CURSO NPTEL - Civil Engineering - Pre-Stressed Concrete Structures

Prestressed Concrete Structures Dr. Amlan K Sengupta and Prof. Devdas Menon

Indian Institute of Technology Madras

1.1 Introduction This section covers the following topics.

• Basic Concept

• Early Attempts of Prestressing

• Brief History

• Development of Building Materials

1.1.1 Basic Concept

A prestressed concrete structure is different from a conventional reinforced concrete

structure due to the application of an initial load on the structure prior to its use.

The initial load or ‘prestress’ is applied to enable the structure to counteract the stresses

arising during its service period.

The prestressing of a structure is not the only instance of prestressing. The concept of

prestressing existed before the applications in concrete. Two examples of prestressing

before the development of prestressed concrete are provided.

Force-fitting of metal bands on wooden barrels The metal bands induce a state of initial hoop compression, to counteract the hoop

tension caused by filling of liquid in the barrels.

Metal bandsMetal bands

Figure 1-1.1 Force-fitting of metal bands on wooden barrels

Pre-tensioning the spokes in a bicycle wheel The pre-tension of a spoke in a bicycle wheel is applied to such an extent that there will

always be a residual tension in the spoke.



Spokes Spokes

Figure 1-1.2 Pre-tensioning the spokes in a bicycle wheel

For concrete, internal stresses are induced (usually, by means of tensioned steel) for

the following reasons.

• The tensile strength of concrete is only about 8% to 14% of its compressive

strength.

• Cracks tend to develop at early stages of loading in flexural members such as

beams and slabs.

• To prevent such cracks, compressive force can be suitably applied in the

perpendicular direction.

• Prestressing enhances the bending, shear and torsional capacities of the flexural

members.

• In pipes and liquid storage tanks, the hoop tensile stresses can be effectively

counteracted by circular prestressing.

1.1.2 Early Attempts of Prestressing

Prestressing of structures was introduced in late nineteenth century. The following

sketch explains the application of prestress.

Place and stretch mild steel rods, prior to concreting

Release the tension and cut the rods after concreting

Place and stretch mild steel rods, prior to concreting

Release the tension and cut the rods after concreting Figure 1-1.3 Prestressing of concrete beams by mild steel rods



Mild steel rods are stretched and concrete is poured around them. After hardening of

concrete, the tension in the rods is released. The rods will try to regain their original

length, but this is prevented by the surrounding concrete to which the steel is bonded.

Thus, the concrete is now effectively in a state of pre-compression. It is capable of

counteracting tensile stress, such as arising from the load shown in the following sketch.

Figure 1-1.4 A prestressed beam under an external load

But, the early attempts of prestressing were not completely successful. It was observed

that the effect of prestress reduced with time. The load resisting capacities of the

members were limited. Under sustained loads, the members were found to fail. This

was due to the following reason.

Concrete shrinks with time. Moreover under sustained load, the strain in concrete

increases with increase in time. This is known as creep strain. The reduction in length

due to creep and shrinkage is also applicable to the embedded steel, resulting in

significant loss in the tensile strain.

In the early applications, the strength of the mild steel and the strain during prestressing

were less. The residual strain and hence, the residual prestress was only about 10% of

the initial value. The following sketches explain the phenomena.



Original length of steel rod (L1)

Original length of concrete beam (L2)

Original length of steel rod (L1)

Original length of concrete beam (L2) a) Beam before applying prestress

Reduced length of concrete beam (L3)Reduced length of concrete beam (L3)

b) Beam at transfer of prestress

Final length of prestressed beam (L4) Final length of prestressed beam (L4)

c) Beam after long-term losses of prestress

Figure 1-1.5 Variation of length in a prestressed beam

The residual strain in steel = original tensile strain in steel – compressive strains

corresponding to short-term and long-term losses.

Original tensile strain in steel = (L2 – L1)/L1

Compressive strain due to elastic shortening of beam = (L2 – L3)/L1

(short-term loss in prestress)

Compressive strain due to creep and shrinkage = (L3 – L4)/L1

(long-term losses in prestress)

Therefore, residual strain in steel = (L4 – L1)/L1

The maximum original tensile strain in mild steel = Allowable stress / elastic

modulus

= 140 MPa / 2×105 MPa

= 0.0007



The total loss in strain due to elastic shortening, creep and shrinkage was also close to

0.0007. Thus, the residual strain was negligible.

The solution to increase the residual strain and the effective prestress are as follows.

• Adopt high strength steel with much higher original strain. This leads to the

scope of high prestressing force.

• Adopt high strength concrete to withstand the high prestressing force.

1.1.3 Brief History

Before the development of prestressed concrete, two significant developments of

reinforced concrete are the invention of Portland cement and introduction of steel in

concrete. These are also mentioned as the part of the history. The key developments

are mentioned next to the corresponding year.

1824 Aspdin, J., (England)

Obtained a patent for the manufacture of Portland cement.

1857 Monier, J., (France)

Introduced steel wires in concrete to make flower pots, pipes, arches and slabs.

The following events were significant in the development of prestressed concrete.

1886 Jackson, P. H., (USA)

Introduced the concept of tightening steel tie rods in artificial stone and concrete

arches.

Figure 1-1.6 Steel tie rods in arches

1888 Doehring, C. E. W., (Germany)

Manufactured concrete slabs and small beams with embedded tensioned steel.



1908 Stainer, C. R., (USA)

Recognised losses due to shrinkage and creep, and suggested retightening the

rods to recover lost prestress.

1923 Emperger, F., (Austria)

Developed a method of winding and pre- tensioning high tensile steel wires

around concrete pipes.

1924 Hewett, W. H., (USA)

Introduced hoop-stressed horizontal reinforcement around walls of concrete

tanks through the use of turnbuckles.

Thousands of liquid storage tanks and concrete pipes were built in the two decades to

follow.

1925 Dill, R. H., (USA)

Used high strength unbonded steel rods. The rods were tensioned and anchored

after hardening of the concrete.

Figure 1-1.7 Portrait of Eugene Freyssinet

(Reference: Collins, M. P. and Mitchell, D.,Prestressed Concrete Structures)

1926 Eugene Freyssinet (France)

Used high tensile steel wires, with ultimate strength as high as 1725 MPa and

yield stress over 1240 MPa. In 1939, he developed conical wedges for end

anchorages for post-tensioning and developed double-acting jacks. He is often

referred to as the Father of Prestressed concrete.



1938 Hoyer, E., (Germany)

Developed ‘long line’ pre-tensioning method.

1940 Magnel, G., (Belgium)

Developed an anchoring system for post-tensioning, using flat wedges.

During the Second World War, applications of prestressed and precast concrete

increased rapidly. The names of a few persons involved in developing prestressed

concrete are mentioned. Guyon, Y., (France) built numerous prestressed concrete

bridges in western and central Europe. Abeles, P. W., (England) introduced the

concept of partial prestressing. Leonhardt, F., (Germany), Mikhailor, V., (Russia) and

Lin, T. Y., (USA) are famous in the field of prestressed concrete.

The International Federation for Prestressing (FIP), a professional organisation in

Europe was established in 1952. The Precast/Prestressed Concrete Institute (PCI) was

established in USA in 1954.

Prestressed concrete was started to be used in building frames, parking structures,

stadiums, railway sleepers, transmission line poles and other types of structures and

elements.

In India, the applications of prestressed concrete diversified over the years. The first

prestressed concrete bridge was built in 1948 under the Assam Rail Link Project.

Among bridges, the Pamban Road Bridge at Rameshwaram, Tamilnadu, remains a

classic example of the use of prestressed concrete girders.



Figure 1-1.8 Pamban Road Bridge at Rameshwaram, Tamilnadu

(Reference: http://www.ramnad.tn.nic.in)

1.1.4 Development of Building Materials

The development of prestressed concrete can be studied in the perspective of

traditional building materials. In the ancient period, stones and bricks were extensively

used. These materials are strong in compression, but weak in tension. For tension,

bamboos and coir ropes were used in bridges. Subsequently iron and steel bars were

used to resist tension. These members tend to buckle under compression. Wood and

structural steel members were effective both in tension and compression.

In reinforced concrete, concrete and steel are combined such that concrete resists

compression and steel resists tension. This is a passive combination of the two

materials. In prestressed concrete high strength concrete and high strength steel are

combined such that the full section is effective in resisting tension and compression.

This is an active combination of the two materials. The following sketch shows the use

of the different materials with the progress of time.



Compression (C) Tension (T) C and T

Stones, Bricks Bamboo, Ropes Timber

Structural steelSteel bars, wires

Reinforced Concrete

PrestressedConcrete

Passive combination

High Strength Steel

High Strength Concrete

Active combination

Concrete

Compression (C) Tension (T) C and T

Stones, Bricks Bamboo, Ropes Timber

Structural steelSteel bars, wires

Reinforced Concrete

PrestressedConcrete

Passive combination

High Strength Steel

High Strength Concrete

Active combination

Concrete

Figure 1-1.9 Development of building materials

(Reference: Lin, T. Y. and Burns, N. H.,

Design of Prestressed Concrete Structures)



1.2 Advantages and Types of Prestressing This section covers the following topics.

• Definitions

• Advantages of Prestressing

• Limitations of Prestressing

• Types of Prestressing

1.2.1 Definitions The terms commonly used in prestressed concrete are explained. The terms are placed

in groups as per usage.

Forms of Prestressing Steel Wires

Prestressing wire is a single unit made of steel.

Strands

Two, three or seven wires are wound to form a prestressing strand.

Tendon

A group of strands or wires are wound to form a prestressing tendon.

Cable

A group of tendons form a prestressing cable.

Bars

A tendon can be made up of a single steel bar. The diameter of a bar is much

larger than that of a wire.

The different types of prestressing steel are further explained in Section 1.7,

Prestressing Steel.

Nature of Concrete-Steel Interface Bonded tendon

When there is adequate bond between the prestressing tendon and concrete, it is called

a bonded tendon. Pre-tensioned and grouted post-tensioned tendons are bonded

tendons.



Unbonded tendon

When there is no bond between the prestressing tendon and concrete, it is called

unbonded tendon. When grout is not applied after post-tensioning, the tendon is an

unbonded tendon.

Stages of Loading The analysis of prestressed members can be different for the different stages of loading.

The stages of loading are as follows.

1) Initial : It can be subdivided into two stages.

a) During tensioning of steel

b) At transfer of prestress to concrete.

2) Intermediate : This includes the loads during transportation of the

prestressed members.

3) Final : It can be subdivided into two stages.

a) At service, during operation.

b) At ultimate, during extreme events.

1.2.2 Advantages of Prestressing

The prestressing of concrete has several advantages as compared to traditional

reinforced concrete (RC) without prestressing. A fully prestressed concrete member is

usually subjected to compression during service life. This rectifies several deficiencies

of concrete.

The following text broadly mentions the advantages of a prestressed concrete member

with an equivalent RC member. For each effect, the benefits are listed.

1) Section remains uncracked under service loads

Reduction of steel corrosion

• Increase in durability.

Full section is utilised

• Higher moment of inertia (higher stiffness)

• Less deformations (improved serviceability).



Increase in shear capacity.

Suitable for use in pressure vessels, liquid retaining structures.

Improved performance (resilience) under dynamic and fatigue loading.

2) High span-to-depth ratios

Larger spans possible with prestressing (bridges, buildings with large column-free

spaces)

Typical values of span-to-depth ratios in slabs are given below.

Non-prestressed slab 28:1

Prestressed slab 45:1

For the same span, less depth compared to RC member.

• Reduction in self weight

• More aesthetic appeal due to slender sections

• More economical sections.

3) Suitable for precast construction

The advantages of precast construction are as follows.

• Rapid construction

• Better quality control

• Reduced maintenance

• Suitable for repetitive construction

• Multiple use of formwork

⇒ Reduction of formwork

• Availability of standard shapes.

The following figure shows the common types of precast sections.



Double T-sectionT-section

Hollow core Piles

Double T-sectionDouble T-sectionT-sectionT-section

Hollow core Piles

L-section Inverted T-section I-girdersL-section Inverted T-section I-girders Figure 1-2.1 Typical precast members

1.2.3 Limitations of Prestressing

Although prestressing has advantages, some aspects need to be carefully addressed.

• Prestressing needs skilled technology. Hence, it is not as common as reinforced

concrete.

• The use of high strength materials is costly.

• There is additional cost in auxiliary equipments.

• There is need for quality control and inspection.

1.2.4 Types of Prestressing Prestressing of concrete can be classified in several ways. The following classifications

are discussed.

Source of prestressing force This classification is based on the method by which the prestressing force is generated.

There are four sources of prestressing force: Mechanical, hydraulic, electrical and

chemical.



External or internal prestressing This classification is based on the location of the prestressing tendon with respect to the

concrete section.

Pre-tensioning or post-tensioning This is the most important classification and is based on the sequence of casting the

concrete and applying tension to the tendons.

Linear or circular prestressing This classification is based on the shape of the member prestressed.

Full, limited or partial prestressing Based on the amount of prestressing force, three types of prestressing are defined.

Uniaxial, biaxial or multi-axial prestressing As the names suggest, the classification is based on the directions of prestressing a

member.

The individual types of prestressing are explained next.

Source of Prestressing Force Hydraulic Prestressing

This is the simplest type of prestressing, producing large prestressing forces. The

hydraulic jack used for the tensioning of tendons, comprises of calibrated pressure

gauges which directly indicate the magnitude of force developed during the tensioning.

Mechanical Prestressing

In this type of prestressing, the devices includes weights with or without lever

transmission, geared transmission in conjunction with pulley blocks, screw jacks with or

without gear drives and wire-winding machines. This type of prestressing is adopted for

mass scale production.



Electrical Prestressing

In this type of prestressing, the steel wires are electrically heated and anchored before

placing concrete in the moulds. This type of prestressing is also known as thermo-

electric prestressing.

External or Internal Prestressing External Prestressing

When the prestressing is achieved by elements located outside the concrete, it is called

external prestressing. The tendons can lie outside the member (for example in I-girders

or walls) or inside the hollow space of a box girder. This technique is adopted in

bridges and strengthening of buildings. In the following figure, the box girder of a bridge

is prestressed with tendons that lie outside the concrete.

Figure 1-2.2 External prestressing of a box girder

(Reference: VSL International Ltd.)

Internal Prestressing

When the prestressing is achieved by elements located inside the concrete member

(commonly, by embedded tendons), it is called internal prestressing. Most of the

applications of prestressing are internal prestressing. In the following figure, concrete

will be cast around the ducts for placing the tendons.



Figure 1-2.3 Internal prestressing of a box girder

(Courtesy: Cochin Port Trust, Kerala)

Pre-tensioning or Post-tensioning Pre-tensioning

The tension is applied to the tendons before casting of the concrete. The pre-

compression is transmitted from steel to concrete through bond over the transmission

length near the ends. The following figure shows manufactured pre-tensioned electric

poles.

Figure 1-2.4 Pre-tensioned electric poles

(Courtesy: The Concrete Products and Construction Company, COPCO, Chennai)



Post-tensioning

The tension is applied to the tendons (located in a duct) after hardening of the concrete.

The pre-compression is transmitted from steel to concrete by the anchorage device (at

the end blocks). The following figure shows a post-tensioned box girder of a bridge.

Figure 1-2.5 Post-tensioning of a box girder


The details of pre-tensioning and post-tensioning are covered under Section 1.3, “Pre-

tensioning Systems and Devices“, and Section 1.4, “Post-tensioning Systems and

Devices”, respectively.

Linear or Circular Prestressing Linear Prestressing

When the prestressed members are straight or flat, in the direction of prestressing, the

prestressing is called linear prestressing. For example, prestressing of beams, piles,

poles and slabs. The profile of the prestressing tendon may be curved. The following

figure shows linearly prestressed railway sleepers.



Figure 1-2.6 Linearly prestressed railway sleepers


Circular Prestressing

When the prestressed members are curved, in the direction of prestressing, the

prestressing is called circular prestressing. For example, circumferential prestressing of

tanks, silos, pipes and similar structures. The following figure shows the containment

structure for a nuclear reactor which is circularly prestressed.

Figure 1-2.7 Circularly prestressed containment structure, Kaiga Atomic Power

Station, Karnataka

(Reference: Larsen & Toubro Ltd, ECC Division, 60 Landmark Years)



Full, Limited or Partial Prestressing Full Prestressing

When the level of prestressing is such that no tensile stress is allowed in concrete under

service loads, it is called Full Prestressing (Type 1, as per IS:1343 - 1980).

Limited Prestressing

When the level of prestressing is such that the tensile stress under service loads is

within the cracking stress of concrete, it is called Limited Prestressing (Type 2).

Partial Prestressing

When the level of prestressing is such that under tensile stresses due to service loads,

the crack width is within the allowable limit, it is called Partial Prestressing (Type 3).

Uniaxial, Biaxial or Multiaxial Prestressing Uniaxial Prestressing

When the prestressing tendons are parallel to one axis, it is called Uniaxial Prestressing.

For example, longitudinal prestressing of beams.

Biaxial Prestressing

When there are prestressing tendons parallel to two axes, it is called Biaxial

Prestressing. The following figure shows the biaxial prestressing of slabs.

Duct for prestressing tendon

Non-prestressed reinforcement

Duct for prestressing tendon


Figure 1-2.8 Biaxial prestressing of a slab

(Courtesy: VSL India Pvt. Ltd., Chennai)



Multiaxial Prestressing

When the prestressing tendons are parallel to more than two axes, it is called Multiaxial

Prestressing. For example, prestressing of domes.



1.3 Pre-tensioning Systems and DevicesThis section covers the following topics.

• Introduction

• Stages of Pre-tensioning

• Advantages of Pre-tensioning

• Disadvantages of Pre-tensioning

• Devices

• Manufacturing of Pre-tensioned Railway Sleepers

1.3.1 Introduction Prestressing systems have developed over the years and various companies have

patented their products. Detailed information of the systems is given in the product

catalogues and brochures published by companies. There are general guidelines of

prestressing in Section 12 of IS:1343 - 1980. The information given in this section is

introductory in nature, with emphasis on the basic concepts of the systems.

The prestressing systems and devices are described for the two types of prestressing,

pre-tensioning and post-tensioning, separately. This section covers pre-tensioning.

Section 1.4, “Post-tensioning Systems and Devices”, covers post-tensioning. In pre-

tensioning, the tension is applied to the tendons before casting of the concrete. The

stages of pre-tensioning are described next.

1.3.2 Stages of Pre-tensioning

In pre-tensioning system, the high-strength steel tendons are pulled between two end

abutments (also called bulkheads) prior to the casting of concrete. The abutments are

fixed at the ends of a prestressing bed.

Once the concrete attains the desired strength for prestressing, the tendons are cut

loose from the abutments.



The prestress is transferred to the concrete from the tendons, due to the bond between

them. During the transfer of prestress, the member undergoes elastic shortening. If the

tendons are located eccentrically, the member is likely to bend and deflect (camber).

The various stages of the pre-tensioning operation are summarised as follows.

1) Anchoring of tendons against the end abutments

2) Placing of jacks

3) Applying tension to the tendons

4) Casting of concrete

5) Cutting of the tendons.

During the cutting of the tendons, the prestress is transferred to the concrete with elastic

shortening and camber of the member.

The stages are shown schematically in the following figures.

Prestressing bed

Steel tendonEnd

abutment

Jack

Prestressing bed

Steel tendonEnd

abutment

Jack

(a) Applying tension to tendons

(b) Casting of concrete

Cutting of tendonCutting of tendon

(c) Transferring of prestress

Figure1-3.1 Stages of pre-tensioning

1.3.3 Advantages of Pre-tensioning

The relative advantages of pre-tensioning as compared to post-tensioning are as

follows. • Pre-tensioning is suitable for precast members produced in bulk.



• In pre-tensioning large anchorage device is not present.

1.3.4 Disadvantages of Pre-tensioning

The relative disadvantages are as follows. • A prestressing bed is required for the pre-tensioning operation.

• There is a waiting period in the prestressing bed, before the concrete attains

sufficient strength.

• There should be good bond between concrete and steel over the transmission

length.

1.3.5 Devices

The essential devices for pre-tensioning are as follows. • Prestressing bed

• End abutments

• Shuttering / mould

• Jack

• Anchoring device

• Harping device (optional)

Prestressing Bed, End Abutments and Mould The following figure shows the devices.

Prestressing bed

MouldEnd

abutment

Jack

Anchoring device

Prestressing bed

MouldEnd

abutment

Jack

Anchoring device

Prestressing bed

MouldEnd

abutment

Jack

Anchoring device

Figure1-3.2 Prestressing bed, end abutment and mould

An extension of the previous system is the Hoyer system. This system is generally

used for mass production. The end abutments are kept sufficient distance apart, and

several members are cast in a single line. The shuttering is provided at the sides and



between the members. This system is also called the Long Line Method. The

following figure is a schematic representation of the Hoyer system

Prestressing bed

A series of moulds

Prestressing bed

A series of moulds

Figure 1-3.3 Schematic representation of Hoyer system

The end abutments have to be sufficiently stiff and have good foundations. This is

usually an expensive proposition, particularly when large prestressing forces are

required. The necessity of stiff and strong foundation can be bypassed by a simpler

solution which can also be a cheaper option. It is possible to avoid transmitting the

heavy loads to foundations, by adopting self-equilibrating systems. This is a common

solution in load-testing. Typically, this is done by means of a ‘tension frame’. The

following figure shows the basic components of a tension frame. The jack and the

specimen tend to push the end members. But the end members are kept in place by

members under tension such as high strength steel rods.

P

Free bodiesPlan or Elevation

Test specimenHigh

strength steel rods

Loadingjack

P

Free bodies

P

Free bodiesPlan or Elevation

Test specimenHigh

strength steel rods

Loadingjack

Plan or Elevation

Test specimenHigh

strength steel rods

Loadingjack

Figure 1-3.4 A tension frame

The frame that is generally adopted in a pre-tensioning system is called a stress bench.

The concrete mould is placed within the frame and the tendons are stretched and

anchored on the booms of the frame. The following figures show the components of a

stress bench.



Jack

Threaded rodElevation

Plan

Mould Strands

Jack


Jack


Plan

Mould Strands

Plan

Mould Strands

Figure 1-3.5 Stress bench – Self straining frame

The following figure shows the free body diagram by replacing the jacks with the applied

forces.

Plan

Load by jack

Tension in strands

Plan

Load by jack

Tension in strands

Figure 1-3.6 Free body diagram of stress bench

The following figure shows the stress bench after casting of the concrete.

Elevation

Plan

Elevation

Plan Figure 1-3.7 The stress bench after casting concrete



Jacks The jacks are used to apply tension to the tendons. Hydraulic jacks are commonly used.

These jacks work on oil pressure generated by a pump. The principle behind the design

of jacks is Pascal’s law. The load applied by a jack is measured by the pressure

reading from a gauge attached to the oil inflow or by a separate load cell. The following

figure shows a double acting hydraulic jack with a load cell.

Figure 1-3.8 A double acting hydraulic jack with a load cell

Anchoring Devices Anchoring devices are often made on the wedge and friction principle. In pre-tensioned

members, the tendons are to be held in tension during the casting and hardening of

concrete. Here simple and cheap quick-release grips are generally adopted. The

following figure provides some examples of anchoring devices.



Figure 1-3.9 Chuck assembly for anchoring tendons

(Reference: Lin, T. Y. and Burns, N. H.,

Design of Prestressed Concrete Structures)

Harping Devices The tendons are frequently bent, except in cases of slabs-on-grade, poles, piles etc.

The tendons are bent (harped) in between the supports with a shallow sag as shown

below.

Harping point Hold up device

a) Before casting of concrete

Harping point Hold up device

a) Before casting of concretea) Before casting of concrete

b) After casting of concreteb) After casting of concrete Figure 1-3.10 Harping of tendons

The tendons are harped using special hold-down devices as shown in the following

figure.



Figure 1-3.11 Hold-down anchor for harping of tendons

(Reference: Nawy, E. G., Prestressed Concrete: A Fundamental Approach)

1.3.6 Manufacturing of Pre-tensioned Railway Sleepers

The following photos show the sequence of manufacturing of pre-tensioned railway

sleepers (Courtesy: The Concrete Products and Construction Company, COPCO,

Chennai). The steel strands are stretched in a stress bench that can be moved on

rollers. The stress bench can hold four moulds in a line. The anchoring device holds

the strands at one end of the stress bench. In the other end, two hydraulic jacks push a

plate where the strands are anchored. The movement of the rams of the jacks and the

oil pressure are monitored by a scale and gauges, respectively. Note that after the

extension of the rams, the gap between the end plate and the adjacent mould has

increased. This shows the stretching of the strands.



Meanwhile the coarse and fine aggregates are batched, mixed with cement, water and

additives in a concrete mixer. The stress bench is moved beneath the concrete mixer.

The concrete is poured through a hopper and the moulds are vibrated. After the

finishing of the surface, the stress bench is placed in a steam curing chamber for a few

hours till the concrete attains a minimum strength.

The stress bench is taken out from the chamber and the strands are cut. The sleepers

are removed from the moulds and stacked for curing in water. After the complete curing,

the sleepers are ready for dispatching.

(a) Travelling pre-tensioning stress bench



Wedge and cylinder assembly at the dead end

Wedge and cylinder assembly at the dead end

(b) Anchoring of strands

Hydraulic jack at stretching end

Initial gap

End plate

Hydraulic jack at stretching end

Initial gap

End plate

(c) Stretching of strands



Extension of ram

Final gap

Threadedrod

Extension of ram

Final gap

Threadedrod

(d) Stretching of strands

Coarse aggregate

Fine aggregate

Coarse aggregate

Fine aggregate

(e) Material storage



Automated batching by weight

Automated batching by weight

(f) Batching of materials

Hopper below concrete mixer Hopper below concrete mixer

(g) Pouring of concrete



(h) Concrete after vibration of mould

(i) Steam curing chamber



(j) Cutting of strands

(k) Demoulding of sleeper



(l) Stacking of sleeper

(m) Water curing



(n) Storage and dispatching of sleepers

Figure 1-3.12 Manufacturing of pre-tensioned railway sleepers



1.4 Post-tensioning Systems and Devices This section covers the following topics

• Introduction

• Stages of Post-tensioning

• Advantages of Post-tensioning

• Disadvantages of Post-tensioning

• Devices

• Manufacturing of a Post-tensioned Bridge Girder

1.4.1 Introduction

Prestressing systems have developed over the years and various companies have

patented their products. Detailed information of the systems is given in the product

catalogues and brochures published by companies. There are general guidelines of

prestressing in Section 12 of IS 1343: 1980. The information given in this section is

introductory in nature, with emphasis on the basic concepts of the systems.

The prestressing systems and devices are described for the two types of prestressing,

pre-tensioning and post-tensioning, separately. This section covers post-tensioning.

Section 1.3, “Pre-tensioning Systems and Devices”, covers pre-tensioning. In post-

tensioning, the tension is applied to the tendons after hardening of the concrete. The

stages of post-tensioning are described next.

1.4.2 Stages of Post-tensioning

In post-tensioning systems, the ducts for the tendons (or strands) are placed along with

the reinforcement before the casting of concrete. The tendons are placed in the ducts

after the casting of concrete. The duct prevents contact between concrete and the

tendons during the tensioning operation.

Unlike pre-tensioning, the tendons are pulled with the reaction acting against the

hardened concrete.



If the ducts are filled with grout, then it is known as bonded post-tensioning. The grout

is a neat cement paste or a sand-cement mortar containing suitable admixture. The

grouting operation is discussed later in the section. The properties of grout are

discussed in Section 1.6, “Concrete (Part-II)”.

In unbonded post-tensioning, as the name suggests, the ducts are never grouted and

the tendon is held in tension solely by the end anchorages. The following sketch shows

a schematic representation of a grouted post-tensioned member. The profile of the duct

depends on the support conditions. For a simply supported member, the duct has a

sagging profile between the ends. For a continuous member, the duct sags in the span

and hogs over the support.

Figure 1-4.1 Post-tensioning (Reference: VSL International Ltd.)

Among the following figures, the first photograph shows the placement of ducts in a box

girder of a simply supported bridge. The second photograph shows the end of the box

girder after the post-tensioning of some tendons.

Figure 1-4.2 Post-tensioning ducts in a box girder




Figure 1-4.3 Post-tensioning of a box girder


The various stages of the post-tensioning operation are summarised as follows.

1) Casting of concrete.

2) Placement of the tendons.

3) Placement of the anchorage block and jack.

4) Applying tension to the tendons.

5) Seating of the wedges.

6) Cutting of the tendons.

The stages are shown schematically in the following figures. After anchoring a tendon

at one end, the tension is applied at the other end by a jack. The tensioning of tendons

and pre-compression of concrete occur simultaneously. A system of self-equilibrating

forces develops after the stretching of the tendons.



Casting bed

Duct

Side viewCasting bed

Duct

Side view (a) Casting of concrete

JackJack

(b) Tensioning of tendons

AnchorAnchor

(c) Anchoring the tendon at the stretching end

Figure 1-4.4 Stages of post-tensioning (shown in elevation)

1.4.3 Advantages of Post-tensioning

The relative advantages of post-tensioning as compared to pre-tensioning are as

follows.

• Post-tensioning is suitable for heavy cast-in-place members.

• The waiting period in the casting bed is less.

• The transfer of prestress is independent of transmission length.

1.4.4 Disadvantage of Post-tensioning

The relative disadvantage of post-tensioning as compared to pre-tensioning is the

requirement of anchorage device and grouting equipment.

1.4.5 Devices

The essential devices for post-tensioning are as follows.

1) Casting bed

2) Mould/Shuttering

3) Ducts



4) Anchoring devices

5) Jacks

6) Couplers (optional)

7) Grouting equipment (optional).

Casting Bed, Mould and Ducts The following figure shows the devices.

Casting bed

Mould

Duct

Casting bed

Mould

Duct

Figure 1-4.5 Casting bed, mould and duct

Anchoring Devices

In post-tensioned members the anchoring devices transfer the prestress to the concrete.

The devices are based on the following principles of anchoring the tendons.

1) Wedge action

2) Direct bearing

3) Looping the wires

Wedge action

The anchoring device based on wedge action consists of an anchorage block and

wedges. The strands are held by frictional grip of the wedges in the anchorage block.

Some examples of systems based on the wedge-action are Freyssinet, Gifford-Udall,

Anderson and Magnel-Blaton anchorages. The following figures show some patented

anchoring devices.



Figure 1-4.6 Freyssinet “T” system anchorage cones

(Reference: Lin, T. Y. and Burns, N. H., Design of Prestressed Concrete Structures)

Figure 1-4.7 Anchoring devices

(Reference: Collins, M. P. and Mitchell, D., Prestressed Concrete Structures)

Figure 1-4.8 Anchoring devices (Reference: VSL International Ltd)



Direct bearing

The rivet or bolt heads or button heads formed at the end of the wires directly bear

against a block. The B.B.R.V post-tensioning system and the Prescon system are

based on this principle. The following figure shows the anchoring by direct bearing.

Figure 1-4.9 Anchoring with button heads


Looping the wires

The Baur-Leonhardt system, Leoba system and also the Dwidag single-bar anchorage

system, work on this principle where the wires are looped around the concrete. The

wires are looped to make a bulb. The following photo shows the anchorage by looping

of the wires in a post-tensioned slab.

Figure 1-4.10 Anchorage by looping the wires in a slab

(Courtesy : VSL India Pvt. Ltd.)

The anchoring devices are tested to calculate their strength. The following photo shows

the testing of an anchorage block.



Figure 1-4.11 Testing of an anchorage device

Sequence of Anchoring

The following figures show the sequence of stressing and anchoring the strands. The

photo of an anchoring device is also provided.



Figure 1-4.12 Sequence of anchoring


Figure 1-4.13 Final form of an anchoring device

(Reference: VSL International Ltd)

Jacks

The working of a jack and measuring the load were discussed in Section 1.3, “Pre-

tensioning Systems and Devices”. The following figure shows an extruded sketch of the

anchoring devices.



Figure 1-4.14 Jacking and anchoring with wedges


Couplers The couplers are used to connect strands or bars. They are located at the junction of

the members, for example at or near columns in post-tensioned slabs, on piers in post-

tensioned bridge decks.

The couplers are tested to transmit the full capacity of the strands or bars. A few types

of couplers are shown.

Figure 1-4.15 Coupler for strands

(Reference: VSL International Ltd)



Figure 1-4.16 Couplers for strands

(Reference: Dywidag – Systems International)

Figure 1-4.17 Couplers for strands

(Reference: Dywidag – Systems International)



Grouting

Grouting can be defined as the filling of duct, with a material that provides an anti-

corrosive alkaline environment to the prestressing steel and also a strong bond between

the tendon and the surrounding grout.

The major part of grout comprises of water and cement, with a water-to-cement ratio of

about 0.5, together with some water-reducing admixtures, expansion agent and

pozzolans. The properties of grout are discussed in Section 1.6, “Concrete (Part-II)”.

The following figure shows a grouting equipment, where the ingredients are mixed and

the grout is pumped.

Figure 1-4.18 Grouting equipment

(Reference: Williams Form Engineering Corp.)

1.4.6 Manufacturing of Post-tensioned Bridge Girders

The following photographs show some steps in the manufacturing of a post-tensioned I-

girder for a bridge (Courtesy: Larsen & Toubro). The first photo shows the fabricated

steel reinforcement with the ducts for the tendons placed inside. Note the parabolic

profiles of the duct for the simply supported girder. After the concrete is cast and cured

to gain sufficient strength, the tendons are passed through the ducts, as shown in the

second photo. The tendons are anchored at one end and stretched at the other end by

a hydraulic jack. This can be observed from the third photo.



(a) Fabrication of reinforcement

(b) Placement of tendons



(c) Stretching and anchoring of tendons

Figure 1-4.19 Manufacturing of a post-tensioned bridge I-girder

The following photos show the construction of post-tensioned box girders for a bridge

(Courtesy: Cochin Port Trust). The first photo shows the fabricated steel reinforcement

with the ducts for the tendons placed inside. The top flange will be constructed later.

The second photo shows the formwork in the pre-casting yard. The formwork for the

inner sides of the webs and the flanges is yet to be placed. In the third photo a girder is

being post-tensioned after adequate curing. The next photo shows a crane on a barge

that transports a girder to the bridge site. The completed bridge can be seen in the last

photo.

(a) Reinforcement cage for box girder



(b) Formwork for box girder

(c) Post-tensioning of box girder



(d) Transporting of box girder

(e) Completed bridge

Figure 1-4.20 Manufacturing of post-tensioned bridge box girders



1.5 Concrete (Part I) This section covers the following topics.

• Constituents of Concrete

• Properties of Hardened Concrete (Part I)

1.5.1 Constituents of Concrete

Introduction Concrete is a composite material composed of gravels or crushed stones (coarse

aggregate), sand (fine aggregate) and hydrated cement (binder). It is expected that the

student of this course is familiar with the basics of concrete technology. Only the

information pertinent to prestressed concrete design is presented here.

The following figure shows a petrographic section of concrete. Note the scattered

coarse aggregates and the matrix surrounding them. The matrix consists of sand,

hydrated cement and tiny voids.

Figure 1-5.1 Petrographic section of hardened concrete

(Reference: Portland Cement Association, Design and Control of Concrete Mixtures)

Aggregate The coarse aggregate are granular materials obtained from rocks and crushed stones.

They may be also obtained from synthetic material like slag, shale, fly ash and clay for

use in light-weight concrete.



The sand obtained from river beds or quarries is used as fine aggregate. The fine

aggregate along with the hydrated cement paste fill the space between the coarse

aggregate.

The important properties of aggregate are as follows.

1) Shape and texture

2) Size gradation

3) Moisture content

4) Specific gravity

5) Unit weight

6) Durability and absence of deleterious materials.

The requirements of aggregate is covered in Section 4.2 of IS:1343 - 1980.

The nominal maximum coarse aggregate size is limited by the lowest of the following

quantities.

1) 1/4 times the minimum thickness of the member

2) Spacing between the tendons/strands minus 5 mm

3) 40 mm.

The deleterious substances that should be limited in aggregate are clay lumps, wood,

coal, chert, silt, rock dust (material finer than 75 microns), organic material, unsound

and friable particles.

Cement In present day concrete, cement is a mixture of lime stone and clay heated in a kiln to

1400 - 1600ºC. The types of cement permitted by IS:1343 - 1980 (Clause 4.1) for

prestressed applications are the following. The information is revised as per IS:456 - 2000, Plain and Reinforced – Concrete Code of Practice.

1) Ordinary Portland cement confirming to IS:269 - 1989, Ordinary Portland Cement,

33 Grade – Specification.

2) Portland slag cement confirming to IS:455 - 1989, Portland Slag Cement –

Specification, but with not more than 50% slag content.

3) Rapid-hardening Portland cement confirming to IS:8041 - 1990, Rapid Hardening

Portland Cement – Specification.



Water The water should satisfy the requirements of Section 5.4 of IS:456 - 2000.

“Water used for mixing and curing shall be clean and free from injurious amounts of oils,

acids, alkalis, salts, sugar, organic materials or other substances that may be

deleterious to concrete and steel”.

Admixtures IS:1343 - 1980 allows to use admixtures that conform to IS:9103 - 1999, Concrete

Admixtures – Specification. The admixtures can be broadly divided into two types:

chemical admixtures and mineral admixtures. The common chemical admixtures are as

follows.

1) Air-entraining admixtures

2) Water reducing admixtures

3) Set retarding admixtures

4) Set accelerating admixtures

5) Water reducing and set retarding admixtures

6) Water reducing and set accelerating admixtures.

The common mineral admixtures are as follows.

1) Fly ash

2) Ground granulated blast-furnace slag

3) Silica fumes

4) Rice husk ash

5) Metakoline

These are cementitious and pozzolanic materials.

1.5.2 Properties of Hardened Concrete (Part I)

The concrete in prestressed applications has to be of good quality. It requires the

following attributes.

1) High strength with low water-to-cement ratio

2) Durability with low permeability, minimum cement content and proper mixing,

compaction and curing



3) Minimum shrinkage and creep by limiting the cement content.

The following topics are discussed.

1) Strength of concrete

2) Stiffness of concrete

3) Durability of concrete

4) High performance concrete

5) Allowable stresses in concrete.

Strength of Concrete

The following sections describe the properties with reference to IS:1343 - 1980. The

strength of concrete is required to calculate the strength of the members. For

prestressed concrete applications, high strength concrete is required for the following

reasons.

1) To sustain the high stresses at anchorage regions.

2) To have higher resistance in compression, tension, shear and bond.

3) To have higher stiffness for reduced deflection.

4) To have reduced shrinkage cracks.

Compressive Strength

The compressive strength of concrete is given in terms of the characteristic compressive strength of 150 mm size cubes tested at 28 days (fck). The characteristic

strength is defined as the strength of the concrete below which not more than 5% of the

test results are expected to fall. This concept assumes a normal distribution of the

strengths of the samples of concrete.

The following sketch shows an idealised distribution of the values of compressive

strength for a sizeable number of test cubes. The horizontal axis represents the values

of compressive strength. The vertical axis represents the number of test samples for a

particular compressive strength. This is also termed as frequency. The average of the

values of compressive strength (mean strength) is represented as fcm. The characteristic

strength (fck) is the value in the x-axis below which 5% of the total area under the curve

falls. The value of fck is lower than fcm by 1.65σ, where σ is the standard deviation of the

normal distribution.



fck fcm

1.65σFrequency

28 day cube compressive strength5% area fck fcm

1.65σFrequency

28 day cube compressive strength5% area

Figure 1-5.2 Idealised normal distribution of concrete strength

(Reference: Pillai, S. U., and Menon, D., Reinforced Concrete Design)

The sampling and strength test of concrete are as per Section 15 of IS:1343 - 1980.

The grades of concrete are explained in Table 1 of the Code.

The minimum grades of concrete for prestressed applications are as follows.

• 30 MPa for post-tensioned members

• 40 MPa for pre-tensioned members.

The maximum grade of concrete is 60 MPa.

Since at the time of publication of IS:1343 in 1980, the properties of higher strength

concrete were not adequately documented, a limit was imposed on the maximum

strength. It is expected that higher strength concrete may be used after proper testing.

The increase in strength with age as given in IS:1343 - 1980, is not observed in present

day concrete that gains substantial strength in 28 days. Hence, the age factor given in

Clause 5.2.1 should not be used. It has been removed from IS:456 - 2000.

Tensile Strength

The tensile strength of concrete can be expressed as follows.

1) Flexural tensile strength: It is measured by testing beams under 2 point loading

(also called 4 point loading including the reactions).

2) Splitting tensile strength: It is measured by testing cylinders under diametral

compression.



3) Direct tensile strength: It is measured by testing rectangular specimens under

direct tension.

In absence of test results, the Code recommends to use an estimate of the flexural

tensile strength from the compressive strength by the following equation.

cr ckf = f0.7 (1-5.1)

Here,

fcr = flexural tensile strength in N/mm2

fck = characteristic compressive strength of cubes in N/mm2.

Stiffness of Concrete The stiffness of concrete is required to estimate the deflection of members. The

stiffness is given by the modulus of elasticity. For a non-linear stress (fc) versus strain

(εc) behaviour of concrete the modulus can be initial, tangential or secant modulus.

IS:1343 - 1980 recommends a secant modulus at a stress level of about 0.3fck. The

modulus is expressed in terms of the characteristic compressive strength and not the

design compressive strength. The following figure shows the secant modulus in the

compressive stress-strain curve for concrete.

εc

fcfck

Ec

fc

εc

fcfck

Ecεc

fcfck

Ec

fcfc

Figure 1-5.3 a) Concrete cube under compression, b) Compressive stress-strain

curve for concrete

The modulus of elasticity for short term loading (neglecting the effect of creep) is given

by the following equation.

c cE = f5000 k

(1-5.2)

Here,

Ec = short-term static modulus of elasticity in N/mm2

fck = characteristic compressive strength of cubes in N/mm2.



The above expression is updated as per IS:456 - 2000.

Durability of Concrete The durability of concrete is of vital importance regarding the life cycle cost of a

structure. The life cycle cost includes not only the initial cost of the materials and labour,

but also the cost of maintenance and repair.

In recent years emphasis has been laid on the durability issues of concrete. This is

reflected in the enhanced section on durability (Section 8) in IS:456 - 2000. It is

expected that the revised version of IS:1343 will also have similar importance on

durability.

The durability of concrete is defined as its ability to resist weathering action, chemical

attack, abrasion, or any other process of deterioration. The common durability

problems in concrete are as follows.

1) Sulphate and other chemical attacks of concrete.

2) Alkali-aggregate reaction.

3) Freezing and thawing damage in cold regions.

4) Corrosion of steel bars or tendons.

The durability of concrete is intrinsically related to its water tightness or permeability.

Hence, the concrete should have low permeability and there should be adequate cover

to reinforcing bars. The selection of proper materials and good quality control are

essential for durability of concrete.

The durability is addressed in IS:1343 - 1980 in Section 7. In Appendix A there are

guidelines on durability. Table 9 specifies the maximum water-to-cement (w-c) ratio

and the minimum cement content for different exposure conditions. The values for

moderate exposure condition are reproduced below.

Table 1-5.1 Maximum water-to-cement (w-c) ratio and the minimum cement content

for moderate exposure conditions (IS:1343 - 1980).

Min. cement content : 300 kg per m3 of concrete

Max w-c ratio* : 0.50

(*The value is updated as per Table 5 of IS:456 - 2000.)



Table 10 provides the values for the above quantities for concrete exposed to sulphate

attack.

To limit the creep and shrinkage, IS:1343 - 1980 specifies a maximum cement content

of 530 kg per m3 of concrete (Clause 8.1.1).

High Performance Concrete

With the advancement of concrete technology, high performance concrete is getting

popular in prestressed applications. The attributes of high performance concrete are as

follows.

1) High strength

2) Minimum shrinkage and creep

3) High durability

4) Easy to cast

5) Cost effective.

Traditionally high performance concrete implied high strength concrete with higher

cement content and low water-to-cement ratio. But higher cement content leads to

autogenous and plastic shrinkage cracking and thermal cracking. At present durability

is also given importance along with strength.

Some special types of high performance concrete are as follows.

1) High strength concrete

2) High workability concrete

3) Self-compacting concrete

4) Reactive powder concrete

5) High volume fly ash concrete

6) Fibre reinforced concrete

In a post-tensioned member, the concrete next to the anchorage blocks (referred to as

end block) is subjected to high stress concentration. The type of concrete at the end

blocks may be different from that at the rest of the member. Fibre reinforced concrete is

used to check the cracking due to the bursting forces.

The following photo shows that the end blocks were cast separately with high strength

concrete.



Figure 1-5.4 End-blocks in a bridge deck


Allowable Stresses in Concrete The allowable stresses are used to analyse and design members under service loads.

IS:1343 - 1980 specifies the maximum allowable compressive stresses for different

grades of concrete under different loading conditions in Section 22.8.

Allowable Compressive Stresses under Flexure

The following sketch shows the variation of allowable compressive stresses for different

grades of concrete at transfer. The cube strength at transfer is denoted as fci.

M30 M60 M40 M60

0.51fci0.44fci

0.54fci0.37fci

Post-tension Pre-tensionM30 M60 M40 M60

0.51fci0.44fci

0.54fci0.37fci

Post-tension Pre-tension

Figure 1-5.5 Variation of allowable compressive stresses at transfer

The following sketch shows the variation of allowable compressive stresses for different

grades of concrete at service loads.



0.34fck

0.41fck

0.27fck

0.35fck

M 30 M 60

Zone I

Zone II

Figure 1-5.6 Variation of allowable compressive stresses at service loads

Here, Zone I represents the locations where the compressive stresses are not likely to

increase. Zone II represents the locations where the compressive stresses are likely to

increase, such as due to transient loads from vehicles in bridge decks.

Allowable Compressive Stresses under Direct Compression

For direct compression, except in the parts immediately behind the anchorage, the

maximum stress is equal to 0.8 times the maximum compressive stress under flexure.

Allowable Tensile Stresses under Flexure

The prestressed members are classified into three different types based on the

allowable tensile stresses. The amount of prestressing varies in the three types. The

allowable tensile stresses for the three types of members are specified in Section 22.7.

The values are reproduced below.

Table 1-5.2 Allowable tensile stresses (IS:1343 - 1980)

Type 1 No tensile stress

Type 2

3 N/mm2.

This value can be increased to 4.5 N/mm2 for temporary loads.

Type 3 Table 8 provides hypothetical values of allowable tensile stresses.

The purpose of providing hypothetical values is to use the elastic analysis method for

Type 3 members even after cracking of concrete.



1.6 Concrete (Part II) This section covers the following topics.

• Properties of Hardened Concrete (Part II)

• Properties of Grout

• Codal Provisions of Concrete

1.6.1 Properties of Hardened Concrete (Part II)

The properties that are discussed are as follows.

1) Stress-strain curves for concrete

2) Creep of concrete

3) Shrinkage of concrete

Stress-strain Curves for Concrete Curve under uniaxial compression

The stress versus strain behaviour of concrete under uniaxial compression is initially

linear (stress is proportional to strain) and elastic (strain is recovered at unloading). With

the generation of micro-cracks, the behaviour becomes nonlinear and inelastic. After the

specimen reaches the peak stress, the resisting stress decreases with increase in strain.

IS:1343 - 1980 recommends a parabolic characteristic stress-strain curve, proposed by

Hognestad, for concrete under uniaxial compression (Figure 3 in the Code).

εcε0 εcu

fc

fckfc

εcε0 εcu

fc

fck

εcε0 εcu

fc

fckfcfc

Figure 1-6.1 a) Concrete cube under compression, b) Design stress-strain curve for

concrete under compression due to flexure



The equation for the design curve under compression due to flexure is as follows.

For εc ≤ ε0

⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

c cck ck

ε εf = f -ε ε

2

0 0

2 (1-6.1)

For εc < εc ≤ εcu

fc = fck (1-6.2) Here,

fc = compressive stress

fck = characteristic compressive strength of cubes

εc = compressive strain

ε0 = strain corresponding to fck = 0.002

εcu = ultimate compressive strain = 0.0035

For concrete under compression due to axial load, the ultimate strain is restricted to

0.002. From the characteristic curve, the design curve is defined by multiplying the

stress with a size factor of 0.67 and dividing the stress by a material safety factor of γm =

1.5. The design curve is used in the calculation of ultimate strength. The following

sketch shows the two curves.

ε0 εcu εc

fcfck

0.447 fck

Characteristic curve

Design curve

ε0 εcu εc

fcfck

0.447 fck

Characteristic curve

Design curve

Figure 1-6.2 Stress-strain curves for concrete under compression due to flexure

In the calculation of deflection at service loads, a linear stress-strain curve is assumed

up to the allowable stress. This curve is given by the following equation.

fc = Ecεc (1-6.3)

Note that, the size factor and the material safety factor are not used in the elastic

modulus Ec.



For high strength concrete (say M100 grade of concrete and above) under uniaxial

compression, the ascending and descending branches are steep.

ε0 εc

fcfck

Es

Eci

ε0 εc

fcfck

Es

Eci

Figure 1-6.3 Stress-strain curves for high strength concrete under compression

The equation proposed by Thorenfeldt, Tomaxzewicz and Jensen is appropriate for high

strength concrete.

⎛ ⎞⎜ ⎟⎝ ⎠⎛ ⎞⎜ ⎟⎝ ⎠

c

c ck nkc

εnε

f = fεn - +ε

0

0

1

(1-6.4)

The variables in the previous equation are as follows.

fc = compressive stress

fck = characteristic compressive strength of cubes in N/mm2

εc = compressive strain

ε0 = strain corresponding to fck

k = 1 for εc ≤ ε0

= 0.67 + (fck / 77.5) for εc > ε0. The value of k should be greater than 1.

n = Eci / (Eci – Es)

Eci = initial modulus

Es = secant modulus at fck = fck / ε0.

The previous equation is applicable for both the ascending and descending branches of

the curve. Also, the parameter k models the slope of the descending branch, which

increases with the characteristic strength fck. To be precise, the value of ε0 can be

considered to vary with the compressive strength of concrete.



Curve under uniaxial tension

The stress versus strain behaviour of concrete under uniaxial tension is linear elastic

initially. Close to cracking nonlinear behaviour is observed.

fc

εc

fcfc

εc

fc

εc

fcfc

(a) (b)

Figure 1-6.4 a) Concrete panel under tension, b) Stress-strain curve for concrete

under tension

In calculation of deflections of flexural members at service loads, the nonlinearity is

neglected and a linear elastic behaviour fc = Ecεc is assumed. In the analysis of ultimate

strength, the tensile strength of concrete is usually neglected.

Creep of Concrete Creep of concrete is defined as the increase in deformation with time under constant

load. Due to the creep of concrete, the prestress in the tendon is reduced with time.

Hence, the study of creep is important in prestressed concrete to calculate the loss in

prestress.

The creep occurs due to two causes.

1. Rearrangement of hydrated cement paste (especially the layered products)

2. Expulsion of water from voids under load

If a concrete specimen is subjected to slow compressive loading, the stress versus

strain curve is elongated along the strain axis as compared to the curve for fast loading.

This can be explained in terms of creep. If the load is sustained at a level, the increase

in strain due to creep will lead to a shift from the fast loading curve to the slow loading

curve (Figure 1-6.5).



εc

fc Fast loading

Slow loading

Effect of creep

εc

fc Fast loading

Slow loading

Effect of creep

Figure 1-6.5 Stress-strain curves for concrete under compression

Creep is quantified in terms of the strain that occurs in addition to the elastic strain due

to the applied loads. If the applied loads are close to the service loads, the creep strain

increases at a decreasing rate with time. The ultimate creep strain is found to be

proportional to the elastic strain. The ratio of the ultimate creep strain to the elastic

strain is called the creep coefficient θ.

For stress in concrete less than about one-third of the characteristic strength, the

ultimate creep strain is given as follows.

cr,ult elε = θε (1-6.5)

The variation of strain with time, under constant axial compressive stress, is

represented in the following figure.

stra

in

Time (linear scale)

εcr, ult = ultimate creep strain

εel = elastic strain

stra

in

Time (linear scale)

εcr, ult = ultimate creep strain

εel = elastic strain

Figure 1-6.6 Variation of strain with time for concrete under compression

If the load is removed, the elastic strain is immediately recovered. However the

recovered elastic strain is less than the initial elastic strain, as the elastic modulus

increases with age.

There is reduction of strain due to creep recovery which is less than the creep strain.

There is some residual strain which cannot be recovered (Figure 1-6.7).



stra

inTime (linear scale)

Residual strain

Creep recoveryElastic recovery

Unloadingstra

inTime (linear scale)

Residual strain

Creep recoveryElastic recovery

Unloading

Figure 1-6.7 Variation of strain with time showing the effect of unloading

The creep strain depends on several factors. It increases with the increase in the

following variables.

1) Cement content (cement paste to aggregate ratio)

2) Water-to-cement ratio

3) Air entrainment

4) Ambient temperature.

The creep strain decreases with the increase in the following variables.

1) Age of concrete at the time of loading.

2) Relative humidity

3) Volume to surface area ratio.

The creep strain also depends on the type of aggregate.

IS:1343 - 1980 gives guidelines to estimate the ultimate creep strain in Section 5.2.5. It

is a simplified estimate where only one factor has been considered. The factor is age of

loading of the prestressed concrete structure. The creep coefficient θ is provided for

three values of age of loading.

Table 1-6.1 Creep coefficient θ for three values of age of loading

Age of Loading Creep Coefficient

7 days 2.2

28 days 1.6

1 year 1.1



It can be observed that if the structure is loaded at 7 days, the creep coefficient is 2.2.

This means that the creep strain is 2.2 times the elastic strain. Thus, the total strain is

more than thrice the elastic strain. Hence, it is necessary to study the effect of creep in

the loss of prestress and deflection of prestressed flexural members. Even if the

structure is loaded at 28 days, the creep strain is substantial. This implies higher loss of

prestress and higher deflection.

Curing the concrete adequately and delaying the application of load provide long term

benefits with regards to durability, loss of prestress and deflection.

In special situations detailed calculations may be necessary to monitor creep strain with

time. Specialised literature or international codes can provide guidelines for such

calculations.

Shrinkage of Concrete Shrinkage of concrete is defined as the contraction due to loss of moisture. The study of

shrinkage is also important in prestressed concrete to calculate the loss in prestress.

The shrinkage occurs due to two causes.

1. Loss of water from voids

2. Reduction of volume during carbonation

The following figure shows the variation of shrinkage strain with time. Here, t0 is the time

at commencement of drying. The shrinkage strain increases at a decreasing rate with

time. The ultimate shrinkage strain (εsh) is estimated to calculate the loss in prestress.

Shrin

kage

str

ain

t0 Time (linear scale)

εsh

Shrin

kage

str

ain

t0 Time (linear scale)

εsh

Figure 1-6.8 Variation of shrinkage strain with time



Like creep, shrinkage also depends on several factors. The shrinkage strain increases

with the increase in the following variables.

1) Ambient temperature

2) Temperature gradient in the members

3) Water-to-cement ratio

4) Cement content.

The shrinkage strain decreases with the increase in the following variables.

1) Age of concrete at commencement of drying

2) Relative humidity

3) Volume to surface area ratio.

The shrinkage strain also depends on the type of aggregate.

IS:1343 - 1980 gives guidelines to estimate the shrinkage strain in Section 5.2.4. It is a

simplified estimate of the ultimate shrinkage strain (εsh).

For pre-tension

εsh = 0.0003 (1-6.6) For post-tension

(1-6.7) ( )shε =

log t +10

0.00022

Here, t is the age at transfer in days. Note that for post-tension, t is the age at transfer

in days which approximates the curing time.

It can be observed that with increasing age at transfer, the shrinkage strain reduces. As

mentioned before, curing the concrete adequately and delaying the application of load

provide long term benefits with regards to durability and loss of prestress.

In special situations detailed calculations may be necessary to monitor shrinkage strain

with time. Specialised literature or international codes can provide guidelines for such

calculations.



1.6.2 Properties of Grout

Grout is a mixture of water, cement and optional materials like sand, water-reducing

admixtures, expansion agent and pozzolans. The water-to-cement ratio is around 0.5.

Fine sand is used to avoid segregation.

The desirable properties of grout are as follows.

1) Fluidity

2) Minimum bleeding and segregation

3) Low shrinkage

4) Adequate strength after hardening

5) No detrimental compounds

6) Durable.

IS:1343 - 1980 specifies the properties of grout in Sections 12.3.1 and Section 12.3.2.

The following specifications are important.

1) The sand should pass 150 µm Indian Standard sieve.

2) The compressive strength of 100 mm cubes of the grout shall not be less than 17

N/mm2 at 7 days.



1.6.5 Codal Provisions of Concrete

The following topics are covered in IS:1343 - 1980 under the respective sections. These

provisions are not duplicated here.

Table 1-6.2 Topics and sections

Workability of concrete Section 6

Concrete mix proportioning Section 8

Production and control of concrete Section 9

Formwork Section 10

Transporting, placing, compacting Section 13

Concrete under special conditions Section 14

Sampling and strength test of concrete Section 15

Acceptance criteria Section 16

Inspection and testing of structures Section 17



1.7 Prestressing Steel This section covers the following topics.

• Forms of Prestressing Steel

• Types of Prestressing Steel

• Properties of Prestressing Steel

• Codal Provisions of Steel

1.7.1 Forms of Prestressing Steel

The development of prestressed concrete was influenced by the invention of high

strength steel. It is an alloy of iron, carbon, manganese and optional materials. The

following material describes the types and properties of prestressing steel.

In addition to prestressing steel, conventional non-prestressed reinforcement is used for

flexural capacity (optional), shear capacity, temperature and shrinkage requirements.

The properties of steel for non-prestressed reinforcement are not covered in this section.

It is expected that the student of this course is familiar with the conventional

reinforcement.

Wires A prestressing wire is a single unit made of steel. The nominal diameters of the wires

are 2.5, 3.0, 4.0, 5.0, 7.0 and 8.0 mm. The different types of wires are as follows.

1) Plain wire: No indentations on the surface.

2) Indented wire: There are circular or elliptical indentations on the surface.

Strands A few wires are spun together in a helical form to form a prestressing strand. The

different types of strands are as follows.

1) Two-wire strand: Two wires are spun together to form the strand.

2) Three-wire strand: Three wires are spun together to form the strand.

3) Seven-wire strand: In this type of strand, six wires are spun around a central wire.

The central wire is larger than the other wires.



Tendons A group of strands or wires are placed together to form a prestressing tendon. The

tendons are used in post-tensioned members. The following figure shows the cross

section of a typical tendon. The strands are placed in a duct which may be filled with

grout after the post-tensioning operation is completed (Figure 1-7.1).

Duct

Grout

Duct

Grout

Figure 1-7.1 Cross-section of a typical tendon

Cables A group of tendons form a prestressing cable. The cables are used in bridges.

Bars A tendon can be made up of a single steel bar. The diameter of a bar is much larger

than that of a wire. Bars are available in the following sizes: 10, 12, 16, 20, 22, 25, 28

and 32 mm.

The following figure shows the different forms of prestressing steel.

Reinforcing barsPrestressing wires, strands and barsReinforcing bars

Prestressing wires, strands and bars

Figure 1-7.2 Forms of reinforcing and prestressing steel



1.7.2 Types of Prestressing Steel

The steel is treated to achieve the desired properties. The following are the treatment

processes.

Cold working (cold drawing) The cold working is done by rolling the bars through a series of dyes. It re-aligns the

crystals and increases the strength.

Stress relieving The stress relieving is done by heating the strand to about 350º C and cooling slowly.

This reduces the plastic deformation of the steel after the onset of yielding.

Strain tempering for low relaxation This process is done by heating the strand to about 350º C while it is under tension.

This also improves the stress-strain behaviour of the steel by reducing the plastic

deformation after the onset of yielding. In addition, the relaxation is reduced. The

relaxation is described later.

IS:1343 - 1980 specifies the material properties of steel in Section 4.5. The following

types of steel are allowed.

1) Plain cold drawn stress relieved wire conforming to IS:1785, Part 1, Specification

for Plain Hard Drawn Steel Wire for Prestressed Concrete, Part I Cold Drawn

Stress Relieved Wire.

2) Plain as-drawn wire conforming to IS:1785, Part 2, Specification for Plain Hard

Drawn Steel Wire for Prestressed Concrete, Part II As Drawn Wire.

3) Indented cold drawn wire conforming to IS:6003, Specification for Indented Wire

for Prestressed Concrete.

4) High tensile steel bar conforming to IS:2090, Specification for High Tensile Steel

Bars used in Prestressed Concrete.

5) Uncoated stress relieved strand conforming to IS:6006. Specification for

Uncoated Stress Relieved Strand for Prestressed Concrete.



1.7.3 Properties of Prestressing Steel

The steel in prestressed applications has to be of good quality. It requires the following

attributes.

1) High strength

2) Adequate ductility

3) Bendability, which is required at the harping points and near the anchorage

4) High bond, required for pre-tensioned members

5) Low relaxation to reduce losses

6) Minimum corrosion.

Strength of Prestressing Steel The tensile strength of prestressing steel is given in terms of the characteristic tensile

strength (fpk).

The characteristic strength is defined as the ultimate tensile strength of the coupon

specimens below which not more than 5% of the test results are expected to fall.

The ultimate tensile strength of a coupon specimen is determined by a testing machine

according to IS:1521 - 1972, Method for Tensile Testing of Steel Wire. The following

figure shows a test setup.

Extensometer

Wedge grips

Coupon specimen

Extensometer

Wedge grips

Coupon specimen

(a) Test set-up



(b) Failure of a strand

Figure 1-7.3 Testing of tensile strength of prestressing strand

The minimum tensile strengths for different types of wires as specified by the codes are

reproduced.

Table 1-7.1 Cold Drawn Stress-Relieved Wires (IS: 1785 Part 1)

Nominal Diameter (mm) 2.50 3.00 4.00 5.00 7.00 8.00

Minimum Tensile Strength fpk

(N/mm2)

2010 1865 1715 1570 1470 1375

The proof stress (defined later) should not be less than 85% of the specified tensile

strength.

Table 1-7.2 As-Drawn wire (IS: 1785 Part 2)

Nominal Diameter (mm) 3.00 4.00 5.00

Minimum Tensile Strength fpk (N/mm2) 1765 1715 1570

The proof stress should not be less than 75% of the specified tensile strength.

Table 1-7.3 Indented wire (IS: 6003)

Nominal Diameter (mm) 3.00 4.00 5.00

Minimum Tensile Strength fpk (N/mm2) 1865 1715 1570

The proof stress should not be less than 85% of the specified tensile strength.

For high tensile steel bars (IS: 2090), the minimum tensile strength is 980 N/mm2. The

proof stress should not be less than 80% of the specified tensile strength.



Stiffness of Prestressing Steel The stiffness of prestressing steel is given by the initial modulus of elasticity. The

modulus of elasticity depends on the form of prestressing steel (wires or strands or

bars).

IS:1343 - 1980 provides the following guidelines which can be used in absence of test

data.

Table 1-7.4 Modulus of elasticity (IS: 1343 - 1980)

Type of steel Modulus of elasticity

Cold-drawn wires 210 kN/mm2

High tensile steel bars 200 kN/mm2

Strands 195 kN/mm2

Allowable Stress in Prestressing Steel As per Clause 18.5.1, the maximum tensile stress during prestressing (fpi) shall not

exceed 80% of the characteristic strength.

≤pi pf 0.8 kf (1-7.1)

There is no upper limit for the stress at transfer (after short term losses) or for the

effective prestress (after long term losses).

Stress-Strain Curves for Prestressing Steel The stress versus strain behaviour of prestressing steel under uniaxial tension is initially

linear (stress is proportional to strain) and elastic (strain is recovered at unloading).

Beyond about 70% of the ultimate strength the behaviour becomes nonlinear and

inelastic. There is no defined yield point.

The yield point is defined in terms of the proof stress or a specified yield strain. IS:1343 - 1980 recommends the yield point at 0.2% proof stress. This stress corresponds to an

inelastic strain of 0.002. This is shown in the following figure.



0.002

Proofstress

εp

fp

0.002

Proofstress

εp

fp

Figure 1-7.4 Proof stress corresponds to inelastic strain of 0.002

The characteristic stress-strain curves are given in Figure 5 of IS:1343 - 1980. The

stress corresponding to a strain can be found out by using these curves as shown next.

0.002 0.005

0.95fpk

0.9fpk

εp

fp

0.002 0.005

0.95fpk

0.85fpk

εp

fp

Stress relieved wires,strands and bars

As-drawn wires

0.002 0.005

0.95fpk

0.9fpk

εp

fp

0.002 0.005

0.95fpk

0.9fpk

εp

fp

0.002 0.005

0.95fpk

0.85fpk

εp

fp

0.002 0.005

0.95fpk

0.85fpk

εp

fp

Stress relieved wires,strands and bars

As-drawn wires

Figure 1-7.5 Characteristic stress-strain curves for prestressing steel

(Figure 5, IS:1343 - 1980)

The stress-strain curves are influenced by the treatment processes. The following figure

shows the variation in the 0.2% proof stress for wires under different treatment

processes.

low relaxation

stress relievedas-drawn

εp

fplow relaxation

stress relievedas-drawn

εp

fp

Figure 1-7.6 Variation in the 0.2% proof stress for wires under different treatment

processes



The design stress-strain curves are calculated by dividing the stress beyond 0.8fpk by a

material safety factor γm =1.15. The following figure shows the characteristic and design

stress-strain curves.

0.8fpk

εp

fpCharacteristic curve

Design curve0.8fpk

εp

fpCharacteristic curve

Design curve

Figure 1-7.7 Characteristic and design stress-strain curves for

prestressing steel

Relaxation of Steel Relaxation of steel is defined as the decrease in stress with time under constant strain.

Due to the relaxation of steel, the prestress in the tendon is reduced with time. Hence,

the study of relaxation is important in prestressed concrete to calculate the loss in

prestress.

The relaxation depends on the type of steel, initial prestress and the temperature. The

following figure shows the effect of relaxation due to different types of loading conditions.

εp

fp

Fast loading

With sustained loadingEffect of relaxation

εp

fp

Fast loading

With sustained loadingEffect of relaxation

Figure 1-7.8 Effect of relaxation due to different types of loading conditions

The following figure shows the variation of stress with time for different levels of

prestressing. Here, the instantaneous stress (fp) is normalised with respect to the initial

prestressing (fpi) in the ordinate. The curves are for different values of fpi/fpy, where fpy is

the yield stress.



10090

80

70

60

5010 100 1000 10,000 100,000

Time (hours)

fpfpi

p i

p y

f=

f0.60.70.80.9

10090

80

70

60

5010 100 1000 10,000 100,000

Time (hours)

fpfpi

p i

p y

f=

f0.60.70.80.9

Figure 1-7.9 Variation of stress with time for different levels of prestressing

It can be observed that there is significant relaxation loss when the applied stress is

more than 70% of the yield stress.

The following photos show the test set-up for relaxation test.

Load cell

Specimen

Load cell

Specimen

(a) Test of a single wire strand



SpecimenSpecimen

(b) Test of a seven-wire strand

Figure 1-7.10 Set-up for relaxation test

The upper limits of relaxation loss are specified as follows.

Table 1-7.5 Relaxation losses at 1000 hours (IS:1785, IS:6003, IS:6006, IS:2090)

Cold drawn stress-relieved wires 5% of initial prestress

Indented wires 5% of initial prestress

Stress-relieved strand 5% of initial prestress

Bars 49 N/mm2

In absence of test data, IS:1343 - 1980 recommends the following estimates of

relaxation losses.

Table 1-7.6 Relaxation losses at 1000 hours at 27°C

Initial Stress Relaxation Loss (N/mm2)

0.5fpk 0

0.6fpk 35

0.7fpk 70

0.8fpk 90

Fatigue Under repeated dynamic loads the strength of a member may reduce with the number

of cycles of applied load. The reduction in strength is referred to as fatigue.



In prestressed applications, the fatigue is negligible in members that do not crack under

service loads. If a member cracks, fatigue may be a concern due to high stress in the

steel at the location of cracks.

Specimens are tested under 2 x 106 cycles of load to observe the fatigue. For steel,

fatigue tests are conducted to develop the stress versus number of cycles for failure (S-

N) diagram. Under a limiting value of stress, the specimen can withstand infinite number

of cycles. This limit is known as the endurance limit.

The prestressed member is designed such that the stress in the steel due to service

loads remains under the endurance limit. The following photo shows a set-up for

fatigue testing of strands.

Figure 1-7.11 Set-up for fatigue testing of strands

Durability Prestressing steel is susceptible to stress corrosion and hydrogen embrittlement in

aggressive environments. Hence, prestressing steel needs to be adequately protected.

For bonded tendons, the alkaline environment of the grout provides adequate protection.

For unbonded tendons, corrosion protection is provided by one or more of the following

methods.



1) Epoxy coating

2) Mastic wrap (grease impregnated tape)

3) Galvanized bars

4) Encasing in tubes.

1.7.4 Codal Provisions of Steel The following topics are covered in IS:1343 - 1980 under the respective sections. These

provisions are not duplicated here.

Table 1-7.7 Topics and sections

Assembly of prestressing and reinforcing steel Section 11

Prestressing Section 12



2.1 Losses in Prestress (Part I) This section covers the following topics.

• Introduction

• Elastic Shortening

The relevant notations are explained first.

Notations Geometric Properties

The commonly used geometric properties of a prestressed member are defined as

follows.

Ac = Area of concrete section

= Net cross-sectional area of concrete excluding the area of

prestressing steel.

Ap = Area of prestressing steel

= Total cross-sectional area of the tendons.

A = Area of prestressed member

= Gross cross-sectional area of prestressed member.

= Ac + Ap

At = Transformed area of prestressed member

= Area of the member when steel is substituted by an equivalent

area of concrete.

= Ac + mAp

= A + (m – 1)Ap

Here,

m = the modular ratio = Ep/Ec

Ec = short-term elastic modulus of concrete

Ep = elastic modulus of steel.

The following figure shows the commonly used areas of the prestressed members.



= + ≡

A Ac Ap At

= + ≡

A Ac Ap At Figure 2-1.1 Areas for prestressed members

CGC = Centroid of concrete

= Centroid of the gross section. The CGC may lie outside the

concrete (Figure 2-1.2).

CGS = Centroid of prestressing steel

= Centroid of the tendons. The CGS may lie outside the tendons or

the concrete (Figure 2-1.2).

I = Moment of inertia of prestressed member

= Second moment of area of the gross section about the CGC.

It = Moment of inertia of transformed section

= Second moment of area of the transformed section about the

centroid of the transformed section.

e = Eccentricity of CGS with respect to CGC

= Vertical distance between CGC and CGS. If CGS lies below CGC,

e will be considered positive and vice versa (Figure 2-1.2).

CGSCGCe

CGS

CGCeCGS

CGCeCGSCGCCGSCGCe

CGS

CGCe

CGS

CGC

CGS

CGC

CGS

CGCe

Figure 2-1.2 CGC, CGS and eccentricity of typical prestressed members

Load Variables

Pi = Initial prestressing force

= The force which is applied to the tendons by the jack.



P0 = Prestressing force after immediate losses

= The reduced value of prestressing force after elastic shortening,

anchorage slip and loss due to friction.

Pe = Effective prestressing force after time-dependent losses

= The final value of prestressing force after the occurrence of creep,

shrinkage and relaxation.

2.1.1 Introduction In prestressed concrete applications, the most important variable is the prestressing

force. In the early days, it was observed that the prestressing force does not stay

constant, but reduces with time. Even during prestressing of the tendons and the

transfer of prestress to the concrete member, there is a drop of the prestressing force

from the recorded value in the jack gauge. The various reductions of the prestressing

force are termed as the losses in prestress.

The losses are broadly classified into two groups, immediate and time-dependent. The

immediate losses occur during prestressing of the tendons and the transfer of prestress

to the concrete member. The time-dependent losses occur during the service life of the

prestressed member. The losses due to elastic shortening of the member, friction at the

tendon-concrete interface and slip of the anchorage are the immediate losses. The

losses due to the shrinkage and creep of the concrete and relaxation of the steel are the

time-dependent losses. The causes of the various losses in prestress are shown in the

following chart.

Losses

Immediate Time dependent

Elastic

shortening Friction Anchorage

slipCreep Shrinkage Relaxation

Figure 2-1.3 Causes of the various losses in prestress



2.1.2 Elastic Shortening

Pre-tensioned Members When the tendons are cut and the prestressing force is transferred to the member, the

concrete undergoes immediate shortening due to the prestress. The tendon also

shortens by the same amount, which leads to the loss of prestress.

Post-tensioned Members If there is only one tendon, there is no loss because the applied prestress is recorded

after the elastic shortening of the member. For more than one tendon, if the tendons

are stretched sequentially, there is loss in a tendon during subsequent stretching of the

other tendons.

The elastic shortening loss is quantified by the drop in prestress (∆fp) in a tendon due to

the change in strain in the tendon (∆εp). It is assumed that the change in strain in the

tendon is equal to the strain in concrete (εc) at the level of the tendon due to the

prestressing force. This assumption is called strain compatibility between concrete

and steel. The strain in concrete at the level of the tendon is calculated from the stress

in concrete (fc) at the same level due to the prestressing force. A linear elastic

relationship is used to calculate the strain from the stress.

The quantification of the losses is explained below.

⎛ ⎞⎜ ⎟⎝ ⎠

p p p

p c

cp

c

p c

∆f = E ∆ε = E ε

f = EE

∆f = mf (2-1.1) For simplicity, the loss in all the tendons can be calculated based on the stress in

concrete at the level of CGS. This simplification cannot be used when tendons are

stretched sequentially in a post-tensioned member. The calculation is illustrated for the

following types of members separately.

• Pre-tensioned Axial Members

• Pre-tensioned Bending Members

• Post-tensioned Axial Members



• Post-tensioned Bending Members

Pre-tensioned Axial Members The following figure shows the changes in length and the prestressing force due to

elastic shortening of a pre-tensioned axial member.

Original length of member at transfer of prestress

Length after elastic shortening

Pi

P0

Original length of member at transfer of prestress

Length after elastic shortening

Pi

P0

Figure 2-1.4 Elastic shortening of a pre-tensioned axial member

The loss can be calculated as per Eqn. (2-1.1) by expressing the stress in concrete in

terms of the prestressing force and area of the section as follows.

(2-1.2)

⎛ ⎞⎜ ⎟⎝ ⎠⎛ ⎞ ⎛ ⎞≈ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

p c

c

i ip

t

∆f = mf

P = mA

P P∆f = m mA A

0

Note that the stress in concrete due to the prestressing force after immediate losses

(P0/Ac) can be equated to the stress in the transformed section due to the initial

prestress (Pi /At). This is derived below. Further, the transformed area At of the

prestressed member can be approximated to the gross area A.

The following figure shows that the strain in concrete due to elastic shortening (εc) is the

difference between the initial strain in steel (εpi) and the residual strain in steel (εp0).



Pi

P0

Length of tendon before stretchingεpi

εp0 εc

Pi

P0

Length of tendon before stretchingεpi

εp0 εc

Figure 2-1.5 Strain variables in elastic shortening

The following equation relates the strain variables.

εc = εpi - εp0 (2-1.3) The strains can be expressed in terms of the prestressing forces as follows.

cc c

Pε =A E

0 (2-1.4)

ipi

p p

Pε =A E

(2-1.5)

pp p

Pε =A E

00

(2-1.6)

Substituting the expressions of the strains in Eqn. (2-1.3)

⎛ ⎞⎜ ⎟⎝ ⎠⎛ ⎞⎜ ⎟⎝ ⎠

i

c c p p p p

i

c c p p p p

i

c p p

i

c p c

P PP= -A E A E A E

P, P + =A E A E A E

Pm 1 P + =A A A

P P =A mA + A

0 0

0

0

0

1 1or

or,

or,

0or i

c t

P P=A A

(2-1.7)

Thus, the stress in concrete due to the prestressing force after immediate losses (P0/Ac)

can be equated to the stress in the transformed section due to the initial prestress (Pi

/At).



The following problem illustrates the calculation of loss due to elastic shortening in an

idealised pre-tensioned railway sleeper.

Example 2-1.1 A prestressed concrete sleeper produced by pre-tensioning method has a rectangular cross-section of 300mm × 250 mm (b × h). It is prestressed with 9 numbers of straight 7mm diameter wires at 0.8 times the ultimate strength of 1570 N/mm2. Estimate the percentage loss of stress due to elastic shortening of concrete. Consider m = 6.

250

40

300

40

Solution

a) Approximate solution considering gross section

The sectional properties are calculated as follows.

Area of a single wire, Aw = π/4 × 72

= 38.48 mm2

Area of total prestressing steel, Ap = 9 × 38.48

= 346.32 mm2

Area of concrete section, A = 300 × 250

= 75 × 103 mm2

Moment of inertia of section, I = 300 × 2503/12

= 3.91 × 108 mm4

Distance of centroid of steel area (CGS) from the soffit,



( )4×38.48× 250 - 40 +5×38.48×40y =9×38.48

=115.5 mm

Prestressing force, Pi = 0.8 × 1570 × 346.32 N

= 435 kN

Eccentricity of prestressing force,

e = (250/2) – 115.5

= 9.5 mm

The stress diagrams due to Pi are shown.

Since the wires are distributed above and below the CGC, the losses are calculated for

the top and bottom wires separately.

Stress at level of top wires (y = yt = 125 – 40)

115.5

e

= +

iP-A

i iP P .e- ± yA I

iP .e± yI

( )

( )3 3

3 8

2

435×10 435×10 ×9.5 = - + × 125 - 4075×10 3.91×10

= -5.8+0.9= -4.9 N/mm

i ic tt

P P .ef = - + yA I



Stress at level of bottom wires (y = yb = 125 – 40),

( )

( )3 3

3 8

2

435×10 435×10 ×9.5 = - - × 125 - 4075×10 3.91×10

= -5.8 - 0.9= -6.7 N/mm

i ic bb

P P .ef = - - yA I

Loss of prestress in top wires = mfcAp

(in terms of force) = 6 × 4.9 × (4 × 38.48)

= 4525.25 N

Loss of prestress in bottom wires = 6 × 6.7 × (5 × 38.48)

= 7734.48 N

Total loss of prestress = 4525 + 7735

= 12259.73 N

≈ 12.3 kN

Percentage loss = (12.3 / 435) × 100%

= 2.83%

b) Accurate solution considering transformed section.

Transformed area of top steel,

A1 = (6 – 1) 4 × 38.48

= 769.6 mm2

Transformed area of bottom steel,

A2 = (6 – 1) 5 × 38.48

= 962.0 mm2

Total area of transformed section,

AT = A + A1 + A2

= 75000.0 + 769.6 + 962.0

= 76731.6 mm2



Centroid of the section (CGC)

A× + A × + A ×y =A

1 2125 (250 - 40) 40

= 124.8 mm from soffit of beam

Moment of inertia of transformed section,

IT = Ig + A(0.2)2 + A1(210 – 124.8)2 + A2(124.8 – 40)2

= 4.02 × 108mm4

Eccentricity of prestressing force,

e = 124.8 – 115.5

= 9.3 mm

Stress at the level of bottom wires, 3 3

3 8

2

435×10 (435×10 ×9.3)84.8= - -76.73×10 4.02×10

= -5.67 - 0.85= -6.52 N/mm

c b(f )

Stress at the level of top wires,

3 3

3 8

2

435×10 (435×10 ×9.3)85.2= - +76.73×10 4.02×10

= -5.67+0.86= -4.81 N/mm

c t(f )

Loss of prestress in top wires = 6 × 4.81 × (4 × 38.48)

= 4442 N

Loss of prestress in bottom wires = 6 × 6.52 × (5 × 38.48)

= 7527 N

Total loss = 4442 + 7527

= 11969 N

≈ 12 kN

Percentage loss = (12 / 435) × 100%

= 2.75 %



It can be observed that the accurate and approximate solutions are close. Hence, the

simpler calculations based on A and I is acceptable.

Pre-tensioned Bending Members The following figure shows the changes in length and the prestressing force due to

elastic shortening of a pre-tensioned bending member.

Pi

wsw (self-weight)

Pi

wsw (self-weight)

Figure 2-1.6 Elastic shortening of a pre-tensioned bending member

Due to the effect of self-weight, the stress in concrete varies along length (Figure 2-1.6).

The loss can be calculated by Eqn. (2-1.1) with a suitable evaluation of the stress in

concrete. To have a conservative estimate of the loss, the maximum stress at the level

of CGS at the mid-span is considered.

(2-1.8)

swi ic

M eP Pe.ef = - - +A I I

Here, Msw is the moment at mid-span due to self-weight. Precise result using At and It in

place of A and I, respectively, is not computationally warranted. In the above

expression, the eccentricity of the CGS (e) was assumed to be constant.

For a large member, the calculation of the loss can be refined by evaluating the strain in

concrete at the level of the CGS accurately from the definition of strain. This is

demonstrated later for post-tensioned bending members.

Post-tensioned Axial Members For more than one tendon, if the tendons are stretched sequentially, there is loss in a

tendon during subsequent stretching of the other tendons. The loss in each tendon can



be calculated in progressive sequence. Else, an approximation can be used to

calculate the losses.

The loss in the first tendon is evaluated precisely and half of that value is used as an

average loss for all the tendons.

(2-1.9)

∑

p p

c

n i,j

j=

∆f = ∆f

mf

P = m

A

1

1

2

121 =212

Here,

Pi,j = initial prestressing force in tendon j

n = number of tendons

The eccentricity of individual tendon is neglected.

Post-tensioned Bending Members The calculation of loss for tendons stretched sequentially, is similar to post-tensioned

axial members. For curved profiles, the eccentricity of the CGS and hence, the stress in

concrete at the level of CGS vary along the length. An average stress in concrete can

be considered.

For a parabolic tendon, the average stress (fc,avg) is given by the following equation.

( )=c,avg c c cf f + f - f1 223 1

(2-1.10)

Here,

fc1 = stress in concrete at the end of the member

fc2 = stress in concrete at the mid-span of the member.

A more rigorous analysis of the loss can be done by evaluating the strain in concrete at

the level of the CGS accurately from the definition of strain. This is demonstrated for a

beam with two parabolic tendons post-tensioned sequentially. In Figure 2-1.7, Tendon

B is stretched after Tendon A. The loss in Tendon A due to elastic shortening during

tensioning of Tendon B is given as follows.

[ ]p p c

p c c

∆f = E ε

= E ε + ε1 2(2-1.11)



Here, εc is the strain at the level of Tendon A. The component of εc due to pure

compression is represented as εc1. The component of εc due to bending is represented

as εc2. The two components are calculated as follows.

∫

∫

Bc

c

c

LB B A

c

LB

B Ac

PεAEδLεL

P .e (x).e (x) = dxL IEP e (x).e (x) dx

E LI

1

2

0

0

=

=

1

=

(2-1.12)

Here,

A = cross-sectional area of beam

PB = prestressing force in Tendon B

Ec = modulus of concrete

L = length of beam

eA(x), eB(x) = eccentricities of Tendons A and B, respectively, at distance x

from left end

I = moment of inertia of beam

δL = change in length of beam

The variations of the eccentricities of the tendons can be expressed as follows.

(2-1.13) ⎛ ⎞+ −⎜ ⎟⎝ ⎠⎛ ⎞+ −⎜ ⎟⎝ ⎠

A A A

B B B

x xe (x) = e ∆eL Lx xe (x)= e ∆eL L

1

1

4 1

4 1

(2-1.14)

−−

2 1

2 1

Where A A A

B B B

, ∆e = e e ∆e = e e

eA1, eA2 = eccentricities of Tendon A at 1 (end) and 2 (centre), respectively.

eB1, eB2 = eccentricities of Tendon B at 1 and 2, respectively.



Substituting the expressions of the eccentricities in Eqn. (2-1.12), the second

component of the strain is given as follows.

(2-1.15)

( )⎡ ⎤+ + +⎢ ⎥⎣ ⎦B

A B A B A B A Bc

P = e e e e e e e eE I 1 1 1 2 2 1 2 2

1 2 85 15 15



2.2 Losses in Prestress (Part II) This section covers the following topics

• Friction

• Anchorage Slip

• Force Variation Diagram

2.2.1 Friction

The friction generated at the interface of concrete and steel during the stretching of a

curved tendon in a post-tensioned member, leads to a drop in the prestress along the

member from the stretching end. The loss due to friction does not occur in pre-

tensioned members because there is no concrete during the stretching of the tendons.

The friction is generated due to the curvature of the tendon and the vertical component

of the prestressing force. The following figure shows a typical profile (laying pattern) of

the tendon in a continuous beam.

Figure 2-2.1 A typical continuous post-tensioned member


In addition to friction, the stretching has to overcome the wobble of the tendon. The

wobble refers to the change in position of the tendon along the duct. The losses due to

friction and wobble are grouped together under friction.

The formulation of the loss due to friction is similar to the problem of belt friction. The

sketch below (Figure 2-2.2) shows the forces acting on the tendon of infinitesimal length

dx.



R

N P + dP

P

dα

dx

P + dP

N

P dα/2

Force triangle

R

N P + dP

P

dα

dxR

N P + dP

P

dα

dx

P + dP

N

P dα/2

Force triangle

P + dP

N

P dα/2

P + dP

N

P dα/2

Force triangle

Figure 2-2.2 Force acting in a tendon of infinitesimal length

In the above sketch,

P = prestressing force at a distance x from the stretching end

R = radius of curvature

dα = subtended angle.

The derivation of the expression of P is based on a circular profile. Although a cable

profile is parabolic based on the bending moment diagram, the error induced is

insignificant.

The friction is proportional to the following variables.

• Coefficient of friction (µ) between concrete and steel.

• The resultant of the vertical reaction from the concrete on the tendon (N)

generated due to curvature.

From the equilibrium of forces in the force triangle, N is given as follows.

dαN = Psin

dα » P = Pdα

22

22

(2-2.1)

The friction over the length dx is equal to µN = µPdα.

Thus the friction (dP) depends on the following variables.

• Coefficient of friction (µ)

• Curvature of the tendon (dα)

• The amount of prestressing force (P)



The wobble in the tendon is effected by the following variables.

• Rigidity of sheathing

• Diameter of sheathing

• Spacing of sheath supports

• Type of tendon

• Type of construction

The friction due to wobble is assumed to be proportional to the following.

• Length of the tendon

• Prestressing force

For a tendon of length dx, the friction due to wobble is expressed as kPdx, where k is

the wobble coefficient or coefficient for wave effect.

Based on the equilibrium of forces in the tendon for the horizontal direction, the

following equation can be written.

P = P + dP + (µPdα + kPdx)

or, dP = – (µPdα + kPdx) (2-2.2)

Thus, the total drop in prestress (dP) over length dx is equal to – (µPdα + kPdx). The

above differential equation can be solved to express P in terms of x.

( )

( )

( )

⎛ ⎞∫ ∫ ∫⎜ ⎟

⎝ ⎠

x

x

P α x

P

PP

x

- µα+kxx

dP = - µ dα+ k dxP

lnP = - µα+ kx

P ln = - µα+ kxP

P = P e

0

0

0 0

0

0

or,

or,

or,

(2-2.3) Here,

P0 = the prestress at the stretching end after any loss due to elastic shortening.

For small values of µα + kx, the above expression can be simplified by the Taylor series

expansion.

Px = P0 (1– µα – kx) (2-2.4)

Thus, for a tendon with single curvature, the variation of the prestressing force is linear

with the distance from the stretching end. The following figure shows the variation of



prestressing force after stretching. The left side is the stretching end and the right side

is the anchored end.

PxP0 PxP0

Figure 2-2.3 Variation of prestressing force after stretching

In the absence of test data, IS:1343 - 1980 provides guidelines for the values of µ and k.

Table 2-2.1 Values of coefficient of friction

Type of interface µ

For steel moving on smooth concrete 0.55.

For steel moving on steel fixed to duct 0.30.

For steel moving on lead 0.25.

The value of k varies from 0.0015 to 0.0050 per meter length of the tendon depending

on the type of tendon. The following problem illustrates the calculation of the loss due

to friction in a post-tensioned beam.

Example 2-2.1

A post-tensioned beam 100 mm × 300 mm (b × h) spanning over 10 m is stressed

by successive tensioning and anchoring of 3 cables A, B, and C respectively as shown in figure. Each cable has cross section area of 200 mm2 and has initial stress of 1200 MPa. If the cables are tensioned from one end, estimate the percentage loss in each cable due to friction at the anchored end. Assume µ = 0.35, k = 0.0015 / m.



Cable ACable BCable C

CL

5050 CGC


CL

5050 CGC

Solution

Prestress in each tendon at stretching end = 1200 × 200

= 240 kN.

To know the value of α(L), the equation for a parabolic profile is required.

md y y= ( L -

d x L 24 2 x )

ym

y

L

x

α(L)

ym

y

L

x

α(L)

Here,

ym = displacement of the CGS at the centre of the beam from the ends

L = length of the beam

x = distance from the stretching end

y = displacement of the CGS at distance x from the ends.

An expression of α(x) can be derived from the change in slope of the profile. The slope

of the profile is given as follows.

md y y= ( L -d x L 2

4 2 x )

At x = 0, the slope dy/dx = 4ym/L. The change in slope α(x) is proportional to x.

The expression of α(x) can be written in terms of x as α(x) = θ.x,

where, θ = 8ym/L2. The variation is shown in the following sketch.



8ym/L

4ym/L

α

0 L/2 L x

θ

8ym/L

4ym/L

α

0 L/2 Lθ

x The total subtended angle over the length L is 8ym/L.

The prestressing force Px at a distance x is given by

Px = P0e–(µα + kx) = P0e–ηx

where, ηx = µα + kx

For cable A, ym = 0.1 m.

For cable B, ym = 0.05 m.

For cable C, ym = 0.0 m.

For all the cables, L = 10 m.

Substituting the values of ym and L

⎧⎪⎨⎪⎩

0.0043x for cable A= 0.0029x for cable B

0.0015x for cable Cηx

The maximum loss for all the cables is at x = L = 10, the anchored end.

⎧⎪⎨⎪⎩

0.958 for cable A = 0.971 for cable B

0.985 for cable C

-ηLe

Percentage loss due to friction = (1 – e–ηL) × 100%

⎧⎪⎨⎪⎩

4.2% for cable A = 2.9% for cable B

1.5% for cable C




CL

CGC

240 kN


CL

CGC

240 kN

Variation of prestressing forces

The loss due to friction can be considerable for long tendons in continuous beams with

changes in curvature. The drop in the prestress is higher around the intermediate

supports where the curvature is high. The remedy to reduce the loss is to apply the

stretching force from both ends of the member in stages.

2-2.2 Anchorage Slip

In a post-tensioned member, when the prestress is transferred to the concrete, the

wedges slip through a little distance before they get properly seated in the conical space.

The anchorage block also moves before it settles on the concrete. There is loss of

prestress due to the consequent reduction in the length of the tendon.

The total anchorage slip depends on the type of anchorage system. In absence of

manufacturer’s data, the following typical values for some systems can be used.

Table 2-2.2 Typical values of anchorage slip

Anchorage System Anchorage Slip (∆s)

Freyssinet system

12 - 5mm Φ strands

12 - 8mm Φ strands

4 mm

6 mm

Magnel system 8 mm

Dywidag system 1 mm

(Reference: Rajagopalan, N., Prestressed Concrete)



Due to the setting of the anchorage block, as the tendon shortens, there is a reverse

friction. Hence, the effect of anchorage slip is present up to a certain length (Figure 2-

2.4). Beyond this setting length, the effect is absent. This length is denoted as lset.

Px

P0

Px

P0

Figure 2-2.4 Variation of prestressing force after anchorage slip

2.2.3 Force Variation Diagram

The magnitude of the prestressing force varies along the length of a post-tensioned

member due to friction losses and setting of the anchorage block. The diagram

representing the variation of prestressing force is called the force variation diagram.

Considering the effect of friction, the magnitude of the prestressing force at a distance x

from the stretching end is given as follows.

(2-2.5) -ηxxP = P e0

Here, ηx = µα + kx denotes the total effect of friction and wobble. The plot of Px gives

the force variation diagram.

The initial part of the force variation diagram, up to length lset is influenced by the setting

of the anchorage block. Let the drop in the prestressing force at the stretching end be

∆P. The determination of ∆P and lset are necessary to plot the force variation diagram

including the effect of the setting of the anchorage block.

Considering the drop in the prestressing force and the effect of reverse friction, the

magnitude of the prestressing force at a distance x from the stretching end is given as

follows.

(2-2.6) ( )' η'xxP = P - ∆P e0



Here, η’ for reverse friction is analogous to η for friction and wobble.

At the end of the setting length (x = lset), Px = P’x

∆P

P0

Px Px’

lset x

Px

Px after stretchingP’x after settingPx beyond lset

∆P

P0

Px Px’

lset x

Px

∆P

P0

Px Px’

lset x

Px

Px after stretchingP’x after settingPx beyond lset

Figure 2-2.5 Force variation diagram near the stretching end

Substituting the expressions of Px and Px’ for x = lset

Since it is difficult to measure η’ separately, η’ is taken equal to η. The expression of

∆P simplifies to the following.

( )( )

( )

( )

⎡ ⎤⎣ ⎦⎛ ⎞⎜ ⎟⎝ ⎠

set set

set

-ηl η'l

- η+η' l

set

set set

P e = P - ∆P e

P e = P - ∆P

P - η+η' l = P - ∆P

η'∆P = P η+η' l = P ηl +η

0 0

0 0

0 0

0 0

1

1

⎛ ⎞⎜ ⎟⎝ ⎠

s setp p

sets set

p p

∆P∆ = lA E

l η'∆ = P ηl +A E η0

12

1 12

⎛ ⎞⎜ ⎟⎝ ⎠

p pset s

s p p

A El = ∆

η'P η +η

∆ A E = η' = η

P η

2

0

0

2

1

for

(2-2.7)

∆P = 2P0ηlset (2-2.8)

The following equation relates lset with the anchorage slip ∆s.

(2-2.9)

Transposing the terms,



Therefore,

s p p

set

∆ A El =

P η0

(2-2.10)

The term P0η represents the loss of prestress per unit length due to friction.

The force variation diagram is used when stretching is done from both the ends. The

tendons are overstressed to counter the drop due to anchorage slip. The stretching from

both the ends can be done simultaneously or in stages. The final force variation is more

uniform than the first stretching.

The following sketch explains the change in the force variation diagram due to

stretching from both the ends in stages.

a) After stretching from right end

b) After anchorage slip at right end

a) After stretching from right end

b) After anchorage slip at right end

c) After stretching from left end

d) After anchorage slip at left end


d) After anchorage slip at left end


d) After anchorage slip at left end Figure 2-2.6 Force variation diagrams for stretching in stages



The force variation diagrams for the various stages are explained.

a) The initial tension at the right end is high to compensate for the anchorage

slip. It corresponds to about 0.8 fpk initial prestress. The force variation

diagram (FVD) is linear.

b) After the anchorage slip, the FVD drops near the right end till the length lset.

c) The initial tension at the left end also corresponds to about 0.8 fpk initial prestress.

The FVD is linear up to the centre line of the beam.

d) After the anchorage slip, the FVD drops near the left end till the length lset. It is

observed that after two stages, the variation of the prestressing force over the length

of the beam is less than after the first stage.

Example 2-2.2 A four span continuous bridge girder is post-tensioned with a tendon consisting of twenty strands with fpk = 1860 MPa. Half of the girder is shown in the figure below. The symmetrical tendon is simultaneously stressed up to 75% fpk from both ends and then anchored. The tendon properties are Ap = 2800 mm2, Ep = 195,000 MPa, µ = 0.20, K = 0.0020/m. The anchorage slip ∆s = 6 mm. Calculate a) The expected elongation of the tendon after stretching, b) The force variation diagrams along the tendon before and after anchorage.

13.7 13.7 3 3.7 15.2 15.2 3.7

0.76 0.6 0.76

All dimensions are in metres

0.6 CL

Inflection points



Solution

Initial force at stretching end

0.75fpk = 1395 MPa

P0 = 0.75fpk Ap

= 3906 kN

The continuous tendon is analys

identified between the points of

inflection points are those where

segments are as follows: 1-2, 2-3,

321

The following properties of parabo

sketch below is used.

The change in slope from the origi

the end of the tendon which is α =

L = length of the segment

e = vertical shift from the or

For segments 2-3 and 3-4 and sub

used.

y

0

ed as segments of parabola. The segments are

maximum eccentricity and inflection points. The

the curvature of the tendon reverses. The different

3-4, 4-5, 5-6, 6-7 and 7-8.

CL

5 764 8

las are used. For segment 1-2, the parabola in the

n to the end of the parabola is same as the slope at

2e/L, where

igin.

sequent pairs of segments, the following property is

e

L α

x



λL

For the two parabolic segments joined at the inflection point as shown in the sketch

above, the slope at the inflection point α = 2(e1 + e2)/λL.

Here,

e1, e2 = eccentricities of the CGS at the span and support respectively

L = length of the span

λL = fractional length between the points of maximum eccentricity

The change in slope between a point of maximum eccentricity and inflection point is

also equal to α.

The change in slope (α) for each segment of the tendon is calculated using the above

expressions. Next the value of µα + kx for each segment is calculated using the given

values of µ, k and x, the horizontal length of the segment. Since the loss in prestress

accrues with each segment, the force at a certain segment is given as follows.

The summation ∑ is for the segments from the stretching end up to the point in the

segment under consideration. Hence, the value of ∑(µα + kx) at the end of each

segment is calculated to evaluate the prestressing force at that point (Px, where x

denotes the point).

e2

e1

α

L

-Σ(µα+kx)xP = P e0



0.163 α 0.111 0.163 0.144 0.144 0.144 0.144

µα +kx 0.0390.050 0.060 0.059 0.059 0.036 0.036

The force variation diagram before anchorage can be plotted with the above values of

Px. A linear variation of the force can be assumed for each segment. Since the

stretching is done at both the ends simultaneously, the diagram is symmetric about the

central line.

a) The expected elongation of the tendon after stretching

First the product of the average force and the length of each segment is summed up to

the centre line.

0.050 Σ(µα +kx) 0.149 k ) 0.110 0.185 0.244 0.303 0.339

e-Σ(µα + kx)

0.738 0.712

0.7830.952 1.000 0.896 0.861

0.831

Px (kN)3906 3718 3500

3363 3246 3058 2883

2781

[ ] [ ]

[ ] [ ]

[ ] [ ]

[ ]

av1 1P L = 3906+3718 ×13.7+ 3718+3500 ×13.72 21 1+ 3500+3363 ×3+ 3363+3246 ×3.72 21 1+ 3246+3058 ×15.2+ 3058+ 2883 ×15.22 21+ 2883+ 2718 ×3.72

= 227612.2 kN



The elongation (∆) at each stretching end is calculated as follows.

3227612×10=2800×195000

= 0.417 m

av

P P

P L∆=A E

b) The force variation diagrams along the tendon before and after anchorage

After anchorage, the effect of anchorage slip is present up to the setting length lset. The

value of lset due to an anchorage slip ∆s = 6 mm is calculated as follows.

6×2800×195000=13.7

=15.46 m

s P Pset

0

∆ A El =P µ

The quantity P0µ is calculated from the loss of prestress per unit length in the first

segment. P0µ = (3906 – 3718) kN /13.7 m = 13.7 N/mm. The drop in the prestressing

force (∆p) at each stretching end is calculated as follows.

02

= 2×13.7×15464= 423.7 kN

p set∆ = P µl

Thus the value of the prestressing force at each stretching end after anchorage slip is

3906 – 424 = 3482 kN. The force variation diagram for lset = 15.46 m is altered to show

the drop due to anchorage slip.

The force variation diagrams before and after anchorage are shown below. Note that

the drop of force per unit length is more over the supports due to change in curvature

over a small distance.



2500

3000

3500

4000

0 20 40 60 8

Distance from end (m)

Pres

tres

sing

forc

e (k

N)

00

After anchorage Before anchorage

2500

3000

3500

4000

0 20 40 60 8

Distance from end (m)

Pres

tres

sing

forc

e (k

N)

After anchorage Before anchorage



2.3 Losses in Prestress (Part III) This section covers the following topics.

• Creep of Concrete

• Shrinkage of Concrete

• Relaxation of Steel

• Total Time Dependent Losses

2.3.1 Creep of Concrete

Creep of concrete is defined as the increase in deformation with time under constant

load. Due to the creep of concrete, the prestress in the tendon is reduced with time.

The creep of concrete is explained in Section 1.6, Concrete (Part II). Here, the

information is summarised. For stress in concrete less than one-third of the

characteristic strength, the ultimate creep strain (εcr,ult) is found to be proportional to the

elastic strain (εel). The ratio of the ultimate creep strain to the elastic strain is defined as

the ultimate creep coefficient or simply creep coefficient θ.

The ultimate creep strain is then given as follows.

(2-3.1) cr,ult elε = θε

IS:1343 - 1980 gives guidelines to estimate the ultimate creep strain in Section 5.2.5. It

is a simplified estimate where only one factor has been considered. The factor is age of

loading of the prestressed concrete structure. The creep coefficient θ is provided for

three values of age of loading.

Curing the concrete adequately and delaying the application of load provide long term

benefits with regards to durability, loss of prestress and deflection.

In special situations detailed calculations may be necessary to monitor creep strain with

time. Specialised literature or international codes can provide guidelines for such

calculations.

The loss in prestress (∆fp ) due to creep is given as follows.

∆fp = Ep εcr, ult (2-3.2)



Here, Ep is the modulus of the prestressing steel.

The following considerations are applicable for calculating the loss of prestress due to

creep.

1) The creep is due to the sustained (permanently applied) loads only.

Temporary loads are not considered in the calculation of creep.

2) Since the prestress may vary along the length of the member, an average value

of the prestress can be considered.

3) The prestress changes due to creep and the creep is related to the

instantaneous prestress. To consider this interaction, the calculation of creep can

be iterated over small time steps.

2.3.2 Shrinkage of Concrete

Shrinkage of concrete is defined as the contraction due to loss of moisture. Due to the

shrinkage of concrete, the prestress in the tendon is reduced with time.

The shrinkage of concrete was explained in details in the Section 1.6, Concrete (Part II).

IS:1343 - 1980 gives guidelines to estimate the shrinkage strain in Section 5.2.4. It is a

simplified estimate of the ultimate shrinkage strain (εsh). Curing the concrete adequately

and delaying the application of load provide long term benefits with regards to durability

and loss of prestress. In special situations detailed calculations may be necessary to

monitor shrinkage strain with time. Specialised literature or international codes can

provide guidelines for such calculations.

The loss in prestress (∆fp ) due to shrinkage is given as follows.

∆fp = Ep εsh (2-3.3) Here, Ep is the modulus of the prestressing steel.

2.3.3 Relaxation of Steel

Relaxation of steel is defined as the decrease in stress with time under constant strain.

Due to the relaxation of steel, the prestress in the tendon is reduced with time. The

relaxation depends on the type of steel, initial prestress (fpi) and the temperature. To



calculate the drop (or loss) in prestress (∆fp), the recommendations of IS:1343 - 1980

can be followed in absence of test data.

Example 2-3.1

A concrete beam of dimension 100 mm × 300 mm is post-tensioned with 5 straight wires of 7mm diameter. The average prestress after short-term losses is 0.7fpk = 1200 N/mm2 and the age of loading is given as 28 days. Given that Ep =

200 × 103 MPa, Ec = 35000 MPa, find out the losses of prestress due to creep,

shrinkage and relaxation. Neglect the weight of the beam in the computation of the stresses.

300

100

50 CGS

300

100

50 CGS

Solution

Area of concrete A = 100 × 300

= 30000 mm2

Moment of inertia of beam section

I = 100 × 3003 / 12

= 225 × 106 mm4

Area of prestressing wires Ap = 5 × (π/4) × 72

= 192.42 mm2

Prestressing force after short-term losses

P0 = Ap.fp0

= 192.4 × 1200

= 230880 N



Modular ratio m = Ep / Ec

= 2 × 105 / 35 × 103

= 5.71

Stress in concrete at the level of CGS

0 0

24 6

230880 230880= - - ×503×10 225×10

cP P ef = - - eA I

= – 7.69 – 2.56

= – 10.25 N/mm2

Loss of prestress due to creep

(∆fp)cr = Ep εcr, ult

= Ep θεel

= Ep θ (fc/Ec)

= m θ fc

= 5.71 × 10.25 × 1.6

= 93.64 N / mm2

Here, θ = 1.6 for loading at 28 days, from Table 2c-1 (Clause 5.2.5.1, IS:1343 - 1980).

Shrinkage strain from Clause 5.2.4.1, IS:1343 - 1980 εsh = 0.0002 / log10(t + 2)

= 0.0002 / log10 (28 + 2)

= 1.354 × 10-4

Loss of prestress due to shrinkage

(∆fp)sh = Epεsh

= 2 × 105 × 1.354 × 10-4

= 27.08 N/mm2



From Table 2c-2 (Table 4, IS:1343 - 1980) Loss of prestress due to relaxation

(∆fp)rl = 70.0 N/mm2

Loss of prestressing force = ∆fp Ap

Therefore,

Loss of prestressing force due to creep = 93.64 × 192.42

= 18018 N

Loss of prestressing force due to shrinkage

= 27.08 × 192.42

= 5211 N

Loss of prestressing force due to relaxation

= 70 × 192.42

= 13469 N

Total long-term loss of prestressing force (neglecting the interaction of the losses and

prestressing force)

= 18018 + 5211 + 13469

= 36698 N

Percentage loss of prestress = 36698 / 230880 × 100%

= 15.9 %

2.3.4 Total Time-dependent Loss

The losses of prestress due to creep and shrinkage of concrete and the relaxation of the

steel are all time-dependent and inter-related to each other. If the losses are calculated

separately and added, the calculated total time-dependent loss is over-estimated. To

consider the inter-relationship of the cause and effect, the calculation can be done for

discrete time steps. The results at the end of each time step are used for the next time

step. This step-by-step procedure was suggested by the Precast / Prestressed



Concrete Institute (PCI) committee and is called the General method (Reference: PCI

Committee, “Recommendations for Estimating Prestress Losses”, PCI Journal, PCI, Vol.

20, No. 4, July-August 1975, pp. 43-75).

In the PCI step-by-step procedure, a minimum of four time steps are considered in the

service life of a prestressed member. The following table provides the definitions of the

time steps (Table 2-3.3).

Table 2-3.3 Time steps in the step-by-step procedure

Step Beginning End

1 Pre-tension: Anchorage of steel

Post-tension: End of curing

Age of prestressing

2 End of Step 1 30 days after prestressing or

when subjected to superimposed

load

3 End of Step 2 1 year of service

4 End of Step 3 End of service life

The step-by-step procedure can be implemented by a computer program, where the

number of time steps can be increased.

There are also approximate methods to calculate lump sum estimates of the total loss.

Since these estimates are not given in IS:1343 - 1980, they are not mentioned here.



3.1 Analysis of Members under Axial Load This section covers the following topics.

• Introduction

• Analysis at Transfer

• Analysis at Service Loads

• Analysis of Ultimate Strength

• Analysis of Behaviour

Notations Geometric Properties

A prestressed axial member may also have non-prestressed reinforcement to carry the

axial force. This type of members is called partially prestressed members. The

commonly used geometric properties of a prestressed member with non-prestressed

reinforcement are defined as follows. A = gross cross-sectional area

Ac = area of concrete

As = area of non-prestressed reinforcement

Ap = area of prestressing tendons

At = transformed area of the section

= Ac + (Es/ Ec) As + (Ep/ Ec) Ap

The following figure shows the commonly used areas of a prestressed member with

non-prestressed reinforcement.



= +

A Ac ApAs

+= +

A Ac ApAs

+

A

≡

AtA

≡

At Figure 3-1.1 Areas for a prestressed member with non-prestressed reinforcement

3.1.1 Introduction

The study of members under axial load gives an insight of the behaviour of a

prestressed member as compared to an equivalent non-prestressed reinforced concrete

member. Prestressed members under axial loads only, are uncommon. Members such

as hangers and ties are subjected to axial tension. Members such as piles may have

bending moment along with axial compression or tension. In this section, no

eccentricity of the CGS with respect to CGC is considered. The definitions of CGS and

CGC are provided in Section 2.1, Losses in Prestress (Part I). The following figure

shows members under axial loads.



Hangers PilesHangers Piles Figure 3-1.2 Members under axial load

The analysis of members refers to the evaluation of the following.

1) Permissible prestress based on allowable stresses at transfer. 2) Stresses under service loads. These are compared with allowable stresses

under service conditions.

3) Ultimate strength. This is compared with the demand under factored loads.

4) The entire axial load versus deformation behaviour.

The stages for loading are explained in Section 1.2, Advantages and Types of

Prestressing

3.1.2 Analysis at Transfer The stress in the concrete (fc) in a member without non-prestressed reinforcement can

be calculated as follows.

c

c

Pf = -A

0

(3-1.1) Here,

P0 = prestress at transfer after short-term losses.

In presence of non-prestressed reinforcement, the stress in the concrete can be

calculated as follows.



cc s c

Pf = -A + (E /E )A

0

(3-1.2)

s

The permissible prestress is determined based on fc to be within the allowable stress at

transfer.

3.1.3 Analysis at Service Loads

The stresses in concrete in a member without non-prestressed reinforcement can be


(3-1.3) ec

c t

P Pf = - ±A A

Here,

P = external axial force (In the equation, + for tensile force and vice

versa.)

Pe = effective prestress.

If there is non-prestressed reinforcement, Ac is to be substituted by (Ac + (Es/Ec) As) and

At is to be calculated including As.

The value of fc should be within the allowable stress under service conditions.

3.1.4 Analysis of Ultimate Strength

The ultimate tensile strength of a section (PuR) can be calculated as per Clause 22.3, IS:1343 - 1980.

In absence of non-prestressed reinforcement,

(3-1.4a) uR Pk pP = f A0.87

In presence of non-prestressed reinforcement,

(3-1.4b) uR y s Pk pP = f A + f A0.87 0.87



In the previous equations,

fy = characteristic yield stress for non-prestressed reinforcement with mild

steel bars

= characteristic 0.2% proof stress for non-prestressed reinforcement

with high yield strength deformed bars.

fpk = characteristic tensile strength of prestressing tendons.

The ultimate tensile strength should be greater than the demand due to factored loads.

The ultimate compressive strength of a section (PuR) can be calculated in presence of

moments by the use of interaction diagrams. For a member under compression with

minimum eccentricity, the ultimate strength is given as follows. Here, the contribution of

prestressing steel is neglected.

PuR = 0.4 fckAc + 0.67 fy As (3-1.5)

3.1.5 Analysis of Behaviour

The analysis of behaviour refers to the determination of the complete axial load versus

deformation behaviour. The analyses at transfer, under service loads and for ultimate

strength correspond to three instants in the above behaviour.

The analysis involves three principles of mechanics (Reference: Collins, M. P. and

Mitchell, D., Prestressed Concrete Structures, Prentice-Hall, Inc., 1991).

1) Equilibrium of internal forces with the external loads at any point of the load

versus deformation behaviour. The internal forces in concrete and steel are

evaluated based on the respective strains, cross-sectional areas and the

constitutive relationships.

2) Compatibility of the strains in concrete and in steel for bonded tendons. This

assumes a perfect bond between the two materials. For unbonded tendons, the

compatibility is in terms of total deformation.

3) Constitutive relationships relating the stresses and the strains in the materials.

The relationships are developed based on the material properties.



Equilibrium Equation

At any instant, the equilibrium is given by the following equation.

P = Acfc + Asfs + Apfp (3-1.6) Here,

fc = stress in concrete

fs = stress in non-prestressed reinforcement

fp = stress in prestressed tendons

P = axial force.

Compatibility Equations

For non-prestressed reinforcement

εs = εc (3-1.7) For prestressed tendons

εp = εc + ∆εp (3-1.8)

Here,

εc = strain in concrete at the level of the steel

εs = strain in non-prestressed reinforcement

εp = strain in prestressed tendons

∆εp = strain difference in prestressed tendons with adjacent concrete

The strain difference (∆εp) is the strain in the prestressed tendons when the concrete

has zero strain (εc = 0). This occurs when the strain due to the external tensile axial

load balances the compressive strain due to prestress. At any load stage,

∆εp = εpe – εce (3-1.9) Here,

εpe = strain in tendons due to Pe, the prestress at service

εce = strain in concrete due to Pe.

The strain difference is further explained in Section 3.4, Analysis of Member under

Flexure (Part III).

Constitutive Relationships

The constitutive relationships can be expressed in the following forms based on the

material stress-strain curves shown in Section 1.6, Concrete (Part II), and Section 1.7,

Prestressing Steel.



For concrete under compression

fc = F1 (εc) (3-1.10)

For prestressing steel

fp = F2 (εp) (3-1.11)

For reinforcing steel

fs = F3 (εs) (3-1.12) The stress versus strain curve for concrete is shown below. The first and third

quadrants represent the behaviour under tension and compression, respectively.

εc

fc

εc

fc

Figure 3-1.3 Stress versus strain for concrete

The stress versus strain curve for prestressing steel is as shown below.

εp

fp

εp

fp

Figure 3-1.4 Stress versus strain for prestressing steel

The following stress versus strain curve is for reinforcing steel.



εs

fs

εs

fs

Figure 3-1.5 Stress versus strain for reinforcing steel

The equilibrium and compatibility equations and the constitutive relationships can be

solved to develop the axial force versus deformation curve. The deformation can be

calculated as εcL, where L is the length of the member.

The following plot shows the axial force versus deformation curves for prestressed and

non-prestressed sections. The two sections are equivalent in their ultimate tensile

strengths.

Deformation

Axial force

Cracking Tensile strengths

Compressive strengths

Prestressed section

Non-prestressed section

Deformation

Axial force



Deformation

Axial force



Prestressed section

Non-prestressed section Figure 3-1.6 Axial force versus deformation curves

From the previous plot, the following can be inferred.

1) Prestressing increases the cracking load.

2) Prestressing shifts the curve from the origin.

• For the prestressed member, there is a compressive deformation in absence

of external axial force.

• A certain amount of external force is required to decompress the member.



3) For a given tensile load, the deformation of the prestressed member is

smaller.

• Prestressing reduces deformation at service loads.

4) For a given compressive load, the deformation of the prestressed member is

larger.

• Prestressing is detrimental for the response under compression.

5) The compressive strength of the prestressed member is lower.

• Prestressing is detrimental for the compressive strength.

6) For a partially prestressed section with the same ultimate strength, the axial load

versus deformation curve will lie in between the curves for prestressed and non-

prestressed sections.

The above conclusions are generic for prestressed members.



3.2 Analysis of Members under Flexure (Part I) This section covers the following topics.

• Introduction

• Analyses at Transfer and at Service

3.2.1 Introduction

Similar to members under axial load, the analysis of members under flexure refers to

the evaluation of the following.

1) Permissible prestress based on allowable stresses at transfer.

2) Stresses under service loads. These are compared with allowable stresses under

service conditions.

3) Ultimate strength. This is compared with the demand under factored loads.

4) The entire load versus deformation behaviour.

The analyses at transfer and under service loads are presented in this section. The

analysis for the ultimate strength is presented separately in Section 3.4, Analysis of

Member under Flexure (Part III). The evaluation of the load versus deformation

behaviour is required in special type of analysis. This analysis will not be covered in

this section.

Assumptions The analysis of members under flexure considers the following.

1) Plane sections remain plane till failure (known as Bernoulli’s hypothesis).

2) Perfect bond between concrete and prestressing steel for bonded tendons.

Principles of Mechanics The analysis involves three principles of mechanics.

1) Equilibrium of internal forces with the external loads. The compression in

concrete (C) is equal to the tension in the tendon (T). The couple of C and T are

equal to the moment due to external loads.

2) Compatibility of the strains in concrete and in steel for bonded tendons. The

formulation also involves the first assumption of plane section remaining plane

after bending. For unbonded tendons, the compatibility is in terms of deformation.




Variation of Internal Forces In reinforced concrete members under flexure, the values of compression in concrete

(C) and tension in the steel (T) increase with increasing external load. The change in

the lever arm (z) is not large.

In prestressed concrete members under flexure, at transfer of prestress C is located

close to T. The couple of C and T balance only the self weight. At service loads, C

shifts up and the lever arm (z) gets large. The variation of C or T is not appreciable.

The following figure explains this difference schematically for a simply supported beam

under uniform load.

Reinforced concrete C2 > C1, z2 ≈ z1

Prestressed concrete C2 ≈ C1, z2 > z1

w2 > w1 w2

z2C2

T2

C1

w1

z1T1

w2

z2C2

T2

w1

C1T1

z1

Reinforced concrete C2 > C1, z2 ≈ z1

Prestressed concrete C2 ≈ C1, z2 > z1

w2 > w1 w2

z2C2

T2

w2

z2C2

T2

C1

w1

z1T1

C1

w1

z1T1

w2

z2C2

T2

w2

z2C2

T2

w1

C1T1

z1

w1

C1T1

z1

Figure 3-2.1 Variations of internal forces and lever arms

In the above figure,

C1, T1 = compression and tension at transfer due to self weight

C2, T2 = compression and tension under service loads

w1 = self weight

w2 = service loads

z1 = lever arm at transfer

z2 = lever arm under service loads.



For the reinforced concrete member C2 is substantially large than C1, but z2 is close to

z1. For the prestressed concrete member C2 is close to C1, but z2 is substantially large

than z1.

3.2.2 Analyses at Transfer and at Service

The analyses at transfer and under service loads are similar. Hence, they are presented

together. A prestressed member usually remains uncracked under service loads. The

concrete and steel are treated as elastic materials. The principle of superposition is

applied. The increase in stress in the prestressing steel due to bending is neglected.

There are three approaches to analyse a prestressed member at transfer and under

service loads. These approaches are based on the following concepts.

a) Based on stress concept.

b) Based on force concept.

c) Based on load balancing concept.

The following material explains the three concepts.

Based on Stress Concept In the approach based on stress concept, the stresses at the edges of the section under

the internal forces in concrete are calculated. The stress concept is used to compare

the calculated stresses with the allowable stresses.

The following figure shows a simply supported beam under a uniformly distributed load

(UDL) and prestressed with constant eccentricity (e) along its length.

eCGCCGS

eCGCCGS

eCGCCGS

Figure 3-2.2 A simply supported beam under UDL

The following sketch shows the internal forces in concrete at a section and the

corresponding stress profiles. The first stress profile is due to the compression P. The

second profile is due to the eccentricity of the compression. The third profile is due to

the moment. At transfer, the moment is due to self weight. At service the moment is

due to service loads.



Internal forces in concrete

–P/A(stress

due to P)

±Pey/ I(stress

due to P.e)

±My/ I(stress

due to M)

Resultant stress profile

M

P+ + =CGC


–P/A(stress

due to P)

±Pey/ I(stress

due to P.e)

±My/ I(stress

due to M)


M

P+ + =CGC

Figure 3-2.3 Stress profiles at a section due to internal forces

The resultant stress at a distance y from the CGC is given by the principle of

superposition as follows.

P Pey Myf = - ± ± A I I

(3-2.1)

For a curved tendon, P can be substituted by its horizontal component. But the effect of

the refinement is negligible.

Based on Force Concept The approach based on force concept is analogous to the study of reinforced concrete.

The tension in prestressing steel (T) and the resultant compression in concrete (C) are

considered to balance the external loads. This approach is used to determine the

dimensions of a section and to check the service load capacity. Of course, the stresses

in concrete calculated by this approach are same as those calculated based on stress

concept. The stresses at the extreme edges are compared with the allowable stresses.

The following figures show the internal forces in the section.

Internal forces at prestressing(neglecting self-weight)

Internal forces after loading

CT

C

T

ec

ez

Internal forces at prestressing(neglecting self-weight)

Internal forces after loading

CT

C

T

ec

ez

CT

C

T

ec

ez

Figure 3-2.4 Internal forces at a section

The equilibrium equations are as follows.

C =T



(3-2.2)

c

M = C.z M = C(e + e) (3-2.3)

The resultant stress in concrete at distance y from the CGC is given as follows.

cCe yCf = - ±

A I(3-2.4)

Substituting C = P and Cec = M – Pe, the expression of stress becomes same as that

given by the stress concept.

(3-2.5) P Pey Myf = - ± ± A I I

Based on Load Balancing Concept The approach based on load balancing concept is used for a member with curved or

harped tendons and in the analysis of indeterminate continuous beams. The moment,

upward thrust and upward deflection (camber) due to the prestress in the tendons are

calculated. The upward thrust balances part of the superimposed load.

The expressions for three profiles of tendons in simply supported beams are given.

a) For a Parabolic Tendon

Bending moment diagram

e

wup

M

L

P

Free body diagram of concrete


e

wup

M

L

P


Figure 3-2.5 Simply supported beam with parabolic tendon

The moment at the centre due to the uniform upward thrust (wup) is given by the

following equation.

upw L

M =2

8



(3-2.6)

The moment at the centre from the prestressing force is given as M = Pe.

The expression of wup is calculated by equating the two expressions of M. The upward

deflection (∆) can be calculated from wup based on elastic analysis.

up

up

Pew = Lw L

∆=EI

2

4

8

5384

(3-2.7)

b) For Singly Harped Tendon

PP

Wup


M


PP

Wup


M


Figure 3-2.6 Simply supported beam with singly harped tendon

The moment at the centre due to the upward thrust (Wup) is given by the following

equation. It is equated to the moment due to the eccentricity of the tendon. As before,

the upward thrust and the deflection can be calculated.

up

up

up

W LM = = Pe

PeW = L

W L ∆=

EI

3

44

48

(3-2.8)



c) For Doubly Harped Tendon

PP

WupWupaL

MBending moment diagram


PP

WupWupaL

MBending moment diagram


Figure 3-2.7 Simply supported beam with doubly harped tendon

The moment at the centre due to the upward thrusts (Wup) is given by the following

equation. It is equated to the moment due to the eccentricity of the tendon. As before,

the upward thrust and the deflection can be calculated.

(3-2.9)

( )

up

up

up

M = W aL = PePeW =aLa - a W L

∆=EI

2 33 424

Example 3-2.1 A concrete beam prestressed with a parabolic tendon is shown in the figure. The prestressing force applied is 1620 kN. The uniformly distributed load includes the self weight. Compute the extreme fibre stress at the mid-span by applying the three concepts. Draw the stress distribution across the section at mid-span.

7.3m

45 kN/m 500

750

145

At end At mid-span

CGC

7.3m

45 kN/m 500

750

500

750

145

At end At mid-span

CGC



Solution

a) Stress concept

Area of concrete, A = 500 × 750

= 375,000 mm2

Moment of inertia, I = (500 × 7503) / 12

= 1.758 × 1010 mm4

Bending moment at mid-span, M = (45 × 7.32) / 8

= 299.7 kNm

Top fibre stress

( )c top toptP Pe Mf = - + y - yA I I

3 3 6

3 10 1

2

1620×10 1620×10 ×145 299.7×10= - + ×375 - ×375375×10 1.758×10 1.758×10

= - 4.32 + 5.01- 6.39= - 5.7 N/mm

0

Bottom fibre stress

( )3 3 6

3 10 1

2

1620×10 1620×10 ×145 299.7×10= - - ×375+ ×375375×10 1.758×10 1.758×10

= -4.32- 5.01+ 6.39= -2.9 N/mm

c bot botbP Pe Mf = - - y + yA I I

0

b) Force concept

CP

C

Pece z

CP

C

Pece z

Applied moment M = 299.7 kN-m



Lever arm z = M / P

= 299.7 × 103 / 1620

= 185 mm

Eccentricity of C ec = z – e

= 185 – 145

= 40 mm

Top fibre stress

( )

3 3

3 10

2

× × ××

× ×

1620 10 1620 10 40= - - 375375 10 1.758 10

= -4.32-1.38= -5.7 N/mm

cc topt

CeCf = - - yA I

Bottom fibre stress

( )

3 3

3 10

2

× × ××

× ×

1620 10 1620 10 40= - + 375375 10 1.758 10

= -4.32+1.38= -2.9 N/mm

cc botb

CeCf = - + yA I

c) Load balancing method

Effective upward load, wup = 8Pe / L2

= 8 × 1620 × 103 × 145 / 73002

= 35.3 kN/m

Residual load wres = 45 – 35.3

= 9.7 kN/m

Residual bending moment Mres = 9.7 × 7.32 / 8

= 64.6 kNm



Residual bending stress (fc)res = 64.6 × 106 × 375 / 1.758×1010

= 1.38 N/mm2

Total top fibre stress (fc)t = – P/A – (fc)res

= – 4.32 – 1.38

= – 5.7 N/mm2

Total bottom fibre stress (fc)b = – P/A + (fc)res

= – 4.32 + 1.38

= – 2.9 N/mm2

The resultant stress distribution at mid-span is shown below.

– 5.7 N/mm2

– 2.9 N/mm2

– 5.7 N/mm2

– 2.9 N/mm2



3.3 Analysis of Members under Flexure (Part II) This section covers the following topics.

• Cracking Moment

• Kern Points

• Pressure Line

Introduction

The analysis of flexural members under service loads involves the calculation of the

following quantities.

a) Cracking moment.

b) Location of kern points.

c) Location of pressure line.

The following material explains each one of them.

3.2.1 Cracking Moment

The cracking moment (Mcr) is defined as the moment due to external loads at which the

first crack occurs in a prestressed flexural member. Considering the variability in stress

at the occurrence of the first crack, the evaluated cracking moment is an estimate.

Nevertheless, the evaluation of cracking moment is important in the analysis of

prestressed members.

Based on the allowable tensile stress the prestress members are classified into three

types as per IS:1343 - 1980. The types are explained in Section 1.2, Advantages and

Types of Prestressing. For Type 1 (full prestressing) and Type 2 (limited prestressing)

members, cracking is not allowed under service loads. Hence, it is imperative to check

that the cracking moment is greater than the moment due to service loads. This is

satisfied when the stress at the edge due to service loads is less than the modulus of

rupture.

The modulus of rupture is the stress at the bottom edge of a simply supported beam

corresponding to the cracking moment (Mcr). The modulus of rupture is a measure of

the flexural tensile strength of concrete. It is measured by testing beams under 2 point



loading (also called 4 point loading including the reactions or middle third loading). The

modulus of rupture (fcr) is expressed in terms of the characteristic compressive strength

(fck) of concrete by the following equation (IS:456 - 2000). Here, fcr and fck are in N/mm2.

(3-3.1) cr ckf = f0.7

The following sketch shows the internal forces and the resultant stress profile at the

instant of cracking.



Mcr

Pe

fcr

yb

CGC



Mcr

Pe

fcr

yb

CGC

Figure 3-3.1 Internal forces and resultant stress profile at cracking

The stress at the edge can be calculated based on the stress concept as follows. The

cracking moment (Mcr) can be evaluated by transposing the terms.

(3-3.2)

e e b cr bcr

cr b e e bcr

cr ecr e

b b

P P ey M y- - + = fA I I

M y P P ey = f + +I A

f I P I M = + + P e y Ay

or,

or,

I

The above equation expresses Mcr in terms of the section and material properties and

prestressing variables.

3.2.2 Kern Points

When the resultant compression (C) is located within a specific zone of a section of a

beam, tensile stresses are not generated. This zone is called the kern zone of a section.

For a section symmetric about a vertical axis, the kern zone is within the levels of the

upper and lower kern points. When the resultant compression (C) under service loads

is located at the upper kern point, the stress at the bottom edge is zero. Similarly, when

C at transfer of prestress is located at the bottom kern point, the stress at the upper



edge is zero. The levels of the upper and lower kern points from CGC are denoted as kt

and kb, respectively.

Based on the stress concept, the stress at the bottom edge corresponding to C at the

upper kern point, is equated to zero. The following sketch shows the location of C and

the resultant stress profile.

Location of resultant compression


C

yb

ktCGC



C

yb

ktCGC

Figure 3-3.2 Resultant stress profile when compression is at upper kern point

The value of kt can be calculated by equating the stress at the bottom to zero as follows.

t b

t b2

tb

Ck yC- + =A I

Ck yC- + =A Ar

r k =y

2

0

or, 0

or,

(3-3.3)

The above equation expresses the location of upper kern point in terms of the section

properties. Here, r is the radius of gyration and yb is the distance of the bottom edge

from CGC.

Similar to the calculation of kt, the location of the bottom kern point can be calculated by

equating the stress at the top edge to zero. The following sketch shows the location of

C and the resultant stress profile.



Cyt

kbCGC



Cyt

kbCGC

Figure 3-3.3 Resultant stress profile when compression is at lower kern point



b t

b t

bt

Ck yC- + =A I

Ck yC - + =A Ar

r k =y

2

2

0

or, 0

or,

(3-3.4)

Here, yt is the distance of the top edge from CGC.

Cracking Moment using Kern Points The kern points can be used to determine the cracking moment (Mcr). The cracking

moment is slightly greater than the moment causing zero stress at the bottom. C is

located above kt to cause a tensile stress fcr at the bottom. The incremental moment is

fcr I/yb. The following sketch shows the shift in C outside the kern to cause cracking and

the corresponding stress profiles.



Cece

∆z

kt

fcr fcr

+ =CGCCGS



Cece

∆z

kt

fcr fcr

+ =CGCCGS

Figure 3-3.4 Resultant stress profile at cracking of the bottom edge

The cracking moment can be expressed as the product of the compression and the

lever arm. The lever arm is the sum of the eccentricity of the CGS (e) and the

eccentricity of the compression (ec). The later is the sum of kt and ∆z, the shift of C

outside the kern.

(3-3.5)

( )( )

( )

cr c

t

crcr t

b

M = C e+ e= C e+ k + ∆z

f I M = C e+ k + y

or,

Substituting C = Pe, kt = r2/yb and r2 = I/A, the above equation becomes same as the

previous expression of Mcr.



⎛ ⎞⎜ ⎟⎝ ⎠

crcr e

b b

cr ecr e

b b

f Ir M =P +e +y y

f I PIM = + +Pe y Ay

2

or,

(3-3.6)

3.2.3 Pressure Line

The pressure line in a beam is the locus of the resultant compression (C) along the

length. It is also called the thrust line or C-line. It is used to check whether C at

transfer and under service loads is falling within the kern zone of the section. The

eccentricity of the pressure line (ec) from CGC should be less than kb or kt to ensure C

in the kern zone.

The pressure line can be located from the lever arm (z) and eccentricity of CGS (e) as

follows. The lever arm is the distance by which C shifts away from T due to the moment.

Subtracting e from z provides the eccentricity of C (ec) with respect to CGC. The

variation of ec along length of the beam provides the pressure line.

c

Mz =C

e = z - e

(3-3.7)

A positive value of ec implies that C acts above the CGC and vice-versa. If ec is

negative and the numerical value is greater than kb (that is |ec| > kb), C lies below the

lower kern point and tension is generated at the top of the member.

If ec> kt, then C lies above the upper kern point and tension is generated at the bottom

of the member.

Pressure Line at Transfer The pressure line is calculated from the moment due to the self weight. The following

sketch shows that the pressure line for a simply supported beam gets shifted from the

CGS with increasing moment towards the centre of the span.



kt

kb

Pressure line

CGCCGS

CL

kt

kb

Pressure line

CGCCGS

CL

Figure 3-3.4 Pressure line at transfer

Pressure Line under Service Loads

The pressure line is calculated from the moment due to the service loads. The following

sketch shows that the pressure line for a simply supported beam gets further shifted

from the CGS at the centre of the span with increased moment under service condition.

kt

kb

Pressure line

CGCCGS

CL

kt

kb

Pressure line

CGCCGS

CL

Figure 3-3.4 Pressure line under service loads

Limiting Zone For fully prestressed members (Type 1), tension is not allowed under service conditions.

If tension is also not allowed at transfer, C always lies within the kern zone. The limiting

zone is defined as the zone for placing the CGS of the tendons such that C always lies

within the kern zone.

For limited prestressed members (Type 2 and Type 3), tension is allowed at transfer

and under service conditions. The limiting zone is defined as the zone for placing the

CGS such that the tensile stresses in the extreme edges are within the allowable values.

The following figure shows the limiting zone (as the shaded region) for a simply

supported beam subjected to uniformly distributed load.



Locus of emin

Locus of emax

CGC

CL

Locus of emin

Locus of emax

CGC

CL

Figure 3-3.4 Limiting zone for a simply supported beam

The determination of limiting zone is given in Section 4.4, Design of Sections for Flexure

(Part III).

Example 3-3.1 For the post-tensioned beam with a flanged section as shown, the profile of the CGS is parabolic, with no eccentricity at the ends. The live load moment due to service loads at mid-span (MLL) is 648 kNm. The prestress after transfer (P0) is 1600 kN. Assume 15% loss at service. Grade of concrete is M30.

18.0m

CGC

18.0m

CGC

200

200

500

1000150

250

150CGS

Values in mm.

Cross-section at mid-span

200

200

500

1000150

250

150CGS

Values in mm.




Evaluate the following quantities. a) Kern levels b) Cracking moment c) Location of pressure line at mid-span at transfer and at service. d) The stresses at the top and bottom fibres at transfer and at service.

Compare the stresses with the following allowable stresses at transfer and at service. For compression, fcc,all = – 18.0 N/mm2 For tension, fct,all = 1.5 N/mm2.

Solution

Calculation of geometric properties

The section is divided into three rectangles for the computation of the geometric

properties. The centroid of each rectangle is located from the soffit.

yt

yb

100

900

1

2

3

+

500y

CGC

Values in mm.

yt

yb

100

900900

11

22

33

+

500500y

CGC

Values in mm.

Area of the section

Area of 1 = A1 = 500 × 200

= 100,000 mm2

Area of 2 = A2 = 600 × 150

= 90,000 mm2

Area of 3 = A3 = 250 × 200

= 50,000 mm2



A = A1 + A2 + A3

= 240,000 mm2

Location of CGC from the soffit

1 2 3×900+ ×500+ ×100=

= 583.3 mm

A A AyA

Therefore, = 583.3 mm=1000.0 - 583.3= 416.7 mm

b

t

yy

Eccentricity of CGS at mid-span

= -150= 583.3 -150 = 433.3 mm

e y

Moment of inertia of 1 about axis through CGC

I A3 21 1

10 4

1= ×500×200 + ×(900 - 583.3)12

=1.036×10 mm


3 22 2

9 4

1= ×150×600 + ×(583.3 - 500)12

= 3.32×10 mm

I A


3 2

3 3

10 4

1= ×250×200 + ×(583.3 -100)12

=1.184×10 mm

I A



Moment of inertia of the section

1 2 3

10

10 4

= + +

= (1.036+0.336+1.184)×10= 2.552×10 mm

I I I I

Square of the radius of gyration

2

10

5 2

=

2.552×10=240,000

=1.063×10 mm

IrA

a) Kern levels of the section

2

51.063×10=583.3

=182.2mm

tb

rk =y

2

51.063×10=416.7

= 255.1mm

bt

rk =y

Values in mm.

255.1

182.2

CGS

CGC

Kern zone+

Values in mm.

255.1

182.2

CGS

CGC

Kern zone+

255.1

182.2

CGS

CGC

Kern zone+

Calculation of moment due to self weight (MDL).

⎛ ⎞⎜ ⎟⎝ ⎠

2 23 2

3 21 m= 24.0 kN/m ×240,000 mm ×

10 mm= 5.76 kN/m

DLw

2

28

5.76×18.0=8

= 233.3 kNm

DLDL

w LM =



b) Calculation of cracking moment

Modulus of rupture

2

= 0.7

= 0.7 30= 3.83kN/mm

cr ckf f

10 3 10

3

3

3.83×2.552×10 0.85×1600×10 ×2.552×10= +583.3 240×10 ×583.3

+0.8×1600×10 ×433.3 Nmm=167.6+ 247.9+554.6= 970.1 kNm

cr ecr e

b b

f I P IM = + + P ey Ay

Live load moment corresponding to cracking

= 970.1- 233.3= 736.8 kNm

LL crM

Since the given live load moment (648.0 kNm) is less than the above value, the section

is uncracked.

⇒ The moment of inertia of the gross section can be used for computation of

stresses.

c) Calculation of location of pressure line at mid-span

At transfer

3233.3×10=1600

=145.8 mm

DLMz =C

=145.8 - 433.3= - 287.5mm

ce = z - e

Since ec is negative, the pressure line is below CGC.

Since the magnitude of ec is greater than kb, there is tension at the top.



Value in mm.

CGS

CGC

Kern zone+

x

287.5

Location of pressure line

Value in mm.

CGS

CGC

Kern zone+

x

287.5

Location of pressure line

At service

3(233.3+ 648.0)×10=0.85×1600

= 648.0 mm

DL+LLMz =C

= 648.0 - 433.3= 214.7mm

ce = z - e

Since ec is positive, the pressure line is above CGC.

Since the magnitude of ec is greater than kt, there is tension at the bottom.

Value in mm.

CGS

CGC

Kern zone+

x

214.7 Location of pressure line

Value in mm.

CGS

CGC

Kern zone+

x

214.7 Location of pressure line

d) Calculation of stresses

The stress is given as follows.


–P/A ±Pey/ I ±My/ I Resultant stress profile

M

P+ + =CGC


M

P+ + =CGC

Calculation of stresses at transfer (P = P0)



30

3

2

1600×10= -240×10

= -6.67 N/mm

PA

Stress at the top fibre

30

10

2

1600×10 ×433.3×416.7=2.552×10

=11.32 N/mm

tP eyI

6

10

2

233.3×10 ×416.7= -2.552×10

= -3.81N/mm

DL tM yI

∴

2

= - 6.67+11.32- 3.81

= 0.84 N/mmtcf

Stress at the bottom fibre

30

10

2

1600×10 ×433.3×583.3= -2.552×10

= -15.85 N/mm

bP eyI

6

10

2

233.3×10 ×583.3=2.552×10

= 5.33 N/mm

DL bM yI

∴2

= - 6.67 -15.85+5.33

= - 17.19 N/mmbcf

Calculation of stresses at service (P = Pe)

0

2

= 0.85

= - 5.67 N/mm

eP PA A

Stress at the top fibre

= 0.85×11.32

= 9.62

f tP eyI

6

10

2

648.0×10 ×416.7= -2.552×10

= - 10.58 N/mm

LL tM yI



∴2

= - 5.67+9.62- 3.81-10.58

= -10.44 N/mmtcf

Stress at the bottom fibre

2

= - 0.85×15.85

= - 13.47 N/mm

f bP eyI

6

10

2

648.0×10 ×583.3=2.552×10

=14.81N/mm

LL bM yI

∴

2

= - 5.67 -13.47+5.33+14.81

=1.0 N/mmbcf

The stress profiles are shown.

17.19

0.84

At transfer1.0

10.44

At service

Numeric values in N/mm2.

17.19

0.84

17.19

0.84

At transfer1.0

10.44

1.0

10.44

At service

Numeric values in N/mm2.

The allowable stresses are as follows.

For compression, fc,comp = – 18.0 N/mm2

For tension, fc,tens = 1.5 N/mm2.

Thus, the stresses are within the allowable limits.



3.4 Analysis of Members under Flexure (Part III) This section covers the following topics

• Analysis for Ultimate Strength

• Analysis of a Rectangular Section

3.4.1 Analysis for Ultimate Strength

Introduction

A prestressed member usually remains uncracked under service loads. The analysis

under service loads assumes the material to be linear elastic.

After cracking, the behaviour of a prestressed member is similar to a non-prestressed

reinforced concrete member. With increasing load, the stress versus strain behaviour of

concrete becomes non-linear. Close to the yielding of the prestressing steel, the stress

versus strain behaviour of steel also becomes non-linear.

The analysis of a prestressed member for ultimate strength is similar to that of a

reinforced concrete member. The analysis aims to calculate the ultimate moment

capacity (ultimate moment of resistance). The capacity is compared with the demand at

ultimate loads.

There is an inconsistency in the traditional analysis at the ultimate state. The force

demand is calculated based on elastic analysis, with superposition for the different load

cases using the load factors. But the capacity is calculated based on the non-linear limit

state analysis. The inconsistency is justified by the following arguments.

1) The moment versus curvature relationship is almost linear till the yielding of the

steel. The moment versus curvature relationship is also referred to as the

behaviour and is explained in Section 3.6, Analysis of Member under Flexure

(Part V).

2) The moment at yield is only slightly lower than the ultimate moment capacity.

Hence the behaviour is practically linear for most of the range of the moment.

3) The calculated moment demand for a load case based on elastic analysis is well

within the moment at yield. Hence, superposition for the load cases is applied to

find out the moment demand under combined loads.



Of course, superposition cannot be used to calculate the deflection under combined

loads.

Variation of Stress in Prestressing Steel In non-prestressed reinforced concrete members, the tension and consequently the

stress in steel increase almost proportionately with increasing moment till yielding. The

lever arm between the resultant compression and tension remains almost constant. In

prestressed concrete members, the tension and consequently the stress in prestressing

steel increase slightly with increasing moment till cracking of concrete. The increase in

moment changes the lever arm significantly. This is explained in Section 3.2, Analysis of

Member under Flexure (Part I). After cracking, the stress in prestressing steel

increases rapidly with moment.

The following sketch explains the variations of the stress in prestressing steel (fp) with

increasing load. The variations are shown for bonded and unbonded tendons. After the

prestress is transferred while the member is supported at the ends, the stress will tend

to increase from the value after losses (fp0) due to the moment under self weight.

Subsequently the stress will tend to drop due to the time dependent losses such as from

creep, shrinkage and relaxation. The losses of prestress are covered in Section 2.3,

Losses in Prestress (Part III). The effective prestress after time dependent losses is

denoted as fpe.

Due to the moment under service loads, the stress in the prestressing steel will slightly

increase from fpe. The increase is more at the section of maximum moment in a bonded

tendon as compared to the increase in average stress for an unbonded tendon. The

stress in a bonded tendon is not uniform along the length. Usually the increase in

stress is neglected in the calculations under service loads. If the loads are further

increased, the stress increases slightly till cracking.

After cracking, there is a jump of the stress in the prestressing steel. Beyond that, the

stress increases rapidly with moment till the ultimate load. At ultimate, the stress is

represented as fpu. Similar to the observation for pre-cracking, the average stress in an

unbonded tendon is less than the stress at the section of maximum moment for a

bonded tendon.



Service load

Load

Bonded

Unbondedfp0

fpe

Self weight

Cracking load

Ultimate load

Losses

fpu

fp

Service load

Load

Bonded

Unbondedfp0

fpe

Self weight

Cracking load

Ultimate load

Losses

fpu

fp

Figure 3-4.1 Variation of stress in prestressing steel

The above sketch assumes that the section is failing in flexure. Other types of failure

are not considered.

Conditions at Ultimate Limit State In the limit states method of analysis, the limit state of collapse (ultimate state) of a

member under flexure is defined as the state when the extreme concrete compressive

strain reaches a value of 0.0035. At ultimate, let the extreme concrete compressive

strain be denoted as εcu. Thus, εcu = 0.0035.

Depending on the amount of prestressing steel, a section can be under-reinforced or

over-reinforced. For an under- reinforced section, the amount of prestressing steel is

less and the steel yields before the extreme concrete strain reaches 0.0035. For an

over-reinforced section, the amount of steel is high and the steel does not yield at

ultimate. The transition situation is called a balanced condition. The strain profiles

across the depths of prestressed flexural members (up to the depth of CGS) for the

three situations are shown below.



Under-reinforced Balanced Over-reinforced

εcuεcu εcu

∆εp

εpu > εpu,bal

∆εp ∆εp

εpu = εpu,bal εpu < εpu,bal

Under-reinforced Balanced Over-reinforced

εcuεcu εcu

∆εp

εpu > εpu,bal

∆εp ∆εp

εpu = εpu,bal εpu < εpu,bal

Figure 3-4.2 Strain profiles along the depths of three prestressed members


εpu = strain in the prestressing steel at the level of CGS at ultimate

condition

εpu,bal = strain in the steel for a balanced section.

The strain difference (∆εp) is the strain in the prestressed tendons when the adjacent

concrete has zero strain (εc = 0). The strain difference gets locked during the transfer of

prestress. The value can be determined as follows.

For pre-tensioned members, the strain difference gets locked when the tendons are cut.

The strain difference at that instant is given as follows.

∆εp = εpi – 0 (3-4.1a)

Here,

εpi = strain in tendons just before transfer

εc = strain in concrete is zero.

For post-tensioned members, the strain difference gets locked when the tendons are

anchored. The strain difference at that instant is given as follows.

∆εp = εp0 – εc0 (3-4.1b)

Here,

εp0 = strain in tendons due to P0, the prestress after transfer

εc0 = strain in concrete due to P0.

In general at any load stage,

∆εp = εpe – εce (3-4.1c)



Here,

εpe = strain in tendons due to Pe, the prestress at service

εce = strain in concrete due to Pe.

As mentioned under material properties, the prestressing steel does not have a definite

yield point. The 0.2% proof stress is defined when the steel reaches an inelastic strain

of 0.2%. Hence, unlike reinforced concrete, the transition from under-reinforced to over-

reinforced section is gradual and there is no definite balanced condition. IS:1343 - 1980

does not explicitly enforce an under-reinforced section. But the IRS Concrete Bridge Code requires that the strain in the outermost tendon should not be less than the

following.

pk

p

f+

E0.87

0.005

The above value can be considered to be the strain in the steel at balanced condition.

Assumptions for Analysis The analysis of members under flexure for ultimate strength considers the following.

1) Plane sections perpendicular to the axis of the member remain plane till the

ultimate state.

2) Perfect bond is retained between concrete and prestressing steel for bonded

tendons.

3) Tension in concrete is neglected.

4) The design stress versus strain curves of concrete and steel are considered.

The methods of analysis will be presented for three types of sections.

1) Rectangular section: A rectangular section is easy to cast, but it is not an efficient

section.

2) Flanged section: A precast flanged section, with flanges either at top or bottom

needs costlier formwork. But the section is efficient in flexure. A flanged section

can also be made of precast web and cast-in-place slab.

3) Partially prestressed section: A section in a member containing both prestressed

and non-prestressed reinforcement.



3.4.4 Analysis of a Rectangular Section The following sketch shows the beam cross section, strain profile, stress diagram and

force couple at the ultimate state.

d

εcu=0.0035

∆εpεpu

fpu

xu

0.447fck0.42xu

Cu

Tu

Strain Stress Force

b

Cross-section

Ap

d

εcu=0.0035

∆εpεpu

fpu

xu

0.447fck0.42xu

Cu

Tu

Strain Stress Force

b

Cross-section

Ap

Figure 3-4.3 Sketches for analysis of a rectangular section

The variables in the above figure are explained.

b = breadth of the section

d = depth of the centroid of prestressing steel (CGS)

Ap = area of the prestressing steel

∆εp = strain difference

xu = depth of the neutral axis at ultimate

εpu = strain in prestressing steel at the level of CGS at ultimate

fpu = stress in prestressing steel at ultimate

The stress block in concrete is derived from the constitutive relationship for concrete.

The relationship is explained in Section 1.6, Concrete (Part II). The compressive force

in concrete can be calculated by integrating the stress block along the depth. The

stress in the tendon is calculated from the constitutive relationship for prestressing steel.

The relationship is explained in Section 1.7, Prestressing Steel.

In the force diagram,

(3-4.2) u ck

u p pu

C = f x bT = A f

0.36 u

(3-4.3) The strengths of the materials are denoted by the following symbols.

fck = characteristic compressive strength of concrete

fpk = characteristic tensile strength of prestressing steel



For analysis of a prestressed section, three principles of mechanics are used. First, the

equilibrium relates the external applied forces with the internal forces. Second, the

compatibility condition relates the strain in the prestressing steel with the strain in

concrete at the level of CGS. This also considers the first two assumptions given in the

previous section. The third principle involves the constitutive relationships of the

materials.

Based on the above principles of mechanics, the following equations are derived.

1) Equations of equilibrium

The first equation states that the resultant axial force is zero. This means that the

compression and the tension in the force couple balance each other.

⇒

⇒

∑

u u

p pu ck u

F =T = CA f = f x b

0

0.36

(3-4.4)

The second equation relates the ultimate moment capacity (MuR) with the internal

couple in the force diagram.

( )uR u u

p pu u

M =T (d - x )= A f d - x

0.420.42

(3-4.5)

2) Equation of compatibility

The depth of the neutral axis is related to the depth of CGS by the similarity of the

triangles in the strain diagram.

u

pu p

x =d + ε - ∆ε

0.00350.0035

(3-4.6)

3) Constitutive relationships

a) Concrete

The constitutive relationship for concrete is considered in the expression Cu =

0.36fckxub. This is based on the area under the design stress-strain curve for concrete

under compression.



b) Prestressing steel

(3-4.7) ( )pu pf = F ε u

The function F(εpu) represents the design stress-strain curve for prestressing steel under

tension.

The known variables in an analysis are: b, d, Ap, ∆εp, fck, fpk.

The unknown quantities are: xu, MuR, εpu, fpu.

The objective of the analysis is to find out MuR, the ultimate moment capacity.

The simultaneous equations 3-4.1 to 3-4.7 can be solved iteratively. This procedure of

analysis is called the strain compatibility method. The steps are as follows.

1) Assume xu .

2) Calculate εpu by rearranging the terms of Eqn. 3-4.6.

3) Calculate fpu from Eqn. 3-4.7.

4) Calculate Tu from Eqn. 3-4.3.

5) Calculate Cu from Eqn. 3-4.2.

If Eqn. 3-4.4 (Tu = Cu) is not satisfied, change xu.

If Tu < Cu decrease xu. If Tu > Cu increase xu.

6) Calculate MuR from Eqn. 3-4.5.

The capacity MuR can be compared with the demand under ultimate loads.

In the strain compatibility method, the difficult step is to calculate xu and fpu. IS:1343 -1980 allows to calculate these variables approximately from Table 11, Appendix B,

based on the amount of prestressing steel. The later is expressed as a prestressed

reinforcement index ωp.

p pk

pck

A fω =

bdf (3-4.8)

Table 11 is reproduced as Table 3-4.1 which is applicable for pre-tensioned and bonded

post-tensioned beams. The values of fpu and xu are given as fpu/(0.87fpk) and xu/d,

respectively.



Table 3-4.1 Values of x and f for pre-tensioned and bonded post-tensioned

rectangular beams (Table 11, IS:1343 - 1980u pu

)

fpu/(0.87fpk) xu/d

ωp Pre-tensioned Bonded post-

tensioned

Pre-tensioned Bonded post-

tensioned

0.025 1.0 1.0 0.054 0.054

0.05 1.0 1.0 0.109 0.109

0.10 1.0 1.0 0.217 0.217

0.15 1.0 1.0 0.326 0.316

0.20 1.0 0.95 0.435 0.414

0.25 1.0 0.90 0.542 0.488

0.30 1.0 0.85 0.655 0.558

0.40 0.9 0.75 0.783 0.653

The values of fpu/(0.87fpk) and xu/d from Table 3-4.1 are plotted in Figures 3-4.4 and 3-

4.5, respectively. It is observed that with increase in ωp, fpu reduces (beyond certain

values of ωp) and xu increases. This is expected because with increase in the amount

and strength of the steel, the stress in steel drops and the depth of the neutral axis

increases to maintain equilibrium.

0

0.2

0.4

0.6

0.8

1

1.2

0 0.1 0.2 0.3 0.4ω p

f pu /

0.8

7 f p

k

Pre-tensioned Post-tensioned (bonded)

Figure 3-4.4 Variation of fpu with respect to ωp (Table 3-4.1)



0

0.3

0.6

0.9

0 0.1 0.2 0.3 0.4

ω p

xu /

d

Pre-tensioned Post-tensioned (bonded)

Figure 3-4.5 Variation of xu with respect to ωp (Table 3-4.1)

Thus given the value of ωp for a section, the values of fpu and xu can be approximately

calculated from the above tables.

Example 3-4.1

A prestressed concrete beam produced by pre-tensioning method has a rectangular cross-section of 100 mm × 160 mm (b × h). It is prestressed with 10 numbers of straight 2.5 mm diameter wires. Each wire is stressed up to a load of 6.8 kN. The design load versus strain curve for each wire is given in a tabular

form. The grade of concrete is M 40. The value of ∆εp is 0.0073.

Estimate the ultimate flexural strength of the member by the strain compatibility method.



100

160

40

CGC

Values in mm

Cross-section of member

100

160

40

CGC

Values in mm

100

160

40

100

160

40

CGC

Values in mm

Cross-section of member Design load (P) versus strain (εp) values for the prestressing wire are given for the range under consideration.

εp P (kN)

0.006 5.4

0.008 7.6

0.010 9.0

0.012 10.0

0.014 10.7

Solution

Strain difference

∆εp = 0.0073

The effective depth of the CGS (d ) is 120 mm.

The strain compatibility method is shown in a tabular form. Here,

Pu = load in a single wire obtained from the table

Tu = 10 × Pu , for the ten wires.



xu

(mm)

xu/d εpu–∆εp

(3-4.6)

εpu Pu

(kN)

(Table)

Tu

(kN)

Cu

(kN)

(3-4.2)

Checking

(3-4.4)

60 0.5 0.0035 0.0108 9.4 94.0 86.4 Tu > Cu

65 0.54 0.0030 0.0103 9.1 91.0 93.6 Tu < Cu

63.5 0.53 0.0031 0.0104 9.15 91.5 91.4 Tu ≈ Cu

The ultimate flexural strength is given as follows.

= ( -0.42 )=91.5 (120.0-0.42×63.5)kNmm=8.5kNm

uR u uM T d x



3.5 Analysis of Members under Flexure (Part IV) This section covers the following topics.

• Analysis of a Flanged Section

3.5.1 Analysis of a Flanged Section

Introduction

A beam can have flanges for flexural efficiency. There can be several types of flanged

section.

1) A precast or cast-in-place flanged section, with flanges either at top or bottom or

at both top and bottom.

2) A composite flanged section is made of precast web and cast-in-place slab.

The following figures show different types of flanged sections.

Double T-sectionT-section

Single box section Double box section

Double T-sectionDouble T-sectionT-sectionT-section

Single box section Double box sectionDouble box section

L-section Inverted T-section I-girderL-section Inverted T-section I-girder Figure 3-5.1 Examples of precast flanged sections



T TT T T T

Box section Composite beam-slab T-section

T TT TT T T TT T

Box section Composite beam-slab T-section Figure 3-5.2 Examples of composite flanged sections

The analysis of a flanged section for ultimate strength is different from a rectangular

section when the flange is in compression. If the depth of the neutral axis from the edge

under compression is greater than the depth of the flange, then the section is treated as

a flanged section. In the following figure, the first strain profile shows that the depth of

the neutral axis (xu) is greater than the depth of the flange (Df). The section is treated

as a flanged section.

The second strain profile shows that xu is less than Df. In this situation, the section can

be treated as a rectangular section.

bf

bw

Df

dxu

Strain profile(xu > Df)

Cross-section

Ap

Strain profile(xu < Df)

xu

bf

bw

Df

dxu

Strain profile(xu > Df)

Cross-section

Ap

Strain profile(xu < Df)

xu

Figure 3-5.3 Two possibilities of strain profile in a flanged section

The effective width or breadth of the flange (bf) is determined from the span of the beam,

breadth of the web (bw) and depth of the flange (Df) as per Clause 23.1.2, IS:456 - 2000.

Analysis of a Flanged Section The following sketch shows the beam cross-section, strain profile, stress diagram and

force couples at the ultimate state. The following conditions are considered.

1) xu > Df : This requires an analysis for a flanged section.



2) Df ≤ (3/7) xu: This ensures that the compressive stress is constant at 0.447fck

along the depth of the flange.

bf

bw

Df

dxu

∆εpεpu

Strain

0.447fck

fpu

0.42xu CuwCuf

TufTuw

0.5Df

+

Stress ForceCross-section

Ap

0.0035bf

bw

Df

dxu

∆εpεpu

Strain

0.447fck

fpu

0.42xu CuwCuf

TufTuw

0.5Df

+

Stress Force

0.447fck

fpu

0.42xu CuwCuf

TufTuw

0.5Df

+

Stress ForceCross-section

Ap

0.0035

Figure 3-5.4 Sketches for analysis of a flanged section


bf = breadth of the flange

bw = breadth of the web

Df = depth of the flange



∆εp = strain difference in the prestressing steel when strain in concrete is zero




The strain difference (∆εp) is further explained in Section 3.4, Analysis of Member under

Flexure (Part III).

In the sketch, the tensile force is decomposed into two components. The first

component (Tuw) balances the compressive force carried by the web, including the

portion of the flange above web (Cuw). Thus Tuw= Cuw. The second component (Tuf)

balances the compressive force carried by the outstanding portion of the flange (Cuf).

Thus Tuf = Cuf.








The expressions of the forces are as follows.

(3-5.1) uw ck u w

uf ck f w f

uw pw pu

uf pf pu

C = f x bC = 7f (b - b )DT = A fT = A f

0.360.44 (3-5.2)

(3-5.3) (3-5.4)

The strengths of the materials are denoted by the following symbols.

Apf = part of Ap that balances compression in the outstanding flanges

Apw = part of Ap that balances compression in the web



Based on the principles of mechanics (as explained under the Analysis of a Rectangular

Section in Section 3.4, Analysis of Member Under Flexure (Part III)), the following

equations are derived.




( ) ( )

⇒

⇒

⇒

∑

u u

uw uf uw uf

pw pf pu ck u w ck f w

F =T = CT +T = C +C

A + A f = f x b + f b - b D

0

0.36 0.447

(3-5.5) f



( ) ( )( ) ( )

uR uw u uf f

pw pu u pf pu f

M =T d - x +T d - D= A f d - x + A f d - D

0.42 0.50.42 0.5

(3-5.6)

From Tuf = Cuf and Eqns. (3-5.2) and (3-5.4), Apf is given as follows. The calculation of

Apw from Ap and Apf is also shown.



(3-5.7) ( )ck f w fpf

pu

pw p pf

f b - b DA =f

A = A - A

0.447

(3-5.8)

2) Equation of compatibility

The depth of the neutral axis is related to the depth of CGS by the similarity of the

triangles in the strain diagram.

u

pu p

x =d + ε - ∆ε

0.00350.0035

(3-5.9)


a) Concrete

The constitutive relationship for concrete is considered in the expressions of Cuw and

Cuf. This is based on the area under the design stress-strain curve for concrete under

compression.


( )pu pf = F ε (3-5.10) u

The function F(εpu) represents the design stress-strain curve for the type of prestressing

steel used.

The known variables in an analysis are: bf, bw, Df, d, Ap, ∆εp, fck and fpk.

The unknown quantities are: Apf, Apw, MuR, xu, εpu and fpu.

The objective of the analysis is to find out MuR , the ultimate moment capacity. The

simultaneous equations 3-5.1 to 3-5.10 can be solved iteratively.

The steps of the strain compatibility method are as follows.

1) Assume xu = Df.

2) The calculations are similar to a rectangular section, with b = bf.

3) If Tu > Cu, increase xu. Treat the section as a flanged section.

4) Calculate εpu from Eqn. (3-5.9).

5) Calculate fpu from Eqn. (3-5.10).

6) Calculate Apf and Apw from Eqn. (3-5.7) and Eqn. (3-5.8), respectively.



7) Calculate Cuw, Cuf, Tuw and Tuf from Eqns. (3-5.1) to (3-5.4). If Eqn. (3-5.5)

(Tu = Cu) is not satisfied, iterate with a new value of xu, till convergence.

8) Calculate MuR from Eqn. (3-5.6).


In the strain compatibility method, the difficult step is to calculate xu and fpu. Similar to

the rectangular section, an approximate analysis can be done based on Table 11 and Table 12, Appendix B, IS:1343-1980. The tables are reproduced in Table 3-4.1 and

Table 3-4.2, respectively, in Section 3.4, Analysis of Member under Flexure (Part III).

The values of xu and fpu are available in terms of a reinforcement index ωpw.

pw pkpw

w ck

A fω =

b df (3-5.11)

Note that the index is calculated based on Apw instead of Ap. The calculation of Apw is

from Eqn. (3-5.8). But Apf depends on fpu, which is unknown. Hence, an iterative

procedure is required.

The steps are as follows.

1) Assume fpu = 0.87fpk.

2) Calculate Apf and Apw from Eqn. (3-5.7) and Eqn. (3-5.8), respectively.

3) Calculate ωpw.

4) Calculate fpu from Table 11 or Table 12.

Compare the calculated value of fpu with the assumed value. Repeat steps 1 to 4 till

convergence.

5) Calculate MuR.

If Df > (3/7) xu, the flange depth is larger than the depth of constant compressive stress.

An equivalent depth of the flange is defined as follows.

yf = 0.15xu + 0.65Df (3-5.12)

The equivalent depth yf is substituted for Df in the expression of MuR.



Example 3-5.1 A bonded post-tensioned concrete beam has a flanged cross-section as shown. It is prestressed with tendons of area 1750 mm2 and effective prestress of 1100 N/mm2. The tensile strength of the tendon is 1860 N/mm2. The grade of concrete is M60. Estimate the ultimate flexural strength of the member by the approximate method of IS:1343 - 1980.

175

460

900140

175

550

460115

Values are in mm.


175

460

900140

175

550

460115

Values are in mm.


Solution

Effective depth d = 900 – 115

= 785 mm

Assume xu = Df = 175 mm. Treat as a rectangular section, with b = bf = 460 mm.

Reinforcement index

P Pk

Pck

A fωbdf

=

1750×1860=460×785×60

= 0.15



Tu = Apfpu

= 1750 × 1618

= 2831.5 kN

Cu = 0.36fckxubf

= 0.36 × 60 × 175 × 460

= 1738.8 kN

Tu > Cu. Hence xu > Df

⇒ Treat as a flanged section

Assume fpu = 0.87fpk

= 1618 N/mm2

Calculate Apf and Apw

ck f w fpf

pu

f b b DAf

2

0.447 ( - )=

0.447×60×(460 -140)×175=1618

= 934 mm

pwA2

=1750 - 934

= 816 mm

Reinforcement index

pw pkpw

w ck

A fω =

b df816×1860=

140×785×60= 0.23

From Table 11,

pu

pk

pu

fff

2

= 0.920.87

= 0.92×0.87×1860

=1489 N/mm



2nd iteration

fpu = 1489 N/mm2


ent index

wpw is same as after 2nd iteration. Hence, the values of fpu, Apf and Apw

have converged.

pfA

2

0.447×60×(460 -140)×175=1489

=1015 mm

pwA2

=1750 -1015

= 735 mm

Reinforcement index

pwω 735×1860=

140×785×60= 0.21

From Table 11,

pu

pk

pu

fff

2

= 0.940.87

= 0.94×0.87×1860

=1521 N/mm

3rd iteration

fpu = 1521N/mm2


pfA

2

0.447×60×(460 -140)×175=1521

= 994 mm

pwA2

=1750 - 994

= 756 mm

Reinforcem

pwω 756×1860=

140×785×60= 0.21

The value of



Ultimate flexural strength

he ultima

(

T te flexural strength is given as follows.

) ( )uR uw u uf fM T x T d D= d- 0.42 + - 0.5

uwT d x A f d x( - 0.42 ) = ( - 0.42 )

=1054.5 kNm

uRM =1054.5+ 739.9

u pw pu u

= 756×1521×(785 - 0.42×337)= 739.9 kNm

uf f pf pu fT d D A f d D( - 0.5 ) = ( - 0.5 )= 994×1521×(785 - 0.5×175)

=1794.4 kNm



3.6 Analysis of Members under Flexure (Part V) This section covers the following topics.

• Analysis of Partially Prestressed Section

• Analysis of Unbonded Post-tensioned Beam

• Analysis of Behaviour

3.6.1 Analysis of Partially Prestressed Section

Introduction

The analyses that are presented in the earlier sections, are for members which do not

have any conventional non-prestressed reinforcement. Usually conventional

reinforcement is provided in addition to the prestressing steel. When this reinforcement

is considered in the flexural capacity, the section is termed as a partially prestressed

section.

The reasons for using a partially prestressed section are as follows.

1) The section is economical.

2) The cambering is less compared to an equivalent section without conventional

reinforcement.

3) The ductility is more in a partially prestressed section.

4) Any reversal of moments (for example, due to earthquake) is not detrimental as

compared to an equivalent section without conventional reinforcement.

Analysis A partially prestressed section can be either rectangular or flanged. A section can be

doubly reinforced with reinforcement near the compression face.

Here, the equations for a doubly reinforced rectangular section are given.

The following sketch shows the beam cross section, strain profile, stress diagram and

force couples at the ultimate state.



0.447fck

fs

b

d xu

0.42xu

∆εp

StressStrain ForceCross-section

dp

d’As’

As

Ap

0.0035

εs’

εpu

εs

fpu

fs’Cs’Cc

Tp

Ts

0.447fck

fs

b

d xu

0.42xu

∆εp


dp

d’As’

As

Ap

0.0035

εs’

εpu

εs

fpu

fs’Cs’Cc

Tp

Ts

Figure 3-6.1 Sketches for analysis of a partially prestressed section



d = depth of the centroid of the reinforcing steel (tension side)

d’ = depth of the centroid of the reinforcing steel (compression side)

dp = depth of the centroid of prestressing steel (CGS)

As = area of the reinforcing steel (tension side)

As’ = area of the reinforcing steel (compression side)


∆εp = strain difference in the prestressing steel when strain in concrete is zero


εs = strain in reinforcing steel (tension side) at ultimate

εs’ = strain in reinforcing steel (compression side) at ultimate


fs = stress in reinforcing steel (tension side) at ultimate

fs’ = stress in reinforcing steel (compression side) at ultimate

fpu = stress in prestressing steel at ultimate.

The strain difference (∆εp) is further explained in Section 3.4, Analysis of Member under

Flexure (Part III).









Cs’ = As’fs’ (3-6.1) Cc = 0.36fckxub (3-6.2) Tp = Apfpu (3-6.3) Ts = Asfs (3-6.4)

The strengths of the materials are denoted by the following symbols.



fy = characteristic yield stress of reinforcing steel

Based on the principles of mechanics (as explained under the Analysis of a Rectangular

Section in Section 3.4, Analysis of Member Under Flexure (Part III)), the following

equations are derived.




∑⇒⇒

⇒

u u

p s c s

' 'p pu s s ck u s s

F =T = CT +T = C +C'

A f + A f = f x b+ A f

0

0.36

(3-6.5)



( ) ( ) ( )( ) ( ) (

puR s p c p u s pA

's s p ck u p u s s p

M =T d - d +C d - x +C' d - d'

= A f d - d + f x b d - x + A' f d - d'

0.42

0.36 0.42

) (3-6.6)

2) Equations of compatibility

For each layer of steel there is a compatibility equation. If there are distributed

reinforcing bars in several layers and the spacing between the layers is large, then the

use of compatibility equation for each layer is more accurate than the use of one

compatibility equation for the centroid of the layers. The following equations are

developed based on the similarity of the triangles in the strain diagram.



(3-6.7) u

p pu p

u s

u'

u s

u

x =d + ε - ∆εd - x ε=

x

x - d' ε=x

0.00350.0035

0.0035

0.0035

(3-6.8)

(3-6.9)


a) Concrete

The constitutive relationship for concrete is considered in the expressions of Cc. This is

based on the area under the design stress-strain curve for concrete under compression.


( )pu pf = F ε1 (3-6.10) u

c) Reinforcing steel ( )( )

s s

' 's s

f = F ε

f = F ε2

3

(3-6.11) (3-6.12)

For mild steel

(3-6.13) 's y

s y

f = ff = f

0.870.87 (3-6.14)

The known variables in an analysis are: b, d, d’, dp, As, As’, Ap, ∆εp, fck, fy and fpk.

The unknown quantities are: MuR, xu, εs, εs’, εpu, fs, fs’ and fpu.

The objective of the analysis is to find out MuR, the ultimate moment capacity.

The previous equations can be solved by the strain compatibility method as

discussed for the fully prestressed rectangular section.

1) Assume xu.

2) Calculate εpu from Eqn. 3-6.7.

3) Calculate fpu from Eqn. 3-6.10.

4) Calculate Tp from Eqn. 3-6.3.

5) Calculate εs from Eqn. 3-6.8.

6) Calculate fs from Eqn. 3-6.11.

7) Calculate Ts from Eqn. 3-6.4.

8) Calculate εs’ from Eqn. 3-6.9.



9) Calculate fs’ from Eqn. 3-6.12.

10) Calculate Cs’ from Eqn. 3-6.1.

11) Calculate Cc from Eqn. 3-6.2.

If Eqn. 3-6.5 (Tu = Cu) is not satisfied, change xu.

If Tu < Cu decrease xu. If Tu > Cu increase xu.

12) Calculate MuR from Eqn. 3-6.6.


3.6.2 Analysis of Unbonded Post-tensioned Beam

In an unbonded post-tensioned beam, the ducts are not grouted. Hence, there is no

strain compatibility between the steel of the tendons and the concrete at a section. The

compatibility is in terms of deformation over the length of the member.

A sectional analysis is not possible. The analysis involves integrating the strain in

concrete to calculate the deformation over the length of the member.

The equation of compatibility is given as follows.

∆p = ∆cp (3-6.15) Here,

∆p = deformation of the tendon

∆cp = deformation of the concrete at the level of prestressing steel (CGS).

The change in stress in steel (∆fp) at ultimate is determined from ∆p. The stress in steel

at ultimate is given by the sum of the effective prestress (fpe) and ∆fp.

fpu = fpe + ∆fp (3-6.16)

The value of fpu is less than that for a bonded tendon. The ultimate moment is given by

the following equation. ( )uR p pu uM = A f d - x0.42 (3-6.17)

The rigorous method of evaluating fpu, based on deformation compatibility, is difficult.

IS:1343 - 1980 allows to calculate fpu and xu approximately from Table 12, Appendix B,

based on the amount of prestressing steel. The later is expressed as the reinforcement

index ωp = Apfpk / bdfck. Table 12 is reproduced as Table 3-6.1 which is applicable for



unbonded post-tensioned beams. The values of fpu and xu are given as fpu/fpe and Xu/d,

respectively. The effective prestress (after the losses) in a tendon is represented as fpe.

Table 3-6.1 Values of x and f for unbonded post-tensioned rectangular beams

(Table 12, IS:1343 - 1980)u pu

fpu/fpe

For values of L/d

xu/d

For values of L/d ωp

30 20 10 30 20 10

0.025 1.23 1.34 1.45 0.10 0.10 0.10

0.05 1.21 1.32 1.45 0.16 0.16 0.18

0.10 1.18 1.26 1.45 0.30 0.32 0.36

0.15 1.14 1.20 1.36 0.44 0.46 0.52

0.20 1.11 1.16 1.27 0.56 0.58 0.64

The values of fpu/fpe and xu/d from Table 3-6.1 are plotted in Figures 3-6.2 and 3-6.3,

respectively. It is observed that with increase in ωp, fpu reduces and xu increases. This

is expected because with increase in the amount and strength in the steel, the stress in

steel drops and the depth of the neutral axis increases to maintain equilibrium.

1

1.25

1.5

0 0.05 0.1 0.15 0.2

ω p

f pu/

f pe 30

2010

l /d

1

1.25

1.5

0 0.05 0.1 0.15 0.2

ω p

f pu/

f pe 30

2010

l /d

Figure 3-6.2 Variation of fpu with respect to wp (Table 3-6.1)



0

0.2

0.4

0.6

0 0.05 0.1 0.15 0.2

ω p

x u /

d

302010

l /d

0

0.2

0.4

0.6

0 0.05 0.1 0.15 0.2

ω p

x u /

d

302010

l /d

Figure 3-6.3 Variation of xu/d with respect to wp (Table 3-6.1)

Thus given the value of ωp for a section, the values of fpu and xu can be approximately

calculated from the above tables.

3.6.3 Analysis of Behaviour

The analysis of behaviour refers to the determination of the complete moment versus

curvature behaviour of the section. The analyses at transfer, under service loads and

for ultimate strength correspond to three instants in the above behaviour.

The curvature (φ) is defined as the gradient of the strain profile.

c cpε + εφ=

d

(3-6.18) Here,

εc = extreme concrete compressive strain

εcp = strain in concrete at the level of prestressing steel (CGS)

d = depth of the CGS.

The following sketch shows the curvature (φ) in the strain profile.



d

εc

εcp

Strain profile

b

Cross-section

Ap

φd

εc

εcp

Strain profile

b

Cross-section

Ap

φ

Figure 3-6.4 Definition of curvature

The analysis of behaviour involves the following three principles of mechanics.

1) Equilibrium of internal forces with the external loads at any point of the

behaviour. There are two equilibrium equations.

a) Force equilibrium equation

b) Moment equilibrium equation.

The internal forces in concrete and steel are evaluated based on the respective

strains, cross-sectional areas and the constitutive relationships.

2) Compatibility of the strains in concrete and in steel for bonded tendons. This

assumes a perfect bond between the two materials. For unbonded tendons, the

compatibility is in terms of deformation.


The relationships are developed based on the material properties.

The equilibrium and compatibility equations and the constitutive relationships can be

solved to develop the moment versus curvature curve for a section.

The following plot shows the curves for a prestressed section and a non-prestressed

section. The two sections are equivalent in their ultimate flexural strengths.



Curvature

Moment

Prestressed section Non-prestressed section

Ultimate strength

Service load level

Cracking load levels

Curvature

Moment

Prestressed section Non-prestressed section

Ultimate strength

Service load level

Cracking load levels

Figure 3-6.5 Moment versus curvature curves

From the previous plot, the following can be inferred.

1) Prestressing increases the cracking load. This leads to the following benefits.

• Reduction of steel corrosion

⇒ Increase in durability.

• Full section is utilised

⇒ Higher moment of inertia (higher stiffness)

⇒ Less deformations (improved serviceability).

• Increase in shear capacity.

2) Prestressing shifts the curve from the origin.

• For the prestressed member, there is a negative curvature causing camber in

absence of external moment.

• A certain amount of external moment is required to straighten the member.

3) For a given moment, the curvature of the prestressed member is smaller.

• Prestressing reduces curvature at service loads.

4) For a given reverse moment, the curvature of the prestressed member is

larger.

• Prestressing is detrimental for the response under reverse moment.

5) The ultimate strength of the prestressed member is lower.

• Prestressing is detrimental under reverse moment.

6) For a partially prestressed section with the same ultimate strength, the

moment versus curvature curve will lie in between the curves for

prestressed and non-prestressed sections.



Ductility

The ductility is a measure of energy absorption. For beams, the curvature ductility (µ) is

defined as

(3-6.19) ϕϕ

= u

y

µ

Here,

φu = curvature at ultimate

φy = curvature at yield.

For prestressed beams, φy can be defined corresponding to a plastic strain of 0.002 in

the prestressing tendons. It has been observed that the ductility of prestressed beams

is less than that in reinforced concrete beams.

In design of members for seismic forces, ductility is an important requirement. In

addition, seismic forces lead to reversal of moments near the supports of beams in a

moment resisting frame. Hence, prestressing of beams in a moment resisting frame is

not recommended in seismic areas.

Experimental Investigation The behaviour of a beam and its ultimate strength can be determined by testing

prototype specimens. The tests can be conducted under static or dynamic loads.

Testing also helps to check the performance of the anchorage units.

The following photo shows the set-up for testing a prototype bridge girder.

Figure 3-6.6 Set-up for testing a bridge girder



4.1 Design of Members This section covers the following topics

• Calculation of Demand

• Design of Sections for Axial Tension

Introduction

The design of prestressed concrete members can be done by the limit states method as

given in Section 4 of IS:1343 - 1980.

First, the force demand in a member under the design loads is determined from a

structural analysis. A preliminary size of the member is assumed for analysis. Next, the

member is designed to meet the demand. If necessary, another cycle of analysis and

design is performed.

The following material explains the calculation of the demand in a member under the

design loads.

4.1.1 Calculation of Demand

In the limit states method, the design loads are calculated from the characteristics loads

by multiplying them with load factors (γf). Several types of loads are considered to act

together under the selected load combinations. The load factors are included in the

load combinations as weightage factors.

The demand in a member for a particular type of load is obtained from the analysis of

the structure subjected to the characteristic value of the load.

The demands for the several load types are then combined under the load combinations,

based on the principle of superposition.

Characteristics Loads For dead loads, a characteristic load is defined as the value which has a 95%

probability of not being exceeded during the life of the structure. This concept assumes

a normal distribution of the values of a particular dead load. In the following figure, the



shaded area above the characteristic value represents 5% probability of exceedance of

the load in the design life of the structure.

Characteristic value

Mean

Frequency

Values of load

5% probabilityof exceedence


Mean

Frequency

Values of load


Figure 4-1.1 Idealised normal distribution for a dead load

For live load, wind load and earthquake load, a characteristic load is defined based on

an extreme value distribution. For example, the characteristic wind load is defined as

the value which has a 98% probability of not being exceeded during a year.


Frequency

Annual maximum mean wind speed



Frequency

Annual maximum mean wind speed


Figure 4-1.2 Extreme value distribution

The characteristics loads can be obtained from IS:875 - 1987 (Code of Practice for

Design Loads for Buildings and Structures) and IS:1893 - 2002 (Criteria for Earthquake

Resistant Design of Structures) as follows.



Table 4-1.1 Codes covering information of loads

Type of load Code

Dead load (DL) IS:875 - 1987 Part 1

Live (imposed) load (LL) IS:875 - 1987 Part 2

Wind load (WL) IS:875 - 1987 Part 3

Snow Load (SL) IS:875 - 1987 Part 4

Earthquake load (EL) IS:1893 - 2002 Part 1

For special loads, there are some guidelines in IS: 875 - 1987, Part 5. In addition,

specialised literature may be referred to for these loads. The special loads are listed

below.

Temperature

Hydrostatic

Soil pressure

Fatigue

Accidental load

Impact and collision

Explosions

Fire

For special situations, the loads are determined from testing of prototype specimens.

Dynamic load tests, wind tunnel tests, shake table tests are some types of tests to

determine the loads on a structure. Finite element analysis is used to determine the

stresses due to concentrated forces and dynamic loads.

Load Factors and Load Combinations The load factors and the combinations of the various types of loads are given in Table 5

of IS:1343 - 1980. The following are the combinations for the ultimate condition.

1.5 (DL + LL)

1.2 (DL + LL ± WL)

1.2 (DL + LL ± EL)

1.5 (DL ± EL)



1.5 (DL ± WL)

0.9 DL ± 1.5 EL

The load combinations for service conditions are as follows.

DL + LL

DL + 0.8 (LL ± EL)

DL ± EL

DL ± WL

Analysis of Structures Regarding analysis of structures, IS:1343 - 1980 recommends the same procedure as

stated in IS:456 - 2000. A structure can be analysed by the linear elastic theory to

calculate the internal forces in a member subjected to a particular type of load.

Design of Members There can be more than one way to design a member. In design, the number of

unknown quantities is larger than the number of available equations. Hence, some

quantities need to be assumed at the beginning. These quantities are subsequently

checked.

The member can be designed either for the service loads or, for the ultimate loads. The

procedure given here is one of the possible procedures. The design is based on

satisfying the allowable stresses under service loads and at transfer. Initially, a

lumpsum estimate of the losses is considered under service loads. After the first round

of design, detailed computations are done to check the conditions of allowable stresses.

Precise values of the losses are computed at this stage. The section is then analysed

for the ultimate capacity. The capacity should be greater than the demand under

ultimate loads to satisfy the limit state of collapse.



4.1.2 Design of Sections for Axial Tension

Introduction Prestressed members under axial loads only, are uncommon. Members such as

hangers and ties are subjected to axial tension. Members such as piles may have

bending moment along with axial compression or tension.

Design of Prestressing Force First, a preliminary dimension of the member is selected based on the architectural

requirement. The prestressing force at transfer (P0) should be such that the

compressive stress in concrete is limited to the allowable value. At service, the

designed prestressing force (Pe) should be such that the tensile stress in concrete

should be within the allowable value. The amount of prestressing steel (Ap) is

determined from the designed prestressing force based on the allowable stress in steel.

At transfer, in absence of non-prestressed reinforcement, the stress in concrete (fc) is

given as follows.

cc

Pf = -A

0

cc s c s

Pf = -A + (E /E )A

0

ec

c t

P Pf = - ±A A

(4-1.1)

Here,

Ac = net area of concrete


In presence of non-prestressed reinforcement, the stress in the concrete (fc) can be


(4-1.2)

Here,


Es = modulus of elasticity of steel

Ec = modulus of elasticity of concrete.

At service, the stress in concrete (fc) can be calculated as follows.

(4-1.3)



Here,

At = transformed area of section

P = external axial force


The external axial force is considered positive if it is tension and negative if it is

compression. In the above expression, non-prestressed reinforcement is not

considered. If there is non-prestressed reinforcement, Ac is to be substituted by (Ac +

(Es/Ec) As) and At is to be calculated including As.

Analysis of Ultimate Strength The ultimate tensile strength of a section (PuR) is calculated as per Clause 22.3,

IS:1343 - 1980. The ultimate strength should be greater than the demand due to

factored loads.

In absence of non-prestressed reinforcement, the ultimate tensile strength of a section

(PuR) is given as follows.

(4-1.4) uR pk pP = f A0.87

In presence of non-prestressed reinforcement,

(4-1.5) uR y s pk pP = f A + f A0.87 0.87

In the previous equations,

fy = characteristic yield stress for non-prestressed reinforcement

with mild steel bars

= characteristic 0.2% proof stress for non-prestressed reinforcement

with high yield strength deformed bars.

fpk = characteristic tensile strength of prestressing tendons.

The following example shows the design of a post-tensioned hanger for tension.



Example 4-1.1

Design a post-tensioned hanger to carry an axial tension of PDL = 300 kN (dead load including self-weight) and PLL = 130 kN. The dimension of the hanger is 250

× 250 mm2.

Design the section without considering non-prestressed reinforcement. Tension is not allowed under service loads. The grade of concrete is M 35. The age at transfer is 28 days. Assume 15% long term losses in the prestress. The following properties of the prestressing strands are available from tests.

Type of prestressing tendon : 7 wire strand Nominal diameter = 12.8 mm Nominal area = 99.3 mm2

Tensile strength fpk = 1860 N/mm2

Modulus of elasticity = 195 kN/mm2.

Solution

Preliminary calculations at transfer

Ac ≈ A = 250 × 250

= 62,500 mm2

Allowable stress for M35 concrete under direct compression at transfer

2

= 0.8×0.51= 0.8×0.51×35=14.3 N/mm

cc,all cif f

Maximum prestressing force at transfer

0 max =

=14.3×62,500= 892,500 N

c cP f A



Preliminary calculations at service

At ≈ A = 250 × 250

= 62,500 mm2

Stress in concrete e

cc t

P Pf = - +A A

Allowable stress at service

2= 0 N/mmct,allfConsidering 15% loss

eP = P00.85

Substituting the values

P P= - +A A

00.850

Preliminary calculations at service (continued…)

Solving, 0

0

0.85 =300+130=

0.85= 506 kN

P P

P

Allowable prestress in tendon

0

2

= 0.8= 0.8×1860=1488 N/mm

p pkf f

Required area of tendon

2

506,000=1488

= 340 mm

pA

Select 4 strands with 2

= 4×99.3

= 397.2 mmpA

Prestress at transfer 0 = 397.2×1488 N= 591kN

P



Final calculations at transfer

2

= 62,500 - 397

= 62103 mmcA

Stress in concrete

OK

0

2

591,000= -62,103

= - 9.5 N/mm

cc

Pf = -A

c cc,af < f ll

Final calculations at service

2=195 kN/mmpE

2

= 5,000 35

= 29,580 N/mmcE

2

195= 62,103+ ×397.229.6

= 64,720 mm

tA

Stress in concrete

e

cc t

P Pf = - +A A

3

2

0.85×591,000 (300+130)×10= - +62,103 64,720

= - 1.4 N/mm

cf

No tensile stress in concrete. OK.

Final calculations for ultimate strength

= 0.87= 0.87×1860×397.2 N= 643.0 kN

uR pk pP f A



Demand under factored loads

=1.5(300+130)= 645.0 kN

uP

PuR ≈ Pu OK

Designed cross-section

250

250 (4) 7-wire strandswith P0 = 591 kN

250

250 (4) 7-wire strandswith P0 = 591 kN

Nominal non-prestressed reinforcement is provided for resisting thermal and shrinkage

cracks.



4.2 Design of Sections for Flexure This section covers the following topics

• Preliminary Design

• Final Design for Type 1 Members

• Special Case

Calculation of Moment Demand For simply supported prestressed beams, the maximum moment at the span is given by

the beam theory. For continuous prestressed beams, the analysis can be done by

moment distribution method. The moment coefficients in Table 12 of IS:456 - 2000 can

be used under conditions of uniform cross-section of the beams, uniform loads and

similar lengths of span.

The design is done for the critical section. For a simply supported beam under uniform

loads, the critical section is at the mid span. For a continuous beam, there are critical

sections at the supports and at the spans.

For design under service loads, the following quantities are known.

MDL = moment due to dead load (excluding self-weight)

MLL = moment due to live load.

The material properties are selected before the design.

The following quantities are unknown.

The member cross-section and its geometric properties,

MSW = moment due to self-weight,

Ap = amount of prestressing steel,

Pe = the effective prestress,

e = the eccentricity.

There are two stages of design.

1) Preliminary: In this stage the cross-section is defined and Pe and Ap are

estimated.



2) Final: The values of e (at the critical section), Pe, Ap and the stresses in concrete

at transfer and under service loads are calculated. The stresses are checked with

the allowable values. The section is modified if required.

4.2.1 Preliminary Design

The steps of preliminary design are as follows.

1) Select the material properties fck and fpk.

2) Determine the total depth of beam (h).

The total depth can be based on architectural requirement or, the following empirical

equation can be used.

h = 0.03 √M to 0.04 √M (4-2.1)

Here, h is in meters and M is in kNm. M is the total moment excluding self-weight.

3) Select the type of section. For a rectangular section, assume the breadth

b = h/2.

4) Calculate the self-weight or, estimate the self-weight to be 10% to 20% of

the load carried.

5) Calculate the total moment MT including self-weight. The moment due to

self-weight is denoted as Msw.

6) Estimate the lever arm (z).

z ≈ 0.65h, if Msw is large (Msw > 0.3MT).

z ≈ 0.5h, if Msw is small.

7) Estimate the effective prestress (Pe)

Pe = MT / z, if Msw is large.

Pe = MI L / z, if Msw is small.

If Msw is small, the design is governed by the moment due to imposed load

(MI L = MT – MSW).

8) Considering fpe = 0.7fpk , calculate area of prestressing steel Ap = Pe / fpe.

9) Check the area of the cross-section (A).

The average stress in concrete at service C/A (= Pe /A) should not be too high as

compared to 50% of the allowable compressive stress fcc,all . If it is so, increase the area

of the section to A = Pe /(0.5fcc,all).



4.2.2 Final Design for Type 1 Members

The code IS:1343 - 1980 defines three types of prestressed members.

Type 1: In this type of members, no tensile stress is allowed in concrete at transfer

or under service loads.

Type 2: In these members, tensile stress is within the cracking stress of concrete.

Type 3: Here, the tensile stress is such that the crack width is within the allowable

limit.

The final design involves the checking of the stresses in concrete at transfer and under

service loads with respect to the allowable stresses. Since the allowable stresses

depend on the type of member (Type 1, Type 2 or Type 3), the equations vary for the

different types. Here, the steps of final design are explained for Type 1 members. The

steps for Type 2 members are explained in Section 4.3, Design of Sections for Flexure

(Part II). The steps for Type 3 members are similar to Type 2, the only difference being

the value of the allowable tensile stress in concrete.

For small moment due to self-weight (Msw ≤ 0.3MT), the steps are as follows.

1) Calculate eccentricity e to locate the centroid of the prestressing steel (CGS).

With increasing load, the compression (C) moves upward from the location of the

tension (T) at CGS. At transfer, under the self-weight, C should lie within the kern zone

to avoid tensile stress at the top. The kern points and kern zone are explained in

Section 3.3, Analysis of Member under Flexure (Part II).

The lowest permissible location of C due to self-weight is at the bottom kern point (at a

depth kb below CGC) to avoid tensile stress at the top. The design procedure based on

the extreme location of C gives an economical section.

The following sketch explains the lowest permissible location of C due to self-weight

moment (Msw) at transfer.



CGCkt

kb eCCGS

T

Internal force in concrete

Stress in concrete

hC

Msw

≡+cb

ct

fb

C/A =P0/A

0

CGCkt

kb eCCGS

T


Stress in concrete

hhC

Msw

≡+cb

ct

fb

C/A =P0/A

0

fb

C/A =P0/A

0

Figure 4-2.1 Stress in concrete due to compression at bottom kern point


A = gross area of cross section

fb = maximum compressive stress in concrete at bottom edge

h = total height of the section

kt, kb = distances of upper and lower kern points, respectively, from CGC

ct, cb = distances of upper and lower edges, respectively, from CGC

P0 = prestress at transfer after initial losses.

The shift of C due to self-weight gives an expression of e.

e = (Msw / P0) + kb (4-2.2)

Here, the magnitude of C or T is equal to P0. The value of P0 can be estimated as

follows.

a) 90% of the initial applied prestress (Pi) for pre-tensioned members.

b) Equal to Pi for post-tensioned members.

The value of Pi can be estimated from the amount of prestressing steel determined in

the preliminary design.

Pi = Ap(0.8fpk) (4-2.3)

Here, the permissible prestress in the steel is 0.8fpk, where fpk is the characteristic

tensile strength.

2) Recompute the effective prestress Pe and the area of prestressing steel Ap.

With increasing load, C further moves up. Under the service loads, C should lie within

the kern zone to avoid tensile stress at the bottom. The highest permissible location of



C due to total load is at the top kern point (at a height kt above CGC) to avoid tensile

stress at the bottom.

The following sketch explains the highest possible location of C due to the total moment

(MT).

ft

CGCkt

kb e

C

CGST

+h


Stress in concrete

MT

≡C C/A =

Pe/A0

cb

ct

ft

CGCkt

kb e

C

CGST

+h


Stress in concrete

MT

≡C C/A =

Pe/A0

cb

ct

Figure 4-2.2 Stress in concrete due to compression at top kern point


ft = maximum compressive stress in concrete at top edge.

The shift of C due to the total moment gives an expression of Pe.

Pe = MT /(e + kt) (4-2.4)

Considering fpe = 0.7fpk , the area of prestressing steel is recomputed as follows.

Ap = Pe / fpe (4-2.5)

3) Recompute eccentricity e

First the value of P0 is updated. The eccentricity e is recomputed with the updated value

of P0.

If the variation of e from the previous value is large, another cycle of computation of the

prestressing variables can be undertaken.

4) Check the compressive stresses in concrete.

The maximum compressive stress in concrete should be limited to the allowable values.

At transfer, the stress at the bottom should be limited to fcc,all , where fcc,all is the

allowable compressive stress in concrete at transfer (available from Figure 8 of IS:1343



- 1980). At service, the stress at the top should be limited to fcc,all , where fcc,all is the

allowable compressive stress in concrete under service loads (available from Figure 7

of IS:1343 - 1980).

a) At Transfer

The stress at the bottom can be calculated from the average stress –P0/A.

bt

P hf = -A c

0 (4-2.6)

To satisfy |fb| ≤ fcc,all, the area of the section (A) is checked as follows.

≥cc,all t

P hAf c

0

(4-2.7)

If A is not adequate then the section has to be redesigned.

b) At Service

The stress at the top can be calculated from the average stress –Pe/A.

e

b

P hf = -t A c(4-2.8)

To satisfy |ft| ≤ fcc,all, the area of the section (A) is checked as follows.

≥ e

cc,all b

P hAf c

(4-2.9)


4.2.3 Special Case

For large moment due to self-weight (Msw > 0.3 MT), the eccentricity e according to

e = (Mw / P0) + kb may violate the cover requirements or, may even lie outside the beam.

In such cases, locate e as per cover requirements. The location of C at transfer will be

within the kern zone without zero stress at the top. The expression of stress at the

bottom is different from that given earlier. The other steps are same as before.



At transfer, the stress at the bottom is calculated using the following stress profile.

e – MSW/P0

C

fb

C/A =P0/ACGC

e – MSW/P0

C

fb

C/A =P0/ACGC

Figure 4-2.3 Stress in concrete due to compression above bottom kern point

⎛ ⎞⎜ ⎟⎝ ⎠

swb

b

MP e - cPPf = - -

A I

000

(4-2.10)

Substituting I = Ar2 and r2/cb = kt

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

sw

bt

Me -P Pf = - +A k

0 01

(4-2.11)


⎛ ⎞⎜ ⎟

≥ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

0 01

sw

cc,all t

Me -P PA +

f k

(4-2.12)

The following example shows the design of a Type 1 prestressed member.

Example 4-2.1 Design a simply supported Type 1 prestressed beam with MT = 435 kNm (including an estimated MSW = 55 kNm). The height of the beam is restricted to 920 mm. The prestress at transfer fp0 = 1035 N/mm2 and the prestress at service fpe = 860 N/mm2. Based on the grade of concrete, the allowable compressive stresses are 12.5 N/mm2 at transfer and 11.0 N/mm2 at service.



The properties of the prestressing strands are given below. Type of prestressing tendon : 7-wire strand Nominal diameter = 12.8 mm Nominal area = 99.3 mm2

Solution

A) Preliminary design

The values of h and MSW are given.

1) Estimate lever arm z. sw

T

MM

55=435

=12.5 %

Since MSW < 0.3 MT ,

Use z = 0.5h

= 0.5 × 920

= 460 mm

2) Estimate the effective prestress.

Moment due to imposed loads

= 435 - 55= 380 kNm

IL T swM = M - M

Effective prestress

3380×10=460

= 826 kN

eP

3) Estimate the area of the prestressing steel.

3

2

826×10=860

= 960 mm

ep

pe

PA =f



4) Estimate the area of the section to have average stress in concrete equal to 0.5 fcc,all .

3

3 2

=0.5

826×10=0.5×11.0

=150×10 mm

e

cc,all

PAf

The following trial section has the required depth and area.

Trial cross-section

100

390

920100

100

Values in mm.

100

390

920100

100

Values in mm. B) Calculation of geometric properties

The section is symmetric about the horizontal axis. Hence, the CGC lies at mid depth.


properties.

410 ct = 460

3

2

1

Values in mm.

CGC410 ct = 460

33

22

11

Values in mm.

CGC

Check area of the section

1 2

2

= 2 + = 2×(390×100)+(720×100)=150,000 mm

A A A



Moment of inertia of the section about axis through CGC

⎡ ⎤⎢ ⎥⎣ ⎦

1 2

3 2

10 4

= 2 +1 1= 2 ×390×100 +(390×100)×410 + ×100×720

12 12=1.6287×10 mm

I I I

3


2

10

2

1.6287×10=150,000

=108,580 mm

Ir = A

Kern levels of the section

2

= =

108,580=460

= 236 mm

t bt

rk kc

Summary after preliminary design

Properties of section

A = 150,000 mm2

I = 1.6287×1010 mm4

ct = cb = 460 mm

kt = kb = 236 mm

Properties of prestressing steel

Ap = 960 mm2

Pe = 826 kN



C) Final design

1) Calculate eccentricity e

≈

swb

Me = + kP0

3

55.0×10= +993.6

290 mm

236

0 0== 960×1035= 993.6 kN

p pP A f

2) Recompute the effective prestress and associated variables.

3435×10=(290+ 236)

= 827 kN

Te

t

MP =e+ k

Since Pe is very close to the previous estimate of 826 kN, Ap, P0 and e remain same.

The tendons are placed in two ducts. The outer diameter of each duct is 54 mm.

Select (10) 7-wire strands with

2

=10×99.3

= 993.0 mmpA


a) At transfer

≥ 0

2

993.6×920=12.5×460

=158,976 mm

cc,all t

P hAf c

b) At service

≥

2

827×920=11.0×460

=150,364 mm

e

cc,all b

P hAf c



The governing value of A is 158,976 mm2. The section needs to be revised. The width

of the flange is increased to 435 mm. The area of the revised section is 159,000 mm2.

Another set of calculations can be done to calculate the geometric properties precisely.

Designed cross-section at mid-span

100

435

920

100

290

100

CGC

CGS

(10) 7-wire strandswith P0 = 994 kN

100

435

920

100

290

100

CGC

CGS

100

435

920

100

290

100

CGC

CGS




4.3 Design of Sections for Flexure (Part II) This section covers the following topics

• Final Design for Type 2 Members

The steps for Type 1 members are explained in Section 4.2, Design of Sections for

Flexure (Part I).

4.3.1 Final Design for Type 2 Members

For Type 2 members, the tensile stress under service loads is within the cracking stress

of concrete. The allowable tensile stress in concrete (fct,all) as per IS:1343 - 1980 is

same for transfer and service load conditions. The value is 3.0 N/mm2, which can be

increased to 4.5 N/mm2 for temporary loads.

The following material provides the steps for sections with small self-weight moment.

For sections with large self- weight moment, the eccentricity e may need to be

determined based on the cover requirements.

1) Calculate eccentricity e to locate the centroid of the prestressing steel (CGS).

Under the self-weight, C may lie outside the kern region. The lowest possible location of

C due to self-weight is determined by the allowable tensile stress at the top.

The following sketch explains the extreme location of C due to self-weight moment (Msw)

at transfer.

fct,all

CGC kbe1C

CGST

+h


Stress in concrete

MSW

≡C e2

kt

fb

cb

ct

fct,all

CGC kbe1C

CGST

+h


Stress in concrete

MSW

≡C e2

kt

fb

cb

ct

Figure 4-3.1 Stress in concrete due to compression outside bottom kern point




fb = maximum compressive stress in concrete at bottom edge

fct,all = allowable tensile stress in concrete at top edge

h = total height of the section

kt, kb = distances of upper and lower kern points, respectively, from CGC

ct, cb = distances of upper and lower edges, respectively, from CGC

e1 = distance between the bottom kern point and the location of

compression

e2 = distance by which the compression travels from CGS due to self

weight

P0 = prestress at transfer after initial losses.

From the previous figure, the shift of C due to self-weight gives an expression of e2. It is

evident that if C is further shifted upwards by a distance e1 to the bottom kern point,

there will be no tensile stress at the top.

swMe =

P20

(4-3.1)

The value of e1 is calculated from the expression of stress corresponding to the moment

due to the shift in C by e1.

tct,all

ct,all

t

ct,all b

P e c = fI

f Ie =

P cf Ak

e =P

0 1

10

10

(4-3.2)

Substituting I = Ar2 and r2/ct = kb

The distance of the CGS below the bottom kern point is given as follows.

sw ct,all bM + f Ak

e + e =P1 2

0(4-3.3)



The eccentricity e is calculated from the following equation.

b

sw ct,all bb

e = e + e + kM + f Ak

e = + kP

1 2

0

(4-3.4)

The above expression can be compared with the expression of Type 1 member e =

(Msw / P0) + kb. Note that the eccentricity has increased for a Type 2 member due to the

allowable tensile stress fct,all.

2) Recompute the effective prestress Pe and the area of prestressing steel Ap.

Under the total load, C may lie outside the kern region. The highest permissible location

of C due to total load is determined by the allowable tensile stress at the bottom.

The following sketch explains the highest possible location of C due to the total moment

(MT).

fct,all

CGCkt

e3C

CGST

+h


Stress in concrete

MT

≡C

ft

kb cb

ct

fct,all

CGCkt

e3C

CGST

+h


Stress in concrete

MT

≡C

ft

kb cb

ct

Figure 4-3.2 Stress in concrete due to compression outside top kern point

From the previous figure, the expression of e3 is obtained by the tensile stress

generated due to the shift of C beyond the upper kern point.

e bct,all

ct,all

e b

ct,all t

e

P e c = fI

f Ie =

P cf Ak

e =P

3

3

3

(4-3.5)

Substituting I = Ar2 and r2/cb = kt



The shift of C due to the total moment gives an expression of Pe.

( )( )

∴

T e t

e t ct,all

T ct,all te

t

M =P e+k +e=P e+k +f Ak

M -f AkP =

e+k

3

t

(4-3.6)

The above expression can be compared with the expression of Type 1 member Pe = MT

/(e + kt). Note that the prestressing force has decreased for a Type 2 member due to

the allowable tensile stress fct,all. This will lead to a decrease in the area of prestressing

steel (Ap). Considering fpe = 0.7fpk, Ap is recomputed as follows.

Ap = Pe/ fpe (4-3.7)


First the value of P0 is updated. The eccentricity e is recomputed with the updated value

of P0.



4) Check the compressive stresses in concrete

The maximum compressive stress in concrete should be limited to the allowable values.

At transfer, the stress at the bottom should be limited to fcc,all , where fcc,all is the

allowable compressive stress in concrete at transfer (available from Figure 8 of IS:1343 - 1980). At service, the stress at the top should be limited to fcc,all , where fcc,all is the

allowable compressive stress in concrete under service loads (available from Figure 7

of IS:1343 - 1980).

a) At Transfer

The stress at the bottom can be calculated from the stress diagram.

( )

⎛ ⎞⎜ ⎟⎝ ⎠

b bb

b b bb

C k +e cCf = - -A I

k c Ce cCf = - 1+ -A r I

1

12

(4-3.8)



From fct,all = Ce1ct / I, substituting Ce1 / I = fct,all /ct

⎛ ⎞⎜ ⎟⎝ ⎠

ct,allbb b

t t

ct,allb b

t t

fcCf = - + - cA c c

fC hf = - - cA c c

1

(4-3.9)


≤

≤

∴ ≥

ct,allb cc,all

t t

cc,all t ct,all b

cc,all t ct,all b

fC h + c fA c cP h f c - f cA

P hAf c - f c

0

0

(4-3.10)


b) At Service

The stress at the top can be calculated from the stress diagram.

( )

⎛ ⎞⎜ ⎟⎝ ⎠

t tt

t t tt

C k + e cCf = - -A I

k c Ce cCf = - + -A r I

3

321

(4-3.11)

From fct,all = Ce3cb / I, substituting Ce3 / I = fct,all /cb

⎛ ⎞⎜ ⎟⎝ ⎠

ct,all ttt

b b

ct,all tt

b b

f ccCf = - + -A c c

f cC hf = - -A c c

1

(4-3.12)

To satisfy |ft| ≤ fcc,all, the area of the section (A) is checked as follows.

≤

≤

∴ ≥

ct,all tcc,all

b b

ecc,all b ct,all t

e

cc,all b ct,all t

f cC h + fA c cP h f c - f cA

P hAf c - f c

(4-3.13)




The following table shows a comparison of equations for Type 1 and Type 2 members.

Table 4-3.1 Comparison of equations for Type 1 and Type 2 members

Type 1 Type 2

Eccentricity

sw

bMe = + kP0

sw ct,all bb

M + f Ake = + k

P0

Effective prestress

T

et

MP =e+ k

T ct,all te

t

M - f AkP =

e+ k

Minimum area based on

stress at bottom at transfer ≥

cc,all t

P hAf c

0

≥cc,all t ct,all b

P hAf c - f c

0

Minimum area based on

stress at top at service ≥ e

cc,all b

P hAf c

≥ e

cc,all b ct,all t

P hAf c - f c

The following example shows the design of a Type 2 prestressed member. The same

section was designed as a Type 1 member in Section 4.2, Design of Sections for

Flexure (Part I). The solutions of the two examples are compared at the end.

Example 4-3.1 Design a simply supported Type 2 prestressed beam with MT = 435 kNm (including an estimated MSW = 55 kNm). The height of the beam is restricted to 920 mm. The prestress at transfer fp0 = 1035 N/mm2 and the prestress at service fpe = 860 N/mm2. Based on the grade of concrete, the allowable compressive stresses are 12.5 N/mm2 at transfer and 11.0 N/mm2 at service. The allowable tensile stresses are 2.1 N/mm2 at transfer and 1.6 N/mm2 at service. The properties of the prestressing strands are given below. Type of prestressing tendon : 7-wire strand Nominal diameter = 12.8 mm Nominal area = 99.3 mm2



Solution

A) Preliminary design

The values of h and MSW are given.

1) Estimate lever arm z. 55=435

=12.5 %

sw

T

MM

Since MSW < 0.3 MT ,

use z = 0.5h

= 0.5 × 920

= 460 mm

2) Estimate the effective prestress.

Moment due to imposed loads

= 435 - 55= 380 kNm

IL T swM = M - M

Effective prestress 3380×10=

460= 826 kN

eP

3) Estimate the area of the prestressing steel.

3

2

826×10=860

= 960 mm

ep

pe

PA =f

4) Estimate the area of the section to have average stress in concrete equal to 0.5 fcc,all .

3

3 2

=0.5

826×10=0.5×11.0

=150×10 mm

e

cc,all

PAf



The following trial section has the required depth and area.

Trial cross-section

100

390

920100

100

Values in mm.

100

390

920100

100

Values in mm. B) Calculation of geometric properties



properties.

410 ct = 460

3

2

1

Values in mm.

CGC410 ct = 460

33

22

11

Values in mm.

CGC

Check area of the section

2

= 2×(390×100)+(720×100)=150,000 mm

A


⎡ ⎤⎢ ⎥⎣ ⎦

1 2

3 2

10 4

= 2 +1 1= 2 ×390×100 +(390×100)×410 + ×100×720

12 12=1.6287×10 mm

I I I

3



Square of the radius of gyration 2

10

2

=

1.6287×10=150,000

=108,580 mm

Ir A

Kern levels of the section 2

108,580=460

= 236 mm

t bt

rk = k =c

Summary after preliminary design

Properties of section

A = 150,000 mm2

I = 1.6287×1010 mm4

ct = cb = 460 mm

kt = kb = 236 mm

Values of prestressing variables

Ap = 960 mm2

Pe = 826 kN

C) Final design

1) Calculate eccentricity e

0 0

= 960×1035= 993.6 kN

p pP = A f

1 20

33

+ =

2.155×10 + ×150,000×23610=

993.6=130mm

sw ct,all bM + f Ake e

P



be = e + e + k=

1 2

1 3 0 + 2 3 6= 36 6 m m

2) Recompute the effective prestress and the area of prestressing steel Ap.

33

1.65435×10 - ×150,000×23610=336+ 236

= 625.6 kN

T ct,all te

t

M - f AkP =

e+ k

Since Pe is substantially lower than the previous estimate of 826 kN, Ap, P0 and e need

to be recalculated.

3

2

625.6×10=860

= 727 mm

ep

pe

PA =f


0 0

= 727×1035= 752.4kN

p pP = A f

→

sw ct,all bb

M + f Ake = + k

P0

33

2.155×10 + ×150,000×23610= + 236

752.4=172+ 236= 408 400 mm

Check the cover requirement

Assuming the outer diameter of duct equal to 54 mm 1= 460 - 400 - ×542

= 33 mm

Clear cover for the duct



The clear cover at the bottom is greater than 30 mm (Clause 11.1.6.2, IS: 1343 - 1980),

which is satisfactory. The side cover in the web is slightly less than 30 mm. The

thickness of the web can be increased to satisfy the requirement.

Since the value of e has changed from 366 mm to 400 mm, prestressing variables are

recomputed.

33

1.65435×10 - ×150,000×23610=400+ 236

= 592.0 kN

T ct,all te

t

M - f AkP =

e+ k

Pe has further reduced from 625.6 kN. Ap and P0 are recalculated.

3

2

592×10=860

= 688.5 mm

pA

Select (7) 7-wire strands with

2

= 7×99.3

= 695.1 mmpA

The tendons can be placed in one duct. The outer diameter of the duct is 54 mm.

0 = 695.1×1035= 719.4kN

P

Since the maximum possible eccentricity is based on cover requirement, the value of e

is not updated.


At transfer ≥

cc,all t ct,all b

P hAf c - f c

0

3

2

719.4×10 ×920=12.5×460 - 2.1×460

=138,352 mm



At service

≥

3

2

592×10 ×920=11×460 -1.65×460

=126,631 mm

e

cc,all b ct,all t

P hAf c - f c

The governing value of A is 138,352 mm2. The section can be revised. The width of

the flange is reduced to 335 mm. The area of the revised section is 139,000 mm2.

Another set of calculations can be done to calculate the geometric properties precisely.

Design cross-section at mid-span

100

335

920

100

400

100

CGC

CGS


100

335

920

100

400

100

CGC

CGS


Comparison of Type 1 and Type 2 sections The solutions from the examples of Type 1 and Type 2 members are placed together in

the next figure for comparison.

100

435

920

100

290

100

CGC

CGS


100

335

920

100

400

100

CGC

CGS


Type 1 Type 2

100

435

920

100

290

100

CGC

CGS


100

435

920

100

290

100

CGC

CGS

100

435

920

100

290

100

CGC

CGS


100

335

920

100

400

100

CGC

CGS


100

335

920

100

400

100

CGC

CGS


Type 1 Type 2 Figure 4-3.3 Sections designed as Type 1 and Type 2 members



The following observations can be made.

1) In Type 2 section, the amount of prestressing steel and the prestressing force

are less than those in a Type 1 section. The area of cross-section is less for Type 2

section.

⇒ Type 2 section is relatively economical.

2) The eccentricity in Type 2 section is larger than in Type 1 section. For unit

prestressing force, the prestressing is more effective in Type 2 section.



4.4 Design of Sections for Flexure (Part III) This section covers the following topics.

• Choice of Sections

• Determination of Limiting Zone

• Post-tensioning in Stages

4.4.1 Choice of Sections

The type of section is selected based on the use of the structure, architectural

requirements, casting and fabrication options, available technology and skilled work

force. Here, a few comments are given for the available types of sections.

1) The section should have large depth below the CGC, so as to have the provision of

large eccentricity. The prestressing force can then be reduced.

2) There should be adequate concrete at the top and bottom to satisfy the allowable

stresses.

3) The end section is usually solid to increase the shear capacity and prevent

anchorage zone failure.

Here, the sections are broadly grouped under rectangular section, T-section, I-section

and inverted T-section. Some variations of each type are shown under the

corresponding broad groups. The sections in each group have similar analysis

procedure. The sections shown are not exclusive.



Table 4-4.1 Types of sections

Broad groups of sections

Rectangular

T-section

I-section

Inverted T-

section

Variations

(a) (b)

Remarks on the sections

(a) More room for

tendons in lower

flange.

(b) Better stability

during erection.

Torsionally stiff

and strong.

Fabrication Easy Easy Expensive form

work Difficult

Space for reinforcement

Adequate Less than

adequate Good Good

Effeciency for non-composite sections

Poor, z ≈ 0.4h Good, z ≈ 0.5h Very good,

z ≈ 0.7h

1) Very

inefficient.

2) Small

ultimate

moment

capacity

Application of non-composite sections

1) Light load and

short span.

2) Msw/MT large.

1) Especially

good for long

span roofs, when

LL « DL.

2) Msw/MT large.

1) Good for

long span and

heavy loads.

2) Msw/MT

small.

Msw/MT large

Effeciency for composite sections

Very good

particularly when

section is shored.

Increases load

capacity only

slightly.

Very good

Very good

with cast-in-

place flange.



Application of composite sections

Good for building

construction

1) Topping

serves to tie all

sections

together.

2) No form

required for

composite pour.

Long span

buildings and

bridges.

Bridges

The different types of sections can be compared by a measure of flexural efficiency η.

The flexural efficiency is defined in terms of the radius of gyration r as follows.

t b

t b

t b

b t

t b

rη=c c

c + cr=c c h

r r+c c=

hk + kη=

h

2

2

2 2

(4-4.1)

Thus for a given value of the depth h, if the

kern zone (kt+kb) is large then the section is efficient.

For a rectangular section, η = 0.33.

For an I-section, η > 0.33.

4.4.2 Determination of Limiting Zone

For full prestressed members (Type 1), tension is not allowed under service conditions.

If tension is also not allowed at transfer, C always lies within the kern zone. The

limiting zone is defined as the zone for placing the CGS of the tendons such that C

always lies within the kern zone. Also, the maximum compressive stresses at transfer

and service should be within the allowable values.




Also, the maximum compressive stresses at transfer and service should be within the



allowable values. Note that the limiting zone is a restriction for the CGS. The individual

tendons may lie outside the limiting zone.

The following figure shows the limiting zone (as the shaded region) for a simply

supported beam subjected to uniformly distributed load.

Locus of emin

Locus of emax

CGC

CL

Locus of emin

Locus of emax

CGC

CL

Figure 4-4.1 Limiting zone for a simply supported beam

The limiting zone is determined from the maximum or minimum eccentricities of the

CGS along the beam corresponding to the extreme positions of C.

The maximum eccentricity (emax) at any section corresponds to the lowest possible

location of C at transfer, that generates allowable tensile stress at the top of the section.

Also, the maximum compressive stress at the bottom should be within the allowable

value.

The minimum eccentricity (emin) at any section corresponds to the highest possible

location of C at service, that generates allowable tensile stress at the bottom of the

section. Also, the maximum compressive stress at the top should be within the

allowable value.

The following material gives the expressions of emax and emin for Type 1 and Type 2

sections. The zone between the loci of emax and emin is the limiting zone of the section

for placing the CGS. The values of emax and emin can be determined by equating the

stresses at the edges of concrete with the allowable values. Else, explicit expressions of

emax and emin can be used. Here, the expressions of emax and emin based on allowable

tensile stress are given.

Type 1 Section At Transfer



The following sketch shows the stress profile in concrete when C is at the lowest

permissible location due to self-weight moment (Msw) at transfer.

Internal forces Stress in concrete

fb

CGCkt

kb emaxC

TCGS

0

cb

ct


fb

CGCkt

kb emaxC

TCGS

0

cb

ct

Figure 4-4.2 Stress in concrete due to compression at bottom kern point

From the shift of C due to self-weight, the following expression can be derived.

swmax b

swmax b

Me - k =P

Me = + kP

0

0

or,

(4-4.2)

Note that since MSW varies, emax varies along the length of the beam. Also, the stress at

the bottom needs to be checked to satisfy the condition |fb| ≤ fcc,all.

At Service

The following sketch shows the stress profile in concrete when C is at the highest

possible location due to the total moment (MT).


ft

CGCkt

kb emin

C

TCGS

0

cb

ct


ft

CGCkt

kb emin

C

TCGS

0

cb

ct

Figure 4-4.3 Stress in concrete due to compression at top kern point

From the shift of C due to total moment, the following expression can be derived.

Tmin t

e

Tmin t

e

Me + k =P

Me = - kP

or,



(4-4.3)

Note that since MT varies, emin varies along the length of the beam. Also, the stress at

the top needs to be checked to satisfy the condition |ft| ≤ fcc,all.

If for a particular section emin is negative, it implies that the CGS can be placed above

CGC. This happens near the supports.

Type 2 Section At Transfer

The following sketch shows the stress profile in concrete when C is at the lowest

permissible location due to self-weight moment (Msw) at transfer.


fct,all

CGC kbe1 CT

CGS e2

kt

fb

cb

ct

emax


fct,all

CGC kbe1 CT

CGS e2

kt

fb

cb

ct

emax

Figure 4-4.4 Stress in concrete due to compression outside bottom kern point

sw ct,all bmax b

sw ct,all bmax b

M + f Ake - k =

PM + f Ak

e = + kP

0

0

or,

(4-4.4)

Note that emax for a Type 2 section is larger than that for a Type 1 section due to the

term fct,all Akb in the numerator. The stress at the bottom needs to be checked to satisfy

the condition |fb| ≤ fcc,all.

At Service

The following sketch shows the stress profile in concrete when C is at the highest

possible location due to the total moment (MT).




fct,all

CGCkt

e3C

TCGS

ft

kb cb

ct

emin


fct,all

CGCkt

e3C

TCGS

ft

kb cb

ct

emin

Figure 4-4.5 Stress in concrete due to compression outside top kern point

T ct,all tmin t

e

T ct,all tmin t

e

M - f Ake + k =

PM - f Ak

e = - kP

or,

(4-4.5)

Note that emin for a Type 2 section is smaller than that for a Type 1 section due to the

term fct,all Akt in the numerator. The stress at the bottom needs be checked to satisfy the

condition |ft| ≤ fcc,all.

The zone between emax and emin is the limiting zone of the section for placing the CGS

for a given loading condition. The values of emax and emin for several sections can be

determined at regular intervals along the length of the beam to get their loci. Note that

the limiting zone for a Type 2 member is larger than the limiting zone for a Type 1

member. The following table shows a comparison of equations for Type 1 and Type 2

members.

Table 4-4.2 Comparison of equations for Type 1 and Type 2 members

Type 1 Type 2

Maximum eccentricity

sw

bMe = + kP0

sw ct,all bb

M + f Ake = + k

P0

Minimum eccentricity

T

te

Me = - kP

T ct,all t

te

M - f Ake = - k

P

The following example shows the calculation of limiting zone based on equating the

stresses at the edges of concrete with the allowable values.



Example 4-4.1 For the Type 2 post-tensioned beam with a flanged section as shown, the span is 18 m. For uniform loads, the profile of the CGS is parabolic. The live load moment at mid-span (MLL) is 648 kNm. The prestress after transfer (P0) is 1600 kN. Assume 15% loss at service. Evaluate the limiting zone of CGS, if the allowable stresses at transfer and at service are as follows. For compression, fcc,all = 18.0 N/mm2 For tension, fct,all = 1.5 N/mm2.

200

200

500

1000150

250

150CGS

Values in mm.

200

200

500

1000150

250

150CGS

Values in mm.

Solution

A) Calculation of geometric properties


properties. The centroid of each rectangle is located from the soffit.



ct

cb

900

100

1

2

3

+

500y

CGC

Values in mm.

ct

cb

900

100

11

22

33

+

500y

CGC

Values in mm. Area of the section

Area of 1 = A1 = 500 × 200 = 100,000 mm2

Area of 2 = A2 = 600 × 150 = 90,000 mm2

Area of 3 = A3 = 250 × 200 = 50,000 mm2

A = A1 + A2 + A3

= 240,000 mm2

Distance of CGC from the soffit

1 2 3×900+ ×500+ ×100=

= 583.3 mm

A A AyA

Therefore, = 583.3 mm=1000.0 - 583.3= 416.7 mm

b

t

cc


3 2

1 1

10 4

1= ×500×200 + A ×(900 - 583.3)12

=1.036×10 mm

I

Moment of inertia of 2

3 2

2 2

9 4

1= ×150×600 + A ×(583.3 - 500)12

= 3.32×10 mm

I



Moment of inertia of 3 3 2

3 3

10 4

1= ×250×200 + A ×(583.3 -100)12

=1.184×10 mm

I

Moment of inertia of the section

1 2 3

10

10 4

= + +

= (1.036+0.336+1.184)×10= 2.552×10 mm

I I I I

Calculation of moment due to self weight.

⎛ ⎞⎜ ⎟⎝ ⎠

2 23 2

3 21 m= 24 kN/m ×240,000 mm ×

10 mm= 5.76 kN/m

SWw

2

2

=8

5.76×18=8

= 233.3 kNm

SWSW

w LM

B) Determination of limiting zone

The values of emax and emin are determined by equating the stresses at the edges of

concrete with the allowable values. The expression of stress is given below.



M

P+ + =CGC


M

P+ + =CGC

Limiting position at mid-span

For emax, consider the load stage at transfer.

i) Calculate e based on fb = – 18.0 N/mm2.



30

3

2

1600×10- = -240×10

= -6.67 N/mm

PA

30

101600×10 × ×583.3= -

2.552×10= - 0.0366e

bP ec eI

6

10

2

233.3×10 ×583.3=2.552×10

= 5.33 N/mm

SW bM cI

2

= - 6.67 - 0.0366 +5.33

= - 18.0 N/mmbf e

18.0 - 6.67+5.33Solving, =

0.0366= 455.2 mm

e

ii) Calculate e based on ft = 1.5 N/mm2.

30

101600×10 × ×416.7=

2.552×10= 0.0261

tP ec eI

e

6

10

2

233.3×10 ×416.7= -2.552×10

= -3.81N/mm

SW tM cI

2

= - 6.67+0.0261 - 3.81

=1.5 N/mmtf e

1.5+ 6.67+3.81Solving, =0.0261

= 460.8 mm

e

Out of the two values of e, the lower value 455.2 mm governs.

∴ = 455.2 mmmaxe



For emin, consider the load stage at service.

i) Calculate e based on ft = – 18.0 N/mm2.

0

2

= 0.85

= - 5.67 N/mm

eP PA A

3

100.85×1600×10 × ×416.7=

2.552×10= 0.022

e tP ec eI

e

6

10

2

648.0×10 ×416.7= -2.552×10

= - 10.58 N/mm

LL tM cI

2

= - 5.67+0.022 - 3.81-10.58

= -18.0 N/mmtf e

-18.0+5.67+3.81+10.58Solving, =

0.022= 93.6 mm

e

ii) Calculate e based on fb = 1.5 N/mm2.

3

100.85×1600×10 × ×583.3= -

2.552×10= - 0.031e

e bP ec eI

6

10

2

648.0×10 ×583.3=2.552×10

=14.81N/mm

LL bM cI

2

= - 5.67 - 0.031 +5.33+14.81

=1.5 N/mmbf e

-1.5 - 5.67+5.33+14.81Solving, =

0.031= 418.4 mm

e

Out of the two values of e, the higher value 418.4 mm governs.

∴ = 418.4 mmmine



Limiting position at end

SW LLM = M = 0.0

At transfer

For fb = – 18.0 N/mm2

2

= - 6.67 - 0.0366

= - 18.0 N/mmbf e

18.0 - 6.67Solving, =0.0366

= 309.6 mm

e

For ft = 1.5 N/mm2

2

= - 6.67+0.026

=1.5 N/mmtf e

1.5+ 6.67Solving, =0.026

= 314.2 mm

e

Selecting the lower value

= 309.6 mmmaxe

At service

For fb = 1.5 N/mm2

2

= - 6.67 - 0.0366

=1.5 N/mmSolving = - 223.0 mm

bf e

e

For ft = – 18.0 N/mm2

2

= - 6.67+0.0261

= - 18.0 N/mmSolving = - 436.0 mm

tf e

e

Since the values of e are negative the CGS lies above CGC. The position of CGS closer

to the CGC is selected.

= - 223.0 mmmine

Similarly, the values of emax and emin can be determined at regular intervals along the

span. The limiting zone is available by joining the points by straight lines.



In the following sketch the limiting zone is shown shaded.

CGC

CL

418.4

455.2

223.0309.6

Values in mm.

CGC

CL

418.4

455.2

223.0309.6

Values in mm. From the sketch of the limiting zone, it is evident that the tendons can be spread out at

the ends. This is necessary to anchor the tendons and reduce the stress concentration

at the ends.

4.4.3 Post-tensioning in Stages

In the previous expressions of emax and emin, the values of P0 and Pe can be for different

levels of prestressing for post-tensioned members. At transfer the member can be

partially prestressed in the casting yard, from which P0 is calculated. After the member

is placed in its permanent location, it can be further prestressed before it is put into

service. The application of prestress in different stages is termed as post-tensioning in stages. The value of Pe is calculated from the revised prestressing force.

With reduced P0 at transfer, emax is increased. Thus, the limiting zone for placing the

CGS and the available zone for the shift in C under service loads are also increased.



4.5 Design of Sections for Flexure (Part IV) This section covers the following topic.

• Magnel’s Graphical Method

Notations The variables used in this section are as follows.

A = area of cross section of member

ct = distance of the top of the section from CGC

cb = distance of the bottom of the section from CGC

e = eccentricity of CGS with respect to CGC

ft = stress at the top of the section

fb = stress at the bottom of the section.

f = allowable compressive stress in concrete cc,all

fct,all = allowable tensile stress in concrete

I = moment of inertia of cross section of member

kt = distance of top kern point from CGC

kb = distance of bottom kern point from CGC

MSW = moment due too self weight

MT = total moment

P0 = prestress at transfer after immediate losses

Pe = prestress at service after long term losses

r = radius of gyration, r2 = I/A

Zt = section modulus corresponding to top of the section = I/ct

Zb = section modulus corresponding to bottom of the section = I/cb

η = ratio of prestressing forces = Pe /P0

4.5.1 Magnel’s Graphical Method

The determination of maximum and minimum eccentricities at the critical section helps

in placing the CGS. But with different types of possible sections, the computations

increase. The graphical method proposed by G. Magnel gives a visual interpretation of

the equations involved.



There are essentially four stress conditions to be checked. These conditions are as

follows.

• At transfer: ft ≤ fct,all and fb ≥ fcc,all

• At service: ft ≥ fcc,all and fb ≤ fct,all

The above expressions are algebraic inequalities where the stresses ft and fb are

positive if tensile and negative if compressive. The allowable tensile stress fct,all is

assigned a positive value and the allowable compressive stress fcc,all is assigned a

negative value. The allowable stresses are explained in the Section 1.5, Concrete (Part

I).

It is to be noted that the values of fcc,all at transfer and at service are different. They are

calculated based on the strength of concrete at transfer and at service, respectively.

Similarly, the values of fct,all at transfer and at service can be different. As per IS:1343 - 1980, the values of fct,all at transfer and service are of course same.

The stresses ft and fb in the four inequalities are expressed in terms of the initial

prestressing force P0, the eccentricity e at the critical section of the member, the section

properties A, Zt, Zb, kt, kb and the load variables Msw and MT.

After transposition, 1/P0 is expressed in terms of e by linear inequality relationships.

For a selected section, these relationships are plotted in the 1/P0 versus e plane. The

acceptable zone shows the possible combinations of 1/P0 and e that satisfy all the four

inequality relationships. A combination of P0 and e can be readily calculated from the

acceptable zone.

The method is explained in a general form. For Type 1, Type 2 and Type 3 members,

the value of allowable tensile stress (fct,all) is properly substituted. For Type 1 members,

fct,all = 0 N/mm2.



At Transfer The following sketch shows the variation of stress in concrete after the transfer of

prestress and due to the self weight.

e P0

MSW

ft

fb

cb

ct

e P0

MSW

ft

fb

cb

ct

Figure 4-5.1 Stress profile in concrete at transfer

The stress at the top is calculated from P0, e, Msw as follows.

⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟⎝ ⎠

t swt

t sw

t

sw

b t

P P ec M cf = - + -A I I

P ec M= - + -A r Z

P Me= - + -A k Z

0 0

02

0

1

1

t

(4-5.1)

The inequality relationship satisfying the stress at the top is expressed in terms of 1/P0

and e as follows.

≤

⎛ ⎞≤⎜ ⎟

⎝ ⎠

≥⎛ ⎞⎜ ⎟⎝ ⎠

t ct,all

swct,all

b t

b

swct,all

t

f f

P Me- + - fA k Z

(- + e/k )P Mf + A

Z

0

0

1

11or,

(4-5.2)

The following sketch shows the plot of inequality relationship. The straight line given by

the above inequality is plotted in the 1/P0 versus e plane and the acceptable zone is

shaded.



bk

0

1P

e

Acceptable

bk

0

1P

ebk

0

1P

e

Acceptable

Figure 4-5.2 Plot based on stress at the top at transfer

The following expression relates the stress at the bottom with the load and section

variables.

⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟⎝ ⎠

b swb

b s

b

sw

t b

P P ec M cf = - - +A I IP ec M= - + +A r Z

P Me= - + +A k Z

0 0

02

0

1

1

b

w

(4-5.3)

The inequality relationship satisfying the stress at the bottom is expressed as follows.

≥

⎛ ⎞≥⎜ ⎟

⎝ ⎠

≥⎛ ⎞⎜ ⎟⎝ ⎠

b cc,all

swcc,all

t b

t

swcc,all

b

f f

P Me- + + fA k Z

( + e/k ) P M-f + A

Z

0

0

1

11or,

(4-5.4)

The following sketch shows the plot of the inequality relationship.

0

1P

tk-e

Acceptable 0

1P

tk-e

Acceptable

Figure 4-5.3 Plot based on stress at the bottom at transfer



At Service The following sketch shows the variation of stress in concrete at service and due to the

total moment.

e ηP0

MT

ft

fb

cb

ct

e ηP0

MT

ft

fb

cb

ct

Figure 4-5.4 Stress profile in concrete at service

Here, Pe is expressed as ηP0, where η is the ratio of effective prestress (Pe) and

prestress at transfer (P0).

The expression of the stress at the top is given below.

⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟⎝ ⎠

t Tt

t T

t

T

b t

ηP ηP ec M cf = - + -A I I

ηP ec M= - + -A r Z

ηP e M= - + -A k Z

0 0

02

0

1

1

t

(4-5.5)

The inequality relationship satisfying the stress at the top is expressed as follows.

≥

⎛ ⎞≥⎜ ⎟

⎝ ⎠⎛ ⎞⎜ ⎟⎝ ⎠≤

⎛ ⎞⎜ ⎟⎝ ⎠

t cc,all

Tcc,all

b t

b

Tcc,all

t

f f

ηP e M- + - fA k Z

e- + ηk

P Mf + AZ

0

0

1

11or,

(4-5.6)

The following sketch shows the plot of inequality relationship. The straight line given by

the above inequality is again plotted in the 1/P0 versus e plane and the acceptable zone

is shaded.



bk

0

1P

e

Acceptable

bk

0

1P

ebk

0

1P

e

Acceptable

Figure 4-5.5 Plot based on stress at the top at service

The following expression relates the stress at the bottom with the load and section

variables.

⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟⎝ ⎠

b Tb

b T

b

T

t b

ηP ηP ec M cf = - - + b

A I IηP ec M= - + +A r Z

ηP e M= - + +A k Z

0 0

02

0

1

1

(4-5.7)

The inequality relationship is expressed as follows.

≤

⎛ ⎞≤⎜ ⎟

⎝ ⎠⎛ ⎞⎜ ⎟⎝ ⎠≤

⎛ ⎞⎜ ⎟⎝ ⎠

b ct,all

Tct,all

t b

t

Tct,all

b

f f

ηP e M- + + fA k Z

e+ ηk

P M-f + AZ

0

0

1

11or,

(4-5.8)

The following sketch shows the plot of the inequality relationship.

0

1P

tk- e

Acceptable

0

1P

tk- e

Acceptable

Figure 4-5.6 Plot based on stress at the bottom at service

Next, the four lines are plotted simultaneously. The common region is the acceptable

zone.



bk

0

1P

tk-

(4-5.4)

(4-5.6) (4-5.2)(4-5.8)

e

Acceptable zone

bk

0

1P

tk-

(4-5.4)

(4-5.6) (4-5.2)(4-5.8)

e

Acceptable zone

Figure 4-5.7 Acceptable zone

A combination of a trial section, prestressing force (P0) and eccentricity (e) at the critical

section, can be plotted in the form of the above graph. If the point lies within the

acceptable zone, then the combination is valid.

The following problem illustrates the use of Magnel’s graphical method.

Example 4-5.1 The section shown is designed as a Type 1 member with MT = 435 kNm (including an estimated MSW = 55 kNm). The height of the beam is restricted to 920 mm. The prestress at transfer fp0 = 1035 N/mm2 and the prestress at service fpe = 860 N/mm2. Based on the grade of concrete, the allowable compressive stresses are 12.5 N/mm2 at transfer and 11.0 N/mm2 at service. The properties of the prestressing strands are given below. Type of prestressing tendon : 7-wire strand Nominal diameter = 12.8 mm Nominal area = 99.3 mm2 For the section, find the acceptable zone by Magnel’s graphical method. Compare the designed values of eccentricity (e) and the inverse of prestressing force at transfer (1/P0) with the acceptable zone.



100

435

920

100

e = 290

100

CGC

CGS


100

435

920

100

e = 290

100

CGC

CGS


Solution

A) Calculation of geometric properties



properties.

410 ct = 460

3

2

1

Values in mm.

CGC410 ct = 460

33

22

11

Values in mm.

CGC

Area of the section

1 2

2

= 2 += 2×(435×100)+(720×100)=159,000 mm

A A A


⎡ ⎤⎢ ⎥⎣ ⎦

1 2

3 2

10 4

= 2 +1 1= 2× ×435×100 +(435×100)×410 + ×100×720

12 12=1.78×10 mm

I I I 3




2

10

2

=

1.7808×10=159,000

=112,000 mm

IrA

Section moduli

3= 38,712,174 mmb t

t

IZ = Z =c

Kern levels

2

= 243.5 mmb tt

rk = k =c

B) Calculation of the inequality relationships of Magnel’s graphical method

Ratio of effective prestress and prestress at transfer

e

pe

p

Pη=Pf

=f

0

0

860=1035

= 0.83

At Transfer

ft ≤ fct,all≥⎛ ⎞⎜ ⎟⎝ ⎠

b

swct,all

t

(- +e/k )P Mf + A

Z0

11

≥⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞≥ ⎜ ⎟⎝ ⎠

eP

e

60

1 -1+ 243.555×100+ ×159,000

38,712,1741 -1+

225,897.9 243.5



The relationship is plotted in the following graph.

0

0.0005

0.001

0.0015

-300 -200 -100 0 100 200 300e (mm)

1/P

0 (1/

kN)

0

0.0005

0.001

0.0015

-300 -200 -100 0 100 200 300e (mm)

1/P

0 (1/

kN)

At Transfer

fb ≥ fcc,all ≥⎛ ⎞⎜ ⎟⎝ ⎠

t

swcc,all

b

( + e/k )P M-f + A

Z0

11

≥⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞≥ ⎜ ⎟⎝ ⎠

60

1 1+ 243.555×1012.5+ ×159,000

38,712,1741 1+

2,213,397.9 243.5

eP

e


0

0.0005

0.001

0.0015

-300 -200 -100 0 100 200 300e (mm)

1/P

0 (1/

kN)

0.0005

0

0.0005

0.001

0.0015

-300 -200 -100 0 100 200 300e (mm)

1/P

0 (1/

kN)

0.0005

At Service

⎛ ⎞⎜ ⎟⎝ ⎠≤

⎛ ⎞⎜ ⎟⎝ ⎠

b

Tcc,all

t

e- + ηk

P Mf +Z

0

11

A

ft ≥ fcc,all

≤⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟⎝ ⎠

60

1 (-1+ 243.5)×0.83435×10-11.0+ ×159,000

38,712,1741£ -1+

45,358.0 243.5

eP

e




0

0.0005

0.001

0.0015

-300 -200 -100 0 100 200 300e (mm)

1/P

0 (1/

kN)

0

0.0005

0.001

0.0015

-300 -200 -100 0 100 200 300e (mm)

1/P

0 (1/

kN)

At Service

⎛ ⎞⎜ ⎟⎝ ⎠≤

⎛ ⎞⎜ ⎟⎝ ⎠

t

Tct,all

b

e+ ηk

P M-f + AZ

0

11

fb ≤ fct,all

≤⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞≤ ⎜ ⎟⎝ ⎠

60

1 (1+ 243.5)×0.83435×100.0+ ×159,000

38,712,1741 1+

2,152,587.1 243.5

eP

e


0

0.0005

0.001

0.0015

-300 -200 -100 0 100 200 300e (mm)

1/P

0 (1/

kN)

00

0.0005

0.001

0.0015

-300 -200 -100 0 100 200 300e (mm)

1/P

0 (1/

kN)

0

The four relationships are plotted in the following graph. The acceptable zone is shown.

The zone is zoomed in the next graph.



0

0.0005

0.001

0.0015

-300 -200 -100 0 100 200 300e (mm)

1/P

0 (1/

kN)

0.0008

0.001

0.0012

225 250 275 300 325e (mm)

1/P

0 (1/

kN)

0.0008

0.001

0.0012

225 250 275 300 325e (mm)

1/P

0 (1/

kN)

The calculated values of e and 1/P0 for the Type 1 section are as follows.

e = 290 mm

1/P0 = 1/(994 kN) = 0.001kN–1.

The solution of the design is shown in the graphs. It lies in the acceptable zone.



4.6 Detailing Requirements for Flexure This section covers the following topic.

• Tendon Profile

• Minimum Amount of Reinforcement

• Miscellaneous Requirements

Introduction The detailing of the prestressing tendons and the reinforcing bars is important to satisfy

the assumptions in the analysis, proper placement of concrete and durability. After the

design calculations, drawings are prepared for construction. These drawings are

referred to as the design drawings which become a part of the construction documents.

The steel fabricator may prepare another set of drawings which are called shop

drawings. These drawings are similar to the design drawings but they contain additional

information such as the bar designations and bar bending schedule. It is essential to

show the detailing in the design drawings so that there is no ambiguity during

construction. It is also necessary to check the details in the shop drawings.

IS:1343 - 1980 specifies some minimum requirements. Here, these requirements are

briefly mentioned. There are requirements for the non-prestressed reinforcement as per

IS:456 - 2000 which are not covered here. The detailing requirements for shear and

torsion are covered in Section 5.3, Design for Shear (Part II) and Section 5.6, Design for

Torsion (Part II), respectively. Of course the detailing is best learned by preparing

drawings for construction projects.

4.6.1 Tendon Profile For a simply supported post-tensioned beam with high uniformly distributed load, a

parabolic profile is selected. The equation of the profile is given as follows.

(4-6.1) ×myy = ( ) ( L - x )

L 24

Here,

L = span of the beam

x = distance from one end

Y = vertical displacement of the tendon (from the level at the ends) at distance x



Ym = vertical displacement of the tendon (from the level at the ends) at the

middle of the beam

The following sketch shows the plot of the equation.

ym

y

L

xym

y

L

x

Figure 4-6.1 Profile of a parabolic tendon

Note that an individual tendon may be displaced from the CGS. Hence, the tendon

need not pass through the CGC at the ends. The figure below shows the parabolic

profiles of the ducts for placing tendon in a simply supported bridge girder.

Figure 4-6.2 Tendon profiles in a simply supported bridge girder

(Courtesy: Larsen & Toubro – Rambøll)

For continuous beams or slabs, parabolic profiles at the spans and at the supports are

connected to get the continuous profile of a tendon. The following sketch shows the

profile of the CGS in a continuous beam. The eccentricities of the CGS at the end span,

first interior support and first interior span are represented as e1, e2 and e3 respectively.



CGCe1

e2e3

Points of contraflexure(inflection points)

Points of maximumeccentricity

CGCe1

e2e3

Points of contraflexure(inflection points)

Points of maximumeccentricity

CGC

Parabola 1 2 3 4 5 6 7× × × × × × × ×

CGC

Parabola 1 2 3 4 5 6 7× × × × × × × ×

Figure 4-6.3 Profile of CGS in a continuous beam

A parabolic segment connects a point of maximum eccentricity with a point of

contraflexure. A point of contraflexure is the location where the curvature of the profile

reverses. It is also known as the inflection point. For varying spans and loading, the

segments on two sides of a point of maximum eccentricity, may not be symmetric. In

the second sketch of the above figure, the different parabolas between the points of

maximum eccentricity and the points of contraflexure are numbered.

The convex segment over a support is required to avoid a kink in the tendon. The length

of a convex segment is determined based on the minimum radius of curvature for the

type of tendon.

A parabolic segment satisfies two conditions.

1) It has zero slope at the point of maximum eccentricity.

2) At a point of contraflexure, the slopes of the parabolic segments on both sides

should match.

The equation of a parabolic segment is given below.

l⎛ ⎞⎜ ⎟⎝ ⎠m

xy = y2

(4-6.2)



xl

yym

xl

yym

Figure 4-6.4 Plot of a parabolic segment

In the previous equation,

l = length of the parabolic segment

x = distance from the point of maximum eccentricity

y = vertical displacement of the profile at distance x

ym = displacement of the point of contraflexure from the point of maximum

eccentricity.

The origin is selected at the point of maximum eccentricity at a critical section. The

equation satisfies the first boundary condition of zero slope at the point of maximum

eccentricity. The length (l) is determined from the requirement of minimum radius of

curvature at the support. The displacement ym is determined from the boundary

condition that at the point of contraflexure, the slopes of the segments on both sides

should match.

The following photo shows the profiles of the tendons in a continuous bridge girder.

Figure 4-6.5 Tendon profiles in a continuous bridge girder

(Courtesy: VSL International Ltd.)

The profile is implemented by the use of hangers or cross bars or chairs of varying

depth at regular intervals. In beams, the duct is supported by hangers from the top bars

or by cross bars attached to the stirrups. The depth of the hanger or cross bar at a



location can be calculated from the equation of the profile. In slabs, the duct is

supported on chairs resting on the form work.

The CGS of the tendon shifts from the centre line of the duct after stretching. The

following sketches show the shifts at the low and high points of the tendon. The shift in

the CGS is available from the type of tendon used and can be accounted for in precise

calculations.

At low point At high point

Shift

Centre line of duct

CGS

CGS


Shift

Centre line of duct

CGS

CGS


Shift

Centre line of duct

CGS

CGS

Figure 4-6.6 Shift in the CGS of a tendon from the centreline of duct

4.6.2 Minimum Amounts of Reinforcement

Minimum Longitudinal Reinforcement A minimum amount of longitudinal reinforcement should be provided to have sufficient

strength after the cracking of concrete.

According to Section 18.6.3.3-a, the minimum amount is as follows.

Minimum (As+ AP) = 0.2% A (4-6.3) Here,

As = area of steel without prestressing,

Ap = area of prestressing steel,

A = total area of cross-section.



ApAsApAs

Figure 4-6.7 Cross-section of a beam showing longitudinal reinforcement

The minimum reinforcement can be reduced to 0.15% A, if high yield strength deformed

bars are used.

Minimum Longitudinal Reinforcement with Unbonded Tendon In a post-tensioned member when the ducts are not grouted, beyond the cracking load,

the number of cracks is small and the crack width is large. To reduce the crack width, a

minimum amount of non-prestressed reinforcement should be provided.

Since the non-prestressed reinforcement is bonded to the concrete, there are several

cracks with small crack width.

UnbondedtendonUnbondedtendonUnbondedtendon

UnbondedtendonNon-prestressed

reinforcement


reinforcement


reinforcement Figure 4-6.8 Crack pattern with and without non-prestressed reinforcement for beams

with unbonded tendon

As per the code of the American Concrete Institute (ACI 318), the minimum amount of

such reinforcement (As) is 0.4% At , where At is the area under tension between the



centroid of the section (CGC) and the tension edge. The above reinforcement is not

intended to provide flexural strength.

As

Unbonded tendon

CGC

At

As

Unbonded tendon

CGC

At

As

Unbonded tendon

CGC

At

Figure 4-6.9 Cross-section of a beam showing longitudinal reinforcement and area

under tension

Minimum Side Face Reinforcement When the depth of the web exceeds 500 mm, a minimum amount of longitudinal

reinforcement should be placed at each face (side face) of the web to check thermal

and shrinkage cracks.

According to Section 18.6.3.3-b, the minimum amount of side face reinforcement (As,sf )

is given as follows.

Minimum As,sf = 0.05% Aw (4-6.4) Here,

Aw = vertical area of the web.

The maximum spacing of the bars is 200 mm.

As,sfAs,sf As,sfAs,sf As,sfAs,sf

Figure 4-6.10 Cross-section of a beam showing side face reinforcement



4.6.3 Miscellaneous Requirements Minimum Cover Requirements A minimum clear cover of concrete is necessary to protect the steel against corrosion

and to develop adequate bond between concrete and steel. The cover is implemented

by chairs or blocks.

Clear coverClear coverClear cover

Figure 4-6.11 Cross-section of a beam showing cover

According to Section 11.1.6, the minimum cover requirements are as follows.

For pre-tensioned members, minimum cover for tendons is 20 mm. For post-tensioned

members, minimum cover for sheathing (duct) is 30 mm or size of the tendon.

The minimum cover should be increased by 10 mm in aggressive environment.

Minimum Spacing Requirements A minimum clear spacing of the tendons or reinforcing bars is necessary for the flow of

concrete during casting and for the bond between concrete and steel.

Clear spacingClear spacingClear spacing

Figure 4-6.12 Cross-section of a beam showing spacing between tendons

According to Section 11.1.7, the minimum spacing requirements are as follows.

For single wires in a pre-tensioned member,

Clear spacing ≥ 3 × wire diameter

≥ 1⅓ × maximum aggregate size.



For large bars or tendons,

Clear spacing ≥ 40 mm

≥ maximum size of tendon / bar

≥ maximum aggregate size + 5 mm.

For grouped tendons (maximum four tendons per group), the requirement is for the

spacing between the groups of tendons.

Vertical spacing

Horizontal spacing

Vertical spacing

Horizontal spacing

Vertical spacing

Horizontal spacing Figure 4-6.13 Cross-section of a beam showing spacing between groups of tendons

According to Section 11.1.8, for grouped tendons the spacing requirements are as

follows.

Horizontal spacing ≥ 40 mm

≥ maximum aggregate size + 5 mm

Vertical spacing ≥ 50 mm.

Anchorage of Reinforcement In a partially prestressed section, where the non-prestressed reinforcement contributes

to flexural strength, the development length of the bars needs to be checked at the

critical section. The bars should be anchored at the supports by hooks to avoid

anchorage failure.

The following photo shows the fabrication of the reinforcement for a post-tensioned box-

girder of a bridge.



Figure 4-6.14 Fabrication of reinforcement




5.1 Analysis for Shear This section covers the following topics.

• Stress in an Uncracked Beam

• Types of Cracks

• Components of Shear Resistance

• Modes of Failure

• Effect of Prestressing Force

Introduction The analysis of reinforced concrete and prestressed concrete members for shear is

more difficult compared to the analyses for axial load or flexure.

The analysis for axial load and flexure are based on the following principles of

mechanics.

1) Equilibrium of internal and external forces

2) Compatibility of strains in concrete and steel

3) Constitutive relationships of materials.

The conventional analysis for shear is based on equilibrium of forces by a simple

equation. The compatibility of strains is not considered. The constitutive relationships

(relating stress and strain) of the materials, concrete or steel, are not used. The strength

of each material corresponds to the ultimate strength.

The strength of concrete under shear although based on test results, is empirical in

nature.

Shear stresses generate in beams due to bending or twisting. The two types of shear

stress are called flexural shear stress and torsional shear stress, respectively. In this

section, the analysis for shear refers to flexural shear stress. The torsional shear stress

is covered in Section 5.4, Analysis for Torsion.

To understand flexural shear stress, the behaviour of a simply supported beam under

uniformly distributed load, without prestressing, will be explained first. The effect of

prestressing force will be subsequently introduced. The presentation will be in the

following sequence.



1) Stresses in an uncracked (homogenous) beam.

2) Types of cracks generated due to the combination of flexure and shear.

3) Components of shear resistance and the modes of failure.

4) Effect of prestressing force.

5.1.1 Stresses in an Uncracked Beam

The following figure shows the variations of shear and moment along the span of a

simply supported beam under a uniformly distributed load. The variations of normal

stress and shear stress along the depth of a section of the beam are also shown.

b

h 12

Variation of normal stress (f)

Variation of shear stress (v)

Shear force diagram

Momentdiagram M

V

b

h 12

Variation of normal stress (f)

Variation of shear stress (v)

Shear force diagram

Momentdiagram M

V

Figure 5-1.1 Variations of forces and stresses in a simply supported beam

Under a general loading, the shear force and the moment vary along the length. The

normal stress and the shear stress vary along the length, as well as along the depth.

The combination of the normal and shear stresses generate a two-dimensional stress

field at a point. At any point in the beam, the state of two-dimensional stresses can be

expressed in terms of the principal stresses. The Mohr’s circle of stress is helpful to

understand the state of stress.

Before cracking, the stress carried by steel is negligible. When the principal tensile

stress exceeds the cracking stress, the concrete cracks and there is redistribution of

stresses between concrete and steel. For a point on the neutral axis (Element 1), the

shear stress is maximum and the normal stress is zero. The principal tensile stress (σ1)

is inclined at 45º to the neutral axis. The following figure shows the state of in-plane

stresses.



σ1 σ2

α = 45o

Principal stresses

σ1σ2

v

2α

Mohr’s circle

v

1

State of pure shear

σ1 σ2

α = 45o

Principal stresses

σ1 σ2

α = 45o

Principal stresses

σ1σ2

v

2α

Mohr’s circle

σ1σ2

v

2α

Mohr’s circle

v

1

State of pure shear

v

1

State of pure shear

Figure 5-1.2 State of stresses at a point on the neutral axis of a beam

At the level of neutral axis, the normal stress is zero and the shear stress is maximum.

An element at that level is under pure shear. A state of pure shear can be conceived as

a state of biaxial tensile-compressive stresses. These principle stresses are inclined at

45° with respect to the axis of the beam. It is necessary to study the principle stresses

to understand the cracking of concrete. The Mohr’s circle is a representation of the

state of in-plane stresses on surfaces of various inclinations passing through a point.

The horizontal and vertical axes represent the normal and shear stresses, respectively.

For a state of pure shear, the centre of the Mohr’s circle coincides with the origin of the

axes. It is expected that the reader is familiar with these concepts from a course in

strength of materials.

Since the shear force is maximum near the supports, cracks due to shear occur near

the supports. The cracks are formed around the neutral axis and perpendicular to the

principal tensile stress (σ1). The cracks are thus inclined at 45° to the axis of the beam.

The following sketch shows the inclination of the cracks forming at the neutral axis.

σ1

σ2

1

v σ1

σ2

σ1

σ2

1

v

1

v

Figure 5-1.3 Inclination of crack at the level of neutral axis

For a point near the bottom edge of the beam (Element 2), the normal stress is

maximum and the shear stress is close to zero. The principal tensile stress (σ1) is

almost parallel to the bottom edge. The angle of inclination of σ1 with respect to the axis

of the beam (α) is much smaller than 45°. The following figure shows the state of in-

plane stresses.



σ2 σ1

(f,v)

2α

Mohr’s circle

σ1

α < 45o

Principal stresses

σ2

2 f

v

Shear stress and normal stress

σ2 σ1

(f,v)

2α

Mohr’s circle

σ2 σ1

(f,v)

2α

Mohr’s circle

σ1

α < 45o

Principal stresses

σ2σ1

α < 45o

Principal stresses

σ2

2 f

v


2 f

v


Figure 5-1.4 State of stresses at a point close to the edge under tension

Adjacent to the bottom edge (edge under tension), the tensile stress due to flexure is

maximum and the shear stress is zero. The state of stress is nearly uniaxial tensile

stress. The principal compressive stress is negligible. The Mohr’s circle is shifted

towards the axis of principal tensile stress.

Since the moment is maximum at mid span, cracks due to flexure occur near mid span.

The cracks are formed at the bottom edge and perpendicular to σ1. Since σ1 is parallel

to the edge, the cracks are perpendicular to the edge.

σ1

σ2

2

v

fσ1

σ2σ1

σ2

2

v

f2

v

f

Figure 5-1.5 Inclination of crack close to the edge under tension

The previous concepts can be used to develop the principal stress trajectories. The

following figure shows the trajectories for a simply supported beam under a uniformly

distributed load. The crack pattern can be predicted from these trajectories.

Figure 5-1.6 Principle stress trajectories

(Courtesy: Pillai, S. U., and Menon, D., Reinforced Concrete Design)



5.1.2 Types of Cracks

The types and formation of cracks depends on the span-to-depth ratio of the beam and

loading. These variables influence the moment and shear along the length of the beam.

For a simply supported beam under uniformly distributed load, without prestressing,

three types of cracks are identified.

1) Flexural cracks: These cracks form at the bottom near the midspan and

propagate upwards.

2) Web shear cracks: These cracks form near the neutral axis close to the support

and propagate inclined to the beam axis.

3) Flexure shear cracks: These cracks form at the bottom due to flexure and

propagate due to both flexure and shear.

In the following figure, the formation of cracks for a beam with large span-to-depth ratio

and uniformly distributed loading is shown.

a) Initiation of flexural cracksa) Initiation of flexural cracks

b) Growth of flexural cracks and formation of flexure shear and web shear cracks.

b) Growth of flexural cracks and formation of flexure shear and web shear cracks.

c) Cracks before failure

Web shearcracks

Web shearcracks

Flexureshearcracks

Flexuralcracks

Flexure shearcracks

c) Cracks before failure

Web shearcracks

Web shearcracks

Flexureshearcracks

Flexuralcracks

Flexure shearcracks

Web shearcracks

Web shearcracks

Flexureshearcracks

Flexuralcracks

Flexure shearcracks

Figure 5-1.7 Formation of cracks in a reinforced concrete beam



5.1.3 Components of Shear Resistance

The components of shear resistance are studied based on the internal forces at a

flexure shear crack. The forces are shown in the following figure.

Vcz

Vs

VPVd

Va

Vcz

Vs

VPVd

Va

Figure 5-1.8 Internal forces at a flexure shear crack

The notations in the previous figure are as follows.

Vcz = Shear carried by uncracked concrete

Va = Shear resistance due to aggregate interlock

Vd = Shear resistance due to dowel action

Vs = Shear carried by stirrups

Vp = Vertical component of prestressing force in inclined tendons

The magnitude and the relative value of each component change with increasing load.

5.1.4 Modes of Failure

For beams with low span-to-depth ratio or inadequate shear reinforcement, the failure

can be due to shear. A failure due to shear is sudden as compared to a failure due to

flexure. The following five modes of failure due to shear are identified.

1) Diagonal tension failure

2) Shear compression failure

3) Shear tension failure

4) Web crushing failure

5) Arch rib failure

The occurrence of a mode of failure depends on the span-to-depth ratio, loading, cross-

section of the beam, amount and anchorage of reinforcement. The modes of failure are

explained next (Courtesy: Pillai, S. U., and Menon, D., Reinforced Concrete Design).



Diagonal tension failure In this mode, an inclined crack propagates rapidly due to inadequate shear

reinforcement.

Figure 5-1.9 Diagonal tension failure

Shear compression failure There is crushing of the concrete near the compression flange above the tip of the

inclined crack.

Figure 5-1.10 Shear compression failure

Shear tension failure Due to inadequate anchorage of the longitudinal bars, the diagonal cracks propagate

horizontally along the bars.

Figure 5-1.11 Shear tension failure



Web crushing failure The concrete in the web crushes due to inadequate web thickness.

Figure 5-1.12 Web crushing failure

Arch rib failure For deep beams, the web may buckle and subsequently crush. There can be

anchorage failure or failure of the bearing.

Figure 5-1.13 Arch rib failure

The objective of design for shear is to avoid shear failure. The beam should fail in

flexure at its ultimate flexural strength. Hence, each mode of failure is addressed in the

design for shear. The design involves not only the design of the stirrups, but also

limiting the average shear stress in concrete, providing adequate thickness of the web

and adequate development length of the longitudinal bars.

5.1.5 Effect of Prestressing Force

In presence of prestressing force, the flexural cracking occurs at a higher load. For

Type 1 and Type 2 sections, there is no flexural crack under service loads. This is

evident from the typical moment versus curvature curve for a prestressed section (refer

to Section 3.6, Analysis of Member under Flexure (Part V)). In presence of prestressing

force, the web shear cracks also generate under higher load.



With increase in the load beyond the cracking load, the cracks generate in a similar

sequence. But, the inclinations of the flexure shear and web shear cracks are reduced

depending on the amount of prestressing and the profile of the tendon.

The effect of prestressing force is explained for a beam with a concentric effective

prestressing force (Pe).

1PePe 1PePe

Figure 5-1.14 A simply supported beam under concentric prestress and uniformly

distributed loads

For a point at the neutral axis (Element 1), there is normal stress due to the prestressing

force (–fpe). The principal tensile stress (σ1) is inclined to the neutral axis at an angle

greater than 45º. With the combination of shear stress, the principal compressive stress

(σ2) is inclined to the neutral axis at an angle much smaller than 45°. The following

figure shows the state of in-plane stresses.

σ2 σ1

(–fpe,v)2α > 90°

Mohr’s circle

σ2

σ1

α > 45o

Principal stresses

v

1fpefpe

Shear stress and prestress

σ2 σ1

(–fpe,v)2α > 90°

Mohr’s circle

σ2 σ1

(–fpe,v)2α > 90°

Mohr’s circle

σ2

σ1

α > 45o

Principal stresses

σ2

σ1

α > 45o

σ2

σ1

α > 45o

Principal stresses

v

1fpefpe


v

1fpefpe

v

1fpefpe


Figure 5-1.15 State of stresses at a point on the neutral axis for a prestressed beam

In the following figure, the formation of cracks for a prestressed beam with large span-

to-depth ratio and uniformly distributed loading is shown. This figure can be compared

with that for a reinforced concrete beam.

PePe PePe

Figure 5-1.16 Formation of cracks in a prestressed beam



After cracking, in presence of prestressing force, the length and crack width of a

diagonal crack are low. Thus, the aggregate interlock and zone of concrete under

compression are larger as compared to a non-prestressed beam under the same load.

Hence, the shear strength of concrete (Vc) increases in presence of prestressing force.

This is accounted for in the expression of Vc.



5.2 Design for Shear (Part I) This section covers the following topics.

• General Comments

• Limit State of Collapse for Shear

5.2.1 General Comments

Calculation of Shear Demand The objective of design is to provide ultimate resistance for shear (VuR) greater than the

shear demand under ultimate loads (Vu). For simply supported prestressed beams, the

maximum shear near the support is given by the beam theory. For continuous

prestressed beams, a rigorous analysis can be done by the moment distribution method.

Else, the shear coefficients in Table 13 of IS:456 - 2000 can be used under conditions

of uniform cross-section of the beams, uniform loads and similar lengths of span.

Design of Stirrups

The design is done for the critical section. The critical section is defined in Clause 22.6.2 of IS:456 - 2000. In general cases, the face of the support is considered as the

critical section.

When the reaction at the support introduces compression at the end of the beam, the

critical section can be selected at a distance effective depth from the face of the support.

The effective depth is selected as the greater of dp or ds.

dp = depth of CGS from the extreme compression fiber

ds = depth of centroid of non-prestressed steel.

Since the CGS is at a higher location near the support, the effective depth will be equal

to ds.

To vary the spacing of stirrups along the span, other sections may be selected for

design. Usually the following scheme is selected for beams under uniform load.

1) Close spacing for quarter of the span adjacent to the supports.

2) Wide spacing for half of the span at the middle.



For large beams, more variation of spacing may be selected. The following sketch

shows the typical variation of spacing of stirrups. The span is represented by L.

L / 4 L / 4L / 2L / 4 L / 4L / 2 Figure 5-2.1 Typical variation of spacing of stirrups

5.2.2 Limit State of Collapse for Shear

The shear is studied based on the capacity of a section which is the limit state of

collapse. The capacity (or ultimate resistance) of a section (VuR) consists of a concrete

contribution (Vc) and the stirrup contribution (VS).

VuR = VC + VS (5-2.1)

Vc includes Vcz (contribution from uncracked concrete), Va (aggregate interlock) and Vd

(dowel action).

The value of Vc depends on whether the section is cracked due to flexure. Section 22.4

of IS:1343 - 1980 gives two expressions of Vc, one for cracked section and the other for

uncracked section. Usually, the expression for the uncracked section will govern near

the support. The expression for the cracked section will govern near the mid span. Of

course, both the expressions need to be evaluated at a particular section. The lower

value obtained from the two expressions is selected.

For uncracked sections,

c co

c t

V =V

V = bD f + f f20.67 0.8

(5-2.2) cp t

Vco is the shear causing web shear cracking at CGC.

In the above expression,


= bw, breadth of the web for flanged sections



D = total depth of the section (h)

ft = tensile strength of concrete = 0.24√fck

fcp = compressive stress in concrete at CGC due to the prestress

= Pe/A.

The value of fcp is taken as positive (numeric value). Note that, a reduced effective

prestress needs to be considered in the transmission length (explained in Section 7.1)

region of a pre-tensioned beam.

The previous equation can be derived based on the expression of the principal tensile

stress (σ1) at CGC.

σ2

σ1

Principal stresses

v

fcpfcp

State of stressat CGC

σ2 σ1

(–fcp,v)

Mohr’s circle

σ2

σ1

Principal stresses

σ2

σ1

Principal stresses

v

fcpfcp


v

fcpfcp


σ2 σ1

(–fcp,v)

Mohr’s circle

σ2 σ1

(–fcp,v)

Mohr’s circle

Figure 5-2.2 State of stresses at a point on the neutral axis for a prestressed beam

The principal tensile stress is equated to the direct tensile strength of concrete (ft).

⎛ ⎞⎜ ⎟⎝ ⎠

cp cp

cp cp c

t

f fσ = - + + v

4

f f V Q= - + +Ib

= f

22

1

2 20

2

2 4

(5-2.3)


I = gross moment of inertia

Q = At y

At = area of section above CGC

y = vertical distance of centroid of At from CGC.



CGC

At

+yCGC

At

+y

Figure 5-2.3 Cross-section of a beam showing the variables for calculating shear

stress in the web

Transposing the terms,

→

20

20.67 0.8

c t cp t

t c

IbV = f + f fQ

bD f + f f

(5-2.4) p t

→c c p

t cp t p

V V +V

= bD f + f f +V

0

20.67 0.8

⎛ ⎞⎜ ⎟⎝ ⎠

≥

c cr

pe uc c

pk u

ck

V =V

f VV = τ bd + Mf M

bd f

01- 0.55

0.1

The term 0.67bD represents Ib/Q for the section. It is exact for a rectangular section and

conservative for other sections.

To be conservative, only 80% of the prestressing force is considered in the term 0.8fcp.

For a flanged section, when the CGC is in the flange, the intersection of web and flange

is considered to be the critical location. The expression of Vc0 is modified by substituting

0.8fcp with 0.8 × (the stress in concrete at the level of the intersection of web and flange).

In presence of inclined tendons or vertical prestress, the vertical component of the

prestressing force (Vp) can be added to Vc0.

(5-2.5)

For cracked sections,

(5-2.6)



Vcr is the shear corresponding to flexure shear cracking. The term (1 – 0.55fpe /fpk)τcbd

is the additional shear that changes a flexural crack to a flexure shear crack.

The notations in the previous equation are as follows.

fpe = effective prestress in the tendon after all losses

≤ 0.6fpk

fpk = characteristic strength of prestressing steel

τc = ultimate shear stress capacity of concrete, obtained from Table 6 of

IS:1343 - 1980. It is given for values of Ap / bd, where d is the depth of

CGS. The values are plotted in the next figure.


= bw , breadth of the web for flanged sections

d = distance from the extreme compression fibre to the centroid of the

tendons at the section considered

M0 = moment initiating a flexural crack

Mu = moment due to ultimate loads at the design section

Vu = shear due to ultimate loads at the design section.

0

0.4

0.8

1.2

0 1 2 3A p /bd x 100

τc(N

/mm

2 )

4

M30 M40 Figure 5-2.4 Variation of shear strength of concrete

The term (M0/Mu)Vu is the shear corresponding to the moment M0, that decompresses

(nullifies the effect of prestress) the tension face and initiates a flexural crack. The

expression of M0 is given below.

ptIM = fy0 0.8

(5-2.7)




fpt = magnitude of the compressive stress in concrete at the level of CGS due to

prestress only.

An equal amount of tensile stress is required to decompress the concrete at the level of

CGS. The corresponding moment is fptI / y.

In the expression of M0,

I = gross moment of inertia

y = depth of the CGS from CGC.

The factor 0.8 implies that M0 is estimated to be 80% of the moment that decompresses

the concrete at the level of CGS. Since the concrete is cracked and the inclination of

tendon is small away from the supports, any vertical component of the prestressing

force is not added to Vcr.

Maximum Permissible Shear Stress To check the crushing of concrete in shear compression failure, the shear stress is

limited to a maximum value (τc,max). The value of τc,max depends on the grade of

concrete and is given in Table 7 of IS:1343 - 1980.

τ≤uc,max

t

Vbd

(5-2.8)

In the previous expression,

dt = greater of dp or ds


ds = depth of centroid of regular steel

Vu = shear force at a section due to ultimate loads.



0

2

4

6

30 40 50 60

f ck (N/mm2)

τc, m

ax (N

/mm

2 )

Figure 5-2.5 Variation of maximum permissible shear stress in concrete



5.3 Design for Shear (Part II) This section covers the following topics.

• Design of Transverse Reinforcement

• Detailing Requirements

• Design Steps

5.3.1 Design of Transverse Reinforcement

When the shear demand (Vu) exceeds the shear capacity of concrete (Vc), transverse

reinforcements in the form of stirrups are required. The stirrups resist the propagation of

diagonal cracks, thus checking diagonal tension failure and shear tension failure.

The stirrups resist a failure due to shear by several ways. The functions of stirrups are

listed below.

1) Stirrups resist part of the applied shear.

2) They restrict the growth of diagonal cracks.

3) The stirrups counteract widening of the diagonal cracks, thus maintaining

aggregate interlock to a certain extent.

4) The splitting of concrete cover is restrained by the stirrups, by reducing dowel

forces in the longitudinal bars.

After cracking, the beam is viewed as a plane truss. The top chord and the diagonals

are made of concrete struts. The bottom chord and the verticals are made of steel

reinforcement ties. Based on this truss analogy, for the ultimate limit state, the total area

of the legs of the stirrups (Asv) is given as follows.

sv u c

v y t

A V -V=s f d0.87

(5-3.1)

The notations in the above equation are explained.

sv = spacing of the stirrups



ds = depth of centroid of non-prestressed steel

fy = yield stress of the stirrups



The grade of steel for stirrups should be restricted to Fe 415 or lower.

Design of Stirrups for Flanges For flanged sections, although the web carries the vertical shear stress, there is shear

stress in the flanges due to the effect of shear lag. Horizontal reinforcement in the form

of single leg or closed stirrups is provided in the flanges. The following figure shows

the shear stress in the flange at the face of the web.

Figure 5-3.1 Shear stress in flange due to shear lag effect

The horizontal reinforcement is calculated based on the shear force in the flange. The

relevant quantities for the calculation based on an elastic analysis are as follows.

1) Shear flow (shear stress × width)

2) Variation of shear stress in a flange (τf)

3) Shear forces in flanges (Vf).

4) Ultimate vertical shear force (Vu)

The following sketch shows the above quantities for an I-section (with flanges of

constant widths).

τf τf max

Shear flow

Vf Vf

Vu

Shear forces

VfVf

bf

Df

τf τf max

Shear flow

τf τf max

Shear flow

Vf Vf

Vu

Shear forces

VfVf

bf

Df

Vf Vf

Vu

Shear forces

VfVf

bf

Df

Figure 5-3.2 Shear flow and shear forces in an I-section



The design shear force in a flange is given as follows.

τ f max ff f

bV = D2 2

(5-3.2)

Here,



τf,max = maximum shear stress in the flange.

The maximum shear stress in the flange is given by an expression similar to that for the

shear stress in web.

τ uf max

f

V A y=I D

1 (5-3.3)

Here,

Vu = ultimate vertical shear force

I = moment of inertia of the section.

A1 = area of half of the flange

= distance of centroid of half of the flange from the neutral axis at CGC.

yA1

yA1

Figure 5-3.3 Cross-section of a beam showing the variables for calculating shear

stress in the flange

The amount of horizontal reinforcement in the flange (Asvf) is calculated from Vf.

fsvf

y

VA =f0.87

(5-3.4)

The yield stress of the reinforcement is denoted as fy .



5.3.2 Detailing Requirements

The detailing requirements for the stirrups in IS:1343 - 1980 are briefly mentioned.

Maximum Spacing of Stirrups The spacing of stirrups (sv) is restricted so that a diagonal crack is intercepted by at

least one stirrup. This is explained by the following sketch.

sv

Elevation

dp ds h

Cross-section

bw

sv

Elevation sv

Elevation

dp ds h

Cross-section

bw

dp ds h

Cross-section

bw

Figure 5-3.4 Cross-section and elevation of a beam showing stirrups

As per Clause 22.4.3.2, the maximum spacing is 0.75dt or 4bw , whichever is smaller.

When Vu is larger than 1.8Vc , the maximum spacing is 0.5dt.

The variables are as follows.

bw = breadth of web



ds = depth of centroid of non-prestressed steel

Vu = shear force at a section due to ultimate loads

Vc = shear capacity of concrete.

Minimum Amount of Stirrups A minimum amount of stirrups is necessary to restrict the growth of diagonal cracks and

subsequent shear failure. For Vu < Vc, minimum amount of transverse reinforcement is

provided based on the following equation.

sv

v y

A =bs f

0.40.87

(5-3.5)




= bw, breadth of the web for flanged sections.

If Vu < 0.5Vc and the member is of minor importance, stirrups may not be provided.

Another provision for minimum amount of stirrups (Asv,min) is given by Clause 18.6.3.2

for beams with thin webs. The minimum amount of stirrups is given in terms of Awh , the

horizontal sectional area of the web in plan. The area is shown in the following sketch.

Awh

Section A - A

Elevation

A A

Awh

Section A - A

Awh

Section A - A

Elevation

A A

Figure 5-3.5 Elevation and horizontal section of a beam showing stirrups

In presence of dynamic load,

Asv,min = 0.3% Awh

= 0.2% Awh , when h ≤ 4bw

With high strength bars,

Asv,min = 0.2% Awh

= 0.15% Awh , when h ≤ 4bw

In absence of dynamic load, when h > 4bw

Asv,min = 0.1% Awh

There is no specification for Asv,min when h ≤ 4bw .



Anchorage of Stirrups The stirrups should be anchored to develop the yield stress in the vertical legs.

1) The stirrups should be bent close to the compression and tension surfaces,

satisfying the minimum cover.

2) Each bend of the stirrups should be around a longitudinal bar. The diameter of

the longitudinal bar should not be less than the diameter of stirrups.

3) The ends of the stirrups should be anchored by standard hooks.

4) There should not be any bend in a re-entrant corner. In a re-entrant corner, the

stirrup under tension has the possibility to straighten, thus breaking the cover

concrete.

The following sketches explain the requirement of avoiding the bend of a stirrup at a re-

entrant corner.

Correct detailing

Re-entrant corner

Incorrect detailing Correct detailing Correct detailing

Re-entrant corner

Incorrect detailing

Re-entrant corner

Incorrect detailing

Re-entrant corner

Incorrect detailing Figure 5-3.6 Cross-section of the bottom flange of a beam showing stirrups

Minimum Thickness (Breadth) of Web To check web crushing failure, The Indian Roads Congress Code IRC:18 - 2000

specifies a minimum thickness of the web for T-sections (Clause 9.3.1.1). The minimum

thickness is 200 mm plus diameter of the duct hole. 5.3.3 Design Steps

The following quantities are known.

Vu = factored shear at ultimate loads. For gravity loads, this is calculated

from VDLand VLL.

VDL = shear due to dead load

VLL = shear due to live load.



After a member is designed for flexure, the self-weight is known. It is included as dead

load.

The grade of concrete is known from flexure design. The grade of steel for stirrups is

selected before the design for shear. As per IS:1343 - 1980, the grade of steel is

limited to Fe 415.

The following quantities are unknown.

Vc = shear carried by concrete

Asv = total area of the legs of stirrups within a distance sv

sv = spacing of stirrups.

The steps for designing stirrups along the length of a beam are given below.

1) Calculate the shear demand Vu at the critical location.

2) Check (Vu / bdt) < τc,max. If it is not satisfied, increase the depth or breadth of the

section. Here, b is the breadth of the web (bw) and dt is larger of dp and ds.

3) Calculate the shear capacity of concrete Vc from the lower of Vco and Vcr. In

presence of inclined tendons or vertical prestress, the vertical component of the

prestressing force (Vp) can be added to Vc0.

4) Calculate the requirement of shear reinforcement through Asv / sv . Compare the

value with the minimum requirement.

5) Calculate the maximum spacing and round it off to a multiple of 5 mm.

6) Calculate the size and number of legs of the stirrups based on the amount

required, type of section and space to accommodate.

Repeat the calculations for other locations of the beam, if the spacing of stirrups needs

to be varied.



Example 5-3.1 Design the stirrups for the Type 1 prestressed beam with the following section (location of tendons shown at mid span).

100

435

920

100

290

100

CGC

CGS

(10) 7-wire strandswith Pe = 826 kN

100

435

920

100

290

100

CGC

CGS

100

435

920

100

290

100

CGC

CGS

(10) 7-wire strandswith Pe = 826 kN

Longitudinal reinforcement of 12 mm diameter is provided to hold the stirrups. The properties of the sections are as follows.

A = 159,000 mm2, I = 1.7808 × 1010 mm4

Ap = 960 mm2

The grade of concrete is M 35 and the characteristic strength of the prestressing steel (fpk) is 1470 N/mm2. The effective prestress (fpe) is 860 N/mm2. The uniformly distributed load including self weight, is wT = 30.2 kN/m. The span of the beam (L) is 10.7 m. The width of the bearings is 400 mm. The clear cover to longitudinal reinforcement is 30 mm.

Solution

1) Calculate Vu at the face of the support (neglecting the effect of compression in

concrete). ⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟⎝ ⎠

=1.5× -2

10.7=1.5×30.2× - 0.22

= 233.3 kN

u TLV w x



Here, x denotes half of the width of bearing. x = 200 mm.

2) Check (Vu / bdt) < τc,max .

Effective depth dt = total depth – cover – diameter of stirrups – ½ diameter of

longitudinal bar.

Assume the diameter of stirrups to be 8 mm.

⎛ ⎞⎜ ⎟⎝ ⎠

1= 920 - 30 - 8 - ×122

= 876 mm

td

3

2

233.3×10=100×876

= 2.7 N/mm

u

w t

Vb d

τc,max for M 35 is 3.7 N/mm2. Hence, (Vu / bdt) < τc,max .

3) Calculate Vc from the lower of Vc0 and Vcr .

2= 0.67bD +0.8co t cp tV f f f

Here,

3

2

826×10=159,000

= 5.19 N/mm

ecp

Pf =A2

= 0.24 35

=1.42 N/mmtf

2t

2

= 0.67 +0.8

= 0.67×100×920 1.42 +0.8×5.19×1.42=173.4 kN

co t cpV bD f f f

The vertical component of the prestressing force can be found out from the equation of

the parabolic tendon.

myy = x ( L - x )

L 24



ym

y

L

xθ ym

y

L

xθ

The following is the expression of the slope of the parabolic tendon.

2

4ta n 2md y yθ = = ( L - xd x L

)

ym

y

L

xθ ym

y

L

xθ

At x = 0.2 m, y = 20 mm, dy/dx = 0.105 and θ = 6.0°.

p eV = P θsin

= 826×0.104= 86.0kN

coV + =173.4+86.0

= 259.4 kNpV

Vp Pe

θ

Vp Pe

θ

⎛ ⎞⎜ ⎟⎝ ⎠

pe ucr c

pk u

f VV = τ bd + Mf M01- 0.55

Here, 860= = 0.1470

pe

pk

ff

58

100 100×960=100×480

= 2.0

pAbd

d = 460 + y

= 460 + 20

= 480 mm



From Table 6, for M 35 concrete, τc = 0.86 N/mm2.

pt

IM fy0 = 0.8

Here,

3 3

10

2

826×10 826×10 ×20= - - ×20159,000 1.7808×10

= -5.19 - 0.02= -5.21N/mm

e ept

P P yf = - - yA I

10

0

6

1.7808×10= 0.8×5.21×20

= 3711.2×10 Nmm= 3711.2 kNm

M

At the critical section,

=1.5 ( - )2

0.2=1.5×30.2× (10.7 - 0.2)2

= 47.6 kNm

u TxM w L x

⎛ ⎞⎜ ⎟⎝ ⎠

pe ucr c

pk u

f VV = τ bd + Mf M01- 0.55Therefore,

crV 30.86 233.3= (1- 0.55×0.58)× ×100×480 + 3711.2×

47.610= 28.1+18204.8=18232.9 kN

The governing value of Vc is 259.4 kN.

⇒ Vu < Vc .

4) Calculate Asv / sv .

Provide minimum steel.

sv

w v y

A =b s f

0.40.87



5) Calculate maximum spacing

sv = 0.75 dt = 0.75 × 876 = 656 mm

sv = 4bw = 4 ×100 = 400 mm

Select sv = 400 mm.

6) Calculate the size and number of legs of the stirrups

Select fy = 250 N/mm2.

2

0.40.87

0.4=100×400×0.87×250

= 73.6 mm

sv w vy

A = b sf

Provide 2 legged stirrups of diameter 8 mm.

2

= 2×50.3

=100.6 mmsv,providedA

Check minimum amount of stirrups.

2

= 0.1%0.1= ×100×400100

= 40 mm

sv,min whA A

Provided amount of stirrups is larger. OK.

Provide same spacing of stirrups throughout the span.

Design of stirrups for flange

1

2

1= ×21= ×435×1002

= 21750 mm

f fA b × D



= 410 mmy

τ 1

3

10

2

233.3×10 ×21750×410=1.7808×10 ×100

=1.17 N/mm

uf max

f

V A y=I D

2 21.17 435= × ×10

2 2=12724 N

f max ff f

τ bV = D

0

2

0.8712724=

0.87×250= 59.0 mm

fsvf

y

VA =f

2

0.40.87

0.4=100×400×0.87×250

= 73.6 mm

svf f vy

A = D sf

For minimum steel

Provide 2 legged stirrups of diameter 8 mm.

Designed section

8 mm diameter stirrups@ 400 mm c/c8 mm diameter stirrups@ 400 mm c/c



5.4 Analysis for Torsion This section covers the following topics.

• Stresses in an Uncracked Beam

• Crack Pattern Under Pure Torsion

• Components of Resistance for Pure Torsion

• Modes of Failure


Introduction The analysis of reinforced concrete and prestressed concrete members for torsion is

more difficult compared to the analyses for axial load or flexure.

The analysis for axial load and flexure are based on the following principles of

mechanics.

1) Equilibrium of internal and external forces

2) Compatibility of strains in concrete and steel

3) Constitutive relationships of materials.

The conventional analysis of reinforced concrete and prestressed concrete members for

torsion is based on equilibrium of forces by simple equations. The compatibility of

strains in concrete and steel reinforcement is not considered.

The strength of each material, concrete or steel, corresponds to the ultimate strength.

The constitutive relationship of each material, relating stress and strain, is not used.

Torsion generated in a member can be classified into two types based on the necessity

of analysis and design for torsion.

1) Equilibrium torsion: This is generated due to loading eccentric to the centroidal

axis. For example, a) in a beam supporting cantilever slab or precast slab or floor

joists on one side, b) in a (curved) bridge deck subjected to eccentric live load

and c) in an electric pole subjected to loads from wires on one side.

The torsion demand is determined by equilibrium condition only. The member needs to

be analysed and designed for torsion. The following figure shows the situations where

eccentric loads are acting on the members.



a) Bridge deck

+ CGC+CGC

b) L-beama) Bridge deck

+ CGC+ CGC+CGC +CGC

b) L-beam Figure 5-4.1 Examples of members under eccentric load

2) Compatibility torsion: This is generated by twisting, to maintain compatibility in

deformation with the connected member. This type of torsion generates in a

primary beam supporting secondary beams.

In compatibility torsion, the torsion demand is determined by both equilibrium and

compatibility conditions. Else, the torsion can be neglected. This implies that primary

beam need not be analysed and designed for torsion, if the secondary beams are

designed as pin supported.

In this section, the emphasis is laid on equilibrium torsion. To understand the behaviour

of a beam under torsion, the presentation will be in the following sequence.

1) Stresses in an uncracked (homogenous) rectangular beam without prestressing

due to pure torsion (in absence of flexure), with constant torque along the span.

2) Crack pattern under pure torsion.

3) Components of resistance for pure torsion.

4) Modes of failure under combined torsion and flexure.

5) Effect of prestressing force.

Although pure torsion is absent in structures, understanding the behaviour of a beam

under pure torsion helps to analyse a beam under combined torsion, flexure and shear.

5.4.1 Stresses in an Uncracked Beam The following figure shows a beam of rectangular cross-section under pure torsion. The

variations of the torsional shear stress (τ) along radial lines in the cross-section are

shown. It can be observed that the maximum shear stress (τmax) occurs at the middle of



the longer side. Hence, the subsequent explanation will refer to the stress condition at

the middle of the longer side.

Variation of torsional shear stress (τ) in the cross-section

1

T

τmax

Variation of torsional shear stress (τ) in the cross-section

1

T

11

T

τmaxτmaxτmax

Figure 5-4.2 Beam subjected to pure torsion

At any point in the beam, the state of two-dimensional stresses can be expressed in

terms of the principal stresses. The Mohr’s circle of stress is helpful to understand the

state of stress.

Before cracking, the stress carried by steel is negligible. When the principal tensile

stress exceeds the cracking stress, the concrete cracks and there is redistribution of

stresses between concrete and steel.

For a point at the middle of the longer side (Element 1), the torsional shear stress is

maximum. The principal tensile stress (σ1) is inclined at 45º to the beam axis.

σ1 σ2

α = 45o

Principal stresses

σ1σ2

τ

2α

Mohr’s circle

τ

1

State of pure shear

σ1 σ2

α = 45o

Principal stresses

σ1 σ2

α = 45o

Principal stresses

σ1σ2

τ

2α

Mohr’s circle

σ1σ2

τ

2α

Mohr’s circle

τ

1

State of pure shear

τ

1

State of pure shear

Figure 5-4.3 State of stresses at the side of a beam

Since the torsion is maximum at middle of the longer side, cracks due to torsion occur

around that location and perpendicular to σ1.



Crack inclination

1

σ1

σ2

τ

Crack inclination

1

σ1

σ2

τ

1

σ1

σ2

τ

Figure 5-4.4 Inclination of crack at the side of a beam

5.4.2 Crack Pattern under Pure Torsion

The cracks generated due to pure torsion follow the principal stress trajectories. The

first cracks are observed at the middle of the longer side. Next, cracks are observed at

the middle of the shorter side. After the cracks connect, they circulate along the

periphery of the beam.



a) Initiation of torsional cracks in longer sidea) Initiation of torsional cracks in longer side

b) Initiation of torsional cracks in shorter sideb) Initiation of torsional cracks in shorter side

c) Spiral torsional cracksc) Spiral torsional cracks Figure 5-4.5 Formation of cracks in a beam subjected to pure torsion

In structures, a beam is not subjected to pure torsion. Along with torsion it is also

subjected to flexure and shear. Hence, the stress condition and the crack pattern are

more complicated than shown before.

5.4.3 Components of Resistance for Pure Torsion

After cracking, the concrete forms struts carrying compression. The reinforcing bars act

as ties carrying tension. This forms a space truss. Since the shear stress is larger near

the sides, the compression in concrete is predominant in the peripheral zone. This is

called the thin-walled tube behaviour. The thickness of the wall is the shear flow zone,



where the shear flow is assumed to be constant. The portion of concrete inside the

shear flow zone can be neglected in calculating the capacity.

The components in vertical and horizontal sections of a beam are shown below.

Figure 5-4.6 Internal forces in a beam

The components can be denoted as below.

Tc = torsion resisted by concrete

Ts = torsion resisted by the longitudinal and transverse reinforcing bars.

The magnitude and the relative value of each component change with increasing torque.

5.4.4 Modes of Failure

For a homogenous beam made of brittle material, subjected to pure torsion, the

observed plane of failure is not perpendicular to the beam axis, but inclined at an angle.

This can be explained by theory of elasticity. A simple example is illustrated by applying

torque to a piece of chalk.



Failure surfaceFailure surface

Figure 5-4.7 Failure of a piece of chalk under torque

For a beam of rectangular section, the plane of failure is further influenced by warping.

Torsional warping is defined as the differential axial displacement of the points in a

section perpendicular to the axis, due to torque.

For a reinforced concrete beam, the length increases after cracking and after yielding of

the bars. For a beam subjected to flexure and torsion simultaneously, the modes of

failure are explained by the Skew Bending Theory. The observed plane of failure is

not perpendicular to the beam axis, but inclined at an angle. The curved plane of failure

is idealised as a planar surface inclined to the axis of the beam.

The skew bending theory explains that the flexural moment (Mu) and torsional moment

(Tu) combine to generate a resultant moment inclined to the axis of the beam. This

moment causes compression and tension in a planar surface inclined to the axis of the

beam. The following figure shows the resultant moment due to flexural moment and

torsion in a beam.

TuMu

Resultant momentTuMu

Resultant moment Figure 5-4.8 Beam subjected to flexural moment and torsion



The modes of failure are explained based on the relative magnitudes of the flexural

moment (Mu) and torsional moment (Tu) at ultimate. Three discrete modes of failure are

defined from a range of failure. The idealised pattern of failure with the plane of failure

and the resultant compression (Cu) and tension (Tu) are shown for each mode

(Courtesy: Pillai, S. U., and Menon, D., Reinforced Concrete Design).

1) Modified bending failure (Mode 1): This occurs when the effect of Mu is larger

than that of Tu.

Zone under Cu

Tu

Zone under Cu

Tu

Zone under Cu

Tu Figure 5-4.9 Idealised pattern for Mode 1 failure

2) Lateral bending failure (Mode 2): This is observed in beams with thin webs

when the effect of Mu and Tu are comparable.

Zone under Cu

TuZone under Cu

Tu

Figure 5-4.10 Idealised pattern for Mode 2 failure

3) Negative bending failure (Mode 3): When the effect of Tu is large and the top

steel is less, this mode of failure occurs.



Zone under Cu

Tu

Zone under Cu

Tu

Figure 5-4.11 Idealised pattern for Mode 3 failure


In presence of prestressing force, the cracking occurs at higher load. This is evident

from the typical torque versus twist curves for sections under pure torsion.

With further increase in load, the crack pattern remains similar but the inclinations of the

cracks change with the amount of prestressing. The following figure shows the

difference in the torque versus twist curves for a non-prestressed beam and a

prestressed beam.

Twist

Torque

Prestressed beam Non-prestressed beam

Ultimate strength

Cracking torque levels

Twist

Torque

Prestressed beam Non-prestressed beam

Ultimate strength

Cracking torque levels

Figure 5-4.12 Torque versus twist curves

The effect of prestressing force is explained for a beam under pure torsion with a

concentric prestressing force (Pe). The following figure shows such a beam.



Pe

TPePe

T

Figure 5-4.13 Beam subjected to pure torsion and prestressing force

For a point at the middle of the longer side (Element 1), there is normal stress due to

the prestressing force (–fpe). The principal tensile stress (σ1) is inclined to the neutral

axis at an angle greater than 45º.

σ2 σ1

(–fpe,τ) 2α > 90°

Mohr’s circle

σ2

σ1

α > 45o

Principal stresses

τ

1fpe


σ2 σ1

(–fpe,τ) 2α > 90°

Mohr’s circle

σ2 σ1

(–fpe,τ) 2α > 90°

Mohr’s circle

σ2

σ1

α > 45o

Principal stresses

σ2

σ1

α > 45o

σ2

σ1

α > 45o

Principal stresses

τ

1fpe


τ

1fpe

Shear stress and prestress Figure 5-4.14 State of stresses at the side of a prestressed beam

In the following figure, the formation of cracks for a prestressed beam under pure

torsion is shown. This figure can be compared with that for a reinforced concrete beam.

Figure 5-4.15 Formation of cracks in a prestressed beam

In presence of prestressing force, the cracking is at a higher torque. After cracking, the

crack width of a spiral crack is low. Thus, the aggregate interlock is larger as compared

to a non-prestressed beam under the same torque. Hence, the torsional strength of



concrete (Tc) increases in presence of prestressing force. This is accounted for in the

expression of Tc.



5.5 Design for Torsion (Part I) This section covers the following topics.

• General Comments

• Limit State of Collapse for Torsion

• Design of Longitudinal Reinforcement

5.5.1 General Comments

Calculation of Torsion Demand The restraint to torsion is provided at the ends of a beam. For beams in a building frame,

the restraint is provided by the columns. Precast beams are connected at the ends by

additional elements like angles to generate the torsional restraint. In bridges,

transverse beams at the ends provide torsional restraint to the primary longitudinal

girders. Box girders are provided with diaphragms at the ends.

For equilibrium torsion in a straight beam with distributed torque (tu), the maximum

torsional moment (Tu) is near the restraint at the support. The following figure shows a

schematic representation of the distributed torque.

tu

Tu

L

tu

Tu

L

Figure 5-6.1 Beam subjected to distributed torque

The torsional moment near the support is given by the following expression.

u

ut LT =2 (5-5.1)

Here,

L = clear span of the beam

tu = distributed torque per unit length.



For a straight beam with a point torque, the maximum torsional moment (Tu) is near the

closer support. If the location of the point torque is variable, Tu is calculated for the

location closest to a support. For a curved beam, Tu is calculated based on structural

analysis.

Design of Torsion Reinforcement The design is done for the critical section. The critical section is defined in Clause 41.2

of IS:456 - 2000. In general cases, the face of the support is considered as the critical

section. When the reaction at the support introduces compression at the end of the

beam, the critical section can be selected at a distance effective depth from the face of

the support.

To vary the amount of reinforcement along the span, other sections may be selected for

design. Usually the following scheme is selected for the stirrup spacing in beams under

uniformly distributed load.

1) Close spacing for quarter of the span adjacent to the supports.

2) Wide spacing for half of the span at the middle.

For large beams, more variation of spacing may be selected. The following sketch

shows the typical variation of spacing of stirrups. The span is represented by L.

L / 4 L / 4L / 2L / 4 L / 4L / 2

Figure 5-6.2 Typical variation of spacing of stirrups

First, an equivalent flexural moment Mt is calculated from Tu. Second, for the design of

primary longitudinal reinforcement, including the prestressed tendon, the total

equivalent ultimate moment (Me1) is calculated from the flexural moment (Mu) and Mt.

Third, the design of longitudinal reinforcement for other faces based on equivalent

ultimate moments Me2 and Me3 is necessary when the equivalent moment Mt is larger

than Mu. The following sketch shows the equivalent ultimate moments for design.



Me1 Me2Me3Me1Me1 Me2Me2Me3Me3

Figure 5-6.3 Equivalent ultimate moments

The design for Me1 is similar to the design of a prestressed section for flexure.

The design for Me2 is similar to the design of a prestressed concrete or reinforced

concrete section. The design for Me3 is similar to the design of a reinforced concrete

section. The design of stirrups including torsion is similar to the design of stirrups in

absence of torsion.

5.5.2 Limit State of Collapse for Torsion

The design for the limit state of collapse for torsion is based on the Skew Bending

Theory. For a beam subjected to simultaneous flexure and torsion, an equivalent

ultimate bending moment at a section is calculated.

The design for torsion involves the design of longitudinal reinforcement as well as the

transverse reinforcement. The longitudinal reinforcement is designed based on the

equivalent ultimate bending moment.

The transverse reinforcement is designed based on the Skew Bending Theory and a

total shear requirement. For the capacity of concrete, to consider the simultaneous

occurrence of flexural and torsional shears, an interaction between the two is

considered.

The equations in IS:1343 - 1980 are applicable for beams of the following sections.

1) Solid rectangular, with D > b.

2) Hollow rectangular, with D > b and t ≥ b/4.

3) Flanged sections like T-beams and I-beams.

The sections are shown in the following sketch.



D

b b

bw

tSolid rectangular Hollow rectangular Flanged

D

b b

bw

tSolid rectangular Hollow rectangular Flanged

Figure 5-6.4 Different sections for torsion design

The variables are as follows.


= bw for flanged section

D = total depth of the section

t = thickness of the section.

The average prestress in a section at the level of CGC, is limited to 0.3fck.

5.5.3 Design of Longitudinal Reinforcement

For the design of the longitudinal reinforcement, there are three expressions of the

equivalent ultimate bending moment for the three modes of failure (Reference: Rangan,

B. V. and Hall, A. S., “Design of Prestressed Concrete Beams Subjected to Combined

Bending, Shear and Torsion”, ACI Journal, American Concrete Institute, March 1975,

Vol. 72, No. 3, pp. 89 – 93). The modes of failure are explained in Section 5.4, Analysis

for Torsion. The figures of the failure pattern are reproduced here for explanation.

Mode 1 Failure



Zone under Cu

Tu

Me1

Zone under Cu

Tu

Me1

Zone under Cu

Tu

Zone under Cu

Tu

Zone under Cu

Tu

Me1Me1

Figure 5-6.5 Idealised pattern and design moment for Mode 1 failure

The equivalent ultimate bending moment for Mode 1 failure (Me1) is given by the

following equation.

(5-5.2) e uM = M + M1 t

The equivalent bending moment for Tu is given as follows.

⎛ ⎞⎜ ⎟⎝ ⎠t u

DM =T +b

21 (5-5.3)


Mu = applied bending moment at ultimate.

Mt = additional equivalent bending moment for torsion.

Tu = applied torsion at ultimate.

Since, the torsion generates tension in the reinforcement irrespective of the sign, the

sign of Mt is same as that of Mu.

Mode 2 Failure



Zone under Cu

Tu Me3Zone under Cu

TuZone under Cu

Tu Me3Me3


The equivalent ultimate transverse bending moment for Mode 2 failure (Me3) is given as

follows.

⎛ ⎞⎜ ⎟⎛ ⎞

⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎜ ⎟⎝ ⎠

e t

b+x DM = M + De +b

21

3

211 22 1

(5-5.4)

In the previous expression

e = Tu/Vu, ratio of ultimate torsion and ultimate shear force at a section.

x1 = smaller dimension of a closed stirrup.

The larger dimension of a closed stirrup is represented as y1. the dimensions are

shown in the following sketch.

x1

y1

x1

y1

Figure 5-6.7 Dimensions of a closed stirrup

The transverse bending moment Me3 is considered when the numerical value of Mu is

less than Mt. Me3 acts about a vertical axis.

Mode 3 Failure



Zone under Cu

Tu

Me2Zone under Cu

Tu

Zone under Cu

Tu

Me2Me2


The equivalent ultimate bending moment for Mode 3 failure (Me2) is given by the

following equation.

Me2 = Mt – Mu (5-5.5) The expression of Mt is same as for Mode 1 failure, given before.

Mode 3 failure is checked when the numerical value of Mu is less than that of Mt. Me2

acts in the opposite sense of that of Mu.

The longitudinal reinforcement is designed for Me1 similar to the flexural reinforcement

for a prestressed beam. The design of flexural reinforcement is covered in Section 4.2,

Design of Sections for Flexure (Part I) and Section 4.3 Design of Sections for Flexure

(Part II). When Me2 is considered, longitudinal reinforcement is designed similar to a

prestressed concrete or reinforced concrete beam. When Me3 is considered,

longitudinal reinforcement is designed similar to a reinforced concrete beam. For a

singly reinforced rectangular section, the amount of longitudinal reinforcement (As) is

solved from the following equation.

(5-5.6)

⎛ ⎞⎜ ⎟⎝ ⎠

y sy s u

ck

f Af A d - = M

f bd0.87 1


d = effective depth of longitudinal reinforcement

fy = characteristic yield stress of longitudinal reinforcement


Mu = one of Me2 and Me3.



5.6 Design for Torsion (Part II) This section covers the following topics.

• Design of Transverse Reinforcement

• Detailing Requirements

• Design Steps

5.6.1 Design of Transverse Reinforcement

For the design of the transverse reinforcement, the capacities of concrete to resist the

torsion and shear need to be determined. To consider the simultaneous occurrence of

flexural and torsional shears, a linear interaction between the two is considered.

The capacity of concrete to resist torsion is reduced from Tc, the capacity under pure

torsion. Similarly, the capacity of concrete to resist shear is reduced from Vc, the

capacity in absence of torsion.

Capacity of Concrete under Pure Torsion The capacity of concrete is determined based on the plastic theory for torsion. The

capacity is equal to the torque generating the first torsional crack (Tcr).

For a reinforced concrete beam, Tcr is estimated by equating the maximum torsional

shear stress (τmax) caused by Tcr to the tensile strength of concrete (0.2√fck). The

estimated tensile strength is less than that under direct tension because the full section

does not plastify as assumed in the plastic theory.

The estimate of the cracking torque (Tcr) for a rectangular section is given below.

⎛ ⎞≈ ⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟⎝ ⎠

cr ck

cr ck

b D bT f -D

bT = b D - fD

2

2

0.2 12 3

0.1 13

(5-6.1)

For flanged sections, the section is treated as a compound section. A compound section

is a summation of rectangular sections.

The cracking torque is estimated as a summation of the capacities of the individual

rectangular sections. Since the interaction between the rectangular sections is



neglected in the summation, the estimate of the cracking torque is a lower bound

estimate.

The following flanged section is shown as a compound section of five rectangles. For an

individual rectangle, the short side is b and the long side is D.

1 3

2

4 5

Figure 5-6.1 Flanged section as a compound section

The estimate of the cracking torque (Tcr) for a compound section is as follows.

⎛ ⎞∑ ⎜ ⎟⎝ ⎠cr ck

bT = b D - fD

20.1 13 (5-6.2)

resist torsion (Tc) for a prestressed concrete beam are th

6.4)

Here, the summation is for the individual rectangles.

For a prestressed concrete beam, the strength of concrete is multiplied by the factor λp,

which is a function of the average effective prestress (fcp).

cpp

ck

fλ = +

f12

1 (5-6.3)

The value of fcp is taken as positive (numeric value). It can be observed that the strength

increases with prestress. The cracking torque (Tcr) and the capacity of concrete to

us estimated as follows.

(5-

⎛ ⎞∑ ⎜ ⎟⎝ ⎠

c cr

c p

T =TbT = b D - λ fD

20.15 13 ck




b = breadth of the individual rectangle

idual rectangle.

InteracIn presence of flexural shear, the torsional capacity of concrete reduces. Similarly, in

r capacity of concrete reduces. This is referred to

D = depth of the indiv

tion of Shear and Torsion

presence of torsion, the flexural shea

as interaction of shear and torsion. The capacity of concrete under shear is explained

in Section 5.2, Design for Shear (Part I). A linear interaction of the shear and torsion

capacities of concrete is considered as shown in the following figure. In the horizontal

axis, the shear demand is normalised with respect to the capacity of concrete under

flexural shear. In the vertical axis, the torsional demand is normalised with respect to

the capacity of concrete under pure torsion.

Tu/Tc

Vu/Vc1.0

1.0

Tu/Tc

Vu/Vc1.0

1.0

Figure 5-6.2 Interaction diagram for shear and torsion

The interaction

(5-6.5) This is a linear interaction equatio


ultimate

ure torsion.

flexural shear.

Based city of concrete to resist torsion

(Tc1) is given below.

equation is given as follows.

u uT Vc c

+ =T V

1

n.

Tu = applied torsion at ultimate

Vu = applied shear at

Tc = capacity of concrete under p

Vc = capacity of concrete under

on the interaction equation, the reduced capa



(5-6.6)

Tc1 is limited to Tu/2 to restrict co reaching its capacity.

t ultimate. The parameter

is the ratio of the corresponding concrete capacities.

(5-6.8) The reduced capacity of concrete to resist

(5-6.9)

Calculation of Transverse ReinThe transverse reinforcement is provided in the form of closed stirrups enclosing the

equal to the higher value determined from

is based on the Skew Bending Theory.

(5-6.10)

The notations are as follows.

b1 = distance between the corner longitudinal bars along the short side

the corner longitudinal bars along the long side.

The di in the following sketch.

ncrete

The parameter e is the ratio of torsion and shear demands a

ec

e = Tu/ Vu (5-6.7) ec = Tc/ Vc

shear is given below.

forcement

corner longitudinal bars. The amount (Asv) is

two expressions.

The first expression

d1 = distance between

Mt = additional bending moment from torsion.

sv = spacing of the stirrups

fy = characteristic yield stress of the stirrups.

mensions b1 and d1 are shown

cc c

c

V =Ve+ e1

e

t vsv

y

M sA =b d f1.5 1 1

⎛ ⎞e⎜ ⎟⎝ ⎠

≤

c1 cc

u

T =Te+ e

T / 2



b1

d1

b1

d1

Figure 5-6.3 Dimensions between the corner bars

The second expression of Asv is based on the concept of total shear.

(5-6.11) sv v TA = A + A2

The first component Av corresponds to the flexural shear to be carried by the stirrups.

The second component At corresponds to the torsional shear to be carried by the

stirrups. The factor 2 considers that the torsional shear is additive to flexural shear in

both the legs.

The following sketch shows the addition of flexural and torsional shears for a hollow

section.

Due to Vu Due to Tu

+

Due to Vu Due to Tu

+

Figure 5-6.4 Distribution of flexural and torsional shears for a hollow section

The two shears are additive in the left web, whereas they are subtractive in the right

web. Since, the stirrups have equal areas in the two legs, the torsional shear is

considered additive to flexural shear in both the legs.

In solid sections, the two shears are not additive throughout the web. The flexural shear

is distributed, whereas the torsional shear is restricted in the shear flow zone. Thus for



solid sections, the expression of Asv is conservative. The following sketch shows the

addition of flexural and torsional shears for a solid section.

Due to Vu Due to TuDue to Vu Due to TuDue to Vu Due to Tu Figure 5-6.5 Distribution of flexural and torsional shears for a solid section

If the breadth of the web is large, the two shears can be designed separately. The

stirrups for flexural shear can be distributed throughout the interior of the web. For

torsional shear, closed stirrups can be provided in the peripheral shear flow zone.

The expressions of Av and At are derived from the truss analogy for the ultimate limit

state.

( )

( )

u c vv

y

u c vT

y

V -V sA =

f d

T -T sA =f b d

1

1

1

1 1

0.87

0.87

(5-6.12)

(5-6.13) The minimum amount of transverse reinforcement is same as that for shear in absence

of torsion.

sv

v y

A =bs f

0.40.87 (5-6.14)

5.6.2 Detailing Requirements

The detailing requirements for torsional reinforcement in Clause 22.5.5, IS: 1343 - 1980

are briefly mentioned.

1) There should be at least one longitudinal bar in each corner. The minimum diameter

of the longitudinal bars is 12 mm.



When any side is larger than 450 mm, provide side face reinforcement (As, sf), as per the following. Minimum amount As,sf,min = 0.1% bD Maximum spacing smax = 300 mm or b, whichever is less.

This amount is sufficient to check thermal and shrinkage cracks.

2) The closed stirrups should be bent close to the tension and compression surfaces

satisfying the minimum cover. The stirrups should be perpendicular to the axis of the

beam. Closed stirrups should not be made of pairs of U-stirrups lapping one another.

This is clarified in the following sketch.

Correct detailingIncorrect detailing Correct detailingCorrect detailingIncorrect detailingIncorrect detailing Figure 5-6.6 Detailing of closed stirrups

3) The maximum spacing is (x1 + y1)/4 or 200 mm, whichever is smaller. Here x1 and y1

are the short and long dimensions of the stirrups respectively.

4) Proper anchorage of stirrups as mentioned under detailing requirements of shear

reinforcement. It is recommended to bend the ends of a stirrup by 135º and have 10

times the diameter of the bar (db) as extension beyond the bend. The following sketch

clarifies the detailing of end hooks.

135°

10db

135°135°

10db10db

Figure 5-6.7 Detailing of end hooks for stirrups



5) The stirrups should be continued till a distance h + bw beyond the point at which it is

no longer required. Here, h is the overall depth and bw is the breadth of the web.

5.6.3 Design Steps

The following quantities are known at the selected section.

Mu = factored flexural moment

Vu = factored shear

Tu = factored torsional moment.

For gravity loads, these are calculated from the dead load and live load.

The grades of concrete and steel are selected before design. As per IS: 1343 - 1980,

the grade of steel for stirrups is limited to Fe 415.

For the design of longitudinal reinforcement, the following quantities are unknown.

The member cross-section.

Me1, Me2, Me3 = total equivalent flexural moment

Ap = amount of prestressing steel,

Pe = the effective prestress,

e = the eccentricity

As = area of longitudinal reinforcement

As’ = area of longitudinal reinforcement in opposite face.

Prestressing steel Ap’ may be provided in the opposite face.

For the design of stirrups, the following quantities are unknown.

Vc1 = shear carried by concrete

Tc1 = torsion carried by concrete

Asv = total area of the legs of stirrups within a distance sv

sv = spacing of stirrups.

The steps for designing longitudinal and transverse reinforcements for beams subjected

to torsion are given.

1) Calculate Mu, Vu and Tu at a selected location. Select a suitable cross-section.

For high value of Tu, as in bridges, a box section is preferred.



For longitudinal reinforcement 2a) Calculate Me1.

2b) Design Ap and As. The design procedure involves preliminary design and final

design, which are explained in the Section 4.2, Design of Sections for Flexure (Part

I) and Section 4.3, Design of Sections for Flexure (Part II)

3a) Calculate Me2 if Mu < Mt.

3b) Design As’. The design procedure is similar for a reinforced concrete section. If

Ap’ is provided, the design is similar to a prestressed concrete section.

4a) Calculate Me3 if Mu < Mt.

4b) Check the adequacy of transverse bending based on the corner bars. If

inadequate, design side face reinforcement (As,sf). As,sf includes the corner bars.

The design is similar to that for a reinforced concrete section.

For transverse reinforcement 5a) Calculate Tc, Eqn. (5-6.4). 5b) Calculate Vc from the lower of Vc0 and Vcr.

5c) Calculate e (if not calculated earlier) and ec.

5d) Calculate Tc1 and Vc1. Limit Tc1 to Tu/2.

6) Calculate Asv / sv from the greater of the values given by Eqns. (5-6.10), (5-6.11), (5-6.12), and (5-6.13). Compare the value with the minimum requirement Eqn. (5-6.14). 7) Calculate maximum spacing and round it off to a multiple of 5 mm.

8) Calculate the size of the stirrups based on the amount required.

Repeat the calculations for other locations of the beam if the spacing of stirrups needs

to be varied.



Example 5-6.1 Design a rectangular section to carry the following ultimate loads.

Tu = 44.5 kNm Mu = 222.5 kNm (including an estimate of self-weight) Vu = 89.0 kN.

The material properties are as follows. fck = 35 N/mm2

fy = 250 N/mm2

fpk = 1720 N/mm2

The prestressing is fpe = 1035 N/mm2.

Solution

1) Calculate Me1.

Let D/b = 2

t uDM =Tb

21+

= 44.5 1+ 2×2= 99.5 kNm

Me1 = Mu + MT

= 222.5 + 99.5

= 322.0 kNm

2) Select section. Design Ap and As.

Select

b = 250 mm

D = 500 mm

d = 450 mm.



Provide (2) 16 mm diameter corner bars. The flexural design results are as follows.

As = 2 × 201

= 402 mm2.

Required amount of prestressing steel with dp = d = 450 mm is Ap = 484 mm2.

Provide 11 mm diameter strands with area = 70 mm2.

Required number of strands = 484 / 70 = 6.8 → 7

Provided amount of prestressing steel

Ap,prov = 7 × 70

= 490 mm2

3) Calculate Me2 .

Since Mu > Mt , design for Me2 is not required.

4) Calculate Me3 .

Since Mu > Mt , design for Me3 is not required.

5a) Calculate Tc .

ecp

pe p

Pf =Af × A

=bD

2

1035×490=250×500

= 4.06 N/mm

cp

pck

fλ

f12

= 1+

12×4.06= 1+35

=1.55



fcp < 0.3 fck . OK

⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟⎝ ⎠

c p ckbT b D λ fD

2

2

= 0.15 1-3

1= 0.15×250 ×500× 1- ×1.55 35 Nmm3×2

= 35.8 kNm

5b) Calculate Vc from the lower of Vco and Vcr .

100 100×490=250×450

= 0.43

pAbd

From Table 6, for M 35 concrete, τc = 0.46 N/mm2.

e e

ptP P ef = - -A I

2

2

9

2

507,150 507,150×200= - -125,000 2.604×10

= -11.85 N/mm

Here,

e = 450 – ½ 500

= 200 mm

I = 250 × 5003 / 12

= 2.604 × 109 mm4.

pt

IM fy0

9

= 0.8

2.604×10= 0.8×11.85×200

=123.43 kNm

τ

⎛ ⎞⎜ ⎟⎝ ⎠

pe ucr c

pk u

f VV = bd + Mf M0

3

1- 0.55

0.46×250×450 89= (1- 0.55×0.6)× +123.43×10 222.5

= 84.0 kN

Here,

fpe/fpk = 1035 / 1720

= 0.6.



co cp tV bD f f2

t

2

= 0.67 f +0.8

= 0.67×250×500 1.42 +0.8×4.06×1.42= 215.6 kN

Here,

tf

2

= 0.24 35

=1.42 N/mm

∴ Vc = Vcr = 84.0 kN

5c) Calculate e and ec .

u

u

Te =V44.5=89.0

= 0.50 m

cc

c

Te =V35.8=84.0

= 0.43 m

5d) Calculate Tc1 and Vc1.

c c

c

eT =Te+ e1

0.50= 35.82×0.50+0.43

=19.26 kNm

cc c

c

eV =Ve+ e1

0.43= 84.0×0.50+0.43

= 38.84 kN

u

cTT 1 < OK2

.

6) Calculate Asv / sv

sv t

v y

A M=s b d f1 1

6

2

1.5

99.5×10=1.5×200×400×250

= 3.3 mm /mm



Estimated values

b1 = 250 – 50

= 200 mm

d1 = 500 – 100

= 400 mm.

u cv

v y

V -VA =s f d

1

1

3

2

0.87

(89.0 - 38.8)×10=0.87×250×400

= 0.58 mm /mm

u cT

v y

T -TA =s f b d

1

1 1

6

2

0.87

(44.5 -19.26)×10=0.87×250×200×400

=1.45 mm /mm

sv V T

v v v

A A As s s

2

= + 2×

= 0.58+ 2×1.45= 3.48 mm /mm

Minimum amount of stirrups

sv

v y

Abs f

0.4=0.87

sv

v

As

2

0.4×250=0.87×250

= 0.46 mm /mm

Select

Asv / sv = 3.48 mm2/mm.



7) Calculate maximum spacing

≤

≤

≤

vx ys 1 1+

4204 + 422

4156 mm

Estimated values

x1 = 250 – 46

= 204 mm

y1 = 500 – 78

= 422 mm.

The other values of sv do not govern.

8) Calculate the size of the stirrups

Select 2 legs of 12 mm diameter stirrups.

Asv = 2 × 113

= 226 mm2

vs 226=

3.48= 65 mm

The spacing can be increased by bundling the stirrup bars.

Designed section

12 mm diameter stirrups @ 65 mm c/c

250

500

200(4) corner bars16 mm diameter

(7) 11 mm diameter strandswith Pe = 507.15 kN

CGC

Dimensions in mm.


250

500



CGC


250

500



CGC

Dimensions in mm.

As D > 450 mm, side face reinforcement is required.



6.1 Calculation of Deflection This section covers the following topics.

• Introduction

• Deflection due to Gravity Loads

• Deflection due to Prestressing Force

• Total Deflection

• Limits of Deflection

• Determination of Moment of Inertia

• Limits of Span-to-effective Depth Ratio

6.1.1 Introduction

The deflection of a flexural member is calculated to satisfy a limit state of serviceability.

Since a prestressed concrete member is smaller in depth than an equivalent reinforced

concrete member, the deflection of a prestressed concrete member tends to be larger.

The total deflection is a resultant of the upward deflection due to prestressing force and

downward deflection due to the gravity loads. Only the flexural deformation is

considered and any shear deformation is neglected in the calculation of deflection.

Shear deformation is included in members such as deep beams and wall type of

structures.

The deflection of a member is calculated at least for two cases.

1) Short term deflection at transfer 2) Long term deflection under service loads

The short term deflection at transfer is due to the prestressing force (before long term

losses) and self-weight. The effect of creep and shrinkage of concrete are not

considered. The long term deflection under service loads is due to the effective

prestressing force (after long term losses) and the gravity loads. The permanent

components of the gravity loads are considered in the effect of creep. These

components are dead load and sustained live load.



6.1.2 Deflection due to Gravity Loads

The methods of calculation of deflection are taught in a course on structural analysis. It

is expected that the students are familiar with the methods. The methods include the

following.

1) Double integration method

2) Moment-area method

3) Conjugate beam method

4) Principle of virtual work

Numerical solution schemes can be developed based on the above methods and

executed in a computer. For members with prismatic cross-sections, common support

conditions and subjected to conventional loading, the deflections are available in tables

of text books on structural analysis.

The expressions of deflection (∆) for a few cases are provided. Here,

I = moment of inertia

E = modulus of elasticity of concrete

Simply supported beams

wL∆=EI

45384

PL∆=EI

3

48

L

w

P at centre

wL∆=EI

45384

PL∆=EI

3

48

L

w

P at centre

L

w

P at centre



Cantilever beams

wL∆=EI

4

8

L

w

PPL∆=EI

3

3

wL∆=EI

4

8

L

w

P

L

w

PPL∆=EI

3

3

6.1.3 Deflection due to Prestressing Force

The prestressing force causes a deflection only if the CGS is eccentric to the CGC. The

deflection due to prestressing force is calculated by the load-balancing method. This

method is explained in Section 3.2, Analysis of Member under Flexure (Part I). The

upward thrust (represented as wup for curved tendons and Wup for bent tendons) and

the upward deflection (also called camber and represented as ∆P) due to the

prestressing forces in typical profiles of tendons are reproduced here.

a) For a Parabolic Tendon

e

wup

L

PP e

wup

L

PP

up

upP

Pew = L

w L∆ =

EI

2

4

8

5384

(6-1.1)



b) For a Singly Harped Tendon

PP

Wup

PP

Wup

up

upP

PeW = L

W L ∆ =

EI

3

4

48

(6-1.2)

b) For a Doubly Harped Tendon

PP

WupWup

aL

PP

WupWup

aL

(6-1.3)

( )up

upP

PeW =aLa a W

∆ =EI

2 33 - 424

L

6.1.4 Total Deflection The total deflection is calculated for the following two cases.

1) Short term deflection at transfer

2) Long term deflection under service loads

The short term deflection at transfer (∆st) is given as follows.

(6-1.4) st P SW∆ = -∆ + ∆0

Here,

∆P0 = magnitude of deflection due to P0

∆sw = deflection due to self-weight

P0 = prestressing force before long term losses.

The long term deflection under service loads is difficult to calculate because the

prestressing force and creep strain influence each other. Creep of concrete is defined

as the increase in deformation with time under constant load. Due to the creep of



concrete, the prestress in the tendon is reduced with time. The ultimate creep strain is

found to be proportional to the elastic strain. The ratio of the ultimate creep strain to the

elastic strain is called the creep coefficient θ. The values of θ as per IS:1343 - 1980 are

given in Section 1.6, Concrete (Part II).

The following expression of the long term deflection under service loads (∆lt) is a

simplified form, where an average prestressing force is considered to generate the

creep strain. The effect of shrinkage on the prestressing force is neglected in the

expression.

( )( )⎛ ⎞⎜ ⎟⎝ ⎠

P Pelt Pe DL SL LL

∆ + ∆∆ = -∆ - θ+ ∆ + ∆ + θ + ∆0 12

(6-1.5)

The notations in the previous equations are as follows.

∆P0 = magnitude of deflection due to P0

∆Pe = magnitude of deflection due to Pe

Pe = effective prestressing force after long term losses.

∆DL = deflection due to dead load (including self-weight)

∆SL = deflection due to sustained live load

∆LL = deflection due to additional live load

A more rigorous calculation of total deflection can be done using the incremental time-step method. It is a step-by-step procedure, where the change in prestressing force

due to creep and shrinkage strains is calculated at the end of each time step. The

results at the end of each time step are used for the next time step. This procedure was

suggested by the Precast / Prestressed Concrete Institute (PCI) committee and is also

called the General method (Reference: PCI Committee, “Recommendations for

Estimating Prestress Losses”, PCI Journal, PCI, Vol. 20, No. 4, July-August 1975, pp.

43-75).

In the PCI step-by-step procedure, a minimum of four time steps are considered in the

service life of a prestressed member. The following table provides the definitions of the

time steps.



Table 6-1.1 Time steps in the step-by-step procedure

Step Beginning End

1. Pre-tension: Anchorage of steel

Post-tension: End of curing Age of prestressing

2. End of Step 1 30 days after prestressing or when

subjected to superimposed load

3. End of Step 2 1 year of service

4. End of Step 3 End of service life

The step-by-step procedure can be implemented in a computer program, where the

number of time steps can be increased.

6.1.5 Limits of Deflection

Clause 19.3.1 of IS:1343 - 1980 specifies limits of deflection such that the efficiency of

the structural element and the appearance of the finishes or partitions are not adversely

affected. The limits of deflection are summarised next.

1) The total deflection due to all loads, including the effects of temperature, creep

and shrinkage, should not exceed span / 250.

2) The deflection after erection of partitions or application of finishes, including the

effects of temperature, creep and shrinkage, should not exceed span/350 or 20

mm, whichever is less.

3) If finishes are applied, total upward deflection due to prestressing force should

not exceed span / 300.

6.1.6 Determination of Moment of Inertia

Type 1 and Type 2 Members These types of members are designed to be uncracked under service loads. The gross

moment of inertia (Ig) can be used to calculate the deflections.

Type 3 Members



This type of members is expected to be cracked under service loads. Strictly, the gross

moment of inertia (Ig) cannot be used in the calculations. IS:1343 - 1980, Clause 22.6.2,

recommends the following.

1) When the permanent load is less than or equal to 25% of the live load, the gross

moment of inertia can be used.

2) When the permanent load is greater than 25% of the live load, the span-to-

effective depth (L/d) ratio should be limited to bypass the calculation of deflection.

If the L/d ratio of a member exceeds the limit, the gross moment of inertia can still be

used if the tensile stress under service loads is within the allowable value. This

recommendation is suggested because the calculation of gross moment of inertia is

simpler as compared to an effective moment of inertia.

6.1.7 Limits of Span-to-Effective Depth Ratio

The calculation of deflection can be bypassed if the span-to-effective depth (L/d) ratio is

within the specified limit.

The limits of L/d ratios as per Clause 22.6.2, IS:1343 – 1980, are as follows.

For L ≤ 10 m

For cantilever beams L /d ≤ 7

For simply supported beams L /d ≤ 20

For continuous beams L /d ≤ 26

For L > 10 m

For simply supported beams L /d ≤ (20 × 10/ L)

For continuous beams L /d ≤ (26 × 10/ L)

Here, L is in metres. Deflection calculations are necessary for cantilevers with L > 10 m.



6.2 Calculation of Crack Width This section covers the following topics.

• Introduction

• Method of Calculation

• Limits of Crack Width

6.2.1 Introduction

The crack width of a flexural member is calculated to satisfy a limit state of serviceability.

Among prestressed concrete members, there is cracking under service loads only for

Type 3 members. Hence the calculation of crack width is relevant only for Type 3

members. The crack width is calculated for the cracks due to bending which occur at

the bottom or top surfaces of a flexural member.

The flexural cracks start from the tension face and propagate perpendicular to the axis

of the member. This type of cracks is illustrated in Section 5.1, Analysis for Shear. If

these cracks are wide, it leads to corrosion of the reinforcing bars and prestressed

tendons. Also, the cracks tend to widen under sustained load or cyclic load. To limit

the crack width, Type 3 members have regular reinforcing bars in the tension zone

close to the surface, in addition to the prestressed tendons.

The crack width of a flexural crack depends on the following quantities.

1) Amount of prestress

2) Tensile stress in the longitudinal bars

3) Thickness of the concrete cover

4) Diameter and spacing of longitudinal bars

5) Depth of member and location of neutral axis

6) Bond strength

7) Tensile strength of concrete.



6.2.2 Method of Calculation

IS:456 - 2000, Annex F, gives a procedure to determine flexural crack width. The

design crack width (Wcr) at a selected level on the surface of the section with maximum

moment is given as follows.

(6-2.1)

( )cr m

crcr min

a εW =a -C+

h - x

321

The notations in the previous equation are as follows.

acr = shortest distance from the selected level on the surface to a

longitudinal bar

Cmin = minimum clear cover to the longitudinal bar

h = total depth of the member

x = depth of the neutral axis

εm = average strain at the selected level.

The values of Cmin and h are obtained from the section of the member. The evaluation

of the other variables is explained.

Evaluation of acr

The location of crack width calculation can be at the soffit or the sides of a beam. The

value of acr depends on the selected level. The following sketch shows the values of acr

at a bottom corner (A), at a point in the soffit (B) and at a point at the side (C).

Neutral axis

acr2

acr3

acr1

B

C

A

Neutral axis

acr2

acr3

acr1

B

C

A

Figure 6-2.1 Cross-section of a beam showing the distances from surface to the

nearest longitudinal bar



Usually the crack width is calculated at a point in the soffit, which is equidistant from two

longitudinal bars. This point is the location of maximum estimated crack width. The

following sketch shows the variables used in computing acr.

s

acrCmin

db

dc

s

acrCmin

db

dc

Figure 6-2.2 Cross-section of a beam showing variables for calculation

Using geometry, the value of acr is obtained from the following equation.

(6-2.2)

⎛ ⎞⎜ ⎟⎝ ⎠

bcr c

dsa = + d -2

2

2 2

Here,

db = diameter of longitudinal bar

dc = effective cover = Cmin + db/2

s = centre-to-centre spacing of longitudinal bars.

The values of db, dc and s are obtained from the section of the member.

Evaluation of x and εm

The values of x and εm are calculated based on a sectional analysis under service loads.

The sectional analysis should consider the tension carried by the uncracked concrete in

between two cracks. The stiffening of a member due to the tension carried by the

concrete is called the tension stiffening effect. The value of εm is considered to be an

average value of the strain at the selected level over the span. The following sketch

illustrates the cracking and the uncracked concrete in a flexural member.



UncrackedconcreteCracked sectionUncrackedconcreteCracked section

Figure 6-2.3 Elevation of a beam showing the cracking and uncracked concrete

The analysis of a Type 3 member should be based on strain compatibility of concrete

and prestressing steel. IS:456 - 2000 recommends two procedures for the sectional

analysis considering the tension stiffening effect.

1) Rigorous procedure with explicit calculation of tension carried by the concrete.

2) Simplified procedure based on the conventional analysis of a cracked section,

neglecting the tension carried by concrete. An approximate estimate of the

tension carried by the concrete is subsequently introduced.

Here, the simplified procedure is explained. For a rectangular zone under tension, the

simplified procedure gives the following expression of εm.

(6-2.3)

( )( )( )m

s s

b h - x a - xε = ε -E A d - x1 3

For a prestressed member, (EpAp + EsAs) is substituted in place of EsAs.

The second term considers the tension carried by the concrete approximately by

reducing the strain (ε1) obtained from the analysis of a cracked section.


a = distance from the compression face to the level at which crack

width is calculated

= h, when the crack width is calculated at the soffit

b = width of the rectangular zone

d = effective depth of the longitudinal reinforcement


Ap = area of prestressing steel.

Es = modulus of elasticity of non-prestressed steel

Ep = modulus of elasticity of prestressed steel



ε1 = strain at the selected level based on a cracked sectional analysis

= εs(a – x)/(d – x)

εs = strain in the longitudinal reinforcement.

The depth of neutral axis (x) can be calculated by a trial and error procedure till the

equilibrium equations are satisfied. The following sketch shows the beam cross section,

strain profile, stress diagram and force couples under service loads. The contribution of

non-prestressed reinforcement is also included.

fs

b

d x0.33x

εdec


dp

As

Ap

εc

εp

εs

fp

fc

C

Tp

Tsfs

b

d x0.33x

εdec


dp

As

Ap

εc

εp

εs

fp

fc

C

Tp

Ts

Figure 6-2.4 Sketches for analysis of a rectangular section


C = 0.5Ecεcxb (6-2.4) Tp = ApEpεp (6-2.5) Ts = AsEsεs (6-2.6)

Based on the principles of mechanics, the following equations are derived.




∑⇒

⇒p s

p p p s s s c c

F =T +T = CA E ε + A E ε = E ε xb

0

0.5

(6-2.7) The second equation relates the moment under service loads (M) with the internal


( ) ( )

( ) (p

s p pA

s s s p c c p

M =T d - d +C d - x

= A E ε d - d + E ε xb d - x

0.33

0.5 0.33

)



(6-2.8)

The value of M should be equal to the moment due to service loads.

2) Equations of compatibility

The depth of the neutral axis is related to the depth of CGS and the depth of non-

prestressed reinforcement by the similarity of the triangles in the strain diagram.

(6-2.9)

c

p c p de

c

c s

εx =d ε + ε - ε

εx =d ε + ε

c

(6-2.10)


Linear elastic constitutive relationships are used in the earlier expressions of C, Ts and

Tp.

The known variables in the analysis are: b, d, Ap, As, εdec, Ec, Ep, Es, M.

The unknown quantities are: x, εc, εp, εs.

The steps for solving the above equations are given below.

1) Assume εc

2) Assume x.

3) Calculate εp and εs from Eqn. (6-2.9) and Eqn. (6-2.10), respectively.

4) Calculate C, Tp and Ts from Eqns. (6-2.4), (6-2.5), (6-2.6), respectively.

5) If Eqn. (6-2.7) is not satisfied, change x. If Tp + Ts < C, decrease x. If Tp

+ Ts > C, increase x.

6) Calculate M from Eqn. (6-2.8). If the value differs from the given value, change

εc and repeat from Step 2.

6.2.3 Limits of Crack Width

Clause 19.3.2 of IS:1343 - 1980 specifies limits of crack width such that the

appearance and durability of the structural element are not affected.



The limits of crack width are as follows.

Crack width ≤ 0.2 mm for moderate and mild environments

≤ 0.1 mm for severe environment.

The types of environments are explained in Table 9, Appendix A of IS:1343 - 1980.



7.1 Transmission of Prestress (Part I) This section covers the following topics.

• Pre-tensioned Members

7.1.1 Pre-tensioned Members

The stretched tendons transfer the prestress to the concrete leading to a self

equilibrating system. The mechanism of the transfer of prestress is different in the pre-

tensioned and post-tensioned members. The transfer or transmission of prestress is

explained for the two types of members separately.

For a pre-tensioned member, usually there is no anchorage device at the ends. The

following photo shows that there is no anchorage device at the ends of the pre-

tensioned railway sleepers.

Figure 7-1.1 End of pre-tensioned railway sleepers


For a pre-tensioned member without any anchorage at the ends, the prestress is

transferred by the bond between the concrete and the tendons. There are three

mechanisms in the bond.

1) Adhesion between concrete and steel

2) Mechanical bond at the concrete and steel interface



3) Friction in presence of transverse compression.

The mechanical bond is the primary mechanism in the bond for indented wires, twisted

strands and deformed bars. The surface deformation enhances the bond. Each of the

type is illustrated below.

Elliptical indentations

Pitch

Circular indentations

Examples of indented wires


Pitch


Pitch

Circular indentationsCircular indentations

Examples of indented wires

Twisted strand Deformed barTwisted strand Deformed bar Figure 7-1.2 Indented wires, twisted strands and deformed bars

The prestress is transferred over a certain length from each end of a member which is

called the transmission length or transfer length (Lt). The stress in the tendon is zero

at the ends of the members. It increases over the transmission length to the effective

prestress (fpe) under service loads and remains practically constant beyond it. The

following figure shows the variation of prestress in the tendon.



fpe

Lt

fpe

Lt

fpe

Lt Figure 7-1.3 Variation of prestress in tendon along transmission length

Hoyer Effect After stretching the tendon, the diameter reduces from the original value due to the

Poisson’s effect. When the prestress is transferred after the hardening of concrete, the

ends of the tendon sink in concrete. The prestress at the ends of the tendon is zero.

The diameter of the tendon regains its original value towards the end over the

transmission length. The change of diameter from the original value (at the end) to the

reduced value (after the transmission length), creates a wedge effect in concrete. This

helps in the transfer of prestress from the tendon to the concrete. This is known as the

Hoyer effect. The following figure shows the sequence of the development of Hoyer

effect.



a) Applying tension to tendon

Originaldiameter

Diameter after stretching

a) Applying tension to tendon

Originaldiameter


Originaldiameter


Originaldiameter


b) Casting of concrete

Originaldiameter


b) Casting of concrete

Originaldiameter


c) Transferring of prestress

fp0

Sinking of tendon

Originaldiameter


c) Transferring of prestress

fp0

Sinking of tendon

Figure 7-1.4 Hoyer effect

Since there is no anchorage device, the tendon is free of stress at the end. The

concrete should be of good quality and adequate compaction for proper transfer of

prestress over the transmission length.

Transmission Length There are several factors that influence the transmission length. These are as follows.

1) Type of tendon

wire, strand or bar

2) Size of tendon

3) Stress in tendon

4) Surface deformations of the tendon



Plain, indented, twisted or deformed

5) Strength of concrete at transfer

6) Pace of cutting of tendons

Abrupt flame cutting or slow release of jack

7) Presence of confining reinforcement

8) Effect of creep

9) Compaction of concrete

10) Amount of concrete cover.

The transmission length needs to be calculated to check the adequacy of prestress in

the tendon over the length. A section with high moment should be outside the

transmission length, so that the tendon attains at least the design effective prestress

(fpe) at the section. The shear capacity at the transmission length region has to be

based on a reduced effective prestress.

IS:1343 - 1980 recommends values of transmission length in absence of test data.

These values are applicable when the concrete is well compacted, its strength is not

less than 35 N/mm2 at transfer and the tendons are released gradually. The

recommended values of transmission length are as follows.

Table 7-1.1 Values of transmission length

For plain and intended wires Lt = 100 Φ

For crimped wire Lt = 65 Φ

For strands Lt = 30 Φ

Here, Φ is the nominal diameter of the wire or strand.

To avoid the transmission length in the clear span of a beam, IS:1343 - 1980

recommends the following.

1) To have an overhang of a simply supported member beyond the support by a

distance of at least ½ Lt.



½ Lt½ Lt Figure 7-1.5 End of a simply supported member

2) If the ends have fixity, then the length of fixity should be at least Lt.

LtLtLt Figure 7-1.6 End of a member with fixity

Development Length The development length needs to be provided at the critical section, the location of

maximum moment. The length is required to develop the ultimate flexural strength of the

member. The development length is the minimum length over which the stress in

tendon can increase from zero to the ultimate prestress (fpu). The development length is

significant to achieve ultimate capacity.

If the bonding of one or more strands does not extend to the end of the member (de-

bonded strand), the sections for checking development of ultimate strength may not be

limited to the location of maximum moment.

The development length (Ld) is the sum of the transmission length (Lt) and the bond

length (Lb).

Ld = Lt + Lb (7-1.1)



The bond length is the minimum length over which, the stress in the tendon can

increase from the effective prestress (fpe) to the ultimate prestress (fpu) at the critical

location.

The following figure shows the variation of prestress in the tendon over the length of a

simply supported beam at ultimate capacity.

Lt

fpu

fpe

Lt

fpu

fpe

Lt

fpu

fpe

Figure 7-1.7 Variation of prestress in tendon at ultimate

The calculation of the bond length is based on an average design bond stress (τbd). A

linear variation of the prestress in the tendon along the bond length is assumed. The

following sketch shows a free body diagram of a tendon along the bond length.

fpuApfpeAp

τbd

Lb

fpu

fpe

fpuApfpeAp

τbd

fpuApfpeAp

τbd

Lb

fpu

fpe

Lb

fpu

fpe

Figure 7-1.8 Assumed variation of prestress in tendon along the bond length

The bond length depends on the following factors.

1) Surface condition of the tendon

2) Size of tendon

3) Stress in tendon



4) Depth of concrete below tendon

From equilibrium of the forces in the above figure, the expression of the bond length is

derived.

( )

τpu pe

bbd

f - f φL =

4 (7-1.2)

Here, Φ is the nominal diameter of the tendon.

The value of the design bond stress (τbd) can be obtained from IS:456 - 2000, Clause

26.2.1.1. The table is reproduced below.

Table 7-1.2 Design bond stress for plain bars

Grade of concrete M30 M35 M40 and above

τbd (N/mm2) 1.5 1.7 1.9

End Zone Reinforcement The prestress and the Hoyer effect cause transverse tensile stress (σt). This is largest

during the transfer of prestress. The following sketch shows the theoretical variation of

σt.

σt

Lt

σt

σt

Lt

σt

Lt

σtσt

Figure 7-1.9 Transverse stress in the end zone of a pre-tensioned beam

To restrict the splitting of concrete, transverse reinforcement (in addition to the

reinforcement for shear) needs to be provided at each end of a member along the



transmission length. This reinforcement is known as end zone reinforcement.

The generation of the transverse tensile stress can be explained by the free body

diagram of the following zone below crack, for a beam with an eccentric tendon.

Tension (T), compression (C) and shear (V) are generated due to the moment acting on

the horizontal plane at the level of the crack. The internal forces along the horizontal

plane are shown in (a) of the following figure. The variation of moment (due to the

couple of the normal forces) at horizontal plane along the depth is shown in (b).

LtLt

Variation of moment at horizontal plane along depth

T C

V

Free body diagram of zone below crack

Lt Variation of moment at horizontal plane along depth

Variation of moment at horizontal plane along depth

T C

V


Lt

T C

V


Lt

Figure 7-1.10 Forces in the end zone

The end zone reinforcement is provided to carry the tension (T) which is generated due

to the moment (M). The value of M is calculated for the horizontal plane at the level of

CGC due to the compressive stress block from the normal stresses in a vertical plane

above CGC. The minimum amount of end zone reinforcement (Ast) is given in terms of

the moment (M) as follows.

sts

MA =f h

2.5(7-1.3)




h = total depth of the section

M = moment at the horizontal plane at the level of CGC due to the

compressive stress block above CGC

fs = allowable stress in end zone reinforcement.

The lever arm for the internal moment is h/2.5. The value of fs is selected based on a

maximum strain.

The end zone reinforcement should be provided in the form of closed stirrups enclosing

all the tendons, to confine the concrete. The first stirrup should be placed as close as

possible to the end face, satisfying the cover requirements. About half the

reinforcement can be provided within a length equal to ⅓Lt from the end. The rest of the

reinforcement can be distributed in the remaining ⅔Lt.

References:

1) Krishnamurthy, D. “A Method of Determining the Tensile Stresses in the End Zones

of Pre-tensioned Beams”, Indian Concrete Journal, Vol. 45, No. 7, July 1971, pp. 286-

297.

2) Krishnamurthy, D. “Design of End Zone Reinforcement to Control Horizontal

Cracking in Pre-tensioned Concrete Members at Transfer”, Indian Concrete Journal, Vol.

47, No. 9, September 1973, pp. 346-349.



Example 7-1.1 Design the end zone reinforcement for the pre-tensioned beam shown in the following figure. The sectional properties of the beam are as follows.

A = 46,400 mm2

I = 8.47 × 108 mm4

Z = 4.23 × 105 mm3

There are 8 prestressing wires of 5 mm diameter.

Ap = 8 × 19.6 = 157 mm2

The initial prestressing is as follows. fp0 = 1280 N/mm2.

Limit the stress in end zone reinforcement (fs) to 140 N/mm2.

400

80

60

90CGC

CGS

200

Cross-section at end

400

80

60

90CGC

CGS

200

400

80

60

90CGC

CGS

200

Cross-section at end Solution

1) Determination of stress block above CGC

Initial prestressing force

P0 = Ap.fpo

= 157 × 1280 N

= 201 kN



Stress in concrete at top

≈

0 0

3 3

5

2

201×10 201×10 ×90= - +46400 4.23×10

0 N/mm

tP P ef = - +A Z

Stress at bottom

0 0

3 3

5

2

201×10 201×10 ×90= - -46400 4.23×10

= -8.60 N/mm

bP P ef = - -A Z

14080

60

CGC

CGS

200

Stress profile Components of compression block

4.30

1.29

8.60

C3

C2

C1

y3 y2 y1

14080

60

CGC

CGS

200

14080

60

CGC

CGS

200

Stress profile Components of compression block

4.30

1.29

8.60

4.30

1.29

8.60

C3

C2

C1

y3 y2 y1

C3

C2

C1

y3 y2 y1

2) Determination of components of compression block

C1 = ½ × 1.29 × 200 × 60

= 7.74 kN

y1 = 140 + ⅓ × 60

= 160 mm

C2 = ½ × 1.29 × 140 × 80

= 7.22 kN

y2 = ⅔ × 140

= 93.3 mm

C3 = ½ × 4.3 × 140 × 80

= 24.08 kN

y3 = ⅓ × 140

= 46.7 mm



14080

60

200

4.30

1.29

8.60

14080

60

200

4.30

1.29

8.60

4.30

1.29

8.60

3) Determination of moment

M = Σ Ci.yi

= C1.y1 + C2.y2 + C3.y3

= (7.74 × 160) + (7.22 × 93.3) + (24.08 × 46.7)

= 3036.6 kN-mm

4) Determination of amount of end zone reinforcement

3

2

2.5

2.5M=140×4002.5×3036.6×10=

140×400=135.6 mm

sts

MA =f h

With 6 mm diameter bars, required number of 2 legged closed stirrups

= 135.6 / (2 × 28.3) ⇒ 3.

For plain wires, transmission length

Lt = 100 Φ

= 500 mm.

Provide 2 stirrups within distance 250 mm (Lt/2) from the end. The third stirrup is in the

next 250 mm.



Designed end zone reinforcement

(3) 6 mm diameter stirrups(3) 6 mm diameter stirrups



7.2 Transmission of Prestress (Part II) This section covers the following topic.

• Post-tensioned Members

7.2.1 Post-tensioned Members

Unlike in a pre-tensioned member without anchorage, the stress in the tendon of a post-

tensioned member attains the prestress at the anchorage block. There is no

requirement of transmission length or development length.

The end zone (or end block) of a post-tensioned member is a flared region which is

subjected to high stress from the bearing plate next to the anchorage block. It needs

special design of transverse reinforcement. The design considerations are bursting

force and bearing stress.

The stress field in the end zone of a post-tensioned member is complicated. The

compressive stress trajectories are not parallel at the ends. The trajectories diverge

from the anchorage block till they become parallel. Based on Saint Venant’s principle,

it is assumed that the trajectories become parallel after a length equal to the larger

transverse dimension of the end zone. The following figure shows the external forces

and the trajectories of tensile and compressive stresses in the end zone.

Stress trajectories in the end zone

yp0

σt y0 = larger transverse dimension ofend zone

y0

Tensile stress trajectories

Compressive stress trajectories

Bearing plate

Stress trajectories in the end zone

yp0


y0



Bearing plate

yp0


y0



Bearing plate

Figure 7-2.1 Stress trajectories in the end zone of a post-tensioned beam

The larger transverse dimension of the end zone is represented as y0. The

corresponding dimension of the bearing plate is represented as yp0. For analysis, the



end zone is divided into a local zone and a general zone as shown in the following

sketch.

Bearing plate

Local zone General zone

y0Bearing plate

Local zone General zone

y0 Figure 7-2.2 Local and general zones in the end zone

The local zone is the region behind the bearing plate and is subjected to high bearing

stress and internal stresses. The behaviour of the local zone is influenced by the

anchorage device and the additional confining spiral reinforcement. The general zone

is the end zone region which is subjected to spalling of concrete. The zone is

strengthened by end zone reinforcement.

The variation of the transverse stress (σt) at the CGC along the length of the end zone

is shown in the next figure. The stress is compressive for a distance 0.1y0 from the end.

Beyond that it is tensile. The tensile stress increases and then drops down to zero

within a distance y0 from the end.

0.1y0 0.9y0

Fbst

σt

Distance alongaxis of beam

0.1y0 0.9y0

Fbst

σt


0.1y0 0.9y0

Fbst

σt


Figure 7-2.3 Transverse stress in the end zone

The transverse tensile stress is known as splitting tensile stress. The resultant of the

tensile stress in a transverse direction is known as the bursting force (Fbst). Compared



to pre-tensioned members, the transverse tensile stress in post-tensioned members is

much higher.

Besides the bursting force there is spalling forces in the general zone.

Spalling force Bursting forceSpalling force Bursting force Figure 7-2.4 Spalling and bursting forces in the end zone

IS:1343 - 1980, Clause 18.6.2.2, provides an expression of the bursting force (Fbst) for

an individual square end zone loaded by a symmetrically placed square bearing plate.

⎡ ⎤⎢ ⎥⎣ ⎦

pbst k

yF = P

y0

0

0.32- 0.3 (7-2.1)

Here,

Pk = prestress in the tendon

yp0 = length of a side of bearing plate

y0 = transverse dimension of the end zone.

The following sketch shows the variation of the bursting force with the parameter yp0 / y0.

The parameter represents the fraction of the transverse dimension covered by the

bearing plate.

bst

k

FP

pyy

0

0

0.02

0.32

1

bst

k

FP

pyy

0

0

0.02

0.32

1 Figure 7-2.5 Variation of bursting force with size of bearing plate



It can be observed that with the increase in size of the bearing plate the bursting force

(Fbst) reduces. The following sketch explains the relative size of the bearing plate with

respect to the end zone.

(1) (2) (3)(1) (2) (3)

Figure 7-2.6 End views of end zones with varying size of the bearing plate

In the above end views of end zones, the bursting force (Fbst) will be largest for Case (1)

and least for Case (3). For a rectangular end zone, Fbst is calculated from the previous

equation for each principle direction. For a circular bearing plate, an equivalent square

loaded area is considered in the calculation of Fbst. For more than one bearing plate, the

end zone is divided into symmetrically loaded prisms. Each prism is analysed by the

previous equation.

End Zone Reinforcement Transverse reinforcement is provided in each principle direction based on the value of

Fbst. This reinforcement is called end zone reinforcement or anchorage zone

reinforcement or bursting links. The reinforcement is distributed within a length from

0.1y0 to y0 from an end of the member.

The amount of end zone reinforcement in each direction (Ast) can be calculated from the

following equation.

bst

sts

FA =f (7-2.2)

The stress in the transverse reinforcement (fs) is limited to 0.87fy. When the cover is

less than 50 mm, fs is limited to a value corresponding to a strain of 0.001.

The end zone reinforcement is provided in several forms, some of which are proprietary



of the construction firms. The forms are closed stirrups, mats or links with loops. A few

types of end zone reinforcement is shown in the following sketches.

Mat LinksMatMat LinksLinks Figure 7-2.7 Types of end zone reinforcement

The local zone is further strengthened by confining the concrete with spiral

reinforcement. The performance of the reinforcement is determined by testing end block

specimens. The following photo shows the spiral reinforcement around the guide of the

tendons.

Figure 7-2.8 Spiral reinforcement in the end zone

(Reference: Dywidag Systems International)

The end zone may be made of high strength concrete. The use of dispersed steel fibres

in the concrete (fibre reinforced concrete) reduces the cracking due to the bursting force.

Proper compaction of concrete is required at the end zone. Any honey-comb of the

concrete leads to settlement of the anchorage device. If the concrete in the end zone is

different from the rest of the member, then the end zone is cast separately.



Bearing Plate High bearing stress is generated in the local zone behind the bearing plate. The

bearing stress (fbr) is calculated as follows.

k

brpun

Pf =A (7-2.3)

Here,

Pk = prestress in the tendon with one bearing plate

Apun = Punching area

= Area of contact of bearing plate.

As per Clause 18.6.2.1, IS:1343 - 1980, the bearing stress in the local zone should be

limited to the following allowable bearing stress (fbr,all).

≤

brbr,all ci

pun

ci

Af = fA

f

0.48

0.8

(7-2.4)

In the above equation,

Apun = Punching area

= Area of contact of bearing plate

Abr = Bearing area

= Maximum transverse area of end block that is geometrically similar

and concentric with punching area

fci = cube strength at transfer.

The expression of allowable bearing stress takes advantage of the dispersion of the

bearing stress in the concrete. The following sketch illustrates the dispersion of bearing

stress in concrete.



Apun

Abr

End view showing bearing plate

Apun

AbrApun

AbrApun

Abr


Apun

Abr


Apun

Abr

Figure 7-2.9 End and isometric views of end zone

The performance of anchorage blocks and end zone reinforcement is critical during the

post-tensioning operation. The performance can be evaluated by testing end block

specimens under compression. The strength of an end block specimen should exceed

the design strength of the prestressing tendons.

The following photos show the manufacturing of an end block specimen.

(a) Fabrication of end zone reinforcement



(b) Anchorage block and guide

(c) End zone reinforcement with guide and duct



(d) End block after casting

Figure 7-2.10 Manufacturing of an end block specimen

Example 7-2.1 Design the bearing plate and the end zone reinforcement for the following bonded post-tensioned beam. The strength of concrete at transfer is 50 N/mm2. A prestressing force of 1055 kN is applied by a single tendon. There is no eccentricity of the tendon at the ends.

Section beyond end zone Section at end zone

100

100

100

400

400 400

600

Section beyond end zone Section at end zone

100

100

100

400

400

100

100

100

400

400 400

600

400

600



Solution

1) Let the bearing plate be 200 mm × 300 mm. The bearing stress is calculated below.

3

2

1055×10=200×300

=17.5 N/mm

kbr

pun

Pf =A

The allowable bearing stress is calculated.

2

0.48

400×600= 0.48×50200×300

= 48 N/mm

brbr,all ci

pun

Af = fA

Limit fbr,all to 0.8 fci = 0.8 × 50 = 40 N/mm2. Bearing stress is less than fbr,all. Hence OK.

2) Calculate bursting force.

In the vertical direction

⎡ ⎤⎢ ⎥⎣ ⎦

⎡ ⎤⎢ ⎥⎣ ⎦

0

0

0.32- 0.3

300=1055 0.32 - 0.3600

=179.3 kN

pbst k

yF = P

y

In the horizontal direction

⎡ ⎤⎢ ⎥⎣ ⎦

⎡ ⎤⎢ ⎥⎣ ⎦

0

0

0.32- 0.3

200=1055 0.32 - 0.3400

=179.3 kN

pbst k

yF = P

y



3) Calculate end zone reinforcement.

3

2

0.87

179.3×10=0.87×250

= 824.6 mm

bstst

y

FA =f

Provide ⅔ Ast = ⅔ × 824.6 = 550 mm2 within 0.1 y0 = 60 mm and 0.5 y0 = 300 mm from

the end.

Select (6) 2 legged 8 mm diameter stirrups.

Provide ⅓ Ast = ⅓ × 824.6 = 275 mm2 within 0.5 y0 = 300 mm and y0 = 600 mm from the

end.

Select (5) 2 legged 6 mm diameter stirrups.

200

300

End view

200

300

200

300

End view



(6) 8 mm stirrups from 60 to 300


End zone reinforcement



End zone reinforcement



8.1 Cantilever Beams This section covers the following topics.

• Analysis

• Determination of Limiting Zone

• Tendon Profile

Introduction Prestressed cantilever beams are present in buildings and bridges. Usually, the

cantilever is provided with a back span (anchor span) to reduce the torsion in the

supporting member. In a building, the cantilever can be an extension of a continuous

beam. In a bridge, the cantilever is a part of the “balanced cantilever” girder. The

construction can be in-situ, where the concrete is cast in formwork that is temporarily

supported. Else, a segmental construction can be adopted, wherein slices of the girder

are placed and temporarily prestressed.

The following figure illustrates the cantilevers in buildings and bridges schematically.

Cantilever spanBack span Cantilever spans

Segment

Cantilever spanBack span Cantilever spanBack span Cantilever spans

Segment

Cantilever spans

Segment

(a) Cantilever in a building frame (b) Cantilevers in a bridge girder

Figure 8-1.1 Cantilevers in buildings and bridges



Figure 8-1.2 Segmental construction, Bridge over Hooghly,

West Bengal (Courtesy: L & T Ramboll)



The analysis of a section of a prestressed cantilever for flexural stresses is similar to

that for a simply supported beam. The difference is that for gravity loads, the bending

moment in cantilever is negative, that is compression is generated at the bottom. Thus,

the CGS is placed above the CGC and the eccentricity e is negative.

The following aspects need to be considered in the analysis and design of a

prestressed cantilever beam.

1) Certain portions of the back span are subjected to both positive and negative

moments. Hence, there will be two design moments at service loads.

2) The beam may be subjected to partial loading and point loading.

3) The sequence of loading is important to design the prestressing force.

4) High values of moment and shear occur simultaneously near the support.

8.1.1 Analysis

The analysis of a cantilever beam with a back span, is illustrated to highlight the aspects

stated earlier. The bending moment diagrams for the following load cases are shown

schematically in the following figure.

1) Dead load (DL)

2) Live load (LL) only on the back span

3) Live load only on the cantilever span

4) Dead load and live load along full length.



Moment diagram for DL

wDL


wDL

Moment diagram for LL on back span

wLL

Moment diagram for LL on back span

wLL

Moment diagram for LL on cantilever span

wLL

Moment diagram for LL on cantilever spanMoment diagram for LL on cantilever span

wLL

Moment diagram for DL + LL along full length

wDL + wLL

Moment diagram for DL + LL along full length

wDL + wLL

Figure 8-1.3 Moment diagrams for dead and live loads



The envelop moment diagrams are calculated from the analysis of each load case and

their combinations. In the following envelop moment diagrams, Mmax and Mmin represent

the highest and lowest values (algebraic values with sign) of the moments at a section,

respectively. Note that certain portions of the beam are subjected to both positive and

negative moments.

Mmax

Mmin

Mmax

Mmin

Mmax

Mmin

Figure 8-1.4 Envelop moment diagrams for dead and live loads

For moving point loads as in bridges, first the influence line diagram is drawn. The

influence line diagram shows the variation of the moment or shear for a particular

location in the girder, due to the variation of the position of a unit point load. The vehicle

load is placed based on the influence line diagram to get the worst effect.

8.1.2 Determination of Limiting Zone

The limiting zone of placing the CGS of the tendons is helpful in selecting a tendon

profile. Note that for a non-prismatic section, the section properties such as A, I, kt, kb, r

change with position along the length.

The limiting zone was explained for a simply supported beam in Section 4.4, Design of

Sections for Flexure (Part III). Here the concept and the equations are first reviewed for

a simply supported beam with positive moment.

For full prestressed members (Type 1), tension is not allowed under service conditions.

If tension is also not allowed at transfer, the compression in concrete (C) always lies

within the kern zone. The limiting zone is defined as the zone for placing the CGS of



the tendons such that C always lies within the kern zone. Also, the maximum

compressive stresses at transfer and service should be within the allowable values.




Also, the maximum compressive stresses at transfer and service should be within the

allowable values.

The limiting zone is determined from the maximum or minimum eccentricities of the

CGS along the beam corresponding to the extreme positions of C. Note that, the

limiting zone is related with the CGS of the tendons. Individual tendons may lie outside

the limiting zone.

For a simply supported beam, the maximum eccentricity (emax) at any section

corresponds to the lowest possible location of C at transfer, that generates allowable

tensile stress at the top of the section. The maximum compressive stress at the bottom

should be within the allowable value. The minimum eccentricity (emin) at any section

corresponds to the highest possible location of C at service, that generates allowable

tensile stress at the bottom of the section. The maximum compressive stress at the top

should be within the allowable value.

The values of emax and emin can be determined by equating the stresses at the edges of

concrete with the allowable values. Else, explicit expressions of emax and emin can be

developed. The following material gives the expressions of emax and emin for Type 1 and

Type 2 sections. The values of emax and emin can be determined at regular intervals

along the length of the beam. The zone between the loci of emax and emin is the limiting

zone of the section for placing the CGS. The equations are summarised

Type 1 Section Based on the stress at the top at transfer, the following expression of emax is derived.

swmax b

swmax b

Me - k =P

Me = + kP

0

0

or,(8-1.1)



Also, the stress at the bottom should be less than the allowable value at transfer.

Based on the stress at the bottom at service, the following expression of emin is derived.

(8-1.2)

Tmin t

e

Tmin t

e

Me + k =PMe = - kP

or,

Also, the stress at the top should be less than the allowable value at service. If for a

particular section emin is negative, it implies that the CGS can be placed above the CGC.

This happens near the supports.

Type 2 Section The corresponding equations for a Type 2 section are given below.

(8-1.3)

sw ct,all bmax b

sw ct,all bmax b

M +f Ake -k =

PM +f Ak

e = +kP

0

0

or,

(8-1.4)

T ct,all tmin t

e

T ct,all tmin t

e

M - f Ake +k =

PM - f Ak

e = - kP

or,

In a simply supported beam, the external moments are always positive. The minimum

moment is due to self weight. The maximum moment is under service loads. For

cantilever beams, the minimum external moment need not be at transfer, when the

moment is due to self weight (MSW). Also, under service loads there are two moments

Mmin and Mmax at a location, obtained from the envelop moment diagrams.

The maximum and minimum eccentricities emax and emin at a particular location are first

determined for service loads from Mmin and Mmax , respectively, at that location and the

effective prestress Pe. Next, another set of emax and emin are calculated at transfer from

MSW and the prestress P0. The final emax is the lower of the two values calculated at

service and at transfer. Similarly, the final emin is the higher of the two values calculated

at service and at transfer.



The expressions of emax and emin for the simply supported beam were developed for

positive moments. For a cantilever, corresponding to a negative moment, the

eccentricity implies that the CGS is located above CGC. The expressions for a

cantilever are given below.

Type 1 section At Service

(8-1.5)

(8-1.6)

minmax b

e

maxmin t

e

Me = + kP

Me = - kP

At Transfer

(8-1.7)

(8-1.8)

swmax b

swmin t

Me = + kP

Me = - kP

0

0

Type 2 section At Service

(8-1.9)

(8-1.10)

min ct,all bmax b

e

max ct,all tmin t

e

M + f Ake = + k

PM - f Ak

e = + kP

At Transfer

(8-1.11) SW ct,all bmax b

SW ct,all tmin t

M + f Ake = + k

PM - f Ak

e = - kP

0

0

(8-1.12)



A typical limiting zone is shown in the following figure.

kt

kb

Limiting zonekt

kb

Limiting zone

Figure 8-1.5 Limiting zone for a non-prismatic beam

8.1.3 Tendon Profile

The tendon profiles for a few beams with cantilever spans are shown schematically in

the following figures. The vertical scale is enlarged to show the location of the CGS

with respect to CGC.

a) Uniform cross section

b) Varying cross section in the cantilever span


b) Varying cross section in the cantilever span Figure 8-1.6 Beams with single cantilever span




b) Varying level of soffit

c) Varying level of top of beam





c) Varying level of top of beamc) Varying level of top of beam Figure 8-1.7 Beams with double cantilever spans

For a prismatic beam with uniform cross section along the length, the tendon profile is

similar to the moment diagram under uniform load. Thus for regions of negative

moment, the CGS is located above the CGC. Since there cannot be a sharp kink in the

tendons and, the supports are not true point supports, the profile is shown curved at the

right support.

For a beam with varying depth, the tendon profile can be adjusted (within emax and emin)

to be straight for convenience of layout of the tendons.



8.2 Continuous Beams (Part I) This section covers the following topics.

• Analysis

• Incorporation of Moment due to Reactions

• Pressure Line due to Prestressing Force

Introduction Beams are made continuous over the supports to increase structural integrity. A

continuous beam provides an alternate load path in the case of failure at a section. In

regions with high seismic risk, continuous beams and frames are preferred in buildings

and bridges. A continuous beam is a statically indeterminate structure.

The advantages of a continuous beam as compared to a simply supported beam are as

follows.

1) For the same span and section, vertical load capacity is more.

2) Mid span deflection is less.

3) The depth at a section can be less than a simply supported beam for the same

span. Else, for the same depth the span can be more than a simply supported

beam.

⇒ The continuous beam is economical in material.

4) There is redundancy in load path.

⇒ Possibility of formation of hinges in case of an extreme event.

5) Requires less number of anchorages of tendons.

6) For bridges, the number of deck joints and bearings are reduced.

⇒ Reduced maintenance

There are of course several disadvantages of a continuous beam as compared to a

simply supported beam.

1) Difficult analysis and design procedures.

2) Difficulties in construction, especially for precast members.

3) Increased frictional loss due to changes of curvature in the tendon profile.

4) Increased shortening of beam, leading to lateral force on the supporting columns.

5) Secondary stresses develop due to time dependent effects like creep and

shrinkage, settlement of support and variation of temperature.



6) The concurrence of maximum moment and shear near the supports needs

proper detailing of reinforcement.

7) Reversal of moments due to seismic force requires proper analysis and design.

Intermediate spanEnd span Intermediate spanEnd span (a) Continuous beam in a building frame

Intermediate spanEnd span Intermediate spanEnd span (b) Continuous beam in a bridge

Figure 8-2.1 Continuous beams in buildings and bridges

8.2.1 Analysis

The analysis of continuous beams is based on elastic theory. This is covered in text

books of structural analysis. For prestressed beams the following aspects are

important.

1) Certain portions of a span are subjected to both positive and negative moments.

These moments are obtained from the envelop moment diagram.

2) The beam may be subjected to partial loading and point loading. The envelop

moment diagrams are developed from “pattern loading”. The pattern loading

refers to the placement of live loads in patches only at the locations with positive

or negative values of the influence line diagram for a moment at a particular

location.

3) For continuous beams, prestressing generates reactions at the supports. These

reactions cause additional moments along the length of a beam.



The analysis of a continuous beam is illustrated to highlight the aspects stated earlier.

The bending moment diagrams for the following load cases are shown schematically in

the following figures.

1) Dead load (DL)

2) Live load (LL) on every span

3) Live load on a single span.


wDL


wDLwDL

wLL

Moment diagram for LL on every span

wLLwLL

Moment diagram for LL on every span

wLL

Moment diagram for LL on one span

wLL

Moment diagram for LL on one span Figure 8-2.2 Moment diagrams for dead and live loads



For moving point loads as in bridges, first the influence line diagram is drawn. The

influence line diagram shows the variation of the moment or shear for a particular

location in the girder, due to the variation of the position of a unit point load. The vehicle

load is placed based on the influence line diagram to get the worst effect. An influence

line diagram is obtained by the Müller-Breslau Principle. This is covered in text books

of structural analysis.

IS:456 - 2000, Clause 22.4.1, recommends the placement of live load as follows.

1) LL in all the spans.

2) LL in adjacent spans of a support for the support moment. The effect of LL in the

alternate spans beyond is neglected.

3) LL in a span and in the alternate spans for the span moment.

The envelop moment diagrams are calculated from the analysis of each load case and

their combinations. The analysis can be done by moment distribution method or by

computer analysis.

In lieu of the analyses, the moment coefficients in Table 12 of IS:456 - 2000 can be

used under conditions of uniform cross-section of the beams in the several spans,

uniform loads and similar lengths of span.

The envelop moment diagrams provide the value of a moment due to the external

loads. It is to be noted that the effect of prestressing force is not included in the envelop

moment diagrams. The following figure shows typical envelop moment diagrams for a

continuous beam.

Mmax

Mmin

Mmax

Mmin

Mmax

Mmin

Figure 8-2.3 Envelop moment diagrams for DL + LL



In the above diagrams, Mmax and Mmin represent the highest and lowest values

(algebraic values with sign) of the moments at a section, respectively. Note that certain

portions of the beam are subjected to both positive and negative moments. The

moment from the envelop moment diagrams will be represented as the M0 diagram.

This diagram does not depend on whether the beam is prestressed or not.

8.2.2 Incorporation of Moment Due to Reactions

As mentioned before, for continuous beams prestressing generates reactions at the

supports. The reactions at the intermediate supports cause moment at a section of the

continuous beam. This moment is linear between the supports and is in addition to the

moment due to the eccentricity of the prestressing force. The concept is explained by a

simple hypothetical two-span beam in the following figure. The beam is prestressed

with a parabolic tendon in each span, with zero eccentricity of the CGS at the supports.

The moment diagram due to the eccentricity of the prestressing force and neglecting the

intermediate support is denoted as the M1 diagram. This diagram is obtained as M1 =

Pe, where, P is the prestressing force (P0 at transfer and Pe at service) and e is the

eccentricity of the CGS with respect to CGC. Neglecting the variation of P along the

length due to frictional losses, the value of M1 is proportional to e. Hence, the shape of

the M1 diagram is similar to the cable profile.



M1 diagram

e

Pe

l l

Profile of the CGS

M1 diagram

e

Pe

l l

Profile of the CGS

w

10wl / 8

3wl / 8 3wl / 8


w

10wl / 8

3wl / 8 3wl / 8w

10wl / 8

3wl / 8 3wl / 8


M2 diagram

wl2/2 = 4Pe

5wl2/8 = 5Pe

+

=

w

Pe

Simplified free body diagram

M2 diagram

wl2/2 = 4Pe

5wl2/8 = 5Pe

+

=

w

Pe

Simplified free body diagram

Figure 8-2.4 Moment diagram due to prestressing force for a two-span beam

Next, the moment diagram due to the prestressing force and including the effect of the

intermediate support is denoted as the M2 diagram. This is obtained by structural

analysis of the continuous beam subjected to the upward thrust. Since the profile of the

tendon is parabolic in each span, the upward thrust is uniform and is given as wup = w =

8Pe/l2. The downward thrust at the location of the central kink is not considered as it

directly goes to the intermediate support. The hold down force at the intermediate

support neglecting the downward thrust is 10wupl/8 = 10Pe/l. The downward forces at

the ends are from the anchorages. The moment diagram due to wup alone (without the



support) is added to that due to the hold down force. The resultant M2 diagram is

similar to the previous M1 diagram, but shifted linearly from an end support to the

intermediate support.

For a general case, the resultant moment (M2) at a location due to the prestressing

force can be written as follows.

M2 = M1 + M1/ (8-2.1)

In the above equation,

M1 = moment due to the eccentricity of the prestressing force neglecting the

intermediate supports

= Pee.

M1/ = moment due to the reactions at intermediate supports.

Pe = effective prestress

e = eccentricity of CGS with respect to CGC.

M1 is the primary moment and M1/ is the secondary moment.

The moment due to the external loads (M0) that is obtained from the envelop moment

diagrams is added to M2 to get the resultant moment (M3) at a location.

M3 = M2 + M0

M3 = M1 + M1/ + M0

(8-2.2)

The variation of M3 along the length of the beam (M3 diagram) can be calculated as

follows.

1) The M0 diagram is available from the envelop moment diagram.

2) Plot M1 diagram which is similar to the profile of the CGS. The variation of Pe

along the length due to friction may be neglected.

3) Plot the shear force (V) diagram corresponding to the M1 diagram from the

relationship V = dM1/dx.

4) Plot the equivalent load (weq) diagram corresponding to the V diagram from the

relationship weq = dV/dx. Note, over the supports weq can be downwards. Also,

a singular moment needs to be included at an end when the eccentricity of the

CGS is not zero at the end.



5) Calculate the values of M2 for the continuous beam (with the intermediate

supports) subjected to weq using a method of elastic analysis (for example,

moment distribution or computer analysis). Plot the M2 diagram.

6) The M3 diagram can be calculated by adding the values of M2 and M0 diagrams

along the length of the beam.

The following figures explain the steps of developing the M2 diagram for a given profile

of the CGS and a value of Pe.

Given profile of the CGS

e

Given profile of the CGS

ee

Step (3) Plotting of V diagram

Step (2) Plotting of M1 diagram

Step (3) Plotting of V diagramStep (3) Plotting of V diagram

Step (2) Plotting of M1 diagramStep (2) Plotting of M1 diagram

Step (4) Plotting of weq diagram

Step (5) Plotting of M2 diagram

Step (4) Plotting of weq diagramStep (4) Plotting of weq diagram

Step (5) Plotting of M2 diagramStep (5) Plotting of M2 diagram Figure 8-2.5 Development of the moment diagram due to prestressing force



The important characteristics of the diagrams are as follows.

1) A positive eccentricity of the CGS creates a negative moment (M1) and an

upward thrust.

2) The M2 diagram has a similar shape to the M1 diagram, which is again similar to

the profile of the CGS. This is because the moment generated due to the

reactions (M1/) is linear between the supports.

8.2.3 Pressure Line due to Prestressing Force

The pressure line (thrust line or C-line) due to the prestressing force only can be

determined from the M2 diagram. It is to be noted that the external loads are not

considered in this pressure line. This is used to select the profile of the CGS.

The calculation of pressure line from the M2 diagram is based on the following

expression. The pressure line can be plotted for the different values of M2 along the

length.

ec= M2/Pe (8-2.3) Here,

ec = distance of the pressure line from the CGC at a location. A positive

value of ec corresponds to a hogging value of M2 and implies that the

pressure line is beneath the CGC.

The following sketch shows the pressure line for a given profile of the CGS.

ece

Pressure line

Profile of the CGS

CGCece

Pressure line

Profile of the CGS

CGC

Figure 8-2.7 Pressure line for a continuous beam

The important characteristics of the pressure line are as follows.

1) The shift of the pressure line from the profile of the CGS is a linear

transformation. It is because M2 diagram has a similar shape to the profile of the

CGS.



⇒ The pressure line will have the same intrinsic shape as the profile of the CGS.

2) Since M2 is proportional to the prestressing force, the eccentricity of the pressure

line (ec) remains constant even when the prestressing force drops from the initial

value P0 to the effective value Pe.

⇒ The location of the pressure line for a given profile of the CGS is fixed,

irrespective of the drop in the prestressing force.

Example 8-2.1 The profile of the CGS for a post-tensioned beam is shown in the sketch. Plot the pressure line due to a prestressing force Pe = 1112 kN.

BA CD 7.5 m7.5 m6 m9 m

0.06 0.240.08 rad

0.12

0.176 rad

CGC0.27

Values of eccentricity in metres.

BA CD 7.5 m7.5 m6 m9 m

0.06 0.240.08 rad

0.12

0.176 rad

CGC0.27

Values of eccentricity in metres.

Solution

1) Plot M1 diagram

The values of M1 are calculated from M1 = Pee.

e (m) M1 (kN m)

0.06 – 66.7

0.24 – 266.9

– 0.12 133.4

0.27 – 300.2



133.4

– 300.2– 266.9– 66.7

M1 diagram (kN m)

Profile of the CGS

0.06 0.24 0.12 0.27

A D B C

133.4

– 300.2– 266.9– 66.7

133.4

– 300.2– 266.9– 66.7

M1 diagram (kN m)

Profile of the CGS

0.06 0.24 0.12 0.27

A D B C

0.06 0.24 0.12 0.27

A D B C

2) Plot V diagram

For AD, 1

-266.9 - (-66.7)=9

= -22.2 kN

dMV =dx

For DB, 1

133.4 - (-266.9)=6

= 66.7 kN

dMV =dx

For BC, to find dM1/dx, an approximate parabolic equation for the M1 diagram can be

used.

( )

( )

e

e

P exM = - L - xL

dMV =dx

P e= - L - xL

1 2

1

2

4

4 2

Pee

M1

L

xPee

M1

L

x



At B, 1

=0

4= -

4×(133.4+300.2)= -15

= -115.6 kN

x

e

dMV =dx

P eL

The exact value of V at B is

V = - 107.0 kN

The difference of V between C and B is given from the change in slope of the M1

diagram.

V|C - V|B = 0.176 × 1112

= 195.7 kN

Therefore, value of V at C is given as follows.

V|C = 195.7 – 107.0

= 89.0 kN

133.4

– 300.2– 266.9– 66.7

M1 diagram (kN m)

66.7– 22.2

– 107.0

89.0

V diagram (kN)

133.4

– 300.2– 266.9– 66.7

M1 diagram (kN m)

133.4

– 300.2– 266.9– 66.7

M1 diagram (kN m)

66.7– 22.2

– 107.0

89.0

V diagram (kN)

66.7– 22.2

– 107.0

89.0

V diagram (kN) 3) Plot equivalent load (weq) diagram

Include moment 66.7 kN-m at A.

Point load at D

W|D = 66.7- (- 22.2 )

= 88.9 kN



Since B is a reaction point, the downward load at B need not be considered.

Distributed load within B and C

BCw 89.0 - (-107.0)=15

=13.0 kN/m

66.7– 22.2

– 107.0

89.0

V diagram (kN)

13.0 kN/m88.9 kN66.7 kN m

Equivalent load diagram

66.7– 22.2

– 107.0

89.0

V diagram (kN)

66.7– 22.2

– 107.0

89.0

V diagram (kN)

13.0 kN/m88.9 kN66.7 kN m


13.0 kN/m88.9 kN66.7 kN m


4) Plot the M2 diagram.

Calculate moment at supports by moment distribution

2

2

88.9×9×615

=128

DF

FEM

Bal

CO

Bal

Total

2

2

88.9×9 ×615

= -192

0.5 0.5213.0×15

12=244

–194.7

–244

–97 122

244

–38.5 –38.5

–66.7 –327.5 327.5 0

66.788.9 13.0

2

2

88.9×9×615

=128

DF

FEM

Bal

CO

Bal

Total

2

2

88.9×9 ×615

=-192

0.5 0.5213.0×15

12=244

–194.7

–244

–97 122

244

–38.5 –38.5

–66.7 –327.5 327.5 0

66.788.9 13.0

In the previous table,

Bal = Balanced

CO = Carry Over moment

DF = Distribution Factor

FEM = Fixed End Moment.



The moment at the spans can be determined from statics. But this is not necessary as

will be evident later.

13.0 kN/m88.9 kN66.7 kN m


327.0

– 66.7

0.0

M2 diagram (kN m)

13.0 kN/m88.9 kN66.7 kN m


13.0 kN/m88.9 kN66.7 kN m


327.0

– 66.7

0.0

M2 diagram (kN m)

327.0

– 66.7

0.0

M2 diagram (kN m) 5) Calculate values of ec at support.

The values of ec are calculated from ec= M2/Pe.

M2 (kN m) ec (m)

– 66.7 0.06

327.0 0.294

0.0 0.184

The deviations of the pressure line from the CGS at the spans can be calculated by

linear interpolation.

Pressure line

0.06

0.1360.294

0.184

Profile of CGS

Pressure line

0.06

0.1360.294

0.184

Profile of CGS



8.3 Continuous Beams (Part II) This section covers the following topics.

• Principle of Linear Transformation

• Concordant Tendon Profile

• Tendon Profiles

• Partially Continuous Beams

• Analysis for Ultimate Strength

• Moment Redistribution

Introduction Before the discussion on the tendon profile (profile of the CGS), the following concepts

are introduced.

1) Principle of linear transformation

2) Concordant tendon profile.

8.3.1 Principle of Linear Transformation

When the profile of the CGS is moved over the interior supports of a continuous beam

without changing the intrinsic shape of the profile within each individual span, the profile

is said to be linearly transformed. In a linear transformation, the curvatures remain

constant and the locations of bends remain unchanged.

The following sketch explains the concept of linear transformation of the profile of the

CGS.

Profile after linear transformation

Profile of the CGS

CGC


Profile of the CGS

CGC

Figure 8-3.1 Linear transformation of the profile of the CGS

Linear transformation cannot involve the movement of the CGS at the ends of a beam

or at the support of a cantilever.



Theorem In a continuous beam, a profile of the CGS can be linearly transformed without

changing the position of the resultant pressure line. This theorem can be proved based

on the requirement that the curvature of the profile of the CGS remains constant under

linear transformation. The following sketch explains that the pressure line remains

constant for linearly transformed profiles of the CGS.

Common pressure line

Profile of the CGS

CGC


Common pressure line

Profile of the CGS

CGC


Figure 8-3.2 Pressure line for linearly transformed profiles of the CGS

8.3.2 Concordant Tendon Profile

A concordant tendon profile in a continuous beam is a profile of the CGS which

produces a pressure line coincident with the profile itself. A concordant tendon profile

does not produce reactions at the supports or secondary moments in the spans. The

upward and downward equivalent loads balance each other.

The following sketch shows a concordant tendon profile which is coincident with the

pressure line.

Concordant profile at the pressure line

CGC

Concordant profile at the pressure line

CGC

Figure 8-3.3 Concordant profile

The advantage of a concordant cable profile is that the calculations become simpler.



1) There is no secondary moment in the spans due to the prestress. The M2

diagram coincides with the M1 diagram.

2) The pressure line due to the prestress coincides with the cable profile. The shift

of the pressure line due to external loads can be measured from the profile

directly.

A concordant profile can be developed from the moment diagram due to external loads

for a certain load combination using the following theorem.

Theorem Every real moment diagram for a continuous beam on non-settling supports produced

by any combination of external loads, whether transverse loads or moments, plotted to

any scale, is one location for a concordant tendon in that beam.

The theorem can be proved based on the condition of no deflection at the supports due

to external loads. Also, for a concordant profile since there is no reaction at any

support, there is no possibility of deflections at the supports. Thus, it is easy to obtain a

concordant profile from the moment diagram of the external loads for a certain load

combination, drawn to a certain scale. The following figure shows the steps of the

development of concordant profile from the moment diagram.

(a) Loading on a continuous beam

(c) Concordant profile

CGC

(b) Moment diagram

(a) Loading on a continuous beam

(c) Concordant profile

CGCCGC

(b) Moment diagram

Figure 8-3.4 Development of concordant profile



Discussion The computation of the concordant profile helps in the layout of the tendon profile. The

tendon profile need not be designed to be a concordant profile. It should be such that

the stresses in concrete at transfer and at service are within the allowable values. If a

concordant profile is selected then the calculations become simpler.

8.3.3 Tendon Profiles

The steps of selecting a tendon profile (profile of the CGS) are based on trials. The

steps are as follows.

1) Assume the section of the beam for calculating self weight. For the preliminary

design, the type and depth (h) of the section can be selected based on

architectural requirement and deflection criteria.

2) Calculate the moment due to self weight (Msw) and the maximum moment (Mmax)

and minimum moment (Mmin) along the length of the beam (envelop moment

diagrams) due to the external loads, including self weight.

3) Compute the required Pe based on the values of Mmax and Mmin, at the critical

locations, similar to the calculations for a simply supported beam. Revise the

section if necessary. If Msw is large,

Pe = MT / z (8-3.1) z ≈ 0.65h (8-3.2)

Here,

MT = Mmax or Mmin

z = estimated lever arm.

4) Considering fpe = 0.7fpk , calculate area of prestressing steel Ap = Pe / fpe.

5) Check the area of the cross-section (A) based on A = Pe /(0.5fcc,all).

6) Calculate the kern distances kb and kt, and the maximum and minimum

eccentricities (emax and emin) along the length. The zone between emax and emin

along the length of the beam is the limiting zone. The equations of emax and emin

are same as that for a simply supported beam.

The value of P0 can be estimated from Pi as follows.

a. 90% of the initial applied prestress (Pi) for pre-tensioned members.



b. Equal to Pi for post-tensioned members.

The value of Pi can be estimated as follows.

Pi = Ap (0.8fpk) (8-3.3) Ap = Pe / 0.7fpk (8-3.4)

7) Select a trial profile of the CGS within the limiting zone. If the profile is a

concordant profile, the pressure line due to prestress coincides with the profile of

the CGS.

Calculate the shift in the pressure line due to external loads. For a Type 1 member, if

the final pressure line lies within the kern zone, then the solution is acceptable. If final

pressure line lies outside the kern zone, try another profile.

For Type 2 and Type 3 members, if the final pressure line lies within a zone such that

the stresses at the edges are within the allowable values, then the solution is

acceptable. If final pressure line lies outside the zone, try another profile.

8) Linearly transform the profile of the CGS to satisfy the cover requirements and

the convenience of prestressing.

For a prismatic beam with uniform cross section along the length, the tendon profile can

be selected similar to the moment diagram under uniform load. Since there cannot be a

sharp kink in the tendons and the supports are not true point supports, the profile needs

to be curved at an intermediate support. For a beam with varying depth, the tendon

profile can be adjusted (within the limiting zone) to be relatively straight for convenience

of layout of the tendons and reduction of losses due to friction. The tendons can be of

segments of single curvature to reduce frictional losses.

The following sketches show the profiles of the CGS for common continuous beams.




b) Varying cross section


b) Varying cross section

d) Uniform cross section with overlapping tendons

c) Combination of a) and b)





Figure 8-3.5 Profiles of CGS for continuous beams

8.3.4 Partially Continuous Beams

Due to the difficulties in construction of continuous beams, an intermediate system

between simply supported beams and continuous beams is adopted. These are called

partially continuous beams.

First, the individual precast members are placed at the site. Next continuity is

introduced by additional prestressing tendons or coupling the existing tendons.

Continuity can also be introduced in a composite construction, where non prestressed

continuity reinforcement is introduced in the cast-in-place topping slab.

A few examples are given in the following sketches. Other innovative schemes are also

used.



Additional tendon

a) With additional tendon

individual tendon

Additional tendon

a) With additional tendon

individual tendon

CouplerJack

b) With coupling of tendons

CouplerCouplerJack

b) With coupling of tendons

Cast-in-place Topping

c) Composite construction with continuity reinforcement

Cast-in-place Topping

c) Composite construction with continuity reinforcement Figure 8-3.6 Partially continuous beams

8.3.5 Analysis for Ultimate Strength

The analysis of continuous beams for ultimate strength is difficult for the following

reasons.

1) Due to non-linear behaviour, superposition of stresses is not valid.

⇒ The concept of load balancing is not truly applicable.

2) The prestressing force varies at the location of cracks.

3) Neglect of the secondary moment due to prestressing is erroneous, unless full

moment redistribution is allowed.

Clause 18.6.4 of IS:1343 - 1980 insists on considering the secondary moment.



8.3.6 Moment Redistribution

It was mentioned in Section 3.4, Analysis of Member under Flexure (Part III), that there

is an inconsistency in the traditional analysis at the ultimate state. The demand is

calculated based on elastic analysis, whereas the capacity is calculated based on the

non-linear limit state analysis. Although the analysis for demand at ultimate is based on

an elastic analysis, IS:1343 - 1980 allows to take advantage of the post-yield

deformation of the highly stressed sections in a continuous beam. The underlying

concept is known as moment redistribution.

Moment redistribution means the transfer of additional moments to the less stressed

sections, as the highly stressed sections with peak moments yield on reaching their

ultimate moment capacities.

To apply moment redistribution, the highly stressed sections are designed for lower

moments and the less stressed sections are designed to carry higher moments than the

values obtained from an elastic analysis. This gives an economical solution.

IS:1343 - 1980, Clause 21.1.1 specifies the following conditions for moment

redistribution.

1) The redistributed moments must be in a state of static equilibrium with the

factored external loads.

2) For serviceability requirements, the ultimate moment of resistance at any section

(MUR) should not be less than 80% of the moment demand from an elastic

analysis (Mu).

3) To limit the demand on post-yield rotation, the reduction in moment at the highly

stressed sections is limited to 20% of the numerically largest moment anywhere

in the beam calculated by an elastic analysis.

4) To ensure ductile behaviour of the highly stressed sections, the following

relationship should be checked.

≤+ 0

100u Mx δ

d.5 (8-3.5)



Here,

xu = depth of neutral axis

d = the effective depth

δM = the percentage reduction in moment.

Example 8-3.1

The prestressed concrete beam shown in the figure, is fixed at the left end and roller supported at the right. It is post-tensioned with a single tendon with a parabolic profile, with indicated eccentricities.

a). Locate the pressure line due to application of a prestress force of 1068 kN. b). Find the primary, secondary and total moments due to prestressing force

at the face of the fixed support. c). What is the magnitude and direction of the reaction produced at the roller

by prestressing force? d). What minor adjustment can be made in the tendon profile to produce a

concordant profile?

6 m 6 m

150250 300300

6 m 6 m

150250 300300



Solution

a) Locate the pressure line.

1) Plot M1 diagram.

The values of M1 are calculated from M1 = Pee.

e (m) M1 (kN m)

– 0.250 267.0

0.150 – 160.2

0.0 0.0

150250

267.0−160.2

Profile of the CGS

M1 diagram (kN m)

150250

267.0−160.2

Profile of the CGS

M1 diagram (kN m)

2) Plot V diagram.

The M1 diagram is made up of two parabolic segments.

267.0−160.2

M1 diagram (kN m)

267.0−160.2

M1 diagram (kN m) For each segment,

( )PexM = - L - x

L1 24



( )

dMV =dx

Pe= - L - xL

1

24 2

Pe

M1

L

xPe

M1

L

x

( )4 267.0+160.2-

12= -142.4 kN

4×160.212

=53.4 kN

V diagram (kN)

x=

PeV = -L0

4x=L

PeV =L

4

( )4 267.0+160.2-

12=-142.4 kN

4×160.212

=53.4 kN

V diagram (kN)

x=

PeV = -L0

4x=L

PeV =L

4

3) Plot equivalent load (weq) diagram.

eq

dVw =dx

= 53.4 +142.412

=16.3 kN/m

16.3 kN/m


16.3 kN/m


4) Plot the M2 diagram.

Calculate moment at supports by moment distribution.



216.3×1212

=195.8

−195.8

195.8

97.9

293.7 0Total

CO

FEM

Bal

216.3×1212

=195.8

−195.8

195.8

97.9

293.7 0Total

CO

FEM

Bal

In the previous table,

Bal = Balanced

CO = Carry Over moment

FEM = Fixed End Moment.

The moment at the span can be determined from statics. But this is not necessary as

will be evident later.

293.7

16.3 kN/m


M2 diagram (kN m)

293.7

16.3 kN/m


M2 diagram (kN m)

5) Calculate values of ec at support.

The values of ec are calculated from ec= M2/Pe.

M2 (kN m) ec (m)

293.7 0.275

The deviations of the pressure line from the CGS at the span can be calculated by

linear interpolation.



150 – 12.5 = 137.5275 – 250

½ (275 – 250) = 12.5

Pressure line

150 – 12.5 = 137.5275 – 250

½ (275 – 250) = 12.5

Pressure line

b) Calculation of primary, secondary and total moments.

M1 = 267.0 kN m primary

M2 = 293.7 kN m total

M1/ = M2 – M1

= 293.7 – 267.0

= 26.7 kN m secondary

c) Calculation of reaction.

L

M2

R2 R2

R1R1

weq

L

M2

R2 R2

R1R1

weq=

1 216.3×12

2= 97.6 kN

eqw LR =

22

293.7=12

= 24.5 kN

MR =L

R1 – R2 = 73.1 kN

Resultant reaction at roller is downwards.

d) The tendon can be shifted to coincide with the pressure line to get a concordant

profile.



275 137.5

Concordant profile

Values in mm.

275 137.5

Concordant profile

Values in mm.



9.1 Composite Sections This section covers the following topics.

• Introduction

• Analysis of Composite Sections

• Design of Composite Sections

• Analysis for Horizontal Shear Transfer

9.1.1 Introduction A composite section in context of prestressed concrete members refers to a section with

a precast member and cast-in-place (CIP) concrete. There can be several types of

innovative composite sections. A few types are sketched below.

T TT T T T

Box section Composite beam-slab T-section

T TT TT T T TT T

Box section Composite beam-slab T-section Figure 9-1.1 Examples of composite sections

The following photos show the reinforcement for the slab of a box girder bridge deck

with precast webs and bottom flange. The slab of the top flange is cast on a stay-in

formwork. The reinforcement of the slab is required for the transverse bending of the

slab. The reinforcement at the top of the web is required for the horizontal shear

transfer.



Figure 9-1.2 Box girder bridge deck with precast webs and bottom flange and CIP

slab; Top: Aerial view, Bottom: Close-up view (Courtesy: José Turmo)

The advantages of composite construction are as follows.

1) Savings in form work

2) Fast-track construction

3) Easy to connect the members and achieve continuity.



The prestressing of composite sections can be done in stages. The precast member

can be first pre-tensioned or post-tensioned at the casting site. After the cast-in-place

(cast-in-situ) concrete achieves strength, the section is further post-tensioned.

The grades of concrete for the precast member and the cast-in-place portion may be

different. In such a case, a transformed section is used to analyse the composite

section.

9.1.2 Analysis of Composite Sections

The analysis of a composite section depends upon the type of composite section, the

stages of prestressing, the type of construction and the loads. The type of construction

refers to whether the precast member is propped or unpropped during the casting of

the CIP portion. If the precast member is supported by props along its length during the

casting, it is considered to be propped. Else, if the precast member is supported only at

the ends during the casting, it is considered to be unpropped.

The following diagrams are for a composite section with precast web and cast-in-place

flange. The web is prestressed before the flange is cast. At transfer and after casting

of the flange (before the section behaves like a composite section), the following are the

stress profiles for the precast web.

At transfer due to

P0 + MSW

+

After casting of flangedue to

Pe + MSW MCIP

Section At transfer due to

P0 + MSW

+

After casting of flangedue to

Pe + MSW MCIP

Section

Figure 9-1.3 Stress profiles for the precast web

Here,

P0 = Prestress at transfer after short term losses

Pe = Effective prestress during casting of flange after long term losses

MSW = Moment due to self weight of the precast web

MCIP = Moment due to weight of the CIP flange.

At transfer, the loads acting on the precast web are P0 and MSW. By the time the flange

is cast, the prestress reduces to Pe due to long term losses. In addition to Pe and MSW,



the web also carries MCIP. The width of the flange is calculated based on the concept of

effective flange width as per Clause 23.1.2, IS:456 - 2000.

At service (after the section behaves like a composite section) the following are the

stress profiles for the full depth of the composite section.

+

At service due to

Pe + MSW MCIP MLL

Section

Unpropped Propped

or +

Figure 9-1.4 Stress profiles for the composite section

Here, MLL is the moment due to live load. If the precast web is unpropped during

casting of the flange, the section does not behave like a composite section to carry the

prestress and self weight. Hence, the stress profile due to Pe + MSW + MCIP is

terminated at the top of the precast web. If the precast web is propped during casting

and hardening of the flange, the section behaves like a composite section to carry the

prestress and self weight after the props are removed. The stress profile is extended up

to the top of the flange. When the member is placed in service, the full section carries

MLL.

From the analyses at transfer and under service loads, the stresses at the extreme

fibres of the section for the various stages of loading are evaluated. These stresses are

compared with the respective allowable stresses.

Stress in precast web at transfer

(9-1.1) SWP P ec M cf = - ± ±

A I I0 0

Stress in precast web after casting of flange

(9-1.2)

e e SW CIPP P ec (M + M )cf = - ± ±A I I



Stress in precast web at service

(a) For unpropped construction

/

e e SW CIP LL/

P P ec (M + M )c M cf = - ± ± ±A I I I

(9-1.3a)

(b) For propped construction

/

e e SW CIP LL/

P P ec M c (M + M )cf = - ± ± ±A I I I

(9-1.3b)

Here,

A = area of the precast web

c = distance of edge from CGC of precast web

c/ = distance of edge from CGC of composite section

e = eccentricity of CGS

I = moment of inertia of the precast web

I / = moment of inertia of the composite section.

From the analysis for ultimate strength, the ultimate moment capacity is calculated.

This is compared with the demand under factored loads. The analysis at ultimate is

simplified by the following assumptions.

1) The small strain discontinuity at the interface of the precast and CIP portions is

ignored.

2) The stress discontinuity at the interface is also ignored.

3) If the CIP portion is of low grade concrete, the weaker CIP concrete is used for

calculating the stress block.

The strain and stress diagrams and the force couples at ultimate are shown below.



∆εp

εpu

Strain

≈xu

Section Stress

0.0035 0.447 fck

fpu

Cuw Cuf

TufTuw

+

Force

bf

bw

Df

d Ap

Figure 9-1.5 Sketches for analysis at ultimate



bw = breadth of the web




∆εp = strain difference for the prestressing steel




fck = characteristic compressive strength of the weaker concrete

Cuw = resultant compression in the web (includes portion of flange

above precast web)

Cuf = resultant compression in the outstanding portion of flange

Tuw = portion of tension in steel balancing Cuw.

Tuf = portion of tension balancing Cuf.


uw ck u wC = f x b0.36 (9-1.4)

uf ck f w fC = f (b - b )D0.447 (9-1.5)

uw pw puT = A f (9-1.6)

uf pf puT = A f (9-1.7)

The symbols for the areas of steel are as follows.



Apf = part of Ap that balances compression in the outstanding flanges

Apw = part of Ap that balances compression in the web

The equilibrium equations are given below. These equations are explained in Section

3.5, Analysis of Members under Flexure (Part IV). The ultimate moment capacity (MuR)

is calculated from the second equation.

( ) ( )⇒

∑pw pf pu ck u w ck f w

F = 0

A + A f = f x b + f b - b D0.36 0.447 f (9-1.8)

( ) ( )uR pw pu u pf pu fM = A f d - x + A f d - D0.42 0.5 (9-1.9)

9.2.3 Design of Composite Sections

The design is based on satisfying the allowable stresses under service loads and at

transfer. The section is then analysed for ultimate loads to satisfy the limit state of

collapse. The member is also checked to satisfy the criteria of limit states of

serviceability, such as deflection and crack width (for Type 3 members only). Before the

calculation of the initial prestressing force (P0) and the eccentricity of the CGS (e) at the

critical section, the type of composite section and the stages of prestressing need to be

decided. Subsequently, a trial and error procedure is adopted for the design.

The following steps explain the design of a composite section with precast web and

cast-in-place flange. The precast web is prestressed before the casting of the flange.

The member is considered to be Type 1 member.

Step 1. Compute e.

With a trial section of the web, the CGS can be located at the maximum eccentricity

(emax). The maximum eccentricity is calculated based on zero stress at the top of the

precast web. This gives an economical solution. The following stress profile is used to

determine emax.



CGCct

cb e

fb

CGS

Web section Stress profile

CGCct

cb e

fb

CGS

CGCct

cb e

fb

CGS

Web section Stress profile Figure 9-1.6 Stress profile for maximum eccentricity of CGS

swmax b

Me = k +P0

Here,

CGC = Centroid of the precast web

kb = Distance of the bottom kern of the precast web from CGC

Msw = Moment due to self weight of the precast web.

P0 = A trial prestressing force at transfer.

Step 2. Compute equivalent moment for the precast web.

A moment acting on the composite section is transformed to an equivalent moment for

the precast web. This is done to compute the stresses in the precast web in terms of

the properties of the precast web itself and not of the composite section.

For a moment Mc which acts after the section behaves like a composite section, the

stresses in the extreme fibres of the precast web are determined from the following

stress profile.

CGC’ ct’

cb’

Composite section

bw

ct’’ ft

fb

Stress profile

CGC’ ct’

cb’

Composite section

bw

ct’’ ft

fb

Stress profile Figure 9-1.7 Stress profile for the composite section

c t

tM c 'f =

I'

c b

bM c 'f =

I'

Here,

CGC’ = centroid of the composite section



ct’ = Distance of the top of the precast web from the CGC’

ct” = Distance of the top of the composite section from the CGC’.

cb’ = Distance of the bottom of the precast web (or composite section)

from the CGC’

I ’ = moment of inertia for the composite section.

The following quantities are defined as the ratios of the properties of the precast web

and composite section.

t

t

t

bb

b

Icm = I'c '

Icm = I'c '

Then the stresses in the extreme fibres of the precast web can be expressed in terms of

mt and mb as follows.

(9-1.10)

t c t t ct

b

m M c m Mf = =I Ak

(9-1.11)

b c b b cb

t

m M c m Mf = =I Ak

Here,

A = Area of the precast web

kb = Distance of the bottom kern of the precast web from CGC

kt = Distance of the top kern of the precast web from CGC

The quantities mt Mc and mb Mc are the equivalent moments. Thus, the stresses in the

precast web due to Mc are expressed in terms of the properties of the precast web itself.

Step 3. Compute Pe

Let MP be the moment acting on the precast web prior to the section behaving like a

composite section. After Mc is applied on the composite section, the total moment for

the precast web is MP + mbMc.



The stress at the bottom for Type 1 member due to service loads is zero.

Therefore,

(9-1.12)

( )

e e P b c

t t

P b ce

t

P P e M + m M- - + =A Ak Ak

M + m MP =

e+ k

0

or,

Note that the prestressing force is acting only on the precast web and hence, e is the

eccentricity of the CGS from the CGC of the precast web.

Step 4. Estimate P0 as follows.

a) 90% of the initial applied prestress (Pi) for pre-tensioned members.

b) Equal to Pi for post-tensioned members.

The value of Pi is estimated as follows.

Pi = Ap(0.8fpk) (9-1.13) Ap = Pe / 0.7fpk (9-1.14)

Revise e, the location of CGS, as given in Step 1 based on the new value of P0.

sw

max bMe = k +P0

(9-1.15)

Step 5. Check for the compressive stresses in the precast web.

At transfer, the stress at the bottom is given as follows.

sw

bt t

P P e Mf = - - +A Ak Ak

0 0(9-1.16)

The stress fb should be limited to fcc,all, where fcc,all is the allowable compressive stress in

concrete at transfer (available from Figure 8 of IS:1343 - 1980).

At service,

( )P t ce e

tb b

M + m MP P ef = - - +A Ak Ak

(9-1.17)



The stress ft should be limited to fcc,all , where fcc,all is the allowable compressive stress in

concrete under service loads (available from Figure 7 of IS:1343-1980). If the stress

conditions are not satisfied, increase A.

Step 6. Check for the compressive stress in the CIP flange.

′ c t

tM c "f =

I' (9-1.18) The stress ft/ should be limited to fcc,all , where fcc,all is the allowable compressive stress

in concrete under service loads.

9.1.4 Analysis for Horizontal Shear Transfer

With increase in the load, the bottom face of the CIP portion tends to slip horizontally

and move upwards with respect to the top face of the precast portion. To prevent this

and to develop the composite action, shear connectors in the form of shear friction reinforcement is provided.

The required shear friction reinforcement (per metre span) is calculated as follows.

τv h

svy

bA =f µ

10000.87

(9-1.19)

The minimum requirements of shear friction reinforcement and spacing are similar to

that for shear reinforcement in the web.


Asv = area of shear friction reinforcement in mm2/m

bv = width of the interface of precast and CIP portions

τh = horizontal shear stress at the interface in N/mm2

fy = yield stress in N/mm2

µ = coefficient of friction

= 1.0 for intentionally roughened interface with normal weight concrete



The shear reinforcement in the web can be extended and anchored in the CIP portion to

act as shear friction reinforcement, as shown below. bv

Intentionally roughened

bv

Intentionally roughened

Figure 9-1.8 Shear reinforcement used for shear transfer

The following example shows the analysis of a composite beam.

Example 9-1.1

The mid-span section of a composite beam is shown in the figure. The precast

web 300 mm × 920 mm (depth) is post-tensioned with an initial force (P0) of 2450

kN. The effective prestress (Pe) is estimated as 2150 kN. Moment due to the self weight of the precast web (MSW) is 270 kNm at mid-span.

After the web is erected in place, the top slab of 150 mm × 920 mm (width) is

casted (unpropped) producing a moment (MCIP) of 135 kNm. After the slab concrete has hardened, the composite section is to carry a maximum live load moment (MLL) of 720 kNm. Compute stresses in the section at various stages.



Section

920

150

300

920

200

Section

920

150

300

920

200

920

150

300

920

200

Solution

1) Calculation of geometric properties.

Precast web

A = 2.76 × 105 mm2

I = 1.95 × 1010 mm2

Distance of CGC from bottom = 460 mm.

Composite section

A/ = 4.14 × 105 mm2

I / = 4.62 × 1010 mm2

Distance of CGC/ from bottom = 638 mm.

638460

CGC/

CGC638

460

CGC/

CGC

2) Calculation of stresses in web at transfer

0 0

3 3 6

5 10

2

2

2450×10 2450×10 ×260×460 270×10 ×460= - ± ±2.76×10 1.95×10 1.95×10

= -0.22 N/mm= -17.54 N/mm

SWP P ec M cf = - ± ±A I I

10



3) Calculation of stresses in web after long term losses

3 3 6

5 10

2

2

2150×10 2150×10 ×260×460 270×10 ×460= - ± ±2.76×10 1.95×10 1.95×10

= -0.97 N/mm= -14.61 N/mm

e e SWP P ec M cf = - ± ±A I I

10

4) Calculation of stresses in web after casting of flange

3 3 6

5 10

2

2

2150×10 2150×10 ×260×460 (270+135)×10 ×460= - ± ±2.76×10 1.95×10 1.95×10

= -4.16 N/mm= -11.42 N/mm

e e SW CIPP P ec (M + M )cf = - ± ±A I I

10

5) Calculation of stresses in the composite section at service

Stress due to MLL

At top fibre At bottom fibre

3

10

2

750×10 ×638= 4.62×10

=10.36 N/mm

/LL b

b /

M cf =I

3

10

2

750×10 ×432= -4.62×10

= -7.01 N/mm

/// LL t

t /

M cf = -I

At top fibre of precast web, the stress due to MLL is calculated from proportionality of

triangles.

2

7.01×282= -432

= -4.57 N/mm

tf



Total stress in precast web

At top fibre At bottom fibre

2

= -4.16 - 4.57= -8.73 N/mm

tf2

= -11.42+10.36= -1.06 N/mm

bf

Total stress in CIP slab

The total stress is due to MLL only.

At top fibre At bottom fibre 2= -7.01 N/mm/

tf 2= - 4.57 N/mm/bf

Stress profiles

– 17.54 – 14.61

–0.22 – 0.97

– 11.42

– 4.16

10.36

– 7.01

– 4.57

– 1.06

– 7.01

– 4.57– 8.73

At transfer

After losses After casting

Due to MLL

At service

– 17.54 – 14.61

–0.22 – 0.97

– 11.42

– 4.16

10.36

– 7.01

– 4.57

– 1.06

– 7.01

– 4.57– 8.73

At transfer

After losses After casting

Due to MLL

At service



9.2 One-way Slabs This section covers the following topics.

• Introduction

• Analysis and Design

9.2.1 Introduction

Slabs are an important structural component where prestressing is applied. With

increase in the demand for fast track, economical and efficient construction, prestressed

slabs are becoming popular. The slabs are presented in two groups: one-way slabs

and two-way slabs. The two-way slabs are presented in details in Sections 9.3 and

9.4.

Rectangular slabs can be divided into the two groups based on the support conditions

and length-to-breadth ratios. The one-way slabs are identified as follows.

1) When a rectangular slab is supported only on two opposite edges, it is a one-way

slab spanning in the direction perpendicular to the edges. Precast planks fall in

this group.

2) When a rectangular slab is supported on all the four edges and the length-to-

breadth (L / B) ratio is equal to or greater than two, the slab is considered to be a

one-way slab. The slab spans predominantly in the direction parallel to the

shorter edge.

The following sketches show the plans of the two cases of one-way slabs. The

spanning direction in each case is shown by the double headed arrow.

B

LL

B

LL (a) (b)

(a) Supported on two opposite edges (b) Supported on all edges (L/B > 2)

Figure 9-2.1 Plans of one-way slabs



A slab in a framed building can be a one-way slab depending upon its length-to-breadth

ratio. A one-way slab is designed for the spanning direction only. For the transverse

direction, a minimum amount of reinforcement is provided. A hollow core slab is also an

example of a one-way slab. A ribbed floor (slab with joists) made of precast double tee

sections, is analysed as a flanged section for one-way bending.

Other types of rectangular slabs and non-rectangular slabs are considered to be two-

way slabs. If a rectangular slab is supported on all the four sides and the length-to-

breadth ratio is less than two, then it is a two-way slab. If a slab is supported on three

edges or two adjacent edges, then also it is a two-way slab. A slab in a framed building

can be a two-way slab depending upon its length-to-breadth ratio. A two-way slab is

designed for both the orthogonal directions.

A slab is prestressed for the following benefits.

1) Increased span-to-depth ratio

Typical values of span-to-depth ratios in slabs are given below.

Non-prestressed slab 28:1

Prestressed slab 45:1

2) Reduction in self-weight


⇒ Increased durability

4) Quick release of formwork

⇒ Fast construction

5) Reduction in fabrication of reinforcement

6) More flexibility in accommodating late design changes.

Precast planks are usually pre-tensioned. Cast-in-situ slabs are post-tensioned. Post-

tensioned slabs are becoming popular in office and commercial buildings and parking

structures, where large column-free spaces are desirable. The maximum length of a

post-tensioned slab is limited to 30 to 40 m to minimise the losses due to elastic

shortening and friction.

Slabs can be composite for the benefits of reduction in form work, cost and time of

construction and quality control. A precast plank can be prestressed and placed in the

final location. A topping slab is overlaid on the precast plank. The grades of concrete in



the two portions can be different. The following sketches show the sections of some

one-way slabs.

Precast and prestressedplank

Cast-in-situ topping

Precast and prestressedplank

Cast-in-situ topping

Figure 9-2.2 Cross-section of a composite slab

9.2.2 Analysis and Design

One-way slabs are analysed and designed for the spanning direction similar to

rectangular beams. The analysis and design is carried out for the width of the plank or

a unit width (say 1 m) of the slab. For continuous slabs, the moment coefficients of

IS:456 - 2000 (Table 12) can be used.

The analysis and design procedures for simply supported rectangular beams are

covered in Sections 3.2 to 3.6 and Sections 4.2 to 4.6, respectively. These materials are

briefly reproduced here.

Preliminary Design

1) Select the material properties fck and fpk. Here, fck is the characteristic

compressive strength of concrete and fpk is the characteristic tensile strength of

prestressing steel.

2) Determine the total depth of slab (h), based on the span to effective depth ratio

(L / d), given in Clause 22.6 of IS:1343 - 1980. Consider d ≈ h – 25 mm. Round

off h to a multiple of 10 mm.

3) Calculate the self weight.

4) Calculate the total moment (MT) including moment due to self weight (Msw).

5) Estimate the lever arm (z).

z = 0.65 h if Msw is large (say Msw > 0.3 MT)

z = 0.5 h if Msw is small.

6) Estimate the effective prestress (Pe).

Pe = MT /z if Msw is large

Pe = MI L/z if Msw is small.

Here, the moment due to imposed loads is given as MI L = MT – Msw. 7) Considering fpe = 0.7 fpk , calculate area of prestressing steel Ap = Pe / fpe .



8) Check the area of cross section (A) A = 1000 mm × h mm. The average stress

C/A should not be too high as compared to 50% fcc,all .

Final Design The final design involves the checking of the stresses in concrete at transfer and under

service loads with respect to the allowable stresses. The allowable stresses depend on

the type of slab (Type 1, Type 2 or Type 3). Here, the steps of final design are

explained for Type 1 slabs only. For Type 1 slabs, no tensile stress is allowed at

transfer or under service loads.

For small moment due to self-weight (Msw ≤ 0.3 MT), the steps are as follows.

1) Calculate eccentricity (e) to locate the centroid of the prestressing steel (CGS).

The lowest permissible location of the compression (C) due to self-weight is at the

bottom kern point (at a depth kb below CGC) to avoid tensile stress at the top. The

design procedure based on the extreme location of C gives an economical section. For

this location of C, the following equation can be derived.

sw

bMe = + kP0

(9-2.1)

The magnitude of C or T is equal to P0, the prestress at transfer after initial losses.

The value of P0 can be estimated as follows.

a) P0 = 0.9 Pi for pre-tensioned slab

b) P0 = Pi for post-tensioned slab

Here, Pi is the initial applied prestress.

Pi = (0.8fpk) Ap (9-2.2) The permissible prestress in the tendon is 0.8fpk.

2) Re-compute the effective prestress Pe and the area of prestressing steel Ap.

For the extreme top location of C under service load, the shift of C due to the total

moment gives an expression of Pe.



T

eT

MP =e+ k (9-2.3)

For solid rectangular slab, kb = kt = h / 6.

Considering fpe = 0.7fpk , the area of prestressing steel is recomputed as follows.

Ap = Pe / fpe (9-2.4)

The number of tendons and their spacing is determined based on Ap. The value of P0 is

updated.

3) Re-compute e with the updated values of Ap and P0.



For large Msw if e violates the cover requirements, e is determined based on cover.

4) Check the compressive stresses in concrete

For the limiting no tension design at transfer, the stress at the bottom (fb) is given as

follows.

(9-2.5)

bt

P h Pf = - = -A c A

0 02

The stress should be less than fcc,all , where fcc,all is the allowable compressive stress in

concrete at transfer. The condition to satisfy can thus be written as | fb | ≤ fcc,all.

For the limiting no tension design at service, the stress at the top (ft) is given as follows.

(9-2.6)

e e

b

P h Pf = - = -t A c A2

The stress should be less than fcc,all , where fcc,all is the allowable compressive stress in

concrete at service. The condition to satisfy can be written as | ft | ≤ fcc,all.

For Type 2 and Type 3 slabs, the tensile stress should be restricted to the allowable

values. For a continuous slab, a suitable profile of the tendons is selected similar to that



in continuous beams. The design of continuous beams is covered in Sections 8.2 and

8.3.

When the value of e is fixed (in either pre-tension or post-tension operations), the

design steps are simpler. If the tendons are placed at the CGC (e = 0), then the uniform

compressive stress due to prestress counteracts the tensile stress due to service loads.

To have zero stress at the bottom under service conditions, the value of Pe can be

directly calculated from the following equation.

e T

b

Teb

P M=A Z

MP = AZ

or,

(9-2.7)

Zb is the section modulus. The above expression is same as Pe = MT / kt, with e = 0.

The stresses at transfer can be checked with an estimate of P0 from Pe.

5) Checking for shear capacity

The shear is analogous to that generates in a beam due to flexure. The calculations can

be for unit width of the slab. The critical section for checking the shear capacity is at a

distance effective depth ‘d’ from the face of the beam, across the entire width of the

slab. The critical section is transverse to the spanning direction. The shear demand

(Vu) in the critical section generates from the gravity loads in the tributary area.

d

Critical section

Tributary area

d

Critical section

Tributary area

Figure 9-2.3 Tributary area and critical section for shear

For adequate shear capacity, VuR ≥ Vu where, VuR = Vc, the shear capacity of uncracked

concrete of unit width of slab. The expression of Vc is given in Section 5.2, Design for

Shear (Part I). If this is not satisfied, it is preferred to increase the depth of the slab to

avoid shear reinforcement.



6) Provide transverse reinforcement based on temperature and shrinkage.

As per IS:456 - 2000, Clause 26.5.2.1, the minimum amount of transverse

reinforcement (Ast,min in mm2) for unit width of slab is given as follows.

Ast,min = 0.15% 1000 h for Fe 250 grade of steel

= 0.12% 1000 h for Fe 415 grade of steel.

Usually the transverse reinforcement is provided by non-prestressed reinforcement.

The minimum reinforcement is sufficient for the transverse moment due to Poisson’s

effect and small point loads. For a heavy point load, transverse reinforcement needs to

be computed explicitly.

The following example shows the design of a simply supported precast prestressed

composite slab.

Reference:

Santhakumar, A.R., Partially Precast Composite PSC Slab, Published by Building

Technology Centre, Anna University, Chennai.

Example 9-2.1

Design a simply supported precast prestressed (Type 1) composite slab for the following data.

Width of the slab = 0.3 m Clear span = 2.9 m Effective span (L) = 3.1 m Thickness of the precast plank = 50 mm Thickness of the cast-in-situ topping slab = 50 mm Grade of concrete in precast plank : M60 Grade of concrete in topping slab : M15

The pre-tensioned tendons are located at mid depth of the precast slab. During the casting of the topping, planks are not propped.



Live load = 2.0 kN/m2 Load due to floor finish = 1.5 kN/m2.

Solution

1) Calculation of moments.

Load per unit area

Weight of precast plank = 1.25 kN/m2

Weight of topping slab = 1.25 kN/m2

Weight of floor finish = 1.50 kN/m2

Live load = 2.00 kN/m2

Total = 6.00 kN/m2

Total moment (MT) along the width of the slab is given as follows.

wBL2 26×0.3×3.1=8 8

= 2.16 kNm

The individual moments are calculated based on the proportionality of the loads.

MSW = moment due to self weight of precast plank

= 2.16 × (1.25 / 6.00) = 0.45 kNm

Mtop = moment due to weight of topping slab

= 2.16 × (1.25 / 6.00) = 0.45 kNm

Mfin = moment due to weight of floor finish

= 2.16 × (1.50 / 6.00) = 0.54 kNm

MLL = moment due to live load

= 2.16 × (2.00 / 6.00) = 0.72 kNm.

2) Calculation of geometric properties.

Precast section



Area

A1 = 300 × 50 = 15,000 mm2

Moment of inertia

300mm

50mm

300mm

50mm

I 31

4

1= ×300×5012

= 3,125,000 mm

Distance to the extreme fibres

b tc = c 50=2

= 25 mm

Section moduli

omposite section

concrete are different for the precast- prestressed (PP) and cast-in-

quivalent area factor = Modulus of CIS / Modulus of PP

b tZ = Z

3

3,125,000=25

=125,000 mm

C

Since the grades of

situ (CIS) portions, an equivalent (transformed) area is calculated. The CIS portion is

assigned a reduced width based on the equivalent area factor (modular ratio).

E

= √(Grade of CIS / Grade of PP)

=√15/60

= 0.5



50mm

50mm

300mm

150mm

Composite section

Equivalent section

50mm

50mm

300mm

150mm

300mm

150mm

Composite section

Equivalent section

Location of CGC from bottom

Atop = 50 × 150 = 7,500 mm2

Abot = 50 × 300 = 15,000 mm2

A2 = Atop + Abot = 22,500 mm2

top botA × + A ×y =

A75 25

937,500=22,500

= 41.7 mm

41.7 mm

58.3 mmCGC

41.7 mm

58.3 mmCGC

Moment of inertia

topI 3 2

4

1= ×(0.5×300)×50 +7500×(75 - 41.7)12

= 9,894,166.8 mm

3 2

4

1= ×300×50 +15000×(41.7 - 25)12

= 7,293,333.5 mm

botI

7 4

= 9,894,166.8+7,293,333.5=17,187,500=1.719×10 mm

I



Distance to the extreme fibres

yb = 41.7 mm

yt = 58.3 mm

Section moduli

b

t

Z

Z

6

3

6

3

=17.19×10 /41.7

= 412,527 mm=17.19×10 /58.3

= 294,703 mm

3) Calculation of prestress

The tendons are located at the mid depth of the precast plank. Hence, e = 0 for the

precast plank. The value of Pe is calculated directly from the following stress profiles.

+

Section MSW + MtopPe Mfin + MLL

+

Stress profiles

+

Section MSW + MtopPe Mfin + MLL

+

Stress profiles To avoid tensile stress at the bottom under service conditions, the resultant stress is

equated to zero.

⎡ ⎤⎢ ⎥⎣ ⎦

SW tope fin LL

b b

SW top fin LLe

b b

M + MP M + M- + + =A Z Z

M + M M + M P = A +Z Z

1 1 2

11 2

0

or,

In the above expression, the first term inside the bracket corresponds to the precast

section. The moments due to self weight and topping slab are resisted by the precast

section alone.

The second term inside the bracket corresponds to the equivalent section. The

moments due to weight of the floor finish and live load are resisted by the equivalent

section.

⎡ ⎤⎢ ⎥⎣ ⎦

P = A 60.45+0.45 0.54+0.72+ ×125,000 412,527

= 50×300× (7.2+3.0)=153,816 N

10



Assuming around 20% loss,

The prestress at transfer (P0) = 1.2 × 153,816

= 184,579 N.

Wires of diameter = 7 mm and ultimate strength (fpk) = 1500 MPa are selected for

prestressing.

Area of one wire (Ap) = 38.48 mm2.

The maximum allowable tension in one wire

= 0.8 fpk × Ap

= 0.8 × 1500 × 38.48

= 46,176 N.

No. wires required = 184,579 / 46,176

= 3.99

→ 4.

Required pull in each wire = 184,579 / 4

= 46,145 N.

Total prestressing force (P0) = 4 × 46,145

= 184,580 N.

Effective prestressing force (Pe) = 0.8 × 184,580

= 147,664 N.

4) Checking of stresses in concrete

a) At transfer

The compressive strength at 7 days (fci ) = 0.7 fck

= 0.7 × 60

= 42 MPa.

Allowable compressive stress (fcc,all) = 0.44 fci



= 0.44 × 42

= 18.5 MPa.

For Type 1 members, the allowable tensile stress (fct,all) is zero.

Stresses at the mid-span of the precast portion

fc = – P0/A1 ± MSW/Z1

= – 12.3 ± (0.45 × 106 / 125,000)

– 8.7

– 15.9

– 8.7

– 15.9

ft = – 15.9 MPa

fb = – 8.7 MPa

∴ |ft| ≤ fcc,all OK

b) After casting of topping slab at 28 days

Allowable compressive stress (fcc,all) = 0.44 fck

= 0.44 × 60

= 26.4 MPa.

The allowable tensile stress (fct,all) is zero.

Stresses at the mid-span of the precast portion

fc = – P0/A1 ± (MSW + Mtop)/Z1

– 5.1

– 19.5

– 5.1

– 19.5 = – 12.31 ± ((0.45 + 0.45) × 106 / 125,000)

ft = – 19.5 MPa

fb = – 5.1 MPa

∴ |ft| ≤ fcc,all OK



c) At service

i) For the precast portion


= 0.35 × 60

= 21 MPa.

The allowable tensile stress (fct,all) is zero.

Stresses at the mid-span of the composite section for unpropped construction

fc = – Pe /A1 ± (MSW + Mtop) / Z1 ± (Mfin + MLL) / Z2

fjunc = – (147,664 / 15,000) – ((0.45 + 0.45) × 106 / 125,000)

– ((0.54 + 0.72) × 106 / 2,063,625)

= – 17.6 MPa

fb = – (147,664 / 15,000) + ((0.45 + 0.45) × 106 / 125,000)

+ ((0.54 + 0.72) × 106 / 412,527)

= 0.4 MPa

≅ 0

|fjung| ≤ fcc,all OK

fb = fct,all OK

ii) For cast-in-situ portion


= 0.35 × 15

= 5.2 MPa.

0.4

– 17.6

– 4.3

– 0.4

0.4

– 17.6

– 4.3

– 0.4

Stresses at the mid-span of the composite section

ft = – (0.54 + 0.72) × 106 / 294,703)

= – 4.3 MPa

fjunc = – (0.54 + 0.72) × 106 / 2,063,625)



= – 0.6 MPa

ft ≤ fcc,all OK

Note that the critical stress at the junction is in the precast portion.

5) Check for shear

VuR = Vc

= Vc0

= 0.67bh√(ft2 + 0.8fcpft)

= 0.67 × 300 × 50 √(1.862 + 0.8 × 9.36 × 1.86)

= 41.9 kN

Vu = wuB L / 2

= 1.5 × 6 × 0.3 × 3.1 / 2 = 4.2 kN

VuR ≥ Vu

Therefore, the shear capacity is adequate.

6) Transverse reinforcement

Using Fe 415 grade of steel, for 1m width

Ast,min = 0.12% 1000 h

= 0.0012 × 1000 × 100

= 120 mm2.

Provide 8 mm diameter bars at 300 mm on centre.

7) Provide nominal reinforcement for shrinkage in the longitudinal direction of the

topping slab.

Using Fe 415 grade of steel, for 1m width

Ast,min = 0.12% 1000 h



= 0.0012 × 1000 × 50

= 60 mm2.

Provide 6 mm diameter bars at 300 mm on centre.



Reinforcement details

(a) Plan of precast plank

(b) Plan of topping slab

(c) Longitudinal Section of composite slab

(4) 7 mm Φ wires

8 mm Φ rebar @ 300 mm c/c


Φ: diameter

(a) Plan of precast plank

(b) Plan of topping slab

(c) Longitudinal Section of composite slab

(4) 7 mm Φ wires



Φ: diameter



9.3 Two-way Slabs (Part I) This section covers the following topics.

• Introduction

• Analysis and Design

• Features in Modeling and Analysis

• Distribution of Moments to Strips

9.3.1 Introduction

The slabs are presented in two groups: one-way slabs and two-way slabs. The one-

way slabs are presented in Section 9.2. When a rectangular slab is supported on all the

sides and the length-to-breadth ratio is less than two, it is considered to be a two-way

slab. The slab spans in both the orthogonal directions. A circular slab is a two-way

slab. In general, a slab which is not falling in the category of one-way slab, is

considered to be a two-way slab.

Rectangular two-way slabs can be divided into the following types.

1) Flat plates: These slabs do not have beams between the columns, drop panels

or column capitals. Usually, there are spandrel beams at the edges.

2) Flat slabs: These slabs do not have beams but have drop panels or column

capitals.

3) Two-way slabs with beams: There are beams between the columns. If the

beams are wide and shallow, they are termed as band beams.

For long span construction, there are ribs in both the spanning directions of the slab.

This type of slabs is called waffle slabs.

The slabs can be cast-in-situ (cast-in-place). Else, the slabs can be precast at ground

level and lifted to the final height. The later type of slabs is called lift slabs. A slab in a

framed building can be a two-way slab depending upon its length-to-breadth (L / B)

ratio. Two-way slabs are also present as mat (raft) foundation.

The following sketches show the plan of various cases of two-way slabs. The spanning

directions in each case are shown by the double headed arrows.



a) Flat plate b) Flat slaba) Flat plate b) Flat slab

d) Waffle slabc) Two-way slab with beams

B

Ld) Waffle slabc) Two-way slab

with beams

B

L

Figure 9-3.1 Plans of two-way slabs

The absence of beams in flat plates and flat slabs lead to the following advantages.

1) Formwork is simpler

2) Reduced obstruction to service conduits

3) More flexibility in interior layout and future refurbishment.

Two-way slabs can be post-tensioned. The main advantage of prestressing a slab is

the increased span-to-depth ratio. As per ACI 318-02 (Building Code Requirements for

Structural Concrete, American Concrete Institute), the limits of span-to-depth ratios are

as follows.

For floors 42

For roofs 48.

The values can be increased to 48 and 52, respectively, if the deflection, camber and

vibration are not objectionable. The following photographs show post-tensioned flat

plate and flat slab.



(a) Flat plate

(b) Flat slab

Figure 9-3.2 Post-tensioned two-way slabs

(Courtesy: VSL India Pvt. Ltd.)

9.3.2 Analysis and Design

Analysis The analysis of two-way slabs is given in Section 31, IS:456 - 2000, under “Flat Slabs”.

The analysis is applicable to flat plates, flat slabs and two-way slabs with deflecting



beams. For two-way slabs with beams, if the beams are sufficiently stiff, then the

method (based on moment coefficients) given in Annex D, IS:456 – 2000, is applicable.

The direct design method of analysing a two-way slab is not recommended for

prestressed slabs. The equivalent frame method is recommended by ACI 318-02. It is

given in Subsection 31.5, IS:456 - 2000. This method is briefly covered in this section

for flat plates and flat slabs.

The slab system is represented by a series of two dimensional equivalent frames for

each spanning direction. An equivalent frame along a column line is a slice of the

building bound by the centre-lines of the bays adjacent to the column line.

The width of the equivalent frame is divided into a column strip and two middle strips.

The column strip (CS) is the central half of the equivalent frame. Each middle strip (MS)

consists of the remaining portions of two adjacent equivalent frames. The following

figure shows the division in to strips along one direction. The direction under

investigation is shown by the double headed arrow.

MS MSCS

l2

l1

1 32MS MSCS

l2

l1

11 332

Figure 9-3.3 Equivalent frame along Column Line 2

In the above figure,

l1 = span of the equivalent frame in a bay

l2 = width of the equivalent frame. This is the tributary width for calculating

the loads.

The following figure shows a typical elevation of an equivalent frame.



Figure 9-3.4 Elevation of an equivalent frame

The analysis is done for each typical equivalent frame. An equivalent frame is modelled

by slab-beam members and equivalent columns. The equivalent frame is analysed for

gravity load and lateral load (if required), by computer or simplified hand calculations.

Next, the negative and positive moments at the critical sections of the slab-beam

members are distributed along the transverse direction. This provides the design

moments per unit width of a slab.

If the analysis is restricted to gravity loads, each floor of the equivalent frame can be

analysed separately with the columns assumed to be fixed at their remote ends, as

shown in the following figure. The pattern loading is applied to calculate the moments

for the critical load cases. This is discussed later.

Figure 9-3.5 Simplified model of an equivalent frame

The steps of analysis of a two-way slab are as follows.

1) Determine the factored negative (Mu–) and positive moment (Mu

+) demands at the

critical sections in a slab-beam member from the analysis of an equivalent frame.

The values of Mu– are calculated at the faces of the columns. The values of Mu

+



are calculated at the spans. The following sketch shows a typical moment

diagram in a level of an equivalent frame due to gravity loads.

Mu–

Mu+

Mu–

Mu+

Figure 9-3.6 Typical moment diagram due to gravity loads

2) Distribute Mu– to the CS and the MS. These components are represented as Mu,

–

CS and Mu,–

MS, respectively. Distribute Mu+ to the CS and the MS. These

components are represented as Mu,+

CS and Mu,+

MS, respectively.

CS

MShalf

MShalf

Mu– Mu

–

Mu+

Mu+

CS

Mu+

MS

Mu–CS Mu

–CS

Mu–MS

CS

MShalf

MShalf

Mu– Mu

–

Mu+

Mu+

CS

Mu+

MS

Mu–CS Mu

–CS

Mu–MS

Figure 9-3.7 Distribution of moments to column strip and middle strips

3) If there is a beam in the column line in the spanning direction, distribute each of

Mu,–

CS and Mu,+

CS between the beam and rest of the CS.

CS

MShalf

MShalf

Mu– Mu

–

Mu+

Mu+

CS

Mu+

MS

Mu–CS Mu

–CS

Mu–MS

Mu–MS

Beamwidth

CS

MShalf

MShalf

Mu– Mu

–

Mu+

Mu+

CS

Mu+

MS

Mu–CS Mu

–CS

Mu–MS

Mu–MS

Beamwidth

Figure 9-3.8 Distribution of moments to beam, column strip and middle strips



4) Add the moments Mu,–

MS and Mu,+

MS for the two portions of the MS (from adjacent

equivalent frames).

5) Calculate the design moments per unit width of the CS and MS.

Design Once the design moments per unit width of the CS and MS are known, the steps of

design for prestressing steel are same as that for one-way slab. The profile of the

tendons is selected similar to that for continuous beams. The flexural capacity of

prestressed slab is controlled by total amount of prestressing steel and prestress rather

than by tendon distribution. But the tendon distribution effects the load balancing.

Some examples of tendon distribution are shown.

100 % banded throughcolumns in both directions

Normal reinforcement

100 % banded in one directionand distributed in otherdirection

100 % banded throughcolumns in both directions

Normal reinforcement

100 % banded in one directionand distributed in otherdirection

75 % in column strip and25 % in middle strip inboth directions

75 % in CS and 25 % in MS in one direction and distributed in other direction

CS MS CS

75 % in column strip and25 % in middle strip inboth directions

75 % in CS and 25 % in MS in one direction and distributed in other direction

CS MS CS

Figure 9-3.9 Typical tendon layouts



Maximum spacing of tendons or groups of tendons should be limited to 8h or 1.5 m,

whichever is less. Here, h is the thickness of the slab. A minimum of two tendons shall

be provided in each direction through the critical section for punching shear around a

column. The critical section for punching shear is described in Section 9.4, Two-way

Slabs (Part II). Grouping of tendons is permitted in band beams.

A minimum amount of non-prestressed reinforcement is provided in each direction

based on temperature and shrinkage requirement. As per IS:456 - 2000, Clause 26.5.2.1, the minimum amount of reinforcement (Ast,min in mm2) for unit width of slab is

given as follows.

Ast,min = 0.15% 1000h for Fe 250 grade of steel

= 0.12% 1000h for Fe 415 grade of steel.

The ducts for placing the individual strands are oval shaped to maintain the eccentricity,

reduce frictional losses and convenient placement of crossing ducts. The ducts are not

commonly grouted as the use of unbonded tendon is not detrimental in buildings.

The following photo shows the ducts for the prestressing tendons and the non-

prestressed reinforcement in a two-way slab.

Oval shaped duct for prestressing tendon

Non-prestressed reinforcementOval shaped duct for prestressing tendon


Figure 9-3.10 Reinforcement in a two-way slab




9.3.3 Features in Modelling and Analysis

The features in modelling and analysing an equivalent frame are discussed.

Gross section versus cracked section For determining the stiffness of a slab-beam member, the gross section can be

considered in place of the cracked section. This is to simplify the calculation of moment

of inertia of the section.

Equivalent column The actual column needs to be replaced by an equivalent column to consider the

flexibility of the transverse beam in the rotation of the slab. The portions of the slab in

the MS rotate more than the portions in the CS because of the torsional deformation of

the transverse beam. In the following figure, the size of the arrows qualitatively

represents the rotation of the slab. Note that the rotation is higher away from the

column.

Upper column

Lower column

c1c2

h

Transverse beam

l2

Upper column

Lower column

c1c2

h

Transverse beam

l2

Figure 9-3.11 Isometric view of a slab-column junction

The transverse beam need not be a visible beam, but a part of the slab in the

transverse direction, bounded by the edges of the column or column capital. In

presence of beam or column capital or in absence of beam, the cross-section of the

modelled transverse beam is taken as shown in the following sketches.



a) In presence of transverse beam

b) In presence of column capital

c) In absence of transverse beam

a) In presence of transverse beam

b) In presence of column capital

c) In absence of transverse beam

Figure 9-3.12 Cross-section of modelled transverse beam

The flexibility of the equivalent column is equal to the sum of the flexibilities of the actual

column and the transverse beam.

(9-3.1) ec c t

= +K ΣK K1 1 1

Here,

Kec = flexural stiffness of the equivalent column

ΣKc = Kc,upper + Kc,lower

Kc,upper = flexural stiffness of the upper column

Kc,lower = flexural stiffness of the lower column

Kt = torsional stiffness of the transverse beam.

An approximate expression for the flexural stiffness of a column (Kc) is given below.

(9-3.2) c cc

E IK =L - h4

2

Here,

Ec = modulus of concrete

L = length of the column

h = thickness of the slab

Ic = moment of inertia of the column.

An approximate expression for torsional stiffness of the transverse beam (Kt) is given

below.

(9-3.3) l

⎛ ⎞⎜ ⎟⎝ ⎠

ct

E CK =cl - 2

22

9

1



Here,

C = equivalent polar moment of inertia of transverse beam

c2 = dimension of column in the transverse direction

l2 = width of equivalent frame.

For a rectangular section, the expression of C is given below.

(9-3.4)

⎛ ⎞⎜ ⎟⎝ ⎠

x x yC = Σy

3

1- 0.633

Here, x and y are the smaller and larger dimensions of the transverse beam. The

expression of C is a lower bound estimate, that is, the calculated value is always lower

than the actual moment of inertia of the transverse beam. For a transverse beam of

compound section, the value of C is the summation of the individual values of the

component rectangles. The splitting of the compound section into component

rectangles should be such, so as to maximise the value of C. For the following two

cases of splitting, select the larger value of C.

Figure 9-3.13 Component rectangles of a compound section

If there is a beam in the column strip in the spanning direction, then Kt is replaced by Kt

(Isb / Is).

Here,

Is = moment of inertia of slab without the projecting portion of the beam

(shaded area in Sketch (a) of the following figure)

Isb = moment of inertia of slab considering it as a T-section

(shaded area in Sketch (b) of the following figure).



(b)(a)

hl2

(b)(a)

hl2

(a)

hl2

Figure 9-3.14 Sections for determining moments of inertia

Slab–beam members The variation of the flexural moment of inertia of a slab-beam member is considered as

follows.

The value of the moment of inertia (I) is constant (say equal to I1) in the prismatic

portion, that is, in between the faces of the columns or column capitals or drop panels.

It is also constant, with a different value (say equal to I2) in the region of a drop panel.

The value varies in the region from the face of the column or column capital to the

center line of the column. But it is approximated to a constant value equal to the

following.

(9-3.5) l

⎛ ⎞⎜ ⎟⎝ ⎠

Ic-

22

2

2

1

Here,

I2 = moment of inertia at the face of the column or column capital

c2 = dimension of column in the transverse direction

l2 = width of equivalent frame.

The following figure shows the variation of the moment of inertia of the slab beam

member.



(a)

(b)I2I1⎛ ⎞⎜ ⎟⎝ ⎠

2Ic

22

21-

l

(a)

(b)I2I1⎛ ⎞⎜ ⎟⎝ ⎠

2Ic

22

21-

l

Figure 9-3.15 a) Elevation of equivalent frame, b) Variation of the moment of inertia of

the slab-beam member

Arrangement of live load Since the factored live load (wu, LL) may not occur uniformly in all the spans in a floor, a

distribution is considered to generate the maximum values of the negative (Mu–) and

positive moments (Mu+) at the critical sections. If the distribution of wu, LL is known, then

the load is applied accordingly. If the distribution is not known, then a pattern loading

is considered based on the value of wu, LL with respect to that of the factored dead load

(wu, DL). Of course, the load case with wu, LL on all the spans should be also analysed.

1) For wu, LL ≤ ¾ wu, DL

The possible variation in wu,LL in the different spans is neglected. wu,LL is applied

uniformly on all the spans.

A B C D E

wu,LL.l2wu,DL.l2

A B C D E

wu,LL.l2wu,DL.l2

Figure 9-3.16 Distribution of live load for wu, LL ≤ ¾ wu, DL

2) For wu LL > ¾ wu,DL

For maximum value of Mu+ in a span, ¾ wu,LL is applied on the span and the alternate

spans. For example, if the maximum value of Mu+ in Span BC of the frame below is to

be determined, then ¾ wu,LL is placed in Spans BC and DE. This distribution will also

give the maximum value of Mu+ in Span DE. For maximum value of Mu

– near the



support, ¾ wu LL is applied on the adjacent spans only. For example, if the maximum

value of Mu– near Support B is to be determined, then ¾ wu,LL is placed in Spans AB

and BC.

A B C D E

¾wu,LL.l2

wu,DL.l2A B C D E

¾wu,LL.l2

wu,DL.l2

Distribution of live load for maximum Mu

+ in Spans BC and DE

A B C D Ewu,DL.l2

¾wu,LL.l2

A B C D Ewu,DL.l2

¾wu,LL.l2

Distribution of live load for maximum Mu

- near Support B

Figure 9-3.17 Distribution of live load for wu LL > ¾ wu,DL

Critical section near a support The critical section is determined as follows.

1) At interior support

At the face of support (column or column capital, if any), but not further than 0.175l1

from the center line of the column.

2) At exterior support

At a distance from the face of column not greater than half the projection of the column

capital (if any).

9.3.4 Distribution of Moments to Strips

In absence of a rigorous analysis (say finite element analysis), the procedure for

reinforced concrete slabs may be used to distribute the moments Mu+ and Mu

– to the

column strip (CS) and middle strip (MS). Of course, ACI 318-02 does not recommend

this procedure to be used for prestressed slabs.



Distribution of Mu– at interior support

Mu,–

CS = 0.75 Mu– (9-3.6)

Mu,–

MS = 0.25 Mu– (9-3.7)

Here,

Mu,–

CS = negative moment in the CS

Mu,–

MS = total negative moment in the two MS at the sides.

Distribution of Mu– at exterior support

If the width of the column or wall support is less than ¾ l2,

Mu,–

CS = Mu– (9-3.8)

Mu,–

MS = 0. (9-3.9) If the width of the column or wall support is greater than ¾ l2, then Mu

– is uniformly

distributed along the width l2.

Distribution of Mu+ at mid span

Mu,+

CS = 0.60 Mu+ (9-3.10)

Mu,+

MS = 0.40 Mu+ (9-3.11)

Here,

Mu,+

CS = positive moment in the CS

Mu,+

MS = total positive moment in the two MS at the sides.

The total moments in MS (Mu,–

MS and Mu,+

MS) are distributed to the two middle strips at

the sides of the equivalent frame, proportional to their widths. The combined MS from

two adjacent equivalent frames is designed for the sum of the moments assigned to its

parts.



9.4 Two-way Slabs (Part II) This section covers the following topics.

• Checking for Shear Capacity

• Spandrel Beams

• Anchorage Devices

• Additional Aspects

9.4.1 Checking for Shear Capacity

The checking of shear capacity of flat plates and flat slabs is of utmost importance. In

absence of beams, the shear is resisted by the slab near the slab-column junction.

The shear capacity of a slab should be adequate to resist the shear from two actions.

1) One-way (beam) shear

2) Two-way (punching) shear.

One-way shear The one-way shear is analogous to that generates in a beam due to flexure. This is

checked in a two-way slab for each spanning direction separately.

The critical section for checking the shear capacity is at a distance effective depth ‘d’

from the face of the column, across the entire width of the frame. The critical section is

transverse to the spanning direction. For gravity loads, the shear demand in the critical

section generates from the loads in the tributary area shown in the next figure. For

lateral loads, the shear demand is calculated from the analysis of the equivalent frame.

In presence of a drop panel two critical sections need to be checked. The first section is

at a distance d1 from the face of the column, where d1 is the effective depth of the drop

panel. The second section is at a distance d2 from the face of the drop panel, where d2

is the effective depth of the slab.



d

CL

l2

Critical section

ln

Plan

Elevation

d

CL

l2

Critical section

ln

d

CLCL

l2

Critical section

ln

Plan

Elevation

(a) Slab without drop panel

d2d1

CL

l2

Critical sections

ln

Plan

Elevation d2

d1 d2d1 d2d1

CLCL

l2

Critical sections

ln

l2

Critical sections

ln

Plan

Elevation

(b) Slab with drop panel

Figure 9-4.1 Critical sections for one-way shear

The calculations for shear can be for unit width of the slab. The shear demand due to

gravity loads per unit width is given as follows.

Vu = wu (0.5ln – d) (9-4.1)

Here, ln is the clear span along the spanning direction.

The shear capacity per unit width is given as follows.

VuR = Vc (9-4.2)



Vc is the shear capacity of uncracked concrete of unit width of slab. The expression of

Vc is given in Section 5.2, Design for Shear (Part I).

For adequate shear capacity

VuR ≥ Vu (9-4.3)

If this is not satisfied, it is preferred to increase the depth of the slab to avoid shear

reinforcement along the width of the slab.

Two-way shear The two-way shear is specific to two-way slabs. If the capacity is inadequate, the slab

may fail due to punching around a column. The punching occurs along a conical

frustum, whose base is geometrically similar and concentric to the column cross-

section. The following figure illustrates the punching shear failure.

Elevation Isometric viewElevation Isometric view Figure 9-4.2 Punching shear failure

Two-way shear is checked for the two spanning directions simultaneously. The critical

section for checking the shear capacity is geometrically similar to the column perimeter

and is at a distance d / 2 from the face of the column. The depth of the critical section is

equal to the average of the effective depths of the slab in the two directions. The

sketches below show the critical section. The tributary area of the column is the area

within the centre-lines of the spans minus the area within the critical section. It is shown

shaded in the third sketch.



b2

b1

c1

c2

Critical section

b2

b1

c1

c2b2

b1

c1

c2

Critical section

(a) Plana

b1

b2

d

b1

b2

d

b1

b2

d

r

(b) Isometric view

CLCL

CLCL

Critical section

CLCL

CLCL

CLCLCLCL

CLCLCLCL

Critical section

(c) Tributary area for shear due to gravity loads

Figure 9-4.3 Critical section for two-way shear

The lengths of the sides of the critical section along axes 1-1 and 2-2 are denoted as b1

and b2, respectively.

b1 = c1 + d (9-4.4) b2 = c2 + d (9-4.5)

Here,

c1 = dimension of the column or column capital along axis 1-1

c2 = dimension of the column or column capital along axis 2-2.

For a non-rectangular column, the critical section consists of the slab edges as per

Figure 13, IS:456 - 2000. For edge and corner columns, the critical section consists of

the slab edges as per Figure 14, IS:456 - 2000.



The demand in terms of shear stress is given as follows.

(9-4.6)

τ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣

2uv uv

uv

b bM MV= + +b d J J

12-2 1-1

0 2-2 1-1

2 2 ⎦

Here,

Vu = shear due to gravity loads from the tributary area

Muv = fraction of moment transferred about an axis

b0 = perimeter of the critical section = 2(b1 + b2).

J = polar moment of inertia of the critical section about an axis

The second and third terms are due to transfer of moments from slab to column. The

moment about an axis is due to the unbalanced gravity loads for the two sides of the

column or due to lateral loads. It is transferred partly by the variation of shear stress in

the critical section and the rest by flexure. The fraction transferred by the variation of

shear stress about an axis is denoted as Muv.

Muv|2-2 = Fraction of moment transferred about axis 2-2

Muv|1-1 = Fraction of moment transferred about axis 1-1

The forces and stresses acting at the critical section are shown below.

Vu Muv|2-2 Muv|1-1

2

2

1 1

Vu Muv|2-2 Muv|1-1

2

2

1 1

(a) Shear and moments acting at the critical section

2

2

1 1

Due to Vu Due to Muv|2-2 Due to Muv|1-1

2

2

1 1

Due to Vu Due to Muv|2-2 Due to Muv|1-1 (b) Stresses acting at the critical section



Due to Vu + Muv|2-2 + Muv|1-1Due to Vu + Muv|2-2 + Muv|1-1 (c) Resultant shear stress diagram at the critical section

Figure 9-4.4 Forces and stresses at the critical section

The fraction of moment transferred by the variation of shear stress about an axis (Muv),

is given in terms of the total moment transferred (Mu) as follows.

Muv = (1- α)Mu (9-4.7)

The value of Mu due to unbalanced gravity loads is calculated by placing live load on

one side of the column only. The value of Mu due to lateral loads is available from the

analysis of the equivalent frame. The parameter α is based on the aspect ratio of the

critical section.

α =bb

1

2

121+3

(9-4.8)

The polar moments of inertia of the critical section, about the axes are given as follows.

(9-4.9)

⎡ ⎤⎛ ⎞⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

23 3 2

2 2 11-11 1= 2 + +

12 12 2bJ b d db b d

(9-4.10)

⎡ ⎤⎛ ⎞⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

23 3 1

1 1 22-21 1= 2 + +

12 12 2bJ b d db b d

For adequate shear capacity

(9-4.11)

τ τ≤v sk c

The shear stress capacity of concrete for a square column is given as follows.

(9-4.12)

τc c= f0.25 k



Here, fck is the characteristic strength of the concrete in the slab. The effect of prestress

is neglected. The factor ks accounts for the reduced shear capacity of non-square

columns.

ks = 0.5 + βc (9-4.13)

The value of ks should be less than 1.0. βc is a parameter based on the aspect ratio of

the column cross-section. It is the ratio of the short side to long side of the column or

column capital.

If τv exceeds ksτc, a drop panel or shear reinforcement needs to be provided at the slab-

to-column junction. The shear reinforcement can be in the form of stirrups or I section

(shear head) or based on shear studs. The reinforcement based on shear studs

reduces congestion for conduits and post-tensioning tendons. If τv exceeds 1.5τc, then

the depth of the slab needs to be increased in the form of drop panels.

The stirrups are designed based on the following equation. Asv is the area of the

vertical legs of stirrups.

(9-4.14)

( )τ τv csv

y

-A =f

0.50.87

The stirrups are provided along the perimeter of the critical section. The first row of

stirrups should be within a distance of 0.5d from the face of the column. They can be

continued in outer rows (concentric and geometrically similar to the critical section) at an

interval of 0.75d, till the section with shear stress τv = 0.5τ .

The different types of reinforcement at the slab-to-column junction are shown in the

following sketches.

References:

1. Bureau of Indian Standards,

Handbook on Concrete Reinforcement and Detailing

(SP 34 : 1987)



2. Khan, S. and Williams, M.

Post-tensioned Concrete Floors

Butterworth-Heinemann Ltd.

Plan at mid-depth of the slab

A A

Elevation of Section A-A

0.75d < 0. 5d


A A


A AA


0.75d < 0. 5d


0.75d < 0. 5d

(a) Stirrups in geometrically similar rows

AA



AA


AAAA


Elevation of Section A-AElevation of Section A-A

(b) Beam cage stirrups



0.5d

~ 2d Elevation of Section A-A

AA


0.5d

~ 2d

0.5d

~ 2d Elevation of Section A-A

AA


AAAA

Plan at mid-depth of the slab (c) Beam cage stirrups with bent-up bars

(d) Shear head reinforcement

(e) Shear stud reinforcement

Figure 9-4.5 Shear reinforcement at slab-column junction



The following photo shows the ducts and reinforcement at the slab-column junction in a

slab with a drop panel.

Figure 9-3.6 Reinforcement at slab-column junction


The residual moment transferred by flexure (Muf), is given in terms of the total moment

transferred (Mu) as follows.

Muf = α Mu (9-4.15)

Additional non-prestressed reinforcement is provided at the top of the slab over a width

c2 + 3h (centred with respect to the column) to transfer Muf.

9.4.2 Spandrel Beams

The flat plates are provided with spandrel beams at the edges. These beams stiffen the

edges against rotation. In turn the beams are subjected to torsion.

The maximum torsion is calculated by assuming a uniform torsional loading along the

width of the equivalent frame (ACI 318-02 recommends a triangular distribution). The

spandrel beams are provided with closed stirrups to resist the torsion. The design for

torsion is given in the Module of “Analysis and Design for Shear and Torsion”.



The following figure shows the distribution of the torsional loading on the spandrel

beam.

Tu,max

l2

Variation of torsion

Tu,max

l2

Variation of torsion

Figure 9-4.7 Torsion in spandrel beams

The maximum torsion (Tu,max) is given as follows. Here, Mu,–e is the moment at the

exterior support of the equivalent frame.

τ-u,e

u,maxMl - c=l

2 2

22 (9-4.16)

9.4.3 Anchorage Devices

In post-tensioned slabs, the anchorage devices transfer the prestress to the concrete.

The device at the stretching end consists of an anchor block and wedges. At the dead

end, the wires are looped to provide the anchorage. Bursting links are provided in the

end zone to resist transverse tensile stresses in concrete. The following photos show

some anchorage devices.

(a) Anchorage device at stretching end



(b) Anchorage device for dead end with bulb

(c) Anchorage device for dead end with plate

Figure 9-4.8 Anchorage devices for slabs




The following photos show the anchorage devices, end zone reinforcement, spandrel

beam before casting and stretching and anchoring of the tendons after casting of

concrete in a slab.

Bursting links

Spandrel Beam

(a) End-zone reinforcement at stretching end

Dead end anchorageBursting links

(b) End-zone reinforcement and anchorage at dead end



(c) Stretching of tendons

(d) Anchorage block at stretching end

Figure 9-4.9 End-zone reinforcement and anchoring of tendons in a slab

(Reference: VSL India Pvt. Ltd.)



9.4.4 Additional Aspects

Restraint from vertical elements Due to the restraint from monolithic columns or walls, the prestressing force in the slab

is reduced. Hence, the stiff columns or walls should be located in such a manner that

they offer least restraint. Alternatively, sliding joints can be introduced which are made

ineffective after post-tensioning of the slab.

Calculation of deflection The deflection of a two-way slab can be approximately calculated by the equivalent

frame method. The deflection at a point is the summation of the deflections of the two

orthogonal strips passing through the point.

For an accurate evaluation, the following models can be adopted.

a) Grillage model

b) Finite element model.

Proportioning of drop panels and column capitals Section 31 of IS:456 - 2000 provide guidelines for proportioning drop panels and

column capitals. A minimum length and a minimum depth (beyond the depth of the

slab) of a drop panel are specified. For column capitals it is preferred to have a conical

flaring at a subtended angle of 90°. The critical sections are shown in Figure 12 of the

code.



9.5 Compression Members This section covers the following topics.

• Introduction

• Analysis

• Development of Interaction Diagram


9.5.1 Introduction

Prestressing is meaningful when the concrete in a member is in tension due to the

external loads. Hence, for a member subjected to compression with minor bending,

prestressing is not necessary. But, when a member is subjected to compression with

substantial moment under high lateral loads, prestressing is applied to counteract the

tensile stresses. Examples of such members are piles, towers and exterior columns of

framed structures.

As the seismic forces are reversible in nature, the prestressing of piles or columns is

concentric with the cross-section. Some typical cross sections are shown below.

Partially prestressed column

Prestressed circular and hexagonal piles

Partially prestressed column

Prestressed circular and hexagonal piles

Figure 9-5.1 Examples of prestressed members subjected to compression



Figure 9-5.2 Stacked prestressed piles

(Reference: Industrial Concrete Products Berhad)

Precast prestressed piles have the following advantages.

1) Large bending and axial tension capacities.

2) Better quality control than bored and CIP piles.


⇒ Increased durability.

4) Use of high strength concrete gives reduced section.

The piles are prestressed by pre-tensioning. The procedure of pre-tensioning is

explained under “Pretensioning Systems and Devices”.

Since a prestressed member is under self equilibrium, there is no buckling of the

member due to internal prestressing with bonded tendons. In a deflected shape, there is

no internal moment due to prestressing.



The justification is explained in the next figure.

C T

b) Under internal prestressing

PP

P

P

a) Under external compression

∆

P∆ C T


C T


PP

P

P


∆

P∆

PP

P

P


∆

P∆

Figure 9-5.3 Internal forces at deflected configuration

In the first free body sketch of the above figure, the external compression P causes an

additional moment due to the deflection of the member. The value of the moment at

mid-height is P∆. This is known as the member stability effect, which is one type of P-∆

effect. If this deflection is not stable, then buckling of the member occurs. In the second

free body sketch, there is no moment due to the deflection of the member and the

prestressing force, since the compression in concrete (C) and the tension in the tendons

(T) balance each other.

When the additional moment due to deflection of the member is negligible, the member

is termed as short member. The additional moment needs to be considered when the

slenderness ratio (ratio of effective length and a lateral dimension) of the member is

high. The member is termed as slender member. In the analysis of a slender member,

the additional moment is calculated by an approximate expression or second order

analysis. In this module only short members will be considered.

9.5.2 Analysis Analysis at Transfer The stress in the section can be calculated as follows.

cPf =A

0 (9-5.1)



Here,

A = Area of concrete


In this equation, it is assumed that the prestressing force is concentric with the cross-

section. For members under compression, a compressive stress is considered to be

positive. The permissible prestress and the cross-section area are determined based on

the stress to be within the allowable stress at transfer (fcc,all).

Analysis at Service Loads The analysis is analogous to members under flexure. The stresses in the extreme fibres

can be calculated as follows.

e

ct t

P N Mcf = + ±A A I

(9-5.2)

In this equation, the external compression for a prestressed member is denoted as N

and is concentric with the cross section. The eccentricity is considered in the external

moment M.


A = area of concrete

At = area of the transformed section

c = distance of the extreme fibre from the centroid (CGC)

It = moment of inertia of the transformed section


The value of fc should be within the allowable stress under service conditions (fcc,all).

Analysis at Ultimate When the average prestress in a member under axial compression and moment is less

than 2.5 N/mm2, Clause 22.2, IS:1343 - 1980, recommends to analyse the member as

a reinforced concrete member, neglecting the effect of prestress. For higher prestress,

the analysis of strength is done by the interaction diagrams.



At the ultimate limit state, an interaction diagram relates the axial force capacity (NuR)

and the moment capacity (MuR). It represents a failure envelop. Any combination of

factored external loads Nu and Mu that fall within the interaction diagram is safe. A

typical interaction diagram is shown below. The area shaded inside gives combinations

of Mu and Nu that are safe.

MuR

NuR

Tension failure

Compression failure

Balanced failureNe1

MuR

NuR

Tension failure

Compression failure

Balanced failureNe1

Figure 9-5.4 A typical interaction diagram for compression and bending

The radial line in the previous sketch represents the load path. Usually the external

loads increase proportionally. At any load stage, M and N are related as follows.

M = N eN (9-5.3)

Here, eN represents the eccentricity of N which generates the same moment M. The

slope of the radial line represents the inverse of the eccentricity (1/eN). At ultimate, the

values of M and N (Mu and Nu, respectively) correspond to the values on the interaction

diagram. For high values of N as compared to M, that is when eN is small, the concrete

in the compression fibre will crush before the steel on the other side yields in tension.

This is called the compression failure.

For high values of M as compared to N, that is when eN is large, the concrete will crush

after the steel yields in tension. This is called the tension failure.

The transition of these two cases is referred to as the balanced failure, when the

crushing of concrete and yielding of steel occur simultaneously. For a prestressed

compression member, since the prestressing steel does not have a definite yield point,

there is no explicit balanced failure.



9.5.3 Development of Interaction Diagram

An interaction diagram can be developed from the first principles using the non-linear

stress-strain curves of concrete under compression and steel under tension. Several

sets of NuR and MuR for given values of eN or xu are calculated. The distance of neutral

axis from the extreme compressive face is denoted as xu. Partial safety factors for

concrete and prestressing steel can be introduced when the interaction diagram is used

for design. Here, the procedure is illustrated for a rectangular section with prestressed

tendons placed at two opposite faces symmetrically, and without non-prestressed

reinforcement.

d2d1Ap1Ap2

B

D

CGC+

d2d1Ap1Ap2

B

D

CGC+

Figure 9-5.5 A rectangular prestressed section

The notations used are as follows.

B = dimension of section transverse to bending

D = dimension of section in the direction of bending

Ap1 = area of prestressing tendons at the tension face

Ap2 = area of prestressing tendons at the compression face

d1, d2 = distances of centres of Ap1 and Ap2, respectively, from the centroid of the

section (CGC).

The strain compatibility equation is necessary to relate the strain in a prestressing

tendon with that of the adjacent concrete. Due to a concentric prestress, the concrete at

a section undergoes a uniform compressive strain. With time, the strain increases due

to the effects of creep and shrinkage. At service, after the long term losses, let the strain

be εce. Also, let the strain in the prestressing steel due to effective prestress be εpe.



εce

εpe εpe

εce

εpe εpe Figure 9-5.6 Strain profile due to effective prestress only

The strain compatibility equation for the prestressed tendons is given below.

(9-5.4) p c p

p pe ce

ε = ε + ∆ε ∆ε = ε - εwhere,

The strain difference of the strain in a prestressing tendon with that of the adjacent

concrete is denoted as ∆εp. The design stress-strain curve for concrete under

compression is used. This curve is described in Section 1.6, Concrete (Part II). The

design stress-strain curve for the prestressed tendon under tension is expressed as fp =

F(εp).

The calculation of NuR and MuR for typical cases of eN or xu are illustrated. The typical

cases are as follows.

1) Pure compression (eN = 0, xu = ∞)

2) Full section under varying compression (0.05D < eN ≤ eN │xu = D , xu ≥ D)

3) Part of section under tension (eN │xu = D < eN ≤ ∞, xu < D)

4) Pure bending (eN = ∞, xu = xu,min)

The above cases are illustrated in the following sketches.

e = 0e

xu

Case 1 Case 2

Elevation

Strain diagram

e = 0e

xu

Case 1 Case 2

Elevation

Strain diagram



e

xu xu

Elevation

Strain diagram

Case 3 Case 4

e

xu xu

Elevation

Strain diagram

Case 3 Case 4 Figure 9-5.7 Typical cases of eccentricity and depth of neutral axis

In addition to the above cases, the case of pure axial tension is also calculated. The

straight line between the points of pure bending and pure axial tension provides the

interaction between the tensile force capacity and the moment capacity.

NuR

MuR

NuR

MuR

Figure 9-5.8 A typical interaction diagram for tension and bending

1. Pure compression (eN = 0, xu = ∞).

The following sketches represent the strain and stress profiles across the section and

the force diagram.

– 0.002

εp1 εp2

Strain profile

fp1 fp2

0.447 fck Stress profile

Tu1 Tu2CuForce diagram

– 0.002

εp1 εp2

Strain profile

fp1 fp2

0.447 fck Stress profile

Tu1 Tu2CuForce diagram

Figure 9-5.9 Sketches for analysis at pure compression

The forces are as follows.



Cu = 0.447fck (Ag – Ap) (9-5.5) Tu1 = Tu2 = Ap1 fp1 (9-5.6)

= Ap1 Ep (– 0.002 + ∆εp)

The steel is in the elastic range. The total area of prestressing steel is Ap = Ap1 + Ap2.

The area of the gross-section Ag = BD. The moment and axial force capacities are as

follows.

MuR = 0 (9-5.7)

NuR = Cu – Tu1 – Tu2

= 0.447fck (Ag – Ap) – Ap Ep (εpe – 0.002 – εce) (9-5.8)

In design, for simplification the interaction diagram is not used for eccentricities eN ≤

0.05D. To approximate the effect of the corresponding moment, the axial force capacity

is reduced by 10%.

∴ NuR = 0.4fck (Ag – Ap) – 0.9Ap Ep (εpe – 0.002 + εce) (9-5.9)

2. Full section under varying compression (0.05D < eN ≤ eN │xu = D , xu ≥ D)


the force diagram.

– 0.002

εp1 εp2

fp1 fp2

0.447 fck

Tu1 Tu2Cu

xu = kD

g Asector

3 / 7 D Strain profile

Stress profile

Force diagram

– 0.002

εp1 εp2

fp1 fp2

0.447 fck

Tu1 Tu2Cu

xu = kD

g Asector

3 / 7 D– 0.002

εp1 εp2

fp1 fp2

0.447 fck

Tu1 Tu2Cu

xu = kD

g Asector

3 / 7 D Strain profile

Stress profile

Force diagram

Figure 9-5.10 Sketches for analysis for section under varying compression

The limiting case for full section under compression corresponds to xu = D, when the

neutral axis lies at the left edge of the section. The strain diagram pivots about a value

of – 0.002 at 3/7D from the extreme compression face. To calculate Cu, first the



reduction of the stress at the edge with lower compression (g) is evaluated. Based on

the second order parabolic curve for concrete under compression, the expression of ‘g’

is as follows.

(9-5.10)

⎛ ⎞⎡ ⎤⎜ ⎟⎢ ⎥⎣ ⎦⎜ ⎟

⎡ ⎤⎜ ⎟⎜ ⎟⎢ ⎥⎣ ⎦⎝ ⎠

⎛ ⎞⎜ ⎟⎝ ⎠

2

2

47= 0.447

3-7

4= 0.4477 - 3

ck

ck

Dg f

kD D

fk

The area of the complementary sector of the stress block is given as follows.

⎛ ⎞⎜ ⎟⎝ ⎠sectorA = g D

= gD

1 43 7421

(9-5.11)

Asector

4 / 7 D

+gx /

Asector

4 / 7 D

+gx /

Figure 9-5.11 Complementary area of the stress block

Distance of centroid from apex (x /) = (3/4)(4/7)D = 3/7 D The forces are as follows.

[ ]⎡ ⎤⎢ ⎥⎣ ⎦

⎡ ⎤⎛ ⎞⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

u ck sector

ck

ck

C = f D - A B

= f D - gD B

= f BD -k -

2

0.44740.44721

4 40.447 121 7 3

(9-5.12)

(9-5.13)

( )⎛ ⎞⎛ ⎞

⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

u p p

p p p

p p c p

u

p p p

u

T = A f= A E ε

= A E ε + ∆ε

Dx - + d= A E - + ∆εDx -

1 1 1

1 1

1 1

1

120.002 3

7



(9-5.14)

( )⎛ ⎞⎛ ⎞

⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

u p p

p p p

p p c p

u

p p p

u

T = A f= A E ε

= A E ε + ∆ε

Dx - - d= A E + ∆εDx -

2 2 2

2 2

2 2

2

22-0.002 3

7

The strains in the concrete at the level of the prestressing steels εc1 and εc2 are

determined from the similarity of triangles of the following strain profile.

εc1

xu – (3 / 7 D)

xu – (D/2 + d1)– 0.002

εc1

xu – (3 / 7 D)

xu – (D/2 + d1)– 0.002

εc2

xu – (3 / 7 D)

xu – (D / 2 – d2)

– 0.002 εc2

xu – (3 / 7 D)

xu – (D / 2 – d2)

– 0.002

Figure 9-5.12 Strain profile across section

The moment and axial force capacities are as follows.

NuR = Cu – Tu1 – Tu2 (9-5.15) MuR = Mc + Mp (9-5.16)

The expressions of Mc and Mp about the centroid are given below. Anticlockwise

moments are considered positive. The lever arms of the tensile forces are shown in the

following sketch.

⎡ ⎤⎢ ⎥⎣ ⎦

/c ck sector

DM = f DB× + A B x + D -

= gD B2

30.447 07 2

10147

(9-5.17)

Mp = Tu1d1 – Tu2d2 (9-5.18)



Tu1 Tu2Cu

d2d1

+Tu1 Tu2Cu

d2d1

Tu1 Tu2Cu

d2d1

+

Figure 9-5.13 Force diagram across the section

3. Part of section under tension (eN │xu = D < , eN ≤ ∞, xu < D)


the force diagram.

– 0.0035

εp1εp2

Tu1Tu2Cu

fp1 fp2

xu

0.447fck

Strain profile

Stress profile

Force diagram

– 0.0035

εp1εp2

Tu1Tu2Cu

fp1 fp2

xu

0.447fck

– 0.0035

εp1εp2

Tu1Tu2Cu

fp1 fp2

xu

0.447fck

Strain profile

Stress profile

Force diagram

Figure 9-5.14 Sketches for analysis for part of section under tension

The forces are as follows. The compression is the resultant of the stress block whose

expression can be derived similar to a reinforced concrete section.

Cu = 0.36fck xu B

Tu1 = Ap1 fp1

= Ap1 F (εp1)

= Ap1 F (εc1+ ∆εp)

Tu2 = Ap2 fp2

= Ap2 Ep εp2

= Ap2 Ep (εc2+ ∆εp)

The strains εc1 and εc2 are calculated from the similarity of triangles of the following

strain diagram.

c

uu

ε =D x+ d - x1

1

0.0035

2(9-5.19)



⎛ ⎞⎜ ⎟⎝ ⎠

c

uu

ε = -D xx - - d

2

2

0.0035

2

(9-5.20)

– 0.0035

xu – (D/2 – d2)(D/2 + d1) – xu

εc2εc1

xu

– 0.0035

xu – (D/2 – d2)(D/2 + d1) – xu

εc2εc1

xu

Figure 9-5.15 Strain profile across section


NuR = Cu – Tu1 – Tu2 (9-5.21) MuR = Mc + Mp (9-5.22)

The expressions of Mc and Mp about the centroid are as follows.

Mc = 0.36fck xu B [ (D / 2) – 0.42 xu ] (9-5.23) Mp = Tu1d1 – Tu2d2 (9-5.24)

The lever arms of the forces are shown in the following sketch. The location of Cu is

similar to that of a reinforced concrete section.

Tu1 Tu2Cu

d2d1

(D / 2) – 0.42 xu

+Tu1 Tu2Cu

d2d1

(D / 2) – 0.42 xu

+

Figure 9-5.16 Force diagram across the section

4. Pure bending (eN = ∞, xu = xu,min)

The value of xu is determined by trial and error from the condition that the sum of the

forces is zero.

Cu – Tu1 – Tu2 = 0

or, 0.36fck xu B – Ap1 fp1 – Ap2 fp2 = 0 (9-5.25)

p p p p pu

ck

A f + A E ε x =

f B1 1 2 2or,

0.36(9-5.26)



The strains εp1 and εp2 are calculated from the strain compatibility equations. The strain

εp2 is within the elastic range, whereas εp1 may be outside the elastic range. The

stresses fp1 and fp2 are calculated accordingly from the stress versus strain relationship

of prestressing steel.

The steps for solving xu are as follows.

1) Assume xu = 0.15 D (say).

2) Determine εp1 and εp2 from strain compatibility.

3) Determine fp1 and fp2 from stress versus strain relationship.

4) Calculate xu from Eqn. (9-5.26).

5) Compare xu with the assumed value. Iterate till convergence.


NuR = 0 (9-5.27) MuR = Mc + Mp (9-5.28)

The expressions of Mc and Mp are same as the previous case.

5. Axial tension

The moment and axial force capacities are as follows. The cracked concrete is

neglected in calculating the axial force capacity.

NuR = – 0.87fpk Ap (9-5.29) MuR = 0 (9-5.30)

The above sets of NuR and MuR are joined to get the interaction diagram.

Example 9-5.1 Calculate the design interaction diagram for the member given below. The member is prestressed using 8 strands of 10 mm diameter. The strands are stress relieved with the following properties.

Tensile strength (fpk) = 1715 N/mm2. Total area of strands = 8 × 51.6



= 413.0 mm2

Effective prestress (fpe) = 1034 N/mm2

Modulus (Ep) = 200 kN/mm2

Strain under fpe (εpe) = 0.0042.

Grade of concrete = M40 Strain under fpe (εce) = – 0.0005.

200 5050

300

300

200 5050

300

300

Solution

Calculation of geometric properties and strain compatibility relationship.

Ag = 300 × 300 = 90,000 mm2

Ap1 = Ap2 = 4 × 51.6 = 206 mm2

d1 = d2 = 100 mm

∆εp = 0.0042 + 0.0005 = 0.0047

∴εp = εc + 0.0047

1. Pure compression (eN = 0, xu = ∞)

MuR = 0 kNm

Cu = 0.447fck (Ag – Ap)

= 0.447 × 40 (90,000 – 413)

= 1601.8 kN

Tu1 = Tu2 = Ap1 Ep (– 0.002 + ∆εp)

= 206.4 × 200 × (0.0047 – 0.002)

= 111.5 kN




= 1601.8 – 2 × 111.5

= 1378.8 kN

With 10% reduction, to bypass the use of interaction diagram for eccentricities

eN ≤ 0.05D

NuR = 1204.9 kN

2. Full section under compression (0.05D < eN ≤ eN │xu = D , xu ≥ D)

400400

Select xu = 400 mm

= (4 / 3) × 300 mm

∴k = 4 / 3

⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟⎝ ⎠

2

2

2

4= 0.447×7 -3

4= 0.447×40 7×(4/3) -3

= 7.13 N/mm

ckg fk

⎛ ⎞⎡ ⎤⎜ ⎟⎢ ⎥⎜ ⎟⎣ ⎦⎝ ⎠

⎛ ⎞⎡ ⎤⎜ ⎟⎢ ⎥⎜ ⎟⎣ ⎦⎝ ⎠

2

22

4 4= 0.447 1-21 7 -3

4 4= 0.447×40×300 1-21 7×(4/3) -3

=1486.9 kN

u ckC f BDk

– 0.002

271.4

εc1

(3/7) 300 = 128.6

150– 0.002

271.4

εc1

(3/7) 300 = 128.6

150 ( )

⎛ ⎞⎜ ⎟⎝ ⎠

1 1 1

150= 206.4×200 -0.002 +0.0047271.4

=148.4 kN

u p p c pT = A E ε + ∆ε



– 0.002

271.4

εc2

350

– 0.002

271.4

εc2

350 ( )

⎛ ⎞⎜ ⎟⎝ ⎠

2 2 2

350= 206.4×200 -0.002 +0.0047271.4

= 87.5 kN

u p p c pT = A E ε + ∆ε


= 1486.9 – 148.4 – 87.5

= 1251.0 kN

Limit NuR to 1240.9 kN to bypass the use of interaction diagram for eccentricities eN ≤

0.05D.

2

2

10=14710= ×7.13×300 ×300

147=13.1 kNm

cM gD B

p u uM = T d - T d1 1 2 2

=148.4×100 -87.5×100= 6.1 kNm

MuR = Mc + Mp

= 13.1 + 6.1

= 19.2 kNm 300300

Select xu = 300 mm

∴k = 1

By similar calculations,

g = 17.9 N/mm2 NuR = 1060.6 kN

Cu = 1304.1 kN Mc = 32.9 kNm

Tu1 = 169.9 kN Mp = 9.6 kNm



Tu2 = 73.6 kN MuR = 42.5 kNm.

3. Part of section under tension (eN │xu = D < eN ≤ ∞, xu < D)

Select xu = 200 mm.

Cu = 0.36fck xu B

= 0.36 × 40 × 200 × 300

= 864.0 kN

1

0.0035= 5200

= 0.0009

cε 0

1 = 0.0009+0.0047= 0.0056

pε

– 0.003550

εc1 200

– 0.003550

εc1 200

Strain corresponding to elastic limit

εpy = 0.87 × 0.8fck / Ep = 0.87 × 1715 / 200 × 103

= 0.0059.

εp1 < εpy

∴fp1 = Ep εp1

= 200 × 103 × 0.0055

= 1115 N/mm2

Tu1 = Ap1 fp1

= 206.4 × 1115

= 230.1 kN

2

0.0035= - 150200

= -0.0026

cε

2 = - 0.0026+0.0047

= 0.0021pε



– 0.0035

150

εc2

200

– 0.0035

150

εc2

200

fp2 = Ep εp2

= 200 × 103 × 0.0021

= 416 N/mm2

Tu2 = Ap2 fp2 NuR = Cu – Tu1 – Tu2

= 206.4 × 416 = 864 – 230.1 – 85.9

= 85.9 kN = 548.0 kN

Mc = 0.36fck xu B [ (D / 2) – 0.42xu ]

= 864 (150 – 0.42 × 200)

= 57.0 kNm

Mp = Tu1d1 – Tu2d2 MuR = Mc + Mp

= 230.1 × 100 – 85.9 × 100 = 57.0 + 14.4

= 14.4 kNm = 71.4 kNm

4. Pure bending (eN = ∞, xu = xu,min)

NuR = 0.0 kN

Try xu = 100 mm.

Cu = 0.36fck xu B

= 0.36 × 40 × 100 × 300

= 432.0 kN

10.0035= 1

100= 0.0052

cε 50

1 = 0.0052+0.0047= 0.0099

pε



– 0.0035150

εc1

100

– 0.0035150

εc1

100

From stress-strain curve

fp1 = 0.87fpk

= 1492 N/mm2

Tu1 = Ap1 fp1

= 206.4 × 1492

= 308.0 kN

– 0.0035

50

εc2

100

– 0.0035

50

εc2

100

2

0.0035= - 50100

= -0.0017

cε

2 = - 0.0017+0.0047

= 0.0029pε

fp2 = Ep εp2

= 200 × 103 × 0.0029

= 580 N/mm2

Tu2 = Ap2 fp2

= 206.4 × 580

= 120.0 kN

Tu1 + Tu2 = 428.0 kN

This is close enough to Cu = 432.0 kN. Hence, the trial value of xu is satisfactory.



Mc = 0.36fck xu B [ (D / 2) – 0.42xu ]

= 0.36 × 40 × 100 × 300 (150 – 0.42 × 100)

= 46.6 kNm

Mp = Tu1d1 – Tu2d2 = 308.0 × 100 – 120.0 × 100

= 18.8 kNm

MuR = 46.6 + 18.8

= 65.4 kNm

5. Axial tension

MuR = 0.0 kNm

NuR = – 0.87fpk Ap

= – 0.87 × 1715 × 413.0

= – 616.2 kN

The above sets of NuR and MuR are joined to get the following interaction diagram. The

limit on axial force capacity to consider the effect of eccentricity less than 0.05D, is not

shown.

-1000

-500

0

500

1000

1500

0 20 40 60 80

M uR (kNm)

NuR

(kN

)




Along with the interaction curve for the prestressed concrete (PC) section, the

interaction curves for two reinforced concrete (RC) sections are plotted. The section

denoted as RC 1 has the same moment capacity at zero axial force. The section

denoted as RC 2 has the same axial force capacity at zero moment. The gross section

of RC 1 is same as that of PC, but the section of RC 2 is smaller.

-1000

-500

0

5001000

1500

2000

2500

0 50 100 150

M uR (kNm)

NuR

(kN

) PCRC 1RC 2

Figure 9-5.17 Interaction diagrams for reinforced and prestressed sections

-1000

-750

-500

-250

00 20 40 60 80

M uR (kNm)

NuR

(kN

)

PC Strength RC 1 StrengthPC Cracking RC 1 Cracking

Figure 9-5.18 Interaction of moment and tension for cracking and strength

Comparing the curves for PC and RC 2, it is observed that if the moment demand is

small, then a smaller reinforced concrete section is adequate to carry the axial force. Of

course with increasing moment, the flexural capacity of the prestressed concrete

section is higher. Comparing the curves for PC and RC 1, it is inferred that for two



sections with same flexural capacities, the axial load capacity of a prestressed concrete

section is less. However if there is tension, the cracking load combination is higher for

PC as compared to RC 1.

Thus, prestressing is beneficial for strength when there is occurrence of:

a) Large moment in addition to compression

b) Moment along with tension.

Such situations arise in piles or columns subjected to seismic forces. In presence of

tension, prestressing is beneficial at service loads due to reduced cracking. Non-

prestressed reinforcement may be used for supplemental capacity.



9.6 Circular Prestressing This section covers the following topics.

• Introduction

• General Analysis and Design

• Prestressed Concrete Pipes

• Liquid Storage Tanks

• Ring Beams

• Conclusion

9.6.1 Introduction

When the prestressed members are curved, in the direction of prestressing, the

prestressing is called circular prestressing. For example, circumferential prestressing in

pipes, tanks, silos, containment structures and similar structures is a type of circular

prestressing. In these structures, there can be prestressing in the longitudinal direction

(parallel to axis) as well. Circular prestressing is also applied in domes and shells.

The circumferential prestressing resists the hoop tension generated due to the internal

pressure. The prestressing is done by wires or tendons placed spirally, or over sectors

of the circumference of the member. The wires or tendons lay outside the concrete

core. Hence, the centre of the prestressing steel (CGS) is outside the core concrete

section.

The hoop compression generated is considered to be uniform across the thickness of a

thin shell. Hence, the pressure line (or C-line) lies at the centre of the core concrete

section (CGC). The following sketch shows the internal forces under service conditions.

The analysis is done for a slice of unit length along the longitudinal direction (parallel to

axis).



CPePe

C

CGC

CGS

Ra) Due to prestress b) Due to internal pressure

T

p

CPePe

C

CGC

CGS

CPePe

C

CGC

CGS

Ra) Due to prestress b) Due to internal pressure

T

p

T

p

Figure 9-6.1 Internal forces under service conditions

To reduce the loss of prestress due to friction, the prestressing can be done over

sectors of the circumference. Buttresses are used for the anchorage of the tendons.

The following sketch shows the buttresses along the circumference.

Buttress Buttress

Figure 9-6.2 Use of buttress in circumferential prestressing

9.6.2 General Analysis and Design

Analysis The basics of analysis and design for circumferential prestressing are provided for a

general understanding. Specific applications such as pipes, liquid storage tanks and

ring beams will be explained later.

Analysis at Transfer The compressive stress can be calculated from the compression C. From equilibrium,

C = P0, where P0 is the prestress at transfer after short-term losses. The compressive

stress (fc) is given as follows

cPf = -A

0 (9-6.1)



Here,

A = area of the longitudinal section of the slice.

The permissible prestress is determined based on fc within the allowable stress at

transfer (fcc,all).

Analysis at Service Loads The tensile stress due to the internal pressure (p) can be calculated from the tension T.

From equilibrium of half of the slice, T = pR where, R is the radius of the mid-surface of

the cylinder. The resultant stress (fc) due to the effective prestress (Pe) and internal

pressure is given as follows.

ec

t

P pRf = - +A A

(9-6.2)

Here,

At = area of the transformed longitudinal section of the slice.

The value of fc should be compressive and within the allowable stress at service loads

(fcc,all). In the previous equation, since Pe = pR and At is greater than A, fc is always

negative. Thus, the concrete will be under compression. To meet the safety standards,

a factor of safety can be further introduced.

Design The internal pressure p and the radius R are given variables. It is assumed that the

prestressing steel alone carries the hoop tension due to internal pressure, that is Pe =

Apfpe = pR.

The steps of design are as follows.

1) Calculate the area of the prestressing steel from the equation Ap = pR / fpe.

2) Calculate the prestress at transfer from an estimate of the permissible initial

stress fp0 and using the equation

P0 = Ap fp0. (9-6.3)

3) Calculate the thickness of concrete shell from the following equation.

A = P0 / fcc,all (9-6.4)



Here, fcc,all is the allowable compressive stress at transfer.

4) Calculate the resultant stress fc at the service conditions using Eqn. (9-6.2). The

value of fc should be within fcc,all, the allowable stress at service conditions.

9.6.3 Prestressed Concrete Pipes

Prestressed concrete pipes are suitable when the internal pressure is within 0.5 to 2.0

N/mm2. There are two types of prestressed concrete pipes: cylinder type and the non-

cylinder type. A cylinder type pipe has a steel cylinder core, over which the concrete is

cast and prestressed. A non-cylinder type of pipe is made of prestressed concrete only.

IS:784 - 2001 (Prestressed Concrete Pipes (Including Specials) - Specification)

provides guidelines for the design of prestressed concrete pipes with the internal

diameter ranging from 200 mm to 2500 mm. The pipes are designed to withstand the

combined effect of internal pressure and external loads. The minimum grade of

concrete in the core should be M40 for non-cylinder type pipes.

First, the core is cast either by the centrifugal method or by the vertical casting method.

In the centrifugal method the mould is subjected to spinning till the concrete is

compacted to a uniform thickness throughout the length of the pipe. In the vertical

casting method, concrete is poured in layers up to a specified height.

After adequate curing of concrete, first the longitudinal wires are prestressed (the wires

can be pre-tensioned). Subsequently, the circumferential prestressing is done by the

wire wound around the core in a helical form. The wire is wound using a counter weight

or a die. Finally a coat of concrete or rich cement mortar is applied over the wire to

prevent from corrosion.

For cylinder type pipes, first the steel cylinder is fabricated and tested. Then the

concrete is cast around it.



Figure 9-6.3 Stacked prestressed pipes (Courtesy: The Indian Hume Pipe Co. Ltd., Mumbai)

The analysis and design of prestressed concrete pipes consider the stresses due to the

different actions. A horizontal layout of the pipe is considered to illustrate them.

Analysis The stresses in the longitudinal direction are due to the following actions.

1. Longitudinal prestressing (fl1)

2. Circumferential prestressing (fl2)

3. Self weight (fl3)

4. Transport and handling (fl4)

5. Weight of fluid (fl5)

6. Weight of soil above (fl6)

Longitudinal prestressing The longitudinal prestressing generates a uniform compression.

l

e

c

Pf = -A1

1 (9-6.5) Here,



Pe = effective prestress

Ac1 = area of concrete in the core.

Circumferential prestressing Due to the Poisson’s effect, the circumferential prestressing generates longitudinal

tensile stress.

le

c

Pf = ×A2 0.284 (9-6.6)

The above expression estimates the Poisson’s effect.

Self weight If the pipe is not continuously supported, then a varying longitudinal stress generates

due to the moment due to self weight (Msw).

ll

sw 3

Mf = ±Z (9-6.7)

Here,

Zl = section modulus about the centroidal axis.

Transport and handling A varying longitudinal stress generates due to the moment during transport and

handling (Mth).

ll

th

Mf = ±Z4 (9-6.8)

Weight of fluid Similar to self weight, the moment due to weight of the fluid inside (Mf) generates

varying longitudinal stress.

ll

f

Mf = ±Z5 (9-6.9)



Weight of soil above The weight of soil above for buried pipes is modelled as an equivalent distributed load.

The expression of stress (fl6) is similar to that for the weight of fluid.

The longitudinal stresses are combined based on the following diagram.

D

fl 1 fl 2 fl 3 + fl 4 + fl 5 + fl 6

+ +

Section of pipe

D

fl 1 fl 2 fl 3 + fl 4 + fl 5 + fl 6

+ +

Section of pipe

Figure 9-6.4 Stress profiles across section

The stresses in the circumferential direction are due to the following actions.

1. Circumferential prestressing (fh1)

2. Self weight (fh2)

3. Weight of fluid (fh3)

4. Weight of soil above (fh4)

5. Live load (fh5)

6. Internal pressure (fh6)

Circumferential prestressing The compressive hoop stress (fh1) is given as follows.

s

h c

s

c

Pf = -AP= -× t

12

1

(9-6.10)

Here,

Ps = tensile force in spiral wire in unit length of pipe

Ac2 = area for longitudinal section of unit length

tc = thickness of the core.



Actions 2. to 5. For each of these actions, first the vertical load per unit length (W) is calculated.

Moment (M) and thrust (T) develop across the thickness owing to distortion of the

section due to W, as shown in the following sketch.

T M

WW

T M

WW

Figure 9-6.5 Forces due to vertical load

The hoop stress at a point is calculated by the following equation.

(9-6.11)

h

h

M Tf = ± +Z A

The expressions of M and T due to W are as follows.

M = CM W R (9-6.12) T = CT W (9-6.13)

Here,

CM = moment coefficient

CT = thrust coefficient

W = vertical load per unit length

R = mean radius of pipe

A = area of longitudinal section for unit length of pipe

Zh = section modulus for hoop stress for same length

= (1/6)t2 × 1000 mm3/m

t = total thickness of core and coat

Values of CM and CT are tabulated in IS:784 - 2001.

The internal pressure is as follows.

h t

pRf =A6 (9-6.14)



The hoop stresses are combined based on the following diagram.

fh 1

fh 2 + fh 3 + fh 4 + fh 5

fh 6

Coat Core

fh 1

fh 2 + fh 3 + fh 4 + fh 5

fh 6

Coat Core

Figure 9-6.6 Stress profiles across the thickness

9.6.4 Liquid Storage Tanks

In the construction of concrete structures for the storage of liquids, the imperviousness

of concrete is an important basic requirement. Hence, the design of such construction

is based on avoidance of cracking in the concrete. The structures are prestressed to

avoid tension in the concrete. In addition, prestressed concrete tanks require low

maintenance. The resistance to seismic forces is also satisfactory.

Prestressed concrete tanks are used in water treatment and distribution systems, waste

water collection and treatment system and storm water management. Other

applications are liquefied natural gas (LNG) containment structures, large industrial

process tanks and bulk storage tanks.

The construction of the tanks is in the following sequence. First, the concrete core is

cast and cured. The surface is prepared by sand or hydro blasting. Next, the

circumferential prestressing is applied by strand wrapping machine. Shotcrete is

applied to provide a coat of concrete over the prestressing strands.



A few photographs are provided for illustration.

(b) Circumferential prestressing (close-up)

(a) Circumferential prestressing

(c) Shotcrete operation

Figure 9-6.7 Construction of a liquid storage tank

(Reference: DYK Incorporated)

IS:3370 - 1967 (Code of Practice for Concrete Structures for the Storage of Liquids)

provides guidelines for the analysis and design of liquid storage tanks. The four

sections of the code are titled as follows.

Part 1: General Requirement

Part 2: Reinforced Concrete Structures



Part 3: Prestressed Concrete Structures

Part 4: Design Tables.

Analysis The analysis of liquid storage tanks can be done by IS:3370 - 1967, Part 4, or by the

finite element method. The Code provides coefficients for bending moment, shear and

hoop tension (for cylindrical tanks), which were developed from the theory of plates and

shells. In Part 4, both rectangular and cylindrical tanks are covered. Since circular

prestressing is applicable to cylindrical tanks, only this type of tank is covered in this

module.

The following types of boundary conditions are considered in the analysis of the

cylindrical wall.

a) For base: fixed or hinged

b) For top: free or hinged or framed.

The applicability of each boundary condition is explained next.

For base Fixed: When the wall is built continuous with its footing, then the base can be

considered to be fixed as the first approximation.

Hinged: If the sub grade is susceptible to settlement, then a hinged base is a

conservative assumption. Since the actual rotational restraint from the footing is

somewhere in between fixed and hinged, a hinged base can be assumed.

The base can be made sliding with appropriate polyvinyl chloride (PVC) water-stops for

liquid tightness.

For top Free: The top of the wall is considered free when there is no restraint in expansion.

Hinged: When the top is connected to the roof slab by dowels for shear transfer, the

boundary condition can be considered to be hinged.



Framed: When the top of the wall and the roof slab are made continuous with moment

transfer, the top is considered to be framed.

The hydrostatic pressure on the wall increases linearly from the top to the bottom of the

liquid of maximum possible depth. If the vapour pressure in the free board is negligible,

then the pressure at the top is zero. Else, it is added to the pressure of the liquid

throughout the depth. The forces generated in the tank due to circumferential prestress

are opposite in nature to that due to hydrostatic pressure. If the tank is built

underground, then the earth pressure needs to be considered.

The hoop tension in the wall, generated due to a triangular hydrostatic pressure is given

as follows.

T = CT w H Ri (9-6.15)

The bending moment in the vertical direction is given as follows.

M = CM w H3 (9-6.16)

The shear at the base is given by the following expression.

V = CV w H2 (9-6.17)

In the previous equations, the notations used are as follows.

CT = coefficient for hoop tension

CM = coefficient for bending moment

CV = coefficient for shear

w = unit weight of liquid

H = height of the liquid

Ri = inner radius of the wall.

The values of the coefficients are tabulated in IS:3370 - 1967, Part 4, for various values

of H2/Dt, at different depths of the liquid. D and t represent the inner diameter and the

thickness of the wall, respectively. The typical variations of CT and CM with depth, for

two sets of boundary conditions are illustrated.



CT CM

H

t

CT CM

H

t

(a) Fixed base – free top

CT CM

H

t

CT CM

H

t

(b) Hinged base – free top

Figure 9-6.8 Variations of coefficients for hoop tension and bending moment

The roof can be made of a dome supported at the edges on the cylindrical wall. Else,

the roof can be a flat slab supported on columns along with the edges. IS:3370 - 1967, Part 4, provides coefficients for the analysis of the floor and roof slabs.

Design IS:3370 - 1967, Part 3, provides design requirements for prestressed tanks. A few of

them are mentioned.

1) The computed stress in the concrete and steel, during transfer, handling and

construction, and under working loads, should be within the permissible values

as specified in IS:1343 - 1980.

2) The liquid retaining face should be checked against cracking with a load factor of

1.2.

σCL/σWL ≥ 1.2 (9-6.18)

Here,

σCL = stress under cracking load

σWL = stress under working load.



Values of limiting tensile strength of concrete for estimating the cracking load are

specified in the Code.

3) The ultimate load at failure should not be less than twice the working load.

4) When the tank is full, there should be compression in the concrete at all points of

at least 0.7 N/mm2. When the tank is empty, there should not be tensile stress

greater than 1.0 N/mm2. Thus, the tank should be analysed both for the full and

empty conditions.

5) There should be provisions to allow for elastic distortion of the structure during

prestressing. Any restraint that may lead to the reduction of the prestressing

force, should be considered.

Detailing Requirements

IS:3370 - 1967, Part 3, also provides detailing requirements. The cover requirement is

as follows. The minimum cover to the prestressing wires should be 35 mm on the liquid

face. For faces away from the liquid, the cover requirements are as per IS:1343 - 1980.

Other requirements from IS:1343 - 1980 are also applicable.

9.6.5 Ring Beams

Ring beams support domes in buildings, tanks, silos and nuclear containment

structures.

Circular prestressing is applied on a dome by a grid of tendons. The cylindrical wall is

prestressed circumferentially and vertically. The ring beam is circumferentially

prestressed. The sketches below show schematic representation of the elements and

the prestressing tendons.



Dome

Ring beam

Cylindrical wall

Raft foundation

Dome

Ring beam

Cylindrical wall

Raft foundation Figure 9-6.9 Cross-section of a nuclear containment structure

Tendon for dome prestressing

Tendon for vertical prestressing of wall

Tendons forprestressing of ring beam

Tendons forprestressing of wall

Tendon for dome prestressing

Tendon for vertical prestressing of wall

Tendons forprestressing of ring beam

Tendons forprestressing of wall

Figure 9-6.10 Typical layout of prestressing tendons at dome and ring beam junction



The following photo shows a prestressed nuclear containment structure.

Figure 9-6.11 Containment Structure, Kaiga Atomic Power Plant, Karnataka

(Reference: Larsen & Toubro)

Analysis The analysis of a ring beam is based on a load symmetric about the vertical axis. Since

the dome is not supposed to carry any moment at the edge, the resultant reaction at the

ring beam is tangential. The following figure shows the forces at the base of dome.

Rd sinθ

H

V

θRd

Rd sinθ

H

V

θRd

Figure 9-6.12 Forces at the base of dome

Let the total vertical load from the dome be W. The vertical reaction per unit length (V)

is given as follows.

d

WV =2πR sinθ

(9-6.19)



Here,

Rd = radius of the dome

θ = half of the angle subtended by the dome.

The horizontal thrust (H) is calculated from the condition of the reaction to be tangential.

The value per unit length is given as follows.

d

H =V cotθW cotθ=πR sinθ2

(9-6.20)

The thrust is resisted by the effective prestressing force (Pe) in the ring beam. Pe can

be estimated from the equilibrium of half of the ring beam as shown in the following

sketch.

Pe

H

Rd sinθ

Plan of ring beam

Pe

H

Rd sinθ

Plan of ring beam Figure 9-6.13 Forces in the ring beam

e dP = H R sinθ

W cotθ=π2

(9-6.21)

9.6.6 Conclusion

Prestressing of concrete is observed in other types of structural elements, such as

bridge decks, shells and folded plates, offshore concrete gravity structures hydraulic

structures, pavements and raft foundations. The analysis of special structures is based

on advanced theory of structural analysis or the finite element method. After the

analysis, the design of such structures follows the basic principles of prestressed

concrete design. It is expected that in future, further innovations in structural form,

prestressing systems and construction technology will promote the application of

prestressed concrete.



A few photos of recent applications follows.

Figure 9-6.14 Cement silo, Jayanthipuram, Tamilnadu

(Reference: Larsen & Toubro)

Figure 9-6.15 Curved box-girder bridge, Jaipur-Kishangarh Highway, Rajasthan

(Reference: L & T Ramboll)



(a) Exterior view

(b) Interior view

Figure 9-6.16 Folded plate

(Department of Ocean Engineering, Indian Institute of Technology Madras)



Figure 9-6.17 Prestressed aqueduct, Gomti, Uttar Pradesh

(Courtesy: Hindustan Construction Company Ltd.)



Bibliography

Books

1. Abeles, P. W., The Principles and Practice of Prestressed Concrete, Crosby

Lockwood and Sons, 1949.

2. Collins, M. P. and Mitchell, D., Prestressed Concrete Structures, Prentice-Hall,

Inc., 1991.

3. Guyon, Y., Prestressed Concrete, John Wiley and Sons, Vol.1, 1953, Vol. 2,

1960.

4. Khan, S. and Williams, M., Post-tensioned Concrete Floors, Butterworth-

Heinemann Ltd., 1995.

5. Krishna Raju, N., Prestressed Concrete, 3rd Edition, Tata McGraw-Hill Publishing

Company Ltd., 1998.

6. Leonhardt, F., Prestressed Concrete – Design and Construction, 2nd Edition,

Wilhelm Ernst and Sohn, 1964.

7. Lin, T. Y. and Burns, N. H., Design of Prestressed Concrete Structures, 3rd

Edition, John Wiley & Sons, 1982.

8. Magnel, G., Prestressed Concrete, Concrete Publications, 1948.

9. Nawy, E. G., Prestressed Concrete – A Fundamental Approach, 5th Edition,

Prentice-Hall, Inc., 2006.

10. Nilson, A., Design of Prestressed Concrete, 2nd Edition, John Wiley & Sons,

1987.

11. Rajagopalan, N., Prestressed Concrete, Narosa Publishing House, 2005.



Codes The codes related with prestressed concrete are listed below according to the

publishing agencies.

Bureau of Indian Standards

IS:784 - 2001 Prestressed Concrete Pipes (Including Fittings) - Specification

IS:1343 - 1980 Code of Practice for Prestressed Concrete

IS:1678 - 1998 Specification for Prestressed Concrete Poles for Overhead Power,

Traction and Telecommunication Lines

IS:1785 - 1983 Specification for Plain Hard Drawn Steel Wire for Prestressed Concrete

Part-1: Cold-drawn Stress-relieved wire

Part-2: As-drawn wire

IS: 2090 - 1983 Specification for High Tensile Steel Bars Used in Prestressed Concrete

IS:2193 - 1986 Specification for Precast Prestressed Concrete Steel Lighting Poles

IS:3370 - 1967 Code of Practice for Concrete Structures for Storage of Liquids

Part-3: Prestressed Concrete Structures

IS:6003 - 1983 Specification for Indented Wire for Prestressed Concrete

IS:6006 - 1983 Specification for Uncoated Stress Relieved Strand for Prestressed

Concrete

IS:6461 - 1973 Glossary of Terms Relating to Cement Concrete

Part 11: Prestressed Concrete

IS:10790 - 1984 Methods of Sampling of Steel for Prestressed and Reinforced Concrete

Part-1: Prestressing Steel

Part-2: Reinforcing Steel



IS:13158 - 1991 Specification for Prestressed Concrete Circular Spun Poles for

Overhead Power, Traction and Telecommunication Lines

IS: 14268 - 1995 Specification for Uncoated Stress Relieved Low Relaxation Seven Ply

Strand for Prestressed Concrete

The Indian Roads Congress

IRC:18 - 2000, Design Criteria for Prestressed Concrete Road Bridges (Post-tensioned

Concrete).

Ministry of Railways, Government of India

IRS Concrete Bridge Code: 1997, Indian Railway Standard Code of Practice for Plain,

Reinforced and Prestressed Concrete for General Bridge Construction.

American Concrete Institute, USA

ACI 318M-05, Building Code Requirements for Structural Concrete and Commentary.

British Standard Institution, UK

BS 8110 : Part 1 : 1997, Structural Use of Concrete : Part 1 Code of Practice for

Design and Construction.

Council of Standards Australia

AS 3600, Concrete Structures, 2001.

European Committee for Standardisation

EN 1992, Design of Concrete Structures, 2005.

Handbook The following handbook is published by the Precast/Prestressed Concrete Institute,

USA.

PCI Design Handbook, 5th Edition.

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