Introduction to Geotechnical Engineering 2nd Edition Holtz Solutions Manual by Scotts77

CHAPTER 2 INDEX AND CLASSIFICATION PROPERTIES OF SOILS

2-1. From memory, draw a phase diagram (like Fig. 2.2, but don’t look first!). The “phases” have a Volume side and Mass side. Label all the parts.

SOLUTION: Refer to Figure 2.2.

2-2. From memory, write out the definitions for water content, void ratio, dry density, wet or moist density, and saturated density.

SOLUTION: Refer to Section 2.2.

2-3. Assuming a value of ρs = 2.7 Mg/m3, take the range of saturated density in Table 2.1 for the six soil types and calculate/estimate the range in void ratios that one might expect for these soils.

SOLUTION: Create a spreadsheet using input values from Table 2.1 and Eq. 2.18.

(Given)(see Eq. 2.18)

ρ' - min ρ' - max emax emin (Mg/m3)(Mg/m3) 0.91.40.890.21 0.41.13.250.55 1.11.40.550.21 0.91.20.890.42 0.00.1 ∞ 16.00 0.30.84.671.13

© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Introduction to Geotechnical Engineering 2nd Edition Holtz Solutions Manual Full Download: http://testbanktip.com/download/introduction-to-geotechnical-engineering-2nd-edition-holtz-solutions-manual/ Download all pages and all chapters at: TestBankTip.com

2-4. Prepare a spreadsheet plot of dry density in Mg/m3 as the ordinate versus water content in percent as the abscissa. Assume ρs = 2.65 Mg/m3 and vary the degree of saturation, S, from 100% to 40% in 10% increments. A maximu m of 50% water content should be adequate.

SOLUTION: Solve Eq. 2.12 and Eq. 2.15 for ρd = f(ρs, w, S, Gs), or use Eq. 5.1.

Index and Classification Properties of Soils Chapter 2

ρ S w ρ = dry ρ w w+S ρs S =100908070605040302010 w ρdry ρdry ρdry ρdry ρdry ρdry ρdry ρdry ρdry ρdry (%) (Mg/m3)(Mg/m3)(Mg/m3)(Mg/m3)(Mg/m3)(Mg/m3)(Mg/m3)(Mg/m3)(Mg/m3)(Mg/m3) 0.02.652.652.652.652.652.652.652.652.652.65 5.02.342.312.272.232.172.091.991.841.591.14 10.02.092.051.991.921.841.731.591.411.140.73 15.01.901.841.771.691.591.481.331.140.890.53 20.01.731.671.591.511.411.291.140.960.730.42 25.01.591.531.451.361.261.141.000.830.610.35 30.01.481.411.331.241.141.020.890.730.530.30 35.01.371.311.231.141.040.930.800.650.470.26 40.01.291.221.141.050.960.850.730.580.420.23 45.01.211.141.060.980.890.780.670.530.380.21 50.01.141.071.000.920.830.730.610.490.350.19 Problem 2-4 0.00 0.50 1.00 1.50 2.00 2.50 3.00 0510152025303540455055 w (%) ρdry (Mg/m3 ) S=100% S=90% S=80% S=70% S=60% S=50% S=40% S=30% S=20% S=10% © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

2-5a Prepare a graph like that in Problem 2.4, only use dry density units of kN/m3 and pounds per cubic feet.

2.12

2.15:

Index and Classification Properties of Soils Chapter 2

SOLUTION: ρ S w From Eq.

and Eq.

ρ = dry ρ w w+S ρs S =100908070605040302010 w γ dry γ dry γdry γ dry γ dry γdry γ dry γ dry γdry γ dry (%) (kN/m3)(kN/m3)(kN/m3)(kN/m3)(kN/m3)(kN/m3)(kN/m3)(kN/m3)(kN/m3)(kN/m3) 0.026.0026.0026.0026.0026.00 26.0026.0026.0026.0026.00 5.022.9522.6622.3021.8621.29 20.5519.5318.0315.6411.18 10.020.5520.0819.5318.8618.03 16.9915.6413.8011.187.12 15.018.6018.0317.3716.5815.6414.4813.0411.188.705.23 20.016.9916.3615.6414.7913.8012.6211.189.407.124.13 25.015.6414.9714.2213.3612.3511.189.798.106.033.41 30.014.4813.8013.0412.1711.1810.048.707.125.232.90 35.013.4912.8012.0411.1810.219.117.836.354.612.53 40.012.6211.9411.1810.349.408.337.125.734.132.24 45.011.8611.1810.449.628.707.686.535.233.732.01 50.011.1810.529.798.998.107.126.034.803.411.82 Problem 2-5a 0.00 5.00 10.00 15.00 20.00 25.00 30.00 0510152025303540455055 w (%) γ dry (kN/m3 ) S=100% S=90% S=80% S=70% S=60% S=50% S=40% S=30% S=20% S=10% © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

2-5b Prepare a graph like that in Problem 2.4, only use dry density units of pounds per cubic feet.

Soils Chapter 2

Index and Classification Properties of

SOLUTION: sw dry s GS SGw γ γ= + S =100908070605040302010 w γ dry γ dry γdry γ dry γ dry γdry γ dry γ dry γdry γ dry (%) (pcf)(pcf)(pcf)(pcf)(pcf)(pcf)(pcf)(pcf)(pcf)(pcf) 0.0165.36165.36165.36165.36165.36165.36165.36165.36165.36165.36 5.0146.01144.14141.86139.04135.45130.72124.21114.7099.4671.12 10.0130.72127.75124.21119.95114.70108.0899.4687.8071.1245.30 15.0118.33114.70110.47105.4799.4692.1282.9471.1255.3533.24 20.0108.08104.0799.4694.1187.80 80.2771.1259.7745.3026.25 25.099.4695.2590.4584.9678.59 71.1262.2551.5438.3421.69 30.092.1287.8082.9477.4371.12 63.8555.3545.3033.2418.48 35.085.7981.4476.5871.1264.95 57.9249.8340.4129.3316.09 40.080.2775.9371.1265.7759.77 53.0045.3036.4826.2514.26 45.075.4271.1266.3961.1655.35 48.8541.5333.2423.7512.79 50.071.1266.8962.2557.1651.54 45.3038.3430.5321.6911.60 Problem 2-5b 0.00 20.00 40.00 60.00 80.00 100.00 120.00 140.00 160.00 180.00 0510152025303540455055 w (%) γ dry (pcf) S=100% S=90% S=80% S=70% S=60% S=50% S=40% S=30% S=20% S=10% © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

2.6. Prepare a graph like that in Problem 2.4, only for S = 100% and vary the density of solids from 2.60 to 2.80 Mg/m3. You decide the size of the increments you need to “satisfactorily” evaluate the relationship as ρs varies. Prepare a concluding statement of your observations.

Note: The relationship between ρdry and w is not overly sensitive to ρs

Index and Classification Properties of Soils Chapter 2

SOLUTION: ρ S w From Eq.

ρ w w+S ρs

2.12 and Eq. 2.15: ρ = dry

ρs = 2.62.652.72.752.8 w ρdry ρdry ρdry ρdry ρdry (%) (Mg/m3)(Mg/m3)(Mg/m3)(Mg/m3)(Mg/m3) 0.02.602.652.702.752.80 5.02.302.342.382.422.46 10.02.062.092.132.162.19 15.01.871.901.921.951.97 20.01.711.731.751.771.79 25.01.581.591.611.631.65 30.01.461.481.491.511.52 35.01.361.371.391.401.41 40.01.271.291.301.311.32 45.01.201.211.221.231.24 50.01.131.141.151.161.17 0.00 0.50 1.00 1.50 2.00 2.50 3.00 0510152025303540455055 w (%) ρ s (Mg/m3 ) © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

2.7. The dry density of a compacted sand is 1.87 Mg/m3 and the density of the solids is 2.67 Mg/m3. What is the water content of the material when saturated?

2.8. A soil that is completely saturated has a total density of 2045 kg/m3 and a water content of 24%. What is the density of the solids? What is the dry density of the soil?

Index and Classification Properties of Soils Chapter 2

SOLUTION: w dry w s 3 w 33 drys S From Eq. 2-12 and Eq. 2-15: ;Note: S = 100% wS 1111 Solving for w: wS(1Mg/m)(100%)16.0% 1.87Mg/m2.67Mg/m ρ ρ= ρ + ρ ⎛⎞ ⎛⎞ =ρ−=−= ⎜⎟ ⎜⎟ ⎜⎟ ρρ ⎝⎠ ⎝⎠

SOLUTION:

Solve using equations or phase diagrams: 3 t dry w satdryw s wdry 3 s drywsat 2045 1649.2kg/m (1w)(10.24) 1 (1000)(1649.2) 2729.6kg/m (1649.210002045) ρ ρ=== ++ ⎛⎞ ρ ρ=−ρ+ρ ⎜⎟ ρ ⎝⎠ ρρ ρ=== ρ+ρ−ρ+−

Solve using phase

relationships: t t w ws s twssss 3 s w ww w 3 s s s 3 s dry t assume V1.0 M2045kg M 0.24M0.24M M MMM20450.24MMM1649.19kg 0.24M M 1000V0.3958m V1000 M 1649.19 2729.6kg/m V10.3958 M 1649.19 1649.2kg/m V1 = = =→= =+→=+→= ρ==→== ρ=== ρ=== © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

diagram

2.9 What is the water content of a fully saturated soil with a dry density of 1.72 Mg/m3? Assume ρs = 2.72 Mg/m3

2-12 and Eq. 2-15: ;Note:

2.10. A dry quartz sand has a density of 1.68 Mg/m3. Determine its density when the degree of saturation is 75%. The density of solids for quartz is 2.65 Mg/m3

2.11. The dry density of a soil is 1.60 Mg/m3 and the solids have a density of 2.65 Mg/m3. Find the (a) water content, (b) void ratio, and (c) total density when the soil is saturated. SOLUTION:

From Eq. 2-12 and Eq. 2-15:

(a) Solving for w: wS(1Mg/m)(100%)24.76% 1.60Mg/m2.65Mg/m w (24.76)(2.65)

(b) From Eq. 2.15:

Index and Classification Properties of Soils Chapter 2

SOLUTION: w dry w s 3 w 33 drys S

Eq.

wS 1111 Solving for w: wS(1Mg/m)(100%)21.4% 1.72Mg/m2.72Mg/m ρ ρ= ρ + ρ ⎛⎞ ⎛⎞ =ρ−=−= ⎜⎟ ⎜⎟ ⎜⎟ ρρ ⎝⎠ ⎝⎠

From

S = 100%

SOLUTION: drydry(final) w dry w s 3 w 33 drys tdry

Eq. 2-12

2-15: wS 1111 Solving for w: wS(1Mg/m)(75%)16.34% 1.68Mg/m2.65Mg/m final (1 ρ=ρ ρ ρ= ρ + ρ ⎛⎞ ⎛⎞ =ρ−=−= ⎜⎟ ⎜⎟ ⎜⎟ ρρ ⎝⎠ ⎝⎠ ρ=ρ+ w)1.68(10.1634)1.95Mg/m3 =+=

Recognize that (initial)(final);S = 75%

From

and Eq.

w dry w s 3 w 33 drys s w Given: S = 100% S

wS 1111

S(100)(1 ρ ρ= ρ + ρ ⎛⎞ ⎛⎞ =ρ−=−= ⎜⎟ ⎜⎟ ⎜⎟ ρρ ⎝⎠ ⎝⎠ ρ == ρ 3 tdry 0.656 .0) (c)

(1w)1.60(10.2476)1.9962.00Mg/m

ρ=ρ+=+==

2.12. A natural deposit of soil is found to have a water content of 20% and to be 90% saturated. What is the void ratio of this soil?

SOLUTION: s s w

w = 20% and S = 90%; assume G2.70

w (20.0)(2.70)

===

(a) WWW625012lb

W50

W 62 (b)

(c)

12 (d) V0.1923ft 62.4

e =−=−= =×== γ=== γ=== === γ === γ =−=−=

wts w s t t t s dry t 3 w w w 3 s s sw vtv

G(2.64)(62.4)

====

V 0.2565 0.84510.84 V0.3035

v s

M100.0649.3150.75g

M113.27100.0613.21g

s w w s s w

M 13.21(100)

(a)w100%26.0326.0% M50.75 w 2.80(26.03)

=−= =−= =×=== ρ

===

Index and Classification Properties of Soils Chapter 2

From Eq. 2.15: e0.60 S(90)(1.0) ρ

2.13. A chunk of soil has a wet weight of 62 lb and a volume of 0.56 ft3. When dried in an oven, the soil weighs 50 lb. If the specific gravity of solids Gs = 2.64, determine the water content, wet unit weight, dry unit weight, and void ratio of the soil.

SOLUTION: Solve using phase diagram relationships.

V0.56 W

W 12(100) w100%24.0%

110.7pcf

89.29pcf

W 50 V0.3035ft

VVV0.560.30350.2565

2.14. In the lab, a container of saturated soil had a mass of 113.27 g before it was placed in the oven and 100.06 g after the soil had dried. The container alone had a mass of 49.31 g. The specific gravity of solids is 2.80. Determine the void ratio and water content of the original soil sample.

SOLUTION: Solve using phase diagram relationships.

ρ © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

(b)FromEq. 2.15: e=0.72880.73 S(1)(100)

2.15. The natural water content of a sample taken from a soil deposit was found to be 12.0%. It has been calculated that the maximum density for the soil will be obtained when the water content reaches 22.0%. Compute how many grams of water must be added to each 1000 g of soil (in its natural state) in order to increase the water content to 22.0%.

SOLUTION:

w0.12M(0.12)M

MMMM0.12M1.12M1000g

M892.857g

M(0.12)(892.857)107.143g

Target state

(Note: M does not change between naturalstate and target state)

MwM(0.22)(89

2.857)196.429g

additional water necessary = 196.429107.14389.28689.29g = −==

2.16. A cubic meter of dry quartz sand (Gs = 2.65) with a porosity of 60% is immersed in an oil bath having a density of 0.92 g/cm3. If the sand contains 0.27 m3 of entrapped air, how much force is required to prevent it from sinking? Assume that a weightless membrane surrounds the specimen. (Prof. C. C. Ladd.)

VVV1.00.600.40m

MGV(2.65)(1000)(0.40m)1060kgM

Index and Classification Properties of Soils Chapter 2

w ws s tswsss s w s ws Natural state M

==→= =+=+== = == =×=

3 3 3 33 t 3 vt 3 stv 3 kg sssst m t kg t m t kg buoytoil m buoybuoy V1m1,000,000cm

SOLUTION:

VnV(0.6)(1.0)0.60m

M 1060 1060 V1.0

== =×== =−=−= =×ρ×=== ρ=== ρ=ρ−ρ=−= γ=ρ×== 3 3 N m 3 N buoy m Force(entrappedair)1373.40.27m370.8N =γ×=×= © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

1060920140 g(140)(9.81)1373.4

2.17. A soil sample taken from a borrow pit has a natural void ratio of 1.15. The soil will be used for a highway project where a total of 100,000 m3 of soil is needed in its compacted state; its compacted void ratio is 0.73. How much volume has to be excavated from the borrow pit to meet the job requirements?

2.18. A sample of moist soil was found to have the following characteristics:

Total volume: 0.01456 m3

Total mass: 25.74 kg

Mass after oven drying: 22.10 kg

Specific gravity of solids: 2.69

Find the density, unit weight, void ratio, porosity, and degree of saturation for the moist soil.

Index and Classification Properties of Soils Chapter 2

SOLUTION: tst v ss 3 t 3 s(emb) t s(borr)s(emb) 3 t(borr)sborr VVV V VV eee1 Embankment V100,000m 100,000 V57,803.47m 0.731 Borrow Pit V VV e1 VV(e1)(57,803.47)(1.151)124,277m ==→= + = == + == + =×+=+=

SOLUTION: 3 33 kg t m NkN tt mm 3 s s sw 3 v 25.74 (a)1767.8571768 0.01456 (b)g(1767.857)(9.81)17,342.6817.34 M 22.10 (c)V0.00822m G(2.69)(1000) V0.014560.008220.006344m 0.006344 e0.77180.77 0.00822 0.7718 (d)n 10.7 ρ=== γ=ρ×=== === ρ =−= === = + w w 10043.5643.6% 718 (e)M25.7422.103.64 3.64 V0.00364 1000 0.00364 S10057.37757.4% 0.006344 ×== =−= == =×== © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

2.19. A gray silty clay (CL) is sampled from a depth of 12.5 feet. The “moist” soil was extruded from a 6-inch-high brass liner with an inside diameter of 2.83 inches and weighed 777 grams. (a) Calculate the wet density in pounds per cubic feet. (b) A small chunk of the original sample had a wet weight of 140.9 grams and weighed 85.2 grams after drying. Compute the water content, using the correct number of significant figures. (c) Compute the dry density in Mg/m3 and the dry unit weight in kN/m3.

2.20. A cylindrical soil specimen is tested in the laboratory. The following properties were obtained:

VVV42.411526.54215.869in0.009184ft

Index and Classification Properties of Soils Chapter 2

SOLUTION: () ( ) () () () 3 2 3 t tt 3 3 t w wts s t dry lb dry ft 1lb 777g 453.6g M (2.83) (a)V637.741in,78.42978.4pcf 4V 1ft 37.741in 12in M 55.7 (b)MMM140.985.255.7g,w100%65.3865.4% M85.2 78.4 (c)47.40pcf (1w)(10.654) 1ft 47.4 0.3 π =×=γ==== =−=−==×=== γ γ=== ++ ρ= () () 33 33 3 kgMg mm NkN dry mm 0.4536kg 759.2880.759 048m1lb (759.288)(9.81)7448.67.45 == γ===

Sample diameter 3 inches Sample length 6 inches Wt. before drying in oven 2.95 lb Wt. after drying in oven 2.54 lb Oven temperature 110°C Drying time 24 hours Specific gravity of solids 2.65 What is the degree of saturation of this specimen? SOLUTION: 2 33 t 33 s s sw 33 vts wts 33 w w w w v (3)

4 W

V642.4115in0.02454ft

2.54 V0.01536ft26.542in G(2.65)(62.4)

WWW2.952.540.41lb

V 1 S100% V π =×== ==== γ =−=−== =−=−= ==== γ =×= 1.354 10071.5% 15.869 ×= © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

W 0.41 V0.00657ft11.354in 62.4

2.21 A sample of saturated silt is 10 cm in diameter and 2.5 cm thick. Its void ratio in this state is 1.35, and the specific gravity of solids is 2.70. The sample is compressed to a 2-cm thickness without a change in diameter.

(a) Find the density of the silt sample, in prior to being compressed.

(b) Find the void ratio after compression and the change in water content that occurred from initial to final state.

V(1.35)(83.553)112.797cmV

MV(1)(112.797)112.797g

MGV(2.70)(83.553)(1)

Index and Classification Properties of Soils Chapter 2

SOLUTION: 3 3 2 3 t w wv v vss 3 tvsssss 3 vw g www cm g sssw cm (10)

4 V S1VV V

(a)V2.5196.350cm

VeV1.35V VVV1.35VV2.35V196.350V83.553cm

π =×= ==→= =×= =+=+==→= === =ρ×== =××ρ= 33 t t gkg t cmm t 2 3 t2 3 s 3 vts final initial 225.594g M112.797225.594338.391g M 338.391 1.7231723 V196.35 (10)

V83.553cm(no change) VVV157.0883.55373.527cm 73.527 e0.88 83.553 112. (c)w = =+= ρ==== π =×= = =−=−= == = 3 wvws final 797 100%50.0% 225.594 final conditions VV73.527cm;M73.527g;M225.594g(no change) 73.527 w100%32.6% 225.594 w50.032.617.4% ×= ==== =×= Δ=−= © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

(b)V2.0157.08cm 4

2.22. A sample of sand has the following properties: total mass Mt = 160 g; total volume Vt =80 cm3; water content w = 20%; specific gravity of solids Gs =2.70. How much would the sample volume have to change to get 100% saturation, assuming the sample mass M t stayed the same? SOLUTION:

M133.33g

M(0.20)(133.33)26.667g;V(1)M26.667cm

M 133.33

V49.383cm;V8049.38330.617cm

G2.70

26.667

S100%87.10%

30.617

Desired condition: S100%

anges,butVandM remain the same

M26.667g;V26.667cm

VV(when S = 100%)

VVV49.38326.66776.053cm

V8076.0533.95cm

Soils Chapter 2

Index and Classification Properties of

3 wss tss s 3 g www cm 33 s sv sw i t

MwM(0.20)M MM0.20M160g

Vch =×= =+= = ==== ====−= ρ =×= = ss 3 ww vw 3 tsv 3

== = =+=+= Δ=−= © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

2.23. Draw a phase diagram and begin to fill in the blanks: A soil specimen has total volume of 80,000 mm3 and weighs 145 g. The dry weight of the specimen is 128 g, and the density of the soil solids is 2.68 Mg/m3. Find the: (a) water content, (b) void ratio, (c) porosity, (d) degree of saturation, (e) wet density, and (f) dry density. Give the answers to parts (e) and (f) in both SI and British engineering units.

SOLUTION:

(b)V47.7612cm47,761.2mm G(2.68)(1)

V80,00047,761.232,238.8mm

Index and Classification Properties of Soils Chapter 2

w w s 33 s s sw 3 v www

M128

(a)M14512817g

17 w100%10013.28113.3%

M 128

32,238.8 e0.67 47,761.2 32,238.8 (c)n100%40.3% 80,000

−= =×=×== ==== ρ =−= == =×= =×ρ= 33 33 3 3 33 gkg t cmm kg lbm t mft kg dry m kg l dry m 17cm17,000mm 17,000 S100%52.7% 32,238.8 145 (e)1.81251812.5 80 1 1812.5113.2(seeAppendixA) 16.018 1812.5 (f)1600.0 (10.13281) 1 1600.099.9 16.018 == =×= ρ=== ⎛⎞ ρ=×= ⎜⎟ ⎝⎠ ρ== + ⎛⎞ ρ=×= ⎜⎟ ⎝⎠ 3 bm ft (seeAppendixA) © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

(d)VM(17)(1)

2.24. A sample of soil plus container weighs 397.6 g when the initial water content is 6.3%. The container weighs 258.7 g. How much water needs to be added to the original specimen if the water content is to be increased by 3.4%? After U.S. Dept. of Interior (1990).

SOLUTION:

M397.6258.7138.9g

M0.063M

M138.9MM0.063MM1.063M

M130.668g

MwM(0.034)(130.668)4.44g

2.25. A water-content test was made on a sample of clayey silt. The weight of the wet soil plus container was 18.46 g, and the weight of the dry soil plus container was 15.03 g. Weight of the empty container was 7.63 g. Calculate the water content of the sample.

SOLUTION:

M15.037.637.40g

M18.4615.033.43g

M 3.43(100)

(a)w100%46.35146.3% M7.40

Index and Classification Properties of Soils Chapter 2

t ws twssss s w s ws

M w0.034

=−= = ==+=+= = Δ Δ== Δ=Δ×==

w w s

=−= =−= =×=== © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

2.26. A soil sample is dried in a microwave oven to determine its water content. From the data below, evaluate the water content and draw conclusions. The oven-dried water content is 23.7%. The mass of the dish is 146.30 grams. After U.S. Dept. of Interior (1990). SOLUTION:

CONCLUSION: The loss of additional water in the soil sample becomes negligible after 8 to 10 minutes in the microwave oven used in the experiment.

Index and Classification Properties of Soils Chapter 2

t ws ss s M231.62146.385.32g

231.62column3 column 5 = 100% 68.9733 =−= = =+ = ×

M(0.237)M 85.320.237MM M68.9733g (this value is constant throughout drying period)

GIVEN CALCULATED Time in Total Oven Mass of Mass of Water Oven Time Soil + Dish Water Content (min) (min) (g) (g) (%) 0 0 231.62 -- -3 3 217.75 13.87 20.11 1 4 216.22 15.40 22.33 1 5 215.72 15.90 23.05 1 6 215.48 16.14 23.40 1 7 215.32 16.30 23.63 1 8 215.22 16.40 23.78 1 9 215.19 16.43 23.82 1 10 215.19 16.43 23.82 19.0 20.0 21.0 22.0 23.0 24.0 25.0 01234567891011 Time in Oven (min) Water Content (% ) © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

2.27. The mass of a sample of silty clay soil plus container is 18.43 g and the weight of the dry soil plus container is 13.67 g. The container we ighs 8.84 g. Compute the water content of the sample.

M13.678.844.83g

M18.4313.674.76g

M 4.76(100) (a)w100%98.5% M4.83

SOLUTION: s w w s

2.28. A specimen of fully saturated clay soil that weighs 1389 g in its natural state weighs 982 g after drying. What is the natural water content of the soil?

SOLUTION:

M1389982407g

w w s

=×==

M 407(100) (a)w100%41.4% M982

2.29. The volume of water in a sample of moist soil is 0.24 m3. The volume of solids Vs is 0.25 m3. Given that the density of soil solids ρs is 2600 kg/m3, find the water content.

SOLUTION:

MV(2600)(0.25)650kg

sss www

MV(1000)(0.24)240kg

240 w100%36.9% 650

2.30. For the soil sample of Problem 2.29, compute (a) the void ratio and (b) the porosity.

SOLUTION: Assume S = 100%

s Gw (2.6)(0.3692) (a)e0.96

=== =×= +

Index and Classification Properties of Soils Chapter 2

=−= =−=

=×==

=−=

=ρ×== =×=

=ρ×==

S(1) 0.96 (b)n10050.0% 10.96 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

2.31. For the soil sample of Problem 2.29, compute (a) the total or wet density and (b) the dry density. Give your answers in Mg/m3, kg/m3, and lbf/ft3

2.32. A 592-cm3 volume of moist sand weighs 1090 g. Its dry weight is 920 g and the density of solids is 2680 kg/m3. Compute the void ratio, porosity, water content, degree of saturation, and

Index and Classification Properties of Soils Chapter 2

SOLUTION: Assume S = 100% 333 333 3 t kgMg lbm t mmft kgMg lbm dry mmft (a)V0.250.240.49m 710 1448.981.4590.4 0.49 650 (b)1326.531.3382.8 0.49 =+= ρ==== ρ====

density in kg/m3 SOLUTION: 33 kgg s mcm 3 s 3 v v s w 3 w w w v 26802.68 920 V343.284cm 2.68 V592343.284248.716cm V 248.716 (a)e0.72450.72 V343.284 0.7245 (b)n100%42.0% 10.7245 (c)M1090920170g 170 V170cm V 170 S100%10068.4% V248.716 (d ρ== == =−= ==== =×= + =−= == ρ =×=×= 33 gkg t cmm 1090 )1.8411841 592 ρ=== © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

total

2.33. The saturated density γsat of a soil is 137 lbf/ft3. Find the buoyant density of this soil in both lbf/ft3 and kg/m3

2.34. A sand is composed of solid constituents having a density of 2.68 Mg/m3. The void ratio is 0.58. Compute the density of the sand when dry and when saturated and compare it with the density when submerged.

of Soils Chapter 2

Index and Classification Properties

SOLUTION: () 3 3 33 3 lbf ft kg m kg lbf ftm lbm ft '13762.474.6 16.018 '74.61195 1 γ=−= ⎛⎞ ρ== ⎜⎟ ⎜⎟ ⎝⎠

SOLUTION: 3 3 3 3 s 3 vt Mg dry m 3 wv w t Mg sat m Mg m Assume V1m V0.58,V10.581.58m 2.68 1.69621.70 1.58 For S = 100%; VV0.58m M(1)(0.58)0.58Mg M2.680.583.26Mg 3.26 2.06 1.58 '2.061.01.06 = ==+= ρ=== == == =+= ρ== ρ=−= © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

2.35. A sample of natural glacial till was taken fr om below the groundwater table. The water content was found to be 52%. Estimate the wet density, dry density, buoyant density, porosity, and void ratio. Clearly state any necessary assumptions.

2.36. A 1-m3 sample of moist soil weighs 2000 kg. The water content is 10%. Assume ρs = 2.70 Mg/m3. With this information, fill in all blanks in the phase diagram of Fig. P2.36.

Index and Classification Properties of Soils Chapter 2

SOLUTION: 3 3 ss ssssws t ww3 wvwv vw 3 t Mg tm dry AssumeV1m,AssumeG2.7 MGV(2.7)(1)(1)2.7Mg,MwM(0.52)(2.7)1.404Mg M2.71.4044.104Mg VM S1VV,V1.404mV V V1.4041.02.404m 4.104 (a)1.71 2.404 2.7 (b)1. 2.404 == =××ρ===×== =+= ==→==== ρ =+= ρ== ρ== 3 3 Mg m Mg m 12 (c)'1.711.00.71 1.404 (d)n100%58.4% 2.404 1.404 (e)e1.4 1.0 ρ=−= =×= ==

SOLUTION: wss tssss w s33 sw s 33 va MMw0.10M M2000M0.10M1.10MM1818.18kg M(0.10)(1818.18)181.82kg M1818.18181.82 V0.673m,V0.181m 27001000 V10.6730.327m,V0.3270.1810.146m =×= ==+=→= == ===== ρ =−==−= air water solid Mw=182 Ms=1818 M Vt=2000 w =0.18 Vs =0.67 Va =0.15 Vv=0.33 Volume(mMass(kg) 3) Vt=1.0 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Introduction to Geotechnical Engineering 2nd Edition Holtz Solutions Manual Full Download: http://testbanktip.com/download/introduction-to-geotechnical-engineering-2nd-edition-holtz-solutions-manual/ Download all pages and all chapters at: TestBankTip.com