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Me211_sol11

Course: ME ME211, Fall 2010
School: Michigan
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Word Count: 553

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The 4- -b H I) 4 \ 4- I + C., &amp; 1 -- -cr -.- ii cJ k d, ., U, cnO Ct, LI) ( *15.20. wooden beam is subjected to a load of 12 kN. If grains of wood in the beam at point A make an angle of 25 with the hothontal as shown, determine the normal and shear stresses that act perpendicular and parallel to the grains due to the loading. friv 12 m 4m F1 Li 1 P &amp;fl/W...

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The 4- -b H I) 4 \ 4- I + C., & 1 -- -cr -.- ii cJ k d, ., U, cnO Ct, LI) ( *15.20. wooden beam is subjected to a load of 12 kN. If grains of wood in the beam at point A make an angle of 25 with the hothontal as shown, determine the normal and shear stresses that act perpendicular and parallel to the grains due to the loading. friv 12 m 4m F1 Li 1 P &fl/W 300mm HH V. 200 mm r4 6 di Lva FiG- FZlJ,M. k.. 6.c7 1w ifr.v V 6.&n kit, ik tcticr 1 1 P Ri7I.-,2?/t: (&rilciv, l2,L. (q3J,j,f (4: (3I4,Vi11. Crt,ssceei%&i. Cciz(OJ)z u2. 0. 1I.2cc1&2(c)oJ) 1k 7(( 7f4L((3)( o7j 4 2..&s-7Mp c45 ((o3) CT) ((7s. 7 & xo!? Q. fST(o ) t.l 22flMI? O , Mpm 6 O.I2S ii .. - 7. 27 2 as2-o 2 2.27-O 1yl2& C 2 - - = 2 (23o ________ 600 psi 15-31. Draw Mohrs circle that describes each of the following states of stress. 2OMPa HO (a) 1 rdc itt 2ksi (b) tiye41 71 (c) Zy? H uvj ftlA$4tK llC ,4 )ft)eskce 7)O (9oo, C((n,O) 7k. rads b. v) c4Je O L... .:: 2 f (C jk it t A ad C 7 7c - Cot4VQW(k1, : - 4ts i , 2 z He&te Qt(2) .2 A:, . UrJa ;Y 1 A(o,c). ((-lou 71-e id C). 1 of 1Cct,rJiq ut out. 7 Adc 0) ccc& Ic f14_of frcI .(ft 4 Zey2oM& 2 LY ortta Au, 1k ie-ftre&ce f FJ ereuc pot 4 C a.e C(C,O) tit P:2o,uMP. cMp) tCMP) jf. 2. *1532. A point on a thin pate is subjected to states two successive of stress as shown. Determine the resulting state of stress with reference to an element oriented as shown at the right. sgktc tie (rc&: Z c tLe pive&, 6% SM P, 6, 8c/(4I?i Z/,L r p y 4 (&t) fevetiie 74 IC C(r. o) 7k . t (oM4. -fr Ca) L4 0 /,4 ?s&1t, & c-ti fe cf Ifrsc J4s of t1p ot%i. 1 at C. 1ie .c 4 j = coMP of (64 c.o 7le ii ak.e ) 7 (r 0. M/ it{e -t,te C) cL f ii . J 4. 61et (c). j oyj 4ci6o) 7 rd4c &sc oi lkcs <f fre p (C tQ) CCo,o tIe Ca) cW fDd The r[ (6;,) C Pof P fIe od z; (Le I 9., c ol o 1 j it . /, \ (d-?q I3Q . ( S;f1= i. irf/p,) cp (MLe the (() Z q 1r fcdc)h cY,>,z (; (d ) 7$5jL(./ 33; 0 T) : :1. 9 ) ( 7 r /37 44 15-39. The thin-walled pipe has an inne r diameter of 0.5 in. and a thickness of 0.025 in. If it is subjected to an internal pressure of 500 psi and the axial tension and torsional loadings shown, determine the princ ipal stresses at a point on the surface of the pipe. 1 4 7.flcj -. 2001b 200 lb 20 lbft 20 Ibft P.e e 77fo.7-- S(.2S7: J skU? /Jg!1 5i .c;: cY. . - I = 2 r). 7 frtI -: (Q.Gf) S .1L frJ7tThc 7 - Ot . - )(o.27S) h / 6Z3rk, :ufr dd 4 A A(7.31, -23.I& 7 I -/r)&1(a cr (&. 2-. 7k s; 6 o 7te rr -/. / 2 ./1L fr ,V / dC ccJ7, LF?/1 t/.e ___7__ 23.I 3 o6 ; Prc Srers j7f+ dL 17S 2 zo65 23.2 o )f (;
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Michigan - ME - ME211
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)Lc,h:=J, ( 1 -.; J )/'1,47( i)~. ~.I- 2.-: /o;l ~= / .;)h2.z:-.(2-P-)JIf ( I -h= (ho o if-) cfw_ 2)71( JfJD):2.-/vJ J;~ )h2.2zRz~ (21&lt;.)- 3-.IIL -td.ACen&gt;GoNoJletlc.t.cfw_ovrnt ~J t&gt;~C .oftVlx:,-1 )S h61JVlJ-&gt;1 rc
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ME 240 13th (and final) Homework Assignment. Due Monday 12/13/10Textbook problems:8.438.628.688.798.89No. 6. Problem contributed by Prof. Scott:The bar shown is undergoing small motions ( &lt; 1) in a horizontal plane (gravity doesnot enter this pro
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Michigan - ME - ME240
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50&gt;pr-.(IIC,0tc-3C)r+3.AV.-.3%.c5. Ashby and Jones II() TLt1i.Jd.Appendix I, Probtem 2.8TrIt111r pin fir ,i 5ntOtn:-,truL11ifli t.uCflOSi1.3iS,Lir,P 1h.iTi. /\j.i4Il(I!C.dLIitOtt2511(E lIeS Iflthc 1% U 1itiitvrt,
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Michigan - ME - ME382
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Michigan - ME - ME382
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Michigan - ME - ME382
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Michigan - ME - ME382
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Michigan - ME - ME382
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ME 211 ADDITIONAL HOMEWORK QUESTIONS1. Figure P1 shows a solid cylindrical shaft of diameter 10 mm and length 1 in, which isbuilt into a wall at one end A. At the other end B, a load of 10 kN is applied at the topand perpendicular to the wall. Also, eq
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LTL5\4.Cr:-2pra-c.-.-_,_j&gt;rUcIF(IPr3C?(iC)4?I-JJL4.I5CrniI1P(ji.(I-flIC?9C)(NThCLc- \d.icC(_3c5-z0C;!19,U9?pC)?\3Lcfw_\0DpDelta LDStressStrainTrue StressTrue StrainTrue Plasti
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Michigan - ME - ME382
GENERAL-PURPOSEAIR VELOCITY TRANSDUCERSFMA-900-V Series822$All modelsOptionalUp to 1.5% AccuracyRemote Probe Model Availablewith 4.6 m (15') CableEach Unit Individually CalibratedDurable Fast-ResponsePlatinum SensorsCompact Solid-State Electr Michigan - ME - ME382 Homework 1 Solutions1.Solution:Solution:Solution:Solution:ME 382 Fall 2010 HW1 Solutions2.Solution:Solution:Solution:Page 2ME 382 Fall 2010 HW1 SolutionsPage 3ME 382 Fall 2010 HW1 Solutions3.[TypeSolution:aquotefromthed Michigan - ME - ME382 ME 382 Fall 2011 HW02 Solutions = 400 106 = 1000 106 = 800 106 = = = 0From Table 5.2: E = 70.3 GPa and =0.345.Problem 1: Dowling 5.163D Hookes Law: =Plugging in known values: =1 + 1 ( + ) 0.004(70,300 ) = 0.3450.0010(70,300 ) = 0.345So Michigan - ME - ME382 VyCI00b.NU04.3(-jIc-.c-NJ--CsCl-cC-I)j1rri.3IIILHIIpLf5)(a-ai0a-qIc5rjIS.xI04NIIriIwS0,c00-IdC9-0p-.4.0SV:1kr0I0C-;*&gt;CiI00I.4-&gt;-_,f-r,)c&gt;%_2)I\JIJ*4-Cd91 Michigan - ME - ME382 Homework 3 Solutions1.Solution:ME 382 Fall 2010 HW3 Solutions2.Solution:Page 2ME 382 Fall 2010 HW3 Solutions3.Solution:Page 3ME 382 Fall 2010 HW3 Solutions4.Solution:Page 4ME 382 Fall 2010 HW3 Solutions5.Solution:Solution:Solution:Page Michigan - ME - ME382 ME 382 Fall 2011 HW04 SolutionsProblem 1: Dowling 7.8Given: = 150 = 30 = 45 SF=3.0 2 + 2 + 1 , 2 = 22Plugging in given values gives:1 = 165 , 2 = 15 , 3 = 0 (a) Tresca (maximum shear stress) criterion: = max(|1 2 |, |2 3 |, |3 1 |)|1 Michigan - ME - ME360 Lab 7: Scaling Laws and ME395, Fall 2011 Prof. Kenn Oldham 11/14/11 Scaling Laws Virtually all physical phenomena depend on dimensional scaling, with respect to length, mass, and Kme. Boeing 747 F16 Beech Bonan Michigan - ME - ME360 International Pumping Systems313 Heinz Rd., Dearborn, MI 48124To:B. Franklin, P.E., Consulting Division ManagerInternational Pumping Systems313 Heinz Rd., Dearborn, MI 48124From: L.A. Larson, Project ManagerDate: 11/14/2011Re:Motor Sizing for Oxy Michigan - ME - ME360 International Pumping Systems313 Heinz Rd., Dearborn, MI 48124To:From:Subject:Date:Dist:L.A. Larson, Project Manager, First Response Technologies91 Greentree Dr., Suite 1A, Chicago, IL 60613Joshua DeBoer, EngineerFlow properties and recommended Michigan - ME - ME360 Model: PX653-25D5VRange: 0-25 in-H2O Michigan - ME - ME360 H)H2&amp;H70I;5:11(IrcII112VIIc.IIIH0bIIIiHcDc-HC,I-Cr-;,(ccICc-vsrdIN\J1t)--14v)0)-IIUD(Oci/_C)0CD3H/Th-4c-I)0S.)c-i z2C;Oj)C-4-cJUI)c-C)vi0-D-c--C)2(%)N-I--IN Michigan - ME - ME360 Pc )ler\cw\Ac.cSsoM/4-c,cCoC(\4LLk-.q.-c4ftQVcc1c rzc.)-/4ev\ 10e Src2,.e,Lv\s%_b,el vwtvX Sx)thaciA(i1pmDP1WJLI-/-Cj.,ILL C)c)c7--4-L.:ccLe:1Lo-L_ -c4 .4-LkLcs-iv5 +YO-C1Cx-CVV\MeL-r\sYVS:X Michigan - ME - ME360 r--r--14_+(1(4Y-q_\(J-J%JIQ:,cq--N.%J4fl%l.JJrri+pt-\_)Ii-pq:*-(_+)-IV\J0c-)q)0rif1I)Ctv(;-1&gt;0(/1HJ-4H41Mc1Ce)3(cH&gt;CIji0I!i.:?:rcHI,HIC_,-+,G(,141k,!0a,(3- Michigan - ME - ME360 tJ (QS&amp;Lf .d_o9i L-Os-t\4YC;o)c_i: o.7pILy\ZyO,;-(a(.tO1b7OZ7SC1(O//IS(,&amp;.4t4k scsfC:100C.cAC,-1-C,NIJf19)0iIcrI-CI,IU%J9-)x,-&gt;E,fl1)\UU04-t41JV.t/-,kcj00N3,-xL5I?_fl%q11 Michigan - ME - ME360 cLEc_)f2t1?-R;L4c:0XZ-vs C r-L:N11qEIf\Sis44L;fYLet4CL t-iC-s L-o);=rAFiey14A_oz:pDbV:-e) Lc.-vsLJ$Fc:L-)o1) TLc-(J0V\k-4-CsVs-.-/IraF2L ri-1(s;i.Lelr-c7(J0/L/-Cs(\Jo)-\J_v -V sV
Michigan - ME - ME360
_i#)\&amp;-teC)C/k L)r,&gt;-N(L)[S)(NKtNS-T3i)1iN--LoLr(NsL r(NS--jLD 2: o:.0///NrsS=-/,K\fSjS1zc2ev,t-sL-49 oPoc -L iL 4 kQve -)1-4isr-Lc9-hLtf4yeiOc_3k t,4-cif- +-1 ,4-s/-svvipre&amp;1C-k\r3?L.1cki
Michigan - ME - ME360
_MEC)ZZZZifbove-\ oc-UL&gt; iSS2PLT)6c2c) /gt-7 7KC)vsji_-FiZTflr-eLx4- VtsL, \4i\ eZv/J#;e4 :+=c&gt;;(S-rr\1A-VtI\-tc4rtsp4-e.LC4101C-Ri\5zLffr_O/-Q_Ij cLJr FLJ-4-000c5-*000 03\r 0000mr0NLr5:(-
Michigan - ME - ME360
*--,JAAI\(119\r-Pk-qL _kLk1L-LS-j&gt;,1V-M,+- Vcc0LcV)_&gt; \f0&amp;sSuw-.SV6K11L&lt;iVzc:t-v&gt;L1-LSJicce-L4_Q_-Lveior,*c)=-zhF-eL:-ivI)-k LIAtzs&amp; +o4pc,L+ Jc-raCc_&amp;S#)Lv-ercorcyit5rs:JA-c0 s
Michigan - ME - ME360
o14RC 3s-:\it)_-(iTC57Z ;pssifczt:&amp;5(e .);kv;;(z)=1gL (A s4f)=fL?,M(4zA,-1o\e4s4LLAks44-e.I)2hL-_-S+SZJ F-ciVcv\LA(.iIq10Pv;4CD(L.-Z1l4Sc1-r cj4So-o:-:uL)c,:L-soQ(C)cO74(cyD.zz
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