Difficulty: Easy
Correct Answer: On the middle fibre along the horizontal (neutral axis) plane
Explanation:
Introduction / Context:
In beams with rectangular or square cross-sections, transverse shear stress varies parabolically over the depth. Understanding where it peaks is important for web sizing and checking shear capacity.
Given Data / Assumptions:
Concept / Approach:
For a rectangular section, τ(y) = (3/2) * V / (b * h) * (1 − (2y/h)^2), which is a parabola with maximum at the neutral axis (y = 0) and zero at the top and bottom fibres.
Step-by-Step Solution:
Recognize the parabolic distribution across depth.At the neutral axis (middle fibre), τ_max = 1.5 * V / (b * h).At extreme fibres, τ = 0.Therefore, the critical shear occurs along the horizontal plane through the neutral axis.
Verification / Alternative check:
Shear flow q = V * Q / I b is maximum where first moment Q is maximum, i.e., at the neutral axis.
Why Other Options Are Wrong:
Lower/top fibres: shear is zero there.Equal at every fibre: false; distribution is parabolic.Corners/diagonals: not where τ is maximum in rectangular sections.
Common Pitfalls:
Confusing shear with bending stress (which is maximum at extreme fibres).
Final Answer:
On the middle fibre along the horizontal (neutral axis) plane.
Discussion & Comments