Elasticity/Sample midterm 2

Sample Midterm Problem 2

Given:

A strain gage rosette provides the following data

${\epsilon }_1=0.01;{\epsilon }_2=0.02;{\epsilon }_{30^o}=0$

where the $X_1$ and $X_2$ directions are perpendicular to each other and ${\epsilon }_{30^o}$ is the extensional strain of a line element at an angle of $30^o$ to the $X_1$ axis (in the counterclockwise direction).

Find:

• (a) Compute ${\epsilon }_{60^o}$.
• (b) Is the result valid if the material is anisotropic ?

Solution

Part (a)

From the previous problem, for an angle of rotation of 30$o^{}$, the rotation matrix $\left[L\right]$ is

$l_{ij}=\left[L\right]=\left[\begin{array}{ccc}\sqrt{3}/2& 1/2& 0\\ -1/2& \sqrt{3}/2& 0\\ 0& 0& 1\end{array}\right]$

Therefore, the components of strain in the rotated co-ordinate system are given by

${\left[\mathit{\epsilon }\right]}^{\text{'}}=\left[L\right]\left[\mathit{\epsilon }\right]{\left[L\right]}^T\text{or,}{\epsilon }_{ij}^{\text{'}}=l_{ip}{l}_{jq}{\epsilon }_{pq}$

Since we are given ${\epsilon }_{30^o}={\epsilon }_{11}^{\text{'}}$, we will calculate the value of this strain in terms of the original components of strain. Thus,

$\begin{array}{cc}{\epsilon }_{11}^{\text{'}}=& l_{1p}{l}_{1q}{\epsilon }_{pq}\\ =& l_{11}{l}_{11}{\epsilon }_{11}+l_{12}{l}_{11}{\epsilon }_{21}+l_{13}{l}_{11}{\epsilon }_{31}+l_{11}{l}_{12}{\epsilon }_{12}+l_{12}{l}_{12}{\epsilon }_{22}+l_{13}{l}_{12}{\epsilon }_{32}+\\ & l_{11}{l}_{13}{\epsilon }_{13}+l_{12}{l}_{13}{\epsilon }_{23}+l_{13}{l}_{13}{\epsilon }_{33}\\ =& l_{11}(l_{11}{\epsilon }_{11}+l_{12}{\epsilon }_{12}+l_{13}{\epsilon }_{13})+l_{12}(l_{11}{\epsilon }_{21}+l_{12}{\epsilon }_{22}+l_{13}{\epsilon }_{23})+\\ & l_{13}(l_{11}{\epsilon }_{31}+l_{12}{\epsilon }_{32}+l_{13}{\epsilon }_{33})\\ =& (\frac{\sqrt{3}}{2})[(\frac{\sqrt{3}}{2})(0.01)+(\frac{1}{2}){\epsilon }_{12}]+(\frac{1}{2})[(\frac{\sqrt{3}}{2}){\epsilon }_{12}+(\frac{1}{2})(0.02)]\\ =& (3/4)(0.01)+(\sqrt{3}/2){\epsilon }_{12}+(1/4)(0.02)\\ =& (5/4)(0.01)+(\sqrt{3}/2){\epsilon }_{12}\end{array}$

Therefore,

$(5/4)(0.01)+(\sqrt{3}/2){\epsilon }_{12}={\epsilon }_{30^o}=0$

Hence,

${\epsilon }_{12}=-(2.5)(0.01)/\sqrt{3}$

Next, for an angle of rotation of 60$o^{}$, the matrix $\left[L\right]$ is

$\begin{array}{cc}\left[L\right]& =\left[\begin{array}{ccc}\cos (60^o)& \sin (60^o)& \cos (90^o)\\ -\sin (60^o)& \cos (60^o)& \cos (90^o)\\ \cos (90^o)& \cos (90^o)& \cos (0^o)\end{array}\right]\\ & =\left[\begin{array}{ccc}1/2& \sqrt{3}/2& 0\\ -\sqrt{3}/2& 1/2& 0\\ 0& 0& 1\end{array}\right]\end{array}$

Therefore, ${\epsilon }_{60^o}={\epsilon }_{11}^{\text{'}}$, is given by

$\begin{array}{cc}{\epsilon }_{11}^{\text{'}}=& l_{1p}{l}_{1q}{\epsilon }_{pq}\\ =& l_{11}{l}_{11}{\epsilon }_{11}+l_{12}{l}_{11}{\epsilon }_{21}+l_{13}{l}_{11}{\epsilon }_{31}+l_{11}{l}_{12}{\epsilon }_{12}+l_{12}{l}_{12}{\epsilon }_{22}+l_{13}{l}_{12}{\epsilon }_{32}+\\ & l_{11}{l}_{13}{\epsilon }_{13}+l_{12}{l}_{13}{\epsilon }_{23}+l_{13}{l}_{13}{\epsilon }_{33}\\ =& l_{11}(l_{11}{\epsilon }_{11}+l_{12}{\epsilon }_{12}+l_{13}{\epsilon }_{13})+l_{12}(l_{11}{\epsilon }_{21}+l_{12}{\epsilon }_{22}+l_{13}{\epsilon }_{23})+\\ & l_{13}(l_{11}{\epsilon }_{31}+l_{12}{\epsilon }_{32}+l_{13}{\epsilon }_{33})\\ =& (\frac{1}{2})[(\frac{1}{2})(0.01)+(\frac{\sqrt{3}}{2}){\epsilon }_{12}]+(\frac{\sqrt{3}}{2})[(\frac{1}{2}){\epsilon }_{12}+(\frac{\sqrt{3}}{2})(0.02)]\\ =& (1/4)(0.01)+(\sqrt{3}/2){\epsilon }_{12}+(3/4)(0.02)\\ =& (7/4)(0.01)+(\sqrt{3}/2)(-(2.5)(0.01)/\sqrt{3})\\ =& (7/4)(0.01)-(5/4)(0.01)=(1/2)(0.01)=0.005\\ \end{array}$

Therefore,

${\epsilon }_{60^o}=0.005$

Part (b)

$\text{The result is valid for all materials.}$