 In the previous video, we demonstrated that shear stress varies linearly with radial position within the cross-section of a circular shaft subjected to torsion. However, we did not directly relate the magnitude of these shear stresses to that torque. In order to do this, we must recall that a resultant force, or moment, is a resultant of a stress state acting over a given domain, in this case the cross-section of the shaft. Let's take a closer look at this. The total internal torque within a shaft is a summation of the torques acting over the cross-sectional area of the shaft. Consider the infinitesimal area on the cross-section, denoted dA, as shown below. The shear stress acting on this area produces a torque equal to the shear force acting on it, multiplied by its moment arm. In this case, the shear force is equal to the shear stress multiplied by dA, and the moment arm is the radial position of dA. Giving us the equation shown here. In the previous video, we found that shear stress as a function of radial position can be expressed as rho over R multiplied by the maximum shear stress at the outer surface of the shaft. Combining these two equations, we can obtain the following integral expression for the total torque. Recognizing that the radius of the shaft and the maximum shear stress are independent of the location within the cross-section, they can be removed from the integral leading to the following expression for the total torque. If we take a closer look at this expression, we can recognize a couple of further simplifications. First, let's take a closer look at the ratio of tau max over the radius R. If we look at the shear stress distribution as a function of radial position as shown here, we observe a triangular distribution where the length of the adjacent edge of the triangle is R, and the opposite side of the triangle is tau max. Due to similar triangles, we can observe that this ratio is preserved at an arbitrary radial position rho. Thus, we can replace tau max over R by tau over rho in our expression. The second simplification that can be made is by recognizing that the remaining integral in our expression is a geometric property of the shaft. Indeed, it is in fact the polar moment of inertia of the cross-sectional area of the shaft about the longitudinal axis. We will denote this property by the capital letter J. Applying these two simplifications, the expression can be rearranged to obtain the following expression for shear stress in terms of resultant torque. This expression is often referred to as the torsion formula. The torsion formula relates shear stress at any radial position rho within a cross-section to the internal torque acting on that cross-section as well as the geometry of that cross-section. By now, you should be very familiar and comfortable with determining the internal loading within a structure by sectioning it, drawing a free body diagram, and applying the principle of equilibrium. The same can be done to determine the internal torsion within a torsional element. The geometry of the shaft comes into play through the polar moment of inertia. Calculating the polar moment of inertia should be reviewed from previous courses and statics, but let's take a quick look at it again to refresh your memory. Let's first take a look at the polar moment of inertia for a solid circular shaft. The polar moment of inertia of a shaft is defined as the integral over the domain of the cross-sectional area of the shaft of the elemental area dA multiplied by its radial position from the axis of the shaft squared. For this derivation, we will designate the radius of the shaft as r and the radial position of the shaft as rho. If we consider a small change in radial position, d rho, and sweep this small change in radial position around the axis of the shaft, we define an element of area shown here in purple. For a small change in radial position, we can define the area of this element as the perimeter multiplied by the radial position change. Or dA is equal to 2 pi rho d rho. Thus, the entire area of the shaft will be covered by increments of d rho in the radial direction. Transferring this result to our integral, we can define a definite integral over the range of radial positions within the shaft. For this solid shaft, the lower limit thus becomes 0 and the outer limit becomes the radius of the shaft r. Substituting our result for the element area dA, we can expand this integral to the following, removing constants from the integral and collecting terms we can then simplify. Integrating rho cubed, we obtain one-quarter rho to the power of 4, resulting in the following. After substituting in our limits of r and 0, we obtain our final expression for the polar moment of inertia of a solid circular shaft. Next, let's consider a hollow circular shaft with inner radius r1 and outer radius r2 as shown here. The definition of polar moment of inertia does not change. Furthermore, a small change in radial position d rho produces the same area element as before with a solid circular shaft. The only difference in transferring this result to our integral is that the limits of integration are changed. The integral is performed from the inner radius r1 to the outer radius r2. Substitution of our element area into the integral, simplification and subsequent integration of the result remain the same as before. However, due to the different limits on the definite integral, the final expression becomes a function of the inner and outer radius as shown here. And there you have it. Now you have all you need in order to determine the shear stress state in a circular torsion shaft. The shear stress is given by the torsion formula. And since the torsion formula only applies to circular shafts, it's also very convenient to keep at hand the solutions for the polar moment of inertia of a solid circular shaft and the polar moment of inertia of a hollow circular shaft. With this formula, you will now be able to begin to look at the design of shafts. What is the required radius of an aluminum shaft to prevent failure given a certain torsional loading? Or what is the maximum torque a given shaft material and dimension can handle? You should be able to calculate these using the torsion formula. However, we still cannot look at statically indeterminate shaft problems because we haven't come up with an expression for the deformation of a shaft due to torsion. And we'll look at that in the next video.