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)

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

8 2 97.63 Moisture content "%" 8.4 10. then by knowing the moisture content for each case we can find Vol.63 Moisture content "%" 8.8 Required: void ratio "e" and degree of saturation "S" at optimum moisture content.15 4.3 14.1 1206.8 124.02 3.24 93.5Text book) Given: Gs = 2.2 12.98 124.6 16.76 105.6 108.24 .4 10. 90.15 4.6 16.67 4.72 Vol.26 4.3 14.9 . of Proctor mold "ft3" 1/30 1/30 1/30 1/30 1/30 Weight of wet soil in mold "lb" 3.26 4.22 112.5 140. 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 .2 12. of Proctor mold "ft3" 1/30 1/30 1/30 1/30 1/30 Weight of wet soil in mold "lb" 3.Soil Mechanics (ECIV 3351) Chapter 5: SOIL COMPACTION Second Term 2011 Problem 2 :( 5.02 3.67 4.

…….98 124.w = 94% .4 10.(2) 3 0 . then e = 0.e = S= .2 12..3 14.24 93..76 105.Soil Mechanics (ECIV 3351) Chapter 5: SOIL COMPACTION Second Term 2011 Moisture content "%" 8.358 ……(1) .5%. 125 = S..8 90..22 112.24 130 125 120 115 110 105 100 95 90 85 80 w% 20 15 10 5 ` From curve: = 125 lb/ft3 @ w% = 12..6 16.

2 0. Vtot = Vv + Vs . then Vs = 2941. we know that Vs is constant and will not change after compaction. 5000 = Vv+Vs .Soil Mechanics (ECIV 3351) Chapter 5: SOIL COMPACTION Second Term 2011 Problem 3: (5.9 Text book) Given that: Embankment fill requires 5000 m3 of compacted soil. Required void ratio = 0. Four borrow pits are available as described in the following table: Borrow Pit A B C D Void ratio 0.Vs …………… (1) e= = 0. Vv = 5000 . Solution: Vtot = 5000 m3 .7 …………(2) from (1) and (2) 0.2 = …….75 Cost "$/m3" 9 6 7 10 Required: Choose the best alternative with the lowest cost .85 1.7.85 = …….7.18m3 Alternative A: e = 0. ereq = 0. …………………………………………………………(B) 4 .7 = . …………………………………………………………(A) Alternative B: e = 1.95 0.

…………………………………………………………(C) Alternative D: e = 0.75 = ……. B with lowest cost = 5 Second Term 2011 .95 = …….Soil Mechanics (ECIV 3351) Chapter 5: SOIL COMPACTION Alternative C: e = 0. …………………………………………………………(D) From alternative costs we can choose Alt.

18 kg.331 kg w% = 11.99 kg mass ( jar + sand + cone ) after filling hole = 2.99 – 2.81 = 3.117 = 3. = = 15. w% 6 8 9 11 12 14 Required: 14.mass of soil to fill the cone = 3. Mass of soil to fill hole only = Mass of soil to fill the hole and cone .8 17. kg/m3.9 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 .9 18..18 – 0.45 18.Soil Mechanics (ECIV 3351) Chapter 5: SOIL COMPACTION Second Term 2011 Problem 4: (5.………………………………… (1) 6 . of the hole = .6% for moist soil from the hole.6 16.13 Text book) Given: = 1667 kg/m3 mass "Ottawa sand" to fill the cone = 0.mass ( jar + sand + cone ) after filling hole = 5. Moist density of soil from the hole = = dry unit weight of soil from the hole = KN/m3.063 kg.81 kg mass of moist soil from the hole = 3.117 kg mass ( jar + sand + cone ) before filling hole = 5.52 18. Vol.…………….94 KN/m3 ……….

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

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