Civil PE Exam AM practice Exam

21 Pages·2015·12.52 MB·English

by  Shahin A. Mansour

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Civil AM Practice Exam- Sorins 2015 Dr. Shahin A. Mansorrr- PF.
he followins information is to be used for oroblems PE#1-L to
The timber formwork and shoring support for a cast-in-place concrete slab is shown below. The 7 in
it
concrete slab is placed on the % plywood that is supported by 2x4 (actual size 1.5" x 3.5") joists
spaced 15 in. o.c. Stringers 4x6 are spaced @45 in. apart. The shores directly under the stringers are
6x6 and are spaced @5 ftintervals along the stringers. The concrete weighs 145 pcf and a construction
live load of 20 psf. Assume the dead load is 3 psf.
4'x8'- 3,4 in. Sheothing
6x6 Shores
I
<t
-Y
PE Exam #1-1
Find: The maximum bending on the joists is most nearly:
(A)
200.00 lb-ft
(B)
26e.00 lb-ft
(c)
2ee.80 lb-ft
(D)
34e.00 lb-ft
Solution: Interior joists are the critical ones because the tributary area will the largest. The tributary
area for an interior joist :15'rx8'. The loads on an interior joist is the summation of:
: :
l/concrete 145 pcf x 7" 1(I2" |ft) 84.58 psf
:
ConstructionLL 20 psf
Dead load: 3 psf
I = 107.58 psf
The tributary width of an interior joist is 15". The load per joist: (107.58 psf ) x 15.1(12"lft):134.48
plf. Using the "Beam Diagrams and Formula" in the Steel Manual, the maximum bending moment will
be at the center support (middle stringer) and its magnitude equals:
M,,*, =-'L8 8:-134'48,x(4',)2 =26g.g6 lb - -fi' Answer:(B).(
PE Exam #1-2
Find: The maximum bending stress on the joists is most nearly:
(A)
880 psi
(B)
980 psi
(C)
1025 psi
D) 1055 psi
Solution: An interior joist is a continuous two equal spans beam
MC
268.961b- frx(12'l fi)x(3.5'12) Answer:(D) {
"rx I = 1053.88ps1
$.5',)(3.5\3lv
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Civil AM Practice Exam- Spring Dr; Shahin A. Mansour. PE
PE Exam #L-3
Given: An excavator has a bucket capacity of 2.8 yd3. Its operation cycle consists of the
: :
following phases: (a) excavation time 45 sec, (b) travel time (two-way) 4 min, and (c)
: :
dumping/transfer time 30 sec. Assume an overall efficiency factor for the excavator 85o/o.
The quantity of excavated material: 50,000 ft3 (bank measure). The material has the following
: : :
properties: Swell 20Yo, Unit weight 1201b/ft3, Water content 30o/o.
Find: The number of days required to complete the job, assuming 8 hr workdays, is most
nearly:
(A) l0
(B) u
(c)
12
(D)
13
Solution:
u,?'?9?fi:,
LCY =5o,ooo xl.Z= =2,222.22yd3
27fr3 I yd3
'p rod.=8dhar *Y9 9h-rt * l"y"!g_ .. ,2'8cyYd3c txe0.g 5 =2-r 7.gld
. daY
(4 min+ # #,
- 2'222'2-2Yd3
#of da'v =lo.zdav Answer:(B){
2l7.8yd' I doy
PE Exam #1-4
Given: A truss is shown below.
12k '116k
Find: The member force (k) in member DI is most nearly:
(A) 22 (compression)
(B) 22 (tension)
(C) 39 (compression)
(D) 39 (tension)
Solution:
DI: interior member :+method of section--+ Consider section 1-1 and take right section of the truss
ZU = 0 CCW ls *, and assume the force in member DI is tension i.e. pulling the joint
"
*)21g1*4(1X16)=0 )
(20)(8)+(16)(16) Fr, =-38.76k (c)
J5\i5
{
Answer:(C)
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CivilAM Practice Exam- Spring Dr. Shahin A. Mansour. PE
PE Exam #1-5
:25 :
Given: A simply supported steel beam (span ft) supports a uniformly distributed load
:
2.75 Wft. The allowable bending stress for the chosen grade of steel 32 ksi. Bending is
ITIiTnIg aaDboouutr tmhee sstlrroonngs aaxxilss i1..ee.. ssttrrcoins-axls bendi
Section A (inz\ d (in) ,S, (inj) S, (inr)
W12 x 65 9.1 2.1 87.9 29.1
W12 x 58 7.0 2.2 78.0 21.4
Wl2 x 53 5.6 2.1 70.6 19.2
W12 x 50 4.6 2.2 64.2 13.9
Find: The lightest W-section among those listed in the above table is:
(A) Wl2 x
50
(B) W12 x 53
(C) W12 x 58
(D) Wl2 x 65
Solution:
M * = W] t s = (2.7 5) (25)' I g = 21 4.g4 l<ft. = 257 g.t3 k - in
k-in
.s ,-2578'13 =80.57in3 Answer:(D){
-
"required 32 kSi
PE Exam #1-6
Given: The department of transportation (DOT) must make a determination about the
feasibility of repairing a bridge. The estimated useful life of the bridge is 25 years. The
MARR is 8%. The relevant cost estimates are shown below. All costs are shown in thousands
of dollars.
Existine Repair
Initial cost 0 75
Annual Cost (maintenance)
Years l-15 10 3
Years 16-25 t4 4
Salvaee Value 5 20
Find: The benefit-cost ratio of making repairs is most nearly:
(A) 0.e0
(B)
O.es
(c)
1.05
(D)
1.10
Solution:
B _7 xrc3 e I 10Y5+t}xro:(p I 48Aoe I Dl? _81,067
c F)Y:rs = 1.1 13
75xto3 -15xto3 (P I 72,870
{
Answer:(D)
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Civil AM Practice Exam- Spring Dr. Shahin A. Mansour. PE
PE Exam #1-7
ft
Given: Water flows in a rectangular open channel at a normal depth of 4.6 for a 12 ft.
channel width. Assume Manning's roughness coefficient (constant with depth) = 0.014.
Find: The longitudinal slope of the channel bed is 0.4%. The flow rate (ft3lsec) is most nearly:
(A) 600
(B) 700
(c)
800
(D) e00
Solution:
Using King's Methods: d lb =# =0.383 , 1G2.68
:
a = K(;)d"' uS (2.6st ffi r+.o,aIa r[.-64 > 7 t2 crs
55.2
Area = 4.6x12=55.2ft.2,perimeterp =12'+2x4.6'=2t.2ft, ond Ro = =2.60 ft
2t.2
e =vA:#ro,),,, J s Gq) = (ss.2) Hf (2.6),,'fi^004 = 702.e crs Answer:(B) {
PE Exam #1-8
Given: Water flows by gravity frorn reservoir A (surface elevation 340 ft) to reservoir B
(surface elevation 270 ft) through a system of cast iron pipes. The characteristics of the pipe
system are given below:
Length: 2,500 ft
:24
Diameter in
:0.02
Friction factor
Minor loss equivalent length:55 ft
Find: The flow rate (ft3lsec) is most nearly:
(A) 42
(B) s5
(c)
6s
(D) 76
Solution:
*
ftr = losses = 340'170' 70'
(hr)(2il@)
(Llqf
h, f
Darcy-Weisbach = =V =
/o
(70)(2)(32.2)(2',)
= 13.28fps
(0.02) (2500 + 55)
e =vA = (L3.zq@fL = 4r.73crs
Answer:(A).(
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Civil AM Practice Exam- Spring Dr. Shahin A. Mansour. PE
PE Exam #l-9
G iven : A cantilever reinforced-concrete retaining
wall is shown. The friction angle between the wall Chesionless bockfi I I
footing and the underlying soil is 20o. uAnnig+l.e w oefi glhntt e=1rn20opl cffricilon =34 dgrees
:257.00 ft
Elevation of the top of the backfill
ft
Elevation of the top of the heal = 243.00
:
ft
Elevation of the bottom of the heal 240.00
4' j' 6'
Find: The factor of safety for sliding of the wall is most nearly:
(A)
1.27
(B) 0,e3
(c)
1.4s
(D)
1.78
Solution:
)
0=34o Ko= 1-tl1 =0.2827
/
1+ sin
Active1arth pressureforce = Fo = 12y' ru'r""= [\12])trro )OT'? e.2827):49021b1ft.
Resisting Force = Friction Force = f = rt'l = tan2l (ZW)
:
= tan20"l6xl4xl20+1x 14x 150 + 1 1x 3 x 1501 6234.81 lblft.
.cr _ Resisting Forces _6234.81 _1 ,,,1
DisturbingForces 4902
{
Answer:(A)
PE Exam #1-10
Given: A parabolic vertical curve is being designed to connect two grades with Gt: -l5o/o and
Gz: *3Yo. The station and elevation of the PVI are 123+64.50 and325.64 respectively.
Find: If PVC is at station ll7 + 50.00, the elevation (ft) of the PVT is most nearly:
(A)
307.21
(B) 30e.4s
(c) rt.t2
3
(D) 313.02
Solution:
Ll2:
(123+64.50) - (117+50) = 6.145 Sta
:
(PW) Etev.: (32s.64) - (s, ) (L I 2) 325.64 - (3%) (6.r4s) :307 .21
{
Answer:(A)
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PE Exam #1-ll
Given: A pavement-based dual loop detector is used to monitor speeds of a stream of vehicles
in one travel lane. The following speeds (mph) were recorded:
45, 51, 58, 47, 50,53, 51, 42, 61, 5l
The traffic flow rate during this period is estimated to be 1,450 vphpl.
Find: If mean headway (sec) is most nearly:
(A) l.e
(B) 2.I
(c)
2.3
(D) 2.s
solution:
r6^o-x. 6eAo
-
Headwav' 2.48 s/veh
--
1450vphpl
{
Answer:(D)
PE Exam #l-Lz
Given: A horizontal circular ourve has PC at coordinates (ft) 4123.64 N, 1064.32 W. Curve
:
radius 1,030 ft, The curye length is 646.35 ft. The tangent at the PC has bearing S42' 30'W.
Find: The coordinates of the PI are most nearly:
(A) 3877.23 N, 1290.12 W
(B) 3897.84 N, 1310.73 W
(c) 4370.05 N, 838.52 W
(D) 4349.44 N, 817.91 W
Solution:
L : 2zrRL/ 360= A = 3z6,0rLR - (360)(646'35) :) A= 35.95"
(2n)(1030)
35'95o
T = RtanA2 2= 1030t* =334.22'
: :
X LW = T sin 4239'= (334.22) (sn42.5") 225.80,
Y = Nr{ =Tc.os42.5o =246.41'
:
PI coordinates = N 4123.68 - y - 4123 .68 - 246.41) = 387 7 .27 N
Plcoordinates = E =1064.32+ x =1064.32+225.80 =1290.12W
{
Answer:(A)
I
'
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Civil AM Practice Exam- Spring Dr. Shahin A. Mansour. PE
PE Exam #1-13
Given: Water flows through a circular conduit of diameter 42 in.
Find: The depth of flow (in) that the pipe conveys maximum flow is most nearly:
(A)
21
(B) 30
(c)
40
(D) 42
Solution:
Using the graph included in the H&H Tables package "Circular Channel Ratios"
Q max will occur at dlD: 0.95
:39.9"
Depth of the flow d: 0.95x 42"
Answer:(C){
PE Exam #l-14
Given: A flow rate: 20 ft3lsec flows through a 30 in diameter concrete pipe (Manning's n
constant with depth: 0.013). The following data was given:
Pipe length: 800 ft
:
ft
Pipe invert elevation at upstream end of pip e 27 5 .64
:270.96
Pipe invert elevation at downstream end of pipe ft.
Find: The depth of flow (in) in the pipe is most nearly:
(A)
14.0
(B)
1s.3
(c)
17.5
(D) 21.8
Solution:
275'64 - 270'96'
slo-p e - = 0.005850 = 0.585%o
800'
Assume the pipe is flowing full:
-ry#(?)"'Jo-*
e..r,= i'!86 ^;,,.1, -> =31.32crs
Using Hydraulics Table og QQtauror' ": 3:0l='3, 2: 0 .64 : (D -d : (0.58) ( 30") :17 .4
{
Answer:(C)
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Civil AM Practice Exam- Spfing Dr. Shahin A. Mansour" PE
en information are to be used for nroblems 15 th 1
Grain size distribution for a soil sample is shown in the curve below. Liquid limit: 34% &plastic limit= l9Yo.
100
g)
C
75
Ua) a
a
t
o-
-+- 50
a
C.
a) a
O ,
(D
25 aq
^
,
2 3 4 5 6789 2 3 4 5 6749 2 3 4 567a9
O.O1 O.t 1.O
1
groin diome+er (mm)
#
Find: The USCS classification is most nearly:
(A) Sw
(B)
SP
(c) ML
(D) cL
Solution:
#4 Sieve:4.75 mm & %o passed:98%o
#200 Sieve: 0.075 mm & % passed #200 =12% i.e. oh retained =88yo, o/o i.e. the soil is coarse
grained S or G
o'55
-
c.".D Duo, , =8.46>6
0.065
D'o 0'182
C,=C..- -
=0.91
Drox Duo 0.065 x 0.55
{
Answer:(A)
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CivilAM Practice Exam- Spring Dr. Shahin A. Mansour" PE
-L6
Find: The AASHTO soil classification is most nearly:
(A) ,4-6
(B) A-2-6
(c) A-2-7
Solution:
Frroxl2Yo
F2oo<35Yo
PI LL- PI =34-lg:l,Yo
=
Also, the group index (GI) is calculated as follows (ust in case if you're asked about it)
GI = 0.01(F200 - 1 5)(P1 - 1 0) = 0.0 1 (12 - I 5)(1 5 - lO) =-v alue i.e. = 0
.
The classification includine the sroup index is: 4-2-6 (0) Answer:(B){
PE Exam #L-17
Given: A concrete mix contains cement, moist sand, and moist coarse aggregate in the
following proportion by weight l:1.8:2.6. The following specifications are given:
Cement
specific gravity: 3.15
:
SSD sand (-.c. 0.5yo) specific gravity :2.70
:0.7Yo)
:2.60
SSD coarse aggregate (m.c. specific gravity
water
Added 5.8 gal per sack cement'
Air
3% (by volume)
The aggregates used for mixing the concrete had the following properties:
sand: :60/o
Wet moisture content
Wet coarse aggregate: moisture content :4Yo
Find: The water content (gallsack) of the concrete is most nearly
(A)
6.1
(B) 7.1
(c)
7.8
(D) 8.s
Solution:
W,,,d= 1.8x94 lb =169.21b drrd Wg,o,,r=2.6x94 lb =244.40lb
Free moisture inrona=6yi (169.2)
^!^1% -8.7792rb
l06Yo
gr*"1=% -W
Free moisture in ' Q44.4) 4.7750 lb
104Yo
lb +7 '7750 lb
Total water content =5.8 gallo, *8'7792 =7.78 gallon / sack
8.341b I gal
{
Answer:(C)
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Civil AM Practice Exam- Spring Dr. Shahin A. Mansour" PE
PE Exam #1-18
Given: A project consists of 10 activities as outlined in the table below. All relationships are
finish-to-start unless otherwise noted.
Activitv Duration (months) Predecessors Successors
A 5 C.D.E
B Ja H
C 3 A F
D 4 A G
E J A H
F 5 C J
G 4 D (FF LAG:5) J
H J B,E I
I 2 H J
J 2 F, G,I
The activity on arrow representation of the project is also shown below.
Find: The critical path for the project is most nearly:
(A) ACFJ
(B) ADGJ
(c) BHrJ
(D) AEHrJ
Solution:
The critical path is the, "continuous, longest path," of activities through a project that
determine the project completion date. The critical path has the lowest, "total float."
The CP is "A-D-G-J".
{
Answer:(B)
l0