Solution4

Solution4 - AE 321 Solution of Homework #4 1. (a) The...

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AE 321 – Solution of Homework #4 1. (a) The equilibrium equations for the case of no body forces is given by the expression σ ij , j = ij x j = 0 ( 1 ) Since the stress is constant everywhere in the body, the equilibrium equations are satisfied everywhere. (b ) The components of the given state of stress in any other Cartesian coordinate system is obtained by using the transformation (rotation) law for 2 nd order tensors, i.e. σ pq = α pi qj ij substituting the hydrostatic state of stress ( ) pq pi qj ij pi qj ij pi qi PP P αα δ ααδ ′ = =− using the property of orthogonality of the transformation (rotation) matrix pi qi = pq ( ) , we obtain pq P pq , i.e. for any Cartesian coordinate system the normal stresses are – P and there are not shear stresses. (c) Since in the hydrostatic state of stress there are no shear stresses, then the principal stresses are 11 = 22 = 33 P (d ) Consider a cube of a material subjected to a uniform external pressure P on all sides as show in Figure 1. Does the above stress state satisfy the boundary conditions (i.e. appropriate stress components conditions) on all external surface of the cube? 1
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Figure 1 Configuration of the loading of the cube of material considered in problem 1(d). The faces of the cube have been numbered to solve the problem. The solution of this problem has two parts: (i) determine the boundary tractions at every surface of the cube, and (ii) determine if the given state of stress satisfy these boundary tractions. (i) The tractions on the surfaces of the cube are determined using the following expression ( ) n T σ n = G G (2) The surface tractions obtained from equation (2) are given in the table below Unit normal vector (from figure) Traction Vector (from figure) Stress on Face (from equation 2) Face 1 [] 0 0 1 = n G ( ) ( ) ( ) 0 , 3 2 1 = = = n n n T T P T 11 =− P , 12 = 13 = 0 Face 2 0 1 0 = n G ( ) ( ) ( ) 0 , 3 1 2 = = = n n n T T P T 22 P , 12 = 23 = 0 Face 3 1 0 0 = n G ( ) ( ) ( ) 0 , 2 1 3 = = = n n n T T P T 33 P , 13 = 23 = 0 Face 4 0 0 1 = n G ( ) ( ) ( ) 0 , 3 2 1 = = = n n n T T P T 11 P , 12 = 13 = 0 2
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Face 5 [] 0 1 0 = n G ( ) ( ) ( ) 0 , 3 1 2 = = = n n n T T P T σ 22 =− P , 12 = 23 = 0 Face 6 1 0 0 = n G ( ) ( ) ( ) 0 , 3 2 3 = = = n n n T T P T 33 P , 13 = 23 = 0 (ii) We check if the given hydrostatic state of stress satisfies the boundary tractions obtained in part (i) . As can be easily verified, the hydrostatic state of stress satisfies the boundary
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Solution4 - AE 321 Solution of Homework #4 1. (a) The...

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