Question

A cable carrying a load of 8 kN per meter run of horizontal span is stretched between the supports 100 m apart. If the supports are at the same level and the central dip is 8 m then the greatest tension in the cable is (in kN)_______.~

• A. 1312.44

Answer 1

The correct option is A 1312.44
$\text{Vertical reaction}=V=\frac{wl}{2}=\frac{8\times 100}{2}=400kN$

$\text{Horizontal reaction}=H=\frac{w{l}^2}{8h}=\frac{8\times {100}^2}{8\times 8}=1250kN$

Max (or) greatest tension

$T_{max}=\sqrt{V^2+H^2}=\sqrt{{400}^2+{1250}^2}$
$=1312.44kN$