The shear stress varies from zero in the axis to a maximum at the outside surface of the shaft. On an element where shear stress is maximum, normal stress is 0. This element where maximum shear stress occursis oriented in such a way that its faces are either parallel or perpendicular to the axis of the shaft as shown in the figure. tau = Tr/J is somewhat analogous to finding the bending stress in a beam, where sigma = M * y / I
Shear stress is zero on the axis passing through the center of a shaft and maximum at the outside surface of a shaft. We want to find the maximum shear stress τ max which occurs in a circular shaft of radius c due to the application of a torque T. The shear stress in a beam is not uniform throughout the cross section, rather it varies from zero at the outer fibres to maximum at the. You overlooked the units of the polar moment of inertia J, which is m^4. As shear rate is under the limits, product defects and problems can be reduced. The shear stress due to an applied torque on a circular shaft is T * r / J. The shear stress in a solid circular shaft in a given position can be expressed as: τ = T r / J (1) where. τ = shear stress (Pa, lb f /ft 2 (psf)) T = twisting moment (Nm, lb f ft) r = distance from center to stressed surface in the given position (m, ft) J = Polar Moment of Inertia of Area (m 4, ft 4) Note So, tau = T*r/J which has units of N-m * m / m^4 = N/m^2 = Pa, a unit of stress, I believe.