ANALYSIS AND DESIGN OF T-BEAM BRIDGE SUPER STRUCTURE USING LIMIT STATE METHOD

BY:
ABHISEK PANDA

3.6 SUPER STRUCTURE DESIGN:
3.6.1 AVAILABLE DATA:
Effective span of T-beam bridge: 18.33m
Total length of bridge: 55m
Clear width of carriage way 7.5m (IRC 5: 1998 & IRC 6: 2014)
Kerb width: 600mm (both side, pedestrians are allowed)
Parapet: 1000 mm 150 mm 150 mm @ 1.5m c/c with 3-cast iron pipes as railing
Camber: 1 in 100 (37.5 mm at center linearly varying to zero at kerbs)
Wearing coat: 80mm
Kerb height above pavement: 200mm (insurmountable type)
Kerb type: full safety ensured
Total kerb height above deck slab: 280 mm
Clear depth of Longitudinal girders: 1400 mm
Width of longitudinal girder: 400mm
Width of cross girder: 300mm
Clear depth of cross girder: 1400 mm
Deck slab thickness: 250 mm
Cantilever slab: 400 mm at girder face; linearly varying @ 0.11 to ends i.e. 200 mm
Total overall depth of the super structure: 1800mm
c/c spacing of longitudinal girders: 2500 mm
c/c spacing of cross girders: 4507.50 mm
clear distance of cantilever span from face of girder: 1800mm
Grade of concrete: M35
Design strength: fcd = 0.67fck/mMPa (Annex – A2 of IRC 112: 2011)
Grade of steel : Fe415 (IS 1786 : 2000)
Design strength of steel : fy/1.15 = 0.87fy MPa (clause-15.2.3.3 of IRC 112:2011)
Poisson's ratio: = 0.2 (Annex-B; B-3-1 of IRC 112:2011 )
Analysis of deck slab: Piegaud's curve
3.6.2 DESIGN OF INTERIOR SLAB PANEL:
The slab is supported on four sides by longitudinal and cross girders which c/c spacing is as shown in the figure above. The slab thickness is 250 mm and breadth and length respectively are 2.5m and 4.5075 m. the effective span will be taken as the clear span of the slabs since these are the continuous slabs. So effective width Beff = 2.1m and effective length Leff = 4.2075m.



Fig-7 (Interior Slab Panel with one wheel of Tracked Vehicle)
3.6.2.1 Bending Moment of the Slab Panel due to Dead Load:
Weight of slab = 0.25 1 1 25 = 6.25 kN/m2
Weight of wearing coat = 0.080 22 1 1 = 1.76 kN/m2
Weight of camber = 0.45 kN/m2
Total dead weight = 8.46 kN/m2
As Pieguad's curve is used in design,
hence K = B/L = 2.5/4.5075 = 0.55
Also dead load moment is to be computed and hence u/B = 1.0 and v/L = 1.0
Using Pieguad's curve for K=0.5, u/B = 1.0 and v/L=1.0,
m1=0.047 and m2=0.004
Also for K=0.6, u/B = 1.0 and v/L = 1.0,
m1=0.047 and m2 = 0.016
After interpolation, we find that m1=0.047 and m2=0.0145
Total dead load on the slab is given by = 8.46 2.54.5075 = 95.33
Short span moment is MB = 95.33 (0.047+0.200.0145) = 4.76 kN-m
Long span moment is ML = 95.33 (0.0145+0.200.047) = 2.28 kN-m
Since these are continuous slabs, a continuity factor of 0.8 will be multiplied. So final end moment is given by MBF = 3.81 kN-m and MBL = 1.824 kN-m
3.6.2.2 Bending Moment due to Live Load:
IRC class AA tracked Vehicle:
The wheel is placed at the center of the slab as shown in the figure previously. For maximum bending moment to be occurred, the load dispersion is taken to be occurring through wearing coat only. The load is disperses at 450 through the wearing coat as per Annex-B of IRC 112:2011.
Hence u = 0.85+20.080 = 1.01m and v = 3.60 + 20.080 = 3.76m
Hence u/B = 0.404 , v/L = 0.8342 and K = B/L = 2.5/4.5075 = 0.55.
Referring to Pieguad's curve,
For K = 0.5, and for above u/B and v/L values,
m1 = 0.084 , m2 = 0.009
Similarly for K = 0.6, we get
m1 = 0.095 and m2 = 0.03
After interpolating, we get
m1 = 0.089 and m2 = 0.019
Short span bending moment is given by, (taking continuity into account)
MB = 0.8350(0.089+0.20.019) = 25.984 kN-m
ML = 0.8 350(0.019+0.20.089) = 10.304 kN-m
As per IRC 6:2010, clause – 208.3, for tracked vehicles, the impact factor is 10% for spans up to 40m.
So MB = 28.58 kN-m and ML = 11.334 kN-m
IRC class-AA wheeled vehicle:
Following different references and guides on bridge engineering, it's clearly understood that though tracked vehicle gives the severest effect along short span/direction but along long span, the wheeled vehicle gives severest effect. By going through "Essentials of Bridge Engineering: D.J Victor" and "R.C.C design: B.C. Punmia, A.K Jain and A.K Jain" we find the following load positioning gives the severest effect among other combinations.

Fig-8 (Loading of wheeled vehicle for severest effect)
Also as per IRC 6:2014 specifications, no other vehicles can come on to the panel during the above arrangement.
Bending Moment calculation due to Wheel Load -2:
Tyre contact dimension : 300 mm 150 mm
So u = 0.3+20.080 = 0.460 m and v = 0.15+20.080 = 0.310 m
The value of u/B = 0.184 , v/L = 0.069 and B/L = 0.55
Using Pieguad's curve, m1 = 0.22 and m2 = 0.20
Hence MB1 = 62.5 (0.22+0.20.2) = 16.25 kN-m and ML1 = 62.5(0.2+0.220.2) = 15.25 kN-m
Bending Moment due to Wheel Load -1
Since Pieguad's curve is applicable to only symmetrical loading pattern, it cannot be used directly for unsymmetrical loading. In order to calculate the moment, a dummy load having equal magnitude will be arranged at equal distance as per real load from the center of slab or from the vertical axis of the slab. The arrangement is shown in the figure.


Fig-9 (Bending moment calculation due to Wheel Load-1)
So u = 2(u1 + x) = 2(0.46+0.37) = 1.66m and v = 0.31m
Load intensity = 37.5/(.31.46) = 262.973 kN/m2
K = B/L = 0.55, u/B = 0.664 and v/L = 0.069
Hence m1 = 0.12 and m2 = 0.14
MB = (0.12+0.20.14)262.9731.660.31 = 20.03 kN-m
Similarly ML = 22.193 kN-m

Now u = 2x = 20.37 = 0.74 and v = 0.31
K = 0.55, u/B = 0.296 and v/L = 0.069
From Pieguad's curve, m1 = 0.184 and m2 = 0.189
MB' = 13.38 kN-m and ML' = 13.622 kN-m
Hence total moment is given by,
MB2 = (20.03 – 13.38) 0.5 kN-m = 3.325 kN-m
ML2 = (22.193 – 13.622) 0.5 kN-m = 4.285 kN-m
Bending Moment due to Wheel Load-3:
Now in this case u = 2(0.46+0.77) = 2.46m , v = 0.31 m
Load intensity is 62.5/(.31.46) = 438.3 kN/m2
K = 0.55, u/B = 0984 and v/L = 0.069
From Pieguad's curve, m1 = 0.09 and m2 = 0.09
MB = 36.10 kN-m and ML = 36.10 kN-m
Taking u = 2x = 20.77 = 1.54 and v = 0.31,
K = 0.55, u/B = 0.616 and v/L = 0.069
Hence m1 = 0.125 and m2 = 0.14
MB' = 32.01 kN-m and ML'= 34.52 kN-m
MB3 = [36.10 - 32.01] 0.5 = 2.045 kN-m
ML3 = [36.10 - 34.52] 0.5 = 0.79 kN-m
Bending Moment due to Wheel Load at - 5:
u =0.46m, v = 2(v1+x) = 2(0.31 + 1.045) = 2.71m
Load intensity = 62.5/(0.460.31) = 438.3 kN/m2
K = 0.55, u/B = 0.184, v/L = 0.60
So m1 = 0.125 and m2 = 0.0306
MB = 71.57 kN-m and ML = 30.35 kN-m
u = 0.46m and v = 2x = 21.045 = 2.09m
K = 0.55, u/B = 0.184 and v/L = 0.464
From Pieguad's curve, m1 = 0.1441 and m2 = 0.043
MB' = 64.30 kN-m and ML' = 30.25 kN-m
So MB = 3.635 kN-m and ML = 0.05 kN-m
Bending Moment due to wheel Load at – 4:
u = 2(u1+x) = 2(0.46+0.37) = 1.66m and v = 2(v1+y) = 2(0.31+1.045) = 2.71m
K = 0.55, u/B = 0.664 and v/L = 0.601
Coefficient m1 = 0.09205 and m2 = 0.027
After multiplying with (u1+x)(v1+y) with both coefficients,
m1' = 0.1015 and m2' = 0.0304
u = 2x = 20.37 = 0.74m, v = 2y = 21.045 = 2.09m
K = 0.588, u/B = 0.296 and v/L = 0.464
So m1 = 0.134 and m2 = 0.0405
Multiplying with xy,
we get m1'=0.052 and m2' = 0.016
u = 0.74m, v = 2.71m
K = 0.55, u/B = 0.296 and v/L = 0.601
m1 = 0.12 and m2 = 0.03 and multiplying with x(v1+y) = 0.5013,
m1' = 0.060 and m2'=0.015
u =1.66 m and v = 2.09m
K = 0.55, u/B = 0.664 and v/L = 0.464
m1 = 0.095 and m2 = 0.038
Multiplying with y(u1+x) = 0.87,
m1'=0.083 and m2' =0.033
So m1 = [(0.1015+0.052)-(0.060+0.083)] = 0.0105
m2 = [(0.0304+0.016)-(0.015+0.033)] 0
MB4 = 37.50.460.31[0.0105+0.20] = 2.761 kN-m
ML4 = 37.50.460.31[0.20.0105+0] = 0.552 kN-m
Bending Moment due to Wheel Load at -6:
1 u = 2[u1+x] = 2[0.46+0.77] = 2.46 m and v = 2[v1+x] = 2[0.31+1.045] = 2.71 m
K = 0.55, u/B = 0.984 and v/L = 0.601
m1 = 0.068, m2 = 0.02
[u1+x][v1+y] = 1.231.355 = 1.67
m1'=0.113 and m2' = 0.0334
2 u = 2x =1.54m and v = 2y = 2.09m
K=0.55, u/B = 0.616 and v/L = 0.464
m1 = 0.097, m2 = 0.038
xy = 0.804
Hence m1' = 0.078 and m2' = 0.0305
3 u = 2[u1+x] = 2.46m and v = 2y =2.09m
K = 0.55, u/B = 0984 and v/L = 0.464
m1 =0.074 and m2 = 0.03
y(u1+x) = 1.285
m1' = 0.095 and m2'=0.038
4 u = 2x = 1.54m and v = 2(v1+y) = 2.71m
K = 0.55, u/B = 0.616 and v/L = 0.6012
m1 = 0.092 and m2 = 0.025
Multiplying x(v1+y) = 1.043 with above coefficients,
m1' = 0.096 and m2' = 0.026
So final coefficients are,
m1 = [(0.113+0.078)-(0.095+0.096)] = 0
m2 = [(0.0334+0.0305)-(0.038+0.026)] 0
The resulting moment is given by,
MB6 = ML6 = 0 kN-m
Total bending moment is given by,
MB= 16.25+3.325+2.045+3.635+2.761 = 28.016 kN-m
ML = 15.25+4.285+0.79+0.05+0.552 = 20.93 kN-m
Applying continuity and impact,
MB = 28.016 0.8 1.18 = 26.45 kN-m
ML = 20.93 0.8 1.18 = 19.76 kN-m
The impact factor is taken as 18% as per clause-208.4 of IRC 6: 2014. It can be seen that the moment along short span for tracked load is greater while the wheel load bending moment along the longer span is severer. Hence the moment for tracked load will be taken along shorter direction and moment along longer direction will be considered from wheel load in the design of deck slab.
3.6.3 WIND LOAD ANALYSIS:
Since the structure is open in longitudinal cross section and it has all possibilities that it will be drowned inside water up to HFL level during storm; as the storm occurs in the tropical region especially in India during rainy season most of the times. Hence there is no need to consider the lateral wind force. But still considering worst condition, let's calculate the uplift force. IS 875 (part-III) : 1987 will be used for this purpose.
As per clause-5.3, IS 875 (part - III):1987 says that
design wind speed is given by , Vz = Vbk1k2k3
Vb = 50 m/sec (Appendix -A) = 180 kmph (for Bhubaneswar zone)
As per clause – 5.3, the wind speed is considered constant up to the height of 10m of any structure.
k1 = 1.08 (Table-1), k2 = 1.00 (Table-2, Category -2, class - A) and k3 = 1 (ɵ = 00, clause – 5.3.3.1)
Hence Vz = 180 1.08 1.00 1.00 = 194.40 kmph 200kmph = 55.56 m/sec
Using clause- 5.4,
pz = 0.6Vz2 = 0.6 (55.56)2 = 1852.15 N/m2 = 1.85 kN/m2
The solidity ratio is equal to 00 in our case. Hence ф = 00
From table-7 (page -19), taking α = 00andф = 00 ,
C1 = (-)1.4 = 1.4 (suction)
As per clause – 6.2.2.7 (page - 27) for overhangs sloping upward, C2 = 0.75 and the positive sign indicates that this will be acting downward. Hence total pressure force per unit run on the slab will be calculated separately for the cantilever and intermediate road way section.



Fig-10 (Wind load on Super Structure)
The pressure distribution is as shown in the figure.
Fuplift/m run = C1pzA = 1.4 .85 4.2 18.33/18.33 = 10.878 kN/m
Fdownward/m run = C2pzA = 0.75 1.85 3.6 18.33/18.33 = 4.995 kN/m 5kN/m
Hence it can be seen that the cantilever section will not be experiencing any uplift force and the wind pressure will be in the same direction as that of the live load.
Also it can be noted that as per clause – 212.5 of IRC 6: 2014, no live load will be considered on the deck when wind speed exceeds 130 kmph.
Now the longitudinal girders can be seen as loaded as below figure for intermediate spans.

Fig-11 (Wind load acting Longitudinally on Super Structure)
Total upward force acting on one slab panel is given by,
10.878 4.2075 = 45.77 kN << 95.33 0.8 = 76.264 kN (dead load of slab panel). Hence it can be concluded that since uplift pressure is lesser than the dead weight of slab panel itself, no danger of negative suction and hence wind pressure need not be considered. Also as the elevated area of the slab panel is quite less, there is no danger of lateral pressure. Hence this can be omitted.
3.6.4 SHEAR FORCE CALCULATION FOR INTERIOR SLAB PANEL:
3.6.4.1 Wheel Load Shear:

Following IRC-112:2011 (B-3.3, Annex- B-3), the dispersion of load through wearing coat & slab will be at 45°.
Hence dispersion of load is given by 0.85+2(0.08+0.25) =1.51m.
For maximum shear to occur , the load dispersion should be within face of girder.
So wheel load will be kept at least 1.51/2=0.755m from the longitudinal girder face; as shown in the figure.


Fig-12 (Load dispersion through deck slab)

Referring IRC-112:2011,clause-B 3.2(page-278),
bef =αa(1-a/l0)+b1
b/l0 =4.20752.1 = 2.003 > 2.00
α =2.6
l0=2.1
a=0.755m
b1=3.6+2× 0.08=3.76m
be=2.6×0.755(1-0.755/2.1)+3.76
=5.017m
Load per meter width is
= 3505.017 = 69.763

Maximum shear force = 69.763×(2.1-0.755)2.1
=44.681 kN/m
Shear force with impact due to tracked vehicle = 49.15 kN/m
3.6.4.2 Dead load shear:

Total dead weight= 8.46 KN/m2
Total dead load shear=8.46×2.12 = 8.883 kN/m
3.6.5 DESIGN BENDING MOMENT AND SHEAR FORCE:

Toal MB = 3.81+28.58=32.39 KN-m
Ml = 1.824+19.76=21.584 KN-m
Total shear force=(8.883+49.15) KN/m
=58.03 KN/m
Since we have considered limit state method the above load values will be 1.5 times that of calculated as per Anex A2 of IRC: 112-2011
So MB = 48.585 50 kN-m
Ml = 32.376 33 kN-m
Vu = 87.045 kN/m

Minimum effective depth as per maximum bending moment is given by
d =50×106(0.36×35×0.48×103×1-0.416×0.48)
= 101.635 mm
As per table – 14.2, clause –14.3.2.1 (page 142) of IRC–112: 2011,
Clear cover =40+8=48 mm
So, d = 250-48 =202 mm > 101.63 mm (ok)
Using 16 mm- bars,cover for short span = 210-8 =202mm and cover for long span =202-16=186 mm.

3.6.6 REINFORCEMENT DETAILS FOR INTEROIR SLAB PANEL:

(i)Area of steel along short direction

(Ast)B =0.5×35415× 1- 1-4.6×50×10635×103×2022×103×202
As per clause -16.6.1.1 of IRC:112-2011
(Ast)min = 0.26×fctmfyk×btd
From table -6.5 (Page-38)of IRC :112-2011
Fctm=2.8 N/mm2
Fyk= 415 N/mm2
Bt=1000 mm
D = 202 mm
(Ast)min=354.351 mm2
Also same clause specified
(Ast)min=0.0013btd
= 262.6mm2
(Ast)max=0.025Ac
=0.025×250×1000=6250mm2
Spacing of bars as per clause-16.6.1.1(4) of IRC-2011 is,
smax< 2h = 2×250 = 500 mm
In our case,spacing of 16 mm bars is
S=1000×π/4×162716.008
=280.811mm> 250 mm
Also as per clause-15.2.1 (2) of IRC:112-2011,
The clear distance between the parallel main reinforcing bars should not be less than dg+10=20+10=30 mm and 20 mm , whichever is greater
Dg= aggregate size = assumed 20mm for slabs, providing 16 mm- bars @ 225mm c/c
((Ast)B)provided = 1000×π/4×162225
=893.61 mm2
Also as per clause -12.2.2, page – 120 of IRC:112-2011, under rare combination of loads,the maximum tensile stress limits to 0.8 fy is to avoid inelastic strain, undesirable cracking/deformation of structure and also to account for long term creep.
We have calculated the steel area taking fyd=0.87fyk ( as per cl-15.2.3.3 IRC:112-2011)
Considering the worst case i.e rare combination of loads, additional steel area required =1-0.80.87 = 0.080 = 8% of required steel area.
So 8% of (Ast)required =8100×716.005 = 57.28 mm2
Ast to be provided=716.005+57.28=773.28 mm2< 893.61 mm2 (ok)
(Ast)Bprov. =893.61 mm2

(ii) Area of steel along long direction
Providing 12mm- bars,
(Ast)L = 0.5×35415×1-1-4.6×33×10635×103×1902×103×190 = 480.53 mm2
(Ast)min=0.26×fctmfyk×bf×d ( cl-16.6.1,IRC:112-2011)
Fctm= 2.8 N/m2 (table-6.5 of IRC:112-2011)
(Ast)min =0.26×2.8415×1000×190 =333.301 mm2
Also as per same clause ,
(Ast)min =0.0013btd=247 mm2
(Ast)max= 0.025Ac=0.025×250×1000 =6250 mm2
Spacing of bars as per clause-16.6.1.1(4 of IRC:112-2011 is,
Smax< 2h=2×250= 500 mm or 250mm ( smaller value is taken)
S =1000×π/4×122480.53 =235.36mm
Hence providing spacing of bars @175 mm c/c & using 12mm bars
((Ast)l)provided =1000×π/4×122175 =646.30 mm2
Also as per clause-15.2.1(2) of IRC:112-2011, the clear distance between the parallel main reinforcing bars should not be less than dg+10=20+10=30 mm or 20 mm (larger value is taken).
Also asper clause-12.2.2,P-120 of loads ,the max tensile stress in steel is limited to 0.8fyk to avoid inelastic stain ,undesirable cracking/deformation of structure & also to account for long term creep.
Hence more area required is = (1-0.80.87)(Ast)required
=0.08×480.53=38.44 mm2
(Astl)required for creep cracking =480.53+38.44 =518.97 mm2< 646.30 mm2
(Ast)L.prov=646.30 mm2
For detailed reinforcement provision, please refer to Appendix B attached with this thesis.

3.6.7 SHEAR CHECK IN INTERIOR DECK SLAB PANEL:
Vu87.045 KN/m = VEd.
Vrdc =0.12k(80ρ1fck)0.33+0.15σcpbwd (clause- 10.3.2, IRC:112-2011)
VRdc =(Vmin+0.15σcp)bwd (minimum)
K=1+200d = 1+200250=1.894 < 2.0 (ok)
Vmin= 0.031K3/2fck1/2 =0.031×1.9953/2×350.5 =0.478 and σcp=0
ρ1 =Astbwd 0.02
Asl =( Ast)B /2 =893.61/2=446.80 mm2
bw =1000, d = 202
ρ1= 2.21×10-3=0.00221 < 0.02 (ok)
VRdc=0.12×1.894×(80×0.00221×35)0.33×1000×250 =103.684 KN/m
(VRdc)min=119.50 KN/m
So minimum shear resistance isVRdc= 119.5 KN/m VEd = 87.045 KN/m
No shear reinforcement is necessary in slabs.
3.6.8 DESIGN OF CANTILEVER SLAB:

Fig-13 (Cantilever portion of Slab)
3.6.8.1 Dead load Moment:
Hand rail (lump sum)=2 KN
Lever arm = 1.725m
Mla = 2×1.725 =3.45 kN-m
R.C.C post = 0.15×0.15×1×25 =0.562 kN
Lever arm = 1.725m
MRP= 0.97 KN-m
Kerb = 075×0.28×1×25 =5.25 kN
Lever arm=1.425 m
Mkerb=7.48 KN-m
R.C.C slab =0.3×1.8×25=13.5 KN
Lever arm=0.9 m
MRS=12.15 KN-m
Wearing coat = 0.080×1.05×22=1.85 KN
Lever arm=0.525m
Mwc=25.02 KN-m
Total dead load moment
Mdc=25.02 KN-m
3.6.8.2 Live load on Kerb:
L.L =400 kg/m2=4 KN/m2
Lateral load due to live load is =750 kg/m=7.5 KN/m
L.L=4×0.6=2.4 KN/m
Lateral L.L =7.5 KN/m
L.L moment =2.4×1.35= 3.24 KN-m
Lateral L.L moment=7.5×0.58 = 4.35 KN-m (cl- 209 of IRC:6-2000)
3.6.8.3 Moment due to wheel load:
As per IRC-6:2010, only IRC-class A & IRC-class-B
Vehicles can come to the cantilever portion,since it can have a minimum distance 150 mm from kerb .


Fig-14 (Wheel load on cantilever slab)
Using IRC-112:2011,
Annexure:B-3(2),Page-279,
effective depth (Bef)= 1.2a+b1
a=0.65 m
b1=0.25+2×0.080 =0.41 m
Bef=1.2×0.65+0.41 =1.19 m
Live load per meter width including impact is
=(0.57×1.5)/1.19=71.85 kN-m
Wheel load moment(Mwc)= 71.85×0.65
= 46.70 KN-m
drequired =119×1060.36×35×0.48×103×(1-0.416×0.48)
=156.80 mm
Providing 40 mm clear cover &16 mm bars,
Effective depth provided is
dprovided =400-(40+8)=352 mm 156.80 (ok)

3.6.9 REINFORCEMENT IN CANTILEVER SLAB:
Main reinforcement is given by
(Ast)main = 0.5×354151-1-4.6×119×10635×103×3522×103×352
=968.4056 mm2
Spacing of 16 mm- bars is given by
S =1000×π4×162 =207.6 mm
Providing 16 mm- bars @ 190 mm c/c,
(Ast)main = 1058.22 mm2
Distribution moment is given by
Mdc = 1.5×0.3×56.043+0.2×25.02
=31.941 kN-m
Distribution reinforcement is
(Ast)dc = 349.2 mm2; Providing 12mm- bars @ 175 mm c/c
[(Ast)dc]provided =646.27 mm2> 349.2 mm2 (ok)
Detailing of reinforcement is done in Appendix-B
3.6.10 CHECK FOR SHEAR IN CANTILEVER PORTION:
Total shear=dead load shear + liveload shear
= 23.162+1.5×5.71.19+2.4=97.411
(where 1.19= bef (Annex-B3,IRC:112-2011))
Design shear = 1.5×97.41= 146.115 kN=VEd
As per clause-10.3.2(2) of IRC:112-2011,
Shear resistance of a structure is given by
VRdc =[0.12k(80ρ1fck) 1/3+(0.15σcp)]bwd
Subject to min VRdc=(Vmin+0.15σcp)bwd
K=1+200d =1+200400 = 1.7< 2.0 (ok)
Vmin=0.031k3/2fck1/2
=0.031×1.713/2×351/2 =0.41
ρ1 =Aslbwd =1058.221000×400 = 2.645×10-3
VRdc=[0.12×1.71×80×2.645×10-3×351/3]×103×400 =158.93 kN
(VRdc)min=0.41×1000×400 =164 kN
VRdc =164 kN >VEd =146.15 kN (ok)
Also IRC:112-2011,cl-10.3.2(5) specified the following criteria.
Ved 0.5bwdvfcd
V=0.6[1-fck310]
=0.6[1-35310]=0.532
So 0.5bwdvfcd= 0.5×1000×400×0.532×0.36×35
=1340.64 KN > VEd (ok)
The live load applied is 0.65 m from the edge of support. The same clause specifies that is the applied load is at av i.e 0.5d to 2d (200mm to 800mm ),then there will be reduction factor multiplied to Ved.So in our case the dead load shear will be as it's i.e 23.162 KN
But live load shear will be (1.5×571.19) ×β
β = reduction factor=av/2d
av= 650 mm
β =6502×400=0.8125
Vls=1.5×571.19×0.8125
=58.377KN
Total shear=84 kN
Ved=126 KN
VRdc=164 KN >126 kN (ok)
It may be noted that the downward wind force will be only 5×1.8=9 KN against live load 86.22 KN. Hence, there is no need of combination of loads taking wind effect in to account.

3.6.11 DEFLECTION CHECK FOR CANTILEVER SLAB:
The deflection will be checked as per is 456:2000.since we have different loading at different positions,we have to consider them separate

3.6.11.1 First Trail (Annex-c of is 456:2000):

Short term deflection
(Igr)end=b×d312=1000×200312=6.67×108mm4
(Igr)mid =22.5×108mm4,Igrat 0.65m=29.35×108mm4
Fcr =0.7fck=4.141 N/mm2
(Mr)end=4.141×6.67×108100=27.6kN-m
(Mr)mid =4.141×22.5×108150=62.115kN-m
(Mr)0.65m=74.158 kN-m
Ec =5000fck=2.958×104 N/mm2
Es =2×105Nmm2,m=EsEc=6.76
Transformed area of compression steel =(m-1)×Asc =3427.93mm2
Transformed area of tension steel=m×Ast =7152.08mm2
Let "x" be the depth of neutral axis
At end
Or 1000×x×x2+3427.93×x-46=7152.08×(152-x)
Or 500x2+10580.01x-1459363.34=0
Or x=44.47mm
At mid, 500x2+10580.01x-1960008.94=0
X=52.92mm
At 0.65m, 500x2+10580.01x-2158693.7=0
X=55.97mm

Fig-15 (Area of Reinforcement in different section of cantilever slab)
(Ir)end=13×1000×44.473+3427.93×44.47-462+7152.08×152-44.472
=1.117×108mm4
(Ir)mid =3.1264×108mm4
(Ir)at 0.65m =5.58×108mm4
z= lever arm iend=0.9d=136mm,zmid=226mm,zat 0.65m=251mm
(M)end=kerb+port&railing+l.due to pedestrian=7.48+3.45+0.97+4.35+3.24=19.49kn-m
(Mmid)=w.c+R.C slab=13.12kn-m
M0.65m=wheel laod=46.70kn-m
(Ieff)end=(ir)end1.2-(mr)end(m)end×zenddend×1-xenddend(bwb)=1.117×1081.2-27.619.49×136152(1-44.47152)=3.67×108mm4
(Ir)end<(Ieff)end<(Igr)end(ok)
(Ieff)mid=-1.455×108>(Ir)mid
So (Ieff)mid=(Ir)mid=3.1264×108mm4
(Ieff )at 0.5 m=88.84×108mm4>(Igr)at 0.65m
So (Ieff )at 0.65 m=29.35×108mm4
(δ1)end=wl33EcIr=10212×180033×2.958×104×(3.67×108)=1.83mm
(δ1)mid =15350×180038×2.958×104×(3.1264×108)=1.21mm
(δ1)At 0.65m =1EcIr1.15×w×l22+wl33=0.64mm
Total deflection due to short tern loading =3.68mm
Deflection due to shrinkage
acs=k3csl2
k3=0.5(for cantilever)
pt=0.7% ,pc=0.4%
k4=0.72pt-pcpt=0.26<1.0
0.25pt-pc=0.3<1.0 (ok)
(cs )end=3.9×10-7
acs=0.63mm
Note-:here at other points,the 'cs' will give much lesser value relating least shrinkage deflection.
Deflection due to creep
Ece =Ec1+θ
θ=1.6(for 28-days strength)
Ece =1.1377×104N/mm2
m =Es/Ec = 17.58
Transformed area for compression steel (m-1)Asc = 9864.7 mm2
Transformed area for tension steel mAst = 18598.93 mm2
Let x-be the depth of neutral axis.
At End:
1000xx2 + 9864.7(x-46) = 18598.93(152-x)
Or 500x2+28463.63x – 3280813.56 = 0
Or x = 57.40 mm
At mid:
500x2 + 28463.63x – 5140706.56 = 0
Or x = 76.853mm
At 0.65 m:
500x2 + 28463.63x – 5657384.835 = 0
Or x = 81.65 mm

Same lever arm as used before will be used.
So = 2.30108 mm4
Similarly Ir.mid = 7.3108 mm4 and Ir at 0.65m = 9.22108 mm4
So
Ir.end< Ieff.end< Igr.end (OK)

Similarly (Ieff)mid=-4.17×108mm4<(Ir)mid
(Ieff)mid =7.3×108mm4
(Ieff)at 0.65m =29.35×108mm4
But in this case ,for calculating perm. Creep using above equations & Ece,only permanent load will be taken care of .no live load will be there
So (W)end=kerb +R.C pol&railing=5.25+2.562=7.812 kN/m
(W)mid =W.C+R.C slab=1.85+13.5=15.35 kn/m
(ai.cc)perm=(Wend)×l33Ece×Ieff+(Wmid)×l38Ece×Ieff
=(2.39+1.35)mm=3.74mm
Short term deflection due to permanent load
ai.perm =7.812×103×180033×2.958×104×3.67×108+15.35×103×180038×2.958×104×3.1264×108=1.4+1.21=2.61mm
So deflection due to creep is given by
3.74-2.61=1.13mm
Total deflection is =3.68+0.63+1.13
=5.44mm
However this deflection will be lesser in practical as more accurate calculations will reveal the result
As per clause 12.4.1 of IRC:112-2011,the deflection should be limited to =cantilever span375
=1800375=4.8mm<5.44mm
Now let's check it as per cl-23.2 of is:456-2000
fs=220.27, pt=0.7%
Modification factor=1.2(fig-4,p-38)
pc=0.4%
Modification factor=1.12
Basic spandepth=7(for cantilever)
Modified spandepth=71.121.2=9.408
Our spandepth= 6<9.408(ok)
But revising the section as providing 16 ф bars 225mm c/c & 12mm ф bars@225mm alternatively&rechecking,the defection criteria is satisfied. Hence total reinforcement provided is given by 1396 mm2.
3.6.11.2 Second Trail- Rechecking of deflection for Cantilever Slab(Annex c of IS-456:2000):
Short term deflection
(Igr)end=b×d312=1000×200312=6.67×108mm4
(Igr)mid =22.5×108mm4,Igr.65m=29.35×108mm4
Fcr =0.7fck=4.141n/mm2
(Mr)end=4.141×6.67×108100=27.6kN-m
(Mr)mid =4.141×22.5×108150=62.115kN-m
(Mr)0.65m=74.158 kn-m

Ec =5000fck=2.958×104N/mm2
Es =2×105nmm2,m=EsEc=6.76
Transformed area of compression steel =(m-1)×Asc =3720.96mm2
Transformed area of tension steel=m×Ast =9436.96mm2
Let "x" be the depth of neutral axis
At end
=1000×x×x2+3720.96×x-46=9436.96×(152-x)
=500x2+13157.92x-1605582.08=0
=x=45.02mm
At mid
=500x2+13157.92x-2549278.08=0
X=59.45mm
At 0.65m
=500x2+13157.92x-2811436.83=0
X=62.97mm
(Ir)end=13×1000×45.023+3720.96×45.02-462+9436.96×152-45.022
=1.38×108mm4
(Ir)mid =4.2×108mm4
(Ir)at 0.65m =5.27×108mm4
z= lever arm. zend=0.9d=136mm,zmid=226mm,zat 0.65m=251mm
(M)end=kerb+port&railing+l.due to pedestrian=7.48+3.45+0.97+4.35+3.24=19.49KN-m
(Mmid)=w.c+R.C slab=13.12KN-m
M0.65m=wheel laod=46.70KN-m
(Ieff)end=(ir)end1.2-(mr)end(m)end×zenddend×1-xenddend=4.5×108mm4
(Ir)end<(Ieff)end<(Igr)end(ok)
(Ieff)mid=-2.05×108mm4;should be greater than (Ir)mid=4.2×108mm4
(Ieff )at 0.65 m=54.44×108mm4<(Igr)at 0.65m=29.35×108mm4
(δ1)end=wl33EcIr=1.5mm, (δ1)mid =0.9mm, (δ1)At 0.65m =1EcIr1.15×w×l22+wl33=0.64mm
Total deflection due to short tern loading =3.04mm
Deflection due to shrinkage
acs=k3csl2
k3=0.5(for cantilever)
pt=0.92% ,pc=0.425%
k4=0.37<1.0
0.25pt-pc=0.495<1.0 (ok)
(cs )end=5.55×10-7
acs=0.9mm
Deflection due to creep-:
Ece =Ec1+θ
θ=1.6(for 28-days strength)
Ece =1.1377×104Nmm2 ,m=EsEce=17.58
Transformed area of compression steel =(m-1)×Asc =10710.68mm2
Transformed area of tension steel=m×Ast =24541.68 mm2
Depth of neutral axis at different section
At end at mid at 0.65m
X=63.2mm x=85.566mm x=91.08mm
(Ir)end=2.80×108mm4, (Ir)mid=9.03×108mm4
(Ir)0.65m=2.80×108mm4
(Ieff)mid=-5.63×108mm4<(Ir)mid
(Ieff)end =6.08×108mm4
(Ieff)at 0.65m =47.80×108mm4; which should not be greater than Igr at 0.65m
(W)end=7.812 kN/m, (W)mid =15.35 kN/m
(ai.cc)perm=(Wend)×l33Ece×Ieff+(Wmid)×l38Ece×Ieff
=(2.2+1.1)mm=3.3mm
Short term deflection due to permanent load
ai.perm =7.812×103×180033×2.958×104×3.67×108+15.35×103×180038×2.958×104×3.1264×108=1.4+1.21=2.61mm
Deflection due to creep = 0.69mm 0.70mm
Hence total deflection = 3.04 + 0.9+0.7 = 4.64 mm < 4.8 mm (ok)
3.6.12 ANCHORAGE LENGTH OF BARS:
3.6.12.1 Intermediate Deck Slab Panel:
In short direction:
main bars = 16 mm @ 225 mm c/c
(lb.net)B=αalb×Ast reqAst provided (page no-152, cl-15.2.4.3, of IRC:112-2011)
Ast reqAst provided =716.005893.61 =0.865
αa =1 (for straight bars)
lb =k
= ×0.87fy4fbd
=16×0.87×4154×30 ( fbd=3.0 N/mm2 table-15.3 of page -150,IRC:112-2011)
=481.4 mm
Also k=30 [ for M35 concrete of table-15.4 of IRC:112-2011]
lb = 30×16 = 480 mm
let's take lb=480mm
(lb.net)B =1×480×0.865
=415.0 mm 420 mm
lb.min =0.3lb =144 mm (for tension anchorage)
lb.min =0.6lb=288 mm ( for compression anchorage)
lb.min=10 =160 mm or 100 mm
lb.net>lb.min (ok)
In long direction:
Ast.reqAst.provided =480.53646.30=0.7430.75
αa =1 (for straight bars)
lb =k
=30×12=360mm
or lb=12×0.87×4154× = 361.05 mm
So lb = 360 mm
lb.net =1×360×0.75 = 270 mm
lb.min =0.3lb=108 mm (tension anchor)
=0.6 lb=216 mm (compression anchor)
lb.min =10 =120 mm or 100 mm
lb.net>lb.min
3.6.12.2 Cantilever slab:
Ast.reqAst.provided =872.561058.22 =0.824 (cl-15.2.4.3 of IRC:112-2011)
αa =1
lb=k =30×12=360 mm
lb.net =0.824×360=296.64 300 mm
3.6.13 CANTILEVER SLAB STIFFENING:

As per clause -16.6.1.4-(1)(a)&(b) of IRC:112-2011,the unsupported edge of a slab parallel to traffic &beyond the clear road width &having varying depth shall be stiffened at any particular point to the resisting moment of 500mm adjacent strip at that particular point.
[(Ast B)0.5m strip]in main slab between girder =93.612=446.805mm2
[(Ast B)0.5m strip]in cantilever section =1396.262=698.132 mm2>446.805mm2(ok)
[(Ast l)0.5m strip]in main slab between girder =646.272=323.135mm2
[(Ast L)0.5m strip]cantilever slab between girder =646.272=323.135mm2
3.6.14 DESIGN OF LONGITUDINAL GIRDER:
Effective span=18.33 m
rid width=400mm
slab thickness=250 mm
Main beam spacing=2.5 m
Depth of rib = 1400 mm
Total depth =1400+250+150=1800mm=D
REACTION FACTOR:
Courbon's method can be applied to our design since span to depth ratio is greater than 2. Arrangement of class-AA tracked load for maximum eccentricity is as given below.


Fig-16 (Arrangement of class-AA tracked load for maximum eccentricity)
As per Courbon's formula ,
Rx= wn[1+Idx2dxe]
Rx=Reaction factor
It may be noted that RA=Rc since the same loading can be reversed in two cases.
RA=2W13[1+3I2×2.52×I×2.5×1.1]
n= no of girders =3
dx=spacing of girders=2.5 m
W1=350 KN
RA=Rc=w3[1+3×1.12×2.5] ( W=700 kN = total axle load )
RB = 2W1/3[1+0]
= 2W1/3 =0.33W
DEAD LOAD:
From cantilever slab:
Weight of parapet railing=2 kN/m
Wearing coat=0.080×1.05× 22=1.848 kN/m
Cantilever slab=0.3×1.8×25 = 13.5 kN/m
Kerb=0.28×0.75×25 = 5.25 kN/m
Parapet = 0.40 kN/m
Total dead load from cantilever slab =23kN/m
Total deck slab load =2×23+8.46×5.4 =91.684 kN/m
Live load due to pedestrian=4× 2=8 kN/m
It is assumed that all loads are equally distributed to 3 girders
So deck slab dead load/girder=99.684/3 =33.228 kN/m
T-beam rib dead load:
0.4×1.4×25 = 15.50 kN/m
Total weight per girder = (33.23+15.5)=48.78 49 kN/m

Cross girder:
Dead weight of one cross girder is=0.3×1.4×25=10.50 kN/m
This load expands up to 2.5 m both sides on intermediate girder
Hence reaction from cross girder = 10.50×2.5=26.25 kN
The end cross beams will not be considered since their loads will be directly transferred to support through piers. Hence end reactions need not be considered.
The presentation is as follows






Fig-17 (Reaction from Cross Girder on Longitudinal Girders)
Maximum dead load bending moment at center of span is,
(Mmax)dead load =49×18.3328+26.25×18.334+26.25×18.334 =2300 kN-m
Total dead load shear at support is
(Vdead)s=49×18.332+26.25×22+26.252=488.46489kN
LIVE LOAD BENDING MOMENT IN GIRDER:
Span =18.33 m, Impact =10%; for maximum bending moment to occur, the live load is placed centrally on the span as shown below.

Fig-18 (Position of Class-AA tracked Load for maximum bending moment)
Live load bending moment is given by 12(3.6825+4.5825)_×700 =2892.75KN-m2893 kN-m
Bending moment with impact & reaction factor is given by
(Ml)outer girder=0.553×1.10×2893=1759.812 KN-m1760 kN-m
(Ml)inner girder=0.333×1.10×2893=1059.71 1060kN-m
LIVE LOAD SHEAR:
For shear force to be maximum,the loading should be arranged as shown below.

Fig-19 (Live load arrangement for maximum shear force )
Wheel load placed at starting will be giving maximum shear i.e. starting point of either end; either 'A' or 'B' . Again one wheel has to be placed on one girder to give severe effect as the distribution load will occur only for one wheel load.
Reaction of 'w2' on girder-B is given by 350×0.452.5=63 kN
Reaction of 'W2' on girder-A is given by 350×2.052.50 =287 kN
Total load on girder –B =350+63 = 413 KN
Maximum reaction on support of girder 'B' is =413×16.5318.33 = 372.44 kN
Maximum live load shear including impact =1.1×372.44=409.68410 kN

TABLE-4(Design Bending Moment and Shear force on Girders)

Bending Moment/Girder
D.L Bending Moment
L.L Bending Moment
Total Bending Moment
Unit
Outer Girder
2300
1760
4060
kN-m
Inner Girder
2300
1060
3360
kN-m
Shear Force/Girder
D.L Shear
L.L Shear
Total Shear
Unit
Outer Girder
489
410
899
kN
Inner Girder
489
410
899
kN








CALCULATION OF BENDING MOMENT USING HENDRY-JAEGAR METHOD:
A =(12π4 )Lh3(nEIrEI)
F =(π22n) (hL)(CJEIr)
c = EI1EI2 =1
L= span of bridge deck = 18.33 m
h = spacing of longitudinal girders = 2.5 m
n = number of cross girders = 5
EI = flexural rigidity of longitudinal girder
CJ = torsional rigidity of longitudinal girder
EI1 & EI2 = flexural rigidities of the outer & inner longitudinal girders
EIr = flexural rigidity of one cross beam .

Fig-20 (Simplified T-beam Longitudinalgirder section)
As per clause – 7.6.1.2 of IRC :112-2011 , the effective flange width will be calculated.
Beff. 1 = 0.2×b1+0.1l0
=0.2×1.05+0.1×18.33=2.043 m
0.2l0=3.66m > 2.043 m (ok)
Bw=0.4 m
Beff=2×2.043+0.4
=4.486 m
B= 2.5 m< 4.488 m
beff=2.5m

Fig-21 (Simplified T-beam Cross girder section)
beff.1=0.2b1+0.1l0
=0.2×2.1037+0.1×0.7×2.5
0.2l0 =0.2×0.7×2.5=0.35>0.596 m
beff.1=0.35 m
beff=2×0.35+0.3=1 m
b =4.5075 m>1m (ok)
Longitudinal girder:
x=A1x1+A2x2A1+A2
=2.5×0.25×0.125+1.55×0.4×1.0252.5×0.25+1.55×0.4
=0.57 m

IL=2.5×0.25312+2.5×0.25×0.4482+0.4×1.55312+0.4×1.35×0.4522 =0.3795 m4
Cross girder:
x=A1X1+A2X2A1+A2
=1×0.25×0.125+0.3×1.55×1.0251×0.25+0.3×1.55 =0.710 m
Ir= 1×0.25312+1×0.25×0.5852+0.3×1.55312+0.3×1.55×0.3152
=0.2261 m4
A = 12π4 ×(18.332.5)3×5×E×0.2261E×0.3795 =144.64
J = Ra3b
ba= 2500250=10
R=0.312 ( Table-7.3 of N. Krishna Raju ,Design of Bridges )
ba= 1550400=3.875
R=0.2787
J = Ra3 b = 0.312×0.253×2.5+0.2787×0.43×1.55 = 0.0398 m4
F = π22×5 ×2.518.33×0.43E×0.0398E×0.2261 =0.0102 0
Two extreme values of 'F' can be taken into analysis i.e. F = 0 or F = .
Taking F = 0 for analysis ,MF= MO+(M - M0)(FA)(3+FA)
MF = MO; MO=0.83 For outer girder,F=0
MO = 0.36 (For inner girder, F = 0)
Design moment for exterior girder :
Dead load moment = 2300 kN-m
Live load moment = 2893×1.1×0.83 = 2641.31 kN-m
Total moment = 4941.00 kN-m >4060 kN-m as calculated using Courbon's method
Interior girder:
Dead load moment = 2300 kN-m
Live load moment = 2893×1.1×0.36 =1145.63 kN-m
Total moment = 3445.63 kN-m 3446 kN-m>3360 kN-m as calculated by Courbon ' s method
MODIFIED COURBON'S METHOD:
Ref : International Journal of scientific & Engineering research Volume 4 , Issue 3 ,March – 2013 (ISSN 2229 -5518)
"Study of Effectiveness of Courbon ' s Theory in the Analysis of T – beam bridges "
By: M.G. Kalyan & hetti &R. P. Shriram
As per journal, they have studied the 4-lane & 6-lane bridge of spans varying 15m-35 for minimum 3-longitudinal girders varying number of girders. They have combined Courbon's method &grillage method (STAAD pro) to get the results.

As per journal,
Pi = × correction factor
Or =
We have calculated
Ra = Rc=ROuter girder =0.553w
Rb=Rinner girder=0.333w
Correction factor is given by,
Y= correction factor,x= span of bridge
X=18.33m

= 0.9214 (rounded up to 4 decimal)
RA=RC= (Router girder) corrected=0.553×0.9214=0.5095
RB=(R inner girder) corrected=0.333×0.9214=0.3068
Total bending moment
Outer girder=2300+2893×1.10×0.5095=3921.40 KN-m
Inner girder=2300+2893×1.10×0.3068=3276.33 KN-m
Out of all 3-analysis, Hendry-Jaegar method gives the highest moment
Design moment-
Outer longitudinal girder=1.5×4771.31=7411.50=7412 KN-m
Intermediate girder=1.5×3276=5169 kN-m
Design shear
Shear force =871×1.5=1348.50 kN

DESIGN OF REINFORCEMENT OF OUTER GIRDER:

Let's assume 32mm ф bars as main reinforcement bars in 4 rows, the clear spacing between bars is 32mm &hence c/c distance is 64mm
Let's take clear cover =40mm( clause-14.3.2,table-14.2,severe condition)
Assuming 10mm dia stirrups& surface reinforcement(as cl-16.5.4)
Effective cover=40+10+12+30+10+15+120=235mm
Hence effective depth= 1800-235=1565mm
Let's take d=1560mm
Df= 250mm
Df /d=0.1602 <0.2
Bf = 2.5m (as calculated earlier)
Xumax =0.48×1600=768mm
Let's check the actual neutral axis depth
Let's take xu =Df =250mm
Mu =0.36× 35× 2500× (1600)2×
=11781kN-m
Imposed moment calculated =7412 kN-m=Ma
Mu >>Ma (hence neutral axis lies inside flange section)
(Ast)required =
=13740.20 mm2
providing 30mm dia bars as main reinforcing bars,
Total bars required =
Let's provide 20 numbers of 30mm-ф bars at 5 rows each containing 4 bars
So (Ast)provided=
The spacing between longitudinal bars is
As per IS 456:2000, clause-26.3.2, the spacing should be minimum of following
Diameter of bars =30mm
5mm more than nominal maximum size of coarse aggregate =20+5=25mm
Our spacing is 35.33 is satisfactory.
For detailing of reinforcement, please refer to Appendix-Battached with this thesis.
DESIGN REINFORCEMENT FOR INNER GIRDER:
d = 1600 mm, Df = 250mm and bf = 2.5m
Assuming neutral axis to be within the flanges, Mu = 1178 kN-m
Applied moment = Mu applied = 5169 kN-m << Mu
So Ast.req = = 9453.60 mm2
Providing 16- 28 mm ф bars in 4- rows, (Ast)provided = 9852.03 mm2
For detailing of reinforcement, please refer to Appendix-B attached with this thesis.
CALCULATION OF ANCHORAGE LENGTH:
Clause -15.2.4.3 of IRC:112-2011
Outer girder
Lb.net=αalb(as) reqd(as)provided lbmin
Let's assume (Ast) reqd(Ast)provided=1 at a section
So αa=1
lb= 4×fydfbd
=304×0.87×4153=902.625 903mm
Or lb=kø=30×30=900mm
lb.net =lb=903mm
lb.min=0.3lb=288.9mm or 100mm or 10ø=300mm
Inner girder
lb=284×0.87×4153=842.45 843mm
lb.net =843mm
Length of bearing = 23lb.net(16.5.1.4. of IRC :112-2011)
=602mm(outer girder)
=562mm(inner girder)

BAR CURTAILMENT:

As we know when the load is divided in the same ratio as the span is divided at a point where B.Moment to be calculated, maximum B.M is obtained. In ILD. Let's consider following cases.


(i) Load at quarter span/ at 4.5825 m from ends of girder

Fig-22.1 (Bar curtailment-Wheel load at quarter span)
The load arrangement is as shown in the figure taking ILD into consideration.
Wheel load moment:
=123.44+2.764×700=2171.40kN-m
Dead load moment:
=488.46×4.5825-49×4.58252×0.5=1724kN-m
As per H-J method, wheel load BM including impact & coefficient
=
Total design moment on outer girder=1.5(1982.5+1724=5560kN-m)
Total design moment on inner girder=1.51724+1.1×2171.40×0.36=3876 kN-m
(ii) Loads at a distance 6.874m from ends of girder

Fig-22.2 (Bar curtailment-Wheel load at 6.874m from end of girder)

Wheel load moment=700×3.45+4.3×0.5=2712.5kN-m
Deadload moment=488.46×6.874-49×0.5×6.8742+26.25×2.2165
=2142kN-m
Total design b.m on outer girder due to impact and H.J coefficient =1.5[2712.5×1.1×0.83+2142=6928kN-m
Total design b.m on inner girders due to impact and H.J coefficient=1.5[2712.5×1.1×0.36+2142=4825kN-m

(iii)Load at a distance 2.29125 m from ends of girder

Fig-22.3 (Bar curtailment-Wheel load at 2.29125m from girder end)

Wheel load moment=700×1.6112+2.005×0.5=1266kN-m
Dead load moment=488.46×2.29125-49×0.5×2.29125=991kN-m
Total design moment on outer girder=1.5991+1.1×0.83×1266=3220.30kN-m
Total design moment on inner girder=1.5991+1.1×0.36×1266=2238.50kN-m

REQUIRED STEEL AREA AS PER BAR CURTAILMENT:

(1)Load at quarter span/ at 4.5825 m from ends of girder
Outer girder:
Providing 16-30mm ф bars ,(Ast.reqd)O.G=
Note-:4-bars are curtailed beyond the section providing development lb.net extension i.e at 3.6795m from ends of girders.
Inner girder:
=(Ast.req)I.G=0.5×35415×1-1-4.6×3876×10635×2500×15602×2500×1560=7035.60mm2
Providing 12-28mmdia bars ,
Note-:4-bars are curtailed beyond the section providing development lb.net extension i.e at 3.7395m from ends of girders
(2)Loads at a distance 6.874m from ends of girder
Outer girder :

Providing 19-30mm dia bars , =(Ast.prov)O.G=13430.31mm2
Note-:1-bars are curtailed at a distance of 5.971m from ends of support
Inner girder:
(Ast.req)I.G=8806.62mm2
Providing 15-28mm dia bars=(Ast.prov)I.G=9236.30mm2
Note-:1-bars are curtailed at a distance of 6.031m from ends of support
(3) Load at a distance 2.29125 m from ends of girder
Outer girder:
(Ast.reqd)O.G=5823.44mm2
Providing 10-30mm dia bars,= (Ast.prov)O.G=7068.60mm2
Note -:10-30mm dia bars are curtailed from 1.388m from ends of support
Inner girder:
(Ast.req)I.G=4025.60mm2
Providing 8mm ф bars , (Ast.prov)I.G=4926.02mm2
Note -:8-28mm ф bars are curtailed from 1.44825m from ends of girder


(OUTER GIRDER)


(INNER GIRDER)

Fig-23 (Bar curtailment Presentation for Longitudinal Girders)

CALCULATION OF INDUCED SHEAR AT DIFFERENT SECTIONS:
Wheel load at starting



Fig-24.1 (Shear force calculation-Wheel load at starting)

Applying impact, total shear =454.3kN
Again reaction at support is =
Dead load shear =
Total shear=899kN 900 kN
Shear at 9.165 from end of girder

Fig-24.2 (Shear force calculation-Wheel load at middle)

Track shear force at middle is =
Dead load shear=
Total shear=196.205 200kN
Shear at 4.5825 from end of girder

Fig-24.3 (Shear force calculation-Wheel load at quarter span)

Track s.f at 4.5825m=120.75+0.5536×413×1.1=296.113kN
Dead load shear=489-49×4.5825=264.46kN
Total shear=561kN
(iv) Shear at 1.95m from end of girder

Fig-24.4 (Shear force calculation-Wheel load at 1.95 m from end)

L.L shear=361.441kN
D.l shear =489-49×1.95=393.45kN
Total shear=755kN



(v) Shear at 3.0m from end of girder

Fig-24.5 (Shear force calculation-Wheel load at 3.0 m from end)

L.L shear=335.34kN
D.L shear=489-(49×3)=342kN
Total shear=678kN
(vi) Shear at 6.87375m from end of girder

Fig-24.6 (Shear force calculation-Wheel load at 6.87375m from end)

L.L shear=239.325kN
D.L shear=489-49×6.87375+26.25=126kN
Total shear=366 kN
(vii) Shear at 4.3825 m from end of girder

Fig-24.7 (Shear force calculation-Wheel load at 4.3825 m from end)

Track shear at 4.3825 m = 0.5 [0.761+0.564]1.10413 = 301 kN
Dead load shear = 489 – (494.3825) = 275 kN
Total shear force = 576 kN

(viii) Assuming bearing =450mm & fan like compression field at an angle 45,total depth up to which shear reinforcement is not necessary is given by,
450+1560=2010=2.01m
Hence shear at 2.01m from end of girder is

Fig-24.8 (Shear force calculation-Wheel load at 2.01m from end)

L.l shear=413×1.1×0.5(0.8903+0.694)=360kN
D.d shear=489-49×2.01=390.51kN
Total shear=750.51KN 751kN
SHEAR RESISTANCE CHECK:
Let's provide 1bent up bar at 4.5825m & at 645mm interval up to 2.00m as shown below
Maximum spacing of bent up bars
[sbmax=0.6d1+cot clause-16.5.28 of IRC:112-2011]
=0.6×1560×2=1.872>0.645(ok)

Now Asw=1×π4×302=706mm2(outer girder)
&1×π4×282=615mm2(inner girder)
As per clause 10.3.3.3 of IRC:112-2011,
Now vrds=asws×z×fywd(cotθ+cotα)sinα
s=0.645m
z=0.9d=0.9×1560=1440mm
fywd=0.8×415=332N/mm2
(VRds)O.G=706645×1404×332×2×12 =721.55kN
VRd.max= cwbwv1fcdz(cotθ+cotα)/(1+cot2θ)
=1×400×0.9×1560×0.6×0.36×35×2/2
=4245.696>>721.55 KN (ok)
(VRds)I.G=615645×1404×332×2×12
=628.543<< 4245.696 kN (ok)
Design shear resistance of member without shear reinforcement is given by
VRdc =[0.12k80ρ1fck0.33+0.15σcp]×bwd(cl-10.3.2 of IRC:112-2011)
σcp=0
vmin=0.031k32fck12
K=1+200d=1+2001560=1.358<2(ok)
vmin=0.29
(VRdc)min=vmin+0.15σcpbwd
=0.29×400×1560=181.105 kN
Now ρ1=Astbwd
Since half reinforcement is always available throughout,
(ρ1)O.G=0.5×14137.167400×1560=0.01133<0.02
(ρ1)I.G =0.5×9852.03400×1560=7.89×10-3<0.02
(VRdc)O.G =0.12×1.35880×0.01133×350.33×400×1560
=318kN>vRdc.min (ok)
(VRdc)I.G=0.12×1.35880×7.89×10-3×350.33×400×1560
=282kN>vRdc.min (ok)

SHEAR REINFORCEMENT DISTRIBUTION ON OUTER GIRDER:

Total shear at face of support= 900kN
However as it's simply supported & a bearing of 450mm is provided,a fan like compression field will exist having steepest angle θ=45 up to which no shear reinforcement will be necessary. so the effective section for shear will be at
0.45+1.560=2.01m from end of girder
Total shear at 2.01m=751kN
Design shear=1.5×751=1126.50 kN
Shear resisted by girder without shear reinforcement = (VRdc)O.G=318 kN
Hence design shear for which shear reinforcement will be provided=VEd
=(1126.50-318)KN=808.50 kN
The shear resisting capacity of available bent up bars at that section is 721.55 kN
But as per clause -10.3.3.3(2)& clause-16.5.2(3)
of IRC :112-2011,only 50% of the shear will be resisted by the bent up bars
Hence shear to be resisted by links/stirrups
By 808.502=404.25kn
[check:(asw.max)bent up bars=0.5×10×0.6×0.36×35×400×64512 ×0.8×415=4154.21mm2>706mm2ok
>615mm2(ok)
Let's provide 4-legged 8mmф vertical stirrups
Asw =4×π4×82=201.062mm2=201mm2
Using clause-10.3.3.2 of IRC:112-2011,
Vrd.max= cwbwv1fcdzfcd(cotθ+tanθ)
=1×400×0.9×1560×0.6×0.36×352
=2122.848kN>>404.25 kN (ok)
As per same clause ,spacing of vertical stirrups given by
S=asw×fywd×z×cotθvrds
=201×0.9×1560×0.8×415×1404.25×103=231.766mm
Let;s provide 4-legged 8mm ф vertical stirrups@ 200mm c/c starting/end of girders.
As per cl-16.5.2 of irc:112-2011,
Min. Shear reinforcement ratio is
ρw.min=0.072×fckfyk=0.072×35415=1.026×10-3
Provided shear reinforcement ratio is
(ρw)=Asws×bw×sin =201200×400=2.125×10-3>( ρw.min)(ok)
Shear force at 4.5825m from ends of girder 561kN. Here no bent up bars are available
Design shear=841.5kN
(Asl)o.g=18×π4×302=12723.45mm2
(ρ1 )o.g=0.02039 > 0.02, (ρ1 )i.g=8620.53400×1560=0.0138 0.02
(vrdc)o.g =0.12×1.358×80×0.02×350.33×400×1560=383.855 383kN
(vrdc)i.g =0.12×1.358×80×0.0138×350.33×400×1560=339.615 339kN
Shear to be resisted by stirrups is
(841.5-383) kN =458.5 kN
VRd.max=2122.848>458.50 kN(ok)
Spacing of stirrups is
S=201×0.9×1560×0.8×415×1458.50×103=204.344mm
Let's provide 4-legged 8mm ф vertical stirrups@190mm c/c up to 6.87375m from girder end on both sides. Shear force at 6.87375m from end of girder is 366kN factored shear=549kN
Shear resistance=383 kN
Total shear to be resisted by stirrups is 549-383=166 kN
Spacing of same 4-legged 8mm vertical stirrups is given by
s=201×0.9×1560×0.8×415×1166×103=564.408mm
As per cl-16.5.2(7) of IRC:112-2011,
Slmax =0.75d(1+cotα)=0.75×1560×1=1170mm>564.408mm(ok)
But As per cl-16.5.3(3,4) of IRC:112-2011, specifies for torsion criteria to be satisfied,the minimum spacing should be 350mm or
18[outer perimeter of the member]=18×3200=400m }(lesser of two)
So provide 4-legged -8mm ф vertical stirrups @360mm c/c in the middle strip band of length 4.5825m
Again it should satisfy (ρw)min
(ρw)min =1.026×10-3
(ρw)prov =201350×400=1.436×10-3>(ρw)min (ok)
SHEAR REINFORCEMENT DISTRIBUTION ON INNER GIRDER:
0.45+1.56=2.01m from girder end
Total design shear=1126.50kN
(VRdc)I.G=282 kN
Design shear for which shear reinforcement will be provided=844.50kn
Bent up bars shear resistance is
(VRds)I.G =628.543 kN
But it will be taking only 844.502=422.25kN
Rest 422.25kn will be taken care of by the vertical stirrups.
Providing 8mm ф -4 legged stirrups,
Asw=201mm2
Cl-10.3.3.2 of IRC:112-2011 specifies
Vrd.max=2122.848kN >> 422.25kN(ok)
Following same clause, spacing of stirrups,
S=201×0.9×1560×0.8×415×1422.25×103=221.88mm
Provide 8mm-4legged stirrups @200mmc/c up to 4.3825m from end of girder on both sides (ρw)min =1.026×10-3

(ρw)prov =201200×400=2.5125×10-3>(ρw)min (ok)

Shear at 4.3825m is given by 576 kN.
Design shear=1.5×576=864kN
No bent up bars are available here.
(VRdc)I.G=339 kN
Shear to be resisted by stirrups=525kN
VR d.max=2122.84kN >>525kN
Spacing of stirrups is given by
S=201×0.9×1560×0.8×415×1302.50×103=178.46mm
Provide 8mm-4legged stirrups @160mmc/c up to 6.87375m
Shear at 6.87375m from end of girder is; Design shear=1.5×360=549kN
(Ast)available=16×π4×282=9852.035mm2
(ρ1 )I.G=9852.035400×1560=0.016<0.02
(VRdc)I.G=356.604 kN
Shear to b resisted by links and/stirrups
(549-356.604)kn=192.396kn 193kN
Providing minimum shear reinforcement spacing of 350mm c/c,
VRds=201×0.9×1560×0.8×415×1350=267.691kN>193kn(ok)
So Provide 8mm-4legged stirrups @350mmc/c in middle strip-band of length 4.5825m
(ρw)min =1.026×10-3

(ρw)prov =201350×400×1=1.436×10-3>(ρw)min (ok)
SURFACE REINFORCEMENT:


Fig-25 (Surface Reinforcement provision)
Surface reinforcement will be provided taking a surface area of the cover portion outside of the stirrups.
Surface area on one face of girder=87×1400=121800mm2
0.01 Act.ext =1218mm2
We have side face reinforcement of approximately 5-12mm dia bars which provide 565.5mm2
Providing 10mm ф bars @150mmc/c,total steel area provided per unit run is
=
Total reinforcement=1274.976>1218mm2(OK)
Surface area on lower end i.e at foot of girder=92×400=36800
=0.01 Act.ext=368mm2
Provision of 4-16mm ф bars will be giving
=
Also stirrups will be giving
=
Total area provided=976.25mm2>368mm2(ok)
All the surface reinforcement bars are perfectly anchored&hence will be serving as shear reinforcement also appear same clause(5) also
Also side face longitudinal bars will be serving as side face reinforcement for stability as per clause-26.5.1.3 of IS-456:2000

CHECK FOR BAR CURTAILMENT ADEQUACY AS PER CURTAILMENT OF GIRDER BARS:
The check will be done as per clause 16.5.1.3 off IRC:-112-2011(page -176)& clause-16.5.1.4 off IRC:112-2011(page -177)
Before checking,it may be noted that the longitudinal bar curtailment rule as per figure-16.2 of IRC:112-2011(page -177) will be taken care of for end sections only since at end of girders,there is maximum shear &minimum reinforcement is acting upon all other sections will be found safer if automatically if end section are found safer
Frs=tensile capacity of reinforcement

n=numbers of bars at ends, d=diameter of bars

=2552.112kN
(Frs)I.G=8×π4×282×0.87×415=2778.54kN
The above Frs should be greater that Fs+

Effective shear will occur at 2.01m from end &its value is 1126.50kN
Fs=0.45×1126.50 507kN
At ends Med 0Medz=0
Ftd=0.5 VEd(cotθ-cotα)
=0.5×1126.50×1=563.25kN
Also another value of Fs appear fig-16.2(A) is
MEdz+NEd=0
Fs+ Ftd =563.25+507=1070.25kn<2552.112 kn=(FRs)o.g
At middle,
Medz=Medmaxz
(Medz)O.G=7412×1060.9×1560=5280kN
(Medz)I.G=5169×1060.9×1560=3682kN
(Fs)O.G=5280kn,(Fs)i.g=3682kN
( VEd=0& also Ftd (FRS)O.G=2552.1122=5104.224kN
(FRS)I.G=1778.542=3557.08kN
As per clause -16.5.4(5) of irc:112-2011,4-16mmф bars which are provided as side face reinforcement will be serving as resisting bars for bending of girder section. So tensile capacity of total 4-16mm ф bars is given by,
So (FRS)O.G=5104.224+290.374=6394.60 kN >(ok)
(FRS)I.G=3847.454kN >3682kN (ok)
3.6.15 DESIGN OF CROSS GIRDER:
Self-weight=0.3×1.4×25=10.5 kN/m
Slab load will be distributed as shown below

Fig-26 (Dead load distribution from slab onCross Girder)
Dead load from slab =2×0.5×2.5×1.25×8.46=26.4375
So udl load =26.43752.5=10.575kN/m
Total load (dead weight) =21.075kN/m
Assuming rigid cross girder,
Reaction on longitudinal girder is =21.075×53=35.125 kN
For max. Bending moment, the loads should be kept at equal distance from intermediate longitudinal girder of both sides of cross girder.












Fig-27 (Wheel load arrangement for maximum bending moment on Cross Girder)
Now for calculating load, following diagram may be provided at beneficial
Load on cross girder from one wheel
=2×3502×(4.5075-0.9)4.5078=280.12kN
Reaction on each longitudinal girder
=280.12×23=186.75kN
Max. Bending moment on cross girder under load =186.75×1.475=275.456 kN-m
In checking impact, Mc =303kN-m
Dead load moment under wheel load ,
=35.125×1.475-21.075×(1.475)22=28.884kN-m
Total design moment=332kN-m
Ultimate moment = Mu =498kN-m
Since the cross girder fixed at intermediate girder,(-)ve moment will exist.
As per D.J victor, Essential Bridges Engineering, referring to Morrice-Little method, distribution coefficient=0.416
So negative moment =0.146× 498 73 kN-m
Since its flange d beam,assuming neutral axis in side flange,
Mu =21241.143kN-m>>>>>498kN-m
(Ast )+ve=498×1060.87×415×1660×(1-0.416×2501600)=886.45mm2
Providing 4-20mm ф bars
(Ast )provided=1256.64mm2
Provision of 4-16-mm ф bars at top will give
(Ast )-ve=804.25mm2>(Ast )required =135.152mm2
Provide 4-12mm ф bars each face uniformly as side reinforcement.

3.6.15.1 SHEAR CHECK FOR CROSS GIRDERS:

Cl-10.3.2 of IRC:112-2011
K=1.33
vmin=0.281
σcp=0
ρ1=2.618×10-3
VRdc=[0.12×1.33×(80×2.618×10-3×35)0.33×300×1600=147.831 kN
VRdc.min=0.281×300×1600=134.880 kN
Design shear VEd =1.5[186.75+35.125]=332.8125 kN
Extra shear to be resisted by stirrups is =184.98 185kN
Providing 8mm dia -4 legged stirrups
Asw =201mm2
Spacing(s)= 519.43mm,8mm ф 4-legged @300mm c/c
Again shear reinforcement ratio
ρw= As×bw×sinα(cl-16.5.2 of IRC 112:2000)
=1.2899×10-3
(ρw)=0.072×35415=1.026×10-3
Assumimg (ρw)min ,
s=201300×1×1.026×10-3=653.021mm>300mm(ok)



Fig-28 (Reinforcement detailing of Cross Girder)

Clause -16.5.2 of IRC:112-2011(6, 7, 8,9,) specifies that
smin=dg+10=30mm
40mm
2øs=2×8=16mm
Smax=0.75d(1+cotα)=0.75×(1+0)=1245mm
As per the Clause -16.5.3 of IRC:112-2011,
The links will not to be of greater spacing of following
1/8 (perimeter of member)=487.5mm
or 350mm
Our provision of 300m (okay)
3.6.16 MODIFIED ANCHORAGE FOR BARS AT THE ENDS OF GIRDERS:
At the ends of girders,bearing available is only 450mm.hence standard 90 bend as per fig 15.2 of irc:112-2011 will be provided.
So modified lb.net will be
lb.net =0.7×lb×1=0.7lb
lb =903mm(for 30mm ф bars)
lb =843mm(for 28mm ф bars)
(lb.net)O.G =632mm
(lb.net)I.G =590mm













REFERENCES:
IRC:112-2011; CODE OF PRACTICE FOR CONCRETE ROAD BRIDGES, INDIAN ROAD CONGRESS
IRC:6-2014; STANDARD SPECIFICATIONS AND CODE OF PRACTICE FOR ROAD BRIDGES; SECTION : II LOADS AND STRESSES
IRC:6-2010; STANDARD SPECIFICATIONS AND CODE OF PRACTICE FOR ROAD BRIDGES; SECTION : II LOADS AND STRESSES
IRC-SP-13; GUIDELINES FOR THE DESIN OF SMALL BRIDGES AND CULVERTS
IS 456 : 2000; INDIAN STANDARD PLAIN AND REINFORCED CONCRETE CODE OF PRACTICE
IS-SP-16; DESIGN AIDS FOR REINFORCED CONCRETE TO IS : 456-1978
IS : 875 – 1987; INDIAN STANDARD CODE OF PRACTICE FOR DESIGN LOADS (OTHER THAN EARTHQUAKE) FOR BUILDINGS AND STRUCUTRES
IS-SP : 34(S&T)-1987; HANDBOOK ON CONCRETE REINFORCEMENT AND DETAILING
ESSENTIALS OF BRIDGE ENGINEERING; VICTOR D.J; OXFORD&IBH PUBLICATION
DESIGN OF BRIDGES; RAJU N.K; OXFORD&IBH PUBLICATION
R.C.C DESIGNS ; PUNMIA B.C, JAIN A.K AND JAIN A.K; LAXMI PUBLICATIONS
REINFORCED CONCRETE VOL.I; SHAH H.J; CHAROTAR PUBLICATIONS
LIMIT STATE DESIGN OF REINFORCED CONCRETE; VARGHESE P.C; PHI LEARNING
THEORY OF STRUCTURES; RAMAMRUTHAM S. AND NARAYAN R.; DHANPATRAI PUBLISHING COMPANY
HIGHWAY ENGINEERING; KHANNA S.K. AND JUSTO C.E.G.
INTERNATIONAL JOURNAL OF SCIENTIFIC & ENGINEERING RESEARCH VOLUME 4 , ISSUE 3 ,MARCH – 2013 (ISSN 2229 -5518) "STUDY OF EFFECTIVENESS OF COURBON ' S THEORY IN THE ANALYSIS OF T – BEAM BRIDGES " BY: M.G. KALYAN & HETTI &R. P. SHRIRAM