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# Lecture notes EML3500

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Date Created: 08/31/15

W EML 3500 Mechanics of Solids ALEX A VOLINSKY VOLINSKYUSFEDU Lecture notes are prepared by the instructor based on the Copyrighted Materials from Pearson Prentice Hall RC Hibbeler Mechanics of Materials and from McGraw Hill FP Beer ER Johnston Jr and JT DeWolf J W Oler Lecture Notes are for studying porpoises only they can not be distributed electronically or sold Grading HomeworkQuiz 15 o15 o getting an A in this class is not possible without Midterm l 20 submitting your homework on time Midterm ll 20 Final Exam 30 A gt90 B gt80 o C gt70 o D gt60 o 0F Professor A Volinsky EML3500 Mechanics of Solids 1 soum momma f Solid Mechanics Mechanics of Materials is a study of the relationship between the external loads on a body and the intensity of the internal loads Within the body The main objective of the study of mechanics of materials is to provide the future engineer With the means of analyzing and designing various machines and load bearing structures 0 Both the analysis and design of a given structure involve the determination of stresses and deformations 0F Professor A Volinsky EML3500 Mechanics of Solids 2 39 39 SOUTH FLORID Forces Co ncetttrated force idealiaatiom 4 g a Surface force WM 39 139 A d force 2 Linear distributed load idealization Copyright 2005 Pearson Prentice Hall ltnc Surface force caused by direct contact of one body with another If the contact area is small it can be idealized as a single concentrated force Linear distributed load force between the ground and the bicycle wheel Loading is represented graphically by a series of arrows along the line 3 The resultant force FR is equivalent to the area under the distributed loading curve and acts though the centroid C or geometric center of the area Body force develops when one body exerts a force on another body without direct physical contact Earth gravitation electromagnetic forces etc UNlVERSlTT Girl SOUTH FLORID yr USP Professor A Volinsky 3 EML3500 Mechanics of Solids Equilibrium of a deformable body Body equilibrium requires both balance of forces and moments so it will not move or rotate ZF0 and ZMO Here F represents all forces along x y and 2 directions and M is the sum of all moments of the forces about any point The best way to account for all forces is to draw a free body diagram 7 0F Professor A Volinsky EML3500 Mechanics of Solids 4 SOUTH FLORlD Support Reactions The surface forces that develop at the supports or points of contact are called reactions If the support prevents translation in a given direction then a force must be developed on the member in that direction Likewise if rotation is prevented a couple moment must be exerted on the member Type of connection Reaction Type of connection Reaction 14 1F C b One unknown F External pin Two unknowne Fx F Roller One unknown F i t m l Pin TWO U k wmi F 3 M F A F Smooth Support AOne unknown F Fined supon Three umkt l w i w M Method of Sections Statics application in Mechanics of Materials is to Determine the resultant force and moment acting within a body necessary to hold the body together when the body is subjected to external loads FR acts through point 0 although its computed value does not depend on the location of this point Moment MR DOES depend on the location of the point 0 which is often chosen as the CENTROID of the sectioned area cross section sectin 9 UNIVERSITY OF SOUTH FLORIDA Professor A Volinsky EML3500 Mechanics of Solids 6 3D Signal Both Inplane and outofplane forces and Mamem T moments need to be considered Fr Mtg 65 Normal Normal force N acts perpendicular to 7 FMS the area Develops when the external loads push or pull on the two segments of a body Shear force V lies in the plane of the area and is developed when the external loads tend to cause sliding Bending V Moment Torsion moment or torque T is developed when the external loads tend to twist one segment of the body with d respect to the other Copyright 2005 Pearson Prentice Hall Inc Bending moment M is caused when the external loads tend to bend the body about an axis lying within the plane of the area I GE PrOfessor A V insky EML3500 Mechanics of Solids 7 SOUTH FLORID Important Points i Meehaaies ef materiais is a study bf the relatidnship between the eaternal leads en a hardy and the intensity bf the internal leads within the badge r Eaterual tarees ean be applied m a bed as distributed er eerieertt rareri strrfitee feedings er as ibe fbrees whieh aet thru ugheut the rulunie at the bedst 1 Linear distributed leadings preduee a resuitartr faree hatring a magnitude equal te the area under the lead diagram and having a Ebeariria that passes thrbugh the eerrrrbiri at this area 9 PL suppert preduees a feree m a partieular direetibn en its attaehed member if it prevents rrarrairrt iua at the member in that direetin and it prbduees a mutate rrtriraear an the member if it presents re re ti art i Ilse equatiens at equilibrium 2 F and E M it must be satisfied in arder m present a bddjr frurn translating with aeeelerated marian and tram retating it When applying the equa tibns at equilibrium it is impbrtant tb first draw the bbdy39is freedaddy diagram in brder tu aeeuunt fur all the terms in the equa39tibns i methnd seetiuns is used tb determine the internal resultant leadings aeting en the surfaee ef the seetidned bbdy In general these resuitants ednsist bf a nbrmal fume shear fbree tbrsieual mement and bending mument F Professor A Volinsky EML3500 Mechanics of Solids SOUTH rroama USP Procedure for Analvsis UNlVERSlTY l SOUTH FLORID The method of sections is used to determine the resultant intarnni loadings at a point located on the section of a body To obtain these resultants application of the method of sections requires the following steps Support Reactions First decide which segment of the l n39rdjtr is to he considered If the segment has a support or connection to another body then before the body is sectioned it will he necessary to determine the reactions acting on the chosen segment of the body Draw the ireeehodr diagram for the entire body and then apt39ilt the necessary equations of equilibrium to obtain these reactions FreeBody Diagram Keep all eaternal distributed loadings couple momenta torques and forces acting on the hodgr in their erect intentions then pass an imaginary section through the hotly at the point where the resultant internal loadings are to be determined If the body represents a member of a structure or mechanical device the section is often taken perpendicninr to the longitudinal aids of the member Draw a treebody diagram of one of the out segments and indi cate the unknown resaltants N ifi l da and T at the section These resultants are normally placed at the point representing the gene metric center or centroid of the sectioned area it the member is subjected to a engineer system of forces only N V and M act at the centroid Establish the t y a coordinate arses with origin at the centroid and show the resultant components acting along the arses Equations of39Enaiiibrinmi Moments shnld be summed at the section about each of the coordinate arses where the resultants act Doing this eliminates the unknown forces N and and allows a direct solution for M and If the solution of the equilibrium equations yields a negative value for a resultant the assumed directionni sense of the resultant is opposite to that shown on the freebody diagram ids 9 Example 11 Determine the resultant internal loadings neting on the eress section at C elf the beam shel wn in Fig Zl ln 2H j 5 A r B is Fig 1 4 Selutien Supper Renet39i ens This problem can he sl ved in the most direct manner by eensidering segment til the llfieaizna sinee then the support reactions at A de net have tn he eempnted OF Professor A Volinsky EML3500 Mechanics of Solids 10 SOUTH FLOHlD i Example 11 Continued FreeBody Diagram Passin an imaginary section perpendiouiar to the 540 N iongitudinal axis of the beam yields the freebody diagram oi segment 1 CB shown in Figs 14h it is important to keep the distributed ioading 1 exactly where it is on the segment untii u ar the section is made Uni V then shouid this loading he reptaoed by a singie resultant force Notice that the intensity of the distributed loading at is found by proportiom 39 ie from FigJi4a In 270 w 180 The viii 2m M 4m magnitue of the resuitant of the distributed ioad is equal to the area under the loading curse triangie and acts throughthe centroid of this areaThuaF 180 Nmiiti 540 N which acts 1 316 n1 2 in in from C as shown in Fig 14b Equations of quuiiibrimm Apptying the equations of eouiiihriutn are have t v 3 ZFx Ut NC t l N iUN C l 39039 Q39U39me if H 539 1 r TZFEUt VCS4UKN0 i215N V5 2 540 N Ans iii Me U MC shing as n Emum MC 2 1nso N u in Am 1 m The negative sign indicates that MC acts in the opposite direction to that shown on the treebody diagram Try soiving this problem using 39 UNiKVJERSiTY Oi a r Professor A Volinsky EML3500 Mechanics of Solids 11 SOUTH FLORIDA Example Review of Statics The structure is designed to support a 30 kN load mi d 20 mm 0 The structure consists of a boom and rod joined by pins zero moment connections at the junctions and supports 600 0 Perform a static analysis to determine the internal force in each structural member and the 30 k1 reaction forces at the supports 800 mm 1 UNIVERSITY F 12 Pr fess r A Volinskv EML3500 Mechanics of Solids EOUTH FLORID Free Body Diagram 0 Structure is detached from supports and the loads and reaction forces are indicated 0 Conditions for static equilibrium ZMC o Ax06rn 30kNO8rn Ax 40kN 2 Fx 2 O Ax Cx Cx Ax 40kN ZFy OAy Cy 30kNO Ay Cy 30kN Ay and Cy can not be determined from 30 these equations Professor A Volinsky EML3500 Mechanics of Solids 13 W Component Free Body Diagram W In addition to the complete structure each component must satisfy the conditions for static equilibrium 0 Consider a freebody diagram for the boom 2MB o Ay08m Ay O substitute into the structure equilibrium equation Cy 30m Results A40kN Cx40kNe Cy30kNT Reaction forces are directed along boom androd UNlVERSlTY Ell I Pr feSS rAV quotnsky EML3500 Mechanics of Solids 14 EOUTH FLORJlD Method of Joints 0 The boom and rod are 2force members ie the members are subjected to only two forces which are applied at member ends For equilibrium the forces must be parallel to to an aXis between the force application points equal in magnitude and in opposite directions Joints must satisfy the conditions for static equilibrium which may be expressed in the s form of a force triangle 30kN 2 FB 0 B FAB FBC 30kN 4 5 3 30 lltN FAB FBC if UNIVERSITY OF Professor A Volinsky EML3500 Mechanics of Solids 1 5 1 SOUTH Rosina Stress Analysis Can the structure safely support the 30 kN load 0 From a statics analysis F AB 40 kN compression F BC 50 kN tension At any section through member BC the r 800mm 30ml internal force is 50 kN With a force intensity or stress of p 50x103N UBCZ Z A 314x10396m2 From the material properties for steel the allowable stress is Gall MP3 0 Conclusion the strength of member BC is adequate UNlVERSlTY l I Pr fess r A Volinskv EML3500 Mechanics of Solids 16 EOUTH FLORID i Design with Materials 0 Design of new structures requires selection of appropriate materials and component dimensions to meet performance requirements For reasons based on cost weight availability etc the choice is made to construct the rod from aluminum Gall 100 MPa What is an appropriate choice for the rod diameter 600 800mm 3 P 30 Uazz5 A P 50X106N 500gtlt1cr6m2 A 01111 100x10 Pa 2 A7zd 4 4A 4500X106m2 2 d 252gtlt10 m252mm 7r 7 0 An aluminum rod 26 mm or more in diameter is adequate OF Professor A Volinsky EML3500 Mechanics of Solids 17 SOUTH FLORlD Example 13 The lmi St in Fig lane consists of the beam A and attached pulleys the cattle and the meter Determine the resultant internal landings acting on the cross section at C it the meter ie lifting the SEMIlb lend with constant velecity Neglect the weight of the pulleys and beam 500 th 13 Example 13 Continued Solution The most direct we to evelee this problem is to section bell the hl 05 ll 5013 lb and the beam at and then ee slder the entire left segment quot quot Freesmly Diagrane See Fig 11 615 Equmfem qf Equilibrium iv 13x 0 500111 NC 0 NC 5am lb Am H 2 Fl 0 5500 1b VC e Vt 500 lb Ame L 2 MC U S lb lt 5001b ft MC 0 ME 239U3900 lb l1 Am Mb 390 AS an exercise try obtaining these same results by considering juet the beam Segment AC Lew remove the pulley 311A frerm the beam and Show the SUDlb fezFee cempenente f the pulley acting on the beam segment AC UNlVEJRSlTY GP i i ProfessorAVoinsky EML3500 Mechanics of Solids 19 SOUTH FLORIDA Normal Stress The resultant internal force for an axially loaded member is normal to a section cut perpendicular to the member aXis The force intensity on that section is defined as the normal stress 039 lim A F 039 5 MAO AA ave A 0 The normal stress at a particular point may not be equal to the average stress but the stress distribution must satisfy the condition P oaWA de jadA A The detailed distribution of stress is statically indeterminate ie can not be found from statics alone UNlVERSlTY l SOUTH FLORJID Professor A Volinsky EML3500 Mechanics of Solids 2 l Shear Stress Forces P and P are applied transversely to the member AB 0 Corresponding internal forces act in the plane of section C and are called shearing forces 0 The internal shear force distribution is defined as the shear of the section and is equal to the load P 0 The corresponding average shear stress is P zIave Z 0 Shear stress distribution varies from zero at the member surfaces to maXimum values that may be much larger than the average value 0 The shear stress distribution cannot be assumed to P39 be uniform F Professor A Volinsky EML3500 Mechanics of Solids 21 SOUTH FLORIDA i General State of Stress Z One normal and two shear stresses acting on each plane of a cube within a body fully define its state of stress Stress SI units units of pressure Pa1Nm2 1KPa1O3Pa 1MPa106Pa 1GPa1O9Pa Uniaxial stress oo Copyright 2005 lPeareen Prentice Hall inc Sign convention Tensile loads are positive compressivenegative Copyright 39 2005 Pearson Prentice Hall lll lC Professor A Volinsky EML3500 Mechanics of Solids 22 Important Points When n berett that is suheeted te en esternel teed s seetiened ther is e diSIfi il tlti ef teree eeting ever the seetened etee whieh helds eeeh segment ef the bed in equilibrium The intet tsitjt eat this nternel three at s peint in the beds is referred tie es stress Stress is the limiting steine ef three per unit area es the eree eppteeehes eetet Fer this definitem the material et the peint is eensitieted t he eentinnens entt enhsive In enetsL there et e sis independent eenipnnents elf stress at eeeh eint in the heady eensisting nf H t i stresst e39m try em and shear isttesst my TF3 TIE magnitude ef these eentpenents depends upnn the type elf leading eeting t1 the heady end the nrientetien ef the element st the peint When s prismatie her is made hens Thinegeneens entt isettepe tttetetisl end is stthjeeted te ssiel free tinting thtengh the eentteitl Let the erssrseetiened eteet then the Hltlt flhl within the her is snhjeetetl net te t t r t f stress This stt ss is essntnett te he nnifrnt nt euereged este1 the eress seetienel street t l some stostns UNlVERSlTY l Professor A Volinsky EML3500 Mechanics of Solids 23 Procedure for Analysis equatien w EA gees the average nnrmal atreee en the erase eeetinnal area et a member when the aeetien ia anhjeeted tel an Eternal reanitant nearmal three P Fer axially leaded memhera N appiieatien at this equatien reniree the tellewing atepa I ternat Leading 1 Seetien the member nerpeettteaier tn ita lengitndinal aaia at the neint where the nermai atreaa ia re he determined an nae the neeeaaary treeheel diagram and eqnatien at three eniiihrinm w te ehtain the internal aaiat teree P at the aeetien eernge Name 5 etermine the memherte ereaeeeetienai area at the aeetien and eempnte the average nermal atreaa er PA 71 it it anggeated that tr he ahewn aeting en a email enlnme element at the material lieeated at a paint en the aeetien where atreae ie ealenlated Te de thia firat draw et en the taee ef the element eeineident with the aeetinned area FL Here a aete in the Etttt tt39f d iaeetine aa the internal teree F ainee all th3 nermal atreaaea en the areas eeetien et in thia direetien tn deaelep thia reanitant The hernia atreae er acting en the eppeaite taee et the element ean be drawn in its apprepriate direetien UNIVERSITY i t iSOUTH FLOEID Professor A Volinsky EML3500 Mechanics of Solids 24 Example 16 The her in Fig 11 11612 has a CO39FISEHWL width of and thickness of M Determine the maximum average norm21 streets in the her when it is suhjeeted t0 the leading ShDW 3 5 It m t leN Q e t2m 2 kN em 30 tth QkN Pm 22 RN 22 m PtkN h IZKN 30 22 12 ct GP Professor A Volinsky EML3500 Mechanics of Solids 25 SOUTH FLORIDA Example 16 Continued Solution farewell Leading By inspemiona the internal axial forces in regions A B Ca and CD are all constant yet have different magnitudes Using the method of sections these loadings are determined in Fig 1 165 and the l 3 W normal force diagram which represents these results graphically is shown in Fig 11160 By inspection the largest loading is in region BC where PEG 2 3D kNi Since the crosssectional area of the bar is C mta l the largest average normal stress also occurs Willth this region oi the ban Average Nermal Stress Applyian Eq l61 we have PRC stir103m The stress distribution aetian on an arbitrary cross section of the bar within region BC is shown in Fig 1 16d Graphically the volume or quot bloek represented by this distribution of stress is equivalent to the load or 3t hN that is 30 kN 857 MPa35 nim li mm GE Professor A Volinsky EML3500 Mechanics of Solids 26 SOUTH FLORIDA i Average Shear Stress Single shear lap joints til I b Vinternal resultant shear force Copyright 2005 Pearson Fret Double shear E a it Shear can arise indirectly due to the action of other types of loading Professor A Volinsky EML3500 Mechanics of Solids 27 BUTH FLELIDA Pure Shear Shear stress can also arise as a f Sectien plane Pure shear Fig 1523 result of other types of loading Equilibrium Censider a relnme element ef material taken at a peint lecaterl en the surface at any sectieneti area en which the average shear stress acts Fig 1523a If we censitier ferce equilibriUm in the y directien then iierce stress area l 7 it 2 a 1 e we as 7 its ay e e i Tar Ta Anti in a similar manner ferce equili h ttm in the tiirectien yields TH Finally talting mements ahetrt the Jr axis mement l l ferce arm l l stress area l l lquot l l 12th ray ea agree he y i Ta Tye se that ray 2 T3 T3 r r In ether wercis ferce and mement equilibrium requires the shear stress acting en the tep face ef the e lementi te he accempaniecl by shear stress acting en three ether facesu Fig 1 235 Here eiiferrr shear stresses must have equal magnitude and be directed either reward er away freer eecii ether at eppesi re edges ef the element This is referred he as the cempiemeemry preperty of shear and under the cenclitiens shewn in Fig 1 23 the material is subjected te pare sheer Altheug h we have censidereci here a case ef simple shear as caused by the direct actien eI a lead in later chapters we will shew that shear stress can alse arise irrdirectiy clue te the actien elf ether types ef leading y USP UNlVERSlTT GE SOUTH FLORlD Professor A Volinsky EML3500 Mechanics of Solids 28 Important Points a If twa parta Whi h ata Haifa aa aata ata jaiaad tagatfari tha aptia laraa aaa aaaaa ahaaaiag at tha matrial with aglifiibla handing If thia ia tha aaaat it ja aara g auitaljla fat agiaaatttg alya ia ta aaaatna that an aaaaaga ahaaa ataaaa aata WEI tha ataaa aaatiaaat araa i D ttaatimaa faataapra aualt aa naila and taaltaa ata arutsajaata ta ahaat t dam Th1 magnitua at a allEatquot fataa aha tha taatattat a gtaataat alaag a ptaaa whiah pasaaa thraagh tha attrfaaaa baiting j i A aata llg lawn fraaahvad g diagram at a aagmnt af the taata ar will aaat a aaaa ta abtaia that magnitu a and djmctin Elf thia faraat 39 UNIVERSITY F I Pr fess r A Volinskv EML3500 Mechanics of Solids 29 SOUTH FLOEID J Procedure for Analysis The eque en THE A is ue ed tie eem pute 011135 the ee eregeeheee ee eee in the materiel pplieetiee requifee the fellewing eiee Iniem i Seem I I eetileum the memberMm pint wiizere the eeerege ehear streee is ten be determined i I Draw the eeeeeeeyfreebedy diagram and ealeulia ite the internal eheerferee quotV Hing atthe eee that is eeeeemy takhell fiche in eqei b um 1 geeemng Sham w i ti mi e the ee e e i area A 11d ee mpute the eeeeege Shear etreee THREE 3 e 11 is eugeeted that Tm he sheen en a small eeleme element ef materiel leeete at e peiet en the eeetien where it is determine T dethiea rst drew THEE e11 the feee ef the elemenneei eidem with the eeetiened elfegg A shear etreee eets in the eeme direetie eye The Sheer etreeeee a i iting Den the three adjeee t ple ee ee then be drawn i11 thei1quot epp repriete dif Eti fellewi g the eehe r e ehewniu Fig 1 23 I UNWERSITY E Professor A Volinsky EML3500 Mechanics of Solids 30 SOUTH FLORJIDE Example 111 three of stnit on red ii The wooden stint shown in Fig i25a is suspended from a iiiInni diameter steei rod which is fastened to the wailt If the strut supports a vertieai ioati of 5 km compute the average shear stress in the rod at the wall and along the two shaded planes of the strut one of which is indicated as shed Solution Internal Sheen As shown on the treehotly diagram in Fig the rod resists a shear force of where it is fastened to the wait A freehotly diagram of the sectioned segment of the strut that is in contact with the ted is shown in Fig i25e Here the shear force acting aiong eaeh shaded piane is Average Shear Stress For the rod V 5008 N 7391an 2 6397 MPa Ana A TT39EUiUUS In For the strata V ZEOUN e H v Tavg 312 MPa Ans l1 Example 111 Continued The aaerageshear stress distributle an the sectioned l d an strith S39egm m SHOW i111 FlgS 1 25d and 11 2512 respectively Also shown with these gures is a typical V lumg element 0f the mammal taken at a plat located 011 the surface of each seetlen Nate carefully how the shear stress must act 011 eael shaded face Of these elements and them on the adjacent faces at the elements V 215 kN V kN 63 Pa feree af rail ea strut 5kN 5 kN BllZ MPa 2 d l a UNlVERSlTY OF 1 Professor A Volinsky EML3500 Mechanics of Solids 32 SOUTH FLORIDA l Allowable Stress Factor of Safety FS cleanc5 or FS tfaHt FS gt1 allow allow Failure stress is determined experimentally for each material In most cases it is the YIELD or FLOW STRESS Design for simple connections AP c or AV I allow allow Copyright 2005 Pearson Prenl f b snow Copyright 2005 Pearson Prentice Halli Inc j UNIVERSITY OF I FLORIDA Professor A Volinsky EML3500 Mechanics of Solids 33 Important Points Daaig a mantbat39 far EtI th ta raaad an alaat39i g all atlwahta attaaa that will aaatala it La aafaly auppatft ita ataadad N quotlaad T hat a at a malty Utili factata that aatt ia aanaa tha aataat attaan in a matahar and aa apaaiag upaa tha iatadad aaaa at tha mamiljaa a faciaa af aafatfy ta apptiad ta ataia tha a xaatabla Innat the mambat aaa aapa ma taut aaaaa illaatt ataa in thia aaatija rapraaaat jaat a fan at the irtaa393t apptaatiana at U aaaraga atmat and ahaar atraa famalaa uaa far atugmaarin daaign anal anjal yata Whaaaaar thaaa aquatiana ara appliadi hawaaaa it 115 imyar tant ta ha a atatt39a tht the attaaa dtatrtbatiaa ta aaauad ta ha aaifraaty attatriaatad a aaaragd Qatar tha aaatia 0F Professor A Volinsky EML3500 Mechanics of Solids 34 SOUTH aoama Procedure for Analysis WED Emitting pl ttall mating the average manual and SHE SJ SIIEES qtttttt Etcartul E n it r ti Shank fjrtf b made at it th gelatin vat whim the ETiTt l EEFESE matting Elma thtt SEEHJI is t EtdEE thus2 mamh l must then he i id t haw a Enf t t area at fEEiE E titJ t I ESDSt the SWEEE that acts tj itth determ thta aren ajaltcatitm t qttir t ttt tttlt tttjtlg aps IntemttILmtdt g 3 i Samth title mam tltmugh amt1 d draw tt f1 EEh y diagram at E1 tatrant t the 1 113111 tt5t The intrttal t tu lta t farce ttt th S ti t l is than at nme tt tl tg tt1 Ettatt nt atquot Equ jb u Regattin rm i Pra vtdedthe alluw ahle SHEEE is knth mi can be datstmi Ett ttm teq im amt1 ItEEElEd it sustain ttm I at the Bastian it than mmuttj ftttttt A 2 Pf a w 31quot A Wfr UKIUNWEESITY E Professor A Volinsky EML3500 Mechanics of Solids 35 SOUTH FLORIDA Example 115 115 The suspehder red is supprted at its end by e fixedeenneeted circular disk as shown in Fig i33e It the red passes through a 4min diameter hele determine the minimum required diameter ef the red and the minimum thickness eat the disk needed to support the teed The allewehte nermel S EI39B SS fer the red is Tatum M1339 and the e39ll39llewehle shear stress fer the disk is THEN 40 m if a Fig 1 33 UNiVERSiTY OF I i Professor A Volinsky EML3500 Mechanics of Solids 36 SOUTH FLORIDA 1 Example 115 Continued Q USF ILlNlVERSlTT SOUTH FLORID Selutien Dimnerer af rm By inspectien the axial three in the red is 20 Thus the required crasssectional area of the red is p 2nv1un339n v E i3333li3m2 Efallllrvar Se that d2 A A 714 U3333ll21r12 a 0206 206 mm Ana Thickness efDisk Sl lOW en the freehotly diagram at the eere sectiert at the Cli Slli 1 3351 the material at the sectioned area must resist shear stress to prevent movement of the lile threngh the hole If this shear stress is assumed te be distributed uniformly ever the seetienedl area then since V 2 2i kNi we have V earthen Tame 106 m2 057110 3 Since the seetienerl area A 2srl2 the required thickness at the disk is narrate 3 m2 r 455153 a 495 p I 2Wm 2 0 m 5 mm as Professor A Volinsky EML3500 Mechanics of Solids 37 Ch2 Strain Applied forces cause the body to change its shape and size deformationstrain Deformation can be highly visible or practically unnoticeable with the naked eye AS AS 861V g AS AS39 2 1 AS Strain is positive if the initial line elongates and is negative if the line contracts Strain is UNITLESS dimensionless quantity SI units are mm Sometimes is expressed in 0o stain of 0022 o Undeformed body Daformed bOdy GE Professor A Volinsky EML3500 Mechanics of Solids 1 SOUTH FLORIDA Normal Strain a Fig 2391 F39ia 23 q H II I StI BSS normal strain Professor A Volinsky EML3500 Mechanics of Solids Shear Strain The change in angle that occurs between two line segments that were originally perpendicular to one another is referred as shear strain 7Z39 7 11m 6 2 BAA along n CeA along t If 6 is smaller than 90 the shear strain is positive if larger negative Undetbrmed body Deformed body OF Professor A Volinsky EML3500 Mechanics of Solids 3 SOUTH FLORlD Cartesian Strain Components I ii 6915 l may Undeformedl Defigrmed Element clement Length of the sides of the deformed Approximate angles between the parallelepiped are Sides Z39nyi 7t2quotny5 Z39sz 1sX Ax 1sy Ay 1sz Az Normal strains cause a change in volume of the rectangular element whereas the shear strains cause a change in its shape Both effects occur during deformation The state of strain at a point in a body is defined by three normal strains ex sy and 82 and three shear strains yxy yyz and yxz I GE PrOfessor A V insky EML3500 Mechanics of Solids 4 SOUTH FLORIDE Important Points Laada will aaaaa all matartal baaiaa ta datarttt and aa a 651111 paints in Illa batty will uadarga diaplaaamaata alt attaagaa ta paattttm NamHat Strata ia a maaaura a tl39la al gatta at aantraatiaa at a small liaa aagmaat in tha hatta wharaaa aaaaa strata ia a maaaura at ttta ahanrga ia aagla that aaa ttra batwaan twat amall lttta aagmanta that am arigittall y parpaadiaular ta aria anathatz The atata at attain at a paint ta attataatariaad by aia attain aampaaaata thraa alarm al att atta ax aw a and thraa ahaar attains aw aw at Thaaa aatapaaaata dapaad upaa ttta ariaatattaa at t ha lina aagmanta and thaif laaa tian in tha badly Strain ia tha gam a tttcal quantity that ta maaauratl uai g aapari tatttal tachaiquaa Claaa atjtattjt t 39aZIa tha a tt aaa in that handy Earl than ba datattttittad tram naat at lal ptapart y ratattaa Mast angittaaring matartala undarga amatl datarmattana and at ttta aarmal attain a at 1 This aaaumpttan at small attain aaa lyata alflawa tha aalaalatiaaa fat nanItal strata ta ha aimpli ad IL a39ttttta titsteatdat appraaimattatta can be ma a abattt that aiaa In most engineering applications e 1 which allows to make certain useful assumptions sin99 cos91 tan99 for small 9 yr USP UNlVERSlT Y Gil Professor A Volinsky SOUTH FLOElD EML3500 Mechanics of Solids 5 Example 21 The Blender rd ShUW in Fig 24 is subjected t0 an increase 0f temperature ateng its exist which creates a nermat strain in the red 0f he 4039ZtU 3ng where Z is given in metere Determine a the displacement at the end B of the red due to the temperature increase and t1 the average normal strain in the red UNIVERSITY OF SOUTH FLORIDA Professor A Volinsky EML3500 Mechanics of Solids 6 Example 21 Continued Salutien Part a Since the nerrrrai strain is reperted at each point aieng the red a vdiifierentiai segment dag iecated at position at Fig 242 has a deformed length that can be determined tram 2 that is a2 1 sorry hag dz The sum tetal Of these segments along the axis yields the defermed length at the red iet It rm 1 z 2 J 1 401U3Izii2 dz 3 z 41300 3 zmjigZm 020239 m The displacement nil the end at the red is therefere B Z 0202391 1 1 in in 2 239 mm JL Arts Parr his The average nernrai strain in the red is determined from eq 2L which assumes that the red er line segment has an original length of 200 min and a change in iength ef 239 mm Hence as as rnrn Eravg I U iig AME Professor A Volinsky EML3500 Mechanics of Solids 7 UNIVERSITY Gt Example 22 22 A farce acting an the grip at the lever arm sheath in Fig 2 511 causes the arm to rotate clockwise through an angle of t9 2 0002 tad Detetmine the average nermat strain devetped in the wire BC a 11 UNIVERSITY OF 7 a 7 Professor A Volinsky EML3500 Mechanics of solids 8 SOUTH FLORIDA Example 22 Continued h Copyright 2005 Pearson Prentice Hall Inc 2 5 Sullutinn Since 3 L002 rad small the Stretch in the Wire CB Fig 2 55 is Bquot 2 605 0002 rad39015 m L001 mThe average normal strain in the Wire is therefme 33 1001 1 I T gg r H mm GE meeSSOVA Volinsky EML3500 Mechanics of Solids 9 SOUTH FLORIDE

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