DESIGN OF A BALANCED CALNTILEVER BRIDGE

Arrangement of Support:X/Y=4 to 5Advantage of Balanced Cantilever Bridge:1. Advantage of continuity2.Reduction in moments3.Not affected by settlement of supports

Railing Post Design:1.5ft1.5ft100*5=500lb150*5=750lb300*5=1500lbIf we design this railing post, as a cantilever beam or as a column, then the required reinforcement is too small.As per ACI code:-Least size of the post= 5x5-Longitudinal bar=4 #5(12mm) dia bar at four corners-Tie bars = #3(10mm) dia bar @ 10 c/cAs per RHD suggestion:- Size of the post= 8x8-Longitudinal bar=4 nos 1 (25 mm) dia bar at four corners-Tie bars = #3(10mm) dia bar @ 10 c/c

Slab design:6.256614Clear Span= 6.25-14/12=5.083t

Specification:-Length of Bridge=285.75 ft-Width of girder =14-c/c spacing of girder = 6.25 ft-Clear span= 6.25-14/12=5.083- Effective span, S = 5.083- 2(1/3*6/12)=4.75 ftA. Load calculation:a) Dead load= Self wt. Of slab + wearing course= 6.5/12*150 + 3/12*120=111.2 PlfAssume slab thickness=6.5b) Live Load= H20 S16

B. Moment calculation:a) Dead load Moment= W *S2/10= = #-ftb) Live Load Moment = 0.8*(S+2)/32*P20= #-ft(Reinforcement perpendicular to the traffic)Due to Continuous support, 0.8c) Impact factor = 50/(S+125) 0.3 =50/(4.75+125)=0.38 Use I.F.=0.3

Moment chart:DLMLLMIMTotal25127008103761C. Depth Calculation:Required Deth, d req= M/Rb = 4.2 t req = 4.2 +1 +0.5*(6/8) = 5.58Check with provided thickness, tpro > t req

D. Steel Calculation:1. Main Steel perpendicular to traffic, As= M/fs*j*d = / fs*j*dHere, d= tpro -1 - 0.5*(6/8) = 0.414 in2 Use 6/8 bar @ 11 c/c2. Distribution Steel (Parallel to traffic) :Percentage=220S 67%= 220/ 4.75 =100.9% >67%So, Distribution steel=0.67*As=0.67*0.414=0.277 in2 Use 5/8 bar @ 11 c/c

Calculation of Girder Profile:285.75142.875d= L/10 to L/12 Using, davg=L/11=114.3/11=10.456.15

A. For parabola (Upper Portion) :y2=c1x,c1=x/y2 =1.432/142.875=0.0142875Here, y=101.42875-100=1.42875x=142.875For x=57.15,y=0.904x=85.725,y=1.1067x=142.875,y=1.42875

B. For parabola (Lower portion) :y2=c2x, c2=y2/x =4.092/85.725 =0.1951Here, y=94.8-90.71=4.09x=57.15+28.575=85.725For, x=28.575,y=2.36

C. For parabola (lower portion at mid span) :y2=c3x,c3=y2/x =5.522/57.15 =0.533Here, y=96.23-90.71=5.52x=114.3/2=57.15For, x=57.15,y=5.52

Interior Girder of Suspended Span :10094.494.0593.07100.64100.917.845.257.15

Dead load Calculation:1. UDL:Slab= (6.5/12)*6.25*0.15=0.508 k/ftWearing course= (3/12)*6.25*0.15=0.156 k/ftFillet= 2*(1/2)*(6*6)/144)*0.15=0.0.038 k/ftGirder= (14/12)*5.2*0.15=0.9748 k/ft145.2Total= 1.677 k/ft47.92 k47.92 k57.151.677 k/ft

2. Upper Parabolic Dead Load (UPDL):Girder= 0.91*(14/12)*0.15=0.159 k /ft57.150.159 k/ft() ab=2.27k(5/12 )ab=3.79k

3. Lower Parabolic Dead Load (LPDL):Girder= 1.73*(14/12)*0.15=0.3028 k /ft57.150.3028 k/ft(1/12) ab=1.442k(1/4) ab=4.326k

4. Concentrated Load (CL):10098.494.0593.07100.647.845.257.15P2P3P16663.923.922100.91P1= [4.7*(10/12)+0.5*(6*6)/144]*6.25*0.15=3.789kP2= [3.92*(34/12)+3.42*(10/12)+0.5*(6*6)/144]*6.25*0.15=13.20k34P3= [6.09*(10/12)+2*0.5*(6*6)/144]*6.25*0.15=4.875k4.710106.09

47.92-1.677x2.44LoadShearBMUDLUPDLLPDLConc.DLShear & BM due to dead load (at a distance x from left support)

Live Load:Effective live load = H20 S16Load factor, S/5 for Smax=10S/5.5 for Smax=14S/7 for Smax=16Where, S= c/c spacing of girderSo, Load factor=6.25/5=1.25

H20 S16 loading,

Design of Interior Girder:Suspended span

Live Load:Effective Live Load = H20 S16 Load factor, S/5 for Smax=10 S/5.5 for Smax=14 S/7 for Smax=16Where, S= c/c spacing of girderLoad factor = 6.25/5=1.25H20 S16 Loading---Modified Loading---

Live Load Shear Calculation:It is proved that maximum shear occur when wheel 2is placed in the section123L1L257.15General equation for LL shear:L1414L2/LL1/L

At L1=0, L2=57.5, V0.0L=V0At L1=5.715, L2=57.5-5.715, V0.1L=V5.715V0.1L = [20*51.435+20 (51.435-14)]-51.43557.15*51.435Impact factor (I.F.) = 50/(125+S)= 50/ (125+57.15)=0.275

Design Shear Chart:

Span

0.0L

=0

0.1L

0.2L

0.3L

0.4L

0.5L

0.6L

0.7L

0.8L

0.9L

1.0L

=57.15

UDL

47.92

0

-47.920

UPDL

2.27

0.128

-3.788

LPDL

1.44

0.721

-4.326

Conc. DL

2.44

-2.44

-2.44

Total DL

54.072

-12.373

-58.474

LL Shear

35.10

-18.33

-35.1

Impact Shear

9.65

-5.041

-9.653

Design Shear

98.82

-35.744

-103.227

Live Moment Moment Calculation:It is provided that maximum moment occur when wheel 2is placed in the section.ab/Lab145 k20 k20 k14General Equation for Live Load Moment (LLM)

Moment of all other section will be determined in same way.a=5.715, b=51.44

Design Moment Chart:

Span

0.0L

=0

0.1L

0.2L

0.3L

0.4L

0.5L

0.6L

0.7L

0.8L

0.9L

1.0L

=57.15

UDL

0

684.65

0

UPDL

0

41.92

0

LPDL

0

36.06

0

Conc. DL

0

69.73

0

Total DL

0

832.36

0

LL moment

0

468.13

0

Impact moment

0

128.74

0

Design moment

0

1429.23

0

5.25.6255.8986.1276.366.5946.827.0597.3047.567.830.0L0.1L0.2L0.3L0.4L0.5L0.6L0.7L0.8L0.9L0.0L100100.286100.404100.495100.572100.636100.699100.756100.808100.857100.904

Steel Calculation:Maximum Moment at 0.5L=1429.225 k-ftM1=Rbd2=213.108*14*73.1282*1/12000=1329.57 k-ftt=79.128, d=79.128-6=73.128756.514t

Effect flange width:1)2)3)As= M/fs(d-t/2)Check T beamDesign beam.