141_Mechanics Homework Mechanics of Materials Solution

141_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: 2.10:7 Each ofthe three 6-m-diameter steel cables is made of an' elastoplastic material for which a, = 345 MPa and E = 200 GPa. A force 1’ is applied to the rigid bar ABC until the bar has moved downward a distance 6 E 2 mm. Knowing that the (rabies were initially taut, determine (a) the maximum value of P, (b) the maximum stress that occurs in eable AD, (6) the final displacement of the bar afier the load is removed. (Hint: In part 6, cable HE is not taut.) PROBLEM 2.108 2.108 Solve Prob. 2.107, assuming that the cables are replaced by rods of the same cross-sectional! area and material. Further assume that the rods are braced so that they ean carry compressive forces. SOLUTION F.» each ma A = H—(aooelz = 23-274 x16“ m‘ Shula at“ ini'i't‘ai YiEIPo’l‘na if: %"M 2 1.7259‘10-3 ' zooxio‘t 3 S‘l'r‘ar‘h in Foals AD anal CF: 5,, =' E“. = E; = lthm - LQS‘xIO' S+Pa§n I'm Y‘oei BE 2 836': 1%: : fin: : 2.50 “(o-3 / Si.“ 5., < 6.. J 6;, = Es” = (Zooxlo‘mzsxro'i) = 250nm" Pa. Since. 835 > Er J '53, = 6,. :- 34$x 10‘ 'P0. For‘cesi RD: 95.: A6“, = (23.2wuo“)(250xro‘) = 7.0;;35xlo’ N P“: A6,. .- (22.274xro")(3%'wo‘) .- 9.7.545Ho’lv For equt'Pu'beOM o‘i2 bar ABC Pun 4' P85 + Pa? - “P = O (a) P = Part?“ + J = (7.0685? 17545 1- 7-0635)*’°3” : 23-? “I "‘ (b3 6,. .- 250u0‘ Pa = 250 HPa. .4 L'efi' '5' = aim-nae rm olr'sppacemem'f Jun'nj Unloading , q -: . pa; . EA 5 W; 5 = 3.534xio‘ ‘5' = Pa: L.» moor-[0‘3 a It) - 0‘6 | I Ft. = “ETA; 5" 200%: £21274" )5 = 7.0685x10‘ 5 G F.» ”when... P': a; + PB; 4 .. .- 14.13? we" 3' as P-P' =0 P‘- P - zaswo’ ~ 5. __ 23.8? 2:03 H.137 K (0" = Lamont)" m Permanen‘i’ Aispzpacenenf o‘P ham send: song-'5': 2H0"s -1.GCIole'3 7 0-3!oxto'3 m '-'—' 0. 310 MM -‘ ...
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