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 $Q>0$ 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)?
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In pin-jointed frames, symmetry of loading and geometry often creates zero-force members. At joints with two non-collinear members and no external load, both members become zero-force members.
In the given frame, all joints are pin-connected and all members are rigid and weightless. The forces are applied at the midpoints of BC and GE. Because members BC and CE have equal lengths, the vertical load at the midpoint of CE creates symmetric moment effects about joint C. Step 1: Check member CD. At joint C, the vertical member BC carries a horizontal force $Q$ at its midpoint. This horizontal load at mid-height produces equal and opposite reactions at B and C. However, the horizontal load is entirely balanced by the member CB and the geometry does not require any load to be transferred into CD. Thus member CD carries no axial force. Step 2: Check member GH. The diagonal member GE carries the effects of the vertical load at the midpoint of CE. The geometry shows that joint G transmits forces only along member GE and the long diagonal to B. Member GH is purely horizontal, and no horizontal reaction is needed at G because the applied loading causes only inclined-force transfer through the diagonal GB–GE assembly. Thus GH remains unloaded. Step 3: Check remaining members AB and EF. Member AB carries load due to the horizontal reaction required to balance the force at the midpoint of BC. Member EF carries load due to the moment equilibrium created by the downward force applied at the midpoint of CE. Hence, both AB and EF are loaded members. Therefore, the only zero-force members are CD and GH. Final Answer: CD and GH
