1. Internal shear force generated in a three hinged arch is always:-
a) 0
b) Infinite
c) Varies
d) Non zero value but remains constant
Explanation: Due to its geometry, shear force always comes out to be zero in 3 hinged arches.
2. What is the degree of indeterminacy of a fixed arch?
a) 1
b) 2
c) 3
d) 4
Explanation: It is indeterminate to 3 degrees.
3. What is the degree of indeterminacy of a two hinged arch?
a) 1
b) 2
c) 3
d) 4
Explanation: It is indeterminate to 2 degrees. It consists of two pin supports at both of its ends.
4. In real life, bending stresses are zero in a three hinge arch.
State whether the above statement is true or false.
a) True
b) False
Explanation: In real life, loads are moving which results in generation of bending stresses.
5. Cable is a tension member.
a) True
b) False
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.
6. The shape of the cable is a funicular polygon.
a) True
b) False
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.
7. 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
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.
8. 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) \(\frac{WL^2}{2H} \)
Explanation:
∑H = 0
HA = HB = 0
∑M = 0
H = \(\frac{WL^2}{8H}. \)
9.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) \(\frac{WL^2}{2H} \)
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 \(\frac{WL^2}{8H}. \)
10. 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) \(\frac{WL}{2}(\sqrt{1+\frac{L^2}{16H^2}}) \)
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 \(\frac{WL}{2}(\sqrt{1+\frac{L^2}{16H^2}}) \)