$$
h _ {0} = h - a _ {\mathrm{s}} = 1 0 0 0 - 4 0 = 9 6 0 \mathrm{mm}
$$
$$
e _ {0} = \frac {M}{N} = \frac {2 4 0 0 \times 1 0 ^ {6}}{7 5 0 \times 1 0 ^ {3}} = 3 2 0 0 \mathrm{mm}, e _ {\mathrm{a}} = \max \left\{2 0, \frac {1 0 0 0}{3 0} \right\} = 3 3. 3 \mathrm{mm}
$$
$$
e _ {\mathrm{i}} = e _ {0} + e _ {\mathrm{a}} = 3 2 0 0 + 3 3. 3 = 3 2 3 3. 3 \mathrm{mm}
$$
$$
e = e _ {\mathrm{i}} + \frac {h}{2} - a _ {\mathrm{s}} = 3 2 3 3. 3 + \frac {1 0 0 0}{2} - 4 0 = 3 6 9 3. 3 \mathrm{mm}
$$
根据《混规》第6.2.17条，受压区高度x所承担的弯矩值 $\mathbf{M}_1$
$$
\begin{array}{r l} M _ {1} & = N e - f _ {\mathrm{y}} ^ {\prime} A _ {\mathrm{s}} ^ {\prime} \left(h _ {0} - a _ {\mathrm{s}} ^ {\prime}\right) \\ & = 7 5 0 \times 1 0 ^ {3} \times 3 6 9 3. 3 - 3 6 0 \times 2 2 8 0 \times (9 6 0 - 4 0) = 2 0 1 4. 8 4 \mathrm{kN} \cdot \mathrm{m} \end{array}
$$
$$
x = h _ {0} - \sqrt {h _ {0} ^ {2} - \frac {2 M _ {1}}{\alpha_ {1} f _ {\mathrm{c}} b}} = 9 6 0 - \sqrt {9 6 0 ^ {2} - \frac {2 \times 2 0 1 4 . 8 4 \times 1 0 ^ {6}}{1 . 0 \times 1 4 . 3 \times 6 0 0}} = 2 8 7. 7 \mathrm{mm} > 2 a _ {\mathrm{s}} ^ {\prime} < \xi_ {\mathrm{b}} h _ {0}
$$
属于大偏压
$$
A _ {\mathrm{s}} = \frac {\alpha_ {1} f _ {\mathrm{c}} b x + f _ {\mathrm{y}} ^ {\prime} A _ {\mathrm{s}} ^ {\prime} - N}{f _ {\mathrm{y}}} = \frac {1 . 0 \times 1 4 . 3 \times 6 0 0 \times 2 8 7 . 7 + 3 6 0 \times 2 2 8 0 - 7 5 0 \times 1 0 ^ {3}}{3 6 0} = 7 0 5 3 \mathrm{mm} ^ {2}
$$
因此选（A）
