To use the torsion formula to relate the torque applied to a rod of circular cross section to the maximum shear stress in the rod a cross section of a solid circular rod is subject to a torque of t = 2.9 kn middot m if the diameter of the rod is d = 5 cm, what is the maximum shear stress? when a torque t is applied to a shaft with a circular cross section, the shaft will deform by twisting about its longitudinal axis. for small deflections, the radial lines will remain straight and the cross sections will remain planar and parallel. the shear strain within the shaft will vary linearly along any radial line. (figure 1) if the material is homogenous and elastic, then the shear stress will also vary linearly along any radial line as long as the maximum shear stress is no greater than the provolone limit for the material. since the shear stress varies nearly, the maximum will occur on the outer boundary of the cross section. the shear stress r at a point on the cross section is related to the torque at the section by tau tp/j, where p is the distance of the point from the central axis and j is called the polar moment of inertia of the cross-sectional area for a solid circular shaft j = pi/2 where c is shaft radium for a circular tube with inner radius c_i and outer radius c_o, the polar moment of inertia is j = pi/2 the maximum stress in a section of a circular tube subject to a torque is tau_max = 34 mpa if the inner diameter is d_i = 5 cm and the outer diameter is d_a = 6.25 cm, what is the torque on the section?