Stress transformation equations and Mohr's circle

Derivation

Consider a wedge cut from a plane-stress element. Let the inclined plane have its normal $n$ at angle $\theta$ (anticlockwise) from the x-axis and area $dA$. The faces on the x and y sides then have areas $dA\cos\theta$ and $dA\sin\theta$.

        y
        |   /  n (normal at angle theta)
   sigma_y  /
  +-----+/
  |     /|
  |    / |-- tau_xy
  +---+--+---- x

Resolve forces along the normal $n$:

$$\sigma_n\, dA = \sigma_x (dA\cos\theta)\cos\theta + \sigma_y (dA\sin\theta)\sin\theta + \tau_{xy}(dA\cos\theta)\sin\theta + \tau_{yx}(dA\sin\theta)\cos\theta$$

Using $\tau_{yx}=\tau_{xy}$ and double-angle identities:

$$\sigma_n = \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta$$

Resolving along the plane gives

$$\tau_{nt} = -\frac{\sigma_x-\sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta$$

Principal planes: $\tau_{nt}=0$ gives $\tan 2\theta_p = \dfrac{2\tau_{xy}}{\sigma_x-\sigma_y}$, and $\sigma_{1,2} = \dfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$.

Mohr's circle

Squaring and adding the two equations eliminates $\theta$:

$$\left(\sigma_n - \sigma_{avg}\right)^2 + \tau_{nt}^2 = R^2$$

This is a circle with centre $(\sigma_{avg}, 0)$ and radius $R$ in the $\sigma$-$\tau$ plane, where $\sigma_{avg} = \dfrac{\sigma_x+\sigma_y}{2}$ and $R = \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$.

Construction

1. Draw $\sigma$ (tension positive) to the right and $\tau$ downward, so that the angle $2\theta$ is measured anticlockwise on the circle. Plot the point for the x-face as $X(\sigma_x, \tau_{xy})$. (In the other common convention, $\tau$ is drawn upward and clockwise shear is positive. State which one you use.)
2. Plot the point for the y-face as $Y(\sigma_y, -\tau_{xy})$.
3. Join $XY$; it cuts the $\sigma$-axis at the centre $C$.
4. Draw the circle with diameter $XY$.

   tau
    |        . S (tau max)
    |     .     .
    |   .         .
 ---+--+----C------+----- sigma
    | sigma2       sigma1
    |   .         .
    |     .     .
    |        . S'

Reading results

• The points where the circle cuts the $\sigma$-axis give $\sigma_1 = OC + R$ and $\sigma_2 = OC - R$.
• The highest and lowest points give $\tau_{max} = \pm R$, with normal stress $\sigma_{avg}$.
• The angle between $CX$ and the $\sigma$-axis (towards $\sigma_1$) is $2\theta_p$. Angles on the circle are twice the physical angles, and they turn in the same sense as the plane turns.
• The point on the circle at angle $2\theta$ from $X$ gives $(\sigma_n, \tau_{nt})$ for the plane at $\theta$.