Given the state of plane stress at a point with the following values:

• $\sigma_x = 90$ MPa
• $\sigma_y = -60$ MPa
• $\tau_{xy} = 20$ MPa

We need to find the normal stress $\sigma_\theta$ acting on a plane inclined at $30^\circ$ counterclockwise (CCW) to the x-axis.

To calculate the normal stress $\sigma_\theta$, we can use the following formula:

$\sigma_\theta = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2} \cos(2\theta) + \tau_{xy} \sin(2\theta)$

Where:

• $\sigma_x = 90$ MPa
• $\sigma_y = -60$ MPa
• $\tau_{xy} = 20$ MPa
• $\theta = 30^\circ$

Let's calculate this step by step.The normal stress $\sigma_\theta$ acting on a plane inclined at $30^\circ$ counterclockwise to the x-axis is approximately $69.82$ MPa.

Would you like more details on how this was calculated or have any questions?

Here are five related questions to expand on this:

1. How do we calculate shear stress on an inclined plane?
2. What is the difference between plane stress and plane strain conditions?
3. How does Mohr's Circle help in visualizing stress transformations?
4. What are the principal stresses for this given state of stress?
5. How can we determine the angle at which maximum normal stress occurs?

Tip: When dealing with stress transformations, always check if the problem involves plane stress or plane strain, as the formulas differ for each condition.