Locate the neutral axis
Q_{\text{below NA}} = Q_{\text{above NA}}
400x(\frac{1}{2}x) = nA_s(d - x)
200x^2 = nA_s(685 - x)
nA_s = \dfrac{200x^2}{685 - x} ← Equation (1)
Moment of inertia
I_{NA} = \dfrac{bx^3}{3}+ nA_s(d –x)^2
I_{NA} = \dfrac{400x^3}{3}+ \dfrac{200x^2}{685 - x}(685 –x)^2
I_{NA} = \frac{400}{3}x^3 + 200x^2(685 –x)
Stress in steel
\dfrac{f_s}{n} = \dfrac{M(d - x)}{I_{NA}}
\dfrac{165}{9} = \dfrac{232.4(685 - x)(1000^2)}{I_{NA}}
I_{NA} = 12\,676\,363.64(685 - x)
Equate the INA
I_{NA} = I_{NA}
\frac{400}{3}x^3 + 200x^2(685 –x) = 12\,676\,363.64(685 - x)
x = 219.6 ~ \text{mm}
From Equation (1)
9A_s = \dfrac{200(219.6^2)}{685 - 219.6}
A_s = 2302.17 ~ \text{mm}^2
Number of bars
N = \dfrac{A_s}{A_b} = \dfrac{2302.17}{\frac{1}{4}\pi (28^2)}
N = 3.74
Use 4 - 28 mm ø bars answer