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Test 1 Examples - th—dJ—dlfllflb lilti 4b LIIVIL L...

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Unformatted text preview: th—dJ—dlfllflb lilti: 4b LIIVIL L l 'v' 1 L IZNL] l NIZIZH l NL: D'KJ titifi' 4'Kti4 l" . Ell/U4 an Ed} .3532]!!! : ' z m. 1:» 5734428931 RYAN MUELLER pm: 34 Question 9 (25 points) ' The rigid barHDE is mom by zwu links}!!! and CD. LiukAEia mud: of aiminum (3 E 70 6P3} and has a crow-madmal area of 500 mm": link CD is made of steal (E =- 2fl0 GPa) and has a cross-sectional area of 600 m“. For the In!“ farce sham determine the deflection {a} or .6, (b) at 13,0?) of E. (52:14pr Frpbkm 2—1, Tcxfi, Fj- 43) i) mam A Fan a! m rigid mméu‘“ BM, EFF-'3 E5” [40 KM _m : PL ,, wgzax 3,3 5.3% E > C5; 2 J13 W "‘" LMH‘FHW) =5 -é,$.5 yin-"fin :3. .. 4.35“.“ 1 {MHz} m‘) ('74:: “97 PA.) :2. :3 - I ' M W EWM‘). ,, =— P .3 A (aim WE‘M-JC 14:: em '3 P4.) E5, B” __ D. Lfl'ihm fiflfifl mm)"‘x DDI - HID arQHfl-l 3L ’ 75,":- 73.7mm . IE ‘ . fimfifa '1 M : nggwa Egg,” ._ HE SE = (wflmfia-(TSITm-K) In.“ ‘ M fl... _.—:— W DB! H , G-‘t'mr'l WEI-TH“ , M 5’ E =ID+LDE EM 9 lg I ;-.. M +rqao*§.atfuu'3):W~+-1.J ‘5 3‘57 "m" ~11 th—dJ—dlfllflb Kit“ 41:! LIITIL, L1U1L I:NL:I1N|:I:H1NL:I D'KJ titid 4'Kt14 1" . [Id/E14 1:12! ngam 61:15 5734428981 R‘MN MUELLER PAGE E15 Qu-tion 10 (25 points) For the aluminum shaft shown (G :- 3.9 x it)" (a) the totqua T mat causes mangle of twist of 5", 5m: torque T in a solid aylindrical shaft of the psi), dietarmine (b) the angle of twist caused by the same lengm and cross-sectional am (PM Parjl‘ 15.1,.) Ted”: SAqfl'IL 11.5 halfanfl). ==> 7” m... L. :erf' fizr‘” fi_ TAJIZQ‘HSJ 5’4! 77" i" 3 L F #5” ‘ {'0 57.2.5" X/fl r44 .7; "‘3' I' I #3 (335215 ”PW" 1"“ 2-33 kip-I‘M £> Sflélfl 5915131371” 39’ - 1"“ £753 muggy-“1w - . fififlfifi “fidflgflfl am *4, grit. £7!“ haiku.) 51mg) A: 1- ; I ‘ 1T“; C: )1, Sat-1b 511A???) A 2 171’” T'- . _ 1,. =- 3-, 55' Ar‘kfi WWI-[r 0- {1L III- flll '_: afrfah (“5""- Cl =5 Laurh” “ {1.55“} In = 'IT' - 0" 4‘51 ’" 13035”)? = marina-Sin" ¢$ (2.,azma')(~is3 . _ Z .3 (3"?XM‘)(153-6’xm”5) «336-? Ma a“: a, a; Mo“ = :9. 31.1.5 in 3' SEP-EE-EE’IES [38:46 LIMC CIUIL ENGINEERING 573 882 4784 HEB/[214 ENGR 2200- Intermediate Strength of Materials WJ. Likes Test #1 F all 2004 Question 9 (25 points). At room temperature (70°F), 3. 0.02-inch gap exists between the ends of the rods shown below. At a later time when the temperature has increased to 320 °F, determine (a) the normal stress in the aluminum rod, and (b) the change in the length of the aluminum rod. f——lfiin.—-’/r—10inu—~lc ml Aluminum Stainless steel A m as ing a = 1.2a.2 E = 10.4 x iospsi E = 23.0 x 105 psi a a 13.3 x 10—31“? a = 9.6 X 10-51“ F r w:- L. m 7L5 fl P<_L: + _.5_.. # _._. J, m - A gr ate Ages, ’37:“ ”35) to > : l (U WUOH‘MO 3 l ll‘l-DliBHDD SEP-EE-EBES 88:47 UMC CTUIL ENGINEERING 573 882 4784 P.E4/E4 ENGR 2200- Intermediate Strength anateriels WJ. Likes Test #1 Fall—2004 Question 8 (25 points). Each of the steel links AB and CD is connected to a support and member BCE by 1&2 inch—diameter steel pins acting in single shear. Knowing that the ultimate normal stress for the steel used in the links is 60 ksi, determine the allowable load P (lbs) if an overall factor of safety of 3.2 is desired. th C I? [‘43 "LL 12 in. FED tort es 2:91 leap; We owe Lom'ud l“ ”3’” “3’ fl 7‘ L“ r ”r"? em 1‘3 DéMB'O \\ 1N m 0‘2. I -'Z. )1 03339115 E\ (V: ”‘ng ‘ 1%k1;3kx F a“ ?= *EF Elms): 3 Sta MP9 Q, J C) @F‘ ‘ lZJP : E) \\\ 3 pg -$ 4?) inc #3 EL \K\\o E)? T?bkhlhs: C).3€5L Rafi ”F- 3; FM k“ 7 EflflflflL ‘ m”- = 9% 15 1'50 rem-two. «Us Amt-75 ”£7 1'an Ail-EA r POLE W A '- AHoot " .- ’I ., I . ”L ‘~ AW : (V‘XDflS) e (Do 1mg ) : size on x ('5’ A 1 "i qufLrg-QXOJ'ZSQWED: 75" MP . “‘ Let“ / Fubl‘wtl . .'-7~5l<'1p6 A Z % . r ‘ "Ht, , err-3.1. - , L, at. -‘-._ TDTRL P.B4 ...
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