Baixe Solucoes resistencia dos materiais e outras Exercícios em PDF para Resistência dos materiais, somente na Docsity!
12n0nãoãa Om Sm aca [mm] CHAPTER | ed => 4 m Lj [a | q e O | 1 [ 9 [ma CJ |! tar tb) PROBLEM LÍ 1.1 Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Knowing thate, = 30 mm and d, = 50 mm, find the average normal stress in the mid section of (a) rod 48, (b) rod BC. SOLUTION rod AB Fome: P= Goxio! N teusiou Area: As Fa = Elson!) = oe serio! GoxI1D? 706. 86 w jo" Sag = 849 MPa = gy. 88x/0* Pa Normo) stress: Ga * x rod BC Force: P= coxo? -tavasxio!) = - ox" Area: À - TA = T (sono) = L 9635 w10"* m . P -Jagxto? Normal stress: Gu = 4: CT Er - 96.77 x10º Pa Gg * 96.3 MPa «a o UR RR o l : Ls Los a 9 OCO 9 ci ca í o Cds ld E) PROBLEM 1.3 1.3 Two solid cylindrical rods AR and BC are welded together at B and loaded as shown. Enowing that d, = 1.2$in and d; = 0 75 in. find the normal stress at the midpoint of (a) rod AB, (b) rod BC. eajê SOLUTION IE (0) rod AB E P= I2+10 = 22 kips & he Falo Tas) o= Lagra im? = Pp - 22 - í «tl lets Sa = A crdaga 17.98 k 25in. | da tb) rod BE le Pp = I0 lips tos A = Ed) = Tlon) = ou mn Po. Jens . Gas = K = soqs * 2R.6 ks e! PROBLEM L.4 1.4 Two solid cylindrical rods AB and BC are welded together at B and loaded as rod BC: 2. d, 10 ips BP O. quo) shown. Knowing that the normal stress must not exceed 25 ksi im either rod, determine the sinallest allowable values of the diamerers d, and d; SOLUTION vod AB: P=Iario = 22 kips & 12%dps Cas * RS ks! Awmsta - Lo ue = Ama O Tá 1. HP. MU) e d, “TS T(as) — 1 1Z0S in . diz |.ost mn P=ibkip Cor aski Aucfa oe as = - , TS” mass 05093 in da: 0.7Mkin, oa I PROBLEM 1,5 4.5 A strain gage located at C on the surface of bane 48 indicates that the average normal stress in the bone is 3 80 MPa when the bone is subjected to two 1200-N forces asshown. Assuming the cross section of the bone at ( Lo be annular and knowing that its outer diametes is 25 mm, determine the inner diameter of the bone's cross section até SOLUTION (4YUzOS) 2. -3 2. dy = (asmo!) — taão vier z azar. G mao! m ] d = 4.98 0107 d= 14.93 mm 8 PROBLEM 1.7 1.7 Link BD consists of a single bar 30 mm wide and 12 mm thick. Knowing that each pin has a 10-mm diameter, determine the maximum value of the average normal stress in link BD if(a) 8 =0,(b) 8=90 SOLUTION Use har ABÉ 20 kn as free body e (a) 60=0º (0.980 sin 30º Xgoxio!) - (0.300 cos 30) Fa = O Fe > I7.324/0* N (b) B=%º (0.450 cos SN Xaoxio!) - (6.800 cus BO) » O Fa * -30 nO N Áreas (a) jensia hocolim, Az (0.030- O.oro)(0.02) - 240 OS m (b) compression A= (ooso)o.or)= a60 xo! mt Stresses . Fe vesauo! “ - to) € 2 * movie 72.2x10 72.2 MPa w c-Je- FÉ - -B3.3x10º - BGMPa a | ll. ] | | í [| —S 3 ra PROBLEM 13 18 Each of the four vertical links has an 8 x 36- mm uniform rectangular cross section and each of the four pins has a t6- mm diameter. Determine the maximum value of the average normal stress in the links connecting (a) points 8 and D, (b) points CandE SOLUTION Use bar ABC asa Free body. ao km É oses fe 0,040 —s ai B) A B a Fas Fes sM,=0 owoF, - (0025 +0.040)(20x10º) = O Fao = 22.4 wo! N Link BD is in fensron ZMg= (o) -(o. 0%) Fis -(ooas row) =o Fe tes Link CE is hm compression. Net area of one Pink For tensim = (0,008X0.036 - 0.04) = Igoxo nt. For tuo paralie) Pia des Amt? 340 “o* mt Tenside ses im funk BD 3 mo So? fe. = eSEi « lol.s6nO! o IOLG MPa Area Pon cne Pink in compressim = (o.c08)(0.03€) = 288x jo mê For two para Mheb Bites, A = 576 IO mt E 12.5 wjo? - do 6 = eo F2Suir. arovio! or -ATMPa ca a co Cl [ a] PROBLEM 1.10 1,10 The frame shown consists oÉ four wooden members, ABC, DEF, BE, and CE. Knowing that each member has a 2» 4.in rectanguiar cross section and that each pin das a - in. diameter, determine the maximum value of she average normal stress (a) in member SE, (b) in member Cf. E É TR SOLUTION a / Add support reuchtons fe + vre so Ih as zhown. =M=o vo D, - (4S+30Yu420) = O Des 900 db. Use member DEF as Free body ZFs =o 4 D r EAD &D, -£D,=0 = y e ET D, = 4Dz = 1200 db. a: É F EMe=Oo +2YfRe) - (3onGID;=o Fer = - 2250 tb. ZMe=0 (30X& Fe) -(5)D =O Fer 240 dk. 225 ta Stress in compressiea member BE Area Az Rinxtin = Bim? (a) Gao = HE» 28660. — — 281 po < Stress im tension member CF Minimum section area vecurs a? pin, Ama 7 QULO-OS = Om” — FR. - Tt . Se RT = IOTIps «a mo = minimum section Using entire Frame as Tree body ( PROBLEM L.II 4.24 For the Pratt bridge truss and loading shown, determine the average normal stress in member RE, knowing that the cross-sectional area of that member is 5.87 ir”, B D F un SOLUTION / VÁ na Use entire tras as Tres body , sb EM=O t E RR RR RE (avo) +(18X30) + (ar)lso) - 36 À, = O gr) or--|-or-) of] by = |Zo kips so bips Sokipe S0kips e E use portioa o truss to the det & a section — cultivy members BD, BE, and CE. q Fur + Zh= 0 AZ > Mo - 30 - EF =O «Fes 50 kps , cs Ger 5 Fase . SO B.S2 es;
Fa Io hps 30 kips Ay = IZ0 lips Use portion & tross to the hedt Pa section Cotting members BD, BE, and CE Dm =o0 2 Fe-(QUnNo)= o A Fes = O kips Gu = Ee es Áee As te - JL: wait -— Ge | 0; | [Ts PROBLEM L.t4 1.44 Two hydrautic cylinders are used to control the position of the robotic arm ABC. Knowing ihat the control rods attached at 4 and DD each have a 20-mm diameter and happen to be parallel in the position shown, determine the average normal stress in (a) member AZ, (5) member DG. dal SOLUTION Use member ABC os Free body. sou DEMO (0.150) É Fi - (0.600800)=0 Fez Helo N Area sf rotio mento AE A = Fdt= EQonS!) = Su ulo É mm! . . - =. pulo? - Stress in rod dE: Sa 2 ÃO = guiéxça = 12-73 x/0t fa (3 Ca = 12.73 MPa — Use com binedl members ABC ond BFD as free body, DzMp=o 5 (o iso XE Fe) — (0.200 (É Fo) - (1.050 - O, 350)(800, = O Foe=-ISdo N Avex ww vod DS é Az Ed*= Flzomo!)" = gmIé Wo im! Stress ia rod DG? Eos z e = Tea = To Pa 6) Goo = TT MPa ati 9 CI Cs a PROBLEM 1.15 1.15 The wooden members 4 and B are to be joned by plywood splice plates which will be fully glued on the surfaces in contact. As part of the design of the joint and knowing that the clearance be:ween the ends af the members is to be 8 mm, determine the smaliest allowable tengrh ifthe average shearing stress in the glue is not to exceed 800 kPa. SOLUTION There are Fou separate aveas S qlue. Each orea must trausmit hab df Ha 204 Load. ThereBor. EF =12 AN = tíx102N Shearina stress in ye €= 800 x10? Pa 3 Let = Jong d qle area and wz width = 100 ma = Ol m -s “As lw as A = Es re = [50 vim = 1580 mm L> 24 + gp = (aXiso)+ 8 = 308 mm a PROBLEM 1,16 SOLUTION A= mdt for Erensing sbress Equadina A Sefving For d: 1.16 Determine the diameter of the largest circular hole which can be punched into a sheet of polystyrene 6-mm thick, knowing that the force exerted by the punch is 45 kN and that à 55-MPa average shearing stress is required to cause the material to fail eylinalrica Pacduvo gurf ara .£L = Pos. AF mdt » £ - E geo - = d FF -* T(o.006X55 x08) 43 dello m d= 43.4 me «a Lo) cd E) Li L [cd lo = PROBLEM 1.19 1.19 The axial force in the column supporting the timber beam shown is P = 75 kN Determine the smailest alowable length Z. of the bearing plate if the bearing stress in the timber is not (o excesd 3.0 MPa SOLUTION «.P. P 8 A Lw La E Expo” Ow Goro Yo.Ito) = 178.6 2/0 m L= 178.6 mm nd Solving For Ls 1.20 An axial load P is supported by a short W250 x.67 column of cross-sectional area 4 = 8580 mm and is distributed to a concrete foundation by a square plate as shown. Knowing that the average normal stress in the column must not exceed 150 MPa and that the bearing stress on the concrete foundation must not exceed 12.5 MPa, determine the side a of the plate which will provide the most economical and safe design. PROBLEM 1,20 SOLUTION Area F codomn: À = 8580mm= S$BO MIO m Normal stress in colomn” 6 = ISOxiOS Pa g= Ê . P= AS: (Essomo VU mo?) = LAB7 IO! N Bearing plate: Ghs= Ê and As O For square plate. As 4 - = q. Polar L agimjo!m = ROS w 10€ ou 32) mm mt =. d — [ Ls L ) [a [| Im 3 PROBLEM 1.21 1.24 Three wooden planks are fastened together by a series of bolts 10 form à solunn. The diameter of each bolt is !4 in. and the inner diameter of each washer is in., which is slightly larger than the diameter of the holes in the planks. Determine the smallest allowable outer diameter « of the washers, knowing that the average normal stress in the bolts is 5 ksi and that the bearing stress between the washers and the planks must not exceed 1.2 kai SOLUTION Bert: Apr Fl = TA): o la6as iu G= És Tensile forceinbolt P= 6,4 =(5 019638) OABNShips Washer: inside cliometer = di = É im, outside diameter = do Bearin area Aw = Ela. AD and Au = E. i & Equeding ERA e 2 Si, z Ni EE .2 do = ds = (5) + COMECE = 1.4393 in do = LtI7 in a PROBLEM 1.22 1.22 Link AB, of width 4 = 2in. and thickness 1 = "4 in., 15 used to support the end ofahorizontal beam. Knowing that the average normel stress in the link is - 20 ksi and that the average shearing stress in each of tas two is 12 ksi, determine (a) the diamezer «? of the pis, (b) the average bearing stress in the link SOLUTION Rod AB is in compression. bt whece bz Aim and bein -5A =-(30X2NKF) = tO kips Pr: too É od Ap= Fo v > uu Ap = - Ea - [EDU = in co d de = TE mts” 1.030 in 382.8 Jesr = É. ca Cl [a | | a p [a 1.8 Each of the four vertical links has an 8 x 36- mm uniform rectangular cross PROBLEM 1,24 section and each of the four pins has a 16- mm diameter 1,24 For the assembly and loading of Prob 1.8, determine (a) the average shearing stress in the pin at C, (b) the average bearing stress at Cin link C£, (c) the average bearing siress at C in member 48C, knowing that this member has a 10 x 50-mm | uniform rectangular cross section. c SOLUTION Use bar ABC as q Free body focas —b p090 — o : ' D) A B E Fao e 2Mg-0 -(ocro)E, - (o.oaskizomo')= 0 Eles AS! (ay Shear in pin at € A= Td* = En (0.016 Y- goL.ogmo wm Double shean T= Fer - RS nto? - 3LIxjo Rà (2NJo1.0€ n19ºº) 31.1 MNPa a [65] Bearing in dink CE at E A=dt= (0.016)0.008)= iagx0"* m” GS. Ate - Sasiliz seio? = U2.8x)0º US MPa 
