Mohr's circle for a plane stress element

Mohr's circle construction

Convention: tension is positive to the right on the $\sigma$ axis; a shear stress that rotates the element clockwise is plotted upward. With this convention, an angle $2\theta$ on the circle is measured in the same sense as the rotation $\theta$ of the plane.

1. Plot X $(\sigma_x,\ \tau) = (100,\ +40)$ for the x face.
2. Plot Y $(\sigma_y,\ \tau) = (20,\ -40)$ for the y face.
3. Join XY; it cuts the $\sigma$ axis at the centre $C$.
4. Draw the circle with centre $C$ and diameter XY.

$$\sigma_{avg} = \frac{100+20}{2} = 60\ \text{MPa}, \qquad R = \sqrt{40^2 + 40^2} = 56.57\ \text{MPa}$$

  tau (cw +)
     |        X(100,40)
     |      .-*-.
     |    .'  |  '.
  ---+---*----C----*---- sigma
     0   B   60    A
     |    '.  |  .'
     |      '-*-'
     |        Y(20,-40)

Principal stresses (points A and B)

$$\sigma_1 = 60 + 56.57 = 116.6\ \text{MPa}, \qquad \sigma_2 = 60 - 56.57 = 3.43\ \text{MPa}$$

Maximum in-plane shear stress

At the top and bottom points of the circle:

$$\tau_{max,\,in\text{-}plane} = R = 56.6\ \text{MPa}, \quad \text{with normal stress } 60\ \text{MPa}$$

Because both principal stresses are tensile, the absolute maximum shear stress (including the third principal stress, $\sigma_3=0$) is $\sigma_1/2 = 58.3$ MPa.

Principal planes

On the circle, the angle XCA is $2\theta_p$ with $\tan2\theta_p = 40/40 = 1$, so $2\theta_p = 45^\circ$, measured clockwise from CX to CA. (By formula, a clockwise shear on the x face is $\tau_{xy} = -40$ MPa, so $\tan2\theta_p = \dfrac{2(-40)}{100-20} = -1$ and $\theta_p = -22.5^\circ$.)

So the plane carrying $\sigma_1$ is $22.5^\circ$ clockwise from the x axis; $\sigma_2$ acts on the plane $90^\circ$ from it.

Stresses on the plane at 20 degrees anticlockwise

Rotate the radius CX by $2\theta = 40^\circ$ anticlockwise. CX starts at $45^\circ$ above the $\sigma$ axis, so the new radius makes $45^\circ + 40^\circ = 85^\circ$ with the $\sigma$ axis:

$$\sigma = 60 + 56.57\cos85^\circ = 64.9\ \text{MPa}$$ $$\tau = 56.57\sin85^\circ = 56.4\ \text{MPa (clockwise sense)}$$

Check by formula with $\tau_{xy} = -40$ MPa: $\sigma = 60 + 40\cos40^\circ + (-40)\sin40^\circ = 64.9$ MPa.

Answer: $\sigma_1 = 116.6$ MPa, $\sigma_2 = 3.43$ MPa; $\tau_{max} = 56.6$ MPa; principal plane $22.5^\circ$ clockwise from x; on the 20 degree plane $\sigma = 64.9$ MPa, $\tau = 56.4$ MPa.