Examen PE Geotecnia

8/12/2019 Examen PE Geotecnia

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Copyright 2008-2012 Pecivilexam.com all rights reserved-GeotechnicalDepth Exam Set #1 1

PE Civil Exam 40-Geotechnical Questions & Answers (pdf Format)

For Depth Exam (Evening Session) Set #1


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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 6

3. Soil Mechanics Analysis -5 Q&A 9

4. Earthquake Engineering -2 Q&A 14

5. Earth Structures -4 Q&A 15

6. Shallow Foundations -6 Q&A 19

7. Earth Retaining Structures -6 Q&A 25

8. Deep Foundations -4 Q&A 31

9. Other Topics- 5 Q&A 35


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Civil Depth (Evening) Questions -Geotechnical Set # 1

I. Subsurface Exploration and Sampling

1. PROBLEM (Drilling and sampling procedures)

Which of the following statement is not true for soil sample?

a. Disturbed soil samples are using for compaction and sub-grade testing.

b. Undisturbed soil samples are using for peak shear, consolidation, permeability, and

density tests.

c. Shelby Tube Sampler is using for collecting undisturbed soil sample.

d. Denison Sampler is using for collecting undisturbed soft clay soil sample.

1. Solution:

Denison Sampler is using for collecting hard cohesive soils, soft rock, cemented soils, and soilscontaining gravel.

Correct Solution is (d)


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2. PROBLEM (In situ testing)

Following figure represent the cone penetration resistance qc = 700 KN at A with an electricfriction cone electrometer. Determine the over consolidation ratio, OCR., where, Nk=15

a. 1.40b. 1.70c. 1.90

d. 1.20

2. Solution:qc = 700 KN

v= Total vertical effect= 3x17+8x19= 203.0 KN/m2v= Effective vertical pressure =3x17+8(19-9.1)= 130.2 KN/m2

OCR=.37((qc -v)/v=.37(700-203.0)/130.2= 1.41

Correct Solution is (a)


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3. PROBLEM (General rock characterization)

During a field exploration, the core barrel was advanced 180 CM during the coring. Thelength of the core recovered in Fig. What was the recovery ratio and Rock quality?

a. .80 Goodb. .48 Poorc. .98 Excellent

d. .58 Fair

3. Solution:

R.Q.D=(30+15+18+38)/180= 0.56, Fair



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6. PROBLEM (Geosynthetics)Which of the following statement is not true for Geosynthetics system?

a. Wall system construction is relatively rapid and it dose not require specialized laboror equipment.

b. Limited foundation preparation is required and its reinforcement is light and easy tohandle.

c. An extremely flexible wall system can not accommodate large total and differentialsettlements without distress. Also, it is not well-suited for applications in regions of

high seismic area.d. Certain Geotextiles may improve drainage characteristics of clayey backfill.

6. Solution:

An extremely flexible wall system can take large total and differential settlements withoutdistress. Also it is well-suited for applications in regions of high seismic area.

Correct Solution is (c)

7. PROBLEM (Frost Susceptibility)

The following soil classification has been determined by laboratory test. What is the highest Frost-susceptibility of following group soil classification?

Types of Soil % finer materials lower than 0.02 mm by weigh t USC System

a. Sands 6% to 15% SM, SW-SM, SP-SM

b. Sands 3% to 10% SW, SP

c. Sands 0% to 3% SW, SP

d. Gravelly Soils 10% to 20% GM, GW-GM, GP-GM

7. Solution:

Group a is highest volume finer soil than other. Therefore, it is highest Frost-susceptibility soil.



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8. PROBLEM (Shear strength properties)

A drained triaxial test has performed for normally consolidated clay with following data.Chamber confining pressure, h=108 KN/m2

Vertical deviator pressure v=158 KN/m2Pore water pressure, U=50 KN/m2

Find the angle of internal friction,

a. 350b. 260c. 250d. 220

8. Solution:

v=158 KN/m2h=108 KN/m2

Pore water Pressure, U=50 KN/m2, will be deducted at drained condition

1= h+v3= h1= 1-U=108+158-U= 266.00-50=216.00 KN/m

2

3= h-U=108.00-50=58.0 KN/m2

Sin= (1- 3)/2 / (1+ 3)/2

=Sin-1{(216-58)/2 / (216+58)/2}=35.210



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III. Soil Mechanics Analysis

9. PROBLEM (Seepage)

The concrete dam 25 m long shown in Figure is embedded 1 m into the ground surface and has asheet pile wall 5 m deep at its heel. The headwater is 7 m deep and the tail water is at groundsurface. The permeability of the soil is k= 20 x 10

-4cm/sec both vertically and horizontally.

What is the quantity of seepage loss under the dam?

a. Qt=2.50x10-5m3/seeb. Qt=8.40x10-5m3/see

c. Qt=4.30x10-5m3/see

d. Qt=5.10x10-5m3/see

9. Solution:

k= 20 x 10-4

cm/sec=2 x 10-5

m/sec,

Dam headwaterheight,h=7m,Impervious layerto Dam bottom height, H=10m

No. of flow channel, Nf =4,No. of potential drop, Nd=13,

Impervious layerto A height, Z=2m

Qt=kh(Nf/Nd)=2 x 10-5x 7 (4/13)=4.30x10-5m3/sec



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10. PROBLEM (Consolidation)

For 50% consolidation of a 25-mm-thick clay layer (drained at both top and bottom) in the

laboratory is 4 min 20 sec. How many days will it take for a 2-m-thick clay layer of the sameclay in the field under the same pressure increment to reach 50% consolidation? In the field,

bottom of the clay have a rock layer.

a. 30.40 daysb. 25.20 daysc. 77.50 daysd. 56.30 days

10. Solution:

Coefficient of Consolidation,

Tv= cvt/ Hd2

Tv= Time factor

Hd(lab)= 25mm = Half thickness of soil layer for two way drainageHd(lab)= 0.025/2=0.0125m

Hd(field2= 2m

tlab= 4min 20 Sec=260 Sec

T50= cvt/ Hd(lab)2= cvt/ Hd(field

2

cvt/ Hd(lab)2= cvt/ Hd(field

2

tlab/ Hd(lab)2= tfield/ Hd(field

2

260/ (0.0125)2= tfield/ 2

2

tfield= 6656000 sec= 77.03 days



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11. PROBLEM (Lateral Earth Pressure)

A retaining wall shown in Figure is at rest, what will be the lateral force per unit length of the wall?

a. 25.1 kN/mb. 67.84 kN/mc. 78.44 kN/m

d. 98.3 kN/m

11 Solution:

=28o= 16.0kN /m3H= 4m

'h= K0v= K0 (q0 +H), q0 = surcharge load=10.0 kN /m2Where, 'h= Effective horizontal pressurev= Effective vertical pressure

K0= 1-sin=0.53'h=.53 x (10+16 x 4= 39.22 kN /m2

AT REST LATERAL EARTH PRESSURETotal Lateral Force, Ph= H/ 2'h= .5 x 4 x 39.22=78.44kN /m



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12. PROBLEM (Lateral Earth Pressure)

A Concrete basement wall shown in Figure, what will be the lateral pressure acting againstthe wall point at 8 ft below the ground level?

a 348 lb/ft2b 570 lb/ft

2

c 538 lb/ft2

d 630 lb/ft2

12. Solution:

Where, 'h= Effective horizontal pressurev= Effective vertical pressure'v= 120 x 4 +60 x 4=720 lb/ft

2

'h= K0v= .4 x 720= 288 lb/ft2Pore water pressure, u=62.4 x 4=250 lb/ft2

Total pressure, h= 'h+u=288+250=538 lb/ft2



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13. PROBLEM (Soil Compaction)

Which compaction method is most effectively applicable for granular soils?a. Pneumatic Rubber-Tired Roller

b. Sheepsfoot Roller

c. Smooth Wheel Roller

d. Vibratory Roller

13. Solution:

Vibratory Rollersare most effective compaction method for granular soils.



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IV. Earthquake Engineering

14. PROBLEM (liquefaction)

Which one of the following soil condition does not meet susceptible liquefactionbehavior?

d Liquidity Index > 0.75

14. Solution:

Liquidity Index should be < 0.75


15. PROBLEM (Earthquake)

An Auditorium building site has the following soil profile peak max acceleration of 0.30g.Estimate the maximum earthquake induce shear stress at 20 depth. Where, stressreduction factor, rd=.95

a 348 lb/ft2

b 570 lb/ft2

c 406 lb/ft2

d 630 lb/ft2

15. Solution:amax=.30gv=10 x 108 + 2 x 108 +112 x 8 =2192 #/ft

2v=2192-10 x 62.4=1568 #/ft

2For 20 depth rd=.95Earthquake induce Shear Stress, f=(.65 amax/g) vrdf=(.65 amax/g) vrd=(.65 x .3g/g)2192 x .95=406.07 lb/ft

2


a Fraction finer than 0.005 mm< 15%b Liquid Limit, LL < 35%c Natural water content > 0.9 LL


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V. Earth Structures

16. PROBLEM (slope stability)

Slope cut in saturated clay as shown Figure The side of the cut slope will make an angle of 56 withthe horizontal. What is the depth cut of slope will have a factor of safety, FS = 3? Given, slopestability number is 0.18.

a. 7.1 mb. 11.2 m

c. 3.0 md. 2.72 m

16. Solution:

= 15.0kN /m3Slope stability number, m=0.18

Cd= Cu/FS=22/3=7.33 kN /m2

Depth of the cut slope, H= Cd / m=7.33/(15 x .18)=2.72 m



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17. PROBLEM (Braced and anchored excavations)

Determine the tension of the tie rods T if they are spaced at 4 meter center of the anchored sheet

pile wall shown in Figure. Unit weight of soil, = 15.0kN /m3.

a 247.00 kNb 345.00 kNc 176.00 kN

d 288.00 kN

17. Solution:Tie rods are spaced at 4 meter center to center= 15.0kN /m3

Pa= 1/2 Ka H2a=1/2 x 15.00 x .33 x 92= 200.48 kN/m (horizontal)Taking moment at tie rod for mobilized passive resistancePp x (9-1-1.5)= Pa x (9-3-1.5)

Pp x 6.5= 200.48 x 4.5Pp= 138.79 kN/m

Tension of the Rods, T= (200.48-138.79) x 4=246.74 kN



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18. PROBLEM (Dam)

The concrete dam of base with 25 m shown in Figure is embedded 1 m into the ground surface andhas a sheet pile wall 5 m deep at its heel. The headwater is 7 m deep and the tail water is at groundsurface. The permeability of the soil is k= 20 x 10

-4cm/sec both vertically and horizontally.

Determine the seepage quantity per meter length of dam per day?

a 15780.00 kg/m2

b 10960.00 kg/m2

c 12800.00 kg/m2

d 12450.00 kg/m

2

18. Solution:

k= 20 x 10-4

cm/sec=2 x 10-5

m/sec,

Dam headwaterheight,h=7m,

Impervious layerto Dam bottom height, H=10mNo. of flow channel, Nf =4,

No. of potential drop, Nd=13,Pressure head per net=h/Nd=7/13=0.54/net

Q=k h Nf/Nd=2 x 10-5

x 7 x 4/13=4.30 x10-5

Q=4.30 x10

-5

x 60 x 60 x 24=3.7 m3/day/m-length

Pressure Head at A, Ph=h+H-(No. of potential drop at A x Pressure head per net)-Z=7+10-(7.5x.54)-2=10.96 m

Pore Water pressure at A, UA=Ph x w=10.96m x 1000 kg/m3

=10960.00 kg/m2

Correct Solution is (b)


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VI. Shallow Foundations

20. PROBLEM (Earthquake bearing capacity)

Determine the gross ultimate earthquake bearing capacity, que

of the following strip

foundation footing. Where, NE =10 & NqE =12.

a 730kN /m2

b 375kN /m2c 250kN /m

2

d 525kN /m2

20. Solution:

=17kN /m3

NE =10 & NqE =12, C=0 for sandy soil

Ultimate EarthquakeBearing Capacity,

que=0.5 BNE+ DNqEque=.5 x 17 x 2 x 10 + 17 x 1.75 x 12=527kN /m

2



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21. PROBLEM (Bearing Capacity)

Determine the ultimate load, Qultof a rectangular footing 6x 4 with eccentric shown inFigure where, Soil Unit Weight, = 118 lb/ft3, Ultimate Bearing Capacity, qu=2000 lb/ft2,eB=.5, eL=.75, L1=1.0 and B1=0.68.

.

a 15.0 Kipb 48.0 Kip

c 28.0 Kipd 31.0 Kip

21. Solution:

Where, eL/L=.75/6= 0.125< 1/6, eB/B=.5/4= 0.125


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Determine the Effective Area of circular footing with eccentric load shown in Figure where,

Soil Unit Weight, = 118 lb/ft3

, Footing Radius of circle, R=5 & e=0.5..

a 54.50 ft2b 65.50 ft

2

c 78.50 ft2

d 63.50 ft2

22. Solution:Where, e=0.50 and R=5

Therefore,Area of ADC, S= R2/2-[e{(R2-e2)+R2Sin-1(e/R)}]S=3.14 x 5

2/2-[0.50{(52-0.52)+52Sin-1(0.5/5)}]

S=3.14 x 52/2-[0.50{(24.75+143.48)}=32.77 ft2Effective Area, A=2S=32.77 x 2= 65.54 ft

2



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23. PROBLEM (Settlement)

Determine the consolidation settlement at center of the clay layer of the mat (40m x 50m)foundation shown in Figure. Given, total effective stress, =200 KN/m2and averageeffective stress increase due to foundation load, =100 KN/m2at the center of the claylayer.

a 345.00 mmb 275.00 mm

c 221.00 mmd 141.00 mm

23. Solution:=200 KN/m2,=100 KN/m2, Cc=0.29 and eo=0.85Settlement, Sc={CcHc/(1+eo)}log {(+)/}Sc={0.29 x 8/(1+.85)}log {(200+ 100)/ 200}=220.83 mm



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The soil is supporting a square footing 5 x 5 shown in Figure Soil unit weight, = 118lb/ft3, c=420 lb/ft2, Nc=17.0, N= 5.0 & Nq=7.0. Determine allowable bearing capacity,where factor of safety is FS=3.

a 6540.00lb/ft2

b 5550.00lb/ft2

c 2520.00lb/ft2

d 4310.00lb/ft2

24. Solution:

Where, D=3 ft,= 118 lb/ft3, c=420 lb/ft2, Nc=17.0, N= 5.0 & Nq=7.0.Ultimate Bearing capacity, qu=0.4 BN+1.3cNc+ DNqqu=0.4 x 118 x 5.0 x 5+1.3 x 420 x 17.0+ 118 x 3 x 7.0= 12940.00 lb/ft

2

qall= 12940.00/3= 4313.33 lb/ft2



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VII. Earth Retaining Structures

26. PROBLEM (Gravity Wall)

Determine theactive earth pressure of the following retaining wall, H=15 feet. Where, unit weight ofsoil,= 118 lb/ft3, = 8oand =30o.

a 11 Kip/ft

b 4 Kip/ftc 6 Kip/ft

d 9 Kip/ft

26. Solution:

KA=[cos30o/(cos8o+{sin(30o+8o) sin8o}]2

KA=[.866 /(.99 +0.292}]2=0.45PA= 1/2 KAH

2=.5 x 118 x .45 x 15

2=5973.75 lb/ft=5.974 Kip/ft



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27. PROBLEM (Cantilever wall)

Determine the total lateral force on wall plus pile cap. Where, equivalent fluid pressure perfeet length of the following wall=64 lb/ft

3.

a 10368 #

b 4608 #c 41470 #

d 52340 #

27. Solution:

Pressure at bottom of the pile cap=64 x18=1152 lb/ft2Fx= 1/2 x 1152 x 18=10368 lb/ft

Total lateral force on wall plus pile cap= 10368 x 4= 41472 #



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29. PROBLEM (Mechanically stabilized earth walls)

Determine the thickness of bottom galvanized steel strip reinforcement of a 6 meter retainingwall in a sandy soil backfill has to be constructed with following value. Assume the corrosion

rate of the strip to be 0.020 mm/year and the lifespan of the structure to be 100 years.

Back fill of sandy soil, = 16 kN/m3, and 1=32o,Foundation soil, = 17 kN/m3, c=40 kN/m2 and =30o,For galvanized steel strip, F.S. =3, Width of strip, w=70 mm, hv=0.4m, hh= 1.0 meter center

to center, Fy=2.5x105kN/m2

a 6.0mm

b 5.0 mmc 4.0 mm

d 3.0mm

29. Solution:

= 16 kN/m3, and 1=32o, w=70 mm=.07m, hv=0.4m, and hh= 1.0m'= Htan2(45-/2)=16 x 6 x tan2(45-32/2)= 29.50 kN/m2Maximum tie force of bottom galvanized steel strip,T= 'hvhh= 29.50x .4 x 1.0= 11.80 kNBreakout ofgalvanized steel strip, Fy=2.5x10

5kN/m2t=T F.S/wfy=11.80 x 3/(.070 x 2.5x10

5)= 0.002m=2.023 mm

Total thickness of bottom steel strip= t + eroded by 100 corrosion

=2.023 + 0.02 x 100=4.023 mm



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30. PROBLEM (Cantilever Retaining wall)

Determine the horizontal pressure of the following cantilever retaining wall shown in Figure.

Back fill of sandy soil, = 19 kN/m3

, and 1=28o

,Foundation soil, = 17 kN/m3, c=40 kN/m2 and =26o,

a 156.50 kN/m2

b 205.50 kN/m2

c 162.00 kN/m2

d 180.00 kN/m2

30. Solution:

= 19 kN/m3, and 1=28oPa= 1/2KaH2=.5 x .40 x 19 x (6+.53)2=162.035 kN/m2Ph=Pa x cos15

o=162.035 x 0.966=156.514 kN/m

2



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31. PROBLEM (Braced and anchored excavations)

Determine the Factor of Safety against heave a braced cut in clay soil shown in Figure.

Length of braced cut, L=15m, Clay soil, = 16 kN/m3

, Nc=5.14, 1=0o

and c=40 kN/m2

.

a 6.0

b 4.0c 3.0

d 5.0

31. Solution:L=15m, B=4m, = 16 kN/m3, Nc=5.14, c=40 kN/m2 and 1=0

o

B/2=42= 2.828> T=2m

Or, T=2m < B/2=2.828Hence, B=T=2m and B=2B=2.828and surcharge, q=0.0F.S=[Nc c{(1+0.2B/L)}+cH/B]/( H+q),F.S=[5.14 x 40{(1+0.2 x 2.83/15)}+40 x 5/2]/( 16 x 5+0.0)F.S= 3.917



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VIII. Deep Foundations

32. PROBLEM (Axial capacity)

A 16 in diameter concrete pile is driven 40 ft into soft clay. Where, C=900 psf. and friction

coefficient, =0.80.Determine the Ultimate pile capacity.

a 132.00 Kip

b 107.00 Kipc 190.00 Kip

d 120.00 Kip

32.Solution:

C=900 psf., D=16, L=40 ft, =0.80,Atip=/4(16/12)2

=1.40 ft2

Qult= Qtip+ QfrictionQtip=9CAtip=9 x 900 x 1.40=11304.00 # =11.30 Kip

Qfriction= CAsurface=0.80 x 900 x (DL)Qfriction=0.80 x 900 x (3.14 x 16/12 x 40)= 120576.00# =120.58 Kip

Qult= Qtip+ Qfriction=11.30+120.58=131.88 Kip



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33. PROBLEM (Axial Capacity)

A 12 in diameter concrete pile is driven 45 ft shown in Figure. Determine the allowable

capacity pile capacity with F.S=3.

a 65.00 Kipb 54.00 Kip

c 38.00 Kipd 48.00 Kip

33. Solution:Qult= Qtip+ Qfriction

Qfriction= CAsurface=C(DL)Qfriction=0.90 x 1100 x (3.14 x 1 x 20)+0.5 x 1800 x (3.14 x 1 x 25)

Qfriction=62172.00+70650.00=132822.00=132.82 KipQtip=9CAtip=9 x 1800 x /4(1)

2= 12717.00# = 12.72 Kip

Qult= Qtip+ Qfriction=12.72+132.82= 145.54 KipQallow= Qult/F.S= 145.54/3= 48.51 Kip



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34. PROBLEM (Pull Capacity)A 12 in diameter concrete pile is driven 45 ft shown in Figure. Determine the pullout

resistance capacity of the pile. Unit Weight of pile, = 150 lb/ft3

a 165.00 Kipb 154.00 Kipc 138.00 Kip

d 133.00 Kip

34. Solution:Pullout resistance capacity, Qpull= Wt. of pile + QfrictionQfriction= CAsurface=C(DL)Qfriction=0.90 x 1100 x (3.14 x 1 x 20)+0.5 x 1800 x (3.14 x 1 x 25)Qfriction=62172.00+70650.00=132822.00=132.82 Kip

Wt. of pile= /4(1)2 x 45 x 150= 5298.75= 5.30 KipPullout resistance capacity, Qpull= Wt. of pile + Qfriction= 138.12 Kip



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IX. Other Topics

36. PROBLEM Cofferdams (systems for temporary excavation support)

A square cofferdam 50 ft each side has placed in a lake to permit dry construction forbridge pier. The coffer dam has consists two square sheet pile enclosure. The innerenclosure separated on all sides from the outer one by 6 ft. The lake bottom waspractically impermeable. The space between sheet piles was filled with a soil whichshowed permeability 0.035 ft/day in the laboratory test. Assuming the sheet pile wasnot water tight and that lake surface was 20 ft above the lake bottom. How much waterper day had to be pumped form the excavation to keep cofferdam dry.

a Q =1645.00 gal/dayb Q =1326.00 gal/dayc Q =678.00 gal/dayd. Q =177.00 gal/day

36 Solution:Coefficient of permeability, k = 0.035 ft /day

Seepage Surface area, A=(50-12) x 4 x20= 3040 ft2

H=20 ftL=6+6=12 fti=H/L=20/12=1.67Q=VA=KiA=.035 x 1.67 x 3040=177.33 ft3/dayQ =177.33 x 7.48=1326.5 gal/day (1 ft3=7.48gal)



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37. PROBLEM(Seepages)

A confined aquifer has shown in Figure with 1.0 mile wide and 50 ft depth with hydraulic

conductivity, K =0.04 ft/sec&Porosity, n=.38. Two observations well are apart at 600 ft andwater elevation deference is 10 ft. What is the seepage velocity, Vs?

a Vs = 0.00105 ft/sec

b Vs = 0.00405 ft/secc Vs = 0.00325 ft/sec

d Vs= 0.00175 ft/sec

37. Solution:

Porosity, n=.38, K =0.04 ft/sec, h=10 ft and L=600 ft

Hydraulic gradient i=h/L=10/600=0.0166Q=KiA= 0.04 x 0.0166 x {50 x (1 mile x 5280)}=176 ft3/secDarcy Velocity, V=Q/A=Ki= 0.04 x 0.0166=0.000666 ft /sec

Seepage Velocity, Vs=V/n=0.000666/0.38=0.00175 ft/sec



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38. PROBLEM(Aquifers)

A well has constructed a confined aquifer with 25 feet depth including hydraulicconductivity, K =0.0035 ft/sec. The two observations well is located at 1200 ft and 2900 ft

form well with water surface elevation is 3.0 ft and 6.0 ft respectively. Determine the waterrate of flow.

a. Q = 3.1246 ft3/secb. Q = 0.1217 ft3/secc. Q = 1.8682 ft3/secd. Q = 2.8012 ft3/sec

38. Solution:

K =0.0035 ft/sec, H=25 ft, h2=6 ft, h1=3 ft, r2=2900 ft, and r1=1200 ft,

Q=2HK{(h2-h1)}/lne(r2/r1) for confined aquiferQ=2 x 3.14 x 25 x 0.0035 (6-3)}/lne(2900/1200)=1.86 ft

3/sec


39. PROBLEM(Safety Management)

Fabricated planks and platforms may be used in lieu of solid sawn wood planks. Fabricatedplanks has recommended by the manufacturer based on the maximum loading capacity 75 lbs

per square foot applied uniformly over the entire span area. What is the name called bymanufactured of the fabricated planks load Capacity?

a. Light-dutyb. Medium-duty

c. Heavy-dutyd. One-person

39. Solution:

Name Called Heavy-duty



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