This lecture we will begin our discussion of forces and moments in slender members. Recall that a slender member is a member whose length is much larger than either of its cross-sectional dimensions. An I-Beam used in building construction is an example of a slender member.
Let's look at the example of a beam, pinned on its left edge and simply supported on its right. A force P acts downward at a distance A from the left support.
We apply the equations of equilibrium to obtain the reaction forces at the supports.
Note that there is no reaction force in the x direction at the left support. We are able to solve for the reaction forces Ray and Rby acting at the edges of the beam.
Now we are interested in finding out what is happening within the beam. To determine the internal forces and moments, we will take a cut of the beam at some arbitrary location x. We take this cut for x less than a because we expect the system to exhibit different behavior to the right and left of the applied load P. We note that there is no axial force acting within the beam.
We now apply the equations of equilibrium to this subsystem. We sum the moments about the point where the cut is taken to determine the internal moment acting at that point. Solving the equations of equilibrium yields the distributions. The shear distribution is constant within this section of the beam. The moment distribution is - as expected - a function of x, the distance from the left support. It is important to note that these distributions are only valid for x less than a.
Now, we take a cut at an arbitrary point x, where x is greater than a. We do so to examine the internal forces and moments to the right of the applied force P. We again apply the equations of equilibrium in order to determine the force and moment distributions as a function of x.
Having determined the shear and moment distributions in all components of the system, it is time to summarize our results. These are the shear force and bending moment diagrams for this beam. We make two very importanty observations. First, there is very different beam behavior to the right and left of point a, the point where the force P acts. Secondly, the derivative of the moment with respect to x results in the shear force distribution. We will examine this derivation in a later lecture.
This problem provides us with a good opportunity to review our sign convention. It is important to keep to this sign convention when making cuts in a beam to ensure consistency. If we are not consistent, we risk solving a problem that is not in proper equilibrium. The distributions we get from such a flawed analysis will be of no use to us as engineers. We must remember to define a positive and a negative face for each cut that we make. On the positive face, positive forces and moments are to be in the positive direction. On the positive face, negative forces and moments are to be in the negative direction. On the negative face, positive forces and moments are to be in the negative direction. On the negative face, negative forces and moments are to be in the positive direction.
This example is of a cantilevered beam subject to an end load. We will apply the same basic procedure here that we did with the previous example.
We apply the equations of equilibrium to the system, obtaining the reaction forces and moment at the wall. A vertical reaction force and a moment act at the wall.
Now, we take a cut of the beam at an arbitrary point x, making sure to maintain the proper sign convention. We again apply the equations of equilibrium to this subsystem and obtain the internal force and moment distributions in the beam. We only take one cut of the beam in this case because for any x (0 less than x, x less than L), the subsytem will have the same external forces acting on it.