Assignment 2

ADVANCED SOIL MECHANICS ASSIGNMENT II

4.1 Following are the results from the liquid an d plastic limit tests for a soil. Liquid Limit test: Plastic limit test: PL = 12.2% Number of blows, N Moisture content (%) a. Draw the flow curve and obtain the liquid 16 36.5 limit. 20 34.1 b. What is the plasticity index of the soil? 28 27 Given: PL = 12.2% Required: a. flow curve & LL =? b. PI =? Solution: a. To draw the flow curve by writing the Moisture content vertically and the number of blows horizontally as follows.

40 ) 36 % ( t n e 32 t n o c e r 28 u t s i o m

24 20 10

15

20

25

30

35

40

Number of blows, blows, N

The moisture content correspond to N = 25, determined from the flow curve, gives the liquid limit of the soil. That is LL = 29.7%. See the following figure. 40 flow curve

36

) % ( t n e 32 t n o c e r u28 t s i o m

LL =29. =29.7% 7%

24 20 10

15

20

25

30

35

40

Number of blows, blows, N

N A M E :

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M E R SH A ;

E NG INE E R IN G

GE O L O GY

( M S c )

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a. Plasticity index of the soil can be calculated from the equation: So,

PI=29.7%12.2%=17.5%

PI=LLPL;

4.2 Determine the liquidity index of the soil described in Problem 4.1 if win situ = 31%. Given: PI = 17.5%, LL = 29.7%, w = 31% & PL = 12.2% Solution:

Required: LI =?

LI = (w – PL)/(LL – PL) = (w  PL)/PI = (31%  12.2%)/(17.5%) = 1.07

4.3 Following are the results from the liquid an d plastic limit tests for a soil. Liquid limit test: Number blows, N 15 20 28

Plastic limit test: PL = 18.7% a. Draw the flow curve and obtain the liquid limit. b. What is the plasticity index of the soil? Given: PL = 18.7% Required: a. flow curve & LL =? b. PI =?

of Moisture content, (%) 42 40.8 39.1

Solution: a. to draw the flow curve by writing the Moisture content vertically and the number of blows horizontally as follows. 50

) %45 ( , t n e t n o 40 c e r u t s i o M35

30 10

15

20

25

30

35

40

Number of blows, N 50

The moisture content corresponding to N = 25, determined from the flow curve, gives the liquid limit of the soil. That is LL = 39.7%

Flow corve

) %45 ( , t n e t n o 40 c e r u t s i o 35 M

LL= 39.7%

30 10

15

20

25

30

35

40

Number of blows, N

N A M E :

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M E R SH A ;

E NG INE E R IN G

GE O L O GY

( M S c )

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b. Plasticity index of the soil can be calculated from the equation:

39.7%  18.7% = 21%

PI=LLPL; that is PI =

4.4 Refer to Problem 4.3. Determine the liquidity index of the soil when the in situ moisture content is 26%. Given: PI = 21%, LL = 39.7%, w = 26% & PL = 18.7% Required: LI =?

Solution:

LI = (w – PL)/(LL – PL) = (w  PL)/PI = (26%  18.7%)/(21%) = 0.35

4.6 A saturated soil has the following characteristics: initial volume (Vi) = 24.6 cm3, final volume (Vf ) = 15.9 cm3, mass of wet soil (M l) = 44 g, and mass of dry soil (M 2) 30.1 g. Determine the shrinkage limit and the shrinkage ratio. Given: Vi = 24.6 cm3, Vf = 15.9 cm3, Ml = 44 g, and M2 30.1 g. Pw = 1 g/cm3 Required: SL & SR =? Solution:   44−. 4.−.    . .   .  (.)() 

 L =  100( )( )(100) = ( = = = 1.89

)(100)(

)(1)(100) = 17.28%

5.2 The sieve analysis of ten soils and the liquid and plastic limits of the fraction passing through the No. 40 sieve are given below. Classify the soils using the AASHTO classification system and give the group indexes. Soil no. Sieve analysis (percent finer) Liquid limit Plastic limit No. 10 No. 40 No. 200 1 98 80 50 38 29 2 100 92 80 56 23 3 100 88 65 37 22 4 85 55 45 28 20 5 92 75 62 43 28 6 48 28 6 NP 7 87 62 30 32 24 8 90 76 34 37 25 9 100 78 8 NP 10 92 74 32 44 35 Solution:Soil no. 1: Using Table 5.1, since 50% of the soil is passing through the No. 200 sieve, it falls under silt-clay classifications — that is, it falls under group A-4, A-5, A-6, or A-7. Proceeding from left to right, it falls under group A-4. From Equation; GI = (F200 - 35) [0.2 + 0.005(LL - 40)] + 0.01(F200 -15) (PI- 10) GI = (50 -35) [0.2 +0.005(38 -40)] +0.01(50 -15) (9-10) = 2.5 ≈ 3 So, the soil will be classified as A-4(3). Soil no. 2: According to Table 5.1, this soil falls under group A-7 (proceed in a manner similar to Soil no. 1). Since, PI > LL – 30 = 33 >56 – 30 = 33 > 26. This is an A-7- 6 soil. GI = (F200 - 35) [0.2 + 0.005(LL - 40)] + 0.01(F200 -15) (PI- 10) GI = (80 – 35) [0.2 + 0.005(56 – 40)] + 0.01(80 – 15) (33 – 10) = 27.55 ≈ 28 So, the soil will be classified as A-7-6(28).

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E NG INE E R IN G

GE O L O GY

( M S c )

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Soil no. 3: According to Table 5.1, this soil falls under group A-6 (proceed in a manner similar to Soil no. 1). GI = (F200 - 35) [0.2 + 0.005(LL - 40)] + 0.01(F200 -15) (PI- 10) GI = (65 – 35) [0.2 + 0.005(37 – 40)] + 0.01(65 – 15) (15 – 10) = 8.05 ≈ 8 Thus, the soil type is A-6(8) Soil no. 4: According to Table 5.1, this soil falls under group A-4 (proceed in a manner similar to Soil no. 1). GI = (F200 - 35) [0.2 + 0.005(LL - 40)] + 0.01(F200 -15) (PI- 10) GI = (45 – 35) [0.2 + 0.005(28 – 40)] + 0.01(45 – 15) (8 – 10) = -4.6≈ 0 Thus, the soil type is A-4(0) Soil no. 5: According to Table 5.1, this soil falls under group A-7 (proceed in a manner similar to Soil no. 1). Since, PI > LL – 30 = 15 > 43 – 30 = 15 > 13. This is an A-7- 6 soil. GI = (F200 - 35) [0.2 + 0.005(LL - 40)] + 0.01(F200 -15) (PI- 10) GI = (62 – 35) [0.2 + 0.005(43 – 40)] + 0.01(62 – 15) (15 – 10) = 8.155 ≈ 8 Thus, the soil type is A-7-6(8) Soil no. 6: Using Table 5.1, since 6% of the soil is passing through the No. 200 sieve, it falls under granular material classification — that is, A-1, A-2, or A-3. Proceeding from left to right, we see that it falls under A-1-a. The group index for A-1-a is zero. So, the soil type is A-1-a (0). Soil no. 7: Using Table 5.1, since 30% of the soil is passing through the No. 200 sieve, it falls under granular material classification — that is, A-1, A-2, or A-3. Proceeding from left to right, we see that it falls under A-2-4. The group index for A-2-4 is zero.

So, the soil type is A-2-4(0) . Soil no. 8: Using Table 5.1, since 34% of the soil is passing through the No. 200 sieve, it falls under granular material classification — that is, A-1, A-2, or A-3. Proceeding from left to right, we see that it falls under A-2-6. GI = 0.01(F -15) (PI -10); Now, F = 34; PI =12; so, GI =0.01(34 -15) (12 -10) = 0.38 ≈ 0 Thus, the soil is A-2-6(0) . Soil no. 9: Using Table 5.1, since 8% of the soil is passing through the No. 200 sieve, it falls under granular material classification — that is, A-1, A-2, or A-3. Proceeding from left to right, we see that it falls under A-3. The group index for A-3 is zero. So, the soil type is A-3(0). Soil no. 10: Using Table 5.1, since 32% of the soil is passing through the No. 200 sieve, it falls under granular material classification — that is, A-1, A-2, or A-3. Proceeding from left to right, we see that it falls under A-2-5. The group index for A-2-5 is zero. So, the soil type is A-2-5(0). 5.4 For an inorganic soil, the following grain-size analysis is given. For this soil, LL = 23 and PL= 19. Classify the soil by using U.S. sieve no. Percent finer a. AASHTO soil classification system 4 100 b. Unified soil classification system 10 90 Give group names and group symbols. 20 64 40 38 80 18 200 13 Solution: N A M E :

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( M S c )

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a. Using Table 5.1, since 13% of the soil is passing through the No. 200 sieve, it falls under granular material classification — that is, A-1, A-2, or A-3. Proceeding from left to right, we see that it falls under A-1-b. The group index for A-1-b is zero. So, the soil type is A-1-b (0). b. Refer to Table 5.2 the percentage passing No. 200 sieve is 13%, which is less than 50%. So it is a coarse grained soil. Thus, coarse fraction = % retained on No. 200 sieve = 100% – 13% = 87% Gravel fraction = % retained on No. 4 sieve = 100% - 100% = 0% Sand fraction = % retained on No.200 sieve - % retained on No.4 sieve = 87 % - 0% = 87% Hence, more than 50% of the coarse fraction is passing No. 4 sieve. Thus, it is a sandy soil. Since more than 12% passing No. 200 sieve, then the group symbol is SC. For the group name, refer to fig 5.4. Since the percentage of gravel is less than 15%, it is clayey sand.

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N A M E :

T I LA H U N

M E R SH A ;

E NG INE E R IN G

GE O L O GY

( M S c )

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Assignment 2

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