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# homework4 - Prob 3.5-4 A steel pipe(E = 200 GPa surrounds a...

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Unformatted text preview: Prob. 3.5-4. A steel pipe (E; = 200 GPa} surrounds a solid aluminum-allay rod (E; *- 70 GPa). and together they are subjected to a compressive force at 200 SIN acting on rigid P_2w m end caps. Determine the shortening of this bimetallic eom- _ pression member, and determine the normal stresses in the. pipe and in the rod. Jim/ﬁlm: ﬂc;{‘rm/}1£ {49/ fifth/>7 ﬁnc/ y/zrrrcf (0," ﬁne/d; P154 and Pas-13 /;;' w; r w/9 m ﬂamcn/ﬁ/mv a/ct/fv/mélé'ah 6544z/xbr.‘ a M. 5'». ) g t : -—_'_ﬁ_ 3 {z} Mitre % (9&4 77[(9.05‘0~)L'{0({yawrjﬂwxwi‘g' 5; =ﬂ 6 = €,:¢;f{r0"‘)ﬁ _ Z1 _ 0,511” 76"— ﬁuﬁ‘i gr/amrmmﬁfmmowﬂn‘) = LeféOK mﬂ-ﬁ §éﬂmrfry til—L o/cﬂ/md .62)” J: C} '—' ﬁ: = —d_' (ﬁféa/frnm‘j) (3’) éwlmcrai anJ/ﬂ h (5/ (‘47)?)6' .6”) (JWJ/nc m and" (3'). ﬂ {/,;%):ep _, 4-: 74% =v/j5747w gauge/9 =—g¢,5?m L 0’3 -C‘: r' 745 = (ax/mm 105-0. r/wmml F —/5ﬁ47£4/ r ‘ 7% = = F , mm _ 0—. *‘ if[/MraMJ-‘a/ap¢mj‘ N 7 «Cf-47. 73% (t) 0: -— —==—— m H lam ma (Ewan/24mg ,9; "' ff/prfmmi" Prob. 3.5-8. A steel pipe With outer dimmer d. and inner diameter d“ and a solid aluminum-alloy rod of diameter at farm a three-segment system that undergoes axial deforma— tionduetouinglelond Pcectinsonnoouaxatpoimc,m shown in Fig. 1315-8. (:1) Calculate the axial stresses 0). ca. and a; in the three elements, and (h) detenm'ne the displace- ments u. and ac. d. -= 2 in.. d, - 1.5 in.. d = 0.75 in. P358, P3_5_9’ P336, PM“, Puma L1 =- L, - 30 in.. L, - so in. P334, and P11J-11 Pc-ukips, E. -30x10’ksi. Eq-E3-10x10’ksi . I ,, f: a 23/“ A“ .- FM ’5 F50 I Efm‘ﬂ‘énam; _ I ﬁrm F190 "5'; 3,—2th E J 57“; =0 (a) 6” H90 2pc, Ii, ' 6 = 1‘2 m) 5/5:- o/gfo mad/fan éeiaw‘or.‘ I “35765 W” =2226(;a");—';; 6;" =ﬂ 6 (2) MA”! - .L 4!; _ J 1,3. (3:745? ﬁ-{ﬁ'EJIrﬁ—c/LEL '6-77/‘(l'0 F e a €——I. H} F :2) FL -—> , .4. 4L; .41. =[dvé z ,1. xix/m W) ﬁCﬂmt/LK: «pf c/éérmaiéhwr {émga (If/x'ﬂfy): 6: f c; r (:3, =0 {3) gamma (,3; Mr m. 746/ {5 rﬂgm 0') J99: .525 m), (m) and {3’} nae/fl: Wear/7, 5m m) 5:»: }m .9 M4) 6 5 570‘ @mérér 0‘} (mi/J”) %é‘4ﬁ£;£/€.€)=o €=WJ€=ZZM£AL ﬁr5=zmérgar , /? sfﬁéhmnm {am/J PAM, Z515 /£r90/z) g‘ :i : 5:245 és; " 5‘; ,L; _ . a; - A: , /é.fai’én 0}. ‘i : r/Of‘ﬂ'ér, ﬂ; d; = /é.j‘Zérr‘ [7”) q; -— /0,84é;£ {Cl (5) ﬂ/y/a/Mcmem/é' M6 ”“J%' 4,19 = f; =ﬁ/f : 5:246ﬂ0ﬁyréu [16 = 4'5; = ﬁg}; = 5¢Z?//O'e)/}a, *hob. 3.5-1.5. A sign whose weight Is W hangs from bolts at points C and D on the rigid. weightless beam in Fig. 3.5- 15. The beam is supported at C and D by wires having enm- ueﬁonal area A and made of material with modulus E and kishingedtoawal] bracketatendB. Beforethesignis attachedtoinbeamBDishorimntalandthenvOsuppon wires are taut but stress-free. Let h = b. (a) Determine expressions tor the tensile stresses a, and 0'; in the two wires. and (1:) determine an expression for the vertical displace- ment. 6;, at D. Assume that the angle of rotation of the beam 30 is very small. Express your answers in term of P. b. A. and E. 919/“ ﬁsh”; I {a} ZEN/r- J/rrrrcx m swkcr. fix/’x‘f‘én'um ,r gag—Mk :0 (has. )4 s erhﬁ, M) F 51/5» gm :0 (is) 578m arr/ﬁrs: -a/cﬁfm¢ﬁ}m Atéew‘ﬂr! [I = W I [L s 4=M £1745 (zJ VA”: 7(= (IL/4'5}! : 4/65 1 ﬁx Pal “I? @m-ﬁ/fy 5 W2. 5/461, -- 14/3,, x {/5631 —- 4/4: jg = 4/35 596mm: 14’ 9/ [ck/armaﬁkn (/mﬂﬁ’ﬂy )5 infar: ‘5; = £374 (5:; 5:: ﬁgﬂhéﬂ a .5156 _ A T 4(3) 5 [a : £5 2/: Z a 24:; ~ 116'; ;o (2') tam/M; (2) ﬁsh] (Zliﬁfﬁ — HUM/[=0 (3’) {Qnﬁhurﬂannex/ps/r) ﬂry a. x:- /r max) M rig/pr (’0 mm! (3') r/r‘w/fmraar/y/ awﬁ c’xprc'n’mn; ,4/ l, M Z; xéffr/LM are (run; of} «71 I? 14’ fp/yr [‘45 Acre 5?.“ Amy). M“ ‘)W 3(4w‘)“/wa=)”"w {M (mg/g) #64545) l, A; g.- *Prob. 3.5-1.6. Solve Prob. 3.5«15 letting it = 2b, rather than k - 6. 5mm ﬂak/242 ﬂ: one/Ma” 5/ 40 5, if—ﬂf {4) fﬂerfcf 0'1 ﬂ: w/mx. Farm ﬁz/ﬁ) m/ /¢!) 9/ PM. eff—a} 3/45: MIMI/ﬂaw W 24 [M‘MWV‘ MW 3 5")”‘1 F; = ———~—Z£f:*:25 M) 6/ ﬁt :ﬂcnn/[aﬂc ’4‘ if, ﬁfth /¢d} ant/[4!) rev/me A - W W _ - 5: ﬂ/‘Y/ﬁ/éKJmen/a/ ﬁrm 5;. (def ﬂaMﬁ/I, M‘wﬁ‘b’ ﬂg’WL E1me 7/— w W) - M rf/fw ;— =zg/7 F“ 6‘4}? 45,5733) #5) ' (/61? m d; =ZJZ P Prob. 3.6-4. The two rod elements in Fig. P364 are stress free when they are assembled together and attached to the rigid walls atA and C. Determine expressions for the stresses v.(AT) and 05(AT) that would result from a uniform temper- ature increase AT. = AT; = EAT. Express your answers in tam at El! E2! Alt A11 L11 [’2’ “1| a}! and aT- 62 = 12F; + «Z L: ma 2; _§°(JL_I -+ ﬁ2%gl AT a L/AEH+ L/AEz __ —E‘7\i1, 'f ﬁzLZDAT ‘ {L/AE: l ‘1’ IL‘/AE 2 L: f. 4%): 6s = L. 1;: — e : 5| 62': O I W + . 'f 2 = (.¥|Fl “ll—II AT!) + ﬁleAT) 9(‘9-“14 ATI Jr 2912 L2 5.1.2.) F; = 1:2 FT ~A -— 9i "'0': O; = E] = LlAzE-z. +1.2 ﬁnal Fa — IELE'Z 'T at + ML) 02:79: LIAzE—Z + LQA"E' Prob. 3.6-1). Steel rods (A = 0.050 in“. E - 30 x 10’ ksi. a- 6.5 x10"PF)mamchodatpolnuA and Ctotlle "rigid" right-angle bracket in Fig. P3612. The bucket is supported by a mouth pin at B, and the rods at A and C are initially stress-tee. (a) Determine the axial am 13de in rods (1} and (2) if the temperature of rod (1) Is Why 50"F. (b) Determine the transverse sheet force 1/; and bending moment M; on the «as: section at D that malt hum this temme (a) (ff/errc'f M “We ﬂ) ””J{z)' Ezqr‘ﬂéfﬂam} +({zm)g =0: 500,3.) r €665. ) : 0 2L; : #24" m 5/5,», 504/ éxec- /¢'M/g£/¢ﬁlfr“ *M/gfmﬂéé’n QM»! a, : f [f f (X, Z, A]? 6: =£ 6 éldé’rﬂcié’ g {3/3 ﬂaw/959p: )z‘ C? i . : ‘ ﬁL /0m. 5’75, 6} -' 26L m Kym/m (x); (2!, mJ/f). (756*vr.1:07?)+27é/-z6):o _ - Kr 4 N? f; t f; {ﬁr/L : Jan f0"/’F)/f:¥1.)(*5z"ﬂ f {a} a/é/r ﬁ’g =/Z/9£’) = xiii/{ova ﬂeﬁr‘mﬁv/{ﬁf ﬂew." {Kg-1?}? {KO—‘3 191%?) : F : ,{é‘foo 0;: ifvén‘ {f} z, I 05 t 7;“? r I. 4500 44f" , 0; = r3, Mir: 01* 5-90 4” m f {4) L, and 5/55” mam/ﬂer gem/mg, E aflﬂK/zhm ,’ U—x pm” 5%; =0 17,, = —/€ : r 77,374, g 4 v3 (2m)? =0 my = Hem}: — far/17'». V9 7 * ‘iz 5/; M} r — 5135/! w}. Prob. 3.7-4- Solve Prob. 3.7-3 with the following data: a—asauxw’ksi, Atsoﬁin‘. gig-0.50m" liq-60in“ L1==901n.. Loom. ﬁ/f’c ﬁbn ,' ﬂt/r/mxiw {ﬂarch a; aha/G. E /Jn., 241 I: 25 =05 6 = f m E/cf’mcq/ 15/5: Lo/c/Q/m 05b»; gtzéayr'ar.‘ {in r s ___-—-————-—'_'_ ._ '4 if" _— M ﬂ {tr/#5) [aznh‘ﬂjox/o’ I2) 'ﬂﬂoo {3} «5’9 for}: ' .1 1 a s ’ If}: Cot £ (/95)‘ Kampaqaam’év) ‘ 0-0062? éeomcfrz e’é C/Cérma ﬁbre -' grit; =dr {1’} [pmér‘qc (/J) (E); (Ma/(j)- ; Mafia” m d 539/; = ammﬂzﬁ = 2.55“?! érﬁr =5 ﬁfe-Harri , ﬂ wrle 07" :9; a azrmz =//""Zi£"‘ _ xi g.gr7xa.;>u . J: 1% : __0_.5’a.a1 Ritz/(FA! ...
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## This note was uploaded on 06/11/2008 for the course AME 20241 taught by Professor Wagner during the Spring '08 term at Notre Dame.

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homework4 - Prob 3.5-4 A steel pipe(E = 200 GPa surrounds a...

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