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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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