Question
Download Solution PDFWhat 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?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFSolution:
Taking. ∑Ma = 0 ⇒ - VB × 300 + 120 × 300 ×
VB = 120 × 150 = 18000 KN
∑Fy = 0 ⇒ VA + VB = 120 × 300
VA = 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.
∴ BMC [From Right Side] = 0
⇒ VB × 150 - HB × 30 - 120 × 150 ×
HB = 45000 KN
∴ Maximum Tension in The cable
Last updated on Jul 2, 2025
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