Concept explainers
A horizontal rigid bar ABC is pinned at end A and supported by two cables at points B and C. A vertical load P = 10 kN acts at end C of the bar. The two cables are made of steel with a modulus elasticity E = 200 GPa and have the same cross-sectional area. Calculate the minimum cross-sectional area of each cable if the yield stress of the cable is 400 MPa and the factor of safely is 2.0. Consider load P only; ignore the weight of bar ABC and the cables.
The minimum cross-section area of each cable.
Answer to Problem 2.4.4P
The minimum cross-section area of each cable is
Explanation of Solution
The vertical load is
The below figure represents the free body diagram of the cable.
Figure-(1)
Here, the length of the portion AB is
Write the expression for the length of the cable 1.
Here, length of the cable 1 is
Write the expression for the length of the cable 2.
Here, length of the cable 2 is
Write the expression for the angle for cable 1 which makes from horizontal.
Write the expression for the angle for cable 1 which makes from horizontal.
Write the expression for the moment about the point A.
Here, the tension in cable 1 is
The below figure represents the displacement diagram.
Figure-(2)
Here, displacement for cable 1 is
Write the expression for the displacement assuming this is a small displacement.
Write the expression for the displacement of portion AB.
Write the expression for the displacement of portion AC.
Write the expression for the deflection in cable 1.
Here, area of the cross-section is
Write the expression for the deflection in cable 2.
Here, the deflection in cable 2 is
Write the expression for the stress equation considering factor of safety for cable 1.
Here, the factor of safety is
Write the expression for the stress equation considering factor of safety for cable 2.
Calculation:
Substitute
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Conclusion:
The minimum cross- section area is calculated by equating modulus of stress, elasticity, cable and factor of safety,height and length of cable.
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Chapter 2 Solutions
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