Deflection of Frames MCQ with answer and solution

Concept

Equivalent beams

Step-by-step solution

1. Analysis of Beam 1 • Boundary Conditions: The left end is securely fixed. The right end features a roller support against a vertical wall. • Behavior under Vertical Loading (UDL): Because the rollers can slide vertically along the wall, this support cannot provide any vertical reaction force (R_y = 0). Being a roller, it also cannot resist any bending moment (M = 0). • Conclusion: For a purely vertical uniformly distributed load (UDL), Beam 1 behaves exactly like a standard cantilever beam of length (L). • (δ_max1 = (wL⁴)/(8EI)) (at the free roller end) • (θ_max1 = (wL³)/(6EI)) (at the free roller end) 2. Analysis of Beam 2 • Symmetry: The structure, support conditions (both outer ends fixed), and the UDL loading are completely symmetric about the central internal hinge. • Internal Forces at the Hinge: • By definition, an internal hinge cannot transmit any bending moment (M_hinge = 0). • Due to the perfect symmetry of the load and geometry, the internal shear force transferred through the hinge must also be zero (V_hinge = 0). Conclusion: Since M = 0 and V = 0 at the center, the internal hinge acts exactly like a free end. This splits Beam 2 into two independent halves, each behaving precisely like a cantilever beam of length L.• δ_max2 = (wL⁴)/(8EI) (at the central hinge) • θ_max2 = (wL³)/(6EI) (at the central hinge)

Explanation

Since both structural configurations simplify to the exact same cantilever behavior under the given loading: • δ_max1 = δ