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T Beam Bridge (Girder Bridge)
Introduction:
1 Tbeam bridges are generally more economical for spans of 12 to 18 m.
2 The girder stem thickness usually varies from 35 to 55 cm and is controlled by the
required horizontal spacing of the positive moment reinforcement.
3 This type of bridges composed of slab supported by girders, and girders supported by
piers and abutments.
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Design of super structure (T – Beam Bridge):
For the above shown bridge design the super structure elements using ultimate strength design
method? Assume any dimensions you don’t know?
Steps of design:
1 Design of slab:
a Load calculations:
Check the minimum depth of superstructure (not only slab) according to table
2.5.2.6.31
t= 0.07*L = 0.07 *15000 = 1050 mm (total depth (slab+girder)
Actual t = 800+300 =1100 mm >1050 mm ok;
Dead load due to weight of slab W
DD
= 0.3*24*1m width of slab =7.2 kN/m
Dead load due to wearing surface W
DW
= 0.075*18 * 1m width =1.35 kN/m
Dead load due to barrier and walk side at cantilever side =
(0.8*0.4*24*1)+(0.2*0.65*24*1) = 10.8 kN/m
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b Design of overhanging part:
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Moment calculations:
Moment due to deck weight and any permanent objects connected to the slab =
M
DD
= 7.2(1.8
2
)/2 + (7.68*1.6)+(3.12*1.08)= 27.32 kN m
M
DW
= 1.35*(0.75
2
)/2= 0.38 kN m
Moment due to truck load, impact and pedestrian load (live load) according to article No
3.6.1.6 and 3.6.1.3.3 =
M
LL
= (1.33*72.5*0.45) + (4.1*1*0.65*1.1) = 46.3 kN m
∑η
i
γ
i
Q
i
≤ R
n
= R
r
η
i =
η
D
η
R
η
I
≥ 0.95
η
i
=0.95 * 1.05 * 1
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η
i
= 0.99 ≥ 0.95 OK
M
u
= ηi * [γ
d
M
D
+ γ
w
M
w
+ γ
l
M
l+im]
With reference to tables 3.4.1.1 and 3.4.1.2 in AASHTO then we can find the values of γ
i
M
u
= 0.99 * [1.25*(27.32) + 1.5*(0.38) + 1.75 * (46.3)] = 114.6 kN m
Shear calculations:
Shear due to deck weight and any permanent objects connected to the slab =
V
DD
= 7.2(1.8) + (7.68)+(3.12)= 23.76 kN
V
DW
= 1.35*(0.75) = 1.01 kN
Shear due to truck load, impact and pedestrian load (live load) according to article No
3.6.1.6 and 3.6.1.3.3 =
V
LL
= (1.33*72.5) + (4.1*1*0.65) = 99 kN
V
u
= 0.99 * [1.25*(23.76) + 1.5*(1.01) + 1.75 * (99)] = 202.4 kN
Flexural design of cantilever part:
d= 30025 – 10 =265 mm
Mu< Ø Mn
Ø Mn = 0.9*0.85*f’c a b (d(a/2)) =0.9* As fy (d(a/2))
114.6*10
6
= 0.9*0.85*35*1000*a (2650.5a)
Solving quadratic equation get a = 16.7 mm
114.6*10
6
= 0.9* As (420) (265(16.7/2))
As= 1182 mm
2
Choose Ø 16 with A
b
= 200 mm
2
No of bars = 1182/200 = 6 Ø 16 / meter long of bridge deck
Spacing = 1000/5 = 200 mm Ø 16@200mm
Check minimum steel and maximum:
As min= 0.25*√35 * 1000* 265/420 = 933 <1182 ok
Check maximum steel:
As
(εs=0.005)
= (0.85*35*1000*a)/420
a =β1 c
c= (0.003/0.008)*265=99.37mm
β1=0.85 (0.05*(3528)/7) = 0.8
As max= 0.85*35*1000*0.8*99.37/420=5630 > 1182 ok
Check shear at cantilever part:
Vc= 0.17√fc’ b d= 0.17*√35 *1000*(300258) *10
3
= 268.5 kN
ØVn= 0.75*268.5 = 201.4 kN ≈ 202 kN
Actually we should check the shear at distance d from face of support (face of girder in
this case) Vu = 146.17 kN (linear interpolation) < 201.4 kN ok;
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**** Don’t forget the shrinkage and distribution steel as per the design of Slab Bridge;
c Live load calculations on slab:
NL = 6/3.6 = 1.67 = 2LANES according to article No. 3.6.1.1.1
The above article (3.6.1.3.3.) means that the lane load will not apply in case of design deck in
transverse direction
M
TL
≈ 0.8* P
72.5
(S+0.6)/32 = 0.8* 72.5 (1.4+0.6)/32 = 3.6 kN m
M
IM
= 0.33*3.6 = 1.18 kN m
M
DD
= 7.2*(
1.4
2
10
) = 1.4 kN m for both positive and negative moment for continuous spans
M
Dw
= 1.35*(
1.4
2
10
) = 0.26 kN m
∑η
i
γ
i
Q
i
≤ R
n
= R
r
η
i =
η
D
η
R
η
I
≥ 0.95
η
i
=0.95 * 1.05 * 1
η
i
= 0.99 ≥ 0.95 OK
M
u
= ηi * [γ
d
M
D
+ γ
w
M
w
+ γ
l
M
l+im]
With reference to tables 3.4.1.1 and 3.4.1.2 in AASHTO then we can find the values of γ
i
M
u
= 0.99 * [1.25*(1.4) + 1.5*(0.26) + 1.75 * (3.6+1.18)] = 10.4 kN m
d= 300 258 = 267 mm
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Ø Mn = Mu = 10.4*10
6
= 0.9*0.85*fc’ a b (d a/2) = 0.9*As fy (d a/2)
10.4*10
6
= 0.9*0.85*35 a (1000) (267 a/2)
Solving quadratic equation get a=1.46mm
As =
10.4∗10
6
420∗(267−0.73)
= 93 mm
2
< As min ; use minimum reinforcement ratio
Use minimum reinforcement ratio As = 0.0025*1000*253 = 632.5 mm
2
As min= 0.25*√35 * 1000* 265/420 = 933 mm
2
choose Ø14 with Ab= 153 mm
2
No of bars = 933/ 153 = 6 bars
Spacing = 1000/ 5 = 200 mm ; choose Ø14 @200 mm top and bottom reinforcement
H W Check the shear capacity of slab in transverse direction?
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2 Design of girders in longitudinal direction (traffic direction):
a Interior girders:
Hints:
**
Mg= m (Mn lanes/NL)
Where m= multi presence factor
M
n lanes
= Moment due to NL No. of lanes
NL = number of lanes
**
Distribution factor:
DF is depends on the following:
a Type of deck
b Spacing between girders or supporting elements
c NL
d Support location
to find D F for moment for interior girders and with reference to table 4.6.2.2.11
in AASHTO specify the type of cross section (a;b;c….etc) in this case the type of
bridge cross section is type (e).
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then according to table 4.6.2.2.2b 1 below and for number of beams (girders) >4 then
the distribution factor can be calculated from equations be low ; you should apply both
equations for one design lane loaded and two or more then take the maximum value of D.F.
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3
In this case n=1
I= 400*
800
= 1.706 x 10
10
mm
4
12
A= 800*400 = 320000 mm
2
e
g
= 400+150 = 550 mm
K
g
= 1 * (1.706 x 10
10
* + (320000 * 550
2
)) = 1.138 x10
11
Ok. Since the value of K
g
is between 4 x10
9
and 3 x10
12
Type of super structure according to table 4.6.2.2.11 is type (e)
Enter to table 4.6.2.2.2.b1 to find the value of D.F
N
b
= 4 girders in our case then
From equation of one design lane loaded:
DF = 0.06+ (
1400
4300
)
0.4
(
1400
15000
)
0.3
(
1.138∗10
11
15000∗300
3
)
0.1
= 0.34
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For interior girders use D.F = 0.84
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NOTE : LEVER RULE
To find the DF using lever rule in case of number of girders = 3 as indicated from table
then we should use the following method (method of lever rule)
1 Find reactions from the deck at the first interior girder by taking the right hand side of
bridge only by assuming internal hinge at the first interior girder
R*1.4  72.5*(1.4) +72.5*(1.81.4) = 0
R = 51.78 kN
2 Divide the reactions by 72.5kN and multiply by m (in this case the right hand side of
bridge capacity = one lane only then m=1.2)
3 DF=
𝑅 𝑚
72.5
To calculate the moment due to tandem or truck load we will use the influence lines tool:
a Truck load
M
truck
= 1.6*35+3.75*175+1.6*175 = 992.25 kN m
b Tandem load
M
tandem
= 110*3.75+3.15*110 = 759 kN m
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Use M
truck
= 992.25 kN m
M
IM
= 0.33*992.25 =327.44 kN m
M
LL
= 9.3 * 15
2
/8 = 261.6 kN m
All of above moment values are per lane;
M
live
per girder = DF (for moment) * 0.5*m*( M
truck +
M
LL+
M
IM
)
=0.84*1*0.5*(992.25+327.44+261.6) = 664.14
kN m
To calculate the value of dead load due to wearing surface and deck weight per girder we
should find the value of flange width of TBeam:
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Then ;
b
f
= the smallest of
150000/4= 3750 mm
or 12*300+400 = 4000mm
or 1400 mm
b
f
= 1400 mm
W
DD
= 24* (0.3*1.4)+24*(0.8*0.4) = 17.76 kN/m
W
Dw
= 18* (0.075*1.4) = 1.89 kN/m
M
DD
= 17.76*15
2
/ 8 = 500 kN.m
M
Dw
= 1.89*15
2
/ 8 = 53.15 kN.m
Mu = 0.99 * [1.25*(500) + 1.5*(53.15) + 1.75 * (664.14)] = 1848.3 kN m
Design of concrete T beam section with flange in compression and span = 15m length
Dimensions of beam (girder) :
bw= 400mm
ts=300 mm
bf= 1400 mm
h= 1100 mm
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d= 1100401015 = 1035mm (expected bar diameter for main Rft = 30 mm)
Note : flexural design of T girder will be explained in class
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To design the shear reinforcement for the interior girders:
1 Find DF from table 4.6.2.2.3.a1 for interior beams
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1
1
V
truck
= 175 *1 + 175*0.71 + 35 *0.43 = 314.3 kN
V
tandem
= 110*1 + 110 * 0.92 =211.2 kN
Design shear according to truck load;
V
IM
= 0.33* 314.3 = 103.7 kN
V
LL
= 9.3*(15)*0.5 = 69.75 kN
V
truck
= 314.3 kN
V
live load
= 314.3+69.75+103.7= 487.75 kN/per lane
V
live load
= DF * m*
∗(V
truck+
V
LL+
V
IM
) = 0.57*1*
2 2
∗( (487.75) =139.0 kN/girder
3 Calculate the shear force per girder for dead load:
b
f
= 1400 mm
V
DD
=
17.76
2
∗ 15 = 133.2 kN
V
DW
=
1.89∗ 15
2
= 14.18 kN
Vu = 0.99 * [1.25*(133.2) + 1.5*(14.18) + 1.75 * (139)] = 426.7 kN m
2 Calculate the shear force due to live load from influence lines of shear:
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Note : shear design of T girder will be explained in class
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