I want to continue our discussion of mechanics of materials today by starting with topic III, Torsion. Here, Torsion, these are the topics we will be looking at in this module. In particular today, we will look at Circular Bars in pure torsion. And Shear stresses and how they are related to the torque, which is the Torsion Formula, and then the formulas which apply to Hollow tubes. Firstly, Torsion is related to the twisting of circular bars and hollow shafts by torsional moments. We'll only be looking at circular bars in the segment. So the twisting moment can be thought of either as a torque, a couple, which is of course a vector whose direction is given by the right hand rule. So, as shown in this diagram here we have this rod subject to torsion by this wrench. And we can have multiple couples or multiple moments acting, as shown here. We have two of them rotating in opposite directions, which we can show schematically like this, which gives the magnitude of the torque, T1 and T2, and their directions. Or alternatively and and conveniently we can show these as vectors. And when we show them as a vector, we usually show them as a vector with two arrowheads. And the direction of the vector of course is the direction of the moment or the couple. And the direction is given by the right hand roll. So in this example. The torque T2 is rotating counter clockwise. If you look end on at that shaft. And so by the right hand rule, it's given by an arrow which is pointing away from the surface. That is a positive torque or a positive couple. Uniform torque is a situation where the torque is constant. In other words, it's constant along the length of the shaft and the shaft is prismatic. In other words, it's constant cross-sectional area and constant properties. We'll also look at cases of non-uniform torque where either the torque and/or the cross-section of the shaft changes along its length. And what we want to derive is some relationship between the applied talk and the amount of twist or deformation of the rod. In other words, we want to find the relationship between the shear stress and the shear strain which is the deformation in the rod. And this relationship is known as the Torsion formula. Now previously, we've only looked at cases of axial loading where the loading is direct along the axis of the member. Now we'll look at the somewhat more complex case of torsion or twisting. So, the analysis will start with the simplest case of a rod in pure torsion. In other words the talk is constant, cross sectional area is constant, and the rod is subject to a torque T at both ends as shown where this torque is positive. And we will only look at round prismatic bars. And from symmetry it follows that a cross-section must remain circular. In other words, it doesn't change shape. The cross-sections are constant in this case. And that follows just from symmetry conditions. In other words, if I imagine a disc being cut out of this rod, a disc would remain a flat disc and any radii or radius that I draw on it would remain straight with no warping. So, in this case, then, we have a line pq here pq as it twists becomes a helical curve, pq prime. And the total angle of twist is phi, where phi is the angle of twist here. So phi is a function of a x, the distance along the rod and this twist varies linearly from one end to the other. We will define the angular strain as being equal to gamma. Where gamma is shown here on this diagram. And gamma the angular strain, we can see from this diagram, increases linearly from 0 at the center to a maximum value at the outer surface of the cylinder. In other words, gamma maximum, the maximum strain, is equal to r, the radius of the tube or shaft, times d phi by dx. Is equal to r theta where theta is the rate of change of angle along the length of the rod d fee by dx. Or gamamax is equal to r phi over l where phi is the total twist is total twist angle and l is the length of the shaft. Now this formula also applies to halo circular tubes. So the angular strain again increases linearly from zero at the center to a maximum at the outside. Therefore, the maximum strain of the outside is equal to r2 over L. Where r2 is the outside diameter and the minimum at the inside of the tube here is equal to r1 phi over L, where r1 is the internal radius. To illustrate that, let's do this example. So we have a rod which is twisted by positive torques T at both ends. Until the angle of rotation between the ends is 3.5 degrees. The length is given as 0.75 meters and the allowable shear strain is .0005 radians. The maximum permissible diameter is most nearly which of these alternatives? So from our previous formula we have that the maximum shear strain, gamma max is r divide by dx and r the radius is half the diameter d over 2 and difi by dx is just phi, the total x divided by the length of the tube, or rod. So therefore rearranging for d which is what we want to calculate here. Is equal to two gamma max L over the fee. So now I can substitute in the given numbers gamma max is 0.0005 the length is 750 millimeters or 0.75 meters, and I have to be careful on the bottom here. The angle phi is 3.5 degrees. However, in that formula phi is radians. So to convert, I multiply by pi/180. And the answer is 12.3 millimeters. And the answer is D. This is maximum diameter. Next, I want to look at shear stresses. And, as we twist a rod here, so I apply some twisting to this, then I would get some shear stresses resulting. In other words, shear stresses which are tangential to the local surface, as shown. So, what we want to do now is find the magnitudes and directions of these shear stresses, and how they're related to torque and the material properties. So, we begin with our shear strain formula again. Gamma max is equal to r theta. And if we look at the details of this element here, we see that an initially square element, ABCD deforms under this twisting action to A prime, B prime, C prime, D prime. In other words, we have an angular strain here of gamma. And as we've previously shown that strain gamma increases linearly from zero at the center to a maximum value at the outer edge. So we could apply Hooke's law in shear in other words the shear stress is directly proportionally to the shear strain. Tau is equal to G times gamma or tau max, therefore is equal to Gr capital theta and the local shear stress tau is equal to row over r, tau max, where row is the local radial distance from the center of the tube. So, in other words, the shear stresses increase linearly from zero of the center to a maximum value at the outer surface as shown in this diagram. Now, in addition, we also generate longitudinal shear stresses because for equilibrium of a material element, the shear stresses are constant around the perimeter. Therefore, shear stresses in this direction also give rise to shear stresses in the longitudinal direction. Next, we want to find the relationship between the torque, the stresses, and the strains. So, we will do this by considering the forces in the moments on an element of area, dA, as shown here at a radial distance row from the center and the shear stress acting on that material element is tau. So, the moment on that element is the force on the element multiplied by its radius arm, and that is therefore equal to tau row dA. The force on the element is stress times area, tau dA, and the moment on is row. So that in turn is equal to tau max over r, times row squared dA, is the moment on that small element of area. And then, the total moment or the total torque acting on that disc is equal to the summation of the moments, in other words, the integral of the moments over the area which is therefore equal to tau max over r, integral over the area of r squared dA. Now, this in turn I can write like this, tau max over r, times ip, where ip is the integral over the area of roe square da is the polar moment of inertia. So therefore, we arrive at our final formula, that the maximum shear stress is equal to tr over ip. And this formula, is known as the torsion formula. And the only shapes that we're interested in here are circles and for a circle, the polar moment of inertia is pi r to the fourth over 2 or in terms of the diameter pi d to the fourth, over thirty two. Therefore, the maximum shear stress tail max is equal to 16T over pi d cubed. From which we notice that the shear stress is inversely proportional to the diameter cubed. Also, the local shear stress in the rod is equal T row divided by Ip, where row is the radial distance from the center. Next, we want to find out what the angle of twist of the rod is. Here we go back to that original diagram. So the angle of twist here, is phi, the total angle of twist along the rod. We know that that is related to tau by this formula. Tau is equal to G, row theta or theta, the rate of change of the angle of twist along the rod, V by the X, is equal to T over G Ip. Therefore, finally, we arrive at this equation that V, the total angle of twist, is equal TL divided by G Ip. And again, a reminder that in all these questions phi, the angles, are given in radians. In addition, this quantity, GIp over L is sometimes denoted by kT, where this is known as the torsional stiffness. In other words, the rigidity, the resistance to twisting under the action of the torque. And here is the relevant section from the handbook in the notation. So, let's do an example on that. So we have a solid aluminum bar. That is twisted by a torque at both ends. The diameter is 52 millimeters and the allowable shear stress is 65 megapascal's. The maximum permissible torque T is most nearly which of these? So we will start from our basic formula for torsion of a circular rod tau max = 16T over pi times the diameter cubed. And in this case, we want to calculate the torque, so we re-cast that formula in terms of T, as shown here. And now, we can substitute in the numbers. The maximum shear stress is 65 megapascals or 65 times 10 to the 6 pascals times pie times the diameter cubed divided by 16. Calculate the numbers that is equal to 1795 newton meters and the closest answer is D. Now lets look at circular tubes. Most tubes for transmission of torque or power are hollow. Because most of the material is where the stress and the moment arm are highest, in other words at the outside. The elements close to the center don't carry much. So hollow shafts are much more efficient than solid rods. In resisting torque for the same weight or mass. The reason for this we can see by considering the moments on elements of area. As we've previously showed, the moment on an element of area DA is proportional to its area. And the area increase outwards as we go outwards. It's also proportional to the moment arm. In other words, proportional to the radial distance r. And finally, the shear stress itself increases linearly outwards. So if you put all of those things together, you find that the moment on an element of area increases very rapidly like the cube of the distance from the center. The cube of the radial distance. Therefore elements in the center here are not carrying much moment, they're not carrying much load. Therefore we can do away with them, and make it hollow. Now, still for a hollow tube, the same formula applies. Tau max is equal to Tr over Ip, and for solid shaft, we have this relationship for the polar moment of inertia, for a hollow shaft, the equation becomes this, in terms of the internal and external diameters or radii. So the same formula applies. But we just use the different formula for the polar moment of inertia. Now remember, everything that we've done so far only applies to linearly elastic material. In other words, Hooke's law in shear applies. And it only applies to circular shafts but these will be the only cases we're interested in. And here again is the corresponding section from the reference handbook. So, let's finish with an example on our hollow aluminum tube. And we're told that this is subject to a torque of 6,200 pound inches applied at the ends. And the length is 24 inches. The internal and external diameters are 1.25 inches and 1.75 inches. And the first question is. The maximum shear stress in the tube is most nearly which of these? So the usual formula applies, TMax is equal to Tr over Ip and the maximum radial distance is half the diameter. D2 over 2. For a hollow shaft, the polar moment of inertia is pi over 32 d2 to the 4th minus d1 to the 4th. So substituting in the numbers, we find that that is equal to 0.681 inches to the 4th. Now our shear stress formula telmax is equal to Td 2 over 2 divided by Ip. Substituting in the numbers that is equal to 7966 pounds per square inch. So the closest answer is C. The next part, if the angle of twist of this hollow shaft is four degrees, the modular elasticity of the aluminum is most nearly which of these? So, first we have to remember to convert that angle to radions by multiplying it by Pi divided by 180, so the twist is 0.0698 radians, and gamma max is equal to r, phi over L, our usual formula. R is half the external diameter. So now I can substitute in the numbers, that is equal to 0.00254 radians is the maximum shear strain of the outer surface. Then I apply Hooke's law in shear in other words tau is equal to G times gamma. So rearranging G is Tau max over gamma max. Substituting in the numbers, we've already computed the maximum shear stress is 7966 divided by gamma max is equal to 3.14x10 to the 6 pounds per square inch, or 3.14 ksi. And the closest answer is B. And this completes our preliminary discussion of torsion.