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E. v1 ∼ ϕi (1) and v2 ∼ ϕi ) finite element model: ∙ [0] [Ae ] [B e ] [De ] ¸½ {∆e } {Λe } ¾ = ½ {f e } {0} ¾ (5) where ([K 11 ] = [0], [K 12 ] = [A], [K 21 ] = [B], [K 22 ] = [C], {F 1 } = {f }, and {F 2 } = {0}) Aeij = e Bij = Z xb 2 (2) (1) d ϕj xa ϕi dx2 Z xb 2 (1) (2) d ϕj xa ϕi dx2 dx, dx, fie = − e Dij = Z xb xa (1) qϕi Z xb (2) (2) xa ϕi ϕj dx dx (6) Note that Hermite cubic interpolations of both w0 and M are implied by Eq. (4a,b), (1) (2) and ϕi = ϕi . The coefficient matrix in Eq. (5) is not symmetric.

C) k3= 70 W/(m. ºC) h1= 50 mm h2= 35 mm h3= 25 mm T∞L = 100o C βL = 10 W/(m2. °K) k1 k2 k3 Fig. 025. The boundary conditions are ³ ´ ³ L R , Q12 + Q21 = 0, Q22 + Q31 = 0, Q32 = −βR U4 − T∞ Q11 = −βL U1 − T∞ ´ L = 100, β = 15 and T R = 35. Thus we have where βL = 10, T∞ R ∞ ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ k1 h1 + βL − hk11 0 0 − hk11 k1 k2 h1 + h2 − hk22 0 0 − hk22 k2 k3 h2 + h3 − hk33 ⎤ ⎧ ⎫ ⎧ 0 L ⎫ 100 ⎪ βL T∞ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎥⎪ ⎥ ⎨ U2 ⎬ ⎨ 0 ⎬ 0 ⎥ = 0 ⎪ U ⎪ ⎪ − hk33 ⎥ ⎪ ⎪ ⎦⎪ ⎩ 3 ⎪ ⎭ ⎪ ⎩ R⎭ k3 U T∞ β 4 R + β R h3 The unknown nodal temperatures can be determined from the above equations.

The boundary conditions are U1 = 0 and Q32 = 1. 1420 The secondary variables can be computed using either the definition or from the element equations. 1 for the natural (or Neumann) boundary conditions µ ¶¯ du ¯¯ = 1, dx ¯x=0 µ ¶¯ du ¯¯ =0 dx ¯x=1 Use the uniform mesh of three linear finite elements to solve the problem. 1. The boundary conditions are Q11 = −1 and Q32 = 0. 10 are solved for the four nodal values ⎡ 52 −55 1 ⎢ −55 104 ⎢ ⎣ 0 −55 18 0 0 PROPRIETARY MATERIAL. 13272 U4 c The McGraw-Hill Companies, Inc.

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