Continuing our discussion of mechanics of materials, stress and strain, the last topic I want to look at here is sheer stress. So, so far we've only been looking at normal stresses, in other words stresses which act perpendicular to the surface. Now I want to look at sheer stresses, stresses which act tangential to the surface. This can arise, for example, if I twist a rod, if I apply a torque and twist this rod, then I can result in sheer stresses acting. Or another way to look at this, suppose I have a connection like this where I have a plate here, A, with a clevis pin here, B, bolting through to another section. The force on both ends of this is P, so this force is being carried by the clevis pin at B. So this bolted connection, if we look at it in side view like this, then we have the bolt passing through here and here's a free body diagram of the bolt and this is a free body diagram of just the part of the clever spin which is in contact with the plate. And in this case we have a shear force on two sides of the pin here, we have a shear force at the top and the bottom, so we say that this pin is in double shear. Now if we look at the free body diagrams here, in particular this free body diagram here at C, we see that the force which is acting on there, the force on this projected area, is the force FB divided the area AB. Where if A/B is the projected area of the bolt, this stress is called the bearing stress, where the projected area is the diameter of the bolt multiplied by tight, or the length of the bolt which is in contact with the plate. And, in this case of course, FB will just be equal to the applied external force, P. Now across this top and bottom part here, where the shear force V is applied, we have a shear stress, and the average shear stress is the shear force divided by the area. In other words, shear stress, which we usually denote by tau, is equal to V over A, or V over pi, d squared over four, where d is the diameter of the bolt. Now, if we look at the distribution of shear stresses within a material element, if I have a small element here as shown, and I have shear stresses around here, tau acting on each face, then I can apply equilibrium of that material element. In other words, the element must be at equilibrium. The sum of the forces on it in any direction must be zero, and the sum of moments must be zero also, for the element not to be rotating. And you can easily show that by equating the sums of the forces and the moments to zero, that the magnitudes of the shear force on all of these faces must be equal. In other words, I must have tau acting on each of these faces. All of those shear forces are equal. And also the directions of the shear forces must be as shown here. So in other words, around this face, the sheer forces must either both be pointing to a corner, such as here, or pointing away from a corner, such as here, and the directions on the opposing faces must always be opposite to each other. And if that's the case, then we say that that element is in pure sheer. Now as a result of these shear stresses, the material element deforms, and it deforms in a sort of diamond shape, like this, and the shear strains look like this, and the deformation in this case is not a linear deformation, it's an angle. And the angular defamation is the change, the decrease in this internal angle here which we denote by gamma. And gamma it's important to remember in the equations that follow, is always expressed in radians. Now again, as for hooks it turns out that for many materials, the amount of angular deformation is directly proportional to the sheer stress. In other words, tau is proportional to gamma. So again we can replace that equation by putting in a constant of proportionality G so that we have tau is equal to G times gamma. This equation is sometimes called Hooke's law in shear, and the constant of proportionality G is called the shear modulus of elasticity. And typical values for steel would be 11,000 pounds per square inch or 28 gigapascals would be typical values. Now we've had three material constants so far. G, E, and nu, where E was the modulus of elasticity, nu is the Poisson's ratio, and now we've had a new one, the sheer modulus of elasticity, and it turns out that these three constants are related to each other. Only two of them are independent, and they are related by this equation here. So if we have the values of any two of those constants, then we can compute the third one from that equation. We also, when weâ€™re dealing with shear stresses and strains, we need a sine convention and the sine convention is given here where a positive face is called positive if it's oriented to the positive axis direction. In other words the outward normal is the same as the coordinate direction. So this face here, the outward normal, is in the same direction as the x axis, so that's a positive face and similarly the top face is a positive face, whereas these two faces, this face and this face are negative faces. And, shear stresses, we define as positive if it's on a positive face and in the same direction as any other coordinate. And similarly if the shear strain is positive if the angle between two positive faces or two negative faces is reduced. So in this diagram here, this shear strain, this is reducing, so that is a positive shear strain. And here are the relative definitions from the reference handbook. So, let's do some examples on that. We have a steel plate weighing 27 kilo newtons and is suspended by a cable sling as shown with a clevis at each end. The clevis pins are 22 millimeters in diameter and 30 millimeters long. And the questions are first of all, the average shear stress in each pin is approximately which of these alternatives? So firstly, we have to apply statics to find tension in the cable here. So simple free body diagram there shows us that the tension in the cable, which I'll call P, is simply equal to half the weight of the steel plate divided by cosine 35 degrees, which is equal to substituting in the numbers 16.5 kilonewtons. Next, we look at a free body diagram of the pin, and in this case, assuming that the pin is in double sheer, in other words, the sheer force is carried on two faces in the pin, such as shown here. So, in this case the shear force is equal to half of the tension force in each cable. So the total force here is P, so the shear force, by symmetry, is half of that force, which is equal to, therefore, 8.24 or .25 kN. So the average shear stress is the shear force divided by the area of the pin, P divided by the area of pin, or P divided by PI by 4 times the diameter squared. The diameter is 22 mm or 0.22 meters is equal to 21,700 kN per square meter or, 21.7 megapascals. So, the closest answer is A. The second part of the same problem is to compute the bearing stress, so the average bearing stress in each pin is approximately which of these alternatives? So, in this case, the bearing stress is the force divided by the projected area of the pin. So that is equal to the force divided by the height of the pin, multiplied by the diameter. The force is 16.5, the height, or the length of the pin is 13 millimeters, or .03 meters, which gives me 25,000 kilo Newtons per square meter, or twenty-five Mega Pascals, and the closest answer is B. The final example, we have a bearing pad here, subjected to a shear force along the top of 16 kN. And we are given that the lateral displacement of the top plate with respect to the bottom plate is 14 mm, so in side view, this elastomer is deforming like this, and the deformation of the top part here, this distance is given as 14 millimeters. So the question is, in this case, we want to calculate the shear modulus of elasticity of the elastomer, it's approximately which of these alternatives? So in this case, the average shear stress of the plate or throughout the elastomer, is the shear force divided by the area of the plate, in other words V divided by A times B. The force we're given is 16 kilonewtons, or 16 times ten cubed, divided by the area a times b which is 474 x 10 cubed Pascalâ€™s. The average shear strain is this angle gamma here, which is equal to the arc tangent of the displacement d divided by this height, t, which is equal to arctangent of 14 over 55, or 14.3 degrees. However, we have to be careful because in our formulas gamma is given in radians. So to convert to radians, we multiply by Pi over 180, so the shear strain is 0.249 radiums. Now we can use our Hook's Law, tau is equal to g times gamma, or rearranging, g is equal to tau divided by gamma, is the shear stress we've calculated is 474 times ten cubed divided by the strain is 0.249 is equal to 1.9 times 10 to the 6th pascals or 1.9 mega pascals and the closest answer is D. And this concludes our discussion of shear stress and strain.