Question

A singly reinforced concrete beam of balanced section is made of M20 grade concrete and Fe415 grade steel bars. The magnitudes of the maximum compressive strain in concrete and the tensile strain in the bars at ultimate state under flexure, as per IS 456: 2000 are \underline{\hspace{2cm}, respectively. (round off to four decimal places)}

• A. 0.0035 and 0.0038
• B. 0.0020 and 0.0018
• C. 0.0035 and 0.0041
• D. 0.0020 and 0.0031

Hint:

Always remember: As per IS 456:2000, $\varepsilon_{c,\text{max}} = 0.0035$ and for Fe415 steel in a balanced section, tensile strain $\approx 0.0038$. These are standard code values useful for design.

Answer 1

The Correct Option is A

Step 1: Maximum strain in concrete.
As per IS 456:2000 (Clause 38.1), the maximum compressive strain in concrete at the extreme fibre in the limit state of flexure is taken as: \[ \varepsilon_{c,\text{max}} = 0.0035 \]

Step 2: Tensile strain in steel (Fe415).
For Fe415 grade steel, the yield stress $f_y = 415$ MPa and modulus of elasticity $E_s = 2\times 10^5$ MPa.
The yield strain is: \[ \varepsilon_y = \frac{f_y}{E_s} = \frac{415}{2 \times 10^5} = 0.002075 \approx 0.0021 \] For a balanced section, the steel reaches just beyond yield at the ultimate state. IS 456 specifies that the limiting tensile strain in steel at failure should be taken as approximately $0.0038$ for Fe415 bars.

Step 3: Match with options.
- Concrete: $0.0035$ - Steel: $0.0038$ These values match option (A).
\[ \boxed{0.0035 \ \text{and} \ 0.0038} \]