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Unformatted text preview: 7650315230! Dgsrou/v 0F MECHANmL, Cmmmg Fat/“’10P?
HUD SEWLI 30W c031 Twﬁs 7503 um
Gm wwm W cﬁm
[: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. manna1‘“): 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) maximumhnorrnalstross theory (1)) maxinnini—si1ear~stress theory ((3) maximum«ﬁxationenergy 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 maximtunnorms}—strcss theory, the (31 v 02 plot shows point a to be
critical. Faiiurc is predicted at. Load a 4 kN (14991349 a: 8 kN I 200 MPH _
2. For maxrmum—shearstress 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 Poim 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 {ma@su§) :. (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’ CLCGCSGI‘CR \VhﬁlC 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 (7423), 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 coidtirawn 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 notfor—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’otomax 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 notfornprofit 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 notfor—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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This homework help was uploaded on 04/01/2008 for the course MECHANICAL 342 taught by Professor Baruh during the Fall '07 term at Rutgers.
 Fall '07
 Baruh

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