A plane stress element has σx= 80 MPa,σy = 20 MPa, and τxy= 50 MPa cw (a)Using Mohr’s circle, find the principal stresses, maximum shear stress and direction.

To solve this problem using Mohr's Circle, we'll follow these steps:

### Given Data:
- \( \sigma_x = 80 \) MPa (stress in the x-direction)
- \( \sigma_y = 20 \) MPa (stress in the y-direction)
- \( \tau_{xy} = 50 \) MPa (shear stress, clockwise)

### Step 1: Determine the Center and Radius of Mohr's Circle
The center \( C \) of Mohr's circle is located at:
\[
C = \left(\frac{\sigma_x + \sigma_y}{2}, 0\right)
\]
\[
C = \left(\frac{80 + 20}{2}, 0\right) = (50, 0) \text{ MPa}
\]

The radius \( R \) of Mohr's circle is given by:
\[
R = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}
\]
\[
R = \sqrt{\left(\frac{80 - 20}{2}\right)^2 + 50^2} = \sqrt{30^2 + 50^2} = \sqrt{900 + 2500} = \sqrt{3400} \approx 58.31 \text{ MPa}
\]

### Step 2: Principal Stresses
The principal stresses \( \sigma_1 \) and \( \sigma_2 \) are located at the points on Mohr's circle along the horizontal axis. They are calculated as:
\[
\sigma_1 = C + R = 50 + 58.31 = 108.31 \text{ MPa}
\]
\[
\sigma_2 = C - R = 50 - 58.31 = -8.31 \text{ MPa}
\]

### Step 3: Maximum Shear Stress
The maximum shear stress \( \tau_{\text{max}} \) occurs at the top (or bottom) of Mohr's circle, which is equal to the radius:
\[
\tau_{\text{max}} = R = 58.31 \text{ MPa}
\]

The corresponding normal stress at the maximum shear stress is the center \( C \) of Mohr's circle:
\[
\sigma_{\text{avg}} = C = 50 \text{ MPa}
\]

### Step 4: Direction of Principal Stresses and Maximum Shear Stress
To find the angle \( \theta_p \) that the principal stresses make with the x-axis (in the physical space), use:
\[
\tan(2\theta_p) = \frac{2\tau_{xy}}{\sigma_x - \sigma_y}
\]
\[
\tan(2\theta_p) = \frac{2 \times 50}{80 - 20} = \frac{100}{60} \approx 1.667
\]

\[
2\theta_p \approx \tan^{-1}(1.667) \approx 59.04^\circ
\]
\[
\theta_p \approx \frac{59.04^\circ}{2} \approx 29.52^\circ
\]

This angle \( \theta_p \) is the orientation of the principal stresses with respect to the original x-axis.

The angle \( \theta_s \) for maximum shear stress (in physical space) is:
\[
\theta_s = \theta_p + 45^\circ \approx 29.52^\circ + 45^\circ = 74.52^\circ
\]

### Summary of Results:
- **Principal stresses:** \( \sigma_1 = 108.31 \) MPa, \( \sigma_2 = -8.31 \) MPa
- **Maximum shear stress:** \( \tau_{\text{max}} = 58.31 \) MPa
- **Direction of principal stresses:** \( \theta_p \approx 29.52^\circ \) (counterclockwise from the x-axis)
- **Direction of maximum shear stress:** \( \theta_s \approx 74.52^\circ \) (counterclockwise from the x-axis)

Would you like a more detailed explanation, or do you have any questions about the process?

### Related Questions:
1. How would the principal stresses change if \( \sigma_y \) was negative?
2. Can you explain how the direction of the principal stresses relates to the original stress element?
3. What is the significance of Mohr's circle in stress analysis?
4. How would the result change if \( \tau_{xy} \) was counterclockwise instead of clockwise?
5. Can Mohr's circle be used to find the principal strains? How?

### Tip:
When working with Mohr's circle, always pay close attention to the sign conventions for stresses, as they directly affect the position and orientation of the circle and the resulting principal stresses.

Learn how to find principal stresses and maximum shear stress of a plane stress element with given stress components using Mohr's Circle. Includes step-by-step calculations and explanations.

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