Difficulty: Easy
Correct Answer: 1400 kg/cm²
Explanation:
Introduction / Context:
In RCC design under the traditional working stress method, permissible (allowable) stresses are specified for different materials. For mild steel (plain bars) used as stirrups or ties, designers use a standard allowable tensile stress to proportion shear reinforcement safely while controlling service-level behavior.
Given Data / Assumptions:
Concept / Approach:
Permissible stress for mild steel in tension under working stress design has been conventionally taken around 1400 kg/cm². This value reflects historical safety factors applied to the characteristic strength to keep service-stress levels within elastic range and to limit crack widths under service loads.
Step-by-Step Solution:
Identify the steel type: mild steel (not high-strength deformed bars).Use standard permissible tensile stress for mild steel in shear reinforcement design = about 1400 kg/cm².Select the matching option.
Verification / Alternative check:
Legacy handbooks and many exam-oriented RCC texts list 1400 kg/cm² as the permissible tensile stress for mild steel in tension for working stress calculations, especially for stirrups and ties. Modern limit-state codes use different approaches (factored capacities), but the question context is clearly working stress in kg/cm².
Why Other Options Are Wrong:
Common Pitfalls:
Mixing up permissible stresses of mild steel with those of high strength bars; confusing limit-state design parameters with working stress permissible values; converting MPa to kg/cm² incorrectly (1 MPa ≈ 10.197 kg/mm² = 101.97 kg/cm²).
Final Answer:
1400 kg/cm²
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