Now, I want to turn to the second topic in Mechanics of Materials axial loadings. So these are the topics we'll look at here. And in particular in this segment, we'll look at axially loaded members under uniform and nonuniform loading conditions. So an axially loaded member refers to a member who's load is directed along the axis of that member and it's parallel to it. And it can only be in tension or compression. And under the action of this applied load in this case P, we assume that the material is extended by an amount delta. So we will be looking at the change in length of a rod like this for uniform and semi-uniform or non-uniform loadings, and we'll also look at statically indeterminate structures. Later on, we'll look at thermal effects and stresses on incline surfaces. Now, a prismatic bar is a bar which is straight and has constant cross section such as shown here, and the definitions we've already had. Strain is the fractional change in the length or delta over L and stress is the force per unit area P over A. And we also know the relationship between those two from Hooke's law, sigma the stress is E times epsilon. So combining these equations together, we find that the elongation is equal to delta is equal to PL over AE. And in particular, this quantity AE is sometimes combined together. And this is known as the axial rigidity of the member. So an example on that, we have an aluminum wire subject to a tensile load P. The modulus of the wire is 10,600 ksi and the permissible elongation of the wire is 8th of an inch. So therefore, the maximum allowable load is which of these alternatives? So the length is 12 feet and the diameter of the wire is a 10th of an inch. So we use our basic equation delta = PL over AE. But in this case we want to compute P, so we rearrange that equation to get P = delta AE over L, which is equal to delta and the area is pi d squared over 4 times E over L. So plugging in the numbers, we have delta is an 8th of an inch, diameter is 0.1 and the modulus of elasticity is 10,600 ksi. So we have to multiply that by 1,000 to change it to PSI, divided by the length is 12 feet and multiplied by 12 to change to inches. And the answer is 72.3 pounds, so the closest answer is B. Now, under uniform loading then, that's fairly simple. We have delta is equal to PL over AE. But for nonuniform loading, we have different types of situations which we'll look at here. Firstly, let's suppose we have bars with intermediate actual loads. So in this case, we have a rod here A B C D. Which is subject to an upward load PB at this point here, a downward load PC and another downward load PD. So this we would say is piece wise constant. In other words, the conditions are constant within each segment AB, BC, CD. The loading is the same cross sectional area material properties are the same. So it's piecewise constant and in this case, we analyze this by separating it, drawing a free body diagram for each segment and finding the force in each segment by statics. So for example, in the segment CD here. Shown here, the internal force N3 in that segment is just equal to the downward load PD and et cetera. We can find it in the other segments as I'll show it in example soon. And in that case, the extension of each segment is now given by Ni Li of Ei Ai with subscript i refers to each different segment and the total elongation we just get by adding them all up, in other words summing of i. And in this case where each segment is the same, the properties Ei times Ai are the same. So that equation just becomes 1 over Ei the summation of NiLi where Ni is the force in each segment and Li is the length of each segment. The second possible case looks like this where we have bars with several prismatic segments. So each segment is constant but we can have different applied forces and in this case the equation is the same. Just the same equation applies although now, the property is the cross sectional area, the modulus of elasticity generally of each segment could be different. And the final case, case three is a situation where we have continuously varying loads or continuously varying dimensions. For example, the rod might be tapering the diameter could be changing and the force could be continuously changing along the member. Well, the analysis here is just the same only now, the analytical segment is an element of the rod dx. And we apply our same basic equation to that element and add them up. So the total elongation is just the sum of the elongations of all the elemental segments. In other words, our summation equation here, just becomes replaced by an integral. So let me illustrate that with some examples. And in this case, we have a brass rod here with modulus elasticity 110 gigapascals constant cross sectional area of 250 squared millimeters and it's loaded by three forces P1, P2, and P3 are shown where the magnitudes are given. The dimensions are given here, the lengths of each segment and the question is, the total change in the length of the bar is most nearly which of these alternatives? So here then, we're going to apply this equation that delta is equal to the sum of the increments of each elemental section. And the first step is to find the force in each segment. So first of all, I'll look at the segment AB here. So if I make a cut here somewhere in that segment and I want to compute the force in there. Let's suppose that the force in the segment AB is NAB. Then applying equilibrium to the rod to the right of that cut. I see that NAB = P1 + P2 minus because it's to the left P3 and substituting in the numbers the force in the segment, AB tensile force is 17 kilonewtons. Next, I turn my attention to the next segment BC and imagine I make a cut through here so the force here, which I'll denote by NBC. Again, applying equilibrium to the segment of the rod to the right of that cut, I see that NBC = P2- P3 = 2.0 kiloneutons also intention. And finally, the last segment CD now, I make my cut somewhere through here and the force in this segment is NCD. So in that segment for equilibrium NCB is equal minus B3 is equal to minus 8 kilonewtons. In other words, that segment is in compression. Now, that I have the forces, I can apply my equation which I can write like this because E and A are constant for each segment. So I can take it out of the summation, so that is equal to 1 over EA force in AB times the length of A et cetera. And this results in a rather lengthy equation, as shown here we know the module as to be elasticity, we know the cross sectional area which I have to multiply by 10 to the minus 6 to change to square meters. And the properties here are shown, here's the force times the length et cetera for each segment. And I add all those up together compute that, the answer is 9.4 x 10 to the- 4 meters or .94 millimeters and the closest answer is therefore A. And because that turned out to be a positive number on the whole this rod elongates. In other words, the point D moves to the right. Another example, we have a nylon bar whose modulus elasticity's 2.1 gigapascals, it weights 11 kilonewtons per cubic meter. That's the specific weight and it's hanging vertically under its own weight. The elongation of the bar at its free mb, is most nearly which of these alternatives? And we're given that the diameter of the rod is 12 millimeters and the initial length is 4.5 meters. So this is a question where we have a continuously varying force along this rod. So we have to use our integral form of the equation. Delta is equal to the integral from 0 to L of N(x) divided by E times A(x) Dx where N(x) is the downward force at any distance X from the top here. So in this case, N of x is obviously varying, because we have gravity operating on the whole length of the rod. So this intend becomes E and A of constant, so I can take that out of the integral. So it becomes the integral from 0 to L of N of X Dx, but the downward force here is just the weight of this section of the rod here. And the weight is the specific weight, which I'll call denote by gamma multiplied by the volume of that length which is the cross sectional area time the length A times x. So substituting that back into the integral equation then, we get 1 over EA integral gamma Axdx from which you see that the cross sectional area is unimportant, it cancels out there. And gamma is constant, so that comes out of the integral. So that becomes gamma over E integral from 0 to L of xdx which is easy to integrate, that is equal to gamma L squared over 2E. So now, I can substitute in the numerical values, gamma the specific way to be given is 11 kilonewtons per cubic meter or 11 x 10 cubed newtons per cubic meter. The length is L 4.5 squared divided by 2, times the modulus of elasticity which is 2.1 gigapasquals or 2.1 x 10 to the 9th, calculating the numbers as 5.3 x 10 to the -5 meters or 0.053 millimeters. So the closest answer is A. Now, a couple of notes on that the total weight of the rod is the specific weight times the volume, which is A times L. Therefore, what would happen if we applied this load simply as a load at the end here, where the load was only applied at that point? Then our formula would have been, delta is the force which is W times L over AE or gamma L squared over E. In other words, the elongation here would have been exactly twice the actual answer. But if we had applied the point, this load at the midpoint at the center of gravity, then we would have gotten the right answer. However, it only works in this case because the cross sectional area of the rod is constant. And this concludes this segment.
