Bridge Chap2 PDF

8/18/2019 Bridge Chap2 PDF

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CHAPTER 2

Bridge Engineering Concept

2.1 INTRODUCTION

In the early development of highways there were very few of structures built

for crossing the roads. Basically, the roads were built to follow the contour of the

ground. In the design of the road, no visibility in vertical and horizontal direction was

considered. When the roads encounter the water, the bridge was constructed. At the

end of the bridges, the level of the road was raised. Since there were no heavy

equipment to carry the components of the bridges, the bridges were light and The

bridges were narrow so that they can be handled by man-power.

The early steel bridge was designed for short span (about 10 m). The type of

the bridges was truss bridge. The truss members were made of steel angles and plates

and they are connected with rivet. The rived angles and the plates were used to support

the floor beams. Often time timber stringers were used for floors.

Since the advancement of construction techniques and the availability of

heavy equipment and also the need of wider and stronger bridges to carry heavy loads,

the bridge engineers started to develop different type of bridges. From the short span

bridges until the long span of bridges. The steel, concrete and prestressed concrete

bridges are the most popular types of bridges. In this note, we will discuss the loads

and the steel bridges.


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. . Bridges Engineering R. Soegiarso4.

2.2 BRIDGE COMPOSITION AND TERMS

Broadly speaking, we can divide the bridges into three parts. The first part isthe super-structure. This part directly supports the loads that come from the

automobiles, trains, pedestrian and the weight of the structure included the pavement.

The second part is the sub-structure. The sub-structure supports the supper structure

and transfers the loads to the ground. The sub-structure can be abutments, piers,

columns and the foundation. The third part is the bearings. The super-structure and the

sub-structure are connected with the bearings. The bearing is a small part of the bridge

structure. However, it may cause a big problem if the bearing does not function

properly particularly after a certain period of time. Schematically, the longitudinal

section of the bridge is shown in Figure 2.1.

Several components of super-structure are as follows:

• Bridge floor . The bridge floor can be made of concrete floor, metal deck

floor or timer floor. This floor directly supports the wheel load of the

automobiles, own weight of the floor slab and the finishing such as,

pavement.

•

The beam system. The beam system can be made of cross beams and

longitudinal beams. The function of the beam system is to transmit the

loads from the floor to the main structure system. The steel plate girder,

box girder, truss structure can considered as the main girder.

•

Cross beams and cross-frames. The cross-beams are constructed to prevent the bending in horizontal direction due to horizontal loads such as

wind or earthquake loads.

•

Support bearings. The support bearings are installed at the contact points

from the super-structure to the sub-structure and used to transmit the loads

from the super-structure to the sub-structure.

Several components of the sub-structure are as follows:

•

Abutments or piers. The abutments and piers are used to transmit to the

loads from the supper structure to the foundations. For multi-span bridges,

the abutments are constructed at the two ends of the bridges and the piers

are constructed between the two ends. The abutments are also used to

support the lateral loads from the soil.


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Clearance

spaneffective

spanstructuresub -

pilesdriven

bearing

length

spaneffective

AbutmentPier

structuresuper

Caison

pilesBored

Figure 2.1: The general cross section

• Foundation. The foundations are used to transmit the loads from the

abutments or piers to the ground. Several types of foundation are

commonly used such as, driven piles, bored piles, caisson etc. The choice

of the type of foundation depends on the soil investigation and the

topography of the area.

Several types of bearings are as follows:

•

Fixed bearings. These bearings are used to transmit the horizontal loads

from the super-structure such as wind loads, earthquake loads, impact

loads etc.

• Movable bearings. The movable bearings are used to accommodate the

possible deformation of the structure such as temperature, shrinkage,

creep, the movement of the foundation etc.

2.3 CLASSIFICATION OF BRIDGES

In the early the development of highways there are a few of crossings are

required to connect between two separate locations. However, since the availability of

heavy equipments and high strength materials several types of bridges were designed

to face the challenge to the need for wider and longer bridges. These days not only the

design but also the construction will play an important role in the decision to choose


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the type of bridges that suitable to a certain condition. Based on several different

criterias, we can make classification of bridges.

2.3.1 CLASSIFICATION OF BRIDGES BASED ON THE MATERIAL

Based on the material used in the super structure the bridges can be classified

as follows:

•

Wooden bridges. The wooden bridge is constructed for short span and

light loads.

•

Stone bridges. The stone-bridge is also constructed for short span and

usually it is constructed with consideration the aesthetic.

•

Steel bridges. The steel bridges are popular in the remote area where thematerial is difficult. It is suitable for short to medium span. The

advantageous of steel bridge for short span are easy to construct and the

quality can be controlled easily.

• Concrete bridges. The concrete bridges are the most popular for

interchanges. The advantageous of concrete bridges are cheaper

particularly for short span and the shape is more flexible than the steel

bridges. However, the dead load of concrete bridges is heavier than the

steel bridges.

• Composite steel bridges. In the composite steel bridge, the concrete slab is

an integral part of the superstructure beams. This type of bridge is popularsince the availability of high strength shear connectors to resist the

horizontal shear between the concrete deck and the steel.

• Aluminum bridges. These type bridges are not popular and very seldom

used in the design.

2.3.2 CLASSIFICATION OF BRIDGES BASED ON THE SHAPE

The bridges based on the shape can be classified as follows:

• Girder bridges. Broadly, these bridges can be classified into concrete

girder bridges and steel girder bridges (plate girder bridges). Most of the

girder bridges are short span bridges. The advantageous of plate girder

bridges and prestressed I girder bridges that they can be fabricated in the


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shops. This type of bridges is commonly used in the highway

intersections.

•

Truss bridges. When the longer span is required, the truss bridges become

the option. The truss bridges more economically for the medium span

however, it is difficult to give a good shape to truss bridges. In Indonesia

particularly, the truss bridges are used to cross small rivers.

• Arch bridges. This type of bridge is occasionally used in the construction.

It has a good shape but the construction is more difficult than the truss or

girder bridges.

• Rigid frame bridges

• Cable stayed bridges

• Suspension bridges

The basic concept of the bridge design is to lay the support system (main girders)

parallel to the road alignment and support the vertical loads. The girder, truss and rigid

frame bridges are popular in Indonesia. One cable stayed bridges constructed in Batam

which is part of Barelang project.

2.4 SURVEYS AND STUDIES

There are many parameters involved before we make the decision of

constructing a bridge. Several surveys and studies have to be conducted to support our

decision. Broadly speaking there are three parameters have to be studied before thedecision is made. The first parameter is called the feasibility study. In this parameter

we evaluate the advantageous and disadvantageous of constructing the bridges. In this

study we will determine whether this bridge is worthwhile to be constructed unless we

have no option. The second parameter, is the design. In the design, we will study the

types of bridges that can support the feasibility studies. The third parameter is the

construction. This is one of the most important aspects in the design of a bridge. For

instance, we design a bridge across a deep valley where there is no possibility of

constructing the middle support, the method of construction becomes a dominant

parameter in making the decision. Eventhough in this studies several different

disciplines are involved, but they have to work together as a team and the results

related each other. When the decision has been made that we are planing to construct

the bridge the next step is to conduct a survey. In order to give a guidance in the

survey, several surveys that can be considered are tabulated in Table (2.1)


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Table 2.1 Types of surveys

TYPE OF SURVEY CONTENT OF SURVEY PURPOSE OF SURVEY

1. Topographical

survey

Drawing up topographical map Selection of bridge position,

length and spacing.

2. Geological survey Collection of geological and soil

history data.

Drawing up geological map and

physical properties

Same as the above.

Selection of position and

Structural plan of substructure

work.

3.

Intersecting roadsetc.

Condition and future plans forwidth, altitude, clearance limit,

structure crossing, longitudinal

grade etc. of intersecting roads

and railways.

Underground installations.

Selection of bridge length,span spacing, clearance,

construction method.

4. Rivers survey Rivers cross section.

Volume and speed of flow.

High and low water level and

river gradient.

Navigating vessels.

Selection of span spacing.

Clearance, pier forms,

foundation, sunken depth of

pier.

Determination of overflow

and prevention section.

Determination of construction

method

Determination of impact load.

5. Survey of sea,

lakes and

marches.

Tidal level, wave height, tide.

Navigating vessels.

Selection of estimated high-

water level.

Selection of water pressure,

construction period and

method.Determination of span

spacing, height under girder,

impact load.


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6. Soil survey. Boring.

Standard penetration test.

Soil test.

Test pit.

Plate bearing test.

Pressiometer test

Measurement of ground water

level.

Selection bearing strata for

use in substructure design.

Determination of allowable

bearing capacity.

Determination of unit weight

of soil, internal friction angle

and viscosity.

Determination of quantity of

settlement due to consoli-

dation.

Selection of constructionmethod.

7. Seismological

survey

Records of earthquakes and

earthquake damages.

Measurement of ground

microtremor.

Confirmation of bedrock.

Determination of design seis-

mic design coefficient.

8. Meteorological

survey

Survey of meteorological

observation records (wind

velocity, temperature, snow and

climate)

Determination of temperature

variation, wind load and snow

load.

Selection of protective

covering, materials, construc-

tion period and method.

9. Survey of

additions to

bridge.

Underground installation at

bridge position.

Plans for construction of

waterpipes, sewers, electricity

and telegraph wires.

Determination of measure-

ment and weight of addition.

10.

Corrosion. Extent of corrosion of existing

structures.

Organic matter, PH.

Exposure test for paints.

Selection of material to be

used.

Determination of rust preven-

tion methods.


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11.

Materials survey. Aggregate and water (quality and

quantity) for concrete.

Selection and calculation of

materials.

12. Construction. Situation of material transporta-

tion routes.

Determination of maximum

transportation load.

Road plans for construction

works.

(JICA, 1977). The purpose of this survey is to give more detail information for

selection the type of the bridge. However, not all surveys have to be conducted but itdepends on the scale of the bridges.

2.5 SELECTION THE BRIDGE TYPE (SUPER-STRUCTURE)

Besides the information from the survey, several considerations in selecting

the type of the supper-structure of the bridge can be listed as follows:

a) The span length. The large portion of the forces in the design of long span

bridges comes from the weight of the structure itself. Usually, the weight

of the concrete structure is much heavier than the steel structure.

b)

The purpose of constructing the bridge. For instance, the type of the

bridge over the road (intersection) might be totally different with the type

of the bridge over the river or deep valley eventhough the span of the

bridge is the same.

c) The soil condition. When it is difficult to find a good bearing strata, we

might consider the light bridge.

d) The layout and the alignment of the bridge. When the layout makes an

angle (skewed bridge) or the curve bridge, special attention should be

given in the design.

e)

The availability of the material. In some areas crush stone is not available

or it is difficult to deliver good quality of concrete, the steel structure may

become a good option.


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2.6 SELECTION SUB-STRUCTURE

The sub-structure can be divided into two parts. The first part is the structureto the resist directly the loads from the upper structure such as abutments or piers.

Several considerations in the selection of this part:

• The location of the substructure. Mostly, the reinforced concrete structure

is used to support the super-structure. Unless for some reasons such as

concreting is not possible, the steel structure will be used.

• Within the rivers, the flow of the water during the normal seasons and the

flood. During flood, the direction of the flow of water might be different

from the normal flow. For normal flow of water the oval might be a good

shape. However, where there is a potential flood and the direction of the

flow of water is uncertain, the circular might be a good shape.

• The connections between the piers and foundation. It depends on the

connection of the super-structure and the first part of the substructure. The

stability of the piers has to be considered.

The second part of the sub-structure is foundation. The foundation is part of the

structure below the ground. The function of the foundation mainly is to transmit the

loads from the substructure to the foundation. Basically, the considerations in making

the selection of the foundation are based on the bearing stratum. The types of

foundation can be listed as follows:

• The bearing stratum within 5 m from the ground. When the bearing

stratum is near the surface the direct foundation can be considered.

• The bearing stratum between 5 – 20 m from the ground, the concrete piles

can be used.

• The bearing stratum over 30 m, the steel piles or the bored piles can be

used.

2.7 COMPOSITE GIRDER BRIDGE

The composite girder bridges can be constructed between prestressed concrete

girder and reinforced concrete slab or between steel girder and reinforced concrete

slab. When the composite construction between steel beam and concrete slab is

considered, creep factor should be considered. In designing the composite beams, there

are four types of loadings have to be considered.


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a) The first type is called long-term loading. Basically the dead load is considered as

long-term loading. When the dead load is working on the steel section, the creep

of the concrete is not considered. However, when the dead load is working on the

composite section such as pavement, curbs and railings the creep of the concrete

has to be considered.

b) The second type is called short-term loading. The live load is considered as short-

term loading. There is no need to consider creep of slab due to live load.

c)

The third type is impact loading. The span contributes to this load and basically

the impact load is considered for live load.

d) The third type is called temporary loading. Earthquake load and wind load are the

temporary loads.

It is important to note that the composite girder is designed based on the assumptionthat the slab and the girder act monolithically. There is no sliding between the steel

and concrete. In the actual construction, there is a stress in concrete after concrete is

setting. This phenomenon has to be considered in the design. Finally, the temperature

difference between the steel and the concrete also contributes to the additional stresses

in the composite section. The stress computation due to the above loads will be

discussed in the examples.

2.7.1 LOADS

The composite girders are designed to resist the loads imposed on the girders

and transfer those loads to the substructure. Based on the sequence of construction, the

loads imposed on the composite girders can be categorized into two categories. The

first category is the loads during construction. Mostly, these loads are acting on the

steel section where the composite action has not been developed. The second category

is the load after construction. These loads are acting after the composite action has

been developed. Based on the duration, the loads can be divided into two categories.

The first category is called permanent loads. The gravity loads are considered as

permanent loads. The second category is called temporary loads. The wind and

earthquake loads are considered as temporary loads. When the creep is considered, the

loads can be divided into two types. The first type is called long-term loading. The

dead loads are considered as long-term loading. The second type is short time loading.

The live loads can be considered as short-time loading. The detail of this load will be

discussed in the examples. Besides these loads, there are many other loads that may


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occur on the structure as shown in Kapsel (Soegiarso, 1999). In this class we will

study the composite girders due to dead and live loads.

a) Dead Loads

Several examples of the dead load in composite girders are as follows:

• The dead load due to the concrete floor or steel floor.

• The dead load due to the weight of the structure such as girders, bracings,

truss structure, gusset plates etc.

• The loads due to the wearing surface such as, pavement, handrails and

curbs.

• Other loads that may added due to the circumstances such as the electric

poles, catwalk etc.

• The loads during construction such as construction equipment.

b)

Live loads

• The loads due to moving vehicles (automobiles, carts, trains). The

magnitude and the distribution of these loads should conform to the

specifications for highway bridges.

• The loads due to pedestrian.

c) Impact coefficient

Since the live loads are the fast moving vehicles, a certain impact allowance due to

live loads has to be considered. The structure elements that are affected to this impact

allowance will discussed in the subsequent section.

2.8 AMERICAN ASSOCIATION OF STATE HIGHWAY AND TRANSPORTATION

OFFICIALS (AASHTO)

2.8.1 Live loads

In designing the bridge components loads imposed on the structure have to be

regulated in the codes. There are many codes that are available and each code isusually used in a particular country. In order to give a reference how the loads are

imposed, in this section we will discuss the AASHTO loads.


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Basically in the design of the bridge, there are two types of loadings have to be

considered. The first type is called truck loading and the second type is the lane

loading. Which ever governs between these two loadings will dictate the design of the

bridge. Based on the intensity of the loads, these two types of loadings (truck loading

and lane loading) can be further divided into two types. The first type is called H-

loadings and the second type is called HS-loadings. The HS loadings are heavier than

the H loadings. Based on the truck axle loads, H loadings are divided into two classes

i.e., H15-44 and H20-44. The affixing 44 indicates the year when the code was

instituted which is in 1944. Similarly, HS loadings are divided into two classes i.e., the

HS20-44 and HS15-44. Hence, there are four standard classes of loading: H15, H20,

HS15 and HS20. The difference between H15 and H20 or HS 15 and HS 20 is the

magnitude of the loads. The loading of H20 is 75 percent of HS20. The magnitude and

the intensity of the lane loads are shown in Figure 2.2. The magnitude of truck loads

for H20-44, H15-44 are shown in Figure 2.3 and the magnitude of truck loads for

HS20-44 and HS15-44 are shown in Figure 2.4.

For continuous span, in computing the maximum negative bending moment,

besides the concentrated load shown in Figure 2.2, an additional concentrated load is

provided. These two loads are placed as such to produce the maximum negative

bending moment.

In AASHTO specifications, the truck loading is assumed to occupy a width of

10 feet ( ≈ 3.0 m) and these loads shall be placed in a 10-foot wide design traffic lanesas shown in Figure 2.2.

2.8.2 Impact coefficient

In considering the impact allowance, the bridge structure is classified into two

groups i.e., group A and group B. Only the structure in group A, the forces obtained

from the live loads have to be multiplied with the impact coefficients. The impact

allowances shall not be applied to the structure in group B. The following are the

structures categorized in groups A and B.

Structure in group A (AASHTO 3.8.1.1)

(1)

Superstructure, including steel or concrete supporting columns, steel towers, legs

of rigid frames, and generally those portions of the structure that extend down to

the main foundation.

(2) The portion above the ground line of concrete or steel piles that are rigidly

connected to the superstructure as in rigid frame or continuous structure.


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P

load ed Concentrat

shear for lbs000,26

momentfor lbs000,18 *

laneload of footlinear lbs640load Uniform

Loading4420HS

Loading4420H

−

−)a

P

load ed Concentrat

shear for lbs500,19

momentfor lbs500,13 *

laneload of footlinear lbs640load Uniform

Loading4415HS

Loading4415H

−

−) b

* For the loading of continuous span involving lane loading refer to Article 3.11.3

which provides for an additional concentrated load.

Figure 2.2: The lane loading

Structure in Group B (AASHTO, 3.8.1.2)

(1) Abutments, retaining walls, piers, piles, except Group A(2)

(2) Foundation pressures and footings

(3)

Timber structures.

(4) Sidewalk loads.

(5) Culverts and structures having 3 feet or more cover.

3.8.2 Impact Formula

3.8.2.1 The amount of the impact allowance or increment is expressed as a fraction of

the live load stress, and shall be determined by the formula:


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125L

50I

+=

where I is impact fraction (maximum 30 percent) and L is length in feet of the portion

of the span that is loaded to produce the maximum stress in the member.

2.9 DESIGN EXAMPLE CONCRETE BRIDGES

In these design examples, we focus our discussion on the concrete part of the

bridges. The first is the design of the concrete slab. And the second is the design of the

slab bridges. The continuous slab bridge is adopted in the second example where the

influence lines have to be drawn in obtaining the maximum moments at particular

cross section.

2.9.1 Bridge slab

In order to be familiar to those loads, let consider the following example. A

multi-stringer composite steel bridge deck has the cross section shown in Figure 2.3.

The slab is continuos over the stringers and half width of flanges is assumed to be 0.15

m. The yield stress of the reinforcement is 24 MPa and the concrete strength is 30

MPa. The deck is cast reinforced concrete slab. The reinforcement is perpendicular to

the traffic. Design the bridge is based on Standard Specifications for Highway Bridges

(AASHTO, 1983) and the specified loading of HS20-44. In the AASHTO

specifications, all formulas are based on US customary units. The future units in theUSA will be metric units.

m40.2 m40.2 m40.2

m90.0 m90.0m20.7

surfacewearingm/kg150 2 m25.0

h

'm/kg50

Figure 2.3: The cross section


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Solution:

a)

Compute the bending moment of the slab.

In order to compute the maximum bending moment in the slab, we have to identify

direction of the reinforcement with respect to the traffic. In AASHTO specification it

is regulated in the following formulas:

The maximum bending moment based on AASHTO specifications 3.24.3.1.

Case A- Main reinforcement perpendicular to traffic (spans 2 to 24 feet inclusive)

HS 20 loading:

=

+

20P32

2s Moment in foot-pounds per foot-width of slab.

HS 15 loading:

=

+15P

32

2s Moment in foot-pounds per foot-width of slab.

In slabs continuous over three or more supports, a continuity of 0.80 shall be applied

to the above formulas for both positive and negative moment. In accurate results can

be obtained from accurate analysis such as using finite element method.

b) Compute impact coefficient

( )30.0

125L

50I ≤+=

In order to use the formulas, the units have to be converted.

c) The thickness of slab

30

10sd

+= (Table 8.9.2)

where s is the span length in feet according to section 8.8 (clear span plus the depth of

the member).


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m40.2

d

)L(spanlearc c

Figure 2.4: Effective span of concrete slab

d Ls c +=

Let assume the span s = 7.5 feet.

"96.6'58.030

105.730

10sd ≈=+=+≥ . Let consider the thickness of the slab is 7.5”

d) The moment due to the dead load:

Own weight = 150.0x12

5.7 = 0.094 Ksf

Weight of overlay (assumed) = 0.030 Ksf

Total DL = 0.124 Ksf

ft/Kft698.08

5.7x124.0x80.0

MM

2

DLDL ===

−+

e) Moment due to lane load

Type of loading HS20-44, the wheel load 20P = 16 Kips (Article 3.24.3)

The main reinforcement is perpendicular to the traffic (Article 3.24.3.1)

( )ft/Kft75.4ft/lbft750,4000,16x

32

25.7P

32

2sM 20 ==

+=

+=

The slab is continuous, the bending moment is multiplied by 0.8 for both positive nad

negative moment (Article 3.24.3.1)

ft/Kft8.375.4x80.0MM LLLL === −+ (Non factored)


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19

f) Impact coefficient and moment (Article 3.8.2.1)

Impact: 377.01255.7

505.12s

50I =+=+=. The maximum impact fraction is 0.3.

Impact moment:

ft/Kft14.180.3x30.0MM II === −+

Factored moment: AASHTO 3.22.1

( )[ ]ILD.MM LD +β+βγ== −+ (Article 3.22, Eq. (3.10))

( )[ ]

( )[ ] ft/Kft63.1114.18.367.1698.03.1

MM67.1M3.1MM ILD

=++=

++== −+

Effective depth (d):

The effective depth can be computed as follows (Figure 2.5):

)clearance(c

hd

2. d steel diaonDistributi

1d entreinforcemMain

)clearance(c

hd

2. d steel diaonDistributi

1d entreinforcemMain

a) Positive moment b) Negative moment

Figure 2.5: Effective depth for positive moment

Positive bending moment

clearanced 5.0hd 1 −−=

Let assume the clearance is 1 in. and 1d is"75.0 , then the effective depth becomes,

( ) "125.6175.05.05.7d """ =−−=

Negative bending moment:


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For negative bending moment we may consider the wearing and particularly when

snow is considered the clearance must be thicker than the area where snow is not

considered. This is due to the salt that can penetrate at the negative moment area. If we

consider the bridge on the snowy area then the effective depth becomes,

wearingclearanced d 5.0hd 21 −−−−=

Let assume the clearance and the wearing is "5.2 , 1d is"75.0 and 2d is "5.0 , then

the effective depth becomes,

( ) "125.45.25.075.05.05.7d """" =−−−=

g) Design reinforcement

In order to make self assessment to the code, let review the following

equations. From Figure 2.6 we obtain the following:

h) To compute bρ

ys b'c f ATand baf 85.0C == (2.1)

ba

'cf 85.0 b

d 2/ad b−

bC

003.0c =ε

T

C

yε cε

sA

Figure 2.6: Stress block diagram

From equilibrium we obtain, bf 85.0

f Aa

'c

ys

b = (2.2)

ba is the balance depth equivalent rectangular stress block.


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21

b1 b Ca β= (2.3)

where from ACI/318R, Article 10.2.7.3 and AASHTO, Article 8.16.2.7,

≥=β

≥

−−=β

≤=β

psii8000f for 65.0

ps4000f for 1000

4000f 05.085.0

psi4000f for 85.0

'c1

'c

'c

1

'c1

(2.4)

Let assume psi000,000,29E s =

000,29/f E/f yysyy =ε→=ε (2.5)

From Figure 2.6 we obtain,

d

C b

yc

c =ε+ε

ε (2.6)

The maximum usable strain at extreme concrete compression fiber shall be assume

equal to 0.003 (ACI/318R-95, Article10.2.4)

Substitute Eq. (2.5) into Eq. (2.6) yields,

( )d f 700,8700,8

Cf 700,8

700,8

d

C

y b

y

b

+=→+= (2.7)

( )d f 700,8

700,8a

y

1 b +β= (2.8)

From Figure (2.26), the equation for the balance failure is as follows:

y bys b'c bdf f A baf 85.0 ρ== (2.9)

( ) by

'c

b ad f

f 85.0=ρ (2.10)

Substitute Eqs. (2.8) into Eq. (2.10) we obtain,


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( )

+

β

=ρ

y

1

y

'c

b

f 000,87

000,87

f

f 85.0 (2.11)

The maximum and minimum reinforcement:

−≤ρ=ρ

−ρ=ρ

)1.5.10Article,108R 318/ACI( f

200 and

f

f 3

)3.3.10Article,108R 318/ACI(75.0

y

min

y

'c

min

bmax

(2.12)

i) To compute

Let consider Figure 2.6, where ba is replaced by a.

( )2/ad Cor TM n −= (2.13)

( )

−=

bf 85.02

f Ad f AM

'c

ys

ysn (2.14)

ρ=→=ρ bd A bd

As

s (2.15)

From Eqs. (2.14) and (2.15) yield,

( )

ρ−ρ=

bf 85.02

bdf d bdf M '

c

yyn (2.16)

( ) bf 85.02

f d bf bd M

'c

22y

22

y2

n

ρ−ρ= (2.17)

0 bd

Mf

f 7.1

f

2

ny'

c

22y =+ρ−ρ

(2.18)

φ=→=φ ununM

MMM (2.19)

Substitute Eq. (2.20) into Eq. (2.18) yields,


21/53

23

0 bd

Mf

f 7.1

f

2

uy'

c

22y =

φ+ρ−

ρ (2.20)

Let2

uu

bd

MR φ= (2.21)

'c

2y

u'c

2y2

yy

f 7.1

f 2

R f 7.1

f 4f f

−±

=ρ

−−==ρ

'c

u

y

'c

f 85.0R 211

f f 85.0 (2.22)

j) Reinforcement due to moment positive:

"125.6d and ft/'K 63.11M ==+

From Eq. (2.21),

( )( )

( )( )( )344.0

125.6129.0

1263.11R

2u ==

Substitute uR into Eq. (2.22) yields,

( ) ( )( )

00602.05.485.0

344.0211

000,60

500,485.0=

−−==ρ

From Eq. (2.11) we obtain,

( )( )03018.0

000,60000,87

000,87

000,60

500,4825.085.0 b =+=ρ

( ) 0226.003018.075.0max ==ρ

OK 00602.0 max →ρ〈=ρ

( )( ) 2s in442.0125.61200602.0A ==+


22/53


Moment negative:

"

125.4d and ft/'K 63.11M ==+

From Eq. (2.21),

( )( )

( )( )( )759.0

125.4129.0

1263.11R

2u ==

Substitute uR into Eq. (2.22) yields,

( ) ( )( ) max

0142.05.485.0

759.0211

000,60

500,485.0ρ〈=

−−==ρ

( )( ) 2s in705.0125.4120142.0A ==−

k) Reinforcement due to moment negative

The reinforcement should be able to carry 1.2 cr M (AASHTO, Article 8.17.1)

Where cr M is the cracking moment.

t

gr

cr y

If M = (AASHTO, Article 8.13.3)

where r f is the modulus of rupture , gI is the moment of inertia of the concrete

section about the centroidal axis, neglecting the reinforcement,t

y is the distance

from the centroidal axis to the extreme fiber. (AASHTO, Article 8.15.2.1)

psi503500,45.7f r == (AASHTO, Article 8.15.2.1.1)

( )"K 58.56

5.75.0

12/)5.7x12(503M

3

cr ==

( ) "K 67.958.562.1M2.1 cr ==

Check for bottom steel ( 2s in442.0A =+ )

( ) ( )( )( )"

'c

ys 578.0125.485.0

60442.0 bf 85.0

f Aa ===

( )2/ad f AM ysn −φ=φ


23/53

25

( )( )( ) cr '

n MK 29.1392/578.0125.660442.09.0M 〉=−=φ

Check for top steel ( 2s in705.0A =− )

( )( )

( )( )"

'c

ys922.0

125.485.0

60705.0

bf 85.0

f Aa ===


( )( )( ) cr '

n MK 49.1392/922.0125.460705.09.0M 〉=−=φ

l) Check Cracking

For psi000,40f y 〉

Ad f z cs= (ACI/318R, Article 10.6.4)

where,

armlinternaisd ,dA

MMMf

s

ILDs αα

++= , cd is the distance from the centroid of the

rebars to the outer surface of the slab and A is the area of concrete in tension zone.

Positive bending moment:

( ) "375.1175.05.0d ""

c =+= Let assume the bar spacing is "8 , the total area of concrete surrounding the bar is

A = 1.375 x 2 x 8 = 2in0.22

Let assume 87.0d =α

( )( )( )

72.284461.0125.687.0

1214.18.3698.0f s =

++= Ksi

( ) ( )( ) Ksi14549.890.22375.172.28z 3 〈==

Negative bending moment:

( ) "375.35.25.075.05.0d """ =++= Let assume the bar spacing is "7 , the total area of concrete surrounding the bar is


24/53


A = 3.375 x 2 x 7.0 = 2in25.47

( )( )( )

74.26705.0125.487.0

1214.18.3698.0f s =++= Ksi

( ) ( )( ) Ksi1450.14525.47375.374.26z 3 〉==

2.9.2 Continuous bridge

This example is a continuous concrete slab bridge with three spans as shown

in Figure 2.7. The first span length is 24 feet, the second span (middle span) is 32 feet

and the third span is 24 feet. The concrete is the normal weight concrete with the

ultimate compressive strength psi500,4f 'c = . The live loads are based on HS20-44. Inthis example we compute the following:

'24 '24'32

h

Figure 2.7: Longitudinal section

a)

Draw the maximum and minimum bending moment diagrams and maximumand minimum shear diagrams.

b) Design the slabs due to the positive and negative bending moments.

c) Check the cracks requirements and the fatigue stress limits.

Solution:

The computation is based on one-way slab with the main reinforcement is parallel to

the traffic.

1. The minimum thickness of the slab:

542.030

10sh ≥

+= (AASHTO, Table 8.9.2)

where s is the clear span plus the depth of the member (AASHTO, Article 8.8).


25/53

27

Let assume the clear span is the 36 feet.

542.04.130

1032h ' ≥=+= . Let h = "17

2. Loads

2.1 Dead loads:

Own weight of the slab: 0.15 x 1.40 x 1 = 0.21 K/ft

Weight of asphalt is assumed = 0.03 K/ft

Weight of the curb is assumed = 0.02 K/ft

Total dead load = 0.26 K/ft

2.2 Live loads:2.2.1 Truck loads

The wheel loads of HS20-44 are shown in Figure 2.8. These wheel loads are

distributed over the slab. The width of slab over which the wheel load is

distributed (E). When the main reinforcement is parallel to the traffic the width E

is computed based on AASHTO, Article 3.24.3.20

E = (4 + 0.06S) feet7≤ where S is the span length

( )

( )

≤=+=→=

≤=+=→=

feet792.53206.04E32S

feet744.52406.04E24S

''

''

'14 '' 3014 −

K 4 K 16 K 16

E

load Wheel

Figure 2.8: The wheel loads of HS20-44


26/53


2.2.2 Lane loads

Qfor lbs000,26

Mfor lbs000,18ft/lbs640

The width of slab over which the wheel load is distributed (2E)

The main reinforcement parallel to the traffic (AASHTO, Article 3.24.3.20

E = (4 + 0.06S) feet7≤

2.3 Impact loads

≤=+=→=

≤=+=→=

3.031.012532

50I32S

3.033.012524

50I24S

'

'

2.4 Influence lines

The influence line is based on the configuration of the supports as shown in Figure 2.9.

This configuration is totally set in a paper with unanimous reference. This is only used

for study. Later on in this notes a computer program is developed to compute the

maximum will be developed.

0 1 122 3 4 5 6 7 8 9 10 11 13 14 15 16 17 18 19 20

L 1.333L L

Figure 2.9: The segments of the bridge

The ratio of middle span length to the first and the third span length is 1.333. The bridge is divided into 20 segments and a unit load is placed from point 1 to 19

successively. Plots of influence diagrams are shown in Figures 2.10-2.12. Based on the

influence line tables, the ordinate in Figures 2.10-2.12 have to be multiplied by L/99


27/53

29

where L is the length of the span. Basically, there are two types of loading have to be

considered. The first type is the truck loading either HS15 or HS20. Only one truck

load is considered for the whole bridge. Therefore, for long span bridge, most probably

the truck loading may not govern. The second type is lane load. In producing the

maximum bending moment, for continuous span, the concentrated loads may be

applied to different spans (AASHTO specifications Article 3.11.3). For maximum

positive moment, only one concentrated load shall be used with uniform loads that

produce the maximum moment.

The bending moments and shear forces at particular section can be obtained by

taking the algebraic sum of the influence line ordinates and multiplied by the length of

each segment. It is important to note that for bending moment only the influence line

ordinates have to be multiplied by L/99. In order to illustrate the computation of

bending moment due to lane load, let consider the influence line of section 1 as shownin Figure 2.9. The algebraic sum of the influence line ordinates between sections 1

through 6 is as follows:

55.3647.163.333.650.972.12T =++++=

( )( )( )96/24d TArea = where d is the distance between points.

( )( )( ) 65.3299/24455.36Area ==

The maximum bending moments and shear forces at each section are as follows:

Table 2.2: Ordinates for influence diagram for bending moments

Moment Unit Load at point

at point 1 2 3 4 5 6 7 8 9

1 12.72 9.50 6.35 3.63 1.47 0.00 -1.18 -1.98 -2.07

2 9.44 19.00 12.70 7.27 2.93 0.00 -2.37 -3.97 -4.13

3 5.90 12.00 19.05 10.90 4.40 0.00 -3.55 -5.95 -6.20

4 2.37 5.00 8.90 14.50 5.87 0.00 -4.73 -7.93 -8.27

5 -1.17 -2.00 -1.25 1.70 7.08 0.00 -5.92 -9.92 -10.30

6 -4.70 -9.00 -11.40 -11.20 -7.70 0.00 -7.10 -11.90 -12.40

7 -3.90 -7.55 -9.56 -9.39 -6.45 0.00 7.84 1.17 -1.738 -3.10 -6.10 -7.73 -7.58 -5.20 0.00 6.28 14.20 8.95

9 -2.30 -4.65 -5.89 -5.76 -3.95 0.00 4.71 10.80 19.60

10 -1.65 -3.20 -4.05 -3.95 -2.70 0.00 3.15 7.35 13.80


28/53


Table 2.2- continued


at point 10 11 12 13 14 15 16 17 18 19

1 -1.93 -1.58 -1.07 -0.52 0.00 0.38 0.55 0.55 0.43 0.23

2 -3.87 -3.17 -2.13 -1.03 0.00 0.77 1.10 1.10 0.87 0.47

3 -5.80 -4.75 -3.20 -1.55 0.00 1.15 1.65 1.65 1.30 0.70

4 -7.73 -6.33 -4.27 -2.07 0.00 1.53 2.20 2.20 1.73 0.93

5 -9.67 -7.92 -5.33 -2.58 0.00 1.92 2.75 2.75 2.17 1.17

6 -11.60 -9.50 -6.40 -3.10 0.00 2.30 3.30 3.30 2.60 1.40

7 -3.35 -3.67 -2.96 -1.54 0.00 1.05 1.49 1.46 1.15 0.64

8 4.90 2.15 0.47 0.03 0.00 -0.20 -0.33 -0.38 -0.30 -0.13

9 13.20 6.97 3.92 1.59 0.00 -1.45 -2.14 -2.21 -1.75 -0.89

10 21.40 13.80 7.35 3.15 0.00 -2.70 -3.95 -4.05 -3.20 -1.65

The bending moment at each point is obtained by multiplying the ordinate of the

particular point in Table 2.2 by L/99, where L is the length based on Figure 2.9

Table 2.3: Ordinates for influence diagram for shears


at point 1 2 3 4 5 6 7 8 91 0.786 0.576 0.385 0.220 0.089 0.00 -0.072 -0.111 -0.215

2 -0.214 0.576 0.385 0.220 0.089 0.00 -0.072 -0.111 -0.215

3 -0.214 -0.424 0.385 0.220 0.089 0.00 -0.072 -0.111 -0.125

4 -0.214 -0.424 -0.615 0.220 0.089 0.00 -0.072 -0.111 -0.125

5 -0.214 -0.424 -0.615 -0.780 0.089 0.00 -0.072 -0.111 -0.125

6L -0.214 -0.424 -0.615 -0.780 -0.911 0.00 -0.072 -0.111 -0.125

7 0.046 0.088 0.111 0.108 0.076 0.00 0.905 0.785 0.647

8 0.046 0.088 0.111 0.108 0.076 0.00 -0.095 0.785 0.647

9 0.046 0.088 0.111 0.108 0.076 0.00 -0.095 -0.215 0.647

10 0.046 0.088 0.111 0.108 0.076 0.00 -0.095 -0.215 -0.352


29/53

31

Table 2.3- continued

Shear Unit Load at point

at point 10 11 12 13 14 15 16 17 18 19

1 -0.117 -0.096 -0.065 -0.031 0.00 0.023 0.033 0.033 0.026 0.014

2 -0.117 -0.096 -0.065 -0.031 0.00 0.023 0.033 0.033 0.026 0.014

3 -0.117 -0.096 -0.065 -0.031 0.00 0.023 0.033 0.033 0.026 0.014

4 -0.117 -0.096 -0.065 -0.031 0.00 0.023 0.033 0.033 0.026 0.014

5 -0.117 -0.096 -0.065 -0.031 0.00 0.023 0.033 0.033 0.026 0.014

6L -0.117 -0.096 -0.065 -0.031 0.00 0.023 0.033 0.033 0.026 -0.046

7 0.500 0.353 0.215 0.095 0.00 -0.076 -0.108 -0.111 -0.088 -0.046

8 0.500 0.353 0.215 0.095 0.00 -0.076 -0.108 -0.111 -0.088 -0.046

9 0.500 0.353 0.215 0.095 0.00 -0.076 -0.108 -0.111 -0.088 -0.046

10 0.500 0.353 0.215 0.095 0.00 -0.076 -0.108 -0.111 -0.088 -0.046

In the concrete design for bridges, the shear force may not be crucial in the

design. But in steel beam, the shear or reaction at the support is carried by the web. In

addition, in the design of shear connectors to resist fatigue, the range of shear has to be

computed. The range of shear is the difference between minimum and maximum shear

envelopes due to live load. The shear force at each point is obtained by multiplying the

ordinate of the particular point in Table 2.3 by the concentrated load. By computing

the maximum and minimum shears at particular section, the range of shear can be

obtained.


30/53


31/53


32/53


0 1 122 3 4 5 6 7 8 9 10 11 13 14 15 16 17 18 19 20

- 0 .

2 1

4

0 .

7 8 0

0 .

0 8 9

- 0 .

0 7

2

0 .

0 2 3

0 .

0 3 3

0 .

0 3 3

0 .

0 2 6

0 .

0 1 4

- 0 .

4 2

4

- 0 .

6 1

5

4sectionfordiagramInfluence

0 .

2 2

0

- 0 .

1 1

1

- 0 .

1 2

5

- 0 .

1 1

7

- 0 .

0 9

6

- 0 .

0 6

5

- 0 .

0 3

1

- 0 .

2 1 4

0 .

7 8 0

0 .

0 8 9

- 0 .

0 7 2

0 .

0 2 3

0 .

0 3 3

0 .

0 3 3

0 .

0 2 6

0 .

0 1 4

- 0 .

4 2 4

- 0 .

6 1 5 5sectionfordiagramInfluence

- 0 .

9 1 1

- 0 .

1 1 1

- 0 .

1 2 5

- 0 .

1 1 7

- 0 .

0 9 6

- 0 .

0 6 5

- 0 .

0 3 1

- 0 .

2 1 4

- 0 .

7 8 0

- 0 .

0 7 2

0 .

0 2 3

0 .

0 3 3

0 .

0 3 3

0 .

0 2 6

0 .

0 1 4

- 0 .

4 2 4

- 0 .

6 1 5

- 0 .

9 1 1

- 0 .

1 1 1

- 0 .

1 2 5

- 0 .

1 1 7

- 0 .

0 9 6

- 0 .

0 6 5

- 0 .

0 3 1

- 1 .

0 0

0 . 9

0 5

- 0 . 0

7 6


- 0 .

0 9 5

0 .

7 8 5

0 .

6 4 7

0 .

5 0 0

0 .

3 5 3

0 .

2 1 5

0 .

0 9 5

- 0 . 1

0 8

- 0 . 1

1 1

- 0 . 0

8 8

- 0 . 0

4 6

- 0 .

2 1 5

- 0 .

0 7 6


- 0 .

0 9 5

0 .

7 8 5

0 .

6 4 7

0 .

5 0 0

0 .

3 5 3

0 .

2 1 5

0 .

0 9 5

- 0 .

1 0 8

- 0 .

1 1 1

- 0 .

0 8 8

- 0 .

0 4 6

- 0 .

3 5 3

- 0 .

0 7 6


- 0 .

0 9 5

- 0 .

2 1 5

0 .

6 4 7

0 .

5 0 0

0 .

3 5 3

0 .

2 1 5

0 .

0 9 5

- 0 .

1 0 8

- 0 .

1 1 1

- 0 .

0 8 8

- 0 .

0 4 6

0.80

6.57

2.47

0.52

9.95

0.18

2.47

0.52

0.52

2.47

13.77

1.72

0.19

12.18

1.72

1.72

8.80

0.81

1.72

1.72

1.96

5.96

1.72

0 .

0 7 6

0 .

1 0 8

0 .

1 1 1

0 .

0 8 8

0 .

0 4 6

0 .

0 7 6

0 .

1 0 8

0 .

1 1 1

0 .

0 8 8

0 .

0 4 6

0 .

0 7 6

0 .

1 0 8

0 .

1 1 1

0 .

0 8 8

0 .

0 4 6

0 .

9 0 5

- 0 .

0 7 6

0 .

7 8 5

0 .

6 4 7

0 .

5 0 0

0 .

3 5 3

0 .

2 1 5

0 .

0 9 5

- 0 .

1 0 8

- 0 .

1 1 1

- 0 .

0 8 8

- 0 .

0 4 6

1.7215.99

1.72 0 .

0 7 6

0 .

1 0 8

0 .

1 1 1

0 .

0 8 8

0 .

0 4 6

6R sectionfordiagram Influence

1 .

0 0

6Lsectionfordiagram Influence

Figure 2.12: Influence diagram for shear


33/53

35

0 1 122 3 4 5 6 7 8 9 10 11 13 14 15 16 17 18 19 20

- 0 .

3 5 3

- 0 .

0 7 6


- 0 .

0 9 5

- 0 .

2 1 5

0 .

5 0 0

0 .

3 5 3

0 .

2 1 5

0 .

0 9 5

- 0 .

1 0 8

- 0 .

1 1 1

- 0 .

0 8 8

- 0 .

0 4 6

- 0 .

5 0 0

1.72

3.65

3.65

1.72

0 .

0 7 6

0 .

1 0 8

0 .

1 1 1

0 .

0 8 8

0 .

0 4 6

Figure 2.12: Influence diagram for shear (-continued)

2.5 Maximum bending moments due to truck loading

The points in the following are refereed to Figure 2.9

• Point 1 of Figure 2.9

( ) ft/K 89.1055.272.1244.5

16

99

24M '=

+=+

( ) ( ) ft/K 10.238.044.5

4

99

2433.198.1

92.5

16

99

24M '−=

+

+−=−

• Point 2

( ) ft/K 59.1446.10.1944.516

9924M '=

+=+

( ) ( ) ft/K 20.477.044.5

4

99

2497.365.2

92.5

16

99

24M '−=

+

+−=−

• Point 3

( ) ft/K 58.13005.1944.5

16

99

24M '=

+=+

( ) ( ) ft/K 30.615.144.5

4

99

2498.395.5

92.5

16

99

24M '−=

+

+−=−


34/53


• Point 4

( ) ft/K 18.11185.15.1444.516

99

24

M'

=

+=

+

( ) ( ) ft/K 40.853.144.5

4

99

243.593.7

92.5

16

99

24M '−=

+

+−=−

• Point 5

( ) ft/K 05.508.744.5

16

99

24M '=

=+

( ) ( ) ft/K 20.1163.692.992.516

0.244.5

4

99

24

M'

−=

++−=

−

• Point 6

( ) ft/K 07.30.23.244.5

16

99

24M '=

+=+

( ) ( ) ( ) ft/K 23.1795.792.5

49.11

92.5

164.11

44.5

16

99

24M '−=

++−=−

• Point 7

( ) ( ) ( ) ft/K 72.515.144.5

453.0

44.5

1684.7

92.5

16

99

24M '=

++=+

( ) ( ) ft/K 12.951.392.5

1656.9

44.5

16

99

24M '−=

+−=−

•

Point 8

( ) ( ) ft/K 13.1020.044.5

4

99

2431.12.14

92.5

16

99

24M '=

−

+=+

( ) ft/K 77.639.610.344.5

16

99

24M '−=

+−=−


35/53


36/53


37/53

39

( ) ( )

( ) ( ) ft/K 17.786.35

92.52

64.02.14

92.52

18

99

24M '=+

=+

( ) ( )

( ) ( ) ft/K 87.430.181.28

44.52

64.073.7

44.52

18

99

24M '−=

++−=−

• Point 9

( ) ( )

( ) ( ) ft/K 41.1095.58

92.52

64.06.19

92.52

18

99

24M '=+

=+

( ) ( )

( ) ( ) ft/K 13.418.887.21

44.52

64.089.5

44.52

18

99

24M '−=

++−=−

•

Point 10

( ) ( )

( ) ( ) ft/K 56.1188.67

92.52

64.04.21

92.52

18

99

24M '=+

=+

( ) ( )

( ) ( ) ft/K 40.308.1508.15

44.52

64.005.4

44.52

18

99

24M '−=+−

−=−

2.7 Shear forces due to truck loading

• Point 0

( ) K 83.3303.00.144.5

16Q =+=+

( ) K 41.0031.0118.092.5

16Q −=−−=−

•

Point 1

( ) K 76.2155.0786.044.5

16Q =+=+

( ) ( ) K 95.0214.044.516

118.092.5

16

Q −=−+−=−

• Point 2


38/53


( ) K 84.105.0576.044.5

16Q =+=+

( ) ( ) K 58.1125.092.5

16424.0

44.5

16Q −=−+−=−

• Point 3

( ) K 13.1385.044.5

16Q ==+

( ) ( ) K 15.2125.092.5

16615.0

44.5

16Q −=−+−=−

•

Point 4

( ) K 65.022.044.5

16Q ==+

( ) ( ) ( ) K 66.2048.092.5

4125.0

92.5

1678.0

44.5

16Q −=−++−=−

• Point 5

( ) K 26.0089.044.5

16Q ==+

( ) ( ) K 71.3118.092.5

4324.0911.0

44.5

16Q −=−+−−=−

• Point 6L

( ) K 12.0014.0028.044.5

16Q =+=+

( ) K 11.452.00.192.5

16Q −=−−=−

•

Point 6R

( ) ( ) K 33.41.044.5

4574.000.1

92.5

16Q =++=+


39/53

41

( ) K 42.0066.0076.044.5

16Q −=−−=−

• Point 7

( ) ( ) K 68.3110.044.5

4427.0905.0

92.5

16Q =++=+

( ) K 42.0066.0076.044.5

16Q −=−−=−

• Point 8

( ) ( )

K 83.244.5

076.04284.0785.0

92.5

16Q =−+=+

( ) ( ) ( ) K 89.0046.044.5

4092.0

44.5

16215.0

92.5

16Q −=−+−+−=−

• Point 9

( ) ( ) K 20.2038.044.5

4155.0647.0

92.5

16Q =++=+

( ) ( ) K 28.1110.044.5

16353.0

92.5

16Q −=−−=−

• Point 10

( ) ( ) ( ) K 71.1111.044.5

16047.0

92.5

4500.0

92.5

16Q =++=+

K 71.1Q =−

2.8 Shear forces due to lane loading

• Point 1

( ) ( )[ ] K 30.2786.0x2652.065.664.0

44.52

1Q =++=+


40/53


( ) ( ) ( )[ ]

( ) ( ) ( )[ ]214.02643.064.0

92.52

1125.02647.264.0

44.52

1Q −+−+−+−=−

K 94.0Q −=−

•

Point 2

( ) ( ) ( )[ ] K 64.1576.02652.093.364.0

44.52

1Q =++=+

( ) ( ) ( )[ ]

( ) ( ) ( )[ ]125.02647.264.0

92.52

1424.02670.164.0

44.52

1Q −+−=−+−=−

K 52.1Q −=−

• Point 3

( ) ( )[ ] K 04.1385.02652.001.264.044.5

1Q =++=+

( ) ( ) ( )[ ]

( ) ( ) ( )[ ]125.02671.264.0

92.52

1615.02678.364.0

44.52

1Q −+−++−=−

K 11.2Q −=−

•

Point 4

( ) ( )[ ] K 60.022.02652.080.064.044.5

1Q =++=+

( ) ( ) ( )[ ]

( ) ( ) ( )[ ] K 66.2125.02647.264.0

92.52

178.02657.664.0

44.52

1Q −=−+−++−=−

• Point 5

( ) ( ) ( )[ ] K 25.0089.02652.018.064.0

44.52

1Q =++=+

( ) ( ) ( )[ ]

( ) ( ) ( )[ ] K 45.3125.02647.264.0

92.52

1911.02695.964.0

44.52

1Q −=−+−++−=−

• Point 6


41/53

43

( )

( ) ( )[ ] K 11.0033.02652.064.044.52

1Q =+=+

( ) ( ) ( )[ ]

( ) ( ) ( )[ ] K 61.3125.02647.264.0

92.52

100.12677.1364.0

44.52

1Q −=−+−++−=−

•

Point 7

( ) ( )[ ]

( ) ( ) ( )[ ] K .752905.02618.1264.0

92.52

172.164.0

44.52

1Q =++=+

( ) ( ) ( )[ ]

( ) ( ) ( )[ ] K 59.0111.02672.164.0

44.52

1095.02619.064.0

92.52

1Q −=−+−+−−=−

• Point 8

( ) ( )[ ]

( ) ( ) ( )[ ] K 30.2785.0268.864.0

92.52

172.164.0

44.52

1Q =++=+

( ) ( ) ( )[ ]

( ) ( ) ( )[ ] K 88.0215.02681.064.0

92.52

1111.02672.164.0

44.52

1Q −=−−+−−=−

• Point 9

( ) ( )[ ]

( ) ( ) ( )[ ] K 84.1647.02696.564.0

92.52

172.164.0

44.52

1Q =++=+

( ) ( ) ( )[ ]

( ) ( ) ( )[ ] K 25.1353.02696.164.0

92.52

1111.02672.164.0

44.52

1Q −=−+−++−=−

• Point 10

( ) ( )[ ]

( ) ( ) ( )[ ] K 40.150.02665.364.0

92.52

172.164.0

44.52

1Q =++=+

( )

( ) ( )[ ]

( )

( ) ( )[ ] K 66.150.02665.364.092.52

1111.02672.164.0

44.52

1Q −=−+−++−=−


42/53


43/53

45

ft/'K 40.22MIL =

−

+

The ultimate moment based on AASHTO 3.10 and group IA is as follows:

( )[ ]ILDu MM2.2M3.1M ++=


44/53


P o

i n t

I L

A

r e a

P

o s i

t i v e

I L

A r e a

N e g a

t i v e

) f t /

K ( M '

D

) f t /

K ( M

'

m a x

L L

+

) f t /

K ( M

'

m a x

L L

−

) f t /

K ( M

'

L L

I m p a c t

−

) f t /

K ( M

'

L L

I m p a c t

+

)

f

t

/ K ( M '

I

L + +

) f t /

K ( M '

I

L − +

) f t /

K ( M

'

I

L

D +

+

+

) f t /

K ( M

'

I

L

D −

+

+

) f t /

K ( r a n g e

M '

1 2 3 4 5 6 7 8 9 1 0

3 4

. 7 3

5 3

. 9 6

5 6

. 9 2

4 3

. 8 6

1 8

. 5 3

1 2

. 5 1

8 . 3

1

3 5

. 8 6

5 8

. 9 5

6 7

. 8 8

1 0

. 0 2

2 0

. 0 2

3 0

. 0 6

4 0

. 0 8

5 4

. 0 6

5 3

. 9 8

3 0

. 1 1

3 0

. 0 5

3 0

. 1 6

1 0 2

. 7 9

6 . 4

2

8 . 8

2

6 . 9

9

0 . 9

8

- 9 . 2

4

- 2 3

. 4 8

- 1 1

. 8 7

1 . 5

0

7 . 5

1

9 . 8

1

1 0

. 8 9

1 4

. 5 9

1 3

. 5 8

1 1

. 1 8

5 . 0

5

3 . 0

7

5 . 7

2

1 0

. 1 3

1 4

. 2 9

1 5

. 3 1

- 2 . 1

0

- 4 . 2

0

- 6 . 3

0

- 8 . 4

0

- 1 1

. 2 0

- 1 7

. 2 3

- 9 . 1

2

- 6 . 7

7

- 5 . 3

0

- 4 . 6

6

3 . 2

7

4 . 3

8

4 . 0

7

3 . 3

5

1 . 5

2

0 . 9

2

3 . 0

4

1 . 7

2

4 . 2

9

4 . 5

9

- 0 . 6

3

- 1 . 2

6

- 1 . 8

9

- 2 . 5

2

- 3 . 3

6

- 5 . 1

7

- 2 . 7

4

- 2 . 0

3

- 1 . 5

9

- 1 . 4

0

1 4 . 1 6

1 8 . 9 7

1 7 . 6 5

1 4 . 5 3

6 . 5 7

3 . 9 9

7 . 4 4

1 3 . 7 7

1 8 . 5 8

1 9 . 9 0

- 2 . 7

3

- 5 . 4

6

- 8 . 1

9

- 1 0

. 9 2

- 1 4

. 5 6

- 2 2

. 4 0

- 8 . 8

0

- 1 1

. 8 6

- 6 . 8

9

- 6 . 0

6

2 0

. 5 8

2 7

. 7 9

2 4

. 6 4

1 5

. 5 1

- 2 . 6

7

- 1 9

. 4 9

- 4 5

. 8 8

- 2 3

. 7 3

- 7 . 3

0

- 4 . 4

3

1 5

. 2 7

2 6

. 0 9

2 9

. 7 1

3 . 6

9

3 . 3

6

- 1 . 2

0

- 9 . 9

4

- 2 3

. 8 0

0 . 6

2

3 . 7

5

1 6 . 2

9

2 4 . 4

3

2 5

. 8 4

2 5 . 4

5

2 1 . 1

3

2 9 . 3

9

1 9

. 3 0

2 2 . 5

7

2 5 . 4

7

2 5 . 9

6

l o a d

L i v e

a n

d

L o a

d

D e a d

t o

d u e

M o m e n

t

:

2 . 4

T a

b l e

(

)

2 6

. 0 x

A

A

M D L

−

+

−

=


45/53


46/53


Table 2.6: The maximum shear forces due to Dead Load and Truck loading

Point +A −A )K (QD

)K (

QL+ )K (

QL− )K (

Q IL++ )K (

Q IL−+ )K (

Q ILD+ ++ )K (

Q ILD− ++

0 10.74 -2.47 2.15 3.83 -0.41 4.98 -0.53 7.13 1.62

1 7.17 -2.90 1.11 2.76 -0.95 3.58 -1.24 4.69 -0.13

2 4.45 -4.17 0.07 1.84 -1.58 2.39 -2.05 2.46 -1.98

3 2.53 -6.49 -1.03 1.13 -2.15 1.47 -2.80 0.44 -3.83

4 1.32 -9.04 -2.00 0.65 -2.66 0.85 -3.46 1.15 -5.46

5 0.70 -12.42 -3.05 0.26 -3.71 0.34 -4.82 -2.71 -7.87

6L 0.52 -16.24 -4.09 0.12 -4.11 0.16 -5.34 -3.93 -9.436R 17.71 -1.72 4.16 4.33 -0.42 5.23 -0.55 9.39 3.74

7 13.9 -1.91 3.12 3.68 -0.42 4.78 -0.55 7.90 2.57

8 10.52 -2.53 2.08 2.83 -0.89 3.67 -1.16 5.75 0.92

9 7.68 -3.68 1.04 2.20 -1.28 2.86 -1.66 3.90 -0.62

10 5.37 -5.37 0.00 1.71 -1.71 2.22 -2.22 2.22 -2.22

Table 2.7: The maximum shear forces due to Dead load and Lane loading

Point +A −A )K (

QD )K (

QL+

)K (

QL−

)K (

Q ILD+

++ )K (

Q ILD−

++

1 7.17 -2.90 1.10 2.30 -0.94 4.09 -0.12

2 4.45 -4.17 0.07 1.64 -1.52 2.20 -1.91

3 2.53 -6.49 -1.03 1.04 -2.11 0.32 -3.77

4 1.32 -9.04 -2.00 0.60 -2.66 -1.22 -5.46

5 0.70 -12.42 -3.05 0.25 -3.45 -2.73 -7.54

6 0.52 -16.24 -4.09 0.11 -3.61 -3.95 -8.78

7 13.9 -1.91 3.12 2.75 -0.59 6.70 2.358 10.52 -2.53 2.08 2.30 -0.88 5.07 0.94

9 7.68 -3.68 1.04 1.84 -1.25 3.43 -0.59

10 5.37 -5.37 0.00 1.40 -1.66 1.82 -2.16


47/53

49

1 102 3 4 5 6 7 8 90

Load Dead toduemomentBendinga)

42.682.8

99.6

98.0

24.9−

48.23−

87.11−

50.1−

51.7

81.9

1 102 3 4 5 6 7 8 90

58.20

82.827.15

69.3 36.320.1−

94.9−

80.23−

88.45−

73.23−

30.7−

62.0

75.3

79.27

64.24

51.15

67.2−

49.19−

43.4−

09.26

71.29

pactImLLDLtoduemomentBendingMin.&Max. b) ++

Figure 2.13: The maximum and minimum bending moments at each section


48/53


1 102 3 4 5 6 7 8 90

Load Dead todue forcesSheara)

10.1

05.3−00.2−

12.3

16.4

08.2

04.1

0.0

1 102 3 4 5 6 7 8 90

13.0−

35.2

69.4

46.2

32.0

55.1−

93.3−

75.5

90.3

59.0−

16.2−83.3−

46.5−

71.2−

90.7

94.0

43.9−

pactImLLDLtoduemomentBendingMin.&Max. b) ++

07.0

03.1−

09.4−

22.2

87.7−

98.1−

62.1

13.7

74.3

39.9

Figure 2.14:Maximum&Minimum shear forces at each section


49/53

51

2.12 Moment positive reinforcement:

The maximum positive moment occurs at section 10:

ft/'K 59.4M,ft/'K 31.15M ,ft/'K 8.9M ILLDL === +++

Based on AASHTO 3.22.1 Group 1 and Article 3.5, the following load combination is

applied:

( )[ ]ILLDLu MM67.1M3.1M ++=

( )[ ] ft/'K 94.5559.431.1567.18.93.1M u =++=

There is no centrifugal force in this problem.

The depth (d) = h – clearance - 2/d 1

Let assume the clearance is "1 and 1d is "75.0 .

"625.15"375.0"1"17d =−−=

Considering Eq. (2.21) we obtain,

( )( )

( )( )( )254.0

625.15129.0

1294.55

bd 9.0

MR

22

uu ===

The required ρ from Eq. (2.11),

( ) ( )( )

0044.05.485.0

254.0211

000,60

500,485.0=

−−==ρ

Check the maxρ from Eq. (2.11),

( )( )0311.0

000,60000,87

000,87

000,60

500,4825.085.0 b =+=ρ

( ) 0233.00311.075.0max ==ρ

OK 0044.0 max →ρ〈=ρ

( )( ) 2s in766.05.14120044.0A ==+

Use bar # 6, the spacing between bars is,


50/53


"89.6"12x766.0

44.0=

The #[email protected]”, the total cross sectional area is 0.81 2in

2.13 The cracking control

For psi000,40f y 〉

Ad f z cs= (ACI/318R, Article 10.6.4)

where,

armlinternaisd ,

dA

MMMf

s

ILLDLs αα

++= and cd is the distance from the centroid of

the rebars to the outer surface of the slab and A is the area of concrete in tension zone.

( ) "25.115.05.0d ""c =+=

Let assume the bar spacing is "5.6 , the total area of concrete surrounding the bar is

A = 1.25 x 2 x 6.5 = 2in25.16


( )( )( ) 22.34766.0625.1587.0

1259.43.158.9

f s =

++

= Ksi

( ) ( )( ) Ksi14538.9625.16375.136.32z 3 〈==

2.14 Moment negative reinforcement

The maximum negative moment occurs at section 6:

ft/'K 17.5M,ft/'K 23.17M ,ft/'K 48.23M ILLDL −=−=−= +++

Similarly with positive moment, based on AASHTO 3.22.1 Group 1 and Article 3.5,

the following load combination is applied:

( )[ ]ILLDLu MM67.1M3.1M ++=

( )[ ] ft/'K 15.7917.523.1767.148.233.1M u −=−−+−=

mailto:#[email protected]:#[email protected]:#[email protected]:#[email protected]


51/53

53

)clearance("5.2

"17"5.13

Distribution steel dia. 0.5"

Main reinforcement #8

The arrangement of the negative reinforcement is assumed based on the above figure,

Let assume the clearance and the wearing is "5.2 , 1d is "75.0 and 2d is "5.0

"625.13"5.0"375.0"5.20.17d =−−−= ( )( )

( )( )( )47.0

625.13129.0

1215.79

bd 9.0

MR

22

uu ===

The required ρ from Eq. (2.11),

( ) ( )( )

0084.05.485.0

47.0211

000,60

500,485.0=

−−==ρ

The maxρ from Eq. (2.11),

( )( ) 0311.0000,60000,87

000,87

000,60

500,4825.085.0 b =+=ρ

( ) 0233.00311.075.0max ==ρ

OK 0084.0 max →ρ〈=ρ

( )( ) 2s in37.1625.13120084.0A ==−

Use steel bar # 8, the spacing is in9.612x37.1

79.0=

The #[email protected]”, the total cross sectional area is 1.45 2in

2.15 The cracking control

In computing cd the wearing surface thickness is reduced.

mailto:#[email protected]:#[email protected]:#[email protected]:#[email protected]


52/53


"375.2"5.0"375.0"5.1d c =++=

Let assume the bar spacing is "5.6 , the total area of concrete surrounding the bar is

A = 2.375 x 2 x 6.5 = 2in875.30


( )( )( )

93.2745.1625.1587.0

1217.523.1748.23f s =

++= Ksi

( ) ( )( ) Ksi14594.130875.30375.393.27z 3 〈==

2.16 Minimum steel requirement

The reinforcement should be able to carry 1.2 cr M (AASHTO, Article 8.17.1)

Where cr M is the cracking moment.

t

gr

cr y

If M = (AASHTO, Article 8.13.3)

where r f is the modulus of rupture , gI is the moment of inertia of the concrete

section about the centroidal axis, neglecting the reinforcement, ty is the distance

from the centroidal axis to the extreme fiber. (AASHTO, Article 8.15.2.1)

psi503500,45.7f r == (AASHTO, Article 8.15.2.1.1)

( )"K 58.56

5.75.0

12/)5.7x12(503M

3

cr ==

( ) "K 67.958.562.1M2.1 cr ==

Check for bottom steel ( 2s in442.0A =+ )

( )( )

( )( )"

'c

ys578.0

125.485.0

60442.0

bf 85.0

f Aa ===


( )( )( ) cr '

n MK 29.1392/578.0125.660442.09.0M 〉=−=φ


53/53

55

Check for top steel ( 2s in705.0A =− )

( )( )

( )( )"

'c

ys922.0

125.485.060705.0

bf 85.0f Aa ===


( )( )( ) cr '

n MK 49.1392/922.0125.460705.09.0M 〉=−=φ

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