# how to find the tension matrix knowing the extensions

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).

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

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

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 $$$$\underline{\underline{\sigma}} = [\sigma] = \sigma_{ij} = \begin{bmatrix} 0.175824 & 0.307692 & 0 \\ 0.307692 & -0.747252 & 0 \\ 0 & 0 & 0 \end{bmatrix}~\mbox{[MPa]}$$$$

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

"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.

• but what did I do wrong in this exercice?
– user28922
Dec 8, 2020 at 10:34
• could you explain me how do i find the tension matrix?
– user28922
Dec 8, 2020 at 12:01