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At a point $ P $ on the surface of a body, the rosette of 45 gauges indicated in the Figure is glued, whose readings are $\varepsilon_1 = 2 \times 10^{-6}$ , $\varepsilon_2 = 1 \times 10^{-6}$ e $\varepsilon_3 = -4 \times 10^{-6}$, respectively in elements 1, 2 and 3. Calculate, using the Mohr circle, the principal stresses and directions main points at point $ P $, knowing that the material of the body presents the following features: $ E = 200 $ GPa and $\nu$ = 0.3. (note: assume a flat state of tension applied to the body).

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I use \begin{equation} \begin{cases} \varepsilon_\mathrm{1} \equiv \varepsilon (\alpha=0^\circ) & = \varepsilon_{xx}\cos^2\alpha_1 + \varepsilon_{yy}\sin^2\alpha_1 + \gamma_{xy}\sin\alpha_1\cos\alpha_1 \\ \varepsilon_\mathrm{2} \equiv \varepsilon (\alpha=+45^\circ) & = \varepsilon_{xx}\cos^2\alpha_2 + \varepsilon_{yy}\sin^2\alpha_2 + \gamma_{xy}\sin\alpha_2\cos\alpha_2 \\ \varepsilon_\mathrm{3} \equiv \varepsilon (\alpha=90^\circ) & = \varepsilon_{xx}\cos^2\alpha_3 + \varepsilon_{yy}\sin^2\alpha_3 + \gamma_{xy}\sin\alpha_3\cos\alpha_3 \end{cases} \end{equation}

and after some calculations i arrive to the deformation matrix at $P$ point: \begin{equation} \underline{\underline{D}} \equiv [D] \equiv D_{ij} = \begin{bmatrix} \varepsilon_{xx} & \varepsilon_{xy} & \varepsilon_{xz} \\ \varepsilon_{yx} & \varepsilon_{yy} & \varepsilon_{yz} \\ \varepsilon_{zx} & \varepsilon_{zy} & \varepsilon_{zz} \\ \end{bmatrix} = \begin{bmatrix} \varepsilon_{xx} & \gamma_{xy}/2 & \gamma_{xz}/2 \\ \gamma_{yx}/2 & \varepsilon_{yy} & \gamma_{yz}/2 \\ \gamma_{zx}/2 & \gamma_{zy}/2 & \varepsilon_{zz} \\ \end{bmatrix} = \begin{bmatrix} 2 & 2 & 0 \\ 2 & -4 & 0 \\ 0 & 0 & 0 \end{bmatrix} \times 10^{-6} \end{equation}

i've done $|\epsilon_{ij}-\delta_{ij}\epsilon|=0$ and got the values $\epsilon_{I}=-1+\sqrt{13}$, $\epsilon_{II}=0$, and $\epsilon_{III}=-1-\sqrt{13}$,

and i use hookes law and got

$\sigma_{I}=\frac{E}{(1+v)(1-2v)}((1-v)\epsilon_{I}+v\epsilon_{II}+v\epsilon_{III})=1.70\times 10^{5}$

However the solutions say that the tension matrix is \begin{equation} \underline{\underline{\sigma}} = [\sigma] = \sigma_{ij} = \begin{bmatrix} 0.175824 & 0.307692 & 0 \\ 0.307692 & -0.747252 & 0 \\ 0 & 0 & 0 \end{bmatrix}~\mbox{[MPa]} \end{equation}

But i don't understand how can i arrive to the tension matrix. I thought is through hookes law but i guess i'm making a mistake somewhere....Could someone explain me how to i get the tension matrix?

In the solutions the deformation matrix is \begin{equation} D = \begin{bmatrix} 2 & 2 & 0 \\ 2 & -4 & 0 \\ 0 & 0 & \varepsilon_{zz} \end{bmatrix} \times 10^{-6} \end{equation} which is different from mine but i also dont understand why $\varepsilon_{zz}$ is not 0

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"A flat state of tension" seems like a computer translation that should be "plane stress."

You seem to have used the formulas for plane strain.

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  • $\begingroup$ but what did I do wrong in this exercice? $\endgroup$
    – user28922
    Dec 8 '20 at 10:34
  • $\begingroup$ could you explain me how do i find the tension matrix? $\endgroup$
    – user28922
    Dec 8 '20 at 12:01

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