Engineering of Foundations Chapter7 Salgado Solution

CHAPTER 7 SITE EXPLORATION Conceptual Problems Problem 7-1 Define all the terms in bold contained in the chapter summary. SOLUTION: No solution provided. End

Problem 7-2 A 55-story, residential building will be built at a beach-front, 60 m × 40 m property. The subsurface is typical of barrier island geology in this area: a very loose sand layer on top of sandstone. Foundations usually consist of piles or piled rafts with the piles bearing on the sandstone. Develop a site investigation plan for this site. SOLUTION:

60m 30m 20m 40m

S-Figure 7-1

The site investigation can be started with eight borings as shown in S-Figure 7-1, six of which would be near the corners of the area. The borings are located schematically,

and corner borings should more properly be enar the corners of the loaded area of the property. All the borings should extend into the sandstone layer. The termination criteria should be decided based on the local standard (or based on the RQD and recovery ratio). Rock cores should be collected and preserved properly for future reference. SPTs and CPTs are suggested for the sand layer. Some CPTs can be performed next to the borings, others in between.

If large scatter or unexepected variations are found between any two

borings, additional borings should be done to reduce the uncertainty for that part of the property. End

Problem 7-3 A 30-story office building will be built at a 30 m × 40 m property. The geology of the area is residual soil of gneiss extending to depths ranging from 10-20 m (this depth can vary significantly across short distances because of the nature of the banding in gneiss). Sound rock (gneiss) is located at that depth. There are occasionally large boulders found at shallow depths. For large buildings, piles to rock are usually used. Develop a site investigation plan for this site. SOLUTION:

20m

20m

15m

15m

S-Figure 7-2 The site investigation can be started with nine borings as shown in the S-Figure 7-2. The borings are located schematically, and corner borings should more properly be enar the corners of the loaded area of the property. All the borings should extend into the sound rock (gneiss) layer. Rock cores should be collected to ensure soundness of the rock if any doubts exist and preserved properly for reference during the pile design stage. SPTs and CPTs are suggested. The locations of the CPTs should be selected to minimize the chances of encountering large boulders. If large scatter or unexepected variations are found between any two borings, additional borings should be done to reduce the uncertainty for that part of the property. End

Quantitative Problems Problem 7-4 A vane shear test was performed at a point within a clay layer. The maximum moment required to rotate the vane, which had a diameter of 60 mm and a

height of 120 mm, was measured as 70N·m. The vane was fully inserted in the soil. What is the undrained shear strength of the clay? SOLUTION: Vane diameter B = 60mm Vane height H = 120 mm = 2B Maximum moment required to rotate the vane T = 70N.m The undrained shear strength su of the clay from a vane shear test (when H = 2B) is given by su =

12T ; n =2, for fully inserted vane πB (12 + n )

su =

12 × 70 ×10−3 kPa = 88.5 ≈ 88 kPa answer 3.14 × (0.06)3 × (12 + 2 )

3

End

Problem 7-5 If a cone penetration test were performed next to the vane shear test of Problem 7-4, which was performed at a depth of 5 m, what value of cone resistance would you expect? The water table is at the surface and the clay has a unit weight of 17 kN/m3. Use Nk = 12. SOLUTION: From Eq. (7.22), we obtain cone resistance: qc = N ksu + σv Now, we calculate total vertical stress at a depth of 5m.

σ v = 17 × 5 = 85kPa

Using Nk=12 for clay layer and su=88.5kPa from Problem 7-4, we can estimate cone resistance at a depth of 5m: q c = 12 × 88.5 + 85 = 1147kPa ≈ 1.15MPa answer End Problem 7-6 A CPT was performed in a deposit of soft clay with the water table at a depth of 1 m. The cone resistance at a depth of 10 m was equal to 0.6 MPa. What is the minimum and maximum value of su of the clay at what depth would expect based on the range of values possible for Nk? The unit weight of this clay is 17 kN/m3. SOLUTION: We will consider the clay layer to be fully saturated due to capillary rise. We can calculate the total stress at 10m as:

σ v = γ sat z = 17 × 10 = 170kPa

Nk values range from 10 to no more than 15. We can calculate su from Eq. (7.22):

qc = N k su + σv

q − σv su = c Nk

Using Nk = 10 and 15, σv = 170kPa and qc = 600kPa in this equation produce maximum and minimum value of su equal to 43 and 29 kPa.

End

Problem 7-7 The results of SPT tests performed with an automatic trip hammer using the standard ASTM split spoon sampler with a liner are shown in the following table. The borehole diameter was within the recommended range. The soil profile consists of a normally consolidated sand with unit weight equal to 20 kN/m3. The water table is located at 4.5 m below the ground surface. Estimate the relative density DR and the peak friction angle φp at depths where SPT measurements are available. Use Eq. (7.6) with Eq. (7.8) to estimate DR, and use Fig. 7-14 to estimate φp. You may use the following table to guide your calculations.

Depth (m)

NSPT

4

25

5

30

6

35

Ch

Cr

N60

σv' (kPa)

SOLUTION: For an automatic trip hammer ERhammer = 80 % Thus, C h =

ER hammer 80 = = 1.33 ER safety 60

DR

φp (deg)

Following Eq. (7.3), for 4m ≤ rod length < 6m, Cr = 0.85; and for rod length = 6m, Cr = 0.95. Standard ASTM split spoon sampler with a liner was used and the borehole diameter was within the recommended range, so Cd = 1 and Cs = 1. Now N60 can be calculated using Eq. (7.1) as N 60 = Ch Cd Cr Cs NSPT at 4m, N60 = (1.33)(1)(0.85)(1)25 = 28.3 at 5m, N60 = (1.33)(1)(0.85)(1)30 = 33.9 at 6m, N60 = (1.33)(1)(0.95)(1)35 = 44.2 Following Eq. (7.6) DR = 100%

N 60 A + BC

σ' v pA

where, A = 36.5, B = 27, pA = 100 kPa. Following Eq. (7.8) C=

K0 ; and so for a normally consolidated sand C = 1. K 0,NC

Unit weight of the sand = 20kN/m3. Water table is at a depth of 4.5m from the ground surface. Now, the effective vertical stresses at different depths will be at 4m, σ′v = 4(20) = 80 kPa at 5m, σ′v = 5(20) – 0.5(9.81) = 95.1 kPa at 6m, σ′v = 6(20) – 1.5(9.81) = 105.3 kPa. From the above mentioned equation, relative densities of the sand deposit at different depths are calculated as 69.8% (at 4m), 73.8% (at 5m), and 82.5% (at 6m).answer

The peak friction angles at those depths are 45.5◦, 46◦, and 47.5◦

answer

(read from

the chart of Figure 7-13). S-Table 7-1

Depth NSPT

Ch

Cr

N60

σ′v (kPa)

DR (%)

φp (deg)

4

25

1.33

0.85

28.3

80

69.8

45.5

5

30

1.33

0.85

33.9

95.1

73.8

46

6

35

1.33

0.95

44.2

105.3

82.5

47.5

(m)

End Problem 7-8 Table 7-8 has SPT blow counts obtained at intervals of 1 m at a sandy site. A donut hammer and a liner sampler without the liner were used. Every care was taken to connect the rod segments firmly and to follow standard procedure. The water table is at a depth of 3 m, and the site is lightly overconsolidated because approximately 2 m of soil of unit weight approximately equal to 17 kN/m3 were removed before the SPT was performed. The K0 of this soil in a normally consolidated state would be 0.48. Calculate the corresponding stress-normalized, energy-corrected blow counts (N1)60. Table 7-8 SPT blow counts for Problem 7-8. Depth (m)

SPT blow count

1

15

2

18

3

22

4

23

5

25

6

28

SOLUTION: For a donut hammer, ERhammer = 45 % Thus, C h =

ER hammer 45 = = 0.75 ER safety 60

Following Eq. (7.3), for rod length < 4m, Cr = 0.75; for 4m ≤ rod length < 6m, Cr = 0.85; and for rod length = 6m, Cr = 0.95. A liner sampler without a liner was used and the borehole diameter was within the recommended range, so Cd = 1 and Cs = 1.2. Now N60 can be calculated using Eq. (7.1) as N 60 = Ch Cd Cr Cs NSPT at 1m, N60 = (0.75)(1)(0.75)(1.2)15 = 10.1 at 2m, N60 = (0.75)(1)(0.75)(1.2)18 = 12.2 at 3m, N60 = (0.75)(1)(0.75)(1.2)22 = 14.9 at 4m, N60 = (0.75)(1)(0.85)(1.2)23 = 17.6 at 5m, N60 = (0.75)(1)(0.85)(1.2)25 = 19.1 at 6m, N60 = (0.75)(1)(0.95)(1.2)28 = 23.9

Now, we need to normalize these values using (7.5):

( N1 )60 = N 60

p A K 0,NC σ'v K 0

For us to do that, we need the K0 value, from which we obtain (4.35):

K 0 = K 0,NC OCR

As an illustrative example, let us calculate (N1)60 at z = 4m.

σ'v = 17kN / m3 × 3m + (17 − 9.81)kN / m 3 × 1m = 58.2kPa σ'v0 = σ'v + Δσ ' = 58.2kPa + 17kN / m3 × 2m = 92.2kPa OCR =

σ'v0 92.2 = = 1.58 σ'v 58.2

K 0 = K 0,NC OCR = 0.48 × 1.58 = 0.60

( N1 )60 = 17.6

100 0.48 = 20.6 58.2 0.60

Similarly, we obtain the following table.

S-Table 7-2

σ'v (kPa) σ'v0 (kPa)

Depth (m)

NSPT

N60

1

15

10.1

17

2

18

12.2

3

22

4

OCR

K0

(N1)60

51

3

0.83

18.6

34

68

2

0.68

17.6

14.9

51

85

1.67

0.62

18.4

23

17.6

58.2

92.2

1.58

0.60

20.6

5

25

19.1

65.4

99.4

1.52

0.59

21.3

6

28

23.9

72.6

106.6

1.47

0.58

25.5

End

Problem 7-9 The cone resistance for a clean sand at 6 m has been measured at 11 MPa. The average total unit weight of the soil column above 6 m is 21 kN/m3. The water table is 3 m below the surface. The soil is normally consolidated, with K0 = 0.45. The soil has

φc = 30˚. Estimate the relative density of the sand at 6 m. SOLUTION: Following Eq. (7.20)

⎛q ⎞ ⎛ σ' ⎞ ln ⎜ c ⎟ − 0.4947 − 0.1041φc − 0.841ln ⎜ h ⎟ p ⎝ pA ⎠ DR = ⎝ A ⎠ ' ⎛σ ⎞ 0.0264 − 0.0002φc − 0.0047 ln ⎜ h ⎟ ⎝ pA ⎠ Effective vertical stress at depth 6m

σ′v = 6(21)-3(9.81) = 96.6kPa. σ′h = 0.45(96.6) = 43.5kPa. pA = reference stress = 100kPa. Given that, for the clean sand, φc = 30◦ ,

⎛ 11 ⎞ ⎛ 43.5 ⎞ ln ⎜ ⎟ − 0.4947 − 0.1041× 30 − 0.841ln ⎜ ⎟ 0.1 ⎠ ⎝ 100 ⎠ = 73% answer DR = ⎝ ⎛ 43.5 ⎞ 0.0264 − 0.0002 × 30 − 0.0047 ln ⎜ ⎟ ⎝ 100 ⎠ End

Problem 7-10 For the sand deposit and conditions of Problem 5-12 and Problem 5-13, estimate and plot the cone resistance qc as a function of depth for the depth range 0-10 m. Plot also the ratio of the small-strain shear modulus G0 to qc. SOLUTION: From the solution of Problem 5-12, we know the variation of relative density along depth. Now cone resistance values can be related to relative densit according to ⎛ σ' ⎞ qc = 1.64 exp ⎡⎣ 0.1041φc + ( 0.0264 − 0.0002φc ) DR ⎤⎦ ⎜ h ⎟ pA ⎝ pA ⎠

0.841− 0.0047D R

For example, at a depth of 1m, DR = 61.5%, and vertical stress

σ'v = 1× 22 = 22kPa . Given that K0 = 0.45 for this deposit, the horizontal stress at 1m σ'h = 0.45 × 22 = 9.9kPa . The reference stress is pA = 100kPa. So the cone resistance value at 1m depth can be calculated as:

0.841− 0.0047×61.5 ⎤ ⎡ ⎛ 9.9 ⎞ ⎥ q c = 100 ⎢1.64 exp ⎣⎡ 0.1041× 30 + ( 0.0264 − 0.0002 × 30 ) 61.5⎤⎦ ⎜ ⎟ ⎝ 100 ⎠ ⎢⎣ ⎥⎦ = 3645kPa

Now from the solution of Problem 5-13, G0 at a depth of 1m is equal to 40970.8kPa. So

G 0 40971 = = 11.2 . qc 3645

The same procedure is repeated for all other depths, and the result is tabulated in S-Table 7-3. S-Table 7-3

Depth(z) 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10

σ'v (kPa) 11 22 33 44 55 66 77 88 99 110 121 132 143 154 165 176 187 198 209 220

σ'h (kPa) 5 9.9 14.9 19.8 24.8 29.7 34.7 39.6 44.6 49.5 54.5 59.4 64.4 69.3 74.3 79.2 84.2 89.1 94.1 99

DR (%) 60.75 61.5 62.25 63 63.75 64.5 65.25 66 66.75 67.5 68.25 69 69.75 70.5 71.25 72 72.75 73.5 74.25 75

qc(kPa) 2436 3644.6 4668.9 5572.5 6428.6 7227.9 8009.8 8756.6 9497.9 10214.5 10931.5 11629.8 12331.8 13019 13711.9 14392.5 15080.2 15757.5 16442.8 17119.2

G0 (kPa) 29918.2 40970.8 49412.1 56593.1 62829.2 68401.7 73755.8 78575.7 83347.2 87744.4 91953 96131.2 100050.9 103998.4 107700.2 111308.3 115049.5 118463.1 122082.7 125377.2

G0/qc 12.3 11.2 10.6 10.2 9.8 9.5 9.2 9 8.8 8.6 8.4 8.3 8.1 8 7.9 7.7 7.6 7.5 7.4 7.3

0

Depth (m)

2

4

6

8

10

6

8

10 G0/qc

12

14

S-Figure 7-3 End

Problem 7-11 For the sand deposit and conditions of Problem 5-14 and Problem 5-15, estimate and plot the cone resistance qc as a function of depth for the depth range 0-10 m. Plot also the ratio of the small-strain shear modulus G0 to qc.

SOLUTION: This solution will follow the same procedure as that of Problem 7-10. In this case due to presence of high water table, the effective stresses at different depths will be less, which in turn will affect the cone resistance values. For example, at a depth of 1m, the vertical stress σ'v = 1× ( 22 − 9.81) = 12.2kPa . Given that K0 = 0.45 for this deposit, horizontal stress at 1m σ'h = 0.45 × 12.2 = 5.5kPa . pA = 100kPa (refernce stress). So the cone resistance value at 1m depth can be calculated as 0.841− 0.0047×61.5 ⎤ ⎡ ⎛ 5.5 ⎞ ⎥ q c = 100 ⎢1.64 exp ⎣⎡ 0.1041× 30 + ( 0.0264 − 0.0002 × 30 ) 61.5⎤⎦ ⎜ ⎟ ⎝ 100 ⎠ ⎢⎣ ⎥⎦ = 2634.8kPa

Now, from the solution of Problem 5-15, G0 at a depth of 1m is equal to 31408 kPa. So

G 0 31407.9 = = 11.9 . qc 2634.8

The same procedure is repeated for all other depths, and the result is tabulated in the following table.

S-Table 7-4

Depth(z) 0.5

σ'v (kPa) 6.1

σ'h (kPa) 2.7

DR (%) 60.75

qc(kPa) 1729.9

G0 (kPa) 22727.1

G0/qc 13.1

1

12.2

5.5

61.5

2634.8

31407.9

11.9

1.5

18.3

8.2

62.25

3364.8

37911.8

11.3

2

24.4

11

63

4045.3

43440.1

10.7

2.5

30.5

13.7

63.75

4662.2

48127.2

10.3

3

36.6

16.5

64.5

5268.8

52481.6

10

3.5

42.7

19.2

65.25

5838.3

56509.5

9.7

4

48.8

22

66

6409.7

60274.6

9.4

4.5

54.9

24.7

66.75

6955.2

63948.6

9.2

5

61

27.5

67.5

7507.9

67334

9

5.5

67

30.2

68.25

8040.8

70499

8.8

6

73.1

32.9

69

8570.3

73685.7

8.6

6.5

79.2

35.6

69.75

9097.4

76637.3

8.4

7

85.3

38.4

70.5

9636.1

79717.1

8.3

7.5

91.4

41.1

71.25

10161.3

82569.5

8.1

8

97.5

43.9

72

10698.9

85349

8

8.5

103.6

46.6

72.75

11224.9

88169.5

7.9

9

109.7

49.4

73.5

11763.9

90832.7

7.7

9.5

115.8

52.1

74.25

12292.7

93562.4

7.6

10

121.9

54.9

75

12835

96132.1

7.5

0

Depth (m)

2

4

6

8

10

6

8

10 G0/qc

12

14

S-Figure 7-4 End

Design Problems Problem 7-12 An SPT log is given in Fig. 7-43. The SPT was performed with a safety hammer using the standard ASTM split spoon sampler with a liner. The borehole

diameter is within the recommended range. The sand is normally consolidated (with a K0 of approximately 0.45) and the water table is very deep. The critical-state friction angle of this sand is approximately 32˚. The unit weights of the sandy clay, silty clay and sand are equal to 17, 15, and 20 kN/m3, respectively. For the sand layer extending from 8.5 to 21 ft: (a) Estimate the relative density DR and the peak friction angle φp that would be obtained in triaxial compression tests performed on ideal, "undisturbed" samples recovered from the following depths where SPT measurements are available: 9.3, 11.5, 14, 16.5 and 19 ft. Use Eq. (7.6) to estimate DR and Eqs. (5.8) and (5.16) to estimate φp. Assume that you are estimating the φp that would be obtained in the laboratory under triaxial compression of a sample consolidated isotropically to the in situ mean effective stress. (b) Estimate φp using Fig. 7-14. Fig. 7-43 SPT log for Problem 7-12. SOLUTION: (a) From Figure 7-43, the NSPT can be calculated at different depths: at 9.3ft, NSPT = (6+6) = 12 at 11.5ft, NSPT = (12+14) = 26 at 14ft, NSPT = (7+10) = 17 at 16.5ft, NSPT = (8+12) = 20 at 19ft, NSPT = (6+7) = 13 For a safety hammer ERhammer = 60 %

Thus, C h =

ER hammer 60 = = 1.0 ER safety 60

Following Eq. (7.3), for rod length < 4m, Cr = 0.75; and for 4m ≤ rod length < 6m, Cr = 0.85. Standard ASTM split spoon sampler with a liner was used and the borehole diameter was within the recommended range, so Cd = 1 and Cs = 1. Now N60 can be calculated using Eq. (7.1) as N 60 = Ch Cd Cr Cs NSPT at 9.3ft (=2.83m), N60 = (1)(1)(0.75)(1)12=9 at 11.5ft (= 3.5m), N60 = (1)(1)(0.75)(1)26 = 19.5 at 14ft (= 4.3m), N60 = (1)(1)(0.85)(1)17 = 14.5 at 16.5ft (= 5.03m), N60 = (1)(1)(0.85)(1)20 = 17 at 19ft (= 5.79), N60 = (1)(1)(0.85)(1)13 = 11.1 Following Eq. (7.6) DR = 100%

N 60 A + BC

σ' v pA

where, A = 36.5, B = 27, pA = 100 kPa. Following Eq. (7.8) C=

K0 ; and so for a normally consolidated sand C = 1. K 0,NC

The unit weights of the sandy clay, silty clay and sand are equal to 17, 15 and 20 kN/m3, respectively. Water table is very deep. Now, the effective vertical stresses at different depths will be at 9.3ft (=2.83m), σ′v = 1.83(17)+0.76(15)+ 0.24(20) = 47.3kPa

at 11.5ft (= 3.5m), σ′v = 1.83(17)+0.76(15)+ 0.91(20) = 60.7kPa at 14ft (= 4.3m), σ′v = 1.83(17)+0.76(15)+ 1.71(20) = 76.7kPa at 16.5ft (= 5.03m), σ′v = 1.83(17)+0.76(15)+ 2.44(20) = 91.3kPa at 19ft (= 5.79m), σ′v = 1.83(17)+0.76(15)+ 3.2(20) = 106.5kPa From the above equation, relative density DR of the sand deposit at 9.3ft can be calculated as

D R = 100

9 47.3 36.5 + 27 ×1× 100

= 42.7%

answer to (a)

Similarly, at 11.5, 14, 16.5, and 19 ft, relative densities are 19.5

at 11.5ft (= 3.5m), D R = 100

at 14ft (= 4.3m), D R = 100

60.7 36.5 + 27 × 1× 100 14.5 76.7 36.5 + 27 ×1× 100

at 16.5ft (= 5.03m), D R = 100

at 19ft (= 5.79m), D R = 100

Calculation of peak friction angle φp Following Eq. (5.8), ⎡ ⎛ 100σ'mp I R = I D ⎢Q − ln ⎜ ⎜ p A ⎝ ⎣⎢

⎞⎤ ⎟⎟ ⎥ − R Q ⎠ ⎦⎥

= 60.7%

= 50.3%

17 91.3 36.5 + 27 ×1× 100 11.1

106.5 36.5 + 27 × 1× 100

answer to (a)

answer to (a)

= 52.7%

= 41.2%

answer to (a)

answer to (a)

where Q = 10 and RQ = 1 for clean silica sand K0 = 0.45 for the sand layer Sample calculation at 9.3ft (=2.83m) At 9.3ft (=2.83m), vertical effective stress σ′v = 1.83(17)+0.76(15)+ 0.24(20) = 47.3kPa knowing K0 = 0.45, the horizontal effective stress

σ 'h = 0.45 × 47.3 = 21.3kPa Thus in situ mean effective stress

σ 'm =

47.3+2 × 21.3 = 30kPa 3

Assume that we are estimating the φp that would be obtained in the laboratory under triaxial compression of a sample consolidated isotropically to the in-situ mean effective stress ' σ 'm = σ3p = 30kPa

Assume φP = 37 o . N=

1 + sin37 o = 4.02 1 - sin37 o

' σ1p = 30 × 4.02 = 120.6kPa

σ 'mp =

σ '1p + 2σ '3p 3

=

120.6 + 2 × 30 = 60.2kPa 3

⎡ ⎛ 100 × 60.2 ⎞ ⎤ I R = 0.427 ⎢10 − ln ⎜ ⎟ ⎥ − 1 = 1.52 ⎝ 100 ⎠ ⎦ ⎣ Following Eq. (5.10), for triaxial condition

φP = φc + 3I R Considering φc = 32◦, φP = 32o + 3 × 1.52 = 36.6o . Clearly our assumption of peak friction angle does not match with the calculated value. Hence we need to iterate again using the obtained value. Finally the value of peak friction angle at this depth converges to φP = 36.6o answer to (a) Similarly the peak friction angles at depths 11.5, 14, 16.5, and 19ft can be calculated as 39.2◦, 37.2◦, 37.3◦ and 35.4◦ respectively. answer to (a)

(b) As shown below, the peak friction angle at depths 9.3, 11.5, 14, 16.5, and 19ft can be obtained from Figure 7-14 as 36◦, 44◦, 39◦, 39.5◦, and 35◦ respectively. answer to (b)

0.0

φ=

0.5

ο 40 φ=

1.5 ο φ=30

ο 5 φ=3

σ'v/pA

ο 45 φ=

1.0

50 ο

2.0

ο φ=25

2.5

3.0

0

10

20

30

40 N60

S-Figure 7-5

The results are summarized in S-Table 7-5.

50

60

70

80

S-Table 7-5

Depth (ft)

NSPT

Cr

N60

σ′v

DR (%)

(kPa)

φp, equation

φp, chart

(deg)

(deg)

9.3

12

0.75

9

47.3

42.7

36.6

36

11.5

26

0.75

19.5

60.7

60.7

39.2

44

14

17

0.85

14.5

76.7

50.3

37.2

39

16.5

20

0.85

17

91.3

52.7

37.3

39.5

19

13

0.85

11.1

106.5

41.2

35.4

35

End

Problem 7-13 Redo Problem 7-12, part (a), assuming consolidation of the sample to the in situ vertical effective stress. SOLUTION: From the solution of Problem 7-12, relative densities at depth 9.3, 11.5, 14, 16.5, and 19ft are 42.7, 60.7, 50.3, 52.7 and 41.2 % respectively. In this problem we are assuming that the sample has been consolidated to the in-situ vertical effective stress in the triaxial test. Thus the values of peak friction angles will be different from those calculated in Problem 7-12. A sample calculation is shown below for depth 9.3 ft.

Calculation of peak friction angle φp Following Eq. (5.8), ⎡ ⎛ 100σ'mp I R = I D ⎢Q − ln ⎜ ⎜ p ⎢⎣ A ⎝

⎞⎤ ⎟⎟ ⎥ − R Q ⎠ ⎥⎦

where Q = 10 and RQ = 1 for clean silica sand Sample calculation at 9.3ft (=2.83m) At 9.3ft (=2.83m), vertical effective stress σ′v = 1.83(17)+0.76(15)+ 0.24(20) = 47.3kPa Assume that we are estimating the φp that would be obtained in the laboratory under triaxial compression of a sample consolidated isotropically to the in-situ vertical effective stress ' σ 'v = σ 3p = 47.3kPa

Assume φP = 37 o . N=

1 + sin37 o = 4.02 1 - sin37 o

' σ1p = 47.3 × 4.02 = 190.1kPa

σ 'mp =

σ '1p + 2σ '3p 3

=

190.1 + 2 × 47.3 = 94.9kPa 3

⎡ ⎛ 100 × 94.9 ⎞ ⎤ I R = 0.427 ⎢10 − ln ⎜ ⎟ ⎥ − 1 = 1.326 ⎝ 100 ⎠ ⎦ ⎣ Following Eq. (5.10), for triaxial condition φP = φc + 3I R Considering φc = 32◦, φP = 32o + 3 × 1.326 = 36o . Clearly our assumption of peak friction angle does not match with the calculated value. Hence we need to iterate again using the obtained value. Finally the value of peak friction angle at this depth converges to φP = 36o answer

Similarly the peak friction angles at depths 11.5, 14, 16.5, and 19ft can be calculated as 38.4◦, 36.5◦, 36.6◦ and 34.8◦ respectively. answer End

Problem 7-14 Redo Problem 7-12, part (a), assuming consolidation of the sample to the in situ horizontal effective stress. SOLUTION: From the solution of Problem 7-12, relative densities at depth 9.3, 11.5, 14, 16.5, and 19ft are 42.7, 60.7, 50.3, 52.7 and 41.2 % respectively. In this problem we are assuming that the sample has been consolidated to the in-situ horizontal effective stress in the triaxial test. Thus the values of peak friction angles will be different from those calculated in Problems 7-12 and 7-13. A sample calculation is shown below for depth 9.3 ft.

Calculation of peak friction angle φp Following Eq. (5.8), ⎡ ⎛ 100σ'mp I R = I D ⎢Q − ln ⎜ ⎜ p ⎢⎣ A ⎝

⎞⎤ ⎟⎟ ⎥ − R Q ⎠ ⎥⎦

where Q = 10 and RQ = 1 for clean silica sand K0 = 0.45 for the sand layer

At 9.3ft (=2.83m) ' σ 3p = 0.45 × 47.3 = 21.3kPa

Assume φP = 37 o . N=

1 + sin37 o = 4.02 1 - sin37 o

' σ1p = 21.3 × 4.02 = 85.6kPa

'

σ mp =

σ '1p + 2σ '3p 3

=

85.6 + 2 × 21.3 = 42.7kPa 3

⎡ ⎛ 100 × 42.7 ⎞ ⎤ I R = 0.427 ⎢10 − ln ⎜ ⎟ ⎥ − 1 = 1.667 ⎝ 100 ⎠ ⎦ ⎣ Following Eq. (5.10), for triaxial condition φP = φc + 3I R Considering φc = 32◦, φP = 32o + 3 ×1.667 = 37.0o . It is same as the assumed value. Therefore, we need not to iterate any further and at this depth (9.3ft) φP = 37.0o answer Similarly the peak friction angles at depths 11.5, 14, 16.5, and 19ft are 39.8◦, 37.7◦, 37.8◦, and 35.8◦ respectively. answer End

Problem 7-15 An SPT log is given in Fig. 7-44 The SPT was performed with a safety hammer using the standard ASTM split spoon sampler with a liner. The borehole diameter is within the recommended range. The sand is normally consolidated, and the water table is very deep. The unit weights of the sandy clay, silty clay, and sand are equal to 17, 15, and 20 kN/m3, respectively. The sand has φc = 30˚. For the sand layer extending from 11 to 21 ft:

(a) Estimate the relative density DR and the peak friction angle φp that would be obtained in triaxial compression tests performed on ideal, undisturbed samples recovered from the following depths where SPT measurements are available: 11.5, 14, 16.5 and 19 ft. Use Eq. (7.6) to estimate DR, and Eqs. (5.8) and (5.16) to estimate φp. Assume that you are estimating the φp that would be obtained in the laboratory under triaxial compression of a sample consolidated isotropically to the in situ mean effective stress. (b) Estimate φp using Fig. 7-14.

Fig. 7-44 SPT log for Problem 7-15 SOLUTION: NSPT can be calculated at different depths at 11.5ft, NSPT = (14+16) = 30 at 14ft, NSPT = (17+22) = 39 at 16.5ft, NSPT = (21+29) = 50 at 19ft, NSPT = (25+27) = 52 For a safety hammer, ERhammer = 60 % Thus, C h =

ER hammer 60 = = 1.0 ER safety 60

Following Eq. (7.3), for rod length < 4m, Cr = 0.75; and for 4m ≤ rod length < 6m, Cr = 0.85. Standard ASTM split spoon sampler with a liner was used and the borehole diameter was within the recommended range, so Cd = 1 and Cs = 1. Now N60 can be calculated using Eq. (7.1) as N 60 = Ch Cd Cr Cs NSPT

at 11.5ft (= 3.5m), N60 = (1)(1)(0.75)(1)30 = 22.5 at 14ft (= 4.3m), N60 = (1)(1)(0.85)(1)39 = 33.2 at 16.5ft (= 5.03m), N60 = (1)(1)(0.85)(1)50 = 42.5 at 19ft (= 5.79m), N60 = (1)(1)(0.85)(1)52 = 44.2 Following Eq. (7.6) DR = 100%

N 60 A + BC

σ' v pA

where, A = 36.5, B = 27, pA = 100 kPa. Following Eq. (7.8) C=

K0 ; and so for a normally consolidated sand C = 1. K 0,NC

The unit weights of the sandy clay, silty clay and sand are equal to 17, 15 and 20 kN/m3, respectively. Water table is very deep. Now, the effective vertical stresses at different depths will be at 11.5ft (= 3.5m), σ′v = 1.07(17)+1.52(20)+0.76(15)+0.15(20) = 63kPa at 14ft (= 4.3m), σ′v = 1.07(17)+1.52(20)+0.76(15)+0.95(20) = 79kPa at 16.5ft (= 5.03m), σ′v = 1.07(17)+1.52(20)+0.76(15)+1.68(20) = 93.6kPa at 19ft (= 5.79m), σ′v = 1.07(17)+1.52(20)+0.76(15)+2.44(20) = 108.8kPa

From the above mentioned equation, relative density DR of the sand deposit at 11.5ft can be calculated as D R = 100

22.5 63 36.5 + 27 ×1× 100

= 64.8%

answer to 1

Similarly, at 14, 16.5, and 19 ft, relative densities are 75.8%, 82.9%, and 81.9% respectively. answer to 1

Calculation of peak friction angle φp Following Eq. (5.8) ⎡ ⎛ 100σ'mp I R = I D ⎢Q − ln ⎜ ⎜ p A ⎝ ⎣⎢

⎞⎤ ⎟⎟ ⎥ − R Q ⎠ ⎦⎥

where Q = 10 and RQ = 1 for clean silica sand Mean stress, σ'm = Considering σ' m =

K0

σ' v + 2σ'h (1 + 2K 0 )σ' v = 3 3 =

0.45

for

the

sand

layer;

at

11.5ft

depth

(1 + 2 × 0.45) × 63 = 39.9kPa 3

Consolidation stress in triaxial test on undisturbed sample σ′c = 39.9kPa = σ′3 Say φp = 36◦, so

φ ⎞ ⎛ N P = tan 2 ⎜ 45o + P ⎟ = tan 2 ( 45o + 18o ) = 3.85 2 ⎠ ⎝ σ'1 = N P σ'3 = 3.85 × 39.9 = 153.6kPa σ'mp =

153.6 + 2 × 39.9 = 77.8kPa 3

⎡ ⎛ 100 × 77.8 ⎞ ⎤ I R ,at11.5ft = 0.648 ⎢10 − ln ⎜ ⎟ ⎥ − 1 = 2.66 ⎝ 100 ⎠ ⎦ ⎣ Following Eq.(5.10), for triaxial condition φP = φc + 3I R

For φc = 30◦, φP at11.5 ft = 30o + 3 × 2.66 = 37.98o . As the calculated value of peak friction angle does not match with the assumed value we need further iterations.

Iteration 2 φp = 37.98◦, NP = 4.2, σ′mp = 82.5kPa, IR = 2.62, φp = 37.86◦

Iteration 3 φp = 37.86◦, NP = 4.18, σ′mp = 82.2kPa, IR = 2.62, φp = 37.86◦ Thus the peak friction angle at depth 11.5ft is 37.9◦. Similarly the peak friction angles at depths 14, 16.5, and 19ft are 39.1◦, 39.8◦, and 39.3◦ respectively.answer to 1 From the chart of Figure 7-14, the peak friction angle at depths 11.5, 14, 16.5, and 19ft are 45◦, 46◦, 48◦, and 47.5◦ respectively.answer to 2 We

summarize

the

results

in

S-Table 7-6. These results illustrate the potential error in using a chart that makes no reference to the intrinsic properties of the soil. If the critical-state friction angle of the soil is low (perhaps because its particles are well rounded and of uniform size), then the chart could overpredict the friction angle by several degrees.

S-Table 7-6

Depth (ft)

NSPT

Cr

N60

σ′v

DR (%)

(kPa)

φp, equation

φp, chart

(deg)

(deg)

11.5

30

0.75

22.5

63

64.8

37.9

45

14

39

0.85

33.2

79

75.7

39.1

46

16.5

50

0.85

42.5

93.6

82.9

39.8

48

19

52

0.85

44.2

108.8

81.9

39.3

47.5

End

Problem 7-16* Two CPTs and one SPT were performed at close proximity. Results are in Table 7-9 and Table 7-10. (a) For the data given, prepare plots of qc, fs and fs/qc vs. depth. (b) Estimate the relative density DR and the peak friction angle φp that would be obtained in triaxial compression tests performed on ideal, undisturbed samples recovered from depths equal to 6.1, 7.6, and 9.10 m using CPT-based methods. The coefficient of lateral earth pressure is equal to 0.4 and the critical-state friction angle is equal to 36°. Use the charts of Fig. 7-26. The sand is normally consolidated and the water table is very deep. (c) Estimate the relative density DR and the peak friction angle φp of the sand at depths equal to 6.1, 7.6, and 9.10 m using an SPT-based method. Use Eq. (7.6) to estimate DR and Eqs. (5.8) and (5.16) to estimate φp. The SPT test was performed with a safety hammer using the standard ASTM split spoon sampler with a liner. The borehole diameter is within the recommended range.

(d) Compare the results obtained in (b) and (c).

Table 7-9 SPT data for Problem 7-16. Depth (m)

Soil Type

Unit

Depth

(from borings)

Weight

(m)

NSPT

(kN/m3) 0-1.5

Clayey Silt

14

6.1

27

1.5-4.3

Sand

15

7.6

25

4.3-5.2

Silty Clay

14.5

9.1

40

5.2-14.3

Sand

19

14.3-16.8

Clayey Silt

15.5

Table 7-10 CPT data for Problem 7-16. depth

qc

fs

fs/qc

qc

fs

fs/qc

(m)

(MPa)

(kPa)

(%)

(MPa)

(kPa)

(%)

0.05

1.29

55.87

4.3

1.5

38.39

2.6

0.1

1.66

60.83

3.7

1.3

51.98

4.0

0.15

1.37

59.92

4.4

1.3

46.35

3.7

0.2

1.37

52.73

3.9

1.3

45.94

3.7

0.25

1.61

53.39

3.3

1.3

46.45

3.5

0.3

1.62

57.92

3.6

1.4

53.51

3.8

0.35

1.86

64.49

3.5

1.4

60.4

4.2

0.4

1.86

70.35

3.8

1.5

64.53

4.3

0.45

1.88

72.81

3.9

1.5

68.28

4.6

0.5

2.14

79.71

3.7

1.6

72.51

4.5

0.55

2.43

87.22

3.6

1.9

95.97

5.1

0.6

2.73

95.76

3.5

2.3

85.23

3.7

0.65

2.94

105.75

3.6

2.1

81.87

3.9

0.7

2.88

113.71

3.9

2.3

83.62

3.7

0.75

2.82

120.79

4.3

2.1

89.88

4.3

0.8

2.67

114.1

4.3

2.0

89.42

4.4

0.85

2.57

108.38

4.2

2.0

89.8

4.5

0.9

2.50

109.09

4.4

2.1

92.35

4.5

0.95

2.57

115.72

4.5

1.9

93.2

4.9

1

2.51

116.5

4.6

1.9

90.43

4.7

1.05

2.48

114.24

4.6

2.0

96.19

4.9

1.1

2.37

117.15

4.9

2.0

103.78

5.2

1.15

2.40

123.25

5.1

2.0

108.85

5.4

1.2

2.36

137.13

5.8

2.0

114.32

5.8

1.25

2.32

148.17

6.4

2.0

123.76

6.2

1.3

2.37

160.71

6.8

2.2

143.25

6.6

1.35

2.37

165.26

7.0

2.3

155.05

6.8

1.4

2.31

152.95

6.6

2.3

160.34

7.1

1.45

2.32

147.83

6.4

2.4

167.56

7.0

1.5

2.32

145.35

6.3

2.5

162.33

6.5

1.55

2.46

141.5

5.8

2.8

149.58

5.4

1.6

2.83

129.19

4.6

3.3

132.36

4.0

1.65

3.28

87.32

2.7

3.8

107.26

2.8

1.7

3.76

56.72

1.5

3.7

67.42

1.8

1.75

3.51

39.06

1.1

3.5

39.27

1.1

1.8

3.25

41.36

1.3

3.3

24.8

0.7

1.85

3.35

47.63

1.4

3.2

23.44

0.7

1.9

3.33

40.95

1.2

3.5

38.68

1.1

1.95

3.38

43.6

1.3

4.1

34.12

0.8

2

4.49

33.41

0.7

3.8

29.54

0.8

2.05

4.60

19.17

0.4

3.4

45.43

1.3

2.1

4.55

48.52

1.1

3.3

45.9

1.4

2.15

3.62

51.84

1.4

2.9

58.8

2.1

2.2

1.33

51.23

3.8

1.2

50.82

4.3

2.25

0.89

25.43

2.9

0.9

41.48

4.7

2.3

1.79

18.96

1.1

1.4

40.1

2.9

2.35

1.95

17.92

0.9

1.8

35.54

2.0

2.4

1.93

31.86

1.7

1.8

30.25

1.7

2.45

1.82

25.59

1.4

1.9

43.09

2.2

2.5

1.64

25.76

1.6

1.7

39.29

2.3

2.55

1.37

18.05

1.3

1.6

22.73

1.4

2.6

1.39

12.53

0.9

1.7

25.09

1.5

2.65

1.43

8.02

0.6

2.0

56.54

2.9

2.7

1.85

1.04

0.1

2.6

7.39

0.3

2.75

2.08

22.01

1.1

2.8

65.69

2.4

2.8

2.91

22.89

0.8

2.0

41.71

2.1

2.85

2.25

27.81

1.2

2.8

32.53

1.2

2.9

2.31

26.98

1.2

3.0

6.23

0.2

2.95

3.16

33.59

1.1

3.4

25.43

0.8

3

4.06

35.36

0.9

4.2

29.46

0.7

3.05

4.22

30.76

0.7

4.2

36.64

0.9

3.1

4.13

21.95

0.5

4.8

32.69

0.7

3.15

3.83

23.21

0.6

4.9

37.05

0.8

3.2

4.15

23.95

0.6

5.0

39.63

0.8

3.25

4.75

27.59

0.6

5.4

37.33

0.7

3.3

5.67

31.23

0.6

6.0

39.37

0.7

3.35

5.70

34.04

0.6

6.4

44.96

0.7

3.4

4.99

28.83

0.6

6.3

50.39

0.8

3.45

4.13

20.18

0.5

5.5

30.07

0.5

3.5

3.45

17.88

0.5

4.2

28.97

0.7

3.55

2.52

19.08

0.8

3.0

18.86

0.6

3.6

1.88

11.21

0.6

2.3

17.44

0.8

3.65

1.43

8.22

0.6

1.9

16.42

0.9

3.7

1.37

7.79

0.6

2.1

20.2

1.0

3.75

1.55

7.95

0.5

3.1

27.22

0.9

3.8

1.97

9.87

0.5

4.5

34.22

0.8

3.85

2.65

12.8

0.5

5.0

32.74

0.7

3.9

3.48

10.62

0.3

4.8

29.58

0.6

3.95

3.51

13.63

0.4

4.3

23.58

0.5

4

3.02

10.74

0.4

3.4

24.11

0.7

4.05

2.61

12.11

0.5

2.9

21.48

0.7

4.1

2.31

24.01

1.0

2.6

33.37

1.3

4.15

2.38

26.57

1.1

3.7

37.37

1.0

4.2

2.70

25.7

1.0

5.7

38.61

0.7

4.25

5.25

37.92

0.7

6.1

85.16

1.4

4.3

4.06

52.78

1.3

3.2

68.95

2.2

4.35

2.12

45.37

2.1

1.9

44.41

2.3

4.4

1.76

49.74

2.8

1.7

37.11

2.1

4.45

3.07

52.31

1.7

2.4

72

3.0

4.5

2.27

62.89

2.8

2.1

76.5

3.7

4.55

2.19

49.26

2.2

2.4

62.01

2.5

4.6

1.93

50.62

2.6

2.0

51.57

2.6

4.65

2.80

76.29

2.7

5.8

56.99

1.0

4.7

3.96

59.79

1.5

3.8

56.7

1.5

4.75

2.49

55.36

2.2

2.3

45.21

2.0

4.8

1.87

37.82

2.0

2.0

32.53

1.7

4.85

1.85

36.54

2.0

1.9

31.27

1.6

4.9

2.01

35.32

1.8

2.1

36.48

1.7

4.95

2.22

51.11

2.3

2.5

69.11

2.7

5

3.37

61.63

1.8

3.0

74.36

2.5

5.05

3.50

92.06

2.6

3.6

72.49

2.0

5.1

5.28

74.73

1.4

16.5

78.55

0.5

5.15

14.03

77.09

0.5

21.6

98.08

0.5

5.2

17.80

83.25

0.5

24.6

121.15

0.5

5.25

22.74

139.02

0.6

27.6

157.1

0.6

5.3

24.36

141.93

0.6

29.3

157.23

0.5

5.35

23.90

130.53

0.5

28.7

143.57

0.5

5.4

22.81

111.92

0.5

28.9

150.45

0.5

5.45

22.65

76.4

0.3

27.9

111.37

0.4

5.5

19.51

92.57

0.5

26.8

94.36

0.4

5.55

22.27

79

0.4

22.6

106.73

0.5

5.6

21.65

72.79

0.3

26.1

138.04

0.5

5.65

16.93

102.34

0.6

26.0

168.5

0.6

5.7

17.07

115.93

0.7

23.7

138.04

0.6

5.75

17.80

109.62

0.6

23.1

105.55

0.5

5.8

19.35

95.36

0.5

23.5

94.06

0.4

5.85

18.30

93.97

0.5

24.0

94.97

0.4

5.9

17.72

68.66

0.4

22.5

106.38

0.5

5.95

17.31

64.68

0.4

20.4

48.48

0.2

6

17.44

70.66

0.4

19.9

65.65

0.3

6.05

17.84

76.01

0.4

19.9

73.28

0.4

6.1

18.68

91.9

0.5

22.2

95.48

0.4

6.15

21.38

113.2

0.5

23.3

112.89

0.5

6.2

21.37

118.71

0.6

23.9

125.65

0.5

6.25

19.14

132.24

0.7

22.7

138.18

0.6

6.3

18.35

112.98

0.6

21.8

146.2

0.7

6.35

17.55

111.63

0.6

21.0

150.07

0.7

6.4

16.97

110.94

0.7

20.6

154.4

0.8

6.45

17.02

110.6

0.6

20.4

-4.29

0.0

6.5

16.92

115.62

0.7

20.3

115.56

0.6

6.55

17.29

97.41

0.6

6.9

139.63

2.0

6.6

17.93

114.6

0.6

19.7

151.12

0.8

6.65

17.65

126.44

0.7

21.3

162.76

0.8

6.7

17.54

123.35

0.7

21.8

167.72

0.8

6.75

16.73

120.34

0.7

21.2

163.98

0.8

6.8

16.06

114.03

0.7

20.6

164.23

0.8

6.85

15.40

106.59

0.7

19.1

148.72

0.8

6.9

14.03

98.82

0.7

16.5

132.57

0.8

6.95

13.00

88.79

0.7

15.9

96.37

0.6

7

13.64

93.93

0.7

15.9

125.85

0.8

7.05

12.81

158.43

1.2

12.4

228.13

1.8

7.1

7.13

180.16

2.5

7.2

212.36

3.0

7.15

3.87

125.51

3.2

3.6

113.34

3.2

7.2

3.12

59.61

1.9

8.3

102.46

1.2

7.25

9.75

48.26

0.5

15.3

105.39

0.7

7.3

13.79

62.58

0.5

17.4

97.15

0.6

7.35

15.19

65.67

0.4

17.6

107.91

0.6

7.4

15.60

83.15

0.5

20.8

119.47

0.6

7.45

17.91

78.71

0.4

22.3

105.65

0.5

7.5

17.03

82.09

0.5

23.0

135.78

0.6

7.55

16.38

78.23

0.5

25.8

151.27

0.6

7.6

17.67

74.14

0.4

25.2

148.78

0.6

7.65

19.09

98.12

0.5

22.6

193.03

0.9

7.7

15.89

105.08

0.7

19.5

190.37

1.0

7.75

13.73

98.14

0.7

17.2

173.36

1.0

7.8

14.03

93.2

0.7

15.9

182.84

1.2

7.85

12.85

157.33

1.2

16.3

189.98

1.2

7.9

5.97

146.85

2.5

14.2

272.85

1.9

7.95

3.74

97.6

2.6

6.1

247.86

4.1

8

3.38

58.84

1.7

3.4

164.55

4.9

8.05

3.88

80.91

2.1

3.4

75.85

2.2

8.1

12.28

89.76

0.7

5.0

98.57

2.0

8.15

15.27

83.11

0.5

17.4

106.55

0.6

8.2

15.61

84.84

0.5

17.1

89.09

0.5

8.25

14.78

84.59

0.6

16.4

91.47

0.6

8.3

16.35

91.04

0.6

15.0

88.01

0.6

8.35

17.78

103.6

0.6

14.1

85.21

0.6

8.4

18.35

111.25

0.6

14.0

88.68

0.6

8.45

16.68

120.83

0.7

15.8

62.09

0.4

8.5

16.19

108.05

0.7

17.5

73.95

0.4

8.55

14.91

101.44

0.7

16.0

97.88

0.6

8.6

14.79

136.64

0.9

17.2

115.82

0.7

8.65

13.59

101.58

0.7

17.3

107.89

0.6

8.7

15.29

91.33

0.6

17.7

86.59

0.5

8.75

19.52

105.59

0.5

19.0

92.12

0.5

8.8

22.41

115.62

0.5

21.0

68.01

0.3

8.85

25.06

113.12

0.5

24.6

-4.39

0.0

8.9

28.08

211.2

0.8

8.95

30.45

168.76

0.6

9

31.94

206.81

0.6

9.05

34.32

187.46

0.5

9.1

34.64

211.51

0.6

9.15

37.37

315.14

0.8

9.2

42.11

344.06

0.8

9.25

38.23

341.31

0.9

9.3

33.55

256.45

0.8

9.35

36.35

195.58

0.5

9.4

34.96

120.52

0.3

9.45

35.90

126.46

0.4

9.5

36.37

124

0.3

9.55

35.85

168.5

0.5

9.6

34.06

293.97

0.9

9.65

34.10

253.82

0.7

9.7

29.22

207.44

0.7

9.75

29.14

145.16

0.5

9.8

32.31

181.05

0.6

9.85

30.29

199.02

0.7

9.9

29.50

-32768

-111.1

9.95

34.62

-32768

-94.6

SOLUTION: (a) qc vs. depth, fs vs. depth, and qc/fs vs. depth plots are given in S-Figure 7-6.

cone resistance, qc (MPa) 0

0

10

20

40

CPT 1 CPT 2

2

depth (m)

30

4

6

8

10

(i)

50

sleeve friction, fs (kPa) 0

0

100

200

300

CPT 1 CPT 2

2

depth (m)

400

4

6

8

10

(ii)

friction ratio, fs/qc (%) 0

0

2

4

6

8

depth (m)

2

CPT 1 CPT 2

4

6

8

10

(iii)

S-Figure 7-6

(b) In order for us to use Figure 7-26, we need to calculate lateral effective stress at each depth. At z = 6.1m qc = 18. 68MPa from CPT 1 and qc = 22.2 MPa from CPT 2 qc, avg = (18. 68+22.2)/2 = 20.44MPa = 20440kPa qc, avg /pA = 20440/100=204.4 σ'v = 14kN / m3 × 1.5m + 15kN / m3 × (4.3 − 1.5)m +14.5kN / m3 × (5.2 − 4.3)m + 19kN / m3 × (6.1 − 5.2)m = 93.2kPa

σ'h = 0.4 × 93.2 = 37.3kPa σ’h/pA = 37.3/100=0.373 From Figure 7-26(d), we can chart off relative density: DR = 80% answer To estimate peak friction angle, let us use Eq. (5.8) and Eq. (5.16). Assuming φp = 42° , N=

1 + sin42o = 5.04 1 - sin42o

σ 'c = σ 'm =

93.2 + 2 × 37.3 = 55.9kPa 3

' σ1p = 55.9 × 5.04 = 281.7 kPa

σ 'mp =

σ '1p + 2σ '3p 3

=

281.7 + 2 × 55.9 = 131.2 kPa 3

⎡ ⎛ 100 ×131.2 ⎞ ⎤ I R = 0.8 ⎢10 − ln ⎜ ⎟ ⎥ − 1 = 3.099 100 ⎝ ⎠⎦ ⎣ φP = φc + 3I R Considering φc = 36◦, φP = 36o + 3 × 3.099 = 45.3o . After iterations, we get

φP = 45o answer

At z = 7.6m qc = 17. 67MPa from CPT 1 and qc = 25.2 MPa from CPT 2 qc, avg = (17. 67+25.2)/2 = 21.435MPa = 21435kPa qc, avg /pA = 21435/100=214.35 σ'v = 14kN / m3 × 1.5m + 15kN / m3 × (4.3 − 1.5)m +14.5kN / m3 × (5.2 − 4.3)m + 19kN / m3 × (7.6 − 5.2)m = 121.7kPa

σ'h = 0.4 ×121.7 = 48.7 kPa σ’h/pA = 48.7/100=0.487 From Figure 7-26(d), we can chart off relative density: DR = 75% answer To estimate peak friction angle, let us use Eq. (5.8) and Eq. (5.16). Assuming φp = 42° , N=

1 + sin42o = 5.04 1 - sin42o

σ 'c = σ 'm =

121.7 + 2 × 48.7 = 73kPa 3

' σ1p = 73 × 5.04 = 367.9 kPa

σ 'mp =

σ '1p + 2σ '3p 3

=

367.9 + 2 × 73 = 171.3kPa 3

⎡ ⎛ 100 × 171.3 ⎞ ⎤ I R = 0.75 ⎢10 − ln ⎜ ⎟ ⎥ − 1 = 2.642 100 ⎝ ⎠⎦ ⎣ φP = φc + 3I R Considering φc = 36◦, φP = 36o + 3 × 2.642 = 43.9o . After iterations, we get

φP = 43.8o answer

At z = 9.1m qc = 34.64MPa from CPT 1 qc /pA = 34640/100=346.4 σ'v = 14kN / m3 × 1.5m + 15kN / m3 × (4.3 − 1.5)m +14.5kN / m3 × (5.2 − 4.3)m + 19kN / m3 × (9.1 − 5.2)m = 150.2kPa

σ'h = 0.4 ×150.2 = 60.1kPa σ’h/pA = 60.1/100=0.601 From Figure 7-26(d), we can chart off relative density: DR = 95% answer To estimate peak friction angle, let us use Eq. (5.8) and Eq. (5.16). Assuming φp = 47° , N=

1 + sin47 o = 6.44 1 - sin47 o

σ 'c = σ 'm =

150.2 + 2 × 60.1 = 90.1kPa 3

' σ1p = 90.1× 6.44 = 580.2 kPa

σ 'mp =

σ '1p + 2σ '3p 3

=

580.2 + 2 × 90.1 = 253.5 kPa 3

⎡ ⎛ 100 × 253.5 ⎞ ⎤ I R = 0.95 ⎢10 − ln ⎜ ⎟ ⎥ − 1 = 3.241 100 ⎝ ⎠⎦ ⎣ φP = φc + 3I R Considering φc = 36◦, φP = 36o + 3 × 3.241 = 45.7o . After iterations, we get

φP = 45.9o answer

(c) Let us first calculate N60. For an safety hammer ERhammer = 60 % Thus, C h =

ER hammer 60 = = 1.0 ER safety 60

Following Eq. (7.3), for 6m ≤ rod length < 10m, Cr = 0.95. Standard ASTM split spoon sampler with a liner was used and the borehole diameter was within the recommended range, so Cd = 1 and Cs = 1. Now N60 can be calculated using Eq. (7.1) as N 60 = Ch Cd Cr Cs NSPT at 6.1m, N60 = (1)(1)(0.95)(1)27=25.7 at 7.6m, N60 = (1)(1)(0.95)(1)25 = 23.8 at 9.1m, N60 = (1)(1)(0.95)(1)40 = 38

and from (b), vertical effective stress at each depth at 6.1m, σ’v = 93.2kPa

at 7.6m, σ’v = 121.7kPa at 9.1m, σ’v = 150.2kPa

Following Eq. (7.6) DR = 100%

N 60 A + BC

σ' v pA

where, A = 36.5, B = 27, pA = 100 kPa. Following Eq. (7.8) C=

K0 ; and so for a normally consolidated sand C = 1. K 0,NC

From the above equations, relative density DR of the sand deposit at 6.1m can be calculated as D R = 100

25.7 95.2 36.5 + 27 ×1× 100

= 64.3%

answer

Similarly, at 7.6m, D R = 100

at 9.1m, D R = 100

23.8 121.7 36.5 + 27 × 1× 100 38 150.2 36.5 + 27 × 1× 100

= 58.6%

answer

= 70.2%

answer

To estimate the peak friction angle, let us assume φp = 42° for a sand deposit at a depth of 6.1m.

At z = 6.1m N=

1 + sin42o = 5.04 1 - sin42o

σ 'c = σ 'm =

93.2 + 2 × 37.3 = 55.9kPa 3

' σ1p = 55.9 × 5.04 = 281.7 kPa

σ 'mp =

σ '1p + 2σ '3p 3

=

281.7 + 2 × 55.9 = 131.2 kPa 3

⎡ ⎛ 100 × 131.2 ⎞ ⎤ I R = 0.643 ⎢10 − ln ⎜ ⎟ ⎥ − 1 = 2.294 100 ⎝ ⎠⎦ ⎣ φP = φc + 3I R Considering φc = 36◦, φP = 36o + 3 × 2.294 = 42.9o . After iterations, we get

φP = 42.8o answer

At z = 7.6m Assuming φp = 42° , N=

1 + sin42o = 5.04 1 - sin42o

σ 'c = σ 'm =

121.7 + 2 × 48.7 = 73kPa 3

' σ1p = 73 × 5.04 = 367.9 kPa

'

σ mp =

σ '1p + 2σ '3p 3

=

367.9 + 2 × 73 = 171.3kPa 3

⎡ ⎛ 100 × 171.3 ⎞ ⎤ I R = 0.586 ⎢10 − ln ⎜ ⎟ ⎥ − 1 = 1.846 100 ⎝ ⎠⎦ ⎣

φP = φc + 3I R Considering φc = 36◦, φP = 36o + 3 × 1.846 = 41.5o . After iterations, we get

φP = 41.6o answer

At z = 9.1m Assuming φp = 47° , N=

1 + sin47 o = 6.44 1 - sin47 o

σ 'c = σ 'm =

150.2 + 2 × 60.1 = 90.1kPa 3

' σ1p = 90.1× 6.44 = 580.2 kPa

σ 'mp =

σ '1p + 2σ '3p 3

=

580.2 + 2 × 90.1 = 253.5 kPa 3

⎡ ⎛ 100 × 253.5 ⎞ ⎤ I R = 0.702 ⎢10 − ln ⎜ ⎟ ⎥ − 1 = 2.134 100 ⎝ ⎠⎦ ⎣ φP = φc + 3I R Considering φc = 36◦, φP = 36o + 3 × 2.134 = 42.4o . After iterations, we get

φP = 42.7o answer (d) Summarized results obtained from (b) and (c) are given in S-Table 7-7

S-Table 7-7 From CPT results (b) Depth (m)

From SPT results (c)

DR (%)

φp (degrees)

DR (%)

φp (degrees)

6.1

80

45

64.3

42.8

7.6

75

43.8

58.6

41.6

9.1

95

45.8

70.2

42.7

As seen in S-Table 7-7, values estimated from both CPT and SPT logs follow similar trends. This trend is in good agreement with cone resistance profile in S-Figure 7-6. However, both relative densities and peak friction angles estimated from CPT logs are larger than those from the SPT logs. This is part a result of the values selected for the constants A and B in Skempton correlations, about which there is considerable uncertainty, accounted for to some extent by the use of values that are on the conservative side. End

Problem 7-17 To estimate the undrained shear strength of a normally consolidated soft clay deposit, vane shear tests were performed at four different depths. In these tests, a rectangular vane with 60 mm diameter and 120 mm height was used. Both ends of the vane were inserted in the soil. The plasticity index of the clay is equal to 65%. The unit weight of the clay is 16 kN/m3, and the water table is at the ground surface. The results are shown in Table 7-11.

Table 7-11 Vane shear test results for Problem 7-17.

DEPTH (m)

Maximum Torque (N.m)

3

7

5

11.1

8

18.5

10

22.4

(a) Estimate the undrained shear strength for each depth and develop a plot of the design undrained shear strength vs. depth (depth on the vertical axis and undrained shear strength on the horizontal axis). To obtain the design undrained shear strength, use Eq. (7.26). (b) Estimate the in-situ undrained shear strength using the correlation of Eq. (6.52). SOLUTION: (a) Vane diameter B = 60mm Vane height H = 120 mm = 2B The undrained shear strength (su)FV of the clay from a field vane shear test (when H = 2B) is given by

( s u )FV =

12T ; n =2, for fully inserted vane πB (12 + n ) 3

Maximum moment required to rotate the vane at 3m depth, T = 7N.m

( s u )FV,at3m =

12 × 7 × 10−3 kPa = 8.8 ≈ 9kPa answer 3.14 × (0.06)3 × (12 + 2 )

similarly for other depths

( s u )FV,at5m =

12 × 11.1×10−3 kPa = 14kPa 3.14 × (0.06)3 × (12 + 2 )

( s u )FV,at8m =

12 ×18.5 × 10−3 kPa = 23.3kPa 3.14 × (0.06)3 × (12 + 2 )

( s u )FV,at10m =

12 × 22.4 × 10−3 kPa = 28.3kPa 3.14 × (0.06)3 × (12 + 2 )

Design undrained shear strength s u = λ ( s u )FV where, λ = 1.18 − 0.0107(PI) + 0.0000513(PI) 2 ≤ 1 . Thus for PI = 65%, λ = 0.7. Design shear strength at depths 3, 5, 8, and 10m are 6.2 (≈ 6), 9.8 (≈ 10), 16.3 (≈1 6), and 19.8 (≈ 20) kPa respectively. answer to (a)

Design Undrained Shear Strength (kPa) 5

10

0

Depth (m)

2 4 6 8 10

S-Figure 7-7

15

20

(b) Using Eq. (6.52), the in-situ undrained shear strength su = 0.11 + 0.0037(PI) σ' v su,at 3m = 3(16-9.81)×[0.11+0.0037(65)] ≈ 7 kPa su,at 5m = 5(16-9.81)×[0.11+0.0037(65)] ≈ 11 kPa su,at 8m = 8(16-9.81)×[0.11+0.0037(65)] ≈ 17 kPa su,at 10m = 10(16-9.81)×[0.11+0.0037(65)] ≈ 22 kPa End