Stress distribution in a transversely loaded rectangular beam When a straight, prismatic rectangular beam is subjected to sagging (positive) bending under transverse loading, where does the maximum compressive normal stress occur?

Introduction / Context:
Bending of beams produces linearly varying normal stress across the depth when Euler–Bernoulli assumptions hold. Recognizing where compression and tension occur is vital for placing reinforcement and checking extreme fiber stresses relative to material limits.

Given Data / Assumptions:

• Prismatic rectangular cross-section, homogeneous isotropic material.
• Sagging (positive) bending moment: concave upward elastic curve in the span.
• Plane sections remain plane; neutral axis passes through the centroid.

Concept / Approach:

Normal stress due to bending is σ = M y / I, where y is measured from the neutral axis. For sagging moment in the usual sign convention, fibers above the neutral axis are in compression and those below are in tension. The magnitude is proportional to the distance |y|, so extremes happen at the outermost fibers.

Step-by-Step Solution:

Verification / Alternative check:

Mohr’s circle for normal stress on fibers across depth shows maxima at the extreme distances from the neutral axis, consistent with σ = M y / I.

Why Other Options Are Wrong:

• Bottom fibre: in sagging, this is tensile, not compressive.
• Neutral axis: σ = 0 by definition.
• “Same at every point”: false for bending; stress varies linearly with y.
• Centroid only: also lies on the neutral axis, so stress there is zero.

Common Pitfalls:

• Reversing the sign convention (for hogging, the answer would flip).
• Confusing shear stress distribution (parabolic) with bending normal stress (linear).

Final Answer:

Top fibre.