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iTwo forces PI and Pz, of magnitude PI::::: 15kN andP2 :::::18leN, are applied as shown to the end A of bar AB,which is welded to a cylindrical member BD of radiusc ::::: 20 mm (Fig. 8.21). Knowing that the distancefrom A tothe axis of memberBD is a ::::: 50 mm and assumingthat allslressesremainbelow the propOrtional]imit of the material,determine(a) the normal and shearingstressesat point K ofthe transversesectionof memberBD located at a distanceb ::::: 60 mm from end B, (b) the principal axesand principalstressesat K, (c) the maximumshearingstressat K.foDu rt'I(jf() F(J'(rtJ.- 6([),- . ,s K N.,.,1-"vtV,'I, ,.".,.,.--M.#./ftO.D61081) N'rnvFFig. 8.22y/d.iff dlUt')(5(t ,, My:: 7.50 N.II),r- KNv 2.-0 T0",. cljf"tct/ct')G0My-PJ.'IS () (VrO P;2 '" H) kNfa15It O. 0 .5".8.21 tP,Fig.",gxO.O.5-.,.,so N 12,.".)l(-, -Itv IS kNFig. 8.23ffy-IA :N. 75. 0,.-'"I ft1 N7 , ()IV ""J\Jr()t-&'jY)( e.-N ft- k(vAdo ('! D J- p(JJAIf.r IJ. t ))".-:.-.';-:- -".',-::-::-.' /"':.'.- :- :'!.:', :.;:-;--:':-',":--- - :I !':\:'; -:'.fU;;,:,:,; ;{ ,:",/ ;-':i:' )-- -- --::-:1'-; :::.I' ;:;.!: '. {!:. 'Jd )" }:::.'i;:;::;': ;j/i. i1.i: : u.S'YH- - -------:---- : \ . ) "i" i.'U .)Y'i !' IkĀ»' \t: , ;? !:,t ;: ;i : (Nt{,

A-7-,-AfrWA -'J -- xiJ,. ,SOP-- f)h---:.-t150CxUh J;"JtAh d.(--V' "A " Qf) i.UI'-.,',q/"( rr:/ "\-v,---I1"1 P (}. o.-7 f.bM po.('1 p

Ne}- x ::- 910eu eA Q i J()Wdij 107 .4(J,;::.f'fPd.r."'J,-. f "6 heCJ.I'y -- l,r()rA((f(\""c-- - -52.5MPaFig. 8.24I PIA52.5t07MPa.8p 22.2 tOJ' 2-&p 15kN'[ "'y( f - fr:Ja;11in -21.4MPa.Fig. 8.26to'( z e s :.)(6'7C-Q' as 22.8 2.-g .,.- 75.1(2 . tOf0 2-MP",--1'1PrA.- -2./.1i. 75.1 MP l'T,n,, MP'1

,"" '" "" ;";, ',; i;;:.;{;:.1 .i; ;f, ",'i:kv:, ,; J';':'! :,;i;\:i.:;:',;,./i;' iD iifi t ;ii.fiii Xl At#wkMi.{ ; .&. ; ;.,{ ;SAMPLESAMPLE PROBLEMPROBLEM 8.48.4AA horizontal 500-lb force acts at point D of crankshaft AB which is held instatic equilibrium by a twisting couple T and by reactions at A and B. Knowing that the bearings are self-aligning and exer no couples on the shaft, determine the normal and shearing stressesat points H, J, K, and L located at theends of the vertical and horizontal diameters of a transverse section located2.5 in. to the left of bearing B.r Ii I.Sin.n. -. . . .5001b.4.5in.SOLUTION4.5 in.Free Body. Entire Crankshaft.A A 2501b%zB 2.50Ib 2:M 0:The geometric properties of the O.9-in.-diameter section areA 7T(O.45in.)2 0.636in26290 psi--LHT 6290 psi,L(0)TT 62-90psi'T JT 524 psi"l(b)T 524psi.,.:::'0Stresses Produced by Twisting Couple T. Using Eq. (3.8), wedeterminethe shearingstressesat points H, J, K, and L and show them inFig. (a).Tc (900Ib . in.)(0.45in.) 6290psi 64.4 X 10 3in4 6290psiKH1 !7T(O.45in.)4 32.2 X 10-3in4J !7T(O.45in.)4 64.4X 10-3in4GT::T 9001b in. /V:250lb O.9-in. diameter.-(500 Ib)(1.8in.) T 0v B 250 IbT 900 Ib in.My (250 Ib)(2.5in.) 6251b . in.Ii 9001b.in. B 250 IbInternal Forcesin TransverseSection. We replacethe reactionB andthe twisting coupleT by an equivalentforce-couplesystemat the centerC ofthe transversesectioncontainjngH. J, K, and L.ill./M!f 62.51b625.-TAKStresses Produced by Shearing Force V. The shearing force V produces no shearing stressesat points J and L. At points Hand K we first compute Q for a semicircle about a vertical diameter and then determine the shearing stress produced by the shear force V 250 lb. These stressesare shownin Fig. (b).Q(1';:0:,.- (T 8730psiT))( 31T4C c3 (O45 in)3 60 7 X 10-3 in3 ,2.!.1Tc233'.(. -VQ :::::(250 Ib)(60.7 X 1O-3in3) 524pSiIt(32.2 X 10-3 in4)(O.9in.),u 8730 psi(c)StressesProduced by the Bending Couple M,. Since the bending couple My acts in a horizontal plane, it produces no stressesat Hand K. Using Eq.(4.15), we detennine the normal stressesat points J and L and show them inFig. (c).(J' -IMylc (625lb . in.)(O.45 in.) 8730PS.I-uL32.2 X 10-3 in4i 8730 psi.,. 6810psi.u "8730psISummary.We add the stressesshown and obtain the total normal andshearingstressesat pointsH. J, K, andL

SAMPLE PROBLEM 8.5.Threeforcesare appliedas shownat pointsA, B, and Jj of a short steelpost.Knowing that the horizontalcrosssectionof the post is a 40 X 140-mmrectangle,determineing stressat pointtheH. principal stresses,principal planes. and maximumshear-SOLUTIONInternal Forces in Section EFG. We replacethe three applied forcesbyan WeequivalentEFG.have. force-couplesystemat the centerC of the rectangularsection.4O mmyMJ :; t).,'5 kN.zIII -30 kN P 50kN V -75 kNMx (50 kN)(0.130m) - (75 kN)(0.200m) -8.5My 0M (30kN)(0.1O0m) 3 kN . mVxG.kN mWe note that thereis no twisting coupleaboutthe y axis. The geometricpropertiesof the rectangularsectionarexA (0.040m)(0.140m) 5.6 X 10-3 m2Ix i2(0.040m)(0.140m? 9.15 X 10-6 m4lz i2(0.140m)(0.040m? 0.747 X10-6 m4Normal Stress at H. We note that normal stresses(J'yare produced bythe centric force P and by the bending couples Mx and M ;.We determine thesign of each stressby carefully examining the sketch of the force-couple system at C.(J'YPIMtla - --AI ,t50 kN UyIMx1bIx(3 kN . m)(0.020 m)(8.5 kN-5.6 X 10-3 m20.747 X 1O-6m4 8.93MPa 80.3MPa- 23.2MPa. m)(O.O25m)9.15 X 10-6 m4cry 66.0MPa. 1Shearing Stress at H. Considering first the shearing force Vx, we notethat Q 0 with respect to the z axis,. since H is on the edge of the cross section. Thus V x produces no shearing stressat H. The shearing force Vz does produce a shearing stress at H and we writeQ A1YJ Tyz (75 kN)(85.5 X 10-6 m3)Ixl (9.15X 1O-6m4)(0.040m) 85.5 X10-6 m3.Tyz 17.52MPa JPrincipal Stresses Principal Planes,and Maximum Shearing Stressat H. We draw Mohr's circle for the stressesat point HT'"mI ;: 8 O"'"iL1VzQ[(0.040 m)(0.045m)](0.0475m)O I