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**Stress Tensor is at a point is define as the state of stress at that point inside a material in the deformed state, placement, or configuration. the state of strress shows all different stress developed on 3-perpendicular plane passing through that point.**

Consider a single crystal in the form of a cube subjected to a combination of normal stresses (\( \sigma_{xx}, \sigma_{yy}, \sigma_{zz} \)) and shear stresses (\( \tau_{xy},\tau_{yz},\tau_{zx} \)). The coordinate axes are chosen parallel to <100> directions if the crystal system is orthogonal (i.e., cubic, tetragonal, or orthorhombic) or to have a specified orientation with respect to the unit cell if it is not. Then the stress tensor for 3D system is given by -

** Bi-axial or plane stress : **

For plane stress condition any one of the mutually perpendicular faces is assumed to be zero. i.e., \( \sigma_{zz} = \tau_{zx} = \tau_{zy} = 0 \)

\( (A) = \begin{bmatrix} 150 & -30 & 0 \\ -30 & 200 & 0\\ 0 & 0 & -80 \\ \end{bmatrix} \) \( (B) = \begin{bmatrix} 200 & 0 & 0 \\ 0 & 150 & -30\\ 0 & -30 & -80 \\ \end{bmatrix} \) \( (C) = \begin{bmatrix} -80 & 30 & 0 \\ 30 & 150 & 0\\ 0 & 0 & 200 \\ \end{bmatrix} \) \( (D) = \begin{bmatrix} 200 & 0 & 0 \\ 0 & 150 & 0\\ 0 & 30 & -80 \\ \end{bmatrix} \)

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\( \sigma \rightarrow e \)[Normal strain]

\( \tau \rightarrow \frac{\gamma}{2} \)[Shear Strain]

**Answer : 800 MPa**

\( G = \frac{Shear Stress}{Shear Strain} = \frac{\tau_{xy}}{\gamma_{xy}} \)

From question \(\frac{\gamma_{xy}}{2} = 0.004 \Rightarrow \gamma_{xy} = 0.008 \)

So , \(G = \frac{\tau_{xy}}{0.008} \) As given G=100GPa

\( \Rightarrow \tau_{xy} = 800MPa \)

The total stress can be divided into a hydrostatic or mean stress tensor \( \sigma_m \), which involves only pure tension or compression , and a deviator stress tensor \( \sigma^{div}_{ij} \), which represents the shear stresses in the total state of stress. The hydrostatic component of the stress tensor produces only volume change. Such as -

\[ \textrm{Total stress } ( \sigma_{ij} ) = \textrm{Hydrostatic stress } ( \sigma_m ) + \textrm{ Deviator stress } (\sigma^{div}_{ij})\]
\[ \sigma_{ij} = \sigma_{ij}^{hydro} + \sigma_{ij}^{div} \]

\[ \begin{bmatrix}
\sigma_{11} & \tau_{12} & \tau_{13} \\
\tau_{21} & \sigma_{22} & \tau_{23}\\
\tau_{31} & \tau_{32} & \sigma_{33} \\
\end{bmatrix} =
\begin{bmatrix}
\sigma_{m} & 0 & 0 \\
0 & \sigma_{m} & 0\\
0 & 0 & \sigma_{m} \\
\end{bmatrix} + \begin{bmatrix}
\sigma_{11} - \sigma_{m} & \tau_{12} & \tau_{13} \\
\tau_{21} & \sigma_{22} - \sigma_{m} & \tau_{23}\\
\tau_{31} & \tau_{32} & \sigma_{33} - \sigma_{m} \\
\end{bmatrix} \]
\[ \textrm{ And, } \sigma_{m} = \frac{\sigma_{11} + \sigma_{22} + \sigma_{33}}{3} \]

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**Answer : 200 GPa**

As \( K = \frac{\sigma}{e_{v}} \)

Also, \( \sigma_{hydo} = \sigma_{m} = \frac{\sigma_{11} + \sigma_{22} + \sigma_{33}}{3} = \frac{90 -20 + 140 }{3} = 70MPa \)

\( \Rightarrow K = \frac{70}{3.5 \times 10^{-4}} \)

\( \Rightarrow K = 200GPa \)

Complex problems can be solve by considering a simplified two-dimension state of stress. Now consider a stress system which consist of two normal stresses \( \sigma_x \) and \( \sigma_y \) and a shear stress \(\tau_{xy} \). The planes at which the acting shear stresses are zero are known as principal planes and the corresponding stress is known as principal stress.

The above figure shows a thin plate with its thickness is normal to the plane of paper. This stress system consist of two normal stresses \( \sigma_x \) and \( \sigma_y \) and a shear stress \(\tau_{xy} \). In order to know the state of stress from the point O on plane AB we have to resolve the normal and shear stress on the plane AB.