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:

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,

40m

60m30m

20m

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

( )u 3

12TsB 12 n

=π +

; n =2, for fully inserted vane

( )

3

u 3

12 70 10s kPa 88.5 88 kPa3.14 (0.06) 12 2

−× ×= = ≈

× × +answer

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:

c k u vq N s= + σ

Now, we calculate total vertical stress at a depth of 5m.

v 17 5 85kPaσ = × =

Using Nk=12 for clay layer and su=88.5kPa from Problem 7-4, we can estimate

cone resistance at a depth of 5m:

cq 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:

v satz 17 10 170kPaσ = γ = × =

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

c k u vq N s= + σ

c vu

k

qsN− σ

=

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 Ch Cr N60 σv' (kPa) DR φp (deg)

4 25

5 30

6 35

SOLUTION:

For an automatic trip hammer ERhammer = 80 %

Thus, hammerh

safety

ER 80C 1.33ER 60

= = =

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

60 h d r s SPTN C C C C N=

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)

60R'v

A

ND100% A BC

p

=σ

+

where, A = 36.5, B = 27, pA = 100 kPa.

Following Eq. (7.8)

0

0,NC

KCK

= ; and so for a normally consolidated sand C = 1.

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

(m) 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

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, hammerh

safety

ER 45C 0.75ER 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

60 h d r s SPTN C C C C N=

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):

( ) 0,NCA1 60 '60

v 0

KpN NK

=σ

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

0 0,NCK K OCR=

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

' 3 3v 17kN / m 3m (17 9.81)kN / m 1m 58.2kPaσ = × + − × =

' ' 3v0 v ' 58.2kPa 17kN / m 2m 92.2kPaσ = σ + Δσ = + × =

'v0'v

92.2OCR 1.5858.2

σ= = =

σ

0 0,NCK K OCR 0.48 1.58 0.60= = × =

( )1 60

100 0.48N 17.6 20.658.2 0.60

= =

Similarly, we obtain the following table.

S-Table 7-2

Depth (m) NSPT N60 σ'v (kPa) σ'v0 (kPa) OCR K0 (N1)60

1 15 10.1 17 51 3 0.83 18.6

2 18 12.2 34 68 2 0.68 17.6

3 22 14.9 51 85 1.67 0.62 18.4

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)

'c h

cA A

R 'h

cA

qln 0.4947 0.1041 0.841lnp p

D0.0264 0.0002 0.0047 ln

p

⎛ ⎞ ⎛ ⎞σ− − φ −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠=⎛ ⎞σ

− φ − ⎜ ⎟⎝ ⎠

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◦ ,

R

11 43.5ln 0.4947 0.1041 30 0.841ln0.1 100D 73%

43.50.0264 0.0002 30 0.0047 ln100

⎛ ⎞ ⎛ ⎞− − × −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠= =

⎛ ⎞− × − ⎜ ⎟⎝ ⎠

answer

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

( )R0.841 0.0047D

c hc c R

A A

q '1.64exp 0.1041 0.0264 0.0002 Dp p

−⎛ ⎞σ

⎡ ⎤= φ + − φ ⎜ ⎟⎣ ⎦⎝ ⎠

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

'v 1 22 22kPaσ = × = . Given that K0 = 0.45 for this deposit, the horizontal stress at 1m

'h 0.45 22 9.9kPaσ = × = . The reference stress is pA = 100kPa. So the cone resistance

value at 1m depth can be calculated as:

( )0.841 0.0047 61.5

c9.9q 100 1.64exp 0.1041 30 0.0264 0.0002 30 61.5100

3645kPa

− ×⎡ ⎤⎛ ⎞⎡ ⎤⎢ ⎥= × + − × ⎜ ⎟⎣ ⎦ ⎝ ⎠⎢ ⎥⎣ ⎦=

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

40970.8kPa. So 0

c

G 40971 11.2q 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) σ'v (kPa) σ'

h (kPa) DR (%) qc(kPa) G0 (kPa) G0/qc 0.5 11 5 60.75 2436 29918.2 12.3 1 22 9.9 61.5 3644.6 40970.8 11.2

1.5 33 14.9 62.25 4668.9 49412.1 10.6 2 44 19.8 63 5572.5 56593.1 10.2

2.5 55 24.8 63.75 6428.6 62829.2 9.8 3 66 29.7 64.5 7227.9 68401.7 9.5

3.5 77 34.7 65.25 8009.8 73755.8 9.2 4 88 39.6 66 8756.6 78575.7 9

4.5 99 44.6 66.75 9497.9 83347.2 8.8 5 110 49.5 67.5 10214.5 87744.4 8.6

5.5 121 54.5 68.25 10931.5 91953 8.4 6 132 59.4 69 11629.8 96131.2 8.3

6.5 143 64.4 69.75 12331.8 100050.9 8.1 7 154 69.3 70.5 13019 103998.4 8

7.5 165 74.3 71.25 13711.9 107700.2 7.9 8 176 79.2 72 14392.5 111308.3 7.7

8.5 187 84.2 72.75 15080.2 115049.5 7.6 9 198 89.1 73.5 15757.5 118463.1 7.5

9.5 209 94.1 74.25 16442.8 122082.7 7.4 10 220 99 75 17119.2 125377.2 7.3

6 8 10 12 14G0/qc

10

8

6

4

2

0D

epth

(m)

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 ( )'v 1 22 9.81 12.2kPaσ = × − = .

Given that K0 = 0.45 for this deposit, horizontal stress at 1m 'h 0.45 12.2 5.5kPaσ = × = .

pA = 100kPa (refernce stress). So the cone resistance value at 1m depth can be calculated

as

( )0.841 0.0047 61.5

c5.5q 100 1.64exp 0.1041 30 0.0264 0.0002 30 61.5100

2634.8kPa

− ×⎡ ⎤⎛ ⎞⎡ ⎤⎢ ⎥= × + − × ⎜ ⎟⎣ ⎦ ⎝ ⎠⎢ ⎥⎣ ⎦=

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

kPa. So 0

c

G 31407.9 11.9q 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) σ'v (kPa) σ'

h (kPa) DR (%) qc(kPa) G0 (kPa) G0/qc 0.5 6.1 2.7 60.75 1729.9 22727.1 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

6 8 10 12 14G0/qc

10

8

6

4

2

0

Dep

th (m

)

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, hammerh

safety

ER 60C 1.0ER 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

60 h d r s SPTN C C C C N=

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)

60R'v

A

ND100% A BC

p

=σ

+

where, A = 36.5, B = 27, pA = 100 kPa.

Following Eq. (7.8)

0

0,NC

KCK

= ; and so for a normally consolidated sand C = 1.

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

R9D 100 42.7%47.336.5 27 1

100

= =+ × ×

answer to (a)

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

at 11.5ft (= 3.5m), R19.5D 100 60.7%60.736.5 27 1

100

= =+ × ×

answer to (a)

at 14ft (= 4.3m), R14.5D 100 50.3%76.736.5 27 1

100

= =+ × ×

answer to (a)

at 16.5ft (= 5.03m), R17D 100 52.7%91.336.5 27 1

100

= =+ × ×

answer to (a)

at 19ft (= 5.79m), R11.1D 100 41.2%106.536.5 27 1

100

= =+ × ×

answer to (a)

Calculation of peak friction angle φp

Following Eq. (5.8),

'mp

R D QA

100I I Q ln R

p

⎡ ⎤⎛ ⎞σ= − −⎢ ⎥⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

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

'hσ = 0.45 47.3 = 21.3kPa×

Thus in situ mean effective stress

'm

47.3+2 21.3σ = = 30kPa3×

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

' 'm 3pσ = σ = 30kPa

Assume oP = 37φ .

o

o

1 + sin37N = = 4.021 - sin37

'1pσ = 30 4.02 = 120.6kPa×

' '1p 3p'

mp

σ + 2σ 120.6 + 2 30σ = = = 60.2kPa3 3

×

R100 60.2I 0.427 10 ln 1 1.52

100⎡ ⎤×⎛ ⎞= − − =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

Following Eq. (5.10), for triaxial condition

P c R3Iφ = φ +

Considering φc = 32◦, P 32 3 1.52 36.6φ = + × =o o . 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 P 36.6φ = o 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 10 20 30 40 50 60 70 80N60

3.0

2.5

2.0

1.5

1.0

0.5

0.0σ'

v/p A

φ=45 ο

φ=50 ο

φ=40 ο

φ=35 οφ=30 ο

φ=25 ο

0 10 20 30 40 50 60 70 80N60

3.0

2.5

2.0

1.5

1.0

0.5

0.0σ'

v/p A

φ=45 ο

φ=50 ο

φ=40 ο

φ=35 οφ=30 ο

φ=25 ο

S-Figure 7-5

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

S-Table 7-5

Depth

(ft) NSPT Cr N60

σ′v

(kPa)DR (%)

φp, equation

(deg)

φp, chart

(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),

'mp

R D QA

100I I Q ln R

p

⎡ ⎤⎛ ⎞σ= − −⎢ ⎥⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

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

' 'v 3pσ = σ = 47.3kPa

Assume oP = 37φ .

o

o

1 + sin37N = = 4.021 - sin37

'1pσ = 47.3 4.02 = 190.1kPa×

' '1p 3p'

mp

σ + 2σ 190.1 + 2 47.3σ = = = 94.9kPa3 3

×

R100 94.9I 0.427 10 ln 1 1.326

100⎡ ⎤×⎛ ⎞= − − =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

Following Eq. (5.10), for triaxial condition

P c R3Iφ = φ +

Considering φc = 32◦, P 32 3 1.326 36φ = + × =o o . 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 P 36φ = o 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),

'mp

R D QA

100I I Q ln R

p

⎡ ⎤⎛ ⎞σ= − −⎢ ⎥⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

where Q = 10 and RQ = 1 for clean silica sand

K0 = 0.45 for the sand layer

At 9.3ft (=2.83m)

'3pσ = 0.45 47.3 = 21.3kPa×

Assume oP = 37φ .

o

o

1 + sin37N = = 4.021 - sin37

'1pσ = 21.3 4.02 = 85.6kPa×

' '1p 3p'

mp

σ + 2σ 85.6 + 2 21.3σ = = = 42.7kPa3 3

×

R100 42.7I 0.427 10 ln 1 1.667

100⎡ ⎤×⎛ ⎞= − − =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

Following Eq. (5.10), for triaxial condition

P c R3Iφ = φ +

Considering φc = 32◦, P 32 3 1.667 37.0φ = + × =o o . It is same as the assumed value.

Therefore, we need not to iterate any further and at this depth (9.3ft) P 37.0φ = o 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, hammerh

safety

ER 60C 1.0ER 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

60 h d r s SPTN C C C C N=

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)

60R'v

A

ND100% A BC

p

=σ

+

where, A = 36.5, B = 27, pA = 100 kPa.

Following Eq. (7.8)

0

0,NC

KCK

= ; and so for a normally consolidated sand C = 1.

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

R22.5D 100 64.8%6336.5 27 1

100

= =+ × ×

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)

'mp

R D QA

100I I Q ln R

p

⎡ ⎤⎛ ⎞σ= − −⎢ ⎥⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

where Q = 10 and RQ = 1 for clean silica sand

Mean stress, ' ' '

' v h 0 vm

2 (1 2K )3 3

σ + σ + σσ = =

Considering K0 = 0.45 for the sand layer; at 11.5ft depth

'm

(1 2 0.45) 63 39.9kPa3

+ × ×σ = =

Consolidation stress in triaxial test on undisturbed sample σ′c = 39.9kPa = σ′3

Say φp = 36◦, so

( )2 2PPN tan 45 tan 45 18 3.85

2φ⎛ ⎞= + = + =⎜ ⎟

⎝ ⎠o o o

' '1 P 3N 3.85 39.9 153.6kPaσ = σ = × =

'mp

153.6 2 39.9 77.8kPa3

+ ×σ = =

R,at11.5ft100 77.8I 0.648 10 ln 1 2.66

100⎡ ⎤×⎛ ⎞= − − =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

Following Eq.(5.10), for triaxial condition

P c R3Iφ = φ +

For φc = 30◦, at11.5ftP 30 3 2.66 37.98φ = + × =o o .

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

(kPa)DR (%)

φp, equation

(deg)

φp, chart

(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

(from borings)

Unit

Weight

(kN/m3)

Depth

(m)

NSPT

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

(m)

qc

(MPa)

fs

(kPa)

fs/qc

(%)

qc

(MPa)

fs

(kPa)

fs/qc

(%)

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

SOLUTION:

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

10

8

6

4

2

0

dept

h (m

)

0 10 20 30 40 50

cone resistance, qc (MPa)

CPT 1CPT 2

(i)

9.95 34.62 -32768 -94.6

10

8

6

4

2

0

dept

h (m

)

0 100 200 300 400

sleeve friction, fs (kPa)

CPT 1CPT 2

(ii)

10

8

6

4

2

0

dept

h (m

)

0 2 4 6 8

friction ratio, fs/qc (%)

CPT 1CPT 2

(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

' 3 3v

3 3

14kN / m 1.5m 15kN / m (4.3 1.5)m

14.5kN / m (5.2 4.3)m 19kN / m (6.1 5.2)m93.2kPa

σ = × + × −

+ × − + × −=

'h 0.4 93.2 37.3kPaσ = × =

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

p 42φ = ° ,

o

o

1 + sin42N = = 5.041 - sin42

' 'c m

93.2 2 37.3σ σ 55.9kPa3

+ ×= = =

'1pσ = 55.9 5.04 = 281.7 kPa×

' '1p 3p'

mp

σ + 2σ 281.7 + 2 55.9σ = = = 131.2kPa3 3

×

R100 131.2I 0.8 10 ln 1 3.099

100⎡ ⎤×⎛ ⎞= − − =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

P c R3Iφ = φ +

Considering φc = 36◦, P 36 3 3.099 45.3φ = + × =o o . After iterations, we get

P 45φ = o 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

' 3 3v

3 3

14kN / m 1.5m 15kN / m (4.3 1.5)m

14.5kN / m (5.2 4.3)m 19kN / m (7.6 5.2)m121.7kPa

σ = × + × −

+ × − + × −=

'h 0.4 121.7 48.7 kPaσ = × =

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

p 42φ = ° ,

o

o

1 + sin42N = = 5.041 - sin42

' 'c m

121.7 2 48.7σ σ 73kPa3

+ ×= = =

'1pσ = 73 5.04 = 367.9 kPa×

' '1p 3p'

mp

σ + 2σ 367.9 + 2 73σ = = = 171.3kPa3 3

×

R100 171.3I 0.75 10 ln 1 2.642

100⎡ ⎤×⎛ ⎞= − − =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

P c R3Iφ = φ +

Considering φc = 36◦, P 36 3 2.642 43.9φ = + × =o o . After iterations, we get

P 43.8φ = o answer

At z = 9.1m

qc = 34.64MPa from CPT 1

qc /pA = 34640/100=346.4

' 3 3v

3 3

14kN / m 1.5m 15kN / m (4.3 1.5)m

14.5kN / m (5.2 4.3)m 19kN / m (9.1 5.2)m150.2kPa

σ = × + × −

+ × − + × −=

'h 0.4 150.2 60.1kPaσ = × =

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

p 47φ = ° ,

o

o

1 + sin47N = = 6.441 - sin47

' 'c m

150.2 2 60.1σ σ 90.1kPa3

+ ×= = =

'1pσ = 90.1 6.44 = 580.2 kPa×

' '1p 3p'

mp

σ + 2σ 580.2 + 2 90.1σ = = = 253.5kPa3 3

×

R100 253.5I 0.95 10 ln 1 3.241

100⎡ ⎤×⎛ ⎞= − − =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

P c R3Iφ = φ +

Considering φc = 36◦, P 36 3 3.241 45.7φ = + × =o o . After iterations, we get

P 45.9φ = o answer

(c) Let us first calculate N60.

For an safety hammer ERhammer = 60 %

Thus, hammerh

safety

ER 60C 1.0ER 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

60 h d r s SPTN C C C C N=

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)

60R'v

A

ND100% A BC

p

=σ

+

where, A = 36.5, B = 27, pA = 100 kPa.

Following Eq. (7.8)

0

0,NC

KCK

= ; and so for a normally consolidated sand C = 1.

From the above equations, relative density DR of the sand deposit at 6.1m can be

calculated as

R25.7D 100 64.3%95.236.5 27 1

100

= =+ × ×

answer

Similarly,

at 7.6m, R23.8D 100 58.6%121.736.5 27 1

100

= =+ × ×

answer

at 9.1m, R38D 100 70.2%150.236.5 27 1

100

= =+ × ×

answer

To estimate the peak friction angle, let us assume p 42φ = ° for a sand deposit at a

depth of 6.1m.

At z = 6.1m

o

o

1 + sin42N = = 5.041 - sin42

' 'c m

93.2 2 37.3σ σ 55.9kPa3

+ ×= = =

'1pσ = 55.9 5.04 = 281.7 kPa×

' '1p 3p'

mp

σ + 2σ 281.7 + 2 55.9σ = = = 131.2kPa3 3

×

R100 131.2I 0.643 10 ln 1 2.294

100⎡ ⎤×⎛ ⎞= − − =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

P c R3Iφ = φ +

Considering φc = 36◦, P 36 3 2.294 42.9φ = + × =o o . After iterations, we get

P 42.8φ = o answer

At z = 7.6m

Assuming p 42φ = ° ,

o

o

1 + sin42N = = 5.041 - sin42

' 'c m

121.7 2 48.7σ σ 73kPa3

+ ×= = =

'1pσ = 73 5.04 = 367.9 kPa×

' '1p 3p'

mp

σ + 2σ 367.9 + 2 73σ = = = 171.3kPa3 3

×

R100 171.3I 0.586 10 ln 1 1.846

100⎡ ⎤×⎛ ⎞= − − =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

P c R3Iφ = φ +

Considering φc = 36◦, P 36 3 1.846 41.5φ = + × =o o . After iterations, we get

P 41.6φ = o answer

At z = 9.1m

Assuming p 47φ = ° ,

o

o

1 + sin47N = = 6.441 - sin47

' 'c m

150.2 2 60.1σ σ 90.1kPa3

+ ×= = =

'1pσ = 90.1 6.44 = 580.2 kPa×

' '1p 3p'

mp

σ + 2σ 580.2 + 2 90.1σ = = = 253.5kPa3 3

×

R100 253.5I 0.702 10 ln 1 2.134

100⎡ ⎤×⎛ ⎞= − − =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

P c R3Iφ = φ +

Considering φc = 36◦, P 36 3 2.134 42.4φ = + × =o o . After iterations, we get

P 42.7φ = o answer

(d) Summarized results obtained from (b) and (c) are given in S-Table 7-7

S-Table 7-7

From CPT results (b) From SPT results (c) Depth (m)

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

( ) ( )u 3FV

12TsB 12 n

=π +

; n =2, for fully inserted vane

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

( ) ( )3

u 3FV,at3m

12 7 10s kPa 8.8 9kPa3.14 (0.06) 12 2

−× ×= = ≈

× × +answer

similarly for other depths

( ) ( )3

u 3FV,at5m

12 11.1 10s kPa 14kPa3.14 (0.06) 12 2

−× ×= =

× × +

( ) ( )3

u 3FV,at8m

12 18.5 10s kPa 23.3kPa3.14 (0.06) 12 2

−× ×= =

× × +

( ) ( )3

u 3FV,at10m

12 22.4 10s kPa 28.3kPa3.14 (0.06) 12 2

−× ×= =

× × +

Design undrained shear strength ( )u u FVs s= λ

where, 21.18 0.0107(PI) 0.0000513(PI) 1λ = − + ≤ . 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)

0

2

4

6

8

10

5 10 15 20

Design Undrained Shear Strength (kPa)

Dep

th (m

)

S-Figure 7-7

(b)

Using Eq. (6.52), the in-situ undrained shear strength

u'v

s 0.11 0.0037(PI)= +σ

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