根据《桩规》第 5.9.10 条
$f_{\mathrm{t}} = 1.43\mathrm{MPa}$ ， $h_{10} = 730\mathrm{mm}$ ， $h_{20} = 400\mathrm{mm}$ ， $h_0 = 1130\mathrm{mm}$
$$
\beta_ {\mathrm{hs}} = \left(\frac {8 0 0}{h _ {0}}\right) ^ {1 / 4} = \left(\frac {8 0 0}{1 1 3 0}\right) ^ {1 / 4} = 0. 9 1 7
$$
$$
b = 0. 8 \times 0. 5 = 0. 4 \mathrm{m}
$$
$$
a _ {\mathrm{x}} = 1 0 0 0 - 2 0 0 - 5 0 0 = 3 0 0 \mathrm{mm}
$$
$$
\lambda_ {\mathrm{x}} = \frac {a _ {\mathrm{x}}}{h _ {0}} = \frac {3 0 0}{1 1 3 0} = 0. 2 6 5, \quad 0. 2 5 < \lambda_ {\mathrm{x}} < 3. 0
$$
$$
\alpha = \frac {1 . 7 5}{\lambda + 1} = \frac {1 . 7 5}{0 . 2 6 5 + 1} = 1. 3 8
$$
$$
b _ {\mathrm{yo}} = \frac {b _ {\mathrm{y1}} h _ {1 0} + b _ {\mathrm{y2}} h _ {2 0}}{h _ {1 0} + h _ {2 0}} = \frac {4 5 0 0 \times 7 3 0 + 2 0 0 0 \times 4 0 0}{7 3 0 + 4 0 0} = 3 6 1 5 \mathrm{mm}
$$
$$
V \leq \frac {1}{\gamma_ {\mathrm{RE}}} \beta_ {\mathrm{hs}} \alpha f _ {\mathrm{t}} b _ {0} h _ {0} = \frac {1}{0 . 8 5} \times 0. 9 1 7 \times 1. 3 8 \times 1. 4 3 \times 3 6 1 5 \times 1 1 3 0 \times 1 0 ^ {- 3} = 8 6 9 7 \mathrm{kN}
$$
因此选（C）
