According to IS 456:2000, for the Fe-500 grade steel having Young's modulus 200 GPa, the minimum strain in tension steel as per the limit state design method is
- A0.0035
- B0.00417Correct
- C0.00217
- D0.00167
Solution
Concept
Under the limit state of collapse, the minimum tensile strain ensures that the steel yields before the concrete crushes, providing adequate ductility.
Formula
εst = (fy / (1.15 × Es)) + 0.002
Step-by-step solution
1. Identify the grade of steel: Fe-500, so characteristic strength fy = 500 MPa. 2. Identify Young's modulus of steel Es = 200 GPa = 200,000 MPa. 3. Use IS 456:2000 formula for minimum strain in tension steel at failure: εst = (fy / (1.15 × Es)) + 0.002. 4. Substitute values: εst = (500 / (1.15 × 200,000)) + 0.002 = 0.002174 + 0.002 = 0.004174.
Shortcut
For Fe-500, remember the yield strain component is ~0.00217 and adding 0.002 gives 0.00417.
Common mistake
Forgetting to add the 0.002 plastic strain component or forgetting the partial safety factor 1.15 for steel.
Correct answer: B) 0.00417