Crowder - Geotechnical 1 Spring 2011

8/11/2019 Crowder - Geotechnical 1 Spring 2011

http://slidepdf.com/reader/full/crowder-geotechnical-1-spring-2011 1/26

Civil Engineering License Exam Review

Geotechnical Engineering- Session 1

Old Dominion University Brian Crowder [email protected]



Geotechnical Engineering Review

Exam Technical Content

The following geotechnical subjects are covered in the Civil Engineering License Exam:

Breadth (Morning) Exam (20%)

Subsurface Exploration and Sampling: drilling and sampling, soil classification, boring log

interpretation

Engineering Properties of Soils: index properties, phase relationships, permeability, and pavement

design criteria

Soil Mechanics Analysis - pressure distribution, lateral earth pressure, consolidation, compaction,

effective and total stresses

Shallow Foundations - bearing capacity, settlement , allowable bearing pressure

Earth Structures - Slope stabi lity, slabs-on-grade

Earth Retaining Structures - Gravity walls, cantilever walls, earth pressure diagrams, stability

analysis, braced and anchored excavations

Breadth Exam Geotechnical Engineering Content

Old Dominion University - Virginia Applied Technology and Professional Development Center



Depth Exam Subject (% points) Depth Exam Detailed Topics

Geotechnical (87.5%) Subsurface Exploration and Sampling

Engineering Properties of Soils and Materials

Soil Mechanics Analysis

Earthquake Engineering

Earth Structures

Shallow Foundations

Deep Foundations

Earth Retaining Structures

Other Topics (12.5%) Groundwater and well fields

Loadings (dead, live, earthquake)

Construction operations and methods

Temporary Structures

Worker Health, Safety, and Environment

Geotechnical Engineering Depth Exam Content

Due to the brevity of this review course, the focus will be on the breadth portion of the exam. Much of the depth exam is

merely a more in-depth coverage of the topics in the breadth exam as well as additional topics from the other sub-

disciplines of civil engineering that are encountered during the practice of geotechnical engineering. For those planning

to take the geotechnical depth exam, it is suggested that additional problems of a more detailed nature be worked in the

geotechnical area, as well as focusing on the other areas as indicated in the table above.

Exam Format Reminder

The entire exam is now in multiple choice format. In the past there was an essay portion of the exam which frequently

included “design” type problems which can have a wide range of acceptable answers. The multiple choice format

means each problem has to be structured such that one answer (with only a small numerical range) can be found. This

should impact how you prepare for the exam as well. You should focus on working sample problems in the same format.

The majority of questions on the exam are completely independent. However, they may be a few questions which pose

a situation and ask two to five follow on questions. You should focus on analysis or general information type problems

and questions. A design type question as found in a textbook can have a range of acceptable answers. For example to

broadly ask what size concrete beam is required for a span of 25 feet and an uniformly applied live load of 100 pounds

per foot, could lead to a range of acceptable concrete beams. The various analysts could choose any range of beam

widths or depths, etc. The exam would be more likely to pose a problem such as for a concrete beam of a given cross




section and reinforcement location, what is the required area of reinforcing steel to support a uniformly applied live load of

100 pounds over a span of 25 feet. In addition, each problem will have all of the required data to reach a solution, since

they can’t require the analyst to look up numerical data that might have a range of acceptable values and which would

impact reaching the examiner’s solution.

In addition, remember that the format of the exam now requires you to work every problem. This means that you never

leave a question blank, always at least make an educated guess. This also means that you have approximately six (6)

minutes to answer each question. Although it may not seem it at times, the examiners assembling the exam have

designed the forty problems in each session to reasonably be completed in the allotted time.

References

There are no geotechnical “codes” that are required for use on the Civil PE exam. However, as indicated in the exam

content review, the geotechnical depth exam does include other areas of civil engineering practice. The main references

that will be may be potentially required for this portion of the exam are from the structural and transportation design

standards. The following summarizes the current design standards related to the coverage areas in the geotechnical

depth exam.

• AASHTO LRFD Bridge Design Specifications, 3rd Edition, 2004, with 2005 and 2006 Interim Revisions, American

Association of State Highway & Transportation Officials, Washington, DC

• ACI 318 Building Code Requirements for Structural Concrete, 2005, American Concrete Institute, Farmington Hills, MI

• ASCE 7 Minimum Design Loads for Buildings and Other Structures, 2005, American Society of Civil Engineers, New

York, NY

• IBC International Building Code, 2006 Edition (without supplements), International Code Council, Falls Church, VA

• AASHTO Guide for Design of Pavement Structures, 1993 edition (4th edition), American Association of State Highway

& Transportation Officials, Washington, DC, plus 1998 supplement.

• AI The Asphalt Handbook, (MS-4), 1989 edition, Asphalt Institute, College Park, MD.

• PCA Design and Control of Concrete Mixtures, 2002, 14th edition, Portland Cement Association, Skokie, IL.

In addition, Lindeburg recommends a copy of NAVFAC Design Manual DM 7.1 and 7.2 from the Department of Navy,

Naval Facilities Engineering Command. Neither of the publications are actively published at this time, however, you will

find businesses publishing copies of the document. In addition, there may be scanned electronic copies of the

documents available.

Coverage of Geotechnical in “Civil Engineering Reference Manual for the

PE Exam”

The purely geotechnical topics are covered as “Topic IV” in Lindeburg’s 10th Edition text or Chapters 35 to 40. Technical

areas that may appear in the depth portion of the exam from other areas in civil engineering include Chapter 54 on walls




and retaining walls, Chapter 55 on reinforced concrete footings, Chapter 76 on flexible pavement design, Chapter 77 on

rigid pavement design, and Chapter 21 on groundwater.

Technical Areas for Geotechnical Engineering Session 1• Soi l Classification

• Boring Log Interpretation

• Engineering Properties of Soils: Mass-Volume Relationships, Atterberg Limits, Permeability, Effective Stress

• Soil Mechanics: Pressure Distribution, Lateral Pressure Distribution, Settlement




Soil Classification

(Lindeburg Page 35-2 through 35-6)

Soil Classification is based on soil particle size distribution, liquid limit, and plasticity index. Particle size distribution is

measured through a sieve analysis. Liquid limit and plasticity index are part of the Atterberg limits. This quantities are

discussed on page 35-21 of Lindeburg. The two main soil classification systems are the AASHTO and Unified SoilClassification System (USCS). Let’s consider a sample soil classification problem that could occur in the morning or

afternoon portion of the exam.

Problem G-1

Given:

Sieve Analysis: % Passing

Soil No. 10 No. 40 No. 100 No. 200 LL PI

A 80 57 25 5 25 10

B 95 90 70 65 55 30

(Source: 101 Solved Civil Engineering Problems, 3rd Edition)

Problem G-1A : According to the AASHTO Soil Classification System, the classification of Sample A is most nearly:

(A) A-1-b

(B) A-3

(C) A-2-5

(D) A-2-4

Problem G-1B: According to the Unified Soil Classification System, the classification of Sample B is most nearly:

(A) CH

(B) MH

(C) OH

(D) OL

Solution G-1A:

Using Table 35.4, Page 35-4, first determine column set to use. Granular material is 35% or less passing No. 200 sieve.

Since only 5% passed the No 200 sieve, Sample A is granular. With 80% passing the No. 10, it exceeds the limit for




A-1-a. WIth 57% passing the No. 40 sieve, Sample A exceeds the limit on A-1-b. With 57% passing the No. 40 and 5%

passing the No. 200, it meets the grain size limits of A-3, however the PI of 10 is not allowable for A-3 which is required

to be non-plastic. With 5% passing the No. 200 sieve, the grain size distribution meets the requirements of the A-2

series of soil classifications which allows a maximum of 35%. Using the LL and PI you can narrow down which subset of

A-2 applies. With a LL of 25 and a PI of 10, it is observed that A-2-4 applies which is a silty or clayey gravel and sand

which provides a good general subgrade rating. ANS. (D)

Solution G-1B:

Using Table 35.5, Page 35-6, first determine if Sample B is coarse-grained for fine-grained. Since over 50% passed the

No. 200 sieve, it is considered fine grained. Since LL is 55 which is more than 50, it is considered highly compressible

and therefore could be MH, CH, or OH. To determine between these classifications the Plasticity chart under Table 35.5

is used. If you plot the point defined by the LL of 55 and PI of 30 it falls within the CH classification. ANS. (A)

Bonus Problem G-1

Another classification system is the USDA textural classification system. (Not in Lindeburg Reference Manual)

Based on

Sand Size - 2.0 to 0.05 mm in diameter

Silt Size - 0.05 to 0.002 mm in diameter

Clay size - smaller than 0.002 mm in diameter.

Although less used, it possible you will see a such a problem on the exam. Using the following data, we will look at an

example of using the USDA classification system

Given - Particle size distribution : 20% gravel,

10% sand, 30% silt, and 40% clay

Note: Chart based on only the fraction of soil

that passes the No. 10 sieve. So, if particle

size distribution includes soil particles that are

larger than 2mm, then correction is necessary.

Modified textural composition

Sand-size = (10 x 100 ) / ( 100 - 20 ) = 12.5%

Silt-size = (30 x 100) / (100-20) = 37.5%

Clay-size = (40 x 100) / (100-20) = 50.0%

Now use USDA classification figure.

Ans. Clay or Gravelly Clay




Boring Log Interpretation

Another typical type of geotechnical problem found on the Civil Engineering license exam is to provide a boring report log

and pose several questions with respect to interpreting the soil properties, the types and results of any standard soil

tests, and the techniques used in conducting the soil study. The standard types of soil tests are covered in Lindeburg on

pages 35-15 to 35-31. The soil tests reviewed are a mixture of field and laboratory tests. For many of the soil tests the

standard abbreviations are listed. It is important to be able to recognize the abbreviations for the tests and geotechnical

equipment as these are frequently used on boring logs and a question might be oriented towards deciphering the

abbreviation. Appendix 35.A of Lindeburg provides a guide to USCS soil boring, well, and geotextile symbols. Let’s

consider a boring log interpretation problem that could appear on the breadth portion of the exam.

Problem G-2

A geotechnical study has been conducted for an upcoming project. A boring log from the completed report is shown

below. Answer the following questions concerning the boring log. (Source: Civil Engineering Sample Examination, 5th

Edition)

Problem G-2A:

The shear strength of the soil between the depth of 1.5 and 2.0 meters is most nearly:

(A) 20 kPa

(B) 5 kPa

(C) 15 kPa

(D) 30 kPa

Problem G-2B:

The abbreviation “SS” in the “No. & Type” column most likely means:

(A) saturated sample

(B) split spoon

(C) saturated surface

(D) submerged sample

Problem G-2C:

How would the soil between depths 1.5 and 2.0 meters be described?

(A) Very soft




(B) Dense

(C) Very stiff

(D) Medium Stiff

Problem G-2D:

The plasticity index of the soil between depths 1.5 and 2.0 meters is most nearly?

(A) 38

(B) 18

(C) 48

(D) 56







Solution G-2A:

The shear strength is plotted near the center of the log on the same scale as the standard penetration blow counts. The

legend shows that the shear strength is indicated by a filled circle. Examining the log, a filled circle is shown at the depth

of approximately 1.6 feet at a value of 30 kPa. Therefore ANS. (D)

Solution G-2B:

The column refers to the number and type of samples taken. The log shows that the standard penetration test (SPT)

was conducted in-situ. As discussed in Lindeburg page 35-17, the SPT measures resistance to the penetration of a

standard split-spoon sampler. Therefore, ANS (B), split spoon is the most likely meaning of the abbreviation “SS.”

Solution G-2C:

Table 35.9 of Lindeburg (page 35-17) shows relationship between relative density and the SPT blow count (N) for sands .

The following table shows relationship for cohesive soils.

The boring log indicates a blow count of 4 in the soil layer in question. As shown in Table 35.9 this is associated with a

medium stiff clay. ANS (D)

Solution G-2D:

As indicated in Lindeburg on page 35-21, the Atterberg limits are sometimes given the symbols wL (liquid limit), wP

(plastic limit), and ws (shrinkage limit). Equation 35.23 (Lindeburg) shows the Plasticity Index, PI = LL - PL . On the boring

log the bar at a depth of approximately 1.7 meters shows the values of the liquid limit, plastic limit, and the in-situ water

content. The liquid limit is approximately 56. The plastic limit is approximately 18. Therefore the plastic limit equals

56-18 or 38. ANS. (A)




Engineering Properties of Soils

Lindeburg covers soil properties and testing in Chapter 35. As discussed there, soil consists of solid soil particles

separated by voids which are filled with either air or water. Therefore the soil can be characterized based on on the

percentages by volume, mass, and weight of soil, water, and air. Table 35.7 on Page 35-9 provides the indexingformulas between all of the related quantities. As part of your exam preparation for the breadth or depth exams, you

should become generally familiar with all of the defining quantities and how to relate them as shown in Table 35.7.

Permeability is discussed on page 35-22 of Lindeburg. The flow of water through a permeable material is given by

Darcy’s Law (Eq. 35.26, 35.27). For certain types of sands the permeability factor can be approximated using Hazen’s

formula (Eq. 35.28). Typical values of permeabilities for UCS soils are provided in Table 35.11. Finally, tests used to

determine the coefficient of permeability more exactly are discussed.

Let’s work a few example problems to help gain some familiari ty with the types of problems you may encounter.

Problem G-3

Given the following soil properties answer the following three questions.

A saturated soil has a void ratio of e=0.50 and a specific gravity of solid solids, Gs = 2.54.

Problem G-3A:

The moisture content of the soil is most nearly:

(A) 19

(B) 20

(C) 21

(D) 25

Problem G-3B:

The saturated unit weight is most nearly:

(A) 110

(B) 126

(C) 130

(D) 140




Problem G-3C:

The dry unit weight is most nearly:

(A) 100

(B) 125

(C) 142

(D) 105

Solution G-3A:

To approach solving these problems you want to first assess the quantities you know. The problem statement will fully

define all of the quantities you need to solve the problem. However, you may have to solve for an additional quantity

using the specified data in order to get your answer. On table 35.7 it can be seen that the degree of saturation, S, is

equal to w(SG)/e . Since the problem specifies that the sample is saturated S=1.0. Therefore the formula simplifies to e =

w(SG). This can be reordered to w = e/(SG). e and SG are given. Therefore, w = 0.50 / (2.54) = 19.7%. You could also

get this formula by looking at the formulae given for the void ratio, e, and substituting the formula for m w as a factor of the

void ratio into it and simplifying terms. Lindeburg has already provided the combined formula at the end of Table 35.7 to

save you that step. ANS. (B)

Solution G-3B:

Table 35.7 provides several equations for the saturated density. To turn the saturated density into a saturated weight,

you simply substitute the unit weight into the density term. Table 35.7 shows that the saturated density = (SG + e)(!w)/

(1+e). Therefore the wet weight is !sat = (SG + e)(!w)/(1+e) = (62.4)(2.54+0.5)/(1+0.5) = 126.5 lb/ft3. ANS. (B)

Solution G-3C:

Table 35.7 provides several formulae for the dry density. As in G-3B, these formulae can be used to find the dry unit

weight. So if dry density = (SG*"w)/(1+e), then the dry unit weight !d = (SG*!water)/(1+e) = (2.54*62.4)/(1+0.5) = 105.6

lb/ft3. ANS. (D)




Problem G-4

A Proctor test is performed on four samples. The mold volume and mass are 1/30 ft3 and 4200 g for each sample. The

follow data are collected. (Source: Civil Engineering Sample Examination, 5th Edition)

Sample Mass of Mold and Soil Water Content

1 6100 g 8.2%

2 6300 g 10.1%

3 6425 g 11.7%

4 6330 g 14.8%

Problem G-4A:

The wet mass density for sample 4 is closest to:

(A) 141 lbm / ft3

(B) 143 lbm / ft3

(C) 145 lbm / ft3

(D) 146 lbm / ft3

Problem G-4B:

The dry mass density for sample 1 is closet to:

(A) 103 lbm / ft3

(B) 116 lbm / ft3

(C) 122 lbm / ft3

(D) 125 lbm / ft3

Problem G-4C:

The relative compaction of sample 2 is approximately:

(A) 90% or less

(B) 92%




(C) 94%

(D) 95% or more

Problem G-4D:

Soils prone to drastic volume changes have which one of the following properties?

(A) Liquid limit below 10%

(B) Moisture content below 10%

(C) Organic content below 10%

(D) Plastic index above 40%

Problem G-4E

One laboratory test used to indicate the grain size of fine-grained soil is the,

(A) sieve analysis

(B) Atterberg limit test

(C) hydrometer analysis

(D) California bearing ratio (CBR) test

Solution G-4A:

m4,wet = (6330 g - 4200 g)*( 0.002205 lbm / 1 g) = 4.697 lbm (conversion of g to lbm inside front cover of Lindeburg)

wet density = "4,w = m4,wet / V = 4.697 lbm / (1/30 ft3) = 140.9 lbm / ft3 . ANS (A)

Solution G-4B:


wet density = "1,w = m1,wet / V = 4.190 lbm / (1/30 ft3) = 125.7 lbm / ft3

dry density (Table 35.7) = "1,dry = "1,w / (1 + w) = 125.7 lbm / ft3 / (1 + 0.082) = 116.2 lbm / ft3 ANS (B)

Solution G-4C:




The term relative compaction refers to the compaction of Sample 2 to the maximum dry density. The maximum dry

density would typically be found by plotting the data on a log plot, curve fitting the data, and identifying the maximum. To

`expedite the solution, you can also look at the data and identify which sample has the maximum dry density and use

that as the maximum value as an approximation. For the data in this case, looking at the mass of the sample, it appears

Sample 3 has the maximum dry density. First, however, we need to calculate the dry density of Sample 2.


wet density = "2,w = m2,wet / V = 4.63 lbm / (1/30 ft3) = 138.9 lbm / ft3

dry density (Table 35.7) = "2,dry = "2,w / (1 + w) = 138.9 lbm / ft3 / (1 + 0.101) = 126.2 lbm / ft3

now, find the dry density of sample 3:

"3,w = (6425 g - 4200 g)*( 0.002205 lbm / 1 g) / (1/30 ft3) = 147.2 lbm / ft3

dry density (Table 35.7) = "3,dry = "3,w / (1 + w) = 147.2 lbm / ft3 / (1 + .117) = 131.8 lbm / ft3

Therefore, the approximate maximum dry density is taken as 131.8 lbm / ft3

The relative compaction of Sample 2 is therefore 126.2 lbm / ft3 / 131.8 lbm / ft3 = 0.958 or 96% ANS (D)

Solution G-4D:

The volumes of soils with high plasticity indices (PI) vary greatly with changes in the moisture content. ANS (D)

Solution G-4E:

A hydrometer analysis is used to determine the particle size distribution for finer particles of soil (see LIndeburg Page

35-2). ANS (C)

Problem G-5

Given the following data from a permeability test on a sample of soil, the coefficient of permeability (meters/year) is most

nearly? (Source: Practice Problems for the Civil Engineering PE Exam, 8th Edition)

Sample Diameter 60 mm

Sample Length 120 mm

Constant Head 225 mm

Flow 1.5 mL in 6.5 min.

(A) 31

(B) 22




(C) 23

(D) 32

Solution G-5:

From Lindeburg, Eqn. 35.29, K = (VL)/(hAt). Therefore K = (1.5 mL x 12 cm ) / ((22.5 cm)*(" /4)*(6 cm)2 * 6.5 min.) =

0.004353 cm / min = 0.00004353 m/min = 22.88 meters / year. ANS. (C)

Problem G-6

Given the following subsurface profile, determine if the effective overburden pressure (psf) at the middle of the clay layer is

most nearly: (Source: NCEES Civill Engineering Sample Questions & Solutions)

(A) 2920

(B) 2046

(C) 900

(D) 1700

SUBSURFACE SECTION (ELEV. 0.0)

SAND

Moist Unit Wt = 115 PCF

Elev. (-)10’-0” # Groundwater Table

SAND

Saturated Unit Weight = 150 PCF

Elev. (-)18’-0”

NORMALLY CONSOLIDATED CLAY


Compression Index - 0.5

Void Ratio = 1.0

Elev. (-) 30’-0”

BEDROCK




Solution G-6:

Effective stress is covered on page 35-14 of Lindeburg. As discussed there, the effective stress is the portion of the total

stress that is supported through soil grain contact. Below the water table, the water applies a hydrostatic or buoyant

force on the soil column. Above the water table, the effective stress and the total stress are the same. Below the water

table the pore water pressure is equal to the weight of the water column. Therefore the solution to problem G-6 can be

found by using buoyant densities below the water table as follows:

$ ’(24ft) = 115 pcf ( 10 ft.) + (150 pcf - 62.4 pcf) (8 ft.) + (95 pcf - 62.4 pcf) (6 ft.) = 2046.4 psf. ANS (B)

Soil Mechanics (Breadth)

For the breadth portion of the exam, you should have general familiarity with the common lateral earth pressure theories

and how to calculate the pressure distribution on a given retaining wall. For each earth pressure theory you should be

familiar with the formulae used to calculate the coefficient of pressure, the magnitude of the resultant forces, and the

location of the resultants of the force components. Lindeburg discusses lateral earth pressures in Chapter 37. In the

2nd Geotechnical review session we will look closer at earth retaining structures.

To ensure appropriate design of foundations and to predict settlement, it is necessary to be able to calculate the increase

in pressure in the various soil layers below a foundation. Lindeburg discusses pressure distribution and settlement in

Chapter 40. Appendix 40.A provides Boussinesq Stress Contour Charts for continuous and square footings.

Problem G-7

Given that a retaining wall will be constructed at a site with sand backfill that has a friction angle of 28 degrees. The

height from the base of the wall to the top of the backfill is 10 ft. Answer the following questions. (Source: NCEES Civill

Engineering Sample Questions & Solutions)

Problem G-7A

According to Rankine’s theory, the active lateral earth pressure coefficient is most nearly:

(A) 0.30

(B) 0.35

(C) 0.40

(D) 0.50

Problem G-7B

The distance from the top of the backfill to the location of the resultant active earth pressure is most nearly:

(A) 3 FT




(B) 5 FT

(C) 7 FT

(D) 9 FT

Solution G-7A:

Really more details are required to correctly find the value of Ka. The most general approach and assumption would be

to assume that the backfill is horizontal and that the wall has a vertical face. With these assumptions, Lindeburg Eq. 37.7

applies.

Ka = (1-sin % )/(1+sin %) = (1 - 0.4695) / ( 1 + 0.4695 ) = 0.36 ANS. (B)

Solution G-7B:

As shown in Eq. 37.9 and discussed on page 37-4, the active pressure is a triangular distribution and therefore the

resultant acts at H/3 above the base of the wall. The question asks for the distance from the top of the backfill, therefore

that distance is 2/3 H = 6.67 feet. ANS (C)

Problem G-8

Problem G-8A:

A 6 ft. x 6 ft. footing is founded 2’ below the surface on a site with the following soil profile and soil properties. Thefooting supports a column with a combined service load of 45 kips. The settlement of the footing, if the settlement of the

sand layers is considered negligible, is most nearly? (Source: 101 Solved Civil Engineering Problems, 3rd Edition)

(A) 2 in.

(B) 1.5 in.

(C) 20 in.

(D) 8.2 in.

Problem G-8B:

Given the situation in Problem G-8A, the size of footing required to support a total service load of 72 kips and have the

same settlement as the 6’ x 6’ footing is most nearly? (Assume a 2:1 vertical to horizontal pressure distribution and no

interaction between the two footings)

(A) 7 ft. x 7 ft.




(B) 7.5 ft. x 7.5 ft.

(C) 8.0 ft. x 8.0 ft.

(D) 9.0 ft. x 9.0 ft.


SAND

Unit Wt = 120 PCF


SAND


Elev. (-)6’-0”

NORMALLY CONSOLIDATED CLAY


Compression Index - 0.38

Original Void Ratio = 0.80

Elev. (-) 10’-0”

BEDROCK

Solution G-8A:

The settlement of a normally consolidated clay is given in Equation 40.16. This equation shows that the original effective

pressure at the middle of the clay layer and the change in pressure at the same point must be found.

$ ’(8 ft.) = 120 pcf ( 4 ft.) + (130 pcf - 62.4 pcf) (2 ft.) + ( 113 pcf - 62.4 pcf) (2 ft.) = 716.4 psf

Use Appendix 40.A to estimate the change in pressure due to the footing load. By looking at ( 0, B) ( i.e. x = 0 and y = 6

ft. from the bottom of the footing to the center of the layer), it is found that the applied pressure at the center of the clay

layer, at the center of the footing is approximately 0.33*p. The applied pressure at the base of the footing is q = (45

kips / 36 sq. ft.) ( 1000 lb. / kip) = 1250 psf. Therefore, pressure increase in clay layer is &p = 0.33 x 1250 psf. = 412.5

psf. Inserting known quantities into Eqn. 40.16, the settlement, S, can be found.

S = (H * Cc * log ( (p’o + &p) / p’o )) / ( 1 + eo ) = ( 4 ft * 0.38 * log ( (716.4 psf + 412.5 psf) / 716.4 psf )) / (1 + 0.80)

S = .17 ft. = 2.0 in. ANS (A)




Solution G-8B:

In order to have the same settlement the change in stress in the middle of the clay layer must be the same for the new

footing as previously calculated, i.e. &p = 412.5 psf. The loaded area at the midpoint of the clay layer can then be

calculated,

A = 72,000 lb. / 412.5 psf = 174.5 sf.

A square footing of this area would be 13.2 ft. Assuming a 2:1 stress distribution, this area can be projected back to the

bearing elevation, i.e. project the area upward 6 ft. Width of new footing would then be 13.2 ft. - 2*0.5*6 ft. = 7.2 ft. For

a footing type problem with the wording “most nearly”, you would typically go with answer nearest value that is equal to

or larger than the calculated value in order for the solution to be acceptable. ANS (B).

Problem G-9

For the following problems, assume a circular mat foundation with a diameter of 40 ft. The maximum service load is

2500 kips. The site has the following soil conditions. (Source: Civil Engineering Sample Examination, 5th Edition)


WELL GRADED SAND

Unit Wt = 130 PCF

Net Bearing Pressure = 3000 psf


SOFT CLAY

Saturated Unit Weight = 132.4 PCF

Compression Index = 0.34

Original void ratio = 1.15

Coefficient of Consolidation = 0.10 sq. ft. / day

Elev. (-)15’-0”

BEDROCK




Problem G-9A

The factor of safety in bearing is most nearly?

(A) 1.2

(B) 1.5

(C) 2.3

(D) 6.0

Problem G-9B

The effective vertical stress at the midpoint of the clay layer prior to installation of the new structure is most nearly?

(A) 700 psf

(B) 1000 psf

(C) 1200 psf

(D) 1400 psf

Problem G-9C

The vertical stress at the midpoint of the clay layer below the center of the foundation is most nearly?

(A) 1000 psf

(B) 2000 psf

(C) 3000 psf

(D) 4000 psf

Problem G-9D

The primary settlement of the foundation is most nearly?

(A) 0.10 ft. or less

(B) 0.25 ft.

(C) 0.50 ft.

(D) 0.75 ft. or more






Solution G-9E:

The time for primary consolidation of a single layer is given by Eqn. 40.20.

t = ( Tv * Hd2 ) / ( Cv )

Since the drainage from the clay layer can only go in one direction, Hd is taken as the full layer thickness.

The value of Tv can be found from Table 40.1. For 80% settlement, Uz = 0.80, and Tv = 0.567.

t = 0.567 * (10 ft)2 / 0.10 ft2 / day = 567 days ANS (D)




BONUS PROBLEM

(Source: NCEES Sample Exam)

The figure shows the basement section of a building. The water table is above the footings. The wall must

support earth for lateral earth pressure. A triangular pressure distribution will be used for the design. The

following equivalent fluid densities apply:

Assume the soil above the water table is saturated.

Considering only one of the basement walls and only that portion between the floor and the top of the

footing, the total applied lateral force (lb/ft) is most nearly:

(A) 3,800

(B) 4,800

(C) 5,300

(D) 5,900


Pressure Above Water

Table

Below Water

Table

Active 27 pcf 21 pcf

At rest 45 pcf 35 pcf

Passive 300 pcf 230 pcf



SOLUTION

Note that the cross section indicates “BACKFILL AFTER THE FLOOR AND SLAB ON GRADE ARE IN

PLACE”. This means the walls are braced before backfill and therefore the wall will not move relative to the

backfill. In that situation you have “at-rest” earth pressure, which is the key element to identify for the

solution.

F1 = 1/2*(!1 h1)(h1)2 = (1/2)(45)(4)2 = 360 lb/ft

F2 = (!1 h1)(h2) = (45)(4)(9) = 1,620 lb/ft

F3 = 1/2*[ (!2 + !H2O) h2]*(h2) = (1/2)(35 + 62.4)(9)2 = 3,945 lb/ft

'F = F1 + F2 + F3 = 5,925 lb/ft

ANSWER (D)


Documents

Crowder - Geotechnical 1 Spring 2011