Civil Engineering PE Sample Exam – AM
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CE SE-01 Project Planning/Quantity Take-Off Methods:
In performing a quantity take-off estimate for a construction project, which of the following information resources is/are used in the estimate:
I. Construction plans
II. Specifications
III. Square foot data
IV. Material and labor costs
a. I, II, & III
b. I, II, & IV
c. II, III, & IV
d. I, II, III, & IV
Solution:
In a quantity take-off estimate, detailed measurements and accounting of materials is assembled from the plans and specifications then multiplied with current material/labor costs to arrive at an overall project estimated cost.
Answer is b
CE SE-02 Project Planning/Project Schedules:
In Critical Path Method – CPM – scheduling, the concept of float is valuable because it allows for scheduling flexibility. Which of the following is not a type of float in CPM scheduling:
a. Flat float
b. Free float
c. Independent float
d. Total float
Solution:
From Fundamental Scheduling Procedures
Free float is the amount of delay which can be assigned to any one activity without delaying subsequent activities.
Independent float is the amount of delay which can be assigned to any one activity without delaying subsequent activities or restricting the scheduling of preceding activities.
Total float is the maximum amount of delay which can be assigned to any activity without delaying the entire project.
Answer is a
CE SE-03 Means and Methods/Construction Loads:
A piece of equipment used to erect a building must be placed midway on a beam. The beam will be designed to carry this construction load and then remain after construction to support a floor. If the beam is 20 feet long and allowed to deflect ¼” maximum, what size W shape should be used? The equipment weighs 25,000 lbs. Neglect the weight of the beam.
a. W8x35
b. W10x26
c. W12x19
d. W14x22
Solution:
Maximum deflection (at the center of the beam) = Δ = (PL3) ÷ (48EI) = 0.25”
P = 25,000 lbs. and L = 20 feet
Esteel = 29,000,000 psi
I = (PL3) ÷ (48EΔ) = (25000 x (10 x 12)3) ÷ (48 x 29,000,000 x 0.25) = 124.14 in4
Using steel tables (like those found in the Manual of Steel Construction):
Find the W shapes with a moment of inertia at least equal to 124.14 in4
A W8x35, W10x26, or W12x19 would meet the criteria. If beam height is not a concern, use the W12x19 since it would weigh the least.
Answer is c
CE SE-04 Means and Methods/Temporary Structures and Facilities:
Shuttering is a type of formwork that uses __________ to form the concrete mold.
a. Metal
b. Plastic
c. Timbers
d. Plywood
Solution:
Shuttering is a type of formwork that uses plywood to form the concrete mold.
Refer to Shuttering and Formwork
Answer is d
CE SE-05 Soil Mechanics/Lateral Earth Pressure:
Using the Rankine method, find the total earth pressure force on a 10 foot high retaining wall. The total unit weight of the soil is 105 pcf and has a friction value of 29 degrees.
a. 1,257.9 lbs.
b. 1,557.2 lbs.
c. 1,618.3 lbs.
d. 1,821.6 lbs.
Solution:
The Active Earth Pressure Coefficient is Ka = [1 – sin (29°)]/[1 + sin (29°)] = 0.347
The Total Earth Pressure Force (Pa) = 0.5 x Ka x γ x H2
Pa = 0.5 x 0.347 x 105 x (10)2 = 1,821.6 lbs.
Answer is d
CE SE-06 Soil Mechanics/Soil Consolidation:
The pressure exerted on the soil directly beneath a 3 ft x 3 ft mat footing that supports a column load of 25 kips is:
a. 18.26 psi
b. 19.29 psi
c. 20.19 psi
d. 22.92 psi
Solution:
25 kips = 25,000 lbs.
Area of footing is 3 ft x 3 ft = 9 sq.ft. = 1,296 sq.in.
Pressure on soil directly beneath footing = 25,000 lbs/1296 sq.in = 19.29 psi
Answer is b
CE SE-07 Soil Mechanics/Slope Stability:
A trench is to be excavated for the installation of large pipes. If the soil cohesion is 250 psf, the stability number is 5.03, and the soil unit weight is 125 pcf, what depth can the trench be excavated before bracing will be required?
a. 4 feet
b. 6 feet
c. 8 feet
d. 10 feet
Solution:
Critical Height = Hc = (c x Ns) ÷ γ
c = soil cohesion = 250 psf
Ns = stability number = 5.03
γ = soil unit weight = 125 pcf
Hc = (250 psf x 5.03) ÷ 125 pcf = 10.06 feet
Answer is d
CE SE-08 Structural Mechanics/Trusses:
When analyzing trusses, which of the following is a fundamental tool:
a. Method of truss analysis
b. Method of section
c. Method of free body
d. Method of stress
Solution:
The two fundamental tools in stress analysis of trusses are the Method of Joint and the Method of Section.
Answer is b
CE SE-09 Structural Mechanics/Axial:
In general, all structural components in building construction will be subject to either tension or compression forces. What is the tensile stress in a ½” diameter steel rod subjected to load of 200 lbs.
a. 200.9 psi
b. 1,018.6 psi
c. 400.1 psi
d. 968.0 psi
Solution:
f = P/A = 200 lbs ÷ (π x r2) = 200 ÷ 0.196 sq.in. = 1,018.6 psi
Answer is b
CE SE-10 Structural Mechanics/Slabs:
A type of foundation typically found in warmer climates where ground freezing and thawing is uncommon is called a slab-on-grade foundation. This type of foundation is also known as a __________.
a. Mat-Slab Foundation
b. Pad Foundation
c. Floating Slab Foundation
d. Spread Foundation
Solution:
Refer to Shallow foundation for more information on shallow foundations.
A Slab-on-Grade foundation is also known as a Floating Slab Foundation.
Answer is c
CE SE-11 Hydraulics and Hydrology/Open-Channel Flow:
An earthen channel with weeds has an average flow velocity of 18.3 ft/s. The hydraulic radius of the channel is 4.75 ft and the linear hydraulic head loss is 8.2 ft. What is the corresponding Gauckler–Manning coefficient?
a. 0.657
b. 0.614
c. 0.687
d. 0.801
Solution:
Refer to Manning formula
V = (k/n) x Rh2/3 x S1/2
V = the cross-sectional average velocity (ft/s)
k = conversion factor (1.4859 ft1/3/s)
n = Gauckler–Manning coefficient
Rh = the hydraulic radius (ft)
S = the slope of the hydraulic grade line or the linear hydraulic head loss (L/L)
V = 18.3 = (1.4859 ft1/3/s ÷ n) x (4.75)2/3 x (8.2)1/2
n = 0.657
Answer is a
CE SE-12 Hydraulics and Hydrology/Storm Characteristics:
Storm frequency analysis and documentation allow meteorologists to predict rainfall intensities so that civil engineers can __________.
I. Design adequate stormwater facilities
II. Design proper retention and/or detention ponds
III. Evaluate existing flood prevention measures
IV. Estimate the need for future runoff controls
a. I, II, & III
b. II, III, & IV
c. I, II, & IV
d. I, II, III, & IV
Solution:
Storm frequency analysis and documentation provides meteorologists, civil engineers, city planners, and others with the historical information to make needed improvements and plan for future installations and modifications.
Answer is d
CE SE-13 Hydraulics and Hydrology/Pressure Conduit:
What pressure is experienced at the bottom of a 12” dia. condenser water pipe that rises from the first floor equipment room to the roof of a 20 story building? Assume each story is 12’-0”.
a. 121 psi
b. 104 psi
c. 193 psi
d. 341 psi
Solution:
The inside diameter of a 12” pipe is 11.94”
The cross-sectional area:
π x r2 = π x (11.94/2)2 = 111.97 in2 = 0.778 ft2
62.4 lb/ft3 x 0.778 ft2 = 48.52 lb/ft
Height = 20 stories x 12 ft/story = 240 ft
Static pressure = 48.52 lb/ft x 240 ft = 11,644.88 lbs
In psi: 11,644.88/111.97 in2 = 104 psi
Answer is b
SE 14 Geometrics/Basic Circular Curve Elements:
A horizontal curve is a transition __________ lanes of roadway that allow vehicles to negotiate turns at gradual rates.
a. between two divergent
b. between two tangent
c. between two convergent
d. between two parallel
Solution:
Refer to Fundamentals of Transportation/Horizontal Curves
Answer is b
CE SE-15 Geometrics/Traffic Volume:
A ramp acceleration lane length is measured from __________ to the point where the ramp lane width becomes less than 12 feet.
a. The Point of Tangency of the last ramp curve
b. The Point of Curve of the last ramp curve
c. The Point of Tangent of the first ramp curve
d. The Point of No Return of the last ramp curve
Solution:
Refer to Geometric Design
Answer is a
CE SE-16 Materials/Soil Classification and Boring Log Interpretation:
In the Unified Soil Classification System (USCS), which letter designates silt?
a. G
b. S
c. M
d. C
Solution:
As per the USCS, G = gravel, S = sand, M = silt, and C = Clay
Answer is c
CE SE-17 Materials/Soil Properties:
Permeability is the measure of a soil’s ability to permit water to flow through its pores or voids. One of the variables in Darcy’s Law, which quantifies the flow of water through porous media, is the coefficient of permeability. Which of the following is another name for coefficient of permeability?
a. Hydraulic Conductivity
b. Coefficient of Laminar Flow
c. Permeability Factor
d. Hydraulic Gradient
Solution:
Refer to Coefficient of Permeability
The coefficient of Permeability is also known as Hydraulic Conductivity
Answer is a
CE SE-18 Materials/Structural Steel:
Steel, as a structural building material, has qualities that make it superior to other materials. One negative quality of structural steel is that it loses structural rigidity at high temperatures like those found in building fires. At what temperature does steel start to lose rigidity?
a. 850°F
b. 1000°F
c. 2000°F
d. 2500°F
Solution:
Refer to Metals – Strength vs. Temperature
Most common steel has lost 10% of its strength at 850°F, 40% of its strength at 1025°F and 90% of its strength at 1500°F.
Answer is a
CE SE-19 Site Development/Excavation and Embankment:
If the fill area at station 1+25 is 200 ft2 and the fill area at station 2+75 is 350 ft2, what is the total net fill between stations 1+25 and 2+75?
a. 275 cubic feet
b. 1,528 cubic yards
c. 1,528 cubic feet
d. 41,250 cubic yards
Solution:
Using the Average End Area Method, V = [(A1 + A2)/2] x L
L (the distance between the two stations) = 275 ft – 125 ft = 150 ft
Volume (V) = [(200 ft2 + 350 ft2)/2] x 150 ft = 41,250 ft3
41,250 ft3/27 = 1,528 cubic yards
Answer is b
CE SE-20 Site Development/ Temporary and Permanent Soil Erosion and Sediment Control:
Erosion is the process by which earth is worn away. The small pieces of rock and soil are moved from one place (eroded) and deposited in another by:
I. Humans
II. Water
III. Wind
IV. Ice
a. I, II, & III
b. I, II, & IV
c. I, III, & IV
d. II, III, & IV
Solution:
Refer to Erosion
Answer is d
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