Explanation:
Refer to the Mohr's Circle section in the Mechanics of Materials chapter of the FE Reference Handbook.
Ultimately, we must determine the angle rotation needed to get the stress element to a principal stress state.
Steps to determine the angle of rotation needed:
1. Draw a quick Mohr's circle based on what's shown in the stress element.
2. Draw a right triangle from the center of the Mohr's circle to a known point along the circle. Determine the Mohr's circle's angle between the known point and the sigma axis (x-axis), which is where the principal stress occurs. That will be the Mohr's circle rotation angle needed to get to a principal stress state.
3. Convert the Mohr's circle's rotation angle needed to get to a principal stress state to the stress element's angle needed to get to a principal stress state.
Step 1. Draw a quick Mohr's circle based on what's shown in the stress element - Identify all points on the Mohr's circle.
Refer to the stress element shown in the FE Reference Handbook to understand its sign notation as well as this problem's explanation image. Play close attention to the sign notation as it's often the number one mistake when drawing Mohr's circles. Normal stresses ($\sigma_x,\sigma_y)$ in tension (going away from the stress element) are positive and normal stresses in compression (going into the stress element) are negative.
For shear stresses ($\tau_{xy}$), look at the shear arrow for the X face of the stress element and for the Y face of the stress element. If a shear arrow causes clockwise rotation to the stress element, then the shear for that stress element face (x or y face) is positive. If a shear arrow causes counterclockwise rotation to the stress element, then the shear for that stress element face (x or y face) is negative. Therefore, two points along the circle can be plotted as:
$$
\text{X face}= (\sigma_x,\tau_{xy})=(+400\si{MPa},+530\si{MPa})
\\
\text{Y face}= (\sigma_y,\tau_{xy})=(-550\si{MPa},-530\si{MPa})
$$
Determine the center of the Mohr's circle.
$$
C=\frac{\sigma_x+\sigma_y}{2}=\frac{400+(-550)}{2}=-75\si{MPa}
$$
Refer to the explanation image to see the final result. Plot the center of the Mohr's circle as well as the X face and Y face of the stress element.
Step 2. Plot a right triangle from the center of the Mohr's circle to the known point along the circumference of the circle.
Use Pythagoras Theorem to solve for the hypotenuse of the triangle, which is also the radius of the circle.
$$
a^2+b^2=hypotenuse^2
\\
(+530)^2+(400-(-75))^2=hypotenuse^2
\\
hypotenuse=radius=711.7\si{MPa}
$$
The angle, 2$\theta$, is the
Mohr's circle's angle to get to a principal stress state. The rotation angle needed to get the
stress element to a principal stress state is $\theta$. This problem asks for the angle to get the
stress element to a principal stress state.
Calculate the Mohr's circle's angle to a principal stress state. Use soh-cah-toa.
$$
\sin{\left[\frac{opposite}{hypotenuse}\right]}^{-1}=angle
\\
\sin{\left[\frac{530}{711.6}\right]}^{-1}=48^\circ
$$
Step 3. Convert the Mohr's circle's rotation angle needed to get to a principal stress state to the stress element's angle needed to get to a principal stress state.
$$
\text{Mohr's Circle's angle}=2\theta=48^\circ
$$
Stress element's rotation angle to a principal stress state $\theta=24^\circ$ clockwise. Refer to blue arrow in the explanation image to understand why the rotation is clockwise to get to a max principal stress point.