A unique property of circular shafts is that under torsion, every cross-section remains plane and undistorted, rotating as a solid rigid slab. To determine the distribution of shearing stress, consider a cylindrical section inside a circular shaft of length L and radius R, fixed at one end. The radius of the cylindrical section is r. Now, consider the small square element formed by two adjacent circles and straight lines on the surface of the cylindrical section before any load is applied. As the torsional load is applied to the shaft, the square element deforms into a rhombus. Since the two sides of the rhombus are fixed, the shearing strain is equal to the angle between lines AB and A'B. Using a small angle approximation and suitable geometry, it can be shown that shearing strain at any given point of a shaft in torsion is proportional to the angle of twist and the distance r from the axis of the shaft. It is maximum at the surface of the shaft.