Question

An unstiffened suspension cable carries a uniformly distributed load of 6.6 kN/m over a span of 20 m as shown in figure. The maximum tension in the cable is ___kN.

• A. 127.9

Answer 1

The correct option is A 127.9
Let the lowest point C be x meter from ‘A’
Cut the cable at supports and at ‘C’ as shown below.

Taking moments about A,
$\Rightarrow H\times 2=wx\times \frac{x}{2}$
$H=\frac{w{x}^2}{4}...\left(i\right)$

Taking moments about B,
$H\times 5=\frac{w(20-x{)}^2}{2}$
$H=\frac{w(20-x{)}^2}{10}...\left(ii\right)$

Equating (i) and (ii) , we get

$\frac{w{x}^2}{4}=\frac{w(20-x{)}^2}{10}$ $\Rightarrow x=7.75m$

$\therefore H=\frac{6.6\times {7.75}^2}{4}=99.1kN$

Maximum tension will be at B

$\therefore T_{max}=\sqrt{H^2+\left[w\right(20-x){]}^2}$

$=\sqrt{(99.1{)}^2+[6.6(20-7.75){]}^2}$

$=127.9kN$