Solutions - ago ”542 ml 1:4“ b I[WK/0 [email protected]

Info icon This preview shows pages 1–10. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
Image of page 7

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 8
Image of page 9

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 10
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ago: ”542 ml 1:4“- b? I [WK/0?, DEQICWN @F' Ngicmi-MCAL erbms:m:g 9W3 fa H410 3&— iii 3 [p 6 q 5......»— wm#...._'_._.__..k,,..w7\___uwwfi% fl 5 -—-- ««««« r‘ ‘ ‘. MD ECDOOON :fi SEED b: mm ‘30 001) N c3“ :- .__E' 3 (:5on WW 300%szth 70x 25 z {1-14 Nipa ..E%— ; L39? ,—; In L132 éfljggzrry; 'Qmflgj 9 Kb “/2, OS “3:: s; I? z“: @302 j M. — iii fiwi‘r ' K ----—v—-M“ SELOJMW"! INN—‘27?“ 150 1W“ K [3 A Cm 6MB To a) @7530 ”(Q—DOOKZQ‘O') r: (:3 a) £44 :3: 65% N QFOLSO 9 I913 $1333“! 8% mwmanjil @ Wk" W‘ «m @i. » oajyfigw ngxggg u 75 (X3 7’: .J O 31 43 0.0 v3 2’ 3 13- “4.333- r: 2 KL”? [0632; 3'01.'t'i"'ri.(""i)N“(“5.12‘i ' ‘ " " ""‘“"‘“' ”WW“W‘“ Known: One and oils spring with known spring constant is attached to at pivoting, rigid fink. Find: Estimate the spring constant: i (a) with respect to a liorizontai force applied at A? (h) with respect to a horizontal force applied at B? (c) with respect to a: horizontal force applied at (2‘? Schematic and Given Data: I; = 5 Nlmm Assumptions: {1) The spring; is linearly elastic, and {2) the lever is rigid. Anaiysis: [. Tm 200 mm Point L! i o————-——-~> ]' Excerpts from this work may be reproduced by instructors for distribution on a not-for—profit basis for testing or instructional purposes onty to students enrolled in courses for which the textbook has been adopted. Any other reproduction or transtation 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. 2. Consider force "l1“ at. a distance “ I" such that the spring is elongated 1 mm. i2 Mn 2 0‘] SNQO 0 mm) =. (1:)(11) 30 F: tees. . . d . .M . . 3. Distance liit‘Ottolt winch is moves Is_(ai’200)1nm. The Spring constant is k’, or ki_ _ Form; i.)00( 200) _ 200,000 Distance“ 21 8-9- 41 For a- - 160 mm 1622-00- 009 - —20 Nlmm 1002 I Fr): 21:20() mm it' = Egg—«099: SNImm 2002 I For a r: 300 mm, 200 000 k' a «mm—’w—m- = 2.22 _\li . . 3002 i mm 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 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 ofthe copyright owner is prohibited. SOLIETION (5.1.7) Known: A steel shaft supported by sclfluligning bearings is subjected to n uniformly distributed load. Find: Using Castigiinno's Method, determine the required diameter, (i, to limit the deflection to 0.2 mm. Schematic and Given Data: \V = 200 thiin. Assumptions: l. The steel shaft remains in the elastic region. 2. The transverse shear deflection is ncgligihle. Analysis: gtitttion l 1. Add dummy load, Q, at point of given deflection. Neglect transverse shear and cousitlcr only bending. Measure ytti'iahlc distance, x, from left bearing. 2. Between i and 2, M]; m {1500 -l- git = lSGOx + 2,; Between 2 anti 3, Mg; 2: (1500 + ‘97)); -- 200(‘x a 5X93“ 2 ..5.) = isnnx ‘t- Q?‘ - 10th? + 100m » 2500 2 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. 'uJ Use "genera? deflection equation“ from Table 5.3.. Due to symmetry, integrate x between 0 and 12.5, and [Eton double. Lot hand Ispertaintolztrge((1)21ucismull (073(1) portions of shaft respectively. 5 , , I 2. 5 , , .3 z 2 [ trttztiwtzssi.d . J. 2 [ Matter-221992. d. l 1 , .i 11; 3 hi1 . ”-2—.” .) l.‘ ' “Q35“; P L531 (130de "ii “2”" . i .12 .‘a + if (isoox + ”if 2 100x? + 1000x - 250mg.) (is I, ' . 4. Now. ict Q a 0. 13.1.. 131”; "m 5 l2.5 s: -2.-. 1399.33. -- ....2 . MM - .1993: - Jiltitttxfi - ZAQOX ,1 t Lt 2 NM _ t 2 2 -| 2 2 Mix .3 s 3 ports: i 1 (250%- was .. 2600a) gm 5 35:5,. ., ‘ iii} 3 4 2 5 . 62,500 764,648 b : Wormw. 4.. _ u ‘ . 11.13 1311 5. Since 13 = 30 x $06 psi (Appendix C—i) EL 2 3““ = 0.0490964 (‘14 4 15 m 39.27%.12... = 0. {)1553 dd ()4 (3. For s =0.21nln : 0.00787 511., 62.500 763,648 0, 0078’? : ..._..._-s.e,,.wm...“WWW + mm-.-“ (39 x rupturnssmn (30 x i.t}‘J)((}.04909::1") 000737 2; 0. 1341.5 + (2.51924 (1“ d“ d“ 2 33.023 (1 2: 3.02 in. 1. Because of symmetry. consider httif the shaft as u cantilever, with the deflection caused by the. 1500 ii) bearing. force being 0.2 mm (0.00787 111.). Excerpts from this work may be reproduced by instructors for distribution on a not-for—profit basis for testing or instructionai 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 ofthe copyright owner is prohibited. Q) N = 200 lhi‘in 0.75:] Neglect transverse shear and consider on! Although F is known to (’iM/OF is taken. Let {S and I}, shat}. t'CSpCCIiVCiy. 3. Between 1 attain/1,2 = F): Between 2 and 3, N133, WW (ix + ‘ 5 N1 ”(61"! 121/0“) (3 2: .V.._._....._“ 17115 0 l 1 .. E 15:5] 5 2313:..3. EIL3 4 .3 0 swim 4.1 ((71? +1 13M "’ ) st, {male 2 0.0553(5. 0. 00787 = w- 45 - "“70 399.)... 30 x 1 (tin). 015530;). it. 00737 e 023.432. .4. t). :3 t9?. (1“ ([4 d4 = 83.019 and Li 2: 3.02 in. ' ' ' ' ' twfor-profit basis for testing or instructions is work ma be reproduced by Instructors for distribution on a no . . . ingirgéigflggued in courses forwhich the textbook has been adopted. Any other reproduction or translation of this War permitted by Sections 107 and 108 of the 1975 United States Copyright Fx — i200tx « 5}j-5-§~§--} = Fx Using the genera! deflection equation in Table 5.3, y bending. Mcnsurc x as shown. inc 1500 lb. it must be. kept as a variable nntii pertain to the small and iatrgc portions of the .. 10(inZ + l000x ~ 2500 r. I23 = WI 17*‘(x)2tix +.f,.§1m[ (Fx .. 100x? + 1000}: — 2500)(x) dx :. k 3 t. -n 5 2.5 +.1.Q.Qttx3 ....z:i1t)-x=" . (609. 371-: - 594nm + 509, 375 — 164,062) Substituting F = 1500 it), a = 0.00787 m, E; = 30 x 106 psi. IL 2 n.04909ns, ,+ KEEN] 500) " 59”” + 50-..2331i:I.§€fe.9§3 30 x §()"(0. 04909L‘i“) t purposes only k beyond that Act without the permission of the copyright owner is prohibited. Stiflli't‘EON (5.25) Knotm: A 37in. Hioztm is made of steel having 8'}: m 42 ksi, Find: Determine the safe axial C()ttt[_)!‘t3$fil\-’LZ load based on a safety .t’actor of 3 for pinned ends and unsupported lengths; of" {a} it) in, (b) 50 in, (c) 100 in, and (ct) 200 in, Schenmtic and Given Data: ;§::1.6:ll§l.2 1,, 3-2.5 in." 123:: 0.40 in"! 55. 3x42 ksi Assumptions: .l. The t—heam does not fall under compression. 2. The l--hen:n is straight. Analysis: Since] = A4712, pm, z: “/1121“ m- “146 = 053 in. ”" it“ TEE q ‘. , i , . t l.“ {gnlfi' :l- , hitler~Joi1nson tangent. pomt [l1({. (5‘ l 3)} as at “55;: [-77:43 \t'ttere it: -: 30 X 30“ ps1 ' lt “ 3' I ‘ , '1 I, -, Lu ”1233(30 x 10-1) :2 m ( (Appellfllk (, l). T’n'HXiM—l m 11.) (a) it"; = fi, = 18-9. Johnson equation flirt. (5.12)] applies: 8 ,3 " r2 . . . s: n: s i - ‘(1 = 42 - .v (18.9}? = 42.5 ksr L" 3 4.21%: _ 51> ,a with SF 2 3,1) 2 (41.500 psi)(l.64 inf-"N3 : 22.67011) I (Note that, in this case, coltnnn action is almost negligible, and Sc;- 5y) ' ' ' ‘ ' -tor—profit basis for testing or instructional purposes only from this work ma be reproduced by Instructors for distribution on a not . . . Exggrgéits enrolled in oourges 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 permtssron ofthe copyright owner is prohtblted. (b) If a 6‘2 2 94. 34. Jt'iimson equation still applies: J ‘ ‘ J . 2 Ser = 42 W $421“ (94,34)? 3 23f} kg; 4n»(30,t)00) P = (28,700) (1.64)}3 = 15,710 H) I W has 100% ,. -- . (L) Mr? w 623-— 153.03. Lulu. ht}. (5.11) Applies. “ 3- Set = "-9131?“ = 35 {730‘0021 2 8.32 m (min? messy P : (832()}(1.64)f3 : 4550 it) i (at) 390139593 = 3"?.36. Eq. (5.11.) eppiies: SC]. e filtfigzimgl x 2. 08 ksi (377. .36) P 2: (2080)(1.64)/3 r: IMO lb fl 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 transiation 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, ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern