05 in load lb for deflection 01 in solid45 load lb

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Unformatted text preview: ¥ 4¡ 8 ¡ #84¡ E ¢ F8  B8 C© D A8 © ¤   ¨¦ #B UIX7SQ©2199(9¦ I"¦   #"   ©¦¦ ©§¥ A U $ W R 3%¦ 0 ) ¦ ' % $  ! ¦  8 ¡8 #© D 8© Geometric Properties = 10 in As = 7 in2 Aa = 12 in2 θ £ D8 4¡ A8 4¡  G8 E8  D4¡ ¡ E8 4¡ ¡ Q¡ ¢ Material Properties Es = 26,875,000 psi σ(yp)s = 86,000 psi Ea = 11,000,000 psi σ(yp)a = 55,000 psi υ = 0.3 θ Loading 1st Load Step: δ = 0.032 in 2nd Load Step: δ = 0.05 in 3rd Load Step: δ = 0.10 in Analysis Assumptions and Modeling Notes The following tube dimensions, which provide the desired cross-sectional areas, are arbitrarily chosen. Inner (steel) tube: inside radius = 1.9781692 in., wall thickness = 0.5 in. Outer (aluminum) tube: inside radius = 3.5697185 in., wall thickness = 0.5 in. The problem can be solved in one of three ways: • using PIPE20 - the plastic straight pipe element • using SOLID45 - the 3-D structural solid element • using SHELL43 - the 2-D plastic large strain shell element In the SOLID45...
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This note was uploaded on 12/09/2010 for the course DEPARTMENT E301 taught by Professor Kulasinghe during the Spring '09 term at University of Peradeniya.

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