Elasticity/Sample midterm 1

Sample Midterm Problem 1

Given:

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

𝐚 = 5 {\hat{𝐞}}_{1} - 3 {\hat{𝐞}}_{2} + 10 \hat{𝐞} 3; 𝐛 = 4 {\hat{𝐞}}_{1} + 6 {\hat{𝐞}}_{2} - 2 {\hat{𝐞}}_{3}; 𝐜 = 10 {\hat{𝐞}}_{1} + 6 {\hat{𝐞}}_{2}

Find:

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

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

d = 128

𝐃 = a_{i} c_{j} = [\begin{matrix} (5) (10) & (5) (6) & (5) (0) \\ (- 3) (10) & (- 3) (6) & (- 3) (0) \\ (10) (10) & (10) (6) & (10) (0) \end{matrix}]

𝐃 = [\begin{matrix} 50 & 30 & 0 \\ - 30 & - 18 & 0 \\ 100 & 60 & 0 \end{matrix}]

𝐃 is a second-order tensor .

δ_{i j} = Kronecker delta

e_{i j k} = Permutation symbol

δ_{i j} = {\begin{matrix} 1 & i f i = j \\ 0 & o t h e r w i s e \end{matrix}

e_{i j k} = {\begin{matrix} 1 & i f i j k = 123, 231, 312 \\ - 1 & i f i j k = 321, 213, 132 \\ 0 & o t h e r w i s e \end{matrix}

g = D_{k k} = D_{11} + D_{22} + D_{33} = 50 - 18 + 0 = 32

g = 32

δ_{i k} e_{i k m} = e_{j j m} = 0

Because $j j m$ cannot be an even or odd permutation of $1, 2, 3$ .

The basis transformation rule for vectors is

v_{i}^{'} = l_{i j} v_{j}

where

l_{i j} = \hat{𝐞} i^{'} ∙ \hat{𝐞} j = \cos (\hat{𝐞} i^{'}, \hat{𝐞} j)

Therefore,

\begin{matrix} [L] & = [\begin{matrix} \cos (3 0^{o}) & \cos (9 0^{o} - 3 0^{o}) & \cos (9 0^{o}) \\ \cos (9 0^{o} + 3 0^{o}) & \cos (3 0^{o}) & \cos (9 0^{o}) \\ \cos (9 0^{o}) & \cos (9 0^{o}) & \cos (0^{o}) \end{matrix}] \\ = [\begin{matrix} \cos (3 0^{o}) & \sin (3 0^{o}) & \cos (9 0^{o}) \\ - \sin (3 0^{o}) & \cos (3 0^{o}) & \cos (9 0^{o}) \\ \cos (9 0^{o}) & \cos (9 0^{o}) & \cos (0^{o}) \end{matrix}] \\ = [\begin{matrix} \sqrt{3} / 2 & 1 / 2 & 0 \\ - 1 / 2 & \sqrt{3} / 2 & 0 \\ 0 & 0 & 1 \end{matrix}] \end{matrix}

Hence,

\begin{matrix} 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{matrix}

Thus,

𝐛^{'} = 6.46 𝐞_{1}^{'} + 3.2 𝐞_{2}^{'} - 2 𝐞_{3}^{'}

The basis transformation rule for second-order tensors is

D_{i j}^{'} = l_{i p} l_{j q} D_{p q}

Therefore,

\begin{matrix} 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}) [(- \frac{1}{2}) (50) + (\frac{\sqrt{3}}{2}) (30) + (0) (0)] + (\frac{1}{2}) [(- \frac{1}{2}) (- 30) + (\frac{\sqrt{3}}{2}) (- 18) + (0) (0)] + \\ (0) [(- \frac{1}{2}) (100) + (\frac{\sqrt{3}}{2}) (60) + (0) (0)] \\ = & (\frac{\sqrt{3}}{2}) [- 25 + 15 \sqrt{3}] + (\frac{1}{2}) [15 - 9 \sqrt{3}] \\ = & - 25 \frac{\sqrt{3}}{2} + \frac{45}{2} + \frac{15}{2} - 9 \frac{\sqrt{3}}{2} \\ = & - 17 \sqrt{3} + 30 \end{matrix}

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