109757634-Civil-Breadth-Mor-Question-2-Some-Pages.pdf

19.

PROBLEM (Member Design )

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

a. b. c. d.

19.

67.00 Kip-ft 145.00 Kip-ft 221.00 Kip-ft 171.00 Kip-ft

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= Wu DLL2/8 2 Moment for triangular live Load, ML=0.0642 WuLL L Mu= WuDLL2/8 + 0.0642 WuLL L2= 47250.00+19645.20= 66895.20#-ft = 66.90Kip-ft Correct Solution is (a)

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

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20.

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 3 footing. Where, the unit weight of concrete is γ= 145 lb/ft , neglect the soil load and footing weight.

a. b. c. d.

20.

1.70 ft 1.50 ft 2.40 ft 1.80 ft

Solution:

Considering, 1ft wide strip of the footing Pu=1.4x4000+1.7x2000=9000.00# Soil pressure, qu=Pu/B qu=2800 lb/ft2 2 2800=9000/B B= 1.79 ft Correct Solution is (d)


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21.

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 3 wt. of concrete is γ= 150 lb/ft .

a. b. c. d.

21.

450 #/ft2 240 #/ft2 2 372 #/ft 2 290 #/ft

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)


23

PROBLEM (Member Design)

22.

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 3 concrete is γ= 150 lb/ft .

a b c d

22.

2

0.11 in 2 0.50 in 0.25 in2 0.31 in2

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 2

2

Mu=WuL /8=320x16 /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.

What is the bottom steel area for the following rectangular simply supported beam? Where, 2 2 Mu=150 ft-kips, fc’=3 kips/in and fy= 60 kips/in .

a. b. c. d.

23.

2.10 in2 2 3.50 in 1.50 in2 1.10 in2

Solution: 2

2

b=12”, d=18”-2”=16”, Mu=150 ft-kips, fc’=3 kips/in and fy= 60 kips/in ., 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.


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. b. c. d.

24.

425.00 Kips 275.00 Kips 490.00 Kips 425.00 Kips

Solution:

Where, b=12”, D=12”, fc’=3 kips/in and fy= 60 kips/in . 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 2

2

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. b. c. d. 25.

360 ft 800 ft 548 ft 558 ft

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. b. c. d. 26.

240 ft 320 ft 260 ft 300 ft

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. b. c. d.

27.

0.40 0.50 0.30 0.20

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. b. c. d.

28.

485 m 410 m 540 m 345 m

Solution:

From the Green book table, the passing sight distance is 485 meters. Design Speed Km/h 30 40 50 60 70 80 90 100 110 120 130

Passing Distance (m) 200 270 345 410 485 540 615 670 730 775 815

Design Speed mph 20 25 30 35 40 45 50 55 60 65 70

Passing Distance (ft) 710 900 1090 1280 1470 1625 1835 1985 2135 2285 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. b. c. d. 29.

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

30.

558.00 ft 458.00 ft 528.00ft 568.00 ft

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. b. c. d.

31.

0.18 ft/ft 0.15 ft/ft 0.26 ft/ft 0.12 ft/ft

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. b. c. d.

32.

-9.00 ft/sec +9.00 ft/sec -6.00 ft/sec +6.00 ft/sec

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. b. c. d. 33.

Circular Semicircular Rectangular Trapezoidal

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.


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. b. c. d.

34.

1170 ft3/sec 1400 ft3/sec 3 980 ft / sec 1260 ft3/sec

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) R 2/3 √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.


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. b. c. d.

3

3.00 ft /sec 1.57 ft3/sec 3.60 ft3/sec 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 .63 ..54 .63 .54 Velocity, V=1.31 C R S =1.31 x 120 x .375 x .001 =2.03 ft/sec 3 Q=VA=1.77 x 2.03=3.60 ft /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. b. c. d.

36.

Lowering the pump Reducing the pump speed Using high suction head pipe Low liquid temperature

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. b. c. d.

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

38.

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

Solution:

Culvert under inlet control will always flow full is no t 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. b. c. d. 39.

0.51 0.45 1.60 0.48 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. b. c. d. 40.

1.5 MGD 3.6 MGD 4.5 MGD 2.2 MGD

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)

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109757634-Civil-Breadth-Mor-Question-2-Some-Pages.pdf

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