97 thoughts on “HOMEWORK 28 - Fall 24”

  1. I was thinking that the middle section would be 0 for both V(x) and M(x) graphs, but then when I broke the bar into segments for the free body diagrams, I wasn't sure.

    2. V(x) and M(x) will not be zero through the middle section. V(x) should give you a constant value, and M(x) should give you a linear equation. V(x) being constant makes intuitive sense because there is no distributed loading in the middle section, therefore the shear force will be the constant value found at the end point of the first section. In other words, V(x) through the section from 2 to 4 will be equal to the shear force found in section 1 evaluated at x = 2.

    3. I am pretty sure you would need to split the bar into the three parts shown above, and none of the V(x) and M(x) would be zero. This is because the downward distributed load is acting on all three segments.

    4. If the mid section is considered alone than yes V(x) and M(x) will be 0 since there are no forces acting there, but when we split it we also consider what came before moving positively along x so that just means that no additional V will not change moving along the mid section.

    5. The reason that the V(x) and M(x) graphs wouldn't have zeros in the middle section is because of the distributed load that you have to consider. the shear will decrease continuously as you move from left to right. Same for the bending moment.

    6. At first it may seem like that however, if you were to split it in the middle section you would find that their is a sheer and moment since the global fbd is acting on the full bar. This means the moment(s) and the force(s) are distributed on the full bar not just a segment of it. However, with the way we are computing for the sheer and moment graphs we can generalize the force acting on it at certain points depending on how we take account of the system. In this scenario we will be splitting the FBD into 3 segments so that we can account for the different forces acting on the bar. Take your first cut somewhere in-between the midpoint of L/3 to the end of L/3 and continue this method to get the rest of your values.

    7. I thought so too, but the first load's shear on the beam creates a shear during the middle section. Therefore, it also affects the M(x) grah.

  2. When a weight force dependent on the length of a segment is present, does v(x) become a function of x, too? Does M(x) then become a squared function?

    2. Yes. The change in the moment can be described as the integral of the shear force on the beam, and taking the integral of a linear function will yield a quadratic function.

    3. V(x) can indeed be a linear function, the distributed load means that the shear can change as the load goes on along the beam. Additionally the moment equation should become non linear because of this

    4. Yes, when the weight force varies along the length (like a distributed load), V(x) becomes a function of x because the shear force is the integral of the load distribution. As a result, M(x), which is the integral of V(x), becomes a quadratic function of x.

      1. Did you make multiple FBDs for the multiple cuts, or did you just make one FBD and made no cuts and solved using moment at the fixed point with the equilibrium equations?

        1. Yes, I made multiple FBDs. I made 1 initial FBD, and then I made a new FBD every time a new force was introduced to the beam. So I had a FBD of the whole beam, a FBD for 0<x<2ft, a FBD for 0<x<4ft, and a FBD for 0<x<6ft.

          1. As a correction, I should have said a change in external forces on the beam instead of new force introduced. From 2ft<x<4ft, there is no new external forces, but there is a force change from 0<x<2ft.

    4. Yes, you still have to solve for the middle section because it changes the moment equation for that part. You have to use the full equivalent force (40) and then create an equation for x for the distances (it won't be written just as x anymore).

    7. Yes, solving for the middle segment is necessary to find the shear and moment. The moment transitions from an exponentially declining value to a linear declining value due to no new forces or moments being applied.

    8. Yes, you do need to solve for the forces and moments in the middle section, taking into account the distributed load that applies specifically in this part. This section would have its own shear force and bending moment equations based on the internal forces and loading.

      1. Yes you will need to split the system into three sections. You will have different shear forces on the middle section then the outer two sections.

      1. I agree with you Kaden, however, there are technically still forces that on the middle section. Because the middle section is attached directly to the first third of the beam, it carries the entire distributed load of 2W. The first section (leftmost) would carry a portion of the distributed load as a function of x, which is an arbitrary point along the first third of the beam. This will eventually lead to the conclusion that the middle section has non-zero shear force resultants.

    3. It will be split into three sections because a new section is required whenever a concentrated force is applied or a distributed force changes. In this case, the distributed force changes at both L/3 and 2L/3, so three sections are needed.

    4. I would split the system into three sections. One from the beginning of the beam to the end of the first distributed load, one for the section where there is no distributed load, and one for the section of the last distributed load

    5. I think this problem should be divided into three sections based on how the distributed load changes along the beam. The first section would cover from 0 to L divided by 3, where the load increases linearly. The second section would cover from L divided by 3 to 2L divided by 3, where the load is constant. The third section would cover from 2L divided by 3 to L, where the load decreases linearly. Splitting it this way makes it easier to calculate the shear force and bending moment for each part of the beam.

  5. How can you tell if quadratic functions for M are concave up or concave down? (Without using a calculator to graph it since it is not allowed on the exam)

    1. You can tell by the sign of the "A" term in y = Ax^2 + Bx + C. If A is positive, the function is concave up. If A is negative, the function is concave down.

    2. Take a look at the number before the squared term in your quadratic function. If it is >0, the graph opens upwards and it is concave up. If it is <0, the graph opens downwards and it is concave down.

    5. If you graph the shear force it is easy to tell the concavity, I watched one of the solutions to a lecture example and it seems that if you are beginning with a large V (large slope of M) and ending with a smaller one then it will be concave down, because slope is decreasing, versus concave up slope is increasing.

    6. To determine if a quadratic function is concave up or down, examine the coefficient of the x^2 term. If the coefficient a is positive, the function is concave up and opens upward. If a is negative, the function is concave down and opens downward.

          1. Yes, you need the respective Feq's in order to find the reactions which are needed to solve for the shear force and bending moments in the different sections of the beam.

      1. I do not think it is possible to split it into two portions because there are different forces acting on the beam in three locations. The first portion has part of the first distributed force, the middle portion contains the entire load of the first distributed force, and the last portion contains part of the second distributed load. Thus, you would need split the beam into three portions to account for all the forces properly.

      2. I think you should just do it in three sections just to be safe, I believe the answer would actually vary depending on how many sections. But even if you're not sure, you should probably do three sections just to be absolutely sure of your answer. Its a little more work but it ensures you have the right equations

      3. If you split into only two parts you are not taking into account that for the first segment (L/3) the downward distributed load changes based on the length of the segment. While for the second segment the value of the downward distributed load the value of L/3 does not change.

      4. I believe it would be possible to split it into two sections, but your answer would be different because the lengths of the sections are different. I think as an approach it's possible, but dividing it into three looks like the approach they want you to use here.

    1. the shear force and bending moment should be nonzero in the middle segment because the applied forces on each creates internal forces in the middle. There are no disrupted loads at segment 2, meaning that the shear force is constant (could be zero or non-zero). But the external force from the previous segment still creates a bending moment dependent on the distance from those forces (linear f(x)).

    1. *To clarify I mean at the second section, it logically makes sense for the shear force to be negative so the you have to flip the arrow on the diagram. Is that because it carries over from the last section?

      1. You can assume for this problem the sheer force is always acting downward. That is because it will always result in clockwise rotation when it is in the negative direction in this scenario.

    1. Correct, I used 4 FBDs Total: One for the whole main system, than one for a cut from 0 to 2, than another cut from 2 to 4, and finally a cut from 4 to 6, which is a total of 4. Hope this helps!

    2. Yes. That is exactly what I did. I have a main FBD for the entire beam and then three FBDs for each of the segments of the beam. I had from 0-2, 2-4, and 4-6.

  12. Do we need to solve for force B_y in the overall FBD? I can see why we need to solve for A_y because it is used in the moment and shear calculations. Is B_y needed for any calculations in this problem?

    2. Yes, you do need to solve for By in the overall FBD because it’s essential for ensuring your shear force graph returns to zero at the end. Solving for By helps maintain equilibrium and confirms that all forces are accounted for in your analysis.

      To find By, you should use the overall FBD and create equivalent force arrows for the two line-loading forces. From there, you can calculate Ay by taking the moment about B, and then use equilibrium equations to solve for By. Even though By is not directly used in the moment and shear calculations, By is necessary to balance the forces and correctly plot the shear graph. Without it, your graph would be incomplete or incorrect.

    1. Yes, the shear force V can be a function of x. When the shear force is a linear function of x, the moment function becomes squared. This is because the shear force is the derivative of the moment, and the moment is the integral of the shear function.

    1. I would recommend using a different FBD for each section, as it may make it easier to understand and grade for the graders. This would also make it less prone to making mistakes.

    1. You can solve this problem by setting up an overall FBD to find the reaction forces and then by splitting the bar into 3 individual sections from 0 to 2, 2 to 4, and 4 to 6 feet. At each section, you use the reaction forces and the distributed load to find the internal resultants.

    2. I believe we do. It helps us and the grader visualize what we are trying to solve for. It helps when summing the forces and moments as well so we don't get confused.

  17. Something interesting to think about before approaching this problem and also to come back to after solving: Since the distributed load acting downward on the left third of the beam is exactly double in height and acting in the opposite direction to the distributed load on the right third of the beam would there be any concrete relationship between any resultant forces/moments caused by this?

  18. When considering Section 2 of the diagram for V(x), should the distributed load be considered up to its extent (integrating p(x) of the left distributed load up to x-L/3) or just simplified to its equivalent force acting at a single point?

    1. In order to analyze section 2 of the beam, you must make a mathematical cut at a point x greater than 2 but less than 4. Therefore, the equivalent force of both distributed loads is not affected by the cut at x. In other words, the distributed loads do not become functions of x. This is not the case when analyzing sections 1 and 3, as the point x will indeed affect the area of the distributed load.

      Hope this helps!

    1. I don't think it's possible as the reaction at the pin joint will be needed when calculating the shear force and bending moment of the third section (x = 2L/3 to x = L)

    2. I'm pretty sure you would still have to solve for By since it would ensure the system stays balanced or else the sheer force might come out incorrect.

    1. When the load on a beam is negative , the moment graph is generally concave down, as the moment decreases along the beam. For a positive load , the moment graph is typically concave up, indicating an increase in the moment. This is because the moment graph is shaped by the integral of the shear force, which is directly affected by the load’s sign.

  22. would the bounds for the integral on the second or third segments be any different? It doesn't feel correct just having X be the bound of integration for them since it doesn't start at zero

    1. I think you just change the lower bounds and you keep the upper bound as x. That is what I did and I'm pretty sure I am correct. I made sure the lower bound was the start of the segment I was working on.

    1. If you are referring to the section from x = 0 to x=L/3 as the A to C region, then yes. This is because the shear force V(x) in this section is a linear function of x. Since the bending moment is the integral of the shear force, and the integral of a linear function is a quadratic (or parabolic) function, M(x) will be parabolic when V(x) is linear.

  25. I'm looking for help on the charts, as I'll still struggling with them. For V(X), I have a negative slope, then a straight line, then a positive slope. For M(X), I have a half-positive parabola, then a negative slope to zero, then a semi-circle back to zero. Does this match what others have?

    1. I'm not sure how much I can tell you but that sounds good. A quick check to see if you are consistent is that the derivative of your M(x) should be what you have for V(x) (or at least the same shape).

  26. For solving for the shear force and bending moment diagrams, how do you calculate the reactions at the supports for this simply-supported beam with a non-uniform distributed load?

  27. In the section of the beam where there is no forces acting does that just mean that the shear force graph is constant at that point? What does it mean for the moment graph?

    1. Yes, when there are no forces acting on a section of the beam, the shear force remains constant in that region because there is no additional load to cause a change in shear. For the moment graph, this means it will be linear in that section, as the moment is the integral of the shear force. A constant shear force results in a straight-line slope on the moment diagram.

  29. Since the distributed load transitions from a uniform to a varying intensity in the first segment, how does this affect the slope of the shear force diagram? I would assume the varying intensity causes the shear force to change non-linearly rather than linearly, as the load distribution isn’t constant. Would this also mean that the bending moment diagram in that section will have a cubic variation instead of the typical quadratic shape seen under uniform loads?

Leave a Reply