Introduction to Elasticity/Sample midterm 2

Sample Midterm Problem 2

Given:

A strain gage rosette provides the following data

$\varepsilon_{1} = 0.01; ~~ \varepsilon_{2} = 0.02; ~~ \varepsilon_{30^o} = 0$

where the $X_1$ and $X_2$ directions are perpendicular to each other and $\varepsilon_{30^o}$ 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 $\varepsilon_{60^o}$.
• (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

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

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

$\left[\boldsymbol{\varepsilon}\right]^{'} = \left[L\right] \left[\boldsymbol{\varepsilon}\right] \left[L\right]^T ~~\text{or,}~~ \varepsilon^{'}_{ij} = l_{ip} l_{jq} \varepsilon_{pq}$

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

$\begin{align} \varepsilon^{'}_{11} = & l_{1p} l_{1q} \varepsilon_{pq} \\ = & l_{11}l_{11}\varepsilon_{11} + l_{12}l_{11}\varepsilon_{21} + l_{13}l_{11}\varepsilon_{31} + l_{11}l_{12}\varepsilon_{12} + l_{12}l_{12}\varepsilon_{22} + l_{13}l_{12}\varepsilon_{32} +\\ & l_{11}l_{13}\varepsilon_{13} + l_{12}l_{13}\varepsilon_{23} + l_{13}l_{13}\varepsilon_{33} \\ = & l_{11}(l_{11}\varepsilon_{11} + l_{12}\varepsilon_{12} + l_{13}\varepsilon_{13}) + l_{12}(l_{11}\varepsilon_{21} + l_{12}\varepsilon_{22} + l_{13}\varepsilon_{23}) + \\ & l_{13}(l_{11}\varepsilon_{31} + l_{12}\varepsilon_{32} + l_{13}\varepsilon_{33}) \\ = & (\frac{\sqrt{3}}{2})\left[(\frac{\sqrt{3}}{2})(0.01) + (\frac{1}{2})\varepsilon_{12}\right] + (\frac{1}{2})\left[(\frac{\sqrt{3}}{2})\varepsilon_{12} + (\frac{1}{2})(0.02)\right] \\ = & (3/4)(0.01) + (\sqrt{3}/2)\varepsilon_{12} + (1/4)(0.02) \\ = & (5/4)(0.01) + (\sqrt{3}/2)\varepsilon_{12} \end{align}$

Therefore,

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

Hence,

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

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

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

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

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

Therefore,

${\varepsilon_{60^o} = 0.005}$

Part (b)

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