ME 211 Homework 5 Solutions

ME 211 Homework 5 Solutions - Problem#18.16...

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Unformatted text preview: Problem#18.16 ME211HomeworkSet#5Solutions 10102008 Twodesignsforashocksupportareshowninthefigure.Thespringhasastiffnessof k=15kN/mandin(a)itisuncompressed,whereasin(b)itisoriginallystretched0.2m. Determinetheaveragenormalstressinthe5mmdiameterboltshankatAwhenthe 6kNloadisapplied.In(b)thebracketBisnotconnectedtothesupport. 1.)ForCase(a),writeyourequilibriumequationtosolveforP,anddeterminethe stressinthespring. F y =0 = P -6 2.)ForCase(b),theloadwillcausethespringtostretch.Theloadisentirely supportedbythespring. F = kx 6kN = (15kN /m)x x = 0.4m P = 6kN P 6kN = = = 306MPa A (0.005) 2 4 P = 6kN P 6kN = = = 306MPa A (0.005) 2 4 Problem#28.80 ThebuiltupshaftconsistsofapipeABandsolidrodBC.Thepipehasaninner diameterof20mmandouterdiameterof28mm.Therodhasadiameterof12mm. DeterminethenormalstressatpointsDandEandrepresentthestressonavolume elementlocatedateachofthesepoints. 1.) DeterminethenominalstressatpointD. 0.028m 2 0.02m 2 Area = - 2 2 P -4 = = = -13.26MPa A .000096 2.)DeterminethenominalstressatpointE. 0.012m 2 Area = = .000113 2 P 8000N = = = 70.7MPa A .000036m 2 Problem#38.85 TheleverisattachedtotheshaftAusingakeythathasawidthof8mmandlength of25mm.Iftheshaftisfixedandaverticalforceof50Nisappliedperpendicularto thehandle,determinetheaverageshearstressdevelopedalongsectionaaofthe key. 1.)DeterminetheforceF,whichisshearforce.Usetheshearforcetodetermine averageshearstress. M o = 0 = F(15) - 50(450) F = 1500N V 1500 = = = 7.50MPa A (0.008)(0.025) Problem#48.27 Thebarsofthetrusseachhaveacrosssectionalareaof1.25in*in.Ifthemaximum averagenormalstressinanybarisnottoexceed20ks,determinethemaximum magnitudePoftheloadsthatcanbeappliedtothetruss. 1.)DrawtheFBDofindividualjointstodeterminethemaximummagnitudePthat canbeapplied. JointA 3 = - P + ( )FAB = 0 y 5 FAB = (1.667)P 4 Fx =FAE + (1.667)P( 5 ) = 0 FAE = (1.333)P F JointE F F y = - P(0.75) + FEB = 0 = - FED + (1.333)P = 0 FEB = (0.75)P x FED = (1.333)P JointB 3 3 = - (0.75)P + ( )FBD -1.667P( ) = 0 y 5 5 FBD = (2.9167)P 4 4 Fx =FBC - (2.9167)P( 5 ) -1.667P( 5 ) = 0 FBC = (3.67)P F 2.)ThehigheststressedmemberisBC BC = (3.67)P = 20 1.25 P = 6.82kip Problem#58.52 Thewoodspecimenissubjectedtothepullof10kNinatensiontestingmachine.If theallowablenormalstressforthewoodis(t)allow=12MPaandtheallowableshear stressisallow=1.2MPa,determinetherequireddimensionsbandtsothatthe specimenreachesthesestressessimultaneously.Thespecimenhasawidthof 25mm. 1.)DrawaFBDofthewoodthatexposestheshearforceoftherod. 2.)Determinetheallowableshearstress.Theareathatisinthedenominatoristhe areathatshearwouldbeactingon,whichisnotthecrosssectionbutthesurface area. allow = 6 V A 5.00(10 3 ) 1.2(10 ) = (0.025)t t = 0.1667m = 167mm 3.)Determinetheallowablenormalstress. allow = 6 P A 10(10 3 ) 12(10 ) = (0.025)b b = 0.03333m = 33.3mm ...
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This note was uploaded on 05/12/2010 for the course MECHENG 211 taught by Professor ? during the Fall '07 term at University of Michigan.

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