EM319.Fall2007.Test1.Solutions

EM319.Fall2007.Test1.Solutions - go Luke-m S EM319 Fall...

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Unformatted text preview: go Luke-m S EM319 Fall 2007 Testl , Tuesday, October 09, 2007 3:03 PM Two cast iron pipes are held together by a bolt, as shown below. The outer diameters of the two pipes are 50 mm and 70 mm and each has a wall thickness of 10 mm. The diameter of the bolt is 15 mm. What is the maximum force P that this assembly can transmit if the allowable stresses in the bolt and the pipes are 200 MPa in shear and 150 MPa intension, respectively. .811: Ila-‘72 V d~ (OOH’ dtO-MA_ In , .3 - '2? l ‘ 25512 Ta?— 3 Li SM! E l 0 A0 fir: M440 Me ftv Ma =7 lMM .— 6-“ I000 } dawolo 5N l —€,0‘.c_(/\ W . 3 lea Tube PM“ : Um AL I i Probleml; Shear Failure ; ta=200*106;d=0.015: M Oct PmaxB: 'Ca*7l'*d2 /2 0mm: 70685.8 70‘ 7 h[6]:= Outer tube failure; i l l i i i ll aa=150*106;d0=0.070;t=0.01;di=dO~—2*t; [WWW ' i E l E l l l l l l l A=7r* (dOZ-diz) /4-2*d*t; Pmaxo=aawA Oui[9]= 237743. inner tube failure I a) mot: ca: 150*106; d0 =o.oso; t=0.01;di =dO-2*t; A=7r*(d02—di2\/4—2*d*t: OK+ Class Title C Page 1 inner tube failure a 7 DW (AC—WV " (JIM-5 h[10];= oa=150*106; d0 =0.050; t: 0.01; di =dO-2*t; A=7r*(d02—diz)/4-2*d*t; Ok+ Pmaxi = oat-A cum: 143496. 5 % Class Title C Page 2 reek " _ E E Mean Mathias Q: 200 Aux): 2 + (3-2) : 2 + X/teo n x ' § 20x12 126:): 1+ ></€o . -. 2cm” 3% Cogs sec-H6 («on are“ t: O‘I‘L€u A0): 2.1 chvc 304mg} i . ‘ (a: [(1406 .1 //// 2 ? Canada ow eiewewf olx 93' X _ N: 80““ + magma 1, w amt m wecm I ‘ g " m QC'tcolA x. __L.. wg‘éjk—fi 96C+cdvx x wx = XQFLfi H+ X/gozoex = 2R15X£x+ b W lmkmdfivm NCX} on A): \é, 3 MLx3= w + Nx a w+ QHtXLX‘r‘xl. Cow‘siaci'w.“ avg Hz“ e‘ewmfi &5( 01 g: N06 dx /(5 Acxfl ‘_ Mwmwi Ase—«E 12501 um zmget ><+7 .— ’— Zfl’t LH- x/flol. Cv'waéc‘éa W} ; E ACXB I i a M“; NZ? A00 ~ methm éH‘ééé (7U (Pier : v; ; a ; ' wwwmmwmwwwwmmeflw 3 , l .1 is mmgflfiwMMMMWW.‘ r HAW“; a E 3. TIM. mechaw'sw 44%; ' I ____._._fi4_:z.ia_(5»9sét 5 $0 7 l Mi ; 2N: + 5N2. -- é-SP=O.®) 30 we see Hindi We [Aowe 3 WkaNA read—ths I ‘ omh +wo ydeuaud' £945 quJru-{ms —.=> s: D. We goons 0M “1: CLon NE {fluod'w-‘A 4” to r (e, - meg—Ct N E ILtAil'mes m bow V > % Em EC, jg: NEL + «AT,L__ {a ' @(owfl L EA I Boa 1;? Se; a Na. * gee), QM“) { NF sham flu: bow EA comazlodch . . 26% A365 rs “NJ, A55uwue ‘4' &e. mg as .SEL 3 ! i M . 3p ’2 _. .__,.-r ‘ N... .. ._V A v 1 ' é o ' ' Z k... ' - g g — E From “42+; , g _ , 6 2 V s. I ,NF +5NE _ . (“NF 0 7 . = P- _ .. .. 73 Mr— .‘ .V '0 . NF WNAWSW£MN “u w~>l~‘e. N Wv‘w WW: Wflmmuwwmmfimfid H l\) \n Z 11 3-3 0! MPa (Eugen-M 3'7 § EMMA”MM“wwmmwfimwwmmmwmwmwmuwwm. V x { I vwmmmmwwfinmmmm> . ' i : n ,WMWW‘WWmmiéuwmmmwmwwmwmmwwwawmzV, ‘ \ I , _ Q . V WW,_WW.MWWW;4mmmwmmm .r ‘ “WWW . womflme *WQ 9-1;. .WWW).WWW».wmwmwnéaemmmmmwmmmwmmammWMWMWM ‘, , , ' g ' , ummmxwmmmm. "WW I (no ‘4 mmmwmmifimeme Wat” 5* ‘ . V . “M www.mmw‘wmw Problem 1 ; Shear Failure; ta = 200*106; d: 0.015; PmaxB: ta*7r*d2/2 70 685.8 Outer tube failure; ca: 150*106; d0 = 0.070; t: 0.01; di =d0-2tt; A=7r~k (dOZ-diz)/4—2*d*t; Pmaxo=aa*A 237 743 . inner tube failure ca: 1501.106,- d0=0.050; t=0.01; di =d0-2*t;A=7r* (do2 ~di2)/4-2*d*t; Pmaxi=aa*A 143 496. 2 EM31 9. Fall 2007. Test1.math.nb |n[56]:= Problemz; a=.;W=.;x=.;t=.;x=.;EM=.; R=1+x/BO; A=2*7r*t*R Wx=y*Integrate[A, x] Nx=W+Wx a=Nx/A 6 = 1 /EM* Integrate [Nat/A, x] a=.;W=80;x=0.1;t=0.125;x=.;EM=11*106; a=Nx/A Plot[a, {x, 0, 240)] 3 6 = l/EM*Integrate[Nx/A, {x, o, 240}] x=0; ‘3 a=leA X Out[59]= 27rt (1 + —) 80 ( l 2‘ Out[60]= [27rtx+8—7rtx y 0 I 1 quen: W+ (27rtx+ —-—ntx2] y 80 W+ (27rtx+—1—7rtx2)7 80 Out[62]= x 2 7ft (1 + E—d) ént (80+x)27—80 (—W+807rtx) Log[80+x] Out[63]= ————————-—————— 2EM7rt 1.27324 (80 + 0.1 (0.785398): + 0.00490874 x2)) owes]: x 1+ E 100 90 80 owes}: 7o ‘ so % ¥ 50 E _ i '50 100 150 200 Out[67]= O . 00120481 Out[69]= 101 . 85 9 EM31 9. Fall 2007. Test1.math.nb 3 problem3; Ne=.;Em=200*109;A=100*10'6;a=12*10"6;AT=—1OO;P=10*103; Nf: (10*Em*A*a*AT+13*P) l29. af=Nf/A Solve[2*Nf+5*Ne = 6.5 P, Ne] Hal. 96 Nel=96 ae=NellA —3793.l —3.7931x1o7 {{Ne—> 14 5112)} {14517.2} {14517.2} {1.45172x108} Problemd; P=.; Lab=.;Em=.;A=.; Lbc=2*La.b; Lac: fisrLab; W=. ; U=.; 6b=.; Nbc=P/Coa[7r/6] Nab=Nbc*Cos[7r/3] Nac=P U=NbcziLbcl (2*Em*A) +Nab2*Lab/ (“mun +Nao2*Lao/ (2*Em*A) W=Pa6bl2 Solve[U=W, 6b] 2P v3— P «3‘ P E 3LabP2 \f3—LabP2 } m*m } Pab 7 i (3 Lab+ \f3—Lab) P H51” “TH 4 EM3I9. Fall 2007. Test1.math.nb Numbers; P =5000.; Lab: 2*12; Em=29*106;A= 0.2; Lbc=2~kLab; Lac: Furled); W=.; U=.; 6b=.; Nbc=P/Cos[7r/6] Nab=Nbc*Cos[7r/3] Nac=P U=Nb02*Lbc/ (2*Em*A) +Nab2-A-Lab/ (2*Em*A) +Nac2*Lac/ (2*Em*A) W=P*6b/2 Solve[U=W, 61)] 5773.5 2886.75 5000. 244 .761 2500. 6b {{5b—> 0.0979045}} ...
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This note was uploaded on 03/19/2008 for the course EM 319 taught by Professor Kennethm.liechti during the Spring '08 term at University of Texas at Austin.

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EM319.Fall2007.Test1.Solutions - go Luke-m S EM319 Fall...

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