Simply supported beam with a symmetric, gradually varying load Load varies from zero at both ends to w per metre at mid-span. Shear force at the centre equals:

Introduction / Context:Shear force V(x) and bending moment M(x) are related to the applied load intensity w(x) by dV/dx = -w(x) and dM/dx = V(x). For symmetric loading on a simply supported beam, the shear diagram is antisymmetric about the mid-span. Identifying where shear is zero helps locate extreme bending moments.

Given Data / Assumptions:

• Simply supported span length l.
• Transverse load intensity increases from 0 at supports to w at mid-span, symmetrically.
• Linear elastic behavior; small deflection theory.

Concept / Approach:By symmetry, reactions at supports are equal and the shear distribution about mid-span must change sign. The point where shear changes sign is exactly where V = 0, which in symmetric cases is at the centre. Since the question asks the shear at the centre, it must be zero.

Step-by-Step Solution:Symmetry ⇒ R_A = R_B.Integrate w(x) over half span to see that left-half resultant equals the left reaction.At mid-span, the net effect of left loads and reaction balances, so V(centre) = 0.Therefore the shear force at the centre is zero.

Verification / Alternative check:Because V = dM/dx, a zero shear indicates an extremum of bending moment at the centre for symmetric loading, consistent with known beam behavior.

Why Other Options Are Wrong:w l / 4 or w l / 2 have dimensions of force but contradict symmetry; w l^2 / 2 has wrong dimensions; the pi-based option is arbitrary.

Common Pitfalls:Confusing resultants of distributed loads with point shear values; forgetting symmetry arguments.

Final Answer:zero