 In order to analyze an engineering system containing torsional loaded members we must first understand a deformation that results from torsion and the associated stress and strain state. We will take a look at these points closer in this video for the case of a circular shaft. Please note that all the formulas derived in this video are only applicable to a circular shaft. However, lucky for us this is the most common shaft geometry used for transmitting torsion in engineering systems. This is actually for good reason, however we will touch upon that in a later video on portion of non-circular shafts. For now let's take a closer look at what happens when we apply a torque to a circular shaft. For a deformable shaft I will be using this foam pool toy that is very flexible and has a circular cross section as you can see here. On the surface of the shaft I have drawn four lines around its circumference, distributed equally along its length. Additionally, a line parallel to the axis of the shaft has also been drawn on the surface. Observe what happens to these lines when an alternating torsional load is applied to the shaft. Two observations can be made. First, the four circumferential lines rotate but do not displace or distort. Second, the line along the axis of the shaft rotates through an angle but remains straight. By adding two more lines to the surface of the shaft, a rectangular element can be defined. Repeating the alternating torsional loading, the nature of this deformation becomes more apparent. As the lengths of the sides of the element do not change it can be concluded that the element does not experience any normal strains. However, the element does rotate through an angle just as a square element subjected to pure shear wood. Let's take a closer look at what is happening on our circular shaft of radius R. Consider a square element A, B, C, D on the surface of the circular shaft as shown here. For illustrative purposes, we have drawn the element large and visible. However, we are going to analyze it as an element of infinitesimal width, dx. Let's first isolate the portion of the shaft containing our infinitesimal element. As we ascertain from observing shear deformation in our form shaft, torsion creates a state of pure shear. Thus, if we apply a torque T to our segment of shaft, the square element A, B, C, D will undergo shear deformation, resulting in infinitesimal rotation of the shaft segment d phi. This angle is equal to the rotation of point B relative to point A along the axis of the shaft. On the surface of the shaft, our infinitesimal element must shear by angle gamma to be compatible with this deformation. This angle gamma is exactly the shear strain experienced by the element. Geometrically, we can relate the tangent of this angle gamma to the ratio of the length B, B prime over length AB. Length B, B prime is the arc length swept by radius R through infinitesimal angle d phi, or r times d phi. And length AB is simply the width of our element dx. Since we expect shear strain in real materials to be small, we can apply a small angle approximation where the tangent of our angle gamma is approximately equal to the angle itself in radians. Applying this approximation to the equation above, we can rearrange it to obtain that the shear strain gamma is equal to the infinitesimal angle of rotation of the segment d phi divided by the infinitesimal width of our segment dx, multiplied by radius R. In the previous slide, we examined the shear deformation on the surface of a segment of our shaft. Let's rename the shear strain at this location to gamma max and take a look at what is happening inside the segment at an arbitrary radius rho. Extracting the interior portion of our original segment at radius rho, we see that the surface of this extracted segment also undergoes shear deformation. Using the same approach as before, it is easy to see that the shear strain gamma at this new radius rho will have a similar relation to d phi and dx. This expression can be rearranged in terms of the outer surface shear strain gamma max resulting in the expression rho over R times gamma max. From this simplified expression, it becomes apparent that shear strain varies linearly with radius rho starting from zero at the center of the shaft and increasing to a maximum at the outermost surface of the shaft. All the previous equations were derived for a solid shaft. However, they are equally applicable to a hollow shaft. In the case of a hollow shaft, strain will vary linearly from a minimum at the inner surface to a maximum strain at the outer surface as shown here. Now that we understand the strain distribution within a shaft subjected to torsion, let's take a closer look at the stress distribution. In order to transform the strain distribution into stress, it is necessary to recall Hooke's law for shear as shown here. It is also necessary to visualize the pure state of shear stress acting on our infinitesimal element that will cause the shear deformation we observe. Finally, we must recall the linear shear strain distribution from the previous slides. Rearranging Hooke's law for shear stress and substituting in our shear strain distribution, we can see that the shear stress distribution also varies linearly from zero at the center of the shaft to a maximum at the outermost surface. Slightly less obvious is that due to the concept of complementary shear stress and the state of pure shear illustrated in our element in the center of the slide, this shear stress distribution describes the shear stress acting perpendicular to the radius over the cross-section of the shaft as well as the shear stress acting in the direction of the axis of the shaft as illustrated on the right. This resulting shear stress distribution again is a result of the principle of complementary shear stress. To recap, we first set out to take a look at the deformation associated with torsion and found that torsion produces twist resulting from a state of pure shear. Next, we took a look at the shear strain distribution compatible with this deformation and found that shear strain varies linearly with radial position within the cross-section of a shaft. Finally, we applied Hooke's law to convert the shear strain into a shear stress and found that shear stress also varies linearly with radial position within the cross-section.