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# Engineering Environmental Sustainability- Take Home FINAL EXAM - Assignment Example

Summary
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…

## Extract of sample "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)
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)

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)
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)
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
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### Take-home Final Exam

...Elegance of Hedgehog Task: The story revolves around Renee Michel and Paloma Josse, residents of an upper-middle Left Bank apartment in Paris France. Widow Renee supervised the building 27 years and is an autodidact in literature and philosophy but conceals to condemnation by the building tenants. She confuses the curiosity of the residents by taking low-quality meals and watching low-quality television but at her room she enjoys reading works, listening to opera and eating quality food. Paloma who is a six-year-old lives on the fifth floor of the building with her parents whom she considers to be snobs and hide her intelligence to avoid exclusion from school. Although the two characters share interest in literature and...
4 Pages(1000 words)Term Paper