 In an earlier video, we examined the stress and strain state in a beam due to moment loading. In it, we examined how moment loading results in curvature deformation, and that an axis exists, known as the neutral axis, along which there are no extensional or normal strains. Using this information, we derived the flexure formula, but we never looked further at the deformation of the beam in terms of its displacement from its original position. Before we jump into looking at the deformation of a beam, we need to recall the coordinate system and sign convention we are using in this course. For a simple horizontal beam, we denote the position along the beam as the z-axis, with the y-axis acting positive downwards as shown here. In this coordinate system, we consider the following internal moments, shear forces, and distributed loads as being positive. Using this convention, we can denote a positive beam deflection by the letter v, which is positive in a downwards direction. All of the formulas derived in this video will be for this sign convention, so please be cautious when alternative coordinate systems and or sign conventions are used. Consider a simple cantilever beam subjected to a moment at its free end. The direction of this moment is such that it will cause the beam to deflect in a positive downwards direction when loaded. We will label this moment negative m, as such a moment would be negative according to our sign convention for moments. The beam as drawn is in its undeformed state. In order to develop a relationship for the deformation, we will redraw this beam in an exaggerated deformed state. If we look at a segment of this deformed beam of width dz, we see that the segment will have a local radius of curvature r that will sweep through an angle d theta. In order to derive the relationship between the moment and the deflection of the beam, we will take a closer look at just this small segment of the deformed beam. Here we see a zoomed in view of the beam. We can see that our radius of curvature sweeps an arc length ds over the angle d theta, giving us that ds is equal to r times d theta. If we look at the left hand side of our element, we can define the displacement at this location as v and the slope of the beam as theta. On the right hand side of our element, both the deflection and slope have become larger by some increment denoted dv and d theta respectively. Recall that the deformation as drawn is exaggerated. In reality, most engineering structures are designed to have a small deformation. Few people enjoy crossing a bridge that deforms a lot. As a result, it is convenient to make a small angle of assumption, which allows us to equate the arc length ds to the undeformed width of our element dz. If we combine these two equations and slightly rearrange, we can see that the inverse of the radius of curvature is equal to the increment in slope divided by the increment in displacement. However, we know that the slope of the curve is the rate of change of the vertical position along the curve with respect to the horizontal position, so we can replace theta by dv over dz. This can be cleaned up to show that the inverse of the radius of curvature is the second derivative of vertical displacement with respect to horizontal position. In order to relate this deformation to loading, we can recall that when deriving the flexure formula for beams, we developed a relationship between the moment loading, radius of curvature, as well as the beam section and material properties. A link to this past video is provided in the top right corner of this video. If we combine this with our previous results, we see that the bending moment is equal to the flexural rigidity EI multiplied by the second derivative of the vertical deflection. Now, we have to be slightly careful here. If we look at the equation in the gray box, we have a small issue. The moment M can be either positive or negative according to our sign convention, but I, E and R are all positive scalars. Thus, the equation should actually include absolute value signs around the moment term. As a result, we have to look at our final equation here and assess what effect the sign of the moment will have. The deformation, slope, and rate and change of slope in our figure here is all in the positive direction, according to our coordinate system. But the deformation is the result of a negative moment. Thus, we need to add a negative sign to our final equation to reflect this. Now that we have the moment curvature relationship, how can we use it? First, we can rearrange it such that we isolate the deformation-related term. Next, we need to realize that this is a differential equation related to horizontal position Z. Thus, the moment term is not a scalar, but a function of Z. Recognizing this, we see that we need to integrate the expression to get our desired deformations. Let's look at this by considering a prismatic beam, a beam where the flexural rigidity is constant along the length. For such a case, if we integrate the moment curvature relationship, we will get the following. Here we can recognize that dv by dz is simply the slope of the deformed beam, which we denoted by theta in previous slides. We can integrate once more and obtain an expression for the deflection of the beam. Both equations for slope and displacement are indefinite integral equations. But what does this mean? Well, it means you will have to dust off your calculus skills and remember that indefinite integrals result in constants of integration. So you will have to be diligent about including and carrying through these constants through your work. But how do we determine what these constants should be? For that, we'll have to look at the boundary conditions of the problem. The first set of boundary conditions we can look at is the support conditions. Roller supports, pin supports and built-in supports each have various constraints on the displacement and slope that they permit. Knowing the location of a support, its z position along the beam, we can equate our displacement equations to the support conditions at this location and solve for the constants of integration. The second type of boundary condition we need to consider is known as a continuity condition. To illustrate this, consider the beam problem shown below. If we draw the moment diagram for this problem, we can see that the moment distribution is a discontinuous function. We will have a moment equation for z is less than a and a different moment equation for z is greater than a. As a result, when we integrate these moment equations to get the deformation, we will have a separate set of deformation equations for z is less than a and z is greater than a. If we look at the expected deformation though, it should be continuous. To ensure this continuous deformation, we need to enforce that the deflection and the slope at z equal a are the same for the two sets of deflection equations. Such a condition is called a continuity condition. The last boundary condition we can use for determining constants of integration is a symmetry condition. If our problem has a plane of symmetry as shown here, then we know that the deformation has to mirror across that plane. So the deflection on either side of the plane have to be equal and the slopes have to be opposite. The beam deformation, however, also needs to be continuous as seen on the previous slide. As a result, the displacement and slope of the beam have to also be equal at the symmetry plane. If we apply both of these sets of constraints, they seem to be conflicting. How can the slopes be opposite but yet the same? Simultaneously applying these constraints leaves only one solution. The slope at a symmetry plane must be zero. This is known as a symmetry condition.