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CHAPTER [| PROBLEM IL1 fit Determine the modulus ofresilience for cach of the following grades oEstractural lá steél; (a) ASTM ATO9 Grade 50: a=50ksi (5) ASTM A9II Grade 65: a=65ksi SOLUTION (c) ASTM ATOS Grade 100: a, 100 ksi Structural steel E = 27x10º psi for all three see be given. to) 6,= S0ksi = goxjo* psi . Sé . Gomot) - : 3 Uy= GE Cojgêrio To 48d indb/in a 0) 6, = 6Sks * GSr)0! ps; 2 GE esmo o ss 7 UGano) T 728 im bb fin a (mm 6= |00ksi = Woxto* ps” SE . Yooxiot) = eg — Ain int à, Ly Ge Uat=100) DIZ inlh in PROBLEM 11,2 11.2 Determine the modulus of resilienoe for cach of the following aluminum alloys: (a) LOOSE =70 GPa, w=S5 MPa (5) 2014-T6: E=72GPa a = 220 MPa SOLUTION (c) 6061-Y6: E =69GPa or = 140 MPa ASominum abloys tw) E= vox Pa, 6 = 55x/0º Pa 2 SÉ . Aston sq. ta 3 Ur = ZE da = RI.6x!6 NemAm 21.6 KW /m — b) E = zanol Pa, 6 * 220x/0º Pa [o à 220x10º + z 2 /, 1 = LL = = e - amd u, REAL = 336»%)0 N tm o 336 [RE Mm e) Es GisloPa, & = Ioulo* Pa gs yo xo*)* : Ss. s 77: dis = OW Nmfm = 420 k/m = E — PROBLEM 11,5 The area unde Fhe 14.5 The stress-strain diagram shown has been drawn from data obtained during a tensile test of an aluminum ailoy. Using E = 72 GPa, (a) determine the modulus of rcsilience of the alloy, (5) the modutus of toughness of the alloy. SOLUTION (0) SG - Es us BE = LEE*= Alaxio ovo À E J296RIO! Nem/n3 = JAIE bT/mS — (6) Modulos OS tovghness = total uvea yudes de stress -strarn Curve The average ovelinade e) the stress-stram curve is oo MPa = S00x10t N/m cuvve is A = (500x/0)(0.18)= Gox/0º N/m* modulos of toughness = Foxtot I/m' = Go MIL = PROBLEM 11.6 116 The stress-strain diagram shown has been drawn from data obtained during the tensile test ofa spevimen of structural steel. Using E = 29 x 108 psi, (a) determine the modulus of resilience of thc stcel, (5) determine the modulus oftoughness of'the steel. SOLUTION tm 6 - Es o) R So. dese Iaquot Ão. 002)" Ur = ZE [E 58.0 imbb/in* — Wb) Modolos of tforghness = total area Under He stress -straru curve À = (57 A0.25-0.002) = 14.44 kips int cs Judy ineleip A int Ar= Hasloas-0.o21)= 3.21 Mplint = 34) inekipoL in3 Ay: Gtroloas-0.005)= 2,83 Kiprlinco 233 inkip in? modotos of toughness = UV, + A, + As +As 8 20 in kip /in3 at Q J ; : 3 [ms o O oa Soaoame pa a La í o = “Gana m (O) PROB 11.7 The load-deformarion diagram shown has been drawn from data obtained during OBLEM 11.7 lhe tensile test of a specimen of structural steel. Knowing that the cross-sectional area of the specimen is 250 mm” and that the defbrmution was measured using à 500-mmi gage length, determine (9) the modulus of resilience of the stos], (6) the modulus of PN) tnughness of the steel, SOLUTION Assomina +hal iebdina ocevr3 al pe càs kv and 5 = 0.6 mm Uy- 4 (esmo Moe!) = 18.75 Nem = 18.95 J Volume S stressed material V= AL: (a50%500)= ias x1o* mamas = 125 * mo mt Ú 13.75 - o 5 - a Uy = Y * agr iso nto = 150 h7/m — A, - (6a. s elo! gem) = Guto* Nom + 6x' J A, E laexiot V76- 8.6)xl5?= Aazuiol Nem = L22í=t0* T As Total energy U- Up+4A+A 4 AÃsE TES* 108 3 3 modulus of tovghness = D - JBSmlO. - cano t T/m V Casa ta - 63 MI/wm* — 2 (160) 61 x 10) = Génio Nom — oelxiotl PROBLEM 11.9 As Using E=29 x 10º psi, determine (a) the strain energy ot the steel rod ABC when 8 kips, (h) the corresponding strain energy density in portions 4B and BC of the rod. SOLUTION Pe 8 kps, E = Qqnio! ks - LJ < « . E A- Ed v- AL, 6: É, us Se U-=vV Portion|) d L A V [= [o] UV wi, m [int tm? use ia-kepfint medir AB lo.e2s| 24 jo.30s [7,303 | 26.08 | tita? ge saxo? ge joas |3e lo.sus [is aliz. | scesmio? | sa aan to? z [e-xeto? ta) Us mex do? indo = téZindh 0 oO Ih AB ori PAsotindiplino= TR in Ab Zon” = se uv: Sesotm kp/mr &6S ind dinê -—a a! To Ed IZ10 A 3U-in, length of aluminum pipe of cross-sectional area 1.85 in” is welded to a fixed support 4 and to a rigid cap 3. “The steel rod EF, 0f0.75-in. diameter, is welded t0capB. Enowing that the modulus of elasticity is 29 x LO psi for steel and 10.6 x 10º for aluminum, determine (a) the total strain energy of the system when P = 1Dkips, (6) the corresponding strain-energy density in the pipe CID and in the rod EF. PROBLEM 11.10 SOLUTION Cm er: ATA: o ums in” mr 4 PEL 4 Qoko) (30) CD: Us 2EA Ulioemot Ya) Fr: no PL - gonotYtus = mL EF: D qEÃ * axaeio MGM) 187.38 im dl, Totadt D= to 4 Ure = 26H indh a . = = 10090 , Et A-guosY 1 . . co: 6:-1S=-StoS psi, U-2g = NENE = 1.378 inodb fin! a - - Joodo . — GÊ. AZ68S n - o .4 Er: = LM = 2268Cpa, U=28" Gini) 2,83 indh/in! us 76.49 ind PROBLEM LLI1 48im, o CD: Mete 1L10 A 30%n. length of aluminum pipe of cross-sectional area 1.85 in? is welded to a fixed support 4 and to a rigid cap B. The steel rod EF, 0f0.75-in. diameter, is welded to cap B. Knowing that the modulus of elasticity is 29 x 10 psi for steel and 10.6 x 109 for aluminum, determine (4) the total stmin energy of the system when P = 10 kips, (6) the corresponding strain-energy density in the pipe CD and in the rod EF, 11.11 Solve Prob. 11.10, when P=8 kips. p SOLUTION For EF: AsEL = omg “(-8000 )*(30) º “sexies ) — 8.985 ind EE: VU ->PL . f8cco)(98) = 09.89 in db ZEA (ZiranioNlo. qa) Total VU: Uo+ Us 168.8 inf = a a CO: S=- Lg = 432 poi, ud SE = 0.882 indb/nt : - B0002.. ; . e QBoB) o m/s EF: O cui 18108 pai, DE tyaariom) — $.65 indbiia? am PROBLEM Í1.14 J1,14 The steel rods AB and BC are made of a steel for which the yield strength is ” O '= 300 MPa and the modulus of clasticity is E = 200 GPa. Determine the maximum strain energy that can be acquired by the assembly without causing any permanent deformation when the length à of'rod 4B is (2) 2 m, (5) dm. 10-mm diameter SOLUTION AT EQ) = 78.54 mi = 7E.S4 96 mt Asc = (E)! R82M ml = 28.174 10! mt êm = Pe GA mn 7 (So0noNas. 27!) = E. 4822 =10º N a U «Ha toy as Am l-a- 6-2 = 4&m 0 - Esso PG teto 0) (ENgoomioGesquos) (ZMacoo nto (as are nto) = 45803 + 25.66 = 30.0 Wm = 30.0 J st (63 a = tm L-u= 6-4 = Am Us (B.u822» CO) (Eu (O . (Bronx 0 KT.samo*) * Bg0o wo ICAO) = 9.606 + J2.72338 = 21.4 Mm = 2.9] a 1.15 Rod 48 is made of a steel for which the yield strength is a = 65 ksi and the PROBLEM 11,15 modulus ofelasticity is E=29 x 10º psi; rod BC is mede of an aluminurh alloy for which & = 40 ksi and E = 10,6 x 10ºpsi. Determine the maximum strain energy that can be acquired by the composite tod 4BC without causing permanent deformation. SOLUTION Ane Elo9)! = O nS6 im" Ez 29000 ks Ao T(o6)= 0282 in" E =IogoO ks Ty = SA Fm cach part 0.6-in. diameter 0,4-in, diameter AB: Pu =(ESAO.I2566 ) = 8.16 hip Bc: Pu: (HO)b.2B214) = 1).3096 kips Use smaller valve P: s.:67% kips nes EL. Ce term tas) à LEMONÊ CEO ZEA (QN2g00oo.izsec) — (2U10600 Ho 28274) =" 256.3x/0* + 467.50" —º ta4xO*inilipe 724 indi a i PROBLEM 11,16 Ur > “ "» a > QU o - e 11.16 Rod AB is made of a steel for which the yield strength is 9, = 300 MPa and the modulus of elasticity is E = 200 GPa. Knowing that a stram energy of 10 J must be acquired by the rod when the axial lond P is applied, determine the diameter of the rod for which the factor of safety with respect to permanent deformation is six. SOLUTION For Puctor of safety of six on He enengy D, = (6X1o)= c0J SE (300m008) s 3 = Sr (S00nt0 2º / Dy 2E —CABooxio*) 225 nloé Tlm - Ih be Emo A Lur (1.8 Raas xo") 148.148 »10 tm 4 = [EA - | entanto) = 13.78006" wm E = 13.73 mm met, PROBLEM 11.17 Ein. diameter 11.17 The rod ABC is made of a steel for which the yield strength is o, = 65 ksi and the modulus of elasticity is E = 29 x 10º psi, Knowing that a strain energy of 90 im « Ib must be acquired by the rod as the axial load P is applied, determine the factor of safety of the rod with respect to permanent deformation when a = 1$ in. SOLUTION tendimetor Ang = (E) = O uigint Ae HMA)» o.19635 in! Pe: Sy An * (65000Y0. 19685 ) - 12763 Ph. PL U = 2 REL Qazes Vis), U,. (raves) (UB 18) “(Rxaa io o. agia) Es = = 4! Dag DO = N Ab. UXarwrios o. Igé8s» 48 in O PROBLEM 11.20 % error F1.19 Show by integration thst the strain energy of the tapered rod AB is U 1 PºL 4 EA im whcre Anis iS the cross-sectional area at end B. 11.20 Solve Prob. 11.19, using the stepped rod shown as an approximation of the tapered rod. What is the percentage error in the answer obtained? SOLUTION Ass Tin Anim + me" 2(, U- papa N ido 2» - EIA > e ey + +qo * mer sms EL - by fossas) - o asse EE a O.2485€ - 0.25 SE xloo% = - 0,575 % -— 11.21 Using E = 10,6 x 10º psi, determine by approximate means the maximum strain energy that can be acquired by the aluminum rod shown if the allowable normal stress is ag = 22 ksi. PROBLEM 11.21 2.B5in. 285in. SOLUTION Ami = TUSV = IGT in” Ca * 22000 psi um Lind) JE (22026) = tias ia (2 MBEBIP) UU. IBIZS) À Tn (1o.Ex ot) ” 2: Dress Pu: Gula t 5887 Bh U ( Pile. PÉ dx . ap? dx ZEA RE ) Et TE d* Dse Simpson 's role + compute He integqual W= 04S im Section din | 402 (| imutitier] om Cat GE) Í 1.50 oaqua ' o.quga R R. to 0.22675 “+ o.q9D70 3 2.58 | 0.15374 2 0.3076 4 2.8s | 0.12311 4 O.4924 & 2,00 | O. MI ! o.) z 2.2625 dr d* U 102.7 in db. ES 1 o PROBLEM 1.24 11.24 7 the truss shown, all members are made of'the same material and have the uniform cross-sectional area indicated. Determine 'tho atrain energy of the truss when the load P is applied, Joint € LF, Do -SE, -P:-o Fe E:-ÊP Fer +2Zk-0 —Fectkç=o Fac = FP Joint D Fe +1 ZF,- O Ko Fo + EF, =O Be Fap = P Member) F L A F'L/A Us Z+EF BC aP 2 A SPRAA .+ EL co “4 22 | A | ÉPLA 2 A eo le lit j A |ieLA - gta GA) 47 . z 4.732 PRA = 2.87 E = ES: 3 E 11.25 Jn the truss shown, all members are made of the same material and have the m PROBLEM 11.25 uniform cross-sectianal area indicated. Determine the strain energy of the truss when ' the load P is applied ma SOLUTION IL . E Ted | +ZR-0 &R-ff*o já a, Faz Fo Es E MZH:o -P-2iFoso Il Ke = Fa=-& P ' Joiat D [1 E +-Zk = 0 N -Re- ÉR, = O o Fao o Fen * -2- PP a 4P — » ! = SEL ds EL If] a . v-2r A pa Member | EF L A | FILA a (oa PS) na es |-Sr| 22] qn |4 PIA = RE (1 A E co |-EP| €1] 24 | 4 PRA - 2 ER | ep» | arl2 | A |ã PRA Es NO z E PA = 0288 A m Lo. L | [a PROBLEM 11.27 11.27 In the truss shown, all members are made of aluminum and have the uniform cross-sectional area indicated. Using E = 72 GPa, determine the strain energy of the truss for the loading shown.. 11.27 Solve Prob. 11.26, assuming that thc 120-kN load is removed, SOLUTION Lec* Ro [13015 = 14Sm Toint 12 - +» Zfç -tEte-bero O Zoo SM fem ess Fo -400 - O Sobving (3 and (2) simultantous dy, Fee = R60 kM Teo = - 260 ky Joint D HZRH- Rot Tã E: 0 Fo = 00 kW “4 Es EL LS EL Ro :Z REA * RE 2 Member) ECEND) LO] A Sim] FÍLIA (e Be 260 tis 1800 73.23 x lo! BD too Es 1208 12.80 x jo” co - 260 1.95 3000 43.99 x do” z 27.67 n to 6 Putos Us 134.67m! 900 Nm = 4900 S at EMTaro?) é 11.28 Taking into account only the effect of normal stresses, determine the strain PROBLEM 11.28 energy ofthe prismatic beam AB for the loading shown. SOLUTION VEM =O aP+ LRg-o Re--9P.. aPj Over portion AD a M= - Px P am d— Ç 2,2 A M Yao * , RETA = 2275, Pxde = ger ER xy . Pla - GEI . aP Over portion DB : Mz-v L L ani Me 2( aPly e Upa - Ç 2ET dv = ml, pv dy E, - Elas ao o Pt pat “aP AEIL” d REJE 31, GEI L Pa” Totah U- Us + Uss T “er (ut) - 11.29 Taking into account only the effect of normal stresses, determino the strain energy of the prismatic beam 4B for the loading shown. PROBLEM 11.29 SOLUTION Symmebric beam anel Aoaclina Fa" Re- HZF,:o R+Bac2P=zo Rato P Over portion AD + M= Rx = Px a a 2 (a a s|a 2,3 - M . 2 tha x) . Pia Us = S, TE MM * ger y X decir) T GEL ado. Over portion DE : M= Pa Do: Pai(L-za) de À AEI 2. E Over portion EB - B, sy muto y Use = Uso = ta Total Dz: Ui + Um + le = La (31 - 4a) —— 39 E 
