In the limit state of collapse for flexure, the strain in tension reinforcement at failure should not be less than (where fy and Es are the characteristic strength and modulus of elasticity of steel respectively. )
- A[fy/(1.5Es) + 0.002]
- B[fy/(1.15Es) + 0.002]Correct
- C[fy/Es + 0.002]
- D[1.15fy/Es + 0.002]
Solution
Concept
Under the limit state of collapse for flexure, IS 456 specifies that the minimum strain in the tension steel at ultimate failure must account for both the yield strain of steel (with a partial safety factor of 1.15) and an additional plastic strain of 0.002 to ensure ductile failure.
Formula
εst = (fy / (1.15 Es)) + 0.002
Step-by-step solution
1. Recall the design stress-strain curve for steel as per IS 456:2000. 2. The design yield stress of steel is taken as fy / 1.15, corresponding to a design yield strain of (fy / (1.15 Es)). 3. To ensure ductile behavior and adequate warning before failure, an additional strain of 0.002 is added to the yield strain. 4. Therefore, the minimum strain in tension reinforcement at failure is given by [fy / (1.15 Es) + 0.002].
Explanation
According to IS 456:2000 (Clause 38.1, note relating to the limit state of collapse for flexure), the strain in tension reinforcement at failure shall not be less than [fy / (1.15 Es) + 0.002]. This ensures that the steel reaches its yield plateau and sufficient ductility is provided before failure.
Why the other options are wrong
A: Uses 1.5 instead of 1.15 (1.5 is for concrete). C: Uses characteristic yield strain without the partial safety factor 1.15. D: Misplaces the partial safety factor 1.15 in the numerator.
Shortcut
Remember that IS 456 incorporates the partial safety factor for steel (1.15) in the denominator of the yield strain term, plus an extra 0.002 plastic strain.
Common mistake
Forgetting the partial safety factor of 1.15 for steel and choosing option (C), or omitting the 0.002 additional strain.
Correct answer: B) [fy/(1.15Es) + 0.002]