geotecnical depth set 2

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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

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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

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

I. Subsurface Exploration and Sampling

1. PROBLEM (Drilling and sampling procedures)

A subsoil investigation has performed which was deposited in 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 Std. 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) 2. PROBLEM (Boring log interpretation)

What will be the minimum boring depth for a 6 (six) story apartment building with a 100 ft wide concrete construction structure?

a. 32.00 Meter b. 11.00 Meter c. 21.00 Meter d. 15.00 Meter

2. Solution:

S=Number of story=6 Depth of Boring, D=6S0.7 (For heavy steel or wide concrete building) ∴ D= 6x60.7

∴ D =21.03 Meter

Correct Solution is (c)

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3. PROBLEM (In situ testing)

What is the Un-drain shear strength of the following Figure represented the cone penetration resistance qc = 700 KN at A with an mechanical friction cone electrometer? Where, Cu = Un-drain shear strength and Nk=20

a. 65.00 KN/m2 b. 13.00 KN/m2 c. 53.00 KN/m2 d. 25.00 KN/m2

3. Solution:

qc = 700 KN

σv = Total vertical effect= 3x17+8x19= 203.0 KN/m2 ∴ Cu = Un-drain shear strength= (qc - σv )/ Nk =(700-203.0)/20= 24.85 KN/m2

Correct Solution is (d)

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II. Engineering Properties of Soils and Materials

4. PROBLEM (Permeability)

What is the flow rate of a permeability of the soil is 3.6 x 10-4

cm/s? Flow takes place through a 120 mm diameter and 300 mm long soil sample, from top to bottom, as shown in the Figure below. The manometers are 130 mm apart, and the water level difference within the two manometers is 110 mm at steady state.

a. 1. 45 cm

3/min

b. 3.20 cm3/min

c. 1.85 cm3/min

d. 2.25 cm3/min

4. Solution: Where, L=120 cm, k=3.6 x 10

-4 cm/s , Y1=110, Y2=130

Hydraulic gradient across the soil specimen, i = 120/130 = 0.923 ∴ velocity of flow, v = k i =(3.6 x 10

-4)(0.923) = 3.322 x 10

-4 cm/s

Cross sectional area of the specimen, A=(∏/4)D2 = 113.04 cm2

∴ Flow rate, Q =vA= (3.322 x 10

-4)(113.04) = 0.0375 cm

3/s = 2.25 cm

3/min


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5. PROBLEM (Phase Relations)

Field density testing (e.g., sand replacement method) has shown bulk density of a compacted road base to be 2.00 t/m

3 with water content of 12%. Specific gravity of the soil grains is 2.70.

Calculate the void ratio.

a. 0.42 b. 0.67 c. 0.51 d. 0.34

5. Solution:

Where, w=12%, Gs= 2.70, γm=2.00 t/m

3

w=Se/Gs= ∴ Se = (0.12)(2.70) = 0.324 γm=(Gs+Se)/(1+e)γw ∴ 2.00=(2.7+0.32)/(1+e) x 1.0 ∴ e = 0.51


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6. PROBLEM (Shear strength properties)

A drained triaxial test has been performed on normally consolidated clay with shown in the following data. Chamber confining pressure, σh=108 KN/m2

Vertical deviator pressure σv=158 KN/m2 Pore water pressure, U=50 KN/m2

Determine the Shear Stress, Τƒ on the failure plane

a. 195 KN/m2 b. 173 KN/m2 c. 115 KN/m2 d. 163 KN/m2

6. Solution:

σv=158 KN/m2, σh=108 KN/m2 Pore water Pressure, U=50 KN/m2, will be deducted at drained condition For normally consolidation soil Effective Normal Stress, σ’=(σ’1+σ’3)/2 +{(σ’1- σ’3)/2}cos2Ф Shear Stress, Τƒ= σ’tanФ+c= σ’tanФ, where C=0 σ1= σh+σv, σ3= σh ∴ σ’1= σ1-U=108+158-U= 266.00-50=216.00 KN/m2

∴ σ’3= σh-U=108.00-50=58.0 KN/m2

SinФ= (σ’1- σ’3)/2 / (σ’1+ σ’3)/2

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

Effective Normal Stress, σ’=(σ’1+σ’3)/2 +{(σ’1- σ’3)/2}cos2Ф

∴ σ’=(216+58)/2 + {(216-58)/2} cos(2x35.210)= 163.47 KN/m2

∴ Shear Stress, Τƒ= σ’tanФ+c= σ’tanФ=163.47x tan35.210=115.36 KN/m2


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A Un-drained triaxial test has been performed on normally consolidated clay shown in the with following data. Chamber confining pressure, σh=108 KN/m2

Vertical deviator pressure σv=158 KN/m2 Pore water pressure, U=50 KN/m2

Find the angle of internal friction, Ф.

a. 290 b. 350 c. 250 d. 220

7. Solution:

Horizontal pressure, σh=108 KN/m2 Vertical Pressure, σv=158 KN/m2 U=50 KN/m2 Pore water Pressure does not reflect at un-drained condition. For normally consolidation soil Shear Stress, Τƒ= σ’tanФ+c= σ’tanФ, where C=0

∴ σ1= σv+σh =158+108= 266.00 KN/m2

σ3= σh=108=108.00 KN/m2

SinФ= (σ1- σ3)/2 / (σ1+ σ3)/2 ∴ Ф=Sin-1{(266-108)/2 / (266+108)/2}= 24.990


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A un-drained triaxial test has been performed on Saturated Cohesionless soil with shown in following data. Total Horizontal pressure, σ3=108 KN/m2

Total Vertical pressure σ1=158 KN/m2 Pore water pressure, U=50 KN/m2

Find the angle of internal friction, Ф

a. 290 b. 240 c. 110 d. 180

8. Solution:

σ3=108 KN/m2 σ1=158 KN/m2 U=50 KN/m2

At Un-drained condition pore water pressure not to be deducted.

σ’1= σ1=158.00= 158.00 KN/m2

σ’3= σ3=108.00 KN/m2

SinФ= (σ’1- σ’3)/2 / (σ’1+ σ’3)/2

∴ Ф=Sin-1{(158-108)/2 / (158+108)/2} = 10.830


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III. Soil Mechanics Analysis

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


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

Following soil profile is shown in the Figure. Calculate the total effective stress at B

a. 282.0 kN /m2 b. 98.1 kN /m2 c. 183.9 kN /m2 d. 102.0 kN /m2

10. Solution:

Total stress at B= 6 x 17 + 10 x 18= 282.0 kN /m2 Since, γw=9.81 kN /m3 Total Pore water pressure at B = 10 x 9.81= 98.1 kN /m2 ∴ Total effective stress at B= 282-98.1=183.9 kN /m2


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11 PROBLEM (Vertical stress, Pore pressure & Effective stress)

Following cross section and plan of a column footing are shown in the Figure. Calculate the increase in stress produced by the column footing at point A. Given, Newmark’s Influence Chart, IV=0.003 & Number of elements inside, M=23, Using Formula, Δσ= IVqM

a. 20.25 kN/m2 b. 15.70 kN/m2 c. 8.60 kN/m2

d. 12.20 kN/m2

11. Solution: P=500 kN, IV=.003, M=23 q= {500/(2x2)}=125 kN/m2 Stress produced by the column footing at point A, Δσ= IVqM ∴ Δσ =.003 x 125 x 23=8.62 kN /m2


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12. PROBLEM (Consolidation)

Following soil profile is shown in the Figure. Normally consolidation laboratory consolidation tests were conducted on a specimen collected from the middle of the clay layer. Calculate the settlement for a surcharge of 25 kN/m2 applied at the ground surface. Given, compression index, Cc=.28.

a. 450 mm b. 399 mm c. 250 mm d. 289 mm

12. Solution:

Since, γw=9.81 kN /m3

γsat=18.0 kN /m3 Cc=.28

Average σo’= 10/2(18.0-9.81)=40.95 kN /m2 ∆σ=25 kN /m3 eo=1.0

Settlement, S= Cc H [log(σo’+∆σ)/ σo’] / (1+ e0), ∴ S = .28 x 10 [log (40.95+25)/ 40.95] / (1+ 1.0),

∴ S = .289m= 289mm


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13. PROBLEM (Lateral Earth Pressure)

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

a. 25.20 kN/m b. 68.80 kN/m c. 81.60 kN/m d. 98.30 kN/m

13. Solution: Φ=28o γ= 16.0 kN /m3 H= 4m

σ'h= K0σ’v = K0 (q0 + γH), q0 = surcharge load=0.0

Where, σ'h = Effective horizontal pressure σ’v = Effective vertical pressure

∴ K0= 1-sinΦ=0.53 ∴ σ'h =.53 x 16 x 4= 33.92 kN /m2

AT REST LATERAL EARTH PRESSURE ∴ Total Lateral Force, Ph= H/ 2 σ'h = .5 x 4 x 33.92=67.84 kN /m

Correct Solution is (b)

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IV. Earthquake Engineering

14. PROBLEM (Pseudo Static Analysis)

Which one of the following statement is not appropriate for Pseudo Static Analysis?

a. In the pseudo static analysis, inertial forces generated by earthquake shaking are represented by an equivalent static horizontal force acting on the slope.

b. The seismic coefficient for this analysis should be the site peak ground acceleration, amax .

c. The vertical component of ground acceleration is normally assumed to be zero during this

representation. d. The factor of safety for a given seismic coefficient can be estimated by using non-traditional slope

stability calculation methods. A factor of safety less than one indicates that the slope is stable for the given lateral force level and further analysis is not required.

14. Solution:

The factor of safety for a given seismic coefficient can be estimated by using traditional slope stability calculation methods. A factor of safety greater than one indicates that the slope is stable for the given lateral force level and further analysis is not required.


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15. PROBLEM (Earthquake)

An Auditorium building site has the following soil profile peak max acceleration of 0.30g. What is the factor of safety at 20 ft depth against liquefaction is mostly? Where, stress reduction factor, rd=.95 and cyclic stress ratio is 0.31. a 1.50 b 1.40 c 0.90 d 1.20

15. Solution:

amax=.30g σv =10 x 108 + 2 x 108 +112 x 8 =2192 #/ft2 σ’v =2192-10 x 62.4=1568 #/ft2 For 20’ depth rd=.95 and Cyclic stress ratio generate for liquefaction, CSRL=.31 Earthquake induce Shear Stress, Τf=(.65 amax/g) σv rd ∴ Τf=(.65 amax/g) σv rd=(.65 x .3g/g)2192 x .95= 406.07 lb/ft2 ∴ Cyclic stress ratio for design Earthquake, CSRE= Τf /σ’v=406.07/1568=0.26 ∴ F.S.=CSRL/CSRE=.31/.26=1.20 Correct Solution is (d)

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V. Earth Structures 16. PROBLEM (Braced and anchored excavations)

Determine the holding capacity of a single anchor slab as shown in the Figure. Where, the unit weight the of the soil, γ= 15.0 kN /m3, Ф=300, width of anchor is 0.2m and c=0.0.

a 45.00 KN b 67.00 KN c 34.00 KN d 78.00 KN

16. Solution:

γ= 15.0 kN /m3, Ф=300, Width of anchor, B= 0.2m , H=2 m, h=0.4m and c=0.0. Pu=(5.4/tanФ)(H2/Bh)0.28(γBhH) ∴ Pu=(5.4/tan300)(22/0.2 x 0.4) 0.28 (15 x 0.2 x 0.4 x 2.0) ∴ Pu= 67.13 KN Correct Solution is (b)

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17. PROBLEM (Slabs-on-grade)

A reinforced concrete floor slab thickness (h), should be determined for a forklift repair shop. The forklift truck axle load is 15 kips, loaded at a frequency of 100 operation per day, which design index is 7. Material Properties: Concrete flexural strength = 485 pounds per square inch Modulus of sub-grade reaction, k = 200 pounds per cubic inch

a Slab Thickness= 5.6” b Slab Thickness= 4.7” c Slab Thickness= 6.7” d Slab Thickness= 7.5”

17. Solution:

Category II, forklift truck axle load is 10 to 15 kips. Using following

1st Considering Concrete flexural strength = 485 psi 2nd Considering K=200 3rd Considering Design index=7 Then Determine the Slab Thickness= 6.7” Correct Solution is (c)

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18. PROBLEM (Dam)

The concrete dam with a base with of 25 m shown in the Figure is embedded 1 m into the ground surface and has a sheet pile wall of 5 m deep at its heel. The headwater is 7 m deep and the tail water is at ground surface. The permeability of the soil is k= 20 x 10-4 cm/sec both vertically and horizontally. Determine the seepage quantity per meter length of dam per day?

a 5.30 m3/day/m-length of dam b 6.20 m3/day/m-length of dam c 2.60 m3/day/m-length of dam d 3.70 m3/day/m-length of dam

18. Solution:

k= 20 x 10-4 cm/sec=2 x 10-5m/sec, Dam headwater height, h=7m, Impervious layer to Dam bottom height, H=10m No. of flow channel, Nf =4, No. of potential drop, Nd=13,

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=3.72 m3/day/m-length of dam


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19. PROBLEM (Slope of earth structure)

Determine the factor of safety against sliding along the soil & rock interface shown infinite slope shown in the Figure.

a FS=1.5 b FS=1.2 c FS=2.0 d FS=2.4

19. Solution:

Where, c=9.0 KN/m2, γ=16.0 KN/m3, H=3 m, β=230 and Ф=160

FS=c/( γH cos2β tanβ) + tan Ф/tanβ ∴ FS=9.0/(16 x 3 x cos223 x tan23) + tan 16/tan23 ∴ FS=0.52+0.68=1.20


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VI. Shallow Foundations

20. PROBLEM (Bearing Capacity)

Determine the ultimate load, Qult of a rectangular footing 6’x 4’ with eccentric shown in the Figure where, the Soil Unit Weight, γ= 118 lb/ft3, Ultimate Bearing Capacity, q’u=3000 lb/ft2, eB=1.5’ and eL=1.75’.

.

a 15.0 Kip b 48.0 Kip c 21.0 Kip d 31.0 Kip

20. Solution:

Where, eL/L=1.75/6= 0.292> 1/6, and eB/B=1.5/4= 0.375>1/6; Therefore, B1=B(1.5-3eB/B)= 4(1.5-3 x 1.5/4)= 3.750 ft L1=L(1.5-3eL/L)= 6(1.5-3 x 1.75/6)= 3.750 ft Effective Area, A’=1/2(L1B1)=1/2 (3.750 x 3.750)= 7.03 ft2 q'u=3000 lb/ft2 ∴ Qult= A’x q'u= 7.03 x 3000= 21093.75=21.09 Kip Correct Solution is (c)

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21. PROBLEM (Load Capacity)

Determine the ultimate load, Qult of circular footing with eccentric load shown in Figure where, the Soil Unit Weight, γ= 118 lb/ft3, Ultimate Bearing Capacity, q’u=2500 lb/ft2, Footing Radius of circle, R=5’ and e=0.5’.

.


21. Solution:

Where, e=0.50 and R=5’ Therefore, Area of ADC, S= πR2/2-[e√{(R2-e2)+R2Sin-1(e/R)}] S=3.14 x 52/2-[0.50√{(52-0.52)+52Sin-1(0.5/5)}] S=3.14 x 52/2-[0.50√{(24.75+143.48)}=32.77 ft2 Effective Area, A’=2S=32.77 x 2= 65.54 ft2 q'u=2500 lb/ft2 ∴ Qult= A’x q'u= 65.54 x 2500= 163850.00= 163.85 Kip Correct Solution is (c)

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22. PROBLEM (Member Design) An isolated 12-inch depth square concrete footing has an axial load of 8000 # as shown in the Figure. Allowable soil bearing pressure is 2800 #/ft2. Determine the size of footing. Unit weight of concrete γ= 145 lb/ft3,

a 1.74 ft b 1.63 ft c 2.44 ft d 1.54 ft

22. Solution:

Soil pressure, q=P/A qallow=2800 lb/ft2 Footing weight= 145 x 12/12=145 lb/ft2 qnet= qallow -Footing weight=2800-145=2655.00 lb/ft2 ∴ qnet=P/A=8000/X2 2655.00=8000/X2 ∴ X= 1.74 ft Correct Solution is (a)

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23. PROBLEM (Member Design) An isolated 12-inch thickness square of concrete footing has axial Dead Load 40 K and Live Load 30 K including Live Load Moment 50 K-ft as shown in the Figure. Net allowable soil bearing pressure is 3000 #/ft2. Determine the minimum size of the footing. Neglect the weight of the soil and footing. a 6.30 ft x 6.30 ft b 4.60 ft x 4.60 ft c 7.50 ft x 7.50 ft d 5.40 ft x 5.40 ft

23. Solution:

Soil pressure, Net qallow=3000 lb/ft2= 3 k/ ft2, B=L, M=50 k-ft P=DL+LL=40+30=70 K Eccentricity, e=M/P=50/75=0.67 Net qallow =4P/{3L(B-2e)} ∴ 3 = 4 x 70/{3 x B(B-2 x 0.67)} 3 = 4 x 70/(3B2-4B) 9B2-12B=280 B=L= 6.30 ft ∴ Footing Size=6.30 ft x 6.30 ft Correct Solution is (a)

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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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25. PROBLEM (Load Capacity)

The soil is supporting a circular footing 5’-0” the diameter is shown in the Figure. The soil unit weight, γ= 115 lb/ft3, c=320 lb/ft2, Nc=17.7, Nγ= 5.0 & Nq=7.4. Determine total allowable load on the footing, where factor of safety is FS=3.

.


24. Solution:

Where, D=3 ft, γ= 115 lb/ft3, c=320 lb/ft2, Nc=17.7, Nγ= 5.0 & Nq=7.4. Ultimate Bearing capacity, qu=0.3 γBNγ+1.3cNc+ γDNq qu=0.3 x 115 x 5.0 x 5+1.3 x 320 x 17.7+ 115 x 3 x 7.4= 10778.70 lb/ft2 qall= 10778.70/3= 3592.90 lb/ft2

Area, A=3.14/4 (5)2=19.62 ft2 ∴ Total allowable load=3592.90 x 19.62=70510.66=70.51 Kip Correct Solution is (d)

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VII. Earth Retaining Structures

26. PROBLEM (Gravity wall)

Determine the passive earth pressure, Pp of the following retaining wall, H=20’. Where, unit weight of soil, γ= 118 lb/ft3, δ= 8o and Φ=30o.

a 81 Kip/ft b 24 Kip/ft c 70 Kip/ft d 36 Kip/ft

26. Solution:

Kp=[cos30o /(√cos8o-√{sin(30o +8o) sin8o}]2 Kp=[.866 /(.99-.292}]2=1.52

∴ Pp= 1/2 γ Kp H2=.5 x 118 x 1.52 x 202= 35872 lb/ft=35.872 Kip/ft


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27. PROBLEM (Cantilever wall)

Determine the total load on the pile group, RA. Where, equivalent fluid pressure per feet length of the following wall =64 lb/ft3.


27. Solution:

Pressure at bottom of the pile cap=64 x18=1152 lb/ft2

Fx= 1/2 x 1152 x 18=10368 lb/ft=10.37 K/ft

Taking moment at RB, ΣRB=0 RA={(12 x 4.75)+(4 x 3.5)+(5 x 1.0)-(4 x 1.0)-(10.37 x 6.0)}/5.5=1.78 K-ft/ft ∴ Total load on pile group, RA=1.78 x 4=7.12 Kip Correct Solution is (b)

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28. PROBLEM (Cantilever wall)

Determine the total lateral load force on the cantilever retaining wall shown in the Figure. The water table is below the foundation. Where, Ka=0.3, unit wt. of back fill soil, γsoil= 115 lb/ft3.

a 2.80 Kip/ft b 5.10 Kip/ft c 6.20 Kip/ft d 1.75 Kip/ft

28. Solution:

Pressure at due to surcharge = 0.3 x 200 = 60.00 lb/ft2 Pressure at due to back fill=.3 x 115 x 12=414.00 lb/ft2

∴ Fx= 1/2 x (60+414.00) x 12=2844.00 lb/ft=2.84 K/ft

Correct Solution is (a)

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29. PROBLEM (Embankment retaining wall)

Determine the lateral active resultant force acting on the facing located from Elv. 0.0’ to 6.0’ on the road embankment retaining wall as shown in the Figure. The water table is below the foundation. Where, Ka=0.3, unit wt. of back fill soil, γsoil= 115 lb/ft3.

a 4.80 Kip/ft b 2.30 Kip/ft c 5.20 Kip/ft d 2.90 Kip/ft

29. Solution:

Pressure at due to surcharge = 0.3 x 250 = 75.00 lb/ft2 Pressure at 0.0’ due to back fill=.3 x 115 x 12=414.00 lb/ft2

Total pressure at 0.0’=414+75=489.00 lb/ft2 Pressure at 6.0’ due to back fill=.3 x 115 x 6= 207.00 lb/ft2 ∴ Total pressure at 6.0’=207+75= 282.00 lb/ft2 Resultant Force between 0.0’ to 6.0’ ∴ F (0.0’ to 6.0’ )= 1/2 x (489.00 +282.00) x 6=2313.00 lb/ft=2.31 K/ft


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30. PROBLEM (Retaining wall)

A cut is made in a saturated clay soil with an un-drained cohesion, c= 600 psf. Determine the lateral earth pressure at the depth of 6 ft as shown in the Figure. Where, unit wt. of clay soil, γsoil= 120 lb/ft3, θ=0 for un-drained condition.

a 90.00 psf, un-Stable Cut b 60.00 psf, un-Stable Cut c -480.00 psf, Stable Cut d -250.00 psf, Stable Cut

30. Solution:

Where, θ=0, γsoil= 120 lb/ft3 , c= 600 psf Paz=Ka σv -2c√ka, ka=tan2(450- θ/2)=1 σv=120 x 6= 720.00 psf ∴ Paz=Ka σv -2c√ka= 1 x 720 -2 x 600 x √1= -480.00 psf, Stable Cut Correct Solution is (c)

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31. PROBLEM (Retaining wall)

A cut is made in a saturated clay soil with an un-drained cohesion, c= 800 psf. What is the depth of the cut, where the lateral earth pressure become zero or positive as shown in the Figure? Where, unit wt. of clay soil, γsoil= 120 lb/ft3, θ=0 for un-drained condition.

a 6.00 ft b 10.50 ft c 15.40 ft d 13.30 ft

31. Solution:

Where, θ=0, γsoil= 120 lb/ft3 , c= 800 psf, Paz=0.0 Paz=Ka σv -2c√ka, ka=tan2(450- θ/2)=1 σv=120 x H Paz=Ka σv -2c√ka= 1 x 120 x H -2 x 800 x √1 0.0= 1 x 120 x H -2 x 800 x √1 ∴ H= 13.33 ft Correct Solution is (d)

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VIII. Deep Foundations

32. PROBLEM (Pile Capacity)

An 18 in diameter concrete pile is driven 50 ft into soft clay. Where, C=800 psf. and friction coefficient, α=0.80. Determine the allowable capacity pile capacity with F.S=3.


32. Solution:

C=800 psf., D=18”, L=50 ft, α=0.80, Atip =π/4(18/12)2=1.76 ft2 Qult= Qtip + Qfriction Qtip =9CAtip=9 x 800 x 1.76=11304.00 # =12.77 Kip

Qfriction = αCAsurface=0.80 x 800 x (πDL) Qfriction =0.80 x 800 x (3.14 x 18/12 x 50)= 150720 # =150.72 Kip Qult= Qtip + Qfriction =12.77+150.72=163.49 Kip

∴ Qallow = Qult /F.S=163.49 /3= 54.50 Kip Correct Solution is (b)

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33. PROBLEM (Pile Group Efficiency)

Determine the pile group efficiency. The 18-inch diameter piles are surrounded in sand shown in the Figure.

a 70% b 90% c 80% d 60%

33. Solution:

Where, n1=4, n2=3, d=3D, D=18” p=∏D=3.14 x 1.5=4.71 ft d=3 x 1.5=4.5 ft Pile efficiency, η={2(n1+n2-2)d + p}/pn1n2 ∴ η={2(4+3-2)4.5 +4.71}/4.71 x 4 x 3=0.80=80% 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 0.60 ft b 0.30 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+(16-2+30/2)(115-62.4)=2074 lb/ ft2=2.07 Kip/ft2 ∆S={CcH1/(1+e0)}log[(δ’+∆δ’)/ δ’] ∴ Settlement, ∆S={0.32 x 30/(1+.85)}log[(2.07+0.65)/ 2.07]=5.19 x 0.12=0.622 ft Correct Solution is (a)

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35. PROBLEM (Pile Drag Force)

Determine the pile drag force. The pile has an 18-inch diameter and it is 40 ft long as shown in the Figure. Where, Φ=300, soil and pile friction angle, δ=.6Φ. a 3.40 kip b 5.50 kip c 1.70 kip d 2.90 kip

35. Solution:

Where, Φ=300, soil and pile friction angle, δ=.6Φ, D=18”=1.5’, γf =118 pcf Hf =8 ft K=1-sinΦ=1-sin300=0.50 p=∏D=3.14 x 1.5=4.71 ft tanδ=tan(.6Φ)=0.324 Pile drag force, Qn=(pKγfH2

f tanδ)/2 ∴ Qn=(pKγfH2

f tanδ)/2=(4.71 x .50 x 118 x 82 x 0.324)/2=2881.11 Ib=2.88 kip Correct Solution is (d)

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IX. Other Topics

36. PROBLEM (Seepage)

A permeable soil layer is underlain by impervious layer as shown in the Figure. Determine the rate of seepage flow per day.

a. Q = 6500.00 ft3/day/ft b. Q = 8600.00 ft3/day/ft c. Q = 7130.00 ft3/day/ft d. Q = 4540.00 ft3/day/ft

36. Solution:

K =0.060 ft/sec, H=10 ft, α=80 Hydraulic Gradient, i=Head loss/Length=Δh/L=L tanα/(L/cosα)=sinα=sin80=0.139 d= 10cosα=9.90 Cross sectional Area, A=d x 1 Seepage, q=kiA=0.060 x 0.139 x 9.90 x1= 0.082566 ft3/sec/ft ∴ Rate of Seepage, q= 0.082566 x 60 x 60 x 24= 7133.70 ft3/day/ft


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37. PROBLEM (Aquifers)

A well has constructed an unconfined aquifer with hydraulic conductivity, K =0.005 ft/sec. An observation well is located at 850 ft form pumping well (rw=3 inch) with a water surface elevation of 12.0 ft and 9.0 ft respectively. Determine the sustainable water rate of flow.

a Q = 3.1246 ft3/sec b Q = 0.1217 ft3/sec c Q = 1.8682 ft3/sec d Q = 2.8012 ft3/sec

37. Solution:

K =0.005 ft/sec, h2 =12 ft, hw =9 ft, r2 =850 ft, and rw =3/12=.25 ft,

Q=∏K{(h22-h2

w)}/lne(r2/rw) for unconfined aquifer ∴ Q= 3.14 x 0.005 (122-92)}/lne(850/.25)=0.1216 ft3/sec

Correct Solution is (b) 38. PROBLEM (Shoring and Re-shoring)

Which statement is not applicable for Shoring and Re-shoring for a building frame construction? a. All vertical shoring equipment shall be plumb. Maximum allowable deviation from

the vertical is one-eighth inch in three feet. If this tolerance is exceeded, the shoring equipment shall not be used until readjusted within this limit.

b. Slabs or beams which are to be re-shored should be allowed to take their actual

permanent deflection before final adjustment of re-shoring equipment is made. c. While the re-shoring is underway, construction loads shall be permitted on the

partially-cured concrete. d. The allowable load on the supporting slab shall not be exceeded when re-shoring.

38. Solution:

No construction loads shall be permitted on the partially-cured concrete.


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39. PROBLEM (Earthquake loads)

A building has to be designed with an earthquake in consideration having its own weight being 7000 kip including the height of building, H=60 ft. Determine the Whiplash Effect due to earthquake of the building that not more than 0.25V where, V is the base shear including following data. Soil Type, S=1.2 Building Type, Ct= =0.075, For R.C. Space Frame I=1, Occupancy Normal Z=0.2, Zone, 2B= Medium hazard Rw=10, Moment resisting R.C. Space Frame


39. Solution:

ΣWx=7000 kip, H=60 ft, Ct= =0.075, S=1.2, I=1, Z=0.2, and Rw=10 T=CtH3/4=1.61 sec Dynamic response, C =1.25S/(T2/3)=1.09 Base Shear, V=(ZIC/Rw)ΣWx Base Shear, V=(0.2 x 1.0 x 1.09 / 10 ) x 7000=152.5 Kip Roof Floor Whiplash Effect, Ft=.07TV= 0.07 x 1.61 x 152.5=17.2 Kip Since, Whiplash Effect, Ft = 17.2 Kip < .25V=38.12 Kip ∴ Whiplash Effect, Ft = 17.2 Kip

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.)


geotecnical depth set 2

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