Neutral axis in bending: stress state at the neutral axis of a beam section Fill the blank with the correct stress condition.

Introduction / Context:
In pure bending of beams, stresses vary linearly across the section depth. The line where the bending stress changes sign is called the neutral axis. Recognizing the stress condition at this axis is fundamental to section design and to locating tension and compression zones.

Given Data / Assumptions:

• Plane sections remain plane (Bernoulli–Euler assumption).
• Linear elastic material and small deflections.
• No axial force superposed, only bending.

Concept / Approach:
The normal bending stress is sigma = (M * y) / I, where y is measured from the neutral axis. At the neutral axis, y = 0, so sigma = 0. Shear stress distribution is separate and has its own neutral line discussion, but in most standard cases maximum shear occurs at the centroidal axis, not necessarily the same as zero bending stress statement.

Step-by-Step Solution:
Use sigma = M * y / I.At y = 0 (neutral axis), sigma = 0.Above the neutral axis: one sign (compression); below it: opposite sign (tension), depending on bending sense.

Verification / Alternative check:
Strain distribution is linear: epsilon = curvature * y; at y = 0, epsilon = 0, hence stress is zero by Hooke's law.

Why Other Options Are Wrong:
Maximum tensile or compressive occurs at extreme fibers (|y| is maximum). Minimum tensile is not a standard descriptor. Maximum shear does not coincide with zero normal stress fact in bending discussion here.

Common Pitfalls:
Confusing shear stress distribution with normal bending stress; assuming neutral axis carries some residual normal stress.

Final Answer:
zero