Continuing our discussion on mechanics of materials axial loadings, in this segment, I want to look at statically indeterminate structures. Now firstly, what do we mean by statically indeterminate structures? Let's first look at a statically determinate structure. For example, this simple situation here, where we have a rod of two different diameters resting on the floor with two downward forces P1 plus P2. Now, I consulted the reaction force here by simply applying statics. Summing the forces in the vertical direction equals to zero, then we see that the reaction force here is equal to P1 plus P2. So very simply we can solve this case by statics alone, and we didn't need the material properties, in other words modualicity lasticity etcetera, to be able to find the forces. However, to find the defamations, we do need the material properties from the equations we had previously. For example, the compression or the elongation of the upper member here, Member 1, is P1L1 / A1E1 and the compression of the lower portion here The force in the lower member is P1 + P2, so the compression or elongation of the lower member is (P1 + P2) L2 / (A2 E2). Now, consider this situation here where we have a rod which is clamped or held firmly between two immovable walls at A and B. And now I have a reaction force here of B at the bottom and RA. How can we solve for those reaction forces? Well, if I apply static equilibrium in this case, in other words, some of the forces in the vertical direction, then I get RA + RB = P. But that's as far as I can go. And now I have two unknowns here, RA and RB. In other words, I cannot solve it because I only have one equation. And this situation we say is statically indeterminate. So to be able to solve it, I need something else. I need some other information. And in this case, the other information is that the total length of the rod from A to B is fixed because it's constrained between these two immovable walls. Or the compression of the lower member here over this region, must be equal to the extension of the upper member. So the total length is fixed. In other words, the total change in length under these imposed forces, or this imposed force B, deltaAB is equal to zero. So the compression of the lower member, which is- Rb * L2 over A2 * E2, because that's in compression so it's negative, plus, because it's in tension, the upper portion, Ra * A1 over E1 Is equal to zero. So now we have two equations here. This equation and this equation, we have two equations and two unknowns. Therefore, we can, in principle, solve it. So in general, to solve these statically indeterminate problems, we have to go through these steps here firstly, we will apply equilibrium. In other words, could be some of the force is equal to zero, some of moments equal zero, etcetera. So in this case the equilibrium equation is this equation, here. Some of the force equals zero. Next, we need a compatibility relationship. In other words, some relationship between the displacements and for this particular case, the relationship was the total length was fixed. In other problems it could be something different. Then we have force-displacement relations, for example the change in length is equal to the force times the length divided by AE. And then the final step is to solve the equations simultaneously. So let's do an example on that. We have a solid steel cylinder encasing a hollow copper cylinder. And on the top we have a solid cap with a force downwards of p applied in a testing machine. So in this case to analyze this, first of all we look at statics. And if we draw a free body diagram of the upper cap here, the free body diagram looks like this where the downward force here is P. And the upward forces are the compressive forces in the copper cylinder, PC, and steel cylinder, PS. So from equilibrium, we see simply some of the forces in the vertical direction is equal to zero, so therefore P is equal to PS plus PC. [COUGH] Our compatibility relationship si that the compression of the two cylinders must be the same. Because the solid disk is moving down at the same rate. So delta s is equal to Delta C. And at deformation equations are Delta is equal to PL over AE so for the steel cylinder we have this equation and similarly for the compass cylinder we have this equation A. So now applying our compatibility equation, we can equate those two together, so now we have two equations in two unknowns, PC and PS, and we can solve them simultaneously for the unknowns. And the answers are PS is given by this expression, The force in the steel cylinder and the force in the copper cylinder by this expression. And the actual deformation we can compute from the deformation of either the steel or the copper cylinder because they're equal. And we get this relationship here. So, let's apply that to a numerical example. So, here, we have a monel outer shell which is enclosing a brass core. We apply a load p to both the shell and the core through a cap plate. The load p required to compress the shell and the core by 0.1 millimeter is most nearly which of these alternatives? So the dimensions are as given here. The original length is 100 millimeters. The diameters are as shown and the modulus of the monel is 170 GPa and of the brass is 96 GPa. So, this is similar to the previous example, but first we need to compute the areas, as shown here. So the area of the monel is pi by four times its outer diameter squared minus inner diameter squared is 62.83 millimeters squared. And similarly the area of the brass, pi by four, times the diameter squared is 28.274 millimeters squared. So, we apply statics here to the cap, sum the forces in the vertical direction is equal to zero. Which tells us that the applied force P is equal to Pm + Pb, where Pm and Pb are the forces in the monel and the brass cylinders, respectively. Next, we have compatibility so that in this case, the compression of both cylinders is the same In other words, Delta M is equal to delta B. Deformation, we have Delta is equal to force times length over AE. So, in other words, rearranging for the force, which is all we wanted to calculate here, is equal to AE over L times Delta. So applying this equation to the monel, we have this, because we know what the compression is, is equal to 10,676 newtons per square meter, or 10.7 kilonewtons per square meter or kilopascals. Similarly, the force in the brass cylinder is equal to this equation, which is equal to 2716 Newtons per square meter, or 2.7 kilonewtons per square meter or kilopascals. And finally, the total force is equal to the sum of the two forces, Pm + Pb is equal to 10.7 + 2.7, or 13.4 kN, and the closest answer is B. So in this case, the monel cylinder carries most of the load. And the brass cylinder carries a much smaller fraction of the load. And this concludes our discussion of statically indeterminate structures.
