civil exam set 3

8/10/2019 Civil Exam Set 3

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Copyright 2008-2012 Pecivilexam.com all rights reserved-Breadth Exam Set #3 1

PE Civil Exam 40-Mix Questions & Answers (pdf Format)

For Breath Exam (Morning Session) Set #-3


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Breadth Exam (morning session): This practice exam contains 40 mixed

questions and answers, each set being from all five areas of civil

engineering:

Table Contents: Page

1. Construction-8 Q & A 3

2. Geotechnical-8 Q & A 11

3. Structural-8 Q & A 21

4. Transportation-8 Q & A 25

5. Water Resources and Environmental-8 Q & A 39


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I. Construction

1. PROBLEM (Earth Work)

A borrow pit contour elevation has been shown in the figure; it has to becut. What is the average volume (yds3) cut from the borrow pit?

.

a. V=11333 yds3b. V=5666 yds3c. V=7536 yds3d. V=2833 yds3

1. Solution:

Area of each grid, A=60x60=3600 ft2V=(1x5+2x7+1x9+2x6+1x5+1x7+3x8+1x9) x [(3600/(4x27)]=2833 yds3

Total Volume of Borrow pit, V=2833 yds3

Correct Answer is (d)


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2 PROBLEM (Earth Work)

AS shown in the Figure, the embankment has to be constructed, the soilsdry unit weight is 106 lb/ft3, moister content is 12.5%. The average areais A1= 625 ft2at station 5+00 and A2=560 ft2at station 7+00. What is

the volume (yd3) of embankment?

a. V=2194 yd3b. V=4389 yd3c. V=3567 yd3d. V=7895 yd3

2

Solution:

L=700-500=200 ftEmbankment volume (yd3), V = {(A1+A2)/2} x L/27V = {(625+560)/2} x 200/27=4389 yd3

Correct Answer is (b)


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3. PROBLEM (Quantity Estimate)

Find the weight for a 26 gage 14" galvanized spiral duct work which is 150'long with a 15% waste for bracing, hangers, waste, and seams.

a. 574 lbsb. 551 lbsc. 633 lbsd. 474 lbs

3. Solution:

Galvanized Sheet, Weight per UnitArea=0.9062 lbs/ft

Duct Length,

L=150 ftDuct diameter, D=14" /12 = 1.17 ft.Duct perimeter, P=D= 3.14 x 1.17=3.68 ft.Total duct area, A=PxL = 3.68 x150=551 sq. ft.Add 15% waste for bracing, hangers,waste, and seamsArea =551 x (1 + 15%)= 633.73 sq. ft.Weight = 633.73x 0.906 = 574 lbs.

Correct Answer is (a)

GaugeNumber

SteelWeight

in

poundsper

squarefoot

US

StandardGauge:

thicknessin inches

Manufactu

rers'Standard:

thicknessin inches

GalvanizedSheet:

weight inlbs/sq ft

StainlessSteel:

weight inlbs/sq ft

10 5.62 0.1406 0.1345 5.7812 5.7937

11 5 0.125 0.1196 5.1562 5.15

12 4.37 0.1094 0.1046 4.5312 4.5063

13 3.75 0.0937 0.0897 3.9062 3.8625

14 3.12 0.0781 0.0747 3.2812 3.2187

15 2.81 0.0703 0.0673 2.9687 2.8968

16 2.5 0.0625 0.0598 2.6562 2.575

17 2.25 0.0562 0.0538 2.4062 2.3175

18 2 0.05 0.0478 2.1562 2.06

19 1.75 0.0437 0.0418 1.9062 1.8025

20 1.5 0.0375 0.0359 1.6562 1.545

21 1.37 0.0344 0.0329 1.5312 1.416

22 1.25 0.0312 0.0299 1.4062 1.2875

23 1.12 0.0281 0.0269 1.2812 1.1587

24 1 0.025 0.0239 1.1562 1.03

25 0.875 0.0219 0.0209 1.0312 0.9013

26 0.75 0.0187 0.0179 0.9062 0.7725

27 0.687 0.0172 0.0164 0.8437 0.7081

28 0.625 0.0156 0.0149 0.7812 0.6438

29 0.562 0.0141 0.0135 0.7187 0.5794

30 0.5 0.0125 0.012 0.6562 0.515


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4. PROBLEM (Material Testing)

Which of the following statements is not true for measuring asphalt bitumentest?

a. The penetration test has performed to measures the consistency(hardness) of asphalt at a specified test condition.

b. The flash point test determines the temperature to which an asphalt canbe safely heated in the presence of an open flame

c. The bitumen content of a bituminous material is measured by means ofits solubility in Carbon Dioxide.

d. The ductility test can measures the distance a standard asphalt samplewill stretch without breaking under a standard testing condition (5cm/min at 25 C).

4. Solution:

C is not true.The bitumen content of a bituminous material is measuredby means of its solubility in Carbon Disulfide.

Correct Answer is (c)


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5. PROBLEM (Quantity Take-Off)

A 5- storied apartment building has to be built, building footings layout with cross-section is shown in the Figure. How much volume of concrete is required forbuilding the foundation up to Ground Level (G.L.) with a 10% wastage?

a. 351 ft3b. 324 ft3c. 342 ft3d. 376 ft3

5. Solution:

Total number of footing =9 Nos.Volume of each footing with extended column= 6x6x1+1x1(3-1) =38ft3Total volume= 38x9x (1+10% wastage) =376.2ft3



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6. PROBLEM (Time-Cost Trade-Off)

A subcontractor has the task of erecting 8000 square meter of metals scaffolds. Thecontractor can use several crews with various costs. It is expected that theproduction will vary with the crew size as given below:

Estimated dailyproduction

(square meter)

Crew size(men)

Crew formation

160 5 1 scaffold set, 2 labors, 2 carpenter, 1foreman

200 6 2 scaffold set, 3 labors, 2 carpenter, 1foreman

240 7 2 scaffold set, 3 labors, 3 carpenter, 1

foreman

Consider the following rates:

Scaffolding $70/day;Labor $90/day;

Carpenter $130/day andForeman $140/day.Determine the least direct cost of this activity considering the different crewsformation.

a. $32400.00b. $32500.00c. $32900.00d. $32300.00

6. Solution:

The calculations are shown in the following table.

Crew size Duration (days) Cost ($)

5 8000/160=50 days 50 x (1x70 + 2x90 + 2x130 + 1x140) =

32500

6 8000/200=40 days 40 x (2x70 + 3x90 + 2x130 + 1x140) =32400

7 8000/230=35 days 35 x (2x70 + 3x90 + 3x130 + 1x140) =32900



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7. PROBLEM(Site Layout And Control)

All residential developments areas to provide two Deep Soil Zones (D.S.Z); one tothe rear and one to the front of the property. Which of the following statements isnot true for the Deep Soil Zones (D.S.Z)?

a. Rear Deep Soil Zones are to have minimum width of 8m or 30% of the averagewidth of the site which ever is the greater and a minimum depth of 18% of thelength of the site up to 8m but not less than 5.5m. Greater than 8m may beprovided if desirable.

b. Deep Soil Zones must be provided for all new developments only, except onlarge lot rural or agriculturally zoned land.

c. Front Deep Soil Zones are to be the width of the site boundary minus thedriveway width and the pathway width by the front setback depth.

d. Deep Soil Zones cannot be covered by impervious surfaces such as concrete,terraces, outbuildings or other structures.

7. Solution:

Deep Soil Zones must be provided for all new developments and existingdevelopment, except on large lot rural or agriculturally zoned land.



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8. PROBLEM (Site Layout And Control)

Which of the following statements is not applicable for Temporary Access RoadsConstruction?

a. Temporary roads shall follow the contour of the natural terrain to the extentpossible. Slopes should not exceed 10 percent.

b. A 6-inch course of Coarse Aggregate shall be applied immediately aftergrading or the completion of utility installation within the right-of-way. Filterfabric may be applied to the roadbed for additional stability.

c. Roadbeds shall be at least 12 feet wide for one-way traffic and 18 feet widefor two-way traffic.

d. All cuts and fills shall be 2:1 or flatter to the extent possible.

8. Solution:

Roadbeds shall be at least 14 feet wide for one-way traffic and 20 feet wide fortwo-way traffic.



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II. Geotechnical

9. PROBLEM (Soil Classification)

A soil sample has the following properties by the Unified Soil Classification System.

LL = 42, PL = 31. What is the soil classification?

a. CHb. MHc. MLd. CL

9. Solution: From plasticity Chart, LL=42, PL=31; PI=42-31=11,ML.



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10. PROBLEM(Boring Log Interpretation)

Which of the following statements is not satisfactory for boring loginterpretation?


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a. The blows/foot is number of times the 140 lbs drive weight dropped 30inches needs to penetrate one foot. The 37 blows per foot confirm thedescription of dense or well-compacted soil.

b. The moisture content range of 1.2 to 5.7% means the soil has enoughmoisture and will not require a lot of water to achieve compaction.

c. The dry density in the top 4 feet is 127.7 pound per cubic foot (PCF). Thisindicates a well-graded soil with a low void ratio. It can be expected to havesignificant cohesion and friction angle

d. The soil between 5 and 8 feet of depth has a dry density of only 96.8 pcf.This indicates the soil is poorly graded and a low percentage of fines. Thissoil will probably have little or no cohesion. That means it may not stand on a

construction slope as steep as 1H to 1V.

10. Solution:

Statement b is not satisfactory

The moisture content range of 1.2 to 5.7% means the soil is very dry and willrequire a lot of water to achieve compaction.



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12. PROBLEM (Lateral Earth Pressure)

Using the Rankine analysis, determine the lateral earth pressures due toearthquakes on a 8 ft rigid concrete retaining wall. The free draining gravel backfillhas a soil unit weight, , of 132 lb/ft3, and an angle of internal friction, , of 33

degrees. The retaining wall will be constructed for passive conditions.

a. 3056 lb/ftb. 5080 lb/ft

c. 7020 lb/ftd. 8078 lb/ft

12. Solution:

Unit weight of soil backfill, = 132 lbs/ft3

Angle of Internal Friction, = 33 degreesWall height, H = 8 ft

Passive case (wall moves toward retained soil)

Coefficient for passive conditions, K

K = KP= (1 + sin ) = (1 + sin 33) = 3.40(1 - sin ) (1 - sin 33)

Lateral earth pressure due to earthquakes,

Pe= 3 KhH2

8

Earthquake coefficient, Kh= 3 K = 3 (3.40) = 2.554 4

Pe= 3 KhH2

8= 3 (2.55) x (132 lb/ft3x (8 ft)2= 8078 lb/ft

8



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13. PROBLEM (Vertical Stress, Pore Pressure & Effective Stress)

Calculate the change in vertical stress at 3 ft below the middle of a 6 ft x 9 ftrectangular foundation. Using the Boussinesq theory and chart. The applied building

load on this foundation is 3200 lb/ft2.

a. 2768 lb/ft2b. 2134 lb/ft2c. 2585 lb/ft2d. 1790 lb/ft2

13. Solution:z = 3 ft

q = 3200 lb/ft2Rectangular footing size, 6 ft x 9 ft

= Pv

Pv= summation of all stress components (i.e. Pv1+ Pv2+ .... + Pvn). In this case,

we analyze the foundation in 4 equal but separate quadrants. Instead of a single 6ft x 9 ft foundation, we have 4 separate 3 ft x 4.5 ft quadrants. This is done so thatone corner of each quadrant is located in the center of the footing.

4Pv= 4q Since the quadrants have equal dimensions with the same applied load,

we simply multiply the equation by 4 (4 quadrants).

= Pv= 4Pv= 4q

m = x = 4.5 ft = 1.5z 3.0 ft

n = y = 3.0 ft = 1.0

z 3.0 ft


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= 0.202, Influence value from Boussinesq chart, where m = 1.5 and n= 1.0.

= 4q= 4(3200 lb/ft2)(0.202) = 2585 lb/ft2



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14. PROBLEM (Settlement)

Determine the consolidation settlement at the center of the clay layer of the mat(30m x 40m) foundation shown in the Figure. Given, the total effective stress,=220 KN/m2and the average effective stress increased due to the foundationload, =90 KN/m2at the center of the clay layer.

a. 345.00 mmb. 275.00 mmc. 187.00 mm

d. 141.00 mm

14. Solution:=220 KN/m2, =90 KN/m2, Cc=0.29 and eo=0.85Settlement, Sc={CcHc/(1+eo)}log {(+ )/ }Sc={0.29 x 8/(1+.85)}log {(220+ 90)/ 220}*1000=186.77 mm

Correct Solution is (c)


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15. PROBLEM (Gravity Wall)

Determine the active earth pressure for the following retaining wall, H=18 feet.Where, the unit weight of the soil is = 120 lb/ft3, = 9oand =32o.

a. 11 Kip/ftb. 4 Kip/ftc. 8 Kip/ftd. 14 Kip/ft

15. Solution:

KA=[cos32o/(cos9o+{sin(32o+9o) sin9o}]2

KA=[0.848 /(0.987 +0.320)]2=0.42

PA= 1/2 KAH2=.5 x 120 x .42 x 182= 8183.45 lb/ft =8.183 Kip/ft

Correct Solution is (c)


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16. PROBLEM (Braced And Anchored Excavations)

Determine the tension of the tie rods T if they are spaced at 2 meters from thecenter of the anchored sheet pile wall shown in Figure. The unit weight of soil is =15.0kN /m

3.

a. 123.00 kNb. 345.00 kNc. 176.00 kNd. 288.00 kN

16. Solution:

Tie rods are spaced at 2 meter center to center= 15.0kN /m

3

Pa= 1/2 Ka H

2

a=1/2 x 15.00 x 0.33 x 9

2

= 200.48 kN/m (horizontal)Taking moment at tie rod for mobilized passive resistancePp x (9-1-1.5)= Pa x (9-3-1.5)Pp x 6.5= 200.48 x 4.5Pp= 138.79 kN/mTension of the Rods, T= (200.48-138.79) x 2=123.38 kN

Correct Solution is (a)


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III. STRUCTURAL

17. PROBLEM (Loadings)

In a building column the axial forces are determined as 180 kips of dead load, 100kips of floor live load, 50 kips from the roof snow load. Determine the requiredstrength of the column without wind & earthquake load. Using the combination loadspecified by AISCs Manual of Steel Construction.

a. 241.00 kipsb. 458.00 kipsc. 346.00 kipsd. 401.00kips

17. Solution:

Given, D=180 kips, L=100 kips, S=50 kips, W=0.0 kips, & E=0.0 Kips

The following load combinations are provided by AISCs Manual of SteelConstruction.

Lr = Roof live load, S = Snow load, R = Rainwater nominal load

Combination of load:

1.2 D + 1.6 L + 0.5 (Lr or S or R)= 1.2 x180 + 1.6 x 100 +0.5 (50)= 401 kips

1.2 D + 1.6 (Lr or S or R) + (0.5 L or 0.8 W)= 1.2 x180 + 1.6 x (50) +0.5 x 100=346.00 kips

1.2 D + 1.6 W + 0.5 L + 0.5 (Lr or S or R)= 1.2 x180 + 0.0 + 0.5 x (50)=241 kips

The required strength for the column is 401 kips.



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18. PROBLEM (Loading)

Determine the dead load acting on the shown in the Figure steel beam withconcrete slab & floor tiles as shown in the Figure. The tiles flooring is 3 thick withcement mortar.

W18x97

6 inch concrete slab

Flooring

6 f t

a. 772 lb/ftb. 630 lb/ftc. 727 lb/ftd. 547 lb/ft ft

18. Solution:

Considering, cement mortar & tiles unit weight is 120 lb/ft3

and

Concrete slab unit weight is 150 lb/ft3

Steel Beam weight = 97 lb/ft

Weight of concrete slab = 150 x 6/12 x 6 = 450 lb/ft

Weight of tiles flooring = 120 x 3/12 x 6 = 180 lb/ftTotal dead weight = 727 lb/ft



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19. PROBLEM (Analysis)

Determine the reaction of YAin the frame as shown in Figure.

a. -6.25 Kb. 6.25 Kc. 31.25 K

d. -12.5 K

19. Solution:

Positive moment is in the clockwise direction

MA=0,

5 x 15 x (15/2) + 25 x15-YDx30=0

YD= 31.25 K

H=0, YD+ YA-25=0,

YA=-31.25+25= -6.25 K



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20. PROBLEM (Member Design)

Determine the ultimate load, Qultof a rectangular footing that is 6x 4 and eccentricshown in the Figure where, soil unit weight is = 120 lb/ft3, and the ultimatebearing capacity is qu=3200 lb/ft

2, eB=1.5 and eL=1.75.

a. 76.8.0 Kipsb. 48.0 Kips

c. 22.5 Kips

d. 25.4 Kips

20. Solution:

Where, eL/L=1.75/6= 0.292> 1/6, and eB/B=1.5/4= 0.375>1/6;Therefore,B1=B(1.5-3eB/B)= 4(1.5-3 x 1.5/4)= 3.750 ftL1=L(1.5-3eL/L)= 6(1.5-3 x 1.75/6)= 3.750 ftEffective Area, A=1/2(L1B1)=1/2 (3.750 x 3.750)= 7.03 ft2q'u=3200 lb/ft

2Therefore, Qult= Ax q'u= 7.03 x 3200= 22496=22.5 Kips



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21. PROBLEM (Mechanics Of Materials)

Which moment is not correct as shown in the following diagrams?

a. -450 K-ft

b. -650 K-ftc. -800 K-ftd. -700 K-ft

21.

Solution:

c is not correct

M at support=0,Moment, M= -(10 x10 x 10/2) + -(20 x 20)=900 K-ft



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Determine the maximum shear for the simply supported reinforced concrete beamshown with a dead load of 1.5 k/ft and a live load of 2.0 k/ft. Assume that the pointof reaction is at the end of the beam.

a. 48.00 k

b. 78.00 kc. 56.00 kd. 72.00 k

22. Solution:

Self weight = (12/12 ft) x (27/12 ft) x 150 lb/ft3= 338 lb/ft = 0.34 k/ftWu = 1.4 (1.5 k/ft + 0.34 k/ft) + 1.7 (2 k/ft) = 5.98 k/ftVu (max) is at the ends = WuL/2 = 5.98 k/ft x (24 ft)/2 = 71.71 k



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Determine the capacity of an 18 x 18 column with 10-#8 bars, tied. Grade 40steel and 3,000 psi concrete.

a 731 kips

b 450 kipsc 340 kipsd 825 kips

23. Solution:

Given,Grade 40 reinforcement has fy = 40,000 psi and fc = 3,000psi

Find Pn, with =0.65 and Pn = 0.80Po for tied columns and

P0= 0.85 fc( Ag Ast) + fsAst

Vertical steel area for #10 bar, A= 3.14/4 x {(8/8)2}=0 .78 in2

Ast = 10 bars (0.78 in2) = 7.8 in2

Concrete area (gross): Ag = 18 in 18 in = 324 in2

Pn = (0.65)(0.80)[0.85x (3000) x (324 7.8) + 40,000 x 7.8]

= 731,281 lb = 731 kips



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24. PROBLEM (Mechanics Of Materials)

Check the adequacy of the shown tension element where, fy=60 ksi. Considering,the Load and Resistance Factor Design (LRFD) method.

Where, Tn, is the nominal strength of the member

Tu, is the Ultimate strength of the member.

tis the resistance factor, t= 0.9 for cross-section yielding

L=100 kips

D=60 kips

Bar 6 in x 0.75 in

a. tTn= 210 kips < Tu= 232 kips

b. tTn= 243 kips > Tu= 232 kips

c. tTn= 243 kips > Tu= 210 kips

d. tTn= 310 kips > Tu= 280 kips

24. Solution:

Given, DL= 60 kips, LL= 100 kipsTn, is the nominal strength of the member

t n uT T

tis the resistance factor taken as: t= 0.9

Fy= Steel yield strength= 60 ksi

Ag= Gross area of section= (6 x 0.75) in2

Tu= 1.2D + 1.6L = 1.2*60 + 1.6*100 = 232 kipsTn= Fyx Ag=60 x 6 x 0.75 = 270 kips

tTn= 0.9*270 = 243 kips > Tu= 232 kips OK



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IV. TRANSPORTATION

25. PROBLEM (Horizontal Curve)

A curving highway has a design speed of 120 km/hr. At one horizontal curve, thesuper-elevation has been set at 8.0% and the coefficient of the side friction is found

to be 0.12. Determine the minimum radius of the curve that will provide a safevehicle.

a. 530 meters

b. 495 meters

c. 567 metersd. 642 meters

25. Solution:

Design speed, V = 120 km/hrSuper-elevation, e = 8%

Coefficient of side friction, f = 0.12

Minimum radius, R

R = V2/(127(e/100+f))=(120)2/(127x(.08+0.12))=566.92 meters



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26. PROBLEM (Horizontal Curve Sight Distance)

A large elevated object is located 46 feet from the centerline of a two-lane highway,

which has 12-foot wide lanes. The elevator is situated on the inside of a horizontalcurve with a radius of 600 feet. Assuming that the elevated object is the only sightrestriction on the curve. What is the minimum sight distance along the curve,

where degree of the curve 12 turns out?

a 351 ft

b 257 ftc 405 ftd 461 ft

26.

Solution

Distance from the center of the inside lane to the object, M = 46-12/2=40 ft.Degree of the curve, D = 12Radius of the curve, R = 600 ftSight distance (ft), S

M = R - Rcos(SD/200)

(SD/200)=cos-1((R-M)/R)= cos-1((600-40)/600)=21.03

S=(21.03x200)/12=350.65 ft.



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27. PROBLEM (Passing Sight Distance)

A vehicle moving at a speed of 50 mph is slowing traffic on a two-lane

highway. What passing sight distance is necessary, in order for a passing maneuverto be carried out safely? Please assume that the passing vehicle accelerates to

passing speed before moving into the left lane.

The following variables have the values given:

Passing vehicle driver's perception/reaction time = 2.5 secPassing vehicle's acceleration rate = 1.47 mph/secInitial speed of passing vehicle = 50 mphPassing speed of passing vehicle = 60 mphSpeed of slow vehicle = 50 mphSpeed of opposing vehicle = 60 mph

Length of passing vehicle = 22 ftLength of slow vehicle = 22 ft

Clearance distance between passing and slow vehicles at lane change = 20 ftClearance distance between passing and slow vehicles at lane re-entry = 20 ftClearance distance between passing and opposing vehicles at lane re-entry = 250 ft

a. 1500 ft

b. 1900 ftc. 1600 ftd. 1800 ft

27. Solution:

Calculate the passing sight distance, D1

V = 73.3 ft/sec (50 mph)T=2.5 secVf = 88 ft/sec (60 mph)

Ui = 73.3 ft/sec (50 mph),A = 2.16 ft/sec/sec (1.47 mph/sec).

S1=VT= 183.3 feet.

Distance D is computed using the equation, Vf2=Ui2+ 2AD

D= (Vf2-Ui2)/2A=(882-73.32)/2x2.16=548.86 ft

D1=S+D=183.3+548.86=732.16 ft


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The passing sight distance is the distance D2, which is defined as the distance that

the passing vehicle travels while in the left lane.

Where, V2 = 14.67 ft/sec (60mph-50mph=10 mph = relative speed of passingvehicle with reference point on the slow vehicle).

S2 = 20 ft + 22 ft + 22 ft. + 20 ft =84 ftS2=V2T,T2=S2/V2=84/14.67=5.73 sec.V = 88 ft./sec. (60 mph)D2= VT2=88x5.73=504.24 ft

The distance, D3=250 ft. is the clearance distance between the passing vehicle andthe opposing vehicle at the moment the passing vehicle returns to the right lane.

The passing sight distance D4 is defined as the distance the opposing vehicletravels during 66% of the time that the passing vehicle is in the left lane.

V = 88 ft./sec. (60 mph) and T4 = 3.7 seconds (5.7*66%).

D4=VT4=88*3.7=325.6 ft

The total passing sight distance, D=D1 + D2 + D3 + D4=1812 ft.



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28. PROBLEM (Ascending Grades)

A highway, with traffic normally runs at 60 mph has an inclined section with a 4%

grade, how much can the elevation of the roadway increase before the speed of thelarger vehicles is reduced to 50 mph?

a. 32 ft

b. 50 ftc. 40 ft

d. 26 ft

28. Solution

From Graph in the Ascending Grades module, we can see that a 4% grade causes areduction in speed of (60 mph-50 mph)=10 mph after 1250 feet.

We can just estimate the elevation increase by multiplying the length of the gradeby the grade.

H= 1250x0.04 = 50 ft.

The elevation of the roadway can only be increased by about 40 feet before heavy

vehicles are reduced to a speed of 50 mph.



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29. PROBLEM (Sag Vertical Curves)

A stopping sight distance of 450 ft. is to be maintained on a sag vertical curve with

tangent grades of -3% and 0%, what should the length of the curve be? Assume aheadlight beam has an upward divergence angle of 1.

a. 465 ft

b. 243ftc. 287 ft

d. 356 ft

29. Solution

Sight distance, S = 450 ft.Beam upward divergence, B = 1Assumed Height of the headlights, H = 2 ft

Change in grade, A = 3% (|G2-G1|

If S > L then

L=2x450-(200 x (2+450 x tan1))/3%=900-657=243 ft

If S < L then (invalid because L < S)

Curve length, L=243 ft



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30. PROBLEM (Vertical Clearances)

A bridge is being designed to pass over a rural two-lane highway with a design

speed of 60 mph. The section of the two-lane highway where the bridge crossesover is an 1800 foot vertical sag curve withA = 3.5. What is the bridge clearance height?

a. 10.50 ft

b. 12.4 ftc. 16.5 ftd. 18.6 ft

30. Solution

DesignSpeed Km/h

PassingDistance (m)

DesignSpeed mph

PassingDistance (ft)

30 200 20 710

40 270 25 90050 345 30 1090

60 410 35 1280

70 485 40 1470

80 540 45 1625

90 615 50 1835

100 670 55 1985

110 730 60 2135

120 775 65 2285

130 815 70 2480

140 75 2580

From Green book table, passing sight distance for a design speed of 60 mph is2,135 feet

Therefore, S=2135 ftL=1800 ftA=3.5Bridge clearance height, H

Here, S> L,S=L/2+400(H-5.75)/A=2273.17 ft

2135=1800/2+400(H-5.75)/3.5Or, 400(M-5.75)=(2135-900)x 3.5H=16.55 ft.

Bridge clearance height is 16.55 feet



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31. PROBLEM (Horizontal Curve)

A horizontal curve is designed with a 600 m radius and is known to have a tangent

length of 52 m. The PI is at station 200+00. Determine the stationing of the PT.

a. PT=200+52

b. PT=200+80c. PT=200+34d. PT=199+48

31. Solution:

PC=PI-T=(200+00)-(0+52)=199+48

PT=PC+L=(199+48)+(1+04)=200+52



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32. PROBLEM (Acceleration And Deceleration)

A racecar is speeding down a level straightaway at 100 km/hr. The car has a

coefficient of drag of 0.3, a frontal area of 1.5 m2, a weight of 10 kN, a wheelbaseof 3 meters, and a center of gravity 0.5 meters above the roadway surface, whichis 1 meter behind the front axle. The air density is 1.054 kg/m3and the coefficient

of road adhesion is 0.6. What is the rate of acceleration for the vehicle?

a. 2.35 m/sce2

b. 2.25 m/sce2c. 1.45 m/sce2d. 1.15 m/sce2

32. Solution:

Use the force balancing equation to solve for a.

Since the straightaway is a level one, the grade is zero,

Aerodynamic resistance is computed:

Rolling resistance is computed:

Tractive Effort is computed:

Looking back to the force balancing equation:

Divide out mass, which can be computed from weight by dividing out gravity.


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Thus, divide mass from the force and acceleration can be found.

Thus, the vehicle is accelerating at a rate of 1.43 meters per second squared.



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V. WATER RESOURCES AND ENVIRONMENT

33. PROBLEM (Energy And/or Continuity Equation)

Determine the height of water in a column that produces a gauge pressure of 16psi.

a. 235 ftb. 32 ftc. 37 ftd. 998 ft

33. Solution:

The relationship between the height of a column of water and the resulting pressureis 2.31 ft of water produces 1 psi.

Ht = P x 2.31 = 16 (psi) x 2.31 (ft/psi) = 36.96 ft.



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35. PROBLEM (Pressure Conduit)

For an 10-inch Class 160 PVC pipeline that is 4000 feet long and has a flow rate of1200 gpm, compare the potential surge pressure caused when a butterfly valve isclosed (in 10 seconds) to a gate valve that requires 30 seconds to close.

Butterfly Valve Gate Valve

a. 135 psi 45 psib. 198 psi 68 psi

c. 375 psi 125 psid. 415 psi 135 psi

35. Solution:

P = 0.028 (Q x L)/(D2x T)

Where,

Q = Flow rate (1200 gpm)D = Pipe Diameter. (10 inches)L = Length of pipeline (4000 feet)Tb = Time to close Butterfly valve (10 seconds)Tg = Time to close Gate valve (30 seconds)P = Surge pressure (psi)?

Surge pressure for Butterfly Valve,

Pbv = 0.028 x (1200 x 4000) / (10

2

x 10)=135 psi

Surge pressure for Gate Valve (Pgv),Pgv = 0.028 x (1200 x 4000) / (102x 30)=45 psi



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36. PROBLEM (Energy Dissipation)

A box culvert, W=2400 mm by L=1800 mm, Q=24.0 m3/sec, supercritical flow inculvert, the normal flow depth = brink depth is yo =1.3 m, the tail water depth isTW=0.90 m. What is length of the energy dissipating pool if d

50/ye =0.45?

a. 20.8 mb. 16.4 mc. 13.6 md. 7.2 m


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36. Solution:

W0=2400mm=2.40m

yo =ye for rectangular section, ye =1.3 m

Vo =Q/A=24.0/(2.40 x 1.3)=7.69 m/s

Fr=Vo /[(9.81)(ye )]1/2=7.69/[(9.81)(1.3)]1/2=2.15

TW/ye =0.9/1.3=0.69, TW/ye


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37. PROBLEM (Open Chanel)

What is the width of a broad-crested weir to convey a river discharge that variesbetween 0.15 and 30.0 m3/sec, y

max=1.75 m, y

min=1.05 m?

a. 29.0 mb. 38.0 mc. 12.0 md. 17.0 m.

37. Solution:



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38. PROBLEM

Determine the 3-hour, 2-year rainfall depth for Fulton County. Where e = 0.767, b =40, and d = 7.6 for the 2-year frequency.

a. 4.30 inchesb. 1.50 inchesc. 3.20 inchesd. 2.16 inches

38. Solution:

Where:

D= rainfall depth (in.)

I= design rainfall intensity (in./hr)

Td = storm duration (min.)

and

e = 0.767, b = 40, and d = 7.6

tc= 3 hours = 180 minutes

Therefore: I = 40/ (180 + 7.6)0.767

I= 0.72 in/hr

D = 0.72 x 180 / 60 = 2.16 inches



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39. PROBLEM (Collection systems -infiltration, inflow)

Which of the following statements is not true for Inflow and infiltration (I/I) insewer systems?

a. Increased operational and capital costs in the sewer network and attreatment plants;

b. Reduced sewer and treatment capacity leading to increased operation ofcombined sewer overflows, flooding and pollution;

c. Increased sewer and treatment capacity restricting for future development;d. Lowering of groundwater levels leading to detrimental effects on local water

resources and loss of soil into sewers causing operational problems andstructural damage.

39. Solution:

C is not true.Reduced sewer and treatment capacity restricting for future development.



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40. PROBLEM (Water Collection)

A community has a population of 40,000. What would be the storage tank capacityfor fire flow?

a. 1.2 MGb. 3.6 MGc. 4.5 MGd. 2.2 MG

40. Solution:

P=40,000

The fire flow is calculated as follows:

Fire flow (gpm) =

Where "P" is the population in 1,000's of people. So, for our community with a

population of 40,000, the fire flow would be:

The required storage capacity for fire flow is calculated as follows:

Capacity, Q = Fire flow DurationCapacity, Q = 6,043 gpm 360 minutes

Capacity, Q = 2,175,480 gal

The storage tank must thus have a fire flow capacity of 2.2 million gallons.


civil exam set 3

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