Introduction to 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 $(\mathbf{e}_1^{'},\mathbf{e}_2^{'},\mathbf{e}_3^{'})$.

Solution

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,

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

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