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Question: A suspension cable is supported at 2 points 2.5m apart. The left support us 2.5m above the right support. The cable is loaded with a uniformly distributed load of 10KN/m throughout the span.
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The maximum dip in the cable from the left support is 4m. Find the maximum and minimum tension in the cable.

Subject : Structural Analysis 1

Topic : Cable & Suspension Bridge

Difficulty : High

sa1(74) • 283 views
 modified 7 months ago by written 9 months ago by
1

To find l:

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

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

$l_{1} = \sqrt{\frac{4}{1.5}} * l_{2}$

$l_{1} = 1.6329 l_{2}$

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

$25=1.6329l_{2}+l_{2}$

$l_{2}= 9.495m$

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

$25=l_{1}+9.45$

$l_{1}=15.505m$

Support reaction calculation:

$V_{A}=w*l_{1}=10*15.505=155.05KN$

$v_{B}=w*l_{2}=10*9.495=94.95KN$

$H_{A}=\frac{w*l_{1}^2}{2*h_{1}}= \frac{10(15.505)^2}{2*4}=300.51KN$

$H_{B}=\frac{w*l_{2}^2}{2*h_{2}}= \frac{10(9.495)^2}{2*1.5}=300.51KN$

$T_{A}=\sqrt{V_{a}^2+H^2}$

$T_{A}=\sqrt{(155.05)^2+(300.51)^2}$

$T_{A}=338.15KN$

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

$T_{B}=\sqrt{(94.95)^2+(300.51)^2}$

$T_{B}=315.15KN$

$T_{max}=338.15KN$

$T_{min}=300.51KN$