Problem 2.7. Strain gauges

Stress, σ’_x [MPa]=

Stress, σ’_y [MPa]=

Stress, τ‘_x [MPa]=

Strain along the thickness, ε’_z  [×10^-4] =

Step 1.  Applying transformation matrices express the measured strains, ${\epsilon }_A, {\epsilon }_B, {\epsilon }_C$ in the function of the unknown strains, $\epsilon {‘}_x, \epsilon {‘}_y, \gamma {‘}_{xy}.$

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See Example 2.4

${\epsilon }_{\mathrm{A}}=\epsilon {‘}_y$

Strain gauge A is placed in y‘ direction thus without any transformation it’s measured strain results directly strain in y‘ direction 

$$\begin{gathered} \mathit{\epsilon }={\mathbf{T}}_{\epsilon }\mathit{\epsilon }\mathbf{‘}=\left[\begin{array}{ccc}{\cos }^{2}45°& {\sin }^{2}45°& \sin 45°\cos 135°\\ {\sin }^{2}45°& {\cos }^{2}45°& –\sin 45°\cos 135°\\ –2\sin 45°\cos 45°& 2\sin 45°\cos 45°& {\cos }^{2}45°–{\sin }^{2}45°\end{array}\right]\left\{\begin{array}{c}\epsilon {‘}_x\\ \epsilon {‘}_y\\ \gamma {‘}_{xy}\end{array}\right\} \\ \to {\epsilon }_B={\epsilon }_x=\epsilon {‘}_x{\cos }^{2}45°+\epsilon {‘}_y{\sin }^{2}45°+\gamma {‘}_{xy}\sin 45°\cos 45°=\frac{\epsilon {‘}_x}{2}+\frac{\epsilon {‘}_y}{2}+\frac{\gamma {‘}_{xy}}{2} \\ \to {\epsilon }_{\mathrm{C}}={\epsilon }_{xy}=\epsilon {‘}_x{\sin }^{2}45°+\epsilon {‘}_y{\cos }^{2}45°–\gamma {‘}_{xy}\sin 45°\cos 45°=\frac{\epsilon {‘}_x}{2}+\frac{\epsilon {‘}_y}{2}–\frac{\gamma {‘}_{xy}}{2} \end{gathered}$$

Step 2.  Transformation results in an equation system. Solve and calculate the three unknown strains in the x’-y‘ coordinate system.

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Step 3. Write the stiffness matrix and determine stresses in the x’-y‘ coordinate system.

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Stresses are obtained by the multiplication of stiffness matrix of the isotropic material and the strain vector.

Eq.(2-58).

$$\begin{gathered} \mathit{\sigma }\mathbf{‘}=\mathit{E}\mathit{\epsilon }{\mathbf{‘}}_{\mathbf{x}}=\frac{1}{1–{\nu }^2}\left[\begin{array}{ccc}E& \nu E& 0\\ \nu E& E& 0\\ 0& 0& \frac{E\left(1–\nu \right)}{2}\end{array}\right]\left\{\begin{array}{c}\epsilon {‘}_x\\ \epsilon {‘}_y\\ \gamma {‘}_{xy}\end{array}\right\}=\left[\begin{array}{ccc}\frac{210}{1–0.3^2}& \frac{0.3\times 210}{1–0.3^2}& 0\\ \frac{0.3\times 210}{1–0.3^2}& \frac{210}{1–0.3^2}& 0\\ 0& 0& 80.77\end{array}\right]\left\{\begin{array}{c}–1.8\\ 6\\ –3.8\end{array}\right\}\times {10}^{–4} \\ \left\{\begin{array}{c}\mathbf{\sigma }{\mathbf{‘}}_{\mathbf{x}}\\ \mathbf{\sigma }{\mathbf{‘}}_{\mathbf{y}}\\ \mathbf{\tau }{\mathbf{‘}}_{\mathbf{x}\mathbf{y}}\end{array}\right\}\mathbf{=}\left\{\begin{array}{c}\mathbf{0}\mathbf{ }\\ \mathbf{126}\\ \mathbf{–}\mathbf{30}\mathbf{.}\mathbf{69}\end{array}\right\}\mathbf{MPa} \end{gathered}$$

Step 4. Give strain, ε_z along the thickness of the plate. 

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The plate is in-plane stress condition, thus no σ_z stress arises along z axis. Strain, ε’_z along the thickness of the plate is caused by the Poisson effect only. See the “generalized” Hooke’s law.

Eq.(2-72).

$$\begin{gathered} \left\{\begin{array}{c}\epsilon {‘}_x\\ \epsilon {‘}_y\\ \epsilon {‘}_z\\ \gamma {‘}_{yz}\\ \gamma {‘}_{xz}\\ \gamma {‘}_{xy}\end{array}\right\}=\left[\begin{array}{cccccc}\frac{1}{E}& –\frac{\nu }{E}& –\frac{\nu }{E}& & & \\ –\frac{\nu }{E}& \frac{1}{E}& –\frac{\nu }{E}& & & \\ –\frac{\nu }{E}& –\frac{\nu }{E}& \frac{1}{E}& & & \\ & & & \frac{2(1+\nu )}{E}& & \\ & & & & \frac{2(1+\nu )}{E}& \\ & & & & & \frac{2(1+\nu )}{E}\end{array}\right]\left\{\begin{array}{c}\sigma {‘}_x\\ \sigma {‘}_y\\ 0\\ 0\\ 0\\ \tau {‘}_{xy}\end{array}\right\} \\ \to \mathit{\epsilon }{\mathbf{‘}}_{\mathbf{z}}=–\frac{\mathrm{\nu \sigma }{‘}_{\mathrm{x}}}{E}–\frac{\mathrm{\nu \sigma }{‘}_y}{E}=0–\frac{0.3\times 0.126}{210}\mathbf{=}\mathbf{–}\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{–}\mathbf{4}} \end{gathered}$$

Applying transformation matrices the measured strains, ${\epsilon }_A, {\epsilon }_B, {\epsilon }_C$ can be expressed in the function of the unknown strains, $\epsilon {‘}_x, \epsilon {‘}_y, \gamma {‘}_{xy}.$ 

See Example 2.4

${\epsilon }_{\mathrm{A}}=\epsilon {‘}_y$

Strain gauge A is placed in y‘ direction thus without any transformation it’s measured strain results directly strain in y‘ direction 

$$\begin{gathered} \mathit{\epsilon }={\mathbf{T}}_{\epsilon }\mathit{\epsilon }\mathbf{‘}=\left[\begin{array}{ccc}{\cos }^{2}45°& {\sin }^{2}45°& \sin 45°\cos 45°\\ \sin 45°& {\cos }^{2}45°& –\sin 45°\cos 45°\\ –2\sin 45°\cos 45°& 2\sin 45°\cos 45°& {\cos }^{2}45°–{\sin }^{2}45°\end{array}\right]\left\{\begin{array}{c}\epsilon {‘}_x\\ \epsilon {‘}_y\\ \gamma {‘}_{xy}\end{array}\right\} \\ \to {\epsilon }_{\mathrm{B}}={\epsilon }_x=\epsilon {‘}_x{\cos }^{2}45°+\epsilon {‘}_y{\sin }^{2}45°+\gamma {‘}_{xy}\sin 45°\cos 45°=\frac{\epsilon {‘}_x}{2}+\frac{\epsilon {‘}_y}{2}+\frac{\gamma {‘}_{xy}}{2} \\ \to {\epsilon }_{\mathrm{C}}={\epsilon }_y=\epsilon {‘}_x{\sin }^{2}45°+\epsilon {‘}_y{\cos }^{2}45°–\gamma {‘}_{xy}\sin 45°\cos 45°=\frac{\epsilon {‘}_x}{2}+\frac{\epsilon {‘}_y}{2}–\frac{\gamma {‘}_{xy}}{2} \end{gathered}$$

Transformation results in an equation system the solution of which results in the three unknown strains in x’-y‘ coordinate system.

$$\begin{gathered} \left\{\begin{array}{c}{\epsilon }_A\\ {\epsilon }_B\\ {\epsilon }_C\end{array}\right\}=\left[\begin{array}{ccc}0& 1& 0\\ \frac{1}{2}& \frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}& –\frac{1}{2}\end{array}\right]\left\{\begin{array}{c}\epsilon {‘}_x\\ \epsilon {‘}_y\\ \gamma {‘}_{xy}\end{array}\right\} \to \left\{\begin{array}{c}\epsilon {‘}_x\\ \epsilon {‘}_y\\ \gamma {‘}_{xy}\end{array}\right\}={\left[\begin{array}{ccc}1& 0& 0\\ \frac{1}{2}& \frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}& –\frac{1}{2}\end{array}\right]}^{–1}\left\{\begin{array}{c}6\times {10}^{–4}\\ 2\times {10}^{–5}\\ 4\times {10}^{–4}\end{array}\right\} \\ \left\{\begin{array}{c}\epsilon {‘}_x\\ \epsilon {‘}_y\\ \gamma {‘}_{xy}\end{array}\right\}= \left\{\begin{array}{c}–1.8\times {10}^{–4}\\ 6.0\times {10}^{–4}\\ –3.8\times {10}^{–4}\end{array}\right\} \end{gathered}$$Stresses in the x’-y‘ coordinate system are obtained by the multiplication of stiffness matrix of the isotropic material and the strain vector.

Eq.(2-58).

$$\begin{gathered} \mathit{\sigma }\mathbf{‘}=\mathit{E}\mathit{\epsilon }{\mathbf{‘}}_{\mathbf{x}}=\frac{1}{1–{\nu }^2}\left[\begin{array}{ccc}E& \nu E& 0\\ \nu E& E& 0\\ 0& 0& \frac{E\left(1–\nu \right)}{2}\end{array}\right]\left\{\begin{array}{c}\epsilon {‘}_x\\ \epsilon {‘}_y\\ \gamma {‘}_{xy}\end{array}\right\}=\left[\begin{array}{ccc}\frac{210}{1–0.3^2}& \frac{0.3\times 210}{1–0.3^2}& 0\\ \frac{0.3\times 210}{1–0.3^2}& \frac{210}{1–0.3^2}& 0\\ 0& 0& 80.77\end{array}\right]\left\{\begin{array}{c}–1.8\\ 6\\ –3.8\end{array}\right\}\times {10}^{–4} \\ \left\{\begin{array}{c}\mathbf{\sigma }{\mathbf{‘}}_{\mathbf{x}}\\ \mathbf{\sigma }{\mathbf{‘}}_{\mathbf{y}}\\ \mathbf{\tau }{\mathbf{‘}}_{\mathbf{x}\mathbf{y}}\end{array}\right\}\mathbf{=}\left\{\begin{array}{c}\mathbf{0}\mathbf{ }\\ \mathbf{126}\\ \mathbf{–}\mathbf{30}\mathbf{.}\mathbf{69}\end{array}\right\}\mathbf{MPa} \end{gathered}$$

The plate is in-plane stress condition, thus no σ_z stress arises along z axis. Strain, ε’_z along the thickness of the plate is caused by the Poisson effect only. See the “generalized” Hooke’s law.

Eq.(2-72).

$$\begin{gathered} \left\{\begin{array}{c}\epsilon {‘}_x\\ \epsilon {‘}_y\\ \epsilon {‘}_z\\ \gamma {‘}_{yz}\\ \gamma {‘}_{xz}\\ \gamma {‘}_{xy}\end{array}\right\}=\left[\begin{array}{cccccc}\frac{1}{E}& –\frac{\nu }{E}& –\frac{\nu }{E}& & & \\ –\frac{\nu }{E}& \frac{1}{E}& –\frac{\nu }{E}& & & \\ –\frac{\nu }{E}& –\frac{\nu }{E}& \frac{1}{E}& & & \\ & & & \frac{2(1+\nu )}{E}& & \\ & & & & \frac{2(1+\nu )}{E}& \\ & & & & & \frac{2(1+\nu )}{E}\end{array}\right]\left\{\begin{array}{c}\sigma {‘}_x\\ \sigma {‘}_y\\ 0\\ 0\\ 0\\ \tau {‘}_{xy}\end{array}\right\} \\ \to \mathit{\epsilon }{\mathbf{‘}}_{\mathbf{z}}=–\frac{\mathrm{\nu \sigma }{‘}_{\mathrm{x}}}{E}–\frac{\mathrm{\nu \sigma }{‘}_y}{E}=0–\frac{0.3\times 0.126}{210}\mathbf{=}\mathbf{–}\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{–}\mathbf{4}} \end{gathered}$$