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: Understanding the loading condition.
A cantilever beam is fixed at one end and free at the other. The uniformly distributed load \( w \) acts over the entire length \( l \) of the beam.
Step 2: Calculating the total load.
The total load acting on the beam is: \[ W = w \times l \]
Step 3: Shear force at the fixed end.
The fixed end must balance the entire applied load. Hence, the shear force at the fixed end is equal to the total load: \[ V = wl \]
Step 4: Conclusion.
The shear force at the fixed end of the cantilever beam is \( wl \).

Answer 2

For a cantilever beam under a uniformly distributed load, the shear force at any section equals the total load carried by the beam segment beyond that section (toward the free end). Since the entire load acts between the fixed end and the free end, the shear right at the wall equals the full load on the beam. Let's test each option:

1. Zero: This would mean the fixed support carries no vertical reaction at all, which is impossible since the entire distributed load \( wl \) has nowhere else to go if not balanced at the wall.
2. \( \dfrac{wl}{4} \): This would only be correct if just a quarter of the beam's length carried the load, or if the load were somehow shared with another support; neither condition applies to a single cantilever fixed at only one end.
3. \( \dfrac{wl}{2} \): This value would apply to a simply supported beam with two supports sharing the load equally, not a cantilever, since a cantilever has only one support carrying the full load.
4. \( wl \): Since the fixed end is the only support, it must supply a reaction equal to the entire distributed load acting over the full length \( l \), giving a shear force of \( wl \) right at the wall. This matches the physical requirement of overall vertical equilibrium of the beam.

Only a reaction (and hence shear force) of \( wl \) satisfies equilibrium for a beam with a single fixed support carrying the entire distributed load.

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