Intermediate-roller beam with equal sagging and hogging: A beam of total length L rests on two intermediate roller supports and carries a uniformly distributed load over its entire length. The beam has equal overhangs at both ends. If the maximum sagging bending moment equals the maximum hogging bending moment, what is the length of each overhang as a fraction of L?

Introduction / Context:
This classic continuous-beam problem tests the relationship between loading, reactions, and internal actions (sagging and hogging moments) for a beam with equal end overhangs supported on two intermediate rollers under a uniformly distributed load. The target is to find the overhang length at which the largest positive (sagging) and negative (hogging) bending moments are equal in magnitude.

Given Data / Assumptions:

• Total beam length = L; equal overhangs = a each side.
• Two intermediate roller supports separated by span (L − 2a).
• Uniformly distributed load of constant intensity across the full length.
• Prismatic member; small deflection; linear elastic behavior.

Concept / Approach:
The internal moments arise from the supported span (hogging over the supports) and the overhanging cantilevers (sagging near free ends). By writing equilibrium and influence relationships (or using standard results/influence lines), one can determine the overhang a/L that equalizes the peak positive and peak negative moments.

Step-by-Step Solution:
Let overhangs = a; supported span = L − 2a.Under full-length UDL, reactions distribute so that hogging peaks over the supports and sagging peaks in the overhangs.Setting |M_hog|max = |M_sag|max and solving the resulting equilibrium/compatibility equation yields a/L ≈ 0.207.Therefore, each overhang length a = 0.207 L.

Verification / Alternative check:
This fraction is a well-known optimum for equalization of moments on a two-span-with-overhang arrangement under uniform load. Numerical checks via influence lines or structural analysis software reproduce a/L ≈ 0.207.

Why Other Options Are Wrong:
0.107 L, 0.307 L, 0.407 L, 0.500 L: these do not equalize the peak hogging and sagging; analysis shows the equality occurs close to 0.207 L.

Common Pitfalls:
Confusing the position of maximum sagging (in the overhang) with that within the supported span, or assuming symmetry automatically implies a = L/4.

Final Answer:
0.207 L.