Problem 2.5. Pressure vessel’s strains

Stress, ${\epsilon }_x$ [×10^-5 MPa]=

Stress, ${\epsilon }_y$ [×10^-5 MPa]=

Stress, ${\gamma }_{xy}$[×10^-5 MPa]=

Principal stress, ${\epsilon }_1$ [×10^-5 MPa]=

Principal stress, ${\epsilon }_2$[×10^-5 MPa]=

Step 1. Strains are related to the stresses due to the compliance matrix of the material. Determine the compliance matrix of the material.

Check compliance matrixHide compliance matrix

Eqs.(2-57) and (2-62).

$$\begin{gathered} \mathbf{S}=\left[\begin{array}{ccc}\frac{1}{E}& –\frac{\nu }{E}& 0\\ –\frac{\nu }{E}& \frac{1}{E}& 0\\ 0& 0& \frac{1}{G}\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{210}& –\frac{0.3}{210}& 0\\ –\frac{0.3}{210}& \frac{1}{210}& 0\\ 0& 0& \frac{1}{80.77}\end{array}\right]\frac{1}{\mathrm{GPa}} \\ \mathrm{where} G=\frac{E}{2\left(1+\nu \right)}=\frac{210}{2\left(1+0.3\right)}=80.77 \mathrm{GPa} \end{gathered}$$

Since the material is isotropic material properties and also the compliance matrix is the same in every directions. 

Step 2. Give strains in x-y coordinate system from the stresses determined in the previous Problem.

Check strainsHide strains

Eqs.(2-57).

$$\begin{gathered} \left\{\begin{array}{c}{\mathrm{\epsilon }}_{\mathrm{x}}\\ {\mathrm{\epsilon }}_{\mathrm{y}}\\ {\mathrm{\gamma }}_{xy}\end{array}\right\}={\mathbf{S\sigma }}_{\mathbf{x}}=\left[\begin{array}{ccc}\frac{1}{E}& –\frac{\nu }{E}& 0\\ –\frac{\nu }{E}& \frac{1}{E}& 0\\ 0& 0& \frac{1}{G}\end{array}\right]\left\{\begin{array}{c}{\sigma }_x\\ {\sigma }_y\\ {\tau }_{xy}\end{array}\right\}=\left[\begin{array}{ccc}\frac{1}{210}& –\frac{0.3}{210}& 0\\ –\frac{0.3}{210}& \frac{1}{210}& 0\\ 0& 0& \frac{1}{80.77}\end{array}\right]\left\{\begin{array}{c}15.87\\ 14.13\\ –4.92\end{array}\right\}\times {10}^{–3} \\ \left\{\begin{array}{c}{\mathbf{\epsilon }}_{\mathbf{x}}\\ {\mathbf{\epsilon }}_{\mathbf{y}}\\ {\mathbf{\gamma }}_{\mathbf{x}\mathbf{y}}\end{array}\right\}\mathbf{=}\left\{\begin{array}{c}\mathbf{5}\mathbf{.}\mathbf{537}\\ \mathbf{4}\mathbf{.}\mathbf{463}\\ \mathbf{–}\mathbf{6}\mathbf{.}\mathbf{096}\end{array}\right\}\mathbf{\times }{\mathbf{10}}^{\mathbf{–}\mathbf{5}} \end{gathered}$$

Step 3. Calculate the principal strains and the principal strain’s directions.

Check principal strainsHide principal strains

Eq.(2-48).

$$\begin{gathered} {\epsilon }_1=\frac{{\epsilon }_x+{\epsilon }_y}{2}+\sqrt{{\left(\frac{{\epsilon }_x–{\epsilon }_y}{2}\right)}^2+{\left(\frac{{\gamma }_{xy}}{2}\right)}^2}=\frac{5.537–4.462}{2}+\sqrt{{\left(\frac{5.537–4.462}{2}\right)}^2+{\left(\frac{–6.069}{2}\right)}^2} \\ {\epsilon }_2=\frac{{\epsilon }_x+{\epsilon }_y}{2}–\sqrt{{\left(\frac{{\epsilon }_x–{\epsilon }_y}{2}\right)}^2+{\left(\frac{{\gamma }_{xy}}{2}\right)}^2}=\frac{5.537–4.462}{2}–\sqrt{{\left(\frac{5.537–4.462}{2}\right)}^2+{\left(\frac{–6.069}{2}\right)}^2} \\ {\mathit{\epsilon }}_{\mathbf{1}}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{095}\mathbf{\times }{\mathbf{10}}^{\mathbf{–}\mathbf{5}} \\ {\mathit{\epsilon }}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{905}\mathbf{\times }{\mathbf{10}}^{\mathbf{–}\mathbf{5}} \end{gathered}$$

Check principal directionsHide principal directions

Eq.(2-47).

${\beta }_0=\frac{1}{2}{\tan }^{–1}\frac{{\epsilon }_x–{\epsilon }_y}{{\gamma }_{xy}}=\frac{1}{2}{\tan }^{–1}\frac{5.537–4.462}{–6.096} \to {\beta }_0=–{40}^{\circ }$

Rotating x-y coordinate system backward by -40^o we arrive at the hoop and axial directions, thus the principal direction of stresses and those of the strains coincide. This statement is true for any isotropic material.

See Footnote l in Section 2.1.4

Step 4. Check the solution by calculating the stresses from the principal strains.

Check principal stressesHide principal stresses

Stresses are obtained by multiplying the stiffness matrix with the strains.

Eqs.(2-58).

$$\begin{gathered} \mathit{\sigma }=\mathit{E}{\mathit{\epsilon }}_{12}=\frac{1}{1–{\nu }^2}\left[\begin{array}{ccc}E& \nu E& 0\\ \nu E& E& 0\\ 0& 0& G\left(1–{\nu }^2\right)\end{array}\right]\left\{\begin{array}{c}{\epsilon }_1\\ {\epsilon }_2\\ 0\end{array}\right\}=\left[\begin{array}{ccc}\frac{210}{1–0.3^2}& \frac{0.3\times 210}{1–0.3^2}& 0\\ \frac{0.3\times 210}{1–0.3^2}& \frac{210}{1–0.3^2}& 0\\ 0& 0& 80.77\end{array}\right]\left\{\begin{array}{c}8.095 \\ 1.905\\ 0\end{array}\right\}\times {10}^{–2} \\ \mathit{\sigma }=\left\{\begin{array}{c}20 \\ 10\\ 0\end{array}\right\}\mathrm{MPa}=\left\{\begin{array}{c}{\sigma }_1 \\ {\sigma }_2\\ 0\end{array}\right\} \end{gathered}$$

The results are equal to the principal stresses which verifies our calculation and meets the expectations.

Strains are related to the stresses due to the compliance matrix of the material. The compliance matrix of the izotrop material is

Eqs.(2-57) and (2-62).

$$\begin{gathered} \mathbf{S}=\left[\begin{array}{ccc}\frac{1}{E}& –\frac{\nu }{E}& 0\\ –\frac{\nu }{E}& \frac{1}{E}& 0\\ 0& 0& \frac{1}{G}\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{210}& –\frac{0.3}{210}& 0\\ –\frac{0.3}{210}& \frac{1}{210}& 0\\ 0& 0& \frac{1}{80.77}\end{array}\right]\frac{1}{\mathrm{GPa}}, \\ \mathrm{where} G=\frac{E}{2\left(1+\nu \right)}=\frac{210}{2\left(1+0.3\right)}=80.77 \mathrm{GPa}. \end{gathered}$$

Since the material is isotropic material properties and also the compliance matrix is the same in every directions. 

The strains in x-y coordinate system are

$$\begin{gathered} \left\{\begin{array}{c}{\mathrm{\epsilon }}_{\mathrm{x}}\\ {\mathrm{\epsilon }}_{\mathrm{y}}\\ {\mathrm{\gamma }}_{xy}\end{array}\right\}={\mathbf{S\sigma }}_{\mathbf{x}}=\left[\begin{array}{ccc}\frac{1}{E}& –\frac{\nu }{E}& 0\\ –\frac{\nu }{E}& \frac{1}{E}& 0\\ 0& 0& \frac{1}{G}\end{array}\right]\left\{\begin{array}{c}{\sigma }_x\\ {\sigma }_y\\ {\tau }_{xy}\end{array}\right\}=\left[\begin{array}{ccc}\frac{1}{210}& –\frac{0.3}{210}& 0\\ –\frac{0.3}{210}& \frac{1}{210}& 0\\ 0& 0& \frac{1}{80.77}\end{array}\right]\left\{\begin{array}{c}15.87\\ 14.13\\ –4.92\end{array}\right\}\times {10}^{–3} \\ \left\{\begin{array}{c}{\mathbf{\epsilon }}_{\mathbf{x}}\\ {\mathbf{\epsilon }}_{\mathbf{y}}\\ {\mathbf{\gamma }}_{\mathbf{x}\mathbf{y}}\end{array}\right\}\mathbf{=}\left\{\begin{array}{c}\mathbf{5}\mathbf{.}\mathbf{537}\\ \mathbf{4}\mathbf{.}\mathbf{463}\\ \mathbf{–}\mathbf{6}\mathbf{.}\mathbf{096}\end{array}\right\}\mathbf{\times }{\mathbf{10}}^{\mathbf{–}\mathbf{5}} \end{gathered}$$

Let us find the principal strains and their directions.

Eqs.(2-48) and (2-47).

$$\begin{gathered} {\epsilon }_1=\frac{{\epsilon }_x+{\epsilon }_y}{2}+\sqrt{{\left(\frac{{\epsilon }_x–{\epsilon }_y}{2}\right)}^2+{\left(\frac{{\gamma }_{xy}}{2}\right)}^2}=\frac{5.537–4.462}{2}+\sqrt{{\left(\frac{5.537–4.462}{2}\right)}^2+{\left(\frac{–6.069}{2}\right)}^2}, \\ {\epsilon }_2=\frac{{\epsilon }_x+{\epsilon }_y}{2}–\sqrt{{\left(\frac{{\epsilon }_x–{\epsilon }_y}{2}\right)}^2+{\left(\frac{{\gamma }_{xy}}{2}\right)}^2}=\frac{5.537–4.462}{2}–\sqrt{{\left(\frac{5.537–4.462}{2}\right)}^2+{\left(\frac{–6.069}{2}\right)}^2}, \\ {\mathit{\epsilon }}_{\mathbf{1}}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{095}\mathbf{\times }{\mathbf{10}}^{\mathbf{–}\mathbf{5}}\mathbf{,} \\ {\mathit{\epsilon }}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{905}\mathbf{\times }{\mathbf{10}}^{\mathbf{–}\mathbf{5}}\mathbf{,} \end{gathered}$$

${\beta }_0=\frac{1}{2}{\tan }^{–1}\frac{{\epsilon }_x–{\epsilon }_y}{{\gamma }_{xy}}=\frac{1}{2}{\tan }^{–1}\frac{5.537–4.462}{–6.096} \to {\beta }_0=–{40}^{\circ }.$

Rotating x-y coordinate system backward by -40^o we arrive at the hoop and axial directions, thus the principal direction of stresses and those of the strains coincide. This statement is true for any isotropic material.

See Footnote l in Section 2.1.4.

We can check the solution by calculating the stresses from the principal strains.

Eq.(2-58).

The results are equal to the principal stresses which verifies our calculation and meets the expectations.