Download - Class room example on T beam bridge
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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
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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
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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
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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
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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.
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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
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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
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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
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= 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
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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
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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)
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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)
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= 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
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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
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= 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
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= [(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)
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= 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)
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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
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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
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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
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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
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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
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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.