Question

The minimum strain at failure in tension steel having yield stress $f_y = 415 \text{ MPa}$ and Young's Modulus $E_s = 200 \text{ GPa}$ as per Limit State Method of design is:

• A. 0.005
• B. 0.0038
• C. 0.0045
• D. 0.0025

Hint:

This requirement ensures that the steel yields significantly before the concrete crushes, which provides ample warning (ductility) before structural failure.

Answer 1

The Correct Option is B

Concept: According to IS 456:2000 for the Limit State of Collapse (Flexure), the maximum strain in tension reinforcement in the section at failure shall not be less than a specific value to ensure ductile failure.

Step 1: Identify the formula for minimum strain.
The formula for the minimum design strain is: \[ \epsilon_{st} \geq \frac{f_y}{1.15 E_s} + 0.002 \] Where:
• $f_y = 415 \text{ N/mm}^2$ (or MPa)
• $E_s = 200 \text{ GPa} = 200 \times 10^3 \text{ N/mm}^2$
• 1.15 is the partial safety factor for steel ($\gamma_s$).

Step 2: Perform the calculation.
\[ \epsilon_{st} \geq \frac{415}{1.15 \times 200000} + 0.002 \] \[ \epsilon_{st} \geq \frac{415}{230000} + 0.002 \] \[ \epsilon_{st} \geq 0.001804 + 0.002 \] \[ \epsilon_{st} \geq 0.003804 \approx 0.0038 \]