How Do You Calculate Principal Stresses and Orientations in a 2D Stress State?

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To calculate principal stresses and orientations in a 2D stress state, the principal stresses were determined using the formula σ₁, σ₂ = (1/2)(σₓ + σᵧ) ± √((σₓ - σᵧ)² + 4τₓᵧ²), yielding values of 96.8 N/mm² and -8.80 N/mm². The angle θ for the principal plane was found using tan(2θ) = (2τₓᵧ)/(σₓ - σᵧ), resulting in θ = 32.69 degrees. The discussion initially focused on calculating normal and shear stresses on planes equally inclined to the axes, which was later clarified. The problem was ultimately resolved, indicating that the calculations for normal and shear stresses were successfully completed. Understanding the orientation and calculation methods is crucial for analyzing stress states in materials.
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Homework Statement



In an element of material, subjected to general two-dimensional stress one axial stress is 66N/mm2 (tensile) and the shear stress is 48N/mm2. Calculate the values and directions of the principal stresses and the normal and shear stresses on planes equally inclined to the axes if the other axial stress of 22N/mm2 is a) tensile and b)compressive.

Homework Equations



\sigma_n=\frac{1}{2}(\sigma_x+\sigma_y)+\frac{1}{2}(\sigma_x-\sigma_y)cos2\theta+\tau_{xy}sin2\theta

\tau_n= \frac{1}{2}(\sigma_x-\sigma_y)sin 2\theta-\tau_{xy}cos2\theta

The Attempt at a Solution



Well I used the fact that the principal stresses are given by

\sigma_1,\sigma_2=\frac{1}{2}(\sigma_x+\sigma_y) \pm \sqrt{(\sigma_x-\sigma_y)^2+4\tau_{xy}^2

and got the values to be 96.8 N/mm2 and -8.80N/mm2

Then I used the fact that

tan2\theta=\frac{2\tau_{xy}}{\sigma_x-\sigma_y}

and got \theta=32.69. That part I got out, but I don't know how to get the normal and shear stresses on planes equally inclined to the axes.

I am not sure about what the question means by equally inclinded to the axes.

EDIT:solved.
 
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