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Unformatted text preview: ago: ”542 ml 1:4“ b? I [WK/0?,
DEQICWN @F' NgicmiMCAL 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 {114 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
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7’:
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43
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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 notfor—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 89
41 For a  160 mm
162200 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 forprofit 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 notforprofit 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 [ Matter221992. 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 notfor—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. ' ' ' ' ' twforprofit 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}j5§~§} = 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,, 32.5 in."
123:: 0.40 in"!
55. 3x42 ksi Assumptions:
.l. The t—heam does not fall under compression.
2. The lhen: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 101) :2 m (
(Appellfllk (, l). T’n'HXiM—l m 11.)
(a) it"; = ﬁ, = 189. 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, ...
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 Fall '07
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