Answer:
Explanation:
Given that:
[tex][\sigma] = \left[\begin{array}{cc}800&300\\ \\300&-400\end{array}\right] (MPa)[/tex]
[tex][\sigma] = \left[\begin{array}{cc}\sigma_{xx}&\sigma_{xy}\\ \\\sigma_{yx}&\sigma_{yy}\end{array}\right][/tex]
Using Determinant method
The principal stress is the maximum or minimum normal stress acting on any plane. For the 2D stress system, the 2-principle plane always carries zero shear stress.
For principal stress [tex]( \sigma_1, \sigma_2)[/tex]
[tex]\sigma_{1,2} = \dfrac{\sigma_x+ \sigma_y}{2} \pm \sqrt{( \dfrac{\sigma_x - \sigma_y}{2} )^2 + \sigma^2_{xy}}[/tex]
[tex]\sigma_{1} = \dfrac{800+(-400)}{2} \pm \sqrt{( \dfrac{800 -(-400)}{2} )^2 + (300)^2}[/tex]
[tex]\sigma_{1} = \dfrac{400}{2} \pm \sqrt{ (600)^2 + (300)^2}[/tex]
[tex]\sigma_{1} = 200+ 670.82 \\ \\ \sigma_{1} = 870.82 \ MPa[/tex]
[tex]\sigma_{2} = \dfrac{800+(-400)}{2} \pm \sqrt{( \dfrac{800 -(-400)}{2} )^2 + (300)^2}[/tex]
[tex]\sigma_{2} = 200- 670.82 \\ \\ \sigma_{1} = - 470.82 \ MPa[/tex]
According to Mohr's circle;
Mohr's circle is the locus provided that the position of the normal stress and the shear stress is acting on any plane.
Center = (a,0)
[tex]a = \dfrac{\sigma_{x}+\sigma_{y}}{2}[/tex]
[tex]a = \dfrac{800+(-400)}{2}[/tex]
a = 200 MPa
radius (r) = [tex]\sqrt{ (\dfrac{\sigma_{x}-\sigma_{y}}{2})^2 + \sigma^2 _{xy}}[/tex]
[tex]=\sqrt{ (\dfrac{800-(-400)}{2})^2 + (300)^2}[/tex]
[tex]=\sqrt{ (600)^2 + (300)^2}[/tex]
r = 670.82 MPa
[tex]\sigma_1 = a +r \\ \\ \sigma_1 = 200 + 670.82 \\ \\ \sigma_1 = 870.82 \ MPa[/tex]
[tex]\sigma_2 =-(r-a)[/tex] (it is negative because of the negative x-axis)
[tex]\sigma_2 =670.82 - 200 \\ \\ \sigma_1 = 470.82 \ MPa[/tex]
[tex]\tau_{max} = radius \ of \ Mohr's \ circle[/tex]
[tex]\tau_{max} = 670.82 \ MPa[/tex]
