Shear deflection of a cantilever under uniformly distributed load: For a cantilever of length L, cross-sectional area A, and shear modulus G, carrying w (force per unit length) uniformly over its entire length, what is the vertical deflection at the free end due solely to shear?

Introduction / Context:
Total deflection of beams consists of bending (flexural) and shear components. For short/deep members, shear deflection may be significant. This question isolates the shear part for a cantilever with uniformly distributed load (UDL).

Given Data / Assumptions:

• Cantilever length L; constant area A and shear modulus G.
• UDL intensity = w (force per unit length).
• Shear strain γ(x) = V(x) / (A G); shear-correction factor neglected as per the statement.

Concept / Approach:
For shear deflection, the incremental rotation due to shear is γ(x) = V(x)/(A G). The deflection at the free end equals the integral of γ(x) along the beam from the fixed end to the free end.

Step-by-Step Solution:
Shear force along the cantilever: V(x) = w (L − x).Shear slope: γ(x) = V(x) / (A G) = w (L − x) / (A G).Shear deflection at free end: δ_s = ∫_0^L γ(x) dx = (1 / (A G)) ∫_0^L w (L − x) dx.Evaluate integral: ∫_0^L (L − x) dx = [L x − x^2/2]_0^L = L^2 − L^2/2 = L^2/2.Therefore, δ_s = w L^2 / (2 A G).

Verification / Alternative check:
Dimensional check: w has units force/length; dividing by A G (force/area) yields length; multiplying by L^2 gives length → consistent.

Why Other Options Are Wrong:
w L^3 / (3 E I): that is a bending-deflection form (flexural, not shear).Other choices misplace coefficients or powers of L.

Common Pitfalls:
Mixing shear deflection with flexural deflection, or forgetting the linear variation of V(x) in a cantilever under UDL.

Final Answer:
w L^2 / (2 A G).