# Introduction to Elasticity/Sample midterm 1

## Sample Midterm Problem 1

Given:

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

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

Find:

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

## Solution

### Part (a)

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

### Part (b)

${\displaystyle \mathbf {D} =a_{i}~c_{j}={\begin{bmatrix}(5)(10)&(5)(6)&(5)(0)\\(-3)(10)&(-3)(6)&(-3)(0)\\(10)(10)&(10)(6)&(10)(0)\end{bmatrix}}}$
${\displaystyle {\mathbf {D} ={\begin{bmatrix}50&30&0\\-30&-18&0\\100&60&0\end{bmatrix}}}}$
${\displaystyle {\mathbf {D} ~{\text{is a second-order tensor}}.}}$

### Part (c)

${\displaystyle {\delta _{ij}={\text{Kronecker delta}}}}$
${\displaystyle {e_{ijk}={\text{Permutation symbol}}}}$
${\displaystyle {\delta _{ij}={\begin{cases}1&{\rm {{if}~i=j}}\\0&{\rm {otherwise}}\end{cases}}}}$
${\displaystyle {e_{ijk}={\begin{cases}1&{\rm {{if}~ijk=123,~231,~312}}\\-1&{\rm {{if}~ijk=321,~213,~132}}\\0&{\rm {otherwise}}\end{cases}}}}$

### Part (d)

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

### Part (e)

${\displaystyle {\delta _{ik}e_{ikm}=e_{jjm}=0}}$

Because ${\displaystyle jjm}$ cannot be an even or odd permutation of ${\displaystyle 1,2,3}$.

### Part (f)

The basis transformation rule for vectors is

${\displaystyle v_{i}^{'}=l_{ij}v_{j}}$

where

${\displaystyle l_{ij}={\widehat {\mathbf {e} }}{i}^{'}\bullet {\widehat {\mathbf {e} }}{j}=\cos({\widehat {\mathbf {e} }}{i}^{'},{\widehat {\mathbf {e} }}{j})}$

Therefore,

{\displaystyle {\begin{aligned}\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{aligned}}}

Hence,

{\displaystyle {\begin{aligned}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{aligned}}}

Thus,

${\displaystyle {\mathbf {b} ^{'}=6.46~\mathbf {e} _{1}^{'}~+~3.2~\mathbf {e} _{2}^{'}~-~2\mathbf {e} _{3}^{'}}}$

### Part (g)

The basis transformation rule for second-order tensors is

${\displaystyle D_{ij}^{'}=l_{ip}l_{jq}D_{pq}\,}$

Therefore,

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