Problem 2.2. Stresses on plate edge

Stress, ${\sigma }_x$ [MPa]=

Stress, ${\sigma }_y$[MPa]=

Stress, $\sigma {‘}_y$[MPa]=

Step 1. Write stress vectors in both x-y and x’-y’ coordinate systems. Find the unknowns.

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Step 2. Rotation of the coordinate system requires the transformation of the stresses. Given and unknown stresses are related to each other through the transformation matrix. Determine transformation matrix.

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${\mathbf{T}}_{\mathbf{\sigma }}=\left[\begin{array}{ccc}{\cos }^{2}45°& {\sin }^{2}45°& 2\sin 45°\cos 45°\\ {\sin }^{2}45°& {\cos }^{2}45°& –2\sin 45°\cos 45°\\ –\sin 45°\cos 45°& \sin 45°\cos 45°& {\cos }^{2}45°–{\sin }^{2}45°\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{2}& 1\\ \frac{1}{2}& \frac{1}{2}& –1\\ –\frac{1}{2}& \frac{1}{2}& 0\end{array}\right]$

Eq.(2-9)

Step 3. Write the transformation parametrically and express the unknown stresses from the equation system of the transformation. 

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Transformation of stress vector from x-y to x’-y’ coordinate system is given by the following equation

Eq.(2-9)

$\left\{\begin{array}{c}\sigma {‘}_x\\ \mathrm{\sigma }{‘}_{\mathrm{y}}\\ \tau {‘}_{xy}\end{array}\right\}\mathbf{=}{\mathbf{T}}_{\sigma }\left\{\begin{array}{c}{\mathrm{\sigma }}_{\mathrm{x}}\\ {\mathrm{\sigma }}_{\mathrm{y}}\\ {\mathrm{\tau }}_{\mathrm{xy}}\end{array}\right\}$

where the transformation matrix is

${\mathbf{T}}_{\sigma }=\left[\begin{array}{ccc}{\cos }^{2}45°& {\sin }^{2}45°& 2\sin 45°\cos 45°\\ {\sin }^{2}45°& {\cos }^{2}45°& –2\sin 45°\cos 45°\\ –\sin 45°\cos 45°& \sin 45°\cos 45°& {\cos }^{2}45°–{\sin }^{2}45°\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{2}& 1\\ \frac{1}{2}& \frac{1}{2}& –1\\ –\frac{1}{2}& \frac{1}{2}& 0\end{array}\right]$

Stresses $\sigma {‘}_x$  and $\tau {‘}_{xy}$ are zero, because there is no surface load to equilibrate them, and ${\tau }_{xy}$ is given in the inital data. In the equation system three unknowns remain (signed red)

$\mathbf{\sigma }\mathbf{‘}\mathbf{=}\left\{\begin{array}{c}0\\ \mathrm{\sigma }{‘}_{\mathrm{y}}\\ 0\end{array}\right\}\mathbf{=}{\mathbf{T}}_{\sigma }\mathit{\sigma }=\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{2}& 1\\ \frac{1}{2}& \frac{1}{2}& –1\\ –\frac{1}{2}& \frac{1}{2}& 0\end{array}\right]\left\{\begin{array}{c}{\sigma }_x\\ {\sigma }_y\\ 25\end{array}\right\}$

Expressing the unknown stresses from the equation system of the transformation the following results are obtained$$\begin{gathered} \frac{{\sigma }_x}{2}+\frac{{\displaystyle {\sigma }_y}}{{\displaystyle 2}}+25=0 \\ \frac{{\displaystyle {\sigma }_x}}{{\displaystyle 2}}+\frac{{\displaystyle {\sigma }_y}}{{\displaystyle 2}}–25=\sigma {‘}_y \to {\mathit{\sigma }}_{\mathbf{x}}\mathbf{=}{\mathit{\sigma }}_{\mathbf{y}}\mathbf{=}\mathbf{–}\mathbf{25}\mathbf{ }\mathbf{MPa}\mathbf{,}\mathbf{ }\mathit{\sigma }{\mathbf{‘}}_{\mathbf{y}}\mathbf{=}\mathbf{–}\mathbf{50}\mathbf{ }\mathbf{MPa} \\ –\frac{{\displaystyle {\sigma }_x}}{{\displaystyle 2}}+\frac{{\displaystyle {\sigma }_y}}{{\displaystyle 2}}=0 \end{gathered}$$