cee166-fa07-mt2-Horvath-soln

cee166-fa07-mt2-Horvath-soln - CE 166 Construction...

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Unformatted text preview: CE 166 Construction Engineering Fall 2007 Exam 2 L Name: Jenna th%= 1. (Paving) Name and briefly describe all the steps in constructing rigid pavements. Name the materials and equipment needed in each step. Transportation should appear several times. Start with aggregate mining. When a step has more than one technological option, list all options (e. g., for paver choice). (30 points) 0-4 Skwe\$, «Alfie's, QWVGWW ben's (move aqojreqfi’d 6‘00“ ‘VW $9,) not «em; qood Mouj“ I. A re NR mmln ~ mme 9mm smurce V at m hmsfifl fl 2 b WW“ 2~ TFQESP§V+~ move R39: source flaggedch urea \I\O\ £cgfj' V V 0 3. wannan = rennng cut, MM, 01— mqou’hc mom—em “+1490; of UJQSh—QQS age log. wag‘hers and clasg'fiflma “MKS Afijrgflqk Li- Feedea: {>sz Malcemds \-o Ck CRUS‘AQK (umwes a $‘eedea matching)" 5. CRUSle : ldecreasw We SUN. 0? Olga/(Pattie 39% breakmgvfi’ MW“ m4“) .... Elm“? Vvoduchm um“ “11 ' I 1...? ananr : Jaw, ewe/swam“, I Impam— Mushea /H-o\mvnek. MM .4: gkwndmgf 3mm eater as ngmwtéf Tekhamfl; Roll, Rod Hill , Ball PM“ (o. Screcmneyz gepam’nm 05— flfiwgaifi m5 31% (in 3011\— m’r\a\r6¢ PlQCflQ L) Vt\mhn% am PQVOWngr- 5Ue% 6W cam use 4,0 warm woman gm vm. ert‘pg ska “W‘s . Tr 3 01+- 4 e r, $320}: M661 re one eeds to be we“ new on S”? t“ tee"P addmwm Wm" W W “m M” "“XWJ ' if “Tam/*3 Commit is DoTChed and Mixed elm using ox ace/ARM baJfl/‘WW (3* WM“ 0R. ‘b bath a . o c e mme Manet J . - g’ < mvek MM 3 Mmem. 6MP“ $-07 Coarse and 9m; aqqrcqme, WARV, and am, qddflnveg “9* such ag our entmmmay Mem tr accewmmg> _ ox. Trampmnng. Concrete to 9H6 or QWek no’r batched Mme) L» m ck mm no. Q6“ "‘9 m" be Cam- Rmva of 0M «wreaked (Pavement) WM 3”“ “Mme/“3 Whacme‘é m we S‘mu‘hneo‘s‘fi 4km” mmmm «wow \Im WUCK} ‘9 “AW “moumsymm Med gmms \oad «tum? :1: *“e “‘90” \2. Mar concrete \9 delwerea, ymoemen’r can begin usm “mm‘ PS wheel barrows, buqqlee, chum, pumps, or cmneg w/ more’re buckefia Meihoé derAs m 9R l \ ' - ll. Credit {grimwmk ‘ 0‘“ WWW 0)‘ wacemem. mph“ \3. EPme POWQ “mgr {pgmvhdin% 9%uspmewl- m 3h"; {Wm Fave): La. MW morale H) we <gzé~\\\{gm‘-\B «t CE166 Fall 2007 Exam 2 ' Page 1 of7 2. (qugipmentm conomics) a) A crawler tractor costs $20,000 as an estimatedmgf $30,000, 1' andgié year life sing the sum-of—the-years’—digits met 0 of depreciation, find r s yeaflTW'w deprec1a ion and book value at the end of each year. Fill in the values in the table. Show your work. (15 points) - D" ~— (200K «30KB = 5 (la/(0L0? .4 .4 - -1_.‘_-_.._~.._-,.‘ ..._..M...._r..._ l D2: "‘ (zoon~z.m<): H5333 L200K,~SDK> = 3"hOOO l5- D6= _1_ 0,00% *5OK3 = “1533 x” is b) If you wanted to get the largest amount of depreciation on your equipment in year 2 of its life, what method would you use? (2 points) IRS Wrescmbed MPH/10d ,-/“ 0) Which method of depreciation allows you to bring the book value of equipment to 0? (2 points) I Rs PreSCRled Hei'md CE166 Fall 2007 Exam 2 Page 2 of 7 /7 Find. WW arcing, 3. (Formwork Design) a) Determine the maximum allowable span of nominal 2” x 4” studs (measured along th studs) for a wall form made from southern/pine carrying a design load of 2,600 lb/ft2. The stud spacing is n center. Assume the studs are continuous over Ws. Limit deflection to 1/3 60 of span length. (14 points) ___—9 \Z " “Bending 1:10.05 as 3 ma (‘WFS‘)(3‘0‘°3‘“3> : lB,-’-\m\ a SMAR” Q: \3-3 FVPv + 2d : 6.3 (\%0@'\)L5.25m‘) + 2 (3.53 : \2.8 m 0 CU: = w 2mm w!er I ° DfilecW‘ Jt= Hm E1)"; : Ho“ U-UWX‘535C‘3 /: zufl'm, (75 21mm ghemfi defovems with a max spacing, of WWW?“ l— wound \ASQ a \2m spacing for mooumm—g \z' b) If in a typical wall formwork design you found out that the steel tie wedge would crush the wooden wales, what remedial design solutions would you recommend? Name 5 solutions. (15 points) a . v I \ l) chomp UJOUA +1899 0f wales Wt “1* lHHCK («UM ;/ K ' 3‘ V} 2) wa\eg ’m SR9‘ H" Kt: ‘l’iLLt‘N‘J-Jné" 3 a) mchge m spae‘in8\ "atrt.,(_my ( ‘3’) ‘0 change 0f Q’rGM fur he wedfixg *1 5) mgrease Slit o; Wale t. CE166 Fall 2007 Exam 2 Page 3 of 7 4. (Tunneling + New Technologies) How would you apply two of the new technologies you have learned about in this course to a tunneling project? Why these technologies? Be specific, but do not write an essay, but rather list and describe things concisely. (22 points) ' \Neambw Cmporem I I This WWU be veng Use’fol m mme MOW '5 cm‘l‘meé‘ M Mam“ PM “%“‘8' m 'cevm'm swarmsams CM be MWB “me Mm ax MM“ °§ “MWS “M” wept/me lines,cab\e, WAKE “YRS. Sewev migems' deemed lines. and gas “flee”; When TunneVrfls it is "mPnrmw‘r not to Virt— oww) 05’ These Majw \ivws because H—nm— emu meafls sewn diGRUP'hM bvl- Possibm tapas. wen Mesa CUMMrSI worKeYS Mfl WV: 0L aback source to U93 and mm,“ mesfl WM‘QS. The “momlr of emqu “Beded to check is less WAR having 11) find jhe blufpemt Loo Kit/13 of) W mf'o,GVld fiolngmck 10 silt 053 wom.Wsa CDMPUKWS Wm «we m mahm W mm: my“ Precise lessening mm of pmblemg. / a Malone \msF-ee’flm Assignier when pm 'msPteor needs to «at WHO 4 WM‘ Tb check weld abuahlg of- le‘mg )3)? example, Mi pevsm has 0x number of TASKS W Go. NH3r “NP‘S‘ vase gnome, mend mmth mam access documents (Specsfimwlnfigmanums), TWS All-ilk, gKel-ch \Ucahms, AM, \Me'rfuc/a WI other mgpemm mm‘°‘°%~ one fichnuoraxa, anemones the mauous was 051 inspect-30w Sou, making 9% were egfmlent W‘AM an \nspechy nears ’rv cover or Mqu amoum- of wnM‘ l“ 0‘ dw' h’e/S‘M defi‘m’rclg warm: to evmmwe wokst ’nme 939A {‘7 347 find Me dmwmflgs,%dr a Gav/hm, am aux/me loam: and foam EXTRACREDIT: (“Q/QM U5Ua‘\\j Conducfit&. Why would the external costs of asphalt concrete pavements differ from portland cement concrete pavements? Name and explain 3 external costs and reasons for differences. (12 points) 6) Repair costs will be higher {P at cm max quhaw \3 used inson of cmmzth CW mix \9 mt as. show an M“ mmwj vesuw m mom remiss/l @ flaoemem— of angw lg aovickek than CAMC/YQK. veezwrse umcHMlnme 0r owner time. Espenauu if WM Use or Gold mm, the amount mt lime a lame (Q closeA is «warm, 1mg use ’hme WM “994 l” Wm M ijdum 053166 or wh‘w‘c mmuem/ _ V ‘ me +0 Cmslwcfim wide“ (3) Salem MSG Cunele \owm Wu cruncegoffimmvafl Angslcvmpamd l‘o V‘VASphalt since it doesn‘t lm/Olve Maw mare/Wk (“Rm flacemm. CE166 Fall 2007 Exam 2 Page 4 of 7 Table Source: Nunnally, Construction Methods and Management, 7th Edition, Pearson Prentice Hall, ISBN 0-13-171685-9, Page 378, Table 13-5 Table 13-5 Concrete form design equations l Support Conditions é Design Condition 1 Span 2 Spans 3 or More Spans l Bending l i l ' 1 Wood 6 = 4.0d‘(—QQ ’5 c = 4.0459)” c = 4.46d(f‘£)é w w w E l c=9.a(fb§ % 6:9.8 (EV (1: 10.95 (52—3- ‘A E I w W w l l Plywood e = 9.8 (fbK—S)‘ e = 9.8 e = 10.95 ( FMS)” l w w \ w r Shear i F A F A F A 1 Wood €=16“§7‘+2d e:12.a-fi+2d case-find 1 Plywood (=24fs-’£/—O+2d 6:19.2fflgiq+2d c=2o§1£59+2d I Deflection e = 5.51 (39)“ e = 6.86 (fly 6 = 6.46 (fly w w w "A: mo 6: 1.72 (5')” e =2.31(15—’)"6 e=2.13 (5)5 w w w |tA=‘A.o e: 1.57 e =2.10(§)% €=1.94(§)% w w w If A = is“ e: 1.37 e :1 .83 e = 1.69 w w w , P Compressron fa or fcl = 7 p Tension f,= 7“ —_—_—_——-—————-—-—-————- Notation: C = length of span. center to center of supports (in.) F,, = allowable unit stress in bending (psi) FbKS = plywood section capacity in bending (ib x in./tt) Fc = allowable unit stress in compression parallel to grain (psi) F“ = allowable unit stress in compression perpendicular to grain (psi) Fslb/O = plywood section capacity in rolling shear (lb/ft) i FV = allowable unit stress in horizontal shear (psi) i fc = actual unit stress in compression parallel to grain (psi) f“ = actual unit stress in compression perpendicular to grain (psi) I, = actual unit stress in tension (psi) A = area of section (in."’)' E = modulus of elasticity (Psi) l = moment of inertia (In. )' El = plywood stiffness capacity (kPamm‘lm) P = applied force (compression or tension) (lb) 5 = section modulus (in.3)' A = deilection (in.) l b = width of member (in.) d = depth of member (in.) w = uniform load per foot of span (lb/it) ‘For a rectangular member: A = bd. S: bd2/6, I= bd3/12 CE166 Fall 2007 Exam 2 Page 5 of 7 Table Source: Nunnally, Construction Methods and Management, 7th Edition, Pearson Prentice Hall, ISBN 0-13-171685-9, Page 382-383, Table 13-7 Table 13—7 Section properties of U.S. standard lumber and timber ([9: width. d= depth) Nominal Size Actual Size Area of Section Section Modulus Moment of Inertia ; (b x d) (545) A S I _______._ _._______ __._____._ t I in. In. mm In.” 103 mm" in.1 1a5 mmJ In.‘ 10‘ mm‘ 1 x 3 0.75 x 2.5 19 x 64 1.875 1.210 0.7812 0.1280 0.9766 0.4065 1 x 4 0.75 x 3.5 19 x 89 2.625 1.694 1.531 0.2509 2.680 1.115 I 1 x 6 0.75 x 5.5 19 x 140 4.125 2.661 3.781 0.6196 10.40 4.328 i 1 x 8 0.75 x 7.25 19 x 184 5.438 3.508 6.570 1.077 23.82 9.913 1 x 10 0.75 x 9.25 19 x 235 6.938 4.476 10.70 1.753 49.47 20.59 1 x 12 0.75 x 11.25 19 x 286 8.438 5.444 15.82 2.592 88.99 37.04 2 x 3 1.5 x 2.5 38 x 4 3.750 2.419 1.563 0.2561 _" 1.053__M*g.§1*2‘0 a 2 x 4 1.5 x ,_.,_§§_>g§g mo; 5.359 2.231 2 x 6 1.5 X 5.5 38 x 140 8.250 5.323 7.563 1.2 ‘ "‘“ébTé'dw‘ ‘ "“"“'8’:656' 2 x 8 1.5 x 7.25 38 x 184 10.88 7.016 13.14 2.153 47.63 19.83 2 x 10 1.5 x 9.25 38 x 235 13.88 8.952 21.39 3.505 98.93 41.18 2 x 12 1.5 x 11.25 38 x 286 16.88 10.89 31.64 5.185 178.0 74.08 i 2 x 14 1.5 x 13.25 38 x 337 19.88 12.82 43.89 7.192 290.8 121.0 3 x 4 2.5 x 3.5 64 x 89 8.750 5.645 5.104 0.8364 8.932 3.718 3 x 6 2.5 x 5.5 64 x 140 13.75 8.871 12.60 2.065 34.66 14.43 3 x 8 2.5 x 7.25 64 x 184 18.12 11.69 21.90 3.589 79.39 33.04 3 x 10 2.5 x 9.25 64 x 235 23.12 14.91 35.65 5.842 164.9 68.63 3 x 12 2.5 x 11.25 64 x 286 28.12 18.14 52.73 8.642 296.6 123.5 3 x 14 2.5 x 13.25 64 x 337 33.12 21.37 73.15 11.99 484.6 201.7 3 x 16 2.5 x 15.25 64 x 387 38.12 24.60 96.90 15.88 738.9 307.5 4 x 4 3.5 X 3.5 89 x 89 12.25 7.903 7.146 1.171 12.50 5.205 4 x 6 3.5 x 5.5 89 x 140 19.25 12.42 17.65 2.892 48.53 20.20 4 x 8 3.5 x 7.25 89 X 184 25.38 16.37 30.66 5.024 111.1 46.26 4 x 10 3.5 x 9.25 89 x 235 32.38 20.89 49.91 8.179 230.8 96.08 4 x 12 3.5 x1125 89 x 286 39.38 25.40 73.83 12.10 415.3 172.8 4 x 14 3.5 x 13.25 89 x 337 46.38 29.92 102.4 16.78 678.5 282.4 ' 4 X 16 3.5 x 15.25 89 x 387 53.38 34.43 135.7 22.23 1034 430.6 6 x 6 5.5 x 5.5 140 x 140 30.25 19.52 27.73 4.543 76.25 19.52 6 x 8 5.5 x 7.5 140 x 191 41.25 26.61 51.56 8.450 193.4 80.48 i 6 x 10 5.5 x 9.5 140 x 241 52.25 33.71 82.73 13.56 393.0 163.6 6 x 12 5.5 x 11.5 140 x 292 63.25 40.81 121.2 19.87 697.1 290.1 6 x 14 5.5 x 13.5 140 x 343 74.25 47.90 167.1 27.38 1128 469.4 6 x 16 5.5 x 15.5 140 x 394 85.25 55.00 220.2 36.09 1707 710.4 CE166 Fall 2007 Exam 2 Page 6 of 7 Table Source: Nunnally, Construction Methods and Management, 7th Edition, Pearson Prentice Hall, ISBN 0-13-171685-9, Page 384, Table 13-8 Table 13—8 Typical values of allowable stress for lumber Allowable Unit Stress (lb/sq in.)[kPa] Species (No. 2 Grade, 4 X 4 (Moisture Content = 19%) [100 x 100 mm] or smaller) Fb F., Fa Fa F, E Douglas fir—larch 1450 185 385 1000 850 1.7 x106 1 [9993} [1276] [2655] [6895] [5861] [11.7 x 106] I Hemlock—fir 1150 150 245 800 675 1.4 x 10‘3 7 V £9291 L1034] [1689] [55115] [4654] [9.7 x 106] L Sgythern pine m 9.2.5 825 1.6 x 10“ I ' [9653] [1241] [2792] [6723] [5688] [11.0 x 106] California redwood 1400 160 425 1000 800 1.3 x 106 [9653} [1103] [2930] [6895] [5516] [9.0 x 106] Eastern spruce 1050 140 255 700 625 1.2 x 106 [7240] [965] [1758] [4027] [4309] [8.3 x 1061 Reduction factor for 0.86 0.97 0.67 0.70 0.84 0.97 wet conditions Load duration factor 1.25 1.25 1.25 1.25 1.25 1.00 (7-day load) _-_————nn_—-———————-— CE166 Fall 2007 Exam 2 Page 7 of 7 ...
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This note was uploaded on 12/01/2011 for the course CE 13972 taught by Professor Chow during the Spring '09 term at University of California, Berkeley.

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cee166-fa07-mt2-Horvath-soln - CE 166 Construction...

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