0:10

These are the topics in the reference manual, and they cover a number of issues.

Â For example, analysis of forces in statically determinate beams,

Â deflection of beams, structural determinacy, etc.

Â 0:23

But I have narrowed this down to some topics which I think are most important,

Â and this will be my outline.

Â First we'll do a statics review, then we'll look at beam deflections, and

Â then we'll look at static determinacy, and indeterminacy.

Â 0:40

So, here is the outline, and here we will be starting with a statics review,

Â and the topics we'll cover here are the conditions for

Â equilibrium of systems of forces.

Â Then we'll look at analysis of trusses, and then beams.

Â And in this segment we'll look at equilibrium.

Â 1:01

So, equilibrium was a topic that was covered in the Montreal Launch Statics,

Â particularly segment 2a, and, basically, this says that all of the forces,

Â and moments acting on a structure are in balance.

Â And furthermore, that there is no deformation.

Â In other words, we're dealing only with rigid bodies.

Â 1:22

So, first we have the concepts of resultants and equilibrium.

Â So, let's suppose that we have an object, or a body,

Â which is subject to three forces, as shown here, F1, F2 and F3.

Â And for example, the resultant of the forces, F2 and

Â F3 here, is the resultant which is denoted by F1.

Â 1:54

Alternatively, we can do it by adding up the scale of components.

Â So, for example, the resultant of three forces here, F1 and F2 and F3,

Â the resultant force in the x direction,

Â is simply the sum of the components of the individual forces in the x direction.

Â And similarly, the resultant force in the y direction,

Â is the sum of the components of the individual forces in the y direction.

Â 2:34

Continuing then, we have these three forces, F1, F2, and F3.

Â And let's suppose that we want to replace this by a single force acting through

Â the point O, and a moment.

Â 3:11

Now, each of these three forces exerts a moment about the point O.

Â For example, M1 is equal to F1, d1, etc.

Â And the sum of these,

Â we suppose that we can replace by the resulting force moment M0.

Â 3:28

So, M0 is just the sum of the moments of the individual forces,

Â cross product of their radius vector, and force vector.

Â So in this case, M0 is R1, cross F1, plus etcetera for the other forces.

Â 3:44

Now, we can also replace this system by a single force only,

Â passing through some point with no moment.

Â And, let's suppose that the resultant force in this case is R,

Â as shown here, which is a distance d from the point O.

Â 4:02

We can fund that distance be taking moments, so the moment of the resultant

Â force, must be equal to the sum of the moment of the individual forces.

Â In other words, R times d must be equal to M0,

Â from which we can calculate the location of the resulting

Â force from d is equal to M0, divided by the resultant force.

Â So, here is the equivalent section from the reference manual.

Â And the conditions for equilibrium, are simply,

Â that the sum of the forces acting on the body must be equal to 0.

Â And the sum of the moments about some point, must also be equal to 0.

Â There were some special cases that we discussed in the static segment.

Â In two dimensions, for example the first one, collinear forces,

Â means that the lines of action of all the forces lie on a common line.

Â And in this case, we have only one independent equilibrium equation,

Â that the sum of the forces, along the line of axis, action of the forces,

Â is equal to 0.

Â The second case, case two, the forces are concurrent at a point.

Â In other words, the lines of action of all the forces,

Â pass through some common point, denoted by O here.

Â 5:27

And, in this case, we have two independent equations.

Â Sum of the forces in the x direction is equal to 0,

Â and the sum of the forces in the y direction is equal to 0.

Â 5:51

And in this case,

Â we have two equations again, the sum of the forces in the x direction,

Â where the x direction is parallel to line of action of the forces, is equal to 0.

Â And also, the sum of the moments about some point is equal to 0.

Â 6:09

The most general case, number four, we have arbitrary forces, and

Â moments or couples acting on the body.

Â And in this case, we have three independent equations,

Â sum of the forces in the x direction is 0, sum of the forces in the y direction is 0,

Â and also sum of the moments about sub point are equal to 0.

Â 6:31

So, in two dimensions then, generally speaking,

Â we have three independent equilibrium equations, which could be these three.

Â Sum FX is 0, sum FY is 0, and sum of the moments about sum point is equal to 0.

Â But they don't have to be those three.

Â They could also, for example, be sum FX equals 0, sum of the moments about sum

Â point A is equal to 0, and sum of the points about sum point B is equal to 0.

Â But in either case, we only have three independent equations in two dimensions.

Â 7:07

So, if we have a system where we have three unknowns, for example,

Â three unknown reactions, we have three equations, therefore we can solve for

Â those unknowns, and we say that that system is statically determinate.

Â 7:21

On the other hand,

Â if we have more than three unknowns, we only have three equations.

Â So, we can not solve it from statics alone, and

Â we say that that situation is statically indeterminate.

Â 7:35

In three dimensions, we have six independent equations,

Â sum of the forces in the component directions are equal to 0, and

Â the sum of the moments in the three component directions are also equal to 0.

Â 7:53

So, let's do an example on that.

Â We have a traffic light pole here,

Â which is supporting three signals, each of which weighs 100 pounds.

Â 8:24

So, summing the forces in the vertical direction, with positive upwards,

Â we have, sum FZ is equal to 0.

Â And the downward components of forces we have are,

Â the 100 pound force of each of these 3 light fixtures.

Â 8:42

So, therefore, we have OZ minus, because all of those forces are acting

Â downwards in a negative Z direction, minus 100, minus 100,

Â minus 100, is equal to 0, from which OZ is equal to 300 pounds force.

Â So, the answer is C.

Â 9:13

So, generally speaking of course, the moment is a vector quantity.

Â It has a magnitude, and direction.

Â So in this case, it's most convenient to calculate the moments

Â about the three component axis, the x, y, and z axis.

Â So, the moment about the axis here is Mx,

Â about the y axis is My, and about the z axis is

Â Mz, where the directions are given by the right hand rule.

Â 9:45

So, first we'll find the moment about the x axis,

Â by summing moments about the x axis passing through O equals 0.

Â And here, we have them, that Mx minus the moment,

Â because the moment on this light fixture is acting in the negative direction,

Â minus 100 multiplies by its momentum, which is 35,

Â which gives us a moment Mx equal to 3500 pound force feet.

Â 10:20

Next we compute the moments about the y axis, and

Â here we have My, and in this case,

Â the direction of the moments

Â of this force and this force, are positive,

Â because they're rotating about the y axis in the positive direction.

Â So, that's 100 times 25, plus 100 times 50.

Â The momentum solving, we have My,

Â the reaction about the y axis is equal to negative 7500 pounds feet.

Â 11:13

In this case, we're asked for the magnitude only of the reaction force.

Â So, the magnitude is the square root of the sum of the squares,

Â of the magnitudes of the individual components.

Â So, substituting in for My and My, we find that M,

Â the magnitude of the moment, is 8,280 pound force feet.

Â So, the answer is B.

Â