CH-1 Structural Design of RCC Building Components

E4-E5 Civil (Technica l) Rev date : 20-04-11

BSNL India For Interna l Circulat ion Only Page: 1

Chapter-1

STRUCTURAL DESIGN

OF

RCC BUILDING

COMPONENTS

Rajendra Mathur Dy. Dir(BS-C) 09412739 232(M) e-mail–[email protected]



Structural Design of RCC Building Components

1.0 Introduction

The procedure fo r ana lysis and design of a given building wil l

depend on the t ype of building, its complexit y, the number of

stories etc. First the architectural drawings of the building are

studied , structural system is f ina lized sizes of structural members

are decided and brought to the knowledge of the concerned

architect. The p rocedure for structural design will invo lve some

steps which will depend on the type of build ing and also its

complexit y and the time ava ilable for structural design. Often, the

work is required to start soon, so the steps in design are to be

arranged in such a wa y the foundation d rawings can be taken up in

hand within a reasonable period of time.

Further, before start ing the structural design, the fo llowing

informat ion o f data are requ ired: ( i) A set of architectural

drawings;( ii) Soil Invest igation report (SIR) of soil data in lieu

thereof; ( iii) Locat ion o f the place o r cit y in order to decide on

wind and seismic loadings;( iv) Data fo r lift s, water tank capacit ies

on top, special roof featu res o r loadings, etc.

Choice of an appropriate structural system for a give n

build ing is vital for its economy and safet y. There are two type of

build ing systems:-

(a) Load Bearing Masonr y Buildings.

(b) Framed Buildings.

(a) Load Bearing Masonry Buildings:-

Small build ings like houses with small spans of beams, s labs

genera lly constructed as load bearing brick walls with reinforced

concrete slab beams. This system is su itable for building up to four

or less stories.(as shown in f ig. below). In such build ings crushing

strength of bricks shall be 100 kg/cm2 mi n i mu m for four stories. This

system is adequate for vert ical loads it also serves to res ists

horizontal loads like wind & earthquake by box act ion. Further, to

ensure its act ion against earthquake , it is necessary to provide RCC

Bands in horizo ntal & vertica l re inforcement in brick wall as per

IS: 4326-1967( Indian Standards Code o f Practice for

Earthquake Resistant Construction of Buildings.) . In some



Buildings, 115mm thick brick walls are provided s ince these walls

are incapable of supporting vert ica l loads, beams have to be provide

alo ng the ir lengths to support adjo ining s lab & the weight of

115mm thick b rick wall of upper

storey. These beams are to rest on 230 mm thick b rick walls or

reinforced concrete columns if required. The design of Load

Bearing Masonr y Build ings are done as per IS:1905-1980 (Indian

Standards Code of Practice for Structura l Safety of Buildings:

Masonry Walls(Second Revision).

Load bearing brick wall

Structural system

(b) Framed Buildings:-

In these t ypes o f bu ildings re inforced concrete frames are provided

in both p rincipal direct ions to resist vertical loads and the vertica l



loads are transmitted to vert ical framing system i.e columns and

Foundations. This t yp e of system is e ffect ive in res ist ing both

vertical & horizontal loads. The brick walls are to be regarded as

non load bearing filler walls only. This system is suitable for multi-

storied build ing which is also effective in res ist ing horizonta l loads

due to earthquake. In this

system the f loor slabs, generally 100-150 mm thick with spans

ranging from 3.0 m to 7.0 m. In certa in earthquake prone areas,

even s ingle or double storey bu ildings a re made framed structures

for safet y reasons. Also the s ingle storey bu ildings of large sto rey

he ights (5 .0m or more ) , like electric substation etc. are made

framed structure as brick walls of large heights are slender and load

carrying capacit y of such walls reduces due to slenderness.

Framed Structura l system



2 .0 Basic Codes for Design . The design should be carried so as to conform to the

following Indian code for reinforced concrete design, published b y

the Bureau o f Indian Standards, New Delhi:

Purpose of Codes

National build ing codes have been formulated in different

countries to la y down gu idelines for the design and construction of

structure. The codes have evo lved from the co llect ive wisdom of

expert structural engineers, ga ined over the years. These codes are

periodica lly revised to bring them in line with current research, and

often, current trends.

The codes serve at least four distinct functions .

First ly, they ensure adequate structural safet y, b y spec ifying certain

essent ia l minimum requ irement for design.

Secondly, the y render the task of the designer relat ively simple;

often, the result o f sophist icate analyses is made ava ilable in the

form of a simple formula or chart.

Thirdly, the codes ensure a measure of consistency among different

designers.

Fina lly, they have some legal validit y in that they protect the

structural designer from any liab ilit y due to structural failures that

are caused by inadequate supervis ion and/or fau lt y mater ial and

construct ion.

( i)IS 456 : 2000 – Plain and reinforced concrete – code of

practice (fourth revision)

( ii) Loading Standards

These loads to be considered fo r structural design are specified in

the fo llowing load ing standards:

IS 875 (Part 1 -5 ) : 1987 – Code of practice for design loads

(other than earthquake) for buildings and structures (second

revision)

Part 1 : Dead loads

Part 2 : Imposed (live) loads

Part 3 : Wind loads



Part 4 : Snow loads

Part 5 : Special loads and load combinations

IS 1893 : 2002 – Criteria for earthquake resistant design of structure (fourth revision) .

IS 13920 : 1993 – Ductile deta iling o f reinforced concrete

structure subject to seismic forces.

Design Handbooks

The Bureau of Indian standards has also published the following

handbooks, which serve as usefu l supplement to the 1978 vers ion of

the codes. Although the handbooks need to be updated to bring them

in line with the recent ly revised (2000 version) o f the Code, many

of the p rovisions continue to be va lid (especially with regard to

structural design provis ions).

SP 16 : 1980 – Design Aids (for Reinforced Concrete) to IS 456 :

1978

SP 24 : 1983 – Explanatory handbook on IS 456 : 1978

SP 34 : 1987 – Handbooks on Concrete Reinforced and Detailing.

General Design Consideration of IS: 456-2000.

The general design and construction of reinforced concrete buildings shall be

governed by the provisions of IS 456 –2000

AIM OF DESIGN

The aim of design is achievement of an acceptable probability that structures

being designed shall, with an appropriate degree of safety –

� Perform satisfactorily during their intended life.

� Sustain all loads and deformations of normal construction & use

� Have adequate durability

� Have adequate resistance to the effects of misuse and fire.

METHOD OF DESIGN –

� Structure and structural elements shall normally be designed by Limit

State Method.

� Where the Limit State Method cannot be conveniently adopted, Working

Stress Method may be used



MINIMUM GRADE OF CONCRETE

The minimum grade of concrete for plain & reinforced concrete shall be as per

table below –

26.4 Nominal Cover to Reinforcement

26.4.1 Nominal Cover

Nominal cover is the design depth of concrete cover to all steel

reinforcements, including links. It is the dimension used in design and

indicated in the drawings. It shall be not less than the diameter of the bar.

26.4.2 Nominal Covers to Meet Durability Requirement

Minimum values for the nominal cover of normal weight aggregate

concrete which should be provided to all reinforcement, including links

depending on the condition of exposure described in 8.2.3 shall be as

given in Table 16.

Table 16 Nominal Cover to Meet Durability Requirements

(Clause 26.4.2)

Exposure Nominal Concrete Cover in mm not Less Than

Mild 20

Moderate 30

Severe 45



Very Severe 50

Extreme 75

NOTES

1. For main reinforcement up to 12 mm diameter bar for mild

exposure the nominal cover may be reduced by 5 mm.

2. Unless specified otherwise, actual concrete cover should not

deviate from the required nominal cover by + 10 mm

3. For exposure condition ‘severe’ and ‘very severe’, reduction of 5

mm may be made, where concrete grade is M35 and above.

26.4.2.1 However for a longitudinal reinforcing bar in a column nominal cover

shall in any case not be less than 40 mm, or less than the diameter of

such bar. In the case of columns of minimum dimension of 200 mm or

under, whose reinforcing bars do not exceed 12 mm, a nominal cover of

25 mm may be used.

26.4.2.2 For footing minimum cover shall be 50 mm.

26.4.3 Nominal Cover to Meet Specified Period of Fire Resistance

Minimum values of nominal cover of normal-weight aggregate concrete to be

provided to all reinforcement including links to meet specified period of fire

resistance shall be as given in Table 16A.



21.4 Minimum Dimensions of RC members for specified Period of Fire

Resistance



DESIGN LOAD

Design load is the load to be taken for use in appropriate method of design. It is –

� Characteristic load in case of working stress method &

� Characteristic load with appropriate partial safety factors for limit

state design.

LOAD COMBINATIONS

As per IS 1893 (Part 1): 2002 Clause no. 6.3.1.2, the following load cases have to

be considered for analysis:

� 1.5 (DL + IL)

� 1.2 (DL + IL ± EL)

� 1.5 (DL ± EL)

� 0.9 DL ± 1.5 EL

� Earthquake load must be considered for +X, -X, +Z and –Z directions.

� Moreover, accidental eccentricity during earthquake can be such that it

causes clockwise or anticlockwise moments. So both clockwise &

anticlockwise torsion is to be considered.

� Thus, ±EL above implies 8 cases, and in all, 25 cases must be considered.

It is possible to reduce the load combinations to 13 instead of 25 by not using

negative torsion considering the symmetry of the building.



STIFFNESS

22.3.1 Relative Stiffness: The relative stiffness of the members may be based

on the moment of inertia of the section determined on the basis of any

one of the following definitions:

a) Gross

Section

The cross-section of the member ignoring reinforcement

b) Transformed

Section

The concrete cross-section plus the area of reinforcement transformed on the basis of modular ratio

c) Cracked

Section

The area of concrete in compression plus the area of reinforcement transformed on the basis of modular ratio

The assumptions made shall be consistent for all the numbers of the

structure throughout any analysis.

22.3.2 For deflection calculations, appropriate values of moment of inertia as

specified in Annexure of IS 456-2000 should be used.

STRUCTURAL FRAMES

22.4 The simplifying assumptions as given in 22.4.1 to 22.4.3 may be used

in the analysis of frames.

ARRANGEMENT OF LIVE LOAD

22.4.1 a) Consideration may be limited to combinations of: 1) Design dead load on all spans with full design live load on two

adjacent spans; and

2) Design dead load on all spans with dull design live load on

alternate spans.

22.4.1 b) When design live load does not exceed three-fourths of the design

dead load, the load arrangement may be design dead load and design

live load on all the spans.

Note: For beams continuous over support 22.4.1 (a) may be assumed. 22.4.2 Substitute Frame: For determining the moments and shears at any

floor or roof level due to gravity loads, the beams at that level together

with columns above and below with their far ends fixed may be

considered to constitute the frame.

22.4.3 For lateral loads, simplified methods may be used to obtain the

moments and shears for structures that are symmetrical. For



unsymmetrical or very tall structures, more rigorous methods should

be used.

MOMENT AND SHEAR COFFICIENTS FOR CONTINUOUS BEAMS 22.5.1 Unless more exact estimates are made, for beams of uniform cross-

section which support substantially uniformly distributed load over

three or more spans which do not differ by more than 15 percent of the

longest, the bending moments and shear forces used in design may be

obtained using the coefficients given in Tables below.

For moments at supports where two unequal spans meet or in case

where the spans are not equally loaded, the average of the two values

for the negative moment at the support may be taken for design.

Where coefficients given in Table below are used for calculation of

bending moments, redistribution referred to in 22.7 shall not be

permitted.

22.5.2 Beams Over Free End Supports

Where a member is built into a masonry wall which develops only

partial restraint, the member shall be designed to resist a negative

moment at the face of the support of W1/24 where W is the total

design load and 1 is the effective span, or such other restraining

moment as may be shown to be applicable. For such a condition shear

coefficient given in Table below at the end support may be increased

by 0.05.

-------------------------------------------------------------------------------------------------------

BENDING MOMENT COFFICIENTS

---------------------------------------------------------------------------------------------------

----

Span Moments Support Moments ------------------------------------------------------------------------------------- Types of Load Near Middle At Middle At Support At Other of End Span of interior next to the Interior span end support Supports ------------------------------------------------------------------------------------------------------- Dead load and 1 1 1 1 imposed load +-- +-- (- )-- (- )-- (fixed) 12 16 10 12



Imposed load 1 1 1 1 (not fixed) +-- +-- (- )-- (- )-- 10 12 9 9 Note:For obtaining the bending moment, the coefficient shall be multiplied by the total design load and effective span. ----------------------------------------------------------------------------------------------------- ---------------------------------------------------------------------------------------------------

----

SHEAR FORCE COFFICIENTS

---------------------------------------------------------------------------------------------------

----

Type of Load At End At Support Next At All Other Support to the end Support Interior Support Outer side Inner Side ------------------------------------------------------------------------------------------------------- Dead load and imposed load 0.40 0.60 0.55 0.50 (fixed) Imposed load 0.45 0.60 0.60 0.60 (not fixed) Note: For obtaining the shear force, the coefficient shall be multiplied by the total design load. -----------------------------------------------------------------------------------------------------

CRITICAL SECTIONS FOR MOMENT AND SHEAR

22.6.1 For monolithic construction, the moments computed at the face of the

supports shall be used in the design of the members at those sections.

For non-monolithic construction the design of the member shall be

done keeping in view 22.2.

22.6.2 Critical Section for Shear The shears computed at the face of the Support shall be used in the

design of the member at that section except as in 22.6.2.1

22.6.2.1 When the reaction in the direction of the applied shear introduces

compression into the end region of the member, sections located at a



distance less than d from the face of the support may be designed for

the same shear as that computed at distance d.

REDISTRIBUTION OF MOMENTS 22.7 Redistribution of moments may be done in accordance with 37.1.1 for

limit state method and in accordance with B-1.2 for working stress method. However, where simplified analysis using coefficients is adopted, redistribution of moments shall not be done.

EFFECTIVE DEPTH

23.0 Effective depth of a beam is the distance between the centroid of the

area of tension reinforcement and the maximum compression fibre,

excluding the thickness of finishing material not placed monolithically

with the member and the thickness of any concrete provided to allow

for wear. This will not apply to deep beams.

CONTROL OF DEFLECTION

23.2 The deflection of a structure or part thereof shall not adversely affect the

appearance or efficiency of the structure or finishes or partitions. The

deflection shall generally be limited to the following:

a) The final deflection due to all loads including the effects of

temperature, creep and shrinkage and measured from the as-cast level

of the supports of floors, roofs and all other horizontal members

should not normally exceed span/250.

b) The deflection including the effects of temperature, creep and

shrinkage occurring after erection of partitions and the application of

finishes should not normally exceed span/350 or 20mm whichever is

less.

23.2.1 For beams, the vertical deflection limits may generally be assumed to be

satisfied provided that the span to depth ratio are not greater than the value

obtained as below:

a) Basic values of span to effective depth ratios for spans up to 10m:

Cantilever 7

Simply supported 20 Continuous 26



b) For spans above 10m, the values in (a) may be multiplied by 10/span

in metres, except for cantilever in which case deflection calculations

should be made.

c) Depending on the area and the type of steel for tension reinforcement,

the value in (a) or (b) shall be modified as per Fig. 4

d) Depending on the area of compression reinforcement, the value of

span to depth ratio be further modified as per Fig. 5

e) For flanged beams, the value of (a) or (b) be modified as per Fig. 6 and

the reinforcement percentage for use in fig. 4 and 5 should be based on

area of section equal to bf d.

Note: When deflections are required to be calculated, the method given

Annexure ‘C’ of IS 456-2000 may be used.



CONTROL OF DEFLECTION – SOLID SLABS

24.1 General

The provisions of 32.2 for beams apply to slabs also. NOTES

1. For slabs spanning in two directions, the shorter of the two spans should

be used for calculating the span to effective depth rations.

2. For two-way slabs of shorter spans (up to 3.5 m) with mild steel

reinforcement, the span to overall depth rations given below may

generally be assumed to satisfy vertical deflection limits for loading class

up to 3 kN/m2.

Simply supported slab 35

Continuous slabs 40

For high strength deformed bars of grade Fe 415,the values given

above should be multiplied by 0.8.

Simply supported slab 28

Continuous slabs 32

23.3 Slabs Continuous Over Supports

Slabs spanning in one direction and continuous over supports shall be

designed according to the provisions applicable to continuous beams.

23.4 Slabs Monolithic with Supports Bending moments in slabs (except flat

slabs) constructed monolithically with the supports shall be calculated by

taking such slabs either as continuous over supports and capable of free, or as



members of a continuous frame work with the supports, taking into account

the stiffness of such support. If such supports are formed due to beams which

justify fixity at the support of slabs, then the effects on the supporting beam,

such as, the bending of the web in the transverse direction of the beam,

wherever applicable, shall also be considered in the design of the beams.

23.4.1 For the purpose of calculation of moment in slabs in a monolithic

structure, it will generally be sufficiently accurate to assumed direct

members connected to the ends of such slab are fixed in position and

direction at the end remote from their connection with the slab.

26.5 REQUIREMENT OF REINFORCEMENT FOR STRUCTURAL

MEMBER

26.5.1 Beams

26.5.1.1 Tension reinforcement

(a) Minimum reinforcement:- The minimum area of tension reinforcement

shall not be less than that given by the following:-

As = 0.85 bd fy

where

As = minimum area of tension reinforcement.

b = breadth of beam or the breadth of the web of T-beam.

d = effective depth, and

fy = characteristic strength of reinforcement in M/mm2

(b) Maximum reinforcement:- the maximum area of tension reinforcement

shall not exceed 0.04bD.

26.5.1.2 Compression reinforcement

The maximum area of comparison reinforcement shall not exceed 0.04 bd.

Comparison reinforcement in beams shall be enclosed by stirrups for effective

lateral restraint.

26.5.1.3 Side face reinforcement

Where the depth of the web in the beam exceeds 750mm, side face reinforcement

shall be provided along the two faces. The total area of such reinforcement shall

be not less than 0.1 % of the web area and shall be distributed on the equally on

the two face at spacing not exceeding 300mm or web thickness whichever is less.

26.5.1.4 Transverse reinforcement in beam for shear torsion



The transverse reinforcement in beam shall be taken around the outer most

tension & compression bars. In T-beams and I-beams, such reinforcement shall

pass around longitudinal bars located close to the outer face of the flange.

26.5.1.5 Maximum spacing of shear reinforcement

Maximum spacing of shear reinforcement means long by axis of the member shall

not exceed 0.75 d for vertical stirrups and d for inclined stirrups at 45” where d is

the effective depth on the section under consideration. In no case shall be spacing

exceed 300mm.

26.5.1.6 Minimum shear reinforcement

Minimum shear reinforcement in the form of stirrups shall be provided such that:

Asv 0.4 bsv 0.87 fy Where Asv = total cross-sectional area of stirrups legs effective in shear.

Sv = stirrups spacing along the length of the member

B = breadth of the beam or breadth of the web of flange beam, and

fy = characteristic strength of the stirrups reinforcement in N/mm2 which shall not

taken greater than 415 N/mm2

Where the maximum shear stress calculated is less than half the permissible value

in member of minor structure importance such as lintels, this provision need not

to be complied with.

26.5.1.7 Distribution of torsion reinforcement

When a member is designed for torsion torsion reinforcement shall be provided as

below:

a) the transverse reinforcement for torsion shall be rectangular closed stirrups

placed perpendicular to the axis of the member. The spacing of the stirrups

shall not exceed the list of x1, x1+y1/4 and 300 mm, where x1, y1 are

respectively the short & long dimensions of the stirrup.



b) Longitudinal reinforcement shall be place as closed as is practicable to the

corner of the cross section & in all cases, there shall be atleast one

longitudinal bar in each corner of the ties. When the cross sectional

dimension of the member exceed 450 mm additional longitudinal bar

shall be provided to satisfied the requirement of minimum reinforcement

& spacing given in 26.5.1.3.

26.3.2 Minimum Distance between Individual Bars

(a) The horizontal distance between two parallel main reinforcing bars shall

usually be not-less than the greatest of the following:

(i) Dia of larger bar and

(ii) 5 mm more than nominal maximum size of coarse aggregate.

(b) When needle vibrators are used it may be reduced to 2/3rd of nominal

maximum size of coarse aggregate,

Sufficient space must be left between bars to enable vibrator to be immersed.

(c) Where there are two or more rows of bars, bars shall be vertically in line

and the minimum vertical distance between bars shall be 15 mm, 2/3rd of

nominal maximum size of aggregate or the maximum size of bars, whichever

is greater.

26.5.2 Slabs

The rule given in 26.5.2.1 and 26.5.2.2 shall apply to slabs in addition to those

given in the appropriate clause.

26.5.2.1 Minimum reinforcement



The mild steel reinforcement in either direction in slabs shall not be less than 0.15

percent of the total cross-sectional area. However, this value can be reduced to

0.12 percent when high strength deformed bars or welded wire fabric are used.

26.5.2.2 Maximum diameter The diameter of reinforcing bars shall not exceed one eight of the total thickness of slab.

26.3.3 Maximum distance between bars - Slabs

1) The horizontal distance between parallel main reinforcement bars shall not be

more than three times the effective depth of solid slab or 300 mm whichever is

smaller.

2) The horizontal distance between parallel reinforcement bars provided against

shrinkage and temperature shall not be more than five times the effective depth of

a solid slab or 300 mm whichever is smaller.

Torsion reinforcement - Slab

Torsion reinforcement is to be provided at any corner where the slab is simply

supported on both edges meeting at that corner. It shall consist of top and bottom

reinforcement, each with layers of bars placed parallel to the sides of the slab and

extending from the edges a minimum distance of one-fifth of the shorter span.

The area of reinforcement in each of these four layers shall be three-quarters of

the area required for the maximum mid-span moment in the slab.

D-l.9 Torsion reinforcement equal to half that described in D-l.8 shall be provided

at a corner contained by edges over only one of which the slab is continuous.

D-1.10 Torsion reinforcements need not be provided at any comer contained by

edges over both of which the slab is continuous.

26.5.3 Columns

A. Longitudinal Reinforcement

a. The cross sectioned area of longitudinal reinforcement shall be not less

than 0.8% nor more than 6% of the gross sectional area of the column.

Although it is recommended that the maximum area of steel should not

exceed 4% to avoid practical difficulties in placing & compacting concrete.



b. In any column that has a larger cross sectional area than that required to

support the load, the minimum percentage steel must be based on the area

of concrete resist the direct stress & not on the actual area.

c. The bar should not be less than 12 mm in diameter so that it is sufficiently

rigid to stand up straight in the column forms during fixing and concerting.

d. The minimum member of longitudinal bars provided in a column shall be

four in rectangular columns & six in circular columns.

e. A reinforced concrete column having helical reinforcement must have at

least six bars of longitudinal reinforcement with the helical reinforcement.

These bars must be in contact with the helical reinforcement & equidistance

around its inner circumference.

f. Spacing of longitudinal should not exceed 300 mm along periphery of a

column.

g. In case of pedestals, in which the longitudinal reinforcement is not taken

into account in strength calculations, nominal reinforcement should be not

be less than 0.15% of cross sectional area.

B. Transverse Reinforcement

a. The diameter of lateral ties should not be less than ¼ of the diameter of

the largest longitudinal bar in no case should not be less than 6 mm.

b. Spacing of lateral ties should not exceed least of the following:-

• Least lateral dimension of the column.

• 16 times the smallest diameter of longitudinal bars to be tied.

• 300mm.

SHEAR

40.1 Nominal Shear Stress

The nominal shear stress in beams of uniform depth shall be obtained by the

following equation:

τv = Vu/ b.d where Vu = shear force due to design loads;

b = breadth of the member, which for flanged section shall be taken as the breadth

of the web, bw; and

d = effective depth.

40.2.3 With Shear Reinforcement

Under no circumstances, even with shear reinforcement, shall the nominal shear

stress in beams should not exceed given in Table 20.

40.2.3.1 For solid slabs, the nominal shear stress shall not exceed half the

appropriate values given in Table 20.



40.3 Minimum Shear Reinforcement

When τv, is less than τc given in Table 19, minimum shear reinforcement shall be provided in accordance with 26.5.1.6.

40.4 Design of Shear Reinforcement

When τv, is exceeds τc , given in Table 19, shear reinforcement shall be provided

in any of the following forms:

a) Vertical stirrups,

b) Bent-up bars along with stirrups, and where bent-up bars are provided, their

contribution towards shear resistance shall not be more than half that of the total

shear reinforcement.

Shear reinforcement shall be provided to carry a shear equal to Vu – τ c b d. the

strength of shear reinforcement Vus shall be calculated as below:

a) For Vertical Stirrups:



0.87 fy Asv d

Vus = ___________ Sv

b) For inclined stirrups or a series of bars bent up at different cross –section:

0.87 fy Asv d

Vus = ___________ (Sin ά + Cos ά) Sv

c) For single bar or single group of parallel bars, all bent up at the same cross sections:

Vus = 0.87 fy Asv Sin ά

Where Asv = total cross –sectional area of stirrups legs or

bent-up bar within a distance Sv,

Sv = spacing of the stirrups or bent-up bars along

the

length of the member.

τ v = nominal shear stress, τ c = design shear strength of the concrete, b = breadth of the member which for flanged

beams, shall be taken as the breadth of the web bw. fy = characteristic strength of the stirrup or bent-

up reinforcement which shall not be taken

greater

than 415 N/mm2,

ά = angle between the inclined stirrup or bent up

bar

and the axis of the member not less than 45o,

and

d = effective depth



DEVELOPMENT LENGTH OF BARS

26.2 Development of Stress in Reinforcement

The calculated tension or compression in any bar at any section shall be

developed on each side of the section by an appropriate development length or

end anchorage or by a combination thereof.

Development length Ld is given by

Ld = φσst /4τbd

φ = nominal diameter of bar, τbd = design bond stress

σst = stress in bar at the section considered at design load

� Design bond stress in limit state method for plain bars in tension is given

in clause 26.2.1.1

� For deformed bars conforming to IS 1786 these values are to be increased

by 60 %.

� For bars in compression, the values of bond stress for bars in tension is to

be increased by 25 percent

B. Shear reinforcement (STIRRUPS)

Development length and anchorage requirement is satisfied, in case of stirrups

and transverse ties, when Bar is bent –

• Through an angle of at least 90 degrees (round a bar of at least its own dia)

& is continued beyond for a length of at least 8 φ, or



• Through an angle of 135 degrees & is continued beyond for a length of at

least 6 φ or

• Through an angle of 180 degrees and is continued beyond for a length of

at least 4 φ

DUCTILE DETAILING AS PER IS: 13920

• Provisions of IS 13920-1993 shall be adopted in all reinforced concrete

structures which are located in seismic zone III, IV or V

The provisions for reinforced concrete construction given in IS 13920-1993

shall apply specifically to monolithic reinforced concrete construction. Precast

and/or prestressed concrete members may be used only if they can provide the

same level of ductility as that of a monolithic reinforced concrete construction

during or after an earthquake.

The definition of seismic zone and importance factor are given in IS 1893-2002.

CODAL PROVISIONS OF IS 13920

5.2 For all buildings which are more than 3 storeys in height, the minimum grade

of concrete shall be M20 (fck = 20 MPa ).

5.3 Steel reinforcements of grade Fe 415 (see IS 1786 : 1985 ) or less only shall

be used. However, high strength deformed steel bars, produced by the thermo-

mechanical treatment process, of grades Fe 500 and Fe 550, having elongation

more than 14.5 percent and conforming to other requirements of IS 1786 : 1985

may also be used for the reinforcement.

Flexure Members

6.1.2 The member shall preferably have a width-to-depth ratio of more than 0.3.

6.1.3 The width of the member shall not be less than 200 mm.

6.1.4 The depth D of the member shall preferably be not more than 1/4 of the

clear span.

6.2 Longitudinal Reinforcement

6.2.1 a) The top as well as bottom reinforcement shall consist of at least two bars

throughout the member length.

b) The tension steel ratio on any face, at any section, shall not be less than ρmin =

0.24(fck)1/2 /fy ; where fck and fy are in MPa.

6.2.2 The maximum steel ratio on any face at any section, shall not exceed ρmax

= 0.025.

6.2.3 The positive steel at a joint face must be at least equal to half the negative

steel at that face.



6.2.4 The steel provided at each of the top and bottom face of the member at any

section along its length shall be at least equal to one-fourth of the maximum

negative moment steel provided at the face of either joint

6.2.6 The longitudinal bars shall be spliced, only if hoops are provided over the

entire splice length, at a spacing not exceeding 150 mm 6.3

Web Reinforcement

6.3.1 Web reinforcement shall consist of vertical hoops. A vertical hoop is a

closed stirrup having a 135° hook with a 10 diameter extension (but not < 75 mm)

at each end that is embedded in the confined core

6.3.2 The minimum diameter of the bar forming a hoop shall be 6 mm. However,

in

beams with clear span exceeding 5 m, the minimum bar diameter shall be 8 mm.

6.3.4 The contribution of bent up bars and inclined hoops to shear resistance of

the section shall not be considered.

6.3.5 The spacing of hoops over a length of 2d at either end of a beam shall not

exceed (a) d/4, and (b) 8 times the diameter of the smallest longitudinal bar;

however, it need not be less than 100 mm. elsewhere, the beam shall have vertical

hoops at a spacing not exceeding d/2.

Columns

7.1.2 The minimum dimension of the member shall not be less than 200 mm.

However, in

frames which have beams with centre to centre span exceeding 5 m or columns of

unsupported length exceeding 4 m, the shortest dimension of the column shall not

be less than 300 mm.

7.1.3 The ratio of the shortest cross sectional dimension to the perpendicular

dimension shall preferably not be less than 0.4.

7.2 Longitudinal Reinforcement

7.2.1 Lap splices shall be provided only in the central half of the member length.

It should be proportioned as a tension splice. Hoops shall be provided over the

entire splice length at spacing not exceeding 150 mm centre to centre. Not more

than 50 percent of the bars shall be spliced at one section.

7.3 Transverse Reinforcement

7.3.1 Transverse reinforcement for circular columns shall consist of spiral or

circular hoops. In rectangular columns, rectangular hoops may be used. A

rectangular hoop is a closed stirrup, having a 135° hook with a 10 diameter

extension (but not < 75 mm) at each end that is embedded in the confined core.

7.3.3 The spacing of hoops shall not exceed half the least lateral dimension of the

column, except where special confining reinforcement is provided, as per 7.4.

7.4 Special Confining Reinforcement

This requirement shall be met with, unless a larger amount of transverse

reinforcement is required from shear strength considerations.



7.4.1 Special confining reinforcement shall be provided over a length lo from

each joint face, towards midspan, and on either side of any section, where flexural

yielding may occur under the effect of earthquake forces. The length ‘lo’ shall not

be less than

(a) larger lateral dimension of the member at the section where yielding occurs,

(b) 1/6 of clear span of the member, and (c) 450 mm.

7.4.2 When a column terminates into a footing or mat, special confining

reinforcement shall extend at least 300 mm into the footing or mat.

7.4.6 The spacing of hoops used as special confining reinforcement shall not

exceed 1/4 of minimum member dimension but need not be less than 75 mm nor

more than 100 mm.

8 JOINTS OF FRAMES

8.1 The special confining reinforcement as required at the end of column shall be

provided through the joint as well, unless the joint is confined as specified by 8.2.

8.2 A joint which has beams framing into all vertical faces of it and where each

beam width is at least 3/4 of the column width, may be provided with half the

special confining reinforcement required at the end of the column. The spacing of

hoops shall not exceed 150 mm.



As per IS-875(Part-1)-1987

0.16-0.23STEEL WORK -ROOFING

0.65MANGLORE TILES

0.15GI SHEET -ROOFING

0.16AC SHEET -ROOFING

78.5STEEL

18LIME -PLASTER

21CEMENT-PLASTER

6-10TIMBER

21-27STONE MASONRY

19-20BRICK MASONRY

25REINFORCED CONCRETE

24PLAIN CONCRETE

kN/m2kN/m3

UNIT WEIGHT

MATERIAL

DEAD LOADS – UNIT WEIGHTS OF SOME MATERIALS/BUILDING COMPONENTS

4.0�STAIRS-NOT LIABLE TO OVER CROWDING

5.0- LIABLE TO OVER CROWDING

7.5–HEAVY VEHICLES

4.0�GARRAGES –LIGHT VEHICLES

10.0�STACK ROOM IN LIBRARIES ,BOOK STORES

5.0-10� FACTORIES & WAREHOUSES

5.0- WITHOUT FIXED SEATING

4.0- WITH FIXED SEATING

� SHOPS,CLASS ROOMS,WAITINGS ROOMS,

RESTAURANTS,WORK ROOMS,THEATRES ETC

4.0– WITHOUT SEPARATE STORAGE

2.5� OFFICIAL – WITH SEPARATE STORAGE

2.0� RESIDENTIAL

LIVE LOAD

(kN/m2)TYPE OF FLOOR USAGE

LIVE LOADS ON FLOORS AS PER IS-875(Part-2)-1987



0.75� ROOF WITHOUT ACCESS

1.5� ROOF WITH ACCESS

LIVE LOADS ON ROOFS

12.0� WEATHER MAKER

10.0� MDF ROOM

6.0� OMC ROOM,DDF ROOM,POWER PLANT,

BATTERY ROOM

6.0� SWITCH ROOM(NEW TECHNOLOGY)

LIVE LOAD (kN/m2)TYPE OF FLOOR USAGE

LIVE LOADS ON FLOORS OF T.E.BLDGS



3.0 Steps for Design of a Multi-Storeyed Building:-

Manual Method of Analysis & Design:-

Step1: Study of architectural Drawings:- Before proceeding for

structural design of any building it is ensure that approved working drawings

are available in the office. All working drawings i.e. each floor plan, elevations,

sections, are studied thoroughly & discrepancy if any brought to the notice of

concern Architect for rectification/correction. The problems coming in

finalization of structural configuration may also be intimated to concern

Architect for rectification/correction if any.

Step2: Finalization of structural Configuration. After receiving

corrected working drawing from the architectural wing, the structural system is

finalized. The structural arrangements of a building is so chosen as to make it

efficient in resisting vertical as well as horizontal loads due to earthquake. The

span of slabs co chosen that thickness of slab 100-150mm and slab panels,

floor beams, and columns, are all marked and numbered on the architectural

plans. Now the building is ready for structural design to start.



ISOMETRIC VIEW OF FRAMED STRUCTURE

Step3: Load Calculation and analysis. For each floor or roof, the

loading intensity of slab is calculated taking into account the dead load of the

slab, finish plaster, etc. including partitions and the live load expected on the

floor, depending on the usage of the floor or roof. The linear loading of beams,

columns, walls, parapets, etc. also calculated.

Step3 (a): Preliminary Sizes of structural members. Before proceeding

for load calculation preliminary sizes of slabs, beams,& columns decided. In

manual load calculation preliminary sizes of structural members should be

judicially fixed as once load calculation & analysis is done it is not easy to revise

the same. But in computer aided analysis & design it can be revised easily.

• Slab:- The thickness of the slab decided on the basis of span/d ratio

assuming appropriate modification factor.

• Beam : The width the beam generally taken as the width of wall

i.e 230 or 300 mm. The width of beam is help full in placement of

reinforcement in one layer & more width is help full in resisting

shear due to torsion. The depth of beam is generally taken as 1/12

th (for Heavy Loads) to 1/15 th (for Lighter Loads) of span.

• Column:- Size of column depends upon the moments from the both

the direction and the axial load. Preliminary Column size may be

finalized by approximately calculation of axial load & moments.

Procedure for vertical load calculation on Columns:

Step(i): First, the load from slab (including Live load & Dead Load) is

transferred on to the adjoining beams using formulas given below|:- For computation of shear force on beams & reactions on columns, an

equivalent uniformly distributed load per linear meter of beam may be taken as : Equivalent u.d.l. on short beam of slab panel = w B/4.0 Equivalent u.d.l. on long beam of slab panel = w B/4 x [2-(B/L)] Where w is the total load on the slab panel in Kn/Sqm & L & B are long span & short spans of slab panel respectively.



Step(ii): Over this load, the weight of wall (if any), self weight of beam etc. are added to get the load on beam (in running metre).

Step(iii):The load (in running metre) on each beam is calculated as in Step 1 &

Step 2.

Step(iv):Then the loads from the beams are transferred to the columns. Step(v):Step (i) to Step (v) is repeated for each floor. Step(vi):These loads at various floors on each column are then added to get the

total loads on each column, footing and the whole building.

Step4: HORIZONTAL (SEISMIC) LOAD CALCULTAION:

The Horizontal Load Calculation or the Load Calculations for Seismic case is

carried out as per the Indian Standard Code IS:1893-2002.

The loads calculated in Para-II above at various floor levels are modified as per

the requirement of Para 7.3.1 of IS:1893-2002.

The Seismic Shear at various floor levels is calculated for the whole Building

using the values from IS 1893-2002.

Calculation o f horizontal loads on buildings (As per is-1893-2002)

Sample example for horizontal load calculation

(I) BUILDING IS ON SEISMIC ZONE-IV

(II) FOUNDATION TYPE ISOLATED FOOTINGS As per clause 7.5.3 o f IS-1893-2002 Design base shear vb

V

b = A h W (F)

Where Ah = Design Horizontal acce lerat ion spectrum value as per

6.4.2 of the code

= (Z/2) (I/R) (Sa/g)

Where Z = Zone factor as per table 2 o f IS Code (1893-2002) = 0.24 (in this case) I = Importance factor as per table 6 of IS-1893-2002)

= 1.5 (Assuming tha t the b ldg. is T.E. Bldg.)



R = Response reduction factor as per table 7 of IS code

= 3.0 (for ordinary R.C. Moment res ist ing frame

(OMRF)

(Sa /g) = Average response accelerat ion coeffic ient for soil

typ e & appropriate natural periods and lamping of the

structure.

For calcu lat ing of (Sa/g) value as above we have to calculate va lue

of T i.e. Fundamental Nat ional Per iod (Seconds) (Clause 7.6 of IS

Code)

T = 0.075 h

0 .7 5 (For RC Frame build ing)

= 0.0 85 h 0 . 7 5 (For Steel frame build ing)

h = Height of building in Meter

In case of building with brick in f ills walls.

T = 0.09 h /d 1 /2

Where h = height of bu ild ing in Meter

and d = Base d imension of the bu ilding at the plinth leve l in

Meter along the considered direction o f the latera l

fo rce.

Value of (Sa/g) is to be read from f ig 2 on page 16 of IS

Code depending upon Soil condit ion & Fundamenta l Natural period

T.

Or the va lue of (Sa/g) may be calcu lated on the basis o f

Following.

Formulas:- (i) For rocky, or hard soil sites

(Sa/g) = 1+15 T if 0 .00≤T≤ 0 .10

= 2.50 if 0 .10≤T≤ 0 .40

=1.00/T if 0 .40≤T≤ 4 .00

(ii) For medium soil sites (Sa/g) = 1+15 T if 0.00≤T≤ 0 .10

= 2.5 0 if 0 .10≤T≤0.55

= 1.36/T if 0 .55≤T≤ 4 .00

( iii) For soft so il sites (Sa/g) = 1+15 T if 0.00≤T≤<0.10



= 2.50 if 0 .10≤T≤0.67

=1.67/T if 0 .67≤T≤ 4 .00

W= Seismic weight of the building as per clause 7.4.2 of the code.

A

B

C

2 b

ays @

7.5

m C

/C

TYPICAL FLOOR PLAN

Bldg. is three storey with Each

storey of 5.0m height

4 bays @ 4.0 m C/C

164.9

157.99

47.01

Frame with EQ Loads


Calculating Seismic weight of building per frame for frame (B)

Length bldg = 16.00 M

C/C distance of frames = 7.50 M

Densit y of R.C.C = 25 KN/m3

Floor slab = 0.15×16.00×7 .50×25 = 450 KN (A)

Co lumn below slab = 0.300×(0.70-0.15)×16.00×25

= 0.30×0 .55×16 .00×25=66 KN (B)

Co lumns = 0.30×0 .60×(5.00+5.00)/2 ×25× 5 = 112.5 KN (C) Live load = 600 kg/m2 = 6.00 kn/m2

As per table 8 of code when live load is above 3.00 kn/m2 50%

of live load to be considered for lamp mass ca lcu lat ion.

Lump mass at First Floor = 0.50×6.00×16×7.50 = 360 KN (D) To tal lamp mass f irst f loor & second floor (Assuming same L.L. on S.F.)

(A)+(B)+(C)+(D) = 450+66+112.50+360

= 988.50 KN

(ii) Wight lamped at terrace Floor slab:-

= 0.13×16.00×7.50×25 = 390 KN (E)

Beam below slab = 0.23×(0.60-0.15)×16.00×25

= 0.23×0.45×16.00×25

= 41 .4.KN (F)

Co lumns = 0.30×0.600×5.00 /2×25×5 = 56.25 KN (G) L.L. = Nil During Earthquake = 0. ( As per the clause 7.3.2 o f the

code the imposed load on roof need not to be considered )

To tal lamped mass at terrace level = (E)+(F)+(G)

= 390+41.40+56.25=487.65 KN

Total weight o f building per framed per inner frame F.F = 988.50 KN S.F. = 988.50 KN Terrace = 487.65 KN 2464.65 KN


Putting a ll values in Formulas (F) Vb = Design base shear = (z/z)(I/R(Sa/g) w Value of T = 0.09 h/Vd

H = Height of bldg. = 15.00 m (3x5.0=15.00m )

d = 16.00 m

T = 0.09×15.00/V 16.00 = 0.3375 = 0.34

For med ium soils For T = 0.34 Sa/g = 2.50 Vb = 0.24 /2×1.50 /3.00×2 .50×2465.65 = 369.85 KN Distribution base shear is done using formula (clause 7.7)

F i = w i h i 2 / ∑ w j h j

2 x Vb

Where Fi = Design lateral force at f loor i

W i = Seismic weight of floor i

hi = he ight of f loor in m from base.

n = number of story’s in the build ing is equal to number of levels at

which masses are located.

Vb = 369.85 KN

Floor W i KN h i (m) W i h i 2 F i

F.F. S.F.

Terrace level

988.50 988.50 487.65

6.00 11.00 16.00

35586 119608.5 124800

47.01 KN 157.99 164.85

∑w i hi 2 = 279994.5 ∑ = 369.85 KN

Step5. VERTICAL LOAD ANALYSIS:

a) GENERAL:

The skeleto n frame work of a mult i sto ried R.C.C. framed

structu re is made up of a system of columns, beams and slabs. It is

presumed that the re inforcements are a lwa ys so arranged that all joints

of the frame are mono lithic.

In view of the uncertain property of materia l creep, shr inkage

and a number of approximate s implifying assumptions made in the

detailed analysis of multi storied framed structures (such as condit ions


of end restraints etc.) it is co nsidered suffic ient to obtain reasonable

accurac y of analysis fo r the design of structu re. If the normal moment

distr ibution is applied to all joints, the work invo lved is enormous.

However with certain assumptions, it is possib le to analyze the frames

and get resu lts which will be adequate for design purposes.

To simplify ana lysis the three dimensional mult isto ried R.C.C.

framed structure are considered as combinat ions of p laner framed in

two d irections. It is assumed that each of these p laner frames act

independent ly o f the frames.

Procedure for Frame analysis for ca lculation of moments in

Columns & beams:

Step(i) : Fir st, the load from slab (including Live load & Dead Load) is

transferred on to the adjo ining beams using formulas given

below|:-

For computation of Bend ing Moments in beams , an equ ivalent

uniformly d istr ibuted load per linear meter of beam may be taken as :

Equiva lent u.d.l. on short beam of slab panel = w B/3.0

Equiva lent u.d.l. on long beam of slab panel = w B/6 x [ 3-(B/L)2 ]

where w is the to tal load on the slab panel in Kn/Sqm & L & B are

long span & short spans of s lab panel respect ive ly.

Step(ii) : Over this load, the weight of wall ( if any), self weight of

beam etc. are added to get the load on beam (in running Meter).

Step(iii) :The load (in running Meter) on each beam is ca lculated as in

Step 1 & Step 2.

Step(iv):Step (i) to Step (iii) is repeated for each floor Step(v):Then these loads are used as u.d.l on a part icular frame for

ana lysis by moment distr ibution method as described in the next

sect ion.

b) METHOD OF ANALYSIS:


Analysis of large framed structures beams too Cumbersome with

the class ical method of structure ana lysis such a Clape yron’s theorem

of three moments, Cast ingiliano’s therefore o f least work, Poison’s

method of vir tual work etc. Therefore, it become necessar y to evolve

simpler methods.

Some of these are :-

a.) Hardy cross method of moment distr ibution.

b.) Kani’s method of iteration.

c) HARDY CROSS METHOD OF MOMENT DISTRIBUTION:

In this method, the ‘balancing’ and ‘carry-over’ const itu te one c ycle

and it has been found the ‘carry-over’ values converge fast enough to

become quite insignificant a fter four c ycle of operation. it is,

therefore, o ften adequate to stop the computation after four cyc les.

The frame is analyzed by this method either:

i. Floor-wise assuming the co lumns to be fixed for ends.

or i i. Taking the frame as a whole. The whole frame analysis can be

carried out for several a lternat ive loading arrangements for

obtaining maximum posit ive and negat ive bending moment.

Generally frames are analyzed floor-wise for the worst

condit ions of loading.

The method is described in the fo llowing steps. Step1: Calculate the st iffness of a ll members. Enter them in the

calculat ion scheme.

Step2: Calculated the d istr ibution factor at a ll joints from the

st iffness. Enter them in the calculat ion scheme.

Step3: Look the jo ints and calculate the f ixed-end moments. Enter

them in the calcu lat ion scheme.

Step4: Unlock the joint one by one by applying imaginary external moments

at each joint which nullifier the unbalanced moment at the joint.

Distribute the imaginary external moment among all members


Meeting at the joint in proportion to the ir re lat ive

st iffness and enter these va lue in the scheme. This

operation is called balancing.

Step5: Enter the carry-over moments at the far in the scheme.

Step6: Repeat steps 4 & 5, t ill the carry-over moments become

insignificant.

Step7: Balance the unbalanced moment obtained from the last

carry-over operation.

Step8: Add the init ial f ixed-end moments, balancing moments and

carry-over moments to get the f inal end moments in beam

& co lumns.

A sample of moment distr ibu tion method is shown on next two pages.

CALCUL ATION OF DISTRIBUTION F ACTOR FOR FRAME ANALYSIS

SNO.

JOIN T MEMBER Size in

Cm

Moment o f Ine rt ia Cm4 ( I)

Le ng th of me

mber

Cm

K=I/L

Sum K D.F.

1

B D

A- II I R igh t beam 30 45 227812. 50 600

379.69

1030. 58

0. 37

Lower Col. 30 45 227812. 50 350 650.

89 0. 63

2

B-II I

Lef t beam 30 45 227812. 50 600 379.

69

1802. 01

0. 21

Righ t beam 30 60 540000. 00 700 771.

43 0. 43

Lower Col. 30 45 227812. 50 350 650.

89 0. 36

3 C-II I

Lef t beam 30 60 540000. 00 700 771.

43

1422. 32

0. 54

Lower Col. 30 45 227812. 50 350 650.

89 0. 46

4

A- II

Upper co l. 30 45 227812. 50 350 650.

89

1681. 47

0. 39

Righ t beam 30 45 227812. 50 600 379.

69 0. 23

Lower Col. 30 45 227812. 50 350 650.

89 0. 39

5

B-II

Lef t beam 30 60 540000. 00 600 900.

00

2973. 21

0. 30

Upper co l. 30 45 227812. 50 350 650.

89 0. 22

Righ t beam 30 60 540000. 00 700 771.

43 0. 26


Lower Col. 30 45 227812. 50 350 650.

89 0. 22

6

C-II

Lef t beam 30 60 540000. 00 700 771.

43

2073. 21

0. 37

Upper co l. 30 45 227812. 50 350 650.

89 0. 31

Lower Col. 30 45 227812. 50 350 650.

89 0. 31

7

A- I

Upper co l. 30 45 227812. 50 350 650.

89

1572. 99

0. 41

Righ t beam 30 45 227812. 50 600 379.

69 0. 24

Lower Col. 30 45 227812. 50 420 542.

41 0. 34

8

B-I

Lef t beam 30 45 227812. 50 600 379.

69

2344. 42

0. 16

Upper co l. 30 45 227812. 50 350 650.

89 0. 28

Righ t beam 30 60 540000. 00 700 771.

43 0. 33

Lower Col. 30 45 227812. 50 420 542.

41 0. 23

9

C-I

Lef t beam 30 60 540000. 00 700 771.

43

1964. 73

0. 39

Upper co l. 30 45 227812. 50 350 650.

89 0. 33

Lower Col. 30 45 227812. 50 420 542.

41 0. 28

FRME ANALYSIS BY MOMENT DISTRIBUTION METHOD

III

U.d . l .

28

6 .00

U.D.L 24

7.00

0 .63 0.37 30X45 0 .21 0.43 0 .36 30X60 0.54

0 .46

0

-84 .0

0 In t .FEM 84.00 0 -98.00 In t .FEM 98.00 0

bal . 52 .92

31 .08 2 .94 6.02 5 .04

-52.92

-45.08

C. O. 14 .63 1.47 15 .54 0.74 -26.46 2.52

-12.66

BAL -10.14 -5.96 2 .14 11.06 3 .66 5.48

4 .66

Tot al 57 .41

-57 .4

1 104.62 17.82 -115.76 53.08

-53.08


Tot al 38 .28 4.06

-34.87

BAL -17.43

BAL -0.42

BAL 12.99

C. O. 26 .46

C.O. 3.01

C.O. -

22.54

29 .25 1.47

-25.32

0 0 0

0 .39 0.22

0 .31

II

25

6 .00

20

7.00

0 .39 0.23 0.3 0.22 0 .26 0.37

0 .31

0

-75 .0

0 In t .FEM 75.00 0 -81.67 In t .FEM 81.67 0

29 .25

17 .25 2 1.47 1 .73

-30.22

-25.32

C. O. 17 .22 1 8 .63 5.39 -15.11 0.87

-20.22

BAL -17.43

-10 .2

8 -0 .58 -0.42 -0 .5 15 .5 12.99

29 .04

-67 .0

3 85 .05 6.44 -95.55 67.82

-32.55

Tot al 41 .81 15.18

-51.01

BAL -7 .26 3.66

2 .08

C. O. 14 .63 0.74

-12.66

34 .44 10.78

-40.43

0 0 0

0 .41 0.28

0 .33

I

28

6 .00

30

7.00


0 .34 0.24 0 .16 0.23 0 .33 0.39 0 .2

8

0

-84 .0

0 In t .FEM 84.00 0 -122.50 In t .FEM

122.50 0

28 .56 20 .1

6 6 .16 8.86 12 .71 -

47.78

-34.

3

C.O. 0 3.08 10 .08 0 -23.89 6.36 0

BAL -6 .02 -4.25 2 .09 3.01 4 .31 2.46 1 .7

6

22 .54

-65 .0

1 102.33 11.87 -129.37 83.54

-32.54

A B C

Step6. HORIZONTAL LOAD ANALYSIS:-

Frame analysis for horizonta l loads calculated in step 4 is carr ied ou t by using :- (a)Approximate Method:-

i) Cant ilever method.

ii) Portal method .

Approximate methods are used for preliminar y designs only. For f ina l

design we may use exact method i.e ( i) Slope deflect ion or matrix

methods (ii) Factor method.

We will no t d iscuss these methods in detail as now modern computer package as STAAD PRO is ava ilable fo r analysis. Step7: DESIGN OF COULMN,FOUNDATIONS, BEAMS & SLABS:

After load calcu lat ion & analysis for vert ica l & horizonta l

loads, design o f Columns ,Foundations, Beams, Slabs and are to be

carried out as per the var ious clauses of IS codes, IS 456-2000,

IS:1893-2002, IS:13920-1993 etc.

The Design of Co lumn, Foundation, Beams and Slabs are discussed in

details in following sect ion.


A. Design o f columns: - With the knowledge of ( i) Vertical load

(ii) Moments due to horizontal loads on either axis;( iii) Moments due

to vertical loads on either axis, act ing on each co lumn, at a ll floor

leve ls of the building, columns are designed by charts of SP-16(Design

Aids) with a load factor of 1.5 for vertical load effect and with a load

factor of 1.2 for the combined effects of the vertical and the horizonta l

loads. The step confirms the s ize of columns assumed in the

architectural drawings. The design o f each column is carr ied out from

the top of foundation to the roof, varying the amount of stee l

reinforcement fo r suitable groups for ease in design. Further,

slenderness effects in each sto rey are considered for each column

group.

Important Considerations in design o f Columns:-

(i)Effective height of column:- The effective height of a column is

defined as the height between the po ints o f contra f lexure of the

buckled co lumn. For effect ive column height refer table 28 (Annexure

E) of IS: 456-2000.

For framed structure effect ive height of column depends

upon re lat ive st iffness of the column & various beams framing into the

column at its two ends. (Refer Annexure E of IS: 456-2000.)

(ii)Unsupported Length: - The unsupported length l, o f a compression

member sha ll be taken as the c lear dis tance between end restraints

except that :-

In beam & slab co nstruct ion, it sha ll be the clear d istance between

the f loor & under side of the shallower beam framing into the columns

in each direct ion at the next higher f loor leve l.

(iii) Slenderness limits for columns: - The unsupported length

between end restraints shall no t exceed 60 times the least latera l

dimensio n of a column.

(iv) Minimum Eccentricity: - All columns sha ll be designed for

minimum eccentric it y equal to unsupported length of column/500 plus

least latera l dimensio n/30, subject to a minimum of 20 mm.

Or emi n ≥ l/500+ D/30 ≥ 20 mm


Where l= unsupported length of column in mm.

D=Lateral dimension o f column in the d irect ion under

considerat ion in mm.

(v)Design Approach: - The design o f column is complex s ince it is

sub jected to axia l loads & moments which may very independently.

Co lumn design requ ired:-

I. Determinat ion o f the cross sect iona l dimension.

II. The area of longitud inal steel & it s distr ibution.

III. Transverse stee l.

The maximum axial load & moments act ing a long the length of

the column are considered for the design of the column sect ion either

by the working stress method or limit state method.

The transverse reinfo rcement is provided to impart effect ive

lateral support aga inst buckling to ever y longitudina l bar. It is either in

the form of circular r ings of polygonal links (lateral t ies) .



B. Design o f foundations: - With the knowledge o f the column loads

and moments at base and the soil data, foundations for columns are

designed

The following is a lis t o f different t ypes of foundations in o rder

to preference with a view to economy: (i) Ind ividual footings ( ii)

Combinat ion of individual and combined footings ( iii) Strip footings

with reta ining wall act ing as str ip beam wherever applicable; ( iv) Raft

foundations of the t ypes (a) Slab (b) beam-slab.

The brick wall footings are also designed at this stage. Often,

plinth beams are provided to support brick walls and also to act as

earthquake t ies in each p rincipal d irect ion. Plinth beams, retaining wal l

if any, are a lso designed at this stage, being considered as part of

foundations.

Important Considerations in design o f Foundations:-

a) Introduction: - Foundations are structural elements that transfer loads from the bu ilding or ind ividual column to the earth. If these loads are to be properly transmitted, foundations must be designed to prevent excessive sett lement or rotation, to minimize differentia l sett lement and to provide adequate safety aga inst slid ing and over turning.

b) Depth of foundation:-

Depth of foundation below ground level may be obtained by using Rankine's

formula

2 p 1 – Sin Ø h = - - - - - - - - - - - - - - - γ 1 + Sin Ø Where h = Minimum dep th of foundation

p = Gross bearing capacit y

γ = Densit y of soil

Ø = Angle of Repose of soil


c) Recommendations o f IS 456 -2000, limit state design, bending,

shear, cracking & development

i) To determine the area required for proper transfer of total load on the soil, the

total load (the combination of dead, live and any other load without multiplying it

with any load factor) need be considered.

To tal Load including Self Weight Plan Area o f foo ting = ------------------------------------- Allowable bearing capacit y of soil

i i) IS 1904 – 1978, Code o f Practice for Structura l Safety of Buildings: shallow foundation, sha ll govern the genera l deta ils.

i i i) Thickness of the edge of footing:-(Reference c lause 34.1 .2) The thickness at the edge shall not be less than 15 cm for footing on soils. iv) Dimension of pedestal:-

In the case of plain Cement Concrete pedestals, the angle between the plane passing through the bottom edge of the pedestal and the corresponding junction edge of the column with pedestal and the horizontal plane shall be governed by the expression.

100 qo Tan α (should not be less than) 0.9 x ----------- + 1 Fck

Where qo = Calculated maximum bearing pressure at the base of the

pedestal/ footing in N/mm2 fck = Character ist ic strength of concrete at 28 days in N/mm2


(v) Bending Moment

(Reference Clauses- 34.2.3 .1 & 34 .2.3.2)

FAC E OF PEDESTA

PLAIN CONCRETE PEDES TAL

α

Co lumn

X Y

X Y

COLUMN

PEDESTAL BASE

FACE OF CO LUMN


ISOLATED COLUMN FOOTING

The bending Moment will be considered at the face of column, Pedestal or wall

and shall be determined by passing through the section a vertical place which

extends completely across the footing, and over the entire area of the footing or,

one side of the said plane.

(vi)Shear

(Reference Clause 33.2.4.1)

The shear strength of footing is governed by the following two factors:-

a) The footing acting essentially as a wide beam, with a potential diagonal

crack intending in a plane across the entire width, the critical section for

this condition shall be assumed as a vertical section located from the face

of the column, pedestal or wall at a distance equal to the effective depth of

the footing in case of footings on soils.

FOR ONE WAY BENDING ACTION

For one way shear action, the nominal shear stress is calculated as follows:-

Vu

τv = -------

b.d

Where

τv = Shear stress

Vu = Factored vertical shear force

b = Breadth of critical section

d = Effective depth

τv < τc ( τc = Design Shear Strength of Concrete Based on % of longitudinal tensile reinforcement refer Table 61 of SP-16)


CRITICAL SECTION FOR ONE -WAY SHEAR

(FOR TWO WAY BENDING ACTION)

For two may bending action, the following should be checked in punching shear. Punching shear shall be around the perimeter 0.5 time the effective depth away from the face of column or pedestal. For two way shear action, the nominal shear stress is calculated in accordance with lause 31.6.2 of the code as follows:-

Vu

τv = ----------

b0.d

Where

τv = Shear stress

b0 = Periphery of the critical section

d = Effective depth

Vu = Factored vertical shear force

B

A

C R IT IC AL SE C T ION

d

d


When shear reinforcement is not provided, the nominal shear stress at the critical section should not exceed [Ks. τc]

Where

Ks = 0.5 + Bc (But not greater than 1)

Short dimension of column or pedestal

Bc = ----------------------------------------------------

Long dimension of column or pedestal

τc = 0.25 fek N/mm2

Note:-It is general practice to make the base deep enough so that shear reinforcement is not required.

(vii)Development Length

(Reference Clause 34.2.4.3)

The critical section for checking the development length in a footing shall be assumed at the same planes as those described for bending moment in clause 34.2.3 of code (as discussed 4.5 of the handout) and also at all other vertical planes where abrupt changes of section occur.

(viii) Reinforcement:- The Min % of steel in footing slab should be 0.12% & max spacing should not be more than 3 times effective depth or 450 mm whichever is less. (Reference Clause 34.3) Only tensile reinforcement is normally provided. The total

reinforcement shall be laid down uniformly in case of square footings. For rectangular footings, there shall be a central band, equal to the width of the footings. The reinforcement in the central band shall be provided in accordance with the following equation.

Reinforcement in central Band width 2

-------------------------------------------------- = ------

Total reinforcement in short direction B + 1

Where Long side of footing

B = --------------------------- Short side of footing


(ix)Transfer of Load at the Base of Column

(Reference Clause 34.4)

The compressive stress in concrete at the base of column or pedestal shall be

transferred by bearing to the top of supporting pedestal or footing.

The bearing pressure on the loaded area shall not exceed the permissible

bearing stress in direct

A1

Compression multiplied by a value equal to ------

A2

but not greater than 2

Where

A1 = Supporting area for bearing of footing, which is sloped or

stepped footing may be taken as the area of the lower base of

the largest frustum of a pyramid or cone contained wholly with

in the footing and having for its upper base, the area actually

loaded and having side slope of one vertical to two horizontal.

A2 = Loaded area at the column base.

For limit state method of design, the permissible bearing stress shall be = 45 fek

4.91 If the permissible bearing stress is exceeded either in column concrete or in

footing concrete, reinforcement must be provided for developing the excess force.

The reinforcement may be provided either by extending the longitudinal bars into

the footing or by providing dowels in accordance with the code as give in the

following:-

1) Minimum area of extended longitudinal bars or dowels must be 0.5% of

cross sectional area of the supported column or pedestal.

2) A minimum of four bars must be provided.

3) If dowels are used their diameter should not exceed the diameter of the

column bars by more than 3 mm.

4) Enough development length should be provided to transfer the

compression or tension to the supporting member.

5) Column bars of diameter larger than 36 mm, in compression only can be

dowelled at the footing with bars of smaller diameters. Te dowel must

extend into the column a distance equal to the development length of the

column bar. At the same time, the dowel must extend vertically into the

footing a distance equal to the development length of the dowel.


C. Design of Floor slabs:- . Design o f f loor s labs and beams is taken up

with the First Floor & upwards .The s labs are designed as one-wa y or

two-way panels, taking the edge condit ions o f the supporting edges in

to account, with the loading alread y dec ided as per functional use of

slab panel.

The design of floor slab is carried out as per clause 24.4 &

37.1.2 & Annexure D o f IS: 456-2000. The Bending moment

coefficients are to be taken from table- 26 of the code depending upon

the support condit ion & bending moment calcu lated & reinforcement

steel may be calculated from the charts of SP-16. The slab design for

particular floor may be done in tabular form as shown below.


SLAB DESIGN

Name of

project:-

Level of s lab

Slab ID

Edge condit io n

Tota l load in

KN/Sq.m w

Short

spa n lx m

long spa n ly m

ly/lx

1.5 *w* lx *lx

sl

ab

th

ic

kne

s

s

in

mm

Shor t span Mome nt KN-M

Steel in short span

Long spa n moment KN-

M

St ee l in

long span

α

x

(

+

)

Α

x

(

-

) mux+

m

u

x

-

S

t

e

e

l

Α

y

(

+

)

α

y

(-

)

m

u

y

+

m

u

y

-

Stee

l

1 2 3 4 5 6 7 8 9 1

0 11= 7 x 9

1

2

=

7

x1

0

1

3 1

4

1

5

1

6

=

7

x1

4

1

7

=

7

x1

5

1 8

S1

Two Adj . Edge. Discont . (Case No.4)

8.50

3.50

5.25 1.5

156.80

120

0.056

0.075

8.78

11.76

0.035

0.047

5.49

7.37

S2


Method of calculation of steel from Tables of SP-16 for slab design

Determine the main reinforcement required for a slab with the following data:

Factored moment Mu 9.60 kN.mper Metre width

Depth of slab 10 cm

Concrete mix M 20

Characteristic strength a) 415 N/mm2

METHOD OF REFERRING TO TABLES FOR SLABS Referring to table 35 (for fck=20 & fy = 415 N/mm2), directly we get the following

reinforcement for a moment of resistance of 9.60 kN.m per Metre width:

8 mm dia at 13 cm spacing or 10 mm dia at 210 cm spacing Reinforcement given in the table is based on a cover of 15 mm or bar diameter which-

ever is greater.

Check for Deflection:-Slab is also checked for control of Deflection as per clause

23.2.1, 24.1 & Fig 4. of the IS:456-2000.

D. Design of floor Beams:-. The beams are designed as continuous beams,

monolithic with reinforced concrete columns with their far ends assumed fixed. The

variation in the live load position is taken into account by following the two-cycle

moment distribution. the moments are applied a face correction to reduce them to the

face of the members. The moments due to horizontal loads are added to the above

moments. Each section of the beam is designed for load factor of 1.5 for vertical load

effect and with a load factor of 1.2 for the combined effects of the vertical and the

horizontal loads.

The effect of the shear due to vertical and horizontal loads is also similarly

taken care of. It may be noted that the shear component due to wind or earthquake

may be significant and it may affect the size and the range of shear stirrups. Bent- up

bars are not effective for earthquake shear due to its alternating nature. The beam

design can be easily done by a computer program which will give reinforcement at

various critical sections along the length of the beam and also shear stirrups required

it saves considerable time and labour of a designer.

In manual method span of a beam is generally designed at three sections i.e

at two supports & at Mid span. The each section is designed for factored Moment,

Shear & equivalent shear for Torsion if any at a section.


Two examples of beam design are given below illustrating calculation of

steel reinforcement with help of SP-16.

Example1.Singly Reinforced Beam

Determine the main tension reinforcement required for a

rectangular beam sect ion with the following data:

Size o f beam 30 X 60 cm

Concrete mix M 20

Character ist ic strength 415 N/mm2 o f reinfo rcement

Facto red moment 170 kN.m

Assuming 20 mm dia bars with 25 mm clear cover, Effect ive depth= 600 – 25 – 20 /2 = 565 mm From Tab le D for fy = 415 N/ mm2

and f c k = 20 N/mm2

Mu, l im/bd2

= 2.76 N/ mm2

= 2.76/1000 X (1000)2

= 2.76 X 103 kN/m2

Mu, l im = 2.76 X 103 bd 2

= 2.76 X 103 X 0.300 X0.565X0.565

= 264.32 kN.m

Actual moment of 170 kN.m is less than Mu, l im. The sect ion is therefo re

to be designed as a s ingly reinfo rced (under-reinforced) rectangu lar

sect ion.

Referring to table 2 of SP-16 we have to ca lculate Mu/bd2 Mu/bd2 = 170 x106 /(300x 565 x565) = 1.78 From Tab le 2. p t =0.556 A st=0.556 x 300x 565/100 =942.42mm2=9.42 cm2

Example2 .Doubly Reinforced Beam (i)Determine the main reinforcements required for a rectangular

with the following data :

Size o f beam 30×60cm

Concrete mix M 20

Character ist ic strength of 415/Nmm2

Reinforcement

Facto red moment 320Kn.m


Assuming 20 mm dia bars with 25 mm clear cover, D=600-25 – 20/ 2 =565 mm From tab le D, fo r fy= 415 N/mm2 and fc k = 20N/mm2 Mu2 lim/bd2= 2.76 N/mm2 = 2.76 × 103 KN/m2 Mu2 lim= 2.76 × 103×0.300×0.565×0.565 = 264.32 KN-M Actual moment of 320 Kn.M is greater than Mu2 lim hence the sect ion

is to be designed as a doubly reinforced section .

Reinforcement from Table 50 Mu/bd2= 320 × 106 / (300×5652 ) = 3.34 N/mm2

d’/d = (25+10)/565 = 0.062

Next higher value of d1 /d = 0.1 , will be used for referr ing to Table 50 For Mu/bd2 = 3.34 and d ’ /d = 0.10, p t

= 1.152, p c = 0.207 Ast = 1.152 x 300x 565 /100 =1952.64 mm2 =19.52 cm2 And Asc = 0.207 x 300x 565/100 =350.86 mm2 =3.51 cm2 (ii) Determine the Shear reinforcement (vertical stirrups ) required

for the same beam section if factored shear force is Vu =250 KN. Shear stress τ v = Vu/bd = 250 × 103 /(300× 565) =1.47 N/mm2

τv < τma x( 2.8 N /mm2) hence sect ion is adequate regarding shear stress. From table 61 for p t=1.15 τc=0.65 N/mm2

Shear capac it y of co ncrete sect ion = τc × b× d = 0.65 × 300 ×565/1000=110 .18 kN Shear to be carried b y st irrups Vu s=Vu - τc × b× d = 250 - 110.18 =139.82 kN Vus/d = 139.82/56.5 = 2.47 kN/cm Referr ing to tab le 62 for steel fy = 415 N/mm2 Provide 8 mm diameter two legged vert ical st irrups at 140 mm spacing.


TABLES

FOR

DESIGN















DETAILING AS PER IS 13920


HOOP SPACING

HOOP SPACING

< d/4 and 8 db

B = BREADTH OF BEAM

db = DIAMETER OF LONGITUDINAL BAR

2d

d

2d

db

50 m max 50 m max

MIN 2 BARS FOR FULL LENGTH

ALONG TOP AND BOTTOM FACE

AS > MIN. Bd

AS < MAX Bd

> d /2

BEAM REINFORCEMENT37

Questions:-

1. Which are the important BIS Codes/handou ts used for structu ral

design of RCC build ings?

2. In which seismic zones provis ions of IS 13920 is to be adopted

fo r all re inforced concrete structu res?

3. (a) What are the basic values o f span to effect ive depth rat ios for

beams as per IS 456 for span upto 10meter for –

( i) Cant ilever

( ii) Simply supported

(iii) Cont inuous

(b) What are the basic values of span to overa ll depth ratios for

two-way slabs upto 3.5 m span & with Fe415 stee l

reinforcement and load ing class upto 3KN/m2?

4. What are the provisions o f IS 456 for nominal cover to meet

durabilit y requirements? As per IS 456 how much minimum

cover should be provided for –

a) Co lumn b) Footing

5. (a) What are the minimum reinforcement provis ion of IS 456 for

beams in respect of:

(i) Tension re inforcement


(ii) Shear re inforcement

(b) What are the IS 456 provisions for maximum reinforcement

in beams for:-

(i) Compression reinfo rcement

(ii) Tension re inforcement

6. What is the maximum permissible spacing for shear

reinforcement in beams? Expla in IS 456 provis ions for side face

reinforcement in beams.

7. How much minimum reinforcement must be provided in s labs?

8. As per IS1893 give formulae for calculating –

a) Design Base Shear (Vb)

b) Design Hor izonta l accelerat ion (Ah)

9. Give formulae for calculat ing t ime period as per IS1893 for –

a) RCC Frame Building b) RCC Building Br ick in f ill walls

10. How vert ical loads on co lumns are ca lculated? Give names of

s impler methods of ana lysis of structures.

11. What is the minimum eccentr icit y for which all columns should

be designed? List out minimum and maximum longitudina l

reinfo rcement required to be provided in co lumns? Give in

brief provisions for maximum spacing of lateral t ies in a

column?

12. What are the cr itical sect ions in isolated footing design fo r the

following:-

a) Bending moment

b ) One way shear

c) Two way shear

13. How many minimum longitudina l reinforcement bars should be

provided in:-

a) Circular co lumn

b) Rectangular column

14. What is the minimum diameter bar that can be used in

longitudina l reinforcement in column?

---------------------

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CH-1 Structural Design of RCC Building Components

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