For a cantilever of length l and constant flexural rigidity E*I carrying a uniformly distributed load w per unit length over the entire span, what is the maximum deflection?

Introduction / Context:
Deflection formulas for standard load cases enable rapid checks of serviceability and stiffness. For a cantilever with full-span UDL, the free-end deflection is the maximum and serves as a benchmark for preliminary sizing and compliance with deflection limits.

Given Data / Assumptions:

• Cantilever length l, uniform load intensity w along 0 to l.
• Constant flexural rigidity E*I.
• Small deflections; linear elastic behavior.

Concept / Approach:
The classic closed-form result for a cantilever under full UDL is maximum deflection at the free end equal to δ_max = w * l^4 / (8 * E * I). This follows from integrating the moment-curvature relationship M / (E * I) = d^2y/dx^2 with boundary conditions y(0) = 0 and dy/dx(0) = 0 at the fixed end.

Step-by-Step Solution:
Moment at a section x from the fixed end: M(x) = w * (l - x)^2 / 2 (taking sign convention).Integrate twice to obtain slope and deflection, applying fixed-end boundary conditions.Evaluate y at x = l to get y(l) = w * l^4 / (8 * E * I).

Verification / Alternative check:
Compare with standard tables: simply supported with UDL gives 5 * w * l^4 / (384 * E * I), confirming that the cantilever case is much larger and specifically equals w * l^4 / (8 * E * I).

Why Other Options Are Wrong:

• 1/384 factors relate to simply supported beams, not cantilevers.
• w * l^3 / (3 * E * I) is a slope-like dimension or end-load deflection form, not correct for full UDL.
• 1/48 corresponds to different loading/support conditions.

Common Pitfalls:
Mixing up simply supported vs. cantilever results; using end-point load formulas for UDL cases; missing the l^4 dependence.

Final Answer:
w * l^4 / (8 * E * I)