Today we are going to start a new subject area, knowledge area eight, Mechanics of Materials. And here is the outline of the materials from the reference handbook, which covers a number of topics. Shear and moment diagrams, stresses and strains, etc. And there are approximately 7 to 11 questions. My outline will be a little bit simpler. First we'll look at stresses and strains, then actual loadings, then torsion, then shear forces and bending moments and beams, and then stresses in beams. So many of the topics covered in their longer outline will be covered here, and some of them will also be covered in other sections, such as statics and structural analysis and structural design. So here's the outline of Mechanics of Materials, and first we'll look at stresses and strains. And here we will look at some basic concepts, equilibrium, etc. And particularly in this section, we'll look at some basic concepts. The concept of equilibrium, and forces, and normal stresses and strains in members. So, Mechanics and Materials is also sometimes called the Mechanics of Deformable Bodies. And here, we're interested in bodies or materials which are not rigid. In other words, the relative distance between any two points or particles changes in the deformation process as a result of applied forces or stresses. This is distinct from statics, where we only talked about rigid bodies, which did not deform at all. Now, the topics we'll be looking at are the relationships between deformations, in other words, strains, and stresses under various types of loading, in particular, actual loading and torsional loading. We want to understand the distribution of internal forces in members such as rods, shafts, beams, columns, etc., and we will also look at bending of beams, and the resulting stresses and strains that result when we bend a beam. Now firstly, some basic concepts of equilibrium, which we already looked at in statics. Suppose we have some body, or an object, which is subject to arbitrary forces of arbitrary magnitudes, directions, lines of actions, etc., F1, F2, and arbitrary moments, M. Then we know that the resultant force is simply the vector sum of all of the forces exerted or are operating on that body. And for the body to be in equilibrium, the resultant, or the sum of the forces must be equal to zero. Similarly, we can have some resultant moment, M naught, which is equal to the sum of the moments of the individual forces, r cross F, plus the sum of any moments or couples added, acting. And also for equilibrium, the sum of the moments must be equal to zero. In two dimensions, as we saw in statics, we have three independent equations. Which could be, for example, some of the forces in the x direction equals zero, some in the y direction equals zero, and some of the moments about some point must be equal to zero. Or equivalently, we could also have some of the forces in x equal to zero, and some of the moments about two separate points equal to zero. But in either case, we have only three independent equilibrium equations. If we have three unknowns then we have three equations, so we can solve for them, and we say that that situation is statically determinate. On the other hand, if we have more than three unknowns, then we say that it is statically indeterminate, and we cannot solve it by statics alone. We need some other information, in particular, the deformation of the material under the action of the applied forces. And that's the topic we'll be looking at in this segment. In three dimensions we have six independent equations. However, mostly here we'll just be looking at two-dimensional problems. The forces which are acting are, generally speaking, external forces, which can be either point forces, for example, here. Or distributed loads, such as here, here, or here, where the distribution can be uniform or some variability. And we also have reaction forces at the supports at A here. And an important part of statics and this subject is the ability to draw free body diagrams which show the forces and the relevant dimensions for the whole structure. And in this case, for example, we've reduced this distributed force to an equivalent force, or resultant force at some location, and similarly with these forces. So this is a free body diagram of the entire structure which we can use to solve for the reaction forces and moments. We're also interested in free bodies of structural elements. For example, this element here, if we redraw the free body diagram just for that element, it looks like that. Where now, the reaction forces at D, which were previously internal forces which didn't appear on the free body diagram of the whole structure, now appear as external forces on the free body of the individual structural part. So that is the same diagram, but now we're also interested in internal forces. For example, the force across this section of this member here. So to do this we make a cut and separate the member into two portions. So for example, this member here. And now the internal forces, which are in this case a moment force, M, and a shear force, V, for example, now become external forces which we can solve. And a lot of what we're looking at in this course will be solving for these internal forces and their effects on the member. We will be dealing extensively with prismatic bars or members. And a prismatic bar is a straight bar with a constant cross-section, and is subject to axial loads at the ends which may be either tensile forces as shown here or compressive. And the forces are directed along the axis of the member. And for a sine convention, we assume that a tensile force, which tries to elongate the bar as positive, and a compressive force, which tends to compress it is negative. Now if we do a free body diagram of the internal forces here. So for example, I make a cut here at MM, and redraw just the left hand portion of that member, we get the free body diagram of the internal forces as shown on the bottom here. And from this we can see that the internal force here, P, is constant along that member and everywhere it's in tension. We can define an average normal stress as the actual force divided by area, so sigma is equal to P over A. And this, of course, has dimensions of force per unit area, or FL to the minus 2, which would be either in British units pound per square foot, or more usually pounds per square inch. Or, in structural engineering here, thousands of pounds per square inch, or ksi. In metric units, the fundamental units are newtons per square meter, which is the same as a pascal. But a pascal is a very small unit, so more commonly here we'll be working with kilopascals, megapascals, or even gigapascals, ten to the ninth pascals. If the bar is elongated by a length of delta, then we define the normal strain, epsilon, as being the fractional increase in the length, or delta over L. The increment divided by the initial length. And this is, of course, a dimensionless quantity. The sine conventions for stresses and strains are the same as for forces. In other words, a tensile stress or strain is positive, and a compressive stress or strain is negative. And here are the the relevant sections from the reference handbook. Now if the rod or the material is not circular, for example, some arbitrary shape as shown here, then again, the average stress is just the force divided by the area, P over A. But you can show that the normal stress is uniform over the cross-section if the resultant force P passes through the centroid of the cross-sectional area. Where the coordinates of the centroid are given by these equations here. Y bar is the integral of ydA over A, and x bar is the integral of xdA over A. In other words, it's the centroid of the area or the center of gravity. Let's do an example on that. So here we have a hollow circular post, which is supporting a load P1 of 16 kN at the top, and a second distributed load P2 at the point B. If the lower part of the post must have the same compressive stress as the upper part, the required magnitude of the load P is most nearly which of these alternatives? And the relevant dimensions, the diameters and the thicknesses of the material are given here. So first, to compute the stresses we'll need the areas, so we'll compute the areas. So the area of the top portion, Ab, is pi by 4, D squared, the outer squared minus the inner diameter squared is equal to that, and substituting in the numbers, we find that that is equal to 679x10^-6 m^2. Similarly, the cross-sectional area of the lower portion BC is pi by 4 outer diameter squared minus inner diameter squared, which is equal to 1442x10^-6 m^2. Now we're ready to compute the stresses. So the stress in the upper portion here, sigma Ab, is the force divided by area. And if I imagine a hypothetical cut through the upper section here, and I apply a free body diagram to that upper section, I see that the compressive force in that upper region is constant, and equal to the applied force, P1. So sigma Ab is P1 over Ab. And the force is given a 16 kilonewtons, 16 times 10 cubed, divided by the area, is 23.6 times 10 to the 6 pascals, or 23.6 megapascals. Next I turn my attention to the lower portion, and if I make a cut through here and do a free body diagram of the portion above there, I see that the downward compressive force there is P1+P2. So the stress there is the force, P1+P2, divided by the area, which is equal to sigma Ab, because we want it to be equal to the stress in the upper portion. So therefore, rearranging for P2, which is what we want to find, we get this expression. And now, substituting in the value for sigma Ab that we found above, we get this, which is equal to 18x10 cubed newtons, or 18 kilonewtons, and the answer is B. This example illustrates the definition of strain, longitudinal strain. And the question is, we have a circular aluminum tube, which is being compressed by forces P at the end, and a strain gauge here, I chose a strain of 400x10^-6. The shortening of the bar is most nearly which of these alternatives? So this is purely a geometrically problem which uses the definition of strain, and the dimensions are as given here. So the normal strain, by definition epsilon, is delta of L, the fractional change in the length of the member. So rearranging the changing length is the strain multiplied by length, plugging in the numbers that is 400x10^-6, times the length of the rod, which is 650 mm, which is 0.26 mm, so the closest answer is B. And in this case, of course, the diameters that were given here is irrelevant to the solution. So this completes our elementary discussion of normal stresses and strains.
