This set of Structural Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Cable Subjected to a Uniform Distributed Load”.
1. Cable is a tension member.
a) True
b) False
View Answer
Explanation: Cable if a flexible member and hence it cannot resist bending moment. Cable cannot bear any compressive loading but it transfers the tensile loading through it. Thus, cable is a tensile member.
2. The shape of the cable is a funicular polygon.
a) True
b) False
View Answer
Explanation: Cable when loaded takes the shape of bending moment diagram that would be formed if the same set of the load is applied on to the simply supported beam of same length and properties. Thus, the shape of the cable is a funicular polygon.
3. The shape of the cable, when loaded with uniformly distributed load throughout the span is _____
a) Linear always
b) Parabolic always
c) Trapezoidal always
d) Linear or parabolic depending upon the intensity of loading
View Answer
Explanation: Cable when loaded takes the shape of bending moment diagram that would be formed if the same set of the load is applied on to the simply supported beam of same length and properties. Thus, the shape of cable loaded with uniformly distributed load throughout is parabolic.
4. The horizontal thrust produced at supports of cable when loaded with uniformly throughout the span is ____
a) \(\frac{WL^2}{32H} \)
b) \(\frac{WL^2}{16H} \)
c) \(\frac{WL^2}{8H} \)
d) \(\frac{WL^2}{2H} \)
View Answer
Explanation:
∑H = 0
HA = HB = 0
∑M = 0
H = \(\frac{WL^2}{8H}. \)
5. Minimum tension in the cable when loaded uniformly throughout the span is ____
a) \(\frac{WL^2}{32H} \)
b) \(\frac{WL^2}{16H} \)
c) \(\frac{WL^2}{8H} \)
d) \(\frac{WL^2}{2H} \)
View Answer
Explanation:
∑H = 0
HA = HB = 0
∑M = 0
H = \(\frac{WL^2}{8H} \)
Minimum tension in the cable equals to the horizontal thrust produced in the cable. Therefore, minimum tension in the cable is \(\frac{WL^2}{8H}. \)
6. Bending Moment at point D for the shown figure is ____
a) 0
b) \(\frac{WL^2}{8} \)
c) \(\frac{WL^2}{2} \)
d) \(\frac{WH^2}{2} \)
View Answer
Explanation: Bending Moment at any point in the cable loaded with uniformly distributed load is zero. Therefore, Bending Moment at point D is zero.
7. Maximum tension in the cable when loaded uniformly throughout the span is ____
a) \(\frac{WL}{2}(\sqrt{1+\frac{L^2}{2H^2}}) \)
b) \(\frac{WL}{2}(\sqrt{1+\frac{L^2}{4H^2}}) \)
c) \(\frac{WL}{2}(\sqrt{1+\frac{L^2}{8H^2}}) \)
d) \(\frac{WL}{2}(\sqrt{1+\frac{L^2}{16H^2}}) \)
View Answer
Explanation: Maximum tension in the cable is the square root of the sum of the squares of maximum vertical and maximum horizontal loads.
Maximum Vertical Load – V = \(\frac{WL}{2} \)
Maximum Horizontal Load – H = \(\frac{WL^2}{8H} \)
Therefore, Maximum Tension is \(\frac{WL}{2}(\sqrt{1+\frac{L^2}{16H^2}}) \)
8. Equilibrium equations used to analyse cable is _____
a) ∑H = 0 only
b) ∑H = 0 and ∑V = 0 only
c) ∑H = 0, ∑V = 0 and ∑M = 0
d) ∑V = 0 only
View Answer
Explanation: To analyse the frame completely, we are required with all the three of equilibrium equations i.e. ∑H = 0, ∑V = 0 and ∑M = 0.
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