Idaho Professional Engineer
Examination Advanced 100-Question
Practice Exam 2026 | Questions &
Answers with Detailed Rationales |
Complete Exam Prep & Study Guide
1. A reinforced concrete beam has a factored flexural demand of 210 kip-ft
and a nominal flexural strength of 280 kip-ft. If the applicable strength
reduction factor is 0.90, what is the design strength?
A. 252 kip-ft
B. 280 kip-ft
C. 310 kip-ft
D. 189 kip-ft
Answer: 252 kip-ft
Rationale: The design strength is φMn = 0.90(280) = 252 kip-ft. The factored
demand of 210 kip-ft is therefore less than the available design strength.
2. A steel tension member has a gross area of 4.50 in² and a yield strength of
50 ksi. Ignoring resistance factors, what is its gross-section yield capacity?
,A. 180 kips
B. 200 kips
C. 225 kips
D. 250 kips
Answer: 225 kips
Rationale: Yield capacity is FyAg = 50(4.50) = 225 kips. Net-section fracture
would require a separate calculation using the net area and tensile strength.
3. A simply supported beam carries a uniformly distributed load of 2.5 kip/ft
over a 24-ft span. What is the maximum bending moment?
A. 150 kip-ft
B. 180 kip-ft
C. 225 kip-ft
D. 300 kip-ft
Answer: 180 kip-ft
Rationale: For a simply supported beam with uniform load, Mmax = wL²/8. Thus
M = 2.5(24²)/8 = 180 kip-ft.
4. A soil has a saturated unit weight of 120 pcf. At a depth of 15 ft below the
groundwater table, what is the effective vertical stress using γw = 62.4 pcf?
A. 936 psf
B. 864 psf
C. 1800 psf
D. 2664 psf
Answer: 864 psf
Rationale: Effective stress below the water table is (γsat − γw)z = (120 −
62.4)(15) = 864 psf.
, 5. A pump delivers water at 1,200 gpm against a total dynamic head of 150 ft.
Assuming water horsepower is calculated from ρgQH, which parameter
most directly increases hydraulic power requirement?
A. Lowering the fluid density
B. Increasing the flow rate
C. Reducing the head
D. Reducing gravitational acceleration
Answer: Increasing the flow rate
Rationale: Hydraulic power is proportional to QH. Increasing either flow rate or
head increases required hydraulic power, while decreasing density would reduce
it.
6. A 480-V three-phase motor draws 35 A at 0.85 power factor. What is its
approximate real electrical power input?
A. 12.1 kW
B. 19.7 kW
C. 24.3 kW
D. 29.1 kW
Answer: 24.7 kW
Rationale: P = √3VIpf = 1.732(480)(35)(0.85) ≈ 24.7 kW.
7. A control system has an open-loop transfer function G(s) = 10/(s+2). What
is its steady-state gain?
A. 0.2
B. 2
C. 5
D. 10
Answer: 5
Rationale: The DC gain is G(0) = 10/(0+2) = 5.
, 8. A steel column is pinned at both ends and has an effective length factor K =
1.0. If its unsupported length is doubled while all other properties remain
unchanged, its Euler critical buckling load becomes:
A. Twice as large
B. Four times as large
C. Half as large
D. One-fourth as large
Answer: One-fourth as large
Rationale: Euler buckling capacity varies inversely with effective length squared:
Pcr ∝ 1/(KL)². Doubling L reduces capacity by a factor of four.
9. A wastewater treatment process receives 5.0 MGD containing 240 mg/L
BOD. What is the approximate BOD loading in lb/day?
A. 5,000 lb/day
B. 8,340 lb/day
C. 10,010 lb/day
D. 12,500 lb/day
Answer: 10,010 lb/day
Rationale: Loading = Q × concentration × 8.34. Thus 5.0 × 240 × 8.34 ≈ 10,008
lb/day.
10.A highway horizontal curve has radius 800 ft and design speed 50 mph. If
superelevation and side friction are represented by the standard
relationship V² = 15R(e+f), what combined value of e+f is required?
A. 0.083
B. 0.156
C. 0.208
D. 0.312
Answer: 0.208
Examination Advanced 100-Question
Practice Exam 2026 | Questions &
Answers with Detailed Rationales |
Complete Exam Prep & Study Guide
1. A reinforced concrete beam has a factored flexural demand of 210 kip-ft
and a nominal flexural strength of 280 kip-ft. If the applicable strength
reduction factor is 0.90, what is the design strength?
A. 252 kip-ft
B. 280 kip-ft
C. 310 kip-ft
D. 189 kip-ft
Answer: 252 kip-ft
Rationale: The design strength is φMn = 0.90(280) = 252 kip-ft. The factored
demand of 210 kip-ft is therefore less than the available design strength.
2. A steel tension member has a gross area of 4.50 in² and a yield strength of
50 ksi. Ignoring resistance factors, what is its gross-section yield capacity?
,A. 180 kips
B. 200 kips
C. 225 kips
D. 250 kips
Answer: 225 kips
Rationale: Yield capacity is FyAg = 50(4.50) = 225 kips. Net-section fracture
would require a separate calculation using the net area and tensile strength.
3. A simply supported beam carries a uniformly distributed load of 2.5 kip/ft
over a 24-ft span. What is the maximum bending moment?
A. 150 kip-ft
B. 180 kip-ft
C. 225 kip-ft
D. 300 kip-ft
Answer: 180 kip-ft
Rationale: For a simply supported beam with uniform load, Mmax = wL²/8. Thus
M = 2.5(24²)/8 = 180 kip-ft.
4. A soil has a saturated unit weight of 120 pcf. At a depth of 15 ft below the
groundwater table, what is the effective vertical stress using γw = 62.4 pcf?
A. 936 psf
B. 864 psf
C. 1800 psf
D. 2664 psf
Answer: 864 psf
Rationale: Effective stress below the water table is (γsat − γw)z = (120 −
62.4)(15) = 864 psf.
, 5. A pump delivers water at 1,200 gpm against a total dynamic head of 150 ft.
Assuming water horsepower is calculated from ρgQH, which parameter
most directly increases hydraulic power requirement?
A. Lowering the fluid density
B. Increasing the flow rate
C. Reducing the head
D. Reducing gravitational acceleration
Answer: Increasing the flow rate
Rationale: Hydraulic power is proportional to QH. Increasing either flow rate or
head increases required hydraulic power, while decreasing density would reduce
it.
6. A 480-V three-phase motor draws 35 A at 0.85 power factor. What is its
approximate real electrical power input?
A. 12.1 kW
B. 19.7 kW
C. 24.3 kW
D. 29.1 kW
Answer: 24.7 kW
Rationale: P = √3VIpf = 1.732(480)(35)(0.85) ≈ 24.7 kW.
7. A control system has an open-loop transfer function G(s) = 10/(s+2). What
is its steady-state gain?
A. 0.2
B. 2
C. 5
D. 10
Answer: 5
Rationale: The DC gain is G(0) = 10/(0+2) = 5.
, 8. A steel column is pinned at both ends and has an effective length factor K =
1.0. If its unsupported length is doubled while all other properties remain
unchanged, its Euler critical buckling load becomes:
A. Twice as large
B. Four times as large
C. Half as large
D. One-fourth as large
Answer: One-fourth as large
Rationale: Euler buckling capacity varies inversely with effective length squared:
Pcr ∝ 1/(KL)². Doubling L reduces capacity by a factor of four.
9. A wastewater treatment process receives 5.0 MGD containing 240 mg/L
BOD. What is the approximate BOD loading in lb/day?
A. 5,000 lb/day
B. 8,340 lb/day
C. 10,010 lb/day
D. 12,500 lb/day
Answer: 10,010 lb/day
Rationale: Loading = Q × concentration × 8.34. Thus 5.0 × 240 × 8.34 ≈ 10,008
lb/day.
10.A highway horizontal curve has radius 800 ft and design speed 50 mph. If
superelevation and side friction are represented by the standard
relationship V² = 15R(e+f), what combined value of e+f is required?
A. 0.083
B. 0.156
C. 0.208
D. 0.312
Answer: 0.208
