Cantilever with uniformly distributed load (UDL): shear force at the fixed end\n\nA cantilever beam of length l carries a uniformly distributed load of intensity w per unit length over the entire span. What is the shear force at the fixed end?

Introduction / Context:
Cantilever beams are common in balconies, crane jibs, and machine tool overhangs. For uniform loading, knowing reactions, shear, and moment at the fixed support is essential to size sections and check safety.

Given Data / Assumptions:

• Cantilever length l, fixed at one end and free at the other.
• Uniformly distributed load w (force per unit length) over the full span.
• Linear elastic beam theory; static equilibrium.

Concept / Approach:
The resultant of a UDL of intensity w on length l is a single force W = w * l acting at the centroid of the distribution, located at midspan from the free end (i.e., l/2 from either end). Shear at the fixed support equals this resultant in magnitude but opposite in direction.

Step-by-Step Solution:
Resultant load: W = w * l.Support shear at fixed end must balance vertical loads: V_fixed + (−W) = 0.Hence V_fixed = W = w * l.(For completeness, fixed-end moment M_fixed = w l^2 / 2.)

Verification / Alternative check:
Construct shear diagram: starting at fixed end with +w l, shear decreases linearly to zero at the free end in the presence of a constant distributed load w; the intercept at the fixed end is indeed w l.

Why Other Options Are Wrong:
Zero contradicts equilibrium. w l / 4 and w l / 2 are fractions appropriate to other problems (e.g., reactions for simply supported beams), not a cantilever’s fixed-end shear. 2 w l is an overestimate.

Common Pitfalls:
Confusing cantilever with simply supported beam; placing the resultant at the wrong location; mixing moment and shear values.

Final Answer:
w l