 # 1. The eigenvectors and eigenvalues for a given stress state are t...

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1. The eigenvectors and eigenvalues for a given stress state are the following: 01 = 10 MPa; = 1 (e1 + e O2 = = 0 n (3) 1 2 e2) - a. What is the maximum shear stress? b. The directions of maximum shear stress bisect the directions ( lines) of maximum and minimum normal stress. (See figure below.) What are the directions of maximum shear stress? c. Assume the matrix of the stress tensor in the eigenvector basis is ordered with (o1,O2,03) respectively ordered in the top, middle, and bottom positions on the diagonal. i. Write the matrix of the stress tensor in the eigenvector basis. ii. Find the traction vector on the plane with unit normal a = 1 with respect to the eigenvector basis. iii. What are the values of (N,S), i.e the values of the normal and shear components of the stress vector? d. Write the Q matrix that is used to transform vector and higher-order tensor components from the e, -basis to the n (i) -basis. (1 pt extra credit: show Q is an orthogonal matrix.) X2 (1) 1 n (e, +22 2 n (3) = 1 2 (e, 1-22 - X1 2 2. Expand the following expressions: a. bkoik, where Sik are the components of the Kronecker delta operator. b. Tij,j + Fi = 0 (3 equations - what is the name of this set of equations?) 1 1 1 -2 c. SijAij, where [Sij] = 1 2 0 , [Aij] = -1 1 0 0 1. 2 -1 (show work for each term) 1 d. Sijuj, using [Sij] from part 2c, and [vi] = 2 . (You can use either matrix 1 notation or indicial notation.) 3. In the absence of body forces, does the following displacement field satisfy the equations of elasticity? What are those equations? u = Bxyz v=y W = sin (ax) where (a,B,y) are constants. (Work must be shown to receive credit.) Tij,j + Fi = 0 equilibrium equations strain-displacement equations 22 22 (1+v)V202 2 + ôx² (o, X to, +02) = 0 (1+v)V2Txy+ X + o, + O2 z ) =0 ôxôy (1+v)V²o, + õy² (onto, + O2) =0 yz ôyôz 22 (OX + of + 02) =0 22 22 (1 + v) V2 T zx + (o, x +oy + O2 =0 ôzôx Beltrami-Michell compatibility equations (for no body force) + = xy ôxôy ôyôz Ox Ox ôy Oz = a de xy ) + = 2 yz 8²ey ôzôx by ôy ôz ôx a den + õe, + de yz ) = 2 õe õe ZX Oz Oz ôx Saint-Venant compatibility equations quite ij - o kk S ij 1+v isotropic Hooke's law a Ow = 0 + ôx Ov + Ow =0 Navier's or Lame's equations Oz (for no body force) HV2, W + ôz + ôv + ôw =0 ôx by Oz = N =T" n = Smar=112/01-03) where 01 202203

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