GRADUATE ENGINEERING DEPARTMENT
GE 5103 Lesson 6 - Quiz
Questions and Answers
A+ Graded – 2026/2027 Update
Official Practice Exam · 2026/2027 Edition
Questions Minutes Passing Score Recertification
50 60 80% Semester
Table of Contents
Section 1: Soil Mechanics and Classification . . . . 14 questions (Q1–Q14)
Section 2: Foundation Design and Analysis . . . . 13 questions (Q15–Q27)
Section 3: Slope Stability and Earth Retaining Structures . . . . 12 questions (Q28–Q39)
Section 4: Geoenvironmental Engineering and Groundwater . . . . 11 questions (Q40–Q50)
Instructions
This practice quiz contains 50 multiple-choice questions divided into 4 sections. You have 60 minutes to
complete the quiz. A passing score of 80% (40 correct answers) is required. Select the single best answer
for each question. All questions are weighted equally. Review the rationale for each answer to strengthen
your understanding of geotechnical engineering concepts covered in GE 5103 Lesson 6.
GE 5103 Lesson 6 — 2026/2027 | Passing Score: 80% | Page 1 of {TOTAL}
, Section 1: Soil Mechanics and Classification — 2026/2027
Q1 Question 1 of 50
A geotechnical engineer performs a sieve analysis on a soil sample collected from a construction site
in San Antonio. The results show that 45% of the soil passes the No. 200 sieve, the liquid limit is 35,
and the plasticity index is 15. According to the Unified Soil Classification System, this soil is classified
as which type?
A. CL, a lean clay, because the plasticity index of 15 and liquid limit of 35 plot below the A-line
on the Casagrande plasticity chart
B. CH, a fat clay, because the liquid limit exceeds 30 and the fines content exceeds 50 percent
C. ML, a silt with low plasticity, because the plasticity index of 15 indicates low plasticity regardless of
the A-line position
D. SC, a clayey sand, because the coarse fraction constitutes 55 percent of the total sample
Correct Answer: A
Rationale:
With 45% fines, LL=35, and PI=15, the point plots below the A-line on the Casagrande chart, classifying the
fines as ML; however, since PI=15 and LL=35 plot above the A-line when checked (PI > 0.73(LL-20) = 0.73(15)
= 10.95), the fines are actually CL. CH requires LL>50, SC requires more than 50% coarse fraction, and ML
plots below the A-line.
Q2 Question 2 of 50
A soil sample has a void ratio of 0.65, a specific gravity of solids of 2.70, and a degree of saturation of
100%. The engineer needs to determine the unit weight of this saturated soil for foundation design
calculations. What is the saturated unit weight?
A. Approximately 108.6 pcf, because the void ratio reduces the effective unit weight below the dry unit
weight
B. Approximately 131.5 pcf, calculated as (Gs + e) times the unit weight of water divided by (1 +
e), which equals (2.70 + 0.65) times 62.4 divided by 1.65
C. Approximately 122.4 pcf, because the saturated unit weight equals the specific gravity times the
unit weight of water alone
D. Approximately 145.8 pcf, because the saturated unit weight includes the weight of solids and water
without accounting for the void ratio
Correct Answer: B
Rationale:
The saturated unit weight formula is (Gs+e)/(1+e) times gamma_w = (2.70+0.65)/1.65 times 62.4 = 3.35/1.65
times 62.4 = 126.7 pcf. Multiplying Gs alone ignores the void space, not accounting for void ratio gives
incorrect values, and the saturated unit weight always exceeds the dry unit weight.
GE 5103 Lesson 6 — 2026/2027 | Passing Score: 80% | Page 2 of {TOTAL}
, Q3 Question 3 of 50
During a standard Proctor compaction test on a clayey soil, the engineer obtains a maximum dry
density of 118 pcf at an optimum moisture content of 16%. A field density test at the construction site
yields a dry density of 112 pcf and a moisture content of 20%. What is the relative compaction, and is
the specification of 95% relative compaction met?
A. Relative compaction is 105.4%, calculated as the ratio of optimum moisture content to field
moisture content, which exceeds the specification
B. Relative compaction is 94.9%, and the over-optimum moisture content of 20% automatically
disqualifies the fill regardless of density
C. Relative compaction is 94.9%, which is slightly below the 95% specification and requires
additional compaction effort before the soil can be accepted
D. Relative compaction is 94.9%, which meets the 95% specification because field values within 1% of
the target are considered acceptable
Correct Answer: C
Rationale:
Relative compaction = field dry density / max lab dry density = 112/118 = 94.9%, just below 95%. Slightly below
specification is not automatically acceptable, the calculation is not based on moisture ratios, and over-optimum
moisture does not automatically disqualify the fill if density requirements are met.
Q4 Question 4 of 50
A consolidated-undrained triaxial test is performed on a saturated clay sample. The total principal
stresses at failure are sigma1 = 120 kPa and sigma3 = 40 kPa, and the pore water pressure at failure
is 25 kPa. What are the effective principal stresses at failure?
A. Effective sigma1 = 145 kPa and effective sigma3 = 65 kPa, obtained by adding the pore water
pressure to each total principal stress
B. Effective sigma1 = 95 kPa and effective sigma3 = 40 kPa, because only the major principal stress
is affected by pore water pressure
C. Effective sigma1 = 120 kPa and effective sigma3 = 15 kPa, because only the minor principal stress
is affected by pore water pressure
D. Effective sigma1 = 95 kPa and effective sigma3 = 15 kPa, obtained by subtracting the pore
water pressure from each total principal stress
Correct Answer: D
Rationale:
Effective stress equals total stress minus pore water pressure for both principal stresses. Adding pore pressure
gives incorrect values, only one stress being affected violates Terzaghi's principle, and both stresses must
have pore pressure subtracted.
GE 5103 Lesson 6 — 2026/2027 | Passing Score: 80% | Page 3 of {TOTAL}
GE 5103 Lesson 6 - Quiz
Questions and Answers
A+ Graded – 2026/2027 Update
Official Practice Exam · 2026/2027 Edition
Questions Minutes Passing Score Recertification
50 60 80% Semester
Table of Contents
Section 1: Soil Mechanics and Classification . . . . 14 questions (Q1–Q14)
Section 2: Foundation Design and Analysis . . . . 13 questions (Q15–Q27)
Section 3: Slope Stability and Earth Retaining Structures . . . . 12 questions (Q28–Q39)
Section 4: Geoenvironmental Engineering and Groundwater . . . . 11 questions (Q40–Q50)
Instructions
This practice quiz contains 50 multiple-choice questions divided into 4 sections. You have 60 minutes to
complete the quiz. A passing score of 80% (40 correct answers) is required. Select the single best answer
for each question. All questions are weighted equally. Review the rationale for each answer to strengthen
your understanding of geotechnical engineering concepts covered in GE 5103 Lesson 6.
GE 5103 Lesson 6 — 2026/2027 | Passing Score: 80% | Page 1 of {TOTAL}
, Section 1: Soil Mechanics and Classification — 2026/2027
Q1 Question 1 of 50
A geotechnical engineer performs a sieve analysis on a soil sample collected from a construction site
in San Antonio. The results show that 45% of the soil passes the No. 200 sieve, the liquid limit is 35,
and the plasticity index is 15. According to the Unified Soil Classification System, this soil is classified
as which type?
A. CL, a lean clay, because the plasticity index of 15 and liquid limit of 35 plot below the A-line
on the Casagrande plasticity chart
B. CH, a fat clay, because the liquid limit exceeds 30 and the fines content exceeds 50 percent
C. ML, a silt with low plasticity, because the plasticity index of 15 indicates low plasticity regardless of
the A-line position
D. SC, a clayey sand, because the coarse fraction constitutes 55 percent of the total sample
Correct Answer: A
Rationale:
With 45% fines, LL=35, and PI=15, the point plots below the A-line on the Casagrande chart, classifying the
fines as ML; however, since PI=15 and LL=35 plot above the A-line when checked (PI > 0.73(LL-20) = 0.73(15)
= 10.95), the fines are actually CL. CH requires LL>50, SC requires more than 50% coarse fraction, and ML
plots below the A-line.
Q2 Question 2 of 50
A soil sample has a void ratio of 0.65, a specific gravity of solids of 2.70, and a degree of saturation of
100%. The engineer needs to determine the unit weight of this saturated soil for foundation design
calculations. What is the saturated unit weight?
A. Approximately 108.6 pcf, because the void ratio reduces the effective unit weight below the dry unit
weight
B. Approximately 131.5 pcf, calculated as (Gs + e) times the unit weight of water divided by (1 +
e), which equals (2.70 + 0.65) times 62.4 divided by 1.65
C. Approximately 122.4 pcf, because the saturated unit weight equals the specific gravity times the
unit weight of water alone
D. Approximately 145.8 pcf, because the saturated unit weight includes the weight of solids and water
without accounting for the void ratio
Correct Answer: B
Rationale:
The saturated unit weight formula is (Gs+e)/(1+e) times gamma_w = (2.70+0.65)/1.65 times 62.4 = 3.35/1.65
times 62.4 = 126.7 pcf. Multiplying Gs alone ignores the void space, not accounting for void ratio gives
incorrect values, and the saturated unit weight always exceeds the dry unit weight.
GE 5103 Lesson 6 — 2026/2027 | Passing Score: 80% | Page 2 of {TOTAL}
, Q3 Question 3 of 50
During a standard Proctor compaction test on a clayey soil, the engineer obtains a maximum dry
density of 118 pcf at an optimum moisture content of 16%. A field density test at the construction site
yields a dry density of 112 pcf and a moisture content of 20%. What is the relative compaction, and is
the specification of 95% relative compaction met?
A. Relative compaction is 105.4%, calculated as the ratio of optimum moisture content to field
moisture content, which exceeds the specification
B. Relative compaction is 94.9%, and the over-optimum moisture content of 20% automatically
disqualifies the fill regardless of density
C. Relative compaction is 94.9%, which is slightly below the 95% specification and requires
additional compaction effort before the soil can be accepted
D. Relative compaction is 94.9%, which meets the 95% specification because field values within 1% of
the target are considered acceptable
Correct Answer: C
Rationale:
Relative compaction = field dry density / max lab dry density = 112/118 = 94.9%, just below 95%. Slightly below
specification is not automatically acceptable, the calculation is not based on moisture ratios, and over-optimum
moisture does not automatically disqualify the fill if density requirements are met.
Q4 Question 4 of 50
A consolidated-undrained triaxial test is performed on a saturated clay sample. The total principal
stresses at failure are sigma1 = 120 kPa and sigma3 = 40 kPa, and the pore water pressure at failure
is 25 kPa. What are the effective principal stresses at failure?
A. Effective sigma1 = 145 kPa and effective sigma3 = 65 kPa, obtained by adding the pore water
pressure to each total principal stress
B. Effective sigma1 = 95 kPa and effective sigma3 = 40 kPa, because only the major principal stress
is affected by pore water pressure
C. Effective sigma1 = 120 kPa and effective sigma3 = 15 kPa, because only the minor principal stress
is affected by pore water pressure
D. Effective sigma1 = 95 kPa and effective sigma3 = 15 kPa, obtained by subtracting the pore
water pressure from each total principal stress
Correct Answer: D
Rationale:
Effective stress equals total stress minus pore water pressure for both principal stresses. Adding pore pressure
gives incorrect values, only one stress being affected violates Terzaghi's principle, and both stresses must
have pore pressure subtracted.
GE 5103 Lesson 6 — 2026/2027 | Passing Score: 80% | Page 3 of {TOTAL}
