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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

15 4. then by knowing the moisture content for each case we can find Vol.26 4.98 124.9 .3 14.Soil Mechanics (ECIV 3351) Chapter 5: SOIL COMPACTION Second Term 2011 Problem 2 :( 5.67 4.8 2 97. 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 .24 93.63 Moisture content "%" 8.5Text book) Given: Gs = 2.6 16. of Proctor mold "ft3" 1/30 1/30 1/30 1/30 1/30 Weight of wet soil in mold "lb" 3.1 1206.2 12.3 14.26 4.63 Moisture content "%" 8.15 4. 90.5 140.67 4.4 10.8 Required: void ratio "e" and degree of saturation "S" at optimum moisture content.6 108.24 .8 124.4 10.02 3.76 105. of Proctor mold "ft3" 1/30 1/30 1/30 1/30 1/30 Weight of wet soil in mold "lb" 3.22 112.6 16.72 Vol.02 3.2 12.

3 14.6 16.w = 94% .5%. 125 = S. …….76 105..22 112.8 90..98 124.358 ……(1) .Soil Mechanics (ECIV 3351) Chapter 5: SOIL COMPACTION Second Term 2011 Moisture content "%" 8.24 93.2 12. then e = 0..24 130 125 120 115 110 105 100 95 90 85 80 w% 20 15 10 5 ` From curve: = 125 lb/ft3 @ w% = 12.4 10..(2) 3 0 .e = S= ..

ereq = 0. Vtot = Vv + Vs . we know that Vs is constant and will not change after compaction.7 = .Vs …………… (1) e= = 0.7.95 0.18m3 Alternative A: e = 0. Four borrow pits are available as described in the following table: Borrow Pit A B C D Void ratio 0. Solution: Vtot = 5000 m3 . then Vs = 2941. …………………………………………………………(B) 4 .2 = …….75 Cost "$/m3" 9 6 7 10 Required: Choose the best alternative with the lowest cost . 5000 = Vv+Vs .7.85 = ……. Vv = 5000 .Soil Mechanics (ECIV 3351) Chapter 5: SOIL COMPACTION Second Term 2011 Problem 3: (5.85 1.7 …………(2) from (1) and (2) 0.9 Text book) Given that: Embankment fill requires 5000 m3 of compacted soil. …………………………………………………………(A) Alternative B: e = 1. Required void ratio = 0.2 0.

B with lowest cost = 5 Second Term 2011 .75 = …….95 = …….Soil Mechanics (ECIV 3351) Chapter 5: SOIL COMPACTION Alternative C: e = 0. …………………………………………………………(C) Alternative D: e = 0. …………………………………………………………(D) From alternative costs we can choose Alt.

Vol. Mass of soil to fill hole only = Mass of soil to fill the hole and cone .52 18.99 kg mass ( jar + sand + cone ) after filling hole = 2.9 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 .mass of soil to fill the cone = 3.9 18.94 KN/m3 ……….331 kg w% = 11.063 kg.8 17.6 16.18 kg.mass ( jar + sand + cone ) after filling hole = 5.Soil Mechanics (ECIV 3351) Chapter 5: SOIL COMPACTION Second Term 2011 Problem 4: (5.………………………………… (1) 6 . = = 15. Moist density of soil from the hole = = dry unit weight of soil from the hole = KN/m3.……………. w% 6 8 9 11 12 14 Required: 14.117 kg mass ( jar + sand + cone ) before filling hole = 5.13 Text book) Given: = 1667 kg/m3 mass "Ottawa sand" to fill the cone = 0. kg/m3.18 – 0.99 – 2.81 = 3.81 kg mass of moist soil from the hole = 3. of the hole = .45 18.117 = 3..6% for moist soil from the hole.

....... ...... 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 ........52 18.45 18....8 17..6 14 16............ so we draw the relationship between w% and w% 6 8 9 11 12 14....(2) 7 ................ 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..9 ............

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