Chapter-17-us-final-solutions.pdf

Chapter 17 17.1

Shelby tube A: Do2 − Di2 (76.2) 2 − (73) 2 Eq. (17.6): AR (%) = (100) = × 100 = 8.96% ≤ 10% (73) 2 Di2 Samples can be considered undisturbed. Suitable for grain size distribution, Atterberg limits, consolidation and unconfined compression tests. Shelby tube B: Do2 − Di2 (3.5) 2 − (3.375) 2 Eq. (17.6): AR (%) = (100) = × 100 = 7.54% ≤ 10% Di2 (3.375) 2 Samples can be considered undisturbed. Suitable for grain size distribution, Atterberg limits, consolidation and unconfined compression tests. Split spoon sampler: Do2 − Di2 (50.8) 2 − (35) 2 Eq. (17.6): AR (%) = (100) = × 100 = 110% ≥ 10% Di2 (35) 2 Samples can be considered as highly disturbed. Suitable for grain size distribution and Atterberg limits tests, but not for consolidation and unconfined compression tests.

17.2 Depth (m) 2 4 6 8 10

17.3

σ 0′ (kN/m2) 34 68 102 136 170

⎡σ ′ ⎤ CN = ⎢ 0 ⎥ ⎣ pa ⎦ 1.71 1.21 0.99 0.857 0.767

−0.5

N60

(N1)60 = CN N60

7 10 11 14 9

≈12 ≈12 ≈11 ≈12 ≈7

⎡ ⎤ ⎢ ⎥ N 60 ⎥ ; pa ≈ 100 kN/m2 φ ′ = tan −1 ⎢ ⎢ ⎛ σ o′ ⎞ ⎥ ⎢12.2 + 20.3⎜⎜ p ⎟⎟ ⎥ ⎝ a ⎠ ⎦⎥ ⎣⎢

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Depth (m) 2 4 6 8 10

17.4

σ o′ 2

(kN/m ) 34 68 102 136 170

po φ′ (deg) N60 2 (kN/m ) [Eq. (17.20)] 100 7 20.1 100 10 21.0 100 11 18.48 100 14 19.37 100 9 10.9 Average φ′ ≈ 18º

1 .7 ⎡ ⎛ 0.06 ⎞ ⎟ ⎢ N 60 ⎜⎜ 0.23 + D50 ⎟⎠ ⎢ ⎝ Eq. (17.18): Dr (%) = ⎢ 9 ⎢ ⎢ ⎣

0 .5

⎤ ⎥ ⎛ 98 ⎞⎥ ⎜⎜ ⎟⎟⎥ (100) ⎝ σ o′ ⎠⎥ ⎥ ⎦

Given γ = 15.7 kN/m3. The following table can now be prepared. Depth z (m) 1.5 3.0 4.5 6.0 7.5

17.5

σ o′ = γ z (kN/m2) 23.55 47.1 70.65 94.2 117.75

⎛ σ′ ⎞ ( N1 )60 = CN N 60 = ⎜⎜ 0 ⎟⎟ ⎝ pa ⎠ Depth (m) 1.5 3 4.5 6 7.5 9

D50 (mm) 0.3 0.3 0.3 0.3 0.3

Dr (%) 99.5 ≈ 100 74.2 ≈ 74 71.7 ≈ 72 70.4 ≈ 70 66.4 ≈ 66

N60 9 10 14 18 20

−0.5 2 N 60 ; φ ′ = 27.1 + 0.3( N1 ) 60 − 0.00054( N 1 ) 60

σ o′ (kN/m2) 1.5 × 18 = 27 3 × 18 = 54 4.5 × 18 = 81 ⎡(5.3 × 18) ⎤ ⎢⎣ + 0.7(18.8 − 9.8)⎥⎦ = 101.7 101.7 + 1.5(18.8 − 9.8) = 115.2 115.2 + 1.5(18.8 − 9.8) = 128.7

N60

CN

(N1)60

φ′

8 9 11

1.924 1.36 1.11

15.4 ≈ 15 12.24 ≈ 12 12.2 ≈ 12

(deg) 31.4 30.6 30.6

12

0.99

11.89 ≈ 12

30.6

15 17

0.931 0.881

13.96 ≈ 14 31.2 14.97 ≈ 15 31.4 Average φ′ ≈ 31°

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17.6

a. The properties should be averaged over a distance of 2B or 4 m below the footing, i.e. up to a depth of 5.5 m. Therefore, design N60 = (8 + 9 + 11) / 3 = 9.33 ≈ 9 and design φ ′ (deg) = (31.4 + 30.6 + 30.6) / 3 = 30.8 ⎛ Df b. Fd = 1 + 0.33⎜⎜ ⎝ B

⎞ 1 .5 ⎟⎟ = 1 + (0.33)⎛⎜ ⎞⎟ = 1.247 ⎝ 2 ⎠ ⎠

N ⎛ B + 0 .3 ⎞ 9 ⎛ 2 + 0 .3 ⎞ ⎛S ⎞ ⎛ 25 ⎞ 2 = 60 ⎜ ⎜ ⎟ (1.247)⎜ ⎟ = 185.53 kN/m ⎟ Fd ⎜ e ⎟ = 0.08 ⎝ B ⎠ 25 0 . 08 2 25 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 2

q net

2

Qnet = qnet × B 2 = 185.53 × 4 = 742 kN

17.7

⎛ qc ⎞ ⎜⎜ ⎟⎟ p 0.26 . Use pa ≈ 100 kN/m2. Eq. (17.39): ⎝ a ⎠ = 7.64 D50 N 60 Depth (m) 1.5 3 4.5 6 7.5 9

17.8

N60 8 9 11 12 15 17

D50 (mm) 0.28 0.28 0.28 0.28 0.28 0.28

qc (kN/m2) 4,390 4,938 6,036 6,585 8,231 9,328

Eq. (17.33): E s = 3qc Using qc from Problem 17.7: Depth (m) 1.5 3 4.5 6 7.5 9

qc (kN/m2) 4,390 4,938 6,036 6,585 8,231 9,328

Es (kN/m2) 13,170 14,814 18,108 19,755 24,693 27,984

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17.9

Eq. (17.35): cu =

cu =

qc − σ c ; Nk ≈ 18.3 Nk

26000 − (22)(118) = 1278.9 lb/ft 2 18.3

17.10 From Eq. (17.43):

⎛ 4.5 ⎞ Recovery ratio, R = ⎜ ⎟(100) = 56.25% ⎝ 8 ⎠

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