Question

I need help on this Geotechnical Engineering its due today

W = V V = W, = -- Air Water Solid Figure 3.13 3.4 During a compaction test in the geotechnical laboratory, the students com- pacted a clayey soil into a cylindrical mold 4 in. in diameter and 4.58 in. in height. The compacted soil in the mold weighed 4 lb, and it had a moisture content of 12%. If G = 2.72, determine the following: a. Dry unit weight los s b. Void ratio c. Degree of saturation d. Additional water (in lb) needed to achieve 100% saturation in the soil sample 3.5 Two undisturbed soil samples, each having a volume of 0.1 ft, are collected

Answer 1

SETE DATE : PRE NO.: 3.41 Given that diameter of Cylindrical mold & 4 in. height of cylindriced mold 4:58 in. ? Weight of compacted Soil - ulb. moisture content 12 %) amb - A 2 2.72 density of water in 26/43 62.4 (W) + 0.03612 lb/in3 lb / 43 Volume of given mold & tr²h O S X 2 X 4.5820 deix oon -= 57.554 in Bulk unit weight of soil (n) y por 57.554 +0,0695 lb/ins polo ma voler ay Dry unit weight (a) Vt d = 0.0695 1to12 = 0.0695 = 0.062 lb/i3. It w 1:12 6 void ratio (e) &d Gw не 0.062 = 2.72x0.63612 Ite te = 1.5832 ez 0.5832

(DATE c Degree of saturation (s) We know that es Go W. S = 2.72 X 12 0.5832 0.5596 55.962 % = s S Additional water 100 7. Saturation needed to achieve es = IN GS a 0-5832 x1 = W x 2-72 w 0:21:44 = 9.44% to wa 21.44 %. additional water = 21:44 - 12 We weight of water Weight of Solid - 9.44 = 100 Weight of water (Vd) x volume of additional watter = 9.44 x 0.062 X 57.554 7.0.336 lb Summary ay Dry unit weight 0.062 lb/in? void ratio 0.5832 Degree of saturation = 55.962 % additional water needed to achieve 100%. saturation a) 0.336 lb. - ole

PAGE NO.: 3:18 niven porosity 1750 kg/m3 that (n)= la 0.36 ar void ratio (e) I we know that e n e = 0.36 do o by bobaan elow by Specific gravity of soil solid Specific gravity =) density of (46) vis density of Solid waiter. . ping 17. 180050 1:75 698) = . summary Tay Noidratio = 0.56 25 W SEC By specific gravity of soil solid 1.75