Question

The shear force at the fixed end of a cantilever beam of length \( l \) carrying a uniformly distributed load \( w \) per unit length is

• A. zero
• B. \( \dfrac{wl}{4} \)
• C. \( \dfrac{wl}{2} \)
• D. \( wl \)

Hint:

For a cantilever with UDL, maximum shear force always occurs at the fixed end.

Answer 1

The Correct Option is D

Step 1: Take a section at distance \( x \) from the free end; the shear force there equals the load carried by the portion of the beam from the free end up to that section.
Step 2: Integrating the uniform load intensity \( w \) over this length gives \( V(x) = \displaystyle\int_0^{x} w\,dx = wx \).
Step 3: At the fixed end, \( x = l \), so the shear force is
\[ V(l) = wl \]

Answer 2

Consider the whole cantilever as a single free body. The uniformly distributed load \( w \) per unit length acting over the full span \( l \) is statically equivalent to one resultant point load of magnitude \( W = wl \) acting at the midpoint of the beam. Since the free end carries no support, the fixed end alone must balance this resultant. Checking each option against this equilibrium requirement:

1. Zero: A zero reaction cannot balance a nonzero resultant downward load \( wl \), so vertical equilibrium of the beam would be violated.
2. \( \dfrac{wl}{4} \): This value is only a quarter of the resultant load \( wl \) that must be balanced, so the sum of vertical forces on the beam would not be zero if the reaction were only this large.
3. \( \dfrac{wl}{2} \): This value is only half of the resultant load \( wl \); a reaction of this size still leaves half the load unbalanced, violating equilibrium.
4. \( wl \): Setting the vertical reaction at the fixed end equal to \( wl \) exactly cancels the downward resultant load, satisfying \( \sum F_y = 0 \) for the whole beam.

Only a fixed-end reaction of \( wl \) keeps the entire cantilever in vertical equilibrium, and this reaction is also the shear force carried at that section.

Therefore, the correct answer is \( wl \).