Difficulty: Easy
Correct Answer: Diagonal tension
Explanation:
Introduction / Context:Concrete is strong in compression and weak in tension. Under shear, inclined tensile stresses (diagonal tension) develop, causing potential cracking. Shear reinforcement (stirrups/bent-up bars) is used to intercept these cracks and carry tensile forces safely, enhancing ductility and preventing brittle web shear failure in beams.
Given Data / Assumptions:
Concept / Approach:
Shear in beams leads to diagonal tension (tensile principal stress on planes approximately at 45 degrees). Concrete alone cannot resist this tension beyond a small limit (diagonal tension capacity). Shear reinforcement functions as tension ties across these potential crack planes, while concrete and aggregate interlock provide residual shear transfer.
Step-by-Step Solution:
Recognize that “vertical shear” is a resultant; the critical failure mode is tension along inclined planes.Provide stirrups (vertical or inclined) to act across the crack as tension members.Ensure spacing and size satisfy code: V_us ≥ V_u − V_c with adequate anchorage.Verification / Alternative check:
Design formulas decompose shear resistance into concrete contribution and steel contribution; the steel leg force is proportional to A_sv * f_y * (d / s) for vertical stirrups, directly addressing diagonal tension.
Why Other Options Are Wrong:
Horizontal shear relates to composite action along interfaces; diagonal compression is typically carried by concrete. “Vertical shear” wording misses the mechanism of crack control; torsion requires special closed stirrups and longitudinal reinforcement beyond ordinary shear design.
Common Pitfalls:
Underestimating shear near supports; exceeding maximum stirrup spacing; inadequate hooks/anchorage for stirrups.
Final Answer:
Diagonal tension
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