91 thoughts on “HOMEWORK 29 - Fall 24”

    2. I would assume that it doesn’t matter, but since the lecture focused on solving the problem graphically, that’s probably the recommended approach. I suggest using the graphical method because, in Exam 2, we were asked to apply two methods for the truss. So, there’s a chance you might need to know both methods for the final exam, practicing now could be beneficial!

    3. What I did for this question was use one method to solve it and then use the other one to verify my answer, that way I was able to practice both and also make sure my answer made sense.

    4. I just did it graphically, but I did solve for By to make sure I had my starting point in the correct spot for shear. The TAs seemed satisfied with just that.

    10. You can use the method of segments if you would Like but the way I approached it was graphing out the Shear graph and then for the bending moment I took the integral of the shear.

    11. You can choose how to approach it—either graphically or using sections. Graphically is often quicker for constructing diagrams, especially if you’re familiar with the relationships between load, shear, and moment. Using sections provides a detailed, step-by-step breakdown and is better if you're asked to show the equations or specific values at key points. Choose based on the problem requirements or your preference.

    1. I'm pretty sure the location doesn't matter. If I remember correctly, an applied moment at one location can be "moved" anywhere on the beam to make solving for reactions simpler-looking.

    2. It doesn't matter when solving for the reaction forces at B because you are looking at the entire beam, but if you're separating the beam into sections, then the location does matter. If the section you're working with doesn't include the moment, which is located at the end of the beam, then don't include it when solving for the shear force and the bending moment.

      1. I agree with Jessica here. In general, when finding the reaction forces at a support, the approach is to view the entire beam/frame/truss as the system, which will encompass all applied forces and moments. And in general, when using sectional analysis, or dividing the beam/truss/frame into parts to make the numerical analysis simpler, you need only consider applied forces/force couples that apply directly to the section you are looking at.

        1. Although there wouldn't be the applied moment at the section nearest the wall, would the reaction moment for the beam on the wall still be included?

          1. The reaction moment for the bean on the wall is still included and is affected by the applied moment. This is because at point B there is a fixed support, so a moment would occur when there is a applied moment.

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    3. Correct the location of the moment does not matter. Whether solving at the location of the moment of at a point away from the moment it is still applied in sum of the moments.

    5. On our calculations it doestn matter, but for the M(x) graph does, to my understanding, the added momentum behaves similarly as the added force does to the V(x) graph.

    2. You would not need to consider the applied moment at point D specifically, no. Though it does affect the moment at point B, since it is a fixed support. Don't forget that there is a moment at point B.

    3. If you do this problem graphically then it is important to consider exactly where the point moment would be, as there would be a discontinuity there on the moment graph. If you do it mathematically through you do not.

  4. To address Abigail's question: no when taking a a moment about another point the location does not matter. I always found it easy to remember if you look at the units of a moment such as KN-m or lb-ft. Since you already have a force times length unit, you don't need to multiply by a length again.

    1. I don't think it matters but I think they want us to use the graphical method. That's what I did and I checked with TAs and they said it was okay.

  7. Can someone explain why we use a different sign convention for the internal stress problems compared to just a regular static equilibrium problem?

    1. I would say its just the way that we are taught. You could change the sign convention if you want to but I would not recommend it because that would mean your opposite convention needs to be switched as well. The convention we learned is just an easy way to execute and less of a worry to determine what sign is what direction. In regards to this specific problem the signs should be the convention as how you did all the other past homework. Its only when you get to the graphing part that signs become a little tricky. For example for moment, if on the diagram it is showing the reaction* moment as positive then on your graph the moment at the point should be negative. It would always follow the opposite of what your reaction moment/force is.

    2. Something that was explained in my section of lecture is to graph the shear force per usual from left to right, as it will mirror the direction of the forces on the beam. However, when doing the moment, it may be best to go from right to left, as the sign convention as you said is different, in the opposite direction for moments applied to the beam. When you do the bending moment from right to left, you can treat it as the standard sign that we have been taught from day 1 as the opposite direction brings the sign back to normal. Hope this helps.

    1. Yes, the concentrated moment at the end of the cantilever beam does affect the bending moment at all of the points on the beam. I made sure to account for it when I made my moment diagram.

    2. Yes, the moment at the free end of the cantilever, where the applied moment M is located, affects the moment at every point along the beam in the moment diagram. In a cantilevered beam, moments are cumulative from the free end toward the fixed end because each force or moment applied along the beam contributes to the bending moment at each preceding point. Thus, the applied moment M at the end adds a constant value to the moment at all points on the beam, influencing the entire bending moment diagram.

  9. When we begin the shear graph we start with a y-intercept equivalent to the force of By correct? Then due to the force w across the entire beam, you would have a linear increase but you'd also have to take into account the force P, yes?

    1. Yes, but I think in this problem By is 0 bc P cancels out the force due to w over L. Moment at B due to the cantilever's "foundation" is nonzero though, so the moment graph starts with that nonzero y-intercept.

      1. I think you forgot to consider the force couple located at D. This would be included in the moment of the entire bar when solving for By. I had a nonzero value for By

    2. Correct you would start with a y intercept equivalent to the force of By but in this case that force is 0. After that it would be a linear increase and once you get to half of the beam you would indeed need to take into account the force P which in this case is negative 40.

    1. The way I solved for this was by first setting up the entire FBD where it is connected to the wall and solving for the reaction forces and the moment, and then from there, you just have to use that value to help solve for the shear and further moment forces.

  11. I feel very confident doing these problems with sections but find the integration method difficult to set up. could someone please explain how to find/conceptualize the inside function?

    1. The inside function for the shear, p(x), is a function that represents the distributed load over whatever segment you are looking at, which is then integrated from the starting point to some generic distance x.
      So, for example, if the segment has no distributed load, p(x) will be 0 and the overall shear will be a constant resulting from the V(0) term that is added on. If the distributed load is a constant, p(x) will also be that constant. If the distributed load has a slope, then p(x) would be an equation in the form Mx + B representing that line.
      It is important to note that the sign of p(x) is dependent on the direction of the distributed load, so even if the distributed load has a positive slope, p(x) could still be negative if the load points down.
      The inner function for the moment is the equation that you found for the shear.

  13. Do the forces essentially cancel each other out, when we account for 10 kN/m over 4m upwards, and 40 kN downwards?
    So only the moment M needs to be accounted for?

    2. Yes. But there is a reaction moment at the fixed support on the other end that is equal and opposite in direction to the moment applied at point D. You have to account for that in your graph for the moment.

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    4. Yes, this is the case except when calculating the forces between the left side and the middle, because there wouldn't be the full force of the distributed load, which would then not fully cancel out. But after that where the distributed load is fully calculated, they should cancel out with the other forces.

    1. It depends on the way you choose to solve the problem. If you choose to use the integration method, one FBD will suffice in your calculations. However, if you choose to do the method of sections (the method used on the last homework), multiple free body diagrams would be better for visualizing the section and showing clearly how you solved for your moment and sheer force equations.

  16. Do we have to actually solve out the equations for each sheer and bending moment, or can we just graph the diagrams based on the forces being applied and what would occur? We learned in lecture how to do it by simply looking at the given picture, and it doesn't say in the question that we specifically need to find the equations, so do we still need to solve for each specifically?

  18. For these types of problems should you always solve them out anyways instead of just doing it graphically? I feel like doing it graphically it is easy to make mistakes.

    1. While it is faster to solve these kinds of problems graphically, a good way to practice using other methods is to solve it out. This way you can make sure you know how to use both methods while also being able to check the answer you got graphically.

    2. You can do it graphically, however, it would be better for content understanding to make the cuts and then make equations from there. For speed on the final exam though, doing it graphically may make more sense as you can just use calculus to figure out how the moment graph will look after making the shear force diagram.

  19. I also did this using the graphic method. I assume because there is a moment at D and the Fy=0, there is a moment at B in the opposite direction at B? So we have a MB and an MD in short.

  20. I saw a comment earlier saying that By should be a nonzero value. Can someone explain how this would be possible? I thought the WL and P forces would cancel out and leave By as the only force acting in the y direction?

    1. I was also confused about this as I have seen these comments you are referring to. I got By to be zero, and does this mean that all the shear forces are equal to zero? Sorry this doesn't make sense to me.

    2. I think that By must be 80kN upwards to balance the total downward force from the distributed load w and the point load P, which add up rather than cancel each other out. This reaction force is necessary to maintain vertical equilibrium for the cantilever beam. I am not sure, but this was my thought process.

      1. This is almost correct! By actually is a value of zero because the distributed force load acts up a total of 40 kN on the beam while force P acts down 40kN, meaning the two forces balance each other out. I think you accidentally reversed the direction of the distributed load.

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  21. Do we need to explicitly solve the equations for each shear and bending moment, or can we just graph the diagrams based on the applied forces and their effects?

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    1. Problem Statement just asks for the diagrams. I added values labeled at important points where there were forces, moments, or any other significant change.

  22. Since the distributed load would have an equivalent load applied in the middle of the beam which would cause an issue where there would be no applied force when breaking the beam apart into sections could you treat the distributed load as two equivalent distributed loads with resultant equivalent forces at L/4 and 3L/4, respectively? If this holds true, could you keep breaking apart the equivalent forces into multiple equivalent forces to suit your needs in future problems?

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    1. Although the distributed load will act in the middle, when solving for the shear and moment you must use the force being at a distance of x/2 because the center of the beam moves as the length x changes. If you were to split the forces and set them at a specific length this would make them concentrated loads rather than the specified distributed load.

  23. For my M(X) graph, I am not sure I did it right. I'm between two: a graph starting at 80 then up to 100 then back down to 80 or a graph, or starting at -80 then up to 100 then down to 80. I'm just not sure if I should start it at 80 or -80, please let me know!

    1. For my graph I started at 80 then went up to 100 then back down to 80 because if you get the moments at each point respectively those are the values that you get so the graph goes correspondingly!

    2. I ended up with the M(x) graph being positive because the moment at the wall is positive using the overall free body diagram but becomes negative when the beam is separated from the wall. This results in positive moments which also makes sense because the moment at point D is positive.

    3. I had the Mx graph starting at -80 going up to -60 with a parabolic curve, then mirroring on the other half to go back down to -80. Just figured this out by doing a moment calculation at the furthest left point and then simply integrating the sheer force curve.

  24. when using the integration method and splitting the beam up into sections, do we have to solve for a new initial moment and shear force when writing the integral for another section.

    1. The points on the M(x) graph were calculated by analyzing the bending moments created by the distributed load w, the concentrated force P at midpoint C, and the applied moment M at the free end D. At C, the combined effects of the distributed load and force P result in a moment of 100kN-m. At D, the total moment sums up to 240kN-m, incorporating all applied forces and moments.

  26. For this question, I solved it using a graphical approach and then verified my answer with the other method. This approach allowed me to practice both techniques and ensured that my solution was accurate.

  29. If the moment on the end didnt equal how much it went down how would the graph look? I think it would stop before the x axis but im not sure how to draw it so it looks like a complete diagram

  30. The combination of the distributed load, concentrated force, and moment makes the shear force and bending moment diagrams unique for each segment of the beam. The distributed load causes a quadratic variation in the moment diagram, while the concentrated force at C creates a jump in the shear diagram. The applied couple at D directly impacts the bending moment diagram but doesn’t affect the shear force.

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