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QUESTION

Write 2 pages thesis on the topic engineering environmental sustainability- take home final exam.

Write 2 pages thesis on the topic engineering environmental sustainability- take home final exam. 13-14, pages 507- 520=13 Both Closed-end steel pipe piles and drilled shafts are under consideration in a project. As shown in Fig. 13-53, the piles will be installed through a 10-m- thick, natural clay layer with ( Su/σ’v)NC= 0.25 into a sand layer with energy corrected, normalized below counts N60 of 25, 29, 32, 40, 42 and 45 for the 6m into unit layer. Sand has unit weight equal to 20 kN/ m3, and the clay has unit weight equal to 17kN/m3. The critical- state friction angle of the sand is 320 and K0= 0.45. The water table is at the surface. In the past the soil profile had been subjected to uniform surcharge of 50kPa applied on the surface of soil deposit, which was later removed causing both the sand and clay layers in the current states to be over consolidated. For a 15-m- long, 500-mm drilled shaft and a geometrically identical closed-end steel pipe pile, calculate (a) the shaft capacity due to the clay layer ( divide the clay into ten sub layers of equal thickness in your calculations), (b) The shaft capacity due to the sand layer, (c) the total shaft capacity, (d) the ultimate base capacity, (e)the ultimate load capacity of the pile, (f) the allowable load based on a suitable factor of safety (without consideration of the strength of the pile cross-section), and (g) the allowable load if the compressive strength of the concrete is 15 MPa.

Figure 13-53 Pile and soil profile for problem 13-14

SOLUTION:

(a) The shaft capacity due to the clay layer

Closed end pipe pile

Let us first divide the clay layer into 10 sub layers. The current vertical effective stress at each layer and past maximum vertical effective stress for each sub layer can be calculated from given data.

For example, at z= 5.5m, the current vertical effective stress σ’v and past maximum vertical effective stress σ’vp are given by

σ’v = (17- 9.81) ˣ 5.5 = 39.5kPa

σ’vp = 50 + 39.5kPa

Therefore, OCR= σ’vp/ σ’v = 89.5/ 39.5 =2.3. In addition, from (6.54), undrained shear strength at a depth of 5.5m below the ground surface is

Su= σ’v ( Su/ σ’v )NC OCR 0.8=39.5ˣ 2.30.8 = 19.2 kPa

Therefore, Su/ σ’v for this layer is given by

Su/ σ’v = 19.2/ 39.5= 0.49 ≤ 1:

Now that α = ( Su/ σ’v )-0.5 NC = (0.25)-0.5= 0.71

Finally, the limited unit shaft resistance is calculated by

qsL ǀat 5.5m for the pipe pile = αSu = 0.71ˣ 19.2kPa = 13.6kPa

Si1milarly, we can finish calculations for other sub- layers and prepare the following table.

S- Table 13-9

Layer

Ztout

(m)

Zbot (m)

Mid

Depth

(m)

Δhi

(m)

σv’

(kPa)

σ vp’

(kPa)

OCR

Su

(kPa)

α

Su/ σ’v

α

qsLi

(kPa)

qsLiΔhi (kN/m)

1

0

1

0.5

1

3.6

53.6

14.9

7.8

2.17

04.1

3.2

3.2

2

1

2

1.5

1

10.8

60.8

5.6

10.7

0.99

0.50

5.4

5.4

3

2

3

2.5

1

18.0

68.0

3.8

13.1

0.73

0.59

7.7

7.7

4

3

4

3.5

1

25.2

75.2

3

15.2

0.49

0.65

9.9

9.9

5

4

5

4.5

1

32.4

82.4

2.5

16.9

0.52

0.69

11.7

11.7

6

5

6

5.5

1

39.5

89.5

2.3

19.2

0.49

0.71

13.6

13.6

7

6

7

6.5

1

46.7

96.7

2.1

21.1

0.45

0.75

15.8

15.8

8

7

8

7.5

1

53.9

103.9

1.9

22.5

0.42

0.77

17.3

17.3

9

8

9

8.5

1

61.1

111.1

1.8

24.4

0.40

0.79

19.3

19.3

10

9

10

9.5

1

68.3

118.3

1.7

26.1

0.38

0.81

21.1

21.1

Σ qsLi Δhi =125.0 kN/m

Therefore, the shaft capacity of closed- ended pipe pile due to clay later is

Qsl= Σ qsLiAsi = ∏BΣ qsLi Δhi= ∏ 0.5mˣ125kNm =196.3 ˣ 196 kN answer to (a)

Drilled shaft

The value α is calculated from (13.50). For example, the value α for the 3rd layer is calculated by

α =0.4 [1-0.12ln(su/ pa)]= 0.4 [1-0.12ln(13.1/1oo)]= 0.50

Now, we can simply prepare the following table:

S- Table 13-10

Layer

Ztot

(m)

Zbot (m)

mid

Depth

(m)

Δhi

(m)

σv’

(kPa)

σ vp’

(kPa)

OCR

Su

(kPa)

α

Su/ σ’v

α

qsLi

(kPa)

qsLiΔhi (kN/m)

1

0

1

0.5

1

3.6

53.6

14.9

7.8

0.52

4.1

4.1

4.1

2

1

2

1.5

1

10.8

60.8

5.6

10.7

0.51

0.50

5.5

5.5

3

2

3

2.5

1

18.0

68.0

3.8

13.1

0.50

0.59

6.6

6.6

4

3

4

3.5

1

25.2

75.2

3

15.2

0.49

0.49

7.4

7.4

5

4

5

4.5

1

32.4

82.4

2.5

16.9

0.52

0.49

8.3

8.3

6

5

6

5.5

1

39.5

89.5

2.3

19.2

0.49

0.48

9.2

9.2

7

6

7

6.5

1

46.7

96.7

2.1

21.1

0.45

0.47

9.9

9.9

8

7

8

7.5

1

53.9

103.9

1.9

22.5

0.42

0.47

10.6

10.6

9

8

9

8.5

1

61.1

111.1

1.8

24.4

0.40

0.47

11.5

11.5

10

9

10

9.5

1

68.3

118.3

1.7

26.1

0.38

0.46

12.0

12.o

Σ qsLi Δhi =85.1 kN/m

Note that the same equation was used regardless of OCR values. The overall result is on the conservation side. Therefore, the shaft capacity of the drilled shaft due to the clay layer is

qsl= Σ qsLiAsi = ∏BΣ qsLi Δhi= ∏ 0.5mˣ85.1Nm =133.7 ͌ 134 kNanswer

(b) Shaft capacity due to the sand layer

Closed- end pipe pile

Let us first calculate the relative density of each soil layer. Using (7.6), for example, the relative density at z= 12.5m is calculated as

σ’v= (17-9.81) ˣ 10+ (20-9.81) ˣ2.5 = 97.4 kPa

σ’v= 97.4 + 50= 147.4 kPa

OCR= 147.4/ 97.4= 1.5

C= (Ko/ KoNC)=√OCR=√1.51= 1.22

DR = 100√ [N60/ A+BC (σ’v/ PA)= 100√ 32/ 36.5 + 27 ˣ1.22 97.4/ 100 = 68.3%]

σ ‘h=Koσ’v= 0.47ˣ 97.4kPa= 43.8kPa

The unit pile base capacity at z= 12.5m is calculated using (13.25):

qbl= 1.64pA exp[0.1041Фc+(0.0264 – 0.0002Фc) DR]( σ ‘h/ PA)0.841- 0.0047DR

=164exp[0.1041 ˣ32+(0.0264- 0.0002 ˣ32) ˣ168.3](43.8/100)0.841-0.0047 68.3

= 11705.3kPa.

Now, taking the interface friction angle between pile and soil as 0.85Фc for the steel pile unit shaft capacity is calculated from (13.35):

V qsL at 12.5m for the pipe pile = 0.02tanδ[ 1.02- 0.0051DR]qbL

= 0.02ˣ tan (0.85ˣ 320)[ 1.02- 0.0051ˣ 68.3] 11705.3

= 80.8kPa

Similarly, we can finish calculation for other sub- layers and prepare the following table.

S- Table 13-11

Σ qsLi Δhi =412.9kN/m

Therefore, the shaft capacity of closed-end steel pipe pile due to sand layer is.

Qsl= Σ qsLiAsi = ∏BΣ qsLi Δhi= ∏ 0.5m 412.9kN/ m 640kN Answer to (b)

The coefficient of lateral earth pressure K is calculated using (13.32).For example, at z= 125

K= 0.7Ko Experise [{o.0114- 0.0022ln [{0.0114- 0.0022ln(σ’v/ PA)} DR]

= 0.7K0exp[{o.o114- 0.0022ln( 97.4]/ 100)} 68.3= 0.69

Taking δ= Фc, the limit unit shaft capacity is calculated by

qsL at 12.5m for drilled shaft =K σ’v tan δ= 0.67ˣ 97.4ˣ tan 32 = 42.kPa

Similarly we can finish calculations on other sub systems and following table:

S-Table 13-12

Σ qsLi Δhi =213.4kN/m

Therefore the fraction of the shaft capacity of the drilled shaft due to the sand layer is

QsL = Σ qsLiAsi= ∏B Σ qsLi Δhi= ∏ˣ 0.5 ˣ 213.4= 335kNanswer to (b)

(c)Total shaft capacity

The total shaft capacity is the sum of the shaft capacity due to the sand and clay layers.

Closed-end pipe pile

QsL= QsL, CLAY + QsL, SAND= 196.3 + 648.6kN= 844.9 ͌845kNanswer to (c)

(d) Based capacity

Closed-end pipe pile

We will calculate the base capacity using soil variables. Let us first calculate relative density of soil layer below pile base. Using (7.6), the relative density at z=15m is calculated as

σ’v= (17-9.81) ˣ 10+ (20-9.81) ˣ5 = 122.9kPa

σ’vp= 122.9 + 50= 172.9kPa

OCR= 172.9/ 122.9= 1.41

C= (Ko/ KoNC)=√OCR=√1.41= 1.19

DR = 100√ [N60/ A+BC (σ’v/ PA)= [100√ 36.5 + 27 ˣ1.19ˣ 122.9/ 100] =77%

σ’h= K0 σ’v0.45ˣ 122.9kPa =55.3kPa

The limit unit pile base capacity at z= 15m is calculated using (13.25):

QbL=164PAexp[0.1041Фc + (0.0264- 0.0002 Фc)DR]( σ’h/PA)0.841-0.0047DR

=164expp0.1041 ˣ 32+ (0.0264- 0.0002 ˣ 32)77](55.3/100) 0.841-0.0047ˣ77

=16110.9kPa

The ultimate unit base resistance of the full displacement pile is obtained by

Qb, 10%=[1.02- 0.0051DR]qbL

=[1.02- 0.005 ˣ 77]16110.9

10106.4kPa

Therefore the ultimate base resistance is calculated as

Qb, ult=qb, 10%Ab=10106.4kPa ˣ ∏/ 4 ˣ (0.5m)2 ͌ 1984kNanswer to (d)

(e) Qult = QsL + Qult =845kN + 1984kN= 2829kNanswer to (e)

Drilled shaft

Qult = QsL + Q ult=469kN +438kN= 907kNanswer to (e)

(f) Allowable load

Closed-ended pipe pile

With a FS= 2.5, the allowable load is found as

Qall= Qult/ FS= 2829/2.5 ͌ 1132kNanswer to (f)

Drilled shaft

Qall= Qult/ FS= 907/2.5 ͌ 363kNanswer to (f)

(g) Allowable load for the drilled shaft if the comprehensive strength of the concrete is 15MPa

We also must check the integrity of the cross section of the drilled shaft. with a FS= 3.0, the design comprehensive strength is obtained from (13.13):

Fcd = f’c/ (FS)LA-2= 15000kPa/3= 5000kPa

Therefore, the allowable shaft load to ensure integrity across section is:

Fall = fcdA= 5000kPaˣ ∏/ 4 ˣ (0.5m)2 ͌ 982kN

The allowable axial load for the drilled shaft from geotechnical consideration is less than the allowable structural load, so, integrity of the cross section is not a concern. Therefore, the final allowable load of drilled shaft is obtained as

Qall = min (geotechnical capacity, structural capacity)= min(363kN, 982kN)

= 363kNanswer to (g)

3-15, pages 520- 524=5

Solution

Calculation of shaft resistance

We will consider the first two layers as clay layers and sand given in table 13-9 and 13-5.

We can now calculate the fundamental soil properties (undrained shear strength for clay and relative density for the sand layers.) Let us first calculate undrained shear strength of the clay layers. Following (7.22),

Su= qc- = σv /Nk

The undrained shear strength at the mid depth of the clay layers are calculated as shown in the following table.

S- Table 13-13

Layer

Average qc (kPa)

Zmid-depth (m)

σv (kPa)

σv (kPa)

Su at the mid-depth (kPa)

Su/ σv

1

300

1.5

25.5

25.5

22.9

0.90

2

500

4

69

59.2

35.9

0.61

Now, relative densities for sand layers can be calculated using Eq. (7.20)

DR = ln(qc/PA)- 0.4947-0.1041Фc- o.841ln (σv/PA)] / [0.0264- 0.0002 Фc – 0.0047ln(σ’h/PA)]

S- Table 13-14

Layer

Average qc (kPa)

Zmid-depth (m)

σv (kPa)

σh(kPa)

DR (%)

3

6500

7

85.8

38.6

50.8

4

10.5

116.4

10.5

52.4

26.2

5

8000

12.5

133.3

60

48.2

Now, from the Table 13-9 and Table 13-5, the unit shaft resistance can be calculated as follows:

For Clay

QsL = αsu,

Where as α=√(Su/ σv)NC/(Su/ σv ) for Su/ σv ≤=√1

For sand

qsL =0.02 tan [1.02- 0.051DR]qbL

Where δ = 0.95 Фc for concrete pile, and q­bL = qc

The values of different unit shaft resistance for different layers are presented 8n the table below:

S- Table 13-14

Layer

Average qc (kPa)

ΔHi

(m)

= α of clay player

DR (%)

For sand layers

qsLi (kPa)

qsLiΔHi

(kN/m)

1

300

3

0.58

-

13.3

39.9

2

500

2

0.70

-

25.1

50.2

3

6500

4

-

50.8

55.9

223.6

4

4000

3

-

26.2

38.5

115.5

5

8000

1

-

48.2

69.9

69.9

∑5i=1qsLi ΔHi=499.1kN/m

QsL=4B∑5i=1qsLi ΔHi=499.1= 598.9 ͌ 599 kN

Calculating the base resistance

At the depth of 13m, σv=137.9, σh=62.1, and qc = 8000kPa Eq. (7.19), DR=47.3%.

Table 13-5

Qb, 10%= [1.02- 0.0051DR]qc

%= [1.02- 0.0051 ˣ 47.3]8000

=6230.2kPa

Q ult = QsL + Qb, ult

= 598.9 + 560.7

= 1159.6kN

The allowable vertical capacity

Qall = Q ult/ FS =1159.6/2.5 = 463.8kN ͌ 464kN

17-8, pages 703 - 705=3

(FS)OMS = ∑Ni=1 [cli + (Wicosαi) tanФ ]/ ∑Ni=1 Wisinαi =893.86/538.02= 1.66 ͌ 1.7

S- Table 17-2

Slice

Y1 (top)

Y2 (top)

Y1 (bottom)

Y2 (bottom)

X1

X2

Wi(kN/m)

αi(rad)

sin αi

1

0.000

3.333

0.000

0.375

0.000 590

3.333 0.112

98.590

0.112

0.112

2

3.333

6.667

0.375

1.563

3.333

6.667

268.787

0.342

0.336

3

6.667

10.000

1.563

3.820

6.667

10.000

376.096

0.595

0.561

4

10.000

10.000

3.820

6.095

10.000

12.071

208.860

0.832

0.739

5

10.000

10.000

6.095

10.000

12.071

14.142

80.873

1.083

0.883

Slice

Wisinαi (kN/m)

Cosαi

N (Wicosαi)

Li

c Li+

NtanФ

1

11.023

0.994

97.972

3.354

128.300

2

90.220

0.942

253.193

3.539

228.999

3

210.879

0.828

311.413

4.025

275.098

4

154.449

0.673

140.599

3.076

149.386

5

71.447

0.469

37.891

4.420

112.081

538.0

893.9

17-9, pages 705- 706=2

(FS)OMS = [∑Ncbi i=1 + (WitanФ ]/ mα]/ [∑Ncbii=1 Wsinαi

Mα= cosαi ( 1+ tanαi tanФ/ FS)

(FS)BSM was determined to be 1.74 by the iteration.

Slice

Wi(kN)

αi ( rad)

Sinαi

Wisinαi

bi

c bi+

WitanФ

Try FS= 1.6

Try FS= 1.7

Try FS= 41.7

c bi+

WitanФ

/ mαi

c bi+

WitanФ/ mαi

c bi+

WitanФ

/ mαi

1

98.590

0.112

0.112

11.o23

3.333

128.266

1.037

1.035

123.949

1.034

124.062

2

268.787

0.324

0.336

90.220

3.334

234.673

1.073

218.659

1.065

220.241

1.063

220.829

3

376.096

0.595

0.561

210.879

3.333

310.671

1.047

288.13

1.034

291.719

1.029

293.062

4

208.860

0.832

0.739

154.449

2.071

171.930

0.962

178.727

0.945

181.94

0.939

183.151

5

80.873

1.083

0.883

71.447

2.071

91.955

0.814

113.028

0.793

115.92

0.786

117.020

538.02

922.19

933.77

938.13

FS

FS

1.71

FS

1.74

FS

1.

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