根据《混规》第 7.1.2 条
CD 为轴心受拉构件:
$$
\rho_ {\mathrm{te}} = \frac {A _ {\mathrm{s}}}{A _ {\mathrm{te}}} = \frac {1 2 \times 4 9 0 . 9}{4 0 0 \times 4 0 0} = 0. 0 3 7 > 0. 0 1
$$
$$
\psi = 1. 1 - 0. 6 5 \frac {f _ {\mathrm{tk}}}{\rho_ {\mathrm{te}} \sigma_ {\mathrm{s}}} = 1. 1 - 0. 6 5 \times \frac {2 . 0 1}{0 . 0 3 7 \times \sigma_ {\mathrm{s}}} = 1. 1 - \frac {3 5 . 3}{\sigma_ {\mathrm{s}}}
$$
$$
\omega_ {\max} \leq 0. 2
$$
$$
\alpha_ {\mathrm{cr}} \psi \frac {\sigma_ {\mathrm{s}}}{E _ {\mathrm{s}}} \left(1. 9 c _ {\mathrm{s}} + 0. 0 8 \frac {d _ {\mathrm{eq}}}{\rho_ {\mathrm{te}}}\right) \leq 0. 2
$$
$$
2. 7 \times \left(1. 1 - \frac {3 5 . 3}{\sigma_ {\mathrm{s}}}\right) \times \frac {\sigma_ {\mathrm{s}}}{2 \times 1 0 ^ {5}} \times \left(1. 9 \times 3 0 + 0. 0 8 \times \frac {2 5}{0 . 0 3 7}\right) \leq 0. 2
$$
得： $\sigma_{\mathrm{s}} = 153.37\mathrm{N / mm}^{2}$
将 $\sigma_{\mathrm{s}} = 153.37\mathrm{N / mm}^2$ 带入 $\psi = 1.1 - \frac{35.3}{153.37} = 0.87_{< 1.0}^{>0.2}$ ，满足要求
根据《混规》第7.1.4条
$$
N _ {\mathrm{q}} = \sigma_ {\mathrm{s}} A _ {\mathrm{s}} = 1 5 3. 3 7 \times 1 2 \times 4 9 0. 9 = 9 0 3. 4 7 \mathrm{kN}
$$
因此选（A）
