Difficulty: Medium
Correct Answer: At mid-height of the section
Explanation:
Introduction / Context:Shear stress in beams is not uniform and depends on the first moment of area Q at the level considered. For non-rectangular sections such as triangles, understanding where the maximum shear occurs is important for checking web stresses and potential shear failures.
Given Data / Assumptions:
Concept / Approach:Shear stress at a level y from a reference is τ = V Q / (I b), where Q is first moment of the area above (or below) the level about the neutral axis, I is second moment of area, and b is width at that level. For a triangle, b varies linearly with y, and Q is a quadratic function of y, so τ becomes a cubic-like function with a single maximum within the depth.
Step-by-Step Reasoning (qualitative):
1) At the extreme fibers (base or apex), Q = 0 or b → 0, giving τ = 0 at both extremes.2) Between extremes, Q increases from zero, while b also changes; the ratio Q / b peaks roughly at the mid-depth.3) Detailed derivations show the maximum shear for a triangular section occurs at mid-height (i.e., halfway between base and apex).Verification / Alternative check:Carrying out the full τ(y) = V Q / (I b) derivation with b(y) linear in y and integrating to find Q(y) leads to a stationary point at mid-depth, which is the maximum τ.
Why Other Options Are Wrong:Apex and base are extreme fibers where τ = 0; centroid is not the location of maximum τ for a triangle; one-third from base is not the maximum for this shape.
Common Pitfalls:Assuming maximum shear always occurs at the neutral axis; for triangles, the neutral axis is at h/3 from the base, but τ_max is at mid-depth, not at the centroidal axis.
Final Answer:At mid-height of the section
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