Cantilever beam of length l with uniformly distributed load w (per unit length): shape of the shear force diagram Choose the correct diagram shape for the entire span.

Introduction / Context:
Shear force diagrams represent the internal transverse shear variation along a beam. For standard loading and support cases, the diagram shape can be inferred directly from load intensity and boundary conditions. Here, a cantilever carries a uniformly distributed load across its full length.

Given Data / Assumptions:

• Beam type: cantilever, fixed at one end and free at the other.
• Load: uniformly distributed load w over 0 to l.
• Linear elastic, prismatic beam.

Concept / Approach:
Relationship: dV/dx = -w(x). For a constant w, V varies linearly with x. Hence the shear force diagram is a straight line, reaching maximum magnitude at the fixed end and zero at the free end.

Step-by-Step Solution:
Take origin at the free end; w is constant.Integrate: V(x) = -∫ w dx = -w x + C.Apply boundary: V = 0 at the free end (x = 0) gives C = 0.Therefore V(x) = -w x, a straight line from 0 at free end to -w l at fixed end.

Verification / Alternative check:
The bending moment diagram is the integral of V and is quadratic (a parabola) peaking at the fixed end. This pairing confirms the linear, triangular SFD.

Why Other Options Are Wrong:
Isosceles or equilateral triangles imply specific side lengths unrelated to load intensity; the correct feature is right angle due to linear variation ending at zero. Rectangle would require constant V. Parabola corresponds to bending moment, not shear force.

Common Pitfalls:
Mixing up shear and moment shapes; forgetting sign conventions; assuming constant shear under UDL, which is incorrect.

Final Answer:
A right-angled triangle