CHAPTER 10
Compressibility and Settlement
QUESTIONS AND PRACTICE PROBLEMS
Section 10.2 Changes in Vertical Effective Stress
10.1 The current σz′ at a certain point in a saturated clay is 181 kPa. This soil is to be covered
with a 2.5 m thick fill that will have a unit weight of 19.3 kN/m3. What will be the value
of σz′ at this point immediately after the fill is placed, before any consolidation has
occurred? What will it be after the consolidation settlement is completed?
Solution
Immediately after the fill is placed, its weight will be carried entirely by the pore water
(i.e., ue=Δσz). Therefore, σ’z at the point will still be 181 kPa.
Once the consolidation settlement is complete, ue=0 and the weight of the fill is
carried entirely by the solid particles, Therefore, at the point of interest
σ´z = σ´z0 + γfill(Hfill) = 181 + (19.3)(2.5) = 229 kPa.
10.2 A circular tank 50 feet in diameter is constructed on top of a saturated soil with a
saturated unit weight of 112 lb/ft3. The total weight of the tank when filled is 6,125 k.
The groundwater table is at the ground surface. Compute σ'z a point 25 feet below the
center of the tank immediately after the tank is placed and filled, before any consolidation
has occurred. What will σ'z be at the same point after consolidation settlement is
completed?
Solution
Before placement of the tank the effective vertical stress is
σ 0′ = γH − u = (112 lb/ft 3 )(25 ft ) − (62.4 lb/ft 3 )(25 ft ) = 4360 lb/ft 3
Immediately after the tank is placed and filled, its weight will be carried entirely by the
pore water (i.e., ue=Δσz). Therefore, σ´z at the point will still be 4360 lb/ft2. However the
total stress and pore water pressure will change.
q=
P 6,125,000 lb
=
= 3119 lb/ft 2
2
A
π (25 ft )
10-1
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10-2
Compressibility and Settlement Chap. 10
Using Figure 9.13
Zf/b = 25/50 =0.5
Iσ = 0.65
(
Δσ z = Iσ q = (0.65) 3119 lb/ft 2
)
= 2027 lb/ft 2
σ ′zf = σ ′z 0 + Δσ z = 4360 lb/ft 2 + 2027 lb/ft 2 = 6387 lb/ft 2
Once the consolidation settlement is complete, ue=0 and the weight of the fill is carried
entirely by the solid particles, Therefore, σ’z at the point is equal to 6387 lb/ft2.
10.3 A 4.0 m thick fill with a unit weight of 20.1 kN/m3 is to be placed on the soil profile
shown in Figure 9.11. Develop plots of σ′z0 and σ′zf vs. depth. The plot should extend
from the original ground surface to a depth of 10.0 m.
Figure 9.11 Soil profile for Example 9.2.
Solution
σ ′zf = σ ′z 0 + γ fill H fill = σ ′z 0 + (20.1 kN/m 3 )(4.0 m ) = σ ′z 0 + 80 kPa
Using the above equation, the initial and final vertical effective stresses are computed as
(z = depth below original ground surface).
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Chap. 10 Compressibility and Settlement
z
(m)
0
2
4.5
10
σz
(kPa)
0
30
72
167
u
(kPa)
0
0
24
78
10-3
σ'z0
(kPa)
0
30
45
89
σ'zf
(kPa)
80
110
128
169
10.4 The groundwater table at a certain site was at a depth of 10 ft below the ground surface,
and the vertical effective stress at a point 30 ft below the ground surface was 2200 lb/ft2.
Then a series of wells were installed, which caused the groundwater table to drop to a
depth of 25 ft below the ground surface. Assuming the unit weight of the soil above and
below the groundwater table are equal, compute the new σz′ at this point.
Solution
Δσ ′z = Δu = (25 − 10 )(62.4 ) = 936 lb/ft 2
σ ′z = 2200 + 936 = 3136 lb/ft 2
10.5 The fill shown in Figure 10.21 was placed many years ago and consolidation in both clay
layers due to the fill placement has is complete. The fill was placed near a river that has
since been dammed. Filling of the reservoir behind the dam has raised the local
groundwater table to a point 2 m below the ground surface. The raising of the water table
has been slow and has not induced any excess pore pressures in the clay soils. Compute
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10-4
Compressibility and Settlement Chap. 10
σz′ at points A and B due to the rise of the groundwater table. Assume the saturated unit
weight of the fill material is 21.1 kN/m3
Figure 10.21 Soil profile for Example 10.6.
Solution
At point A
σ ′z 0 =
∑ γH − u
= (20.3 )(2 ) + (21.1 )(6.5) + (19.0 )(4.0 ) − (9.8 )(10.5)
= 151 kPa
At point B
σ ′z 0 =
∑ γH − u
= (20.3 )(2 ) + (21.1 )(6.5) + (19.0 )(9.0 )
+ (18.0 )(10.0 ) − (9.8 )(25.5)
= 279 kPa
10.6 A 1.00 m3 element of soil is located below the groundwater table. When a new
compressive load was applied, this element consolidated, producing a vertical strain, εz ,
of 8.5%. The horizontal strain was zero. Compute the volume of water squeezed out of
this soil during consolidation and express your answer in liters.
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Chap. 10 Compressibility and Settlement
10-5
Solution
Since the soil is below the groundwater table, it is saturated (S=100%). Thus, the volume
of water squeezed out of the element is equal to the total change in volume.
(
3
)
⎛ 100 cm ⎞ ⎛ 1 liter ⎞
ΔV = ε zV = (0.085) 1.00 m 3 ⎜
= 85 liters
⎟ ⎜
3 ⎟
⎝ m ⎠ ⎝ 1000 cm ⎠
Section 10.5 Consolidation Test
10.7 A consolidation test is being performed on a 3.50 in diameter saturated soil sample that
had an initial height of 0.750 in and an initial moisture content of 38.8%.
(a) Using Gs = 2.69, compute the initial void ratio, e0.
(b) At a certain stage of the test, the normal load P was 300 lb. After the consolidation at
this load was completed, the sample height was 0.690 in. Compute σz′ (expressed in
lb/ft2), εz, and e.
Solution
(a)
wG s (0.388)(2.69)
=
= 1.04
S
1
e0 =
(b)
A=
π (3.5 / 12)2
4
= 0.0668 ft 2
σ ′z =
P
300
=
= 4490 lb/ft 2
A 0.0668
εz =
ΔH 0.0750 − 0.0690
=
= 8.00%
H
0.0750
e = (1 − ε z )(1 + e0 ) − 1 = (1 − 0.0800 )(1 + 1.04 ) − 1 = 0.876
10.8 A consolidation test of a soft marine silty clay produced the following data
σz′
(lb/ft2)
εz
50
250
500
1000
2000
4000
8000
16,000
4000
500
0.007
0.028
0.059
0.097
0.145
0.189
0.235
0.278
0.271
0.253
The initial void ratio of the sample was 1.24 and σz0′ at the sample depth was 270 lb/ft2
(a) Plot these data on an arithmetic diagram similar to that in Figure 10.10(a)
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10-6
(b)
(c)
(d)
(e)
(f)
Compressibility and Settlement Chap. 10
Plot these data on a semilogarithmic diagram similar to that in Figure 10.10(b)
Using Casagrande’s method, find σ'c
Using Schmertmann’s method adjust the test results
Determine Cc and Cr
The soil has a plasticity index of 17. Based on Kulhawy and Mayne’s correlations, do
the consolidation test results seem reasonable?
Solution
(a)
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Chap. 10 Compressibility and Settlement
10-7
(b)
(c) See Casagrande construction on plot – σ’c = 525 psf
(d) See Schmertmann construction on plot
(e) Using the reconstructed curve:
(ε z )b − (ε z )a
Cc
=
1 + e0
log σ z' b − log σ z'
(
) (
)
=
a
0.39 − 0.27
= 0.12
log(100,000) − log(10,000)
Cc = 0.12(1 + e0 ) = 0.12(2.24 ) = 0.27
Using the original decompression curve:
(ε z )b − (ε z )a
Cr
=
1 + e0
log σ z' b − log σ z'
(
) (
)
a
=
0.038 − 0.008
= 0.015
log(100,000) − log(1000)
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10-8
Compressibility and Settlement Chap. 10
Cc = 0.015(1 + e0 ) = 0.015(2.24 ) = 0.033
(f) Per Kulhawy and Mayne’s correlation:
Cc = Ip/74= 17/74 = 0.23
Cr = Ip/370= 17/370 = 0.046
The test results are fairly close to these values – Therefore, they appear to be reasonable
10.9 A consolidation test on a sample of clay produced the following data:
σz′ (kPa)
εz
8
0.032
16
0.041
32
0.051
64
0.069
128
0.109
256
0.173
512
0.240
1024
0.301
16
0.220
The initial void ratio of the sample was 1.21, and σz0′ at the sample depth was 40 kPa.
(a)
(b)
(c)
(d)
(e)
(f)
Plot these data on an arithmetic diagram similar to that in Figure 10.10(a)
Plot these data on a semilogarithmic diagram similar to that in Figure 10.10(b).
Using Casagrande’s method, find σc′.
Using Schmertmann’s method, adjust the test results.
Determine Cc and Cr.
The soil has a plasticity index of 23. Based on Kulhawy and Mayne’s correlations,
do the consolidation test results seem reasonable?
Solution
(a)
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Chap. 10 Compressibility and Settlement
10-9
(b)
(c) See Casagrande construction on plot – σ’c = 525 psf
(d) See Schmertmann construction on plot
(e) Using the reconstructed curve:
(ε z )b − (ε z )a
Cc
=
1 + e0
log σ z' b − log σ z'
(
) (
)
=
a
0.27 − 0.015
= 0.26
log(1000) − log(100)
Cc = 0.26(1 + e0 ) = 0.26(2.21) = 0.57
Using the original decompression curve:
(ε z )b − (ε z )a
Cr
=
1 + e0
log σ z' b − log σ z'
(
) (
)
a
=
0.065 − 0.015
= 0.050
log(1000) − log(100)
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10-10
Compressibility and Settlement Chap. 10
Cr = 0.050(1 + e0 ) = 0.050(2.21) = 0.11
(f) Per Kulhawy and Mayne’s correlation:
Cc = Ip/74= 23/74 = 0.31
Cr = Ip/370= 23/370 = 0.044
The test values are greater than Kulhawy and Mayne’s correlation by about a factor of 2.
This indicates the soil has an unusually high compressibility.
Section 10.6 Consolidation Status in the Field
10.10 Before the placement of any fill, a consolidation test was performed on a soil sample
obtained from Point B in Figure 10.20. The measured preconsolidation stress was 88 kPa.
Determine whether the soil is normally consolidated or overconsolidated, then compute
the overconsolidation margin and overconsolidation ratio at Point B.
Note: These computations are based on the initial conditions, and thus should not
include the weight of the proposed fill.
Figure 10.20 Soil profile for Example 10.5
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Chap. 10 Compressibility and Settlement
10-11
Solution
At point B:
σ ′z 0 =
∑ γH − u
= (18.5)(1.5) + (19.5)(2.0 ) + (16.0 )(3.0 ) − (9.8)(5.0 )
= 66 kPa
σ´z0 < σ´c, so the soil is Overconsolidated
σ m′ = σ c′ − σ ′z = 88 − 66 = 22 kPa
OCR =
σ c′
88
=
= 1.3
66
σ ′z
10.11 A saturated, normally consolidated, 1000-year-old fine-to-medium sand has an SPT
N60 = 2 at a depth where the vertical effective stress is about 1000 lb/ft2 and D50 = 0.5
mm. Using the techniques described in Chapters 3 and 4, determine the relative density
of this soil, then estimate Cc/(1+e0) based on Table 10.4.
Solution
Using Equation 3.2:
N1, 60 = N 60
2000 lb/ft 2
2000
= 12
= 17
1000
σ ′z
Using Equation 4.21 – 4.24:
C P = 60 + 25 log D50 = 60 + 25 log(0.5) = 52
⎛ t ⎞
⎛ 1000 ⎞
C A = 1.2 + 0.05 log⎜
⎟ = 1.2 + 0.05 log⎜
⎟ = 1.25
⎝ 100 ⎠
⎝ 100 ⎠
COCR = OCR 0.18 = 10.18 = 1
Dr =
N 1, 60
C p C A C OCR
× 100% =
17
× 100% = 51%
(52)(1.25)(1)
10.12 A consolidation test has been performed on a sample obtained from Point A in
Figure 10.3. The measured preconsolidation stress was 1500 lb/ft2.
(a) Determine if the soil is normally consolidated or overconsolidated
(b) Compute the overconsolidation margin and the overconsolidation ratio
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10-12
Compressibility and Settlement Chap. 10
(c) Compute σc′ at Point B
Figure 10.3 Soil profile for Example 10.1. el. = elevation.
Solution
(a) At Point A:
σ ′z 0 =
∑ γH − u
= (98)(1.6) + (100)(4.4) − (62.4)(4.4)
= 322 lb/ft 2
(b)
σ´z0 < σ´c, so the soil is overconsolidated.
σ m = σ c′ − σ ′z = 1500 − 322 = 1178 lb/ft 2
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Chap. 10 Compressibility and Settlement
OCR =
10-13
σ c′ 1500
=
= 4.7
322
σ ′z
(c) At Point B:
σ ′z 0 =
∑ γH − u
= (98)(1.6 ) + (100)(16.4) − (62.4)(16.4)
= 773 lb/ft 2
σ c′ = σ ′z 0 + σ m = 773 − 1178 = 1951 lb/ft 2
10.13 A consolidation test has been performed on a sample obtained from a saturated clay at a
point 6.5 m below the ground surface. The groundwater table is at the ground surface and
the unit weight of the clay is 18.5 kN/m3. The measured preconsolidation stress was
260 kPa.
(a) Determine if the soil is normally consolidated or overconsolidated
(b) Compute the overconsolidation margin and the overconsolidation ratio
(c) Compute σc′ at depth of 12 m in the same soil.
Solution
(a) At a point 6.5m below ground surface:
σ ′z 0 = ∑ γH − u
σ ′z 0 = (18.5)(6.5) − (9.8)(6.5)
σ ′z 0 = 56.6 kPa
(b)
σ´z0 < σ´c, so the soil is overconsolidated.
σ m = σ c′ − σ ′z = 260 − 56 .6 = 203 kPa
OCR =
260
σ c′
=
= 4.6
σ ′z 56.6
(c) At a point 12 m below the ground surface:
σ ′z 0 = ∑ γH − u
σ ′z 0 = (18.5)(12) − (9.8)(12)
σ ′z 0 = 104 kPa
σ c′ = σ ′z 0 + σ m = 104 − 203 = 307 kPa
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10-14
Compressibility and Settlement Chap. 10
10.14 Laboratory consolidation test for all soils have both a recompression curve and a virgin
curve. This is true even for normally consolidated soils. Explain how the lab test for a
normally consolidated soil can have a recompression curve even though the soil in the
field has never been preloaded.
Solution
During the process of sampling the soil removed from the field is unloaded from its in
situ condition. Even the highest quality field samples undergo this unloading. Therefore
even normally consolidated soils will exhibit a recompression curve during laboratory
testing.
Section 10.8 Consolidation Settlement Predictions
10.15 A 2 m thick fill with an in place unit weight of 19.2 kN/m3 is to be placed on top of the
soil profile shown in Figure 10.25. Both the sand and the clay are normally consolidated.
The sand is fine, poorly graded and medium dense with Cc/(1+e0) estimated to be 0.009.
Lab consolidation tests on the clay produced Cc/(1+e0) = 0.18. Compute the total
consolidation settlement due to the placement of the fill. Would it be acceptable to ignore
the settlement generated by consolidation of the sand layer?
Figure 10.25 Soil Profile for Problem 10.15.
Solution
Applied stress from the proposed fill
Δσ z = γ
fill H fill
(
)
= 19.2 kN/m 3 (2 m )
For this computation we will divide the sand stratum and the clay stratum each into two
layers.
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Chap. 10 Compressibility and Settlement
Layer
1
2
3
4
H (m)
1
1
0.5
0.5
zf
0.5
1.5
2.25
2.75
At midpoint of layer
σ'z0
Δσz
σ'zf
(kPa) (kPa) (kPa)
8.6 38.4
47.0
21.3 38.4
59.7
27.2 38.4
65.6
33.9 38.4
72.3
10-15
Cc/(1+e0)
0.009
0.009
0.18
0.18
Eqn.
10.23
10.23
10.23
10.23
Σ=
δc,ult
(mm)
6.66
4.03
34.43
29.62
74.74
δc,ult = 75 mm
It would be acceptable to ignore settlement in the sand layer particularly since it’s a dense
sand. Note that approximately 10 mm of settlement occurs in the sand layer compared to
65 mm in the clay layer.
10.16 A 5.0 ft thick fill is to be placed on the soil profile as shown in Figure 10.3. A
consolidation test performed on a sample obtained from Point B produced the following
Compute the ultimate
results:
Cc = 0.27, Cr = 0.10, e0 = 1.09, σc′ = 760 lb/ft2.
consolidation settlement due to the weight of this fill and determine the ground surface
elevation after the consolidation is complete.
Note: The first layer in your analysis should extend from the original ground
surface to the groundwater table.
Solution
At sample B:
σ ′z 0 = ∑ γH − u = (98)(1.6 ) + (100)(16.4 ) − (62.4)(16.4) = 773 lb/ft 2
σ c′ = 760 lb/ft 2
σ z 0 ≈ σ c′ Therefore soil is normally consolidated
Proposed fill:
σ ′zf = σ ′z 0 + γ fill H fill = σ ′z 0 + (122)(5.0 ) = σ ′z 0 + 610 lb/ft 2
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10-16
Layer
1
2
3
4
5
Compressibility and Settlement Chap. 10
H
(ft)
1.6
4.0
6.0
10.0
10.0
zf
0.8
3.6
7.0
16.6
26.6
At midpoint of layer
σ'z0
Δσz
σ'zf
(psf) (psf) (psf) Cc/(1+e0)
78.0 610
688.0
0.129
232.0 610
842.0
0.129
420.0 610 1030.0
0.129
721.0 610 1331.0
0.129
1097.0 610 1707.0
0.129
Eqn.
10.23
10.23
10.23
10.23
10.23
Σ=
δc,ult
(ft)
0.20
0.29
0.30
0.34
0.25
1.38
δc,ult = 1.38 ft. Therefore, final ground surface elevation is 10.6 + 5 – 1.38 = 14.2 feet.
10.17 A 4.0 m thick fill is to be made of a soil with a Proctor maximum dry unit weight of
19.4 kN/m3 and an optimum moisture content of 13.0%. This fill will be compacted at
optimum moisture content to an average relative compaction of 92%. The underlying
soils are as shown in Figure 10.26. Consolidation tests were performed at Points A and B,
with the following results:
Sample
A
B
Cc
0.59
0.37
Cr
0.19
0.14
e0
1.90
1.21
σc′ (kPa)
75
100
The silty sand is normally consolidated. Using hand computations, determine the
ultimate consolidation settlement due to the weight of this fill.
Figure 10.26 Soil profile for Problems 10.17 and 10.30.
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Chap. 10 Compressibility and Settlement
10-17
Solution
At sample A:
σ ′z 0 = ∑ γH − u = (14.0)(3.0) + (17.5)(2.1) + (13.5)(4.2) − (9.8)(6.3) = 74 kPa
σ c′ = 75 kPa
σ z 0 ≈ σ c′ Therefore soil is normally consolidated
At sample B:
σ ′z 0 =
∑ γH − u = (14.0)(3.0) + (17.5)(2.1) + (13.5)(6.0)
+(15.0 )(2.0 ) − (9.8)(10.1)
= 91 kPa
σ c′ = 100 kPa
σ z 0 ≈ σ c′ Therefore soil is normally consolidated
Proposed fill:
γ f = C R γ d ,max (1 + w) = (0.92)(19.4)(1 + 0.13) = 20.2 kN/m 3
σ ′zf = σ ′z 0 + γ fill H fill = σ ′z 0 + (20.2)(4.0 ) = σ ′z 0 + 81 kPa
Layer
1
2
3
4
5
H
(m)
3.0
2.1
3.0
3.0
3.1
zf
1.5
4.1
6.6
9.6
12.7
At midpoint of layer
σ'z0
Δσz
σ'zf
(kPa) (kPa) (kPa)
21.0
81
102.0
50.0
81
131.0
64.0
81
145.0
75.0
81
156.0
88.0
81
169.0
Cc/(1+e0)
0.013
0.013
0.200
0.200
0.170
Eqn.
10.23
10.23
10.23
10.23
10.23
Σ=
δc,ult
(m)
0.03
0.01
0.21
0.19
0.15
0.59
δc,ult = 59 cm
10.18 The owner of the land shown in the profile in Example 10.5 has decided not to build the
proposed fill. Instead, the land will be used for farming. To provide irrigation water, a
series of shallow wells will be drilled into the sand, and these wells will cause the
groundwater table to drop to the bottom of the sand layer (i.e., 2.0 m below its current
position). Compute the ultimate consolidation settlement due to this drop in groundwater.
Do you think such a settlement will adversely affect the farming?
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10-18
Compressibility and Settlement Chap. 10
Figure 10.20 Soil profile for Example 10.5
Solution
Compute σ’z0 using the original groundwater table (same value as used in Example 10.5)
and σ’zf using the final groundwater table. The unit weight of the zone between the two
groundwater tables will drop from 19.5 kN/m3 to 18.5 kN/m3.
Layer
1
2
3
4
5
H (m)
1.5
2.0
3.0
3.0
4.0
zf
0.8
2.5
5.0
8.0
11.5
At midpoint of layer
σ'z0
Δ σz
σ'zf
(kPa)
(kPa)
(kPa)
13.9
0
13.9
37.4
8.8
46.2
56.4
17.6
74.0
75.0
17.6
92.6
96.7
17.6
114.3
Cc/(1+e0)
0.013
0.013
0.200
0.200
0.170
Eqn.
10.23
10.23
10.23
10.23
10.23
Σ=
δc,ult
(m)
0.00
0.00
0.07
0.05
0.05
0.18
δc,ult = 18 cm, this would not adversely affect farming.
10.19 A certain site is underlain by the soil profile shown in Figure 10.22 with the groundwater
table at the initial location. The groundwater table will remain at this location, but a
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Chap. 10 Compressibility and Settlement
10-19
20.0-ft deep fill with a unit weight of 119 lb/ft3 is to be placed. The only consolidation
data available is from a test conducted on a sample from Point B. The test results are as
follows: e0 = 1.22, Cc = 0.31, Cr = 0.09, σc′ = 3800 lb/ft2. Determine the ultimate
consolidation settlement due to the weight of this proposed fill. Assume the sands are
normally consolidated.
Figure 10.22 Soil profile for Example 10.7.
Solution
At sample B:
σ ′z 0 = ∑ γH − u
= (121)(7.0) + (125)(6.0) + (127)(18) + (110)(611) − (62.4)(35) = 2909 lb/ft2
σ m = σ c′ − σ ′z 0 = 3800 − 2909 = 891 lb/ft 2
σ ′zf = σ ′z 0 + γ fill H fill = σ ′z 0 + (119)(20) = σ ′z 0 + 2380 lb/ft 2
Using the computations in Example 10.7 with appropriate modifications:
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10-20
Layer
1
2
3
4
5
6
Compressibility and Settlement Chap. 10
H
(ft)
7.0
6.0
9.0
9.0
7.0
8.0
zf
3.5
10.0
10.5
26.5
34.5
42.0
At midpoint of layer
σ'z0
Δσz
σ'c
σ'zf
(psf)
(psf) (psf)
(psf) Cc/(1+e0) Cr/(1+e0)
423.0 2380
2803.0
0.011
1035.0 2380
3415.0
0.011
1513.0 2380
3893.0
0.011
2095.0 2380
4475.0
0.011
2552.0 2380 3443.0 4932.0
0.14
0.04
2909.0 2380 3800.0 5289.0
0.14
0.04
Σ=
δc,ult
(ft)
0.06
0.03
0.04
0.03
0.19
0.20
0.54
δc,ult = 0.54 ft = 6.5 in.
10.20 Using the data in Example 10.8, compute the consolidation settlement at the edge of the
tank. Then compute the differential settlement, which is the difference between the
settlement at the center and the edge.
Hint: Compute Δσz using Figure 9.13.
Solution
Use figure 9.13 to compute σz,induced
Layer
1
2
3
4
5
6
H
(m)
3.0
1.5
2.0
3.0
3.0
4.0
zf
1.5
3.8
2.5
8.0
11.0
14.5
At midpoint of layer
σ'z0
Δσz
σ'zf
(kPa) (kPa) (kPa)
28.8
51
79.8
71.5
42
113.5
95.0
41
136.0
114.0
39
153.0
132.6
34
166.6
154.3
29
183.3
Cc/(1+e0) Cr/(1+e0)
0.002
0.008
0.008
0.19
0.19
0.19
Σ=
δc,ult
(mm)
3
2
2
73
57
57
194
Round off to: δc,ult = 190 mm
Differential settlement = 320 -190 = 130 mm
Section 10.9 Secondary Compression Settlement
10.21 Using the data from Example 10.10, develop a plot of secondary compression settlement
vs. time for the period 40 to 100 years after completion of the fill. Is the rate of
secondary compression settlement increasing or decreasing with time?
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Chap. 10 Compressibility and Settlement
10-21
Figure 10.20 Soil profile for Example 10.5
Solution
Using Equation 10.28:
t (yr)
40
δs
(mm)
0
δs/yr
(mm/yr)
1.7
50
17
60
32
70
44
80
54
90
63
100
72
1.5
1.2
1.0
0.9
0.9
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10-22
Compressibility and Settlement Chap. 10
The rate of secondary compression settlement is slowly decreasing with time.
Section 10.12 Heaving Due to Unloading
10.22 Point C in Figure 10.20 was originally at elevation 12.00 m, but it dropped to elevation
11.52 m as a result of the consolidation settlement described in Example 10.5. Now that
the consolidation is complete, the fill is to be removed. Compute the new elevation of
Point C after the natural soils rebound in response to the fill removal. Ignore any
secondary compression settlement.
Solution
Clay: Cr/(1+e0) = 0.08/(1+1.10) = 0.038
Sand: Per page 393, Cr is about one-third of Cc, so Cr/(1+e0) = (1/3)(0.008) = 0.003
Layer
1
2
3
4
5
H (m)
1.5
2.0
3.0
3.0
4.0
zf
0.8
2.5
5.0
8.0
11.5
At midpoint of layer
Δσz
σ'z0
σ'zf
(kPa)
(kPa)
(kPa)
71.5
-57.6
13.9
95.0
-57.6
37.4
114.0
-57.6
56.4
132.0
-57.6
74.4
154.3
-57.6
96.7
Cr/(1+e0)
0.003
0.003
0.038
0.038
0.038
δc,ult
Eqn.
(mm)
10.24
-3.20
10.24
-2.43
10.24 -34.84
10.24 -28.39
10.24 -30.85
Σ= -99.70
Final elevation of Point C is 11.52 + 0.099 = 11.62 m
10.23 In Example 10.7 the total settlement due to the lowering of the groundwater table was
computed to be 4.6 in. It has been several years after the groundwater table was lowered
and consolidation is complete. Now the upper 30 ft of the soil profile is to be excavated
to create underground parking for a mid-rise building. How much will the base of the
excavation heave due to removal of the soil? If the total mass of the structure being
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Chap. 10 Compressibility and Settlement
10-23
constructed is 65 % of the mass of the soil removed, what will be the net settlement or
heave of the structure?
Figure 10.22 Soil profile for Example 10.7
Solution
Heaving due to removal of the soil
Layer
1
2
3
4
5
6
H (ft)
1.0
3.0
3.0
3.0
3.0
3.0
zf
30.5
32.5
34.5
38.5
41.5
44.5
At midpoint of layer
σ'z0
Δσz
σ'zf
(psf)
(psf)
(psf)
3632.9 -3600
32.9
3736.6 -3600
136.6
3831.8 -3600
231.8
4022.2 -3600
422.2
4165.0 -3600
565.0
4307.8 -3600
707.8
Cr/(1+e0)
0.0467
0.0467
0.0467
0.0467
0.0467
0.0467
Eqn.
10.24
10.24
10.24
10.24
10.24
10.24
Σ=
δc,ult
(ft)
-0.10
-0.20
-0.17
-0.14
-0.12
-0.11
-0.84
δheaving = 0.84 ft = 10 in.
If 65% of the weight of the soil removed is replaced by the building, then the net change
in vertical stress is 0.65(–3600) = –1260 lb/ft2.
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10-24
Layer
1
2
3
4
5
6
Compressibility and Settlement Chap. 10
H (ft)
1.0
3.0
3.0
3.0
3.0
3.0
zf
30.5
32.5
34.5
38.5
41.5
44.5
At midpoint of layer
σ'z0
Δσz
σ'zf
(psf)
(psf)
(psf)
3632.9
-1260 2372.9
3736.6
-1260 2476.6
3831.8
-1260 2571.8
4022.2
-1260 2762.2
4165.0
-1260 2905.0
4307.8
-1260 3047.8
Cr/(1+e0)
0.0467
0.0467
0.0467
0.0467
0.0467
0.0467
Eqn.
10.24
10.24
10.24
10.24
10.24
10.24
Σ=
δc,ult
(ft)
-0.01
-0.03
-0.02
-0.02
-0.02
-0.02
-0.12
There would be a net heave of 0.12 ft = 1.4 in.
Comprehensive
10.24 A road embankment is being placed across a shallow section of a bay. The existing
profile consists of 1 m of water over a 5-m thick normally consolidated clay soil which
overlies a very dense and stiff gravelly sand. A consolidation test on the clay generated
the following results: Cc = 0.21, e0 = 1.21. The embankment material is expected to be
place at a unit weight of 18.1 kN/m3. Determine the thickness of the embankment such
that the final elevation of the embankment is 2 m above the water level. This will require
an iterative solution.
Solution
Assume a unit weight of 15.0 kN/m3 for the clay per table 4.1
Through trial and error use 3.4 m:
H
Layer (m)
1
0.50
2
0.50
3
1.00
4
1.00
5
1.00
6
1.00
zf
1.25
1.75
2.50
3.50
4.50
5.50
At midpoint of layer
σ'z0
Δσz
σ'zf
(kPa) (kPa) (kPa)
1.3
53.9
55.2
3.9
53.9
57.8
7.8
53.9
61.7
13.0
53.9
66.9
18.2
53.9
72.1
23.4
53.9
77.3
Cc/(1+e0)
0.095
0.095
0.095
0.095
0.095
0.095
Eqn.
10.23
10.23
10.23
10.23
10.23
10.23
Σ=
δc,ult
(m)
0.077
0.056
0.085
0.068
0.057
0.049
0.392
Hfill = 3.4 m, this fill height will compensate for the settlement, to achieve the required
elevation of 2 ft above the water level.
10.25 Develop a spreadsheet that can compute one-dimensional consolidation settlement due to
the weight of a fill. This spreadsheet should be able to accommodate a fill of any unit
weight and thickness, underlain by multiple compressible soil strata. It also should be
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Chap. 10 Compressibility and Settlement
10-25
able to accommodate both normally consolidated and overconsolidated soils. Since the
computer does all of the computations, the spreadsheet should use at least 50 layers.
Once the spreadsheet is completed, use it to solve Examples 10.5 and 10.6. Submit
printouts of both analyses.
Solution
A variety of solutions are possible. The spreadsheet Settlement.xlsx contains an example
of one solution.
10.26 A cross-section through a tidal mud flat area is shown in Figure 10.24. This site is
adjacent to a bay, is subject to varying water levels according to tides, and is occasionally
submerged when heavy runoff from nearby rivers raises the elevation of the water in the
bay. For analysis purposes, use a groundwater table at the ground surface, as shown. A
crust has formed in the upper 0.8 m of soil due to dessication (drying) and is stiffer than
the underlying soil. This crust is overconsolidated case I, and the soils below are
normally consolidated. The proposed fill is required to protect the site from future
flooding, and thus permit construction of a commercial development. Use the
spreadsheet developed in Problem 10.17 to determine the ultimate consolidation
settlement due to the weight of this proposed fill. Then, consider the possibility that the
crust was not recognized in the site characterization program, and perform another
analysis using Cc /(1 + e0) = 0.20 and γ = 14.0 kN/m3 for the entire 5.0 m of clay.
Compare the results of these two analyses and comment on the importance of recognizing
the presence of crusts.
Figure 10.24
Typical crust near the ground surface in an otherwise normally
consolidated clay. el. = elevation.
Solution
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10-26
Compressibility and Settlement Chap. 10
Using a spreadsheet solution the settlement accounting for the desiccated crust is
approximately 640 mm. If the desiccated crust is ignored the computed settlement
increases to approximately 850 mm. Clearly it is important to include the desiccated
crust in the computation if one exists.
10.27 Explain the difference between normally consolidated soil and overconsolidated soils,
and give examples of geologic conditions that would form each type.
Solution
A normally consolidated soil is one for which the current vertical effective stress, σ´z0, is
the greatest vertical effective stress the soil has ever experienced. Example of such a soil
include: a) a lacustrine clay at the bottom of a lake, and b) a recently deposited alluvial
soil.
An overconsolidated soil is one which the current vertical effective stress is less
than the vertical which existed sometime in the past. Examples of this kind of soil
include: a) those that were once covered by a glacier that has since melted, and b) a soil
below an area where the overlying soils were excavated or removed by erosion.
10.28 What types of natural soils are best suited for testing in a consolidometer? Why? Which
are not well suited? Why?
Solution
Consolidation test results are very sensitive to the quality of the sample, so poor samples
do not produce good results. Soft and medium clays are best suited for testing in a
consolidometer because we can obtain high quality “undisturbed” samples at a reasonable
cost. Conversely, clean sands and gravels are difficult to sample, and thus are poor
candidates for consolidation tests.
10.29 According to the results from a consolidation test, the preconsolidation stress for a certain
soil sample is 850 lb/ft2. The in-situ vertical effective stress at the sample location is
797 lb/ft2, and the proposed load will cause σz to increase by 500 lb/ft2. Which equation
should be used to compute the consolidation settlement, 10.23, 10.24, or 10.25? Why?
Solution
In this case, σ´c is less than 20% greater than σ´z0, indicating that the soil is only slightly
overconsolidated. In this case, the soil also could be analyzed as being normally
consolidated using Equation 10.23. However, the stresses at this point satisfy the
conditions for overconsolidated case II (σ´z0 < σ´c < σ´zf), so Equation 10.25 could also
be used to compute the consolidation settlement.
10.30 A 3.0 m thick fill with a unit weight of 18.1 kN/m3 is to be placed on the soil profile
shown in Figure 10.26. Consolidation test results at Points A and B are as stated in
Problem 10.17, except that σc′ at point B is now 200 kPa. Using Equation 10.21 with Cc
or Cr as appropriate, develop a plot of vertical strain, εz vs. depth from the original
ground surface to the top of the glacial till. How does this curve vary within a given soil
stratum? Why? Does it suddenly change at the strata interfaces? Why?
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Chap. 10 Compressibility and Settlement
10-27
Solution
The soil represented by the sample from point B is now overconsolidated Case I
At midpoint of layer
σ'z0
Δσz
σ'zf
(kPa) (kPa) (kPa) Cc/(1+e0) Cr/(1+e0)
Layer
zf
0.0
0.0 54.3 54.3
0.013
1.0
14.0 54.3 68.3
0.013
Silty
2.0
28.0 54.3 82.3
0.013
Sand
3.0
42.0 54.3 96.3
0.013
Saturated 3.0
42.0 54.3 96.3
0.013
Silty
4.0
50.0 54.3 104.3
0.013
Sand
5.1
58.0 54.3 112.3
0.013
5.1
58.0 54.3 112.3
0.200
6.6
64.0 54.3 118.3
0.200
Soft
8.1
69.0 54.3 123.3
0.200
Clay
9.6
75.0 54.3 129.3
0.200
11.1 80.0 54.3 134.3
0.200
0.06
Medium 11.1 80.0 54.3 134.3
Clay
14.2 96.0 54.3 150.3
0.06
Eqn.
10.23
10.23
10.23
10.23
10.23
10.23
10.23
10.23
10.23
10.23
10.23
10.23
10.24
10.24
δc,ult
(m)
0.0000
0.0089
0.0122
0.0141
0.0141
0.0166
0.0190
0.2927
0.3522
0.4084
0.4542
0.4995
0.1498
0.1659
εz
0.000
0.009
0.006
0.005
0.005
0.004
0.004
0.057
0.053
0.050
0.047
0.045
0.013
0.012
εz
0.000
0.050
0.100
0.0
2.0
4.0
z (m)
6.0
8.0
10.0
12.0
14.0
16.0
The strain decreases with depth within any given soil strata, because the ratio σ´zf/ σ´z0
decreases with depth. However, the strain suddenly changes at the strata interfaces
because the compressibility changes there.
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10-28
Compressibility and Settlement Chap. 10
10.31 Considering the variation of strain with depth, as found in Problem 10.30, does a 1-m
thick layer near the top of a stratum contribute more or less to the consolidation
settlement than a 1-m thick stratum near the bottom? Explain. Does this finding support
the statement in Section 10.10 that “The presence of crusts has a significant impact on
settlement computations, even if they are much thinner than the underlying compressible
soils?” Explain.
Solution
A 1 m thick layer near the top of a stratum contributes more to the consolidation
settlement than does a 1 m thick layer near the bottom of the same strata because the
strain is greater near the top. Thus, the compressibility near the ground surface is
especially significant. If a crust is present, its compressibility will be less than that of the
underlying soils, so the resulting consolidation will be correspondingly less. Therefore,
this finding does support the statement in Section 10.10.
10.32 A shopping center is to be built on a site adjacent to a tidal mud flat. The ground surface
elevation is +0.2 m, and the groundwater table is at the ground surface. The underlying
soils consist of 7.3 m of medium clay with Cc /(1+e0 ) = 0.18, Cr /(1+e0 ) = 0.06, σm′ = 0,
and γ = 15.1 kN/m3. The clay stratum is underlain by relatively incompressible stiff soils.
In order to provide sufficient flood control protection, a fill must be placed on this
site before the shopping center is built, thus maintaining the entire site above the highest
flood level. This fill will have a unit weight of 19.0 lb/ft3. According to a hydrologic
study, the fill must be thick enough so that the ground surface elevation is at least +1.8 m
after all of the consolidation settlement is complete. Use the spreadsheet developed in
Problem 10.17 to determine the required ground surface elevation at the end of
construction. Assume all of the settlement occurs after construction.
Solution
This requires an iterative process, through trial and error, try H = 2.5m
Layer
1
2
3
4
5
6
H
(m)
1.30
1.25
1.25
1.25
1.25
1.00
zf
(m)
0.7
1.9
3.2
4.4
5.7
6.8
At midpoint of layer
σ'z0 Δ σz
σ'zf
(kPa) (kPa) (kPa) Cc/(1+e0) Cr/(1+e0)
3.4 47.5 50.9
0.18
0.06
10.2 47.5 57.7
0.18
0.06
9.9 47.5 57.4
0.18
0.06
23.5 47.5 71.0
0.18
0.06
30.1 47.5 77.6
0.18
0.06
36.0 47.5 83.5
0.18
0.06
Eqn.
10.23
10.23
10.23
10.23
10.23
10.23
Σ=
δc,ult
(m)
0.274
0.169
0.131
0.108
0.093
0.066
0.841
The total amount of fill needed is H = 2.5, Elev = (+0.2) + (+2.5) - .9 = (+1.8)
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Chap. 10 Compressibility and Settlement
10-29
10.33 A consolidation test has been performed on a sample of lodgement till from a region that
was once covered with a glacier. The current vertical effective stress at the sample
location is 1800 lb/ft2 and the measured preconsolidation stress is 32,500 lb/ft2.
(a) Assuming the glacier was in place long enough for complete consolidation to occur,
and assuming the ground surface and groundwater table elevations have remained
unchanged, compute the maximum thickness of the glacier. The specific gravity of
glacial ice is about 0.87.
(b) Although glacial ice was present for a very long time, it also extended over very large
areas, so the required drainage distance for the excess pore water was very long. As a
result, all of the excess pore water pressures may not have dissipated (Chung and
Finno, 1992). Therefore, our assumption that complete consolidation occurred may
not be accurate. If so, would our computed thickness be too large or too small?
Explain.
Solution
(a)
σ m′ = σ c′ − σ ′z 0 = 32,500 − 1,800 = 30,700 lb/ft 2
H ice =
σ m′
σ′
30,700
= m =
= 570 ft
γ ice G s γ w (0.87 )(62.4)
(b) If all the excess pore water pressures had not dissipated while the glacier was present,
then the vertical effective stress did not reflect the entire weight of the ice. Thus, the
value of σ’c is not as high as it would have been if all the excess pore water pressures
had dissipated. Therefore, our computed value of Hf is too small.
10.34 A highway is to be built across a wetlands with the soil profile shown in Figure 10.27
below. These wetlands are subject to flooding, so a fill must be placed to keep the
pavement above the highest flood level. According to a hydrologic analysis, the roadway
must be at elevation 7.0 ft or higher to satisfy this requirement. Sandy fill material that
has a compacted unit weight of 122 lb/ft3 is available from a nearby borrow site.
A subsurface exploration program has been completed at this site, and laboratory
tests have been performed. The results of this program are tabulated below:
Moisture
Dry unit
content
Depth
weight
σc′
3
(%)
Cc/(1+e0) Cr/(1+e0) (lb/ft2)
(ft)
(lb/ft )
2.0
95
28.6
0.13
0.06
3000
7.5
89
33.0
0.16
0.06
550
13.0
92
30.5
0.12
0.05
850
24.0
93
29.9
0.14
0.07
4800
All depths are measured from the original ground surface.
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10-30
Compressibility and Settlement Chap. 10
Figure 10.27 Soil profile for Problem 10.34. el. = elevation.
The natural soils will settle under the weight of the proposed fill. Approximately
25 years will pass before this settlement is complete. Therefore, the road must be built at
an elevation higher than 7.0 ft so that after the settlement is complete it is at 7.0 ft. The
pavement thickness is 0.5 ft, so the top of the fill must remain at or above elevation 6.5 ft.
(a) Use the spreadsheet developed in Problem 10.25 to determine the required elevation
of the roadway immediately after construction. Assume that no settlement occurs
during construction, and the pavement has the same unit weight as the fill.
Hint: As the fill thickness becomes greater, the settlement increases. Thus, this
problem requires a trial-and-error solution. You will need to estimate the required fill
thickness, then compute the settlement and final roadway elevation. Try to have one
trial that produces a road elevation that is too high, and another that produces one too
low. Then interpolate to find the required fill thickness.
Note 1: As the fill settles, the lower portion will become submerged below the
groundwater table, so σ´zf will be less than predicted by Equation 10.2. However, we
will ignore this effect.
Note 2: Laboratory tests have been performed on two samples of the soft clay.
Combine these two sets of test results, then assign γ, Cc/(1+e0), Cr/(1+e0), and σ´m
values that apply to the entire stratum.
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Chap. 10 Compressibility and Settlement
10-31
Note 3: We do not have any unit weight data for the portion of the crust above the
groundwater table. Therefore, assume it is the same as that below the groundwater
table. In this case, this assumption should introduce very little error.
(b) Is our assumption regarding the submergence of the fill (per note 1 in part a)
conservative or unconservative? Is this a reasonable assumption? Explain.
Solution
(a)
@ 2.0 feet
σ ′z 0 =
∑ γH − u
= (95)(2 ) − (62.4 )(1.5) = 95.4 lb/ft 2
σ m′ = σ c′ − σ ′z 0 = 3000 − 96 .4 = 2903.6 lb/ft 2
Average soft clay values:
γ = 90.5lb / ft 3
Cc
= 0.14
1 + e0
Cr
= 0.055
1 + e0
σ c' = 700 lb/ft 2
Assume average values occur @ 10.25 feet
σ ′z 0 =
∑ γH − u
= (90.5)(3.25) + (95)(7 ) − (62.4 )(9.75) = 350.7 lb/ft 2
σ m′ = σ c′ − σ ′z 0 = 700 − 350 .7 = 349.3 lb/ft 2
@ 24.0 feet
σ ′z 0 =
∑ γH − u
= (95)(7 ) + (90.5)(16 ) + (93)(3) − (62.4 )(23.5) = 925.6 lb/ft 2
σ m′ = σ c′ − σ ′z 0 = 4800 − 925 .6 = 3874 .4 lb/ft 2
Through trial error use 7.25 ft
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10-32
Layer
1
2
3
4
5
6
7
Compressibility and Settlement Chap. 10
H
(ft)
3.5
3.5
5.0
6.0
5.0
4.0
4.0
zf
(ft)
1.75
5.25
9.50
15.0
20.5
25.0
29.0
σ'z0
(psf)
88.3
202.4
329.7
484.2
638.8
1029.6
1152.0
At midpoint of layer
Δ σz
σ'c
σ'zf
(psf)
(psf)
(psf)
884.5 2991.9
972.8
884.5 3106.0 1086.9
884.5
679.0 1214.2
884.5
833.5 1368.7
884.5
988.1 1523.3
884.5 4904.0 1914.1
884.5 5026.4 2036.5
Cc/(1+e0) Cr/(1+e0)
0.13
0.06
0.13
0.06
0.14
0.055
0.14
0.055
0.14
0.055
0.14
0.07
0.14
0.07
Σ=
δc,ult
(ft)
0.22
0.15
0.26
0.26
0.18
0.08
0.07
1.22
Hfill = 7.25 ft, this fill height will compensate for the settlement, to achieve the
required elevation of +7.0 ft or higher. Therefore, the roadway elevation immediately
after construction should be 1.0 + 7.25 = 8.25 ft.
(b) This assumption is conservative, because the actual value of σ´zf will be less than that
used in the analysis
10.35 An engineer has suggested an alternative design for the proposed highway in Problem
10.34. This design consists of using geofoam for the lower part of the fill, as shown in
Figure 6.45. It will be covered with at least 1.0 ft of soil to provide a buffer between the
pavement and the geofoam. Compute the minimum required geofoam thickness so that
the roadway will always be at elevation 7.0 or higher.
Hint: The geofoam is an extra “hidden” layer between the fill and the natural
ground surface with γ = 0.
Solution
Through trial and error, use H=1.1 ft
Layer
1
2
3
4
5
6
7
H
(ft)
3.5
3.5
5.0
6.0
5.0
4.0
4.0
zf
1.75
5.25
9.50
15.00
20.50
25.00
29.00
At midpoint of layer
σ'z0
Δσz
σ'c
σ'zf
(psf) (psf)
(psf)
(psf)
88.3 134.2 2991.9 222.5
202.4 134.2 3106.0 336.6
329.7 134.2 679.0 463.9
484.2 134.2 833.5 618.4
638.8 134.2 988.1 773.0
1029.6 134.2 4904.0 1163.8
1152.0 134.2 5026.4 1286.2
δc,ult
Cc/(1+e0) Cr/(1+e0)
(ft)
0.13
0.06
0.08
0.13
0.06
0.05
0.14
0.055
-0.03
0.14
0.055
-0.03
0.14
0.055
-0.02
0.14
0.07
0.01
0.14
0.07
0.01
Σ= 0.08
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Chap. 10 Compressibility and Settlement
10-33
The required minimum fill of Hfill = 1.1 ft is needed, this fill height will compensate for
the settlement to achieve the required elevation of +7.0 ft or higher, with a combination
of 5 ft of geofoam. Therefore, the roadway elevation immediately after construction
should be 1.0 + 1.1 + 5.0 = 7.1 ft.
10.36 A series of prefabricated dual-bore steel tubes similar to the one in Figure 10.28 are to be
installed in an underwater trench to form a tunnel. The trench will be in seawater, which
has a unit weight of 64.0 lb/ft3. The tubes will be floated into position, and sunk into
place by temporarily flooding the interior. Then, non-structural concrete will be placed
into chambers along the tube to act as ballast, and the inside will be pumped dry. The
completed tube will be 80 ft wide, 300 ft long, and 40 ft tall, and weigh 32,000 tons
exclusive of buoyant forces. Finally, the tube will be covered with soil, producing the
cross-section shown in Figure 10.29.
Figure 10.28 This prefabricated tunnel section is part of the Central Artery Project in
Boston. It was floated into position, and then sunk to the bottom of the bay. (Photograph
by Peter Vanderwarker, courtesy of the Central Artery/Tunnel Project.)
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10-34
Compressibility and Settlement Chap. 10
Figure 10.29 Final cross-section for underwater tunnel as described in Problem 10.36.
(a) The interior of the tube will be dewatered after the concrete is placed, but before the
trench is backfilled. Once this is done, will the tube remain at the bottom of the
trench, or will it float up to the water surface?
(b) After the trench is backfilled, what will be the net Δσz in the soft clay? Assume Δσz is
constant with depth.
(c) Using the final cross-section, compute the ultimate consolidation settlement or heave
of the tube due to Δσz′ in the soft clay. Assume no heave occurs during construction.
(d) The weakest parts of the completed tunnel will be the connections between the tube
sections. In order to avoid excessive flexural stresses at these connections, the
structural engineer has specified a maximum allowable differential settlement or
differential heave of 5 in along the length of the tube (the term “allowable” indicates
this value already includes a factor of safety). An evaluation of the soil profile
suggests the differential settlement or heave will be no more than 50 percent of the
total. Has the structural engineer’s criteria been met?
Solution
(a)
32,000 ton =
γ =
(b)
2,000 lb
= 64,000,000 lb
1 ton
64,000,000
P
=
= 66.7 lb/ft 3
V (80)(40)(300)
The tube will remain at the bottom of the trench because γw< γtunnel.
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Chap. 10 Compressibility and Settlement
10-35
Δσ ′z, net = σ soil − σ tunnel
= γH − γ tunnel H tunnel
= (100 − 64.0 )(40 ) − (66.7 − 64 )(40 )
= −1332 lb/ft 2
(c)
H
Layer (ft)
1
5.0
2
5.0
3
5.0
4
5.0
5
5.0
6
5.0
zf
97.5
102.5
107.5
112.5
117.5
122.5
At midpoint of layer
σ'z0
Δ σz
σ'zf
(psf)
(psf)
(psf) Cc/(1+e0) Cr/(1+e0)
2250.0 -1332
918.0
0.19
0.06
2430.0 -1332 1098.0
0.19
0.06
2610.0 -1332 1278.0
0.19
0.06
2790.0 -1332 1458.0
0.19
0.06
2970.0 -1332 1638.0
0.19
0.06
3150.0 -1332 1818.0
0.19
0.06
Eqn.
10.24
10.24
10.24
10.24
10.24
10.24
Σ=
δc,ult
(ft)
-0.11
-0.09
-0.09
-0.08
-0.07
-0.07
-0.50
δheave = 0.50 ft = 6 inches
(d) Yes, the structural engineer’s criteria have been met because the differential will be 3
inches.
10.37 A proposed building is to have three levels of underground parking, as shown in Figure
10.30. To construct this building, it will be necessary to make a 10.0 m excavation,
which will need to be temporarily dewatered. The natural and dewatered groundwater
tables are as shown, and the medium clay is normally consolidated. The chief
geotechnical engineer is concerned that this dewatering operation may cause excessive
differential settlements in the adjacent building and has asked you to compute the
anticipated differential settlement across the width of this building. Assume the wall is
perfectly rigid, and thus does not contribute to any settlement problems, and that the
maximum allowable differential settlement from one side of the building to the opposite
side is 50 mm. Neglect any loss in σz′ below the existing building due to the removal of
soil from the excavation. Discuss the implications of your answer.
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10-36
Compressibility and Settlement Chap. 10
Figure 10.30 Soil profile for Problem 10.37.
Solution
Near side:
Layer
1
2
3
4
5
H
(m)
2.00
2.00
2.00
2.00
2.30
zf
13.50
15.50
17.50
19.50
21.65
At midpoint of layer
σ'z0
Δ σz
σ'zf
(psf)
(kPa)
(kPa)
148.2
87.55
235.7
161.6
87.55
249.1
175.0
87.55
262.5
188.4
87.55
275.9
202.8
87.55
290.3
Cc/(1+e0)
0.26
0.26
0.26
0.26
0.26
Σ=
δc,ult
(m)
0.105
0.098
0.092
0.086
0.093
0.474
Far side:
H
Layer (m)
1
2.00
2
2.00
3
2.00
4
2.00
5
2.30
zf
13.50
15.50
17.50
19.50
21.65
At midpoint of layer
σ'z0
Δσz
σ'zf
(kPa) (psf) (kPa) Cc/(1+e0)
148.2 36.05 184.2
0.26
161.6 36.05 197.6
0.26
175.0 36.05 211.0
0.26
188.4 36.05 224.4
0.26
202.8 36.05 238.8
0.26
Σ=
δc,ult
(m)
0.049
0.045
0.042
0.040
0.043
0.219
Differential settlement across building = 0.47-0.22 = 0.25 m = 250 mm
This is well in excess of the allowable differential settlement, and therefore is
unacceptable
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Chap. 10 Compressibility and Settlement
10-37
10.38 The Palacio de las Bellas Artes in Mexico City, shown in Figure 10.1, is an interesting
example of large consolidation settlement. It is supported on a 1.8 to 3.0 m thick mat
foundation which is approximately 65 m wide and 115 m long. The average bearing
pressure between the bottom of this mat and the supporting soil is 115 kPa (Ledesma,
1936).
The soil conditions beneath the palace are too complex to describe in detail here.
However, we can conduct an approximate analysis using the following simplified profile:
Cc
Depth
γ
Description
3
1 + e0
(m)
(kN/m )
0-5
Sandy fill
17.5
0
5 - 45
Normally consolidated soft clays
11.5
0.53
> 45
Stiff soils
0
For our simplified analysis, use a groundwater table at a depth of 5 m, and assume
the bottom of the mat is at the original ground surface. In addition, assume the fill has
been in pace for a very long time, so the consolidation settlement due to the weight of the
fill is complete.
Divide the soft clay zone beneath the center of the building into five layers of
equal thickness. Then, compute Δσz at the midpoint of each layer using the methods
described in Chapter 9. Finally, compute the ultimate consolidation settlement beneath
the center of the palace due to its own weight.
Solution
Layer
1
2
3
4
5
H
(m)
8.0
8.0
8.0
8.0
8.0
zf
4.0
12.0
20.0
28.0
36.0
At midpoint of layer
σ'z0
σz
σ'zf
(kPa) (kPa) (kPa)
94.0
114
208.0
108.0
109
217.0
121.0
101
222.0
135.0
90
225.0
149.0
80
229.0
Cc/(1+e0)
0.53
0.53
0.53
0.53
0.53
Eqn.
10.23
10.23
10.23
10.23
10.23
Σ=
δc,ult
(m)
1.46
1.28
1.12
0.94
0.79
5.60
δc,ult = 5.6 m
Commentary: This is an extremely large settlement. It is due to the wide, heavy load and
the very soft soil.
10.39 A fill is to be placed at a proposed construction site, and you need to determine the
ultimate consolidation settlement due to its weight. Write a 200–300 word essay
describing the kinds of field exploration, soil sampling, and laboratory testing you will
need to perform to generate the information needed for this analysis. Your essay should
describe specific things that need to be done, and what information will be gained from
each activity.
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10-38
Compressibility and Settlement Chap. 10
Solution
The field exploration will need to include drilling exploratory borings at the site of the
proposed fill. These borings should extend through any soils that are soft or loose enough
to produce significant settlement, and into firm underlying strata. High quality
undisturbed samples need to be obtained in all clay and silt strata, and appropriate in-situ
test, such as the standard penetration test, in sandy strata. The undisturbed samples need
to be brought to a soil mechanics laboratory where their moisture content and dry unit
weight would be measured, and consolidation test performed.
The field exploration program also must determine the location of the
groundwater table. This will probably be done by installing observation wells in one or
more of the exploratory borings, and monitoring these wells.
The laboratory consolidation test data will be used to determine Cc, Cr, and σ´m
for each clay or silt stratum, and the SPT test results will be used to determine the relative
density and compressibility of the sand strata. The moisture content and unit weight tests
are used to determine w, γ, and e.
The laboratory test program also must include a Proctor compaction test on the
soils to be used for the fill. This is needed to determine the unit weight of the fill.
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