Composite Bridge Design

STEP 1: DESIGN DATA
1.1 Proposed Structural System

1.1.2 Superstructure Type: R.C Deck and
Steel Pratt Truss Girders
1.1.3 Number of spans: 1
1.1.4 Road importance: Trunk Road

1.1.5 Preliminary Member Sizing
Effective span = 30m
Total carriage way width = 7.50m
Pedestrian kerb ( on each side) width = 1.50 m
Surfacing slope = 3%
Deck Slab thickness (AASHO Table 2.5.2.6.3-1)
1.2(𝑆+3000)
Minimum thickness, tmin = = 𝟒𝟐𝟎𝐦𝐦
30
Depth of truss girders, h:

𝑠𝑝𝑎𝑛 𝑠𝑝𝑎𝑛 30000
ℎ~ 𝑠𝑎𝑦 =
(15 𝑡𝑜 25) 20 20
= 𝟏, 𝟓𝟎𝟎𝐦𝐦
Spacing of vertical truss members
For simplifying design, take this spacing = depth of
truss girders = 1,500mm
Bracing members to the trusses to be provided at
interval of 3.00m along the span of the bridge
STEP 2: STRUCTURAL LAYOUT
STEP 3: LOADS ANALYSIS ON DECK
DEAD LOAD (ℽfL = 1.15)
Permanent Dead Load
S.W of slab in carriageway: = 0.42 x 24 x 1.15 = 11.60kN/m2
Kerb 1: = 0.10 x 24 x 1.15 = 2.76 kN/m2

Kerb 2: = ½ x 0.42 x 24 x 1.15 = 5.80 kN/m2
Superimposed Dead Load (ℽfL = 1.75)

S.W of wearing course = 0.030 x 20 x 1.75 = 1.05kN/m2
(at the edge)
= 0.1125 x 20 x 1.75 = 3.94 kN/m2
(at apex)
Parapet = 5.00 x 1.75 = 8.75kN/m2
LIVE LOAD
HA Load
Reference Calculations
Carriageway width = 7.5m
Loaded length = 30m
Cl 3.2.9.3.1. Number of notional lanes
Number of notional lanes =2
Notional lane width (bL) = 7.5/2 = 3.75m
Cl 6.2.1. Nominal HA UDL
Table 13 Loaded length = 30m
HA UDL = 30 kN/m (per notional lane)
Cl 6.2.2. Knife Edge Load (KEL)
KEL = 120 kN (per notional lane)
LIVE LOAD
HA Load
HA alone (L = 1.50)

UDL: 30 kN/m/lane : 30 x 1.5/3.75 = 12.00 kN/m2
KEL: 120 kN/lane: 120 x 1.5/3.75 = 48.00 kN/m
(transversal line load)
HA with HB (L = 1.30)

HA Loading
UDL: 30 x 1.3/3.75 = 10.40 kN/m2
KEL: 120 x 1.3/3.75 = 41.60 kN/m
(transversal line load)
HB Loading (taking 45 Units)
Total wheels: 16 wheels in 4 axles and (L = 1.30)
Load per wheel: 45 x 2.5 x 1.3 = 146.25 kN
Load per axle : 146.25 x 4 = 585.00 kN
Total vehicle load: 585 x 4 = 2,340 kN
Pedestrian (L = 1.5)

Pedestrian Load: 5 x 1.5 = 7.50 kN/m2
ANALYSIS OF DECK SLAB
Dead Load (DL)
Consider 1.0 m width along the slab
Slab VL= 11.60 x 0.15 = 1.74kN ML = -11.60 x 0.152/2 = - 0.13k Nm

VR = 11.60 x7.5/2 = 43.5kN MR = 11.6 x 7.52/8-0.13 = 81.43kNm
Kerb VL1 = 2.76 x 1.50 = 4.14kN ML1 = -2.76 x1.52/2 = -3.11kNm
VL2 = 5.80 x 1.35 = 7.83kN ML2 = - (5.80 x1.35) x (1.35/3 + 0.15)

= - 4.70kNm
Surfacing: VR = 0.5 x (1.05+4.99) x 7.5/2 = MR = 11.33 x 3.75 – 1.05 x 3.752/2
11.33 kN – 0.5 x 3.94 x 3.752/3 = 25.87kNm
Parapet: VL = 8.75 kN ML = 8.75 x 1.5 = - 13.13 kNm
Live Load
Case 1: HA on both lanes
Due to VR = 12.00 x 7.5/2 = 45kN MR = 12.00 x 7.52/8 =

UDL 84.38 kNm
Due to KEL VR = 48.00 x 7.5/2 = MR = 48.00 x 7.52/8 =
180 kN 337.5 kNm
Case 2: HA on One Lane Only
Due to VR = ⅜ x12.0 x 7.5 = MR = 12.00 (⅜x7.5)2/2 =

UDL 33.75kN 47.46kNm
Due to VR = ⅜ x 48 x 7.5 = MR= 48(⅜x 7.5)2/2 = 190 kNm

KEL 135 kN
Pedestrian Load
VL = 7.5x 1.50 = 11.25 kN ML = 7.5 x 1.52/2 = 8.44 kNm

Case 3: HB alone
Load per wheel = 146.25 kN
VR 
P
4 L  4 a  6  = 146.25/7.5 (4 x 7.5 – 4 x 0.25 – 6) = 448.5kN
L
Case 3: HB alone
a = 3.75 – 0.25 – 1 = 2.50m
M R  4 L  4a  6  a  1  ( Px1)
P
L
MR = (146.25/7.5(14)(3.5)– (146.25 x 1)
= 809.3kNm
Case 4: HA + HB
VR,HB = 448.5 kN
VR,HA,UDL = (1/8)WL = 10.40 x 7.5/8 = 9.75 kN
VR,HA,KEL = 41.60 x 7.5/8 = 39.0 kN
MHB = 146.25x (2 + 0.25) -146.25 x 2 – 146.25 x 1 = 517.59 kNm
MHA,UDL = 47.46x 1.3/1.5 = 41.13 kNm (from HA on one lane )
MHA,KEL = 190 x 1.3/1.5 = 164.67 kNm (from HA on one lane)
Summary of Design Loads for the Deck
Load type - Case Shear Force Reaction Bending Moment
in kN in kN in kNm
Outer Inner Hogging Sagging
Dead Load (DL)
Slab 1.74 43.5 0.13 81.43
Kerb 1 4.14 4.14 3.11
Kerb 2 7.83 7.83 4.70
Surfacing Dead Load
Surfacing 11.33 11.33 25.87
Parapet 8.75 8.75 13.13
Sub Total 22.46 54.83 77.29 21.07 107.3
Live Load
Case 1: HA on both
lanes
HA UDL 45.00 45.00 84.38
HA KEL 180.00 180.00 337.50
Sub-Total 225.0 225.0 421.88
Bridges © 2017
Summary of Design Loads for the Deck (ctd)

Case 2: HA on one lane
HA UDL 33.75 33.75 47.46
HA KEL 135.0 135.0 190.0

Sub-Total 168.75 168.75 237.46
Case 3: HB alone 448.5 448.5 809.3

Case 4: HA + HB
HA UDL 9.75 9.75 41.13
HA KEL 39.00 39.0 164.67
HB 448.5 448.5 517.59
Sub-Total 497.25 497.25 723.39
Pedestrian 11.25 11.25 8.44
Design Values 33.71 552.08 585.79 29.51 916.60
17
Bridges © 2017
DESIGN OF DECK
Total thickness of slab = 420 mm
Concrete cover = 35 mm
Bar diameter = 20 mm
Concrete class = C40
Steel grade = 460
Effective depth: d = 420 – 35 – 10 = 375 mm
Design moment = 916.6 kNm
M 916.6 106
K 2
 2
 0.217  K ' (0.156)
f cu bd 40 x1000 x375
Doubly reinforced section
Design of Compression Steel

( K  K ' ) fcubd 2 0.217  0.156 40 1000 3752 
AS 1  
0.87 fy ( d  d ' ) 0.87 x 460 x375  45
 1,891.3mm 2 / m 18
Bridges © 2017
DESIGN OF DECK
Design of Tension Steel
AS  AS 1  AS 2
 AS 1 
0.156  fcubd 2
0.87 fy 0.775d 
 AS 1 

0.15640 1000  3752 
0.87 x 4600.775 x375
 1,891.3  7,544.6
 9,436mm 2 / m 19
Bridges © 2017
DESIGN OF DECK
Provide Y20 – 150 mm c/c (As = 2,094mm2/m) on top
within deck transverse direction.
Provide Y40– 125 mm c/c (As = 10,053mm2/m) at bottom

within deck transverse direction. 50% of tension steel has to
be provided at the support for taking unforeseen negative
moments.
Secondary reinforcement
As= 0.13%bh = 0.13/100(1000x420) = 546mm2/m
Provide Y16– 300 mm c/c (As = 670mm2/m) both at

bottom and top within deck longitudinal direction.
20
Bridges © 2017
Design of Kerb:
The design moment is 29.51 kNm

Using the same effective depth,
M 29.51106
K 2
 2
 0.0052  0.156
f cu bd 40 x1000 x375
z = 0.95d = 356.3 mm
As = 29.51 x 106/(0.87 x 460 x 356.3) = 207mm2/m
As,min = 0.13 x 1000 x 450/100 = 585 mm2/m --- Provide minimum

reinforcement
Provide Y16 – 300 c/c (As = 670 mm2/m)
21
Bridges © 2017
Design for Shear

V = 552.08kN
v = V/bd = 552080/(1000 x 375) = 1.472 N/mm2 < 5.0 N/mm2 OK
Critical shear stress:
vc = 0.79x (100 x 10053/(1000x375))^0.333x(40/25)^0.5/1.25 = 1.111 N/mm2
Provide shear reinforcement:

0.4 + vc = v < 5.0 N/mm2
Asv b( v  v c )

sv 0.87 f yv
Assume bar diameter = 16 mm
Asv 2-legs = 402 mm2,
Sv = 0.87 x 460 x 402/(1000 x (1.29-0.89) = 402 mm c/c, say 400 mm c/c
Take the minimum reinforcement of Y16 – 300 c/c
22
Bridges © 2017
23

Composite Bridge Design

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What are the steps involved in the design of the bridge deck?

What are the steps involved in the design of the bridge deck?

What loads are considered in the design of the bridge deck?

What loads are considered in the design of the bridge deck?

STEP 1: DESIGN DATA

1.1 Proposed Structural System

1.1.4 Road importance: Trunk Road

Depth of truss girders, h:

Kerb 1: = 0.10 x 24 x 1.15 = 2.76 kN/m2

Superimposed Dead Load (ℽfL = 1.75)

HA alone (L = 1.50)

HA with HB (L = 1.30)

Total wheels: 16 wheels in 4 axles and (L = 1.30)

Load per wheel: 45 x 2.5 x 1.3 = 146.25 kN

Load per axle : 146.25 x 4 = 585.00 kN

Total vehicle load: 585 x 4 = 2,340 kN

Pedestrian (L = 1.5)

Slab VL= 11.60 x 0.15 = 1.74kN ML = -11.60 x 0.152/2 = - 0.13k Nm

VL2 = 5.80 x 1.35 = 7.83kN ML2 = - (5.80 x1.35) x (1.35/3 + 0.15)

Due to VR = 12.00 x 7.5/2 = 45kN MR = 12.00 x 7.52/8 =

Due to VR = ⅜ x12.0 x 7.5 = MR = 12.00 (⅜x7.5)2/2 =

Due to VR = ⅜ x 48 x 7.5 = MR= 48(⅜x 7.5)2/2 = 190 kNm

VL = 7.5x 1.50 = 11.25 kN ML = 7.5 x 1.52/2 = 8.44 kNm

Load per wheel = 146.25 kN

a = 3.75 – 0.25 – 1 = 2.50m

Summary of Design Loads for the Deck (ctd)

HA KEL 135.0 135.0 190.0

Case 3: HB alone 448.5 448.5 809.3

Design Values 33.71 552.08 585.79 29.51 916.60

Design of Compression Steel

Provide Y40– 125 mm c/c (As = 10,053mm2/m) at bottom

Provide Y16– 300 mm c/c (As = 670mm2/m) both at

The design moment is 29.51 kNm

As = 29.51 x 106/(0.87 x 460 x 356.3) = 207mm2/m

As,min = 0.13 x 1000 x 450/100 = 585 mm2/m --- Provide minimum

Provide Y16 – 300 c/c (As = 670 mm2/m)

Design for Shear

Provide shear reinforcement:

Take the minimum reinforcement of Y16 – 300 c/c

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