Deflection of a cantilever with an applied end moment A cantilever of length L is subjected to a constant bending moment M at its free end (no transverse load). If the flexural rigidity is EI, what is the vertical deflection of the free end?

Introduction / Context:The slope-deflection behavior of a cantilever under an applied end moment is a basic case in elastic beam theory. With pure end moment and no transverse load, the shear force is zero along the span, and curvature is constant, allowing a straightforward integration to obtain slope and deflection.

Given Data / Assumptions:

• Prismatic beam with constant E and I.
• Cantilever length = L, fixed at one end, free at the other.
• Applied end moment = M (clockwise or counterclockwise sign as per convention).

Concept / Approach:From elastic curve: M(x) = constant = M. Curvature κ = M / (E * I). Then dy' = M / (E * I) under standard sign convention. Integrating the curvature twice with boundary conditions at the fixed end (y = 0 and dy/dx = 0 at x = 0) yields closed-form expressions for slope and deflection.

Step-by-Step Solution:Write differential equation: E * I * d^2y/dx^2 = M.Integrate once: E * I * dy/dx = M * x + C1; apply slope at fixed end (x = 0): dy/dx = 0 ⇒ C1 = 0.Integrate again: E * I * y = (M * x^2)/2 + C2; apply deflection at fixed end (x = 0): y = 0 ⇒ C2 = 0.At free end x = L: y_free = (M * L^2) / (2 * E * I).

Verification / Alternative check:Deflection varies with L^2, and slope at the free end is θ_free = (M * L) / (E * I). Units: M has forcelength, so y_free has length, confirming dimensional consistency.

Why Other Options Are Wrong:(M * L)/(E * I): This is the slope, not the deflection.(M * L^3)/(3 * E * I) and other powers: Correspond to different load cases (e.g., end point load or UDL), not a pure moment.

Common Pitfalls:

• Confusing the pure moment case with an end load or uniformly distributed load.
• Forgetting the boundary conditions at the fixed end, leading to wrong constants.

Final Answer:y_free = (M * L^2) / (2 * E * I)