 In the previous videos, we established that torsion creates a state of pure shear within a circular shaft. We then derive the torsion formula that relates the internal torque to the magnitude of that shear stress. But how does the shaft deform and how do we relate this to the internal torque? We have not quite looked at this and it is needed in order to solve statically indeterminate problems, so let's take a closer look at relating the internal torque to the deformation of a shaft. Here we have a solid circular shaft of radius r and length l. If you recall from earlier videos, we observed that if we subject this shaft to a uniform torque T, the right-hand side of the shaft will rotate through an angle phi relative to the left-hand side. We will refer to this angle as the angle of twist of the shaft. If we once again look at a small increment of length of the shaft, dx, we can see once again our element of pure shear deformation. Associated with this pure shear deformation is a small angle of twist. The cross-sectional plane containing side Cb rotates through an angle of twist relative to the cross-sectional plane containing side Ad. This angle of twist is denoted d phi in the figure shown here. If you subdivide the shaft along its length into many of these elements, each element will have its own angle of twist along its length. The angle of twist of the overall shaft would thus be a summation of each of these individual angles of twist. If we divide the shaft into an infinite number of elements, we obtain an infinite sum or an integration of angles of twist. If we focus in on that segment of shaft with length dx, we can recall that the angle of twist associated with this segment can be related to the shear strain state. In previous videos, we already established that the angle of twist of this segment d phi over dx is the rate of change of the shear strain with respect to the radial position within the shaft cross-section. It is the slope of the linear shear stress distribution. We can rearrange this equation in terms of d phi as shown below. Now that we have the angle of twist in terms of shear strain, we need to apply Hooke's law in order to convert from strains to stresses. In an earlier unit on material properties, we established that shear stress and shear strain are linearly related through the shear modulus g. We can use Hooke's law to rewrite the equation for angle of twist in terms of shear stress as shown below. Next, we can apply the torsion formula relating stress to the torque derived earlier. Substituting this into the formula below, we obtain an expression for the angle of twist d phi. Collecting terms and simplifying, we can clean up this result a little to obtain the following expression for d phi. Finally, we need to remember that the overall angle of twist for a shaft is the infinite sum or integration of all the d phi's along the length of the shaft. Thus, the overall angle of twist for a circular shaft is the integration over the length of the shaft of the internal torque divided by the product of the shear modulus and polar moment of inertia of the shaft multiplied by the infinitesimal segment length dx. Now we have an integral solution for the angle of twist of our shaft. Taking a look at this integral solution, you may be tempted to remove the torque, shear modulus, and polar moment of inertia outside of the integral as our shaft had a constant radius, was made of a single material, and had a constant internal torque. But is this really the case for general shafts? Real shafts may have variations in cross-section as the compound conical cylindrical shaft shown here. This variation in geometry means that the polar moment of inertia will vary as a function of x. Other shafts may have variations in material properties as shown here. This would result in a variation in the shear modulus of the shaft material along its length. And finally, shafts may have a complex torsional loading applied to them, with not just torques at the end but intermediate torques applied along its length. This will result in a variable internal torque distribution. Thus, strictly speaking, the torsion deformation formula is an integral solution where torque, shear modulus, and polar moment of inertia all can be functions of x. Lucky for us, there are several special cases where some of these are constant, and we'll take a closer look at these in the following slides. The first special case we will look at is that of a circular shaft with uniform cross-section and internal torque along its length. This is the simplest case we can consider where the internal torque and shear modulus and polar moment of inertia are all constants. This allows us to remove these quantities from the integral. The integral of dx over the length of the shaft is precisely the shaft length, allowing us to simplify further, resulting in the expression that the angle of twist is the internal torque times the shaft length divided by the product of the shear modulus and the polar moment of inertia. The second special case we will look at is a uniform circular shaft subjected to multiple torques. Consider the uniform shaft shown here. If torques are applied at locations a, b, and c, as illustrated here, the internal torque is no longer constant along the length of the shaft. However, if we draw the internal torque diagram for the shaft, we can see that the shaft becomes segmented into sections with uniform torque. In this particular case, section a, b, and section b, c. We can thus consider the shaft as a series of uniform loaded shafts and use a solution from the previous special case, adding the result for each segment. This equation can be further generalized as a summation of angles of twist from individual segments with uniform internal torque, material, and cross-section. Now that you have a generalized force displacement relationship for a circular shaft subjected to torsion, as well as two simplified force displacement relationships, you are now well equipped to start calculating the deformation of circular shafts subjected to torsion. Furthermore, you are now able to solve statically indeterminate problems containing torsional loading.