Shear in UDL case:\nFor a simply supported beam carrying a uniformly distributed load of intensity w per unit length over the full span, what is the shear force at the midspan (centre) of the beam?

Introduction / Context:
Shear force and bending moment distributions under simple load cases are staple knowledge in structural analysis. For a uniformly distributed load (UDL), recognizing the symmetry of shear lets you answer midspan values instantly—useful for quick checks in design and exams.

Given Data / Assumptions:

• Simply supported beam with span l.
• UDL of intensity w acts along the entire span.
• Static, linear elastic behavior; supports provide vertical reactions only.

Concept / Approach:
For a symmetric UDL, reactions at supports are equal: R_A = R_B = w l / 2. Shear force V(x) at a distance x from the left support equals R_A − w x, a straight line decreasing from +w l / 2 to −w l / 2. The point where V = 0 is the midspan.

Step-by-Step Solution:

Verification / Alternative check:
The shear diagram is a straight line crossing zero at midspan. The maximum bending moment occurs where shear is zero, i.e., at the midspan, confirming internal consistency (M_max = w l^2 / 8).

Why Other Options Are Wrong:

• w l^2 / 2, w l^2 / 4, w l^2 / 8 are bending-moment-like terms (units of force*length), not shear.
• w l / 2 is the end shear (support reaction), not the midspan shear.

Common Pitfalls:
Mixing up shear and moment formulas; always check dimensional consistency to avoid such errors.

Final Answer:
Zero