Question

An L-shaped elastic member ABC with slender arms AB and BC of uniform cross-section is clamped at end A and connected to a pin at end C. The pin remains in continuous contact with and is constrained to move in a smooth horizontal slot. The section modulus of the member is same in both the arms. The end C is subjected to a horizontal force P and all the deflections are in the plane of the figure. Given the length AB is 4a and length BC is a (BC vertical, hanging down from B, with C at the bottom subjected to horizontal force P), the magnitude and direction of the normal force on the pin from the slot, respectively, are

Accepted Answer

The correct answer is (A) 3P/8, and downwards. The slot provides one unknown vertical reaction R_C at C, making this a 1°-indeterminate frame. Taking moments of the loads at C about a cut at distance x up from C along BC gives bending moment M_BC=Px (R_C is collinear with BC, contributing no moment there). Continuing into AB (distance y from B): M_AB=aP+yR_C. Applying Castigliano's theorem (∂U/∂R_C=0, since vertical deflection at C must be zero): ∫₀^4a(aP+yR_C)y dy=0 → 8a³P+(64/3)a³R_C=0 → R_C=−3P/8. The negative sign means the actual reaction is 3P/8 directed downward (opposite the assumed-upward direction).

Explanation