homework4 - Prob. 3.5-4. A steel pipe (E; = 200 GPa}...

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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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homework4 - Prob. 3.5-4. A steel pipe (E; = 200 GPa}...

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