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Solution 3.1 Prove the following: (a) e 1 G w s d + γ = γ Start with fundamental definitions: s s s s w s s s d V and W for subsitute ; ) e l ( V V ; G V W ; V W + = γ = = γ d w s s s w s d e l G ) e l ( V G V γ = + γ = + γ = γ (b) e = n 1 n S = n ) n 1 ( wG e wG s s =

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### Transcript of G γ s w γ - TEST BANK 360testbank360.eu/...soil-mechanics-and-foundations... · G s w d min s w d...

Solution 3.1 Prove the following:

(a) e1

G wsd +

γ=γ

sssswsss

d VandWforsubsitute;)el(VV;GVW;V

W+=γ==γ

dws

s

swsd el

G)el(V

GVγ=

=+

γ=γ

(b) e = n1

n−

∴ S = n

)n1(wGe

wG ss −=

Solution 3.2

d d dr

d d

D min max

max min

γ − γ γ=γ − γ γd

Dr = relative density which is usually defined in terms of min, max and current void ratio (emin, emax, e, respectively). The corresponding dry densities are:

e1G

;e1

G;

e1G ws

dmin

wsd

max

wsd maxmin +

γ=γ

=γ+γ

Relative density in terms of void ratios is:

re eD

e emax

max min

−=

Solving for the appropriate void ratios and substituting into :

1G

e;1G

e;1G

ed

ws

d

wsmin

d

wsmax

maxmin

−=−=−=γγ

γγ

γγ

s w s w

d dr

s w s w

d d

G G1 1D G G1 1

min

min max

γ γ− − +

γ γ=

γ γ− − +

γ γ

=

maxmin

min

dd

dd

11

11

γ−

γ

γ−

γ

=

maxmin

min

maxmax

max

min

min

min

dd

d

dd

d

dd

d

dd

d

11

11

γγ

γ−

γγ

γ

γγ

γ−

γγγ

=

minmax

minmax

min

min

dd

dd

dd

dd

γγ

γ−γ

γγ

γ−γ

= minmax

minmax

min

min

dd

dd

dd

dd

γ−γ

γγ

γγ

γγ

rD∴ = ⎟⎟⎠

⎞⎜⎜⎝

γ

γ

γ−γ

γ−γ

d

d

dd

dd max

minmax

min

Solution 3.3 Given Gs = 2.70, w = 0.65, and e = 1.96, Plot S vs. w. Use S = (w Gs) / e and realize that S will equal 1 at about 71.6% water content. Assume that up to this point, the void ratio will remain constant. After this point, the void ratio will increase and the saturation will remain at 1.0. The relationship is linear up to saturation. Notice that there are no data points shown as this is generated data, not observed data.

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0 20 40 60 80 100

w, water content (%)

S, s

atur

atio

n

Solution 3.4 Strategy It is easiest to consider that each sphere occupies a unit volume or else you many calculate the volume of N number of spheres, or if D is the diameter of the sphere, the volume occupied by it in the array is D3 for the

cubic (loose) array and 3 1D2

for the dense array.

Loose array Step 1. Calculate volume of sphere of diameter D:

6

3DVsphereπ

=

Step 2. Calculate solid volume ratio occupied by sphere

66

3

3

ππ

=⎟⎟⎠

⎞⎜⎜⎝

=D

D

VS

Step 3. Calculate the porosity

4764.06

1 =−=πn

Step 4. Calculate the void ratio

91.05236.04764.0

1==

−=

nne

Dense Array Step 1. Calculate volume of sphere of diameter D:

6

3DVsphereπ

=

Step 2. Calculate volume of space occupied by sphere. The height of space occupied is a tetrahedron.

Height of tetrahedron is2D3

. Space occupied is: 33 2D D D D4 3

× × =12

3

3

D6

VS1 1D2

⎛ ⎞π⎜ ⎟

8π⎝ ⎠= =

Step 3. Calculate Porosity

2595.018

1 =−=πn

Step 4. Calculate void ratio

35.02595.01

2595.01

=−

=−

=n

ne

Solution 3.5 D = 6πrηv

W = s3r

34 γπ

W ′ = Effective weight of particle = )(r34

wsat2 γ−γπ

For equilibrium, D = W ′

( )wsat3r

34vr6 γ−γπ=ηπ

( ) v2

9rwsat γ−γ

η=

[To show that D = vr6 ηπ ]

D = CD γw A2

v2

CD = drag coefficient of sphere = RN

24

NR = Reynold’s number = vdηγ w

d = diameter of sphere

A = 4d 2π

(projected area of sphere)

∴D = vr6vd32vA

N24 2

wR

ηπηπγ ==

Solution 3.6 Md = mass of dry soil = 100 grams Mwe = mass of water of equivalent volume to dry soil = volume of water displaced x density of water (1 gram/cm3) = (537.5 – 500) x 1 = 37.5 grams

ds

we

M 100G 2M 37.5

= = = .67

Solution 3.7 Given: S = 1.0, MT = 500g, Ms = 400 g (a) Water Content (w)

( )T sw

s s

500 400M MMw 0.25 25%

M M 400

−−= = = = =

(b) Void Ratio (e)

e = ( )sG w2.7 0.25 0.675

S= =

Alternatively: Gs = 2.7, 1 gm/ (mass density of water) =ρw

3cm

3ss

s s

M 400V 1

G 2.7 1= = =

ρ ×48.1cm

Because S = 1, Vv = Vw

Vw = Vv = ( ) 3w

w

500 400M100 cm

1

−= =

ρ

3v

3s

V 100 cme 0.675

V 148.1 cm∴ = = =

(c) Saturated Unit Weight (γsat)

ssat w

G e 2.7 0.6759.8

1 e 1 0.675+ +⎛ ⎞ ⎛ ⎞

γ = γ =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠= 19.7 kN/ 3m

(d) Effective Unit Weight )(γ ′

wsat γγγ −=′ = 19.7 – 9.8 = 9.9 kN/ 3m

Solution 3.8 (a) Void Ratio (e)

w1d +γ

=γ = 1.16.19 = 17.8 kN/ 3m

e1G ws

d +γ

=γ ; 8.17

8.97.2e

×=∴ - 1 = 0.49

(b) Degress of Saturation (S)

Se = w , S = sGe

wG s = 49.0

7.21.0 × = 0.55 or 55%

(c) % of Air voids 3

ws

ss m67.0

8.97.28.17

GW

V =×

=

sv VVV −= = 1 – 0.67 = 0.33 3m

Weight of water per 3m = =×=γ 8.171.0w d 1.78 kN

8.978.1Vw = = 0.18 3m

wvair VVV −= = 0.33 – 0.18 = 0.15 3m

% of air voids = 10033.015.0

VV

v

air ×= = 45%

Solution 3.9 Since the soil is saturated S = 1 Find void ratio Se = ; e = 2.7 = 0.62 sWG 23.0×Find dry unit weight

e1G ws

d +γ

=γ = 3m/kN3.1662.1

8.97.2=

×

Find bulk unit weight ( w1d + )γ=γ = 16.3 = 20.1 kN/( 23.01+ ) 3m

Find porosity

n = 62.162.0

e1e

=+

= 0.38 or 38%

Solution 3.10 The total volume of wet sand, VT is equal to 4.64 x 10-4 m3. The wet weight is 8N and after drying, the dry weight is 7.5N. First, find the water content:

%7.6%100xN5.7

)N5.7N8(www

solids

water =−

=

Next, find the dry unit weight:

334

T

solidsd m/kN2.16

m10x64.4N5.7

Vw

==γ −

Then, using the relationship:

ws

d e1G

γ+

=γ , solve it for the void ratio, e.

64.01m/kN81.9m/kN2.16

70.270.21Ge 33w

d

s =−⎟⎟⎠

⎞⎜⎜⎝

⎛=−γ

γ=

Finally, use the relationship:

sGweS = to obtain the saturation, S = 0.28 = 28%

Solution 3.11 Given: w = 0.10, γ = 18.5 kN/ 3m , Gs = 2.7, emax = 0.87 (loose state), emin = 0.51 (dense state) (a) Relative Density (Dr)

Dr = minmax

max

eeee

−−

x 100

Determine e by first calculating dγ

3d

18.516.8 kN /m

1 w 1.1γ

γ = = =+

s w s w

dd

G G 2.7 9.8e 1 1; e 0.575

1 e 16.8γ γ ×

γ = ⇒ = − = − =+ γ

Calculate Dr: Dr = 0.87 0.575

100 81.9%0.87 0.51

−× =

(b) Degree of Saturation

swG 0.1 2.7S 4

e 0.575×

= = = 7%

Solution 3.12 Given: e = 1.2 Determine bulk (γ) and effective (γ′) unit weights for the following degrees of saturation: (a) S = 0.75

3sw

G Se 2.7 0.75 1.29.8 16 kN/m

1 e 1 1.2+ + ×⎛ ⎞ ⎛ ⎞

γ = γ = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

3s

sat w

G e 2.7 1.29.8 17.4 kN /m

1 e 1 1.2+ +⎛ ⎞ ⎛ ⎞γ = γ = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠

wsat γ−γ=γ′ = 17.4 – 9.8 = 7.6 kN/ 3m (b) S = 0.95

2.7 0.95 1.29.8

1 1.2+ ×⎛

γ = ⎜ +⎝ ⎠

⎞⎟ = 17.1 kN/ 3m

wsat γ−γ=γ′ = 17.1 – 9.8 = 7.3 kN/ 3m

(c) For S = 1, satγ = 17.4 kN/ 3m % Error in γ if S = 0.95 but assumed to be 1.0:

% Error = 17.4 17.1100

17.1−

× = 1.8%

Solution 3.13 Sieve size Weight % Rtd Cum.% Rtd % finer

4.75 3.1 12.8 12.8 87.22 5.8 24.0 36.8 63.2

0.85 3.8 15.7 52.5 47.50.25 2.6 10.7 63.2 36.8

0.075 6.8 28.1 91.3 8.7Pan 2.1Sum 24.2 Grain Size Distribution:

Problem 3.13

0

20

40

60

80

100

0.010.1110

Limiting Diameter (mm)

Perc

ent F

iner

(%)

From the grain size distribution, d10 = 0.085 mm, d30 = 0.18 mm, and d60 = 1.70 mm. Strictly speaking, for a sand to be graded SW, it must have a Cu greater than 6 and 1 < Cc < 3.

60u

10

d 1.70C 20.0

d 0.085= = =

2 2

30c

60 10

d 0.18C 0

d d (1.70) (0.085)= = = .22

Hence, by strict application of the classification system, this is an SP. However, there is only about 10% fines and Cu (20) is much greater than 6. Recognize that some gap-grading is happening here. Better to call it an internally unstable well-graded (silty) sand. This soil would be a poor choice to use as a filter.

Solution 3.14

0

20

40

60

80

100

0.010.1110

Limiting Diameter (mm)

Perc

ent F

iner

by

Wei

ght Soil A

Soil BSoil C

Soil A: More than 50% of the sample is greater than 0.075 mm (no. 200 sieve) and all of the soil is smaller than 4.75 mm (no. 4 sieve). Hence, the first letter will be S. For the approximate 37% of the sample that is finer than the no. 200 sieve, there is a ratio of 4:1 clay to silt. This amount of clay should produce significant plasticity in the soil and plot above the A-line. Hence, the soil should be classified as SC, a clayey sand. Soil B: 100% of the sample passes the no. 4 sieve and 0% passes the no. 200 sieve. Hence, the first letter will be S. Look at the grain size distribution. From the grain size distribution, d10 = 0.12 mm, d30 = 0.40 mm, and d60 = 0.75 mm.

60u

10

d 0 75C 6d 0 12

. .

.= = = 25

2 2

30c

60 10

d 0 40C 1d d 0 75 0 12

. .( . ) ( . )

= = = 78

This soil satisfies all criteria for a well-graded sand, SW. Soil C: 100% of the sample passes the no. 4 sieve and 0% passes the no. 200 sieve. Hence, the first letter will be S. Look at the grain size distribution. From the grain size distribution, d10 = 0.10 mm, d30 = 0.19 mm, and d60 = 0.43 mm.

60u

10

d 0 43C 4d 0 10

. .

.= = = 3

2 2

30

60 10

d 0 19Cc 0 84d d 0 43 0 10

. .( . ) ( . )

= = =

This soil does not satisfy Cc (CC = 0.84 which is less than 1) criteria and does not satisfy Cu (Cu = 4.3 < 6) criteria for a well-graded sand and hence, is a poorly-graded sand, SP.

Solution 3.15 . ( Dr. Budhu check if this solution is correct?)

Percent Finer sieve

opening(mm) Soil A B C 4.75 100.00 100.00 100.00

2 (No. 10) 89.89 74.28 97.00 0.85 77.03 63.82 77.40

0.425(No. 40) 56.81 31.64 59.40 0.15 47.75 11.79 23.00

0.075(No. 200) 34.13 0.00 1.20

Step1: % passing No 200 sieve. Soil A: 34.13% passing i.e. < 35%, Soil B: 0% passing i.e. < 35%, Soil C: 1.2 % passing i.e. < 35% Soils A, B and C are granular Step2: Make a table according to Table 3.7

%Passing

Sieve No Soil A B C No. 10 89.89 74.28 97.00No. 40 56.81 31.64 59.40

No. 200 34.13 0.00 1.20LL 23% - - PL 8% - - PL 15% - -

Group Index Soil A: GI = 0.01(F-15)(PI-10) = 0.96 =1 Soil B: GI = 0 (non plastic) Soil C: GI = 0 (non plastic) Step3: Use Table 3.7 with the values in step 2 to classify soils according to AASHTO Soil A: A-2-6 (1) Soil B: A-1-b Soil C: A-3

Solution 3.16 (a) Determine LLLL = 40%

30354045505560

1 10 100

Number of blows

Wat

er c

onte

nt (%

)

25

(b) Liquidity Index (LI) PI = 0.40 – 0.23 = 0.17

LI = 17.0

23.038.0I

ww

p

pl −=

− = 0.88

(c) Since LI is within the range 0 < < 1, the soil is plastic and brittle failure is unlikely. LI

Solution 3.17

y = 22.972x0.3024

R2 = 0.9999

10

100

1 10 100

Penetration (mm)

Wat

er c

onte

nt (%

)

LL = 58% c = 22.97, m = .3024

m 0.3024PL c(2) 22.97(2) 28.3= = = PI = LL – PL = 58 – 28.3 = 29.7%

Solution 3.18

w(%) W γd (kN/m3) 12 17 16.08 13 18 16.87 14 18.7 17.38 16 19.3 17.62 19 18.9 16.82

15.00

16.00

17.00

18.00

19.00

20.00

21.00

10 12 14 16 18 20

water content (%)

dry

unit

wei

ght (

kN/m

3)

zero air voids

Optimum Moisture Content = 15.5%, Maximum dry unit weight = 17.7 kN/m3

e = 5.017.178.97.21G

d

ws =−⎟

⎠⎞

⎜⎝⎛ ×=−⎟⎟

⎞⎜⎜⎝

⎛γγ

S = sw G 0.155 (2.7)0.84

e 0.5= = = 84%

Zero air voids – use ( )d s w sG /(1 wGγ = γ +

w(%) γd

12 19.98 13 19.59 14 19.20 16 18.48 19 17.49

Solution 3.19 Se = wGs

eembankment = 455.095.0

7.216.0=

×

fill

embankment

fill

embankment

e1e1

VV

++

=

vfill = ( ) 3m2.316151455.01

3.11000,200=

++×

Solution 3.20 (a)

s s ssat w s

s s ssat w w

s s

s

G e G w G w; e G w1 e S 1G G w G (1 w)1 G w 1 G w

G 2.7

+γ = γ = = =

++ +

∴γ = γ = γ+ +

=

Depth (m) 1 2 3 4 5 6

w (%) 21.3 23.6 6.1 32.7 41.5 42

γsat(kN/m3) 20.4 20.0 24.1 18.6 17.7 17.6

0

1

2

3

4

5

6

7

0 10 20 30 40 5w (%); (kPa)

Dep

th (m

)

0

saturated unit weight

water content

(b) The water content at depth = 3m is very low in comparison with the other values and is rather suspicious (c) The soil at depth = 3m is likely a thin sand layer.

Solution 3.21

(a) Fill : 3d m/kN9.15

082.012.17

=+

Embankment: ( ) 3

d m/kN1.181995.0 ==γ

Minimum fill needed: per 3m of embankment: 9.151.18

= = 1.14 3m

(b) and water in fill = 0.82 = 1.3 kN sW wWW = 9.15× water needed = .11 = 2 kN 1.18×

Additional water = 2 – 1.3 = 0.7 kN per 3m (c) water needed = 0.12 = 2.17 kN 1.18×

additional water = 2.17 – 1.3 = 0.87 kN

(d) Number of truck loads = 400,11m10

m000,1143

3=

(e) Find Cost: Distance: 10 km Extra charge = (10-2) 3(10 2) 0.5 \$4 / m− × =

Purchase, load, and spread and compact per 3m = \$(15 + 4 + 1.02) = \$20.02 Cost = 114,000 × \$20.02 = \$2,282,280

Solution 3.22 I. Suitability of Soils ITEM PIT 1 PIT 2 % Fines CuCc

5 7 1.3

22 --- ---

Pit 2 contains too many fines (and may not be suitable) Pit 1 has few fines and is well graded ⇒ it will compact to higher densities, which implies it will have higher shear strength and lower compressibility. Pit 1 contains a better soil for the embankment than Pit 2. Cost

Volume of embankment = 3102.242.22122.213 ×⎥

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ ×××+× x 10 = 47.96 34 m10×

Extra mileage charge/ 3m

Pit 1: = 190 km (∑=

+10

1nn28 )

)Pit 2: = 150 km (∑=

+10

1nn24

Pit 1 : ( ) = \$51,374,752 410 190 0.5 1.02 1.1 47.96 10+ × + + × ×

Pit 2: = \$46,334,156 ( ) 412 150 0.55 1.26 0.85 47.96 10 1.1+ × + + × × ×The actual cost is likely to be more because we have calculated the compacted volume. The volume of pit material required is likely to be different for each pit.