02

取黏土试样测得：质量密度$\rho=1.80g/cm^3$，土粒比重$G_s = 2.70$，含水量 $\omega=30\%$。 拟使用该黏士制造比重为1.2的泥浆，计算制造$1m^3$泥浆所需的黏士质量为下列哪个选项？

(A) 0.41t

(B) 0.67t

(C) 0.75t

(D) 0.90t

解答一

$$V_s + V_w = 1 \Rarr \cfrac{m_s}{2.7} + \cfrac{1.2 × 1 - m_s}{1} = 1 \Rarr m_s = 0.318t \\~\\ m = 0.318 × ( 1 + 0.3) = 0.413t$$

解答二

《工程地质手册》公式2-6-1

注意

只适用完全饱和泥浆

$$m_s = V\rho_s \cfrac{\rho_2 - \rho_w}{\rho_s - \rho_w} = 1 × 2.7 × \cfrac{1.2 - 1}{2.7 - 1} = 0.45t \\~\\ m = \rho V = \rho_d(1 + \omega)V = 0.318 × (1 + 0.3) × 1 = 0.413t$$

解答三

$$1.2 = \cfrac{2.7 + e}{1 + e} \Rarr e = 7.5 \\~\\ \rho_d = \cfrac{2.7}{1 + 7.5} = 0.318t/m^3 \\~\\ m = \rho V = \rho_d(1 + \omega)V = 0.318 × (1 + 0.3) × 1 = 0.413t$$