ME211solution_6 correction

ME211solution_6 correction - By the assumption of a small...

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By the assumption of a small angle, theta, 't a n A AC A C A θ ±± '3 2 AA CA B AB A ε = == 1.5 1.5(2 ) 0.0524 = 180 π εθ ⎛⎞ ==° ⎜⎟ ⎝⎠ A
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1 200 '' tan 0.9273 150 rad θ == 1 197 ' tan 0.92 150 rad ( '' ') (0.9273 0.92) 0.0073 rad γ θθ =− Note: Since is larger than 2 π , shear strain is negative.
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--- 7.' & .. = tll\-'(~) ... o. <J2.?3 (9-, ~ t"'-(~) = 0. Cfl. 0.: &-,. . - th ': 00 73 :t b = -t"11\ B'~: ~o () fJ1. (fo . .'. AT. .: Jto~) = (12. S. )..QV "I"{'=-PE.:3 (.: All po·nrt~ 0)\ ~e Y\8~t 3iol.e. a.p -ti-te. n.ct_"sw,-»- motH. cJ~ ~~ l\ H = Al t I.1i = 11 ~'~'" 3 ~ II ~.5" (3'1 ~ t~,,"'(!iE\; t#~-(I \ A"') lI~S) ::: 0." 4.& @:: (}-a. - fJ s ::. O. q ~?3 - 0. q I If-' :: ~ OI'2. .? Y'~ tb\~(-l 8';= ~ f d-~. r",~, ~-lT~ ~ -~-+<i-~)=-01~ -o,oo')Jto.o/2.?':' o.ooty, _r 7-',' ; oot q. nJl
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For small angles, 2 0.00662252 302 rad αψ == = 2 0.00496278 403 rad βθ = Shear strain: 3 ( ) 0.0116 11.6(10 ) Ax y rad rad γα β =+= = 3 ( ) ( ) 0.0116 11.6(10 ) Bx y rad rad γθ ψ =− +
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From the curve of the elastic region (below), 40 ksi σ = when 0.001 ε = 3 40 0 40.0(10 ) 0.001 0 E ksi == From the stress-strain diagram, 40.0 Y ksi = 2 40.0 (0.5 ) 7.85 4 YY PA k π ⎡⎤ ⎛⎞ = ⎜⎟ ⎢⎥ ⎝⎠ ⎣⎦ i p From the stress-strain diagram, 76.5 u ksi = 2 76.5 (0.5 ) 15 4 uu k = i p
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Allowable normal stress: Y allow FS σ = 57.5 3 allow = 19.17 allow ksi = allow P A = 4 19.17 A = 2 0.209 A in =
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This homework help was uploaded on 04/04/2008 for the course ME 211 taught by Professor Thouless during the Fall '08 term at University of Michigan.

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ME211solution_6 correction - By the assumption of a small...

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