Question

Atterberg limits and natural water contents for five fine grained soils are listed in the following table:

Soil No.	LL (%)	PL (%)	Natural Water Content (%)
1	10	10	4
2	33	18	20
3	46	34	12
4	74	31	32
5	74	49	16

Classify the soils in the Unified Soil Classification System, estimate their optimum water contents and maximum dry densities, and indicate approximately how much each soil must be wetted or dried if it is to be efficiently compacted.

Answer 1

SOLUTLON N we have given fine grained soil. . 3011 Somple 1: Liquid limit (LL) = 10%. Plastic limit (PL) - 10% P.Inder (PL)- Lil-P.L-10-10=0 Natural water content = 4%. SOLL Sample 2 : liquid limit (LL) = 33 % Plastic limit (PL) = 18% Plasticily Index PL = 33-18 = 15% Naderat water coment - 20. soll Sample 8: liquid limit (LL) = 464. Plastic limit (ll): 34). DE = 46-34 = 12 % similarly for soil sample 4 ll= 944; PLE 312; PF = 43 Natural water content = 32%. soil sample s LL2 74% ; PL= 49%; PR = 25 Nechiral water content = 16%.

on plotting the given and obtained data in atter being graph or chand we get: A-line PE=0.73/11-20) Plasticity index 90 (PI) S ME | MH 105 H ( 4 CLAMLEDLI 10 20 30 40 50 to do so Liquid limit 01.) → from the table it is obtained that given soil samples can be classified as: somple 1 = MC : low plasticity sult ☺ sample 2. : (L = tow plastiüty clay 0 Sample 3 = MC = low plasticity sår © sample 4 : cn = High Plasticity clay. samiple S = mx = sugu plastify gilt. Now as per Isser "Corndation Between compaction characteristics and Afterbery limits of fine grained soils" we have OMC MDD 0.916*PL -0.030* PL. -0.875. -0.18 *PL -0.027 * PC +21-182 R-807 R20-835.

Samples: OMC = 0.910*10 -0.030 * 0 -0.875 = 0.29% MOD = -0.18 X 10 -0.027* 0+21182 = 19 since N. water cont. = 44 pamble in to be welted in order to seach Lombocł su il efficiently Samyled OMC = 0.916* 18- 0.030% 15 -0.875 = 15:16 %. MOD = -0.18 X18 - 0.023x15 +21.189 = 17:54 19 1943 1. Sample should be dried. OMC = 0.918 * 34 -0.030* 12 -0.875 - 29.41 %. MOD = -0.18x34-0.027x12 + $1.182 = 14.74 A sample should be wetted. Sample 4. OMC = 0.916* 31 -0.030* 43-0.875 = 26.435. MOD = -0.18X31 -0.09 7x 43 +91.182 - 14 3. Jumple should be doned. Sample s. OMC = 0.416*44 -0.030 *25 -0.375 = 43.26 MDD = -0.18 49 -0.029X35 +21:18 2 = 11.6 Sample should be welled. .