A cantilever beam with point load at free end and udl on entire span, deflection will be

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Examoogle · Accepted Answer

Option B: $\frac{Wl^3}{3EI} + \frac{wl^4}{8EI}$

Explanation & Concepts:
The total deflection of a cantilever beam under multiple loads is calculated by the principle of superposition. The deflection at the free end due to a point load $W$ is $\frac{Wl^3}{3EI}$ and the deflection due to a uniformly distributed load (udl) $w$ is $\frac{wl^4}{8EI}$.

Working Principle: According to the principle of superposition, when a beam is subjected to multiple independent loads, the resultant deflection at any point is the algebraic sum of the individual deflections produced by each load acting independently. For a cantilever beam, the free-end slope and deflection depend on the load intensity, span, and flexural rigidity $EI$.

Key Formulas: $\delta_{point} = \frac{Wl^3}{3EI}$ — Deflection due to point load at free end; $\delta_{udl} = \frac{wl^4}{8EI}$ — Deflection due to udl over entire span

Key Points: The cantilever beam follows the Euler-Bernoulli beam theory.; The principle of superposition is valid for linear elastic materials.; The maximum deflection for both load cases occurs at the free end.

Advantages: Simplifies analysis of complex loading patterns.; Allows modular calculation of structural responses.

Disadvantages: Valid only within the limit of proportionality (Hooke's Law).; Does not account for non-linear material behavior or large deformations.

Applications: Structural design of building overhangs.; Analysis of bracket supports and machine components.

Additional Info: The deflection formula $\frac{wl^4}{8EI}$ for UDL represents the total deflection at the tip of the cantilever.. Option D is the formula for a simply supported beam with central point load and UDL, which is not applicable to a cantilever.

Answer: B is correct — The total deflection is the sum of the tip deflection caused by the point load and the tip deflection caused by the uniformly distributed load.

Exam Tip: Always remember that for simply supported beams, denominators are usually larger (e.g., $48$ and $384$) compared to cantilevers ($3$ and $8$) due to boundary conditions.

Examoogle · Answer

Option A: $\frac{Wl^2}{16EI} + \frac{wl^3}{24EI}$

Examoogle · Answer

Option C: $\frac{Wl^2}{2EI} + \frac{wl^3}{6EI}$

Examoogle · Answer

Option D: $\frac{Wl^3}{48EI} + \frac{5wl^4}{384EI}$

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