MIT2_092F09_sol5

# MIT2_092F09_sol5 - 2.092/2.093 FINITE ELEMENT OF SOLIDS AND...

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1 2 x ⎞⎛ 2 y h 1 = 1 + ⎟⎜ 1 + 4 a ⎠⎝ b 1 2 x ⎞⎛ 2 y h 2 = 1 ⎟⎜ 1 + 4 a ⎠⎝ b 1 2 x ⎞⎛ 2 y h 3 = 1 ⎟⎜ 1 4 a ⎠⎝ b 1 2 x ⎞⎛ 2 y h 4 = 1 + ⎟⎜ 1 4 a ⎠⎝ b u 1 v ⎢ ⎥ 1 2 u = [ h h h ] u v 1 2 h 3 4 , v = [ h h h h ] 2 u 1 2 3 4 3 v 3 u 4 v 4 u ε xx = = h 1, x 0 h 2, x 0 h 3, x 0 h 4, x 0 U x 2.092/2.093± F INITE E LEMENT OF S OLIDS AND F LUIDS F ALL 2009± Homework 5- solution Instructor: Prof. K. J. Bathe Assigned: TA: Seounghyun Ham Due: Problem 1 (20 points):± x y a b 1 2 3 4 Session10 Session12

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v ε yy = = 0 h 1, y 0 h 2, y 0 h 3, y 0 h 4, y U y where U T = [ u 1 v 1 u 2 v 2 u 3 v 3 u 4 v 4 ] ; ε zz = 0 As ε V = ε xx + ε yy ε V = B ε V U = h 1, x h 1, y h 2, x h 2, y h 3, x h 3, y h 4, x h 4, y U Hence Joint 1 Joint 2 t P 2 R 2 t ө t P t 1 ө t P 2 b a 2 2 K = t ∫∫ β B T ε V B dxdy ε V b a 2 2 Problem 2 (20 points): R 2 u 2 u 1 1 2 1 L 2 L t ө Assuming tension in the bar as positive, the equilibrium of the joints gives:
t t t P 1 P 2 cos θ = 0 (1) R t t 2 P 2 sin θ = 0 (2) t EA t t EA P 1 = u 1 , P 2 = δ L 2 (3) L 1 L 2 where

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Unformatted text preview: + (0.5 + t u 2 ) 2 − L 2 (4) 0.5 + t u From the geometry, tan t = 2 , t u t 2 = −Δ , R = − P (5) 5 − u 2 1 Eq. (1) and (2) are the force equilibrium equations. We use them by assuming a Δ , t t solving from equation (1) for u 1 , then substituting u 1 and Δ into the equation (2) to obtain the corresponding P. We can also solve them in different way. We first assume t t ө and then calculate u 1 and Δ . P ×10 4 Vs. Δ EA 3± P ×10 4 Vs. Δ Using ADINA EA Displacement u 1 Vs. Δ 4± MIT OpenCourseWare 2.092 / 2.093 Finite Element Analysis of Solids and Fluids I Fall 2009 For information about citing these materials or our Terms of Use, visit: ....
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