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Soil Mechanics (ECIV 3351

)
Chapter 5: SOIL COMPACTION

Second Term
2011

Problem 1: (5.1 Text book)
Given that: Gs = 2.72
Required:

= ?? "lb/ft3" @ w = 5%, 8%, 10%, 12% and 15%

Solution:
Note:
"zero air voids unit weight" which means that voids contain no air but only water, i.e.
saturation case S=100% .

,

.
w%
5
8
10
12
15

149.4
139.39
133.43
127.96
120.55

1

02 3.8 Required: void ratio "e" and degree of saturation "S" at optimum moisture content.2 12.5 140.8 124.6 16.15 4.3 14.4 10. 90.24 93.9 . of Proctor mold "ft3" 1/30 1/30 1/30 1/30 1/30 Weight of wet soil in mold "lb" 3.72 Vol.3 14.26 4.98 124. Solution: To determine the optimum moisture content we have to sketch the relationship between water content "w%" and dry unit weight To find for each case we have to find due to given data .67 4.Soil Mechanics (ECIV 3351) Chapter 5: SOIL COMPACTION Second Term 2011 Problem 2 :( 5.15 4.02 3.4 10.2 12.76 105. of Proctor mold "ft3" 1/30 1/30 1/30 1/30 1/30 Weight of wet soil in mold "lb" 3.22 112.1 1206.26 4.67 4.6 16.63 Moisture content "%" 8.24 .6 108.8 2 97.5Text book) Given: Gs = 2. then by knowing the moisture content for each case we can find Vol.63 Moisture content "%" 8.

8 90.24 93.e = S= .Soil Mechanics (ECIV 3351) Chapter 5: SOIL COMPACTION Second Term 2011 Moisture content "%" 8..5%.2 12.98 124.76 105. then e = 0.358 ……(1) .4 10..6 16.22 112.3 14. 125 = S.w = 94% .. ……...24 130 125 120 115 110 105 100 95 90 85 80 w% 20 15 10 5 ` From curve: = 125 lb/ft3 @ w% = 12.(2) 3 0 .

2 = …….Soil Mechanics (ECIV 3351) Chapter 5: SOIL COMPACTION Second Term 2011 Problem 3: (5. …………………………………………………………(B) 4 . 5000 = Vv+Vs .85 1.85 = …….9 Text book) Given that: Embankment fill requires 5000 m3 of compacted soil.7.7 = . Required void ratio = 0.18m3 Alternative A: e = 0.Vs …………… (1) e= = 0.2 0.7. Vv = 5000 . Vtot = Vv + Vs . we know that Vs is constant and will not change after compaction.75 Cost "$/m3" 9 6 7 10 Required: Choose the best alternative with the lowest cost .95 0. Solution: Vtot = 5000 m3 .7 …………(2) from (1) and (2) 0. Four borrow pits are available as described in the following table: Borrow Pit A B C D Void ratio 0. then Vs = 2941. …………………………………………………………(A) Alternative B: e = 1. ereq = 0.

B with lowest cost = 5 Second Term 2011 .Soil Mechanics (ECIV 3351) Chapter 5: SOIL COMPACTION Alternative C: e = 0.75 = ……. …………………………………………………………(D) From alternative costs we can choose Alt. …………………………………………………………(C) Alternative D: e = 0.95 = …….

w% 6 8 9 11 12 14 Required: 14.mass of soil to fill the cone = 3.52 18.6% for moist soil from the hole.13 Text book) Given: = 1667 kg/m3 mass "Ottawa sand" to fill the cone = 0. kg/m3. Moist density of soil from the hole = = dry unit weight of soil from the hole = KN/m3.Soil Mechanics (ECIV 3351) Chapter 5: SOIL COMPACTION Second Term 2011 Problem 4: (5.8 17.331 kg w% = 11.18 – 0.117 kg mass ( jar + sand + cone ) before filling hole = 5.063 kg.45 18.mass ( jar + sand + cone ) after filling hole = 5.94 KN/m3 ………. = = 15. Vol..81 kg mass of moist soil from the hole = 3.99 kg mass ( jar + sand + cone ) after filling hole = 2.81 = 3.9 of compaction in the field Relative compaction in the field Solution: Mass of soil to fill the hole and cone = mass ( jar + sand + cone ) before filling hole .…………….18 kg. Mass of soil to fill hole only = Mass of soil to fill the hole and cone . of the hole = .117 = 3.6 16.………………………………… (1) 6 .99 – 2.9 18.

...................... .......(2) 7 .......6 14 16.........9 .45 18. 20 19 18 17 16 15 14 13 12 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 From curve Then........52 18. so we draw the relationship between w% and w% 6 8 9 11 12 14.... R = = 19KN/m3 ……………............9 18...Soil Mechanics (ECIV 3351) Chapter 5: SOIL COMPACTION Relative compaction in the field (R) = We have to find Second Term 2011 .....8 17..