Modulus of elasticity of concrete
$E_c = 4700\sqrt{f'_c} = 4700\sqrt{21}$
$E_c = 21\,538.11 ~ \text{MPa}$
Modular ratio
$n = \dfrac{E_s}{E_c} = \dfrac{200\,000}{21\,538.11}$
$n = 9$
Allowable stresses
$f_s = 140 ~ \text{MPa}$ for Grade 275
$f_c = 0.45f'_c = 0.45(21) = 9.45 ~ \text{MPa}$
Steel area
$A_s = 4 \times \dfrac{1}{4}\pi(32^2) = 1024\pi ~ \text{mm}^2$
$nA_s = 9(1024\pi) = 9216\pi ~ \text{mm}^2$
Moment of area
$Q_{\text{above NA}} = Q_{\text{below NA}}$
$300x(\frac{1}{2}x) = nA_s(d - x)$
$150x^2 = 9216\pi(600 - x)$
$150x^2 + 9216\pi x - 5\,529\,600\pi = 0$
$x = 257.22 \, \text{ and } \, -450.24$
Use $x = 257.22 ~ \text{mm}$
Moment of inertia
$I_{NA} = \dfrac{300x^3}{3} + nA_s(d - x)^2$
$I_{NA} = \dfrac{300(257.22^3)}{3} + 9216\pi(600 - 257.22)^2$
$I_{NA} = 5\,103\,735\,931 ~ \text{mm}^4$
Moment capacity
$f_b = \dfrac{Mc}{I}$
Based on concrete
$f_c = \dfrac{Mx}{I_{NA}}$
$9.45 = \dfrac{M(257.22)(1000^2)}{5\,103\,735\,931}$
$M = 187.51 ~ \text{kN}\cdot\text{m}$
Based on steel
$\dfrac{f_s}{n} = \dfrac{M(d - x)}{I_{NA}}$
$\dfrac{140}{9} = \dfrac{M(600 - 257.22)(1000^2)}{5\,103\,735\,931}$
$M = 231.61 ~ \text{kN}\cdot\text{m}$
Use the safe value of M
$M = 187.51 ~ \text{kN}\cdot\text{m}$ answer
