31. Breadth-Practice-Exam-1-pdf

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Civil PE Breadth Practice Exam #1 Please visit path-to-pe.com for more study resources. Do you have questions or comments about the exam? Contact Mari at path.to.pe@gmail.com .
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1 . A clay loam soil would likely be placed in which SCS hydrologic soil group? A. Group A B. Group B C. Group C D. Group D 2 . What is the USCS group symbol of the non-plastic soil with the following gradation characteristics? A. GW B. GM C. SP D. SM path-to-pe.com Civil PE Breadth Practice Exam #1 Page 3
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3 . What is the most likely type of slope failure for the slope shown below? A. Rotational failure B. Transitional failure C. Infinite slope failure D. The slope is not likely to fail 4 . Which of the following statements is true regarding soil plasticity? A. The shrinkage limit is greater than the liquid limit. B. The plastic limit is less than the shrinkage limit. C. The plasticity index is equal to the plastic limit minus the liquid limit. D. Soils with water contents greater than the liquid limit behave as a viscous liquid material. path-to-pe.com Civil PE Breadth Practice Exam #1 Page 4
5 . A stormwater outfall pipe is being designed to transmit stormwater from a catch basin to the base of a steep slope. The measured slope dimensions are shown below. Design guidelines require the pipe to outlet from a diffuser tee pipe section placed a minimum of 3 horizontal feet from the ordinary high water mark (OHWM). What is most nearly the length of pipe [ft] needed? A. 400 B. 390 C. 590 D. 594 path-to-pe.com Civil PE Breadth Practice Exam #1 Page 5
6 . What is most nearly the magnitude of the reaction moment [in-kip] at the fixed connection at Point A? Assume an unfactored load combination. A. 9980 B. 830 C. 870 D. 10400 7 . A soil profile is shown in the figure. Which of the following statements is true? A. The total vertical stress at Point A > the total vertical stress at Point B. B. The effective vertical stress at Point A > the effective vertical stress at Point B. C. The effective vertical stress at Point B > the effective vertical stress at Point A. D. The effective vertical stress at Point B > the total vertical stress at Point B. path-to-pe.com Civil PE Breadth Practice Exam #1 Page 6
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8 . A graph of a modified Proctor test on a silty sand is shown below. Compaction specifications indicate this soil should be compacted to a minimum 95% of the maximum dry density. What is most nearly the target range of moisture contents to meet this specification during compaction? A. 11-13% B. 13-16% C. 11-16% D. 8-18% path-to-pe.com Civil PE Breadth Practice Exam #1 Page 7
9 . A pipe carries a fluid with density of 1100 kg/m 3 through a transition as shown in the diagram below. What is most nearly the gauge pressure [Pa] at Point 2? A. 5211 B. 10899 C. 2083 D. 27607 10 . A cylindrical retention tank has a diameter of 24 feet and a maximum inflow rate of 100 gpm. The retention tank is 10 feet tall and is equipped with a spillway outlet 8 feet above its base. Assuming the tank is empty at the start of a storm event that discharges 80 gpm into the tank, what is most nearly the time [hours] that water will begin to spill from the tank? A. 4.5 B. 1.4 C. 8.5 D. 5.6 path-to-pe.com Civil PE Breadth Practice Exam #1 Page 8
11 . A 230-foot by 120-foot mat footing foundation is to be poured. The mat footing will be 2 feet thick, with a 6-inch thick bearing pad of crushed gravel beneath it. What is most nearly the volume [yd 3 ] of concrete that must be delivered to the site, allowing for 8% waste? A. 59600 B. 55200 C. 2200 D. 2050 12 . A 6-foot-tall embankment is to be built atop a normally consolidated deposit of non-plastic silt. The silt has a unit weight of 120 pcf, initial void ratio of 0.5, a degree of saturation of 20 percent, and an organic content of 1 percent. What will be the primary mechanism of settlement triggered by the placement of the embankment? A. Immediate settlement B. Consolidation settlement C. Secondary compression D. All of the above 13 . A 20-foot-tall internally braced, shored excavation will be advanced through the soil profile shown below, then a mat footing foundation will be placed atop the exposed subgrade soils. The mat footing is designed to support a building load of 2 kips per square foot. What is most nearly the anticipated consolidation settlement of the building [inches] triggered by the building load? Ignore the dead load of the footing. A. -1 B. 0 C. 0.5 D. 1 path-to-pe.com Civil PE Breadth Practice Exam #1 Page 9
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14. A footing will be subjected to a load of 1000 psf. The footing was preliminarily designed to have dimensions of 4-feet wide by 4-feet long, and placed at the ground surface atop clay soils with a cohesion of 250 psf (angle of internal friction = 0 degrees). However, preliminary design did not satisfy the minimum required factor of safety for bearing capacity of 2.0. What modification to the design can be performed to increase the factor of safety for bearing capacity? Assume groundwater is well below the footing and does not affect design. A. Increase the length of the footing A. Increase the load on the footing B. Increase the depth of the footing C. None of the above 15 . A 1200-foot-long equal tangent vertical curve starts at PVC station 34+50 at elevation 600 feet. The initial grade of the curve is -3% and the final grade is 2%. What is most nearly the elevation [feet] of the lowest part of the curve? A. 480.2 B. 632.4 C. 561.3 D. 589.2 16 . An earthworks contractor would like to sell a 12-year-old excavator and use those funds to buy a new excavator for $100,000. The 12-year old excavator initially cost $60,000, with a salvage value of $8,000. When it was purchased, it had an expected lifespan of 20 years. Assuming the old excavator is sold at its current book value (assuming straight-line depreciation), how much more funding [$] does the contractor need to purchase the new excavator? A. $44,300 B. $28,800 C. $42,600 D. $71,200 17 . A 75-foot-long, 16-inch-diameter ductile iron watermain has a capacity of 12 cfs and is sloped at 1%. What is the head loss [ft] experienced over the length of the pipe? A. 0 B. 0.9 C. 0.4 D. 0.6 path-to-pe.com Civil PE Breadth Practice Exam #1 Page 10
18 . In the truss shown below, which members are zero-force members? A. CD, CF, DE, DF, FG B. AG, BC, BH, CF, FG C. AG, BH, CD, DE, EF D. AB, BC, CF, DE, EF 19 . A streambed is measured to have an average uniform flow velocity of 84 ft/min over a 20-day period. The average depth of flow is approximately 5 feet, and the streambed dimensions are shown below. The average slope of the stream is 0.1%. Estimate the Manning’s roughness coefficient [unitless] of the streambed. A. 0.05 B. 0.1 C. 0.07 D. 0.001 path-to-pe.com Civil PE Breadth Practice Exam #1 Page 11
20 . A monitoring well will take 2 laborers 48 hours to install, or 3 laborers 28 hours to install. The hourly labor cost is $60 per worker, and each worker receives a $100 per diem stipend for lodging and meals. A typical workday is 8 hours long. What is the lowest cost [$] for the work? A. $6,240 B. $5,040 C. $6,960 D. $5,760 21 . A pipe installation adjacent to a busy roadway is planned. The pipe will be installed 6 feet below the ground surface, and the excavation will be unshored. The excavated soil will be a clay with an unconfined compressive strength of 2 tsf. What is the maximum slope angle of the excavation that meets OSHA excavation requirements? A. 2H:1V B. 1.5H:1V C. 1H:1V D. 0.75H:1V path-to-pe.com Civil PE Breadth Practice Exam #1 Page 12
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22 . A concrete gravity retaining wall is being designed. The wall will support granular (cohesionless) backfill with an internal friction angle of 34 degrees and a unit weight of 120 pcf. What is most nearly the thickness of the wall [ft] required to maintain a factor of safety against overturning (measured about Point O) of 3.0? Assume Rankine earth pressures and the geometry shown below. Neglect wall friction. A. 1 B. 2 C. 3 D. 5 23 . What is most nearly the gripping force, R B, [lbf] on the pipe shown below when a 90 lbf force is applied to the handles of the slip-joint pliers? The joint at Point A is a pin. A. 450 B. 232 C. 200 D. 300 path-to-pe.com Civil PE Breadth Practice Exam #1 Page 13
24 . Which of the following is not a typical temporary erosion and sedimentation control method? A. Stabilized construction access B. Silt fence C. Storm drain inlet protection D. Mechanically stabilized earth walls 25 . Given a horizontal curve with the following parameters, determine the station at the start of curve (PVC). PT station = 26+34.72 Radius = 1200 feet Intersection Angle = 15°15’ A. 23+17.42 B. 23+15.33 C. 26+29.15 D. 26+29.18 26 . An aging storm sewer made of vitrified clay will be replaced by slip-lining a new plastic pipe into it. The new plastic pipe must be able to support an equivalent flow rate as the old pipe. The following pipe sizes are commercially available. The minimum diameter [inch] of the new plastic pipe is most nearly: A. 17 B. 14 C. 16 D. 15 path-to-pe.com Civil PE Breadth Practice Exam #1 Page 14
27 . Two steel lightposts are subjected to equal lateral loads as shown. Each lightpost is made of solid rectangular steel and they are the same length. Which lightpost will experience greater max deflection from the applied load? Assume both lightposts will deflect along their x axis. A. Lightpost 1 will experience greater max deflection. B. Lightpost 2 will experience greater max deflection. C. Each lightpost will experience the same max deflection. D. There is not enough information to answer the question. path-to-pe.com Civil PE Breadth Practice Exam #1 Page 15
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28 . What is most nearly the maximum line load, w o [kip/ft], that produces the same maximum moment in the beam as a uniform load of 25 kip/ft? A. 50 B. 25 C. 85 D. 30 29 . The results of three compression tests on concrete cylinders are provided below. Each concrete sample has a unit weight of 155 pcf. What is most nearly the modulus of elasticity (ksi) of the concrete? Test # Compressive Strength of Concrete (psi) 1 3,550 2 3,150 3 3,200 A. 3,600 B. 3,800 C. 3,570 D. 3,650 path-to-pe.com Civil PE Breadth Practice Exam #1 Page 16
30 . A flexible spread footing is 4 feet wide, 6 feet long, and 2 feet thick. What allowable bearing load [psf] will result in a maximum soil settlement of 0.5 inches, given a modulus of subgrade reaction of 100 pci and a minimum FS of 2.0? A. 25 B. 50 C. 3,600 D. 7,200 31 . An outfall pipe will be supported by a concrete deadman as shown in the figure. What is the planned invert of the pipe at the center of the deadman [ft]? The slope of the pipe between the catch basin and crest of the bluff is constant. A. 145.93 B. 145.43 C. 145.87 D. 146.13 path-to-pe.com Civil PE Breadth Practice Exam #1 Page 17
32 . A 600-foot-long embankment section is planned over an alignment with varying ground surface.The geotechnical engineer has stated that the on-site soils are not appropriate for re-use along this section. What is most nearly the volume of imported embankment [yd 3 ] material needed? A. 296,250 B. 14,970 C. 10,970 D. 404,250 33 . A steel water main is to be replaced with an HDPE pipe. The design requires a pipe that can withstand a minimum water pressure of 820 kPa. The available HDPE pipes are rated in psi. The minimum HDPE pipe rating required is: A. 160 psi B. 115 psi C. 130 psi D. 60 psi 34 . A 250-acre watershed has the following breakdown of soil types: 60 percent soil A, 25 percent soil B, and 15 percent soil C. Curve numbers for each soil type are provided in the table below. Using the NRCS method, what is most nearly the runoff depth (in) for a 25-year frequency, 24-hour storm with a total rainfall of 6 inches? Soil Curve Number, CN A 74 B 47 C 61 A. 2.4 B. 2.0 C. 1.5 D. 2.8 path-to-pe.com Civil PE Breadth Practice Exam #1 Page 18
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35 . A rectangular reinforced concrete beam requires a minimum total steel reinforcement area of 1.6 in 2 . Choose the most appropriate cross section to meet this requirement. 36 . A concrete slab is poured and kept at a constant temperature of 60 ℉ . How long [hrs] will it take for the concrete to mature to 100 ℉ -hr? A. 3.6 B. 4.2 C. 1.7 D. 2.5 37 . Per the following CPM diagram, how long [days] after project start will shoring installation be complete? A. 13 B. 17 C. 6 D. 7 path-to-pe.com Civil PE Breadth Practice Exam #1 Page 19
38 . 10 years of measured peak streamflow are provided in the table below. What is most nearly the estimated future peak streamflow [cfs] over a 10-year return period? Assume normal distribution. A. 22,500 B. 19,600 C. 26,000 D. 20,000 39 . A driver slams on the brakes on a level asphalt road, coming to a stop just before a stop sign. The vehicle travels 60 feet in 3 seconds during deceleration. What is most nearly the vehicle’s initial traveling speed [mph]? Assume a friction factor of 0.9 between the car’s tires and the road surface. A. 40 B. 60 C. 20 D. 120 path-to-pe.com Civil PE Breadth Practice Exam #1 Page 20
40 . A construction project involves building a new bridge over a river. The bridge will be 1000 feet long and 40 feet wide, and will have a total of 10 spans. The project will involve the following activities: ● Excavation: 135000 cubic feet ● Pile driving: 1000 linear feet ● Concrete pour: 4000 cubic yards ● Structural steel: 9420 linear feet ● Asphalt paving: 45000 square feet What is most nearly the estimated cost of the construction project, given the following unit prices? ● Excavation: $20 per cubic yard ● Pile driving: $100 per linear foot ● Concrete pour: $85 per cubic yard ● Structural steel: $5 per linear foot ● Asphalt paving: $10 per square yard A. $3,637,000 B. $637,000 C. $1,037,100 D. $947,000 path-to-pe.com Civil PE Breadth Practice Exam #1 Page 21
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Civil PE Breadth Practice Exam #1 Solution Key Please visit path-to-pe.com for more study resources. Do you have questions or comments about the exam? Contact Mari at path.to.pe@gmail.com .
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path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 2
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1. A clay loam soil would likely be placed in which SCS hydrologic soil group? A. Group A B. Group B C. Group C D. Group D NCEES Reference Handbook: pg 379 (keyword: “SCS hydrologic soils”) Solution: Per the description on page 379, clay loam soils fit within SCS hydrologic soil group C. The correct answer is C. 2. What is the USCS group symbol of the non-plastic soil with the following gradation characteristics? A. GW B. GM C. SP D. SM NCEES Reference Handbook: pg 116 (keyword: “USCS”, “soil classification”) Solution: Use the table on pg 116 and process of elimination. 1. Since % passing the #200 sieve < 50% (i.e. percent retained on #200 sieve > 50%), the soil is coarse-grained (either a sand or a gravel). The group symbol will start either with either an “S” or a “G” 2. Compare % gravel to % sand. a. % gravel = % retained on the #4 sieve = 100% - 86.9% = 13.1% path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 3
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b. % sand = % passing #4 sieve but retained on #200 sieve = 86.9% - 32.6%) = 54.3% c. % sand > % gravel, so the soil is a sand. The group symbol will start with an “S”. 3. % fines = % passing #200 sieve = 32.6% > 12%. The solution has to be either an SM or an SC. Since the problem states the soil is non-plastic, SM is the most likely soil type. The correct answer is D. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 4
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3. What is the most likely type of slope failure for the slope shown below? A. Rotational failure B. Translational failure C. Infinite slope failure D. The slope is not likely to fail NCEES Reference Handbook: Section 3.6; pgs 106-113 (keyword: “slope failure”) Solution: The solution comes from recognizing that a failure surface is not likely to extend through the impermeable sandstone layer. Due to the shape of the slope (the partially saturated weathered sandstone layer is thin), an infinite slope failure is the most likely type of slope failure in this case. The correct answer is C. 4. Which of the following statements is true regarding soil plasticity? A. The shrinkage limit is greater than the liquid limit. B. The plastic limit is less than the shrinkage limit. C. The plasticity index is equal to the plastic limit minus the liquid limit. D. Soils with water contents greater than the liquid limit behave as a viscous liquid material. NCEES Reference Handbook: pg 138 (keywords “Atterberg Limits”, “Shrinkage Limit”, “Plastic Limit”, “Liquid Limit”) Solution: By studying the Atterberg limit figure on pg 138, it is apparent that choices A through C are not correct. The correct answer is confirmed in the table at the base of the page - the definition of the liquid limit is the water content at which a soil begins to behave as a viscous fluid material. The correct answer is D. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 5
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5. A stormwater outfall pipe is being designed to transmit stormwater from a catch basin to the base of a steep slope. The measured slope dimensions are shown below. Design guidelines require the pipe to outlet from a diffuser tee pipe section placed a minimum of 3 horizontal feet from the ordinary high water mark (OHWM). What is most nearly the length of pipe [ft] needed? A. 400 B. 390 C. 590 D. 594 NCEES Reference Handbook: Section 1.3 (keyword: “distance between two points”) Solution: Length = Length of Section 1 + Length of Section 2 + Length of Section 3 Know: ?????[%] = ???? ??? * 100 → ?𝑖?? = ?????[%] * ??? 100 And ?????ℎ = ?𝑖?? 2 + ??? 2 path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 6
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Length of Section 1 = x1 = 90 ft Length of Section 2: ?𝑖?? 2 = 30 * 65?? 100 = 19. 5?? ?????ℎ 2 = 19. 5 2 + 65 2 = 67. 86?? Length of Section 3: ?𝑖?? 3 = 173.2 * (120−3)?? 100 = 202. 64?? ?????ℎ 3 = 202. 64 2 + (120 − 3) 2 = 234?? L = L1 + L2 + L3 = 90 ft + 67.86 ft + 234 ft =391.86 ft L is most nearly 390 feet. The correct answer is B. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 7
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6. What is most nearly the magnitude of the reaction moment [in-kip] at the fixed connection at Point A? Assume an unfactored load combination. A. 9980 B. 830 C. 870 D. 10400 NCEES Reference Handbook: pg 35 (keyword: “cantilevered beam slopes”) Solution: Step 1: Recognize that, in addition to the live load point load shown, there is also a uniform dead load of the beam (W beams are named by size x weight [lbf] per linear foot). The uniform dead load of the beam is 67 lbf / ft. Step 2: Recognize that the solution calls for units of in-kips. For simplicity, find the length of the beam in inches = 32ft * 12 in/ft = 384 in. Also find the uniform dead load of the beam in kip/in: 67 ??? ?? * ( ??? 1000 ??? ) * ( ?? 12 ?? ) = 0. 006 ??? ?? Step 3: Use the first two rows of the “Cantilevered Beam Slopes and Deflections” table in the NCEES reference handbook to find the maximum moment at Point A. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 8
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? ? = ? ???? + ? ???? = [26 ?𝑖? * 384𝑖?] + [ 0.006 ???/?? * (384??) 2 2 ] = 10426 𝑖? * ?𝑖? The answer is most nearly 10400 in-kips. The correct answer is D. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 9
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7. A soil profile is shown in the figure. Which of the following statements is true? A. The total vertical stress at Point A > the total vertical stress at Point B. B. The effective vertical stress at Point A > the effective vertical stress at Point B. C. The effective vertical stress at Point B > the effective vertical stress at Point A. D. The effective vertical stress at Point B > the total vertical stress at Point B. NCEES Reference Handbook: The difference between effective and total vertical stresses in soils does not seem to be in the handbook, but it is an important concept to know. ??????𝑖?? ????𝑖??? ?????? = ????? ????𝑖??? ?????? − ???? ???????? An equivalent expression: σ' = σ − ? An equivalent expression: γ'? = γ? − γ ? ? ? where z = depth, = unit weight, ’ = buoyant unit weight, γ γ γ ? = ??𝑖? ??𝑖?ℎ? ?? ????? = 62. 4 ??? path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 10
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Solution: Knowing this concept, we can solve this problem by finding the total and effective vertical stresses at Points A and B. Point A: σ ? = γ ???? ? ???? + γ ???? ? ????(????? 𝑃???? ?) = (120 ??? * 10 ??) + (105??? * 2??) = 1410 ??? ? ? = γ ? ? ?(????? 𝑃???? ?) = (62. 4??? * 0??) = 0 ??? σ ? ' = σ ? − ? ? = 1410 ??? Point B: σ ? = γ ???? ? ???? + γ ???? ? ????(????? 𝑃???? ?) = (120 ??? * 10 ??) + (105??? * 5??) = 1725 ??? ? ? = γ ? ? ?(????? 𝑃???? ?) = (62. 4??? * 3??) = 187. 2 ??? σ ? ' = σ ? − ? ? = 1725 ??? − 187. 2 ??? = 1537. 8 ??? The correct answer is C. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 11
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8. A graph of a modified Proctor test on a silty sand is shown below. Compaction specifications indicate this soil should be compacted to a minimum 95% of the maximum dry density. What is most nearly the target range of moisture contents to meet this specification during compaction? A. 11-13% B. 13-16% C. 11-16% D. 8-18% NCEES Reference Handbook: pg 152, Section 3.9.1 (keyword: “modified proctor”) Solution: ● The maximum dry density from the chart is approximately 125 pcf. ● 95% of the maximum dry density = 125 pcf * 95% = 119 pcf (approximately) ● If you draw a horizontal line through the graph at 119 pcf, it intersects the compaction curve at moisture contents of approximately 11% and 15.5%. This represents the range of moisture contents where dry density above 119 pcf can be achieved. ● The answer is most nearly between 11% and 16%. The correct answer is C. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 12
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9. A pipe carries a fluid with density of 1100 kg/m 3 through a transition as shown in the diagram below. What is most nearly the gauge pressure [Pa] at Point 2? A. 5211 B. 10899 C. 2083 D. 27607 NCEES Reference Handbook : pg 307 (keword: “Bernoulli Equation”) Solution: Recognize that there is continuity of the pipe between Points 1 and 2, so the Bernoulli equation of constant energy applies. ? 1 + 𝑃 1 γ + ? 1 2 2? = ? 2 + 𝑃 2 γ + ? 2 2 2? Rearrange to find P 2 ? 2 = γ * [(? 1 − ? 2 ) + 𝑃 1 γ + ( ? 1 2 2? − ? 2 2 2? )] Plug and chug, knowing that g = 9.807 m/s 2 (Section 1.1.2 of handbook), and γ = ρ * ? (?ℎ??? ρ = ????𝑖?? ?? ???𝑖?) ? 2 = 1100 * 9. 807 * [(162 − 163. 2) + 11900 1100*9.807 + ( 2.5 2 2? − 0.75 2 2? )] = 2083 ?? Double check the units work out (reference Section 1.2 Conversion Factors): ?? = ?? ? 3 * ? ? 2 * [(?) + 𝑃? ?? ? 3 * ? ? 2 + ( ? 2 ? 2 * ? 2 ? )] ?? = ?? ?*? 2 + ?? + ?? ?*? 2 = ?? + ?? + ?? → ??𝑖?? ??? ???? The correct answer is C. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 13
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10. A cylindrical retention tank has a diameter of 24 feet and a maximum inflow rate of 100 gpm. The retention tank is 10 feet tall and is equipped with a spillway outlet 8 feet above its base. Assuming the tank is empty at the start of a storm event that discharges 80 gpm into the tank, what is most nearly the time [hours] that water will begin to spill from the tank? A. 4.5 B. 1.4 C. 8.5 D. 5.6 NCEES Reference Handbook : pg 10 Section 1.3.7.12 - volume of a cylinder (keyword: “cylinder”) Solution: Simplify the question down to the key concepts. We are given a rate of inflow and are looking for the number of hours for the tank to fill up to the spillway elevation, 8 feet above the tank’s base. The volume of the tank between its base and the spillway can be represented with h = 8 ft, d = 24 ft. ? = π*(24??) 2 *(8??) 4 = 3619. 115 ?? 3 Convert ft 3 to gallons (reference Section 1.2 Conversion Factors) 3619. 115 ?? 3 * 7. 481 = 27074. 599 ??? Time to fill the tank to this volume = V / inflow ?𝑖?? = ? / ? = 27074.599 ??? 80 ?𝑎? ?𝑖? = 338. 4?𝑖? = 5. 6 ℎ? The correct answer is D. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 14
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11. A 230-foot by 120-foot mat footing foundation is to be poured. The mat footing will be 2 feet thick, with a 6-inch thick bearing pad of crushed gravel beneath it. What is most nearly the volume [yd 3 ] of concrete that must be delivered to the site, allowing for 8% waste? A. 59600 B. 55200 C. 2200 D. 2050 NCEES Reference Handbook : n/a Solution: Volume of footing = 230 ft x 120 ft x 2 ft = 55200 ft 3 Volume, accounting for waste = 55200 ft 3 * 1.08 = 59616 ft 3 Convert volume to cubic yards (3 feet = 1 yard) 2208 yd 3 59616 ?? 3 * ( ?? 3?? ) 3 = The correct answer is C. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 15
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12. A 6-foot-tall embankment is to be built atop a normally consolidated deposit of non-plastic silt. The silt has a unit weight of 120 pcf, initial void ratio of 0.5, a degree of saturation of 20 percent, and an organic content of 1 percent. What will be the primary mechanism of settlement triggered by the placement of the embankment? A. Immediate settlement B. Consolidation settlement C. Secondary compression D. All of the above NCEES Reference Handbook: n/a Solution: This question requires a general knowledge of soil settlement, and can be solved through process of elimination. Total soil settlement = immediate settlement + consolidation settlement + secondary compression (or creep). Immediate settlement is typically caused by soil particles rearranging themselves in reaction to an applied load. Immediate settlement is a typical source of settlement for most soils, especially granular and non-plastic soils. It is not a huge source of settlement in overconsolidated soils. This type of settlement applies to the soil described in this problem. Consolidation settlement is caused by excess water gradually being squeezed out of void spaces in the soil matrix after a load is applied. Consolidation settlement is a typical source of settlement for saturated (S > 90%), fine-grained soils. Consolidation settlement in granular soils with high permeability (sands and gravels) is not easily distinguishable from immediate settlement, so it is not typically considered for granular soils. Since S < 90% for the soil described in this problem, it is not likely to be a source of settlement. Secondary compression, or creep, occurs under a constant load and can be caused by gradual degradation of the soil particles over time. Secondary compression is a typical source of settlement for organic soils The correct answer is A. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 16
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13. A 20-foot-tall internally braced, shored excavation will be advanced through the soil profile shown below, then a mat footing foundation will be placed atop the exposed subgrade soils. The mat footing is designed to support a building load of 2 kips per square foot. What is most nearly the anticipated consolidation settlement of the building [inches] triggered by the building load? Ignore the dead load of the footing. A. -1 B. 0 C. 0.5 D. 1 NCEES Reference Handbook: pgs 89-90 (keyword: “normally consolidated”, “overconsolidated”) Solution: Step 1: Determine the magnitude of unloading from the excavation. Note that effective (not) total vertical stress is needed. ??????𝑖?? ????𝑖??? ?????? = ????? ????𝑖??? ?????? − ???? ???????? γ'? = γ? − γ ? ? ? where z = depth, = unit weight, ’ = buoyant unit weight = γ γ γ ??? −γ ? γ ? = ??𝑖? ??𝑖?ℎ? ?? ????? = 62. 4 ??? ?????? = γ ???,???? * ℎ ??? + γ' ???? * ℎ ??? = (110??? * 17??) + (120??? − 62. 4???) * 3?? = 2043??? path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 17
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Step 2: Recognize the unload value is slightly greater than the applied load of 2000 psf, so the clay soils will experience a net unloading effect. Knowing this, we can eliminate options C and D, since the additional load will not trigger downward settlement. By observation, we can deduce the most likely solution to this problem is B -> 0 inches of settlement, since it would take a significant amount of unloading to result in -1 inches of expansion in the soils. To be sure, we can use the equation for settlement in overconsolidated soils (pg 90), modified to recognize that no normal consolidation will occur in this scenario. ? = 16?? 1+0.6 (0. 24??? 10 ( 2000??? 2043??? )) =− 0. 02?? = − 0. 3 𝑖? **Note that the equation above is more properly done using effective stresses at the midpoint of the consolidating layer. However, since we are doing the numerical solution as a quick check, we have simplified our solution for this problem for time management’s sake. The correct answer is B. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 18
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14. A footing will be subjected to a load of 1000 psf. The footing was preliminarily designed to have dimensions of 4-feet wide by 4-feet long, and placed at the ground surface atop clay soils with a cohesion of 250 psf (angle of internal friction = 0 degrees). However, preliminary design did not satisfy the minimum required factor of safety for bearing capacity of 2.0. What modification to the design can be performed to increase the factor of safety for bearing capacity? Assume groundwater is well below the footing and does not affect design. A. Increase the length of the footing B. Increase the load on the footing C. Increase the depth of the footing D. None of the above NCEES Reference Handbook: pg 96-96 Section 3.4.2.1 (keyword: “bearing capacity”) Solution: The ultimate bearing capacity of soils for a concentrically loaded square footing is shown in the following equation. Before getting too deep into plugging and chugging, first realize that some values will zero out since the bearing soil is purely cohesive. First consider the factors Nc, Nq, and N Ɣ , using the bearing capacity factors table on pg 97. When , Nc= 5.14, Nq = 1, and N Ɣ = 0. The third ϕ = 0 term in the bearing capacity equation is now zero, and the second term has been simplified.. ? ??? = 5. 14(?)(? ? ) + ?(? ? ) + 0 Now consider the shape factors s c and s q , using the Shape Correction Factors table on pg 96. Since = 0, s q is simplified to 1, and the bearing capacity equation becomes: ϕ ? ??? = 5. 14(250???)(1 + ? 5*? ) + ? Know that q = surcharge due to soil above the footing = γ * ? ??????? By inspection, we can see that increasing the width of the footing and load on the footing will not increase the factor of safety. Increasing the depth of the footing will increase the factor of safety. The correct answer is C. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 19
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15. A 1200-foot-long equal tangent vertical curve starts at PVC station 34+50 at elevation 600 feet. The initial grade of the curve is -3% and the final grade is 2%. What is most nearly the elevation [feet] of the lowest part of the curve? A. 480.2 B. 632.4 C. 561.3 D. 589.2 NCEES Reference Handbook: pg 278, Section 5.3.1 (keyword: “vertical curve”) Solution: Use the ‘curve elevation’ equation in Section 5.3.1. Solving the first derivative of the curve equation will give you the horizontal location (“x”) of the flat part of the curve. ????? ??????𝑖?? = ? 𝑃?? + ? 1 ? + ?? 2 Curve elevation, first derivative: 0 = ? 1 + 2?? Solve for x: , where a is defined in Section 5.3.1 as ? = − ? 1 2? ? = ? 2 −? 1 2? So ? = −? 1 2* ? 2 −? 1 2*𝐿 = 720?? Plug x = 720ft into the curve elevation equation: 589.2 ft ? = 600 + (− 0. 03) * (720) + (0. 000021) * 720 2 = The correct answer is D. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 20
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16. An earthworks contractor would like to sell a 12-year-old excavator and use those funds to buy a new excavator for $100,000. The 12-year old excavator initially cost $60,000, with a salvage value of $8,000. When it was purchased, it had an expected lifespan of 20 years. Assuming the old excavator is sold at its current book value (assuming straight-line depreciation), how much more funding [$] does the contractor need to purchase the new excavator? A. $44,300 B. $28,800 C. $42,600 D. $71,200 NCEES Reference Handbook: pg 38, Section 1.7.5 and 1.7.6 (keyword: “straight line”, “book value” Solution: We are solving for: x = additional funds needed = cost of new excavator - selling price of old excavator Selling price = current book value = initial cost - ⅀ Dj ? ? = ?−? ? ? , ?ℎ??? ? = 𝑖?𝑖?𝑖?? ???? ?? ??? ????????? = $60, 000 ? ? = ???????? ??????? ????? 𝑖? ???? ? = $8000 ? 𝑖? ?ℎ? ???????? ?𝑖?? ?? ?ℎ? ????? = 20 ????? ? ? = $60,000−$8,000 20 = $2, 600 Σ? ? = ? ? * 12 = $31, 200 → ??????𝑖??𝑖?? ?? ?ℎ? ??? ????????? Selling price = $60,000 - $31,200 = $28,800 X = $100,000 - $28,800 = $71,200 The correct answer is D. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 21
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17. A 75-foot-long, 16-inch-diameter ductile iron watermain has a capacity of 12 cfs and is sloped at 1%. What is the head loss [ft] experienced over the length of the pipe? A. 0 B. 0.9 C. 0.4 D. 0.6 NCEES Reference Handbook: pg 328 Section 6.3.1.2 (keyword: “head loss”) Solution: First make sure all the given values are in the correct units. L [ft] = 75ft D [ft] = 16in *(1ft / 12in) = 1.333ft Q [cfs] = 12cfs The Hazen-Williams coefficient can be found on pg 329, Section 6.3.1.4. For a ductile iron pipe, C = 140. Plug and chug. ℎ ? = 4.73*? ? 1.852 *? 4.87 * ? 1.852 = 0. 924?? The correct answer is B. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 22
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18. In the truss shown below, which members are zero-force members? A. CD, CF, DE, DF, FG B. AG, BC, BH, CF, FG C. AG, BH, CD, DE, EF D. AB, BC, CF, DE, EF NCEES Reference Handbook: The NCEES reference handbook does not cover the rules for determining zero-force members of a truss, but they are worth memorizing/practicing. The rules are: path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 23
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Solution: Step 1: Draw in the reaction forces.We don’t need to try to solve for them, knowing where the reaction forces are is enough for this problem. Step 2: Apply the first rule of thumb for finding zero force members: path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 24
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Step 3: Apply the second rule of thumb for finding zero force members: path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 25
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Step 4: Apply the third rule of thumb for finding zero force members: Note: since we’ve already determined that member ED is a zero-force member, we can treat Joint E as a joint that satisfies the requirements for the third rule of thumb. The correct answer is C (AG, BH, CD, DE, EF). path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 26
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19. A streambed is measured to have an average uniform flow velocity of 84 ft/min over a 20-day period. The average depth of flow is approximately 5 feet, and the streambed dimensions are shown below. The average slope of the stream is 0.1%. Estimate the Manning’s roughness coefficient [unitless] of the streambed. A. 0.05 B. 0.1 C. 0.07 D. 0.001 NCEES Reference Handbook: pg 345, Section 6.4.5.1 (keyword: “Manning’s”) Solution: We can use the flow velocity version of Manning’s equation to solve for n. ? = 1.486 ? * ? 𝐻 2/3 * ? 1/2 Where R H = cross-sectional area of flow [ft 2 ] / wetted perimeter [ft] = A / P There are readily available equations for these values in Section 6.4.5.4 (search “trapezoid”). 𝐴 = (? + ??) * ? = (10 + 2. 5 * 5) * 5 = 112. 5 ?? 2 ? = ? + 2? 1 + ? 2 = 10 + 2 * 5 * 1 + 2. 5 2 = 36. 926?? path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 27
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? 𝐻 = 𝐴/? = 3. 047?? Solve for n: ? = 1.486 ? * ? 𝐻 2/3 * ? 1/2 (make sure v has units of ft/s. ) ? = 84 ?? ??? * ( ??? 60? ) = 1. 4 ?? ? ? = 1.486 1.4 * 3. 047 2/3 * 0. 001 1/2 = 0. 071 The correct answer is C. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 28
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20. A monitoring well will take 2 laborers 48 hours to install, or 3 laborers 28 hours to install. The hourly labor cost is $60 per worker, and each worker receives a $100 per diem stipend for lodging and meals. A typical workday is 8 hours long. What is the lowest cost [$] for the work? A. $6,240 B. $5,040 C. $6,960 D. $5,760 NCEES Reference Handbook: n/a Solution: Compare the cost of Scenario A (2 laborers) to Scenario B (3 laborers): Scenario A ????? ???? = 2 ??????? * 48 ℎ???? * $60 ??????*ℎ??? = $5, 760 # ???? ?? ???? = 48 ℎ???? * ( ??? 8 ℎ???? ) = 6 ???? $1,200 ????? ??? ?𝑖?? = 2 ??????? * $100 ???*?????? * 6 ???? = ????? ???? = ????? ???? + ????? ??? ?𝑖?? = $5, 760 + $1, 200 = $6, 960 Scenario B ????? ???? = 3 ??????? * 28 ℎ???? * $60 ??????*ℎ??? = $5, 040 # ???? ?? ???? = 28 ℎ???? * ( ??? 8 ℎ???? ) = 3. 5 ???? (????? ?? ?? 4 ????) $1,200 ????? ??? ?𝑖?? = 3 ??????? * $100 ???*?????? * 4 ???? = ????? ???? = ????? ???? + ????? ??? ?𝑖?? = $5, 040 + $1, 200 = $6, 240 It is cheaper to hire 3 workers for this task. The correct answer is A. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 29
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21. A pipe installation adjacent to a busy roadway is planned. The pipe will be installed 6 feet below the ground surface, and the excavation will be unshored. The excavated soil will be a clay with an unconfined compressive strength of 2 tsf. What is the maximum slope angle of the excavation that meets OSHA excavation requirements? A. 2H:1V B. 1.5H:1V C. 1H:1V D. 0.75H:1V NCEES Reference Handbook: Pg 157-158, Section 3.10.1 and 3.10.2 (keyword: “OSHA”, “excavation”) Solution: Read the descriptions of the OSHA soil types. A clay soil with unconfined compressive strength of 2 tsf would typically be classified as a Type A soil. However, since the excavation will be next to a busy roadway, it will be subject to vibrations. The appropriate OSHA soil type for this scenario is Type B. Per Section 3.10.2, the maximum unshored sideslope angle for Type B soils is 1H:1V. The correct answer is C. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 30
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22. A concrete gravity retaining wall is being designed. The wall will support granular (cohesionless) backfill with an internal friction angle of 34 degrees and a unit weight of 120 pcf. What is most nearly the thickness of the wall [ft] required to maintain a factor of safety against overturning (measured about Point O) of 3.0? Assume Rankine earth pressures and the geometry shown below. Neglect wall friction. A. 1 B. 2 C. 3 D. 5 NCEES Reference Handbook: pg 79-80, Section 3.1.2 (keyword: “Rankine”) Solution: Step 1: Draw in the forces acting on the wall and their moment arms. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 31
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Step 2: The minimum factor of safety for overturning about Point O is 3. Use the factor of safety equation to solve for t. Know that FSo = resisting moments / overturning moments. 𝐹? ? = ?'?'+?? 𝑃?*? ?𝑎 3 = 1500 ???*1.33??+1500?*(2??+0.5?) 1696.3 ???*3.33?? 3 = 2000+3000?+750? 2 5654.3 = 0. 354 + 0. 531? + 0. 133? 2 0 = − 2. 646 + 0. 531? + 0. 133? 2 0 = 0. 133? 2 + 0. 531? − 2. 646 Solve for t using the quadratic equation (search “quadratic equation” to find it in the NCEES reference handbook on pg 4) ? = −(0.531)± 0.531 2 −4(0.133)(−2.646) 2*0.133 ? = 2. 9, − 6. 9 Use the positive root, 2.9 ft ~ 3ft. The correct answer is C. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 32
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23. What is most nearly the gripping force, R B, [lbf] on the pipe shown below when a 90 lbf force is applied to the handles of the slip-joint pliers? The joint at Point A is a pin. A. 450 B. 232 C. 200 D. 300 NCEES Reference Handbook: no specific reference, except knowing that the moment at a pin connection is zero Solution: Step 1: Draw the free body diagram for one of the plier arms. Step 2: This problem can be solved simply by realizing that the summation of moments about A should be zero. ? ? = 0 = 90??? * 5𝑖? − ?? * 1. 5𝑖? (???𝑖?? ??????𝑖?? ??????? ?? ???𝑖?𝑖??) ?? = 90???*5?? 1.5?? = 300??? The correct answer is D. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 33
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24. Which of the following is not a typical temporary erosion and sedimentation control method? A. Stabilized construction access B. Silt fence C. Storm drain inlet protection D. Mechanically stabilized earth walls NCEES Reference Handbook: n/a Solution: By inspection, a mechanically stabilized earth wall is a structural wall type and not a typical temporary erosion and sedimentation control method. The correct answer is D. 25. Given a horizontal curve with the following parameters, determine the station at the start of curve (PVC). PT station = 26+34.72 Radius = 1200 feet Intersection Angle = 15°15’ A. 23+17.42 B. 23+15.33 C. 26+29.15 D. 26+29.18 NCEES Reference Handbook: pgs 271-273, Section 5.2.1 (keyword: “horizontal”) Solution: We are given STA PT = 26+34.72 (use 2634.72 in calculations), radius R = 1200ft, and intersection angle = 15°15’, which we must convert to decimal degrees (15° + 15’/60 = ∆ 15.25°). To find STA PC, we need to subtract the length of curve from the STA PT: ??𝐴 ?? = ??𝐴 ?? − ? From pg 273, we know ? = ?∆π 180 = 1200*15.25*π 180 = 319. 395?? ??𝐴 ?? = ??𝐴 ?? − ? = 2634. 72 − 319. 395 = 2315. 33 = 23 + 15. 33 The correct answer is B. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 34
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26. An aging storm sewer made of vitrified clay will be replaced by slip-lining a new plastic pipe into it. The new plastic pipe must be able to support an equivalent flow rate as the old pipe. The following pipe sizes are commercially available. The minimum diameter [inch] of the new plastic pipe is most nearly: A. 17 B. 14 C. 16 D. 15 NCEES Reference Handbook: pg 238, Section 6.3.1.1 (keyword: “circular pipe flow”) Solution: Know that we can apply the equation for circular pipe flow using the given parameters, and know that the new pipe must be able to handle the same (or greater) amount of flow as the old pipe (Q new >= Q old ). First make sure the parameters we’re given are in the correct units as meant to be used in the circular pipe flow equation, D old =18”=1.5ft C = Hazen-Williams Coefficient [unitless] S old = S new = S = 2% = 0.02 Solve for D new. ? ??? ≥ ? ??? 0. 432 * ? ??? * ? ??? 2.63 * ? ≥ 0. 432 * ? ??? * ? ??? 2.63 * ? Eliminate like terms. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 35
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? ??? ≥( ? ??? *? ??? 2.63 ? ??? ) 1 2.63 ≥ 1. 333?? ≥ 16𝑖? The correct answer is C. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 36
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27. Two steel lightposts are subjected to equal lateral loads as shown. Each lightpost is made of solid rectangular steel and they are the same length. Which lightpost will experience greater max deflection from the applied load? Assume both lightposts will deflect along their x axis. A. Lightpost 1 will experience greater max deflection. B. Lightpost 2 will experience greater max deflection. C. Each lightpost will experience the same max deflection. D. There is not enough information to answer the question. NCEES Reference Handbook: pg 245, Shears Moments, and Deflections Table, Row 21 (keyword: “cantilevered beam”) Solution: From the table on pg 245, use the equation for max ∆ path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 37
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The maximum deflection of the posts are a function of the load (P), the distance of the load from the fixed end (b), the length of the poles (l), the modulus of elasticity of the poles (E), and the moment of inertia of the poles (I). Search “moment of inertia” to find the quick reference to moment of inertias for rectangular sections on pg 24. Since both lightposts will deflect along their x axis, we take the moment of inertia equation about the x axis. The simplest way to compare deflections for each beam is to simplify the deflection equations for each lightpost so they use the same terms. The point loads are the same: P1 = P2 = P The moduli of elasticity are the same: E1 = E2 = E The length of the lightposts are the same: L1 = L2 = L The point of application for each load can be put in terms of L: b1 = 0.5*L, b2 = L Since we know all the dimensions for the beams, we can simply calculate Ix for both beams: path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 38
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𝐼 ?,1 = 16*20 3 3 = 42667𝑖? 3 𝐼 ?,2 = 18*24 3 3 = 82944𝑖? 3 Plug the values in to the equation for maximum deflection and compare: ∆ ???,1 = 𝑃(0.5?) 2 6*?*42667?? 3 * (3? − 0. 5?) = 2. 4?10 −6 * 𝑃? 3 ? ∆ ???,2 = 𝑃(?) 2 6*?*82944?? 3 * (3? − ?) = 4. 0?10 −6 * 𝑃? 3 ? Lightpost 2 will experience more deflection. The correct answer is B. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 39
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28. What is most nearly the maximum line load, w o [kip/ft], that produces the same maximum moment in the beam as a uniform load of 25 kip/ft? A. 50 B. 25 C. 85 D. 30 NCEES Reference Handbook: pg 34 Simply Supported Beams Slopes and Deflections table (keyword: “maximum moment”) Solution: The table on pg 34 has quick reference equations for both a uniform load and a linearly increasing line load. ? ???,??????? = ? ???,???? = ? ? ??𝑖???? *? 2 8 = ? ? *? 2 9 3 Solve for w o . ? ? = 9 3 * ? ??𝑖???? 8 = 1. 949 * 25?𝑖?/?? = 48. 7 ?𝑖?/?? ~ 50?𝑖?/?? The correct answer is A. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 40
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29. The results of three compression tests on concrete cylinders are provided below. Each concrete sample has a unit weight of 155 pcf. What is most nearly the modulus of elasticity (ksi) of the concrete? Test # Compressive Strength of Concrete (psi) 1 3,550 2 3,150 3 3,200 A. 3,600 B. 3,800 C. 3,570 D. 3,650 NCEES Reference Handbook: pg 261, Section 4.3.2.1 (keyword: “modulus of elasticity”) Solution: The modulus of elasticity is defined as E c = 33w c 1.5 ?' ? f’ c is defined as the compressive strength of concrete (psi). w c is not defined, but know that it refers to the unit weight of concrete (pcf). The most accurate solution would likely be to calculate the modulus of elasticity for each test, then average the results. A quicker solution that gives a fairly close result is to average the compressive strengths of each test, then calculate the modulus of elasticity with the average compressive strength. ?' ?,??? = (3550 + 3150 + 3200)/3 = 3300 ??𝑖 E [ksi] = 33*155 1.5 = 3658 psi ~ most nearly 3650 psi. 3300 The correct answer is D. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 41
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30. A flexible spread footing is 4 feet wide, 6 feet long, and 2 feet thick. What allowable bearing load [psf] will result in a maximum soil settlement of 0.5 inches, given a modulus of subgrade reaction of 100 pci and a minimum FS of 2.0? A. 25 B. 50 C. 3,600 D. 7,200 NCEES Reference Handbook: pg 229 (keyword: “modulus of subgrade reaction”) Solution: k = P / where P may be taken as the ultimate bearing capacity, q u = q allowable *FS ∆ q allowable = = 25 psi ?∆ ?? = 100???*0.5 2 Convert q allowable to psf ? ????????? = 25 ??? ?? 2 * ( 12 2 ?? 2 ?? 2 ) = 3600 ??? The correct answer is C. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 42
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31. An outfall pipe will be supported by a concrete deadman as shown in the figure. What is the planned invert of the pipe at the center of the deadman [ft]? The slope of the pipe between the catch basin and crest of the bluff is constant. A. 145.93 B. 145.43 C. 145.87 D. 146.13 NCEES Reference Handbook: n/a Solution: Determine the slope of the pipe. ? = ???? ??? = 146.23−145.33 72−105 =− 0. 027 Determine the pipe invert at the center of the deadman (i.e. STA 0+82 + ½*2ft = STA 0+83) ? = ???? ??? = 146.23−? 72−83 Solve for y: ? = 146. 23 − (72 − 83) * ? = 145. 93 The correct answer is A. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 43
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32. A 600-foot-long embankment section is planned over an alignment with varying ground surface.The geotechnical engineer has stated that the on-site soils are not appropriate for re-use along this section. What is most nearly the volume of imported embankment [yd 3 ] material needed? A. 296,250 B. 14,970 C. 10,970 D. 404,250 NCEES Reference Handbook: pg 49, Section 2.1.2.1 (keyword: “earthwork volumes”) Solution: The average end area method can be used to solve this problem. Note that this equation can be applied as a summation across multiple cross sections. ? = Σ(? * ? 1 +? 2 2 ) The distance between cross sections: L1 = 300 ft L2 = 300 ft The area of the embankment for each cross section: A1 = 500 sq.ft A2 = 675 sq.ft A3 = 125 sq.ft path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 44
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Note that this problem gives you extraneous information in providing cut areas - since the problem states re-use of on-site soils are not appropriate, the area of imported material needed for each cross section is simply the area of the embankment. ? = ? 1 * ? 1 +? 2 2 + ? 2 * ? 2 +? 3 2 ? = 300?? * 500?? 2 +675?? 2 2 + 300?? * 675?? 2 +125?? 2 2 V = 296,250 ft3 = 10,972 yd3 (conversion factor: 1yd = 3ft, 1yd3 = 27ft3) The correct answer is C. 33. A steel water main is to be replaced with an HDPE pipe. The design requires a pipe that can withstand a minimum water pressure of 820 kPa. The available HDPE pipes are rated in psi. The minimum HDPE pipe rating required is: A. 160 psi B. 115 psi C. 130 psi D. 60 psi NCEES Reference Handbook: pg 3, Section 1.2 (keyword: n/a) Solution: This is simply a unit conversion problem. Refer to the Section 1.2 Conversion Factor table. Per the table, multiply kPa * 0.145 to convert to psi. 820 kPa * 0.145 = 118 psi Note that 118 psi is the minimum pressure rating. We need to use the next pressure rating, which is 130 psi. The correct answer is C. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 45
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34. A 250-acre watershed has the following breakdown of soil types: 60 percent soil A, 25 percent soil B, and 15 percent soil C. Curve numbers for each soil type are provided in the table below. Using the NRCS method, what is most nearly the runoff depth (in) for a 25-year frequency, 24-hour storm with a total rainfall of 6 inches? Soil Curve Number, CN A 74 B 47 C 61 A. 2.4 B. 2.0 C. 1.5 D. 2.8 NCEES Reference Handbook: pg 378, Section 6.5.2.2 (keyword: “NRCS”, “runoff depth”) Solution: Solve for runoff depth, Q. Q depends on total precipitation, P (given = 6 inches) and maximum basin retention, S (dependent on the curve number). Calculate the weighted average curve number. ?? ??? = Σ(?? * % ????) Σ% ???? = 6530 100 = 65. 3 Calculate ? = 1000 ?? 𝑎?? − 10 = 5. 3 Calculate ~ 2.4 in ? = (𝑃−0.2?) 2 𝑃+0.8? = (6−0.2*5.3) 2 6+0.8*5.3 = 2. 383 𝑖? The correct answer is A. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 46
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35. A rectangular reinforced concrete beam requires a minimum total steel reinforcement area of 1.6 in 2 . Choose the most appropriate cross section to meet this requirement. NCEES Reference Handbook: pg 258, Section 4.3.1 (keyword: “reinforcing bars”, “#3”) Solution: The minimum As = 1.6 in 2 Calculate the total cross sectional area of steel for each of the given options. Use the given area of steel bars given in the “ASTM Standard Reinforcing Bars” table. Options C and D meet the minimum requirement for total steel reinforcement. However, Option D has much greater steel reinforcement than needed. Therefore, Option C is the most appropriate option. The correct answer is C. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 47
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36. A concrete slab is poured and kept at a constant temperature of 60 ℉ . How long [hrs] will it take for the concrete to mature to 100 ℉ -hr? A. 3.6 B. 4.2 C. 1.7 D. 2.5 NCEES Reference Handbook: pg 74, Section 2.5.3.1 (keyword: “concrete maturity”) Solution: Simply solve the concrete maturity equation for t and plug and chug. ? = (? − ? ? ) * ? ? = ? ?−? ? = 100 60−32 = 3. 57ℎ???? ~ 3. 6 ℎ???? The correct answer is A. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 48
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37. Per the following CPM diagram, how long [days] after project start will shoring installation be complete? A. 13 B. 17 C. 6 D. 7 NCEES Reference Handbook: pg 70, Section 2.4.1.1 (keyword: “CPM”) Solution: Note the question is asking for number of days after project start that shoring installation will be complete. From Node A to Node D, the shortest timeframe will be 10 days (Node A - C) + 7 days (Node C - D). Therefore the amount of days before shoring will be done will be 17 days. The correct answer is B. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 49
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38. 10 years of measured peak streamflow are provided in the table below. What is most nearly the estimated future peak streamflow [cfs] over a 10-year return period? Assume normal distribution. A. 22,500 B. 19,600 C. 26,000 D. 20,000 NCEES Reference Handbook : pg 376, Section 6.5.1.6 (keyword: “return period”) Solution: Use the simplified flood frequency equation to get X T = estimated event magnitude. K can be read from the “Frequency Factor for Normal Distribution” table on pg 376, for a 10-year return period. K = 1.282 and s can both be determined using the data sample (reference pg 13 - 14, Section ? 1.4.1) path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 50
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=19625 cfs ? ?=1 ? ∑ (? ? − ? ) 2 = 220801250 ? = 1/9 * 220801250 = 4953 ??? ? ? = ? + ? ? ? = 19625 + 1. 282 * 4953 = 25975 ??? ~ 26000 ??? The correct answer is C. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 51
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39. A driver slams on the brakes on a level asphalt road, coming to a stop just before a stop sign. The vehicle travels 60 feet in 3 seconds during deceleration. What is most nearly the vehicle’s initial traveling speed [mph]? Assume a friction factor of 0.9 between the car’s tires and the road surface. A. 40 B. 60 C. 20 D. 120 NCEES Reference Handbook : pg 270, Section 5.1.4.3 (keyword: “deceleration”) Solution: Where db is given (60 feet), f is given (0.9), G is zero since the roadway is level, and v2 is zero since the vehicle has come to a full stop. 40.2 mph ? 1 = (60)(30)(0. 9) = The correct answer is A. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 52
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40. A construction project involves building a new bridge over a river. The bridge will be 1000 feet long and 40 feet wide, and will have a total of 10 spans. The project will involve the following activities: ● Excavation: 135000 cubic feet ● Pile driving: 1000 linear feet ● Concrete pour: 4000 cubic yards ● Structural steel: 9420 linear feet ● Asphalt paving: 45000 square feet What is most nearly the estimated cost of the construction project, given the following unit prices? ● Excavation: $20 per cubic yard ● Pile driving: $100 per linear foot ● Concrete pour: $85 per cubic yard ● Structural steel: $5 per linear foot ● Asphalt paving: $10 per square yard A. $3,637,000 B. $637,000 C. $1,037,100 D. $947,000 NCEES Reference Handbook: pg 3, Section 1.2 (keyword: n/a) Solution: Convert the excavation and asphalt paving quantities into the correct units and multiply quantity * unit cost to get the cost of each activity. Then sum the cost of each activity to get the cost of the project. The correct answer is B. path-to-pe.com Civil PE Breadth Practice Exam #1 - Solutions Page 53
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