BRIDGE DATA 3:EFFECTIVE SPAN OF Tee BEAM= 10 m Width of carriage way= 7.5 mThickness of wearing coat =80 mmSpacing of main girders =2.5 mWidth of kerb =0.5 mWidth of footpath =1 mThickness of deck slab =250 mmModular ratio =10Number of main Girders =4Stress in concrete (compression) =10 Stress in steel (tension) = 200 M30 Grade and Fe-415 Grade HYSD bars.

As width of carriage way is 7.5m, number of proposed lanes are 2.Therefore LIVE LOAD COMBINATION: ONE LANE OF 70R OR TWO LANES OF CLASS A

6.5 DESIGN BASED ON IRC: 21-20006.5.1 Load calculation for grillage model

-1 (i) Dead load: Self weight

(ii)SIDL a) Wearing coat (80 mm) = -1.84 kN/m2 b) Weight of kerb = -7.8 kN/m2 c) Weight of crash barrier = -14.86 kN/m2 d) Pedestrian load = -3.889 kN/m26.5.2 Sectional properties of longitudinal and transverse members:

Fig 6.33: Cross Section of End Longitudinal girder A=1.037*106 mm2, Yc=914 mm, Ixx =0.2466*10 12 mm4, Iyy=108 mm4 Izz=0.1233 *10 12 mm4

Fig 6.34: Cross Section of Intermediate Longitudinal girder A=0.975*106 mm2, Yc=906 mm, Ixx =0.240*10 12 mm4, Iyy=108 mm4 Izz=0.120 *10 12 mm4

6.5.3 Design of reinforcement for external girder at L/2 6.5.3.1 Design of B.M reinforcement for external girder at mid span:Table 6.35: B.M due to DL and SIDL for external longitudinal girder at mid spanLoad type BENDING MOMENT (kN-m)

DEAD LOAD243

SIDL282

Table 6.36: B.M due to Live load for external longitudinal girder at mid spanLoad typeBENDING MOMENT(kN-m)BENDING MOMENT WITH IMPACT FACTOR (kN-m)

CLASS A 327397

70R TRACKED9831081

70 R WHEELED690862

Mmax = (Dead load +SIDL) B.M +max. of (class A or 70R Tracked or 70R Wheeled)B.M =243+282+1081 =1606 kN-mDesign of section: Effective depth d = = = 1041mm.Ast = = = 8676 mm2 Provide 12 # of 32 ASt provided =9650 mm2 Effective depth provided= 1140 mm.Stress check: Calculation of neutral axis

hc = 273 mm Bf *Df*(hc-Df) +Bw*(hc-Df)*(hc - ) =m*Ast*(d-hc) 1) Compressive stress in concrete: fc = * = * fc = 6.29 < 10 2) tensile stress in steel fst = = fst = 141< (200 )6.5.3.2 Design of Shear reinforcement for external girder at L/2Table 6.37: S.F due to DL and SIDL for external longitudinal girder at mid spanLoad type SHEAR FORCE (kN)

DEAD LOAD54

SIDL92

Table 6.38: S.F due to Live load for external longitudinal girder at mid spanLoad typeSHEAR FORCE (kN)SHEAR FORCE WITH IMPACT FACTOR (kN)

CLASS A 107129

70R TRACKED341375

70 R WHEELED240300

Vmax = (DEAD LOAD +SIDL)S.F +Max. OF (CLASS A OR 70RTRACKED OR 70R WHEELED)S.F =54+92+375 = 521KNCheck for shear stress = = = > = < =Design of shear reinforcement for Vs.:Vs =Vu - *b*d = 512*103- 0.589*350*1140 =276 kN = = =242 mm2 Adopt 4 legged 10 @ 200 mm c/c. 6.5.4 Design of reinforcement for external girder at l/4:6.5.4.1 Design of B.M reinforcement for external girder at quarter span Table 6.39: B.M due to Dead load and SIDL for external longitudinal girder at quarter spanLoad type BENDING MOMENT (kN-m)

DEAD LOAD181

SIDL215

Table 6.40: B.M due to Live load for external longitudinal girder at quarter spanLoad typeBENDING MOMENT(kN-m)BENDING MOMENT WITH IMPACT FACTOR (kN-m)

CLASS A 253307

70R TRACKED785863

70 R WHEELED553691

Mmax = (Dead Load +SIDL)B.M +Max. OF (class A OR 70RTracked OR 70R Wheeled)BM =181+215+863 =1259 kN-mDesign of section: Effective depth d = = = 922 mm.Ast = = = 6211 mm2 Provide 8 # of 36 ASt provided =8143 mm2 Effective depth provided= 1140 mm.

Stress check: Calculation of neutral axis

hc = 204 mm Bf *Df*(hc-Df) +Bw*(hc-Df)*(hc - ) =m*Ast*(d-hc) 1) Compressive stress in concrete: fc = * = * fc = 4.35 < 10 2) tensile stress in steel fst = = fst = 125< (200 )6.5.4.2 Design of Shear reinforcement for external girder Table 6.41: S.F due to Dead load and SIDL for external longitudinal girder at quarter spanLoad type SHEAR FORCE (kN)

DEAD LOAD109

SIDL104

Table 6.42: S.F due to Live load for external longitudinal girder at quarter spanLoad typeSHEAR FORCE (kN)SHEAR FORCE WITH IMPACT FACTOR (kN)

CLASS A 119144

70R TRACKED344378

70 R WHEELED253316

Vmax = (DEAD LOAD +SIDL)S.F +Max. OF (CLASS A OR 70RTRACKED OR 70R WHEELED)S.F =109+104+378 = 591KNCheck for shear stress = = = > = < =Design of shear reinforcement for Vs.Vs =Vu - *b*d = 591*103- 0.589*350*1140 =355 kN = = =253mm2Adopt 4 legged 10 @ 150 mm c/c.

6.6 DESIGN BASED ON IRC: 112-20116.6.1 LOAD CALCULATION FOR GRILLAGE MODEL

-1 (i) Dead load: Self weight (ii)SIDL a) Wearing coat (80 mm) = -1.84 kN/m2 b) Weight of kerb = -7.8 kN/m2 c) Weight of crash barrier = -14.86 kN/m2 d) Pedestrian load = -3.889 kN/m2LOAD COMBINATION =1.35*(DL) +1.75 *(SIDL)+1.5*(LIVE LOAD)6.6.2 Sectional properties of longitudinal and transverse members:

Fig 6.35: Cross Section of End Longitudinal girder A=1.037*106 mm2, Yc=914 mm, Ixx =0.2466*10 12 mm4, Iyy=108 mm4 Izz=0.1233 *10 12 mm4

Fig 6.36: Cross Section of Intermediate Longitudinal girder A=0.975*106 mm2, Yc=906 mm, Ixx =0.240*10 12 mm4, Iyy=108 mm4 Izz=0.120 *10 12 mm4

6.6.3 Design of reinforcement for external girder at L/2 6.6.3.1 Design of B.M reinforcement for external girder:Table 6.43: B.M due to Dead load and SIDL for external longitudinal girder at mid spanLoad type BENDING MOMENT (kN-m)

DEAD LOAD327

SIDL494

Table 6.44: B.M due to Live load for external longitudinal girder at mid spanLoad typeBENDING MOMENT(kN-m)BENDING MOMENT WITH IMPACT FACTOR (kN-m)

CLASS A 491596

70R TRACKED14701617

70 R WHEELED9381172

Mmax = (Dead load +SIDL) B.M +max. of (class A or 70R Tracked or 70R Wheeled)B.M =327+494+1617 =2438 kN-mDesign of section: Step1: Assume neutral axis lies in the flange =1.2 - =1.2 - =0.07x = 77 mm