What is the maximum tension (approximately) in the cable as shown in figure, if it carries a uniform horizontally distributed load of intesity 120 kN/m?

 

Solution:

Taking. ∑Ma = 0 ⇒ - V_B × 300 + 120 × 300 × \(\frac{300}{2}\) = 0

V_B = 120 × 150 = 18000 KN

∑F_y = 0 ⇒ V_A + V_B = 120 × 300

V_A = 18000 kN

As we know that cable is always subjected to axial tension, Bending moment at each point along the length of the cable is zero.

∴ BM_C [From Right Side] = 0

⇒ VB × 150 - H_B × 30 - 120 × 150 ×  \(\frac{150}{2}\) = 0

H_B = 45000 KN

∴ Maximum Tension in The cable 

\({T_{\max }} = \sqrt {V_B^2 + H_B^2} \)

\(T = \sqrt {{{\left( {18000} \right)}^2} + {{\left( {45000} \right)}^2}} = 48466.48KN = 48.5MN\)