Elasticity/Sample midterm 1

Sample Midterm Problem 1
Given:

The vectors $$\mathbf{a}\,$$, $$\mathbf{b}\,$$, and $$\mathbf{c}\,$$ are given, with respect to an orthonormal basis $$(\widehat{\mathbf{e}}_{1},\widehat{\mathbf{e}}_{2},\widehat{\mathbf{e}}_{3})$$, by

\mathbf{a} = 5~\widehat{\mathbf{e}}_{1} - 3~\widehat{\mathbf{e}}_{2} + 10~\widehat{\mathbf{e}}{3}~; \mathbf{b} = 4~\widehat{\mathbf{e}}_{1} + 6~\widehat{\mathbf{e}}_{2} - 2~\widehat{\mathbf{e}}_{3}~; \mathbf{c} = 10~\widehat{\mathbf{e}}_{1} + 6~\widehat{\mathbf{e}}_{2} $$

Find:


 * (a) Evaluate $$ d = a_m~c_m~b_1$$.
 * (b) Evaluate $$ \mathbf{D} = \mathbf{a}\otimes\mathbf{c}$$. Is $$\mathbf{D}\,$$ a tensor?  If not, why not? If yes, what is the order of the tensor?
 * (c) Name and define $$\delta_{ij}\,$$ and $$e_{ijk}\,$$.
 * (d) Evaluate $$ g = D_{ij} \delta_{ij} \,$$.
 * (e) Show that $$ \delta_{ik} e_{ikm} = 0 \,$$.
 * (f) Rotate the basis $$(\widehat{\mathbf{e}}_{1},\widehat{\mathbf{e}}_{2},\widehat{\mathbf{e}}_{3})$$ by 30 degrees in the counterclockwise direction around $$\widehat{\mathbf{e}}_{3}$$ to obtain a new basis $$(\mathbf{e}_1^{'},\mathbf{e}_2^{'},\mathbf{e}_3^{'})$$. Find the components of the vector $$\mathbf{b}\,$$ in the new basis $$(\mathbf{e}_1^{'},\mathbf{e}_2^{'},\mathbf{e}_3^{'})$$.
 * (g) Find the component $$D_{12}\,$$ of $$\mathbf{D}\,$$ in the new basis $$(\mathbf{e}_1^{'},\mathbf{e}_2^{'},\mathbf{e}_3^{'})$$.

Part (a)


d = [(5)(10) + (-3)(6) + (10)(0)](4) = 128 $$

{d = 128} $$

Part (b)


\mathbf{D} = a_i~c_j = \begin{bmatrix} (5)(10) & (5)(6) & (5)(0) \\              (-3)(10) & (-3)(6) & (-3)(0) \\               (10)(10) & (10)(6) & (10)(0)              \end{bmatrix} $$

{\mathbf{D}   = \begin{bmatrix} 50 & 30 & 0 \\              -30 & -18 & 0 \\               100 & 60 & 0              \end{bmatrix} } $$

{\mathbf{D}~\text{is a second-order tensor}.} $$

Part (c)


{\delta_{ij} = \text{Kronecker delta}} $$

{e_{ijk} = \text{Permutation symbol}} $$

{        \delta_{ij} = \begin{cases} 1 & \rm{if}~ i=j \\ 0 & \rm{otherwise} \end{cases} }   $$

{        e_{ijk} = \begin{cases} 1 & \rm{if}~ ijk = 123,~231,~312 \\ -1 & \rm{if}~ ijk = 321,~213,~132 \\ 0 & \rm{otherwise} \end{cases} }   $$

Part (d)


g = D_{kk} = D_{11} + D_{22} + D_{33} = 50 - 18 + 0 = 32 \, $$

{g = 32}\, $$

Part (e)


{ \delta_{ik} e_{ikm} = e_{jjm} = 0} $$ Because $$jjm$$ cannot be an even or odd permutation of $$1,2,3$$.

Part (f)
The basis transformation rule for vectors is

v_i^{'} = l_{ij} v_j $$ where

l_{ij} = \widehat{\mathbf{e}}{i}^{'}\bullet\widehat{\mathbf{e}}{j} = \cos(\widehat{\mathbf{e}}{i}^{'},\widehat{\mathbf{e}}{j}) $$ Therefore,
 * $$\begin{align}

\left[L\right] &= \begin{bmatrix} \cos(30^o) & \cos(90^o-30^o) & \cos(90^o) \\ \cos(90^o+30^o) & \cos(30^o) & \cos(90^o) \\ \cos(90^o) & \cos(90^o) & \cos(0^o) \end{bmatrix} \\ &= \begin{bmatrix} \cos(30^o) & \sin(30^o) & \cos(90^o) \\ -\sin(30^o) & \cos(30^o) & \cos(90^o) \\ \cos(90^o) & \cos(90^o) & \cos(0^o) \end{bmatrix} \\ &= \begin{bmatrix} \sqrt{3}/2 & 1/2 & 0 \\ -1/2 & \sqrt{3}/2 & 0 \\ 0 & 0 & 1             \end{bmatrix} \end{align}$$ Hence,
 * $$\begin{align}

b_1^{'} & = l_{11} b_1 + l_{12} b_2 + l_{13} b_3 = (\sqrt{3}/2)(4) + (1/2)(6) + (0)(-2) = 2\sqrt{3} + 3 = 6.46 \\ b_2^{'} & = l_{21} b_1 + l_{22} b_2 + l_{23} b_3 = (-1/2)(4) + (\sqrt{3}/2)(6) + (0)(-2) = -2 + 3\sqrt{3} = 3.2\\ b_3^{'} & = l_{31} b_1 + l_{32} b_2 + l_{33} b_3 = (0)(4) + (0)(6) + (1)(-2) = -2     \end{align}$$ Thus,

{\mathbf{b}^{'} = 6.46~\mathbf{e}_1^{'}~ +~ 3.2~\mathbf{e}_2^{'}~ -~ 2\mathbf{e}_3^{'} } $$

Part (g)
The basis transformation rule for second-order tensors is

D_{ij}^{'} = l_{ip} l_{jq} D_{pq} \, $$ Therefore,
 * $$\begin{align}

D_{12}^{'} = & l_{11} l_{21} D_{11} + l_{12} l_{21} D_{21} + l_{13} l_{21} D_{31} + l_{11} l_{22} D_{12} + l_{12} l_{22} D_{22} + l_{13} l_{22} D_{32} +\\ & l_{11} l_{23} D_{13} + l_{12} l_{23} D_{23} + l_{13} l_{23} D_{33} \\ = & l_{11} (l_{21} D_{11} + l_{22} D_{12} + l_{23} D_{13}) + l_{12} (l_{21} D_{21} + l_{22} D_{22} + l_{23} D_{23}) +\\ & l_{13} (l_{21} D_{31} + l_{22} D_{32} + l_{23} D_{33}) \\ = & (\frac{\sqrt{3}}{2})\left[(-\frac{1}{2})(50)+(\frac{\sqrt{3}}{2})(30)+(0)(0)\right] + (\frac{1}{2})\left[(-\frac{1}{2})(-30)+(\frac{\sqrt{3}}{2})(-18)+(0)(0)\right] + \\ & (0)\left[(-\frac{1}{2})(100)+(\frac{\sqrt{3}}{2})(60)+(0)(0)\right]\\ = & (\frac{\sqrt{3}}{2})\left[-25+ 15\sqrt{3}\right] + (\frac{1}{2})\left[15 - 9\sqrt{3}\right] \\ = & -25\frac{\sqrt{3}}{2} + \frac{45}{2} + \frac{15}{2} - 9\frac{\sqrt{3}}{2} \\ = & -17\sqrt{3} + 30 \end{align}$$

{D_{12}^{'} = -17\sqrt{3} + 30 = 0.55} $$