**Maximum Moment**
$M_{max} = 29.05(4)(2) = 232.4 ~ \text{kN}\cdot\text{m}$

**For Balanced Capacity**

$f_c = 0.45f_c' = 0.45(21) = 9.45 ~ \text{MPa}$

$f_s = 165 ~ \text{MPa}$

$\dfrac{x_{bal}}{f_c} = \dfrac{d}{f_c + f_s/n}$

$\dfrac{x_{bal}}{9.45} = \dfrac{685}{9.45 + 165/9}$

$x_{bal} = 233 ~ \text{mm}$

$C_{bal} = \frac{1}{2}f_c \, b x_{bal} = \frac{1}{2}(9.45)(400)(233)$

$C_{bal} = 440.37 ~ \text{kN}$

$M_{bal} = C_{bal}(d - \frac{1}{3}x_{bal}) = 440.37[ \, 0.685 – \frac{1}{3}(0.233) \, ]$

$M_{bal} = 267.45 ~ \text{kN}\cdot\text{m}$

*M*_{max} < *M*_{bal}, thus, the beam is singly-reinforced.

**For Singly Reinforced Section**

$f_s = 165 ~ \text{MPa}$

$f_c = ~ ?$

Accurate Solution for finding *A*_{s} (Not recommended)

Approximate Solution for finding *A*_{s} (Recommended)

$M_{max} = T(d - \frac{1}{3}x_{bal})$

$M_{max} = A_sf_s(d - \frac{1}{3}x_{bal})$

$A_s = \dfrac{M_{max}}{f_s(d - \frac{1}{3}x_{bal})}$

$A_s = \dfrac{232.4(1000^2)}{165[ \, 685 - \frac{1}{3}(233) \, ]}$

$A_s = 2319.13 ~ \text{mm}^2$

Number of bars

$N = \dfrac{A_s}{A_b} = \dfrac{2319.13}{\frac{1}{4}\pi (28^2)}$

$N = 3.76$

Use 4 - 28 mm ø bars *answer*