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Copyright © 2008-2013 Pecivilexam.com all rights reserved PE Civil Breadth Exam Set #2 1







PE Civil Exam 40-Mix Questions & Answers (pdf Format)
For Breath Exam (Morning Session) Set #-2

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Copyright © 2008-2013 Pecivilexam.com all rights reserved PE Civil Breadth Exam Set #2 2

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 19

4. Transportation-8 Q&A 27

5. Water Resources and Environmental-8 Q&A 34

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I. Construction

1. PROBLEM (Earth work)


An embankment cross section is shown in the Figure, the soil in the embankment has to be
compacted, and the soil dry unit weight is106 lb/ft3, moisture content is12.5%. The borrowed
soil will be brought in a 10 cubic yard truck at haul with a void ratio of 0.7 and a moisture
content of 10%. The borrowed site soil has a void ratio of 1.0 & specific gravities, Gs=2.67.
What will be the number of haul truck loads for 100 ft length of embankment construction?

a. 314 numbers of trucks
b. 414 numbers of trucks
c. 360 numbers of trucks
d. 276 numbers of trucks


1. Solution:


Vol. of Embankment per 100 ft length= {(24+74)/2} x 16 x 100=78400 ft3


Moist unit weight of truck γ =(1+w)Gs γw /(1+e)=2.67x62.4(1+0.1)/(1+.70)=107.80 lb/ft3


The dry unit weight of truck γd = γ /(1+w)=107.8)/(1+.1)=98.0 lb/ft3


Weight of dry soil per truck= 10 x 27 x 98=26460 lb


Dry weight of Embankment compacted soil= 78400 x 106=8310400 lb


Number of truck= 8310400/26460=314 number of trucks


Correct Solution is (a)

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II. Geotechnical

9. PROBLEM (Soil Classification)


A soil sample has been given in the following value. What is the AASHTO soil classification?


Sieve Passing
No-10 100%
No- 4 80%
No-200 37%
L.L 44
PI 12


a. A-7-6
b. A-7-5
c. A-7
d. A-4




9. Solution:


Soil passing #200 sieves is 37% which is greater than 35%; therefore it is a silty clay soil. So it
falls under A-7 on AASHTO classification chart.
Also, PI=12<(LL-30=14), Therefore, it is A-7-5.


Correct Solution is (b)

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III. STRUCTURAL



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
Considering, DL factor=1.4 and LL factor=1.7
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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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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