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For the M 20 concrete and the Fe 415 steel, the balanced moment of resistance is:

  • a)
    3.15 bd2

  • b)
    3.73 bd2

  • c)
    2.76 bd2

  • d)
    5.80 bd2

Correct answer is option 'C'. Can you explain this answer?
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Limiting Moment of Resistance Factor for M 15 Grade Concrete and Fe 415 Grade Steel

The limiting moment of resistance factor for M 15 grade concrete and Fe 415 grade steel is given by the expression:

$Z_{\text{lim}} = \frac{0.87f_yA_s(d - \frac{A_s}{2})}{0.36f_ckb(d - \frac{0.42x_{\text{u}})}{x_{\text{u}}})}$

where,

$A_s$ = area of steel reinforcement

$d$ = effective depth of the beam

$f_y$ = characteristic strength of steel

$f_ck$ = characteristic strength of concrete

$b$ = width of the beam

$x_{\text{u}}$ = lever arm distance

Calculating the Limiting Moment of Resistance Factor

For M 15 grade concrete, the characteristic strength of concrete is 15 N/mm².

For Fe 415 grade steel, the characteristic strength of steel is 415 N/mm².

Assuming a rectangular beam with a width of 230 mm and an effective depth of 300 mm, the area of steel reinforcement can be calculated as:

$A_s = \frac{\pi}{4}d^2\frac{1}{s}$

where $s$ is the spacing of the steel bars.

Assuming a spacing of 150 mm, the area of steel reinforcement can be calculated as:

$A_s = \frac{\pi}{4}(300)^2\frac{1}{150} = 1767.15 mm^2$

The lever arm distance can be calculated as:

$x_{\text{u}} = 0.48d$

$x_{\text{u}} = 0.48(300) = 144 mm$

Substituting the values in the expression for the limiting moment of resistance factor, we get:

$Z_{\text{lim}} = \frac{0.87(415)(1767.15)(300 - \frac{1767.15}{2})}{0.36(15)(230)(300 - \frac{0.42(144)}{144})} = 2.07$

Therefore, the limiting moment of resistance factor for M 15 grade concrete and Fe 415 grade steel is 2.07.
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