Question

A cable carrying a load of 15 kN per metre run of horizontal span, is stretched between two supports 150 m apart. The supports are at the same level and the central dip is 10 m. The difference between the greatest and the least tensions in the cable will be

• A. 184.43 kN
• B. 147.42 kN
• C. 198.88 kN
• D. 166.31 kN

Answer 1

The correct option is B 147.42 kN

$Verticalreaction;V=\frac{wl}{2}=\frac{15\times 150}{2}=1125kN$

$Horizontalreaction;H=\frac{w{l}^2}{8h}=\frac{15\times {150}^2}{8\times 10}=4218.75kN$
$\therefore Maximumtension=T_{max}=\sqrt{V^2+H^2}=\sqrt{{1125}^2+{4218.75}^2}=4366.17kN$

$Minimumtension=T_{min}=H=4218.75kN$

$\therefore T_{max}-T_{min}=147.42kN$