GATE ME 2022 Set 2 — Question 47
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Engineering Mechanics → Statics → Frames & Machines
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Question
The lengths of members BC and CE in the frame shown in the figure are equal. All the members are rigid and lightweight, and the friction at the joints is negligible. Two forces of magnitude are applied as shown, each at the mid-length of the respective member on which it acts.Which one or more of the following members do not carry any load (force)?
Correct answer
(B) CD; (D) GH
Solution
To determine which members carry no load, we analyze the equilibrium of the frame.
1.Member Analysis: Members BC and CE are subjected to external forces at their midpoints. This means they act as beams and will transmit forces and moments to the joints B, C, and E.
2.Joint C: Joint C connects members BC, CD, and CE. Member CD is a two-force member connected to a pin support at D. For member CD to carry a load, there must be a vertical reaction at support D.
3.Joint G: Joint G connects members BG, EG, and GH. Member GH is a two-force member connected to a pin support at H. For member GH to carry a load, there must be a horizontal reaction at support H.
4.Symmetry and Loading: Given that the lengths and the forces are applied identically (one horizontally on the vertical member BC and one vertically on the horizontal member CE), the structure exhibits a specific anti-symmetry. By analyzing the global equilibrium and the sub-assemblies, it can be shown that the reactions at supports D and H are zero ( and ).
5.Conclusion: Since the reactions at D and H are zero, the internal forces in members CD and GH must also be zero. Members AB and EF, being connected to the primary supports A and F that balance the external loads , will carry non-zero loads.
Therefore, members CD and GH do not carry any load. Correct options are (B) and (D).Continue learning with Success Tracker
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