Problem 2.18. Hole on a plate

Original stress, σ [N/mm²]=

Tangental strain ε_y [μ]=

Step 1. Writing the material law give the relation between the original stresses and strains (without the hole) in hydrostatic stress state.

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Step 2. Give stresses and strains at a point 50mm from the middle after drilling a hole.

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Stresses around the hole are given in Figure 2.30c.

$$\begin{gathered} \mathbf{\epsilon }=\left\{\begin{array}{c}{\epsilon }_x^{}\\ {\epsilon }_y^{}\\ {\gamma }_{xy}^{}\end{array}\right\}=\mathbf{C} \mathit{\sigma }=\left[\begin{array}{ccc}\frac{1}{E}& 0& 0\\ 0& \frac{1}{E}& 0\\ 0& 0& \frac{2}{E}\end{array}\right]\left\{\begin{array}{c}{\sigma }_x^{}\\ {\sigma }_y^{}\\ {\tau }_{xy}^{}\end{array}\right\}=\left[\begin{array}{ccc}\frac{1}{210\times {10}^6}& 0& 0\\ 0& \frac{1}{210\times {10}^6}& 0\\ 0& 0& \frac{2}{E}\end{array}\right]\left\{\begin{array}{c}\sigma \left(1–\frac{R^2}{x^2}\right)\\ \sigma \left(1+\frac{R^2}{x^2}\right)\\ 0\end{array}\right\} \\ \mathbf{\epsilon }\mathbf{ }(x=50\mathrm{mm})=\left[\begin{array}{ccc}\frac{1}{210\times {10}^6}& 0& 0\\ 0& \frac{1}{210\times {10}^6}& 0\\ 0& 0& \frac{2}{E}\end{array}\right]\left\{\begin{array}{c}\sigma \left(1–\frac{{15}^2}{{50}^2}\right)\\ \sigma \left(1+\frac{{15}^2}{{50}^2}\right)\\ 0\end{array}\right\}, \mathrm{where} R=15\mathrm{mm} \end{gathered}$$

Step 3. Express original stress from the measured strain difference.

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Step 4. Calculate the expected hoop strain.

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Step 5. Check the calculation by expressing the measured strain difference from the results.

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The material law gives the relation is between the original stresses and strains (without the hole) in hydrostatic stress state:

${\mathit{\epsilon }}_{}^{\mathbf{0}}=\left\{\begin{array}{c}{\epsilon }_x^0\\ {\epsilon }_y^0\\ {\gamma }_{xy}^0\end{array}\right\}=\mathbf{S} {\mathit{\sigma }}^{\mathbf{0}}=\left[\begin{array}{ccc}\frac{1}{E}& 0& 0\\ 0& \frac{1}{E}& 0\\ 0& 0& \frac{2}{E}\end{array}\right]\left\{\begin{array}{c}{\sigma }_x^0\\ {\sigma }_y^0\\ {\tau }_{xy}^0\end{array}\right\}=\left[\begin{array}{ccc}\frac{1}{210\times {10}^3}& 0& 0\\ 0& \frac{1}{210\times {10}^3}& 0\\ 0& 0& \frac{2}{210\times {10}^3}\end{array}\right]\left\{\begin{array}{c}\sigma \\ \sigma \\ 0\end{array}\right\}$

After drilling a hole stresses and strains at a point 50 mm from the middle are

Stresses around the hole are given in Figure 2.30c.

The original stress can be expressed from the measured strain difference:

$$\begin{gathered} {\epsilon }_x–{\epsilon }_x^0=–20\mu =\frac{1}{E}\left(\sigma \left(1–\frac{15^2}{50^2}\right)–\sigma \right)=–\frac{\sigma }{E}\frac{15^2}{50^2} \\ \to \mathit{\sigma }=\left({\epsilon }_x^0–{\epsilon }_x\right)E\frac{50^2}{15^2}=–20\times {10}^{–6}\times 210\times {10}^3\frac{50^2}{15^2}\mathbf{=}\mathbf{46}\mathbf{.}\mathbf{7}\frac{\mathbf{N}}{{\mathbf{mm}}^{\mathbf{2}}} \end{gathered}$$

The expected hoop strain is

$\mathbf{\epsilon }=\left\{\begin{array}{c}{\epsilon }_x\\ {\epsilon }_y\\ {\gamma }_{xy}\end{array}\right\}=\left[\begin{array}{ccc}\frac{1}{E}& 0& 0\\ 0& \frac{1}{E}& 0\\ 0& 0& \frac{2}{E}\end{array}\right]\left\{\begin{array}{c}\sigma \left(1–\frac{R^2}{x^2}\right)\\ \sigma \left(1+\frac{R^2}{x^2}\right)\\ 0\end{array}\right\}=\left[\begin{array}{ccc}\frac{1}{210\times {10}^3}& 0& 0\\ 0& \frac{1}{210\times {10}^3}& 0\\ 0& 0& \frac{2}{210\times {10}^3}\end{array}\right]\left\{\begin{array}{c}46.7\left(1–\frac{{15}^2}{{50}^2}\right)\\ 46.7\left(1+\frac{{15}^2}{{50}^2}\right)\\ 0\end{array}\right\}=\left\{\begin{array}{c}202.2\mu \\ \mathbf{242}\mathbf{.}\mathbf{2}\mathit{\mu }\\ 0\end{array}\right\}$The calculation can be checked by expressing the measured strain difference from the results

${\epsilon }_x^0–{\epsilon }_x=\frac{\sigma }{E}–202.22\mu =\frac{46.667}{210\times {10}^3}–202.22\mu =20\mu$