In this module, we're going to start a new topic, structural analysis. Number 12 in the FE reference manual. These are the topics in the reference manual, and they cover a number of issues. For example, analysis of forces in statically determinate beams, deflection of beams, structural determinacy, etc. But I have narrowed this down to some topics which I think are most important, and this will be my outline. First we'll do a statics review, then we'll look at beam deflections, and then we'll look at static determinacy, and indeterminacy. So, here is the outline, and here we will be starting with a statics review, and the topics we'll cover here are the conditions for equilibrium of systems of forces. Then we'll look at analysis of trusses, and then beams. And in this segment we'll look at equilibrium. So, equilibrium was a topic that was covered in the Montreal Launch Statics, particularly segment 2a, and, basically, this says that all of the forces, and moments acting on a structure are in balance. And furthermore, that there is no deformation. In other words, we're dealing only with rigid bodies. So, first we have the concepts of resultants and equilibrium. So, let's suppose that we have an object, or a body, which is subject to three forces, as shown here, F1, F2 and F3. And for example, the resultant of the forces, F2 and F3 here, is the resultant which is denoted by F1. And to find the resultant, we simply add all of the individual forces vectorially. Alternatively, we can do it by adding up the scale of components. So, for example, the resultant of three forces here, F1 and F2 and F3, the resultant force in the x direction, is simply the sum of the components of the individual forces in the x direction. And similarly, the resultant force in the y direction, is the sum of the components of the individual forces in the y direction. So, continuing with this example, we can replace any system of forces by a single force acting through some point, and a moment. Continuing then, we have these three forces, F1, F2, and F3. And let's suppose that we want to replace this by a single force acting through the point O, and a moment. So, to do this we can simply move the forces, their lines of action to the point O. And the result end of these forces, is simply the vector sum of those individual forces. So, in this case the resultant R is simple F1 plus F2 plus F3, added vectorially. Now, each of these three forces exerts a moment about the point O. For example, M1 is equal to F1, d1, etc. And the sum of these, we suppose that we can replace by the resulting force moment M0. So, M0 is just the sum of the moments of the individual forces, cross product of their radius vector, and force vector. So in this case, M0 is R1, cross F1, plus etcetera for the other forces. Now, we can also replace this system by a single force only, passing through some point with no moment. And, let's suppose that the resultant force in this case is R, as shown here, which is a distance d from the point O. We can fund that distance be taking moments, so the moment of the resultant force, must be equal to the sum of the moment of the individual forces. In other words, R times d must be equal to M0, from which we can calculate the location of the resulting force from d is equal to M0, divided by the resultant force. So, here is the equivalent section from the reference manual. And the conditions for equilibrium, are simply, that the sum of the forces acting on the body must be equal to 0. And the sum of the moments about some point, must also be equal to 0. There were some special cases that we discussed in the static segment. In two dimensions, for example the first one, collinear forces, means that the lines of action of all the forces lie on a common line. And in this case, we have only one independent equilibrium equation, that the sum of the forces, along the line of axis, action of the forces, is equal to 0. The second case, case two, the forces are concurrent at a point. In other words, the lines of action of all the forces, pass through some common point, denoted by O here. And, in this case, we have two independent equations. Sum of the forces in the x direction is equal to 0, and the sum of the forces in the y direction is equal to 0. Case number three, where all the lines of the forces are parallel to each other, and this can occur commonly under systems, under the action of gravity. And in this case, we have two equations again, the sum of the forces in the x direction, where the x direction is parallel to line of action of the forces, is equal to 0. And also, the sum of the moments about some point is equal to 0. The most general case, number four, we have arbitrary forces, and moments or couples acting on the body. And in this case, we have three independent equations, sum of the forces in the x direction is 0, sum of the forces in the y direction is 0, and also sum of the moments about sub point are equal to 0. So, in two dimensions then, generally speaking, we have three independent equilibrium equations, which could be these three. Sum FX is 0, sum FY is 0, and sum of the moments about sum point is equal to 0. But they don't have to be those three. They could also, for example, be sum FX equals 0, sum of the moments about sum point A is equal to 0, and sum of the points about sum point B is equal to 0. But in either case, we only have three independent equations in two dimensions. So, if we have a system where we have three unknowns, for example, three unknown reactions, we have three equations, therefore we can solve for those unknowns, and we say that that system is statically determinate. On the other hand, if we have more than three unknowns, we only have three equations. So, we can not solve it from statics alone, and we say that that situation is statically indeterminate. In three dimensions, we have six independent equations, sum of the forces in the component directions are equal to 0, and the sum of the moments in the three component directions are also equal to 0. So, let's do an example on that. We have a traffic light pole here, which is supporting three signals, each of which weighs 100 pounds. We can neglect the weight of the horizontal, and vertical poles. And the first question is, the vertical component of the reaction force at the support at O is most nearly which of these alternatives? So, let's denote the vertical component of the force here by OZ, in the Z direction. So, summing the forces in the vertical direction, with positive upwards, we have, sum FZ is equal to 0. And the downward components of forces we have are, the 100 pound force of each of these 3 light fixtures. So, therefore, we have OZ minus, because all of those forces are acting downwards in a negative Z direction, minus 100, minus 100, minus 100, is equal to 0, from which OZ is equal to 300 pounds force. So, the answer is C. The next part, we want to calculate the magnitude of the momentary action at O, it's most nearly which of these alternatives? So, generally speaking of course, the moment is a vector quantity. It has a magnitude, and direction. So in this case, it's most convenient to calculate the moments about the three component axis, the x, y, and z axis. So, the moment about the axis here is Mx, about the y axis is My, and about the z axis is Mz, where the directions are given by the right hand rule. So, first we'll find the moment about the x axis, by summing moments about the x axis passing through O equals 0. And here, we have them, that Mx minus the moment, because the moment on this light fixture is acting in the negative direction, minus 100 multiplies by its momentum, which is 35, which gives us a moment Mx equal to 3500 pound force feet. Next we compute the moments about the y axis, and here we have My, and in this case, the direction of the moments of this force and this force, are positive, because they're rotating about the y axis in the positive direction. So, that's 100 times 25, plus 100 times 50. The momentum solving, we have My, the reaction about the y axis is equal to negative 7500 pounds feet. And Mz, the moment about the z axis, we don't need to worry about, that is 0, because none of those light fixtures have a force component in the Xy plane. In this case, we're asked for the magnitude only of the reaction force. So, the magnitude is the square root of the sum of the squares, of the magnitudes of the individual components. So, substituting in for My and My, we find that M, the magnitude of the moment, is 8,280 pound force feet. So, the answer is B. So, for further similar examples, see that statics module segment IIb. And this concludes our preliminary discussion of statics.
