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ABHISEK PANDA

3.6 SUPER STRUCTURE DESIGN:

3.6.1 AVAILABLE DATA:

1. Effective span of T-beam bridge: 18.33m2. Total length of bridge: 55m3. Clear width of carriage way 7.5m (IRC 5: 1998 & IRC 6: 2014)4. Kerb width: 600mm (both side, pedestrians are allowed)5. Parapet: 1000 mm 150 mm 150 mm @ 1.5m c/c with 3-cast iron pipes as railing6. Camber: 1 in 100 (37.5 mm at center linearly varying to zero at kerbs)7. Wearing coat: 80mm8. Kerb height above pavement: 200mm (insurmountable type)9. Kerb type: full safety ensured10. Total kerb height above deck slab: 280 mm11. Clear depth of Longitudinal girders: 1400 mm12. Width of longitudinal girder: 400mm13. Width of cross girder: 300mm14. Clear depth of cross girder: 1400 mm15. Deck slab thickness: 250 mm16. Cantilever slab: 400 mm at girder face; linearly varying @ 0.11 to ends i.e. 200 mm17. Total overall depth of the super structure: 1800mm18. c/c spacing of longitudinal girders: 2500 mm19. c/c spacing of cross girders: 4507.50 mm20. clear distance of cantilever span from face of girder: 1800mm21. Grade of concrete: M35

22. Design strength: fcd = 0.67fck/mMPa (Annex – A2 of IRC 112: 2011) 23. Grade of steel : Fe415 (IS 1786 : 2000)24. Design strength of steel : fy/1.15 = 0.87fy MPa (clause-15.2.3.3 of IRC 112:2011)25. Poisson’s ratio: = 0.2 (Annex-B; B-3-1 of IRC 112:2011 )26. 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.

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ABHISEK PANDA

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-mLong span moment is ML = 95.33 (0.0145+0.200.047) = 2.28 kN-mSince 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:

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ABHISEK PANDA

1. 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 getm1 = 0.095 and m2 = 0.03

After interpolating, we getm1 = 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-mML = 0.8 350(0.019+0.20.089) = 10.304 kN-mAs 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

2. 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.

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ABHISEK PANDA

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 mmSo u = 0.3+20.080 = 0.460 m and v = 0.15+20.080 = 0.310 mThe value of u/B = 0.184 , v/L = 0.069 and B/L = 0.55Using Pieguad’s curve, m1 = 0.22 and m2 = 0.20Hence 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.

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ABHISEK PANDA

Fig-9 (Bending moment calculation due to Wheel Load-1)

So u = 2(u1 + x) = 2(0.46+0.37) = 1.66m and v = 0.31mLoad intensity = 37.5/(.31.46) = 262.973 kN/m2

K = B/L = 0.55, u/B = 0.664 and v/L = 0.069Hence m1 = 0.12 and m2 = 0.14MB = (0.12+0.20.14)262.9731.660.31 = 20.03 kN-mSimilarly ML = 22.193 kN-m

1. Now u = 2x = 20.37 = 0.74 and v = 0.31K = 0.55, u/B = 0.296 and v/L = 0.069From Pieguad’s curve, m1 = 0.184 and m2 = 0.189MB

’ = 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-mML2 = (22.193 – 13.622) 0.5 kN-m = 4.285 kN-m

Bending Moment due to Wheel Load-3:

2. 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

3. Taking u = 2x = 20.77 = 1.54 and v = 0.31,

K = 0.55, u/B = 0.616 and v/L = 0.069

BY:

ABHISEK PANDA

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:

4. u =0.46m, v = 2(v1+x) = 2(0.31 + 1.045) = 2.71mLoad intensity = 62.5/(0.460.31) = 438.3 kN/m2

K = 0.55, u/B = 0.184, v/L = 0.60So m1 = 0.125 and m2 = 0.0306MB = 71.57 kN-m and ML = 30.35 kN-m5. u = 0.46m and v = 2x = 21.045 = 2.09mK = 0.55, u/B = 0.184 and v/L = 0.464From Pieguad’s curve, m1 = 0.1441 and m2 = 0.043MB

’ = 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:

6. 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

7. 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

8. u = 0.74m, v = 2.71m

BY:

ABHISEK PANDA

K = 0.55, u/B = 0.296 and v/L = 0.601m1 = 0.12 and m2 = 0.03 and multiplying with x(v1+y) = 0.5013,m1

’ = 0.060 and m2’=0.015

9. u =1.66 m and v = 2.09m

K = 0.55, u/B = 0.664 and v/L = 0.464m1 = 0.095 and m2 = 0.038Multiplying 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 = 37.5

0.460.31 [0.0105+0.20] = 2.761 kN-m

ML4 = 37.5

0.460.31 [0.20.0105+0] = 0.552 kN-m

Bending Moment due to Wheel Load at -6:

10. 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 mK = 0.55, u/B = 0.984 and v/L = 0.601m1 = 0.068, m2 = 0.02[u1+x][v1+y] = 1.231.355 = 1.67m1

’=0.113 and m2’ = 0.0334

11. 2 u = 2x =1.54m and v = 2y = 2.09mK=0.55, u/B = 0.616 and v/L = 0.464m1 = 0.097, m2 = 0.038xy = 0.804Hence m1

’ = 0.078 and m2’ = 0.0305

12. 3 u = 2[u1+x] = 2.46m and v = 2y =2.09mK = 0.55, u/B = 0984 and v/L = 0.464m1 =0.074 and m2 = 0.03y(u1+x) = 1.285m1

’ = 0.095 and m2’=0.038

13. 4 u = 2x = 1.54m and v = 2(v1+y) = 2.71mK = 0.55, u/B = 0.616 and v/L = 0.6012m1 = 0.092 and m2 = 0.025Multiplying x(v1+y) = 1.043 with above coefficients,m1

’ = 0.096 and m2’ = 0.026

BY:

ABHISEK PANDA

So final coefficients are, m1 = [(0.113+0.078)-(0.095+0.096)] = 0m2 = [(0.0334+0.0305)-(0.038+0.026)] 0The 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-mML = 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-mML = 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 thatdesign 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 = 0.6Vz

2 = 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

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ABHISEK PANDA

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/mFdownward/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.

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ABHISEK PANDA

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 =2.6l0=2.1a=0.755mb1=3.6+2× 0.08=3.76mbe=2.6×0.755(1−0.755/2.1)+3.76 =5.017mLoad per meter width is

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ABHISEK PANDA

= 3505.017 = 69.763

Maximum shear force = 69.763×(2.1−0.755)

2.1 =44.681 kN/mShear force with impact due to tracked vehicle = 49.15 kN/m3.6.4.2 Dead load shear:

Total dead weight= 8.46 KN/m2

Total dead load shear=8.46×2.1

2 = 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-mTotal shear force=(8.883+49.15) KN/m

=58.03 KN/mSince we have considered limit state method the above load values will be 1.5 times that of calculated as per Anex A2 of IRC: 112-2011So 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 mmAs per table – 14.2, clause –14.3.2.1 (page 142) of IRC–112: 2011, Clear cover =40+8=48 mmSo, 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− 4.6×50×106

35×103×2022⌋×103×202

As per clause -16.6.1.1 of IRC:112-2011

(Ast)min = 0.26×f ctmf yk

×btd

BY:

ABHISEK PANDA

From table -6.5 (Page-38)of IRC :112-2011Fctm=2.8 N/mm2

Fyk= 415 N/mm2

Bt=1000 mmD = 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 mmIn our case,spacing of 16 mm ∅ bars is

S¿ 1000×π /4×162

716.008 =280.811mm> 250 mmAlso 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 greaterDg= aggregate size = assumed 20mm for slabs, providing 16 mm-∅ bars @ 225mm c/c

((Ast)B)provided = 1000×π /4×162

225

=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

BY:

ABHISEK PANDA

Providing 12mm-∅ bars,

(Ast)L = 0.5×35415

×[1−√1− 4.6×33×106

35×103×1902 ]×103×190 = 480.53 mm2

(Ast)min=0.26×f ctmf yk

×b f ×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 /4×122

480.53 =235.36mm

Hence providing spacing of bars @175 mm c/c & using 12mm ∅ bars

((Ast)l)provided =1000×π /4×122

175 =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 0.8f yk 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

BY:

ABHISEK PANDA

1. 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 =[0.12 k (80 ρ1 f ck)0.33+0.15σcp ]bwd (clause- 10.3.2, IRC:112-2011)

VRdc =(Vmin+0.15σ cp)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 σ cp=0

ρ1=A st

bwd≤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:

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ABHISEK PANDA

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

BY:

ABHISEK PANDA

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 .

BY:

ABHISEK PANDA

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 =√ 119×106

0.36×35×0.48×103×(1−0.416×0.48)

=156.80 mm

Providing 40 mm clear cover &16 mm ∅ bars,

Effective depth provided is

BY:

ABHISEK PANDA

dprovided =400-(40+8)=352 mm ≫156.80 (ok)

3.6.9 REINFORCEMENT IN CANTILEVER SLAB:

Main reinforcement is given by

(Ast)main = 0.5×35415 [1−√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)

2. 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

BY:

ABHISEK PANDA

As per clause-10.3.2(2) of IRC:112-2011,

Shear resistance of a structure is given by

VRdc =[0.12k(80 ρ1 f ck)1 /3+(0.15σcp)¿bw d

Subject to min VRdc=(Vmin+0.15σ cp)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 =Aslbw d

=1058.221000×400 = 2.645×10−3

VRdc=[0.12×1.71× (80×2.645×10−3×35 )1/3]×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−f ck

310]

=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 V ed.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

BY:

ABHISEK PANDA

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)

1. 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×d3

12=1000×200

3

12=6.67×108mm4

(Igr)mid =22.5×108mm4 , ( I gr )at 0.65m=29.35×108mm4

Fcr =0.7√ f ck=4.141N /mm2

(Mr)end=4.141×6.67×108

100=27.6kN−m

(Mr)mid =4.141×22.5×108

150=62.115 kN−m

(Mr)0.65m=74.158 kN-mEc =5000√ f ck=2.958×104 N /mm2

Es =2×105 Nmm2 ,m= Es

E c=6.76

Transformed area of compression steel =(m-1)×Asc =3427.93mm2

BY:

ABHISEK PANDA

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 500 x2+10580.01 x−1459363.34=0Or x=44.47mmAt mid, 500 x2+10580.01 x−1960008.94=0X=52.92mmAt 0.65m, 500 x2+10580.01 x−2158693.7=0X=55.97mm

Fig-15 (Area of Reinforcement in different section of cantilever slab)

(Ir)end=13×1000× (44.47 )3+3427.93× (44.47−46 )2+7152.08× (152−44.47 )2

=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-mM0.65m=wheel laod=46.70kn-m

(Ieff)end=

(i¿¿ r )end

1.2−(m¿¿ r )end

(m)end×zenddend

×(1− xend

dend )(bw

b)¿

¿=

1.117×108

1.2− 27.619.49

× 136152

(1−44.47152

)=3.67×108mm4

( I ¿¿r )end<(I ¿¿eff )end<(I ¿¿ gr )end(ok )¿¿¿

BY:

ABHISEK PANDA

(I ¿¿eff )mid=¿¿So ( I ¿¿eff )mid=( I ¿¿r)mid=3.1264×10

8mm4¿¿(Ieff )at 0.5 m=88.84×108mm4>¿(Igr)at 0.65m

So (Ieff )at 0.65 m=29.35×108mm4

(δ 1)end=w l3

3Ec I r= 10212× (1800 )3

3× (2.958×104 )×(3.67×108)=1.83mm

(δ 1)mid =15350× (1800 )3

8× (2.958×104 )×(3.1264×108)=1.21mm

(δ 1)At 0.65m =1

Ec I r [1.15×w×l2

2+w l3

3 ]=0.64mmTotal deflection due to short tern loading =3.68mmDeflection due to shrinkageacs=k3csl2

k3=0.5(for cantilever)pt=0.7% ,pc=0.4%

k4=0.72p t−pc

√ p t

=0.26<1.0

0.25pt−pc=0.3<1.0(ok )(cs )end=3.9×10−7

acs=0.63mmNote-:here at other points,the ‘cs’ will give much lesser value relating least shrinkage deflection.Deflection due to creep

Ece =E c

1+θθ=1.6(for 28-days strength)Ece =1.1377×104N /mm2

m =Es/Ec = 17.58Transformed 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 = 0Or x = 57.40 mmAt mid:500x2 + 28463.63x – 5140706.56 = 0

BY:

ABHISEK PANDA

Or x = 76.853mmAt 0.65 m:500x2 + 28463.63x – 5657384.835 = 0Or 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 thereSo (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=(W end)×l3

3 Ece×I eff+(Wmid)×l3

8 Ece×I eff=(2.39+1.35)mm=3.74mmShort term deflection due to permanent load

ai.perm =7 .812×103×18003

3×2 .958×104×3 .67×108+ 15 .35×103×18003

8×2.958×104×3 .1264×108=1.4+1.21=2.61mm

So deflection due to creep is given by3.74-2.61=1.13mmTotal deflection is =3.68+0.63+1.13=5.44mmHowever 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 span

375

=1800375

=4.8mm<5.44mm

2. Now let’s check it as per cl-23.2 of is:456-2000fs=220.27, pt=0.7%

BY:

ABHISEK PANDA

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×d3

12=1000×200

3

12=6.67×108mm4

(Igr)mid =22.5×108mm4 , ( I gr ) .65m=29.35×108mm4

Fcr =0.7√ f ck=4.141n /mm2

(Mr)end=4.141×6.67×108

100=27.6kN−m

(Mr)mid =4.141×22.5×108

150=62.115 kN−m

(Mr)0.65m=74.158 kn-m

Ec =5000√ f ck=2.958×104 N /mm2

Es =2×105 nmm2 ,m= Es

E c=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 )

=500 x2+13157.92 x−1605582.08=0=x=45.02mmAt mid=500 x2+13157.92 x−2549278.08=0X=59.45mm

BY:

ABHISEK PANDA

At 0.65m=500 x2+13157.92 x−2811436.83=0X=62.97mm

(Ir)end=13×1000× (45.02 )3+3720.96× (45.02−46 )2+9436.96× (152−45.02 )2

=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-mM0.65m=wheel laod=46.70KN-m

(Ieff)end=

(i¿¿r )end

1.2−(m¿¿ r )end

(m)end×zenddend

×(1− xend

dend)¿

¿=4.5×108mm4

( I ¿¿r )end<(I ¿¿eff )end<(I ¿¿ gr )end(ok )¿¿¿(I ¿¿eff )mid=¿¿(Ieff )at 0.65 m=54.44×108mm4<¿(Igr)at 0.65m=29.35×108mm4

(δ 1)end=w l3

3Ec I r=1.5mm,(δ 1)mid =0.9mm, (δ 1)At 0.65m =

1Ec I r [ 1.15×w×l2

2+w l3

3 ]=0.64mmTotal deflection due to short tern loading =3.04mmDeflection due to shrinkageacs=k3csl2

k3=0.5(for cantilever)pt=0.92% ,pc=0.425%k4¿0.37<1.00.25pt−pc=0.495<1.0(ok )(cs )end=5.55×10−7

acs=0.9mmDeflection due to creep-:

Ece =E c

1+θθ=1.6(for 28-days strength)

Ece =1.1377×104 Nmm2

,m=E s

Ece=17.58

Transformed area of compression steel =(m-1)×Asc =10710.68mm2

Transformed area of tension steel=m×Ast =24541.68mm2

BY:

ABHISEK PANDA

Depth of neutral axis at different sectionAt end at mid at 0.65mX=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=(W end)×l3

3 Ece×I eff+(Wmid)×l3

8E ce×I eff=(2.2+1.1)mm=3.3mmShort term deflection due to permanent load

ai.perm =7.812×103×18003

3×2.958×104×3.67×108+ 15.35×103×18003

8×2.958×104×3.1264×108=1.4+1.21=2.61mm

Deflection due to creep = 0.69mm 0.70mmHence 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:

1. In short direction:

main bars = 16 mm∅@ 225 mm c/c

(lb.net)B=α a lb×A st req

A st provided (page no-152, cl-15.2.4.3, of IRC:112-2011)

A st req

A st provided =716.005893.61 =0.865

α a =1 (for straight bars)

lb =k∅

=∅×0.87 f y4 f bd

=16×0.87×415

4×30 ( fbd=3.0 N/mm2 table-15.3 of page -150,IRC:112-2011)

BY:

ABHISEK PANDA

=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)

2. In long direction:

A st .req

A st . provided =480.53646.30=0.7430.75

α a =1 (for straight bars)

lb =k∅

=30×12=360mm

or lb=12×0.87×415

4× = 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:

BY:

ABHISEK PANDA

A st .req

A st . provided =872.561058.22 =0.824 (cl-15.2.4.3 of IRC:112-2011)

α a =1

lb=k∅ =30×12=360 mm

lb.net =0.824×360=296.64 300 mm3.6.13 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

=698.132mm2>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

1. 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.

BY:

ABHISEK PANDA

Fig-16 (Arrangement of class-AA tracked load for maximum eccentricity)

As per Courbon’s formula ,

Rx= ∑ wn

[1+∑ I

∑ dx2dxe]

Rx=Reaction factor

It may be noted that RA=Rc since the same loading can be reversed in two cases.

RA=2W 1

3[1+

3 I2×2.52×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

2. 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

BY:

ABHISEK PANDA

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 49kN/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

BY:

ABHISEK PANDA

Fig-17 (Reaction from Cross Girder on Longitudinal Girders)

Maximum dead load bending moment at center of span is,

(Mmax)dead load =49×18.332

8+ 26.25×18.33

4+ 26.25×18.33

4 =2300 kN-m

Total dead load shear at support is

(Vdead)s=49×18.33

2+26.25×2

2+ 26.25

2 =488.46489kN

3. 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.

BY:

ABHISEK PANDA

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=2892.75 KN-m2893 kN−m

Bending moment with impact & reaction factor is given by (Ml)outer girder=0.553×1.10×2893=1759.812KN−m1760kN−m(Ml)inner girder=0.333×1.10×2893=1059.711060 kN−m4. LIVE LOAD SHEAR:

For shear force to be maximum,the loading should be arranged as shown below.

BY:

ABHISEK PANDA

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)

5. CALCULATION OF BENDING MOMENT USING HENDRY-JAEGAR METHOD:

A =(12π4

)( Lh )3

¿)

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

BY:

ABHISEK PANDA

F =¿) (hL)(

CJE I r

)

c = E I 1E I 2

=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

EI 1& EI 2= flexural rigidities of the outer & inner longitudinal girders

EI r = 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=2.043m

0.2l0=3.66m > 2.043 m (ok)

Bw=0.4 m

Beff=2×2.043+0.4

BY:

ABHISEK PANDA

=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+0.1×0.7×2.50.2l0 =0.2×0.7×2.5=0.35>0.596m∴beff.1=0.35 mbeff=2×0.35+0.3=1mb =4.5075 m>1m (ok)Longitudinal girder:

x=A1 x1+A2 x2A1+¿A 2

¿

=2.5×0.25×0.125+1.55×0.4×1.025

2.5×0.25+1.55×0.4

=0.57 m

IL=2.5×0.253

12+ (2.5×0.25 )× (0.448 )2+ 0.4×1.55

3

12+(0.4×1.35 )×0.4522=0.3795m4

BY:

ABHISEK PANDA

Cross girder:

x=A1 X1+A2 X2

A1+A2

¿ 1×0.25×0.125+0.3×1.55×1.0251×0.25+0.3×1.55 =0.710 m

I r=1×0.253

12+1×0.25×0.5852+ 0.3×1.55

3

12+0.3×1.55×0.3152

¿0.2261m4

A = 12π4

׿)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 = π2

2×5× 2.518.33

× 0.43 E×0.0398E×0.2261

=0.01020

Two extreme values of ‘F’ can be taken into analysis i.e. F = 0 or F = ∞ .

Taking F = 0 for analysis ,M F=MO+(M¿¿∞−M 0)√ (F √A)(3+F√ A)

¿

❑⇒M F

= MO¿; MO=0.83 (For outer girder , F=0 ) ¿

MO=¿ 0.36 (For inner girder, F = 0)

Design moment for exterior girder :

BY:

ABHISEK PANDA

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−mas calculated by Courbon ’ s method

6. 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. ShriramAs 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 calculatedRa = Rc=ROuter girder =0.553wRb=Rinner girder=0.333wCorrection factor is given by,Y= correction factor,x= span of bridgeX=18.33m

= 0.9214 (rounded up to 4 decimal)RA=RC= (Router girder) corrected=0.553×0.9214=0.5095RB=(R inner girder) corrected=0.333×0.9214=0.3068

BY:

ABHISEK PANDA

Total bending momentOuter girder=2300+2893×1.10×0.5095=3921.40 KN-mInner girder=2300+2893×1.10×0.3068=3276.33 KN-mOut of all 3-analysis, Hendry-Jaegar method gives the highest momentDesign moment-Outer longitudinal girder=1.5×4771.31=7411.50=7412 KN-mIntermediate girder=1.5×3276=5169 kN-mDesign shear Shear force =871×1.5=1348.50 kN

7. 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 64mmLet’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=235mmHence effective depth= 1800-235=1565mmLet’s take d=1560mmDf= 250mmDf /d=0.1602 <0.2Bf = 2.5m (as calculated earlier)Xumax =0.48×1600=768mmLet’s check the actual neutral axis depthLet’s take xu =Df =250mm

Mu =0.36× 35× 2500× (1600)2× =11781kN-mImposed 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

BY:

ABHISEK PANDA

So (Ast)provided= The spacing between longitudinal bars is As per IS 456:2000, clause-26.3.2, the spacing should be minimum of followingDiameter of bars =30mm5mm more than nominal maximum size of coarse aggregate =20+5=25mmOur spacing is 35.33 is satisfactory.8. For detailing of reinforcement, please refer to Appendix-Battached with this thesis.9. DESIGN REINFORCEMENT FOR INNER GIRDER: d = 1600 mm, Df = 250mm and bf = 2.5mAssuming neutral axis to be within the flanges, Mu = 1178 kN-mApplied 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

10. For detailing of reinforcement, please refer to Appendix-B attached with this thesis.11. CALCULATION OF ANCHORAGE LENGTH: Clause -15.2.4.3 of IRC:112-2011Outer girder

Lb.net=α a lb(a¿¿ s)reqd

(a¿¿s ) provided≥ lbmin¿¿

Let’s assume (A¿¿st)reqd

(A¿¿ st) provided¿¿=1 at a section

So α a=1

lb=(∅4 )×( f ydf bd )=304× 0.87×415

3 =902.625≈903mm

Or lb=kø=30×30=900mmlb.net =lb=903mmlb.min=0.3lb=288.9mm or 100mm or 10ø=300mmInner girder

lb=284× 0.87×415

3=842.45≈843mm

lb.net =843mm

Length of bearing = 23 lb.net(16.5.1.4. of IRC :112-2011)

BY:

ABHISEK PANDA

=602mm(outer girder)=562mm(inner girder)

12. 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:

=12

[3.44+2.764 ]×700=2171.40 kN−m

Dead load moment:=488.46×4.5825−49×4.58252×0.5=1724 kN−mAs per H-J method, wheel load BM including impact & coefficient

=Total design moment on outer girder=1.5(1982.5+1724=5560 kN−m)Total design moment on inner girder=1.5 (1724+1.1×2171.40×0.36 )=3876 kN-m(ii) Loads at a distance 6.874m from ends of girder

BY:

ABHISEK PANDA

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

Wheel load moment=700× [3.45+4.3 ]×0.5=2712.5 kN−mDeadload moment=488.46×6.874−[49×0.5×6.8742+26.25×2.2165 ]=2142kN-mTotal design b.m on outer girder due to impact and H.J coefficient =1.5 ¿Total design b.m on inner girders due to impact and H.J coefficient=1.5¿

(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-mDead load moment=488.46×2.29125−49×0.5× (2.29125 )=991kN−mTotal design moment on outer girder=1.5 (991+1.1×0.83×1266 )=3220.30kN−mTotal design moment on inner girder=1.5 (991+1.1×0.36×1266 )=2238.50 kN−m

13. REQUIRED STEEL AREA AS PER BAR CURTAILMENT:

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

Outer gird er:

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:=¿

Providing 12-28mmdia bars ,

BY:

ABHISEK PANDA

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 , =¿Note-:1-bars are curtailed at a distance of 5.971m from ends of supportInner girder:¿Providing 15-28mm dia bars=¿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 girderOuter girder:¿Providing 10-30mm dia bars,=¿Note -:10-30mm dia bars are curtailed from 1.388m from ends of supportInner girder:¿Providing 8mm ф bars , ¿Note -:8-28mm ф bars are curtailed from 1.44825m from ends of girder

(OUTER GIRDER)

BY:

ABHISEK PANDA

(INNER GIRDER)

Fig-23 (Bar curtailment Presentation for Longitudinal Girders)

14. CALCULATION OF INDUCED SHEAR AT DIFFERENT SECTIONS: 1. Wheel load at starting

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

Applying impact, total shear =454.3kN

BY:

ABHISEK PANDA

Again reaction at support is =

Dead load shear =Total shear=899kN ≈ 900 kN2. 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≈200kN3. 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=12

[0.75+0.5536 ]×413×1.1=296.113kN

Dead load shear=489−(49×4.5825 )=264.46 kNTotal 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)

BY:

ABHISEK PANDA

L.L shear=361.441kND.l shear =489−(49×1.95 )=393.45 kNTotal 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=335.34kND.L shear=489-(49×3)=342kNTotal 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=239.325kND.L shear=489−(49×6.87375+26.25 )=126 kNTotal 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)

BY:

ABHISEK PANDA

Track shear at 4.3825 m = 0.5 [0.761+0.564]1.10413 = 301 kNDead load shear = 489 – (494.3825) = 275 kNTotal 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=2.01mHence 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)=360 kND.d shear=489−(49×2.01 )=390.51 kNTotal shear=750.51KN≈751 kN4. SHEAR RESISTANCE CHECK: Let’s provide 1bent up bar at 4.5825m & at 645mm interval up to 2.00m as shown belowMaximum spacing of bent up bars ¿]=0.6×1560×2=1.872>0.645(ok)

Now A sw=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=asw

s× z× f ywd(cotθ+cotα)sinα

s=0.645mz=0.9d=0.9×1560=1440mmf ywd=0.8×415=332N /mm2

(V Rds)O.G=706645

×1404×332×2× 1√2

=721.55 kN

V Rd .max=∝cwbw v1 f cd z(cotθ+cotα) /(1+cot2θ)❑=1×400×0.9×1560×0.6×0.36×35×2/2

BY:

ABHISEK PANDA

=4245.696>>721.55 KN (ok)

(V Rds)I . G=615645

×1404×332×2× 1√2

=628.543<< 4245.696 kN (ok)Design shear resistance of member without shear reinforcement is given by

VRdc =[0.12k (80 ρ1 f ck )0.33+0.15σ cp]×bw d(cl−10.3 .2of IRC :112−2011)σ cp=0

vmin=0.031 k32 f ck

12

K=1+√ 200d =1+√ 2001560=1.358<2(ok )

vmin=0.29(VRdc)min=(vmin+0.15σ cp )bw d=0.29×400×1560=181.105 kN

Now ρ1=A st

bwdSince 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 =[0.12×1.358 (80×0.01133×35 )0.33 ]×400×1560¿318 kN>vRdc .min (ok )

(VRdc)I.G=[0.12×1.358 (80×7.89×10−3×35 )0.33 ]×400×1560=282 kN>vRdc. min (ok )

5. SHEAR REINFORCEMENT DISTRIBUTION ON OUTER GIRDER:

Total shear at face of support= 900kNHowever 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 girderTotal shear at 2.01m=751kNDesign shear=1.5×751=1126.50 kNShear resisted by girder without shear reinforcement = (VRdc)O.G=318 kNHence design shear for which shear reinforcement will be provided=VEd

=(1126.50-318)KN=808.50 kN

BY:

ABHISEK PANDA

The shear resisting capacity of available bent up bars at that section is 721.55 kNBut 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

=404.25 kn

[check:(asw.max)bent up bars=0.5×10×0.6×0.36×35×400×645

1√2

×0.8×415=4154.21mm2>706mm2 (ok )

>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=∝cwbw v1 f cd zf cd

(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×f ywd× z×cotθ

vrds

=201×0.9×1560×0.8×415×1

404.25×103=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×√ f ck

f yk=0.072×√35

415=1.026×10−3

Provided shear reinforcement ratio is

(ρw)=A sw

s×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

¿1 )o.g=0.02039 > 0.02, ¿1 )i.g=8620.53400×1560

=0.0138≈0.02

BY:

ABHISEK PANDA

(vrdc)o.g =0.12×1.358× (80×0.02×35 )0.33×400×1560=383.855≈383 kN(vrdc)i.g =0.12×1.358× (80×0.0138×35 )0.33×400×1560=339.615≈339 kNShear to be resisted by stirrups is(841.5-383) kN =458.5 kNVRd.max=2122.848>458.50 kN(ok)Spacing of stirrups is

S=201×0.9×1560×0.8×415×1

458.50×103=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=549kNShear resistance=383 kNTotal shear to be resisted by stirrups is 549-383=166 kNSpacing of same 4-legged 8mm vertical stirrups is given by

s=201×0.9×1560×0.8×415×1

166×103=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 or18 [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 4.5825mAgain it should satisfy (ρw)min

(ρw)min =1.026×10−3

(ρw)prov =201

350×400=1.436×10−3>( ρw)min (ok )

6. SHEAR REINFORCEMENT DISTRIBUTION ON INNER GIRDER: 0.45+1.56=2.01m from girder end Total design shear=1126.50kN(VRdc)I.G=282 kNDesign shear for which shear reinforcement will be provided=844.50knBent up bars shear resistance is(VRds)I.G =628.543 kN

But it will be taking only 844.502

=422.25 kN

Rest 422.25kn will be taken care of by the vertical stirrups.Providing 8mm ф -4 legged stirrups,Asw=201mm2

BY:

ABHISEK PANDA

Cl-10.3.3.2 of IRC:112-2011 specifiesVrd.max=2122.848kN >> 422.25kN(ok)Following same clause, spacing of stirrups,

S=201×0.9×1560×0.8×415×1

422.25×103=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 =201

200×400=2.5125×10−3>( ρw)min (ok )

Shear at 4.3825m is given by 576 kN.Design shear=1.5×576=864kNNo bent up bars are available here.(VRdc)I.G=339 kNShear to be resisted by stirrups=525kNVR d.max=2122.84kN >>525kNSpacing of stirrups is given by

S=201×0.9×1560×0.8×415×1

302.50×103=178.46mm

Provide 8mm-4legged stirrups @160mmc/c up to 6.87375mShear at 6.87375m from end of girder is; Design shear=1.5×360=549kN

(Ast)available=16×π4×282=9852.035mm2

¿1 )I.G=9852.035400×1560

=0.016<0.02

(VRdc)I.G=356.604 kNShear to b resisted by links and/stirrups(549-356.604)kn=192.396kn≈193 kNProviding minimum shear reinforcement spacing of 350mm c/c,

VRds=201×0.9×1560×0.8×415×1

350=267.691 kN>193 kn(ok)

So Provide 8mm-4legged stirrups @350mmc/c in middle strip-band of length 4.5825m(ρw)min =1.026×10−3

(ρw)prov =201

350×400×1=1.436×10−3>( ρw)min (ok )

7. SURFACE REINFORCEMENT:

BY:

ABHISEK PANDA

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)1. All the surface reinforcement bars are perfectly anchored&hence will be serving as shear

reinforcement also appear same clause(5) also2. 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

3. CHECK FOR BAR CURTAILMENT ADEQUACY AS PER CURTAILMENT OF GIRDER BARS:

BY:

ABHISEK PANDA

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 saferFrs=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 1126.50kNFs=0.45×1126.50 ≈ 507kN

At ends Med≈0M ed

z=0

Ftd=0.5 VEd(cotθ−cotα)=0.5×1126.50×1=563.25kNAlso another value of Fs appear fig-16.2(A) is

M Ed

z+NEd=0

Fs+ Ftd =563.25+507=1070.25kn<2552.112 kn=(FRs)o.g

At middle,

BY:

ABHISEK PANDA

M ed

z=M edmax

z

(M ed

z)O.G=7412×10

6

0.9×1560=5280 kN

(M ed

z)I.G=5169×10

6

0.9×1560=3682 kN

(Fs)O.G=5280kn,(Fs)i.g=3682kN

( VEd=0& also Ftd<M Edmax

z )

(FRS)O.G=2552.1122=5104.224kN(FRS)I.G=1778.542=3557.08kNAs 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/mSlab 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

=10.575 kN /m

Total load (dead weight) =21.075kN/mAssuming rigid cross girder,

BY:

ABHISEK PANDA

Reaction on longitudinal girder is =21.075×5

3=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)

BY:

ABHISEK PANDA

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

=280.12 kN

Reaction on each longitudinal girder

=280.12×2

3=186.75kN

Max. Bending moment on cross girder under load =186.75×1.475=275.456 kN-mIn checking impact, Mc =303kN-mDead load moment under wheel load ,

=35.125×1.475−21.075× (1.475)2

2=28.884 kN−m

Total design moment=332kN-mUltimate moment = Mu =498kN-mSince 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=0.416So negative moment =0.146× 498≈73 kN-mSince its flange d beam,assuming neutral axis in side flange,Mu =21241.143kN-m>>>>>498kN-m

(Ast )+ve=498×106

0.87×415×1660×(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:112-2011K=1.33vmin=0.281σcp=0ρ1=2.618×10-3

VRdc=[0.12×1.33×(80×2.618×10-3×35)0.33×300×1600=147.831 kNVRdc.min=0.281×300×1600=134.880 kNDesign shear VEd =1.5[186.75+35.125]=332.8125 kN

BY:

ABHISEK PANDA

Extra shear to be resisted by stirrups is =184.98≈185 kNProviding 8mm dia -4 legged stirrups Asw =201mm2

Spacing(s)= 519.43mm,8mm ф 4-legged @300mm c/cAgain shear reinforcement ratio

ρw= A

s×bw×sinα¿

¿ (cl-16.5.2 of IRC 112:2000)

=1.2899×10-3

(ρ¿¿w)¿=0.072×√35415

=1.026×10−3

Assumimg ¿¿)min ,

s=201

300×1×1.026×10−3=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=30mm40mm2øs=2×8=16mmSmax=0.75d(1+cotα)=0.75×(1+0)=1245mmAs per the Clause -16.5.3 of IRC:112-2011,The links will not to be of greater spacing of following1/8 (perimeter of member)=487.5mmor 350mmOur provision of 300m (okay)

BY:

ABHISEK PANDA

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 belb.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:

1. IRC:112-2011; CODE OF PRACTICE FOR CONCRETE ROAD BRIDGES, INDIAN ROAD CONGRESS

2. IRC:6-2014; STANDARD SPECIFICATIONS AND CODE OF PRACTICE FOR ROAD BRIDGES; SECTION : II LOADS AND STRESSES

3. IRC:6-2010; STANDARD SPECIFICATIONS AND CODE OF PRACTICE FOR ROAD BRIDGES; SECTION : II LOADS AND STRESSES

4. IRC-SP-13; GUIDELINES FOR THE DESIN OF SMALL BRIDGES AND CULVERTS

BY:

ABHISEK PANDA

5. IS 456 : 2000; INDIAN STANDARD PLAIN AND REINFORCED CONCRETE CODE OF PRACTICE

6. IS-SP-16; DESIGN AIDS FOR REINFORCED CONCRETE TO IS : 456-1978

7. IS : 875 – 1987; INDIAN STANDARD CODE OF PRACTICE FOR DESIGN LOADS (OTHER THAN EARTHQUAKE) FOR BUILDINGS AND STRUCUTRES

8. IS-SP : 34(S&T)-1987; HANDBOOK ON CONCRETE REINFORCEMENT AND DETAILING

9. ESSENTIALS OF BRIDGE ENGINEERING; VICTOR D.J; OXFORD&IBH PUBLICATION

10. DESIGN OF BRIDGES; RAJU N.K; OXFORD&IBH PUBLICATION

11. R.C.C DESIGNS ; PUNMIA B.C, JAIN A.K AND JAIN A.K; LAXMI PUBLICATIONS

12. REINFORCED CONCRETE VOL.I; SHAH H.J; CHAROTAR PUBLICATIONS

13. LIMIT STATE DESIGN OF REINFORCED CONCRETE; VARGHESE P.C; PHI LEARNING

14. THEORY OF STRUCTURES; RAMAMRUTHAM S. AND NARAYAN R.; DHANPATRAI PUBLISHING COMPANY

15. HIGHWAY ENGINEERING; KHANNA S.K. AND JUSTO C.E.G.

16. 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