Find the SF and BM for the girder at a distance of 40m from the left end. The supporting cable has a central dip of 12m. Find, also the maximum tension in the cable and draw the BMD for the girder.

Subject : Structural Analysis 1

Topic : Cable & Suspension Bridge

Difficulty : High

1.To find support reaction:

Considering the stiffening girder as a suspension beam supporting the given external load system,

$ \Sigma M_{a}=0 $

$240 *25+300*80-V_{B}*120=0 $

$V_{B}=250KN $

$\Sigma F_{Y}=0 $

$V_{A}-240-300+250=0 $

$V_{A}=290KN $

2.To find H:

Beam moment at C$= BM_{c}=290*60-240*35) $

$BM_{c}=9000KNm $

Beam moment under the 240KN load=$(290*25)=7250KNm $

Beam moment under the 300KN load= $(250*40)=1000KNm $

Horizontal reaction at each end of the cable,$ H= \frac{M_{c}}{h} $

$H=\frac{9000}{12}=750KN $

3.Find equivalent UDL:

Let $ w_{e} / unit run $= UDL transferred to the cable.

$H=\frac{W_{e}*l^2}{8*h} = \frac{W_{e}*120^2}{8*12}= 750 $

$W_{e}=5KN/m $

Each vertical reaction for the cable= $V= \frac{W_{e}*l}{2} = \frac{5*120}{2} =300KN $

Max tension in the cable=$ T_{max}= \sqrt{V^2+H^2}= \sqrt{300^2+750^2} $

$T_{max}=807.8KN $

4. S.F calculation :

For the girder, S.F at any section = $SF_{x} = (Beam shear- H tan\theta) $

For the cable at any point,$ tan\theta=\frac{4h}{l^2}*(l-2x) $

At 40m from the left end,$ tan\theta = \frac{4*12}{120^2}*(120-2*40)= \frac{2}{15}=0.1333 $

Beam shear at 40m from left end $= (290-240) =50KN $

Actual SF at 40m from the left end $ =(50-750* \frac{2}{15})= -50KN$

5.BM calculation:

For the girder, BM at any section,$ M=(Beam moment-H_{mom}) = Beam moment – H_{y} $

Beam moment at 40m from the left end $= (290*40-240*15)=8000KNm $

At 40m from the left end, for the cable, $y=\frac{4h}{l^2}*x(l-x)=\frac{4*12}{120^2}*40*80= \frac{32}{3}m $

Actual BM at 40m from left end $=(Beam moment-H_{moment})=(8000-750*\frac{32}{3})=0 $

6.Actua lBMD for the girder:

Dip of the cable 25m from left end $=\frac{4h}{l^2}*x(l-x)=\ frac{4*12}{120^2}*25*95=7.92m $

Actual BM at 25m from left end $=(Beam moment-H_{moment})=(7250-750*7.92)=1310KNm $

Dip of the cable 80m from the left end $= \frac{4*12}{120^2}*80*40 =\frac{32}{3}m $

Actual BM at 80m from left end $=(1000-750* \frac{32}{3})=2000KNm $

7.Actual BMD for the stiffening girder: