Homework_Solution__2 - 1.33 Member ABC, which is supported...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1.33 Member ABC, which is supported by a pin and bracket a1 Canal 5 cable ED, Problem 1 '35 “'35 dfifiifimd TB suppml lllfi Efide load P as fihfl'fl'fl. KLIDWiJIg lhIIL lI'LI: uilimale load [Dr Gable 3D is WU kN, dumminc LEI: factor ufsafcw with respch 10 cable failure. p Lise member ABC as II. * 'Frflt 5.9va and ha}: Hund- mem ber "E‘D is at. inuev'Feufiee men-when“. (Liar; 4gEM: : '3': ELEM 52.11.3105 40010-2.) mm W HM) 41am msoajfoe‘r-iapsm so"-'Eo-L1‘r= O Lgauqa P - 0.7mm? F3; = CI Fm, = LEI‘E'SS P = {1.3I%3§i(1é*f9$3= f-fiflqu'fls“ FE” : mono! N . 3. {2' rr E Fuzr- _ IC-‘ICZIHO, 3 ES‘T 3.‘1t5- "" - - Hm :fiomvro Problem 1 _43 1.43 The two wooden members showrs, which support a lfi-kN load, are joined by plywood splices fully glued on the surfaces in ccmtact. The ultimate shearing stress in the glue is 2.5 MPa and the ciearance between the members is 6 mm. Determine lhe required length l. of each splice if a factor of safety of 2.75 is to be achieved. There awe. '4- sepaufiau'i'xt area): chcafiue. Each qflue when, mus'fl Wmsmll'f' 8 MN O‘F 5 ifieaw peach P= 8w 2 swam Requfi'ea‘ uf‘f'r'ma‘lte fleece”. PD : CF53}? - (2.751(3vto‘): 52:27::93 N Qaunv¥3afl (Pew.ij 0‘]? Ear/91 3,993 area, PL; - - fl“ fuw ‘— ‘— W) -3 J L- 29+C = {23(70Axto'3h 0.0% = omsgm w LBonj‘Hw mt SFQI‘LE; w L3 l4é.8 MM 4'11 1.53 In [In shat-I structure shown, a IS-mm—dimnclcr pin diameter pins 33% used at E and D. The ultimate shearln connections, and the ultimate normal straw is dflfl MFa in link EL}. Knowing tlm a factor ofsafe’ry '0le is desimd. dawn-ulna- the lzrgest lead 1' 411m can be applied aid. Note that link ED is not taint"ch mum! the pin hoies. Problem 1.53 is Hard at C and |0-mm~ 3 aims: is Jfitl MPa atall Use 'Free Basia! EEC- HEMC = D: 0.23:2: P — 0.12:: PM = o P E 73f can If” +DEHB= {lJEDP- (3.12::- C = O P = %C (:13 Tamale-v- um hef' San-$13M bl; .pu'MII-C ED. E. 1': Cl FM=EAM+=Fs—fimi ____ (Hill;whirl";I : 5, quiz? N (ta-«IE‘Juflfihwlfldsl SEER!" in Pins B Quasi D- FIB: = fAPM "' {,Ta': *{I—'I5aflm J(E}flbfft53)1 - Smdflw ml“ oF FED is awn-we" N. Fm“ m P: élfggflhxmii = Less .0: in“ H She-av I'm FFM :L'J' C in C = 271%.. r 23; if" 1vl2lij‘5-‘g'i1tfg‘1féxld’llar 23:14:10" M Fm“ (2“; "P2 Ell/23214250“) = 2.121119%! Smafififr‘ Lit-)0? or? IP 4:5. Jfimmflt Uni-ve- P: LK'QBKJD? M ‘Pztcsa law a! ggwgthjztge 0.45_ in. When a terasilc force P is applied. Kn‘owing Funding“ mien mm”: of“? the magmtude of the {me P, (b) the 8?. 03? no“ h. 1. {DL IE: C r -3. {} =. ___, =. a... _ ’{LOS'IIXID o‘- q' 3 #9 P AL‘ ' 33:?“ “0"”? .. I Problem 2.4 24 A zs-m L é 22 Ft = 33:? in. A = $4‘ - firs-2511‘: — |.°IO7>4103}L P: L907 5. _ H 55 (tha‘ficomgy ? "a 03‘! 6"- EE - "L'" = H‘g—a—r—H .- sa‘gxerT-Ffi Pmbiem 2.9 M A blag. arm-mm Ienglh M50 x 40 mm mm. semi mummtmmmmmmfiwminkngmunheblockshwldbcal A -_ {5916;119:1g 2,9510 mm“ muslfiJifinfimm-igimllml. -3. 1 :: 1x10 w. 50: 30 NFL-.1 Squuu‘Pa. E: 15m" F... CflflSIIaFEVFIH‘J ijfawygfig Eli-rigs; 5': E P: A5 = (ind‘lfismm‘) = IEOKJDE’ N can h'JEi‘r‘n? d9aumLh JE'PMML‘h-bw : 8 = $5" P: hEEL‘é“: = {Quro'3JC?5x!D?JCO-Dflf13 ? 2132*!03“ Tha wafftw limit-M fiauarns. P: [EOHf—‘n‘aw 13:16:52.0 It“ I ———h________. —-——————____________________ Problem 2.14 2.14 m 4—mm-diameler cablefiCis made orasteer wifll £— 200 GM. lint-wing that the maximum ms in the: cable nmst nut amccd 190 MP3 and mm the clungafio-n ufthe cable mustnot exceed 6 mm,find the maximum load Pmalcan b; appliedasaflmun. L3: = 1161+ ‘+" = "Li"! m U5? lam" A3 0.5 n'Pr"?e baa} .. F Fm». 4 -— " r 1+ .. EMA - 3.3.: F -{G}Lm1mfi;a)- O ‘P = 0.?5‘9? FE: A! Cansr'd'em'nj ijouméfie 5+M5$i 5‘: Norm“ Ft: fir A = EA“ r. yam-mt: J'Lé'fiévgo" if)" E? = If“ a; 5A = tiara #:0‘301546xm'fi) = 2.3mm" H Canfid‘wiu: airpouméje efipnja‘ffi‘uni 5 = G £154.». F L:- gas (£2556): "lunar ‘r g .= H - :0 to lfaxtcfl_ 3 5 AE Fa: Lu “HEW—a— 2-Wifir'EON SMaEwa W539; fiaflE/ns. lac: 2,.cjfflxloa N F: 0.?5m I?“ :- Lcnflyofivimmmnm‘)Hnazawo‘u PrL‘IS‘E W m:- I ...
View Full Document

This document was uploaded on 10/28/2011 for the course MECHANICAL 222 at Michigan State University.

Page1 / 4

Homework_Solution__2 - 1.33 Member ABC, which is supported...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online