Strength of Materials – Simply supported beam with a central point load Consider a simply supported beam of span L carrying a single concentrated load P exactly at mid-span. State whether the shear force diagram changes sign at the midpoint under this loading condition.

Introduction / Context:In strength of materials, understanding how shear force varies along a beam under specific loads is fundamental. A simply supported beam with a central point load is a classic case used to teach sign conventions and the relationship between loads, shear force, and bending moment.

Given Data / Assumptions:

• Beam is simply supported at two ends.
• A single point load P acts at mid-span (x = L/2).
• Beam is prismatic, linearly elastic, and weightless for simplicity.
• Upward reactions at supports are equal due to symmetry, each = P/2.

Concept / Approach:Shear force V(x) is the algebraic sum of vertical forces to the left or right of a section. For a point load, V has a jump (discontinuity) at the load location. With the load at mid-span, the shear changes from positive to negative (or vice versa depending on sign convention) at that exact point.

Step-by-Step Solution:Take sections to the left of mid-span: V_left = +P/2 (upward reaction).Take sections immediately to the right of the point load: V_right = +P/2 − P = −P/2.The sudden change from +P/2 to −P/2 at mid-span shows a sign change in V(x).Bending moment is maximum where shear crosses zero; here it occurs at mid-span, consistent with theory.

Verification / Alternative check:The bending moment diagram is triangular on each half with peak M_max = (P/2)(L/2) = PL/4 at mid-span. Maximum moment coincides with V = 0, confirming the shear sign change at the center.

Why Other Options Are Wrong:“It changes sign only at the supports” is false; supports are reaction points, not zero-shear locations for this load case. “It never changes sign for a point load” contradicts the jump at the load. “Depends on beam material only” is incorrect; material does not affect static equilibrium of internal forces.

Common Pitfalls:Mixing sign conventions; forgetting that a concentrated load introduces a shear force jump; assuming zero shear only for distributed loads.

Final Answer:Correct