Question
Download Solution PDFA cable of span 120 m and dip 10 m carries a load of 6 kN/m of horizontal span. The maximum tension in the cable is ______.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
VA + VB = 120 x 6 = 720 kn
Bending moment at point A, ∑MA = 0
VB × 120 – (6 × 120 × 120/2 ) = 0
VB = 360 KN
So,
VA = 360 KN
We know that bending moment in cable is zero everywhere.
In figure we assume a point C which is at bottom most part of cable.
Stand at C and look left,
Bending moment at C = 0
VA × 60 – HA × 10 – (6 × 60 × 60/2) = 0
360 × 60 - HA × 10 – (6 × 60 × 60/2) = 0
HA = 1080 KN
The maximum tension in the cable Tmax = \(\sqrt {V{a^2} + H{a^2}} \)
Tmax = \(\sqrt {{{360}^2} + {{1080}^2}} \)
∴ Tmax = 1138.42 kN
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