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Soil Mechanics Principles and Practice Fourth Edition
Solutions Manual
Graham Barnes
Contents
2 3 4 5 6 7 8 9 10 11 12 13
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 Fourth Edition
Graham Barnes
Exerci se
Solution
Chapter 2
Soil Description and Classification
Volume of gravel = (m4 – m1) – (m3 – m2)
2.1
= (1870.6 – 845.2) – (2346.0 – 1608.7) = 1025.4 – 737.3 = 288.1 cm3 mass of gravel = m2 – m1 = 1608.7 – 845.2 = 763.5 g s = 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
2.3
ms = 537.5 g
d30 = 0.23 mm
d50 = 0.42 mm
V0 = 250ml
moderately uniformly graded
wax G s = 0.90
volume of soil lump = 250 – 544.4 – 537.5 = 242.33 cm3 0.90 bulk density b = ms = 537.5 × 10 6 = 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 Water content = 537.5 – 479.2 × 100 = 12.17 %, say 12.2 % 479.2
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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
Cc = 0.232 = 0.89 0.54 × 0.11
mw = 544.4 g
% passing
Soil Mechanics Principles and Practice Fourth Edition
Graham Barnes
Exerci se
Solution Dry densityd = b = 2.218 = 1.977 Mg/m3, say 1.98 Mg/m3 1 + w 1 + 0.122 Dry weight density = 1.977 × 1000 × 9.81 = 19.39 kN/m3 1000 From Table 2.19
b = Gsw (1 + w ) (1 + e)
2.4
1 + e = 2.72 × 1.0 × 1.122 = 1.376 2.218
void ratio e = 0.376, say 0.38
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) × 100 = 3.2 %
The degree of saturation would be 100 % Sr = wGs = 100 e
2.5
w = e × 100 = 0.376 × 100 = 13.82 %, say 13.8 % Gs 2.72
b = Gsw (1 + w) = 2.72 × 1.0 × 1.1382 = 2.25 Mg/m3 (1 + e) 1.376
Av = n (1 Sr) with n = e and Sr = wGs Av = e – w Gs 1 + e e 1 + e
2.6
e = Av + w Gs = 0.05 + 0.135 × 2.68 = 0.4335 1 Av 1 – 0.05
b = 2.68 × 1.0 × (1 + 0.135) = 2.122 Mg/m3 , say 2.12 Mg/m3 1 + 0.4335
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Soil Mechanics Principles and Practice Fourth Edition
Graham Barnes
Exerci se
Solution
Plastic limit = 26.6 + 27.3 = 26.95, say 27 % 2
2.7
Plasticity index = 55 – 27 = 28 % Classification is CH
2.8
Liquidity index = 35 – 27 = 0.29 28
Volumetric shrinkage
2.9
Consistency index = 55 – 35 = 0.71 28
35.5 21.7 100 38.9% 35.5
Assuming the soil at the liquid limit is fully saturated 57.20 37.39 Mass of solids 37.39g Volume of solids 13.90ml 1 0.52 2.69 At the shrinkage limit mass of water 37.39 0.16 5.98g Volume air 21.7 - 13.90 - 5.98 1.82 ml 1.82 Av 100 8.39% 21.7
Chapter 3
Permeability and Seepage
Q = 200 ml
3.1
3.2
3.3
l = 100 mm
h = 234 mm
t = 225 s
k = Q l = 200 × 10 3 × 100 × 4 = 8.6 × 102 mm/s = 8.6 × 105 m/s 2 Aht × 75 × 234 × 225 a = × 4 2 = 12.57 mm2 A = × 100 2 = 7854.0 mm2 4 4 k = 2.3 × 12.57 × 120 log10 (950/740) = 2.7 × 105 mm/s = 2.7 × 108 m/s 7854 30 × 60
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volume of water 5.98ml
Soil Mechanics Principles and Practice Fourth Edition
Graham Barnes
Exerci se
Solution koverall = Q l = 100 × 10 3 × 100 × 4 = 4.5 × 103 mm/s = 4.5 × 106 m/s 2 A h t × 75 × 672 × 745 kv = Lsand + Lsilt d Lsand = 95 mm Lsand /ksand + Lsilt/ksilt
Lsilt = 5 mm
ksand = 8.6 × 105 m/s
4.5 = 95 + 5 d ksilt = 2.4 × 104 mm/s = 2.4 × 107 m/s 3 2 10 95 × 10 /8.6 + 5/ksilt
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 Hc = 4.0 – 0.5 = 3.5 m Case B f = 2.75
3.5
k = q = 2.5 d f d Hc 1000 × 60 × 2.75 × 0.15 × 3.5 = 2.9 × 105 m/s l = 4.5 m
3.6
H1 = 2.20 m
r1 = 11.0 m
H2 = 2.65 m
r2 = 37.0 m
q = 540 litres/min k = 540 loge (37.0/11.0) = 8.6 × 104 m/s 1000 × 60 × 2 × × 4.5 × 0.45
3.7 Figure 3.43
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Soil Mechanics Principles and Practice Fourth Edition
Graham Barnes
Exerci se
Solution
datum at base of gravel h = 7.5 m
nf = 3 nd = 8
q = 3 × 7.5 × 3 × 1000 × 60 = 5.1 litres/min/m, say 5 litres/min/m 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
3.7
3.8
Shape factor is the same 3/8 k overall
3 6 1.34 10 4 m/s 10 5 10 4
q = 1.34 × 7.5 × 3 × 1000 × 60 = 22.6 litres/min/m, say 23 litres/min/m
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Soil Mechanics Principles and Practice Fourth Edition
Graham Barnes
Exerci se
Solution 104
8
Consider last block for boiling condition l = 1.8 m h = 7.5/8 = 0.938 m Upward seepage force = w × h × l × 1 = 9.8 × 0.938 × 1.8 × 1 = 16.55 kN
3.9 boiling
Downward seepage force = bgravel × l × 1.0 × 1 +subsand × 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