$$\begin{aligned}\phi M_n &= 0.850 \times 546.6 = \boxed{\mathbf{464.6}\ \text{kN}\cdot\text{m}} \\ \frac{M_u}{\phi M_n} &= \frac{250.0}{464.6} = \mathbf{0.538}\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$

$$\begin{aligned}\phi M_n &= 0.650 \times 525.5 = \boxed{\mathbf{341.6}\ \text{kN}\cdot\text{m}} \\ \frac{M_u}{\phi M_n} &= \frac{250.0}{341.6} = \mathbf{0.732}\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$

$$\begin{aligned}\eta &= 1.00,\quad \beta_1 = 0.80,\quad \varepsilon_{cu} = 0.0033 \\ a &= \beta_1 c = 0.80 \times 149.0 = \boxed{\mathbf{119.2}\ \text{mm}}\end{aligned}$$

$$\begin{aligned}\eta &= 1.00,\quad \beta_1 = 0.80,\quad \varepsilon_{cu} = 0.0033 \\ a &= \beta_1 c = 0.80 \times 142.4 = \boxed{\mathbf{113.9}\ \text{mm}}\end{aligned}$$

$$\begin{aligned}\varepsilon_t &= 8.53\text{‰},\quad \varepsilon_y = 2.00\text{‰},\quad \varepsilon_{t,tcl} = 5.00\text{‰} \\ &\to\ \text{Tension Controlled},\quad \text{철근 항복}\end{aligned}$$

$$\begin{aligned}\rho_f &= \frac{3{,}040}{400 \times 509.1} = 1.493\%,\quad \rho_{fb} = 0.357\% \\ \rho_f &> \rho_{fb}\ \to\ \text{콘크리트 압축파괴}\end{aligned}$$

$$\begin{aligned}\rho &= 1.493\%,\quad \rho_b = \boxed{\mathbf{3.175}\ \text{\%}},\quad \rho_{max} = 2.420\%\end{aligned}$$

$$\begin{aligned}\rho_{fb} &= 1.00 \times 0.85 \times 0.80 \times \frac{30}{850} \cdot \frac{E_f\varepsilon_{cu}}{E_f\varepsilon_{cu}+f_{fu}} = \boxed{\mathbf{0.357}\ \text{\%}} \\ \rho_f &= 1.493\%\quad (\rho_f/\rho_{fb} = 4.18)\end{aligned}$$

$$\begin{aligned}\varepsilon_s &= 7.97\text{‰}\ \to\ f_s = \min(E_s\varepsilon_s,\ f_y) = \boxed{\mathbf{400}\ \text{MPa}}\end{aligned}$$

$$\begin{aligned}f_{fu} &= C_E f^*_{fu} = 0.85 \times 1{,}000 = 850\ \text{MPa} \\ f_f &= \boxed{\mathbf{382}\ \text{MPa}}\quad (f_f/f_{fu} = 0.45)\end{aligned}$$

$$\begin{aligned}M_n &= A_s f_y\left(d-\frac{a}{2}\right) = 3{,}040 \times 400 \times \left(509.1-\frac{119.2}{2}\right) = \boxed{\mathbf{546.6}\ \text{kN}\cdot\text{m}}\end{aligned}$$

$$\begin{aligned}c &= 142.4\ \text{mm},\quad a = \beta_1 c = 113.9\ \text{mm} \\ M_n &= A_f f_f\left(d-\frac{\beta_1 c}{2}\right) = 3{,}040 \times 382 \times \left(509.1-\frac{113.9}{2}\right) = \boxed{\mathbf{525.5}\ \text{kN}\cdot\text{m}}\end{aligned}$$

$$\begin{aligned}\varepsilon_t &= 8.53\text{‰}\ \to\ \phi = \boxed{\mathbf{0.850}\ \text{}}\quad (\text{Tension Controlled})\end{aligned}$$

$$\begin{aligned}\varepsilon_{ft} &= 9.08\text{‰},\quad \varepsilon_{fu} = 18.89\text{‰},\quad \frac{\varepsilon_{ft}}{\varepsilon_{fu}} = 0.481 \\ \phi &= \boxed{\mathbf{0.650}\ \text{}}\quad (\text{Compression Controlled})\end{aligned}$$

$$\begin{aligned}\rho &= 1.493\% \le \rho_{max} = 2.420\%\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$

$$\begin{aligned}\text{상한 없음}\quad (\rho_f &= 1.493\%,\ \rho_{fb} = 0.357\%)\end{aligned}$$

$$\begin{aligned}1.2 M_{cr} &= 1.2 \times 82.8 = 99.4\ \text{kN}\cdot\text{m} \\ \phi M_n &= 464.6 \ge 1.2M_{cr}\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$

$$\begin{aligned}A_{f,min} &= \max\left(\frac{0.41\sqrt{30}}{850},\ \frac{2.3}{850}\right) b_w d = \boxed{\mathbf{551}\ \text{mm²}} \\ \rho_f &> \rho_{fb}\ \to\ \text{적용 대상 아님 (압축파괴 구간)}\end{aligned}$$