Question

A plane stress state (2D) of material is defined by: [σ] = [800300300−400] (MPa). Define the principal stresses σ1, σ2 using:

• Determinant Method
• Mohr’s Circle (determine also ????max)

Answer 1

Step-by-step explanation:

Given that:

`[\sigma] = \left[\begin{array}{cc}800&300\\ \\300&-400\end{array}\right] (MPa)`

`[\sigma] = \left[\begin{array}{cc}\sigma_(xx)&\sigma_(xy)\\ \\\sigma_(yx)&\sigma_(yy)\end{array}\right]`

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
`( \sigma_1, \sigma_2)`

`\sigma_(1,2) = (\sigma_x+ \sigma_y)/(2) \pm \sqrt{( (\sigma_x - \sigma_y)/(2) )^2 + \sigma^2_(xy)}`

`\sigma_(1) = (800+(-400))/(2) \pm \sqrt{( (800 -(-400))/(2) )^2 + (300)^2}`

`\sigma_(1) = (400)/(2) \pm √( (600)^2 + (300)^2)`

`\sigma_(1) = 200+ 670.82 \\ \\ \sigma_(1) = 870.82 \ MPa`

`\sigma_(2) = (800+(-400))/(2) \pm \sqrt{( (800 -(-400))/(2) )^2 + (300)^2}`

`\sigma_(2) = 200- 670.82 \\ \\ \sigma_(1) = - 470.82 \ MPa`

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)

`a = (\sigma_(x)+\sigma_(y))/(2)`

`a = (800+(-400))/(2)`

a = 200 MPa

radius (r) =
`\sqrt{ ((\sigma_(x)-\sigma_(y))/(2))^2 + \sigma^2 _(xy)}`

`=\sqrt{ ((800-(-400))/(2))^2 + (300)^2}`

`=√( (600)^2 + (300)^2)`

r = 670.82 MPa

`\sigma_1 = a +r \\ \\ \sigma_1 = 200 + 670.82 \\ \\ \sigma_1 = 870.82 \ MPa`

`\sigma_2 =-(r-a)` (it is negative because of the negative x-axis)

`\sigma_2 =670.82 - 200 \\ \\ \sigma_1 = 470.82 \ MPa`

`\tau_(max) = radius \ of \ Mohr's \ circle`

`\tau_(max) = 670.82 \ MPa`