Question

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.

Answer 1

Here, we don't have 3^rd 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

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

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

$\theta _{p} = -2.45$