By J. Reddy

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Extra info for An Intro. to the Finite Element Method [SOLUTIONS]

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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.

Dx cj d2 φj d2 φj d2 φi + λ φ + φj i dx2 dx2 dx2 ! 2 + λ φi φj # ) dx cj (1) which is a quadratic (matrix) eigenvalue problem, and it is more diﬃcult (but not impossible) to solve. 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.

We have ¡ ¢ 1 π (76)4 − (60)4 10−12 = 108, 161 N-m k1 = (27 × 10 ) 32 500 × 10−3 1 π(50)4 × 10−12 = 94, 493 N-m k2 = (77 × 109 ) 32 500 × 10−3 9 The boundary conditions are θ1 = 0, T21 + T22 = 6, 325 N-m The condensed equations are (108, 161 + 94, 493) θ2 = 6, 325; TL = − (108, 161 + 94, 493) θ2 = −6, 325 PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. ° All rights reserved. 1: One-dimensional heat conduction/convection: µ ¶ du d a + cu = q − dx dx for 0 < x < L NBC: specify nx a EBC: specify u, du + β(u − u∞ ) = Q dx where nx = −1 at x = xa and nx = 1 at x = xb .