By J Reddy
J.N. Reddy's, An advent to the Finite aspect procedure, 3rd version is an replace of 1 of the most well-liked FEM textbooks to be had. The ebook keeps its robust conceptual strategy, essentially analyzing the mathematical underpinnings of FEM, and supplying a normal process of engineering software areas.
Known for its targeted, rigorously chosen instance difficulties and large collection of homework difficulties, the writer has comprehensively lined quite a lot of engineering parts making the ebook approriate for all engineering majors, and underscores the wide variety of use FEM has within the expert world.
A supplementary textual content site situated at http://www.mhhe.com/reddy3e includes password-protected ideas to end-of-chapter difficulties, basic textbook details, supplementary chapters at the FEM1D and FEM2D laptop courses, and extra!
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Extra info for An Introduction to The Finite Element Method[Solutions]
20264 Again, the exact solution is the same as the finite element solution at the nodes. 13 using the Neumann boundary conditions µ ¶¯ µ ¶¯ du ¯¯ du ¯¯ = 0, =0 ¯ dx x=0 dx ¯x=1 Use the uniform mesh of three linear elements to solve the problem and compare against the exact solution cos πx u(x) = π2 Solution: For this case, the boundary conditions require Q11 = 0 and Q32 = 0. SInce none of the UI are specified, the condensed equations are the same as the assembled equations. However, the coeﬃcient matrix of the assembled equations is singular and the solution can be determined by specifying one of the UI .
C The McGraw-Hill Companies, Inc. ° (2a) All rights reserved. 22 using the Ritz method. 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
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.
An Introduction to The Finite Element Method[Solutions] by J Reddy