Question

A suspension cable of 100 m span has a dip of 10 m. It is stiffened by a two-hinged girder whose weight is 20 kN/m. Determine the maximum tension in the cable if a point load of 500 kN rolls over the girder.

• A. 3125 kN
• B. 3365.73 kN
• C. None of the above
• D. 1250 kN

Answer 1

The correct option is B 3365.73 kN

To find w on the cable
We know that in a two-hinged stiffening girder the live load is transmitted to the cable as equivalent u.d.l over the entire span.

Therefore, total weight on the cable per meter length.

$w=\frac{Total(D.L+L.L)}{Span}$

$=\frac{20\times 100+500}{100}=25kN/m$

Step 2: To find H, the horizontal pull

$H=\frac{w{l}^2}{8y_c}=\frac{25\times 100\times 100}{8\times 10}=3125kN$

Vertical reaction at supports

$V_A=V_B=\frac{25\times 100}{2}=1250kN$

Maximum tension in the table
$T_{max}=\sqrt{H^2+V^2}=\sqrt{{3125}^2+{1250}^2}=3365.73kN$