By J Reddy
J.N. Reddy's, An advent to the Finite point technique, 3rd variation is an replace of 1 of the preferred FEM textbooks to be had. The booklet keeps its robust conceptual procedure, in actual fact analyzing the mathematical underpinnings of FEM, and supplying a common strategy of engineering software areas.
Known for its special, conscientiously chosen instance difficulties and vast choice of homework difficulties, the writer has comprehensively lined quite a lot of engineering parts making the publication approriate for all engineering majors, and underscores the wide variety of use FEM has within the specialist world.
A supplementary textual content site situated at http://www.mhhe.com/reddy3e includes password-protected strategies to end-of-chapter difficulties, basic textbook info, supplementary chapters at the FEM1D and FEM2D machine courses, and extra!
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Extra info for An Introduction to The Finite Element Method[Solutions]
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The solution of these equations is P1 = 39 12 15 Qa , P2 = Qa , P3 = Qa 14 7 14 PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. ° All rights reserved. 6: Consider the hydraulic pipe network (the flow is assumed to be laminar) shown in Fig. 6. 5 cm 3 4 • L = 55 m D = 5 cm 4 L = 60 m D = 8 cm 5 • • L = 70 m D = 5 cm 5 6• 6 Fig. 6 Solution: The assembled equations are ⎡ 1 R1 ⎢− 1 ⎢ R1 ⎢ ⎢ 0 ⎢ ⎢ 0 ⎢ ⎢ ⎣ 0 0 1 R1 − R11 + R12 + − R12 − R13 0 0 1 R3 0 − R12 1 1 R2 + R4 0 − R14 0 0 − R13 0 1 1 R3 + R5 − R15 0 = PROPRIETARY MATERIAL.
SOLUTIONS MANUAL 35 where u and v are the dependent varibales, a, b, f and q are known functions of x. Also identify the primary and secondary variables of the formulation. 1: Consider the following differential equations governing bending of a beam using the Euler—Bernoulli beam theory: − d2 w M = 0, − dx2 EI − d2 M =q dx2 (1) where w denotes the transverse deflection, M the bending moment and q the distributed transverse load. Develop the weak forms of the above pair of coupled second-order differential equations over a typical element (xa , xb ).
K4 k2 1 5 k6 2 1 P k5 k1 4 3 k3 Fig. 2 Solution: The assembled stiffness matrix is ⎡ k1 ⎢ −k1 ⎢ [K] = ⎢ ⎢ 0 ⎣ 0 0 −k1 k1 + k2 + k3 + k4 −k3 −k2 −k4 0 −k3 k3 + k5 −k5 0 0 −k2 −k5 k2 + k5 + k6 −k6 ⎤ 0 −k4 ⎥ ⎥ 0 ⎥ ⎥ −k6 ⎦ k4 + k6 The boundary conditions are: U1 = 0, Q62 + Q42 = P , and the equilibrium requires that the sums of all Q’s be zero. Hence, the condensed set of equations is ⎡ k1 + k2 + k3 + k4 ⎢ −k3 ⎢ ⎣ −k2 −k4 −k3 k3 + k5 −k5 0 PROPRIETARY MATERIAL. −k2 −k5 k2 + k5 + k6 −k6 ⎤⎧ ⎫ ⎧ ⎫ −k4 U2 ⎪ ⎪0⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎨ ⎪ ⎨ 0⎬ 0 ⎥ U 3 ⎥ = ⎦ ⎪ 0⎪ −k6 U ⎪ ⎪ ⎪ ⎭ ⎪ ⎩ ⎪ ⎭ ⎩ 4⎪ P k4 + k6 U5 c The McGraw-Hill Companies, Inc.