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Copyright © 2008-2011 pe-exam.org all rights reserved PE Civil Breadth Exam Set #2 20 18. PROBLEM (Member Design) A cantilever beam is shown in the Figure. Determine the ultimate bending moment. a. 18.00 Kip-ft b. 8.00 Kip-ft c. 5.50 Kip-ft d. 34.00 Kip-ft 18. Solution: L=6 ft W LL =100 #/ft W DL =200 #/ft Wu= 1.7 x 100 + 1.4 x 200=450.00#/ft Mu=WuL 2 /2=8100.00#-ft = 8.1 Kip-ft Correct Solution is (b)

Civil Breadth Mor Question 2(Some Pages)

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Copyright © 2008-2011 pe-exam.org all rights reserved PE Civil Breadth Exam Set #2 20

18. PROBLEM (Member Design)

A cantilever beam is shown in the Figure. Determine the ultimate bending moment.

a. 18.00 Kip-ft b. 8.00 Kip-ft c. 5.50 Kip-ft d. 34.00 Kip-ft

18. Solution:

L=6 ft WLL=100 #/ft WDL=200 #/ft Wu= 1.7 x 100 + 1.4 x 200=450.00#/ft Mu=WuL2/2=8100.00#-ft = 8.1 Kip-ft Correct Solution is (b)

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19. PROBLEM (Member Design)

A simply supported beam is shown in the Figure. Determine the ultimate bending moment.

a. 67.00 Kip-ft b. 145.00 Kip-ft c. 221.00 Kip-ft d. 171.00 Kip-ft

19. Solution:

L=30 ft WLL=200 #/ft WDL=300 #/ft WuDL= 1.4 x 300=420.00 #/ft WuLL= 1.7 x 200 =340.00 #/ft Moment for uniformly Dead Load, MD= WuDLL2/8 Moment for triangular live Load, ML=0.0642 WuLL L2 Mu= WuDLL2/8 + 0.0642 WuLL L2= 47250.00+19645.20= 66895.20#-ft = 66.90Kip-ft Correct Solution is (a)

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20. PROBLEM (Member Design)

A continuous 12-inch thick concrete wall footing has axial DL load of 4000 #/ft and LL=2000 #/ft as shown in the Figure. The soil bearing capacity is 2800 #/ft2. Determine the width of the footing. Where, the unit weight of concrete is γ= 145 lb/ft3, neglect the soil load and footing weight.

a. 1.70 ft b. 1.50 ft c. 2.40 ft d. 1.80 ft

20. Solution:

Considering, 1ft wide strip of the footing

Pu=1.4x4000+1.7x2000=9000.00# Soil pressure, qu=Pu/B qu=2800 lb/ft2 2800=9000/B2 B= 1.79 ft

Correct Solution is (d)

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21. PROBLEM (Member Design)

Determine the factored design load for deflection control one way uniformly loaded slab. The slab is rest on16 ft apart simply supported as shown in the Figure. Considering dead weight of the slab and it is carrying a live load of 120 lb/ft2. Where, fc’=3 kips/in2, fy= 60 kips/in2 and unit wt. of concrete is γ= 150 lb/ft3.

a. 450 #/ft2

b. 240 #/ft2 c. 372 #/ft2 d. 290 #/ft2

21. Solution:

Considering 1-ft wide strip Determine the depth of slab that deflection control. For one way slab, h=L/20=16x12/20=9.60 in Dead load of slab, Wd=150x9.60/12=120 #/ft2 Factor design load, Wu=1.4xWd+1.7xLL=1.4x120+1.7x120=372 #/ft2 Correct Solution is (c)

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22. PROBLEM (Member Design)

Determine the steel area per feet of one way uniformly loaded slab. The slab is rest on16 ft apart simply supported as shown in the Figure. Considering, slab dead weight is 120 lb/ft2 and it is carrying a live load of 120 lb/ft2. Where, fc’=3 kips/in2, fy= 60 kips/in2 and unit weight of concrete is γ= 150 lb/ft3.

a 0.11 in2

b 0.50 in2 c 0.25 in2 d 0.31 in2

22. Solution:

Considering 1-ft wide strip Slab has been controlled by deflection, Therefore, D=L/20=16x12/20= 9.60 in, d=9.6-.75= 8.85 in Factor design load, Wu=1.4xWd+1.7xLL

=1.4x120+1.7x120 =372.00 #/ft2

Mu=WuL2/8=320x162/8=11904.00=11.90 k-ft Considering, a=.1d= 8.85 x.1=.89 As=Mu/Φfy(d-a/2)=(11.9 x12)/0.85x60(9.6-.89/2)=0.31 in2

Correct Solution is (d)

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23. PROBLEM (Member Design)

What is the bottom steel area for the following rectangular simply supported beam? Where, Mu=150 ft-kips, fc’=3 kips/in2 and fy= 60 kips/in2.

a. 2.10 in2

b. 3.50 in2 c. 1.50 in2 d. 1.10 in2

23. Solution:

b=12”, d=18”-2”=16”, Mu=150 ft-kips, fc’=3 kips/in2 and fy= 60 kips/in2., Φ=0.85 Assume, a=.15d=.15x16=2.4” As=Mu/Φfy(d-a/2)=(150 x12)/0.85x60(18-2.4/2)=2.10 in2

Correct Solution is (a)

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24. PROBLEM (Member Design)

What is the design strength of the following spiral Column section? Where, Mu=100 ft-kips, fc’=3 kips/in2 and fy= 60 kips/in2.

a. 425.00 Kips

b. 275.00 Kips c. 490.00 Kips d. 425.00 Kips

24. Solution:

Where, b=12”, D=12”, fc’=3 kips/in2 and fy= 60 kips/in2. Steel area, Ast= 6*.44= 2.65 in2 , Column area, Ag=π/4d2=(3.14/4) * (12 x12)= 113.04in2 For spiral colum, Φ=.75 and β= .85 Po=0.85fc’(Ag-Ast)+fyAst =0.85 x 3 (113.04-2.65) + 60 x 2.65= 432.84 kips Design Strength, Pu= ΦβPo=0.75 x 0.85 x 432.84= 275.94 Kips Correct Solution is (b)

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IV. TRANSPORTATION

25. PROBLEM (Horizontal Curve)

A Horizontal curve has an interior angle of Δ= 40o shown in the Figure. What is the Length of Curve, Where, R=800 ft?

a. 360 ft b. 800 ft c. 548 ft d. 558 ft

25. Solution:

Δ= 40o R=800 ft , D=2R Length of Arc, L= ΔπD/360=40 x 3.14 x 2 x 800 /360= 558.22 ft

Correct Solution is (d)

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26. PROBLEM (Sight Distance)

A two lane highway has a center line shown in the Figure. A car was passing near a building and building would be 22 ft from the center of the highway. Determine the sight distance for the driver in outside lane, where each lane is 12 ft wide.

a. 240 ft b. 320 ft c. 260 ft d. 300 ft

26. Solution:

BC=450-22=428 ft AC=450 + 6=456 ft (For the driver in outside lane) CosΔ/2=BC/AC=428/456 Δ/2= 20.18 Δ= 2 x 20.18= 40.37 D=2 x AC= 2 x 456= 912 Arc/ Δ=πD/360 Arc= ΔπD/360= 40.37 x 3.14 x 912 / 360= 321.13 ft Sight Distance, S= 321.13 ft Correct Solution is (b)

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27. PROBLEM (Stopping Sight Distance)

A driver sees a body lying across the road and needs to stop. If his vehicle was initially traveling at 100 km/h and skids to a stop on a 2.5% upgrade, taking 75 m to do so, what was the coefficient of friction on this surface?

a. 0.40 b. 0.50 c. 0.30 d. 0.20

27. Solution:

Correct Solution is (b)

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28. PROBLEM (Passing Sight Distance)

A vehicle moving at a speed of 70 Km/h is slowing traffic on a two-lane highway. What is the passing sight distance in order for a passing maneuver to be carried out safely?

a. 485 m b. 410 m c. 540 m d. 345 m

28. Solution:

From the Green book table, the passing sight distance is 485 meters.

Design Speed Km/h

Passing Distance (m)

Design Speed mph

Passing Distance (ft)

30 200 20 710 40 270 25 900 50 345 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

Correct Solution is (a)

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29. PROBLEM (Vertical Curve)

A vertical curve has a length of L=600 ft & P1, station 102+25.0 shown in the Figure. Calculate the grade on the curve at the midpoint.

a. 2.50% b. 0.75% c. 1.75% d. -0.75%

29. Solution:

A=G2-G1= -1.75% –(+3.25%)= -5% X=L/2=600/2=300 S=G1+ XA/L=(3.25%)+ 300(-5%)/600=3.25%- 2.5%=0.75%

Correct Solution is (b)

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30. PROBLEM (Vertical Curve for Stopping Sight Distance)

A vertical curve is required between ascending 3.5% and descending -1.5% grade, the design speed is 55 mph & the stopping sight distance is S=495 ft. Calculate the length of the vertical curve for required stopping sight distance.

a. 558.00 ft b. 458.00 ft c. 528.00ft d. 568.00 ft

30. Solution:

A=G2-G1= -1. 5% –(+3.5%)= -5%=5% Assume S>L, S=495 ft L=2S-2158/A=2 x 495-2158/ 5= 558.4 ft and L>S, L=AS2/2158=5 x 4952/2158= 567.71 ft it is correct length.

Correct Solution is (d)

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31. PROBLEM (Super Elevation)

A vehicle weighs 3500 lb and travels at 35 mph on curve road with radius of 450 ft. What is the super elevation, where tire friction is not needed to prevent the vehicle from sliding?

a. 0.18 ft/ft b. 0.15 ft/ft c. 0.26 ft/ft d. 0.12 ft/ft

31. Solution:

V=35 mph, R=450 ft, f=0

Super elevation, e= V2/15R –f={352/(15 x 450)}-0= 0.18 ft/ft

Correct Solution is (a)

32. PROBLEM (Acceleration and Deceleration)

A vehicle weighs of 3500 lb travels 65 mph brake and decelerates at constant rate and slides 500 ft on a level road before stopping. What is the acceleration of the vehicle?

a. -9.00 ft/sec b. +9.00 ft/sec c. -6.00 ft/sec d. +6.00 ft/sec

32. Solution:

Vo=65 mph=65 x 5280/(60 x 60)= 95.33 ft /sec, V=0 ft/sec, s= 500 ft

Perception-reaction time, t=2s/Vo+V=2 x 500 / (95.33-0) = 10.49 Sec Acceleration, a= V-Vo/t=0-95.33/10.49= -9.09 ft/sec

Correct Solution is (a)

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V. WATER RESOURCES AND ENVIRONMENT

33. PROBLEM (Open Channel)

Which of the following has the most efficient cross section of a channel section?

a. Circular b. Semicircular c. Rectangular d. Trapezoidal

33. Solution:

The most effective efficient cross section is a semicircular channel section, because it has a larger wetted perimeter than any other section.

Correct Solution is (b)

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34. PROBLEM (Open Channel) A triangular concrete channel has been constructed with a slope of 0.003. The water depth is 6 ft. What is the flow rate? Side slope 3:1, n = .014.

a. 1170 ft3/sec b. 1400 ft3/sec c. 980 ft3/ sec d. 1260 ft3/sec

34. Solution: Area, A=3 x 62= 108 ft2 Wetted perimeter, P= (2 x 6 √1+32)=37.94 ft Hydraulic Radius, R=A/P=108/37.94=2.85 ft Velocity, V=(1.49/n) R2/3 √S=(1.49/0.014) x 2.852/3√.003=11.70 ft/sec Q=VA=11.7 x 108=1264 ft3/ sec

Correct Solution is (d) 35. PROBLEM (Open Channel)

A circular sewer line has an 18 in diameter with slop of S = .001. The flow is uniform and steady. What is the flow of the sewer line? Using Hazen Williams full flow velocity with C = 120.

a. 3.00 ft3/sec b. 1.57 ft3/sec c. 3.60 ft3/sec d. 1.76 ft3/sec

35. Solution:

D=18”=1.5ft, S = .001, C=120 Area, A=(∏/4)d2 =(∏/4)1.52 = 1.77 ft2 For Circular pipe, Hydraulic Radius, R=D/4=1.5/4=.375 Velocity, V=1.31 C R.63 S..54=1.31 x 120 x .375.63 x .001.54=2.03 ft/sec Q=VA=1.77 x 2.03=3.60 ft3/sec Correct Solution is (c)

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36. PROBLEM (Pump)

Which of the following statements is not applicable for the increase of net positive suction head (NPSH) to eliminate cavitations?

a. Lowering the pump b. Reducing the pump speed c. Using high suction head pipe d. Low liquid temperature

36. Solution:

Using high suction head pipe Correct Solution is (c)

37. PROBLEM (Water Distribution)

Which of the following statements is not applicable for two or more parallel water distribution loops?

a. The flow divides in parallel pipe make the head loss in each branch the same b. The head loss between the two junctions is the same as the head loss in each branch. c. The total flow rate the sum of the flow rates in the two or more branches. d. Size and Length of parallel line should be the same.

37. Solution:

Size and Length of parallel line should be the same are not applicable

Correct Solution is (d)

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38. PROBLEM (Culverts) Which of the following statement is not true for culverts?

a. If water can flow through and out of the culvert faster than it can enter control b. If water can enter and flow through the culvert faster than it can out, outlet controls. c. Culvert under inlet control will always flow full d. Culvert under outlet control can flow either partially full or full.

38. Solution:

Culvert under inlet control will always flow full is not correct.

Correct Solution is (c)

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39. PROBLEM (Hydrology Rainfall)

Determined Average Runoff Coefficient, “C” as shown in the Figure, which is the runoff, contributes to a collector.

a. 0.51 b. 0.45 c. 1.60 d. 0.48

39. Solution:

A1 = 4 ac ,C1 = 0.38 , t1 = 20 min , A2 = 2ac , C2 = 0.58 , t2 = 12 min, A3=3ac, C3=.64, t3=15 min C=(C1A1+C2A2+C3A3)/(A1+A2+A3)=(4 x .38 + 2 x .58+3 x.64 )/(4+2+3)=0.51

Correct Solution is (a)

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40. PROBLEM (Wastewater Collection)

Estimate the wastewater flow of a community has a population of 15,000.

a. 1.5 MGD b. 3.6 MGD c. 4.5 MGD d. 2.2 MGD

40. Solution:

P=15,000, water use q=100 gal per capita/day, Peak hourly factor=3 Peak hourly flow, Q=15000 x 100 x 3=4.5 MGD

Correct Solution is (c) --------------------END---------------------------------------