Compressibility of Soil and Rock
Chapter 8
CHAPTER 8 COMPRESSIBIL COMPRESSIBILITY ITY OF SOIL SOIL AND A ND ROCK 8-3. Determine the overconsolidation ratio (OCR) for the five fine-grained soils of Fig. 8.9a. SOLUTION: Eq. Eq. (8.2 (8.2)) : OCR OCR = Soil #8: OCR = Soil #9: OCR =
σ 'p σ 'vo
200 kPa 160 kPa 250 kPa 170 kPa
= 1.25 = 1.47
Soil #10: OCR =
350 kPa kPa
Soil #11: OCR =
350 kPa kPa
Soil #13: OCR =
290 kPa
230 kPa 280 kPa 340 kPa kPa
= 1.52 = 1.25 = 0.85
8-5. What is the OCR of the clay till in Fig. 8.9c? SOLUTION: Eq. (8.2) : OCR =
σ 'p 420 kPa = OCR = = 11.4 σ 'vo 36.7 kPa
8-6. Estimate the the preconsolidation stress for: (a) the undisturbed undisturbed Leda clay in Fig. 8.9d, (b) undisturbed Mexico City clay in Fig. Fig. 8.9e, (c) undisturbed Chicago clay in Fig. 8.9f, 8.9f, and (d) the swelling clays from Texas in Fig. 8.9g. SOLUTION: (a) (a) Leda Leda clay clay (undi undist stur urbe bed) d)::
σ'p = 220 220 to 280 280 kPa kPa
(b) (b) Mexi Mexico co City City clay clay (sam (sampl ple e CP2) CP2):: (c) Chic Chicag ago o clay clay (sam (sampl ple e CP1) CP1):: (d) (d) Tex Texas clay clay (sam (sampl ple e #1) #1):
σ'p = 90 to120 kPa kPa
σ'p = 110 110 to140 kPa kPa
σ'p = 300 300 to 400 400 kPa kPa
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Compressibility of Soil and Rock
Chapter 8
8-7. 8-7. Determine the compression indices for the four soils of Problem 8.6. SOLUTION: Eq. 8.7: Compression index = Cc
Recompression index = Cr =
=
e1 − e2 σ' log 2 σ '1
e1 − e2 σ' log 2 σ '1
(a) Leda clay (undisturbed): Cc
=
1.39 − 0.6 5000 log 1000
(b) Mexico City clay (sample CP2): Cc
(c ) Chicago clay (sample CP1): Cc
(d) Texas clay (sample #1): Cc
=
=
=
= 1.13;
14.5 − 2.15 1000 log 100
1.13 − 0.42 5000 log 2
0.87 − 0.63 3000 log 300
Cr =
= 12.35;
= 0.21;
= 0.24;
2.2 − 2.08 10, 000 log 1.0 Cr =
Cr =
Cr =
= 0.03
14.1 − 13.6 5000 log 2
0.95 − 0.85 5000 log 2
0.9 − 0.86 3000 log 10
= 0.15
= 0.03
= 0.016
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Compressibility of Soil and Rock
Chapter 8
8-8. The pressure versus void ratio data determined from a consolidation test on an undisturbed clay specimen are as follows: (a) Plot the pressure versus void ratio curve on both arithmetic and semilogarithmic graphs. (b) Determine the equations for the virgin compression curve and for the rebound curve for unloading, starting at 1280 kPa. (c) What are the corresponding modified compression and recompression indices for this soil? (d) Estimate the stress to which this clay has been preconsolidated. (After A. Casagrande.)
SOLUTION: (a) Arithmetic graph. 0.9 0.85 0.8 0.75 o 0.7 i t a R0.65 d i o V 0.6
0.55 0.5 0.45 0.4 0
200
400
600
800
1000
1200
1400
Effective Stress (kPa)
continu ed on next page
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Compressibility of Soil and Rock
Chapter 8
Problem 8-8 continued.
Semilogarithmic graph.
0.9 0.85 0.8 0.75 o 0.7 i t a R0.65 d i o V 0.6
0.55 0.5 0.45 0.4 1
10
100
1000
10000
Effective Stress (kPa)
(b) Eq. 8.7: Compression index = Cc
Recompression index = Cr =
e1 − e2 σ' log 2 σ '1
=
=
e1 − e2 σ' log 2 σ '1
=
0.8 − 0.54 10,000 log 1.0
(c) Eq. 8.9: Modified compression index = Ccε
=
Eq. 8.15: Modified recompression index = Cr ε
=
(d) From the semillogarithmic plot: σ'p
1.0 − 0.3 10,000 log 60
= 0.32
= 0.065
Cc 1 + eo Cr 1 + eo
=
0.315 1 + 0.864
= 0.17
=
0.065
= 0.035
1 + 0.864
≈ 330 kPa
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Compressibility of Soil and Rock
Chapter 8
8-9. A building is to be constructed on a stratum of the clay 7 m thick for which consolidation data are given in Problem 8.8. The average existing effective overburden pressure on this clay stratum is 126 kPa. The average applied pressure on the clay after construction of the building is 285 kPa. (a) Estimate the decrease in thickness of the clay stratum caused by full consolidation under the building load. Estimate the decrease in thickness due to the bu ilding load if the clay had never been preconsolidated under a load greater than the existing overburden. (c) Show on the e versus log σ plot of Problem 8.8 the values of Δe used. SOLUTION:
σ 'p 310 kPa = = 2.5 σ 'vo 126 kPa σ 'p σ ' + Δσv use Eq. 8.19b: sc = CrεHo log + CcεHo log vo σ 'vo σ 'p 310 (126 + 285) sc = (0.035)(7 m)log + (0.17)(7 m)log = (7 m)(0.0137) + (7 m)(0.0208) (a) Eq. (8.2) : OCR
=
126 310 sc = 0.09579 + 0.1457 = 0.2415 m = 241mm
Δe1 = 0.0137 → Δe1 = (0.0137)(1 + 0.864) = 0.0255 1 + eo Δe2 = 0.0208 → Δe2 = (0.0208)(1+ 0.864) = 0.0388 1 + eo (b) σ 'p = σ 'vo = 126 kPa σ ' + Δσv (126 + 285) = (0.17)(7 m)log use Eq. 8.13: sc = CcεHo log vo 126 σ 'p sc = Ho × ε = (7 m)(0.0873) = 0.611m = 611mm ΔeNC = 0.0873 → ΔeNC = (0.0873)(1 + 0.0873) = 0.0949 1 + eo ε = 0.0873
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Compressibility of Soil and Rock
Chapter 8
8-10. The compression curve for a certain clay is a straight line on the semilogarithmic plot, and it passes through the point e = 1.15, σ ’v = 65 kPa and e = 0.76, σ ’v = 825 kPa. Determine an equation for this relationship. (After Taylor, 1948.) SOLUTION: Eq. 8.7: Compression index = Cc
=
e1 − e2 σ' log 2 σ '1
=
1.15 − 0.76 825 log 65
= 0.353
e = ( −0.1535)Ln(σ 'v ) + 1.7907
1.4
1.2
1 o i t 0.8 a R d i o 0.6 V
e = -0.1535Ln( σ') + 1.7907 Cc=0.353
0.4
0.2
0 10
100
1000
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Compressibility of Soil and Rock
Chapter 8
8-12. The following consolidation test data were obtained from undisturbed San Francisco Bay 3 Mud. For this clay, LL = 88, PL = 43, ρ s = 2.70 Mg/m and w = 105.7%. Initially, the specimen 3 height was 2.54 cm and its volume was 75.14 cm . Plot the data as percent consolidation versus log pressure. Evaluate the preconsolidation pressure and the modified virgin compression index. SOLUTION: Strain
a
Strain
b
% Strain
Stress
Dial Reading
Void
(kPa)
(mm)
Ratio
0
12.700
2.765
0.000
0.000
0.00
5
12.352
2.712
0.014
0.014
1.39
10
12.294
2.703
0.016
0.016
1.62
20
12.131
2.679
0.023
0.022
2.26
40
11.224
2.541
0.059
0.058
5.88
80
9.053
2.211
0.147
0.144
14.54
160
6.665
1.849
0.243
0.238
24.04
320
4.272
1.486
0.340
0.332
33.58
480
2.548
1.224
0.409
0.400
40.45
160
2.951
1.285
0.393
0.384
38.85
40
3.533
1.374
0.369
0.361
36.52
5
4.350
1.499
0.336
0.329
33.25
c
e −e Ro − Ri Δe = o i; Strainb : ε = 1 + eo 1 + eo 25.4 mm ⎛ Straina + Strainb ⎞ Strainc : % Strain = ⎜ ⎟ × 100 2 ⎝ ⎠
ε=
Straina :
ε2 − ε1 0.45 − 0.053 = = 0.32 σ '2 700 log log 40 σ '1 Eq. 8.9: Compression index = Cc = (1 + eo )Ccε = (1 + 2.765) × (0.32) = 1.20 Modified Compression index = Ccε
=
From the semillogarithmic plot: σ'p
≈ 38 kPa
Effective Stress (kPa) 1
10
100
1000
0.00
5.00
10.00
15.00
n i a r t S20.00 t n e 25.00 c r e P 30.00
35.00
40.00
45.00
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Compressibility of Soil and Rock
Chapter 8
8-13. Plot the data of Problem 8.12, on a void ratio versus log pressure graph. Evaluate the preconsolidation pressure and the virgin compression index. Do these values agree with what you found in Problem 8.12? SOLUTIONS: 3.5
3
2.5
o i t a R 2 d i o V 1.5
1
0.5 1
10
100
1000
Effective Stress (kPa)
σ 'p ≈ 39 kPa;
Eq. 8.7: Compression index = Cc
=
e1 − e2 σ' log 2 σ '1
=
2.1 − 1.486 320 log 100
= 1.21
These values agree with the solutions in Problem 8-12.
8-14. The initial water content of the sample in Problem 8.12 is 105.7%, and the density of the 3 solids ρ s = 2.70 Mg/m . Compute the wet and dry density and degree of saturation of the consolidation test sample if the dry weight of the sample is 52.8 g. If the final water content is 59.6%, compute the degree of saturation and dry density at the end of consolidation. SOLUTION: (a) w i
= 105.7%, Ms = 52.8 g
w
=
Mw
Vs
=
Ms
→ Mw = (1.057)(52.8 g) = 55.81g → Mt = 55.81+ 52.8 = 108.61g
Ms
=
52.8 g
= 19.55 cm3 ;
Vw
=
Mw
=
55.81g
= 55.81cm3
ρs 2.70 g cm ρw 1.0 g cm Vv = e × Vs = (2.765)(19.55 cm3 ) = 54.071cm3 ; Vt = 19.55 cm3 + 54.071cm3 = 73.621cm3 ρt = S=
Mt Vt Vw Vv
=
3
108.61g 3
73.621cm
× 100 =
= 1.47 Mg m3 ; ρd = 3
55.81cm
54.071cm3
(b) w f = 59.6%;
S=
Ms Vt
=
3
52.8 g 3
73.621cm
= 0.717 Mg m3
× 100 = 103.2% (not possible)
ρs w (2.70)(0.596) × 100 = × 100 = 107% (not possible) ρw e (1.0)(1.499)
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Compressibility of Soil and Rock
Chapter 8
8-15. A 7.8 m thick layer of soft San Francisco Bay Mud is to be loaded with a granular fill 3.2 m 3 thick, on the average. The total density of the fill is about 1.8 Mg/m . Assume that the test data in Problem 8.12 is typical of the clay layer, and that the layer is normally consolidated. What consolidation settlement will take place due to the weight of the fill? SOLUTION:
Δσv = (3.2 m )(1.8 Mg m3 )(9.81 m s2 ) = 56.50 kPa Clay is NC, thus: σ' vo = σ 'p = 38 kPa From Problem 8-12: Ccε = 0.32 σ ' + Δσv 38 + 56.5 = (0.32)(7.8 m)log = 0.9875 m = 99 mm sc = CcεHo log vo σ 'p 38
8-16. Assume the laboratory test results in Problem 8.12 are typical of another San Francisco Bay Mud site, but where the clay is slightly overconsolidated. The present vertical effective overburden stress is calculated to be about 15 kPa, and the thickness of the clay is 3.9 m. At this 3 location, the granular fill ( ρ = 1.8 Mg/m ) will be only about 1.2 m thick. Estimate the consolidation settlement due to the weight of the fill. SOLUTION:
Δσv = (1.2 m)(1.8 Mg m3 )(9.81m s2 ) = 21.9 kPa;
σ' vo = 15 kPa; σ 'p = 38 kPa 0.0226 − 0.01 = 0.0097 From Problem 8-12 plot: Ccε = 0.32 and Cr ε = log Eq. 8.17: sc sc
σ 'vo + Δσv σ 'vo 15 + 21.9
20 1.0
= CrεHo log
= (0.0097)(3.9 m)log
15
= (3.9 m)(0.00379) = 0.0148 m = 1.5 mm
8-17. What settlement would you expect at the overconsolidated site in Problem 8.16 if the fill to be constructed were 4 m thick? SOLUTION:
Δσv = (4 m)(1.8 Mg m3 )(9.81m s2 ) = 70.63 kPa;
σ'vo = 15 kPa; σ 'p = 38 kPa 0.0226 − 0.01 From Problem 8-12 plot: Ccε = 0.32 and Cr ε = = 0.0097 log use Eq. 8.19b: sc
= CrεHo log
= (0.0097)(3.9 m) log
38
σ 'p σ ' + Δσv + CcεHo log vo σ 'vo σ 'p (15 + 70.63)
+ (0.32)(3.9 m)log 15 sc = 0.0153 + 0.440 = 0.456 m = 456 mm sc
20 1.0
38
= (3.9 m)(0.0039) + (3.9 m)(0.11291)
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Compressibility of Soil and Rock
Chapter 8
8-18. Plot the following data and determine the preconsolidation pressure and the modified 3 compression index. Specimen height is 25.4 mm, w n = 32.5%, ρ d = 1.45 Mg/m . Sample is from a depth of -11.5 m.
% Str ai n
Press ure (kPa)
0.09
5
0.11
10
0.12
20
0.26
40
0.98
80
1.91
160
4.19
320
8.05
640
8.03
320
7.83
160
7.21
80
7.34
160
7.60
320
8.35
640
12.65
1280
17.41
2560
22.18
5120
21.65
1280
20.63
160
19.26
40
15.35
5
Effective Stress (kPa) 1
10
100
1000
10000
0.00
5.00
n i a 10.00 r t S t n e c r 15.00 e P
20.00
25.00
SOLUTION:
ε2 − ε1 0.25 − 0.031 = = 0.154 σ '2 8000 log log 300 σ '1
Modified Compression index = Ccε
=
From the semillogarithmic plot: σ'p
≈ 260 kPa
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Compressibility of Soil and Rock
Chapter 8
8-19. At the site where the sample of Problem 8.18 was taken, the soil profile consists of about 6.5 m of sand and rubble fill and then 9.1 m of clay. The water table is about 1.8 m below the 3 ground surface. Average densities of the sand and rubble fill are 1.45 Mg/m above the water 3 table and 1.70 Mg/m below the water table. Estimate the consolidation settlement if the average stress increase in the compressible layer is: (a) 50 kPa, (b) 100 kPa, and (c) 250 kPa. SOLUTION: (a)
Δσv = 50 kPa;
For the clay: ρsat
σ 'p = 260 kPa (from Problem 8-18)
= ρdry × (1 + w ) = (1.45 Mg m3 )(1 + 0.325) = 1.92 Mg m3
σ'vo = ⎡⎣(1.8 m)(1.45 Mg m3 ) + (4.7 m)(1.70 − 1.0 Mg m3 ) + (4.55 m)(1.92 − 1.0 Mg m3 )⎤⎦ × 9.81m s2 σ'vo = (10.086 Mg m2 ) × 9.81m s2 = 98.9 kPa (at the center of the clay layer) From Problem 8-18 plot: Ccε
0.0815 − 0.05 640 log 1.0 98.9 + 50 = (0.0112)(9.1m)log 98.9
= 0.154 and Cr ε =
= 0.0112
σ 'vo + Δσv σ 'vo sc = (9.1m)(0.00199) = 0.0181m = 1.8 mm
Eq. 8.17: s c
(b)
= CrεHo log
Δσv = 100 kPa;
σ 'p = 260 kPa; σ'vo = 98.9 kPa
σ 'vo + Δσv 98.9 + 100 = (0.0112)(9.1m)log σ 'vo 98.9 sc = (9.1 m)(0.003398) = 0.0309 m = 3.1mm
Eq. 8.17: s c
(c)
Δσv = 250 kPa;
Eq. 8.19b: sc sc
= CrεHo log
σ 'p = 260 kPa; σ'vo = 98.9 kPa
= CrεHo log
= (0.0112)(9.1m)log
σ 'p σ ' + Δσv + CcεHo log vo σ 'vo σ 'p
260
+ (0.154)(9.1m)log
98.9 sc = 0.0428 + 0.179 = 0.222 m = 22.2 mm
(98.9 + 250) 260
= (9.1m)(0.0047) + (9.1m)(0.01967)
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Compressibility of Soil and Rock
Chapter 8
8-20. Plot the following void ratio versus pressure data, and evaluate the compression index and the recompression index. Determine the preconsolidation stress. Void
Pressure
Ratio
(kPa)
1.025
0
1.006
10
0.997
20
0.978
40
0.950
80
0.911
160
0.893
200
0.837
300
0.780
400
0.655
800
0.504
2000
0.542
500
0.589
160
0.681
20
1.100 1.000 0.900 0.800
o i t a R0.700 d i o V
0.600 0.500 0.400 0.300 1
SOLUTION: From the plot:
10
Recompression index = Cr =
1000
10000
Effective Stress (kPa)
σ 'p ≈ 220 kPa
Eq. 8.7: Compression index
100
= Cc = e1 − e2 σ' log 2 σ '1
e1 − e2 σ' log 2 σ '1
=
=
1.1 − 0.30 6200 log 71
0.8 − 0.504 2000 log 71
= 0.412
= 0.090
8-21. Use the consolidation data from Problem 8.20 to compute the settlement of a structure that adds 175 kPa to the already existing overburden pressure of 130 kPa at the middle of a 6 m thick layer. SOLUTION:
Δσv = 175 kPa;
σ'vo = 130 kPa;
σ 'p = 220 kPa;
OCR =
220 130
= 1.7
= 0.412, Cr = 0.09, eo = 1.025 σ 'p Ho H σ ' + Δσv use Eq. 8.18b: sc = Cr log + Cc o log vo σ 'vo σ 'p 1 + eo 1 + eo
From Problem 8-20: C c
⎛ 6 m ⎞ 220 ⎛ 6 m ⎞ (130 + 175) = (0.09) ⎜ + (0.412) ⎜ = 0.0609 + 0.1732 log ⎟ ⎟ log 220 ⎝ 1 + 1.025 ⎠ 130 ⎝ 1+ 1.025 ⎠ sc = 0.234 m = 234 mm sc
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Compressibility of Soil and Rock
Chapter 8
8-22. What would be the settlement of the same structure in Problem 8.21 if the overconsolidation ratio of the clay were 1.0 and σ ’vo + Δσ v = 305 kPa at the middepth of the clay layer? Show your work and assumptions on the e versus log s curve of Problem 8.20. SOLUTION:
Δσv = 175 kPa; σ'vo + Δσv = 305 kPa; OCR = 1.0 → σ 'p = σ'vo = 220 kPa From Problem 8-20: Cc = 0.412, Cr = 0.09, eo = 1.025 σ ' + Δσv Ho use Eq. 8.11: sc = Cc log vo σ 'vo 1 + eo ⎛ 6 m ⎞ (305) = 0.173 m = 173 mm sc = (0.412) ⎜ ⎟ log ⎝ 1 + 1.025 ⎠ (220)
8-23. The consolidation curve of Fig. Ex. 8.9 is typical of a compressible layer 5 m thick. If the existing overburden pressure is 50 kPa, compute the settlement due to an additional stress of 150 kPa added by a structure. SOLUTION:
Δσv = 150 kPa;
σ' vo = 50 kPa;
σ 'p = 120 kPa;
120 50 eo = 0.87 OCR =
= 1.7; Ho = 5 m
= 0.44, Cr = 0.03, σ 'p σ ' + Δσv Ho H + Cc o log vo Eq. 8.18b: sc = Cr log σ 'vo σ 'p 1 + eo 1 + eo From Examples 8.9 and 8.12: C c
⎛ 5 m ⎞ 120 ⎛ 5 m ⎞ (50 + 150) = (0.03) ⎜ + (0.44) ⎜ = 0.0305 + 0.261 log ⎟ ⎟ log 120 50 ⎝ 1 + 0.87 ⎠ ⎝ 1 + 0.87 ⎠ sc = 0.291m = 291mm sc
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Compressibility of Soil and Rock
Chapter 8
8-24. For the test data of Problem 8.12, construct the field virgin compression curve using the Schmertmann procedure for an OCR of unity. SOLUTION: 3.5
3 39.000, 2.765 2.5 o i t a R 2 d i o V
1.5 590.000, 1.161 1
0.5 1
10
100
1000
Effective Stress (kPa)
Field virgin compression curve based on Schmertmann’s (1955) procedure. Field compression index = Cc
This compares to Cc
=
e1 − e2 σ' log 2 σ '1
=
2.765 − 1.161 = 1.36 590 log 39
= 1.21 using the un-adjusted lab compression curve (see Problem 8-13).
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Compressibility of Soil and Rock
Chapter 8
8-25. Do Problem 8.24 for an OCR = 2.5. SOLUTION: 3.5
3 39, 2.71 2.5 o i t a R 2 d i o V
1.5 570, 1.1613 1
0.5 1
10
100
1000
Effective Stress (kPa)
Field virgin compression curve based on Schmertmann’s (1955) procedure. Field compression index (OCR = 2.5): Cc
=
e1 − e2 σ' log 2 σ '1
=
2.71 − 1.161 = 1.33 570 log 39
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Compressibility of Soil and Rock
Chapter 8
8-26. At the midpoint of a 7.5 m thick soil layer, the void ratio is 1.9. Find this point on the field virgin compression curve determined in Problem 8.24. What is the corresponding pressure? If this pressure is doubled over the entire site, compute the consolidation settlement of the layer. SOLUTION: 3.5
3
2.5 o i t a R 2 d i o V
170, 1.9
1.5
340, 1.49
1
0.5 1
10
100
1000
Effective Stress (kPa)
= 7.5 m, e1 = 1.9 From the plot: p1 = 170kPa p2 = 2po = 340 kPa → e2 = 1.49 (from plot) ⎛ Δe ⎞ ⎛ 1.9 − 1.49 ⎞ sc = Ho ( ε) = Ho ⎜ ⎟ = (7.5 m) ⎜ ⎟ = 1.06 m = 106 mm 1 e 1 1.9 + + ⎝ ⎠ o ⎝ ⎠ Ho
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Compressibility of Soil and Rock
Chapter 8
8-31. Figure P8.31 shows a proposed foundation site, with 10 ft of sand overlying 15 ft of clay with consolidation properties shown. The clay is normally consolidated. Assume 1-D conditions. (a) Compute the initial σ ’v at the middle of the clay layer prior to excavation and construction. (b) After excavation and during construction, the foundation area will be heavily loaded with the structure and equipment so that σ ’v at the middle of the clay layer will be increased to 3900 psf. Determine the settlement that will occur under these conditions. (c) After construction is completed, the equipment will be removed and the final σ ’v at the middle of the clay layer will be 3200 psf.
SOLUTION:
σ 'vo = (10 ft)(110 pcf ) + (7.5 ft)(120 − 62.4 pcf ) = 1532 psf (b) Δσv = 3900 − 1532 = 2368 psf Given: Ccε = 0.165, Crε = 0.033, Ho = 15 ft σ ' + Δσv Eq. 8.13: sc = CcεHo log vo σ 'vo (1532 + 2368) = 1.0 ft sc = (0.165)(15 ft)log (a)
(1532)
(c)
Δσv = 3200 − 1532 = 1668 psf
Compute the swell that occurs after equipment is removed (part b to part c): sswell
= CrεHo (log σb − log σc )
= (0.033)(15) (log3900 − log 3200) = 0.0425 ft The final settlement = 1.0 − 0.0425 = 0.96 ft sswell
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Compressibility of Soil and Rock
Chapter 8
8-32. As part of a construction project, a 7.5 m thick layer of clay is to be loaded with a temporary 3 m thick sand layer (refer to Fig. P8.32).The figure shows the water-table location, soil unit weights, and the compression curve properties for the clay. Assume the sand layer remains dry. (a) Calculate the value of σ ’v in the middle of the clay layer (at 3.75 m below the water table) before the sand layer is applied, and after consolidation is complete. (b) Based on your answer in part (a), and the compression curve characteristics, calculate the settlement that will occur under these conditions. (c) How much will the clay layer heave when the 3 m sand layer is removed?
SOLUTION: (a)
σ 'vo = (3.75 m)(20.5 − 9.81kN m3 ) = 40.09 kN m2
(b)
Δσv = (3 m)(16 kN m3 ) = 48.0 kN m2 , σ 'p = 74 kN m2 , OCR =
use Eq. 8.19b: sc
= CrεHo log 74
+ (0.18)(7.5 m)log 40.09 sc = 0.05992 + 0.1022 = 0.162 m = 162 mm sc
= (0.03)(7.5 m)log
σ 'p σ ' + Δσv + CcεHo log vo σ 'vo σ 'p (40.09 + 48)
(c) sswell sswell
= CsεHo log
74
σ 'p 74 kPa = = 1.85 σ 'vo 40.09 kPa
= (7.5 m)(0.00799) + (7.5 m)(0.01362)
σ 'vf ( 40.09 + 48.0) = (0.03)(7.5 m)log = (7.5 m)(0.01026) σ 'vo 40.09
= 0.077 m = 77.0 mm
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Compressibility of Soil and Rock
Chapter 8
8-33. Refer to Fig. 8.5a. Determine: (a) Using log interpolation between 100 and 1000, determine the σ v value at a vertical strain, ε v = 20%. (b) If the initial void ratio, eo = 2.6, determine Cr and Cc for this soil. For C c use the portion of the curve between σ ’v = 100 and 500 kPa. (c) If the original clay layer thickness is 9.5 m, determine the settlement that occurs in the layer when it is loaded from 200 to 400 kPa. [Note: You don’t need the results from part (b) to do this.]
SOLUTION: (a) There are various approaches for interpolating logarithmic scales. One approach for interpolating the desired x (or σ 'v ) value is as follows:
( ) × ( x11− f );
x = x 2f
where, f =
a
a+b In this case, the x data point ( at ε v Thus, f
= 20%) is 18.2% of the way between 100 and 1000 kPa.
= 0.182 and x = (10000.182 ) × (1000.182 ) = (3.5156)(43.2514) = 152 kPa
∴ at εv = 20%, σ'v = 152 kPa ε2 − ε1 0.34 − 0.13 = = 0.300 500 σ '2 log log 100 σ '1 Compression index = Cc = Ccε (1 + eo ) = (0.30)(1 + 2.6) = 1.08
(b) Modified Compression index = Ccε
=
ε2 − ε1 0.07 − 0.0 = = 0.035 σ '2 1000 log log 10 σ '1 Recompression index = Cr = Crε (1 + eo ) = (0.035)(1+ 2.6) = 0.126 Modified Recompression index = Cr ε
=
(c) For the load increment from 200 to 400 kPa, sc
Δε = 32 - 23 = 9.0% (From plot in Fig. 8.5a)
= ( Δε)(Ho ) = (0.09)(9.5 m) = 0.855 m = 855 mm
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Compressibility of Soil and Rock
Chapter 8
8-34. A large embankment is to be built on the surface of a 15-ft clay layer. Before the embankment is built, the initial σ ’v at the middle of the clay layer is 480 psf. The results from a 1-D consolidation test on the clay from the middle of the layer are as follows: σ ’p = 1800 psf
Cr ε = 0.0352
Ccε = 0.180
If the final σ ’v at the middle of the layer after the embankment loading is 2100 psf, what is the settlement, in inches, of the clay layer resulting from this loading? SOLUTION:
σ 'p 1800 psf = = 3.75 σ 'vo 480 psf σ 'p σ ' + Δσv + CcεHo log vo use Eq. 8.19b: sc = CrεHo log σ 'vo σ 'p (a)
psf, OCR = σ 'vo = 480 psf, σ 'vf = 2100
sc
= (0.0352)(15 ft)log
sc
= 0.4839 ft = 5.8 in
1800 480
+ (0.18)(15 ft)log
(2100) 1800
= 0.3031+ 0.1808
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Compressibility of Soil and Rock
Chapter 8
8-35. Figure P8.35 shows a proposed site where an excavation will be made. The 10 ft layer of sand will be removed, so that the top of the 24 ft normally consolidated clay layer will be exposed. Assume full capillarity in the clay only. (a) Assume that the water-table location remains the same during excavation. Compute the σ v. σ ’v, and u values at the middle of the clay layer before and after the excavation. (b) Assuming 1-D conditions, compute how much the clay layer will deform due to this excavation, in inches. Specify whether this is settlement or heave.
SOLUTION: (a) Before excavation
σv = (10 ft)(110 pcf) + (12 ft)(120 pcf) = 2540 psf u = (12 − 3 ft)(62.4 pcf ) = 561.6 psf
σ 'v = 2540 − 561.6 = 1978 psf After excavation
σv = (12 ft)(120pcf) = 1440 psf u = (12 − 3 ft)(62.4 pcf ) = 561.6 psf
σ 'v = 1440 − 561.6 = 878 psf Δσ = 1978 − 878 = 1100 psf, σ 'p = σ 'v = 1978 psf σ 'p (1978) sswell = CsεHo log = (0.035)(24 ft)log = (24 ft)(0.01234) σ 'vf (878) sswell = 0.2963 ft = 3.5 in (heave)
(b)
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Compressibility of Soil and Rock
Chapter 8
8-36. Figure P8.36 shows the soil profile at a site where you plan to lower the water table. You have results from two consolidation tests, one from the upper 12 ft thick overconsolidated crust, and another from the lower 32 ft thick normally consolidated zone. You plan to lower the water table from its current 12 ft depth to 20 ft below ground surface. The consolidation properties for each layer are shown. Assume full capillarity. a) Compute the σ ’v in the middle of each layer before and after the water table is lowered. (b) Determine the total settlement that will result from lowering the water table.
SOLUTION: (a) Center of layer 1: original water table at 12 ft
σv = (6 ft)(120 pcf) = 720 psf; u = 0; σ 'vo = 720 psf Center of layer 2: original water table at 12 ft
σv = (12 ft)(120 pcf) + (16 ft)(118 pcf) = 3328 psf u = (28 − 12 ft)(62.4 pcf ) = 998.4 psf σ 'vo = 3328 − 998.4 = 2329.6 psf Center of layer 1: water table lowered to 20 ft
σv = (6 ft)(120 pcf) = 720 psf; u = 0; σ 'vf = 720 psf (no change) Center of layer 2: water table lowered to 20 ft
σv = (12 ft)(120 pcf) + (16 ft)(118 pcf) = 3328 psf u = (28 − 20 ft)(62.4 pcf ) = 499.2 psf σ 'vf = 3328 − 499.2 = 2828.8 psf (b) Consolidation settlement will occur in layer 2.
Δσ = 2828.8 − 2329.6 = 499.2 psf, Layer 2: σ 'p2 = σ 'vo = 2329.6 psf Given: Ccε = 0.185, Cr ε = 0.034, Ho = 32 ft σ ' + Δσv Eq. 8.13: sc = CcεHo log vo σ 'vo (2329.6 + 499.2) = 0.499 ft = 6.0 in sc = (0.185)(32 ft)log (2329.6)
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Compressibility of Soil and Rock
Chapter 8
8-37. When a consolidation test is performed on some soils, the virgin compression region is not linear, but bilinear. Figure P8.37 shows such a compression curve from a 15 ft thick layer. Required: (a) What vertical strain, occurs when the soil is loaded from an initial σ ’v1 = 560 psf to σ ’v2 = 3000 psf? (b) If you load the soil further, to σ ’v3 = 4000 psf, how much additional settlement occurs? (c) Finally, if you unload from 4000 psf back to σ ’v4 = 3000 psf, what additional deformation (in ft) occurs?
SOLUTION:
⎛ 980 ⎞ ⎛ 3000 ⎞ εv = (0.032)log ⎜ + (0.14)log ⎜ ⎟ ⎟ = 0.00778 + 0.068 = 0.0758 = 7.58% ⎝ 560 ⎠ ⎝ 980 ⎠ sc = (15 ft)(0.0758) = 1.14 ft
(a)
⎛ 4000 ⎞ εv = (0.17)log ⎜ ⎟ = 0.0212 = 2.12%; sc = (15 ft)(0.0212) = 0.318 ft ⎝ 3000 ⎠ sc − total = 1.14 + 0.318 = 1.46 ft
(b)
⎛ 4000 ⎞ εswell = (0.032)log ⎜ ⎟ = 0.004 = 0.4% (heave) ⎝ 3000 ⎠ sswell = (15 ft)(0.004) = 0.06 ft = 0.72 in
(c)
Aside : net settlement (snet ) for loading and unloading described in parts a, b, and c: snet
= 1.146 − 0.06 = 1.09 ft; or about 1 ft
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Compressibility of Soil and Rock
Chapter 8
8-38. Refer to the compression curve marked Soil 13 in Fig. 8.9a. Disregard the small unloading cycle in the middle of the curve. The initial void ratio for this soil is 1.17, and the preconsolidation pressure is 290 kPa. Note: the right-hand vertical boundary of this graph is at σ ’vc = 2000 kPa. Required: (a) Determine the C r and Cc for this soil based on the compression curve. (b) If a layer of this soil 12 m thick is loaded from 50 to 800 kPa (two of the data points shown on the curve), what settlement will result, in m? SOLUTION: (a) Cr
(b)
=
1.17 − 1.135 = 0.0106; ⎛ 2000 ⎞ log ⎜ ⎟ ⎝ 1 ⎠
Δσv = 750 kPa;
OCR =
290 50
Cc
=
σ'vo = 50 kPa;
1.35 − 0.67 ⎛ 2000 ⎞ log ⎜ ⎟ ⎝ 100 ⎠
= 0.523
σ 'p = 290 kPa;
eo
= 1.17; Ho = 12 m
= 5.8
Eq. 8.18b: sc
= Cr
Ho 1 + eo
log
σ 'p σ ' + Δσv H + Cc o log vo 1 + eo σ 'vo σ 'p
⎛ 12 m ⎞ 120 ⎛ 12 m ⎞ (50 + 750) = (0.0106) ⎜ + (0.523) ⎜ log ⎟ ⎟ log 290 = 0.0223 + 1.2745 50 ⎝ 1 + 1.17 ⎠ ⎝ 1+ 1.17 ⎠ sc = 1.30 m sc
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