Cantilever under pure end moment: A cantilever of length L is subjected to a constant bending moment M applied at the free end (no shear). What is the vertical deflection at the free end?
-
AM L / (E I)
-
BM L^2 / (2 E I)
-
CM L^2 / (3 E I)
-
DM L^3 / (3 E I)
-
EZero
Answer
Correct Answer: M L^2 / (2 E I)
Explanation
Introduction / Context:Different loadings on cantilevers produce characteristic slope and deflection formulas. A pure moment at the free end creates a uniform bending moment along the length, simplifying integration.
Given Data / Assumptions:
- Cantilever length L; flexural rigidity E I constant.
- Applied end moment M at the free end; no transverse load.
- Small deflection; Euler–Bernoulli beam theory (plane sections remain plane).
Concept / Approach:The differential equation is E I * y'(x) = M (constant). Integrating twice with fixed-end boundary conditions (zero slope and zero deflection at the built-in end) yields the slope and deflection distributions.
Step-by-Step Solution:E I y' = M → y' = M / (E I).Integrate: y' → y' dx → y' = (M / (E I)) x + C1.Boundary at fixed end x = 0: slope y'(0) = 0 ⇒ C1 = 0.Integrate: y = (M / (E I)) x^2 / 2 + C2.Boundary at fixed end x = 0: y(0) = 0 ⇒ C2 = 0.Deflection at free end: y(L) = M L^2 / (2 E I).
Verification / Alternative check:The slope at the free end is θ(L) = M L / (E I); differentiating the deflection confirms consistency.
Why Other Options Are Wrong:M L / (E I): that is the slope, not the deflection.Other denominators (3 E I) arise for different load cases (e.g., point load).Zero: a pure end moment definitely causes rotation and deflection.
Common Pitfalls:Applying the point-load formulas instead of the constant-moment case.
Final Answer:M L^2 / (2 E I).