In the previous discussions of beam deflections, we looked at the differential equations and solutions to the differential equations. Now, I'll give some examples on how to apply them. So, the first one, we're given that the deflection curve of this simple beam which is under a distributed load is given by this expression. V is equal to minus q not L to the fourth over pie to the fourth EI, sine pie x over L. So, it's a simply supported beam of length L and we have some kind of a distributed load q(x). But we don't know the equation for that. So, the possible solutions are given here. Which one is the correct one? So, to solve this, we use the load equation which was this form of the differential equation EI d4 x by dx to the fourth is equal to minus q. So, in this case, I can rearrange that then to get d4 dx to the fourth is minus q over EI, the same equation. And in this case, we're given the equation for the curve. We're given the deflection curve which is this, so what I'm going to have to do is differentiate this four times to find out the function, q of x. So, I differentiate it once, I get dv by dx is minus q0 L cubed over pi cubed E I cosine pi x over L. Then, differentiate again, d2v by dx squared is equal to set etc. So, the cosines say it changes to a sine and I multiply by pi over L. Do it again. D3v by dx cubed is equal to this, sine changes to a cosine and finally the d4x by dx to the fourth is minus q0 over EI sine pi x over L. And now, comparing these two equations here. Comparing this equation with this equation, we see that q is equal to q0 sine pi x over L and so, the correct answer is B. This is the same problem, only now, we want to find the reaction at the support A. Is it which of these four alternatives? So, to solve this, I use the shear force equation form of the differential equation which is this, V is equal to EI d3v by dx cubed. And we've already solved for D3v by dx cubed from the previous problem, so we don't need to do that again. It's equal to this, q0 L over pi EI, cosine pi x over L. So, comparing these two equations, we see that V, the sheer force, is equal to q0 L pi, divided by pi, multiplied by cosine pi x over L, is the variation in the sheer force along the beam. Now, in this case, the question was to find the reaction at the support A here. And from statics, the reaction at that point is just equal to the sheer force at that point. In other words, at x equals zero, the vertical reaction Ra here. Ra the vertical reaction is equal to the shear force. So, substituting x equals zero into this equation, we see that the shear force at that point which is equal to the reaction is equal to qL over pi. So, the answer is C. The next problem is the same one yet again, only now, we want to find the maximum bending moment in the beam. It is which of these alternatives is it? So, here, we use the bending moment equation, which is this one, EI d2v by dx squared, is equal to the local bending moment, M. And again, from the previous problems we already know, d2v by dx squared is equal to this, q0L squared, over pi squared EI, etc. So, comparing these two equations here, we see that M, the local bending moment in the beam, is given by this expression. Now, the question though, is to find the maximum bending moment. And the maximum bending moment, we can either see its location from elementary statics or by drawing the bending moment diagram, etc. But in this case, it's fairly easy to show that the maximum bending moment occurs in the middle which we could also see by just looking at this equation. The maximum bending moment occurs in the middle when sine pi x over L is equal to one, in other words, x over L is equal to X is equal to L over two, the midpoint. So, substituting in, we have the maximum occurs at x equals L over two, so substituting in is equal to qL squared over pi squared. And the correct answer is A. Another example, similar one, but now the shape of the deflection curve is given by this different expression here. And again, we want to find the distributed load. Which of these alternatives is it? So, this proceeds similar to the solution in the last case. We start from the load equation in this case because it's a distributed load. EI d4v by dx to the fourth is equal to minus q, or rearranging, we get this. And we proceed the same way we did last time. We differentiate this expression here with respect to x four times. And I won't go through those. The final answer is d4v by dx to the fourth, is equal to minus q0x over LEI. So, comparing these two equations, we see that q must be equal to q0, x over L. And so, the answer is B. Another example, here we have a simply supported beam which is loaded as shown with a distributed load of 1.8 kips/ft or a 1000 pounds per foot. We're given the moment of inertia of the beam is 285 inches to the fourth. Modulicity, elasticity and the length are as shown. So, first question, the maximum deflection of the beam is most nearly which of these? Now, to solve this, we have to see which case we have to look up the case in the reference handbook. And this turns out to be case number four, where remembering that their symbol for q here is w in the reference handbook. So, from that case, we find that the maximum deflection, v max, is given by this expression 5qL to the fourth over 384 EI. So, now, we can just substitute in 5Q is 1.8 kips per foot, or 1.8 times 10 cubed pounds per foot. The length is 14 raised to the 4th power, 384, the moment of the modules of elasticity is 30 times 10 to the 6th tons per square foot. Multiply that by 144 to change to pounds per square foot. And the moment of inertia is 285 inches to the fourth. Divided by 12 to the fourth to change that to foot to the fourth. And computing that we get 0.0152 feet or 0.182 inches, and the answer is C. And obviously, we realize that the maximum deflection occurs in the center of the beam here at this point. The second part, we want to calculate the angle of rotation of the supports, it's most nearly which of these angles? So, again, we look up case four in the reference handbook from which we find that the maximum slope, theta max, is equal to ql cubed over 24EI. And again, we recognize that the maximum slope here occurs at the supports at both ends here. So, again, substituting in the given numbers, we get this and we get that. That is equal to .00347 and recognize, of course, that that is in radians. So, to convert that to degrees, multiplied by 180 over pi, we get .199 degrees and the closest answer is B. Another example, this cantilever beam is loaded by a concentrated load P, equal to 6900 newtons are shown. The moment of inertia is given, modulus of elasticity and length. And first question, the maximum deflection of the beam is most nearly which of these? So, this case, we look up in the reference handbook, this is case number seven, from which we find that the maximum deflection is given by this expression. PL cubed over 3EI. So, now it's just a matter of substituting in the numbers. The force, P, is 6900 Newtons. Length, L, is 1.5 meters. Modulus of elasticity is 210 gigapascals, or 210 times ten to the ninth. And the moment of inertia is 4.91 times ten to the minus six meters to the fourth, which gives us a deflection of 00.00753 meters or 7.53 millimeters. And the answer is D. So, the beam in this case is deflecting like this, and of course, the maximum deflection is occurring at the end here, V max. The second part of this, the angle of rotation at the free end, is most nearly which of these angles? So, again, from the same case in handbook, we find that the maximum slope or rotation theta max which also occurs at the end right here is equal to PL cubed over PL squared, I'm sorry, over 8EI. Substituting in the numbers, we find that that is equal to .00502 radians or .288 degrees. So, the answer is B. And this concludes our discussion of examples of solutions of the differential equations of beam deflections.
