$$\begin{aligned}\phi M_n &= 0.850 \times 546.6 = \boxed{\mathbf{464.6}\ \text{kN}\cdot\text{m}} \\ \frac{M_u}{\phi M_n} &= \frac{350.0}{464.6} = \mathbf{0.753}\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$

$$\begin{aligned}\phi M_n &= 0.650 \times 525.5 = \boxed{\mathbf{341.6}\ \text{kN}\cdot\text{m}} \\ \frac{M_u}{\phi M_n} &= \frac{350.0}{341.6} = \mathbf{1.025}\ \ \color{#d40000}{\times\ \textbf{NG}}\end{aligned}$$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$\begin{aligned}f_r &= 0.63\sqrt{30} = 3.45\ \text{MPa} \\ I_g &= \tfrac{400\times600^3}{12} = 7{,}200\times10^6\ \text{mm}^4 \\ M_{cr} &= \frac{f_r I_g}{y_t} = \frac{3.45\times7{,}200\times10^6}{300} = \boxed{\mathbf{82.8}\ \text{kN}\cdot\text{m}} \\ M_s &= 120\ >\ M_{cr} = 82.8\ \to\ \text{균열 단면 → 균열 및 처짐 검토}\end{aligned}$$

$$\begin{aligned}f_r &= 0.63\sqrt{30} = 3.45\ \text{MPa} \\ I_g &= \tfrac{400\times600^3}{12} = 7{,}200\times10^6\ \text{mm}^4 \\ M_{cr} &= \frac{f_r I_g}{y_t} = \frac{3.45\times7{,}200\times10^6}{300} = \boxed{\mathbf{82.8}\ \text{kN}\cdot\text{m}} \\ M_s &= 120\ >\ 0.8M_{cr} = 66.3\ \to\ \text{균열 단면 → 균열 및 처짐 검토}\end{aligned}$$

$$\begin{aligned}E_c &= 27.5\ \text{GPa} \\ n &= E/E_c = 200{,}000/27{,}537 = 7.26 \\ \rho &= \frac{A}{bd} = \frac{3{,}040}{400\times509.1} = 0.0149 \\ k &= \sqrt{2\rho n+(\rho n)^2}-\rho n = 0.370 \\ I_{cr} &= \tfrac{bd^3}{3}k^3 + nAd^2(1-k)^2 = 3{,}163\times10^6\ \text{mm}^4 \\ f_{s} &= \frac{M_s\,n\,d(1-k)}{I_{cr}} = \frac{120\times10^6\times7.26\times509.1\times(1-0.370)}{3{,}163\times10^6} = \boxed{\mathbf{88}\ \text{MPa}}\end{aligned}$$

$$\begin{aligned}E_c &= 27.5\ \text{GPa} \\ n &= E/E_c = 45{,}000/27{,}537 = 1.63 \\ \rho &= \frac{A}{bd} = \frac{3{,}040}{400\times509.1} = 0.0149 \\ k &= \sqrt{2\rho n+(\rho n)^2}-\rho n = 0.198 \\ I_{cr} &= \tfrac{bd^3}{3}k^3 + nAd^2(1-k)^2 = 965\times10^6\ \text{mm}^4 \\ f_{fs} &= \frac{M_s\,n\,d(1-k)}{I_{cr}} = \frac{120\times10^6\times1.63\times509.1\times(1-0.198)}{965\times10^6} = \boxed{\mathbf{83}\ \text{MPa}}\end{aligned}$$

$$\begin{aligned}s_1 &= 375\frac{\kappa_{cr}}{f_s}-2.5c_c = 375\times\frac{280}{88}-2.5\times53 = 1{,}055 \\ s_2 &= 300\frac{\kappa_{cr}}{f_s} = 300\times\frac{280}{88} = 950 \\ s_{max} &= \min(s_1,\,s_2) = \boxed{\mathbf{950}\ \text{mm}} \\ s &= \frac{b-2c_c-d_b}{n-1} = \frac{400-2\times53-25.4}{3-1} = 134\ \le\ s_{max}\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$

$$\begin{aligned}\frac{E_fw}{f_{fs}k_b} &= \frac{45{,}000\times0.7}{83\times1.20} = 316.3 \\ s_1 &= 1.15\times316.3-2.5\times53 = 231 \\ s_2 &= 0.94\times316.3 = 297 \\ s_{max} &= \min(s_1,\,s_2) = \boxed{\mathbf{231}\ \text{mm}} \\ s &= \frac{b-2c_c-d_b}{n-1} = \frac{400-2\times53-25.4}{3-1} = 134\ \le\ s_{max}\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$

$$\begin{aligned}\beta_{cr} &= \frac{h-kd}{d-kd} = \frac{600-100.7}{509.1-100.7} = 1.223 \\ d_c &= c_c+d_b/2 = 65.7\ \text{mm} \\ f_{fs,lim} &= \frac{0.36E_f}{d_c\beta_{cr}k_b} = \frac{0.36\times45{,}000}{65.7\times1.223\times1.20} = \boxed{\mathbf{168}\ \text{MPa}} \\ f_{fs} &= 83\ \le\ 168\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$

$$\begin{aligned}\kappa_{cr} &= 280\ (\text{건조환경})\end{aligned}$$

$$\begin{aligned}k_b &= 1.20 \\ w &= 0.7\ \text{mm}\end{aligned}$$

$$\begin{aligned}\left(\tfrac{M_{cr}}{M_a}\right)^3 &= \left(\tfrac{82.8}{120}\right)^3 = 0.3287 \\ I_e &= 0.3287\times7{,}200 + (1-0.3287)\times3{,}163 = \boxed{\mathbf{4{,}490}\ \text{×10⁶ mm⁴}} \\ \frac{I_e}{I_g} &= \mathbf{0.624}\end{aligned}$$

$$\begin{aligned}\gamma &= 1.72-0.72\left(\tfrac{0.8M_{cr}}{M_a}\right) = 1.72-0.72\times0.552 = 1.322 \\ I_e &= \frac{I_{cr}}{1-\gamma\left(\tfrac{0.8M_{cr}}{M_a}\right)^2\left[1-\tfrac{I_{cr}}{I_g}\right]} = \frac{965}{1-1.322\times0.552^2\times(1-0.134)} \\ &= \boxed{\mathbf{1{,}482}\ \text{×10⁶ mm⁴}} \\ \frac{I_e}{I_g} &= \mathbf{0.206}\end{aligned}$$

$$\begin{aligned}\Delta_i &= K\frac{M_a L^2}{E_c I_e} = 0.1042\times\frac{120\times10^6\times6000^2}{27{,}537\times4{,}490\times10^6} = 3.64\ \text{mm} \\ \Delta_{sus} &= 0.60\times3.64 = 2.18 \\ \Delta_L &= \Delta_i-\Delta_{sus} = \boxed{\mathbf{1.46}\ \text{mm}} \\ \Delta_L &\ \le\ L/360 = 16.7\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$

$$\begin{aligned}\Delta_i &= K\frac{M_a L^2}{E_c I_e} = 0.1042\times\frac{120\times10^6\times6000^2}{27{,}537\times1{,}482\times10^6} = 11.02\ \text{mm} \\ \Delta_{sus} &= 0.60\times11.02 = 6.61 \\ \Delta_L &= \Delta_i-\Delta_{sus} = \boxed{\mathbf{4.41}\ \text{mm}} \\ \Delta_L &\ \le\ L/360 = 16.7\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$

$$\begin{aligned}\lambda_\Delta &= \frac{\xi}{1+50\rho'} = \frac{2}{1+50\times0} = 2.00 \\ \Delta_{lt} &= \lambda_\Delta\Delta_{sus} = 2.00\times2.18 = 4.37\ \text{mm} \\ \Delta_L+\Delta_{lt} &= 1.46+4.37 = \boxed{\mathbf{5.82}\ \text{mm}}\ \le\ L/240 = 25.0\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$

$$\begin{aligned}\lambda_\Delta &= 0.6\,\xi = 0.6\times2 = 1.20 \\ \Delta_{lt} &= \lambda_\Delta\Delta_{sus} = 1.20\times6.61 = 7.94\ \text{mm} \\ \Delta_L+\Delta_{lt} &= 4.41+7.94 = \boxed{\mathbf{12.35}\ \text{mm}}\ \le\ L/240 = 25.0\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$

$$\begin{aligned}M_{s,sus} &= 0.60\times120 = 72.0\ \text{kN}\cdot\text{m} \\ f_{fs,sus} &= \frac{M_{s,sus}\,n_f\,d(1-k)}{I_{cr}} = \boxed{\mathbf{50}\ \text{MPa}} \\ &\ \le\ 0.30f_{fu} = 0.30\times850 = 255\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$

$$\begin{aligned}h_{min} &= \frac{L}{16} = 375\ \text{mm}\ \le\ h = 600\ \to\ \text{처짐 계산 생략 가능}\end{aligned}$$

$$\begin{aligned}l_d &= \frac{0.90\,d_b\,f_y}{\lambda\sqrt{f_{ck}}}\cdot\frac{\alpha\beta\gamma}{(c+K_{tr})/d_b} = \frac{0.90\times25.4\times400}{1.0\sqrt{30}}\times\frac{1\times1\times1}{2.50} \\ &= \boxed{\mathbf{668}\ \text{mm}}\ (26.3d_b) \\ \text{약산식 }l_{db} &= \frac{0.6\times25.4\times400}{\sqrt{30}} = 1113\ \to\ 1113\ /\ 1669\end{aligned}$$

$$\begin{aligned}c_b &= \min(c+d_{bv}+d_b/2,\ s/2) = \min(65.7,\ 67.2) = 65.7 \\ c_b/d_b &= \min(2.59,\,3.5) = 2.59 \\ l_d &= \frac{d_b\left(\dfrac{f_{fr}}{0.083\sqrt{f_{ck}}}-340\right)\psi_t}{13.6+c_b/d_b} = \frac{25.4\left(\dfrac{850}{0.083\sqrt{30}}-340\right)\times1}{13.6+2.59} \\ &= \boxed{\mathbf{2{,}400}\ \text{mm}}\ (94.5d_b)\end{aligned}$$

$$\begin{aligned}\alpha &= 1\ (일반),\quad \beta = 1,\quad \gamma = 1\ (D22 이상) \\ c &= \min(65.7,\ 67.2) = 65.7 \\ \frac{c+K_{tr}}{d_b} &= \min\!\left(\frac{65.7+0}{25.4},\,2.5\right) = 2.50\end{aligned}$$

$$\begin{aligned}\psi_t &= 1\ (일반) \\ c_b/d_b &= 2.59\ (\le3.5)\end{aligned}$$

$$\begin{aligned}l_d &= \max(668\times0.753292,\ 300) = \boxed{\mathbf{503}\ \text{mm}}\ (19.8d_b)\end{aligned}$$

$$\begin{aligned}l_d &= \max(2400\times1,\ 20d_b = 508,\ 300) = \boxed{\mathbf{2{,}400}\ \text{mm}}\ (94.5d_b)\end{aligned}$$

$$\begin{aligned}l_{db} &= \max\!\left(\frac{0.25\times25.4\times400}{\sqrt{30}},\ 0.043\times25.4\times400\right) = 464 \\ l_d &= \max(464,\ 200) = \boxed{\mathbf{464}\ \text{mm}}\end{aligned}$$

$$\begin{aligned}l_{d,c} &= l_d = \boxed{\mathbf{2{,}400}\ \text{mm}}\end{aligned}$$

$$\begin{aligned}l_{hb} &= \frac{0.24\times1\times25.4\times400}{\sqrt{30}} = 445 \\ l_{dh} &= \max(445\times0.70,\ 8d_b = 203,\ 150) = \boxed{\mathbf{312}\ \text{mm}}\ (\text{꼬리 }12d_b = 305)\end{aligned}$$

$$\begin{aligned}f_{fu} &= 850\ \text{MPa}\ \to\ \text{구간 }520 < f_{fu} < 1040 \\ l_{bhf} &= 850\times25.4/(3.1\sqrt{30}) = 1272 \\ l_{dh} &= \max(1272\times0.7,\ 12d_b = 305,\ 230) = \boxed{\mathbf{890}\ \text{mm}}\ (\text{꼬리 }12d_b,\ r\ge3d_b = 76)\end{aligned}$$

$$\begin{aligned}l_{dt} &= \frac{0.24\times1\times25.4\times400}{\sqrt{30}} = 445\ \to\ \max(\cdot,\ 8d_b,\ 150) = \boxed{\mathbf{445}\ \text{mm}}\ (\text{그 외 부위 (4.1-7)})\end{aligned}$$

$$\begin{aligned}l_d^{(4.5.2)} &= 668\ (\text{300 최솟값 및 감소 미적용}) \\ l_{lap} &= \max(1.3\times668,\ 300) = \boxed{\mathbf{868}\ \text{mm}}\ (B급)\end{aligned}$$

$$\begin{aligned}l_d(f_{fu}) &= 2400\ (\text{최솟값 및 감소 미적용}) \\ l_{lap} &= \max(1.3\times2400,\ 20d_b = 508,\ 300) = \boxed{\mathbf{3{,}121}\ \text{mm}}\ (B급)\end{aligned}$$

$$\begin{aligned}l_{s1} &= \left(\frac{1.4\times400}{\sqrt{30}}-52\right)\times25.4 = 1276 \\ l_{s2} &= 0.072\times400\times25.4 = 732 \\ l_s &= \max\big(\min(l_{s1},l_{s2}),\ 300\big) = \boxed{\mathbf{732}\ \text{mm}}\end{aligned}$$

$$\begin{aligned}l_{lap,c} &= l_{lap} = \boxed{\mathbf{3{,}121}\ \text{mm}}\end{aligned}$$

$$\begin{aligned}\frac{\phi M_n}{V_u}+l_a &= \frac{464.628\times10^6}{180\times10^3}+150 = \boxed{\mathbf{2{,}731}\ \text{mm}} \\ l_d &= 503\ \le\ 2731\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$

$$\begin{aligned}\frac{\phi M_n}{V_u}+l_a &= \frac{341.564\times10^6}{180\times10^3}+150 = \boxed{\mathbf{2{,}048}\ \text{mm}} \\ l_d &= 2400\ >\ 2048\ \ \color{#d40000}{\times\ \textbf{NG}}\end{aligned}$$