$$\begin{aligned}\phi V_n &= 0.75\,(V_c + V_s) = 0.75\,(185.9 + 360.4) = \boxed{\mathbf{409.7}\ \text{kN}} \\ \frac{V_u}{\phi V_n} &= \frac{150}{409.7} = \mathbf{0.37}\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$

$$\begin{aligned}\phi V_n &= 0.75\,(V_c + V_f) = 0.75\,(92.9 + 202.7) = \boxed{\mathbf{221.8}\ \text{kN}} \\ \frac{V_u}{\phi V_n} &= \frac{150}{221.8} = \mathbf{0.68}\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$

$$\begin{aligned}V_c &= \tfrac{1}{6}\Big(1+\tfrac{N_u}{14A_g}\Big)\lambda\sqrt{f_{ck}}\,b_w d \\ &= \tfrac{1}{6}\Big(1+\tfrac{0}{14\times240{,}000}\Big)(1.0)\sqrt{30}\times400\times509.1 \\ &= \boxed{\mathbf{185.9}\ \text{kN}}\end{aligned}$$

$$\begin{aligned}k_p &= N_u/(14A_g) = 0/(14\times240{,}000) = 0.000 \\ k_s &= (0.3/d)^{0.25} = (0.3/0.509)^{0.25} = 0.88 \\ &\to [0.75,\,1.1] \to 0.88\ \to\ 1.00\ (\text{최소보강 만족}) \\ V_c &= \tfrac{1}{6}(1+0.000)(1.00)(\beta_v\,0.5)\sqrt{30}\times400\times509.1 = \boxed{\mathbf{92.9}\ \text{kN}}\end{aligned}$$

$$\begin{aligned}A_v &= 2\times\pi\times13^2/4 = 265\ \text{mm}^2 \\ V_s &= \frac{A_v f_{yt} d}{s} = \frac{265\times400\times509.1}{150} = \boxed{\mathbf{360.4}\ \text{kN}} \\ V_{s,max} &= \tfrac{2}{3}\sqrt{30}\times400\times509.1 = 743.6\ \ge V_s\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$

$$\begin{aligned}A_v &= 2\times\pi\times13^2/4 = 265\ \text{mm}^2 \\ V_f &= \frac{A_{fv} f_{ft} d}{s} = \frac{265\times225\times509}{150} = \boxed{\mathbf{202.7}\ \text{kN}} \\ V_{f,max} &= 0.25\,\xi_v f_{ck} b_w z = 0.25\times0.528\times30\times400\times433 = 685.5\ \ge V_f\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$

$$\begin{aligned}f_{yt} &= 400\ \text{MPa} \le 500\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$

$$\begin{aligned}f_{fb} &= (0.05\,r_b/d_b + 0.3)\,f_{fu} \\ &= (0.05\times3+0.3)\times850 = 382\ \text{MPa} \\ 0.005E_f &= 0.005\times45{,}000 = 225\ \text{MPa} \\ f_{ft} &= \min(225,\ 382) = \boxed{\mathbf{225}\ \text{MPa}}\end{aligned}$$

$$\begin{aligned}\phi V_c &= 0.75\times185.9 = 139.4 < V_u = 150\ \to\ \text{스터럽 필요} \\ \frac{A_v}{s} &= \frac{V_u-\phi V_c}{\phi f d} = \frac{(150-139.4)\times10^3}{0.75\times400\times509} \\ &= \boxed{\mathbf{0.069}\ \text{mm²/mm}} \\ s_{req} &= A_v\big/\tfrac{A_v}{s} = 265/0.069 = 3{,}833\ \text{mm}\end{aligned}$$

$$\begin{aligned}\phi V_c &= 0.75\times92.9 = 69.7 < V_u = 150\ \to\ \text{스터럽 필요} \\ \frac{A_v}{s} &= \frac{V_u-\phi V_c}{\phi f d} = \frac{(150-69.7)\times10^3}{0.75\times225\times509} \\ &= \boxed{\mathbf{0.935}\ \text{mm²/mm}} \\ s_{req} &= A_v\big/\tfrac{A_v}{s} = 265/0.935 = 284\ \text{mm}\end{aligned}$$

$$\begin{aligned}A_{v,min} &= \max(0.0625\sqrt{30},\,0.35)\,\frac{b_w s}{f_{yt}} \\ &= \max(0.342,\,0.35)\times\frac{400\times150}{400} = \boxed{\mathbf{52}\ \text{mm²}} \\ A_v &= 265 \ge 52\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$

$$\begin{aligned}A_{fv,min} &= \max(0.0625\sqrt{30},\,0.35)\,\frac{b_w s}{f_{ft}} \\ &= 0.350\times\frac{400\times150}{225} = \boxed{\mathbf{93}\ \text{mm²}} \\ A_{fv} &= 265 \ge 93\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$

$$\begin{aligned}s_{max} &= \min(d/2,\,600) = \min(255,\,600) = 255\ \text{mm} \\ V_s &= 360.4 \le \tfrac{1}{3}\sqrt{f_{ck}}\,b_w d = 371.8\ \to\ \text{절반 규정 미적용} \\ s_{allow} &= \min(s_{req},\,s_{min},\,s_{max}) = \min(3{,}833,\,758,\,255) \\ &= \boxed{\mathbf{255}\ \text{mm}} \ \ge s = 150\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$

$$\begin{aligned}s_{max} &= \min(d/2,\,600) = \min(255,\,600) = 255\ \text{mm} \\ V_f &= 202.7 \le \tfrac{1}{3}\sqrt{f_{ck}}\,b_w d = 371.8\ \to\ \text{절반 규정 미적용} \\ s_{allow} &= \min(s_{req},\,s_{min},\,s_{max}) = \min(284,\,427,\,255) \\ &= \boxed{\mathbf{255}\ \text{mm}} \ \ge s = 150\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$

$$\begin{aligned}f_{fu} &= C_E\,f_{fu}^* = 0.85\times1000 = \boxed{\mathbf{850}\ \text{MPa}}\end{aligned}$$