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Problem 1: The equations of motion of anaxisym metric elasticity problem are given by∂σrr∂r+∂σrz∂z+1r(σrr−σθθ) +fr=ρ∂2ur∂t2,∂σrz∂r+∂σzz∂z+σrzr+fz=ρ∂2uz∂t2.(1.1)where uranduzare the displacement components along the rand z coordinates, respect-lively (due to symmetry about theθ-axis, the displacement component= 0 and all variables are independent ofθ). The stresses are related to the strains∗(Hooke’s law foran isotropic material)σij= 2μεij+λδijεkk; (σ11=σrr, σ12=σrθ, σ33=σzz,and so on)(1.2)and the nonzero strains are related to the displacements (ur,uz) byεrr=∂ur∂r, εrz=12(∂ur∂z+∂uz∂r), εθθ=urr, εzz=∂uz∂z(1.3)Develop: (a) the weak form of the equations with displacements (ur,uz) as the unknowns,(b) the semi discrete finite element model of the equations, and (c) the fully discretized finite element model by completing the time approximation step.∗Those who are not familiar with the index notation used in Eq. (1.2), the explicit stress-strain relations areσrr= 2μεrr+λ(εrr+εθθ+εzz) ;σθθ= 2μεθθ+λ(εrr+εθθ+εzz)σzz= 2μεzz+λ(εrr+εθθ+εzz) ;σrθ= 2μεrθ, σrz= 2μεrz(1.4)Problem 2: For the plane elasticity problem shown in Figure 1, give (a) global stiffnesscoefficientsK99,K9(32), and K(10)(28); and (b) the specified boundary degrees of freedom and compute the contribution of the specified forces to the nodes.Problem 3: Analyze the plane elasticity problem in Example 11.7.1of the text book using meshes of linear (a) triangular and (b) rectangular elements given in Table 11.7.312 cm4 cmcm1GPa,260.333,GPa,6921=====gee stress••••••••••••••12345611161514131281091•ht0 = 3 kN/cm2•74323•••••4 cm4 cm12•435678123465Fig A8-3Figure 1: A plane elasticity problem with a mesh of quadratic elements.(attached). You must submit tabulated results of the maximum displacements and stressesfor various meshes shown in Table 11.7.3 (i.e., duplicate the results of Table 11.7.3).••y7in.120=axin.lb100/p=in.160=bin.0360.h=61230 10 psi,0.25EEn== ́ =Mesh shown is for 4×4 Figure 2: A plane elasticity Problem.2PLANE ELASTICITY667Table 11.7.2Finite element results for a thin plate (plane stress assumption)using various meshes of triangular and rectangular elements andmaterial properties†U3U4U7U8MeshMaterial(×10−4) (×10−4) (×10−4) (×10−4)1×1Isotropic:11.2911.96410.113−1.080E= 30×106psi10.8532.32610.853−2.326ν= 0.25G=E/[2(1 +ν)]1×1Orthotropic:E1= 31×106psiE2= 2.7×106psi10.7671.66610.651−1.579G12= 0.75×106psi10.7282.67510.728−2.675ν12= 0.28†For each mesh, the first row corresponds to triangular elements and the second row to one rectangular element.Table 11.7.3Deflections and stresses in an isotropic plate subjected to uniform edge load (Example 11.7.1).ux(120,0)uy(120,0)Mesh(×10−4) (×10−4)σxxσyyσxy1×111.2911.964285.967.4210.80(80, 53.33)†(40, 106.7) (80, 53.33)2×211.3722.175294.169.3623.20(40, 26.67) (20, 53.33) (40, 26.67)Triangles4×411.2842.126306.273.7535.93(20, 13.33) (10, 146.7) (20, 13.33)8×811.2092.054331.681.3148.04(10, 6.67)(5, 153.3) (10. 6.67)16×1611.1792.014372.591.9458.90(5, 3.33)(2.5, 156.7) (5, 3.33)1×110.8532.326277.825.840.0(60, 80)(60, 80)(60, 80)2×211.0782.021277.837.4613.23(30, 40)(30, 40)(30, 40)Rectangles*4×411.1502.009288.149.7427.73(15, 20)(15, 60)(15, 20)8×811.1621.997308.056.5340.97(7.5, 10)(7.5, 50) (7.5, 10)16×1611.1661.992339.561.253.14(3.75, 5)(3.75, 75) (3.75, 5)†Location of the stress. *Meshes of rectangular elements reflect solution symmetries present in the problem but meshes of triangular elements do not. The8×8mesh of nine-node elements gives the results: 11.901,1.781, 346.0 (3.17, 4.226), 61.81 (3.17, 84.23), and 53.49 (3.17, 4.226).

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