PE Civil Exam 40- Geotechnical Questions & Answers (pdf ... · PDF fileThe sand layer...

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PECivilExam.com Copyright © 2008-2012 Pecivilexam.com all rights reserved- Geotechnical Depth Exam Set #2 PE Civil Exam 40- Geotechnical Questions & Answers (pdf Format) For Depth Exam (Evening Session) Set #2

Transcript of PE Civil Exam 40- Geotechnical Questions & Answers (pdf ... · PDF fileThe sand layer...

Page 1: PE Civil Exam 40- Geotechnical Questions & Answers (pdf ... · PDF fileThe sand layer underlying the clay and it was under artesian pressure. ... Unit weight of concrete γ= 145 lb/ft3and

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Copyright © 2008-2012 Pecivilexam.com all rights reserved- Geotechnical Depth Exam Set #2

PE Civil Exam 40- Geotechnical Questions & Answers (pdf Format) For Depth Exam (Evening Session) Set #2

Page 2: PE Civil Exam 40- Geotechnical Questions & Answers (pdf ... · PDF fileThe sand layer underlying the clay and it was under artesian pressure. ... Unit weight of concrete γ= 145 lb/ft3and

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PE Civil Depth Exam (Evening Session): This practice exam contains 40-Geotechnical questions and answers each set from all Geotechnical & Soil Foundation Engineering:

Table Contents: Page

1. Subsurface Exploration and Sampling -3 Q&A 3 2. Engineering Properties of Soils and Materials-5 Q&A 5

3. Soil Mechanics Analysis -5 Q&A 10

4. Earthquake Engineering -2 Q&A 15

5. Earth Structures -4 Q&A 17

6. Shallow Foundations -6 Q&A 21

7. Earth Retaining Structures -6 Q&A 27

8. Deep Foundations -4 Q&A 33

9. Other Topics- 5 Q&A 38

Page 3: PE Civil Exam 40- Geotechnical Questions & Answers (pdf ... · PDF fileThe sand layer underlying the clay and it was under artesian pressure. ... Unit weight of concrete γ= 145 lb/ft3and

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Civil Depth (Evening) Questions- Geotechnical Set # -2

1. PROBLEM (Drilling and sampling procedures)

A subsoil investigation has performed which was deposited in the sand. Calculate the relative density of the soil at 30 ft depth. Where, uniformity coefficient, Cu=3.1, Effective Overburden Pressure, σ’=2200 psi and Field Standerd Penetration Number, N60=13.

a. 30% b. 65% c. 55% d. 45%

1. Solution: Relative Density, Dr=11.7 + +0.76(222 N60+1600-53 σ’-50C2u)0.5 Relative Density, Dr at the depth 30 ft,

∴ Dr =11.7 + +0.76(222 x 13+1600-53 x 2200/144-50 x 3.12)0.5 ∴ Dr = 54.66%

Correct Solution is (c)

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9. PROBLEM (Vertical stress, Pore pressure & Effective stress)

Following soil profile is shown an exploratory drill in saturated stiff clay shown in the Figure. The sand layer underlying the clay and it was under artesian pressure. Calculate maximum depth of open cut before the bottom heaves in clay. a. 8.80 ft b. 11.50 ft c. 10.30 ft d. 13.20 ft

9. Solution:

γsat=110 pcf & γw =62.4

Consider heaving the interface point between sand & clay where effective stress, σ’= 0.00. σ= σ’ +u, u is the pour water pressure σ= u (H-Hexc) γsat=H1 γw,

Where, H=Total depth of clay, Hexc=Depth of excavation, H1=20-8=12 ft ∴ (20-Hexc) 110=12 x 62.4 ∴ Hexc=13.20 ft

Correct Solution is (d)

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24. PROBLEM (Member Design) An isolated 12-inch thickness square of concrete footing has an axial Dead Load 40 K and Live Load 30 K including Live Load Moment 150 K-ft shown in the Figure. Determine the Maximum soil bearing capacity. Unit weight of concrete γ= 145 lb/ft3and Unit weight of soil, γ= 120 lb/ft3.

a 2.40 Ksf b 3.80 Ksf c 3.00 Ksf d 4.00 Ksf

24. Solution:

Footing weight = 145 x 8.5 x 8.5 x 1.0= 10.48 Kip Soil weight = 120 x 8.5 x 8.5 x (4-1)= 26.01 Kip P1=10.48+26.10= 36.49 Kip B=L=8.5, M=150 K-ft P2=DL+LL=40+30=70 K P=P1+P2= 106.49 Kip Eccentricity, e=M/P=150/ 106.49= 1.41 Soil Bearing Capacity, qmax =4P/{3L(B-2e)} = 4 x 106.49/{3 x 8.5(8.5-2 x 1.41)} ∴ Soil Bearing Capacity, qmax = 2.94 Ksf Correct Solution is (c)

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34. PROBLEM (Pile Group Settlement)

Determine the consolidation settlement in the clay layer of the pile group. Each pile has an 18-inch diameter and 60 ft long as shown in the Figure.

a 0.40 ft b 0.60 ft c 0.90 ft d 1.10 ft

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34. Solution: Where, Qg=500Kip, n1=4, n2=3, d=3D, D=18”=1.5’, Cc=0.32, e0=.85 d=3 x 1.5=4.5 ft Lg=(n1-1)d+2(D/2)=(4-1)4.5+2(1.5/2)=15 Bg=(n2-1)d+2(D/2)=(3-1)4.5+2(1.5/2)=10.5 H1=30 Z=H1/2=30/2=15 ft ∆δ’=Qg/(Lg+z)(Bg+z)=500/(15+15)(10.5+15)=0.65 Kip/ft2 ∴ δ’=5 x110+(40-2+30/2)(115-62.4)=550+(38+15)x52.5=3332.5 lb/ ft2=3.33 Kip/ft2 ∆S={CcH1/(1+e0)}log[(δ’+∆δ’)/ δ’] ∴ Settlement, ∆S={0.32 x 30/(1+.85)}log[(3.33+0.65)/3.33]=5.19 x 0.08=0.40 ft Correct Solution is (a)

40. PROBLEM (OSHA Regulations)

A “B” type soil has to be excavated less than 20 ft deep. What will be Maximum Allowable Slope (H:V) of the excavation?

a. VERTICAL (90 Deg.) b. 3/4:1 (53 Deg.) c. 1:1 (45 Deg.) d. 1 1/2:1 (34 Deg.)

40. Solution:

Maximum Allowable Slope is 1:1 (45 Deg.)

Correct Solution is (c)