# When doing Mohr's circle, why do we rotate by twice the angle?

On page-3 of these MIT OCW notes, it is said that if we want to find the stresses when the square element is rotated by $$\theta$$ , then we should rotate diamter of mohr circle by $$2 \theta$$ and read of the numbers from the ends of diameter.

Is there any intuitive reason as to why we have to use twice the angle?

One intuitive reason is that if you want to rotate by 90 degrees then $$\sigma_x$$ would be equal to $$\sigma_y$$, and $$\sigma_y\rightarrow \sigma_x$$. In order for that to happen you'd have to travel 180 degrees. This is better evidenced for principal stresses ($$\sigma_1, \sigma2$$, however, it can be seen and in the generic stress state When one switches to resolving along a different set of mutually perpendicular axes, rotated by $$\theta$$ from the original set, one gets one factor of $$\cos\left(\theta\right)$$ in the stress component from projecting the force onto the new set of axes, and another factor of $$\cos\left(\theta\right)$$ in the stress component from projecting the surface area onto the new set of axes, making a factor of $$\cos^2\left(\theta\right)$$ altogether. By a well-known trigonometric identity, that factor is the same as $$\left(1/2\right)+\left(1/2\right)\cos\left(2\theta\right)$$. The first $$(1/2)$$ is why the centre of Mohr's circle is offset from the origin of a longitudinal stress-shear stress graph by $$(1/2)$$ of the sum of principal stresses; the second $$(1/2)$$ is why the radius of Mohr's circle is $$(1/2)$$ of the difference between the principal stresses; and the $$\left(2\theta\right)$$ is why one turns $$\left(2\theta\right)$$ around Mohr's circle from the longitudinal stress axis of the graph.