Shear in concrete — according to IS 456 working-stress traditions for M 150 concrete, the safe diagonal tensile stress (permissible shear stress in concrete without shear reinforcement) is approximately:
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A5 kg/cm²
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B10 kg/cm²
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C15 kg/cm²
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D20 kg/cm²
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E25 kg/cm²
Answer
Correct Answer: 25 kg/cm²
Explanation
Introduction / Context:Diagonal tension (shear cracking) capacity of plain concrete governs the minimum shear resistance of beams before shear reinforcement contributes. Traditional working-stress values (and their limit-state analogs) scale with concrete strength. For M 150 concrete, the safe diagonal tensile stress is a standard memorization point in many exams.
Given Data / Assumptions:
- Concrete grade: M 150 (characteristic strength about 15 N/mm²).
- No shear reinforcement considered at the section.
- Objective: choose permissible diagonal tension (shear) stress.
Concept / Approach:
Permissible diagonal tension increases with concrete strength. A limit-state expression is commonly written as τ_c ≈ 0.62 * sqrt(fck) N/mm². For fck ≈ 15 N/mm², τ_c ≈ 0.62 * 3.873 ≈ 2.4 N/mm² ≈ 24 kg/cm². Rounded to the nearest option, this aligns with 25 kg/cm² for quick design and exam purposes.
Step-by-Step Solution:
Compute sqrt(15) ≈ 3.873.Multiply by 0.62 → ≈ 2.4 N/mm².Convert to kg/cm² (1 N/mm² ≈ 10.197 kg/cm²) → ≈ 24.5 kg/cm² → choose 25 kg/cm².Verification / Alternative check (if short method exists):
Cross-check with traditional tables for M 150; the result falls in the same ballpark used in educational references.
Why Other Options Are Wrong:
5–20 kg/cm² understate M 150 capacity; 25 kg/cm² best matches the calculated value and common tabulations.
Common Pitfalls (misconceptions, mistakes):
Confusing τ_c (concrete shear stress) with τ_v (applied shear); ignoring size effect and minimum shear reinforcement rules even when τ_v < τ_c.
Final Answer:
25 kg/cm²