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2.2-10 b

# 2.2-10 b - 93 GHAPIER 2 Axially Loaded Members 2FV=0 F12 W...

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Unformatted text preview: 93 GHAPIER 2 Axially Loaded Members 2FV=0 F12 W. W '> F. 27+ 13(1 ~%) + 13—132 use constraint equation to deﬁne horiz. position, then solve for location It I:1 . F2 L] + W 2 L3 “t“ h +_ l k: substitute expressions for F1 & F2 above into constraint equ. (SE soive for x —2L1Lk|k3 __, k3WL — akin +‘2L3Lk1k3 + 211Lk1k3 + kIWL x = —2P(1;, + k2} x = 134.7 mm (W (b) NEXT REMOVE-P AND FIND NEW VALUE OF SPRING Part (c) - continued CONSTANT K1 so THAT BAR 15 1101112. UNDER WEtGI—[T W bfﬂEILS W ' L W I W _ Now, Ft = 7 F: = 7 since P = 0 “(,1 b) ‘ " E M}; x 1:, 2 @ l _. same constraint equation as above but now P = 0: L _ b Em=o W E! :3— 2 F] = W — F2 L + ; — L1 11 - = 0 I kt ( ‘ ) k2 _ . , L solve for k. ' “(if " b) — '1 F = w — ' k! = {at [L Lm: h')]] w I I L — b 4.11 ** [ 3 ‘ "- . _ t WL N n I:1 = W kI m 0.204—~ +— ~( L- 1-H mm constraint equation - substitute above expressions (-2) USE K] = 0.300 N/mm BUT RELOCA‘I‘E for Fl & F3 and solve for b spouse Kl (x = b) so THAT BAR ENDS up ,7 J t! r. 1 ' F Fa LN Homz. rostTlm UNDER “EICHI'W LI + Wt _ (L3 + h) _ __ z 0 kl k3 U243 use the following data N ‘ N k, = 0300““ kg = 0.4— L1 2 250 mm mm mm L; = 200 mm L =1 350 mm WLMb FBD ﬂ ZlelkzL + Win:2 — 2L3k,k3L - lltklkgL w Wle 1 ’ (gm-1kg; w zen-[kg ~— 211Mb — 2Wk. b = 74.1 mm <— ...
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