FE Civil Practice Exam 2027 Edition |
Full-Length 2026–27 Questions, Correct
Answers & Rationales
1. A reinforced concrete column carries a factored axial load of 1,200 kN. If
the load is increased by 15%, what is the new factored load?
A. 1,260 kN
B. 1,320 kN
C. 1,380 kN
D. 1,410 kN
Answer: 1,380 kN
Rationale: The increased load is 1,200(1.15) = 1,380 kN. Percentage increases
should be applied to the original quantity rather than added as an absolute
value.
2. A soil sample has a wet mass of 182 g and a dry mass of 156 g. If the mass
of water is 26 g, what is the water content?
A. 12.5%
B. 14.3%
,C. 16.7%
D. 18.2%
Answer: 16.7%
Rationale: Water content is w = Mw/Ms × 100 = 26/156 × 100 = 16.7%. The
denominator is the dry mass of soil solids.
3. A simply supported beam spans 8 m and carries a uniform load of 20 kN/m
over its entire span. What is the maximum bending moment?
A. 80 kN·m
B. 120 kN·m
C. 160 kN·m
D. 320 kN·m
Answer: 160 kN·m
Rationale: For a simply supported beam under a uniform load, Mmax = wL²/8.
Thus Mmax = 20(8²)/8 = 160 kN·m.
4. A concrete cylinder has a diameter of 150 mm and fails under a
compressive load of 530 kN. What is its approximate compressive stress?
A. 20.1 MPa
B. 25.0 MPa
C. 30.0 MPa
D. 35.4 MPa
Answer: 30.0 MPa
Rationale: The cross-sectional area is π(150²)/4 = 17,671 mm². Therefore, stress
= 530,000/17,671 ≈ 30.0 MPa.
5. A survey line has a horizontal distance of 420 m and an elevation difference
of 31.5 m. What is the approximate slope expressed as a percentage?
,A. 5.5%
B. 7.5%
C. 8.2%
D. 13.3%
Answer: 7.5%
Rationale: Percent slope = elevation difference/horizontal distance × 100 =
31.5/420 × 100 = 7.5%.
6. A project has an initial cost of $250,000 and produces annual net savings of
$40,000. Ignoring salvage value, what is the simple payback period?
A. 4.25 years
B. 5.00 years
C. 6.25 years
D. 8.00 years
Answer: 6.25 years
Rationale: Simple payback = initial investment/annual savings = 250,000/40,000
= 6.25 years.
7. For a normally distributed population with mean 72 and standard deviation
8, approximately what percentage of observations lie between 56 and 88?
A. 68%
B. 90%
C. 95%
D. 99.7%
Answer: 95%
Rationale: The interval 56 to 88 is ±2 standard deviations from the mean. For a
normal distribution, approximately 95% of observations fall within ±2ς.
8. A 4-m-long steel bar has an axial tensile force of 80 kN, cross-sectional area
of 1,000 mm², and modulus of elasticity of 200 GPa. What is its elongation?
, A. 0.8 mm
B. 1.2 mm
C. 1.6 mm
D. 2.4 mm
Answer: 1.6 mm
Rationale: Stress = P/A = 80 MPa. Strain = ς/E = 80/200,000 = 0.0004. Elongation
= εL = 0.0004(4,000) = 1.6 mm.
9. A force of 12 kN acts perpendicular to a 3-m lever arm. What is the
resulting moment?
A. 4 kN·m
B. 15 kN·m
C. 36 kN·m
D. 48 kN·m
Answer: 36 kN·m
Rationale: Moment equals force multiplied by perpendicular distance: M = Fr =
12(3) = 36 kN·m.
10.A project has a present benefit of $850,000 and a present cost of $700,000.
What is the benefit-cost ratio?
A. 0.82
B. 1.00
C. 1.21
D. 1.55
Answer: 1.21
Rationale: Benefit-cost ratio = present benefits/present costs = 850,000/700,000
= 1.214. A ratio greater than 1 indicates benefits exceed costs.
11.A fluid flows through a circular pipe with diameter 0.20 m at a velocity of
2.5 m/s. What is the approximate discharge?
Full-Length 2026–27 Questions, Correct
Answers & Rationales
1. A reinforced concrete column carries a factored axial load of 1,200 kN. If
the load is increased by 15%, what is the new factored load?
A. 1,260 kN
B. 1,320 kN
C. 1,380 kN
D. 1,410 kN
Answer: 1,380 kN
Rationale: The increased load is 1,200(1.15) = 1,380 kN. Percentage increases
should be applied to the original quantity rather than added as an absolute
value.
2. A soil sample has a wet mass of 182 g and a dry mass of 156 g. If the mass
of water is 26 g, what is the water content?
A. 12.5%
B. 14.3%
,C. 16.7%
D. 18.2%
Answer: 16.7%
Rationale: Water content is w = Mw/Ms × 100 = 26/156 × 100 = 16.7%. The
denominator is the dry mass of soil solids.
3. A simply supported beam spans 8 m and carries a uniform load of 20 kN/m
over its entire span. What is the maximum bending moment?
A. 80 kN·m
B. 120 kN·m
C. 160 kN·m
D. 320 kN·m
Answer: 160 kN·m
Rationale: For a simply supported beam under a uniform load, Mmax = wL²/8.
Thus Mmax = 20(8²)/8 = 160 kN·m.
4. A concrete cylinder has a diameter of 150 mm and fails under a
compressive load of 530 kN. What is its approximate compressive stress?
A. 20.1 MPa
B. 25.0 MPa
C. 30.0 MPa
D. 35.4 MPa
Answer: 30.0 MPa
Rationale: The cross-sectional area is π(150²)/4 = 17,671 mm². Therefore, stress
= 530,000/17,671 ≈ 30.0 MPa.
5. A survey line has a horizontal distance of 420 m and an elevation difference
of 31.5 m. What is the approximate slope expressed as a percentage?
,A. 5.5%
B. 7.5%
C. 8.2%
D. 13.3%
Answer: 7.5%
Rationale: Percent slope = elevation difference/horizontal distance × 100 =
31.5/420 × 100 = 7.5%.
6. A project has an initial cost of $250,000 and produces annual net savings of
$40,000. Ignoring salvage value, what is the simple payback period?
A. 4.25 years
B. 5.00 years
C. 6.25 years
D. 8.00 years
Answer: 6.25 years
Rationale: Simple payback = initial investment/annual savings = 250,000/40,000
= 6.25 years.
7. For a normally distributed population with mean 72 and standard deviation
8, approximately what percentage of observations lie between 56 and 88?
A. 68%
B. 90%
C. 95%
D. 99.7%
Answer: 95%
Rationale: The interval 56 to 88 is ±2 standard deviations from the mean. For a
normal distribution, approximately 95% of observations fall within ±2ς.
8. A 4-m-long steel bar has an axial tensile force of 80 kN, cross-sectional area
of 1,000 mm², and modulus of elasticity of 200 GPa. What is its elongation?
, A. 0.8 mm
B. 1.2 mm
C. 1.6 mm
D. 2.4 mm
Answer: 1.6 mm
Rationale: Stress = P/A = 80 MPa. Strain = ς/E = 80/200,000 = 0.0004. Elongation
= εL = 0.0004(4,000) = 1.6 mm.
9. A force of 12 kN acts perpendicular to a 3-m lever arm. What is the
resulting moment?
A. 4 kN·m
B. 15 kN·m
C. 36 kN·m
D. 48 kN·m
Answer: 36 kN·m
Rationale: Moment equals force multiplied by perpendicular distance: M = Fr =
12(3) = 36 kN·m.
10.A project has a present benefit of $850,000 and a present cost of $700,000.
What is the benefit-cost ratio?
A. 0.82
B. 1.00
C. 1.21
D. 1.55
Answer: 1.21
Rationale: Benefit-cost ratio = present benefits/present costs = 850,000/700,000
= 1.214. A ratio greater than 1 indicates benefits exceed costs.
11.A fluid flows through a circular pipe with diameter 0.20 m at a velocity of
2.5 m/s. What is the approximate discharge?
