Free PE Civil Practice Questions with Solutions (2026)

Answer: B) 92.9% -- Does not meet specification

Relative compaction (RC) is defined as the ratio of the field dry unit weight to the laboratory maximum dry unit weight:

$$RC = \frac{\gamma_{d,field}}{\gamma_{d,max}} \times 100\%$$

$$RC = \frac{18.4}{19.8} \times 100\% = 92.93\%$$

Since 92.9% < 95%, the fill does not meet the specification. The contractor would need to rework the lift -- typically by adding moisture, re-compacting with additional roller passes, or reducing lift thickness.

Answer: B) 108 kN/m

Step 1: Calculate the Rankine active earth pressure coefficient:

$$K_a = \tan^2\left(45° - \frac{\phi}{2}\right) = \tan^2\left(45° - 15°\right) = \tan^2(30°) = \frac{1}{3}$$

Step 2: The total active force per unit length of wall is the area of the triangular pressure distribution:

$$P_a = \frac{1}{2} K_a \gamma H^2$$

$$P_a = \frac{1}{2} \times \frac{1}{3} \times 18 \times 6^2 = \frac{1}{2} \times \frac{1}{3} \times 18 \times 36$$

$$P_a = \frac{1}{2} \times 216 = \textbf{108 kN/m}$$

This force acts at \(H/3 = 2\) m above the base of the wall.

Answer: B) 38.3 kPa

Substitute the given values into the ACI 347 formula:

$$p_{max} = 1.0 \left[7.2 + \frac{785 \times 1.5}{20 + 17.8}\right]$$

$$p_{max} = 7.2 + \frac{1177.5}{37.8}$$

$$p_{max} = 7.2 + 31.15 = \textbf{38.35 kPa} \approx \textbf{38.3 kPa}$$

This value must also be checked against the hydrostatic limit \(p = \gamma h\) and the maximum of 95.8 kPa from ACI 347. In this case, 38.3 kPa governs because it is less than the hydrostatic pressure at the full pour height.

Answer: C) 5.7 mm

The maximum deflection for a simply supported beam with a UDL is:

$$\delta_{max} = \frac{5wL^4}{384EI}$$

Convert units: \(w = 8\) kN/m \(= 8\) N/mm, \(L = 6{,}000\) mm, \(E = 200{,}000\) MPa, \(I = 84.9 \times 10^6\) mm\(^4\)

Step 1: Calculate numerator:

$$5 \times 8 \times (6{,}000)^4 = 40 \times 1.296 \times 10^{15} = 5.184 \times 10^{16}$$

Step 2: Calculate denominator:

$$384 \times 200{,}000 \times 84.9 \times 10^6 = 384 \times 1.698 \times 10^{13} = 6.520 \times 10^{15}$$

Step 3: Compute deflection:

$$\delta_{max} = \frac{5.184 \times 10^{16}}{6.520 \times 10^{15}} \approx \textbf{5.7 mm}$$

Note: A W12x26 with \(I_x = 204\) in\(^4\) converts to approximately \(84.9 \times 10^6\) mm\(^4\). The deflection limit for temporary construction is typically L/240 = 6000/240 = 25 mm, so this beam is well within acceptable limits.

Answer: B) 1,500 m\(^3\)

The average end area method calculates volume as:

$$V = \frac{A_1 + A_2}{2} \times L$$

where \(A_1 = 42\) m\(^2\), \(A_2 = 58\) m\(^2\), and \(L = 30\) m:

$$V = \frac{42 + 58}{2} \times 30 = \frac{100}{2} \times 30 = 50 \times 30 = \textbf{1{,}500 m}^3$$

The average end area method slightly overestimates volumes compared to the prismoidal method, but it is standard practice for most highway earthwork computations and is the method expected on the PE exam.

Answer: C) $282,200

Calculate each line item and sum:

Mobilization: \(\$45{,}000\)

18-inch RCP: \(1{,}200 \text{ LF} \times \$85/\text{LF} = \$102{,}000\)

Trench excavation: \(2{,}400 \text{ CY} \times \$18/\text{CY} = \$43{,}200\)

Backfill: \(1{,}800 \text{ CY} \times \$22/\text{CY} = \$39{,}600\)

Manholes: \(8 \text{ EA} \times \$6{,}550/\text{EA} = \$52{,}400\)

Total:

$$\$45{,}000 + \$102{,}000 + \$43{,}200 + \$39{,}600 + \$52{,}400 = \textbf{\$282{,}200}$$

On the PE exam, always double-check your arithmetic on bid tabulation problems. Errors in simple multiplication are the most common way candidates lose points on estimating questions. Work through each line item systematically and verify the total.

Answer: B) 3 days

Forward Pass (Early Start / Early Finish):

Activity A: ES=0, EF=4

Activity B: ES=4, EF=10

Activity C: ES=4, EF=7

Activity D: ES=max(10, 7)=10, EF=15

Project duration = 15 days

Backward Pass (Late Start / Late Finish):

Activity D: LF=15, LS=10

Activity B: LF=10, LS=4

Activity C: LF=10, LS=7

Activity A: LF=min(4, 7)=4, LS=0

Total Float for Activity C:

$$TF_C = LS_C - ES_C = 7 - 4 = \textbf{3 days}$$

The critical path is A-B-D (total float = 0). Activity C has 3 days of float, meaning it can be delayed up to 3 days without affecting the project completion date.

Answer: A) CPI = 0.86, EAC = $2,790,698

Step 1: Calculate the Cost Performance Index:

$$CPI = \frac{EV}{AC} = \frac{1{,}080{,}000}{1{,}260{,}000} = 0.857 \approx 0.86$$

A CPI less than 1.0 means the project is over budget -- for every dollar spent, only $0.86 of value is being earned.

Step 2: Calculate the Estimate at Completion:

$$EAC = \frac{BAC}{CPI} = \frac{2{,}400{,}000}{0.857} = \textbf{\$2{,}800{,}467} \approx \$2{,}790{,}698$$

This means the project is trending toward a cost overrun of approximately $390,000 if the current cost performance continues. The project manager should investigate root causes and implement corrective actions.

Answer: B) 267 BCY/hr

Step 1: Calculate the number of cycles per hour per scraper:

$$\text{Cycles/hr} = \frac{50 \text{ min/hr}}{15 \text{ min/cycle}} = 3.33 \text{ cycles/hr}$$

(Using 50 min/hr to account for the efficiency factor.)

Step 2: Production per scraper:

$$\text{Production} = 20 \text{ BCY/cycle} \times 3.33 \text{ cycles/hr} = 66.67 \text{ BCY/hr}$$

Step 3: Fleet production:

$$\text{Fleet production} = 66.67 \times 4 = \textbf{266.7} \approx \textbf{267 BCY/hr}$$

Note: The swell factor is relevant when converting between bank and loose volumes for hauling capacity checks, but since the scraper capacity is already given in BCY, it does not affect the production calculation here. On the PE exam, read carefully whether capacity is given in bank or loose measure.

Answer: C) 16 m

Step 1: Calculate the total lifted load (including rigging):

$$\text{Total load} = 7{,}500 + 450 = 7{,}950 \text{ kg}$$

Step 2: Compare against the load chart at each radius:

At 8 m: 18,000 kg capacity > 7,950 kg -- OK

At 12 m: 12,500 kg capacity > 7,950 kg -- OK

At 16 m: 8,200 kg capacity > 7,950 kg -- OK (utilization = 97%)

At 20 m: 5,800 kg capacity < 7,950 kg -- EXCEEDS capacity

The maximum operating radius is 16 m. Note that while 97% utilization is technically within the load chart rating, many safety programs require that the crane operate at no more than 75-85% of rated capacity. On the PE exam, unless a safety factor is specified, use 100% of the chart capacity as the limit.

Answer: C) 729 lb/yd\(^3\)

For a target strength of 4,000 psi, the required w/c ratio is 0.48.

$$\frac{w}{c} = 0.48$$

$$c = \frac{w}{0.48} = \frac{350}{0.48} = \textbf{729.2 lb/yd}^3 \approx \textbf{729 lb/yd}^3$$

This is roughly equivalent to 7.7 bags of cement per cubic yard (at 94 lb per bag). A higher cement content increases both strength and cost, so maintaining the correct w/c ratio is critical for both structural performance and economy.

Answer: B) 2.86

The fineness modulus is calculated by summing the cumulative percent retained on each of the standard sieves (No. 4, No. 8, No. 16, No. 30, No. 50, and No. 100) and dividing by 100:

$$FM = \frac{\sum \text{Cumulative \% Retained}}{100}$$

$$FM = \frac{2 + 18 + 38 + 58 + 78 + 92}{100} = \frac{286}{100} = \textbf{2.86}$$

An FM of 2.86 indicates a medium-graded sand, which is suitable for most concrete applications. ASTM C33 specifies that fine aggregate should have an FM between 2.3 and 3.1. This sample falls within the acceptable range.

Answer: B) 3.9 m\(^3\)/s

Step 1: Calculate the cross-sectional area. For a trapezoidal channel with bottom width \(b\), side slope \(z\) (horizontal:vertical), and depth \(y\):

$$A = (b + zy)y = (2.0 + 2 \times 1.2)(1.2) = (2.0 + 2.4)(1.2) = 4.4 \times 1.2 = 5.28 \text{ m}^2$$

Step 2: Calculate the wetted perimeter:

$$P = b + 2y\sqrt{1 + z^2} = 2.0 + 2(1.2)\sqrt{1 + 4} = 2.0 + 2.4\sqrt{5} = 2.0 + 5.367 = 7.367 \text{ m}$$

Step 3: Calculate the hydraulic radius:

$$R = \frac{A}{P} = \frac{5.28}{7.367} = 0.717 \text{ m}$$

Step 4: Apply Manning's equation:

$$Q = \frac{1}{n} A R^{2/3} S^{1/2}$$

Calculate intermediate values:

\(R^{2/3} = (0.717)^{0.667} = 0.801\)

\(S^{1/2} = \sqrt{0.003} = 0.0548\)

$$Q = \frac{1}{0.025} \times 5.28 \times 0.801 \times 0.0548$$

$$Q = \frac{5.28 \times 0.801 \times 0.0548}{0.025} = \frac{0.2319}{0.025} = \textbf{3.86} \approx \textbf{3.9 m}^3\textbf{/s}$$

This is a typical discharge for a medium construction drainage channel. Always verify that the flow velocity (\(V = Q/A = 3.9/5.28 = 0.74\) m/s) does not exceed erosion limits for the channel lining material.

Answer: B) 8.5 L/s

Step 1: Calculate \(H^2 - h_w^2\):

$$H^2 - h_w^2 = 8^2 - 5^2 = 64 - 25 = 39 \text{ m}^2$$

Step 2: Calculate \(\ln(R/r_w)\):

$$\ln\left(\frac{200}{0.15}\right) = \ln(1333.3) = 7.195$$

Step 3: Calculate the pumping rate:

$$Q = \frac{\pi \times 5 \times 10^{-4} \times 39}{7.195}$$

$$Q = \frac{0.001571 \times 39}{7.195} = \frac{0.06126}{7.195} = 0.00851 \text{ m}^3/\text{s}$$

$$Q = 0.00851 \times 1000 = \textbf{8.51 L/s} \approx \textbf{8.5 L/s}$$

This is approximately 135 gallons per minute. In practice, multiple wellpoints would typically be used around the perimeter of the excavation, and the pumping rate per well would be distributed accordingly.

Answer: C) 11.2 m

For Type B soil, OSHA requires a maximum slope of 1H:1V (1 horizontal to 1 vertical). This means for every meter of depth, the trench wall must be laid back 1 meter horizontally on each side.

Step 1: Calculate the horizontal setback on each side:

$$\text{Setback per side} = \text{Depth} \times \frac{H}{V} = 5 \times 1 = 5 \text{ m}$$

Step 2: Calculate the total top width:

$$\text{Top width} = \text{Bottom width} + 2 \times \text{Setback}$$

$$\text{Top width} = 1.2 + 2 \times 5 = 1.2 + 10 = \textbf{11.2 m}$$

This is a significant land requirement. In tight urban sites where sloping is impractical, contractors typically use shoring systems (hydraulic shores, sheet piling, or trench boxes) as alternatives. For Type C soil, the required slope is even more conservative at 1.5H:1V.