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

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Examoogle · Accepted Answer

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

Explanation & Concepts:
The slope of a cantilever beam at the free end is determined by the principle of superposition, where the total slope is the sum of the slope due to a point load at the free end and the slope due to a uniformly distributed load (UDL) on the entire span. For a point load W, the slope at the free end is $\frac{Wl^2}{2EI}$, and for a UDL of intensity w, it is $\frac{wl^3}{6EI}$.

Working Principle: According to the Euler-Bernoulli beam theory, the slope $	heta$ is the integral of the curvature $\frac{M(x)}{EI}$. By evaluating the moment equations for a point load $M_P = W(l-x)$ and a UDL $M_u = \frac{w(l-x)^2}{2}$, and integrating over the beam length, the superposition of these individual slopes results in the total rotation at the free end.

Key Formulas: $	heta_P = \frac{Wl^2}{2EI}$ — Slope due to point load; $	heta_u = \frac{wl^3}{6EI}$ — Slope due to UDL

Key Points: Superposition applies because the structure behaves linearly elastically.; The slope is measured relative to the fixed end (where slope is zero).; Both loadings produce maximum slope at the free end of the cantilever.

Advantages: Simplifies complex problems into solvable elementary components.; Provides a direct analytical result for standard load cases.

Disadvantages: Applicable only within the linear elastic limit of the material.; Assumes small deflection theory.

Applications: Design of cantilever brackets and overhang beams.; Verification of structural software simulations.

Additional Info: Option B represents the deflection (not slope) at the free end.. Option D relates to a simply supported beam with UDL, where the coefficient 5/384 is used for deflection.. Ensure consistent units for load (N or kN), length (m or mm), and flexural rigidity (EI).

Answer: C is correct — The total slope at the free end is the summation of the individual slopes caused by the point load and the UDL, defined as $\frac{Wl^2}{2EI} + \frac{wl^3}{6EI}$.

Exam Tip: Always distinguish between slope (	heta) and deflection (\delta). Slopes involve $l^2$ and $l^3$, whereas deflections involve $l^3$ and $l^4$ for point and UDL respectively.

Examoogle · Answer

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

Examoogle · Answer

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

Examoogle · Answer

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

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