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A suspension cable of horizontal span 95m is supported at two different levels. The right support is higher than the left support by 4m.

The dip to the lowest point of the cable below the left support is 5m. The cross section area of the cable is $3500\ mm^2$. Find the uniformly distributed load that can be carried by the cable if the maximum stress is limited to $600\ N/mm^2$.


Subject : Structural Analysis 1

Topic : Cable & Suspension Bridge

Difficulty : Low

1 Answer
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diagram

To find length:

$\frac{l_{1}}{l_{2}}= \frac{h_{1}}{h_{2}} $

$l_{1} = \sqrt{\frac{h_{1}}{h_{2}}} * l_{2} $

$ l_{1} = \sqrt{\frac{5}{9}} * l_{2} $

$ l_{1} = 0.7453 l_{2} $

$l=l_{1}+l_{2} $

$95=0.7453l_{2}+l_{2} $

$l_{2}= 54.43m $

$l=l_{1}+l_{2} $

$95=l_{1}+54.43 $

$l_{1}=40.57m $

$V_{A}=w*l_{1}=P*40.57=40.57PN $

$V_{B}=w*l_{2}=P*54.43=54.43PN $

$H=\frac{w*l_{2}^2}{2*h_{2}}= \frac{P(54.44)^2}{2*9}=164.65P $

$T_{max}=T_{B}=\sqrt{V_{B}^2+H^2} $

$T_{max}=\sqrt{(54.43P)^2+(40.57P)^2} $

$T_{max}=173.413P $

$\sigma=\frac{P}{A} $

$P=\sigma * A $

$173.413P=600*3500 $

$P=\frac{600*3500}{173.413} $

$P=12109N/m $

$P=12.10KN/m $

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