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1. The nonzero stress components relative to axes (x, y, z) are Oxx = -90 MPa,...

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1. The nonzero stress components relative to axes (x, y, z) are Oxx = -90 MPa, Oyy = 50 MPa, and Oxy= 6 MPa. a. Determine the
1. The nonzero stress components relative to axes (x, y, z) are Oxx = -90 MPa, Oyy = 50 MPa, and Oxy= 6 MPa. a. Determine the principal stresses, 01202203 b. Calculate the maximum shear stress. c. Calculate the octahedral shear stress. d. Determine the angle between the x axis and the X axis, where X is in the direction of the principal stress 01.
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Here, we don't have 3rd component in the direction of z axis, so we won't be having principle stress in the same direction.

σ3 = 0MPa

Now, applying 2D principle stress equation.

\sigma _{1} = \frac{\sigma _{xx}+ \sigma _{yy}}{2} + \sqrt{\left (\frac{\sigma _{xx}- \sigma _{yy}}{2} \right )^{2} + \tau _{xy}^{2}}

\sigma _{2} = \frac{\sigma _{xx}+ \sigma _{yy}}{2} - \sqrt{\left (\frac{\sigma _{xx}- \sigma _{yy}}{2} \right )^{2} + \tau _{xy}^{2}}

Here, Ory = Try

1. Principle Stresses

\sigma _{1} = \frac{\sigma _{xx}+ \sigma _{yy}}{2} + \sqrt{\left (\frac{\sigma _{xx}- \sigma _{yy}}{2} \right )^{2} + \tau _{xy}^{2}}

* * *==

\sigma _{1} = 50.2566 MPa

\sigma _{2} = \frac{\sigma _{xx}+ \sigma _{yy}}{2} - \sqrt{\left (\frac{\sigma _{xx}- \sigma _{yy}}{2} \right )^{2} + \tau _{xy}^{2}}

+ ( += =

\sigma _{2} = -90.2566 MPa

Here, -ve sign shows compressive stress in nature.

So, as per question, \sigma _{1} = -90.2566 MPa and 02= -50.2566 M Pa

2. Maximum Shear Stress

\tau _{max} = \sqrt{\left (\frac{\sigma _{xx}- \sigma _{yy}}{2} \right )^{2} + \tau _{xy}^{2}}

\tau _{max} = \sqrt{\left (\frac{-90-50}{2} \right )^{2} + 6^{2}}

\tau _{max} = 70.2566 MPa

3. Octahedral Shear Stress

Toet = 5V2 (024 +02y – OrzOyu) + 672,

\tau _{oct} = \frac{1}{3}\sqrt{2\left ((-90)^{2} + 50^{2} -(-90)50 \right ) +6 *6^{2} }

\tau _{oct} = 58.13 MPa

4. Angle

\tan2 \theta _{p} = \frac{2\tau _{xy}}{\sigma_{xx} - \sigma _{yy}}

\tan2 \theta _{p} = \frac{2*6}{-90 - 50}

\theta _{p} = -2.45

.

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