 In this video, we will relate resultant torque to the shear flow distribution in an arbitrary thinwalled closed section shaft. In the drawn showing here, the wall thickness has been exaggerated so that we can more clearly annotate it during our derivation. However, it is important that the conditions for a thinwalled shaft discussed in the previous video still apply. If torsion is being transmitted by this shaft, there will be an internal torque which we will label T acting on the cross section. This torque is a resultant of the shear flow acting along the perimeter of the cross section. To relate these, let's consider an infinitesimal element of the shaft perimeter with a length ds. The shear flow acting on this element will produce a resultant force which we will call df. Because the element is infinitesimal, the curvature of the path ds is negligible and the resultant force is precisely equal to the shear flow q multiplied by the element length ds. This resultant shear force will generate a moment, dt, that contributes to the overall resultant torque of the cross section. To define this moment, we will define a datum point, o, about which to sum all of the torsional moments acting on the cross section. Denoting the moment arm for df about point o as r, we can express the moment due to df as dt equals r times df. Substituting in our result for df, we get that dt is equal to r times q times ds. If we integrate this result over the entire perimeter of the cross section, we can obtain the total resultant torque T. If we recall from previous video, shear flow is constant in the case of torsion and thus can be removed from the integral, resulting in the following integral equation. What does this integral represent physically? Well r times ds defines a rectangular area. If we look at our section closely, we see that r sweeps out a triangular area along ds. This triangular area is precisely one-half the rectangular area r times ds. Recognizing that the area swept by r around the entire perimeter is equal to the area enclosed by the median line, we can see that the integral of r times ds must be equal to twice the area enclosed by the median line. Taking this result, we can forget about needing to solve integrals and simply replace the integral with twice the enclosed area. Typically, we are interested in determining the shear flow from the resultant torque, so the equation can be rearranged as follows. This analog to the torsion formula for thin walled closed sections is commonly referred to as Brett's formula.