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The 3-hinged stiffening girder of a suspension bridge of span 120m is subjected to two point loads of 240KN and 300KN at distance 25m and 80m from the left end.

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 Answer
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diagram


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:

dgrm

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:

dgrm

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