Question

What is the maximum tension (kN) on a cable carrying 2.5 kN/m uniform load if the lowest point of the cable is 40m and 3m away from the one of the support along the horizontal and vertical direction respectively? Round off to 3 decimal places.

Answer 1

Given a cable on which Uniformly distributed load (w) =  2.5KN/m is acting.

Lowest point on cable (h) = 40m and 3m away from one of the supports

As the lowest point is at midpoint on the cable, so 3m would be halfway on the cable, so the total length of the cable is (2 * 3) = 6m

therefore, Length of cable (L) = 6m

We need to find maximum tension on the cable

The maximum tension in the cable is the Resultant reaction at any support

So, first we need to determine vertical reaction and horizontal reaction at any of the support

• Vertical reaction

as the loading is uniform , so the support reaction will be:

V_A = V_C = (wL)/2 = (2.5 * 6)/2 = 7.5KN

• Horizontal reaction

As there is no horizontal force , so, H_A = H_C

Taking moment about point B (lowest point), and considering forces on left side, we get:

  

H_A = 0.28125 KN

Maximum tension in cable (T_max) = Resultant reaction at any support

Resultant reaction at A = R_A

R_A = 7.505 KN

Therefore, the maximum tension in the cable (T_max) = 7.505 KN