Question

A cable is supported between two points 30 m horizontally apart. The left support is 3 m above the right support . The cable carries a load of 2 kN/m on the horizontal span . The lowest point of the cable is 6 m below the left support. The maximum tension in cable (in kN)

• A. 86.6 kN
• B. 51.4 kN
• C. 35.2 kN
• D. 62.3 kN

Answer 1

The correct option is D 62.3 kN


Let X = horizontal distance between A and C.
Consider equilibrium of part AC of the cable

$Resolvingvertically,V_a=2x$

$TakingmomentaboutA,H\times 6=\frac{2x^2}{2}$

$\Rightarrow H=\frac{x^2}{6}$

Now consider equilibrium of whole cable :

Taking moment about B, $V_a\times 30=H\times 3+2\times \frac{{30}^2}{2}$

$\Rightarrow 30V_a=3H+900$

$\Rightarrow 10V_a=H+300...\left(1\right)$

$Substituting{V}_{a}andH,$

$10\left(2x\right)=\frac{x^2}{6}+300$

$\Rightarrow x^2-120x+1800=0$

$\Rightarrow x=17.5736m$

$\Rightarrow V_a=2x=2\times 17.5736=35.1472kN$

$Also,H=\frac{x^2}{6}=\frac{{17.5736}^2}{6}=51.4719kN$

Hence, maximum tension,

$T_{max}=\sqrt{{35.1472}^2+{51.4719}^2}$

$T_{mzx}=62.327kN$