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

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 $