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All rights reserved. 60 AN INTRODUCTION TO THE FINITE ELEMENT METHOD h1 h2 h3 Air at temperature, T∞R = 35o C Film coefficient, βR = 15 W/(m2. °K) k1= 50 W/(m. ºC) k2= 30 W/(m. ºC) 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.

2h 27 )(x − h), ψ2e (h/3) = 1 gives C = 3 3 2h c The McGraw-Hill Companies, Inc. ° All rights reserved. 18a,b). Use the local coordinate x ¯ for simplicity. , x ¯ = 0: ψ1 (0) = 1 → c1 = − Thus we have µ 3¯ x ψ1 (¯ x) = 1 − h ¶µ 3¯ x 1− 2h 9 2h3 (2) ¶µ x ¯ 1− h ¶ (3) Similarly, the Lagrange interpolation function for node 2 of a cubic element with equally-spaced nodes should be of the form, because it must vanish at x ¯ = 0, x ¯ = 2h/3 and x ¯ = h, where x ¯ is the local coordinate with the origin at node 1, ψ2 (¯ x) = c2 (¯ x − 0)(¯ x− 2h )(¯ x − h) 3 (4) The constant c2 is determined from the condition that ψ2 (h/3) = 1: c2 = we have µ ¶µ ¶ x ¯ 3¯ x x ¯ ψ2 (¯ x) = 9 1− 1− h 2h h 27 2h3 .

139 (4) The weighted-residual solutions are more accurate than the Ritz solution because they use higher-order polynomials that satisfy all boundary conditions. 24: Consider the Laplace equation −∇2 u = 0, 0 < x < 1, 0 0 u(x, 0) = x(1 − x), PROPRIETARY MATERIAL. u(x, ∞) = 0, 0≤x≤1 c The McGraw-Hill Companies, Inc. ° All rights reserved. SOLUTIONS MANUAL 29 Assuming an approximation of the form U1 (x, y) = c1 (y)x(1 − x) find the differential equation for c1 (y) and solve it exactly.

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An Intro. to the Finite Element Method [SOLUTIONS] by J. Reddy


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