Solutions Manual

Soil Mechanics Principles and Practice Third Edition

Solutions Manual

Graham Barnes

Contents

2 3 4 5 6 7 9 10 11 12 13 14

Soil description and classification Permeability and seepage Effective stress and pore pressure Contact pressure and stress distribution Compressibility and consolidation Shear strength Shallow foundations – stability Shallow foundations – settlement Pile foundations Lateral earth pressure and retaining structures Slope stability Earthworks and soil compaction

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e Chapter 2

Solution Soil Description and Classification Volume of gravel = (m4 – m1) – (m3 – m2) = (1870.6 – 845.2) – (2346.0 – 1608.7) = 1025.4 – 737.3 = 288.1 cm3

2.1

mass of gravel = m2 – m1 = 1608.7 – 845.2 = 763.5 g

ρs = 763.5 = 2.65 g/ cm3 or Mg/m3 288.1

Sieve size

Mass retained g

Mass passing g

20 14 10 6.3 3.35 2 1.18 600 425 300 212 150 63 tray

0 18.9 67.4 44.2 75.8 122.1 193.7 240.0 282.2 242.1 233.7 265.3 240.0 80.0

2105.4 2086.5 2019.1 1974.9 1899.1 1777.0 1583.3 1343.3 1061.1 819.0 585.3 320.0 80.0 -

2.2

d10 = 0.11 mm U = 0.54 = 4.9 0.11

d30 = 0.23 mm Cc =

d50 = 0.42 mm

mw = 544.4 g

∴moderately uniformly graded

0.232 = 0.89 0.54 × 0.11

V0 = 250ml

wax Gs = 0.90

volume of soil lump = 250 – 544.4 – 537.5 = 242.33 cm3 0.90 bulk density ρb = ms = 537.5 × 106 = 2.218 Mg/m3, say 2.22 Mg/m3 V 242.33 × 106 Bulk weight density = 2.218 × 1000 × 9.81 = 21.76 kN/m3, say 21.8 kN/m3 1000 Moisture content = 537.5 – 479.2 × 100 = 12.17 %, say 12.2 % 479.2

ρd = ρb = = 2.218 = 1.977 Mg/m3, say 1.98 Mg/m3 1+w 1 + 0.122 Bulk weight density = 1.977 × 1000 × 9.81 = 19.39 kN/m3, say 19.4 kN/m3

Graham Barnes 2010

1

100 99.1 95.9 93.8 90.2 84.4 75.2 63.8 50.4 38.9 27.8 15.2 3.8

d60 = 0.54 mm

2.3 ms = 537.5 g

% passing

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution 1000

From Table 2.19

ρb = Gsρw (1 + w) (1 + e) 1 + e = 2.72 × 1.0 × 1.122 = 1.376 2.218

void ratio e = 0.376, say 0.38

2.4 n = e = 0.376 = 0.273, say 0.27 1 + e 1.376 Sr = w Gs × 100 = 0.122 × 2.72 × 100 = 88.3 %, say 88% e 0.376 Av = n (1 - Sr) = 0.273 (1 – 0.883) = 3.2 %

The degree of saturation would be 100 % Sr = w Gs = 100 e

2.5

w = e × 100 = 0.376 × 100 = 13.82 %, say 13.8 % Gs 2.72

ρb = Gsρw (1 + w) = 2.72 × 1.0 × 1.1382 = 2.25 Mg/m3 (1 + e) 1.376

Av = n (1 - Sr) with n =

2.6

e 1+e

and Sr = wGs e

Av = e – w Gs 1+e

e = Av + w Gs = 0.05 + 0.135 × 2.68 = 0.4335 1 - Av 1 – 0.05

ρb = 2.68 × 1.0 × (1 + 0.135) = 2.122 Mg/m3 , say 2.12 Mg/m3 1 + 0.4335

Liquid limit Plastic limit = 26.6 + 27.3 = 26.95, say 27 % 2

2.7

Plasticity index = 55 – 27 = 28 % Classification is CH

Graham Barnes 2010

2

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution Liquidity index = 35 – 27 = 0.29 28

2.8 Consistency index = 55 – 35 = 0.71 28

Chapter 3

Permeability and Seepage Q = 200 ml

3.1

l = 100 mm

h = 234 mm

t = 225 s

k = Q l = 200 × 103 × 100 × 4 = 8.6 × 10-2 mm/s = 8.6 × 10-5 m/s Aht π × 752 × 234 × 225 a = π × 42 = 12.57 mm2 4

A = π × 1002 = 7854.0 mm2 4

3.2 k = 2.3 × 12.57 × 120 log10 (950/740) = 2.7 × 10-5 mm/s = 2.7 × 10-8 m/s 7854 30 × 60

koverall = Q l = 100 × 103 × 100 × 4 = 4.5 × 10-3 mm/s = 4.5 × 10-6 m/s A h t π × 752 × 672 × 745 kv =

3.3

Lsand + Lsilt d Lsand /ksand + Lsilt/ksilt

Lsand = 95 mm

Lsilt = 5 mm

ksand = 8.6 × 10-5 m/s 4.5 = 95 + 5 d 103 95 × 102 /8.6 + 5/ksilt ksilt = 2.4 × 10-4 mm/s = 2.4 × 10-7 m/s

At top of clay total head = 5.0 + 5.0 = 10.0 m At base of clay total head = 3.0 m ∴Head difference = 10.0 – 3.0 = 7.0 m

3.4

i = 7.0 = 1.4 5.0 q = 100 × 100 × 5 × 1.4 × 1000 × 60 × 60 × 24 108 = 60480 litres/day = 60.48 m3/day

Graham Barnes 2010

3

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution Hc = 7.5 – 4.0 = 3.5 m f = 2.75

3.5

k=

q f d Hc

=

2.5 d 1000 × 60 × 2.75 × 0.15 × 3.5

= 2.9 × 10-5 m/s

Figure 3.41

3.7

datum at base of gravel h = 7.5 m

nf = 3

nd = 8

q = 3 × 7.5 × 3 × 1000 × 60 = 5.1 litres/min, say 5 litres/min 105 8 Location

Total head m

Elevation head m

Pressure head m

pore pressure kPa

0 1 2 3 4 5 6 7 8

7.5 7.5 × 7/8 = 6.56 7.5 × 6/8 = 5.63 7.5 × 5/8 = 4.69 7.5 × 4/8 = 3.75 7.5 × 3/8 = 2.81 7.5 × 2/8 = 1.88 7.5 × 1/8 = 0.94 7.5 × 0/8 = 0

5.5 2.56 0.1 -1.9 -3.0 -3.0 -3.0 -1.8 0

2.0 4.0 5.53 6.59 6.75 5.81 4.88 2.74 0

19.6 39.2 54.2 64.6 66.2 57.0 47.8 26.8 0

Graham Barnes 2010

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Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution

3.7

Shape factor is the same 3/8

k overall =

3 10

5

×

6 10 4

= 1.34 ×10 −4 m/s

3.8 q = 1.34 × 7.5 × 3 × 1000 × 60 = 22.6 litres/min, say 23 litres/min 104 8 Consider last block for boiling condition ∆l = 1.8 m ∆h = 7.5/8 = 0.938 m Upward seepage force = γw × ∆h × ∆l × 1 = 9.8 × 0.938 × 1.8 × 1 = 16.55 kN

3.9

Downward seepage force = γbgravel × ∆l × 1.0 × 1 + γsubsand × ∆l × ∆l × 1 = 20.5 × 1.8 × 1.0 × 1 + (19 – 9.8) × 1.8 × 1.8 × 1 = 36.9 + 29.8 = 66.7 kN 16.55 kN < 66.7 kN ∴ ULS verified. See Worked example 3.12 for comments on this inequality.

Consider block 2.0 m × 4.0 m for heaving mean head hm = 1.41 + 2.81 = 2.11 m 2 upward force = 9.8 × 2.11 × 2.0 × 1 = 41.4kN Downward force = 20.5 × 1.0 × 2.0 × 1 + (19 – 9.8) × 3.0 × 2.0 × 1 = 41.0 + 55.2 = 96.2 kN 41.4 kN < 96.2 kN ∴ultimate limit state (ULS) verified

3.9

Applying partial factor γG;dst on destabilising action of 1.335 in the UK National Annex Design upward force = 41.36 × 1.335 = 55.2 kN partial factor γG;stb on stabilising action is 0.90 in the UK National Annex Design downward force = 96.2 × 0.90 = 86.6 kN 55.2 kN < 86.6 kN ∴ ULS verified. See Worked example 3.11 for comments on this inequality. Note: without gravel

Graham Barnes 2010

55.2 kN > 49.7 kN (55.2 × 0.90) kN ∴ULS is not verified.

5

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution head difference h = 8.0 m

3.10

nf = 2.5 nd = 16 q = 5 × 8.0 × 2.5 × 60 × 60 × 1000 106 16 = 22.5 litres/hour

3.11

Along the slip surface

D15soil = 0.15 mm

3.12

Location

Total head m

Elevation head m

Pressure head m

Pore pressure kPa

1

5.6

5.6

0

0

2

8.0 × 11/16 = 5.50

5.2

0.3

2.9

3

8.0 × 10/16 = 5.00

3.9

1.1

10.8

4

8.0 × 9/16 = 4.50

2.7

1.8

17.6

5

8.0 × 8/16 = 4.00

1.9

2.1

20.6

6

8.0 × 7/16 = 3.50

1.25

2.25

22.1

7

8.0 × 6/16 = 3.00

0.90

2.1

20.6

8

8.0 × 5/16 = 2.50

0.6

1.9

18.6

9

8.0 × 4/16 = 2.00

0.4

1.6

15.7

10

8.0 × 3/16 = 1.50

0.2

1.3

12.7

11

8.0 × 2/16 = 1.00

0.1

0.9

8.8

12

8.0 × 1/16 = 0.50

0.05

0.45

4.4

13

8.0 × 0/16 = 0

0

0

0

D85soil = 2.1 mm

Minimum D15filter = 5 × D15soil = 5 × 0.15 = 0.75 mm Maximum D15filter = 5 × D85soil = 5 × 2.1 = 10.5 mm < 5% fines

Graham Barnes 2010

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Soil Mechanics Principles and Practice Third Edition

Exercis e Chapter 4

Graham Barnes

Solution Effective Stress and Pore pressure Total stress = 5 × 9.8 + 5 × 18 = 139 kPa

4.1

Pore pressure = 10 × 9.8 = 98 kPa Effective stress = 139 – 98 = 41 kPa

Total stress = 5 × 18 = 90 kPa

4.2

Pore pressure = 5 × 9.8 = 49 kPa Effective stress = 90 – 49 = 41 kPa

Top of sand Total stress = 5 × 21 = 105 kPa Pore pressure = 3 × 9.8 = 29.4 kPa Effective stress = 105 – 29.4 = 75.6 kPa

4.3 Bottom of sand Total stress = 5 × 21 + 5 × 18 = 195 kPa Pore pressure = 8 × 9.8 = 78.4 kPa Effective stress = 195 – 78.4 = 116.6 kPa

At 5 m Total stress = 2 × 16.5 + 3 × 18 = 87 kPa Pore pressure = 2 × 9.8 = 19.6 kPa Effective stress = 87 – 19.6 = 67.4 kPa

4.4 At 2 m Total stress = 2 × 16.5 = 33 kPa Pore pressure = –1 × 9.8 = –9.8 kPa Effective stress = 33 – – 9.8 = 42.8 kPa

4.5 Before pumping a)

top of clay

b) middle

Graham Barnes 2010

1.0 × 16.5 + 3.0 × 18 – 3 × 9.8 = 41.1 kPa 1.0 × 16.5 + 3.0 × 18 + 2.5 × 21 – 5.5 × 9.8 = 69.1 kPa

7

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution c)

bottom of clay

1.0 × 16.5 + 3.0 × 18 + 5.0 × 21 – 8 × 9.8 = 97.1 kPa

Immediately after pumping a) top of clay 3.0 × 16.5 + 1.0 × 18 – 1.0 × 9.8 = 57.7 kPa b) middle

69.1 kPa (no change)

c) bottom of clay

97.1 kPa (no change)

Long-term a)

top of clay

57.7 kPa

b) middle

3 × 16.5 + 1 × 18 + 2.5 × 21 – 3.5 × 9.8 = 85.7 kPa

c)

3 × 16.5 + 1 × 18 + 5 × 21 – 6 × 9.8 = 113.7 kPa

bottom of clay

Note: change in effective stress = 16.6 kPa throughout = 2 × 9.8 – 2 × (18 – 16.5) = 19.6 – 3 = 16.6 kPa

4.6

Change in effective stress = 2 × 9.8 = 19.6 kPa

At 2 m Present σv´ = 2 × 19.5 = 39.0 kPa

Past maximum σv´ = 2 × 19.5 = 39.0 kPa

OCR = 39.0 = 1.0 39.0 At 6 m Present σv´ = 6 × 19.5 – 4 × 9.8 = 77.8 kPa

4.7

Past maximum σv´ = 6 × 19.5 = 117.0 kPa

OCR = 117.0 = 1.50 77.8 At 20 m Present σv´ = 20 × 19.5 – 18 × 9.8 = 213.6 kPa Past maximum σv´ = 20 × 19.5 – 14 × 9.8 = 252.8 kPa OCR = 252.8 = 1.18 213.6

4.8

Vertical effective stress σV´ = 6 × 20.5 – 4.5 × 9.8 = 78.9 kPa Horizontal effective stress σH´ = 120 – 4.5 × 9.8 = 75.9 kPa K0 = 75.9 = 0.96 78.9 For normally consolidated clay K0≈1 – sinφ´ = 1 – sin25o = 0.577

Graham Barnes 2010

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Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution ∴ clay is overconsolidated

Cell pressure kPa 0 50 50 100 100 150 150 200 200 300 300 400

4.9

– 40

– 90

– 140

– 190

– 290

–

σ1 – σ3 0 68 117 146 171 190 198

4.10

Chapter 5

Back pressure kPa 0

Pore pressure kPa –5 12 38 67 89 126 140 184 190 285 290 388

∆u

∆σ3

B

–

–

–

17

50

0.34

–

–

–

29

50

0.58

–

–

–

37

50

0.74

–

–

–

44

50

0.88

–

–

–

95

100

0.95

–

–

–

98

100

0.98

∆u 0 15 19 12 1 –13 –25

u 300 315 319 312 301 287 275

∆σ1 – ∆σ3 0 68 117 146 171 190 198

A 0 0.22 0.16 0.08 0.01 –0.07 –0.13

Contact Pressure and Stress Distribution From middle load

σv =

3Pz 3 2πR 5

z = R = 2.0 m

5.1

σv = 3 × 500 × 2.03 = 59.7 kPa 2 × π × 2.05

from side loads z = 2.0 m

R2 = 2.02 + 4.02 = 20

R = 4.47 m

σv = 3 × 500 × 2.03 = 1.1 kPa 2 × π × 4.475

Stress beneath middle load = 59.7 + 1.1 + 1.1 = 61.8 kPa, say 62 kPa

Graham Barnes 2010

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Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution

From middle load

z = R = 2.0 m

5.2

σv =

2Pz 3

πR 4

σv = 2 × 500 × 2.03 = 159.2 kPa π × 2.04

from side loads z = 2.0 m

σv = 2 × 500 × 2.03 = 6.4 kPa π × 4.474

R = 4.47 m (from 5.1)

Stress beneath middle load = 159.2 + 6.4 + 6.4 = 172 kPa

Beneath centre ½α = tan–1(4/5) = 38.7o

α = 77.3o = 1.349 radians

β = –½α = – 38.7o σv = 80/π [1.349 + sin77.3 × cos0] = 59.2 kPa Beneath edge β = 0 α = tan–1(8/5) = 58.0o = 1.012 radians

σv = 80/π [1.012 + sin58 × cos58] = 37.2 kPa

5.3 4m from edge β = tan–1(4/5) = 38.66o

ω = tan–1(12/5) = 67.38o α = 67.38 – 38.66 = 28.72o = 0.501 radians σv = 80/π [0.501 + sin28.72 × cos(28.72 + 2 × 38.66)] = 9.4 kPa

Beneath centre z = 6.0 m

B = 3.0 m

L = 6.0 m

L/B = 2

z/B = 2

I = 0.12

L = 12.0 m

L/B = 2

z/B = 1

I = 0.20

σv´ = 0.12 × 105 × 4 = 50.4 kPa

5.4 Beneath corner z = 6.0 m

B = 6.0 m

σv´ = 0.20 × 105 = 21.0 kPa

5.5 At A Stress at corner of 3No. 10 × 10 m

z/B = 8/10 = 0.8 L/B = 1 I = 0.197 (Giroud)

σv = 80 × 0.197 × 3 = 47.3 kPa, say 48 kPa

Graham Barnes 2010

10

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution

At B Stress at corner of 2No. 10 × 20 m minus 1No. 10 × 10 m

σv = 80 × (2 × 0.217 – 0.197) = 19 kPa

z/B = 8/10 = 0.8 L/B = 2 I = 0.217

At centre z = 6.0 m

B = 3.0 m

z/h = 6/12 = 0.5 L/B = 6/3 = 2 From Figure 5.10

5.6

L = 6.0 m

h = 12 m

h/B = 12/3 = 4

I = 0.136 by interpolation

σv = 0.136 × 105 × 4 = 57.1 kPa, say 57 kPa At corner z = 6.0 m

B = 6.0 m

L = 12.0 m

h = 12.0 m

L/B = 2

z/h = 6/12 = 0.5 h/B = 12/6 = 2

I = 0.223

σv´ = 0.223 × 105 = 23.4 kPa, say 23 kPa

At centre z = 6.0 m

B = 6.0 m

L = 12.0 m

L/B = 2

z/B = 6/6 = 1

5.7 From Figure 5.12 I = 0.43

σv´ = 0.43 × 105 = 45.2 kPa, say 45 kPa

Chapter 6

Compressibility and Consolidation Final void ratio ef = 0.273 × 2.70 = 0.737

∆e =

6.1

∆H ( 1 + e) H

Pressure kPa 0 25 50 100 200 400 800

Graham Barnes 2010

Thickness mm 19.0 18.959 18.918 18.836 18.457 17.946 17.444

11

∆H mm

∆e

ef

–

–

0.041 0.082 0.164 0.543 1.054 1.556

0.004 0.008 0.016 0.053 0.103 0.152

0.856 0.852 0.848 0.840 0.803 0.753 0.704

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution 200 25 0

17.526 17.669 17.782

1.474 1.331 1.218

6.1

pc´≈125 kPa

6.2

p0´= 5 × (21.2 – 9.8) = 57 kPa

OCR = 125 = 2.2 56

Graham Barnes 2010

12

0.144 0.130 0.119

0.712 0.726 0.737

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution Pressure increment kPa 0 – 25 25 – 50 50 – 100 100 – 200 200 – 400 400 – 800

6.3

∆H

Hi

∆σ

mv m2/MN

0.041 0.041 0.082 0.379 0.511 0.502

19.0 18.959 18.918 18.836 18.457 17.916

25 25 50 100 200 400

0.086 0.087 0.087 0.201 0.138 0.070

From the graph At p0´ = 57 kPa

e = 0.847

At σv´ = 107 kPa

e = 0.839

At σv´ = 157 kPa

e = 0.820

For increment p0´ + 50

mv = 0.847 – 0.839 × 1 × 1000 = 0.087 m2/MN 1.847 50

For increment p0´ + 100

mv = 0.847 – 0.820 × 1 × 1000 = 0.146 m2/MN 1.847 100

At σv´ = 200 kPa

e = 0.803

At σv´ = 400 kPa

e = 0.753

At σv´ = 800 kPa

e = 0.704 Pressure increment kPa

Compression index Cc

200 – 400

Cc = 0.803 – 0.753 = 0.166 log10(400/200)

400 – 800

Cc = 0.753 – 0.704 = 0.163 log10(800/400)

200 – 800

Cc = 0.803 – 0.704 = 0.164 log10(800/200)

6.4

∴ say Cc = 0.164

6.5

Graham Barnes 2010

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Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution

H0 = 18.50 mm Hf = 17.76 mm

d=

H=

18.50 + 17.76 = 18.13mm 2

18.13 = 9.065mm 2

t 90 = 3.35

t90 = 11.22 mins

cv = 0.848 × 9.0652 × 60 × 24 × 365 = 3.26 m2/yr, say 3.3 m2/yr 11.22 1000 × 1000 mv = (18.50 – 17.76) × 1 × 1000 = 0.20 m2/MN 18.50 200 k = cv mv γw =

3.26 365 × 24 × 60 × 60

× 0.20 × 9.8 = 2.0 × 10–10 m/s 1000

Split the clay into 2 m thick sub-layers mv = 0.24 – 0.02z

∆σ = 2.5 × 20 = 50 kPa Depth m 1 3 5 7 9

6.6

a) Tv = cv t = 12 × 6 = 0.67 d2 32 × 12

δH = mv × ∆σ × H

mv

δH

0.22 0.18 0.14 0.1 0.06

22 18 14 10 6 Σ = 70 mm

From Figure 6.15

Z=1

Initial excess pore pressure = 50 kPa Remaining pore pressure = 50 × (1 – 0.76) = 12 kPa

6.7 b)

Tv = 12 × 6 = 0.167 62 × 12

Z=1

Uv = 0.173

Remaining pore pressure = 50 × (1 – 0.173) = 41.4 kPa, say 42 kPa

Graham Barnes 2010

14

Uv = 0.755, say 0.76

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution ∆σv = 3 × 9.8 = 29.4 kPa

δH = 0.45 × 29.4 × 5 = 66.1 mm

d = 5/2 = 2.5 m

time t = 6/2 + 20 = 23 weeks Tv = 5.5 × 23 = 0.389 2.52 × 52

6.8

0.389 = π U v 4

2

Uv > 60%∴use log10 (100

U v = 0.704

(

0.389 = 1.781 − 0.933log10 100 − Uv

– Uv) = 1.781 – 0.389 = 1.492

)

U = 69%

0.933 ∴ settlement = 0.69 × 66 = 46 mm

U = 90%

6.9

∴Tv = 0.848

0.848 = 5.5 × t 2.52

t = 0.848 × 2.52 = 0.96 years = 50.1 weeks 5.5 ∴ time from start of pumping = 50.1 + 6/2 = 53.1 weeks

For 1-dimensional consolidation alone d = 10/2 = 5.0 m 0.071 = cv t d2

6.10

∴Tv = 0.071

Uv = 30% ∴ cv t = 0.071 × 52

∴cH t = 3.5 × 0.071 × 52

Tr = cH t = 3.5 × 0.071 × 52 = 0.399 4R2 4 × (0.564 × 3.5)2 r = 0.10 m

n = 0.564 × 3.5 = 19.74 0.10

∴Ur = 0.755

1 – Uoverall = (1 – Uv )(1 – Ur) = (1 – 0.3) (1 – 0.755) = 0.172

Graham Barnes 2010

15

∴Uoverall = 0.828, or 83%

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution Height of embankment = 6 + δH Consolidation settlement = δH = 0.30 × (6 + δH) × 20.5 × 13 = 0.080 × (6 + δH) 1000

δH = 0.48/0.92 = 0.521 m check: = 0.30

× 6.521 × 20.5 × 13 = 521 mm

required Uoverall = 521 – 50 = 0.904 521 time required = ½ + 2 = 2.5 months Tv = 2.5 × 2.5 = 0.0123 6.52 × 12

6.11

∴Uv =

4Tv = 0.125 π 2πrw = 2

R = 0.525s

n = R = 0.525s = 21.15s rw 0.0248

(1 – 0.904) = (1 – 0.125) (1 – Ur) ∴Ur = 0.890

× (70 + 8) ∴s=

rw = 24.8 mm

n 21.15

1.323 Tr = cH t = 7.0 × 2.5 =∴ 1.323 s= 2 2 2 Tr 4R 4× (0525s) × 12 s 1.323 Tr

n

Tr

15 20 25 30 40

0.547 0.626 0.687 0.737 0.817

n 21.15 0.709 0.946 1.182 1.418 1.891

1.555 1.454 1.388 1.34 1.273

From graph s = 1.36 m, say 1.35 m

Chapter 7

Shear Strength Peak strength τf = cp´ + σN´ tanφp´

= 5 + (185 – 90) × tan 29o = 57.7 kPa

Residual strength τf = cr´ + σN´ tanφr´

= 0 + (185 – 90) × tan 12o = 20.2 kPa

7.1

7.2

σv´ = 122 – 32 = 93 kPa

σH´ = 97 – 32 = 65 kPa

Apex of Mohr circle

τ = σv´ – σH´ = 93 – 65 = 14 kPa

Graham Barnes 2010

16

∴s =

1.323 Tr

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution 2

2

σv´ + σH´ = 93 + 65 = 79 kPa 2 2

at the normal stress of

at failure τf = c´ + σN´ tanφ´

= 8 + 79 tan 28o = 50 kPa

50 >>14 ∴soil is not at failure

σf´ = 93 + σ3f´ + (93 – σ3f´) cos (90 + 28) 2 2 = 46.5 + 0.5σ3f´ – 21.8 + 0.235σ3f´ = 24.7 + 0.735σ3f´ 8 + (24.7 + 0.735σ3f´) tan28o = (93

– σ3f´ ) sin (90 + 28) = 41.1 + 0.441σ3f´ 2

7.3

21.1 + 0.391σ3f´ = 41.1 – 0.441σ3f´ OR σ3f´ = Kaσv´ – 2c´√Ka

σ3f´ = 20.0 = 24.0 kPa 0.832

Ka = 1 – sin28o = 0.361 1+ sin28o

σ3f´ = 0.361 × 93 – 2 × 8 × √0.361 = 24 kPa

7.4

σ3´ = σ3 – u = 450 – u

σ1´ = σ1´ – σ3´ + σ3´

s = σ1´ + σ3´ 2

t = σ1´ – σ3´ ´2

p´ = σ1´

u

σ1

σ 1´

σ3´

0 68 117 146 171 190 198

300 315 319 312 301 287 275

450 518 567 596 621 640 648

150 203 248 284 320 353 373

150 135 131 138 149 163 175

Maximum sinφ´ with maximum σ1´/σ3´ = 2.17

Graham Barnes 2010

17

q = σ1´ – σ3

3

σ1 – σ3

M = (q/p´)max = 198 = 0.822 241

+ 2σ3´

σ1´ σ3 ´ 1.00 1.50 1.89 2.06 2.15 2.17 2.13

s´

t´

p´

q

150 169 190 211 235 258 274

0 34 59 73 86 95 99

150 158 170 187 206 226 241

0 68 117 146 171 190 198

∴sin φ´ = 95 = 0.368 258

φ´ = 21.6o

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution

7.5

7.6

at σN´ = 100 kPa

τ = 15 + 100 tan39 = 96 kPa

ε = ∆l = ∆l l0 76

A0 = π × 382 = 1134.1 mm2 4

A = A0 (1 – ε)

σ1 – σ3 = F A

7.7

Cell pressure

kPa 100 200 400

σ3 350 400 500

∆l mm

Force N

ε %

Area mm2

11.4 9.6 8.5

331 309 312

15.0 12.6 11.2

1334.2 1297.6 1277.1

σ1 – σ3 143 208 312

σ1 493 608 812

u 326 354 418

7.8

Graham Barnes 2010

τ = 64 kPa

18

σ1´ 167 254 394

∴F = 96 = 1.50 64

σ1 – σ3 kPa 248.1 238.1 244.3

σ 3´ 24 46 82

s´ 96 150 238

cu kPa 124 119 122

t´ 72 104 156

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution σ3 100 200 300

σ1 344 644 940

σ1 – σ3 244 444 640

t´ 122 222 320

7.9

1.75 = N – λln150 = N – λ × 5.011

7.10

1.65 = N – λln300 = N – λ × 5.704

0.10 = λ × (5.704 – 5.011)

λ = 0.10 = 0.144, say 0.14 0.693

qf´ = 368 kPa

7.11

N = 1.75 + 0.144 ln150 = 2.47

σ1´ = 368 + 300 = 668 kPa

pf´ = 1/3(668 + 2 × 300) = 422.7 kPa M = 368 = 0.87 422.7

7.12

σ3´ = 150 kPa

M = 0.87

qf´ = Mpf´

σ1´ – 150 = 0.87 × 1/3 (σ1´ + 2 × 150) = 0.29 σ1´ + 87 σ1´ = 237 = 333.8 kPa 0.71

Graham Barnes 2010

σ1´ – σ3´ = 333.8 – 150 = 183.8 kPa

19

s´ 222 422 620

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution OR ∴pf´ = 450 = 211.3 kPa 2.13

qf´ = 3 = 0.87 pf´ pf´ – p0´ pf´ – 150

σ3´= 150 kPa

7.13

pf´ = 211.3 kPa

v = 2.37 – 0.144 × ln211.3 = 1.599

σ3´= 300 kPa

∴e = 0.599, say 0.60

pf´ = 422.7 kPa

v = 2.37 – 0.144 × ln422.7 = 1.499

7.14

∴qf´ 0.87 × 211.3 = 183.8 kPa

∴e = 0.499, say 0.50

σ3´= 150 kPa vf = v0 = 1.75

vf = Γ + λ ln pf´ (on CSL)

ln pf´ = 2.37 – 1.75 = 4.306 0.144

pf´ = 74.1 kPa

qf´ = M qf´ = 0.87 × 74.1 = 64.5 kPa, say 65 kPa

σ1´ – σ3´ = 64.5 kPa

1/3σ1´ + 2/3σ3´ = 74.1 kPa

σ1´ = 64.5 + σ3´ kPa

1/3(σ3´+ 64.5) + 2/3σ3´ = 74.1 kPa

σ3´ = 74.1 – 64.5 = 52.6 kPa 3

u =σ3

– σ3´ = 150 – 52.6 = 97.4 kPa

u = pf

– pf´ = 171.5 – 74.1 = 97.4 kPa

OR pf = p0´ + 1/3qf = 150 + 64.5 = 171.5 kPa 3

σ3´= 300 kPa

vf = v0 = 1.65

vf = Γ + λ ln pf´ (on CSL)

ln pf´ = 2.37 – 1.65 = 5.000 0.144

pf´ = 148.4 kPa

qf = M pf´ = 0.87 × 148.4 = 129.1 kPa, say 129 kPa

σ1´ – σ3´ = 129.1 kPa σ1´ = 129.1 + σ3´ kPa

pf´ = 1/3(σ3´+ 129.1) + 2/3σ3´ = 148.4 kPa

σ3´ = 148.4 – 129.1 = 105.4 kPa 3

uf =σ3

OR pf = p0´ + 1/3 qf =300 + 1/3 × 129.1 = 343.0 kPa

Graham Barnes 2010

20

– σ3´ = 300 – 105.4 = 194.6 kPa, say 195 kPa

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution u = pf

Chapter 9

9.1

– pf´ = 343.0 – 148.4 = 194.6 kPa

Shallow Foundations - Stability Net Rd = cukNcsc sc = 1.2 A´ γcu Combination 1 Rd = 3.52 × 95 × 5.14 × 1.2 = 7178 kN 1.0

Nc= 5.14 Combination 2 Rd = 3.52 × 95 × 5.14 × 1.2 = 5127.2 kN 1.4

Combination 1 Net Ed = 1.35 × 2925 + 1.5 × 1225 + q – q = 5786.25 kN Ed < Rd (5786.3 < 7178)

9.2

Combination 2 Net Ed = 1.0 × 2925 + 1.3 × 1225 + q – q = 4517.5 kN Ed < Rd (4517.5 < 5127.2) ∴ ULS is verified

Combination 1 5786 = B2 × 95 × 5.14 × 1.2 1.0

Combination 2 4517.2 = B2 × 95 × 5.14 × 1.2 1.4

B2 = 9.87

B2 = 10.79

9.3 B = 3.14 m

B = 3.29 m

Combination 2 B = 3.29 m, say 3.3 m

9.4 Combination 1 φ´ = 25o

Nc = 20.7

sq = 1 + sin 25o = 1.423

Nq = 10.7

Nγ = 9.1

sγ = 0.7

sc = (1.423 × 10.7 – 1) = 1.467 (10.7 – 1)

Net Rd = c´ Nc sc + q´ Nq sq + 0.5γ´ B´ Nγ sγ + γw hw A´ q´ = 1.5 × 20.5 – 1.0 × 9.8 = 20.95 kPa

–q γ´ = 20.5 – 9.8 = 10.7 kN/m3

Net Rd = 4 × 20.7 × 1.467 + 20.95 × 10.7 × 1.423 + 0.5 × 10.7 × 3.3 × 9.1 × 0.7 + 1.0 × 9.8 – 1.5 × 20.5 iiiiiiA´ 1.0

= 121.5 + 319.0 + 112.5 + 9.8 – 30.75 = 532.0 kPa Net Rd = 532.0 × 3.32 = 5794.0 kN

5786 < 5794 ∴ULS verified, but see below Combination 2

Graham Barnes 2010

21

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution tan25o/1.25 = 0.373 φd = 20.46o

Nc = 15.3

Nq = 6.7

sq = 1 + sin 20.46o = 1.35

sγ = 0.7

sc = (1.35 × 6.7 – 1) = 1.41 (6.7 – 1)

Nγ = 4.3

Net Rd = 4 × 15.3 × 1.41 + 20.95 × 6.7 × 1.35 + 0.5 × 10.7 × 3.3 × 4.3 × 0.7 + 1.0 × 9.8 – 1.5 × 20.5 iiiiiiA´ 1.25

= 69.0 + 189.5 + 53.1 + 9.8 – 30.75 = 290.65 kPa Net Rd = 290.65 × 3.32 = 3165.2 kN

4517.5 > 3165.2 ∴ULS not verified q´ = 1.5 × 20.5 = 30.75 kPa

γ´ = 20.5 kN/m3

Combination 1 Net Rd = 4 × 20.7 × 1.467 + 30.75 × 10.7 × 1.423 + 0.5 × 20.5 × 3.3 × 9.1 × 0.7 – 1.5 × 20.5 iiiiiiA´ 1.0

= 121.5 + 468.2 + 215.5 – 30.75 = 774.45 kPa Net Rd = 774.45 × 3.32 = 8433.8 kN

9.5

5786 < 8434 ∴ULS verified Combination 2 Net Rd = 4 × 15.3 × 1.41 + 30.75 × 6.7 × 1.35 + 0.5 × 20.5 × 3.3 × 4.3 × 0.7 – 1.5 × 20.5 iiiiiiA´ 1.25

= 69.0 + 278.1 + 101.8 – 30.75 = 418.15 kPa Net Rd = 418.15 × 3.32 = 4553.7 kN

4517.5 < 4553.7 ∴ULS verified

Graham Barnes 2010

22

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution Combination 1 γφ = 1.0

φ´ = 38o

Nq = 48.9

q´ = 2.0 × 19.5 – 1.5 × 9.8 = 24.3 kPa sq = 1 + 3.0 sin 38o = 1.41 4.5

Nγ = 74.9

γ´ = 19.5 – 9.8 = 9.7 kN/m3

sγ = 1

– 0.3 ×3.0 = 0.80 4.5

Net Rd = 24.3 × 48.9 × 1.41 + 0.5 × 9.7 × 3.0 × 74.9 × 0.8 + 1.5 × 9.8 – 2.0 × 19.5 iiiiiiA´

9.6

= 1675.5 + 871.8 + 14.7 – 39.0 = 2523.0 kPa Combination 2 γφ = 1.25

φd = 32.0o

tan38o/1.25 = 0.625

sq = 1 + 3.0 sin 32o = 1.353 4.5

sγ = 1

Nq = 23.2

Nγ = 27.7

– 0.3 ×3.0 = 0.80 4.5

Net Rd = 24.3 × 23.2 × 1.353 + 0.5 × 9.7 × 3.0 × 27.7 × 0.8 + 1.5 × 9.8 – 2.0 × 19.5 iiiiiiA´ = 762.8 + 322.4 + 14.7 – 39.0 = 1060.9 kPa

Combination 1 Ed = 1.35 × 2150 + 1.5 × 350 = 3427.5 kN Net Rd = 2523.0 × 3.0 × 4.5 = 34060.5 kN

γR = Rd = 34060.5 = 9.94 Ed 3427.5

9.7 Combination 2 Ed = 1.0 × 2150 + 1.3 × 350 = 2605.0 kN Net Rd = 1060.9 × 3.0 × 4.5 = 14322.2 kN

γR = Rd = 14322.2 = 5.50 Ed 2605.0 WF = 2.5 × 0.5 × 25 WB1 = 1.55 × 1.0 × 21 WB2 = 0.65 × 1.0 × 21

9.8

= 31.25 kN = 32.55 kN = 13.65 kN 77.45 kN

Combination 1 Design vertical load =125 × 1.35 + 45 × 1.5 = 236.25 kN Total vertical load = 77.45 + 236.25 = 313.7 kN 313.7 × e = 236.25 × 0.45 + 13.65 × 0.925 – 32.55 × 0.475 = 103.48 kNm

Graham Barnes 2010

23

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution e = 103.48 = 0.330 m 313.7 B´ = 2.5 – 2 × 0.33 = 1.84 m

sc = sq = sγ = 1.0

q = q´ = 1.5 × 22 = 33 kPa

γ´ = 22 – 9.8 = 12.2 kN/m3

γc = γφ = 1.0

Nc = 18.1

φ´ = 23o

Nq = 8.7

Nγ = 6.5

Net Rd = 4 × 18.1 + 33 × 8.7 + 0.5 ×12.2 × 1.84 × 6.5 – 33 = 399.46 kPa iiiiiiA´ 1.0 Rd = 399.46 ×1.84 = 735.0 kN Net design load = 313.7 –1.84 × 22 = 273.2 kN 273.2 < 735.0 ∴ULS verified Combination 2 Design vertical load =125 × 1.0 + 45 × 1.3 = 183.5 kN Total vertical load = 77.45 + 183.5 = 260.95 kN 260.95 × e = 183.5 × 0.45 + 13.65 × 0.925 – 32.55 × 0.475 = 79.74 kNm e = 79.74 = 0.306 m 260.95

9.8

γc = γφ = 1.25

B´ = 2.5 – 2 × 0.306 = 1.89 m

sc = sq = sγ = 1.0

φd = 18.76o Nc = 13.7 Nq = 5.7

tan23o/1.25 = 0.340

Net Rd = 4 × 13.7 + 33 × 5.7 + 0.5 ×12.2 × 1.89 × 3.2 – 33 = 235.83 kPa iiiiiiA´ 1.25 Rd = 235.83 ×1.89 = 445.7 kN Net design load = 260.95 – 1.89 × 22 = 219.4 kN 219.4 < 445.7 ∴ULS verified

9.9

WF = 3.3 × 3.8 × 0.6 × 25 = 188.10 kN WB = 3.3 × 3.8 ×1.2 × 20 = 300.96 kN 489.06 kN Combination 1 Design vertical load = 2925 × 1.35 + 1225 × 1.5 = 5786.25 kN Design horizontal load = 150 × 1.35 + 50 × 1.5 = 277.50 kN Total vertical load = 489.06 + 5786.25 = 6275.31 kN e = 277.50 × 3.3 = 0.146 m 6275.31

Graham Barnes 2010

24

Nγ = 3.2

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution L´ = 3.8 – 2 × 0.146 = 3.51 m

B´ = 3.30 m

sc = 1 + 0.2 × 3.30 = 1.188 3.51

Net vertical load = 6275.31 – 3.3 ×3.51 × 21.5 = 6026.3 kN 277.50   i c = 0.5+ 0.5 1−  = 0.932 3.3×3.51×95  

Net Rd = 95 × 5.14 × 1.188 ×0.932 = 540.65 kPa iA´ 1.0

iiiii

Rd = 540.65 ×3.3 ×3.51 = 6262.35 kN 6026.3 < 6262.35 ∴ULS verified Combination 2 Design vertical load = 2925 × 1.0 + 1225 × 1.3 = 4517.50 kN Design horizontal load = 150 × 1.0 + 50 × 1.3 = 215.0 kN Total vertical load = 489.06 + 4517.50 = 5006.56 kN e = 215.0 × 3.3 = 0.142 m 5006.56 L´ = 3.8 – 2 × 0.142 = 3.52 m

B´ = 3.30 m

sc = 1 + 0.2 × 3.30 = 1.188 3.52

Net vertical load = 5006.56 – 3.3 ×3.52 × 21.5 = 4756.8 kN 215.0   i c = 0.5 + 0.5 1−  = 0.949 3.3 × 3.52 × 95   Net Rd = 95 × 5.14 × 1.188 × 0.949 = 393.2 kPa iA´ 1.4

iiiii

Rd = 393.2 × 3.3 × 3.52 = 4567.4 kN 4756.8 > 4567.4 ∴ULS not verified

9.10 Combination 2 Design vertical load = 2925 × 1.0 + 1225 × 1.3 = 4517.50 kN Design horizontal load = 150 × 1.0 + 0 × 1.3 = 150.0 kN Total vertical load = 489.06 + 4517.50 = 5006.56 kN e = 150.0 × 3.3 = 0.099 m 5006.56 L´ = 3.8 – 2 × 0.099 = 3.60 m

B´ = 3.30 m

sc = 1 + 0.2 × 3.30 = 1.183 3.60

Net vertical load = 5006.56 – 3.3 × 3.6 × 21.5 = 4751.1 kN

Graham Barnes 2010

25

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution

150.0   i c = 0.5 + 0.5 1−  = 0.966 3.3 × 3.6 × 95   Net Rd = 95 × 5.14 × 1.183 ×0.966 = 398.6 kPa iA´ 1.4

iiiii

Rd = 398.6 × 3.3 × 3.6 = 4735.4 kN 4751.1 > 4735.4 ∴ULS still not verified; change size of foundation (width or length)

Combination 1 Design horizontal load = 277.50 kN Design resistance = A´ cud = 3.3 × 3.51 × 95 = 1100.4 kN 1.0 277.50 < 1100.4 ∴ULS verified

9.11 Combination 2 Design horizontal load = 215.0 kN Design resistance = A´ cud = 3.3 × 3.52 × 95 = 788.2 kN 1.4 215.0 < 788.2 ∴ULS verified

Chapter 10

Shallow foundations - Settlements

For centre L/B = 10/5 = 2

H/B = 20/2.5 = 8

From Figure 10.1

I = 0.465

B = 2.5 m

ρi = s0 = 120 × 2.5 × 0.465 × 4 × 1000 = 27.9 mm, say 28 mm 20 × 1000

10.1 For corner L/B = 2

H/B = 20/5 = 4

From Figure 10.1

B=5m

I = 0.375

ρi = s0 = 120 × 5 × 0.375 × 1000 = 11.25 mm, say 11 mm 20 × 1000

10.2 For centre L/B = 2

H/B = 5/2.5 = 2

B = 2.5 m

H/B = 20/2.5 = 8

Graham Barnes 2010

I = 0.230 I = 0.465

26

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution ρi = s0 = 120 × 2.5 × (0.465 – 0.230) × 4 × 1000 = 14.1 mm, say 14 mm 20 × 1000

For corner L/B = 2

H/B = 5/5 = 1

B=5m

I = 0.085

H/B = 20/5 = 4

I = 0.375

ρi = s0 = 120 × 5 × (0.375 – 0.085) × 1000 = 8.7 mm, say 9 mm 20 × 1000 ∴differential settlement of pipe = 14 – 9 = 5 mm From Table 10.2 for L/B = 2 µr ≈0.75

ρrigid = 0.75 × 27.9 = 20.9 mm, say 21 mm

10.3

For average settlement for L/B = 2 H/B = 20/5 = 4 From Figure 10.4 µ1 = 0.76 µ0 = 1.0

ρaverage = 1.0 × 0.76 × 120 × 5 × 1000 = 22.8 mm, say 23 mm 20 × 1000

For centre

L/B = 2

k = 30 – 10 × 2.5 = 0.25 10 20

H/B = 20/2.5 = 8 From Figure 10.5

B = 2.5 m I = 0.35

ρi = s0 = 120 × 5 × 0.35 × 4 × 1000 = 42.0 mm 10 × 1000

10.4 For corner

L/B = 2

H/B = 20/5 = 4

k = 30 – 10 × 5 = 0.50 From Figure 10.5 10 20

B=5m

I = 0.20

ρi = s0 = 120 × 5 × 0.20 × 1000 = 12.0 mm 10 × 1000

10.5 a)

elastic

L/B = 2

H/B = 16/2 = 8

B=2m

ρi = s0 = 40 × 2 × 0.465 × 4 × 1000 = 18.6 mm, say 19 mm 8 × 1000

Graham Barnes 2010

27

From Figure 10.1, I = 0.465

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution b) with yield From Figure 10.7, F = 3

fy = 1.5

ρy = 1.5 × 18.6 = 27.9 mm, say 28 mm

Assume 2 m thick layers Layer 1 2 3 4

10.6

Depth m 1 3 5 7

p0´

∆σ

P

w0

e0

∆e

δH

18.5 55.5 82.7 100.1

70 40 25 15

0.680 0.236 0.115 0.061

40 35 31 28

1.080 0.945 0.837 0.756

0.102 0.035 0.017 0.009 Σ

98.0 36.4 18.7 10.4 163.5

Say 164 mm

Layer 1 2 3 4

10.7

Depth m 1 3 5 7

∆σ

mv

δH

70 40 25 15

0.76 0.68 0.60 0.52

106.4 54.4 30.0 15.6 206.4

Σ 206.4 × 0.8 = 165.1 mm, say 165 mm

Graham Barnes 2010

28

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution Ef = 4 + 4 × (28 – 4) = 8 MPa 24 For centre L/B = 2

H/B = 20/4 × 1.2 = 6

k = 28 – 8 × 4 = 0.50 8 20

B = 2.5 m

From Figure 10.13

I = 0.33

D/B = 4/8 = 0.50 From Figure 10.13 FD = 0.93 H/B = 20/8 = 2.5 From Figure 10.14 FB = 0.96

10.8

ρT = s1 = 70 × 4 × 0.33 × 0.93 × 0.96 × 4 × 1000 = 41.2 mm, say 41 mm 8 × 1000 For corner L/B = 2

H/B = 20/8 × 1.2 = 3

k = 28 – 8 × 8 = 1.00 8 20

B = 2.5 m

From Figure 10.13

I = 0.225

D/B = 4/8 = 0.50 From Figure 10.13 FD = 0.93 H/B = 20/8 = 2.5 From Figure 10.14 FB = 0.96

ρT = s1 = 70 × 8 × 0.225 × 0.93 × 0.96 × 1000 = 14.1 mm, say 14 mm 8 × 1000

First 5 years

ρs = s2 = 8 × 1000 × 1.5 × log10 (5/1) = 83.9 mm, say 84 mm 100

After 30 years

ρs = s2 = 8 × 1000 × 1.5 × log10 (30/1) = 177.3 mm, say 177 mm 100

10.9

Graham Barnes 2010

29

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution Schertmann

Immediate

p0´ = 2 × 19.5 = 39 kPa

∆p = p – p0´ = 90 – 39 = 51 kPa

σv´ = 7 × 19.5 – 3 × 9.8 = 107.1 kPa

10.10

C1 = 1 – 0.5 × 39 = 0.918 51

C2 = 1.0 (short-term)

Izmax = 0.5 + 0.1 (51/107.1)0.5 = 0.569

qc = constant

ΣIz ∆z = 0.569 × 1.5 × B × ½ + (0.569 – 0.1) × 0.5 × B × ½ + 0.1 × 0.5 × B = 0.594B = 0.594 × 10 = 5.94 m ρi = 0.618 × 1.0 × 51 × 5.94 × 1000 = 18.7 mm, say 19 mm 2.5 4 × 1000 long-term C2 = 1 + 0.2 log10 (10 × 30) = 1.495

ρT = 1.495 × 18.7 = 28.0 mm

Burland and Burbridge ZI = 5.8 m

10.11

Hs > ZI ∴fs = 1.0

q´ = 90 kPa

Ic = 1.71 = 0.068 101.4 Immediate

ρi = 1.0 × 1.0 × (90 – 2/3 × 39) × 100.7 × 0.068 = 21.8 mm, say 22 mm

Long-term

ft = 1 + 0.3 + 0.2log10 (30/3) = 1.5

ρT = 1.5 × 21.8 = 32.7 mm, say 33 mm

Chapter 11

Pile Foundations Rb;k = π × 0.752 × 9 × 1.0 × 120 = 477.1 kN 4

11.1

σv0´ = 2 × 19.5 = 39 kPa

Rs;1,k = π × 0.75 × 7.0 × 1.0 × 50

= 824.7 kN

Rs;2,k = π × 0.75 × 4.0 × 0.5 × 120 = 565.5 kN ΣRs;k = 1390.2 kN

Graham Barnes 2010

30

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution Combination 1 Fd = 1250 × 1.35 + 215 × 1.50 = 2010.0 kN Alternative 1 From the UK National Annex for n = 4 ξ3 = 1.38 ξ4 = 1.29 Rc;dmean = 477.1× 134 × 1.1 + 824.7 × 63 × 1.1 + π × 0.75 × L × 0.5 × 134 × 1.1 120 1.38 50 1.38 1.38 = 424.7 + 828.3 + 125.8L Rc;dmin = 477.1× 103 × 1.1 + 824.7 × 42 × 1.1 + π × 0.75 × L × 0.5 × 103 × 1.1 120 1.29 50 1.29 1.29 = 349.2 + 590.7 + 103.5L use the minimum values 103.5L = 2010.0 – 349.2 – 590.7 L = 10.34 m, say 10.35 m Alternative 2

11.2

Rc;d =

Model factor = 1.4

477.1 + 824.7 + π × 0.75 × L × 0.5 × 120 1.4 × 1.0 1.4 × 1.0 1.4 × 1.0

101.0L = 2010.0 – 340.8 – 589.1

≤ 2010.1

∴L = 10.7 m

Combination 2 Fd = 1250 × 1.0 + 215 × 1.30 = 1529.5 kN Alternative 1 use the minimum values Rc;dmin = 349.2 + 590.7 + 103.5L 2.0 1.6 1.6 64.7L = 1529.5 – 174.6 – 369.2 Alternative 2 Rc;d =

≤ 1529.5 ∴L > 15.23 m, say 15.25 m

Model factor = 1.4

477.1 + 824.7 + π × 0.75 × L × 0.5 × 120 1.4 × 2.0 1.4 × 1.6 1.4 × 1.6

170.4 + 368.2 + 63.1L < 1529.5

≤ 2010.1

∴L ≥15.7 m

(Rc;m)mean = (2160 + 2040 + 2280 + 2310 + 2450)/5 = 2248 kN (Rc;m)min = 2040 kN

11.3

Rc;k = lowest of 2248 × 1.1 and 2040 × 1.1 = 1831.7 and 2077.8 kN 1.35 1.08 Rc;k = 1831.7 kN

Graham Barnes 2010

31

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution Operative shaft length L = 18 – 2 = 16 m Characteristic cuk at base = 75 + 4 × 18 = 147 kPa Characteristic base resistance = 9 × 147 × 0.352 = 162.1 kN

11.4

Characteristic shaft resistance = 1831.7 – 162.1 = 1669.6 kN Average cuk at mid-length of shaft = 75 + 4 × 10 = 115 kPa

α × 115 × 16 × 4 × 0.35 = 1669.6

α = 0.648, say 0.65

Minimum strength cuz = 68 + 2.5z Mean strength cuz = 88 + 5z

At 1 m bgl At 8 m bgl At 15 m bgl

Characteristic cuk kPa 75 + 4 = 79 75 + 8 × 4 = 107 75 + 15 × 4 = 135

Minimum cuk kPa 68 + 2.5 = 70.5 68 + 8 × 2.5 = 88.0 68 + 15 × 2.5 = 105.5

Mean cuk kPa 88 + 5 = 93.0 88 + 8 × 5 = 128.0 88 + 15 × 5 = 163.0

Alternative 1

11.5

Rc;dmean = 163.0 × 9 × π × 0.452 + 128.0 × 0.65 × 14 × π × 0.45 = 117.0 + 952.4 = 1069.4 kN 1.33 × 1.5 × 4 1.33 × 1.3 Rc;dmin = 105.5 × 9 × π × 0.452 + 88.0 × 0.65 × 14 × π × 0.45 = 83.9 + 725.7 = 809.6 kN 1.20 × 1.5 × 4 1.20 × 1.3 Alternative 2

model factor = 1.2

Rc;d = 135 × 9 × π × 0.452 + 107 × 0.65 × 14 × π × 0.45 = 107.4 + 882.4 = 989.8 kN 1.20 × 1.5 × 4 1.2 × 1.3 Rc;d = 809.6 kN

Graham Barnes 2010

32

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution σv´ at 1.5 m = 1.5 × 19.5 = 29.25 kPa For zc/d φ´after = 3/4 φ´ + 10 = 3/4 × 34 + 10 = 35.5o

zc/d = 11.3

zc = 11.3 × 0.45 = 5.1 m σv´ at 5.1 m = 5.1 × 19.5 – 3.5 × 9.8 = 64.2 kPa kstanδ = 0.60

For Nq φ´after = 34 + 40 = 37o 2

Rb;k = 110 × 64.2 × π 4

11.6

Nq = 110

× 0.452 = 1123.2 kN

0 to 1.5 m bgl Rs,1,k = 29.25

× 0.5 × 1.5 × π × 0.45 × 0.60

= 18.6 kN

1.5 to 5.1 m bgl Rs,2,k = (64.2 + 29.25) × 0.5 × 3.6 × π × 0.45 × 0.60 = 142.7 kN 5.1 to 15 m bgl Rs,3,k = 64.2

× 9.9 × π × 0.45 × 0.60

= 539.1 kN Rs,k = 700.4 kN

Combination 1 Rc;d = 1123.2 + 700.4 = 1823.6 kN, say 1824 kN 1.0 1.0 Combination 2 Rc;d =

1123.2 + 700.4 = 471.9 + 333.5 = 805.4 kN, say 805 kN 1.4 × 1.7 1.4 × 1.5

11.7 Combination 1 Fd = 450 × 1.35 + 120 × 1.5 = 787.5 κΝ for the downdrag load Dd = 233.9 × 1.0 ∴for the ultimate limit state Fd + Dd ≤Rc;d Rc;d = Rc;k = 1044.8 = 1044.8 kN γt 1.0 787.5 + 233.9 = 1021.4 kN

1021.4 < 1044.8 ∴ULS verified, but see below

Combination 2 Fd = 450 × 1.0 + 120 × 1.3 = 606.0 κΝ for the downdrag load Dd = 233.9 × 1.25 = 292.4 kN ∴for the ultimate limit state Fd + Dd ≤Rc;d Rc;d = Rc;k = 1044.8 = 522.4 kN γt 2.0

Graham Barnes 2010

33

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution 606.0 + 292.4 = 898.4 kN

Chapter 12

898.4 > 522.4 ∴ULS not verified

Lateral Earth Pressures and Retaining Structures

Combination 1

Combination 2

γG = 1.35 γφ = 1.0 γR = 1.0

γG = 1.0 γφ = 1.25 γR = 1.0

∴ φd = φk = 36o

∴ tanφd = tan36o = 0.581 φd = 30.17o 1.25

Ka = 1– sin36 = 0.260 1+ sin36

Ka = 1– sin30.17 = 0.331 1+ sin30.17

12.1 Assume water table is the characteristic Assume water table is the characteristic

Total thrust = 9.88 + 29.64 + 11.94 + 44.1 = 95.56 kN, say 95.6 kN moment = 9.88 × 3.67 + 29.64 × 1.5 + 11.94 × 1.0 + 44.1 × 1.0 = 136.76 kNm point of application = 136.76 = 1.43 m 95.56

Total thrust = 12.58 + 37.74 + 15.20 + 44.1 = 109.62 kN, say 109.6 kN moment = 12.58 × 3.67 + 37.74 × 1.5 + 15.20 44.1 × 1.0 = 162.08 kNm

point of application = 162.08 = 1.48 m 109.62

Combination 1

Combination 1

Kp = 1 + sin36 = 3.852 1 – sin36

Kp = 1 + sin30.17 = 3.021 1 – sin30.17

12.2

Graham Barnes 2010

34

× 1.0 +

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution Total thrust

Total thrust

= 146.38 + 439.14 + 176.81 + 44.1 = 806.43 kN, say 806.4 kN

= 114.80 + 344.40 + 138.66 + 44.1 = 641.96kN, say 642 kN

moment

moment

= 146.38 × 3.67 + 439.14 × 1.5 + 176.81 × 1.0 + 44.1 × 1.0 = 1416.83 kNm

= 114.80 × 3.67 + 344.40 × 1.5 + 138.66 × 1.0 + 44.1 × 1.0 = 1120.68 kNm

point of application = 1416.83 = 1.76 m 806.43

point of application = 1120.68 = 1.75 m 641.96

12.2

Combination 1 as before because c´ and φ´ are unfactored Combination 2 Sand

tanφd = tan36o = 0.581 φd = 30.17o 1.25

Ka = 0.331

Clay

tanφd = tan26o = 0.390 φd = 21.32o 1.25

Ka = 0.467

cd = 12 = 9.6 kPa 1.25

12.3

point of application = 107.66 = 1.07 m 100.66

12.4 Combination 2

φd = 30.17o

δ/φ = 0.66

Ka = 0.285

Total thrust = 100.50 kN

Moment = 10.83 × 3.67 + 32.49 × 1.50 + 13.08 × 1.0 + 44.1 × 1.0 = 145.66 kNm

Graham Barnes 2010

35

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution Point of application = 145.66 = 1.45 m 100.5 Shear force = (10.83 + 32.49 + 13.08) × tan(30.17 × 2/3) = 20.65 kN

Combination 1

γG = 1.35 γQ = 1.5 unfavourable

design surcharge = 10 × 15 = 15 kPa behind line

γG = 1.0 γQ = 0 favourable design surcharge = 10 × 0 = 0 kPa behind wall γφ = 1.0 Weight Base Stem Backfill

12.5

1.0 × 3.8 × 25 0.8 × 5.0 × 25 1.5 × 5.0 × 21

95.0 100.0 157.5

Distance from O 1.90 1.90 3.05

Moment 180.5 190.0 480.4

Overturning

Total horizontal thrust = 17.1 + 71.82 = 88.9 kN Shear force = 88.9 × tan37 = 67.0 kN Shear force moment = 67.0 × 3.8 = 254.6 kNm

Overturning moment = 17.1 × 3.0 + 71.8 × 2.0 = 194.9 kNm Resisting moments = 180.5 + 190.0 + 480.4 + 254.6 = 1105.5 kNm 194.9 < 1105.5 ∴ ULS verified Sliding Total V = 95.0 + 100.0 + 157.5 + 67.0 = 419.5 kN Resisting force = 419.5 × tan25 = 195.6 kN 88.9 kN < 195.6 kN ∴ ULS verified Bearing pressure (including surcharge) Moment from surcharge = 10 × 1.5 × 1.5 × 3.05 = 68.6 kNm Total V = 419.5 + 10 × 1.5 × 1.5 = 442.0 kN Resultant moment = 180.5 + 190.0 + 480.4 + 254.6 + 68.6 – 194.9 = 979.2 kNm Distance of V from O = 979.2 = 2.215 m

Graham Barnes 2010

36

Eccentricity e = 2.215 – 1.9 = 0.315 m

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution 442.0 qmax = 442.0 × (1 + 6 × 0.315) = 174.2 kPa 1 × 3.8 3.8 B´ = 3.8 – 2 × 0.315 = 3.17 m

qmin = 442.0 × (1 – 6 × 0.315) = 58.5 kPa 1 × 3.8 3.8

qmean = 442.0 = 139.4 kPa 3.17

Combination 2

γG = 1.0

12.5

γQ = 1.3

γφ = 1.25

unfavourable

φk = 37o

tan37o = 0.603 1.25

φd = 31.08o

δk = 25o

tan25o = 0.373 1.25

φd = 20.46o

Ka = 0.26

δ = 31.08o

Overturning

Total horizontal thrust = 20.28 + 98.28 = 118.56 kN Shear force = 118.56 × tan31.08 = 71.46 kN Shear force moment = 71.46 × 3.8 = 271.56 kNm Overturning moment = 20.28 × 3.0 + 98.28 × 2.0 = 257.4 kNm Resisting moments = 180.5 + 190.0 + 480.4 + 271.6 = 1122.5 kNm 257.4 kNm < 1122.5 kNm ∴ ULS verified Sliding Total V = 95.0 + 100.0 + 157.5 + 71.5 = 424.0 kN Resisting force = 424.0 × tan20.46 = 158.2 kN 118.6 kN < 158.2 kN ∴ ULS verified Bearing pressure (including surcharge) Moment from surcharge = 10 × 1.3 × 1.5 × 3.05 = 59.5 kNm Total V = 424.0 + 10 × 1.3 × 1.5 = 443.5 kN Resultant moment = 180.5 + 190.0 + 480.4 + 271.6 + 59.5 – 257.4 = 924.6 kNm Distance of V from O = 924.6 = 2.085 m 443.5 Eccentricity e = 2.085 – 1.9 = 0.185 m

Graham Barnes 2010

37

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution qmax = 443.5 × (1 + 6 × 0.185) = 150.8 kPa 1 × 3.8 3.8

qmin = 443.5 × (1 – 6 × 0.185) = 82.6 kPa 1 × 3.8 3.8

B´ = 3.8 – 2 × 0.185 = 3.43 m qmean = 443.5 = 129.3 kPa 3.43

Fφ = 1.2 (CIRIA 104, Table 5) worst credible parameters, permanent works Design tanφ = tan27o = 0.425 1.2 Table 12.1

δ /φ active = 2/3

Figure 12.11

Ka = 0.38

design φ = 23.0o

δ /φ passive = 0.50 Kp = 3.0

12.6

Zone

Force kN

Lever arm from anchor m

Moment kNm

Active 1 0 – 2.0 m 2 2.0m – toe 3 2.0m – toe

4 water

Passive 5 soil 6 water

2 × 20 × 0.38 × 2.0× ½ = 15.20

0.33

5.06

40.0 × 0.38 × (4.0 + d) = 60.8 + 15.2 d [20 × (6.0 + d) – (4.0 + d) × 9.8] × 0.38 – 4.0 × 0.38 = 15.50 + 3.88d force (15.50 + 3.88d) × (4.0 + d) ½ = 31.0 + 1.94d2 + 15.51d (4.0 + d) × 9.8 × (4.0 + d) ½ = 78.4 + 39.2d + 4.9d2

½ (4.0 + d) + 1.0 = 3.0 + 0.5d 2/3 (4.0 + d) + 1.0 =3.67 + 0.67d

182.4 + 7.6d2 + 76.0d

2/3 (4.0 + d) + 1.0 = 3.67 + 0.67d

(78.4 + 39.2d + 4.9d2) × (3.67 + 0.67d) = 287.73 + 196.39d + 44.24d2 + 3.28d3

10.2d × 3.0 × d × ½ = 15.3d2

2/3d + 5.0 = 0.67d + 5.0

10.25d3 + 76.5d2

9.8 × d × d × ½ = 4.9d2

0.67d + 5.0

3.28d3 + 24.5d2

(31.0 + 1.94d2 + 15.51d) × (3.67 + 0.67d) = 113.77 + 77.69d + 17.51d2 + 1.30d3

ΣMoments Force

Graham Barnes 2010

0

d

38

d2

d3

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution 1 2 3 4 5 6 totals

5.06 182.40 113.77 287.73 588.96

76.0 77.69 196.39 350.08d

7.60 17.51 44.24 – 76.5 – 24.5 – 31.65d2

65.81 + 39.12d – 3.54d2 – d3 = 0 by trial d = 5.58 m, say 5.6 m Check: Force kN 15.2 145.62 177.86 449.70 476.39 152.57

No. 1 2 3 4 5 6

Lever arm m 0.33 5.79 7.39 7.39 8.72 8.72

12.6

Moment kNm 5.06 843.14 1314.39 3323.28 – 4154.12 – 1330.41 Σ = 1.34 ∴OK

Total thrust Pa = 1 + 2 + 3 + 4 = 15.2 + 145.6 + 177.9 + 449.7 = 788.4 kN Total thrust Pp = 5 + 6 = 476.4 + 152.6 = 629.0 kN Anchor force T = (788.4 – 629.0)

12.7 Design tensile strength = 300 1.1 fn = 1.1

× 2.5 = 398.5 kN, say 399 kN

× 60 × 3 = 49.09 kN 1000

TDj = 49.09 = 44.63 kN fn 1.1

fms = 1.0 ∴φd = φp = 37o

length of reinforecement in active zone = (7.0

– 0.7) × tan(45 – 37/2) = 3.14 m

Lej = 8.0 – 3.14 = 4.86 m tanδ = 0.6 × tan37o = 0.452 equation 12.68

TAj = 4.86 × 2 1.3

δ = 24.3o, say 25o

× 0.06 × [tan25 × 1.0 × 19.0 × 0.7] = 2.78kN 1.0

for inextensible reinforcement – coherent gravity method equation 12.74 K0 = 1 – sinφ = 1

– sin37o = 0.398

Ka = 1 – sin37o = 0.249 1 + sin37o at 0.7 m

Graham Barnes 2010

K = 0.398 – 0.7 × (0.398 – 0.249) = 0.381 6.0

39

1.30 3.28 – 10.25 – 3.28 – 8.95d3 =0

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution equation 12.76 Tj = 0.381 × 0.7 × 19 × 0.6 = 3.04 kN 3.04 kN > 2.78 kN ∴state is breached by pull-out determine required length of reinforcement TAj = Tj Lej

× 2 × 0.06 × (tan25 × 1.0 × 19 × 0.7) = 3.04 1.3

Lej = 5.31 m

1.0

length of reinforecement = 5.31 + 3.14 = 8.45 m, say 8.5 m

Chapter 13

Slope Stability

With hw = 0

F = c´ + tanφ´cos2β γsub z γ z sinβ cosβ

At 2.0 m

F = 5 + tan 26o × cos214o × 9.2 × 2.0 = 1.51 19 × 2.0 × sin14o cos14o

At 4.0 m

F = 5 + tan 26o × cos214o × 9.2 × 4.0 = 1.23 19 × 4.0 × sin14o cos14o

At 6.0 m

F = 5 + tan 26o × cos214o × 9.2 × 6.0 = 1.13 19 × 6.0 × sin14o cos14o

13.1

At 2.0 m γod = 1.51 = 1.21 1.25

13.2

At 4.0 m γod = 1.23 = 0.98 1.25 At 6.0 m γod = 1.13 = 0.90 1.25

Graham Barnes 2010

40

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution At 2.0 m with seepage through the suction zone F = c´ + tanφ´cos2β (γz +γwhs) γ z sinβ cosβ F = 5 + tan 26o cos214o × (19 × 2.0 + 9.8 × 2.0) = 3.53 19 × 2.0 × sin14o cos14o At 4.0 m ‘dry’ slope with z = hw

13.3

F = c´ + tanφ´cos2β γz γ z sinβ cosβ F = 5 + tan 26o × cos214o × 19 × 4.0 = 2.24 19 × 4.0 × sin14o cos14o At 6.0 m wet slope with hw = 4.0 m F = 5 + tan 26o × cos214o × (19 × 6.0 – 9.8 × 6.0 + 9.8 × 4.0) = 1.81 19 × 6.0 × sin14o cos14o 1.0 = 0 + tan φr´ cos210o × (21 × 2.5 – 9.8 × 2.5 + 9.8 × 0.5) = 3.554 tan φr 21 × 2.5 × sin10o cos10o

13.4 tan φr´ = 1.0 = 0.281 3.554

Graham Barnes 2010

φr´ = 15.7o

41

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution

13.5 Slice 1 2 3 4 5 6 7 8 9 10 11

b m 0.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5

F = cu × Σ bsecα ΣW sinα

h m 0.14 1.05 2.55 3.70 4.60 5.50 5.90 5.80 4.85 3.50 1.50

W kN 1.3 29.1 70.8 102.7 127.6 152.6 163.7 161.0 134.6 97.1 41.6

cumin = 0.157 × 1.0 × 18.5 × 6.0 = 17.4 kPa, say 17 kPa

Graham Barnes 2010

bsecα

Wsinα

-25 -20 -12 -4 +4 11.5 19 28 37 47 60.5 Σ

0.55 1.60 1.53 1.50 1.50 1.53 1.59 1.70 1.88 2.22 3.05 18.65

-0.55 -9.95 -14.7 -7.2 8.9 30.4 53.3 75.6 81.0 71.0 36.2 324.0

cu = F × ΣW sinα = 1.0 × 324.0 = 17.4 kPa, say 17 kPa Σ bsecα 18.65

From Taylor’s curves Ns = 0.157

13.6

αο

42

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution

Slice 1 2 3 4 5 6 7 8 9 10

13.7

b m 2.3 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0

h m 0.85 2.60 4.20 5.50 6.50 7.15 7.45 7.4 6.8 4.9

W kN 39.1 104.0 168.0 220.0 260.0 286.0 296.0 296.0 272.0 196.0

αο

bsec α

-33 -23 -13 -5 +5 13 23 33 44 59

2.74 2.17 2.05 2.01 2.01 2.05 2.17 2.38 2.78 3.88 24.24

Σ Pw = ½ γw zc 2 = ½ × 9.8 × 3.02 = 44.1 kN at yc = 2.0 + 2.0 = 4.0 m Radius R = 13.0 m F=

13.0 × 35 × 24.24 = 11029 = 1.378, say 1.38 13.0 × 601.9 + 44.1 × 4.0 8001

β = 21.8o

13.8

13.9

D = 14/7 = 2.0

Ns = 0.165

γod =

cuk = 55 = 1.79 Ns γcu γk H 0.165 × 1.4 × 19 × 7.0

Nsd =

38 = 0.162 1.4 × 21 × 8

From chart slope angle = 30o

13.10

A = c´b + (W – ub) tanφ B=

secα 1 + tanα tanφ F

F = A × Bccc ΣWsinα

Graham Barnes 2010

43

Wsin α -21.3 -40.6 -37.8 -19.2 22.7 64.3 115.7 161.2 188.9 168.0 601.9

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution

F = 1.30 Slice

b

h

W

α

1 2 3 4 5 6 7 8 9 10 11

3.0 3.0 3.0 3.0 3.0 3.0 3.0 3.0 3.0 3.0 1.45

0.4 1.55 2.85 3.90 4.55 49.5 5.05 4.70 3.87 2.52 0.85

26.6 103.2 189.8 259.7 303.0 329.7 336.3 311.7 257.7 167.8 56.6

-14.8 -8.8 -2.9 +2.9 8.8 14.8 20.9 27.3 34.1 41.5 47.8

Σ

Wsin α -6.8 -15.8 -9.6 +13.1 46.4 84.2 120.0 143.0 144.5 111.2 41.9 672.1

Initial F = 1.30

F = 889.7 = 1.324 672.1

Initial F = 1.40

F = 898.7 = 1.337 say F = 1.33 672.1

u

A

B

A×B

B

A×B

3.9 13.8 23.7 30.6 34.4 35.5 33.3 27.2 15.9 0 0

16.3 39.1 66.8 87.9 106.4 117.9 124.3 122.0 111.4 90.8 32.0

1.15 1.07 1.02 0.98 0.96 0.94 0.94 0.94 0.96 1.00 1.05

18.8 41.8 68.1 86.1 102.1 110.8 116.9 114.7 106.9 90.8 33.6 889.7

1.14 1.07 1.02 0.98 0.96 0.95 0.95 0.95 0.98 1.02 1.07

18.6 41.8 68.1 86.1 102.1 112.0 118.1 115.9 109.2 92.6 34.2 898.7

Σ

c´ = 3 = 0.015 γH 22.2 × 9

hw = 2.7 = 0.30 H 9.0

c´ = 0.005 γH

a = 0.07

b = 2.31

F = 0.07 + 2.31 × tan26o = 1.20

c´ = 0.025 γH

a = 0.25

b = 2.35

F = 0.25 + 2.35 × tan26o = 1.40

13.11

For c´ = 0.015 γH

Graham Barnes 2010

F = 1.20 + 0.01 (1.40 – 1.20) = 1.30 0.02

44

F = 1.40

Σ

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution γc = γφ = 1.25

γG = 1.0 γγ = 1.0

let C = 3.0 × b + tan26o (W – ub) and D = secα 1.25 1.25 1 + tanα tanφ γφ Slice

b

h

W

α

1 2 3 4 5 6 7 8 9 10 11

3.0 3.0 3.0 3.0 3.0 3.0 3.0 3.0 3.0 3.0 1.45

0.4 1.55 2.85 3.90 4.55 49.5 5.05 4.70 3.87 2.52 0.85

26.6 103.2 189.8 259.7 303.0 329.7 336.3 311.7 257.7 167.8 56.6

-14.8 -8.8 -2.9 +2.9 8.8 14.8 20.9 27.3 34.1 41.5 47.8

13.12

Σ

Wsin α -6.8 -15.8 -9.6 +13.1 46.4 84.2 120.0 143.0 144.5 111.2 41.9 672.1

u

C

D

3.9 13.8 23.7 30.6 34.4 35.5 33.3 27.2 15.9 0 0

13.0 31.3 53.5 72.7 85.2 94.3 99.4 97.0 89.1 72.7 14.2

1.153 1.077 1.021 0.982 0.954 0.937 0.932 0.937 0.955 0.993 1.041

Σ

C× D 15.0 33.7 54.6 74.0 81.3 88.4 92.6 90.9 85.1 72.2 14.8 687.8

γod = 687.8 = 1.023 672.1

F = 1.20 Slice

b

h

W

α

1 2 3 4 5 6

8.0 8.0 8.0 8.0 8.0 6.0

2.03 5.04 6.70 7.00 5.75 2.35

308.6 766.1 1018.4 1064.0 874.0 267.9

-12.0 -2.4 7.2 16.9 27.2 37.1

Wsin α -64.2 -32.1 127.6 309.3 399.5 161.6

Σ

901.7

13.13 Initial F = 1.20

F = 1205.1 = 1.34 901.7

Initial F = 1.30

F = 1209.6 = 1.34 901.7

For γM = γc = γφ = = 1.25 γ1 = 1.34 = 1.072 1.25

Chapter 14

Earthworks and Soil compaction

Graham Barnes 2010

45

F = 1.30

u

A

B

A×B

B

A×B

19.9 49.4 65.7 63.7 41.7 3.4

103.4 197.4 249.2 275.3 269.4 135.1

1.105 1.016 0.965 0.944 0.951 0.989

114.3 200.6 240.5 259.9 256.2 133.6

1.099 1.002 0.968 0.951 0.963 1.005

Σ

1205.1

Σ

113.6 197.8 241.2 261.8 259.4 135.8 1209. 6

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution Number of blows 1 2 3 4 6 8 12 16 24 32 48 64 96 128

Penetration mm 43.0 56.0 63.0 68.0 75.0 80.5 87.0 91.0 95.5 98.5 100.5 101.5 102.0 102.0

Change n to 4n 25.0 24.5 24.0 23.0 20.5 18.0 13.5 10.5 6.5 3.5 1.5 0.5

14.1

At 5 mm

Number of blows = 27.5

MCV = 10

× log10 27.5 = 14.4

a = 31.6, say 32 b = 5/6.4 = 0.78

14.2

Graham Barnes 2010

Table 14.4 sensitivity is moderate

46

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Solution a) shrinkage factor SF = γb = 1.86 = 0.94 or 94% γc 1.98 required volume of cutting = 15450 = 16447 m3 0.94 b)

14.3

load factor LF = γm = 1.67 = 0.898 γb 1.86 machine load in bank m3 = 15 × 0.898 = 13.47 m3 total number of loads = 16447 = 1221 13.47

c)

Hourly production = 13.47 × 4 × 60 × 0.75 = 303.1 bankm3/hour 8 Total time required = 16447 = 54.3 hours 303.1

14.4

w%

ρb Mg/m3 ρd Mg/m3

14.3

15.8

17.7

19.2

20.9

22.6

1.967

2.008

2.065

2.088

2.077

2.072

1.721

1.734

1.754

1.752

1.718

1.690

ρdmax = 1.76 Mg/m3

wopt = 18.5 %

ρb = ρd (1 + w) = 1.76 × 1.185 = 2.086 Mg/m3 2.086 = 2.72 × (1 + 0.185) 1+e

e = 0.545

14.5 n=

0.545 = 0.353 1 + 0.545

Sr = w Gs × 100 = 0.185 × 2.72 × 100 = 92.3 % e 0.545

Av = 0.353 × (1 – 0.923) × 100 = 2.7 %

Graham Barnes 2010

47

Soil Mechanics Principles and Practice Third Edition

Graham Barnes

Exercis e

Graham Barnes 2010

Solution

48