Pure bending – stress distribution:\nIn a beam subjected to pure bending, the intensity of normal stress in any fibre is __________ the distance of that fibre from the neutral axis.

Introduction / Context:
Understanding how bending stress varies across a beam’s depth is central to safe and economical design. Under pure bending (constant bending moment, zero shear), the normal stress varies linearly with the distance from the neutral axis, reaching zero at the neutral axis and maximum at the extreme fibres.

Given Data / Assumptions:

• Euler–Bernoulli beam theory applies.
• Plane sections remain plane and perpendicular to the neutral axis after bending.
• Material behavior is linear elastic and homogeneous.

Concept / Approach:
The flexure formula provides the relationship between bending stress and the fibre distance y from the neutral axis: sigma = M * y / I, where M is the bending moment and I is the second moment of area about the neutral axis. This shows a direct proportionality to y.

Step-by-Step Solution:

Verification / Alternative check:
The linear stress distribution is consistent with zero stress at the neutral axis and equal magnitudes of tension and compression at symmetric extreme fibres for symmetric sections under pure bending.

Why Other Options Are Wrong:

• Equal to / less than / more than: these are not general relationships; the correct dependence is proportional, not a fixed inequality.

Common Pitfalls:
Forgetting sign convention: fibres on one side are in tension (positive), and on the other in compression (negative). Magnitudes still scale linearly with |y|.

Final Answer:
Directly proportional to