EEE aFE CIVIL PRACTICE EXAM 58. The frame in the figure below is — PININ MEMBLR i ITYPICAL) RIGID BEAM-COLUMN A CONNECTION (T¥PICAL) PIN BOUNDARY * cryPIcaL) OA unstable © B stable and determinate © C. indeterminate one degree © D._ indeterminate two degrees Copyright© 2017 by Noes 42FE CIVIL PRACTICE EXAM jow with ts correct description ‘March each of the beams diagrammed bel CONCENTRATE == UN ee D LIVE LOAD UNIFORM LIVE LOAD FORM DEAD LOAD D UNIFORM DEAD LOA IVE LOAD. DAD UNIFORM LIVE LOAP | rr NCENTRATED | fried 7 [sa UNIFORM DEAD LOAD | CONCENTRATED LIVE LOAD ford UNIFORM DEAD LOAD CONCENTRATED LIVE LOAD Sopra DOH by nceES 43 | eam Types Fixed-end [indeterminate lr Multispan ‘Simpl pan Next60. A concentrated load of 20 kips mov ‘truss, the maximum force (kips) mn Member C 20 ips 4 oA BS oB 2 oc 375 oD 8 ‘copyright© 2017 by NCES FE CIVIL PRACTICE EXAM } sr the truss shown. NEBlecting the, es throug! 5 the moving foad is Most nea D due to i anFE CIVIL PRACTICE EXAM Which ofthe vertical-toad influence lines shown below i correct for Member UU of shown below? ea ol. PINNED JOINT uy u Uy Us (ypieal) 6@ 18 f= 108ft Option A option B Option € Option D OA Option A 2B Option B OC Option c 2D Option D Porrent 2017 by Ncees 45 init ~FE CIVIL PRACTICE EXAM 62. The beam shown below is known as # P { A B indeterminate beam simply supported beam D. continuously loaded beam oo000 figure below has its compress length, Assume /, = 50 ksi and 63, The W2I » $7 steel beam shown at the one-third points over its segment The maximum factored load w. (kips/ft) that the beam can carry ji nearly ‘Copyignt © 2017 by NceES 46FE CiviL py town below isthe ero ‘ding applied in the 64. The figure sh transverse lo, RaCTice Exam 85 Section of, Yeditection Mark the location on the cross sect SIDE View CROSS SECTION a7 a “Seman © 2017 by noees oeFE CIVIL PRACTICE EXAM ross section of a steel rectangular beam, ‘The beam i subjected t shown below is the ct The figure loading applied in the ¥-direction transverse 1s section that experiences maximum normal stress t * SIDE VIEW Mark the location on the ros CROSS SECTION S>mron © 2047 Next> Pe “7o™ FE CIVIL PRACTICE EXAM 65. According to ACI 318-11, the value of @ that should be used in computing y strength 6M, for the beam section shown below 18 most nearly Bea 7 CLOSED TIED Lb STIRRUPS 218m foo ksi Osi 4-10 A, 5.08 an? ; >00 voxFE CIVIL PRACTICE EXAM A reinforced concrete beam is subjected to a factored mom Moment M, = 638 hips. For 66. A rete /1_ = 4,000 psi, and for ste! f= 60,000 psi. The beam rit am i reinfor tao rows, positioned as shown in the figue below: It may be Noe Wah gh 18 bars 10.90, The minimum adequate overall with (othe nearest ahake thet ¥etfcaton most nearly’ inch) fr this beam is, #8 BARS win pin, COVERTO_,| py in OSttEL "| po © et 2 in (CENTER-CENTER SPACING) ever 3 sin (CENTER BOTTOM SPACING) _y Sie ee 1 Sin, CLEAR SPACING | oA 10 OB 12 oc 2B op 15 next Snr: M2017 by NoEES “0FE CIVIL PRACTICE EXAM 87. The fl lexural desigi reed concret a sign strength ({t-kips) of the reinforeed eonerete beam section shy, By, 4 4 isin din eno oy eee ’ be td in f= 4.000 psi f~ 60 ksi A 267 OB (297 oc 9 oD 354 | ! copyright © 2017 by NCEES 50FE CIVIL PRACTICE Exam 68. The flexural design strength (fi-kips) of the reinforced concrete nearly T i | 20in 24in ow | | eooe L be 16 in f= 4,000 psi fy = 60 ksi OA 258 OB 289 oc B4 oD #9 next> >Praht © 2917 by NOFF@ 51FE CIVIL PRACTICE EXAM ¥ sy associted waa lS 000, 69. Mad Mark the area ofthe Arbor chart prowess a. Diseth ae ; =" We Ss Z0 ih Then — if } E if VV oA Sy | le o8 5 ae itor OH oc 2 ay 0 ? Kit ot 7 7 — 1620 0 ee Liquip LIMIT LL) no, An andinusbed spe of sl 8 3 SPE So of solids of 2-70, a moisture cote ; An andi gran of0 63 The eee of saturation MES nearly ms ' oA 2% SB ast oc ° oD 8% ° copynnt © 2017 by NEES a 52 7 vot2. FE CIVIL PRACTICE Exam Direct shear test data of a sand are shown bel Area of sample = 16 in® Normal load at failure = $12 Ib Shear stress at failure ~ 16 psi The angle of friction is most nearly 0° OA oB 27 oc 30° oD 83 has a 10-ft upper layer of sand t of the sand 1s | loration indicates that a level site ‘ground surface The unit wei £10 fi 1s most nearly Subsurface expl groundwater table is at the effective overburden stress (psf) at a depth o A. B c 0000 dD, Next>| a cTICE EXAM seamutar soit wth an antermal ANTS OF, seach pressure (Ib f°) possible oni FE CIVIL PRA! 73, A 12-fthigh retaining wall has backfill of and a unit weight of 125 pet. The Rankine most neatly | OA 2280 OB 3.000 oc 9.000 oD 27.000 | 74. A strip footing having a wich JF ~2 fais to be constructed at ground surface (Dj ~ 0} Uncen: the footing is sand having the following bearing capacity factors: N;= 0, Ny = 2, and. The unit weight of sand (7) 120 pet. Recalling that qui ~eN DAN, + 0.5 ¥BN, the ls | bearing capacity guy (psf) of the Footing is most nearly | OA 1200 © B 2400 oc 3.000 oD 4800 | Copyright © 2017 by NCEES ey -FE CIVIL PRACTICE EXAM nsolidated 10-ft clay layer is surcharged, which caus , €8 a deere ‘A normally cot ase thickness, 13. ‘The coefficient of consolidation is 0.16 das diay layer 18 confined between two layers of day andthe time factor 2 or” Sn Consolidation is most nearly time (ays) required for Soh, OA s o B. 38 oc 188 5D 750 ‘a dam that is 600 ft long (measured low net for the cross section of the total seepage (7 day) 16. The figure below shows the fl perpendicular to the cross section). Assuming & uniform cross section. tinder the dam would be most nearly 30M, joo ne Copyright © 2017 by NCEES 58 2FE CIVIL PRACTICE EXAM imerseetion is being reconstructed to address safety problems, and ty S, and itis est estimated that a3, Anurban an ur nly enlusive countermeasures havea crash elution fat the ly the expected umber of eases per eu 1 and Po Significant rote o5 respemeapated, the expected number of SVeraHe crashes per year alter econsincct ae is am 10M is most ease oB 4 oc 6 oD 64 as a gravel surface. A deer suddenly leaps into the ros 1g at SO mph when the deer appears. T ‘gravel surface is 0.65. The 84. A person is driving a car on a road that h: erred is on a 10% downgrade, and the car is travelin Fee en tion time is 1-5 sce, and the coefficient of friction on the deceleration rate, @ coefficient of friction f = 32.2 ft/sec? ‘The total distance (ft) required to stop 1s most nearly OA 153 om 2 oe 3 OD 262 oe "IOh ©2017 by NcEES next ae 59FE CIVILPRACTICE EXAM , ne has a volume of 1400 vehicleshr and an av EEEFE CIVIL PRACTICE EXAM 89. Consider the equation. ‘AsiOs +373. COF 2 AS 1s 75 for arsenic: | 1 standard gram vor oxygen, and 12 £8 caFb0n. ox ‘Atomic weights may be taken a 6 for ONYEER. GS ith carbon wil nd the equation above, the reaction mmole of AS: ‘ah formation of oA Lgram mole of AS OB. 28grams of CO oC -150 grams of AS © D___ agreater amount By weight of CO than of AS fe filters, Each filter has a surface area of 1005" Ts ‘min. The quant int has multipl nd the backwash duration i812 90, A surface water treatment pla ost nearly fier backwash rate is 15 gpmft, a sires required for backwash of each Filter 1S m oA 4,800 OB 13,200 oc 18,000 oD $76,000 ‘Copyright © 2017 by NCEES 62 veFE CIVIL PRACTICE EXAM waste flow wi 8 S.day HOD of tewater treatment plant is. proces: oj, A municipal wast Jo0 mp/L_ at 20°C. If the BOD rate constant fy (bi 2p ng/L) ofthe raw wastewater at 20°C is Se rece ean nearly 3 day ' the ultimate oA o B. oc oD. 92. You are designing a7 aerobic system 10 biodegrade benzene (Colle) ‘The biodegradation follows the chemical reaction below (note that you must balance this equation). The benzene concentration 1S 500 mg/L. (C= 12,H=1,0> 16) (CoHe + O2 > CO: + H20 “The amount of oxygen (mg/L) that must be consumed to completely biodegrade ‘PE benzene is most nearly oA 200 °B 500 oc 800 2D. 1,600 ‘eeram © 2017 wy s NEXT> ReFE CIVIL PRACTICE EXAM 93. Operational manuals, warranties, guarantees, and as-built drawings are generally pros | owner during which phase ofthe project” Construction Procurement Close-out 0000 game Feasibility 94, A loader has a fllbucket capacity of 3 yd and the average time required to pe tached af soil nto a truck is Tun, The loader is supported by four trucks witha volun: Te a each anda cycle time of 12 min plus the time to Toad the truck “The ideal productivity (yd) of this system is most nearly OA 180 oB 22 oc m7 oD 30 | oprah © 2017 by ncees 64FE CIVIL PRACTICE EXAM luction of the excavator 1s the controlling factor in & highway dite J average prod vce gs, The aver eavators with four different bucket si7es are avilable as renal uns Te cont Niet proportional tothe bucket capacty of the excavator Aun nal te choose vd‘ hr) 18 equal to (excavator eycles per hour) - (average bucket payout aosreveeyele) The excavator characteristics are as follows pad in ; y ‘Average Bucket Payload Excavator | ey) 1 | 050 2 | 1.00 3 17s 4 2.00 ‘The optimally efficient excavator is © A Excavator! © B Excavator? © C Excavator 3 © D Excavator 4 Coomom6 2017 by neers os a a_iFE CIVIL PRACTICE EXAM 96. An embankment having a volume of 320.000 yd’ is to be nee a borrow he, Unit weight and moisture content of the borrow material were determined 10 be 106, 18.2% respectively. ‘The embankment material has a otal unit NBM OF I? Pet anda na, content of 16.7% The volume of borrow (yd) needed to co mel nearly OA 274,100 OB 315500 OC 324.500 © D 373.600 The value of Angle A in the following figure is most nearly 7 gt Wy NOrTOSCALE OA 30°18 47 OB 324750" OC 390538 OD 42°3509 ‘Copyright© 2017 by NCEES 6 NextFE CIVIL PRACTICE EXAM gg, The cross-sectional areas to be excavated (cut) at certain sect i sections ofa road Projet are a follows Station Area (ft2) | [3x00 | 247 400 | 269 4035 322 5x00 | 395 | | 5+65 418 | 6-00 293 L_t00 | 168 Using the prismoidal method, the earth to be excavated (yd°) between Sections 4-35 and 5-65 is ‘most nearly oA 1,460 OB 1,840 oc 1,860 oD 1,900 next> Iyright © 2097 by NCBEA 6799, FE CIVIL practice EXAM ee « bearing and distance fo “a5. {UM from Point B wo Pout The Beasne an oo Pomtk ae Samar aml ea7 I The, coudinates orp E The coordinates of Pont K are 1ESITISN and 10.5679°8 "HE () in attude 1 most meals SA on SB on oC Oe oD oa A 100. The area inside the quadrilateral, PC. PI, PT and O. equals 83,164 ft" The shaded area (i between the circular curve and the tangents most neces Pl Ph 36:48 Pee Pr OA 2.879 OB 3577 oc S407 OD 82K6 ‘© Copyright 2017 by NCEES 68SOLUTIONSFE CIVIL SOLUTIONS Detailed solutions for each question begin nthe neNE PIES Jo] alole|>|o ° | alojo | | o|>|=|0|> \ >FE CIVIL SOLUTIONS fer to the Mathematics section of the FE Reference H reference Handbook ae [3ren* eiI7 He CORREC T ANSWER IS: C ext feo! HE CORRECT ANSWER rhe roots of a function 3S defined as points where F=0 inthis case, aivide 2 polynomials sx2 +5 6x46 6xt6 ° 2and* eho $+ 6 factors to (x +2008 +3) rherefore, the roots oF! ares? Tie CORRECT ANSWER IS: C nmFE CIVIL SOLUTIONS Refer to the Mathematics section ofthe 1 Reference Handbook (x AY + (y= AY + (= my? =F with center at (Ht, k m0) (r-07 +Q- P+ e-CP-P rey THE CORRECT ANSWER IS! © “the cross product of vectors A and B is a vector perpendicular A and B fii k Lg Olea) —f-2-0)¢K(2— = Ai + 25-2 divide by the magnitude To obtain a unit vector THE CORRECT ANSWER IS:as FE CIVIL SOLUTIONS refer to the Engineering Probability and Statisties section of the / of the Hi: Reference He Handbook N _ 801 _ go 95 a 90 50 | Mode median C90 88 87 85 85 e sample is 89. The median of the sample is 90 The mode of the sample 1s 90 ‘The mean of th “dian and the mode are equal ‘Therefore, the m« fer to the Engineering Probability and Statistics section ofthe FE Reference Handboo Binomial distribution 05 (chance of getting a head) Pe 05 (chance of not getting a head) n= 10 (number of trials) x= 4 (number of heads) (10191817) (9 5)!” Prot) DIDO J os4 0.5% He0s*0.5%) = =0.2081 THE CORRECT ANSWER I 7310. FE CIVIL SOLUTIONS section of the FE Reference Hayy, Rete Hes to the Engineering Probability and Stats! Th \ 1. ahat sun to 12 There are 36 possible rolls oy, fee Is only one throw, 6 and 6, that sul therefore 35/36 will have a sum fess than 12 3536-0972 D THE CORRECT ANSWER I eyes Px)= 38+ 2s xGx*2)=0 %(x) negative below x = 1/3 1G) positive above x= 13 Since f"(x) = O and f%(«) changes sign atx = —1/3, the inflection point is at x THE CORRECT ANSWER IS: 5 | The following formulas are in the is Fie rows of Column B | 1 SaV3e asia 2 SAD3+ASI9—3 3 Sass + Asiao3 4 SASS ASIO 8 3 Savaranac3 In spreadsheet equation fo the formula in Cell BS iy SASS + ASI'2 3 THE CORRECE ANSWER IS; ¢ 74FE CIVIL SOLUTIONS Column Value ot A Rew 0 0 wy sas An SAM aw SAM pn CORRECT ANSWER IS: ge oggwes AS ITB SS K igpwves A 85/3 28 K gna A284 071 Hie CORRECT ANSWERS! A VAR x 4 5 6 pxIT LOOP wo step 1 Atihe conclusion of the routine, VAR 6 CORRECT ANSWER IS: D Professional Conduct states: 7 on : 7 = Jedd by education or expenienee quatiti ke assignments only whe involved Hicensees shall unde weil technical fields of engine 4g OF SUEVEY IN HE CORRECT ANSWER IS: € —_—e 7816, 17. 18, 19, FE CIVIL SOLUTIONS i fection B, in the Ethics sec Refer to the NCEES Rules of Professional Conduct, Section B, in the Ethics section, FE Reference Handbook ‘THE CORRECT ANSWERS ARE: A AND E- Refer to the NCES Rules of Professional Conduct im the Ethics section FE Reference Handbook Licensees may express a professional opinion publicly only he Founded on adequate knowledge of the facts and a competent evaluation of the subject mater THE CORRECT ANSWER IS: C Black's Law Dictionary defines lien as a claim on property for payment of debt ‘THE CORRECT ANSWERIS: A Refer to the Engineering Economies section ofthe FE Reference Handbook A=PUAP. Pm) = 100,000 (4 P. 1%, 60) ~ 100,000 0.0222) = $2,220/month THE CORRECT ANSWER IS: BFE CIVIL SOLUTIO ais INS rhe easiest way 10 solwe ts problem is (0100 athe presen wor fx 0. Of each option ; esent worth values are all given By 0 mr pirst Cost + Annual Cost » (4, 12%, 8) —s, ) Salvage Valu ? inst Cost + Annual Cost « 4.9676 Salvage Vitae 2 le 0.40390 on PA) = $63,731 hen) = $63,392 p(C) = $63.901 P(D) = $63,222 ows are all costs, so the {wo most preferable projects, those with te D, and the difference between them is $170, lowest present ‘The eash flows are 3) worth costs, are .CT ANSWER IS: B ‘THE CORRE! 2. _ Refer to the Engineering Economics section of the FE Reference Handbook $5,000 i= 10% EEE IEEE 2 ce fecreey cure ’ 500/yr $7500 4 A A= A+ P41, AG. A (4 a) v(4. ) = 500+7, so0( 4 10%, 10) -5.000( 4 10%, 10) = 500 + 7,500(0.16275) — 5,000( 0.06275) = 1,407 per year IE CORRECT ANSWER IS: C ~~ yy ia co2, 23, FE CIVIL SOLUTIONS $1.50 (5,000) = $7,500 $80.50 (5,000) = $2,500 ‘Annual savings ~ $7,500 ~ $2,500 ~ $5.00 Additional investment = $15,000 ~ $1,000 Payback ~ $14,000/S5,000 = 2.8 years $14,000 THE CORRECT ANSWER IS: B Refer to Resolution of a Force inthe Statics setion of the Re R= Dh Re Dhr 1 212-129=083N 123 Ry =2.12+ Seos 105 124483=695N Ry =212+ Ssin 105° In. +R,? =Vo. ‘THE CORRECT ANSWER I 6.999 R 22a = 3(a) 9-37 3 4 (260) + 3(300)= F(260) + $(300)=140 peo fre WE 5707 1407 R= 396N D THE CORRECT ANSWER I 78 zference HandbookFE CIVIL SOLUTIONS ems of Forces i the States section ofthe 1 Reference Handbnk. Ux 10 the Syst ody diagram w | 4 4 f I B x sy Since free £ vets =0 rd ’ ' ’ THE CORRECT ANSWER IS: C ”FE CIVIL SOLUTIONS 26. Zero-force members usually occur at joints where members are aligned as Follows, 4 or tc That is joints where two members are afong the same line (O% and OC) and the third mens, sense ray angle erate a zero-orce member. Tat member (OB) 16a 760-ere nea because the forces in OA and OC must be equal and opposite ‘lem, we immediately examine Joints B and E For this specific pro rp B , BE ¢ Cy, A “ \G 4p BG is zero-force member CE isa zero-force member ow, examine Joint G. Since BG 8 zero-frce member, the joint effectively looks like d 4c ka ¢ therefore, CG is another zero-force member and, Finally, examine Joint C. Since both CG and CE are zero-foree members, the ovat effective looks like >, & 8 - and, therefore, CF i another zero-force member Thus, BG, CE, CG, and CF are the 220° members ‘THE CORRECT ANS 80FE CIVIL SOLUTIONS ne twaton of econ am mesa the direction parallel t the ans is given 1 fxd where d4=(22 — 4 4 Ls \an fil jane. Normal to the Pl i mN (at impending sli Tangent sy <0: _ mg sind+ HN = amg sin g png €08 © =O 2108 tang =H cose tang = 025 THE CORRECT ANSWER IS: D Poe a1Saga nEn nara FE CIVIL SOLUTIONS (60 fusec)? — 2 (12 fU/sec*V(140 3,360 f?/sec” = 3,600 ft/sec" =240 ft/sec? vy = 15.5 fsee ‘THE CORRECT ANSWER IS: C 430. Let subscript 1 refer tothe ball and subscript 2 refer to the block. mm tmvy = my a""2 mz =2m, ‘my (10 ft/sec) +0 = m1 fusec + 2m (4 flsee) _(=¥)_ 4-2) €= y=) (10-0) ‘s1. Refer to Impact in the Dynamics section of the FE Reference Handbook 1 Energy may or may not be conserved Itis rs ray may or may not be conserved. It is conserved only if the coefficient of restitution S Momentum is always conserved Is: D THE CORRECT ANSWEPerrier ae err een ee ae AAT Eneagy yy Ne Dynan Mamieg etn vty thee hy NY Wie iad : ‘ Wray 0 oo z 1 : bmw Uh= mghy 1 Urea Sw nay i. Sm m2 8)=35 Ye JOR 7g ns THE CORRECT ANSWER IS: B om x “ IOkKN 6KN : , 6m : Area | 1302) 26kNem : SHEAR DIAGRAM Area 2 (6) IS kN Area} 444) 10 kNem MOMENT DIACKAY Maximum mapnitude of the bending momedt ts 26 KN THE CORRECE ANSWER IS: D 8334, 35. 36. FE ci, SOLUTIONS al Press ence Handbook ne Vessel in the Mechanics of Materials sec The eyling c mm. Thus “2” bE considered thin-walled ty 402 Io this 86.1 12 mm ang Pr 1 cre r= litfe 35043 el eee ee (1.680 MPa\356 5 6 SENN 56 my sre 12mm THE CORRECT ANSWER IS: B mM nd _ x(0.2)*(840 10 eo 1,319Nm THE CORRECT ANSWER IS: C EP =PA=(I4e10") (* rn Figg = 275 KN = 04 = 6810" A= 404104 4 P= LAMPs SS 1 ‘THE CORRECT ANSWER IS: A3 8 » FE CIVIL SOLUTIONS fo Torsion mn the Mechantes of Materials section of ‘ection of the I. ef ference Handhiok Refer t 1 7 jax twhen P=" mi 121m 0.196 in? p= 1.30010 5 =0.009in PE. searranged gives aE 300X12 (1,300K12)__ g 84 10° psi = 8.840kst PL Fa ~ (@.009K0.196) que CORRECT ANSWER IS: C mies of Material section ofthe Weference Handhos sefer to Mohr's Circle in the Mecha is Cirele, the maximum inplane shear stress 1S Ts R From a constructed Moh a-V500 JOA T si n ME CORRECT ANSWER I 8540. a. 2 8 44 FE CIVIL SOLUTIONS The Euler formula is used for elastic stabiit cof elavely Fong columns Subjected 0 con, axial loads in compression : THE CORRECT ANSWER IS: D ce at the jobsite) Was as r It is assumed that the conerete wie ratio at the plant (and hence at the J ) WAS AS Fequired fi, proper strength THE CORRECT ANSWER IS: C ‘The moisture content of each aggregate cludes (1) water that would pe needed to b aggregates to SSD condition (the absorbed vate?) snd (2) the excess water that 1 fee toad, Tere water Since the as-used moisture content greater than the absorption for er to the mix, thus reducing the water aggregate, each aggregate contributes the excess VA nae ded wo mix. The water added 1 the mix the Water computed for over-dry ageremrs {G05 Ib yd’) plus the excess water in each ageregate Final water = 305 ~ [(2.0% ~ 0.5%6)/100] » 1,674 ~ (6.0% - 0.7%)/100] = 1,100 = 221 6 hyd ‘THE CORRECT ANSWER IS: B eter to the defiion of Charpy’ Impact Fest the Materials Science Structure of section of the FE Reference Handbook. — THE CORRECT ANSWER IS: D ‘By definition, a metal with high har ith high hardness has high tensile strength and a high yield st ‘THE CORRECT ANSWER IS: DFE CIVIL SOLUTIONS = 0.204 m rie CORREC ‘erro the Fluid Mechanies section of the M1 Reference Hundbon yo Refertoth Handbook dynamic viscosity (H) mits of absolute are ky/(m-s) Units of kinematic viscosity (9) are mis rhe relationship between the 10 Is yop where pis the density m kg/m” = 1 511263(1,000) = 0.001188 = 1.19 10° THE CORRECT ANSWER IS: A sforence Handbook P. Referto the Fluid Mechanics section of the IE: Re THE CORRECT ANSWER IS: D ~~ *48, 49. FE CIVIL SOLUTIONS waluated at the mean height, an The mean pressure ofthe fluid acting on the gate 1s evaluated af MF mean heighy, ang, of pressure is 2/3 of the height from the top. thus, the total 10,610 N 3 Fy = pe (11) =1,600(9 807)3(3) = and its point of application is 1.00 m above the hinge A moment balance about the hin F(3)- F,()=0 _ Fr _ 70,610 23,537N THE CORRECT ANSWER Use the rational formula Q- CIA Ocfs Q= (0.10)(1.5 in /hr\20 acres THE CORRECT ANSWER IS: A (AE) (200° 2 2 A 2-322 (2001 Inne = 008 2 gs} orson 026)= 0052 n how =hian ~My yo = 0186-0052 =0.134 8 Since the head loss from A to B is larger than the head loss f __- than the head at B This means there i low omc roe OM ® £4: the head at C8 ‘THE CORRECT ANSWER IS: B BRsi FE CIVIL SOLUTIONS nd velocity, 21) is proportional tof Forth ally turbulent flow conditions. Referring to the M fox ane same FTE covesporaing th mnamber 86 OT ne Fluid Mechames 6c of the / Reference Handi in Da pe el me PE smn os se CORRECT ANSITERIS A 238 (4 o1(0.625)™(000878) o-F0r8 Q-25 12h THE CORRECT ANSWER TS: vation difference The 34h THE CORRECT ANSWER IS: B tt eee55. FE CIVIL SOLUTIONS Pressure difference between the sixth floor and ground level (2416) 12 )/_1n? =[ 5 | 6 floors « 1A? )o312 wi Bay) st Pressure on sixth floor ~ 85 ~ 31.2 ~ $3.8 psi THE CORRECT ANSWER IS: B Solve for h. 10 fi/see = 1.318 = 150 » (12/12/4)" "(9.000 ft Solve for H. 10 = 82.5 (x/9,000) hh 180.7 THE CORRECT ANSWER I Q =ClA 4 = 2-2 = 226 acres 1 065«6 Catchment Area = 2 26 acres x 43,560 fA /acre = 98,400 ft? 100% 98,400 _ og f 100 4 THE CORRECT ANSWER IS: 4 iFE CIVIL SOLUTIONS senard Tip Had (unit fad) a Frnt Cand compu Fores in a ty seing each F force by 40) fan bars. (This can easily ve dom H svi eh emember = change in length AF = 2 by its fee Fim the table bso (be sue w FL ned values of AL =“ and f) use the silt to get the displacement at Joint C sen sum pth = + 1.0464 in, down SAE 7 an i FL vember | F¢Kips) Lain) f SE —aB 30.0 240 0.1034 125 01 BC 492 473 0.2008 1231 02472 | ow -750 480 03103 1875 0s8i8 AD -30.0 288 0.0745 00559 BD -250 240 | -00517 0.0323 Y= 10864 THE CORRECT ANSWER I: aFE CIVIL SOLUTIONS Unknown reactions and internal forces at internal pins. 6 « 2 ~ 12 etd body components (shown as FBDs) @ 3 equations per component 4 Number of equations: 4 x 3 = 12 ‘There are 12 equations and 12 unknowns -» determinate (stable By member sang, g - ' { wont = 6 @ +@+- ~ ze u (10) (7) Rtat sony 1) UNKNOWN FORCE ‘THE CORRECT ANSWER I 92FE CIVIL SOLUTIONS RATED LIVE LOAD UNIFORM LIVE LOAD. GE UiroRs! DEAD LOAD CONCENTI —Simplespan | esrorttTVEE ” | epee ied a of ‘Multispan UNIFORM DEAD LOAD, | PiONCENTRATED LIVE LOAD UNIFORM DEAD LOAD ‘CONCENTRATED LIVE | Bi [ featend—] LOAD EE SHOWN ABOVE. IIE CORRECTLY LABELED BEAMS AR a os60, 61. | FE CIVIL SOLUTIONS The maximum force in CD occurs when the wheel 1s placed at the the peak of the i ocati influence line they, Max Foy = 1.875 x 20 ips ~ 375 kips MN VAININVAVA| Influence line for CD THE CORRECT ANSWER IS: C “w oO =FE CIVIL SOLUTIONS ofa cantilever beam, it 18 mot statically ind Indeterminate, it compietely 1y supported, definition py defining only at a specific point 6 gd tis 108 ECO! ga Meso! computation) For bracing @ 15 [100 Mant Cs 10 (given) from the Za 12le, the Civil Engineering section of the / Refer From 157. p= 484 fees Tp 477 L143 BE free Hn since Lp <28= 10

-@i)M2L)].° 24 00-+ (6) «$+ 14) (-6)/ 2 8S = 209.63 fi THE CORRECT ANSWER IS Try equation for $<. as? 100[ 2h + Jame : 9600" 1o0[ y2x330+ J3=1.00) = 1,966 ft S'S Las assumed. Therefore, the correct equation was used ‘THE CORRECT ANSWER IS: B 40.8333° T= Roan (£2) = 600 tan 5 = 22334 x =p = 133° = = = 427.61 1= RIE = 600 = 40.8333" ~ 765 PC= P11 = 20+00 00 ~ 223.34 = 1747666 PI=PC +L = 17+76 66 + 427 61 = 2240427 ‘THE CORRECT ANSWER I 1028. 8 FE CIVIL SOLUTIONS angles are based on a vehicle st ible gaps. topped at the stop har and | gaure sieht part cles for accept a aor sireet THE coRRECT ANS 5(4) + 0.1009) 44(2) + 0 ation with asphalt willbe thicker under identical condito alt or any combin: asph ‘qHE CORRECT ANSWER IS: C mbined CR are not additive, 80 60% sRs from highest 10 lowes!) Crash factors cR=CR) + (1- CR) CRz (order C 25+ (1 -0.25)0-15 036 1 singg no change n ADT Crashes prevented 3610-036 10-36-64 THE CORRECT ANSWER IS: D — 103FE CIVIL SOLUTIONS H Dismee- sya] tee 5of06s—0 10] 151.52 + 110.25 =2618 +1.47(50%15) THE CORRECT ANSWER IS: D 3,600 sec/hr oo 2.57 sec/vehic PCIE eicesTe em ‘THE CORRECT ANSWER IS: 4 Pvc wrix-wtnptcranspect-ston eet 6+ 12535-1580 THE CORRECT ANSWER IS: B sr. Refer the Chemisty section of he erence Handbook forthe equilibrium conse! chemical reaction. 4a+Beor2C72D ‘THE CORRECT ANSWER IS: D 104FE CIVIL SOLUTIONS pesiden® 8 79.639 tater vote =~] ( swab ae view -[ agent CN day af} 79,6391 br 2.86 hr eadens© 155 957 A HE CORRECT ANSWER IS: € gp 2 moles of AS 75 gimole of As = 150 8 of AS THE CORRECT ANSWER IS: C r= 100 #215 gpm A (12 min) 18,000 gal THE CORRECT ANSWER IS: C BOD; = 200 mg/L 1S days 4-023 day 10D, = BOD,» {| Wome. = BOD yy Wb =20/(1-« 8) pops Dj, = 292 7 mgd m4 HF CORRECE ANSWER IS: ( ce hd93, 94, (£0min’ he yy {Omin'h \Smin/ truck Jl Soap = 18034 hr truck J THE CORRECT Answer 60 min/he Production of Excavator 3 *1 75 LCY capacity 050 min eyele Yew Similar cal ther three. 2OLCY he culations for the o exeava ee ie bee THE Corr ECT ANSWER IS: ¢ 106FE CIVIL SOLUTIONS ght of embankment material = 122 it wel pry unit 108 5590 7 5(320.000) = 315,470 ya HE CORREC T ANSWER IS: B aw of Cosines Uses 2be cos A gehre-2 480? +7207 - 391 08 A= 4804720) A =30° 18°47" THE CORRECT ANSWER IS: & st Refer to Eanhwork Formulas in the Civil Engineering section of the FE Reference Handboos Volume to be excavated = 130[322 + (4395) + 418} [(6)27)] = 1862 »¢ THE CORRECT ANSWER IS: C 2 107 Blc sare 2 | 27 79:16 io LL SILISN \sF 1050723 F oF oo KepSmas wy Bun tisttas hasnt = 2% 8155328 ig Bg ~ 1080723 ~ 8758.82 = 175191 Tat (yy | Dep(e) | __Coordiante_| (eos6)| (Lsin® | EC) | NO) 75532 | 1125061 908.94 | 1171330, GOT 14 | 116TH 10,507.23 list) 15 =0.12 ALat= 11,511.27 THE CORRECT ANSWER IS: 108oases Po rr the fractional part of the circle jee wt ye = 30788! _ 36.88 gration of cHrCle 300° 360! = 0.1022 sgeqof fastion of eitele = 0.1022 x m x (500)? = 80,285 92 shaded area = total atea ~ fraction of circle = 83.164 ~ 80.285 =2,879 7 THE CORRECT ANSWER IS: A. 109PREPARATION MATERIAL ei EXAM BLISHED BY NCEES PE Reference Handbook PE Practice Exams for all modules. Chemical civil Electrical and Computer Environmental Industrial and Systems ‘Mechanical Other Disciplines ‘or more information about these and other NCES publications and services, Visit us at www ncees org of contact Client Services at (800) 250-3196."cees.orglexams