30m Box Girder Design Taiwan Bridge

SIMPLE SPAN R.C.BOX GIRDER BRIDGE

1. DESIGN DATA AND SPECIFICATIONS

Subject Information:
Superstructure type:- Simple span R.C.Box Girder
Clear span : 30.00 m
Clear Roadway 10.00 m
L 3.50 m
Sidewalk
R 0.39 m
Material properties:
Concrete :- Class A Concrete (C-30)
fc'= = 24 MPa ( fc' cylinder )
fc=0.4*fc' = 10 MPa
Ec=4800sqrt(fc') = 23,515 MPa
Reinforcement steel:
Grade 420 steel: For rebars diam. 20mm and above
fy = 420 MPa
fs = 165 MPa
Es = 200,000 MPa
Grade 300 steel: For rebars less than diam. 20
fy = 300 MPa
fs = 140 MPa
Es = 200,000 MPa
Modular ratio n = Es / Ec = 8.51 Use n = 9
Live Loading: (1) Design Truck : HL-93
(2) Design Tandem
DESIGN TRUCK
P/4 P P P P
6ft
(4.3m) (4.3- 9.2m) (1.80m)

P = 145KN/2 72.5 KN
LONGITUDINAL ARRANGEMENT TRANSVERSE ARRANGEMENT
DESIGN TANDEM
P P P P
1.20m 1.80m

P = wheel load = 1/2*110KN = 55 KN

LONGITUDINAL ARRANGEMENT TRANSVERSE ARRANGEMENT

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 30m Box Girder Design Taiwan Bridge

Dynamic Load Allowance

Section 3.13, the vehicular dynamic load allowance IM
IM = 33% For Strength Limit State There fore IM = 33%
IM = 15% For Service Limit State There fore IM = 15%
The live loads shall be factored by 1+IM/100 = 1.33

Design method: Load and Resistance Factor Design (LRFD)

Whole-width design specified in AASHTO LRFD 3rd Edition 2004 Article 4.6.2.2.1
Reference: ERA's Bridge Design Manual 2002
AASHTO Standard Specifications for Highway Bridges, 1998
2. Bridge Cross Section and preliminary dimensions
13.490

3.50 9.99 0.385

Ø30 pvc
0.20

0.79
1.90
1

0.30 2.10 0.30 2.10 0.30 2.10 0.30 2.10 0.30 2.10 0.30 0.79
0.20

12.30

Given Data Clear Rdwy width, RW= 10.00 m

Sidewalk width, SW-L= 3.50 m
Sidewalk width, SW-R= 0.39 m
Total bridge width, WT= 13.89 m
No. of lane
lane= 4
Skew Angle= 0 Degree
3. PRELIMINARY SUPERSTRUCTURE DIMENSIONS
No. of box= 5
No. of web= 6
c/c of support= 30.50 m
Top slab thickness, T = 0.22 m
Bottom slab thickness, T = 0.20 m
Multiple presence factor for one lane loaded = 1.20 (Section 3.8.1)
Multiple presence factor for two lande loaded = 1.00
Recommended min. girder depth=0.06*S= 1.83 m ( Tab. 8.9.2 - Art. 8.9.2)
Use Min. Girder depth D= 1.90 m
c/c girder spacing, a= 2.40 m
End of slab to center of exterior girder, c= 0.94 m
Web width recommended, bw>= 0.20 m
Use web width, bw= 0.30 m
Box clear width= 2.10 m
Top slab thickness= 0 22 m
0.22
Overhang slab thickness near ext. girder= 0.30 m
fillet= 0.10 m
Width of diaphragm= 0.25 m
Total number of diaphragm= 3
c/c spacing of diaphragms= 15.100 m
3
unit weight of concrete= 24 KN/m
Recommended minimum depth ( Tab. 8.9.2 - Art. 8.9.2)
For simple spans of Box
Box-Girders
Girders the minimum depth in feet is given as:
Dmin = 0.06*S

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where:
S span length as defined in Art. 8.8 in feet.
S=
i.e. S= clear span plus depth of member but need not exceed the c/c of supports
Clear span plus depth of member = 31.90 m
c/c of supports= 30.50 m
Design span length, S= 30.50 m
Crown and Wearing surface
Use 2.0% crown in transverse direction and assume 10cm asphalt wearing surface
Posts and Railings
Use 0.9m high concrete parapet

4. DESIGN OF OVER HANG

4.1 LOADS
4.1.1 DEAD LOADS

Dead loads Moment arm ablout face MDL

(KN/m) of exterior girder (KNm/m)
W1(railing)= 5.76 0.6425 3.701
W4(overhang slab)= 4.95 0.471 2.330
Sum WDL = 10.71 MDL = 6.031

b) LIVE LOADS
i ) Design Truck Load
For the design of deck overhangs with a cantilever, not exceeding 1.8 m from the centerline of the exterior girder to the face of
a structurally continuous concrete railing, the outside row of wheel loads shall be replaced with a uniformly distributed line load of
15 kN/m intensity, located 0.3 m from the face of the railing.
x = 0.00 m c-bw/2= 0.79 m
Live load moment per linear meter width of slab PLL= 15.00
15 00 KN/m
MLL= PLL*x= 0.00 KNm/m
Impact Factor IM = 33%
Live Load plus impact: MLL+IM = 0.00 KN/m

ii) Railling Load

Railing loads shall be applied on an effective length of E =1140+0.833X, X= 0.49 m E = 1.55 m
where: X- is the distance in mm from the center of the post to the point under investigation
According to Art. 2.7 of the AASHTO 1996, the design load, P is , P = 44.51 KN
height of top of rail from top of curb = 0.90 m
Moment arm = 1.10 m
Therefore the railing live load is as follows:
MRLL = 31.58 KNm/m

iii ) Pedestrian Live Load

Pedestrian live load = 85 lb/ft2 = 4.00 KN/m2
Moment due to pedestrian load = 0.18 KNm/m

Total Design Moment

a) Dead load plus rail live load MTR = 1.25MDL+1.75MRLL = 62.81
b) Dead load plus truck live load MTT = 1.25MDL+1.75MLL = 7.54
c) Dead load plus pedestrian live load MTP = 1.25MDL+1.75MLL = 7.85
MTOT = 1.25*MDL + 1.75*MLL+IM = 62.81

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4.2 Design for flexure

Main reinforcement perpendicular to traffic
Mu = 62.81 KNm/m
Minimum Reinforcement Ø= 0.90
fr = 0.63 * SQRT(fc') = 3.09 N/mm2 b= 1000 mm
Icr = bh3/12 = 2.25E+09 mm4 fy = 300 N/mm2
Mcr = fr * Icr / (yt) = 46.30 KNm/m yt = h/2 fc' = 24 N/mm2
1.2 * Mcr = 55.55 KNm/m D= 300 mm
1.33 times Factored Strength Load= 83.54 KNm/m diam = 16 mm
Mdesign = 62.81 KNm/m cover = 50 mm
d= 242 mm
Rn 
Mu
 fc1   2Rn 
 Smax = 450 mm
  0.85   1  1 
mom b  d eff
2
 c1 
 fy
 0.85f

A s    d eff
R n = 1.19 N/mm2 = 0.00410
A s = 991.10 N/mm2 min= 0.03*(f'c/fy) = 0.0024
OK !
Requiredspacing= 202.87 mm
Use diameter 16 mm bars c/c 200 mm

5.1 Interior span slab

5.1.1 Loadings
a) Dead Loads
Dead loads computation
22.00 cm thick slab = 5.28 KN/m2
8.00 mm thick asphalt = 1.80 KN/m2
wDL = 7.08 KN/m2

MDL= (WDL*S2 *0.80)/8 where 0.80 is a cotinuity factor

Span length S = Clear span = 2.10 m (Art. 3.24.1.2)
MDL = 1/8( wDL*S2*0.80) = 3.12 KNm/m

b) Live Load
2400 mm = 29140
i) Unfactored Negative Live Load moment for a girder spacing of 2400 mm = 29140
2500 mm = 29720
MLL = 29.14 KNm/m S= 2.10 m
Note: The value from table A4-1(AASHTO LRFD 1998) is dynamic impact included

Factored Design moment

Total Design moment
MTOTAL =1.75*MLL= 51.00 KNm/m overhang dead load and spand dead load are assumed to balance out for
negative moment conservatively

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5.2.2 a Design for flexure Mu = 51.00 KNm/m

Ø= 0.90
R n = 1.62 N/mm2 = 0.00563 b= 1000 mm

fy = 300 N/mm2
min= 0.03*(f'/fy) = 0.0024 OK ! fc' = 24 N/mm2
A s = 1053.66 mm2/m D= 220 mm

diam = 16 mm
Required As = 1053.66 mm2/m cover = 25 mm
Spacing, s = 190.82 mm2/m d eff = 187 mm
S max = 330 mm
Use diameter 16mm bars c/c 190 mm (Top reinf.-transverse) of interior slab
As provided = 1,058 mm2/m

2400 mm = 25500
ii) Unfactored Positive Live Load moment for a girder spacing of 2400 mm = 25500
2500 mm = 26310
MLL = 25.50 KNm/m
Note: The value from table A4-1 is impact included
i)
Total Design moment
MTOTAL =1.25* MLL+1.75*MLL = 48.53 KNm/m

5.2.2 b Design for flexure Mu = 48.53 KNm/m

Ø= 1
R n = 1.54 N/mm2 = 0.00535 b= 1000 mm

fy = 300 N/mm2
min= 0.03*(f'/fy)
( y) = 0.0024 OK ! fc' = 24 N/mm2
A s = 1000.50 mm2/m D= 220 mm

diam = 16 mm
Required As = 1000.50 mm2/m cover = 25 mm
Spacing, s = 200.96 mm2/m d= 187 mm
S max = 330 mm
Use diameter 16 mm bars c/c 200 mm (Bottom reinf.-transverse)
As provided = 1005.31 mm2/m

Distribution Reinforcements (Art. 3.24.10.2)

For main reinforcement perpendicular to traffic, the bottom distribution reinf. is given as percentage of the main slab reinf. as given below:
As(distr.) % = 1750/sqrt(S) <= 50%
Therefore, % As dist. =36% S= 2400.00 mm
As dist.= 359.11 mm2/m
spacing = 314.93 mm diam. of bar= 12 mm
Use diam. 12.00 mm mm bars c/c 310 mm (bottom reinf. -longitudinal.)
As provided = 365 mm2

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Temperature and shrinkage reinforcements (Art. 8.20)

As (temp and shrink.) is not less than 1/8 square inch per feet.
As (temp and shrink.)= 264.58 mm2/m
Using diam. 12 bar
spacing = 427 mm diam. of bar = 12 mm
Use diam. 12 c/c 400 mm (top. Reinf -both direction. )

6. DESIGN OF LONGITUDINAL GIRDERS

6.1. Loads
6.1.1 Dead Loads
a) Exterior Girder

C= 0.94 m
Uniform dead loads per linear meter span (KN/m) a= 2.40 m
a
railing= 5.76 KN/m bw= 0.30 m
overhang slab= 4.95 KN/m Dg= 1.90 m
deck slab= 7.13 KN/m Tt= 0.22 m
bottom slab= 6.48 KN/m Tb= 0.20 m
girder = 10.66 KN/m
sum 40.00 KN/m

Weight of Diaphragms:(at two extreme supports and at Center)

Conc. Load P1 = 9.32 KN
Conc. Load P2 = 9.32 KN

P1 P2 P1 ( KN)
L/2 L/2
WDL

L
WDL= 40.00 KN/m P1= 9.32 KN P2= 9.32 KN
L= 30.50 m L1= 15.10 m L2= 15.10 m

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Shear Forces and Bending Moments due to Dead Loads on Exterior Girder
VDL(x)= (P1+P2/2+ wL/2) - (P1+ W*X) = P2/2+W*L/2-W*X if 0<= x <L/2
MDL(x)= (P1+P2/2+W*L/2)*X - P1*X - W*X2/2 = (P2/2+WL/2)*X - WX2/2 if 0<= x <L/2
x VDL MDL
(m) (KN) (KNm)
0.00 614.66 0.00
1.525 553.66 890.85
3.050 492.66 1688.67
4.575 431.66 2393.47
6 100
6.100 370 66
370.66 3005 24
3005.24
7.625 309.66 3523.99
9.150 248.66 3949.71
10.675 187.66 4282.40
12.200 126.66 4522.08
13.725 65.66 4668.72
15.250 0.00 4722.35

b) Interior Girder

C= 0.94 m
a= 2.40 m
bw= 0.30 m
Dg= 1.90 m
Tt= 0.22 m
Tb= 0.20 m

Uniform dead loads per linear meter span (KN/m)

slabs = 24.19 KN/m
girder = 10.66 KN/m
Wearing surface = 4.32 KN/m
Fillet = 0.48 KN/m
sum WDL = 39.65 KN/m

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Diaphragms: Conc. Load P1 = 18.65 KN Conc. Load P2 = 18.65 KN

P1 P2 P1 ( KN)
L/2 L/2
WDL

L
P1= 18.65 KN WDL= 39.65 KN/m

P2= 18.65 KN L= 30.50 m

Shear Forces and Bending Moments due to Deal Loads on Interior Girder
VDL(x)= (P1+P2/2+ wL/2) - (P1+ W*X) = P2/2+W*L/2-W*X if 0<= x <L/2
MDL(x)= (P1+P2/2+W*L/2)*X - P1*X - W*X2/2 = (P2/2+WL/2)*X - WX2/2 if 0<= x <L/2

x VDL MDL
(m) (KN) (KNm)
0.00 613.96 0.00
1.525 553.49 890.18
3.050 493.03 1688.15
4.575 432.57 2393.92
6.100 372.10 3007.48
7.625 311.64 3528.83
9.150 251.18 3957.98
10.675 190.71 4294.92

12.200 130.25 4539.66

13.725 69.79 4692.19
15.250 0.00 4752.51

Whole Width Dead Load

x VDL MDL
(m) (KN) (KNm)
0.00 3685.15 0.00
1.53 3321.30 5342.41
3.05 2957.44 10129.95
4.58 2593.59 14362.61
6.10 2229.74 18040.40
7.63 1865.88 21163.31
9.15 1502.03 23731.34

10.68 1138.18 25744.50

12.20 774.33 27202.79
13.73 410.47 28106.20
15.25 0.00 28454.73

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6.1.2 Live Loads

6.1.2.2 Transverse Load Distribution

The design truck or tandem shall be positioned transversely such that the center of
wheel load is not closer than300 mm from the face of the anycurb or railing,

Distribution Factor for moment

Interior Girder:
Case-1: One Design lane loaded where 2100 <= S<= 4000
gmom_int_1 = [1.75+S/1100]X[300/L]0.35X[1/NC]0.45 where 18000 <= L<= 73000
where Nc > 3
L= 30.50 m ts= 0.22 m
S= 2.40 m Nc= 5

gmom_int_1 = 0.38 Lane

Case-2: Two or more design lanes loaded

0.3 0.25
gmom_int = [13/Nc] x[S/430]X[1/L]
gmom_int_2
mom int 2 = 0.56
0 56 Lane
a e

There fore, gmom_ext_2 is maximum of the above two values,gmom_ext_1 or gmom_ext_2 with the following reduction
factor for skew angle
Cf= 1.05-0.25xtan  = 0.00 Degree 0 << 60

= 0.00 Rad
Cf= 1.00 gmom_int = 0.56 Lane

Whole Width Distribution For Moment= 4.00 Lane

Distribution Factor for Shear

Interior Girder:
Case-1: One Design lane loaded
The distribution of live load per lane for shear in interior girder is determined according to the formulas given in Table 13-7.
where 1800<=S<=4900

6000<=L<=73000
890<=D<=2800
0.6 0.1
shr int 1= [S/2900] x[D/L] =
gshr_int_1 0.68 Lane L= 30.50 m

S= 2.40 m
Case-2: Two or more design lanes are loaded D= 1.90 m

The distribution of live load per lane for shear in interior girder
gshr_int_2= [S/2200]0.9x[D/L]0.1= 0.82 Lane

There fore, gshr_int maximum of the above two values, gshr_int_2 or gshr_int_2 with correction factor for skew applied
Cf= 1+[0.25+L/70D)tan 
= 0.00 Degree 0.00 Rad
Cf= 1.00

gshr_int= 0.82 Lane

Shole Width Distribution For Shear= 4.00 Lane

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 30m Box Girder Design Taiwan Bridge

6.1.3 FACTORED LOADS

Load Factors and Load Combinations
The load factors and load combinations are according to ERA's Bridge Design Manual 2002, section 3.3
The load combination to be used for design is Strenght - I Limit state, Table 3-1.

Factored Load = gb *DL + 1.75* ( LL + IM ) = 1.25*DL+1.75*(LL+IM) gb = 1.25

Distribution factors shear are
IM= 33% Srength & Service
gshr= 4.00 Lane IM= 15% fatigue

Factored Shear Forces

Live Load Shear Shear
x VLLTrack VLLTandem VLane Load VDL max max max
(m) (KN) (KN) (KN) (KN) (KN) (KN) (KN)
0.00 384.26 280.37 141.83 3685.15 2611.57 6296.72 1751.18
1.53 363.06 266.07 127.64 3321.30 8425.16 5763.32 1635.47
3.05 341.85 251.77 113.46 2957.44 7673.64 5229.92 1519.76
4.58 320.64 237.47 99.28 2593.59 6922.12 4696.52 1404.05
6.10 299.44 223.17 85.10 2229.74 6170.59 4163.12 1288.34
7.63 278.23 208.87 70.91 1865.88 5419.07 3629.72 1172.63
9.15 257.02 194.57 56.73 1502.03 4667.55 3096.32 1056.92
10.68 235.82 180.27 42.55 1138.18 3916.02 2562.92 941.22
12.20 214.61 165.97 28.37 774.33 3164.50 2029.52 825.51
13.73 193.41 151.67 14.18 410.47 2412.97 1496.12 709.80
15.25 172.20 137.37 0.00 0.00 1603.18 916.10 594.09

gmom= 4.00 Lane

Live Load Moment g Moment g
x MLL,Track MLL,Tandem MLane Load MDL max max max
(m) (KN) (KN) (KN) (KNm) (KNm) (KNm) (KNm)
0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
1.53 553.66 405.76 205.47 5342.41 13270.87 9109.76 2825.51
3.05 1042.64 767.91 389.31 10129.95 25094.59 17234.04 5328.06
4.58 1466.94 1086.44 551.52 14362.61 35471.15 24372.83 7507.67
6.10 1826.57 1361.36 692.11 18040.40 44400.56 30526.15 9364.31
7.63 2121.51 1592.66 811.06 21163.31 51882.81 35693.98 10898.00
9.15 2351.77 1780.35 908.39 23731.34 57917.91 39876.33 12108.74
10.68 2527.49 1924.42 984.09 25744.50 62600.18 43127.11 13036.95
12.20 2658.79 2024.88 1038.16 27202.79 66023.96 45500.20 13723.06
13.73 2725.42 2081.72 1070.60 28106.20 68000.58 46887.81 14086.21
15.25 2727.36 2094.95 1081.42 28454.73 68530.04 47289.95 14126.41

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6.2. DESIGN FOR FLEXURE

a) Design Loads
The factored loads in the above tables are the design loads and the structure shall be designed to carry the expected design loads as calculated
above.
b) Effective Compression Flange Width, beff, ( AASHTO Art. 8.10.1.1)
Exterior Girder
The total width of slab effective as T-girder flange shall not exceed one-fourth of the span length of the girder. The efffective flange
width overhanging on either side shall not exceed six times the thickness of the slab or one half the clear distance to the next web.
Therefore, the effective compression flange width, beff, is the minimum of the following:
a. 1/4 span = 7.63 m L = 30.50 m
b. 2*6hf+bw = 2.94 m hf = 0.22 m (average)
c. bw+2*1/2(clear spacing b/n webs ) = 2.40 m bw = 0.30 m
bw+(Ext. cantilever + 1/2*clear spacing b/n girders) = 2.14 m Overhang L = 0.79 m
Effective compression flange width, beff = 2.14 m Clear S b/n gir.=2.10 m
Bottom slab width = 1.35 m
Interior Girder
The effective compression flange width, beff, is the minimum of the following:
a 1/4 span
a. = 7 63 m
7.63 L = 30.50
30 50 m
b. 2*6hf+bw = 2.94 m hf = 0.22 m
c. bw+2*1/2(clear spacing b/n webs ) = 2.40 m bw = 0.30 m
Effective compression flange width, beff = 2.40 m Clear S b/n gir.=2.10 m
Use compression flange width, beff = 14.14 m

c) Spacing Limits for Reinforcements ( Art. 8.21.1 & 8.21.3)

The clear distance between parallel bars in a layer shall not be less than 1.5 bar diameter , maximum size of aggregate or 1.1/2 inches.
1.5 diam. =
1.5*diam. 48.00 mm d
Assume d= 32 mm
max. size of aggrergate = 1 inch = 25.40 mm
1.1/2 inch = 38.10 mm
Hence the minimum clear distance between parallel bars in
a layer with no lapping is = 48.00 mm = 0.05 mm
Assuming vertical lapping of bars, one over the other,
The minimum clear distance between parallel bars in a layer is = 48.00 mm
Therefore, the minimum center to center spacing of bars in a layer is = 80.00 mm
Use center to center spacing of bars in a layer = 80.00 mm

Minimum clear distance between two layers of bars is 1 inch = 25.00 mm

Thus, minimum center to center spacing of bars when there is lap = 89.00 mm
Use vertical center to center spacing of bars b/n layers = 90.00 mm

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d) Section Capacity and Reinforcement

The maximum design moment is:
Mmax = 68,530.04 KNm

Try 150 No. diam. a di ai*di d' d

1st row = 50 32 804.25 0.05 m 42.63 0.05 1.85
25 32 804.25 0.05 m 42.63 0.05 1.85
2nd row = 25 32 804.25 0.09 m 68.36 0.06 1.84
25 32 804.25 0.09 m 68.36 0.08 1.82
25 32 804.25 0.09 m 68.36 0.08 1.82
0 32 804.25 0.09 m 68.36 0.08 1.82
As= 120637.158
D = 1.90 m
d' = sum ai*di = 0.08 m
sum ai
d = D - d' = 1.82 m
beff = 14.14 m

For rectangular beam analysis (a < hf)

hf), use the following formulas:
r= As/bd

a = As*fy/(0.85*fc'*b)
ØMn =Ø*As*fy*(d-a/2)
For T-beam analysis (a > hf ), use the following formulas:
r= As/bd

a = As*fy/(0.85*fc'*b)
The steel area, Asf, required to balance the longitudinal compression force in the overhang
Asf = 0.85 fc (b bw) hf/fy
0.85*fc'*(b-bw)*hf/fy
The resisting moment provided by the force Asf*fy
M1 =Asf*fy*(d - hf/2)
The remaining steel area (As-Asf), at a stress fy, is balanced by the compression in the rectangular portion of the beam.
The depth of the equivalent rectangular stress block in the zone is found from the horizontal equilibruim
a = (As - Asf)*fy/(0.85*fc'*bw)
An additional resisting moment, M2, is thus provided by the force (As - Asf)*fy and 0.85*fc'a*bw at lever arm of (d-a/2)
M2 =(As - Asf)*fy*(d-a/2)
The total resisting moment, Mu, is the sum of M1 amd M2
Mu = M1 + M1
ØMu = Ø ( M1 + M2 )
Assume rectangular beam analysis As= 120,637 mm2
b= 13.885 m
r= As/bd = 0.01223 d= 1.817 m

a = As*fy/(0.85*fc'*b) 178.88 mm fy = 420.0 MPa

It is rectangular beam as assumed fc' = 24.00 Mpa
ØMn =Ø*As*fy*(d-a/2) 78,787.39 KNm bw = 1.800 m
hf = 0 22 m
0.22
Ø= 0.90

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The following computations for T-beam analysis do not apply

Asf = 0.85 fc (b bw) hf/fy
0.85*fc'*(b-bw)*hf/fy = - mm2
M1=Asf*fy*(d-hf/2) = - KNm
a = (As - Asf)*fy/(0.85*fc'*bw) = - mm2
M2 =(As - Asf)*fy*(d-a/2) = - KNm
Mu = M1 + M1 = - KNm
ØMu = Ø ( M1 + M2 ) = - KNm
There fore, the section capacity is
ØMn = 78,787.39 KNm >> Mmax= 68,530.04 KNm
OK!!
e) Checking maximum steel area
rmax = 0.75*rb = 0.75*0.85*b1*(fc'/fy)*600/(600+fy) b1= 0.85
= 0.0182 fc' = 24 N/mm2
rprovided = 0.0122 << max fy = 420 N/mm2
OK !! bbottom= 13.885

f) Check for minimum reinforcement (Section 9.4.5)

ØMn > = 1.2Mcr.
Mcr = fr * Ig / yt
fr = 0.63 * SQRT( fc' ) = 3.086 N/mm2
Centroid of cross section
y1=Sum(Ai*yi)/(sumAi) = btop= 14.14 m
Sum Ai*yi = 8.351 m3 bbottom= 13.89
Sum Ai = 8.552 m2 bw= 1.80 m
Yt=Sum(Ai*yi)/(sumAi) = 0.976 m htop flang= 0.22 m
Gross moment of inertia, Ig, about centroid D= 1.90 m
p 3 + A1*d12 =
I1=1/12*btop*hf 2.0717 m4 hbottom flang= 0.20
I2=1/12*bw*(D - ht.f )3 + A2*d22= 0.4898 m4
I3=bb*hbf^3/12+bb*(hbf/2)^2 = 2.1426
Sum Ig = 4.7041 m4
The crackimg moment, Mcr,
Mcr = fr * Ig / yt = 14,868.26 KNm
1.2Mcr = 17,841.91 KNm
1.33Mmax = 91,144.95
Min of the above two= 17,841.91
Mmax = 68,530.04 KNm >> 1.2Mcr.
OK!
g) Serviceability Requirements

MS Consulatancy 13
 30m Box Girder Design Taiwan Bridge

Fatigue stress limits ( Section 9.6.2 )

Fatigue stress limits will be checked for the service load conditions. The permissible stress range is given by Eq. 9.19
ff = 145 - 0.33 fmin + 55 (r/h) r/h= 0.30
MDL= 28454.73
MLL (Fatigue) min= 0.00 KNm
k = [pn+0.5(ts/d)2] / (pn+(ts/d) = 0.508 MLL+IM (Fatigue) max= 14126.41 KNm
kd = 0.923 m = 0.01223
T-beam btf= 13.89 m
If it is rectangular beam d= 1.82 m
j = 1 - k/3 = - hf= 0.22 m
n= 9
If it is T-beam As= 120637.16 mm2
j = 6 - 6(hf/d)+2(hf/d)2 - (hf/d)3/(2pn) = 0.942
6 - 3 (hf/d)
The minimum stress, fmin, is :
fmin = MDL/(As*j*d) = 137.76 N/mm2
The maximim stress, fmax, is caused by the total load ( MDL+LL+IM )
fmax = (MDL+LL+IM) /(As*j*d) = 189.06 N/mm2
The actual stress range
range, delta ff, is:
Delta ff = fmax - fmin = 51.29 N/mm2
The fatigue stress limit, ff, is
ff = 145 - 0.33* fmin + 55 (r/h) = 116.04 N/mm2 >> 51.29 N/mm2
OK !!

Control of Cracking by Distribution of Reinforcement (Sec. 9.4.5)

To control flexural cracking of the concrete, tension reinforcement shall be well distributed within maximum flexural zones.
Components shall be so proportioned that the tensile stress in the steel reinforcement at service limit state, fs, does not exceed
fs = Z / (dc*A)1/3 < = 0.6*fy tb 0.200 m

A = Effective tension area/ No. of bars=bw*(2y')/N = 0.01533 m2 y' = d' = 0.083 m

dc = distance measured from extreme tension fiber to center of
the closest bar = 0.050 m bb.f= 13.89 m
Z = crack width parameter, assumed = 30.00 KN/mm Msrvervice-I= 47,289.95
There fore, for crack control the maximum allowable stress is No. of bars= 150
fsa = Z / (dc*A)1/3 = 327.81 N/mm2
0.6*fy = 252.00 N/mm2
Th maximim
The i i stress,
t ffmax, att service
i lload
d iis
fmax = Msrvice-I /(As*j*d) = 228.95 N/mm2 << fsa = 252.000 N/mm2

OK!

MS Consulatancy 14
 30m Box Girder Design Taiwan Bridge

h) Bar Cutting
Development of Reinforcement (Sec. 9.4.3)
Positive moment reinforcement: Atleast one-third of the positive moment reinforcement in imple span members shall extend along
the same face of the member beyond the centerline of the support.

The basic development length, ldb, in mm is

For bars diam. 35 and smaller, ldb = 0.02Ab*fy/sqrt(fc') >=0.06db*fy db= 32 mm
ldb = 0.02Ab*fy/sqrt(fc') = 1379 mm fy= 420 Mpa
fc'= 24.0 Mpa
The tension development length,ld, is 0.06db*fy= 806.4 mm
ld = ldb * modification factor = 1379 mm mod. Factor = 1.00

Standard Hook Detail

e
e = 4*db = 4*32mm = 128mm
d = 10*db =320mm
d

a = 1/2*circumf.+4*db = 630mm

12db= 384.00

5db= 160.00

Lap splices of Reinforcement in Tension

The length of lap for tension lap splices shall not be less than either 300mm or the 1.3 times the development length
Lap splices for diam. 32 bar = 1,793 mm

Flexural Reinforcement Extension Length (Sec. 9.4.5)

Except at supports of simple spans and at the free ends of cantilevers, reinforcement shall be extended beyond the point at which

it is no longer required to resist flexure for a distance not less than:

a) the effective depth of the member = 1.82 m
b) 15 times the nominal diameter of bar = 0.48 m
c) 1/20 of the clear span = 1.50 m
There fore, extension length is max.of the above = 1.82 m

x Mx
MOMENT ENVELOPE
(m) (KNm)
y = -703.5x2 + 15280x - 14482
0.00 0.00 80000
1.53 13270.87 70000
60000
3.05 25094.59
nt (KNm)

50000
4.58 35471.15
40000
Momen

6.10 44400.56 30000

7.63 51882.81 20000
10000
9.15 57917.91
0
10.68 62600.18
0.00 1.53 3.05 4.58 6.10 7.63 9.15 10.68 12.20 13.73 15.25
12.20 66023.96 Distance (m)

13.73 68000.58
15.25 68530.04

MS Consulatancy 15
 30m Box Girder Design Taiwan Bridge

Bar Cutting and Resisting moment

D = 1.90 m m hf = 0.22 m
btop= 14.1425 m fy = 420.00 Mpa
bbottom= 13.885 fc' = 24.00 MPa
bw = 1.80 m Ø = 0.90

Bars T0tal No of Bars Diam. As d' d =D-d' a=

p=As/bd Asfy/(0.85fc'b)
(mm) (mm2) (m) (m)
(mm)

G-2_ 25 150 32 120637 0.083 1.817 0.01202 176

G 2 25
G-2_ 125 32 100531 0.082
0 082 1.818
1 818 0.01000
0 01000 146
G-2_ 25 100 32 80425 0.061 1.839 0.00778 117
G-2_ 25 75 32 60319 0.053 1.847 0.00577 88
G-1_ 50 50 32 40212 0.053 1.847 0.00385 59
Rect. Beam T-beam
Rect. or ØMn
T-beam Asf=0.85fc' M1=Asf.fy(d-hf/2) a= (As - M2 ØMn
=Ø*Asfy(d-a/2)
(b-bw)hf/fy (KNm) Asf)*fy/(0.85*fc' =(As-Asf)fy =Ø ( M1 + M2
(KNm)
Rect. 78861.6 - - - - -
Rect. 66295.1 - - - - -
Rect. 54127.0 - - - - -
Rect. 41111.3 - - - - -
Rect. 27630.0 - - - - -
Bar Schedule
Total No Mn(KN-m) Length
G-5_ 25 150 78862 7.56 G-5a_ 0
G-4_ 25 125 66295 17.06 25094.59 3.05 G-3a_ 0
G-3_ 25 100 54127 23.89 27630.01 4.20 25.73 G-4a_ 0
G-2 25
G-2_ 75 41111 25.73 35471.15 4.58 G-2a 0
G-2a_
G-1_ 50 50 27630 31.45

i ) Skin Reinforcement
If the depth of side face of a member exceeds 3ft, longitudinal skin reinforcement shall be uniformly distributed
along both side faces of the member for a distance d/2 nearest the flexural tension reinforcement.
Area of skin reinforcement, Ask > = 0.012*(d - 30) in2 / ft (on each side face)
Depth of side face = D - hf = 1680 mm >> 3ft = 914.40 mm
Skin Reinf. shall be provided!
Ask > = 0.012*(d - 30) in2 / ft = 0.49852 in2/ft d = 1.82 mm
= 1055 mm2/m diam. = 16 mm
Spacing = as/As * 1000 = 191 mm
Maximum spacing, Smax, lesser of d/6 or 12" = 303 mm
Use diam. 16 mm mm bars c/c 300 mm

MS Consulatancy 16
 30m Box Girder Design Taiwan Bridge

6.3 DESIGN FOR SHEAR

Shear strength
Design of cross sections subject to shear shall be based on
Vu < = ØVn = Ø (Vc + Vs ) Ø= 0.900
where Vu = factored shear forces at the section
Vn = the nominal shear strength, determined as the lesser of Vn = Vc + Vs or Vn=0.25*fc'*bv*dv
Vc = the nominal shear strength provided by the concrete, determined by:
Vc = 0.083*b*SQRT(fc')*bv*dv, where = 2.00

Vs = the nominal shear strength provided by the shear reinforcement

Vs = Av
Av*fv*d
fv d / s
Shear strength provided by concrete
Shear stress provided by concrete 0.16600
vc = 0.083*b*sqrt(fc') = 0.81 MPa

The shear strength carried by concrete, Vc=vc*bw*d= 0.81 bd (N)

Shear strength provided by shear reinforcement

Where the factored shear force, Vu, exceeds shear strength ØVc, shear reinforcement shall be provided to satisfy
the equation, Vu < = ØVn = Ø ( Vc + Vs )
When shear reinforcement perpendicular to the axis of the member is used
Vs = Vu/Ø - Vc = Av*fy*d / s
There fore, spacing of shear reinforcement, s, is :
S = Av*fy*d / Vs = Av*fy*d/(Vu/Ø-Vc)=Av*fy*d/(Vu/Ø - 0.813*bw*d )
Minimum shear reinforcement ( Eq.12.34 ) bw = 1.80 m
Av = 0.083*sqrt(fc')*bw*S / fy fc' = 24.00 MPa

Smax = Av*fy / ( 0.083*sqrt(fc')*bw ) fy = 300.00 MPa

Maximum spacing of shear reinforcement Av =diam. 12 1357.17 mm2
If Vu < 0.10*fc' * bw * d, S <= 0.8d <= 600mm 0.10*fc' * bw= 4320 Mpa
If Vu >= 0.10*fc' * bw * d, S <= 0.4d <= 300mm
Sections located less than a distance d from support may be designed for the same shear as that omputed at a distance d

10000.00
FACTORED SHEAR FORCE DIAGRAM

8000.00

6000.00

4000.00

2000.00

0.00
0.00 1.53 1.82 3.05 4.58 6.10 7.63 9.15 10.68 12.20 13.73

MS Consulatancy 17
30m Box Girder Design Taiwan Bridge

Vcc=
x Vu
V d Vs=Vu-Ø
Vs V Ø Vc 0.10*fc'*bw*d
0 10*fc'*b *d
0.88bw.d*10-3
(m) (KN) (m) (KN) (KN)
(KN)

0.00 2611.57 1.85 2,925.65 0,000.00 250,668.93

1.53 8425.16 1.85 2,925.65 6,435.65 250,668.93
1.82 7817.64 1.85 2,925.65 5,760.62 250,668.93
3.05 7673.64 1.85 2,925.65 5,600.62 250,668.93
4.58 6922.12 1.84 2,912.98 4,778.27 249,583.20
6.10 6170.59 1.84 2,912.98 3,943.24 249,583.20
7 63
7.63 5419 07
5419.07 1 82
1.82 2 879 32
2,879.32 3 141 87
3,141.87 246 699 22
246,699.22
9.15 4667.55 1.82 2,879.32 2,306.85 246,699.22
10.68 3916.02 1.82 2,878.44 1,472.69 246,624.57
12.20 3164.50 1.82 2,878.44 0,637.66 246,624.57

13.73 2412.97 1.82 2,878.44 0,000.00 246,624.57

15.25 1603.18 1.82 2,878.44 0,000.00 246,624.57

Smax-2= Spacing by analysis

Smax-1 Spacing
x y( q ( )
Av*fy/(0.083*sqrt(fc')* y
S=Av*fy*d/
(mm) provided
(m) bw) (Vu/Ø-Vc)
(mm)
(mm) (mm)

0.00 600 556 - 130

1.53 600 556 117 130
0.00 600 556 131 130
3.05 600 556 134 130
4.58 600 556 157 130
6.10 600 556 190 190
7.63 600 556 236 190
9.15 600 556 321 300
10.68 600 556 502 300
12.20 600 556 1160 400
13.73 600 556 - 400
15.25 600 556 - 400

MS Consulatancy 18
 30m Box Girder Design Taiwan Bridge

7 DEFLECTION
a) Computation of Gross Moment of Inertia
Centroid of cross section
For simplicity of calculations,Area,
the slab
Ai surface is assumed
Centroid, level,
yi i.e., withoutAicrossfall.
yi The center of gravity is calculated from bottom of girder
Part (m2) (m) (m3) Centroid of area

Top Slab 3.05 1.79 5.467913 y1=sum Ai*yi/(sum Ai)= 1.00 m

Bottom Slab 2.46 0.10 0.246
Girder 2.66 0.94 2.504
Sum 8.18 8.22

Gross moment of Inertia

Area, Ai yi ycg = yi-y1 Ig Ai*(ycg)2 Ig+Ai*(ycg)2
Part (m2) (m) (m) (m4) (m3) (m4)
Top Slab 3.05 1.79 0.79 0.012321 1.883 1.8956
Bottom Slab 2.46 0.10 0.90 0.010914 2.014 2.0249
Girder 2.66 0.94 0.06 0.243134 0.011 0.2543
Sum Ig = 4.1748

b) Computation of Effective Moment of Inertia

Ie = [ Mcr/Ma ]3*Ig + [ 1 - (Mcr/Ma)3] Icr < = Ig
fr = 0.63*sqrt(fc') = 3.09 N/mm2
Mcr = fr*Ig / yt = 12823.22 KNm

Weight of superstructure W W
Part (KN) (KN/m)
Top Slab 2236.04 72.36
B tt
Bottom Slab
Sl b 1800
1800.72
72 58.28
58 28
Girder 975.02 31.97
Sum W (KN/m) 162.61
Ma = W*L2/8 = 19,407 KNm 28,454.73
Transformed section

beff
fc
c

C
kd

kd
d

d
jd

nAs fs
T

bw

kd = 0.923 m d= 1.817 m

beff= 14.14 m d'= 0.083 m

bw= 1.80 m D= 1.900 m
n= 9 hf= 0.220 m

As= 120,637 mm2 j= 0.942 m

MS Consulatancy 19
 30m Box Girder Design Taiwan Bridge

Centroid of cross section Area,

( y1 measured
Ai from bottom)
yi Ai yi
Part (m2) (m) (m3)

Top Slab 2.20 1.79 3.938 y1=sum Ai*yi/(sum Ai)= 1.25 m

Girder
Transformed steel, 0.91 1.33 1.203
nAs 1.09 0.08 0.090

Sum 4.19 5.23

Computation of Moment of Inertia of Cracked section, Icr

Area, Ai y g = yi-y1
ycg y y g
Icg (y g)2
Ai*(ycg) (y g)2
Ig+Ai*(ycg)
g
Part (m2) (m) (m4) (m3) (m4)
Slab 2.20 0.5419 0.012549 0.6462 0.6587
Girder
Transformed steel, 0.91 0.0805 0.052092 0.0059 0.0580
nAs 1.09 1.165 - 1.4742 1.4742
Sum = 0.7167 m4
Effective moment of Inertia, Ie
Ie = [ Mcr / Ma]3 * Ig + [ 1 - (Mcr/Ma)3] Icr = 1.714 m4 <= Ig = 4.175 m4

Computation of Dead Load Deflection using effective moment of Inertia,

Inertia Ie.
Ie
The maximum deflection is calculated by using the formula;
Deflection, max. = 5*W*L4/(384*E*Ie) = 45.45 mm E = Ec= 23,515 Mpa
W= 162.61 KN/m
Long term deflection = Instantaneous deflection * factor L= 30.50 m
Factor = 3 - 1.2*(As'/As)> =1.6 3.00 As' = 0
Long term deflection = 136.36 mm

Calculation of live load deflection using the effective moment of inertia

The live load deflection at middspan for different loading conditions that is analogous to the moving
load pattern is given as follows:

P P P/4

- When investigating the maximum absolute deflection, all design lanes should be loaded; and all
pp g components
supporting p q y
should be assumed to deflect equally.
- The live load portion of load combination Service I of Table 3-3( i.e, load factors for both live
and dead load equal to 1.0) with dynamic load allowance factor should be applied.

MS Consulatancy 20
30m Box Girder Design Taiwan Bridge

P with IM = 580.00 kN w lane load = 37.20 N/mm

x =l/2
l/2 = 15.25 5wl4/(384EI) =
def. Lane =5wl 10.40 mm

-2
Px/(6eIl) = 1.20E-12 mm
According to Chapter 3, .
- The live load deflection should be taken as the larger of :
* That resulting from the design truck alone, or
* That resulting from 25 percent of the design truck together with the design lane load.

p
(0.25*Axle
Def at midspan + Lane) Max Def
x x1 x2 x3 (axle) [mm] [mm]
0.00 30.50 26.23 21.97 -2.54 9.76 9.76
1.53 28.98 24.71 20.44 4.51 11.53 11.53
3.05 27.45 23.18 18.92 10.33 12.98 12.98
4.58 25.93 21.66 17.39 14.97 14.14 14.97
6.10 24.40 20.13 15.87 18.54 15.03 18.54
7.63 22.88 18.61 14.34 21.09 15.67 21.09
9.15 21.35 17.08 12.82 22.70 16.07 22.70
10.68 19.83 15.56 11.29 23.46 16.26 23.46
12.20 18.30 14.03 9.77 23.44 16.26 23.44
13.73 16.78 12.51 8.24 22.71 16.08 22.71
15.25 15.25 10.98 6.72 21.36 15.74 21.36
Absolute Maximum Deflection= 23.46 mm
Therefore, the maximum live load deflection is the maximum of the tabulated values above.

Def Live load Max =

Def. 23 46 mm
23.46
For Vehicular load, the deflection limit is equal to Span/800.= 38.13 mm OK!

MS Consulatancy 21