0132497468-Ch10_ISM.pdf

Stress Distribution and Settlement Analysis

Chapter 10

CHAPTER 10 STRESS STRESS DISTRIBUTION DISTRIBUTION AND A ND SETTLEMENT ANALYSIS ANA LYSIS

10-1. Compare the stress distribution with depth for (a) a point load of 1200 kN and (b) a 1200 kN load applied over an area of 3 x3 m. Plot the results. results. 10-2. If you used the Boussinesq (or Westergaard) theory for Problem 10.1, do the problem again but use the Westergaard (or Boussinesq) theory instead. Comment on the differences between the two theories. theories. SOLUTION: Rec t an g u l ar L o ad Co rn rn er er

Po i n t L o ad

Cen te ter 2

Depth, z (m )

z

Q/z

z

(k Pa)

(k Pa)

NBouss

Nwest

z -

Bouss (k Pa)

(k Pa)

115.02

z -

West (k Pa)

1

28.76

1200.00

0.477

0.318

57 5 72.96

381.97

5

4.98

19.92

48 4 8.00

0.477

0.318

22 22.92

15.28

10

1.38

5.52

12 1 2.00

0.477

0.318

5. 5.73

3.82

15

0.63

2.50

5.33

0.477

0.318

2. 2.55

1.70

20

0.35

1.42

3.00

0.477

0.318

1. 1.43

0.95

25

0.23

0.91

1.92

0.477

0.318

0. 0.92

0.61

30

0.16

0.63

1.33

0.477

0.318

0. 0.64

0.42

40

0.09

0.36

0.75

0.477

0.318

0. 0.36

0.24

45

0.07

0.28

0.59

0.477

0.318

0. 0.28

0.19

50

0.06

0.23

0.48

0.477

0.318

0. 0.23

0.15

Stress Increase (kPa) 0

10

20

30

40

50

0 10

) m20 ( z , 30 h t p e 40 D

Boussinesq Westergaard Rectangle Center

50 60

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Chapter 10

10-3. Compute the data and draw a curve of σ z/Q versus depth for points directly below a point load Q. On the same plot draw curves of σ z/Q versus depth for points directly below the center of square footings with breadths of 6.5 m and 20 m, respectively, each carrying a uniformly distributed load Q. On the basis of this plot, make a statement relative to the range within which loaded areas may be considered to act as po int loads. (After Taylor, 1948.) SOLUTION: As can be viewed in the plot, at a depth of about two times the loaded area, the σz value is not significantly different than the point load solution. (A Q value of 1000 was was used in this solution.)

6.5 m x 6.5 m L o ad Co r n er Depth, z (m )

Po i n t L o ad

20 m x 20 m L o ad

Cen t er

z

Co r n er Q/z2

z

(k Pa)

(k Pa)

1

6.28

25.10

NBouss

Nwest

z -

Bouss (k Pa)

(k Pa) 1000.00

0.477

0.318

z -

West (k Pa)

Cen t er

z

z

(k Pa)

(k Pa)

47 477.46

318.31

0.62

2.50

5

3.01

12.04

40 40.00

0.477

0.318

19 19.10

12.73

0.58

2.32

10

1.10

4.39

10 10.00

0.477

0.318

4. 4.77

3.18

0.44

1.75

15

0.53

2.13

4.44

0.477

0.318

2. 2.12

1.41

0.30

1.21

20

0.31

1.24

2.50

0.477

0.318

1. 1.19

0.80

0.21

0.84

25

0.20

0.80

1.60

0.477

0.318

0. 0.76

0.51

0.15

0.60

30

0.14

0.56

1.11

0.477

0.318

0. 0.53

0.35

0.11

0.45

40

0.08

0.32

0.63

0.477

0.318

0. 0.30

0.20

0.07

0.27

45

0.06

0.25

0.49

0.477

0.318

0. 0.24

0.16

0.05

0.22

50

0.05

0.21

0.40

0.477

0.318

0. 0.19

0.13

0.04

0.18

Stress Increase (kPa) 0

10

20

30

40

50

0 10

) m20 ( z , 30 h t p e 40 D

Boussinesq Westergaard 6.5m x 6.5m square load 20m x 20m square load

50 60



Chapter 10

10-4. The center of a rectangular area at ground surface has Cartesian coordinates (0, 0), and the corners have coordinates (7, 18). All dimensions are in meters. The area carries a uniform pressure of 150 kPa. Estimate the stresses at a depth of 20 m below ground surface at each of the following locations using the Boussinesq approach: (0, 0), (0, 18), (7, 0), (7, 18), and (12, 28). SOLUTION: (12,28)

+

qo = 150 kPa and z = 20 m

(0, 18)

(7, 18)

+

Determine the stress increase using Fig. 10.4 (or Eq. 10.6) for the vertical stress under the corner of a uniformly loaded rectangular area. Use superposition as necessary. (Influence values presented below were determined using the Boussinesq solution as given by Eq. 10.6.)

+(0, 0) + (7,0)

(a) (0,0) x = 7, y = 9

→ I = 0.0592 (multiply I by 4)

σz = 4qoI = 35.49 kPa (b) (0,18) x = 7, y = 36


σz = 2qoI = 30.21 kPa (c) (7,0) x = 14, y = 18


σz = 2qoI = 26.25 kPa (d) (7,18) x = 14, y = 36

→ I = 0.1672 (multiply by 1)

σz = qoI = 25.08 kPa (e) (12,28) (x1 = 36, y1 = 19),

σz = qo

(x 2

∑ I = q (I − I i

0

1

2

= 19, y2 = 10), (x3 = 46, y3 = 5), (x 4 = 5, y 4 = 10) − I3 + I4 ) = (150 kPa) × (0.1942 − 0.1181− 0.0756 + 0.0475)

σz = 7.2 kPa



Chapter 10

10-5. Compare the results of Problem 10.4 with those of the 2:1 method. Comments? SOLUTION: (Eq.10.2)

σz =

qoBL (B + z)(L + z)

=

(150 kPa)(14 m)(36 m) (36 + 20)(14 + 20) m2

σz = 39.7 kPa



Chapter 10

10-6. Calculate the stress distribution with depth at a point 3.5 m from the corner (along the longest side) of a rectangularly loaded area 15 by 35 m with a uniform load of 75 kPa.

+ (15, 38.5) qo = 75 kPa and z varies Determine the stress increase using Fig. 10.4 (or Eq. 10.6) for the vertical stress under the corner of a uniformly loaded rectangular area. Use superposition as necessary. (Influence values presented below were determined using the Boussinesq solution as given by Eq. 10.6.)

35.0 m.

+

(0, 0)

SOLUTION: Boussinesq

Depth

1

2

2:1 Method

v

(m)

v

(kPa)

(kPa)

1

0.2500

0.2477

0.17

68.36

5

0.2464

0.1711

5.65

49.22

10

0.2292

0.0989

9.78

35.00

15

0.2026

0.0632

10.45

26.25

35

0.1100

0.0177

6.92

11.25

50

0.0718

0.0093

4.69

7.13

100

0.0242

0.0025

1.63

2.54

Boussinesq x = 38.5, y = 15

→ I1 x = 3.5, y = 15 → I2 σz = qo (I1 − I2 ) (see tabulated results) (2: 1 Method, Eq.10.2)

σz =

qoBL (B + z)(L + z)

=

(75 kPa)(15 m)(35 m) (35 + z)(15 + z) m2

Stress Increase (kPa) 0.00 0 ) m ( h t p e D

20.00

40.00

60.00

80.00

20 40 60 80 100

Boussinesq 2:1 Method

120



Chapter 10

10-7. How far apart must two 18 m diameter tanks be placed such that their stress overlap is not greater than 10% of the contact stress at depths of 10, 20, and 30 m? SOLUTION: Use Fig. 10.5. Determine x for I values of 5% At z = 10 m: z At z = 20 m: z At z = 30 m: z

r

= 1.11, x r = 2, x = 18 m

r

= 2.22, x r = 2.4, x = 21.6 m

r

= 3.33, x r = 2.3, x = 20.7 m

10-9. Work Example 10.5, using superposition of the results of Figs. 10.7 and 10.4. How does your answer compare with the solution for Example 10.5?

SOLUTION: Scan fig from p.471 Fig. 10.7 − Corner of triangular load: Assume L is very large in comparison to z; thus, m For z = 3 m :

m = 10,

n = 2, I = 0.179

For z = 6 m :

m = 10,

n = 1, I = 0.125

= 10 + .

Fig. 10.4 − Corner of rectangular loaded area: Assume y is very large in comparison to x; thus, n = 10 + . For z = 3 m :

m = 1.67,

n = 10, I = 0.234

For z = 6 m :

m = 0.83,

n = 10, I = 0.189

Apply superposition: For z = 3 m :

σz = qo (I1 + I2 ) = 2 × (59 kPa)(0.179 + 0.234) = 48.7 kPa

For z = 6 m :

σz = qo (I1 + I2 ) = 2 × (59 kPa)(0.125 + 0.189) = 37.1 kPa



Chapter 10

10-10. Given the data of Example 10.6. Instead of a load on the surface, compute the depth of an excavation to cause a reduction in stress a t the bottom of the excavation of 200 kPa if ρ = 2.1 3 Mg/m . The excavation plan area is shown in Fig. Ex. 10.6a.

SOLUTION: Find z for

Δσ = 200 kPa at point O’ using the Boussinesq method (Fig. 10.4 or Eq. 10.6)

ρ = 2.1 Mg

3;

m

(

γ = 2.1 Mg

⎛ 9.81 m ⎞ = 20.6 kN ⎜ ⎟ ) m ⎝ s ⎠ m 3

2

3

γ × z = Δσ v 20.6 kN

× z = −200 kPa m3 z = 9.71m excavation

Now, determine

σz 9.71 m below point O', for qo = −200 kPa using superposition.

Find σz at z = 9.71 m, for q o = -200 kPa. (Influence values presented in the table below were determined using the Boussinesq solution as gi ven by Eq. 10.6.) Rectangle

x

y

I

1

60

100

0.25

2

60

20

0.24

3

10

20

0.203

4

10

40

0.206

i

0.899

σz = qo (

∑I =

∑ I ) = (−200 kPa)(0.899) = −179.8 kPa 1



Chapter 10

10-11. For the excavation of Problem 10.10, estimate the stress change at a depth of 50 m below the bottom of the excavation at point O’. qo = -200 kPa, z = 50 m Determine the stress decrease using Fig. 10.4 (or Eq. 10.6) for the vertical stress under the corner of a uniformly loaded rectangular area. Use superposition by adding I values for the 4 rectangular areas as tabulated below. SOLUTION: Find σz at z = 50 m, for q o = -200 kPa. (Influence values presented in the table below were determined using the Boussinesq solution as gi ven by Eq. 10.6.) Rectangle

x

y

I

1

60

100

0.212

2

60

20

0.106

3

10

20

0.033

4

10

40

0.050

i

0.401

σz = qo (

∑I =

∑ I ) = (−200 kPa)(0.401) = −80.2 kPa 1



Chapter 10

10-13. A strip footing 2.5 m wide is loaded on the ground surface with a pressure equal to 175 kPa. Calculate the stress distribution at depths of 2.5, 7.5 and 12.5 m under the center of the footing. If the footing rested on a normally consolidated cohesive layer whose LL was 78 and whose PL was 47, estimate the settlement of the footing. Assume wn = 50%, S = 100%, γ’ = 7.5 3 kN/m , and the total clay l ayer thickness beneath the footing = 15 m. SOLUTION: Find σz for qo = 175 kPa. Use ½ the footing width and multiply by 4 (superposition) to determine the maximum value of σz at the midpoint of the strip footing. Use this value for settlement determinations. (Influence values presented below were determined using the Boussinesq solution as given by Eq. 10.6.) Assume x is very large in comparison to y; thus, m = ∞. For z = 2.5 m :

n=

1.25

= 1.0, I = 0.137, σ z − 2.5 = qo × I = 4(175 kPa)(0.137) = 95.9 kPa 2.5 1.25 For z = 7.5 m : n = = 0.167, I = 0.052, σz −5 = qo × I = 4(175 kPa)(0.052) = 36.4 kPa 7.5 1.25 For z = 12.5 m : n = = 0.10, I = 0.032, σz −10 = qo × I = 4(175 kPa)(0.032) = 22.4 kPa 12.5 Estimate Cc using Eq. 8.28: C c Estiamte eo . Assume w n

= 0.009(LL − 10) = (0.009)(78 − 10) = 0.612

≈ 50%, S ≈ 100%, and Gs ≈ 2.7 → e =

Gs w S

=

(2.7)(50) (100)

= 1.35

Determine sc by dividing clay profile into 3 layers with centers at z = 2.5, 5, and 10 m. Use Eq. 10.15 for NC clay: s c

=

Cc

n

∑ H log σ +σΔσ

1 + eo i =1

i

i

i

i

⎡ ⎛ 18.8 + 95.9 ⎞ ⎛ 56.3 + 36.4 ⎞ ⎛ 93.8 + 22.4 ⎞ ⎤ 5 × log ⎜ + 5 × log ⎜ + 5 × log ⎜ ⎢ ⎟ ⎟ ⎟⎥ 1 + 1.35 ⎣ ⎝ 18.8 ⎠ ⎝ 56.3 ⎠ ⎝ 93.8 ⎠ ⎦ sc = (0.260) ⎡⎣3.927 + 1.083 + 0.465⎤⎦ sc = 1.42 m sc

=

0.612



Chapter 10

10-16. A large oil storage tank 90 m in diameter is to be constructed on the soil profile shown in Fig. P10.16. Average depth of the oil in the tank is 18 m, and the specific gravity of the oil is 0.92. Consolidation tests from the clay layer are similar to those given in Problem 8.18. Estimate the maximum total and differential consolidation settlement of the tank. Neglect any settlements in the sand. Work this problem: (a) assuming conditions at the middepth of the clay are typical of the entire clay layer, and (b) dividing the clay layer into four or five thinner layers, computing the settlement of each thin layer and summing up by Eq. (8.14). Hint: See Example 10.8.

SOLUTION: Gs

= 0.92, ρoil = (0.92) (1 Mg m3 ) = 0.92 Mg m3

γ oil = ( 0.92 Mg m3 ) ( 9.81 m s2 ) = 9.025 kN m3 qo

= ( 9.025 kN m3 ) (18 m) = 162.4 kPa

From Fig. 10.5: for r = 45 m and z = 30 m, consolidation indices from Problem 8.18: clay layer properties:

w

σz = Δσv = 135 kPa Ccε

= 0.154, Cr ε = 0.0112, σ'p = 260 kPa

= 29.3%, ρd = 1.5 Mg m3 , eo = 0.79

clay:

ρt = (1.5 Mg m3 ) (1 + 0.293) = 1.94 Mg m3 , ρ ' = 1.94 − 1.0 = 0.94 Mg m3

sand:

ρt = 1.81 Mg m3 , ρ ' = 1.81 − 1.0 = 0.81 Mg m3 , estimate ρdry =

At the center of the clay layer:

σ'vo

1.81

= 1.7 Mg m3 0.06 + 1 = ⎡⎣(2)(1.7) + (18)(0.81) + (10)(0.94)⎤⎦ × 9.81 = 268.6 kPa

σ'vo ≈ σ'p ∴ clay is NC σ 'vo + Δσv σ 'vo (268.6 + 135) sc = (0.154)(20 m)log = (0.154)(20 m)(0.1768)

(a) Eq. 8.13: sc

= CcεHo log

268.6 sc = 0.5447 m = 544.7 mm

solutio n continued on next page



Chapter 10

10-16 contin ued.

n

(b) For multiple layers: sc

= Ccε

∑ H log σ ' σ+' Δσ vo

i

i =1

vo

See tabulated values below. When σ 'vo Solution: sc

> σ'p , assume σ'p = σ 'vo

= 0.3572 m = 357 m m

Depth Below Clay Surface To p

v

Bo tt om

' vo

Cen ter of

Su bl ay er Th ic kn es s

Ef fec ti ve Ov er bu rd en

Sublayer

Ho

Pressure

'p

v

' vf

Pr ec on so l. Pr es su re

Pr es su re Ch an ge

F in al Pr es su re

C om p re ss i on R at i o Rec om p. Cu rv e C

r

Vi rg in Cu rv e C

C ha ng e i n T hi ck nes s H

c

(m)

(m)

(m)

(m)

(kPa)

(kPa)

(kPa)

(kPa)

0.0

5.0

2.50

5.00

199.40

260.0

148.0

347.40

0.0112

0.154

0.1034

5.0

10.0

7.50

5.00

245.54

260.0

139.0

384.54

0.0112

0.154

0.1323

10.0

15.0

12.50

5.00

291.65

291.7

130.0

421.65

0.0112

0.154

0.1233

15.0

20.0

17.50

5.00

337.76

337.8

120.0

457.76

0.0112

0.154

0.1017

20.0

27.5

SUM =

15.00

#VALUE!

SUM =

0.3572

#VALUE! #VALUE!

#VALUE!

(m)



Chapter 10

10-17. Estimate the ultimate consolidation settlement under the centerline of a 17 x 17 m mat foundation. The mat is 1.2 m thick reinforced concrete, and the average stress on the surface of the slab is 80 kPa. The soil profile is shown in Fig. P10.17. Oedometer tests on samples of the clay provide these average values: Neglect any settlements due to the sand layer. Cc = 0.40, Cr = 0.03, clay is NC

SOLUTION: Estimate γ conc qo

= 23.6 kN m3

= ( 23.6 kN m3 ) (1.2 m) + 80 = 108.3 kPa

qnet

= 108.3 − (1.9 Mg m3 ) ( 9.81 m s2 ) (1.2 m) = 85.9 kPa

From Fig. 10.4 (Eq. 10.6): for m = consolidation properties: clay layer properties: Gs w

=

,

z Cc = 0.40, Cr

n=

8.5

Determine I at center of each sublayer.

z = 0.03,

σ'p = σ 'vo for NC clay

= 42%, ρsat = 1.8 Mg m3 , ρ ' = 1.8 − 1.0 = 0.8 Mg m3

(2.7)(42)

= 1.13 → Ccε = 0.187, Cr ε S 100 Break clay into 4 sublayers as shown in the table below. estimate eo

=

w

8.5

Depth Below Clay Surface To p

Bo tto m

' vo

Cen ter o f

Su bl ay er Th ic kn es s

Ef fec ti ve Ov erb ur den

Sublayer

Ho

Pressure

'p Pr ec on so l. Pres su re

= 0.014

v


' vf Fi nal Pres sur e

Compression Ratio Rec om p. Cu rv e C

r

Vi rg in Cu rv e C

Change in T hi ck nes s H

c

(m)

(m)

(m)

(m)

(kPa)

(kPa)

(kPa)

0.0

2.0

1.00

2.00

101.05

101.05

85.8

186.85

0.187

0.014

0.0075

2.0

4.0

3.00

2.00

116.74

116.74

83.5

200.24

0.187

0.014

0.0066

4.0

7.0

5.50

3.00

136.36

136.36

74.8

211.16

0.187

0.014

0.0080

7.0

10.0

8.50

3.00

159.91

159.91

60.2

220.11

0.187

0.014

0.0058

20.0

27.5

SUM =

10.00

#VALUE!

#VALUE! #VALUE!

(kPa)

#VALUE!

(m)

SUM =

0. 02 04

solutio n continu ed on next page



Chapter 10

10-17 contin ued.

Use Eq. 10.15 for NC clay: s c

=

Cc

n

∑ H log σ +σΔσ

1 + eo i =1

At the center of each clay sublayer:

i

i

i

i

σ'vo = (1.9 Mg m3 ) ( 9.81 m s2 ) ( 5 m) + ( 0.8 Mg m3 ) ( 9.81 m s2 ) z

σ'vo = 93.2 kPa + (7.848)z; where, z = depth below the clay surface use superposition to determine Δσv = σ z =4 ( qnet ) (I) RESULTS : sc

= 0.0204 m = 20 mm



Chapter 10

10-18. Three uniformly distributed loads of 100 kPa each are applied to 10 x 10 m square areas on the soil profile shown in Fig. P10.18. Undisturbed samples of the clay were taken prior to construction, and consolidation tests indicated that the average preconsolidation stress is about 110 kPa, the average compression index is 0.50, and the a verage recompression index is 0.02. Estimate the total consolidation settlement for the clay layer only under the center of the middle loaded area.

SOLUTION: qo

= 100 kPa, assume eo = 0.9

consolidation indices:

Ccε

0.50

=

= 0.55, Crε =

0.9 clay: γ ' = (1.83 − 1.0)(9.81) = 8.14 kN m3

0.02 0.9

= 0.022, σ 'p = 110 kPa

γ ' = (2.0 − 1.0)(9.81) = 9.81 kN m3 At the center of the clay layer: σ' vo = (10)(9.81) + (1.5)(8.14) = 110.3 kPa sand:

σ'vo ≈ σ'p ∴ clay is NC Eq. 8.13: sc

σ 'vo + Δσv σ 'vo

= CcεHo log

Use Fig. 10.4 (Eq. 10.6) to calculate Δσv Δσv

= σz = qo

∑ I = 4q (I − I i

Δσv = 42.76 kPa

0

1

2

= σz at the center of the 3 mats, at depth z = 11.5 m.

+ I3 ) = ( 4)(100 kPa) × (0.1226 − 0.1200 + 0.1043)

(see tabulated values below)

Rectangle

x

y

I

1

30

5

0.1226

2

20

5

0.1200

3

5

5

0.1043

(110.3 + 42.76)

sc

= (0.55)(3 m)log

sc

= 0.235 m = 235 mm

110.3

= (0.55)(3 m)(0.1423)



Chapter 10

10-19. A series of oil storage tanks are to be constructed near Mystic River power station in Boston, MA. The typical tank is 22 m in diameter, and it exerts an average foundation stress of about 125 kPa. The soil profile at the site is very similar to that shown in Fig. 8.19(a), see next page. Estimate both the total and differential consolidation settlement under the average tank. SOLUTION: Eq. 8.19b: sc

⎡ = Ho ⎢Crε ⎣⎢

σ 'p

∑ log σ '

+ Ccε

vo

∑ log σ ' σ+' Δσ vo

⎤ ⎥ ⎦⎥

v

p

Calculate settlement of the silt and clay layers from depth 7 to 32 m. Break region into 4 sublayers. Estimate σ'vo and σ'p by interpolating values from the plot in Fig. 8.19a. Use empirical expressions from Chpt. 8 to estimate Cc and Cr . Use Fig. 10.5 to determine

Δσv beneath the center and the edge of the tank,

at the center of each sublayer. The differential settlement is the difference between these two sc values. ......................................................................................................................... ..... Assume Gs

= 2.7

Estimate eo for the upper organic and silty layers. eo Gs w

Gs w

=

(2.7)(0.3)

S (2.7)(0.4)

1.0

= 0.81

= = 1.08 S 1.0 Use Table 8.3 to estimate Cc for the upper organic and silty layers. Estimate eo for the lower blue clay layer. eo

Cc

= 1.15 × 10−2 × 30 = 0.345, Ccε =

1 + eo

=

0.345 1.81

= 0.191

0.191

= 0.0191 10 Use Table 8.4 to estimate Cc for the lower blue clay layer. estimate Cr ε

assume Cc

=

Cc

=

=

= 0.4, C cε =

Cc 1 + eo

=

0.4 2.08

= 0.192, Cr ε =

0.192 10

= 0.0192

See table below for Δσv values determined from Fig. 10.5. Depth below tank (z), m

z/r

Icenter

(kPa)

Icenter

(kPa)

10

0.91

0.696

87.0

0.35

43.8

215

1.36

0.476

59.4

0.28

35.0

21

1.91

0.305

38.1

0.21

26.25

28.5

2.59

0.188

23.5

0.13

16.25

center

center

solutio n continued on next page



Chapter 10

10-19 contin ued. Table below summarizes consolidation settlement calculation for the tank center. 10-19: Tank Center Depth Below Clay Surface

' vo Su bl ay er Thi ck nes s

Ef fec ti ve Ov erb urd en

Sublayer

Ho

Pressure

(m)

(m)

6.0

3.00

6.0

10.0

10.0

'p

v

' vf

Compression Ratio

Pr ec on so l. Pr es sur e

Pr es su re Chan ge

Fi nal Pr es sur e

(kPa)

(kPa)

(kPa)

(kPa)

6.00

8.93

80.00

87.0

95.88

0.191

0.0191

1.1003

8.00

4.00

14.10

30.00

59.5

73.55

0.191

0.0191

0.2803

18.0

14.00

8.00

19.74

19.74

38.1

57.85

0.192

0.0192

0.0717

18.0

25.0

21.50

7.00

26.79

26.79

23.5

50.29

0.192

0.0192

0.0368

20.0

27.5

SUM =

25.00

#VALUE!

Top

Bo tto m

(m)

(m)

0.0

Cen ter o f

Rec om p. Cur ve C

#VALUE! #VALUE!

r

Vi rg in Curv e

Change in

C

Th ic kn es s H

c

(m)

#VALUE!

SUM =

0.38 88

Table below summarizes consolidation settlement calculation for the tank edge. 10-19: Tank Edge Depth Below Clay Surface

' vo Su bl ay er Th ic knes s

Ef fec ti ve Ov er bu rden

Sublayer

Ho

Pressure

(m)

(m)

6.0

3.00

6.0

10.0

10.0

'p

v

' vf

Compression Ratio

Pr ec on so l. Pr ess ur e


Fi nal Press ur e

(kPa)

(kPa)

(kPa)

(kPa)

6.00

8.93

80.00

43.8

52.73

0.191

0.0191

0.8838

8.00

4.00

14.10

30.00

35.0

49.10

0.191

0.0191

0.2669

18.0

14.00

8.00

19.74

19.74

26.3

45.99

0.192

0.0192

0.0564

18.0

25.0

21.50

7.00

26.79

26.79

16.3

43.04

0.192

0.0192

0.0277

20.0

27.5

SUM =

25.00

#VALUE!

To p

Bo ttom

(m)

(m)

0.0

Cen ter of

Rec om p. Cur ve C

#VALUE! #VALUE!

#VALUE!

r

Vi rg in Cu rv e

Change in

C

Th ic kn es s H

c

(m)

S UM =

0 .3 51 0

10-19. Solution Summary Total maximum conso lidation settlement = 389 mm Differential settlement = 389 – 351 = 38 mm



Chapter 10

10-20. A new highway to Siracha, Thailand, is to be constructed east of Bangkok, across a region of deep deposits of very soft marine clay. A typical soil profile is shown in Fig. 8.21(a). The average Cc = 0.8 below the drying crust. The proposed embankment is 17 m wide at the top, has three horizontal to one vertical side slope, and is 2.5 m high. Estimate the ultimate consolidation settlement of the centerline of the embankment. SOLUTION: Eq. 8.19b: sc

⎡ = Ho ⎢Crε ⎢⎣

∑

log

σ 'p + Ccε σ 'vo

∑

log

σ 'vo + Δσv ⎤ ⎥ σ 'p ⎥⎦

Calculate settlement of the silt and clay layers from depth 0 to 10 m. Break upper crust (0 to 4 m) into 2 sublayers, and lower clay (4 to 10 m ) into 2 sublayers. Estimate σ'vo and σ'p by interpolating values from the plot in Fig. 8.21a. Use Fig. 10.6 to determine

Δσv beneath the center of the embankment,

at the center of each sublayer. ..............................................................................................................................

= 2.7, w = 15% (upper crust)

Assume Gs

Estimate eo for the lower green clay sublayers. eo

=

Gs w

=

(2.7)(1.0)

S 1.0 Use Table 8.4 from Chpt. 8 to estimate Cc for upper crust sublayers upper crust:

Ccε

green clay:

Ccε

=

=

Cc

=

1 + eo Cc

1 + eo

=

0.4 3.7

0.8 3.7

= 0.108; estimate Cr ε =

= 0.216; estimate C r ε =

0.108 10

0.216 10

= 2.7 → Cc = 0.4

= 0.0108

= 0.0216

See table below for Δσv values determined from Fig. 10.6. Estimate γ fill

= 20.4 kN m3 , a = 7.5 m, b = 17 m, qo = (2.5 m) ( 20.4 kN m3 ) = 51kPa

z below embankment (m)

a/z

b/z

I

2 (kPa)

1

7.5

17

0.499

50.9

3

2.5

5.7

0.499

50.9

5.5

1.4

3.1

0.49

50.0

8.5

0.88

2.0

0.475

48.4

v =

From profile in Fig. 8.21a:

z

σ 'vo = 3.3z + 2

Estimate σ'p values from plot in Fig. 8.21a

solution continued on next page



Chapter 10

10-20 contin ued. 10-20: Embankm ent Center Depth Below Clay Surface Top

Bo tto m

' vo

Cen ter o f

Su bl ay er Thi ck nes s

Ef fec ti ve Ov erb urd en

Sublayer

Ho

Pressure

'p Pr ec on so l. Pr es sur e

v

Pr es su re Chan ge

' vf Fi nal Pr es sur e

Compression Ratio Rec om p. Cur ve C

r

Vi rg in Curv e C

Change in Th ic kn es s H

c

(m)

(m)

(m)

(m)

(kPa)

(kPa)

(kPa)

(kPa)

0.0

2.0

1.00

2.00

5.30

34.00

50.9

56.20

0.0108

0.108

0.0646

2.0

4.0

3.00

2.00

11.90

30.00

50.9

62.80

0.0108

0 .108

0.0780

4.0

7.0

5.50

3.00

20.15

36.00

50.0

70.15

0.0216

0 .216

0.2041

7.0

10.0

8.50

3.00

30.05

50.00

48.4

78.45

0.0216

0.216

0.1411

20.0

27.5

SUM =

10.00

#VALUE!

#VALUE! #VALUE!

#VALUE!

(m)

SUM =

0.42 31

10-20. Solution Summary Consolidation s ettlement at embankment center = 423 mm



Chapter 10

10-21. Figure P10.21 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: (a) At the center of the clay layer: (b) Eq. 8.13: sc

= CcεHo log

σ 'vo + Δσv σ 'vo

3900

= (2.475)(0.406) = 1.0 ft 1532 σ' 3200 = (0.495)(−0.0859) = −0.0425 ft (c) sc = CrεHo log vf = (0.033)(15)log 3900 σ 'vo sc

= (0.165)(15)log

σ'vo = (10 ft)(110 pcf) + (7.5 ft)(120 − 62.4 pcf) = 1532 psf

(0.52 in)

Heave = 0.52 inches occurs after the equipment is removed (part b to part c). Net movement = 12 − 0.52 = 11.5 in

(settlement )



Chapter 10

10-22. 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. The figure below 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) At the center of the clay layer:

σ'vo = (3.75 m)(20.5 − 9.81 kN m3 ) = 40.09 kPa

σ 'p = 74 kPa (b) σ'vo < σ 'p ∴ clay is OC σ'vf = (3 m)(16 kN m3 ) + 40.09 = 88.09 kPa Eq. 8.19b: sc sc

= CrεHo log

= (0.03)(7.5)log

74

40.09 sc = 0.162 m = 162 mm (c) sc

= CrεHo log

σ 'p σ ' + Δσv + CcεHo log vo + σ 'vo σ 'p + (0.18)(7.5)log

88.09 74

= 0.05989 + 0.1022

σ 'vf 40.09 = (0.03)(7.5)log = −0.0769 m = 77 mm (heave) 88.09 σ 'vo



Chapter 10

10-23. The figure shows the 1-D compression curve for a clay. (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 = 0.846, determine Cr and Cc for this soil. For C c, use the portion of the curve between σ ’v = 200 and 800 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 70 to 200 kPa.

SOLUTION:

σ 'v = 140 kPa at ε v = 20% (∼ 60% of the way between 100 and 200) 0.40 − 0 (b) Ccε = = 0.248 → Cc = Ccε (1 + eo ) = (0.248)(1 + 0.846) = 0.458 (a)

900 22 0.32 − 0.30 Crε = = 0.01 → Cr = Crε (1 + eo ) = (0.01)(1 + 0.846) = 0.0185 1000 log 10 s (c) ε v = c ; From the consolidation curve, Δεv = 0.20 − 0.05 = 0.15 Ho log

sc

= (0.15)(9.5 m) = 1.42 m



Chapter 10

10-24. 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: At the center of the clay layer:

σ'vo = 480 kPa, σ 'p = 1800 kPa

σ'vo < σ 'p ∴ clay is OC σ'vf = 2100 kPa Eq. 8.19b: sc sc

= CrεHo log

= (0.0352)(15)log

σ 'p σ ' + Δσv + CcεHo log vo σ 'vo σ 'p

1800

480 sc = 0.484 ft = 5.8 inches

+ (0.18)(15)log

2100 1800

= 0.3031+ 0.1808



Chapter 10

10-25. The figure 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 ) + (3 ft)(120 p cf ) + (9 ft)(120 p cf ) = 1100 + 360 + 1080 = 2540 p sf u = (9 ft)(62.4 pcf ) = 561.6 psf

σ 'v = σv − u = 2540 − 561.6 = 1978.4 psf after excavation

σv = (3 ft)(120 pcf) + (9 ft)(120 pcf) = 1440 psf u = (9 ft)(62.4 pcf) = 561.6 psf

σ 'v = σv − u = 1440 − 561.6 = 878.4 psf

(b) sc sc

= CrεHo log

σ 'vf 878.4 = (0.035)(24)log = −0.2962 ft 1978.4 σ 'vo

= 3.55 inches of heave



Chapter 10

10-26. The figure 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 be low ground surface. The consolidation properties for each layer are shown. Assume full capillarity. (a) Compute σ ’v the 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) water table at 12 ft

σv = (12 ft)(120 p cf ) + (16 ft)(118 pcf ) = 1200 + 1888 = 3088 psf u = (16 ft)(62.4 p cf ) = 998.4 p sf σ 'v = σv − u = 3088 − 998.4 = 2089.6 psf water table at 20 ft

σv = (12 ft)(120 p cf ) + (16 ft)(118 pcf ) = 1200 + 1888 = 3088 psf u = (8 ft)(62.4 pcf) = 499.2 psf σ 'v = σv − u = 3088 − 499.2 = 2588.8 psf Consolidtaion settlement will occur in the lower, soft clay, layer. Eq. 8.19b: sc sc

= CcεHo log

= (0.185)(32)log

σ 'vo + Δσv σ 'p

2588.8 2089.6

= 0.551 ft = 6.6 in



Chapter 10

10-27. When a consolidation test is performed on some soils, the virgin compression region is not linear, but bilinear. The figure shows such a compression curve from a 15 ft thick layer. (a) What vertical strain, ε v, 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 feet) occurs?

SOLUTION:

⎡ σ 'vf ⎤ 980 3000 + (0.17)log ⎢Ci log ⎥ = (0.032)log 560 980 σ 'vi ⎦ ⎣ εv = 0.00778 + 0.0826 = 0.090 = 9.0%

(a)

εv =

∑

4000

(b) sc

= (15 ft)(0.14)log

(c) sc

= (15 ft)(0.032)log

sc

3000

= (15)(0.01749) = 0.26 ft (3.1in)

3000 4000

= (15)( −0.004)

= −0.06 ft (−0.72 in) → heave



Chapter 10

10-28. The figure shows a soil profile where a clay layer will consolidate under an embankment loading of 150 kPa. There is no capillarity. Your firm performed two consolidation tests: i) one test indicated that the soil is overconsolidated, with σ ’p = 110 kPa. ii) one test indicated that the soil is normally consolidated. Both tests gave the same Cr ε and Ccε values. Assume Ccε = 0.25. (a) Determine the initial σ ’v at the middle of the clay layer (i.e., at depth 5.5 m). (b) Compute the settlement due to the embankment loading, assuming that the overconsolidated assumption is correct (σ ’p = 110 kPa). (c) Compute the settlement again, this time assuming that the soil is normally consolidated.

SOLUTION: (a) At the center of the clay layer:

σ'vo = (2.5 m)(16 kN m3 ) + (3 m)(20.5 − 9.81 kN m3 ) = 72.07 kPa

σ 'p = 110 kPa, σ'vf = 150 + 72.1 = 222.1kPa σ'vo < σ 'p ∴ clay is OC σ 'p σ ' + Δσv Eq. 8.19b: sc = CrεHo log + CcεHo log vo σ 'vo σ 'p (b)

sc

= (0.025)(6)log

110

+ (0.25)(6)log

72.1 sc = 0.4852 m = 485.2 mm

222.1 110

= 0.0275 + 0.4577

σ'vo = σ 'p = 72.1kPa σ ' + Δσv 222.1 sc = CcεHo log vo = (0.025)(6)log = 0.073 = 73.3 mm 72.1 σ 'p (c) Assume


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