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1- Introduction _ Internal Resultant Loadings (1).pdf
1- Introduction _ Internal Resultant Loadings (1).pdf
Yusfarijerjis
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Mar 20, 2023
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About This Presentation
strength
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882.38 KB
Language:
en
Added:
Mar 20, 2023
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20 pages
Slide Content
Slide 2
Strength of Materials
Strengthofmaterialsisabranchofmechanicsthatstudiestheinternal
effectsofstressandstraininasolidbodythatissubjectedtoan
externalloading.Stressisassociatedwiththestrengthofthematerial
fromwhichthebodyismade,whilestrainisameasureofthe
deformationofthebody.Inadditiontothis,mechanicsofmaterials
includesthestudyofthebody’sstabilitywhenabodysuchasacolumn
issubjectedtocompressiveloading.Athoroughunderstandingofthe
fundamentalsofthissubjectisofvitalimportancebecausemanyofthe
formulasandrulesofdesigncitedinengineeringcodesarebasedupon
theprinciplesofthissubject.
Slide 3
Historical Development
Theoriginofmechanicsofmaterialsdatesbacktothebeginningofthe
seventeenthcentury,whenGalileoperformedexperimentstostudythe
effectsofloadsonrodsandbeamsmadeofvariousmaterials.However,
atthebeginningoftheeighteenthcentury,experimentalmethodsfor
testingmaterialswerevastlyimproved,andatthattimemany
experimentalandtheoreticalstudiesinthissubjectwereundertaken
primarilyinFrance,bysuchnotablesasSaint-Venant,Poisson,Lamé,
andNavier.
Overtheyears,aftermanyofthefundamentalproblemsofmechanics
ofmaterialshadbeensolved,itbecamenecessarytouseadvanced
mathematicalandcomputertechniquestosolvemorecomplex
problems.Asaresult,thissubjectexpandedintootherareasof
mechanics,suchasthetheoryofelasticityandthetheoryofplasticity.
Researchinthesefieldsisongoing,inordertomeetthedemandsfor
solvingmoreadvancedproblemsinengineering.
Slide 4
Equilibrium of a Deformable Body
ExternalLoads.
Abodyissubjectedtoonlytwotypes
ofexternalloads:
•Surfaceforcesarecausedbythe
directcontactofonebodywiththe
surfaceofanother.
•Bodyforceisdevelopedwhenone
bodyexertsaforceonanotherbody
withoutdirectphysicalcontact
betweenthebodies.Examples
includetheeffectscausedbythe
earth’sgravitationorits
electromagneticfield
Slide 5
SupportReactions.
Thesurfaceforcesthatdevelopatthesupportsorpointsofcontact
betweenbodiesarecalledreactions.Fortwo-dimensionalproblems,
i.e.,bodiessubjectedtocoplanarforcesystems,thesupportsmost
commonlyencounteredareshowninthetablebelow.Notecarefullythe
symbolusedtorepresenteachsupportandthetypeofreactionsit
exertsonitscontactingmember.Asageneralrule,ifthesupport
preventstranslationinagivendirection,thenaforcemustbe
developedonthememberinthatdirection.Likewise,ifrotationis
prevented,acouplemomentmustbeexertedonthemember.
Slide 6
EquationsofEquilibrium
Equilibriumofabodyrequiresbothabalanceofforces,topreventthe
bodyfromtranslatingorhavingacceleratedmotionalongastraightor
curvedpath,andabalanceofmoments,topreventthebodyfrom
rotating.Theseconditionscanbeexpressedmathematicallybytwo
vectorequations:
Ofteninengineeringpracticetheloadingonabodycanberepresented
asasystemofcoplanarforces.
Successfulapplicationoftheequationsofequilibriumrequires
completespecificationofalltheknownandunknownforcesthatacton
thebody,andsothebestwaytoaccountforalltheseforcesistodraw
thebody’sfree-bodydiagram.
Slide 7
Inmechanicsofmaterials,staticsisprimarilyusedtodeterminetheresultant
loadingsthatactwithinabody
•Normalforce,N.Thisforceacts
perpendiculartothearea.Itisdeveloped
whenevertheexternalloadstendtopushor
pullonthetwosegmentsofthebody.
•Shearforce,V.Theshearforceliesinthe
planeoftheareaanditisdevelopedwhen
theexternalloadstendtocausethetwo
segmentsofthebodytoslideoverone
another.
•Torsionalmomentortorque,T.Thiseffect
isdevelopedwhentheexternalloadstendto
twistonesegmentofthebodywithrespectto
theotheraboutanaxisperpendiculartothe
area.
•Bendingmoment,M.Thebendingmoment
iscausedbytheexternalloadsthattendto
bendthebodyaboutanaxislyingwithinthe
planeofthearea.
Internal Resultant Loadings
Slide 8
Coplanar Loadings.
Ifthebodyissubjectedtoacoplanarsystemofforces,thenonly
normal-force(N),shear-force(V)andbending-moment(M)
componentswillexistatthesection.Ifweusethex,y,zcoordinate
axes,thenNcanbeobtainedbyapplyingFx=0,andVcanbe
obtainedfromFy=0,Finally,thebendingmoment(Mo)canbe
determinedbysummingmomentsabout
pointO(thezaxis),Mo=0
Slide 9
Procedure for Analysis
Theresultantinternalloadingsatapointlocatedonthesectionofabody
canbeobtainedusingthemethodofsections.Thisrequiresthefollowing
steps.
SupportReactions.
•Firstdecidewhichsegmentofthebodyistobeconsidered.Ifthe
segmenthasasupportorconnectiontoanotherbody,thenbeforethe
bodyissectioned,itwillbenecessarytodeterminethereactions
actingonthechosensegment.Todothisdrawthefree-bodydiagram
oftheentirebodyandthenapplythenecessaryequationsof
equilibriumtoobtainthesereactions.
Slide 10
Free-BodyDiagram.
•Keepallexternaldistributedloadings,couplemoments,torques,and
forcesintheirexactlocations,beforepassinganimaginarysection
throughthebodyatthepointwheretheresultantinternalloadings
aretobedetermined.
•Drawafree-bodydiagramofoneofthe“cut”segmentsandindicate
theunknownresultantsN,V,M,andTatthesection.These
resultantsarenormallyplacedatthepointrepresentingthe
geometriccenterorcentroidofthesectionedarea.
•Ifthememberissubjectedtoacoplanarsystemofforces,onlyN,V,
andMactatthecentroid.
•Establishthex,y,zcoordinateaxeswithoriginatthecentroidand
showtheresultantinternalloadingsactingalongtheaxes.
Slide 11
Equations of Equilibrium.
•Momentsshouldbesummedatthesection,abouteachofthe
coordinateaxeswheretheresultantsact.Doingthiseliminatesthe
unknownforcesNandVandallowsadirectsolutionforM(andT).
•Ifthesolutionoftheequilibriumequationsyieldsanegativevalue
foraresultant,theassumeddirectionalsenseoftheresultantis
oppositetothatshownonthefree-bodydiagram.
Slide 12
EXAMPLE1.Determinetheresultantinternalloadingsactingonthe
crosssectionatCofthecantileveredbeamshowninthefigurebelow.
Slide 13
Solution:
ThesupportreactionsatAdonothavetobedeterminedifsegment
CBisconsidered.
Theintensityofthedistributedloading
atCisfoundbyproportion
270
9
=
�
6
w=180N/m
Themagnitudeoftheresultantofthedistributedloadisequaltothearea
undertheloadingcurve(triangle)andactsthroughthecentroidofthis
area(i.e??????=0.5×6??????×180�/??????=540�)actat(
1
3
6??????=2??????)
Slide 14
Equations of Equilibrium.
՜
+
??????
�=0 −�
??????=0
+??????
�=0 ??????
??????−540=0 ??????
??????=540�
+ �
??????=0 −�
??????−540×2=0 �
??????=−1080�.??????
Slide 15
EXAMPLE2.Determinetheresultantinternalloadingsactingonthecross
sectionatCofthemachineshaftshowninthefigurebelow.Theshaftis
supportedbyjournalbearingsatAandB,whichonlyexertverticalforceson
theshaft.
Slide 16
Solution:
+ �
??????=0
−??????
�×0.4+120×0.125−225×0.1=0
??????
�=−18.75�
Slide 17
Equations of Equilibrium.
՜
+
??????
�=0 −�
??????=0
+??????
�=0 −18.75−40−??????
??????=0 ??????
??????=−58.8�
+ �
??????=0 �
??????+40×0.025+18.75×0.25=0
�
??????=−5.69�.??????
Slide 18
EXAMPLE 3. The 500-kg engine is suspended from the crane boom in the
figure below. Determine the resultant internal loadings acting on the cross
section of the boom at point E.
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