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:

Difficulty: Easy

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

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