HW3 - —| 2.35 Compressive centric forces 01‘40 kips are...

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Unformatted text preview: —| 2.35 Compressive centric forces 01‘40 kips are applied at both ends of the assembly shown by means of rigid plates. Knowing that E, = 29 x 10" psi and 5., = 10.1 x 10“ psi, determine (a) the normal stresses in the steel core and the aluminum shell, (5) the deformation of the assembly. Le+ Tra- = For-ll-[on 64‘ 6.96%»? ‘Fmrce m-luh-sni-l b?! Ethel?)- Pfi -.. PorJf—l'o“ Janka" 'szz cavn‘eJ if cor-e. Sleelcore — 13$]:— — git-£6: 5' m. a ‘ L S -‘ PS'L :- Efififi § ESP: PS '— g ___J°___. EaAnfi“ EsAs .— fiflf: 0-7854541 : _tl-le'03 (19.: xro‘Mr-L r233) + C7»?>'ID°]{O.785‘!\J — —szo.m no" 6'5 = E52 : L'ZQxIO‘UQROflliO") “Jamxao‘fi: 48.6“ ks? A [3“: Egg = (IOJHOQKGIO‘WHD'G) nemwo‘ps: —s,27 inf-4! 5 = Le - (lowszoar no“ )=-G.4¢l.w!o" 5: — 6.2Jx:d’.'n. 4 Problem 2.41 Dimensions in mm 40-min diam. 30-min diam. EA SUE : EA P = Fa.l - eoxgoa _ PL (P.—eowto3)§o.r2m _ Sac“ _'—'" _ gA ' 251.321.10‘ P = 9,. — gone“ . (a —50xao3§Co.roa3 7%.onx {0‘ L. Sec: ET ' 3+1: E. :9: {2‘ — iomgcfi 3r: _ (En- Iooxlo‘XoJooW 7%.‘210 x 10‘ L 4:: E : 5” : 5.3+ Sac—$3“ +505 = I'LISI’hce Poih+ E canno+ Mow: r224};ve +o A, Eco.) anaszawgo“q F?It — 292.52%“? : o $3,; = 9,—100xgoz‘ r 62.833031—{005'103 = - 37.2“!03 U 3c: = 8m“' SM. 7 2.41 Two cylindrical rods. one of meel and the other of brass, are joined at C and restrained by rigid supports an! and E. For 111.: loading shown and knowing that E; = 200 GPa and E5 = 105 GPa, determine (a) the reactions at A and E, (b) the deflection of point C. Bio—C'- E : 200*!0“ ‘Pa. A = 15m“ -—- Luamxgol‘ rm": LZSsCMzo" m‘ EA: 25:-327No‘ N C+..E: E: Losym“Pq A _- 1.}csoj‘r 706.86 m: : ?oe.2e.x;o“m‘ EA= 7"}.220xlo‘ N L: “SO-mm: OJSOM - T’L - _ . 5"“ ' 251.3272106 ‘ “6"”“0 2" L T120 mm: OJQDM - Lil-17.47xzo‘” a - 26.2q3sro“ ;' [001mm = 0.300;... = Lamas-“53?, -‘ gagwxgo" L: {00 M = 0.100 m = Lawssx Jo”?A — 13%.?35Ho“ 3.25357xzo“ PA w 242.42%:0“ 5A5 = 9 62.8kM-=— 4!! I2, =. 62. Ssixroa N 57.2 Hue ad Lissa??va PA — 26.8%:‘(04 = (Ligaemloqflszssax :03] - 26.8‘32xlo" é 46.3HD- EM 5:: 1%.B‘prn—i 4-. Problem 2.46 Use tam" as a, '11:”: 19:33.?“ flair/Igor -2LT, a LT: + LTD = o (3» +Tfi§=m Tg+T3+n+Tg—F>=D (an, sug54l'+u+.‘mfi (1'1 :nJm (3) M1 Jl‘ut-AQ-ne L, L). “‘an + 2T5 = 0 TE: 27:. (3‘) SuLE+f+d+[n-a m) (2*); Martin 9149(4)) T's-123+ 3Tn+4T.—P=O 1073:? T» = $13 "' T9 = 21; = (21%}? Teri—P 4 E= mam—(551°) 1:733? 4 TD = (31(52)- mcé P3 1-D: g p 4 2.46 The rigid bar ABCD is suspended from f0 ur identical wires tension in each wire caused by the load P - Dclc ' ' Show. rm Di‘Fof‘qufl'OM-S LEJI' 9 Be HM? wo+fi+iur 9F Ea.» ABCD am! 8;) 5a, a; am? '5; be +1” Aa¥wmw+.m 0? mm A1 8,5; M D FM.“ Samar” e= 85 = 5, + L6 gc— g“ 2L8 = 253 — sa ‘53: SA+3L9 333-285 m Sl'hcc 0.” wires. am L‘OEBVI'FI'CGHPJ 'H-e gum“ in Jrle val—was are. Fmrjcr‘hbwqj +9 +Le (Je‘FLN‘w' sch-omen T;= 273—»: TD .~ 373—223 2.5] A rod consisting of two cyiindrica] portions AB and BC is restrained at both ends. Portion AB is made of steel (E, = 200 GPa, a, = l 1.7 x 10‘6I°C) and portion BC is made of brass (Eb: ms GPa, ab = 209 x mire; Knowing that the rod is initially unstressed, determine the compressive force induced in ABC when 111cm is a temperature rise of 50°C. m.mmd1nmeter _ ‘1 81’“! AM; 2 Hi 5)“ : Ell-{3.95 .‘ : i.c‘iG-3.§Ki03vn:= 1.96351‘2'0-3m l _c z jcé.gémm = ?OG_86‘K!O M 1| '4. ‘ 30.51111 diameter has: FNC. 'H'mrmaj eannsFom. 5T- LABOISfifiJ—i + Laid-4&1") : (0.250% I I, 7 x io")(so‘) + (0_300)(‘20.q x (0")C503 = 459.75 x to“ m .I' Shari-warm. pine +0 thinLtci CoMflWcS'yvt 1Com ’P. g :- + $.— -.° EEAM Eb Au : {3.250 P + o_ 300 ‘P (?0mio“‘1(106.3eflo“3 {Iosxio“1(|.%35'*to'53 = 3.2135w0‘q P For 25w ne‘i oie'Flie-c‘i'iond 5f; 3 5-.- sxzzzssxro‘”? = HS‘LTSHD‘G P: N’LGQMOB N P: “+7.6ij 4 2.60 At room temperature (20°C) a [LS-mm gap exists between the ends of the rods shown. At a later time when the temperature has reached 140°C, determine (a) the manual stress in the aluminum rod, (b) the change in length of the aluminum rod. 331‘ ‘: Ho -20 = #2073 5,? Laotimfr-‘H- LSDQCAT} Stainless steel 2 = (o. 500 )(2 3x : o“)(r20) + (0.25030 7- 3 ' f0" 31:” D} A = 800 mm E=190I3Pn = |-_347xjo"m (1 =17.3 X 10‘EPC “1mg ' ' - CPa ,_ __ -1 £- . -s. '3 a? - LEI-l7“ l0 -- O. HO '= 0.897’40 .44 PL“ PLs _ Lt. .L—s ea c» 25A: "(9A3 am“) Q3053 0.250 “F 4.. = 3.6%“! NO qdfizooowto-Q (:‘toxfquSOOHO'qu 7 P seaward“ P = 0.2q7vto“ P= ZSE-BWIDE N ‘5 -E 7 _ EZ’Z-EL’Qfl T — Hefixgo‘ Pa. —H6.2MP«. 4 AK ’1000 x to" P L“ 5.x "' LadaiA-r—N! _ EMA“ (232.34 x ro"*J(0-S°°3 _ ._ K '5‘ _. .....— A ‘C m _ (0-500193 to 1020} (75“09M90003m4} 563:“: 0.363 MM "'" Pr oblem 2.6! [r1 astandard tensilctest.analuminumrodof20- - Eiageilhsirl] force of Pr:- 30 RN. Knowing that v = t] 3smgmgaer is st. - p on am 111 ' _ ' E ' ' Oflhe rod. g no on)de150mmgagelength’QMhechaggfin T P : 30:: IO‘ N . __ 2 _ l :1. 150mm 20-mmchameter A - E — H (20‘! T 3"}.1‘5‘? men-L : 314.157); (0"M- ___ E _ SOKIOB _ ‘ L 5:! A ' smmwa‘ ‘ “fly” P“ 6" ‘?5 99331:?" E : —-Y— 7 _;__ __.. -5 3 E .70 x we LSG‘H‘Z x30 P‘ . __ _ — _ 033) Q} - LEJ T ClbDKFDBMLSK—L‘HSHO3) : 205,104 M O'QOSMn‘ Er = -’U§» T —(0.35)(L364:3x;o'3) r —'- 47144 no“ at 8, 1 (2m I0"31’u‘t77.%x(o“) r _ figs-Mo" M H- MD. thlnm 0 n“: Problem 2.69 2.69 A fabric used in air-inflated stmcmres is subjected to a biaxial I _ results in normal stressm: a, = 120 MP2: and a, = 160 MPa. Knowifiadml properties of the fabric can be approximated as E = 87 GPa and gullh determine the change in length of (a) side A3, (I?) side BC, (5) diagonal Mt 6n = nowo‘ Pa, 53: 0, 6; = Monmfi a“: é—(sg— v63 _ v62) _ l _ 873m)fir = 75102 x lo" [J20 Ho" — (0.341(“0 uro‘fl l'l s2 = gM-vs; 46;. + 6;) = f.3?o.' we“ W): 40-30020 mm") + I60 was] (an Saw/75332,; = (100m)(759.02xlo")= 0.0154 mm (b) Sa,¢=CFEv-;)E2 = (.75MM)CI.37OIX Io“): 0,1523 “m .‘ A “L 3 Lake) 5%»: angLJr hang 112 ABC. as ob L) “A c. b C1 '—' 0:4 in." Ob+afn A.‘¥Fe¢e«+rafis L»; cafeJIUS. 2c aiC = 20.0%. + .25 a“) dc. = «3: can + i;— an, But (12:00 mmJ b: lac:th crm = :1; 1mm dc».e Sm = 0.0754 hm db = as; C II 0. I370 1mm _ _ loo 7:5 _ g3 SA: - pic. — .125 (0.075%)+ I25, (DAMS) — o.r22a m4 ...
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This note was uploaded on 02/25/2010 for the course ENGR 2530 taught by Professor Lindaschadlerfeist during the Spring '08 term at Rensselaer Polytechnic Institute.

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HW3 - —| 2.35 Compressive centric forces 01‘40 kips are...

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