civil pe exam sample problems

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CIVIL PE EXAM SAMPLE PROBLEMS

GEOTECHNICAL DEPTH

PREPARED BY

CIVILPEEXAMS.COM

Civil PE Exam: Geotechnical Depth Sample Problems 1

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PROBLEM #1

A 4-inch diameter double-tube barrel is used to obtain rock core specimens from a natural rock

formation. The 5-foot long core run yielded a total of 43 inches of specimens. The rock core

recovery is most nearly:

A) 55%

B) 9%

C) 72%

D) 43%

PROBLEM #2

Samples of an organic soil deposit were obtained with the use of a split-spoon sampler.

Laboratory classification testing yielded the following results:

Percent Passing the No. 4 Sieve: 98%



Liquid Limit: 64

Plastic Limit: 44

The correct Unified Soil Classification System (USCS) symbol for this soil deposit is:

A) GM

B) OL

C) ML

D) OH



Problem #3

A geosynthetic will be utilized at a site with very soft fine-grained soils. Which geosynthetic will

be most appropriate to provide a primary function of separation of fine-grained soils from

coarse-grained soils?

A) Geogrid

B) Geotextile

C) Geomembrane

D) Geofoam



PROBLEM #4

A continuous footing has a width (B) of 4 feet and is embedded 2 feet into a clay material. The

groundwater table is at the bottom of the footing. The clay material has the following

properties:

Moist soil unit weight (γm): 115 pcf

Saturated soil unit weight (γsat): 120 pcf

Cohesion (c): 500 psf

Angle of internal friction (φ): 20 degrees

Assuming a Factor of Safety of 2.0 and utilizing the bearing capacity factors listed in the table

above, the allowable bearing capacity of the footing is most nearly:

A) 3,500 psf

B) 9,400 psf

C) 4,700 psf

D) 5,500 psf



PROBLEM #5

A site contains the stratigraphy presented below.

It has been determined that the vertical total and effective stresses at the middle of the clay

layer are 855 psf and 387 psf, respectively. The water table is at the ground surface. If the water

table is raised 10 feet above the existing ground surface, calculate the change in vertical

effective stress at the center of the clay layer.

A) 387 psf

B) 855 psf

C) 1,479 psf

D) No change



PROBLEM #6

A 20-foot high retaining wall is to be constructed over soft clay materials and it has been

determined (from external stability evaluations) that sliding will control the retaining wall

design. The retaining wall weighs a total of 3,000 lb/ft. Based on the information presented in

the diagram below, the minimum width (B) of the retaining wall that will provide a Factor of

Safety of at least 1.5 is most nearly:

A) 12 feet

B) 12.5 feet

C) 13 feet

D) 13.5 feet

© 2011 CivilPEExams.com

CIVIL PE EXAM SAMPLE PROBLEMS

GEOTECHNICAL DEPTH SOLUTIONS

PREPARED BY

CIVILPEEXAMS.COM

Civil PE Exam: Geotechnical Depth Sample Problem Solutions 1


PROBLEM #1: SOLUTION

Determine the rock core recovery.

Given:

1) 4-inch diameter barrel

2) 5-foot long core run

3) 43 inches of recovery

Solution:

The rock core recovery is the percentage of rock specimens recovered during the run and is defined as

follows:

�� %� = �� ℎ � �� ℎ � �� × 100%

The most important part of this problem is to remember to use units that are consistent. So, plugging

into the above equation we have:

�� %� = 43 ��ℎ��5 �� × 12 ��ℎ��

��× 100% = 71.67%

Answer is C




Determine the Unified Soil Classification System (USCS) symbol.

Given:

Percent Passing the No. 4 Sieve: 98% Percent Passing the No. 40 Sieve: 96%

Percent Passing the No. 200 Sieve: 90% Liquid Limit: 64

Plastic Limit: 44

Solution:

The No. 4 sieve separates sand from gravel. Therefore, since 98% passes the No. 4 sieve, we know that

we have a soil sample with 2% gravel-sized fragments. Therefore, the gravel (“G”) symbol does not

apply. So, answer “A” can be eliminated. Also, since the problem states that the soil sample is organic,

then the letter “O” will be the first symbol in the classification. Therefore, we can also eliminate answer

“C”. In order to determine the final answer, we will need the following graph:

Source: Naval and Facilities Design Manual 7.1

In order to determine the plasticity of the sample, we need to calculate the plasticity index (PI) as

follows:

)* = �� − )� = 64 − 44 = 20

Now, we can plot the point of LL = 64 and PI = 20 on the above figure to determine the second letter of

the classification. Since the dot falls on the right side of the “A” line as well as on the right side of the LL

= 50 line, the sample can be classified as “OH.”

Answer is D




Which geosynthetic will be most appropriate to provide to provide a primary function of separation of

fine-grained soils from coarse-grained soils?

Given:

A) Geogrid

B) Geotextile

C) Geomembrane

D) Geofoam

Solution:

Geogrids have relatively large apertures that will allow fine-grained soils to mitigate upwards to the

coarse-grained soils during dynamic loading (i.e. traffic, compaction equipment, etc.). Therefore,

geogrids are not appropriate. Geomembranes are used in landfill operations to prevent liquids and

solids from migrating to the natural groundwater table. Thus, this material is not applicable. Geofoam is

typically used as a light weight fill in order to reduce the contact stresses on compressible soils. So, this

geosynthetic is not appropriate. Geotextiles have very, very small openings that generally only allow

water to flow through for drainage purposes and their primary function is generally to provide

separation of dissimilar materials.

Answer is B




The allowable bearing capacity of the footing is most nearly:

Given:

Moist soil unit weight (γm): 115 pcf

Saturated soil unit weight (γsat): 120 pcf

Cohesion (c): 500 psf

Angle of internal friction (φ): 20 degrees

Solution:

Since this is a continuous footing, Terzaghi’s bearing capacity equation can be used and no corrections

for size, depth or inclination are required. The bearing capacity equation can be written as follows:

,-./ = �01 + 34506 + 12 7308

where,

qult = ultimate bearing capacity (psf)

c = 500 psf

γ = 115 pcf (above water) & 120 pcf (below water)

Df = footing embedment = 2 feet

B = footing width = 4 feet

Nc, Nq, Nγ = Bearing capacity factors = 14.8, 6.4, & 5.4 (from above table for φ = 20 deg.)



Since the water table is at the bottom of the footing, a submerged unit weight should be used for the

last term of the equation. The moist unit weight can be used for the soil above the water table.

Therefore, the ultimate bearing capacity can be calculated as follows:

,-./ = �01 + 34506 + 12 7308

= 500 9��: �14.8� + 115 9

��< �2 ��6.4� + 12 �4 �� =120 9

��< − 62.4 9��<> �5.4�

= 9,494 ��

The allowable bearing capacity can be calculated as follows:

,A.. = ,-./B�

Where,

qall = allowable bearing capacity

qult = ultimate bearing capacity = 9,494 psf

FS = factor of safety = 2.0

Therefore,

,A.. = ,-./B� = 9,494 ��

2 = 4747 ��

Answer is C




Determine the change in vertical effective stress by raising the water level 10 feet above the existing

ground surface:

Given:

Solution:

The PE exam contains questions that are very simple in concept and the questions may provide data that

is irrelevant to the question, thus confusing the test taker. This is one of those questions. If you

understand soil mechanics and theory of stresses, by inspection, you should understand that there is no

change in vertical effective stresses by raising the water level by any depth above the ground surface

given that the water level is already at the ground surface. The proof of this is demonstrated as

hereafter.



The following diagram presents the problem after the water table is raised.

Let’s calculate the effective stress before raising the groundwater level. The vertical effective stress at

the middle of the clay layer can be calculated as follows:

CDE = CD − ∆�

CD = 2�� 130 �� + 3��115 �� + 2.5��100 �� = 855 ��

∆� = 7.5��62.4 �� = 468 ��



Therefore,

CDE = 855 �� − 468 �� = 387 ��

Now, the vertical effective stress at the middle of the clay layer after raising the water level is as follows:

CDE = CD − ∆�

CD = 10��62.4 �� + 2�� 130 �� + 3��115 �� + 2.5��100 �� = 1,479 ��

∆� = 17.5��62.4 �� = 1,092 ��

Therefore,

CDE = 1,479 �� − 1,092 �� = 387 ��

Thus, there is no change in the vertical effective stress by raising the groundwater table. This is a very

important concept to understand.

Answer is D




Based on the information presented in the diagram below, the minimum width (B) of the retaining wall

that will provide a Factor of Safety of at least 1.5 is most nearly:

Given:

Solution:

The following free body diagram applies to this problem:



The factor of safety against sliding can be written as follows:

B� = �� B�� *�� = G��H + 7�I

)A

The only force inducing sliding is the active soil pressure. The soil resistance is provided by the clay layer.

Rearranging the above equation and solving for B, we have:

7 = )AB� − G��H�I

where,

B = wall width

Pa = 0.5γH2Ka = 0.5(110pcf)(20ft)

2(0.238) = 5,236 lb/ft

W = weight of wall = 3,000 lbs/ft

FS = 1.5

δ = 15 degrees (base friction angle of clay)

Cα = α*c = 0.55*1,000 psf

Substituting the above values in the equation and solving for B we have

7 = )AB� − G��H�I

= 5,236 9

�� 1.5� − 3,000 9�� tan �15�

0.55 ∗ 1,000 ��

7 = 12.82 ��

Answer is C