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

By J. Reddy

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Alternatively, we identify the operator A of the problem to be A = −d2 /dx2 so that it does not include the unknown, λ (not consistent with the definition of the method). Then 0= Z 1 0 = = A(φi )R dx = n ½Z 1 X j=1 "Z Ã n 1 d2 φ d2 φ X i j j=1 n X j=1 0 dx2 0 ¾ A(φi ) [A(φj ) − λφj ] dx cj d2 φi + λ φj dx2 dx2 ! # dx cj (Kij − λMij ) cj PROPRIETARY MATERIAL. 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.

011181 rad, PROPRIETARY MATERIAL. 010825 rad c The McGraw-Hill Companies, Inc. ° All rights reserved. 8: A steel (Gs = 77 GPa) shaft and an aluminum (Ga = 27 GPa) tube are connected to a fixed support and to a rigid disk, as shown in Fig. 8. If the torque applied at the end is equal to T = 6, 325 N-m, determine the shear stresses in the steel shaft and aluminum tube. 8 mm 76 mm Aluminum tube 50 mm Steel shaft 500 mm Fig. 8 Solution: The assembled system of equations for the two-element mesh is ∙ k1 + k2 − (k1 + k2 ) − (k1 + k2 ) k1 + k2 ¸½ θ1 θ2 ¾ = ½ T11 + T12 T21 + T22 ¾ where ki are the shear stiffnesses ki = Gi Ji /hi and h1 = h2 = 500 × 10−3 m.

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.

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