Difficulty: Easy
Correct Answer: y_free = (M * L^2) / (2 * E * I)
Explanation:
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:
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:
Final Answer:
y_free = (M * L^2) / (2 * E * I)
Discussion & Comments