Solutions - 7650315230 Dgsrou/v 0F MECHANm-L Cmmmg...

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Unformatted text preview: 7650315230! Dgsrou/v 0F MECHANm-L, Cmmmg Fat/“’10P? HUD SEWLI 30W c031 Twfis 7503 um Gm wwm W cfi-m [:1 - Q5: 66 PQ‘L 1* 66000 03/“?— _7L,_ Emu Cb F9: 261‘ T __E:_~. \H CW: '77— : I67— : me (“*— J’ "frag 77d PS “jg‘fl” : £45“ , CMOLXIM ML mwfiipmfiD 3 (2°11 : ééQQQ _. - dfi Qoéme'TSOO Ln [5 X75012 6mg???” ng : | ll‘mflw‘o .M __ s ‘ ‘ CED kg P ' fig" Cmdmmm 94W Wmeflj CW => gem. gawk” 9&:W 16%7300 2200:3pr "W “:i‘ 1.40 m Fg : 3% =3 ’2 9 2 W 7mg .5 0L : Q°HXJZX\(X'7$OO :2 {5me 426000er SOLUTION (6.9) Known: A thin piste of known material is loaded in tension and has a centrai crack of given length perpendicular to the direction of the applied load. Find: Estimate the highest tensile load that the plate will support. Schematic and Given Data: Leaded beryllium copper su =9s ksi S},=117 ksi KI . = 70 ksi infl's C. 2w = 8 in. t m 0.05 in. EC = l.5 in. P = ‘? Assumptions: 1.. The crack length is a small fraction of the piale width. 2. The tensile stress based or} the net area (minus the area of the crack) is less than the yield strength. 3. Yielding has occurred within one small volume of the material at the crack root. 4. Crack propagation to total fracture occurs instantaneously when the limiting value of the stress intensity factor K] equals or excoeds the fracture toughness Km for the material. Analysis: 199“: wwwlflsm = 44. 9 so ‘i'Tsrc' 1,. SW75 2. Since the area equals 2M. P :- (T3(2\Vt) : 44,900(8)({}.05) = 17,960 lb I i. From liq. (6.2), (SE = Comment: The PIA stress based on the net area, {(ZW — 20), is 55.26 ksi which is less than S‘. = 1.17 ksi. Hence the second assumption is satisfied. Excerpts from this work may be reproduced by instructors for distribution on a not—for- to students enroiled in courses forwhieh the textbook has been adopted. Any other if permitted by Sections 107 and 108 of the 1976 United States Copyright Act without t profit basis for testing or instructional purposes only eproduction or translation of this Work beyond that he permnssron ofthe copyright owner is prohibited. manna-1‘“): K m “J um SOLUTION (6.19} I Known: A machine frame made of steel having, known 8}» and SSy is loaded in a test fixture. The principal stresses at two critical points on the surface are known. Find: Compute the test: load at which the frame will experience initial yieicling according to the (a) maximumhnorrnal-stross theory (1)) maxinnini—si1ear~stress theory ((3) maximum«fixation-energy theory Discuss the relative validity of“ each theory for this application. Compute the. value of test load at which yielding would crnnmcnoo. Schematic and Given Data: (12: 100 MPa 1 ud— (,i+—~ a l we a] e 200 Ml’a i Test toad : 4 RN 02.. S), a 466 Ml’n 3‘ : 250 Mi’a by Excerpts from this work may be reproduced by instructors for distribution on a not—for—profit basis for testing or instructional purposes only to students enroiled in courses for which the textbook has been adopted Any other reproduction or translation of this Work beyond that permitted by Sections 107 and 108 of the 1975 United States Copyright Act without the permission ofthe copyright owrier is prohibited. (31 (Mill’s) 400 (‘5 Theory Load Line _200 v - 7 for 13 0.5773), £30 85),: “250/” 600 , T r 133(31‘)’ - Sheer DE. Theory Diagonal \‘h—w— Moln‘ Them 1 Assumption: The material is homogeneous. Analysis: 1. For the maximtun-norms}—strcss theory, the (31 v 02 plot shows point a to be critical. Faiiurc is predicted at. Load a 4 kN (1499-13-49 a: 8 kN I 200 MPH _ 2. For maxrmum—shear-stress theory. the or ~ 02 plot shows pgjnt t) to be critical, Failure is predicted at Excerpts from this work may be reproduced by instructors for distribution on a not—for—profit basis for testing or instructions: purposes only to students enrotled in courses for which the textbook has been adopted. Any other reproduction or translation of this Work beyond that permitted by Sections 107 and 108 of the 1976 United States Copyright Act without the permission of the copyright owner is prohibited. Load =4kN =6.4 kit . 3. For maximnmhdistortionenergy theory, the (it rsngm Shows Poi-m b {0 b6 critical. Failure is predicted at lead “him MP?! 7..) m I More precisely, from Bo. (6.7), 09. = [(1.50)2 +0100)?— -(150){-i0{})]1i2 2 218 MPa Thus, failure is predicted at ' l 400 MPa 1': fl“, Loni 4kNl‘218Mpa '7 MN I Comment: 1. Maximum normal stress theory should not be used for this application since it gives good results only for brittle fractures. Maximum shear stress theory may he used but is not very accurate. Maximum distortion energy theory will give the best results for this application. Yieiding is expected to begin at. a load of 7.3 kN. sew Excerpts from this work may be reproduced by instructors for distribution on a not~for~profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this Work beyond that permitted by Sections 107 and 108 of the 1976 United States Copyright Act without the permission of the cepyright owner is prohibited. W9 W w‘wgw m (WM M W Uflfirnaté, flaw} Qu : 930 Mm, :qgo N/mmz' \{ludi 2 5 600 {la/90““ 1 606 N/mm?’ 23W 6M9 W w W W QR 990 MP» 2: I32 real, 15mm fig '84?» Cg :0er gm can: or? CAw'cLO 2 mm. QT )CR :{ (m1 ’— U%CIOG>’%Q® r" (3‘3 9 3 0L7, 5 [935006,] v {[email protected]§) :. (rag—{336g .4230. 5': :- vlfiflnogv — [‘DBLI'ECfl ._-_ LE); 1 “" LQfSCEESO) : {£63 £332») 3321229??? W 0901 w (69, | “0696 [Ego "sofit’f't‘tON (8.28) Known: A11 unnotched bflt' and 21 notched bar ot known material have the some min imum cross section. Find: For each bar. estimate (at) the value of static tensile load P causing fracture {13) the value of automating axial. load x P that would be just on the verge of producing eventual fatigue fracture {after perhaps } -5 mitiiou cycles). Schematic and Given Data: .m P 4—m ' i 30 mm 30 mm T 35 mm 30 mm :' = 2.5 mm Machined surface A181 IGSO normalized steel Assumption: The bat is manufactured as specifier] with regard to the enticed fillet: geometry and the bar surface finish. Analysis: 1. For a static fracture of it ductile material, the notch has little effect. Hence, for both bars, 3 P g A‘Su where Su 3 748.] MP2: (Appendix 04:1) P m(30111111)2(748.1 M ’3) 2 673 X 103 N 13 = (no kN I 2- so = 311’ CLCG-CSGI‘CR \Vhfil-C Sn: 3-“ = Mpii CL: C]- 2 CR 2 1, Ct; m 0.8 (Table 8.1) CS 2 0.74 (Fig. 8.13} Excerpts from this work may be reproduced by instructors for distribution on a not—for ' ' ' ' ' I _ “profit beats for testin or instructional or 0553 on! to students enrolted In courses for which the textbook has been adopted. Any other reproduction or translagtion of this Workpbegond that y permitted by Sections 107 and 108 of the 1976 United States Copyright Act without the permission of the copyright owner is prohibited. 8,, = 0.5(748.1)(§ )(t).8)(0.74)ti )(It) : 22] MPH From Fi g. 4.39, K = 2.50 7 Assuming Bhn : 217 (Appendix (7-423), using Fig 8.24, L] m 0.86 Thus, Kr: 1 + (Kl - Uq [‘Eq. (8.2)] Kr = t + (1.50)(0.86) = 2.29 3. For the unnotched bar. P = AS" 2 (30 mth (221 M‘Pa) =199xm3 N =199 kN 4. For the notched bar, P: A‘Sn/Kf :3 £99 kN/2.29 = 8'? kN I Excerpts from this work may be reproduced by instructors for distribution on a not—for—profit basis for testing or instructional purposes only to students enrolied in courses for which the textbook has been adopted. Any other reproduction or translation of this Work beyond that permitted by Sections 107 and 108 of the 1976 United States Copyr§ght Act without the permission ofthe copyright owner is prohibited. St'.)l,li'l‘l0i\i (8.36) Known: A coid-tirawn rectangular steet bar has knowu hardness value and dimensions and is to have infinite life with 90% reliability and a safety factor oi“ 1.3. Find: Estimate the nurximum tensile force 113M001} be zippiied to the ends: (a) if the force is con'ipleteiy reversed, (b) if the. force varies between zero and a maximum vaiue. Schematic and Given Data: Ht} film h. m 10 mm I) 2 (at) mm (1 = I2 mm Assumption: The hole is symmetrically machined in the plate. Anniysis: 1. For 1.40 131111, S“ x 0.5(140) = 70 ksi or Su : 6.890(70) : 482.3 MPn {Appendix A—l) ’2. From liq. (3. 1.0a), S): m 525 Elm w 30,000 m 42,800 psi : 295 MW (May be higher for cold drawn, in any case, problem is not affected) 3. 3H m sn’ C1,C(;C,;(}r(2n this. (8.1)] S.{ = 0.53” (Fig. 8.5) (71,2 CT m Ck : 1 (Table 8.1.) Cti = 08 (Table 8.1) :8 s 0.78 (Fig. 8.13) Sn = (0.5){482.3)(I)(0.8)(0.78)(i){l) = 150 M1321 I But this is (conservativeiy) for 50% reliability. For 90% reliability back off 1.3 standard deviations (Fig. 6.1.?) of 8% or 10.4%. Therefore, 8;; (90% reliability) = 150((.i.896) = 134 MPH. 4. From Fig. 4.40, Ki m 2.5 From. Fig. 8.24, q = 0.85 (by extrapolation) Thus, Kr: l + (Ki —1)q [$118.11)] =1 + (l.5)(0.85) 2 2.28 - A i?) F , or F = 0.210 Excerpts from this work may be reproduced by instructors for distribution on a not-for—profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this Work beyond that permitted by Sections 107 and 108 of the 1976 United States Copyright Act without the permission of the copyright owner is prohibited. 6. For eompietety reversed load, Gum = 1’34 MPa; Fum = 0.31 (Tom); on with a safety factor of 1.3, For mrodoanax load “in l “a On (MM) 100 3“ “2% miifc.,‘£?0% retinbility 1 ()0 200 30!) 0“. (Ml-‘11) 400 8. From the figure above. tor xet’o-to-max toad, i onm m 105 + 105 z 210 Ml’a. ’i‘horot’orc, with a safety factor of 1.3. Puma = {f.).21)(21.0)/i..3 = 34 kN ! Excerpts from this work may be reproduced by instructors for distribution on a not-fornprofit basis for testing or instructional purposes onty to students enrolted in courses for which the textbook has been adopted. Any other reproduction or translation of this Work beyond that permitted by Sections 107 and 108 of the 1976 United States Copyright Act without the permission ofthe copyright owner is prohibited. SOLEi’I‘lON (8.39) Known: A ,l {Z—ln. pitch roller chain plate is made of carbon stccl hostwircniccl to give known values of Sn and 83:. All surfaces are comparable to tho "machincc " category. The link is loaded in repeated axial tension by pins that go through the two holes. The safety factor is 1.2. F ind: Esiinmtc the value of maxinnnn tensile force that would give infinite fatigue: life. Schematic and Given Data: i 0.050 in. ’4: (l. :25 in. din. Su 3 I40 ksi s? = I In its Assumption: The roller chain plate manufactured as Specified with regard to the. surface finish and critical hole geometries. Analysis: 1. The not tensile area in a section through the hole axis is (0.382 h 0.125)(0.(J50) = 0.0129 in.2 2. From Fig. 4.40, with (Nb m 0.33, Kr m 3.3 From Fig. 8.24, q = 0.37 From liq. (8.2), Kr m i + (K wl)q Kr: i. + (2.3)(0.87) = 3.00 _ _ I 2((l.0l29) — “(3.31 Al. Sn = Sill CLCGCSCchR 3. (Fm : Ga 3—" = Snr W 0.58“ C1, 3 01‘ a CR : 1 (Table 8.1) Ck; m 0.85 (Table 8. i) C35 = 0.70 (Fig. 8.13) ...‘n :10.5(l4l})(l)(l).85)(0.70)(l)(l) :- 42 ksl I Excerpts from this work may be reproduced by instructors for distribution on a notwfor—profit basis for testing or instructionai purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this Work beyond that permitted by Sections 107 and 108 of the 1976 United States Copyright Act without the permission of the copyright owner is prohibited. =10 60 30 [60 $270 140 0m 6. From the graph, 0m 3 (fax 32 ksi. With SF = 1.2, (i .2)(t 1113);“ = 32,000 Therefore, I: = 229 ID I Excerpts from this work may be reproduced by instructors for distribution on a not-for—profit basis for testing or instructionai purposes onty to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this Work beyond that permitted by Sections 107 and 108 of the 1976 United States Copyright Act without the permission of the copyright owner is prohibited. ...
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