A.
Design of Superstructure 1.0 Design Data 1.1.
Materials and its Properties:
M25 Fe415 Characteristics Strengthof Concrete fck Permissible direct compressive stress, σc Permissible flexural compressive stress, σcbc Maximum Permissible shear stress, τmax ( 0.07 0.07*fck) fck)
=
25 MPa
=
6.2 MPa 8.3 MPa 1.75 MPa
= =
Basic Permissible Stresses of Reinforcing Bars as per IRC : 21-1987, Section III: Permissible Flexural Tensile stress, σst = 200 MPa Permissible direct compressive stress, σco = 170 MPa Self weight of materials as per IRC : 6-2000: Concrete (cement-Reinforced) Macadam (binder premix) 1.2.
= =
24 kN/m3 22 kN/m3
Geometrical Properties:
Effective Span of Bridge Total length of span Numbers of span Width of expantion Joint Total length g of Bridge g Nos. of longitudinal Girder Spacing of Girder Rib width of main girder Overall depth of main girder Depth of kerb above deck slab Nos. of cross girder Spacing of cross girder Rib width of cross girder Overall depth of cross girder Deck slab thickness Deck slab thickness at edge Thickness of wearing coat Fillet size (horizontal) Fillet size (vertical) Bridge Width: Carriageway width Footpath width Kerb width Outer Kerb width Inner Total Width of Deck Slab Total depth of Kerb Outer Total depth of Kerb Inner
= = = = = = = = = = = = = = = = = = = = = = = = = =
24.00 24.56 2 40 49.2 3 2.4 400 2000 225 6 4.8 300 1500 220 150 80 150 150
m
6 0.45 0.15 0 7.2 0.375 0
m
m mm m m mm mm mm m mm mm mm mm mm mm mm
m m m m m m
Page 1 Fig 1.BRIDGE CROSS-SECTION
Fig 1.BRIDGE CROSS-SECTION
2.0
Design of Slab 2.1 Design of Cantilever slab: The h cantilever il slab l b iis ddesigned i d bby effective i width id h method. h d
Thickness of slab =
300 mm at junction with rib 150 mm at free end 0.5 kN/m (assumed) (for IRC class A loading) 54 %
Self weight of Railing = Impact factor =
25 %
(for ( IRC class AA loading) g)
Dead Load Bending Moment and Shear Force: S.No.
Item
Width
Depth
Unit Wt
Assumed
1 Railing/Parap et 2 Kerb (outer) 3 Kerb (inner) 4 Wearing Coat 5 Slab
0.2 0 0.4 1 1
0.225 0 0.08 0.15 0.075 7.684
Total
24 24 22 24 24
Load / m run (kN) 0.5 kN 1.08 0 0.704 3.6 1.8
Distance 1.525-0.100=
kN kN kN kN kN
1.525-0.100= 0.15+0.175/2=
0.150/2= 0. 50/
1.525/2= 1.525/3=
0.9 m 0.9 0.2375 0.075 0.5 0.3333
kN
Dead load Shear force at theface of rib = Dead load Bending Moment at the face of rib =
57KN
Moment
m m m m m 3.875 7.684 3.875
IRC Class AA Track Loading
0.97 0 0.05 1.8 0.6
kN.m kN.m kN.m kN.m kN.m kN.m kN kN.m
37.5KN
350KN
IRC Class A Loading
0.45 kN.m
IRC Class AA Wheel Loading
Fig. 2.a, 2.b & 2.c
Live Load Bending Moment and Shear Force: IRC Class AA will not operate on the cantilever slab that shown in fig 2.b & 2.c above and Class A Loading is to be considered and the load will be as shown in fig 2.a above. Effective width of dispersion be is computed by equation be
=
1.2X+ bw
Here X= bw=
0.125 m 0.41 m
Hence be=
0.56 m
IRC Class A Loading Load Live Load per m width including impact Maximum Moment due to live load Average thickness of cantilever slab Taking pedestrain load (LL) Eff ti width Effective idth off slab l b Cantilever length of slab Maximum Bending moment Shear force at the face of slab Total Design Shear Force Total Design Bending Moment Design of Section: Modular Ratio, m
=
= = =
280 3 σ cbc
28.5 kN 76.339 kN 9.5424 kNm = 225 = 5.0 = 0 0.45 45 = 1 = 1.406 = 2.250 = 86.3 = 14.82
mm kN/m2
m m kN.m kN kN kN.m
=
11.245 Page 2
280 3 σ cbc m σ cbc
=
Neutral axis factor, k
=
Lever arm factor, j
1 −
k 3
1
Moment of resistance coefficient, R =
2
=
0.3182
=
0.8939
=
1.1804
=
112.06
+ σ st
m σ cbc
× k × j × σ st
Therefore, required effective depth of slab= M Rb
d=
Effective depth p of slab,, pprovided
Provide φ
10
254 mm
=
M
=
Area of steel required, Ast
σ
st .
mm bars @
=
j .d
=
> d reqO.K. q mm2
326.42
mm c/c, giving area of steel
200
mm
=
393
mm2
> required, Ok.
Distribution Steel: Distribution steel is to be provided for 0.3 times live load moment plus 0.2 times dead load moment. 4.06 kN.m Moment = Effective depth 244 mm =
Area of steel required, Ast
σ
M st . j . d
2 93.057 mm
=
Half reinforcement is to be provided at top and half at bottom. Provide ø 10 mm bars 200 mm c/c at both top and bottom, giving area of steel t l=
mm2 > required required, OK OK.
392 5 392.5
Check for min. area of Steel: Min. area of steel @ 0.12 % =
2 360 mm
Design for Shear: Dead load shear Live load Shear Total tanβ =
=
7.68
= =
2.250 kN 9.93 kN
=
0.005
Provided. O.K.
kN
0.150 τv =
Shear stress,
V-
M × tan β d bd
Percentage area of tension steel, pt Allowable shear stress as per code is given by
τ c = k 1 .k 2 .τ co
k 1 = 1 . 14 − 0 . 7 × d ≥ 0 . 5
=
k
Ad t Adopt
co
%
0.986
k 2 = 0.5 + 0.25 ρ ≥ 1 (where
τ
0.13
=
N/mm2
( d being in m)
=
Value ot
<
=
ρ
0.500446
2
A bd
=
≥
=
s
)
1.00 1
for M25 grade of concrete from code = 0.4 N/mm2
Allowable shear stress
τ =
c
= K1 × K
2
× 0 .4
0.3944 N/mm2
>
τ
v
, Hence Safe
Page 3
2.2 Design of Interior Panels: The slab panel is designed by Pigeaud’s method. D
B
C
A Fig : 3 Bridge Plan
Short span of slab, Long span of slab,
Bs Ls
Calculation of Bending moments a) Due to Dead load: Self weight of wearing coat Self weight of deck slab
= =
= =
2
m
4.5
m
1.76 5.28 7.04
kN/m2 kN/m2 2
kN/m Total = Since the slab is supported on all four sides and is continuous, Piegaud’s curves are used to calculate bending moments. Ratio k = Bs/Ls Ratio, = 0 44 0.44 As the panel is loaded with UDL, u/Bs = 1 v/Ls = 1 Where, u & v are the dimensions of the loaded area. From the Pigeaud’s curve, m1 = 0.0457 m2 = 0.0086 Total dead load W = 63.36 kN Moment along short span, M1 = W (m1 +0.15m2) = 2.98 kN-m Moment along long span, M2 = W (0.15m1 +m2) = 0.98 kN-m Considering effects of continuity, 0.8 Moment along short span, M1 = 2.38 kN-m Moment along long span, M2 = 0.78 kN-m
b) Due to Live load: Class AA Tracked Vehicle For maximum bending moment one wheel is placed at the center of panel. Tyre contact length along short span, x = 0.85 Tyre contact length along long span, y = 3.6 Loaded length, u = 1.034 m Loaded width, v = 3.766 m
m
350 kN
Ratio, k = Bs/Ls = 0.44 u/Bs /B = 0 517 0.517 v/Ls = 0.837 From the Pigeaud’s curve, m1 = 0.0813 m2 = 0.0147 Moment along short span, M1= W(m1+0.15m2) = Moment along long span, M2= W(0.15m1+m2) =
W1=350kN N
Wheel load, W =
m
29.227 kN-m 9.413 kN-m Page 4
Fig: 4
Bending moment including impact and continuity, M1 = 29.227 kN-m M2 = 9.413 kN-m
Fig: 4
c) Due to Live load: Class AA Wheeled Vehicle Case-I: When two loads of 37.5 kN each and four loads of 62.5kN are placed such that two loads of 62.5kN lies at center line of pannel. = m Tyre contact width (along short span), 0.30 Tyre contact length (along long span), = m 0.15 Disperced width along short span, u = m 0.510 Disperced width along long span, V = m 0.380
62.5kN 62.5kN
W4
37.5kN
W1
W5
X
37.5kN
X
62.5kN
W2
0 44 0.44
=
Y
62.5kN
K
W3
W6 Y
Fig: 5
Bending moment due to load W1:
62.5 kN
Ratio, k = Bs/Ls = 0.44 u/Bs = 0.255 v/Ls = 0.084 From the Pigeaud’s curve, m1 = 0.1965 m2 = 0.1383 Moment along short span, M1= W(m1+0.15m2) = 13.578 kN-m Moment along long span, span M2= W(0.15m1+m2) = 10.486 kN-m Bending moment including impact and continuity, 13.578 kN-m M1 = M2 = 10.486 kN-m Bending moment due to load W2: 62.5 kN Wheel load is placed unsymmetrical wrt the X-X Intensity of loading, q = 322.45
kN/m2
Considering loaded area 2.000 x 0.380 Loaded area = 0.760 m2 Total applied load = q x area = 245 kN Ratio, k = Bs/Ls = 0.44 u/Bs = 1.000 v/Ls = 0.084 From the Pigeaud Pigeaud’ss curve, m1 = 0.0935 m2 = 0.0742 Moment along short span, M1= W(m1+0.15m2) = 25.650 kN-m Moment along long span, M2= W(0.15m1+m2) = 21.628 kN-m Bending moment including impact and continuity, M1 = 25.650 kN-m
m
Page 5
M2 =
21.628 kN-m
Next, Consider the area between the real and the dummy load i.e., Loaded area = 0.566 m2 Total applied load = q x area = 183 kN Ratio, k = Bs/Ls = 0.44 u/Bs = 0.745 v/Ls = 0.084
From the Pigeaud’s curve, m1 = 0.1157 m2 = 0.0944 Moment along short span, M1= W(m1+0 W(m1+0.15m2) 15m2) = Moment along long span, M2= W(0.15m1+m2) = Bending moment including impact and continuity, M1 = 23.718 kN-m M2 = 20.412 kN-m Final Moment M1 = 0.966 kN-m M2 = 0.608 kN-m Bending moment due to load W3: Intensity of loading, q = 193.47 kN/m2 Considering loaded area 1.710 Loaded area = 0.650 m2 Total applied load = q x area = Ratio, k = Bs/Ls = 0.44 u/Bs = 0.855 v/Ls = 0.084 From the Pigeaud’s curve, m1 = 0.1036 m2 = 0.08325 Moment along short span, M1= W(m1+0.15m2) = Moment along long span, M2= W(0.15m1+m2) W(0 15m1+m2) = Bending moment including impact and continuity, 14.598 kN-m M1 = M2 = 12.423 kN-m
1.490 m
X
0.380 m
23 718 kN-m 23.718 20.412 kN-m
37.5 kN Wheel load is placed unsymmetrical wrt the X-X 0.380
x
m
126 kN
14.598 kN-m 12 423 kN-m 12.423
Next, Consider the area between the real and the dummy load i.e., Loaded area = 0.262 m2 Total applied load = q x area = 51 kN Ratio, k = Bs/Ls = 0.44 u/Bs /B = 0 345 0.345 v/Ls = 0.084
From the Pigeaud’s curve, m1 = 0.1765 m2 = 0.1312 M1= W(m1+0.15m2) = 9.957 Moment along long span, M2= W(0.15m1+m2) ( ) = 8.002 Bending moment including impact and continuity, M1 = 9.957 kN-m M2 = 8.002 kN-m Final Moment 2.321 kN-m M1 = M2 = 2.210 kN-m Bending moment due to load W4: 62.5 Intensity of loading, q = 322.45
0.69 m
X
0.380 m
kN-m kN-m
kN Wheel load is placed unsymmetrical wrt the Y-Y kN/m2
Page 6
Considering loaded area 2.780 x Loaded area = 1.418 m2 Total applied load = q x area = 457 kN Ratio, k = Bs/Ls = 0.44 u/Bs = 0.618 v/Ls = 0.255 From the Pigeaud’s curve, 0.1168 m1 = m2 = 0.0648 Moment along short span, M1= W(m1+0.15m2) = 57.832 kN-m Moment along long span, span M2= W(0.15m1+m2) = 37.628 kN-m Bending moment including impact and continuity, M1 = 57.832 kN-m M2 = 37.628 kN-m
0.510
m
2.02 m
Next, Consider the area between the real and the dummy load i.e., Loaded area = 1.030 m2 Total applied load = q x area = 332 kN R i k=B Ratio, Bs/Ls /L = 0 44 0.44 u/Bs = 0.449 v/Ls = 0.255
X
0.510 m
From the Pigeaud’s curve, m1 = 0.1353 m2 = 0.0712 Moment along short span, M1= W(m1+0.15m2) ( ) = 48.480 kN-m Moment along long span, = M2= W(0.15m1+m2) 30.386 kN-m Bending moment including impact and continuity, M1 = 48.480 kN-m M2 = 30.386 kN-m Final Moment M1 = 4.676 kN-m M2 = 3 621 kN-m 3.621 Bending moment due to load W5: Wheel load is placed unsymmetrical wrt the Both X-X and Y-Y For X- X Axis
W
=
62.5 kN
Intensity of loading, q = 322.45 kN/m2 Considering loaded area 2.000 x Loaded area = 0.760 m2 Total applied load = q x area = 245 kN Ratio, k = Bs/Ls = 0.44 u/Bs = 1.000 v/Ls = 0.084 From the Pigeaud’s curve, m1 = 0.0935 m2 = 0.0742 Moment along short span, M1= W(m1+0.15m2) = 25.650 kN-m Moment along long span, M2= W(0.15m1+m2) = 21.628 kN-m Bending moment including impact and continuity, M1 = 25.650 kN-m M2 = 21.628 kN-m Next, Consider the area between the real and the dummy load i.e., Loaded area = 0.566 m2
0.380 m
1.490 m
X
0.380 m
Page 7
Total applied load = q x area = 183 kN Ratio, k = Bs/Ls = 0.44 u/Bs = 0.745 v/Ls = 0.084 From the Pigeaud’s curve, m1 = 0.1157 m2 = 0.0944 Moment along short span, M1= W(m1+0.15m2) = 23.718 kN-m Moment along long span, M2= W(0.15m1+m2) = 20.412 kN-m Bending moment including impact and continuity, continuity M1 = 23.718 kN-m M2 = 20.412 kN-m Moment Along X-X M1 = 0.966 kN-m M2 = 0.608 kN-m For Y- Y Axis W
=
62.5 kN
Intensity of loading, q = 322.45 kN/m2 Considering loaded area 2.780 x 0.510 Loaded area = 1.418 m2 Total applied load = q x area = 457 kN Ratio, k = Bs/Ls = 0.44 u/Bs = 0.618 v/Ls = 0.255 From the Pigeaud Pigeaud’ss curve, m1 = 0.1168 m2 = 0.0648 Moment along short span, M1= W(m1+0.15m2) = 57.832 kN-m Moment along long span, = 37.628 kN-m M2= W(0.15m1+m2) Bending moment including impact and continuity, M1 = 57 832 kN-m 57.832 kN m M2 = 37.628 kN-m Next, Consider the area between the real and the dummy load i.e., 2.020 m Loaded area = 1.030 m2 Total applied load = q x area = 332 kN Ratio, k = Bs/Ls = 0.44 u/Bs = 0.449 v/Ls = 0.255 From the Pigeaud’s curve, 0.1353 m1 = m2 = 0.0712 Moment along short span, M1= W(m1+0.15m2) = 48.480 kN-m Moment along long span, M2= W(0.15m1+m2) = 30.386 kN-m Bending moment including impact and continuity, M1 = 48.480 kN-m M2 = 30.386 kN-m Final Moment 4.676 kN-m M1 = M2 = 3.621 kN-m Resultent Moment M1 = 5.642 kN-m M2 = 4.230 kN-m
m
X
0.51 m
Page 8
Bending moment due to load W6: Wheel load is placed unsymmetrical wrt the Both X-X and Y-Y For X X Axis X-X
W
=
Intensity of loading, q =
37.5 kN 193.47
kN/m2
Considering loaded area 1.710 x Loaded area = 0.650 m2 Total applied load = q x area = 126 kN Ratio, k = Bs/Ls = 0.44 u/Bs = 0 855 0.855 v/Ls = 0.084 From the Pigeaud’s curve, m1 = 0.1036 m2 = 0.08325 Moment along short span, M1= W(m1+0.15m2) = 14.598 kN-m Moment along long span, M2 W(0.15m1+m2) M2= W(0 15 1 2) = 12 423 kN-m 12.423 Bending moment including impact and continuity, 14.598 kN-m M1 = M2 = 12.423 kN-m Next, Consider the area between the real and the dummy load i.e., Loaded area = 0.262 m2 Total applied load = q x area = 51 kN Ratio, k = Bs/Ls = 0.44 u/Bs = 0.345 v/Ls = 0.084
0.380 m
0.69 m
X
0.380 m
From the Pigeaud’s curve, 0.1765 m1 = m2 = 0.1312 Moment along short span, M1= W(m1+0.15m2) = 9.957 kN-m Moment along long span, span M2= W(0.15m1+m2) = 8.002 kN-m Bending moment including impact and continuity, M1 = 9.957 kN-m M2 = 8.002 kN-m Moment Along Y-Y 2.321 kN-m M1 = M2 = 2.210 kN-m For Y- Y Axis
W
=
37.5 kN
Intensity of loading, q = 193.47 kN/m2 Considering loaded area 2.780 x Loaded area = 1.418 m2 Total applied load = q x area = 274 kN Ratio, k = Bs/Ls = 0.44 u/Bs = 0.618 v/Ls = 0.255 From the Pigeaud’s curve, m1 = 0.1168 m2 = 0.0648 Moment along short span, M1= W(m1+0.15m2) = 34.699 kN-m Moment along long span, = 22.577 kN-m M2= W(0.15m1+m2)
0.510
m
Page 9
Bending moment including impact and continuity, M1 = 34.699 kN-m M2 = 22.577 kN-m Next, Consider the area between the real and the dummy load i.e., Loaded area = 1.030 m2 Total applied load = q x area = 199 kN Ratio, k = Bs/Ls = 0.44 u/Bs = 0.449 v/Ls = 0.255
From the Pigeaud’s curve, m1 = 0.1353 m2 = 0 0712 0.0712 Moment along short span, M1= W(m1+0.15m2) = 29.088 kN-m Moment along long span, M2= W(0.15m1+m2) = 18.231 kN-m Bending moment including impact and continuity, M1 = 29.088 kN-m M2 = 18.231 kN-m Final Moment M1 = 2.806 kN-m M2 = 2.173 kN-m Resultent Moment 5.127 M1 = M2 = 4.383 Total Moment Due to IRC Class AA Wheeled Vechicle Moment along short span,M1 Moment along long span,M2 c) Due to Live load: Class A Loading
2.020 m
= =
X
0.510 m
32.309 kN kN-m m 25.539 kN-m
IRC Class A Loading: For maximum bending moment one wheel of 57kN should be placed at the centre of span and other at 1.2 m from it as shown. Neglecting small eccentricity of 80mm. Tyre contact length along short span, Y = 0.5 m
Tyre contact length along long span, X = 0.25 m Imaginary load W3 = W2 is placed on the other side of W1 to make loading symmetrical. Due to loads W2 & W3 Bending moment at center of panel will be that due to load W1 and half.
Page 10
W1
57kN
X
57kN
Y
X
W2
Y Fig: 6
Bending moment due to load W1: Wheel load, W1 =
57 kN
Loaded length, u = 0.696 m Loaded width, v = 0.465 m Ratio, k = Bs/Ls = 0.44 u/Bs = 0.35 v/Ls = 0.10 From the Pigeaud Pigeaud’ss curve, m1 = 0.1717 m2 = 0.1245 Moment along short span, M1= W(m1+0.15m2) = 10.851 kN-m Moment along long span, M2= W(0.15m1+m2) = 8.565 kN-m Bending moment including impact and continuity, M1 = 10 851 kN-m 10.851 kN m M2 = 8.565 kN-m Bending moment due to load W2: Wheel load is placed unsymmetrical wrt the Y-Y 57 kN Intensity of loading, q = 176.09 kN/m2 Wheel load, W2 =
Considering loaded area 2.865 x 0.696 m Loaded area = 1.993 m2 Total applied load = q x area = 351 kN Ratio, k = Bs/Ls = 0.44 u/Bs = 0.637 v/Ls = 0.348 From the Pigeaud’s curve, m1 = 0.10843 m2 = 0.0497 Moment along short span, M1= W(m1+0.15m2) = 40.676 kN-m Moment along long span, M2= W(0.15m1+m2) = 23.154 kN-m Bending moment including impact and continuity, 40.676 kN-m M1 = M2 = 23 154 kN-m 23.154 kN m Next, Consider the area between the real and the dummy load i.e., 1.935 m Loaded area = 1.346 m2 Total applied load = q x area = 237 kN Ratio, k = Bs/Ls = 0.44 u/Bs = 0.430 v/Ls = 0.348 From the Pigeaud’s curve, m1 = 0.1313
X
0.696 m
Page 11
m2 = 0.0552 Moment along short span, M1= W(m1+0.15m2) = 33.081 kN-m Moment along long span, M2= W(0.15m1+m2) = 17.751 kN-m Bending moment including impact and continuity, M1 = 33.081 kN-m M2 = 17.751 kN-m Final Moment M1 = 3.798 kN-m M2 = 2.702 kN-m Total Bending Moment due to load W1 & W2 will be, M1 = 14.6 kN-m M2 = 11.3 kN-m Design Bending Moment due to LL: M1 = 32.3 kN-m M2 = 25.5 kN-m Calculation of Shear Force a) Due to Dead load: Dead load shear force =
7.04 kN
b) Due to Live load: Class AA Tracked Vehicle Load of Tracked Vehicle=
350 kN
350kN
350kN
Fig 7.a
Dispertion in the direction of span, = 1.45 m For maximum shear, load is kept such that whole dispersion is in the span. That is at from the edge of beam.
0.725 m 0.3
Effective width of slab = αx 1 − x + bw l Span Ratio (L/B) = a for continuous slab = x = bw = Therefore effective width of slab = Load per meter width = Shear force at left edge = Shear force including impact & continuity =
2.25 2.60 0.725 3.76 4.96 70.54 44.97
m m kN kN 44.97 kN
c) Due to Live load: Class AA Wheeled Vehicle
37.5kN
62.5kN
62.5kN
37.5kN
Page 12
37.5kN
62.5kN
37.5kN
62.5kN
Fig 7 7.b b
Dispersion width in the direction of span = Loads are placed such that outermost load is at distance of x bw
Effective width for first wheel
= =
0.900
m
0.450 0.31
m from edge of the beam.
=
=
1.217 m
Wheel Load But the center to center distance of two axel are Average effective width for one wheel = Portion of load in span = Load per meter width of slab =
=
62.50 kN 1.2 m, thus effective width will overlap. 1.208 m 1.000 m 51.72 kN
For second wheel
=
62.50 kN m 0.550 1.347 m 1.2 m, thus effective width will overlap. 1.273 m 49.1 kN
Wheel Load x
= =
Effective width for second wheel But the center to center distance of two axel are Average effective width for one wheel = Load per meter width of slab = For third wheel
Wheel Load X
Effective width for third wheel Load Acting on Span Acting at 0.2 Effective width for third wheel Load per meter width of slab Shear force at left edge Shear force including impact and continuity
= = = = m from Support = = = =
37.50 kN -0.050 0 400 0.400 16.67
m kN
0.918 18 37.3
m kN kN
37.318
kN
< 1.2 m
b) Due to Live load: IRC Class A loading
57kN
57kN
Page 13
Fig 7.c
Fig 7.c
Shear force due to load W1: 57 kN Dispersion width in the direction of short span = For maximum shear force, the load should be placed at distance of of girder. In this position second load will be as shown.
m
x αx 1 − + bw l
Effective width for first wheel = Where,
1.1 0.55 m from web
x= bw= bw L/B=
0.55 0.41 m 2.3
Therefore, Effective width = But distance between axels is Average effective width / wheel = Load W1 = Load per meter width of slab = And shear force = Shear force including impact & continuity =
m α= 2.6 1.447 m 1.2 m and hence effective width overlaps. 1.32 m 57.00 kN 43.07 kN/m 32.30 kN 32 30 32.30 kN
Shear force due to load W2:
x αx 1 − + bw l
Effective width for second wheel = Where,
0.350 m 0.41 m 2.4 1.161 m
x= bw= L/B=
Therefore, Effective width = = =
Load W2
Effective load
α=
2.6 < 1.2 m
57.0 kN 18.1 kN 15.6 kN
Load per meter width of slab = And shear force
Shear force including impact & continuity Total shear force
= = =
12.89 kN 12.89 kN 45.19 kN
Design of Section: Total Design Bending Moments, 34.69 kN-m M1 = M2 = 26.32 kN-m Total Design Shear force, 52.23 kN S.F. = Effective depth required, d=
M = Rb
171 4 mm 171.4
Use 30 mm Effective depth available = Area of steel required along short span, Ast =
M = σ st × j × d
1055
Clear cover 184 mm
&
12 mm
dia
O.K.
mm2
Check for minimum area of steel: Page 14
Min. area of steel @
Provide ø steel =
264 12 mm bars
1130.4
Effective depth for long span = Area of steel required along long span,
M = σ st × j × d
Ast = Provide ø steel =
851
12 mm bars
1130.4
mm2
100 mm c/c at both top and bottom, giving area of mm2 > required. 173 mm
mm2
100 mm c/c at both top and bottom, giving area of mm2 > required.
Check for shear: Nomin
τv =
Vu = b× d
2 0.284 N/mm
Provided percentage area of tensile steel = Permissible shear stress, Κ×τc = 1.16 x
0.51 % 0.363 N/mm2
0.313 =
> O.K.
3.0 Design of Longitudinal Girder Effective Span of Bridge Slab thickness Width of Rib Spacing of main Beam Over all depth of Beam
= = = = =
24 0.22 0.4 2.4 2
m m m m m
c/c
3.1 Calculation of dead load moment and shear force on longitudinal girder: Let the over all depth of the longitudinal girder be = Weight of Rib per m =
2000 mm, the depth of its rib will 1.78 m 17.09 kN/m
Dead load due to deck Dead load from each cantilever portion (refer design of cantilever slab) = 7.68 kN/m 2 Dead load of slab & Wearing coat = 7.04 kN/m Total Dead load per m from deck = 51.98 kN/m This load is borne by all the three girders Dead Load per girder due to Deck Slab = 17.33 kN/m Let the Depth of rib of cross girder to be = 1.28 m let its width be = 0.3 m Weight of rib of cross girder = 9.216 kN/m Length of each cross girder = 4m It is assumed that the weight of each cross girder is equally borne by the entire three longitudinal girders. This weight acts as point load on each girder its value being = 12.29 kN 3 kN
12.3 kN
12.3 kN
12.3 kN
12.3 kN
12.3 kN
34.413 kN/m
RA
Fig : 8 Total UDL RA= RB
= =
RB
34.413333 kN/m 449.824 kN
Bending Moment (BM) Page 15
BM at Centre of span BM at ¼ th of span th
Span BM att 1/8 Span BM at 3/8
th
Shear Force (SF) SF at Support SF at 1/8th Span th
Span SF at 3/8 Span SF at 1/4
th
SF at Center of span
Distance from Support At Center of Span At 3/8th Span
= =
2654.707 kNm 2064.758 kNm
=
2492.474 kNm
=
1157 748 kNm 1157.748 kN
= =
437.536 kN 334.296 kN
=
218.768 kN
= =
115.528 kN 0 kN
BM
SF
2654.71 2492.47 2064.76 0.00
At 1/4th Span At Support
0.00 115.53 218.77 437.54
3.2 Calculation of live load moment and shear force on longitudinal girder: Impact factor for: 6
4 .5 + L
=
0.150
IRC Class AA Tracked Vehicle =
0 100 0.100
IRC Class AA Wheeled Vehicle =
0.215
Distribution of live loads on longitudinal girder for bending moment: IRC Class AA Tracked Vehicle: Reaction on the girder will be maximum when the eccentricity is maximum. Eccentricity will be maximum when the loads are very near to the kerb. Position of loads for maximum eccentricity is shown in figure. All the th girders i d are assumedd to t have h the th same momentt off inertia. i ti
W1
W1
Fig 9 Fig.
Reaction factor for Outer Girder, RA =
2 W1 3I 1 + 3 2I × (2.4
2 W1 = 3
Reaction factor for Inner Girder, RB = If W be the axel load, then wheel load W1 = W/2 Then reaction factor, RA = RB =
)2
× 2 . 4 × 0 . 35 =
0.81 W1
0.667 W1
0.406 W 0.333 W
IRC Class A Loading: Position of loads for maximum eccentricity is shown in figure.
W1
W1
W1
W1
Page 16
W1
W1
W1
W1
Fig. 10
Reaction on Outer Girder RA,
4 W1 3
RA=
RB =
3I × 2 .4 × 0 .1 = 1 + 2 2I × (2.4 )
4W 1 [1 + 0 ] = 3
1.333
1.625
W2
W2
If W be the axel load, then wheel load W2 = W/2 Then reaction factor, RA =
0.813 W
RB =
0.667 W
Bending g Moment due to Live load: IRC Class AA Tracked Vehicle The influence line diagram for bending moment is shown in figure. Effective span of girder, le = 24.0 m ITC Class AA Tracked Vehicle Load: = 700 kN 1 −
Ordinate of Bending Moment at considered section, M x =
x x L
Calculation of bending moment at L/2.
Ordinate of Influence line at mid span =
6.0 m Leff=
12
24 m Leff 3.6m
700 kN
5.1
6 RB
RA
Fig. 11a: ILD for BM at L/ 2
3885.0
kN-m Bending Moment including impact and rection factor for outer Girder Bending Moment including impact and rection factor for Inner Girder
Bending Moment =
Calculation of bending moment at 3L/8.
= = 9m
=
Ordinate of Influence line at mid span =
Leff=
1736 kN-m 1425 kN-m 24 m 9
5.625 m Leff 3.6m
700 kN
4.5
4.95 5.63
3*Leff / 8
RB
RA
Fig. 11b: ILD for BM at 3L/ 8
1855 7 1855.7
kNm Bending Moment including impact and rection factor for Outer Girder Bending Moment including impact and rection factor for Inner Girder
Bending Moment =
Calculation of bending moment at L/4.
Ordinate of Influence line at mid span =
= = 6
=
4.500 m
m
Leff=
829.3 kNm 680.4 kNm 24 m 6
Leff 3.6m
700 kN
Page 17 Leff / 4
700 kN
3.15 4.500
Leff / 4
4.05 RB
RA
Fig. 11c: ILD for BM at L/ 4
1500.1
= Bending Moment including impact and rection factor for Outer Girder Bending Moment including impact and rection factor for Inner Girder
Bending Moment
Calculation of bending moment at L/8.
3
=
Ordinate of Influence line at mid span =
m
kNm = =
670.3460 kNm 550.0275 kNm
Leff=
24 m 3
2.625 m Leff
3.6m
700 kN
1.050 Leff / 8
2.400 2.625 RB
RA
Fig. 11d: ILD for BM at L/ 8
881.2 = kNm Bending Moment including impact and rection factor for Outer Girder Bending Moment including impact and rection factor for Inner Girder
Bending Moment
393.792 323.11
= =
kNm kNm
Bending Moment due to Live load: IRC Class A Loading The influence line diagram for bending moment is shown in figure. 24 m Effective span of girder, le = Unit Loads Values Loads Values W1 27 W5 68 kN kN W2 27 W6 68 kN kN W3 114 W7 68 kN kN W4 114 W8 68 kN kN Distances X X1 X2 X3 X4
Values Varies Varies 1.1 3.2 1.2
Unit
Distances X5 X6 X7 X8 X9
m m m
1 −
Ordinate of Bending Moment at considered section, M x = Calculation of bending moment at L/2
Values 4.3 m 3 m 3 m 3 m Varies
=
Ordinate of Influence line at mid span = W3=144kN
x
12 m, when load W4 is at L/2 Leff= 24 m
6m
W1=27kN W2=27kN
x L
W4=144kN
W5=68kN
W6=68kN
W7=68kN
x1= 6.5
RA
6.00 X= 12
RB
Fig. 12a: ILD for BM at L/ 2
Load Nos.
Load Values kN
W1 W2 W3 W4 W5 W6 W7
27 27 114 114 68 68 68
Position from Maximum
IL Ordinate
-5.50 -4.40 -1.20 0.00 4.30 7.30 10.30
Total = Total Bending Moment including impact for Outer Girder =
3.250 3.800 5.400 6.000 3.850 2.350 0.850
Moment Component 87.75 102.60 615.60 684.00 261.80 159.80 57.80 1969.35 kN-m 1840.11 kN-m
Page 18
Total Bending Moment including impact for Inner Girder = Calculation of bending moment at 3L/8
=
Maximum Ordinate of Influence line = W1=27kN
x1= 4.7
5.625 m
W2=27kN
W3=144kN W4=144kN
1509.84 kN-m
9.0 m, when load W3 is at 3L/8 Leff= 24 m W5=68kN
W6=68kN
W7=68kN
W8=68kN
5.625 X= 9.0
RA
RB
Fig. 12b: ILD for BM at 3L/ 8
Load Nos Nos.
Load Values kN
Position from Maximum
IL Ordinate
W1 W2 W3 W4 W5 W6 W7 W8
27 27 114 114 68 68 68 68
-4.30 -3.20 0.00 1.20 5.50 8.50 11 50 11.50 14.50
2.938 3.625 5.625 5.175 3.563 2.438 1 313 1.313 0.188
Moment Component 79.31 97.88 641.25 589.95 242.25 165.75 89 25 89.25 12.75
Total = Total Bending Moment including impact for Outer Girder = Total Bending Moment including impact for Inner Girder =
Calculation of bending moment at L/4
=
Maximum Ordinate of Influence line = W1=27kN W2=27kN
4.500 m
W3=144kN W4=144kN
1918.39 kN-m 1792.49 kN-m 1470.76 kN-m
6 m, when load W3 is at L/4 Leff= 24 m
W5=68kN
W6=68kN
W7=68kN
W8=68kN
X1= 1.7 4.50
X= 6
RA
RB
Fig 12c: ILD for BM at L/ 4 Fig.
Load Nos.
Load Values kN
Position from Maximum
IL Ordinate
W1 W2 W3 W4 W5 W6 W7 W8
27 114 114 68 68 68 68 68
-4.30 -3.20 0.00 1 20 1.20 5.50 8.50 11.50 14.50
1.275 2.100 4.500 4 200 4.200 3.125 2.375 1.625 0.000
Moment Component
Total = Total Bending Moment including impact for Outer Girder = Total Bending Moment including impact for Inner Girder = Calculation of bending moment at L/8
=
Maximum Ordinate of Influence line = W3=144kN W4=144kN
2.625 m W5=68kN
W6=68kN
34.43 239.40 513.00 285 60 285.60 212.50 161.50 110.50 0.00 1556.93 kN-m 1454.75 kN-m 1193.64 kN-m
3 m, when load W4 is at L/8 Leff= 24 m W7=68kN
W8=68kN
2.625
RA
X= 3
Fig. 12d: ILD for BM at L/ 8
RB
Page 19
Fig. 12d: ILD for BM at L/ 8
Load Nos.
Load Values kN
W1 W2 W3 W4 W5 W6 W7 W8
27 114 114 68 68 68 68 68
Position from Maximum O 0.00 di t
IL Ordinate
Moment Component
0.000 0.000 2.625 2.475 1.938 1.563 1.188 0.813
0.00 0.00 1.20 5.50 8.50 11.50 14.50
0.00 0.00 299.25 168.30 131.75 106.25 80.75 55.25
Total = Total Bending Moment including impact for Outer Girder = Total Bending Moment including impact for Inner Girder =
841.55 kN-m 786.32 kN-m 645.19 kN-m
Absolute Maximum BM Calc lation of bending moment at the load point which Calculation hich is eq equidistance idistance from res resultant ltant Ordinate of Influence line at mid span = 6m W1=27kN W2=27kN
W3=144kN W4=144kN
R
Leff= W5=68kN
24 m W6=68kN
W7=68kN
6 RA
x= 12
RB
Fig. 12e: ILD for BM at L/ 2 / Absolute Maximum BM
C.G of the Load system from outer 27 kN Wheel Load = 6.420 m The heavier wheel load near C.G. of load System is 114kN which lies at a distance of
6.42-(1.1+3.2+1.2)= X
=
0.92 m from CG 0.46 m
Load Nos.
Load Values kN
Position from Left support
IL Ordinate
W1 W2 W3 W4 W5 W6 W7 W8
27 114 114 68 68 68 68 68
6.04 7.14 10.34 11.54 15.84 18.84 21.84 0.00
3.020 3.570 5.170 5.770 4.080 2.580 0.000 0.000
Total = Total Bending for Outer Girder = g Moment including g impact p Total Bending Moment including impact for Inner Girder =
Moment Component 81.54 406.98 589.38 392.36 277.44 175.44 0.00 0.00 1923.14 kN-m 1796.93 kN-m 1474.41 kN-m
Page 20
Shear Force due to Live load: IRC Class AA Tracked Vehicle. At Support Effective span of girder, le = Load Class AA Tracked vehicle W1=
24 m 350 kN
RC =
RQ'
Q
f
C/C
c
b
h
C
CL of outer L Girder kerb Line
X
R
RR'
e
CL of inner L Girder
B
a
RS =
0.35
R
C/C
RR =
W1 W1 W1 W1 W1 W1
i
RB =
0.358 0.215 0.716 0.430 0 176 0.176 0.105
C/C
RQ =
Q
1.025
g
Reaction at support, RA =
CL of outer L Girder
A
Interm mediateX -G irder
W1 W1
d
PC =
1.146 0.281
Ist X- Girder
PB =
a
For maximum shear at support, load should be as near the support as possible. The length of the load is 3.6m, the SF will be max. when the C. G. of the load is placed at a distance of 1/2*3.6= 1.8m From the support along its length, thus the load will lies between the support & the Ist Intermediate X-girder, the width of track being 0.85m, the CG of load will thus lie at a distance of 1.2+0.85/2=1.625 m from kerb of footpath Load act at a distance of 1.8 m from support A, B and C L= 4.8 m C/C Distance of L Girder= 2.4 m 1.8 X= X1= 3.0 a= 0.6 b= C= 1.625 d= 2.05 f= 2.050 e= g= 1.375 h= 0.675 i= 1.725 L Loads on Girders, kerb Line PA= 0.573 W1
S
RS'
S
X1
Fig. 13 The loads on the cross girder i.e. RQ, RR & RS are to be distributed by normal Courbon's theory. Total load, ∑W = W1 = 0.750 2.05 m C.G. of loads from Q = Eccentricity, e = 0.35 m Reaction factor for outer girder, FQ = 0.305 W1 in this case xi=2.4 m and ∑xi2=(2.4)2+(0)2+(2.4)2= 2 x (2.4)2
Reaction factor for inner girder, FR =
0.250
W1
in this case xi=0 m It may be seen that the reaction FQ and FR act as load at 1/3 span of outer longitudinal girder and inner longitudinal girder respectively. The reactions at support A and B due to those loads are RA due to FQ = 0.24375 W1 RB due to FR =
0.200 W1
`
0.602 Total reaction on outer Girder = 0.916 Total reaction on inner Girder = Max shear at support including impact for outer girder Max shear at support including impact for inner girder At Intermediate Section Effective span of girder, le =
W1 W1 = =
231.7 kN 352.7 kN
24.0 m
Ordinate of Bending SF at considered section, SF x At Left
=
At Right
x = 1 − 1 − L
1 −
x L
Page 21
Shear at 1/8th span Calculation of bending moment at L/8
=
Ordinate of Influence line at Left O di t off Influence Ordinate I fl line li att Right Ri ht x= 3
3 m, when load Placed at Just Right of L/8
= =
0.875 m 0.125 0 125 m
a'= 0.125
a= 0.875
Leff= L ff 24 b= 0.725
m
3.6m a' 350 kN a
b
RA
RB
Fig. 14a: ILD of SF at L/ 8 of Span
S.F.
=
280 kN 125.13 kN 102.667 kN
S.F. including impact for outer girder = S.F. including impact for inner girder = Shear at 1/4th span Calculation of bending moment at L/4
=
Ordinate of Influence line at Left Ordinate of Influence line at Right 6 x=
6 m, when load Placed at Just Right of L/4
= =
0.75 m 0.25 m
a'=
0.25
a=
0.750
Leff= 24 m b= 0.600
3.6m
350 kN
a'
b
a
RA
RB
Fig. 14b ILD of SF at L/ 4 of Span
S.F. =
236.25 kN 105.574 kN 86.625 kN
S.F. including impact for outer girder = S.F. including impact for inner girder = Shear at 3/8th span Calculation of bending moment at 3L/8
=
Ordinate of Influence line at Left Ordinate of Influence line at Right x= 9
9 m, when load Placed at Just Right of 3L/8
= = a'=
0.375
0.625 m 0.375 m a= 0.625
Leff= 24 m b= 0.475
3.6m
350 kN
a'
b
a
RA
RB
Fig. 14c 14cILDof ILD of SF SFat at 3L/ 8 of Span
S.F. =
192.5 kN 86.023 kN 70.583 kN
S.F. including impact for outer girder = S.F. including impact for inner girder = Shear at 1/2th span Calculation of bending moment at L/2
Ordinate of Influence line at Left Ordinate of Influence line at Right x= 12
=
12 m, when load Placed at Just Right of L/2
= =
0.5 m 0.5 m
a'= 0.5
a= 0.5
Leff= Leff 24 m b= 0.350
3.6m a' 350 kN a
RA
Fig. 14d ILD of SF at L/ 2 of Span
b
RB Page 22
Shear Force due to Live load: IRC Class A Load. The influence line diagram for shear force is shown in figure. Effective span of girder, le = 24.0 m Loads Values Loads Values W1 27 W5 68 kN W2 27 W6 68 kN W3 114 W7 68 kN W4 114 W8 68 kN Distances X X1 1 X2 X3 X4
Values Varies Varies 1.1 3.2 1.2
m m m
Ordinate of Bending SF at considered section, SF x At Left = 1 −
Distances Values X5 4.3 X6 6 3 X7 3 X8 3 X9 Varies x L
At Right
kN kN kN kN
m m m m
=
x 1 − 1 − L
Page 23
Calculation of Shear Force at Support
=
Ordinate of Influence line y3= W3=144kN W4=144kN
Y3
Y4
24.0 m, when load W3 Leff=
1 W5=68kN
W6=68kN
W7=68kN
W8=68kN
Y5
Y6
Y7
Y8
RA
24.0 m
RB
Fig. 15a: ILD for SF at Support
Load Nos.
Load Values kN
W3 W4 W5 W6 W7 W8
114 114 68 68 68 68
Position from Left support 0 1.20 5.50 8.50 11.50 14.50
IL Ordinate
Moment Component
114.00 108.30 52.42 43.92 35.42 26.92
1.000 0.950 0.771 0.646 0.521 0.396
Y3= Y4= Y5= Y6= Y7= Y8=
Total = Total SF including impact for Outer Girder = Total SF including impact for Inner Girder = Calculation of SF at L/8
=
Ordinate of Influence line
3.000 m, when load W3
0.875
At Right At Left
0.125
W3=144kN W4=144kN
Y3
380.97 kN 355.97 kN 292.07 kN
Y4
Leff=
x=
24.0 m
W5=68kN
W6=68kN
W7=68kN
W8=68kN
Y5
Y6
Y7
Y8
Fig. 15b: ILD for SF at L/ 8
RA Load Nos.
Load Values V l kN
W3 W4 W5 W6 W7 W8
114 114 68 68 68 68
Position from f L Left ft support 0 1.20 5.50 8.50 11.50 14.50
RB
IL Ordinate
Y3= Y4= Y5= Y6= Y7= Y8=
Moment Component
0.875 0.825 0.646 0.521 0.396 0.271
99.75 94.05 43.92 35.42 26.92 18.42
Total = Total SF including impact for Outer Girder = Total SF including impact for Inner Girder =
Calculation of SF at L/4
Ordinate of Influence line W1=27kN W2=27kN
Y1
RA
=
6 000 m, 6.000 m when load W3 0.25
W3=144kN W4=144kN
Y3
300.05 300 05 kN 280.36 kN 230.04 kN
0.75
At Right At Left
Leff=
Y4
W6=68kN
W7=68kN
Y5
Y6
Y7
Fig. 15c: ILD for SF at L/ 4
x=
24.0 m
W5=68kN
Y3' Y2
3
6.00
W8=68kN
Y8
RB Page 24
Load Nos.
W1 W2
Load Values kN 27 27
Position from Left/ Right at X 4.3 Y1= 3.2 Y2=
W3 W4 W5 W6 W7 W8
114 114 68 68 68 68
0 1.20 5.50 8.50 11 50 11.50 14.50
IL Ordinate
Moment Component
-0.071 -0.117 0.750 0.700 0.521 0.396 0 271 0.271 0.146
Y3= Y4= Y5= Y6= Y7= Y8=
-1.91 -3.15 85.50 79.80 35.42 26.92 18 42 18.42 9.92
Total = Total SF including impact for Outer Girder = Total SF including impact for Inner Girder = Calculation of SF at 3L/8
Ordinate of Influence line
=
9.000 m, when load W3
0.625
At Right At Left
0.375
W1=27kN W2=27kN
Y1
250.90 kN 234.44 kN 192.36 kN
Leff=
W3=144kN W4=144kN
W5=68kN
W6=68kN
W7=68kN
Y5
Y6
Y7
Y3' Y2
Y4
Y3
24.0 m
x=
9.0 W8=68kN
Y8
Fig. 15d: ILD for SF at 3L/ 8
RA Load L d Nos. N
W1 W2 W3 W4 W5 W6 W7 W8
Position P ii from Left/ Right at X 4.3 3.2
L d Load Values kN 27 27 114 114 68 68 68 68
0 1.20 5.50 8.50 11.50 14.50
RB
IL Ordinate O di
Y1= Y2= Y3= Y4= Y5= Y5 Y6= Y7= Y8=
Moment Component M C
-0.196 -0.242 0.625 0.575 0.396 0.271 0.146 0.021
-5.29 -6.53 71.25 65.55 26.92 18.42 9.92 1.42
Total = Total SF including impact for Outer Girder = Total SF including impact for Inner Girder = Calculation of SF at L/2
Ordinate of Influence line
=
181.65 kN 169.73 kN 139.27 kN
12 000 m, 12.000 m when load W3
0.5
At Right At Left
0.5 W1=27kN W2=27kN
Y1
Leff= W3=144kN W4=144kN Y3'
Y2
Y3
W5=68kN
Y5
Y4
24.0 m
x=
W6=68kN
Y7
Y6
Fig. 15e: ILD for SF at L/ 2
RA Load Nos.
Load Values kN
W1 W2
27 27
Position from Left/ Right at X 4.3 Y1= 3.2 Y2=
12.0
RB
IL Ordinate
-0.321 -0.367
Moment Component
-8.66 -9.90 Page 25
114 114 68 68 68
W3 W4 W5 W6 W7
0 1.20 5.50 8.50 11.50
0.500 0.450 0.271 0.146 0.021
Y3= Y4= Y5= Y6= Y6 Y7=
57.00 51.30 18.42 9.92 1.42
Total = Total SF including impact for Outer Girder = Total SF including impact for Inner Girder =
Design of Section: Total Design Bending Moments and Shear Forces for Outer Girder: Bending Moment (kN (kN-m) m) Section Due to Due to LL Total DL 0.00 X=0 0.00 0.000 1157.75 X = L/8 1551.540 393.79 2064.76 670.35 2735.104 X = L/4 2492.47 829.26 X = 3L/8 3321.735 X = L/2 1736.11 4390.817 2654.71 Total Design Bending Moments and Shear Forces for Inner Girder: Bending Moment (kN-m) Section Total Due to Due to LL DL 0.000 0.000 X=0 0.00 1480.859 X = L/8 323.111 1157.75 2614.79 X = L/4 550.028 2064.76 680.419 3172.89 X = 3L/8 2492.47 X = L/2 1424.50 4079.21 2654.71
119.49 kN 111.65 kN 91.61 kN
Due to DL
Shear Forces (kN) Due to LL Total
437.536 334.296 218.768 115.528 0.000
Due to DL
355.966 280.359 234.439 169.733 111.646
793.502 614.655 453.207 285.261 111.646
Shear Forces (kN) Due to LL Total
437.54 334.30 218.77 115.53 0.00
352.72 230.04 192.36 139.27 91.61
790.252 564.334 411.128 254.796 91.607
Design of Outer Girder:
Overall depth of beam, D = Rib width, bw = Flange width of T-beam will be, bf = bw + 1/5 x lo Therefore width of flange, bf =
2000 mm 400 mm 2.5 m 2400 mm
2.4 m
>
170 mm from bottom of T-beam to the centre of gravity of rod rod, 1830
d=
Area of steel required, Ast= Provide
M
σ st × j × d
=
13420 mm2
Provided area of steel = 13843 Total Provided area of steel = Number of bars in bottom row Width of beam Side and bottom clear cover to bars C.G. of the bottom row of bars from bottom Clear distance between vertical bars C G off th C.G. the S Second d row off bbars ffrom bbottom tt C.G. of the third row of bars from bottom C.G. of the fourth row of bars from bottom C.G. of the fifth row of bars from bottom C.G. of the bar group from bottom
16 nos. of φ 2 nos. of φ
32 25
mm bars+
mm2
13843 mm2 = = = = = = = = = =
4 400 40 56 32 120 184 248 308.5 163.1
nos.
= =
400
mm mm
Fig. 16 : Cross Section of Girde
mm
mm mm mm mm
mm mm
≤
Effective Depth Df
2000
mm bars
170
mm
O.K.
1830 mm 220 mm Page 26
Check for stresses: Calculation of depth of neutral axis: Assuming that the effective area in of compression and tension sides about neutral axis, we get ½ × bw × xa2 + (bf – bw) × Df × (xa – Df/2) = m × Ast × (d – xa) xa = 482 Solving, we get, mm Let compressive stress in concrete at top of flange = σ And compressive stress in concrete at bottom of flange = σ' σ’ =
Then,
x
a
− D xa
× σ = 0.543 σ
f
Position of C.G. of compressive stress in flange from top, x1 =
σ + 2σ ' D f × = 3 σ +σ'
99.1 mm
½ x(σ +σ ')x Bf x Df
Compressive force in flange, C1 =
407402 σ
=
½ x σ 'xx (Xa – Df)x Bw
Compressive force in rib, C2 =
28420 σ
=
C.G. of compressive force in rib from top, x2 307.2 mm 435822.3 σ
Total compressive force, C = C.G. of total compressive force from top
=
Therefore, lever arm, jd =
C1 × x1 + C2 × x2 = C1 + C2
1717.3 mm d
Critical Neutral axis depth, nd =
σ 1 +
=
st
m σ
Moment of resistance of the section is given by Mr =
112.7 mm
582.3 mm
<
xa
c
− σ +σ 1 bf × d f × = × d − y 2
699626722.86 σ
Equating Mr to external B.M we get 4390816575.00 Mr = σ= 6.28 < 8.3 O.K. Stress Developed in Steel Reinforcement is given by t
=
d − xa m × σ × xa
=
197.6 <
200 O.K.
Check for minimum area of steel Minimum area of tension steel in beam @ 0.2 % of web area 2 13843 mm
1600 mm2 < Ast provided =
O.K.
Design for shear: τ
Assuming
v
=
V B × d
=
####### nos. of φ
percentage area of tension steel =
32 1.00 % 0.420 N/mm2
1.08 N/mm2 will be continued up to support, then provided
Permissible shear stress, τc = < τv Shear reinforcement is required. Shear reinforcement shall be provided to carry a shear of, Vs = V - tc . bw . D = 457195 N Assuming φ 12 mm 2-legged vertical stirrups having area of steel, Asv = Spacing, S =
Asv × σ st × d = Vs
226 mm2
181 mm c/c Page 27
Vs As per minimum shear reinforcement requirements, maximum spacing,
σ st × Asv 0.4 × b w
Smax =
=
283 mm c/c.
Hence provide φ 12 mm, 2-legged vertical stirrups @ 100 mm c/c at support. Design summary: Area of steel required and provided at different sections of Outer girder are given in below: Tension Reinforcement (Fe 415): Section BM Area of Steel Required Area of Steel Provided Area No. Dia. L/2
3L/8 L/4 L/8 Support
4390.82
KNm
13420
mm2
3321.73 KNm 2735.10 KNm 1551.54 KNm
10153 8360 4742
mm2 mm2 mm2
0.00 KNm
0
mm2
16 2 14 12 10 10
32 25 32 32 32 32
13843 11254 9646 8038 8038
mm2 mm2 mm2 mm2 mm2
Shear Reinforcement ((Fe 415): ) Section SF 2-legged vertical Stirrups required 2-legged vertical Stirrups provided 1029.4 φ@ φ@ L/2 111.646 kN 10 mm 10 250 mm 402.87 3L/8 285.261 φ@ φ@ kN 10 mm 10 250 mm 253.58 L/4 453.207 φ@ φ@ kN 10 mm 10 200 mm 269.24 L/8 614.655 φ @ φ @ kN 12 mm 12 150 mm kN mm 12 φ@ 208.56 12 φ@ 100 Support 793.502 mm Area of steel required and provided at different sections of Inner girder are given in below: Tension Reinforcement (Fe 415): Section BM Area of Steel Required Area of Steel Provided No. Dia. Area L/2 3L/8 L/4 L/8 Support
4079 3172.89 2614.79 1480.86 0 000 0.000
KNm KNm KNm KNm KNm
Shear Reinforcement (Fe 415): Section SF L/2 91.607 kN 3L/8 254.796 kN L/4 411.128 kN L/8 564.334 kN Support pp 790.252 kN
12468
mm2
9698 7992 4526 0
mm2 mm2 mm2 mm2
2-legged vertical Stirrups required 10 10 10 12 12
φ@ φ@ φ@ φ@ φ@
150 150 150 150 100
mm mm mm mm mm
16 2 14 12 10 10
32 25 32 32 32 32
13843 11254 9646 8038 8038
mm2 mm2 mm2 mm2 mm2
2-legged vertical Stirrups provided 10 10 10 12 12
mm
φ@
250 250 200 150
φ@
100
mm
φ@ φ@ φ@
mm mm mm
Page 28
4 0 Design of Cross Girder: 4.0 Dead Load Overall Depth of cross girder = Width of cross girder = Self weight of cross girder=
1.5 m 0.3 m 9.216 kN/m
2.4 4.8
Fig. 17
Dead load from slab = 20.2752 kN Thi load This l d iis assumed d as uniformly if l di distributed ib d lload d per meter run = Total Dead load per meter run = 17.664 kN/m Assuming, cross girder as rigid, reaction on main girder =
8 448 kN/m 8.448 kN/
13.52 kN
Live Load: IRC Class AA Tracked Vehicle Maximum bending moment occurs when one wheel of a vehicle lies near center of span. Position for maximum bending moment is shown in figure. Deck Slab is assumed to p y supported. pp pp be simply The critical supported between two cross ggirder.
W
Effective load coming on cross girder =
l − 1 .8 / 2 = l
W1
569
kN
W1
Fig. 19 Reaction on each longitudinal girder Maximum B.M. occurs under the load, Bending moment including impact Dead Load Bending Moment at the section, Total bending moment =
=
189.58 kN
=
260.68 kNm 286.74 kNm 1.8876 kNm
=
=
288.6 kNm
LL Shear force including impact = Total shear force =
208.5 kN 222.1 kN
Therefore, Design Moment = Design Shear Force =
288.6 kNm 222.1 kN
C Cross girder i d is i designed d i d as T-Beam. TB Assuming effective depth, d =
1450.00 mm
M
Area of tension steel required =
σ st × j × d
=
1113.37
mm2
Minimum area of tension steel in beam @ 0.2 % of web area = 900 3 nos. of φ
Hence provide bars having area of steel =
1472
25
mm2 mm bars +
< 0
1113.4 mm2 20
nos.of φ
mm2 .
Page 29
D i for Design f shear: h Nominal shear stress, τv =
V B × d
=
2 2 0.51 N/mm < τmax = 1.9 N/mm O.K.
Provided percentage area of tension steel, p = 0.33 % 2 Permissible shear stress, τc = 0.275 N/mm2 < 0.51 N/mm O.K. Shear reinforcement is required. Shear reinforcement shall be provided to carry a shear force of Vus = Vu - tc . bw . D = 139433 N Assuming φ 10 mm 2-legged vertical stirrups having area of steel, Asv = Spacing, S =
Asv × σ st × d = Vus
As per minimum shear reinforcement requirements, maximum spa vertical stirrups,
Hence provide φ end cross girder and of intermediate cross girder.
327 mm c/c 10
0.87fy × Asv = 0.4 × b w 10 10
157 mm2
mm 2-legged vertical stirrups @ mm 2-legged vertical stirrups @
mm 2 legged
472 mm 150 150
mm c/c through out the length of mm c/c through out the length
Elastomeric Bearing On Bridge Used According IRC 83-part II it is reccomended use of elastomeric bearins of size (250X400X50 ) mm embedding 5 plates of 3mm thickness and 6 mm clearance in plan G= 1 kN/mm2 For Vertical Load 793 50 kN 793.50
Page 30
