Saturated Unit Weight

Take a unit volume of solids: it weighs $G_s \gamma_w$ and occupies 1, while the voids occupy e and, if saturated, hold $e\gamma_w$ of water. Total weight $(G_s + e)\gamma_w$ over total volume (1 + e) gives the saturated unit weight directly. For $G_s$ = 2.68, e = 0.70 and $\gamma_w$ = 9.81 kN/m³ that is (3.38 × 9.81) ÷ 1.70 = 19.50 kN/m³. In imperial units $G_s$ = 2.65 with e = 0.55 and $\gamma_w$ = 62.4 pcf gives (3.20 × 62.4) ÷ 1.55 = 128.8 pcf.

The trap is applying $\gamma_{\text{sat}}$ above the water table. Soil there is moist, not saturated, and its unit weight is lower: using $\gamma_{\text{sat}}$ for a whole profile overstates total stress and, worse, understates the effective stress ratio the strength depends on. The second trap is arithmetic laziness: $\gamma_{\text{sat}}$ is not $\gamma_d + \gamma_w$. A denser soil holds less water, so as e falls $\gamma_{\text{sat}}$ rises while the water carried per unit volume falls. Run the numbers at e = 0.4 and e = 1.0 and watch the two effects part company.