Question

for the given sand, the maximum and minimum dry unit weights are 108 lb/ft3 and 92 lb/ft3, respectively. given that Gs=2.65, determine the moist unit weight of the soil when the relative density is 60% and moisture content is 8%.

Answer 1

The moist unit weight of compaction = 109.05 lb/ft3

Step-by-step explanation:

In order to determine the moist unit weight, the dry unit weight has to be evaluated first. If Y is the moist unit weight, then:

Y = Yd (1 + m)

Where:

Yd = dry unit weight

m = moisture content of soil = 8% = 0.08

But the dry unit weight is unknown. In order to calculate the dry unit weight, we will make use of the formula for relative density R;

R = [(Yd — Ydmin) ÷ (Ydmax — Ydmin)] × [Ydmax ÷ Yd]

Where:

R = relative density = 60% = 0.6

Yd = dry unit weight

Ydmin = minimum dry weight = 92 lb/ft3

Ydmax = maximum dry weight = 108 lb/ft3

Therefore R = 0.6 = [(Yd — 92) ÷ (108 — 92)] × [108/Yd]

0.6 = [(Yd — 92)/16] × [108/Yd], or

0.6 = (0.0625Yd — 5.75) × [108/Yd]

0.6Yd = 6.75Yd — 621

6.75Yd — 0.6 Yd = 621

6.15Yd = 621

And Yd = 100.98 lb/ft3 = dry unit weight

But we are asked to find the moist unit weight = Y = Yd (1 + m)

where Yd = dry unit weight and m = moisture content of soil = 8% = 0.08

Therefore, Y = 100.98 (1 + 0.08) = 109.05 lb/ft3.