以 CD 为研究对象， $F_{Cy}=\frac{1}{2}\times48\times2.5=60kN$

以 ABC 为研究对象，对 B 点取矩，力矩平衡

$$ \sum \mathrm{M} _ {\mathrm{B}} = 0, $$

$$ F _ {\mathrm{A}} \cdot 1 0 = P \cdot 5 + \frac {1}{2} q \cdot 1 0 ^ {2} - F _ {\mathrm{Cy}} \cdot 2. 5 - \frac {1}{2} q \cdot 2. 5 ^ {2} $$

$$ 1 0 F _ {\mathrm{A}} = 6 0 0 \times 5 + \frac {1}{2} \times 4 8 \times 1 0 ^ {2} - 6 0 \times 2. 5 - \frac {1}{2} \times 4 8 \times 2. 5 ^ {2} $$

$$ F _ {\mathrm{A}} = 5 1 0 \mathrm{kN} $$

$$ V _ {\mathrm{B}} = 6 0 0 + 4 8 \times 1 0 - 5 1 0 = 5 7 0 \mathrm{kN} $$

根据《混规》第6.3.1条:

$$ \frac {h _ {\mathrm{w}}}{b} = \frac {6 1 0}{3 0 0} = 2. 0 3 < 4, V \leq 0. 2 5 \beta_ {\mathrm{c}} f _ {\mathrm{c}} b h _ {0} = 0. 2 5 \times 1. 0 \times 1 6. 7 \times 3 0 0 \times 6 1 0 \times 1 0 ^ {- 3} = 7 6 4 \mathrm{kN} $$

第6.3.4条

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

$$ \frac {A _ {\mathrm{sv}}}{s} = \frac {5 7 0 \times 1 0 ^ {3} - 0 . 7 \times 1 . 5 7 \times 3 0 0 \times (6 5 0 - 4 0)}{3 6 0 \times (6 5 0 - 4 0)} = 1. 7 \mathrm{mm} ^ {2} / \mathrm{mm} $$

因此选（C）。