Craig's Soil Mechanics (Solution's Manual) PDF

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Craig’s Soil Mechanics Seventh Edition Solutions Manual Craig’s Soil Mechanics Seventh Edition Solutions Manual R.F. Craig Formerly Department of Civil Engineering University of Dundee UK First published 1992 by E & FN Spon, an imprint of Thomson Professional Second edition 1997 Third edition 2...

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Craig’s Soil Mechanics Seventh Edition Solutions Manual

Craig’s Soil Mechanics Seventh Edition Solutions Manual

R.F. Craig Formerly Department of Civil Engineering University of Dundee UK

First published 1992 by E & FN Spon, an imprint of Thomson Professional Second edition 1997 Third edition 2004 11 New Fetter Lane, London EC4P 4EE Simultaneously published in the USA and Canada by Spon Press 29 West 35th Street, New York, NY 10001 This edition published in the Taylor & Francis e-Library, 2004. “To purchase your own copy of this or any of Taylor & Francis or Routledge’s collection of thousands of eBooks please go to www.eBookstore.tandf.co.uk.”

Spon Press is an imprint of the Taylor & Francis Group ª 1992, 1997, 2004 R.F. Craig All rights reserved. No part of this book may be reprinted or reproduced or utilised in any form or by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying and recording, or in any information storage or retrieval system, without permission in writing from the publishers. British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library Library of Congress Cataloging in Publication Data A catalog record for this book has been requested ISBN 0-203-31104-3 Master e-book ISBN

ISBN 0-203-67167-8 (Adobe eReader Format) ISBN 0–415–33294–X (Print edition)

Contents

1 Basic characteristics of soils

1

2 Seepage

6

3 Effective stress

14

4 Shear strength

22

5 Stresses and displacements

28

6 Lateral earth pressure

34

7 Consolidation theory

50

8 Bearing capacity

60

9 Stability of slopes

74

Author’s note In order not to short-circuit the learning process it is vital that the reader should attempt the problems before referring to the solutions in this manual.

Chapter 1

Basic characteristics of soils

1.1 Soil E consists of 98% coarse material (31% gravel size; 67% sand size) and 2% fines. It is classified as SW: well-graded gravelly SAND or, in greater detail, well-graded slightly silty very gravelly SAND. Soil F consists of 63% coarse material (2% gravel size; 61% sand size) and 37% non-plastic fines (i.e. between 35 and 65% fines); therefore, the soil is classified as MS: sandy SILT. Soil G consists of 73% fine material (i.e. between 65 and 100% fines) and 27% sand size. The liquid limit is 32 and the plasticity index is 8 (i.e. 32
24), plotting marginally below the A-line in the ML zone on the plasticity chart. Thus the classification is ML: SILT (M-SOIL) of low plasticity. (The plasticity chart is given in Figure 1.7.)

Figure Q1.1

2

Basic characteristics of soils

Soil H consists of 99% fine material (58% clay size; 47% silt size). The liquid limit is 78 and the plasticity index is 47 (i.e. 78
31), plotting above the A-line in the CV zone on the plasticity chart. Thus the classification is CV: CLAY of very high plasticity.

1.2 From Equation 1.17
w 1:00 ¼ 1:55 ¼ 2:70  1:095  1:91
; e ¼ 0:55

1 þ e ¼ Gs ð1 þ wÞ

Using Equation 1.13 Sr ¼

wGs 0:095  2:70 ¼ ¼ 0:466 0:55 e

ð46:6%Þ

Using Equation 1.19
sat ¼

Gs þ e 3:25
w ¼  1:00 ¼ 2:10 Mg=m3 1þe 1:55

From Equation 1.14 w¼

e 0:55 ¼ 0:204 ¼ Gs 2:70

ð20:4%Þ

1.3 Equations similar to 1.17–1.20 apply in the case of unit weights; thus, d ¼

Gs 2:72  9:8 ¼ 15:7 kN=m3 w ¼ 1:70 1þe

sat ¼

Gs þ e 3:42 w ¼  9:8 ¼ 19:7 kN=m3 1þe 1:70

Using Equation 1.21 0 ¼

Gs
1 1:72 w ¼  9:8 ¼ 9:9 kN=m3 1þe 1:70

Using Equation 1.18a with Sr ¼ 0.75 ¼

Gs þ Sr e 3:245 w ¼  9:8 ¼ 18:7 kN=m3 1þe 1:70

Basic characteristics of soils

3

Using Equation 1.13 w¼

Sr e 0:75  0:70 ¼ 0:193 ¼ Gs 2:72

ð19:3%Þ

The reader should not attempt to memorize the above equations. Figure 1.10(b) should be drawn and, from a knowledge of the definitions, relevant expressions can be written by inspection.

1.4  382 76 ¼ 86 200 mm3 4 Mass 168:0 ¼ ¼ 1:95 Mg=m3 Bulk density ð
Þ ¼ Volume 86 200  10
3

Volume of specimen ¼

Water content ðwÞ ¼

168:0
130:5 ¼ 0:287 130:5

ð28:7%Þ

From Equation 1.17
w 1:00 ¼ 1:80 ¼ 2:73  1:287  1:95
; e ¼ 0:80

1 þ e ¼ Gs ð1 þ wÞ

Using Equation 1.13 Sr ¼

wGs 0:287  2:73 ¼ 0:98 ¼ 0:80 e

ð98%Þ

1.5 Using Equation 1.24
d ¼


2:15 ¼ ¼ 1:92 Mg=m3 1 þ w 1:12

From Equation 1.17
w 1:00 ¼ 1:38 ¼ 2:65  1:12  2:15
; e ¼ 0:38

1 þ e ¼ Gs ð1 þ wÞ

Using Equation 1.13 Sr ¼

wGs 0:12  2:65 ¼ 0:837 ¼ 0:38 e

ð83:7%Þ

4

Basic characteristics of soils

Using Equation 1.15 A¼

e
wGs 0:38
0:318 ¼ 0:045 ¼ 1:38 1þe

ð4:5%Þ

The zero air voids dry density is given by Equation 1.25
d ¼

Gs 2:65  1:00 ¼ 1:95 Mg=m3
w ¼ 1 þ ð0:135  2:65Þ 1 þ wGs

i.e. a dry density of 2.00 Mg/m3 would not be possible.

1.6 Mass (g)


(Mg/m3)

w


d (Mg/m3)


d0 (Mg/m3)


d5 (Mg/m3)


d10 (Mg/m3)

2010 2092 2114 2100 2055

2.010 2.092 2.114 2.100 2.055

0.128 0.145 0.156 0.168 0.192

1.782 1.827 1.829 1.798 1.724

1.990 1.925 1.884 1.843 1.765

1.890 1.829 1.790 1.751 1.676

1.791 1.733 1.696 1.658 1.588

In each case the bulk density (
) is equal to the mass of compacted soil divided by the volume of the mould. The corresponding value of dry density (
d ) is obtained from Equation 1.24. The dry density–water content curve is plotted, from which wopt ¼ 15%

Figure Q1.6

and


dmax ¼ 1:83 Mg=m3

Basic characteristics of soils

5

Equation 1.26, with A equal, in turn, to 0, 0.05 and 0.10, is used to calculate values of dry density (
d0,
d5,
d10 respectively) for use in plotting the air content curves. The experimental values of w have been used in these calculations; however, any series of w values within the relevant range could be used. By inspection, the value of air content at maximum dry density is 3.5%.

1.7 From Equation 1.20 e¼

G s
w
1
d

The maximum and minimum values of void ratio are given by Gs
w
1
dmin Gs
w ¼
1
dmax

emax ¼ emin

From Equation 1.23 ID ¼

Gs
w ð1=
dmin
1=
d Þ Gs
w ð1=
dmin
1=
dmax Þ

½1
ð
dmin =
d Þ 1=
dmin ½1
ð
dmin =
dmax Þ 1=
dmin

d

dmin
dmax ¼
dmax

dmin
d
 1:72
1:54 1:81 ¼ 1:81
1:54 1:72

¼

¼ 0:70 ð70%Þ

Chapter 2

Seepage

2.1 The coefficient of permeability is determined from the equation k ¼ 2:3

al h0 log At1 h1

where   0:0052 m2 ; l ¼ 0:2 m 4  A ¼  0:12 m2 ; t1 ¼ 3  602 s 4 h0 1:00 ¼ 0:456 log ¼ log 0:35 h1 2:3  0:0052  0:2  0:456 ;k¼ ¼ 4:9  10
8 m=s 0:12  3  602 a¼

2.2 The flow net is drawn in Figure Q2.2. In the flow net there are 3.7 flow channels and 11 equipotential drops, i.e. Nf ¼ 3.7 and Nd ¼ 11. The overall loss in total head is 4.00 m. The quantity of seepage is calculated by using Equation 2.16: Nf 3:7 ¼ 1:3  10
6 m3 =s per m q ¼ kh ¼ 10
6  4:00  11 Nd

Figure Q2.2

Seepage

7

2.3 The flow net is drawn in Figure Q2.3, from which Nf ¼ 3.5 and Nd ¼ 9. The overall loss in total head is 3.00 m. Then, q ¼ kh

Nf 3:5 ¼ 5:8  10
5 m3 =s per m ¼ 5  10
5  3:00  9 Nd

The pore water pressure is determined at the points of intersection of the equipotentials with the base of the structure. The total head (h) at each point is obtained from the flow net. The elevation head (z) at each point on the base of the structure is
2.50 m. The calculations are tabulated below and the distribution of pressure (u) is plotted to scale in the figure. Point

h (m)

h
z (m)

u ¼ w (h
z) (kN/m2)

1 2 3 4 5 6

2.33 2.00 1.67 1.33 1.00 0.67

4.83 4.50 4.17 3.83 3.50 3.17

47 44 41 37 34 31

e.g. for Point 1: 7  3:00 ¼ 2:33 m 9 h1
z1 ¼ 2:33
ð
2:50Þ ¼ 4:83 m h1 ¼

Figure Q2.3

8

Seepage

u1 ¼ 9:8  4:83 ¼ 47 kN=m2 The uplift force on the base of the structure is equal to the area of the pressure diagram and is 316 kN per unit length.

2.4 The flow net is drawn in Figure Q2.4, from which Nf ¼ 10.0 and Nd ¼ 11. The overall loss in total head is 5.50 m. Then, q ¼ kh

Nf 10 ¼ 2:0  10
6 m3 =s per m ¼ 4:0  10
7  5:50  11 Nd

Figure Q2.4

2.5 The flow net is drawn in Figure Q2.5, from which Nf ¼ 4.2 and Nd ¼ 9. The overall loss in total head is 5.00 m. Then, q ¼ kh

Nf 4:2 ¼ 4:7  10
6 m3 =s per m ¼ 2:0  10
6  5:00  9 Nd

Seepage

9

Figure Q2.5

2.6 The scale transformation factor in the x direction is given by Equation 2.21, i.e. pffiffiffiffiffiffiffi pffiffiffiffiffi 1:8 kz xt ¼ x pffiffiffiffiffi ¼ x pffiffiffiffiffiffiffi ¼ 0:60x kx 5:0 Thus in the transformed section the horizontal dimension 33.00 m becomes (33.00  0.60), i.e. 19.80 m, and the slope 1:5 becomes 1:3. All dimensions in the vertical direction are unchanged. The transformed section is shown in Figure Q2.6 and the flow net is drawn as for the isotropic case. From the flow net, Nf ¼ 3.25 and Nd ¼ 12. The overall loss in total head is 14.00 m. The equivalent isotropic permeability applying to the transformed section is given by Equation 2.23, i.e. k0 ¼

pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðkx kz Þ ¼ ð5:0  1:8Þ  10
7 ¼ 3:0  10
7 m=s

Thus the quantity of seepage is given by q ¼ k0 h

Nf 3:25 ¼ 1:1  10
6 m3 =s per m ¼ 3:0  10
7  14:00  12 Nd

10

Seepage

Figure Q2.6

2.7 The scale transformation factor in the x direction is pffiffiffiffiffiffiffi pffiffiffiffiffi kz 2:7 xt ¼ x pffiffiffiffiffi ¼ x pffiffiffiffiffiffiffi ¼ 0:60x kx 7:5

Thus all dimensions in the x direction are multipled by 0.60. All dimensions in the z direction are unchanged. The transformed section is shown in Figure Q2.7. The equivalent isotropic permeability is pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k0 ¼ ðkx kz Þ ¼ ð7:5  2:7Þ  10
6 ¼ 4:5  10
6 m=s

The focus of the basic parabola is at point A. The parabola passes through point G such that GC ¼ 0:3 HC ¼ 0:3  30 ¼ 9:0 m Thus the coordinates of G are x ¼
48:0

and

z ¼ þ20:0

Substituting these coordinates in Equation 2.34
48:0 ¼ x0


20:02 4x0

Seepage

11

Figure Q2.7

Hence, x0 ¼ 2:0 m Using Equation 2.34, with x0 ¼ 2.0 m, the coordinates of a number of points on the basic parabola are calculated, i.e. x ¼ 2:0
x z

2.0 0

z2 8:0 0 4.00


5:0 7.48


10:0 9.80


20:0 13.27


30:0 16.00

The basic parabola is plotted in Figure Q2.7. The upstream correction is drawn using personal judgement. No downstream correction is required in this case since  ¼ 180 . If required, the top flow line can be plotted back onto the natural section, the x coordinates above being divided by the scale transformation factor. The quantity of seepage can be calculated using Equation 2.33, i.e. q ¼ 2k0 x0 ¼ 2  4:5  10
6  2:0 ¼ 1:8  10
5 m3 =s per m 2.8 The flow net is drawn in Figure Q2.8, from which Nf ¼ 3.3 and Nd ¼ 7. The overall loss in total head is 2.8 m. Then,

12

Seepage

Figure Q2.8

q ¼ kh

Nf 3:3 ¼ 4:5  10
5  2:8  7 Nd
5 3 ¼ 5:9  10 m =s per m

2.9 The two isotropic soil layers, each 5 m thick, can be considered as a single homogeneous anisotropic layer of thickness 10 m in which the coefficients of permeability in the horizontal and vertical directions, respectively, are given by Equations 2.24 and 2.25, i.e. kx ¼ kz ¼

H1 k1 þ H2 k2 10
6 fð5  2:0Þ þ ð5  16Þg ¼ 9:0  10
6 m=s ¼ H1 þ H2 10 H1 þ H2 10 ¼ H1 H2 5 5 þ þ
6 k1 k2 ð2  10 Þ ð16  10
6 Þ

¼ 3:6  10
6 m=s

Then the scale transformation factor is given by pffiffiffiffiffi pffiffiffiffiffiffiffi kz 3:6 xt ¼ x pffiffiffiffiffi ¼ x pffiffiffiffiffiffiffi ¼ 0:63x 9:0 kx Thus in the transformed section the dimension 10.00 m becomes 6.30 m; vertical dimensions are unchanged. The transformed section is shown in Figure Q2.9 and the flow net is drawn as for a single isotropic layer. From the flow net, Nf ¼ 5.6 and Nd ¼ 11. The overall loss in total head is 3.50 m. The equivalent isotropic permeability is

Seepage

Figure Q2.9

k0 ¼

qffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðkx kz Þ ¼ ð9:0  3:6Þ  10
6 ¼ 5:7  10
6 m=s

Then the quantity of seepage is given by q ¼ k0 h

Nf 5:6 ¼ 5:7  10
6  3:50  11 Nd
5 3 ¼ 1:0  10 m =s per m

13

Chapter 3

Effective stress

3.1 Buoyant unit weight:  0 ¼ sat
w ¼ 20
9:8 ¼ 10:2 kN=m3 Effective vertical stress: 0v ¼ 5  10:2 ¼ 51 kN=m2

or

Total vertical stress: v ¼ ð2  9:8Þ þ ð5  20Þ ¼ 119:6 kN=m2 Pore water pressure: u ¼ 7  9:8 ¼ 68:6 kN=m2 Effective vertical stress: 0v ¼ v
u ¼ 119:6
68:6 ¼ 51 kN=m2

3.2 Buoyant unit weight:  0 ¼ sat
w ¼ 20
9:8 ¼ 10:2 kN=m3 Effective vertical stress: 0v ¼ 5  10:2 ¼ 51 kN=m2

or

Effective stress

15

Figure Q3.1/3.2

Total vertical stress: v ¼ ð200  9:8Þ þ ð5  20Þ ¼ 2060 kN=m2 Pore water pressure: u ¼ 205  9:8 ¼ 2009 kN=m2 Effective vertical stress: 0v ¼ v
u ¼ 2060
2009 ¼ 51 kN=m2

3.3 At top of the clay: v ¼ ð2  16:5Þ þ ð2  19Þ ¼ 71:0 kN=m2 u ¼ 2  9:8 ¼ 19:6 kN=m2

0v ¼ v
u ¼ 71:0
19:6 ¼ 51:4 kN=m2 Alternatively,  0 ðsandÞ ¼ 19
9:8 ¼ 9:2 kN=m3

0v ¼ ð2  16:5Þ þ ð2  9:2Þ ¼ 51:4 kN=m2

At bottom of the clay: v ¼ ð2  16:5Þ þ ð2  19Þ þ ð4  20Þ ¼ 151:0 kN=m2 u ¼ 12  9:8 ¼ 117:6 kN=m2

0v ¼ v
u ¼ 151:0
117:6 ¼ 33:4 kN=m2 NB The alternative method of calculation is not applicable because of the artesian condition.

16

Effective stress

Figure Q3.3

Figure Q3.4

3.4  0 ¼ 20
9:8 ¼ 10:2 kN=m3 At 8 m depth: 0v ¼ ð2:5  16Þ þ ð1:0  20Þ þ ð4:5  10:2Þ ¼ 105:9 kN=m2

3.5  0 ðsandÞ ¼ 19
9:8 ¼ 9:2 kN=m3

 0 ðclayÞ ¼ 20
9:8 ¼ 10:2 kN=m3

Effective stress

17

Figure Q3.5

(a) Immediately after WT rise: At 8 m depth, pore water pressure is governed by the new WT level because the permeability of the sand is high. ; 0v ¼ ð3  16Þ þ ð5  9:2Þ ¼ 94:0 kN=m2 At 12 m depth, pore water pressure is governed by the old WT level because the permeability of the clay is very low. (However, there will be an increase in total stress of 9 kN/m2 due to the increase in unit weight from 16 to 19 kN/m2 between 3 and 6 m depth: this is accompanied by an immediate increase of 9 kN/m2 in pore water pressure.) ; 0v ¼ ð6  16Þ þ ð3  9:2Þ þ ð3  10:2Þ ¼ 154:2 kN=m2 (b) Several years after WT rise: At both depths, pore water pressure is governed by the new WT level, it being assumed that swelling of the clay is complete. At 8 m depth: 0v ¼ 94:0 kN=m2

(as above)

At 12 m depth: 0v ¼ ð3  16Þ þ ð6  9:2Þ þ ð3  10:2Þ ¼ 133:8 kN=m2

18

Effective stress

3.6 Total weight: ab ¼ 21:0 kN Effective weight: ac ¼ 11:2 kN Resultant boundary water force: be ¼ 11:9 kN Seepage force: ce ¼ 3:4 kN Resultant body force: ae ¼ 9:9 kN ð73 to horizontalÞ (Refer to Figure Q3.6.)

Figure Q3.6

Effective stress

3.7 Situation (1): (a)  ¼ 3w þ 2sat ¼ ð3  9:8Þ þ ð2  20Þ ¼ 69:4 kN=m2 u ¼ w ðh
zÞ ¼ 9:8f1
ð
3Þg ¼ 39:2 kN=m2

0 ¼ 
u ¼ 69:4
39:2 ¼ 30:2 kN=m2 (b) 2 ¼ 0:5 4 j ¼ iw ¼ 0:5  9:8 ¼ 4:9 kN=m3 # i¼

0 ¼ 2ð 0 þ jÞ ¼ 2ð10:2 þ 4:9Þ ¼ 30:2 kN=m2 Situation (2): (a)  ¼ 1w þ 2sat ¼ ð1  9:8Þ þ ð2  20Þ ¼ 49:8 kN=m2 u ¼ w ðh
zÞ ¼ 9:8f1
ð
3Þg ¼ 39:2 kN=m2

0 ¼ 
u ¼ 49:8
39:2 ¼ 10:6 kN=m2 (b) 2 ¼ 0:5 4 j ¼ iw ¼ 0:5  9:8 ¼ 4:9 kN=m3 " i¼

0 ¼ 2ð 0
jÞ ¼ 2ð10:2
4:9Þ ¼ 10:6 kN=m2 3.8 The flow net is drawn in Figure Q2.4. Loss in total head between adjacent equipotentials:
h ¼

5:50 5:50 ¼ 0:50 m ¼ Nd 11

Exit hydraulic gradient: ie ¼


h 0:50 ¼ ¼ 0:71
s 0:70

19

20

Effective stress

The critical hydraulic gradient is given by Equation 3.9: ic ¼

0 10:2 ¼ 1:04 ¼ 9:8 w

Therefore, factor of safety against ‘boiling’ (Equation 3.11): F¼

ic 1:04 ¼ 1:5 ¼ ie 0:71

Total head at C: hC ¼

nd 2:4  5:50 ¼ 1:20 m h¼ 11 Nd

Elevation head at C: zC ¼
2:50 m Pore water pressure at C: uC ¼ 9:8ð1:20 þ 2:50Þ ¼ 36 kN=m2 Therefore, effective vertical stress at C: 0C ¼ C
uC ¼ ð2:5  20Þ
36 ¼ 14 kN=m2 For point D: hD ¼

7:3  5:50 ¼ 3:65 m 11

zD ¼
4:50 m uD ¼ 9:8ð3:65 þ 4:50Þ ¼ 80 kN=m2 0D ¼ D
uD ¼ ð3  9:8Þ þ ð7  20Þ
80 ¼ 90 kN=m2 3.9 The flow net is drawn in Figure Q2.5. For a soil prism 1.50  3.00 m, adjacent to the piling: hm ¼

2:6  5:00 ¼ 1:45 m 9

Effective stress

Factor of safety against ‘heaving’ (Equation 3.10): F¼

ic 0d 9:7  3:00 ¼ 2:0 ¼ ¼ im w hm 9:8  1:45

With a filter: 0d þ w  w hm ð9:7  3:00Þ þ w ;3¼ 9:8  1:45 ; w ¼ 13:5 kN=m2 F¼

Depth of filter ¼ 13.5/21 ¼ 0.65 m (if above water level).

21

Chapter 4

Shear strength

4.1  ¼ 295 kN=m2 u ¼ 120 kN=m2

0 ¼ 
u ¼ 295
120 ¼ 175 kN=m2

f ¼ c0 þ 0 tan 0 ¼ 12 þ 175 tan 30 ¼ 113 kN=m2

4.2 03 (kN/m2)

1
3 (kN/m2)

01 (kN/m2)

100 200 400 800

452 908 1810 3624

552 1108 2210 4424

The Mohr circles are drawn in Figure Q4.2, together with the failure envelope from which 0 ¼ 44 .

Figure Q4.2

Shear strength

23

4.3 3 (kN/m2)

1
3 (kN/m2)

1 (kN/m2)

200 400 600

222 218 220

422 618 820

The Mohr circles and failure envelope are drawn in Figure Q4.3, from which cu ¼ 110 kN/m2 and u ¼ 0.

Figure Q4.3

4.4 The modified shear strength parameters are 0 ¼ tan
1 ðsin 0 Þ ¼ tan
1 ðsin 29 Þ ¼ 26 a0 ¼ c0 cos 0 ¼ 15 cos 29 ¼ 13 kN=m2

The coordinates of the stress point representing failure conditions in the test are 1 1 ð1
2 Þ ¼  170 ¼ 85 kN=m2 2 2 1 1 ð1 þ 3 Þ ¼ ð270 þ 100Þ ¼ 185 kN=m2 2 2 The pore water pressure at failure is given by the horizontal distance between this stress point and the modified failure envelope. Thus from Figure Q4.4 uf ¼ 36 kN=m2

Figure Q4.4

24

Shear strength

4.5 3 (kN/m2)

1
3 (kN/m2)

1 (kN/m2)

u (kN/m2)

03 (kN/m2)

01 (kN/m2)

150 300 450 600

103 202 305 410

253 502 755 1010

82 169 252...