Slab Bridge Design Example

Slab Bridge - Design Example
Assumptions
 Main reinforcements are placed parallel to the center line
of road way
 The bottom of the slab is assumed level
 Center to center support is assumed perpendicular to the
supports
 Additional amount of concrete thickness for cross fall is
not considered for flexural design
Preliminary dimensions:
 Clear span = 7.2 m  Skewness angle = 0˚ 00'
 Width of abutment shelf 00"
= 0.5 m  Railing depth = 0.35 m
 Clear width of super  Railing width = 0.25 m
structure = 4.2 m  Post cross section = 0.25
 Total width of super m * 0.25 m
structure = 5 m  Post height = 0.55 m
 No. of lane = 1  Average post spacing =
 Total length of super 1.4 m
structure = 7.2 + 0.5 + 0.5
= 8.2 m
1.4 m
7.2 m
0.5 m
0.5 m
Effective Span = 7.7 m
Longitudinal Elevation
0.25 m
Railing
0.35 m
4.2 m
0.55 m
5m Post
Cross sectional view Curb 0.4 m
0.4 m
Details of Curb, Post and Railing
Materials:
 Concrete: f'c = 30 MPa  For bar size <= 20 mm
 Unit weight of concrete = diameter, Grade:40: fyk =
24 kN/m3 300 MPa
 Steel: For bar size > 20  Es = 2 * 105 MPa; Ec =
mm diameter,Grade:60: 24000 MPa;
fyk = 420 MPa  Modulus of rupture =
f  0.63 f ' = 3.45 MPa
r c
Preliminary thickness of slab:

 Depth of slab = 6% of the span length = (6/100)* 7700 =
462 mm
 Consider overall depth of slab = 470 mm
Live loads:
Equivalent width of longitudinal strips per lane for both shear
and moment with one lane =
o E1 = 250 + 0.42 √L1W1 L1 = 7700 mm
W1 = 5000 mm
o E1 = 250 + 0.42 √7700 * 5000 = 2856 mm = 2.86 m
 Impact factor (Dynamic load allowance): (1 + IM/100) =
(1 + 33/100) = 1.33
Design truck load:
Front axle load = 35 kN; Load dispersed per m in
35
transverse direction = 2.86 = 12.24 kN/m
Other axle loads = 145 kN Load dispersed per m in

145
transverse direction = 2.86 = 50.7 kN/m
Design tandem load:

Axle load = 110 kN; Load dispersed per m in transverse
110
direction = 2.86 = 38.46 kN/m
Lane load:
Load = 9.3 kN/m; Load dispersed per m in transverse
9.3
direction = 3 = 3.1 kN/m
Shear force calculation:
Loads on span:
50.7 kN 50.7 kN
4.3 m
Truck load
38.46 kN
38.46 kN Tandem load
1.2 m Lane load = 3.1 kN/m
A B
7.7 m
0.8442
1 0.4416
IL for maximum reaction at A

Maximum shear at A due to truck loads = 50.7 * 1 + 50.7 *
0.4416 = 73.09 kN
Maximum shear at A due to tandem loads = 38.46 * 1 +
38.46 * 0.8442 = 70.93 kN
Maximum shear at A due to lane load = 3.1 [½ * 7.7 * 1] =
11.94 kN
Design truck load governs.
Shear with dynamic load allowance = (73.09 * 1.33) +
11.94 = 109.15 kN
Bending moment calculation:
Absolute maximum bending moment due to moving loads
is found by placing one of the axle loads and the resultant
of the system at equidistant from the center.
For design truck load:
50.7 kN 50.7 kN
4.3 m
A C D B
2.15 m
0.625 m 1.075 m
m R
3.85 m 2.775 m
7.7 m
For this moving load of design truck, absolute maximum

bending moment occurs at section D, for the load position
shown .
(4.925 * 2.775)/7.7
= 1.775
0.225
ILD for bending moment at D

BM at D = 50.7 (1.775 + 0.225) = 101.4 kN m
For Design Tandem Load:
Load position for absolute maximum BM to occur at D, a
section 0.3 m from mid span section is shown below.
Maximum BM at D = 38.46 * (1.9133 + 1.36) = 125.89 kN m
Hence the BM is governed by design tandem load
38.46 kN 38.46 kN
1.2 m
A B
C D
2.95 m
R 0.3 m
3.85 m 3.55 m
4.15 m
7.7 m
(4.15 * 3.55)/7.7 = 1.9133

1.36
ILD for BM at D
For Design Lane Load:
Lane load BM is determined at the section of maximum
BM, section D, of design tandem loading
Lane load BM at D = 3.1*(7.7 * 1.9133)/2 = 22.84 kN m
Total maximum BM
Total maximum BM with IM = (125.89*1.33) + 22.84 =
190.27 kN m
Dead Loads
Assume depth of wearing surface = 0.075 m
Weight of wearing surface = 0.075 * 22.5 = 1.6875 kN/m2
Weight of deck slab = 0.47 * 24 = 11.28 kN/m2
Weight of railing, post and curb
=[ { (0.35 * 0.25 * 24) + 0.7143 * (0.25 * 0.25 * 0.55 *
24) + (0.4 * 0.4 * 24)} * 2] ÷ 5
= 2.6 kN/m2
Total weight of components DC = 11.28 + 2.6 = 13.88
kN/m2
Maximum shear due to DC = VDC = 13.88 [½ * 7.7 * 1] =
53.44 kN/m width
Maximum Shear due to DW = VDW = 1.6875 [½ * 7.7 * 1] =
6.5 kN/m width
Maximum bending moment due to DC at the section of
maximum LL BM
13.88 kN/m
4.15 m 3.55 m
4.15 * 3.55 / 7.7 = 1.9133

BM at the section D due to the load DC = = 13.88 [½ * 7.7 *
1.9133] = 102.24 kNm
Maximum BM due to DW at the section of maximum LL BM = =
1.6875 [½ * 7.7 * 1.9133] = 12.43 kNm
Design load combination for strength I limit state
Q = ηiγiQi
Consider the design to be of conventional one, non-redundant
and typical bridge, for these assumptions,
ηD = 1.00 ηR = 1.05 ηI = 1.00
Factored resistant moment Mu = 1.05[1.25 MDC + 1.5MDW +
1.75MLL+IM]
= 1.05[1.25 (102.24 ) + 1.5(12.43 ) + 1.75(190.27)] = 479.4 kNm
As per ERA bridge design manual LRF design procedure,
factored resistant moment Mu = фMn
Where, Mn = Nominal resistant moment,
Ф = Resistance factor
Mn = Mu/ф, has to be used in the equation for steel ratio ρ
Steel ratio ρ can be written as,
2𝑀𝑢 𝑓𝑐′ 2∗479.4∗10 6 30
ρ = 1 − 1 − ∅𝑏𝑑 2 𝑓 ′ = 1 − 1 − 0.9∗1000∗433 2 ∗30
𝑐 𝑓𝑦 420
= 0.00712
Effective depth d = 470 – 25 – 24/2 = 433 mm; assuming
24 mm dia reinforcement to be used
ρmin = 0.03 * (30/420) = 0.00241 < 0.00712
As = 0.00712 * 1000 * 433 = 3082.96 mm2
𝜋∗12 2 ∗1000
Spacing of 24 mm diameter rods = = 146.74 mm
3082 .96
Provide 24 mm diameter rods at 140 mm c/c as main

flexural reinforcement along the span.
 * (12)2 *1000
As provided =  3231mm2
140
Distribution of main reinforcement to resist against cracking:

Concrete is considered cracked if tensile stress in concrete >=
80% of the modulus of rupture
fr = Modulus of rupture = 3.45 MPa (as determined in the
earlier steps)
Service load moment Mu = 1.05[1.00(102.24 ) + 1.00(12.43 ) +
1.00(190.27)] = 320.187 kNm
320.187∗10 6
fct = tensile stress in concrete = M/Z = 1000 ∗470 2 /6 = 8.69 MPa
0.8fr = 0.8 * 3.45 = 2.76; fct > 0.8fr Therefore the section has to
be taken as a cracked section
Steel stress: To find the steel stress, for cracked section,
consider the equivalent transformed section of concrete.
Modular ratio = m = Es/Ec = 2 * 105/24000 = 8.33;
mAs = 8.33 * 3231 = 26914.23 mm2
x can be determined by taking moments of areas above and
below NA
1000 * x * x/2 = 26914.23 (433 – x)
It can be reduced to a quadratic equation,
x2 + 53.8x – 23307.7 = 0 X = 128.12 mm
Moment of inertia I = 1/3(1000 * 128.123) + 26914.23 (433 –
128.12)2 = 3.2 * 109 mm4
320.187∗10 6 ∗(433−128.12)
fs/m = My/I = = 30.5;
3.2∗10 9
fs = 8.33 * 30.5 = 254.11 MPa

𝑍
fsa = Where Z = 30,000 for moderate exposure;
(𝑑 𝑐 𝐴)1/3
dc = 25 + 24/2 = 37 mm
A = 2dcS/1 = 2 * 37 * 140/1 = 10360 mm2
30000
fsa = [37 = 412.98 MPa
10360 ]1/3
𝑍
But fsa = (𝑑  0.6 fy ; 0.6(420) = 252 MPa
𝑐 𝐴)1/3
fs > fsa Hence the provision of spacing of main reinforcement is

not safe against cracking.
The spacing of reinforcement shall be adjusted to satisfy the
above criterion
Consider spacing of 24 mm diameter rods at 120 mm center to
center.
 * (12) 2 *1000
As provided =  3770mm 2
120
mAs = 8.33 * 3770 = 31404.1 mm2
x can be determined by taking moments of areas above and
below NA
1000 * x * x/2 = 31404.1(433 – x)
It can be reduced to a quadratic equation,
x2 + 62.81x – 27195.95 = 0 X = 136.47 mm
Moment of inertia I =
1/3(1000 * 136.473) + 31404.1 (433 – 136.47)2 = 3.6 * 109 mm4
320.187∗10 6 ∗(433−136.47)
fs/m = My/I = = 26.37;
3.6∗10 9
fs = 8.33 * 26.37 = 219.69 MPa

𝑍
fsa = Where Z = 30,000 for moderate exposure; dc =
(𝑑 𝑐 𝐴)1/3
25 + 24/2 = 37 mm
A = 2dcS/1 = 2 * 37 * 120/1 = 8880 mm2
30000
fsa = [37 = 435 MPa
8880 ]1/3
𝑍
But fsa = (𝑑  0.6 fy ; 0.6(420) = 252 MPa
𝑐 𝐴)1/3
fs < fsa Hence the provision of spacing of main reinforcement is
safe against cracking.
Distributor reinforcement
For primary reinforcement parallel to traffic, % of main
reinforcement as distributor reinforcement =
1750 / s  50%
Where S = Span length = 7.7m = 7700 mm

1750 / 7700  20%
20% of main reinforcement = (20/100)*3231 = 646.2 mm2
𝜋(82 )1000
Spacing of 16 mm dia rods = = 311.15 mm
646.2
Provide 16 mm diameter rods at 300 mm c/c as distribution

reinforcement
Temperature and shrinkage reinforcement
Temperature and shrinkage reinforcement As  0.75 Ag/fy
As = 0.75 * 1000*470/420 = 839.3 mm2 for both direction
For top layer alone: As = 839.3/2 = 419.6 mm2
𝜋(62 )1000
Spacing of 12 mm dia rods = = 269.5 mm
419.6
Provide 12 mm dia rods at 260 mm C/C in both direction at the

top as temperature and shrinkage reinforcements.

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Slab Bridge - Design Example

Effective Span = 7.7 m

Cross sectional view Curb 0.4 m

Details of Curb, Post and Railing

Preliminary thickness of slab:

Other axle loads = 145 kN Load dispersed per m in

Design tandem load:

IL for maximum reaction at A

For this moving load of design truck, absolute maximum

ILD for bending moment at D

(4.15 * 3.55)/7.7 = 1.9133

4.15 * 3.55 / 7.7 = 1.9133

Provide 24 mm diameter rods at 140 mm c/c as main

Distribution of main reinforcement to resist against cracking:

fs = 8.33 * 30.5 = 254.11 MPa

fs > fsa Hence the provision of spacing of main reinforcement is

fs = 8.33 * 26.37 = 219.69 MPa

Where S = Span length = 7.7m = 7700 mm

Provide 16 mm diameter rods at 300 mm c/c as distribution

Provide 12 mm dia rods at 260 mm C/C in both direction at the

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