Beams with equal overhangs under uniformly distributed load A simply supported beam has a clear span l and equal overhangs a on both sides (total length l + 2a). It carries a uniformly distributed load over the whole length including the overhangs. The bending moment within the main span changes sign (i.e., a point of contraflexure occurs in the span) if:

Introduction / Context:
Beams with overhangs exhibit both sagging and hogging bending moments under a uniformly distributed load (UDL). Whether the moment within the supported span changes sign depends on the relative magnitude of the overhangs compared to the span. This determines if negative (hogging) effects from the overhangs penetrate into the main span sufficiently to create a contraflexure point.

Given Data / Assumptions:

• Equal overhangs of length a at both ends; clear span between supports is l.
• Uniformly distributed load of intensity w acting along the entire length (l + 2a).
• Linear elastic, small deflection theory.

Concept / Approach:

Because of symmetry, the reactions at the supports are equal. The overhangs produce negative moments at the supports which compete with the positive (sagging) moments in the main span. A sign change within the span occurs when the negative influence from the overhangs is strong enough, which happens when the overhangs are relatively long compared to the span.

Step-by-Step Solution (qualitative):

Verification / Alternative check:

Detailed shear and moment equations show zero-moment locations moving into the span once the overhang load effect dominates, consistent with the rule l < 2a.

Why Other Options Are Wrong:

• l > 2a: overhangs are too short; the span moment remains of one sign (sagging).
• l = 2a is the limiting case; the question asks when a sign change occurs, which is strictly when l is less than 2a.
• l = 3a or 4a again correspond to shorter overhangs relative to the span.

Common Pitfalls:

• Confusing the condition for zero support moment with the condition for a span contraflexure point.

Final Answer:

l < 2a.