17 thoughts on “HOMEWORK 29.A - SUM 24”

    1. The way I solved it does involve calculations by segments but I did not draw out the individual segments. Rather I solved for sheer force and bending moment about B, C and D using equilibrium equations and by dividing the distributed load over each segment. The graphs I got do make sense but I am still not 100% sure if my way of doing it was right. I think one way of checking the validity of our sheer force diagram is by subtracting the maximum and minimum sheer forces (Vmax - Vmin) and ensuring that it's equal to the external force applied at the point where the maximum and minimum sheer forces coincide. In this case, that would be point C and the external force on it is 40kN.

      1. I found the rate of change of V using the distributed load and adjusted the graph/equations to account for P. Then I took the integrals of V to find rate of change of M and adjusted for the moments at B and D.

        I'm not sure what you mean with the validity check and where min/max coincide. Also, wouldn't the external force on C not be 40 kN since you have both P and the distributed load?

  2. No, I didn't show my section calculations for the bending moment. I did for shear force, but that is pretty easy as we know it starts at 0 then climbs until just before the applied load P where it drops then climbs up again because the distributed load. Then since we know that bending moment is just the integral of shear force we can find the correct graph for bending moment. However, as there is a plus C in an integral we have to shift it up or down a certain amount which is where the concentrated couple comes into play.

      1. In my calculation I found By to be zero. If you sum the forces in the y direction you have 40 up and 40 down which cancels out leaving By to equal zero.

  3. How does the force P = 40 at C affect the shear force diagram of the beam with the load w = 10 kN/m over its length L = 4m. My understanding is that cause a straight decrease after the segment BC. That is the way I am approaching it. However, after the straight decrease, does it go up linearly? Just confused on that part

    1. Fort the shear force I found that it started at zero and then increased linearly until point C and then it dropped straight down at that point and continued to climb linearly again. I believe the way that you were thinking is the correct way.

      1. I'm fairly certain this is correct, the concentrated force at C causes a discontinuity on the shear force diagram, then after that point the linear climbing continues, just from the point it drops to back up to zero

    2. The way you can think about it is that the distributed loads cause lines and curves that are not just horizontal on the shear force diagram. The distributed load will cause the shear force to continually increase throughout the member. However, at the point L/2 the point P causes a large decrease but after the point it continues to increase because of the distributed load. So yes, it will go up linearly with the same slope as before the force C.

  5. If you find shear force and moment equations for each section through integration only, what is the most reliable way to know the added constant? It's easy for shear force because it's easy to visualize what V is at the beginning of each section. Is there a simple way to do this for the moment too?

    1. It's easy to see the moment reaction at point B (this is where the graph starts) and the moment on the end (this is where the graph ends) but I'm mostly asking for the middle of the beam

      1. There is no added constant for the moment in the middle because the forces in the middle of the beam cancel out. However you can tell that slope of the moment graph in the middle is zero because the shear force graph intersects zero at the middle of the beam.

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