MIT2_094S11_hw_9_sol

MIT2_094S11_hw_9_sol - 2.094 FINITE ELEMENT ANALYSIS OF...

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2.094 F INITE E LEMENT A NALYSIS OF S OLIDS AND F LUIDS S PRING 2008 Homework 9 - Solution Instructor: Prof. K. J. Bathe Assigned: 04/17/2008 Due: 04/24/2008 Problem 1 (10 points): [] 1 (1 )(1 ) ) ) ) 4 Hr s r s r s r = + +− −+ s 1 ) ) 00 22 S s rr HH =− ⎡⎤ == ⎢⎥ ⎣⎦ 2 2 5 0 2 s J r = + from and 2(6 ) yr =+ 1 (6 5 ) 2 z sr s + det 5 Jr 111 1 5 1 444 4 11 1 1 5 02 44 4 4 sss s rs B r r ++− + −− = r + + + Page 1 of 6
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0 0 k k k ⎡⎤ = ⎢⎥ ⎣⎦ The conductivity matrix is {} 11 (5 ) 2(6 ) kT T V K B k B ydydz B k B r r drds ++ −− ==+ + ∫∫ The convection matrix is 1 1 17 2(6 ) 2 c TT cs s s s S K hH H dS H r dr + ⎛⎞ == + ⎜⎟ ⎝⎠ because det s dS ydl y J dr where 1/2 22 1 17 det 2 s s yz J rr =− ∂∂ ⎛⎞⎛⎞ =+ = ⎜⎟⎜⎟ ⎝⎠⎝⎠ with 2 y r = and 2 z s r = The heat flow load vector is e B QQ Q where 2(6 ) (5 ) BB B V QH q d V H q d r + + d s 1 1 17 2( ˆ 6 2 ˆ ) e es T s s T s ee S Q H dS H r dr θθ + + Page 2 of 6
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Problem 2 (10 points): Note that φ is zero on the boundary of the shaft. Therefore we have only one degree of freedom at node 1,
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MIT2_094S11_hw_9_sol - 2.094 FINITE ELEMENT ANALYSIS OF...

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