Calculate the design stress for a steel grade Fe-415 after applying necessary factor of safety, in case of design of RCC members in accordance with limit state design.
- A415 N/mm²
- B435 N/mm²
- C217.5 N/mm²
- D361.05 N/mm²Correct
Solution
Concept
In limit state design, the design stress of steel is fy divided by the partial safety factor 1.15, which equals 0.87 fy.
Formula
fd = fy / γm = fy / 1.15 ≈ 0.87 fy
Step-by-step solution
1. Given steel grade Fe-415, so fy = 415 N/mm². 2. Partial safety factor for steel, γm = 1.15. 3. Design stress = fy / 1.15 ≈ 0.87 × fy. 4. Design stress = 0.87 × 415 = 361.05 N/mm².
Explanation
In limit state design we do not use the full yield strength of steel. We divide it by the partial safety factor 1.15, so design stress = 0.87 fy. For Fe-415 this is 0.87 × 415 = 361.05 N/mm².
Why the other options are wrong
A: 415 N/mm² is the characteristic yield strength, with no safety factor applied. B: 435 N/mm² is the design stress (0.87 fy) for Fe-500, not Fe-415. C: 217.5 N/mm² is not the limit state design stress of Fe-415.
Shortcut
Design stress of steel = 0.87 × fy.
Common mistake
Using fy directly, or using the Fe-500 value of 435 N/mm².
Correct answer: D) 361.05 N/mm²