Livro Beer - 5ª ed Mecanica dos Materiais (soluções), Notas de estudo de Mecatrônica

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Problem Solutions By Dean Updike 1.1 Iwo solid ylindrical rods 45 and BC are wolded together at B and loaded as Problem 1.1 shown, Determine the magnitude of the force P for which the tensile stress in rad A! eee erro ja dice fomaguitude-pf the-compressive-stressipaod:BC seas “ amem e Ve 50 mm 75 mm Bag CEO cs T6S-Simpat Po. , Sag EF Ana = S07-3x0 *P qt [eme WO amo 2 750 vam e] ha = FUSO 447 mt Saes <2limol — + Ag 6 260 — go E. = B.0543 — 23b-43 P PTN EANA Equating Eu to 2 Ga - ut so7.3 wioº Pe R(posAs — grbifexio P) Pena Problem 1.2 1.210 Prob. £.1, knowing at P = 160 kN, determine the average normal stress at the midsection af (a) rod 46, (b) rod 1. 1 Tio solid cyfindrical rods 48 and BC are welded together at B and loaded as shown, Determine the magnitude of the force E for which the tensile stress iu rod AB is Lwice the magnitude of he compressive stress in rad RC. ta) Rod AB. 50 uma Bois Tum : P= sto te (tension) ho td Aq = ES TES rstgsmm po TRT 4 pa 3 120 kN I Ga o bee orem má Am 1963-5 -—— FORO e 750 um (1) Eed BC. F-vso-(Qzo) = Bo kn ve. Bo EM compression, z 2 Age = Tee > lr. ELG G ma cu E tome Cear 180 MPa " Et ds = Efe = 106.86 mul= 706. BEMIG! ml Gu. E. doxto Cl A 706, B6rio = Ario Pa Geo HRAAMPa Proprietary Materisi. € 2069 Fhe MeGraw-Hitl Companies, Ine, All cighes rescrved. No part of this Manual may be displayed, reproduced, or distsfbated ia any forms ve by day umeans, without the prior written permission of the publisher, or used beyond the limited distributioa to tenchers and educntors permittod by Mefirave-Hál for their individual course preparation. A student using this manual is using it veios permission. Problem 1.7 1,7 Each of the fow vertical links hás am 8 x 36-mm uniforin rectangular cross Scctioy and cach of the four pins has a Hó-mm diameter. Determine fhe maximum valug ot the average normal stress in the links connectina (4). points B anil D.(6) points C and desafio Ê B Use bar ABC asa Free body, 20 4M 0.0RS e poda PRaRo Il. Faso [a EM, =0" (tomo — (o. vas + o. ogo)lzond) = o Fo = ses Link BD is m fension. EMp= 0: -(o04)Es «(o o2s)aomd) = o Ee = RS = N Link Cos ia compressio, Ned area Ff one Sink For fension = (o cos)lo.c36- 0.0) = 160x40 "mt. For tuo paraltel fintes, Ag: SAO m Tênside svess in Pink BD. Ga + fio - SESNOÊ (o seuo! Ghpz ILCMPa a 2 tam Ane 320 xj0o"* Area For one Pink in com press = (9.008 )(0.036) oa = pasx 10 mê Fortwo parabel Bates, Az SIA m e 1? e db SE fe = aii Fator o Gs MlTMA Proprictary Material. 92009 The MeGraw-HBl Companies, Ene. ALL rights reserved, No part of this Mamial raay be displayed, repraducei, or distobued in any forma or by any means, vwithont Ih prior wpiiten permission of the publisher, or used beyond the liniteilistribulioa te tenolhera and educators permítted by MoGraw-Hilt for their individual course preparation. A student neing this manual is usia if witltoi permission Problem 1,8 18 Knowing that the echtral portion of the link 80 bas a unitorm crass-sectional area of 800 imin/, determine the magnitude of the load P for which the normal stress dn that portion Of ED is $0 MPa. Fa = SA =(Sox108)(g00x 15 = 400 N Bo = fis, Can 2.00 m ea Dse Free Body AC For safras, o.56m mx 3 IR ERVA Tas Uibr IO AL 4) 4 aii (gone) CH) - Plovsid4)y=s o P= 38.1xl00N Peas un a Problem 1.9 1.9 Knowing that Bok DE is 25 mm vide and 3 mm thick, derenmíne the nostmal stress in the central portion of that Bak wien (4) 6 - O, (bj 8 — 96º 109 100 mm N fone cosmo emahe D] Use membey CEF os o Prez body ! Sum | 260 ima DEM o 0) Feto U)(agosinB)-fo 4)l24o cosa) = O For = Ibo sinO -Zz0cos 6 N A ve * tor025)(01003) = 75 %46ºm* Soe* Ee Aoc O 6-0: Fe=-B26 N -—320 cen DÊ 2 CL q47M maes Sos * some E 457 1Pe, 6) 06-40: Fer -!0N -t6o 75 8d Gor ” = —adaMPa -—a 1.11 Thesigid bar EFO is supported by the truss system shown. Enaw- ing that the member CG is à solid circular rod of 18 mom dimnetor, determine e the normal-stress-in-CGu= ecmemensemineceem anamnese rea Problem 1.11 Using porkios EFGCB «s a Free body + 5 -0:f% Fa-/5=0 Fa = 25 EN LE | EB reef 2-5 a Using beam EFG as « Free bol Tas Y OM -o: =[uajad Ee + 04) (ad E) = o Fc = Fu = 25kN Cross Sectional area E member CG ” z . z Am = Edo Elooby= 254.4 x10Pm Normal cress iu CG. - 25 Ger fm. ZE 0. rp [a > Da asda é & «at Problem 1.12 T.12 The rigid bar EFG is supported ly the truss system showg. Deter- mine the cross-sectional area of member AE Tor which the normal stress in He member is 105 MPa. Using portion EFGCB as à Free body HZK -o:fi Fe ts o Far = Z5EM 13 ace B af DS ro Stress in mcaber AE Gar * /05 4183 Em co Fe gs Soc Age Breno > E E 3 “e fm, F, 25x0 -b oz Ce Ane E e 2B3$eixio m at the time of failure. 15 mm a, Area being sheared N : À= JO mm x (Smm = 1350mm 2 [850 =/0" m E TA stect laica. sgoad Force Pe Seo! N a ” Shearing shress Lo Tc qiaiO a = SGSmgO? Pe 5.08 MPa a Proprictacy Material, O 2009 The McGraw-Hill €. distributed in any fosmt or by soy teams, wvilhout fe pri antes, Jec. AH nights reserved. No part of this Manual may be displayed, reproduced, or wrillen permission oF the publisher, or used beyond the limited distribution to teacher and educalors permilted by Melirav-Hi) for their individua! course preparation. À student using this magual is using il without permission. Problem 1.17 1.17 Two wooden planhs, cach 12 mas Uhick and 275 mm wide, are joined by the dry tnortise joint shown. Knowing fhat the wood used shears offalong its grain when the Som CaNErARE Si stress veachos 8 MPa: determine the magnitude “of the axial loud e um which will cause the joint to fail, fe 16 eum Soy areas must be sheared oPÊ wlhes Hhe quiul Fails, Each of fhese areas has alimensioas LG um X 12 mm, its ave being A-UEVIZ 192 mad = taguto we? At Foitore ht Foro E carried by cock of cuens ia F- TA -lexto liga xl) = Is3€ N= 536 kb Since Hheve am six Faisuve queas Ps GE c(GXi53c) = QIkN Problem 1.18 1.18 A load P is applied to a stecl rod itpported as showo by sn aluminum plate into which a 12-min-diamiter hole has heen drilled, Knowing, that the shearing stress must not exceed 180 MPa in fe steel rod and 70 MPa in the atuminum plate, determine the largest load E that cam be applied to the rod. < 4mm vam 4 | em For He steeb rod, A = Tah = (myb.elayo.o10) = B76.GI MIO im" q — RE%A, P = (180m0%37e.49>16") = E7.Berio' N Sum For the afominum plate, bz Td, t, = (mYo.omoYX0.008) = Loosanto” m” Ta E E me LA = Tha PB = Coro YU.cossrioS = vo. 372 m10º N a The Aimitina válse For + Dãad Pis tle smalior or f amd Pr P-eusbuio! N P=677kN à Problem 1.21 1.25 A 40-KN axial load js applied to a short wooden post that is supported by à concrete fuoting resting on uadisturbed soil, Determine (a) the maximum bearing . Stress on the concrete footing, (A) the size of the footing For which the average. s dt lhe soil is 145 kPa 2 O mm ta) Bearing stress on concrete Focting. Pe yo kN = Horto N A 6 - " (ipoizoi e axioma = JAxi0 im" “ = 2,.288x)p* Pa 2.33 MPa «ti 6) Footing arec. P= 4x0! N S= 45 kPa = Eoxio! Pa ou. Pp 2 Po. go N 6-7 hd fendas T O 27586 m Since Jhe area is square, A= b=Jà = JO giss, = OSS m be SS mm Problem 1.22 122 An axial load Pis anppored by a short WZ00 x 59 column of vross-seetional uva À = 7560 min” and is distributed to à concrete toundation by a square plate as shown. Khowing that the average normal in the cor uma must noi exceed 200 MPa and flat the bearing stress on the concrete foundation must not exceed 20 MPa, determine the side q of the plate that wii provide the most economica! and sufe design. For the column =| ow Pe GA (ox NIELS sia LA For lhe axa ph.te , Ss: 20MA; A: + LR. gos tm Since the pÍude is sguave Aco a JR AsiEE = ou 75m =275 mm ww itt Companies, Jue, Att sig teserved. No part of this Manual may he displayed, repsnduced. or Proprietary Material, 2009 The Mel distributed in any form ar by any means, without the prior written permission of lhe publisher, or use beyonul die invited eistributiva to teachers and educators perminted by McGraw-Hill for their individual course preparation, À stadent using Uia manual is using without permission. 1.23 A 6-min-diameter pin is used at connection € of the pedal shown. Rnowing thal Problem 1,23 P=S00 N, determine (a) the average shcaring stress in the pin, (h) the nominal bearingstressin the pedal at º, (ch the nominal bearing stress in each support bracket EUA - eme ” 75 eum - on mo] Draw Tyee bo? | Piaquam al Ado. B 15 mm Since ACD is « Q-fore member, the reuation al E is aivected tover? poini E, He intesccatiou o the Dines of action of lhe other two forces. From geomelry, ce-fiods ts = 325mm. HM ZE -0: EcC-P=o (= 2CP = Qels0o)-1800N to to [a 2Y 300) c - (0) War Se cd Sã Gio orono' Pa Quea8o MPa ca o 5 - a [o tZ00 e : = - = 2 to -27 tt 6 A OD TIA Fa Gr 2.1 MPa a ( Ge dl. + tãom = 2Tulos Pa 6: 217M Po a ho Zdi ICMS) O 1.24 Knowing that a force P of magnitude 750 N is applied to the pedal shown, Problem 1.24 determine (a) lhe diameter ofthe pin at E forwhich the average shearing stress ir the pin is-48 MPa, (6) die corresponding beaciag stress ix he pedal at C, (e) the comespanding beating stress in Me each suppor bracket at C. mm Dem VEM > vd Draw Free body eliagram of ACD. Since ACD is a B- Force member, the reaction at É is divected tomava poiet E, 4e intrystet ion of +he Lines oP achou of the other two Tovces. Smm From qeomelry, CE = faosF+ past = 325 mm +tz6-0: BEC-P=o C-26P-(260750)= 1950 N = AC. À€ FC. IMIAsSO Vo = . Br" ca) Tom E à EAR Eldon ID É ESP m dz G57mm -. e E se . e ss a db & da dt (asno) (Gnjo 5 ss. 9xiot Pa Gus 38.9MPa o g-3L.& iaso - somo Pa Gr8SoMPa «a E Zdt - QMESI IS seio) Problem 1.26 1.26 Two identical linkage-and-hydrantic-cylinder systems control the “o -—position-ofthe-forks oba fork-lift truck: The load supported:by the one system shown is 6 kN. Knowing dei (he thickness of member BD is 16 mm, deter- : muinto (4) the average shearing stress in the 12.mm-diameter pin al 8, (b) lh - bearing stress at B im member BD, f 300 mm SO0 amu 8, GEN | 4 > Bs 4D2Mg= O: 380 e ! obE -tos)têj= o Use one Fork as a Free Body. E = 8 eN — LIRrOo E+RB,= O Be: -E B, = GEN sem “go man 400 vam! bt =0 B,- GEN B-/8/48/- [5440 qareN Ae Edo Efe) peseprad tao GIRA A Tonto 6Imfa (1) Beocing stress «À B. 6- B imel Lorem Bo dt loror)to 06 Proprivtary Material. € 2809 The MeGraw-lilt Companies, Ino. All ights reserved. No par ofihis Manval may be displayed, reproduced, ot distributed in any form or by any scans, withomt the prior wailten permission of the publisher, or used beyond the limited distrilmaim ta teachers and educators permittod by Mutiraw-Hill for their individual course preparation. À student using this manual is using it wihout permission. Fi y ú | ermine shearing Problem 1.27 1.27 For the assembly and loading of Prob. 1.7, determine (a) the average 8] stress in the piu at B, (6) lhe average bearing stress at 2 in member 2D, (c) the average bearing stress at Bin member ARC, kngwing fhatthis member hasa FO, 5U- mm uniform rectanpular cross section. 1.7 Each of the four vertical links has an 8 x 36-mm uaiform rectangular cross section uud each of the four pins has a 16-mm diameter. Determine the maximum value of the average nóemal stress in the links connectitg (2) points 3 and D, (b) points C and É. Use bar ABC as o Free body. Cold 0,05 o.oto pe Fe» Fes DZM = o: (ooo), -(o. 05x 0.040 (A0u/0)) = O Foo * 82.5x10º N (2) Shear pin at B. Ts se Tor double shear, where Az Tdts Eloow) = golcexo tm 8 En) 80.8 10º L=80.8MPa at 0 Bearing: Pink BD. A = dt = (o.06)/0.008): 28x m* S = Eras - LOS NILE. casulo Gr 270 MPa «) Bearing m ABC + B. A- dt = (o ouMooo)=- Igorxio! mt 6 De (ueReo = 208 x10º 6 - 208 MPa a Proprictary Material. O 2009 The MeGraw-HII Companies, Inc. AU rights reserved, No part of this Manual may be displayed, repreduced, or distributed in any form or by any means, vwithout the prior written pormissioa of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student usíng this manual is using it without permission.