（1）支座截面，P 作用在 A 支座边，

$$ V _ {\mathrm{A}} = P + 0. 5 q l = P + 0. 5 \times 4 0 \times 8 = P + 1 6 0 $$

根据《混规》第6.3.4条

$$ h _ {0} = 6 5 0 - 7 0 = 5 8 0 \mathrm{mm} $$

$$ V _ {\mathrm{cs}} = \alpha_ {\mathrm{cv}} f _ {\mathrm{t}} b h _ {0} + f _ {\mathrm{yv}} \frac {A _ {\mathrm{sv}}}{s} h _ {0} $$

$$ V _ {\mathrm{cs}} = 0. 7 \times 1. 4 3 \times 2 5 0 \times 5 8 0 + 3 6 0 \times \frac {2 \times 5 0 . 3}{1 0 0} \times 5 8 0 = 3 5 5. 2 \mathrm{kN} $$

根据《混规》第6.3.1条

$$ h _ {\mathrm{w}} / b \leq 4, 0. 2 5 \beta_ {\mathrm{c}} f _ {\mathrm{c}} b h _ {0} = 0. 2 5 \times 1 4. 3 \times 2 5 0 \times 5 8 0 \times 1 0 ^ {- 3} = 5 1 8. 4 \mathrm{kN} > V _ {\mathrm{cs}} $$

$V_{\mathrm{A}} = P + 160 = 355.2\mathrm{kN}$ 得 $P = 195.2\mathrm{kN}$

(2) C 截面, P 作用在 C 右侧,

验算最小配箍率要求

$$ \frac {2 \times 50.3}{250 \times 250} = 0.16 \% > 0.24 \frac {f _ {\mathrm{t}}}{f _ {\mathrm{yv}}} = 0.24 \times \frac {1.43}{360} = 0.095 \% $$

$$ V _ {\mathrm{C}} = \frac {6 . 5}{8} P + 4 0 \times 2. 5 = \frac {6 . 5}{8} P + 1 0 0 $$

$$ V _ {\mathrm{cs}} = 0. 7 \times 1. 4 3 \times 2 5 0 \times 5 8 0 + 3 6 0 \times \frac {2 \times 5 0 . 3}{2 5 0} \times 5 8 0 = 2 2 9. 2 \mathrm{kN} < 5 1 8. 4 \mathrm{kN} $$

$V_{A}=\frac{6.5}{8}P+100=229.2kN$ 得 P=159.02kN

因此选（B）。