User:Eas4200c.f08.aeris.guan/hw5

Strain Displacement Relationship and Equilibrium Equations (Stresses)
With the theory of elasticity, we first considered the kinematic assumptions (21.3) which resulted in realizing that the four strain components in the xx, yy, zz, and yz directions were equal to zero (21.3)and consequently the stresses in these directions were zero (22.4). Comparing this to the equations in page 70 of the book, we find that since the book has used the convention of an (x,y) coordinate plane in order to analyze the strain and stress, however we used a (y,z) coordinate system. These expressions are easily equated by using a cyclic permutations of the variables. Then, the stress strain relationship (tensorial or engineering) were established (23.2). Now, we will rewrite the stress strain relationship as shown below. $$ [\varepsilon_{ij}]_{6x1}= \begin{bmatrix} A_{3x3} & 0_{3x3}\\ 0_{3x3} & B_{3x3}\\

\end{bmatrix}_{6x6} [\sigma_{ij}]_{6x1}= C[\sigma_{ij}]_{6x1} $$ Where A and B are matrices with the values shown below and B is a diagonal matrix. The 6x6 block diagonal matrix will be referred to as the C matrix. $$A= \begin{bmatrix} \frac{1}{E} &\frac{-\nu }{E} &\frac{-\nu}{E} \\ \frac{-\nu }{E} &\frac{1}{E} &\frac{-\nu }{E}  \\ \frac{-\nu }{E} &\frac{-\nu }{E} &\frac{1}{E} \\ \end{bmatrix}, B= \begin{bmatrix} \frac{1}{2G} &0  &0 \\ 0 &\frac{1}{2G}  &0 \\ 0 &0  &\frac{1}{2G} \end{bmatrix} $$ However, we will keep the C in the block diagonal form. To solve for the stress, we multiply both sides by the inverse of C. To invert a diagonal matrix, one must take the inverses of each value of the diagonal. It is the same for a block diagonal matrix as seen below. Invert C by inverting the diagonal matrices. $$ [\sigma_{ij}]_{6x1}= \begin{bmatrix} A_{3x3}^{-1} & 0_{3x3}\\ 0_{3x3} & B_{3x3}^{-1}\\

\end{bmatrix}_{6x6} [\varepsilon_{ij}]_{6x1}= C^{-1}[\varepsilon_{ij}]_{6x1}

$$ $$ C^{-1}C= \begin{bmatrix} AA^{-1} & 0\\ 0 & BB^{-1}\\

\end{bmatrix} =I

$$ Where I is the identity matrix, a diagonal matrix with the diagonal values all equaling one. Then, we will continue on proving $$\displaystyle \sigma_{xx}=\sigma_{yy}=\sigma_{zz}=\tau_{yz}=0$$. First, we will apply the zero strain relations in which the strains in the xx, yy, zz, and yz directions are zero. Knowing that there are essentially six strain components since the strain tensor is a 3 by 3 symmetrical matrix, a 6 by 1 column matrix can be constructed where the first four rows are zero and the last two are non zero. Subscripts x, y, and z will be replaced with indices 1, 2, and 3 respectively. This gives the following expression. $$ [\sigma_{ij}]= \begin{bmatrix} A^{-1} & 0\\ 0 & B^{-1}\\ \end{bmatrix} \begin{bmatrix} 0\\ 0\\ 0\\ 0\\ \varepsilon _{31}\\ \varepsilon _{12}\\ \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\\ 0\\ \sigma _{31}\\ \sigma _{12}\\ \end{bmatrix}

$$ Multiplying the top three rows of the 6 x 6 matrix to the top three rows of the strain tensor results in $$\displaystyle \sigma_{11}=\sigma_{22}=\sigma_{33}$$. The resulting last three rows are derived in the collapsible box below. (16.2) Recall the road map step: equilibrium equation for stresses (section 4.2) in a nonuniform stress field. To understand the nonuniform stress state, let's consider a 1 dimensional case (a cantilever beam) as a model:



It is necessary to point out that the axial stress in the uniform case is the same for every value of x (i.e. $$\displaystyle \sigma(x)=\sigma(x+dx)$$) and F(x)=0. For the nonuniform case, axial stress DOES vary with x (i.e. $$\sigma(x)\ne\sigma(x+dx)$$) and F(x) is either a nonuniform function or constant. We will use this concept to derive the equilibrium equations for the stresses.