Deflection of a cantilever with uniformly distributed load: Maximum deflection in terms of W = w * l

Difficulty: Medium

Correct Answer: W * l^3 / (8 * E * I)

Explanation:


Introduction / Context:
Deflection formulas are central to serviceability checks. For a cantilever of length l with a uniformly distributed load w (per unit length), maximum deflection occurs at the free end. The problem defines W = w * l for convenience, and asks for the deflection in that form.


Given Data / Assumptions:

  • Cantilever length = l.
  • UDL intensity = w, total load W = w * l.
  • Young’s modulus = E, second moment of area = I.
  • Small deflection theory, prismatic beam.


Concept / Approach:
Standard result for a cantilever with UDL: y_max = w * l^4 / (8 * E * I). Substituting W = w * l gives y_max = (W * l^3) / (8 * E * I). This is obtained by integrating beam differential equations or using tabulated results.


Step-by-Step Solution:

Start with y_max = w * l^4 / (8 * E * I).Given W = w * l → w = W / l.Substitute: y_max = (W / l) * l^4 / (8 * E * I) = W * l^3 / (8 * E * I).Hence, the correct expression is W * l^3 / (8 * E * I).


Verification / Alternative check:
Dimensional check: numerator has force * length^3; denominator E * I has force * length^2; result is length, as required.


Why Other Options Are Wrong:
W * l^3 / (3 * E * I) is the cantilever tip load case factor; the 48 denominators belong to simply supported beam formulae; W * l^2 / (8 * E * I) has wrong power of l.


Common Pitfalls:
Mixing up the 8 vs 3 vs 48 factors; forgetting to substitute W = w * l correctly; using simply supported results for cantilever problems.


Final Answer:
W * l^3 / (8 * E * I)

More Questions from Strength of Materials

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion