27

TEE BEAM AND SLAB BRIDGE DECK

INTRODUCTION:

This is the most common type of bridge. This type of bridge is economical for

spans between 10 to 25 m. The floor arrangement of a girder bridge depends

on the width of the road way. For a two lane or wider bridges the road way is

supported on the number of longitudinal girders generally with transverse

beams or diaphragms. The spacing of girders affects to a considerable extent

the cost of a bridge. As the number of girders are more form work is costly

and the number of bearings required are more. Wide spacing between

girders gives thicker slab but the number of girders are less, resulting in

less cost or form work and less number of bearings. Cross girders are

provided to connect main girders at suitable intervals. Maximum spacing of

cross girders is 10 m. By providing cross girders at closer intervals the slab

can be designed as two-way slab thus resulting in thinner section of the

slab. The cross girders in that case are to be designed for live load in the

most unfavorable position. When the slab is designed as two-way reinforced

slab, cross beam is intended to distribute the loading internally. The cross

beam should be intended to distribute the loading internally. The cross

beam should be designed to resist the moments produced by worst position

of loads. When the function of the cross beams is to act as the stiffening

girders, it should be provided with bottom reinforcement is equal to 0.3% of

the effective cross section of the cross beam. If the cross beam is monolithic

with the slab no top reinforcement is required as the reinforcement in the

slab above the cross beam will take any tension that may develop. In case

cross beam is not touching the slab nominal reinforcement is provided at

the top.

COMPONENTS OF A T-BEAM BRIDGE:

The T-BEAM superstructure consists of the following components.

1) Deck slab

2) Cantilever portion

28

3) Footpaths, if provided, kerbs and rails.

4) Longitudinal girders, considered in design to be of T section

5) Cross beams or diaphragms.

6) Wearing coat.

Standard details are used for kerbs and hand rails. The width of the kerb

may vary from 475mm to 600mm. Wearing coat can be asphaltic concrete of

average thickness 56mm or of cement concrete of M30 grade for an average

thickness of 75mm. Footpaths of about 1.5m width are to be provided on

either sides of the bridge located in municipal areas.

NUMBER AND SPACING OF MAIN GIRDERS:

The bridge having three main girders which is applicable for a two-lane

carriage way of 7.5m width. If the width of the bridge is adopted as 12.0m,

at least 4 main girders are necessary. The lateral spacing of the longitudinal

girders will affect the cost of the bridge. Hence in any particular design, the

comparative estimates of several alternative arrangements of girders should

be studied before adopting the final design. With closer spacing, the number

of girders will be increased, but the thickness of deck slab will be increased.

Usually this may result in less cost of materials. But the cost of form work

will increase due to large number of girder forms, as also cost of vertical

supports and bearing. Relative economy of two arrangements with different

girder spacings depends upon the relation between the unit cost of

materials and the unit cost of form work. The aim of the design should be to

adopt a system which will give a minimum cost. For the conditions in India

a three girder system is usually more economical than a four girder system

for a bridge width of 8.7m catering to two lane carriageway.

GENERAL FEATURES:

A typical tee beam deck slab generally comprises the longitudinal girder,

continuous deck slab between the tee beams and cross the girders to

provide lateral rigidity to the bridge deck. The longitudinal girders are

29

spaced at intervals of 2 to 2.5 m and the cross girders are provided at 4 to 5

m interval. The distribution of live loads among the longitudinal girders can

be estimated by any one of the following rational methods.

1. Courbons method.

2. Guyon Massonet method.

3. Hendry Jaegar method.

COURBONS METHOD:

Among these methods, Courbons method is the simplest and is applicable

when the following conditions are satisfied.

1) The ratio of span to width of deck is greater than 2 but less than 4

2) The longitudinal girders are interconnected by at least five symmetrically

placed longitudinal girders.

3) The cross girders extends up to a depth of at least 0.75 times the depth

of the longitudinal girders.

Courbons method is popular due to the simplicity of computations as

detailed below:

When the live loads are proportioned nearer to the kerb of the bridge as

shown in the figure (1) the centre of gravity of live load acts eccentrically

with the centre of gravity of the girder system. Due to this, eccentricity the

loads shared by each girder is increased or decreased depending upon the

position of the girder. This is calculated by Courbons theory by a reaction

factor given by.

RX = (W/ n)[1 + I/d2X .I) dX . e]

Where RX = Reaction factor for the girder under consideration

I = Moment of inertia of each longitudinal girder

30

dX = Distance of the girder under consideration from the central

axis of the bridge

W = Total concentrated live load

n = Number of longitudinal girders

e = Eccentricity of live load with respect to the axis of the bridge

The live load bending moments and the shear forces are computed for each

of the girders. The maximum design moments and shear forces are obtained

by adding the live and dead load bending moments. The reinforcements in

the main longitudinal girders are designed fo the maximum moments and

shears developed in the girders.

An approximate method may be used for the computation of bending

moments and shear forces in the cross girders. The cross girders are

assumed to be equally rigid so that the reactions due to the dead and live

loads are assumed to be equally shared by the cross girders. This

assumption will simplify the computation of bending moments and shear

forces on the cross girders.

K E R B

WW

e

A X IS O F T H E B R ID G E

1 .2 m

0 .8 5 0 .8 51 .2 m

d x

F IG .(1 ) P O S IT IO N O F L IV E L O A D F O R M A X IM U M B .M IN G IR D E R A

A B

31

DESIGN OF CROSS BEAMS: Since the purpose of the cross beams is to

stiffen the longitudinal beams and to provide a stiff superstructure, an

appropriate design usually adequate. However, if a more rigorous design

design in required, the following procedure may be adopted. Dead load

bending moment is computed considering a trapezoidal distribution of the

weight of the deck slab and wearing coat, besides including the self weight.

The cross beam is considered continuous over the spans. Suitable weighted

moment factors may be computed, considering different dispositions of the

standard loading over the cross beams. Similarly shears may also be

computed. Reinforcements may be provided to suit the values of moment

and shear. Additional cranked bars are usually two bars of 20 dia or 22 dia

may be provided to cater the diagonal tension. Using the approximate

method, the depth of the intermediate cross beam may be arranged such

that the bottom of the cross beam is at the top of the bottom flange of the

longitudinal beam when a bulb is provided or to a depth atleast 0.75 of

overall depth when straight T-ribs are adopted. Width of cross beam may be

adopted nominally as 250mm. The reinforcements may be provided at 0.5%

of gross area at bottom and 0.25% of gross area at top. The same

reinforcement may also be used for the end cross beam. Nominal shear

reinforcement consisting of 12 dia two-legged stirrups or 10 dia four-legged

stirrups at 150mm centers will usually be adequate. The design of the end

cross beam may be performed on the same lines as for the intermediate

cross beams. In earlier days the depth of the end cross beam had never been

reduced to about 0.6 of that for intermediate beam. With the use of

elastometric bearings provision has to be made for the possibility of lifting

the deck to replace the bearings. Hence the present practice is to keep the

depth of the end cross beam the same as for the intermediate cross beam.

The bottom reinforcement may be taken as half the bottom reinforcement for

the corresponding intermediate cross beam. The top reinforcement is kept

the same in addition, two bars of 20 dia or 22 dia are provided at top as

cranked bars to cater to diagonal tension occurring during the lifting

operation. The locations of jacks for lifting have to be indicated, and

additional mesh reinforcement should be provided at these locations.

32

DESIGN OF R.C.C T-BEAM BRIDGE DECK FOR THE GIVEN DATA:

GIVEN DATA:

Clear width of road way = 7.5 m

Span (c/c of bearings) = 18 m

Live load = IRC class AA tracked vehicle.

Average thickness of wearing coat = 80 mm

Materials = M 20 grade concrete

Fe 415 grade of steel

Spacing of cross girders = 3 m

DATA:

Effective span of T-beams = 18 m

Width of carriage way = 7.5m

Thickness of wearing coat = 80 mm

Materials = M 20 grade concrete

Fe 415 grade of steel

Spacing of cross girders = 3 m

PERMISISSIBLE STRESSES:

cb = 6.7 N/mm2 , m=10, st = 190 N/mm2, j = 0.91, Q = 0.762

CROSS-SECTION DETAILS:

Three main girders are spaced at 2.5 m centres.

Thickness of wearing coat = 80 mm

Thickness of deck slab = 250 mm

33

Kerbs 600 mm wide by 300 mm deep are provided. Cross girders are

provided at every 3m interval.

Breadth of cross girder = 300 mm

Depth of main girder = 160 cm at the rate of 10 cm per metre of span.

The depth of cross girder is taken as equal to the depth of main girder to

simplify the computations. The cross section of the deck and the plan

showing the spacing of cross girders are shown in figure (2).

DESIGN OF INTERIOR SLAB PANELS:

BENDING MOMENTS

Dead weight of slab = (110.224) = 6.00 KN/m2

Dead weight of wearing coat = (0.0822) = 1.76 KN/m2

Total load = 7.76 KN/m2

Live load is class AA tracked vehicle. One wheel is placed at the centre of the

panel as shown in fig (3).

u = (0.85 + 2 0.08) = 1.01 m

v = (3.6 + 2 0.08) = 3.76 m

(u/B) = (1.01/2.5) = 0.404

(v/L) = (3/3) = 1 (since dispersion exceeding the panel V=L)

K = (B/L) = (2.5/3) = 0.83

Referring to Pigeauds curves: m1 = 0.079 m2 = 0.041

MB = W (m1 + 0.15 m2)

= 350 (0.079 + 0.15 0.041)

= 29.8 kN-m.

As the slab is continuous design B.M = 0.8 MB

34

Design B.M including impact and continuity factor is given by

MB (short span) = (1.25 0.8 29.8)

= 29.8 kN-m

Similarly long span moment

ML = 350 (0.041 + 0.15 0.079)

= 18.5 kN-m

ML (long span) = (1.25 0.8 18.5)

= 18.5 kN-m

SHEAR FORCES:

Dispersion in the direction of span

= [0.85 + 2 (0.08 + 2)]

= 1.41 m

For maximum shear, the load is kept such that the whole dispersion is in

span. The load is kept at (1.41/2) = 0.705 m from the edge of the beams as

shown in figure (4)

Effective width of slab = KX (1-X/L) + bW

Breadth of the cross girder = 300 mm

Clear width of the panel = (3 0.3) = 2.7

Therefore (B/L) = (2.7/ 2.2) =1.2

From table given in the first chapter K for continuous slab is given as 2.36

Effective width of slab = 2.36 0.705 [1 (0.705/2.2)] + 3

= 4.13 m

Load per meter width = 350/4.13

35

= 84.75 kN

Shear force = 84.75 [(2.2 0.705)/2.2]

= 57.6 kN

DEAD LOAD B.M AND SHEARS:

Dead load = 7.76 kN/mm2

Total load on panel = (3 2.5 7.76)

= 58.2 kN

(u/B) = 1

(v/L) = 1

K = (B/L) = (2.5/3) = 0.83

(1/k) = 1.2

Referring to Pigeauds curves: m1 = 0.042 m2 = 0.029

MB = 58.2 (0.042 + 0.15 0.029)

= 2.7 kN-m

Taking continuity into effect

MB = (0.8 2.28) = 2.16 kN-m

ML = 0.8 58.2 (0.029 + 0.15 0.042)

= 1.64 kN-m

Dead load shear force = (7.76 2.2)/2

= 8.536 kN

DESIGN MOMENTS AND SHEARS:

Total bending moment MB = (29.8 + 2.16) = 31.96 kN-m

36

Total bending moment ML = (18.5 + 1.64) = 21.14 kN-m

Total shear forces = (57.6 + 7.216) = 64.816 kN

DESIGN SECTION:

Effective depth

d =

= . .

= 204.5 mm

Adopt overall depth = 250 mm

Effective depth = 230 mm

Ast (short span) = (32.08 106)/ (190 0.91 230)

= 804 mm2

Use 12 mm dia bars at 120 mm centers (Ast = 905 mm2)

Eff depth for long span using 10 mm dia bars = (23065)=219 mm

Ast (long span) = (21.14 106)/ (190 0.91 219)

= 559 mm2

Use 10 mm dia bars at 120 mm centers (Ast = 629 mm2)

CHECK FOR SHEAR STRESS:

Normal shear stress V = (64.816 103)/ (1000 230)

= 0.281 N/mm2

(100 Ast / bd) = (100 905) / (1000 230)

= 0.39

37

From IS: 456 2000 design strength of concrete is

C = 0.43 N/mm2

Since V < C shear stresses are permissible within the limits.

DESIGN OF LONGITUDINAL GIRDERS:

REACTIONS: Using Courbons theory, the I.R.C class AA loads are

arranged for maximum eccentricity as shown in figure (5)

Reaction factor for the outer girder is

RA = (2W1/3) [1 + (3I 2.5 1.1) / (2I 2.52)]

= 1.107W1

RB = (2W1/3) [1 +0]

= 0.667W1

If W = axle load = 700 kN

Therefore W1 = 0.5W

RA = (1.107 0.5W)

= 0.5536W

RB = (0.667 0.5W) = 0.333W

DEAD LOAD FROM SLAB PER GIRDER:

The dead load of deck slab is calculated with reference to figure (6) is

shown below

Weight of

1) Parapet railing = 0.7 kN/m

2) Wearing coat = (0.081.122) = 1.936 kN/m

3) Deck slab = (0.251.124) = 6.600 kN/m

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4) Kerb = (0.500.624) = 7.200 kN/m

Total = 16.436 kN/m

Total dead load of the deck = (2 16.436) + (7.76 5.3)

= 74 kN/m

It is assumed that the dead load is shared equally by all the girders

Dead load per girder = (74/3)

= 24.67 kN/m.

LIVE LOAD BENDING MOMENTS IN GIRDERS:

Span of the girder = 18 m

Impact factor (for class AA) = 10%

The live load is placed centrally on the span as shown in figure (7)

Bending moment = (3509) (350 (1.8/2))

= 2835 kN m

Bending moment including impact and reaction factor for outer girder

= 2835 1.1 0.5536

= 1726.4 kN/m

Bending moment including impact and reaction factor for inner girder

= 2835 1.1 0.333

= 1038.46 kN-m

LIVE LOAD SHEAR:

For estimating the maximum live load shear in the girders, the IRC class

AA loads are placed as shown in figure (8)

39

Reaction of W2 on girder B = (350 0.45) / 2.5

= 63 kN

Reaction of W2 on girder A = (350 2.05) / 2.5

= 287 kN

Total load on the girder B = (350 + 63)

= 413 kN

Maximum reaction in girder B = (413 16.2)/18

= 371.7 kN

Maximum reaction in girder A = (287 16.2)/ 18

= 258.3 kN

Maximum live load shear with impact factors in

Inner girder = (371.7 1.1)

= 408.87 kN

Outer girder = (258.3 1.1)

= 284.13 kN

Maximum live load shear with impact factors in inner girder is

408.87 KN and outer girder is 284.13 KN.

DEAD LOAD B.M AND S.F IN MAIN GIRDER:

The depth of the girder is assumed as 1800 mm (100 mm for every meter

of span)

Depth of rib = (1.8 0.25) = 1.55 m

Width = 0.3 m

Weight of rib per meter = (1 0.3 1.55 24)

40

= 11.16 kN/m

The cross girder is assumed to have the same cross sectional dimensions

as the main girder

Weight of cross girder = 11.16 kN/m

Reaction on main girder = (11.16 2.5)

= 27.9 kN

Reaction from deck slab on each girder = 24.67 kN/m

Therefore total dead load per meter on girder = (24.67+11.16)

= 35.83 kN/m

Maximum bending moments are computed from the figure (9)

Maximum B.M at centre of span is obtained as

Mmax=(420.129) ((27.9/2)9) (27.96) (27.93) (35.83 (81/2))

= 1829 kN-m

Maximum shear at support

MAX.S.F = [(35.8318)/2] + [(27.97)/2] 27.9 = = 393 kN

DESIGN B.M AND S.F

The design B.M and shears are given in table below

BENDING MOMENTS (kN-m)

DLBM LLBM TOTAL BM

OUTER

GIRDER 1829 1727 3556

INNER GIRDER 1829 1039 2868

41

SHEAR FORCE (kN)

DLSF LLSF TOTAL SF

OUTER

GIRDER 393 284.13 677.13

INNER GIRDER 393 408.87 801.87

The beam is designed as a T-section. Assuming an effective depth

d = 1900 mm

Approximate lever arm = [(1900 (250/2)] = 1775 mm

Ast = (3556 106) / (190 1775)

= 10545 mm2

Provide 16 bars of 32 mm diameter in three rows (Ast=12864 mm2).

Shear reinforcements are designed to resist the maximum shear at

supports.

Nominal shear stress V = (V/bd)

= (801.87103) / (3001900)

= 1.407 N/mm2

Assuming 2 bars of 36 mm diameter to be bent up at support section

resisted by bent up bars is given by

Vus = sv Asv sin = (15028041)/ (1000 2) = 171 kN

Shear to be resisted by vertical stirrups

42

= 801.87 171

= 631 kN

Using 10 mm dia 4-legged stirrups, spacing

Sv = (sv Asv d) / V

= (1504791900) / 631103

= 143 mm

Provide 10 mm diameter 4-legged stirrups at 150 mm c/c

DESIGN OF CROSS GIRDER

Self weight of the cross girder = 11.16 kN/m

Loads on the cross girder are as shown in fig (10)

Dead load from slab = (20.52.51.257.76)

= 24.25 kN

Uniformly distributed load = (24.5/2.5)

= 9.8 kN/m

Total load on cross girder = (11.16 + 9.8)

= 20.96 kN/m

Assuming the cross girder to be rigid, reaction on each cross girder

= (20.96 5)/ 3 = 34.94 kN

For maximum B.M in the cross girder, the loads of I.R.C class AA should

be placed as shown in figure (11)

Load on the cross girder = [(350 (3 - 0.9)/3] = 245 kN

Assuming the cross girder as rigid, reaction on each longitudinal girder

is

43

= [(2245)/3]

= 163.33 kN

Maximum B.M in cross girder under the load

= 163.331.475 = 241 kN-m

Live load B.M including impact = 1.1 241 = 265 kN-m

Dead load B.M at 1.475 m from support

= (34.941.475 20.961.4752/2)

= 29 kN-m

Total design B.M = (265 + 29)

= 294 kN-m

Live load shear including impact

= (2245/3) 1.1

= 180 kN

Dead load shear = 34.94 kN

Total design shear = (180 + 34.94)

= 215 kN

Assuming an effective depth for cross girder = 1970 mm

Ast = [(394106)/(2000.911970)]

= 1099 mm2

Provide 4 bars of 20 mm diameter (Ast = 1256 mm2)

Shear stress V = (V/bd)

= (215103) / (3001970)

44

= 0.36 N/mm2

Using 10 mm diameter 2 legged stirrups,

Spacing Sv = [(2002791970) / (215103)]

= 290 mm

Adopt 10 mm diameter 2 legged stirrups at 150 mm c/c throughout the

length of the cross girder.

The details of reinforcement are shown in figure (12.1), figure (12.2) and

figure (13)

DESIGN OF BEARINGS:

Vertical reactions from girders

Dead load reaction from girder is calculated using the figure (9) is as

follows.

Rd = (35.83 18 + 27.9 6)/2 = 406.17 kN

Live load reaction including impact factor is calculated from the figure(7)

and figure (8)

Rl= 408.87 kN

Total vertical reaction from the girders = Rd+Rl

=406.17+408.87=814.87kN.

Assuming horizontal reaction = 100 kN.

Permissible compressive stresses in concrete bed = 4kN/mm2

Permissible bearing stress in steel plate = 185 N/mm2

Permissible shear stress in steel = 105 N/mm2

(Above values are obtained from IRC: 83-1982)

45

DESIGN DETAILS:

1) BED PLATE:

Area of bed plate = (814.87 1000)/4

= 20.37 104 mm2

Provide a bed plate of overall size 40065040 mm and top plate of

overall size 40065040 mm.

2) ROCKER DIAMETER:

Let R = Radius of rocker surface in contact with the flat surface of

bed plate.

Vertical design load per unit length (170R33)/ E2 = Nominal ultimate tensile strength of material (250 N/mm2) R = Radius of the rocker spherical surface.

E = Modulus of elasticity of material (200 kN/mm2)

Design load per unit length = (814.87 1000)/650

= 1253.33 N/mm2

Hence = 1253.33 =

R = 137.38 mm

Provide a radius of 200 mm for rocker surface.

3) ROCKER PIN:

Providing 2 rocker pins the horizontal shear force to be resisted by

each pin = 100/2 = 50 kN

If d = diameter of rocker pin

=

105=50103

d = 24.6 mm

46

Adopt a tapering pin with a diameter of 25 mm at top and bottom

diameter of 30 mm and height 55 mm.

4) THICKNESS OF BASE PLATE:

Maximum bending moment about the central axis of base plate

= (814.87/2) 1000 100

= 4.074 107 Nmm

If t = thickness of base plate required

Section modulus Z = (bt2)/6 = (M/)

t =

= 48.48 mm

Provide an overall thickness of 72 mm for the central portion of the

base plate.

5) CHECK FOR BEARING STRESS:

Assuming 50% contact area between top and bottom plates

Bearing stress = (814.87 1000)/ (650 100)

= 12.53 N/mm2 < 185 N/mm2

Hence bearing stresses are within permissible limits. Figure (14) shows

the dimensional details of the rocker bearing.

PIER:

Design of pier: The salient dimensions of the pier like the height, pier

width and batter are determined as follows

1) HEIGHT: The top level of pier is fixed 1 to 1.5 m above the high

flood level, depending upon the depth of water on the upstream

side. Sufficient gap between the high floods level and top of pier is

essential to protect the bearings from flooding.

2) PIER WIDTH: The top width of pier should be sufficient to

accommodate the two bearings. It is usually kept at a minimum of

600 mm more than the outer to outer dimension of the bearing

plates.

3) PIER BATTER: Generally the sides are provided with a batter of 1

47

in 12 to 1 in 24. Short pier have vertical sides. The increased

bottom width is required to restrict the stresses developed under

loads within safe permissible values.

4) CUT AND EASE WATERS: The pier ends are shaped for

streamlining the passage of water. Normally the cut and ease

waters are either shaped circular or triangular.

STABILITY ANALYSIS OF PIER:

A pier as shown in figure (15) supports the deck forming a simple

highway. The various forces acting on the pier are listed below.

Dead load from each span= ((35.818+27.96)/2) 3 = 1217.7

1218 kN

Reaction due to live load on one span

= 408.87 3

= 1226.61 kN 1227 kN

Assuming,

Breaking force = 140 kN

Wind pressure on pier = 2.4 kN/mm2

Material of pier = 1:3:6 cement concrete

Density = 24 kN/mm2

DESIGN COMPUTATIONS:

1) STRESS DUE TO DEAD LOADS AND SELF WEIGHT OF PIER

From figure (15)

Dead load from superstructure = (2 1218) = 2346 kN

Self-weight of pier = 8.50.5(2+3)10 24 = 5100 kN

Total direct load = 7446 kN

Compressive stress at base of pier = (7446/(8.53))=292 kN/mm2

48

2) EFFECT OF BUOYANCY:

Width of pier at H.F.L = 19 m

Submerged volume of pier = 8.50.5(1.9+3.0)9 = 187.4 m3

Reduction in weight of pier due to buoyancy = (187.410)

=1874 kN

Tensile stress at the base due to buoyancy = - (1874/(8.53))

= - 73.5 kN/mm2

3) STRESS DUE TO ECCENTRICITY OF LIVE LOAD:

Reaction due to live from one span is 1227 kN acting at an

eccentricity of e=0.5 m

Moment about base = M = 12270.5 = 613.6 kN-m

Section modulus = Z = (8.532)/6

= 12.75 m3

Stresses developed at the base of pier due to the eccentricity of the

live load = (1227/ (8.53)) (613.6/12.75)

= 48.12 48.12

= 96.24 kN/mm2

= 0

4) STRESSES DUE TO LONGITUDINAL BREAKING FORCES:

Assuming Breaking force at bearing level = 100 kN

Moment about the base of pier = (100 10) = 1000 kNm

Stresses at base = (M/Z) = (1000/12.75)

= 78.43 kN/mm2

5) STRESSES DUE TO WIND PRESSURE:

Total wind pressure on pier = (area) (wind intensity)

= ((2+3)/2)10 2.4

= 60 kN

Assuming the wind load to act at the mid height of the pier,

moment about the base of the pier = (605) = 300 kNm

Modulus of section at base = Z = (38.53)/6 = 361.25 m3

Stresses developed at base due to wind load

= (M/Z) = (300/361.25) = 0.83 kN/mm2

49

The maximum and minimum stresses developed are computed by

combining the stresses due to the various load combinations as

shown in the below table.

S.No Type of load

Stress (kN/mm2)

When dry During Floods

1 Dead load and

self-weight 292 292

2 Buoyancy -- -73.5

3 Eccentric live

load 96.24 96.24

4 Braking force 78.43 78.43

5 Wind pressure 0.83 0.83

Maximum stresses 467.5 394

Minimum stresses 308.98 235.48

The material of the pier being 1:3:6 cement concrete the maximum

permissible compressive stress in concrete is 2 N/mm2 or 2000 kN/mm2

Hence the stresses developed at the base of the pier are within safe

permissible.

FOUNDATION:

Total load on the foundation is calculated as follows

Dead load of girder on each pier = 2346 kN

Dead load of pier = 5100 kN

Live load = 12272 = 2454 kN

Total load = 9900kN

50

Assuming width of pier = 1 m

Length of pier = 9 m

Size of pile = 300 mm by 300 mm

Spacing of pile = 1.5 m

Materials M-20 grade of concrete and Fe415 grade of steel.

Assuming hard strata is available at a depth of 6 ma below the ground

level at bridge site.

1) ARRANGEMENT OF PILE AND PILE CAP:

Fourteen pile are arranged at a spacing of 1.5 m as shown in

figure(16)

Load on each pile = (9900/14) = 707 kN

2) PILE REINFORCEMENTS:

a) Longitudinal reinforcement

Length of pile above ground level = 0.6 m

Total length of pile = (6+0.6) = 6.6 m

Size of pile = 300 mm by 300 mm

L = 6.6 m B = 0.3 m

Ratio (L/B) = 6.6/0.3 = 22 > 12

Hence the pile is designed as long column

Reduction coefficient = (1.25-(L/48B))

= (1.25 (22/48)) = 0.792

Safe permissible stress in concrete = = 0.795 5

= 3.96 N/mm2

Safe permissible stress in steel = (0.792 190) = 150 N/mm2

Load carrying capacity of pile is expressed as

P = + 707103 = 3.96[(300300)- ] + 150 Solving = 2400 mm2

According to IRC 78:1893 the longitudinal reinforcement A

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should not be less than 1.25 percent of gross cross section for

pile with a length less than 30 times the least width.

Hence A (0.0125300300) 1125 mm2 Adopt 8 bars of 20 mm diameter providing an area of A=2513 mm2

b) Lateral reinforcement

In the body of the pile the lateral reinforcement should be not

less than 0.2 percent of the gross volume.

Using 8 mm diameter ties

Volume of tie = 50[4(300-80)] = 44000 mm3

If p=pitch of the ties

Volume of pile per pitch length = (300300)p mm3

= 90000 p mm3

Hence equating we have 44000 = 0.00290000p

P = 244 mm

Maximum permissible pitch = 0.5300 = 150 mm

Hence provide 8 mm diameter ties at 150 mm centres in the

main body of the pile.

c) Lateral reinforcement near pile head:

Lateral reinforcement is of particular importance in resisting

driving stresses near the pile head provided for a length of

3B = 3 300

= 900 mm

Spiral reinforcement is provided near pile head using 8 mm

diameter helical ties

Volume of spiral per mm length = (0.6/1000)[(3003001)]

= 540 mm2

If p= pitch of the spiral

p= (circumference of spiral AS)/540

Provide a clear cover of 40 mm to the main longitudinal

reinforcement of 20 mm diameter bars and using 8 mm

diameter spiral ties inside the main reinforcements.

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Diameter of spiral = [300-(240)-2(20)-8] = 172 mm

p = ( 172 50)/50 = 50 mm

Adopt 8 mm diameter spirals at a pitch of 50 mm for a length of

900 mm at the top of pile.

d) Lateral reinforcement near pile ends

Lateral reinforcement of 0.6 percent of gross volume is provided

in form of ties for a distance of 3 times the least lateral

dimension both at top and bottom of the pile.

Volume of the ties = 0.6 percent of gross volume for a length of

(3300) = 900 mm

Using 8 mm diameter ties

Volume of each tie = 50[4(300-80)] = 44000 mm3

If p= pitch of the ties

Volume of piles per pitch length = (300300p)

= 90000p

44000 = (0.6/100)(90000p)

Solving p = 81.48 mm

Adopt 8 mm diameter ties at 80 mm centres for a length of 900

mm from the ends of the pile both at top and bottom.

The longitudinal and cross-sections of the pile with

reinforcement details is shown in figure (16.1)

3) PILE CAP:

The arrangement of pile with pile cap is shown in figure (16).

Referring to figure (16.2) the maximum bending moment in pile

cap is computed as follows.

MZZ = (0.5W1-0.5W0.25) = 0.375W kNm

Where W= 7072 = 1414 kN

MZZ = 531 kNm

The effective depth required is given by

531 100.874 1500 d = 636.42 mm

Adopt overall depth of 690 mm

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ASt = (531 106)/(2300.9636) = 4033 mm2 per 1.5m width

Using 25 mm diameter bars spacing is given by

S= (1500491)/4033 = 182 mm

Adopt 25 mm diameter bars at 180 mm centres

Distribution steel = 0.12% of gross area

= 0.00126901000

= 828 mm2/m

Adopt 16 mm diameter bars at 250 mm centres as distribution

steel.

Maximum shear force = V = 707 kN

Shear stress = V = (V/bd) = (707 1000)/(1500 630) = 0.748

C = 0.28 N/mm2

Vus = 707 [(0.281500600)/1000] = 455 kN

Using 10 mm diameter stirrups (8 legged) spacing is given by

S = (879230630)/(4551000) = 201 mm

Dopt 10 mm diameter stirrups at 200 mm centres in a width of 1500

mm.