Unformatted text preview: 2.10:7 Each ofthe three 6mdiameter 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
crosssectional! area and material. Further assume that the rods are braced so that
they ean carry compressive forces. SOLUTION
F.» each ma A = H—(aooelz = 23274 x16“ m‘ Shula at“ ini'i't‘ai YiEIPo’l‘na if: %"M 2 1.7259‘103 ' 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 “(o3 / 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 70635)*’°3” : 23? “I "‘ (b3 6,. . 250u0‘ Pa = 250 HPa. .4 L'efi' '5' = aimnae 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 PP' =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 03!oxto'3 m ''—' 0. 310 MM ‘ ...
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 Summer '10
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