Answered: A cable with supports at the same…

A cable with supports at the same elevation has a span of 550 ft. The cable weighs 4 lb / ft and the sag at midspan is 100 ft. Determine The tension in the cable at midspan.

International Edition---engineering Mechanics: Statics, 4th Edition

4th Edition

ISBN:9781305501607

Author:Andrew Pytel And Jaan Kiusalaas

Publisher:Andrew Pytel And Jaan Kiusalaas

Chapter6: Beams And Cables

Section: Chapter Questions

Problem 6.72P: The cable AOB weighs 24 N/m. Determine the sag H and the maximum tension in the cable.

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Concept explainers

Combined Loading

Any functioning component, such as machine parts, structural members, pressure vessels, and so on, are all under the influence of more than one force. When anybody is under the influence of more than one force such as shear forces, bending moments, axial forces, torsion, and so on, then such body is said to be under combined loading. A very practical example of combined loading is the crankshaft of an internal combustion engine (IC engine). The crankshaft of the IC engine is under high rpm, which is caused due to the torsional forces. Due to the presence of piston, counterbalances, and internal imbalances, the crankshaft experiences lateral forces and due to the rise in temperature, the crankshaft undergoes thermal expansion, which pulls the crankshaft along the axial direction and lateral direction. The presence of unbalanced loads along the periphery of the crankshaft also induces a bending moment.

Question

A cable with supports at the same elevation has a span of 550 ft. The cable weighs 4 lb / ft and the sag at midspan is 100 ft. Determine The tension in the cable at midspan.

Expert Solution

Step 1

Given data:

Weight of the cable, W = 4 lb/ft.
Sag at the midspan of the cable, h = 100 ft.
Distance between the cable's support, a = 550 ft.

Draw the diagram of the cable,

The expression for the sag of the cable is,

$\begin{array}{rcl} h & = & (\frac{T_{o}}{W}) [\cosh (\frac{\frac{a}{2}}{\frac{T_{o}}{W}}) - 1] \\ 100 ft & = & (\frac{T_{o}}{W}) [\cosh (\frac{\frac{550 ft}{2}}{\frac{T_{o}}{W}}) - 1] \\ 100 ft & = & (\frac{T_{o}}{W}) [\cosh (\frac{275 ft}{\frac{T_{o}}{W}}) - 1] . . . . . . . . . . . . . . . . . . . . . . . . . (1) \end{array}$

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