Midterm2

# Midterm2 - try/— ‘*2.90 The composite cube of Prob 2.89...

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Unformatted text preview: try/— ‘ *2.90 The composite cube of Prob. 2.89 is constrained against deformation in the z ' . _\ . f PROBLEM 2.9a ‘ . , dlrectlon and elongated in the x direction by 0.035 mm due to a tensile load in the x direction.. Determine (a) the stresses ax, 0;, and a” (b) the change in the dimension in they direction. E, = 50 CPI: v“ = 0.254 Es. = 15.2 CPS. v,Er = 0.254 E, = 15.2 CH: v = 0.423 SOLUTION - Coms‘i‘t‘aim'i‘ Condi‘iﬁbn Loni conoiE‘Hd-t 63 = O From equw'ifiw (3} 0 "' " a” 6‘;- ‘f E; 6.: (o.25ﬂ(1&2) = 0,0772% 5;: \$1: vﬂEz 6" = 50 Ex con'i'inueal Pf'o‘a Fem 2-90 CCM+I'|1 L) ed FV‘OM equw‘kau (I) wf-Hn = o I _ Egg- =- 0-7333 5K _ 'E’, a, 6“ " 033037 = (50:10“! )g 375x :0“) _ 3 z m) 6.; I 0.38039“ - thzsxto 1%. we M90, as 6‘1=_(o.ov7m'a)(q~r.casuo‘) =- 334W”; 1:“ = 3.45- ma 4 . A E 313.1 J; - 2% Frran(21 83': Ex 6", .9.E.-_)(6")f is: I = _W+b_ ,(o.tsiii)1'3-wemg:1 50“” 1512x101 1* - 32.3 -73 x-io“ _ ‘8‘] 2 L35! 3 (whm)(“ 323335404.) =' _Q‘o'2q MM d 33w” 1': in ' ,E _ E r 4 3: 21" Fla-K J— J- C “Thwv‘ 9 an AB: a...___c2‘m25° # " «e» = 4 ° ‘3 mama.) ‘ '5'9'5Y’0 '" C : 25.15110 W: = 25.!5MM KC - 50.3 than i : 3 .__. (Z E [150 I ._ M - Rod BC. (2 1T (3x106) 3:. 83: we "‘ m} c = 31.;9xto"m = 3km um cat: 2:; = 63.9....“ «In. 3.19 Theﬂbmle stressisSO WammebraserodABandﬁ MPainthealuuﬁnum red BC. Knowmg that a torque T = 1250 N-m is applied at A, determine the required diameter of (a) rod AB, (b) for BC. .351 (1', .‘-1_-:': q: ’. RQBEEM-3i19 SOLUTION beam shown is made ofa high—strength, low-alloy mmmm Using a factor of safety of 3.0, detennine, ghe largest; beam when it is bent about the z axis. Neglect the 4.3 The wide-ﬂange 0&=345 MPa and our—450 MPa. couple that can be applied to the effect of ﬁllets. PROBLEM 4.3 SOLUTION I, -. r;- w + A A‘ Ti; (250)033‘ +(250)(J8){m )1 ISL 70:; Ho‘ m. 4 6;.er : (Isawto‘)(2=u.7sxlo“) : 243m? NW C OJSO 243 kN-m 4.53 Aconuetcbcamisreinfomd “new 1mg laced _ .. PROBLEM 4.58 magmas ofelasﬁcity is 3 x 106 psi ﬁx a}; wring 30px 10‘; Usmg an allowable 3111333 of 1350_p3i ﬁx- the mm and 20 ksi fm. the. 51:11: determine the largest allowable posnive bcndjm moment in the beam_ ._ . . J. .— Ah 3 3c!" = BQK‘é): = 1.30% ;..Z ‘l. nAa = [8.040 1‘." LocoJre. ned‘l‘mﬁ gun's. 3x33: - (I3.ow)(M-x\ = 0 4x3 name at — 252.5: = o -J3-oqo +4 13.0451+(4)(4)(25'2.5c) (270+) - IL! —x = 7.?QJ i‘I’I Solve 479» x x= -‘-' 5,005 8 x3 + nASO‘f-XY‘ = éisxaoosds + (:3.oqo)(7.9as)" .. 6‘]: M - “j n: 1.0) U]: 6.005 In) 16]: I350 P5? = (1350)U730.\$) '— ‘ 3 ‘ . \ (1-0 )(éJDOS') - 339 {o ‘ND‘W‘ 2 33‘? k'p-IV} h'—' I0 'lj'= 7.9‘15' J 6‘ = 20 x103 Par M = {201103)(1730-§) - X 5 _' .. u . (IO )(‘7Aﬁ53a #53 lo .99 m — 433 k‘p,” aLoOSQ 'Hne Smajler‘ VOL/pa): 38‘! lap-{n 1' 32.? kCP ...
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## This note was uploaded on 05/07/2008 for the course MAE 131a taught by Professor Nesteranko during the Spring '07 term at UCSD.

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Midterm2 - try/— ‘*2.90 The composite cube of Prob 2.89...

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