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## 53 thoughts on “Homework 27.B”

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Watch your signs on Ay when calculating the Moment for each section.

Reminder that given moments in a moment equation don't depend on how far away the moment is on the arm.

For this problem, you do not need to make a cut between A and C, since there is no force or moment acting at C. You can solve the problem by cutting between A and B, as well as cutting between B and D.

Does this mean that the M(x) and V(x) are constant from A to B?

Yes Antonio, M(x) and V(x) are constant from A to B.

Hello Antonio,

No, since there are shear forces acting on the beam on this interval from A to B, but the internal shear force is constant at this regionn

Or do you have to make a cut between A and B AND B and D?

Hi Antonio, yes, the two cuts you should make are between A and B and B and D.

Does the weight of the beam not matter?

Hi Alexander, the weight of the beam is negligible in this case. In cases where it is not, it is usually considered as a uniform loading.

Hello Alexander,

Since no sufficient information about the beam weight and mass provided by the problem statement we can neglect its effect

I don't think we need to account for the weight of beam since they didn't give us that information.

Does not matter, the reaction forces only work against the force P and create moments to work against the applied moments M and 2M at A and B, respectively.

Since there are only 3 different areas of the beam with a force applied, are there only two different sections of the beam we need to consider?

Christian,

That is correct. The only sections that need to be considered are A to B and B to D. There is no need to consider the segments of A to C or C to B because there are no forces acting at C.

That is correct, for the problem given, we would only have to consider that there would be two different sections for the beam that we would have to take into account.

Be very careful with the signs on each of the answers.

Are we considering the moments M and 2M as the moments caused by the reactions at points A and B? Or do we still have to calculate the moments caused by the reactions at those points?

Yes, I believe that the moments M and 2M are at point A and point B respectively. Therefore, you don't need to do the extra moment on these points.

If I sum the moments about B, how do I account for the moment about A? Do I just add it? Do I multiply it by the distance between A and B?

Hello Zachary, in the moment equation for sum, moment is just added, unlike force which is multiplied with its distance.

Moments in this case are treated as constants, so you just add it at the beginning.

Is there any significance to cutting the beam at C? Would the calculations at that point not be the same as if there was only a cut between A and B.

Hi Tessa, since there are no forces acting at point C, it would not be significant to make a cut there. The only cuts that are significant in this problem are between A and B, and B and D.

The given moments at A and B in this problem are negative, correct?

Hi Lukas, yes. Both of the concentrated couples (moments) of M and 2M at A and B respectively are in a clockwise direction, making them both negative.

Yes, since it is counterclockwise, the moments are negative.

Sorry. "counterclockwise" should be changed to "clockwise" above.

when taking the moment about point A, do i still add in the M that is already on the arm at A?

Hi Noah, yes. The concentrated couple (moment) that is already on the arm at A still needs to be included when taking the moment from point A.

As point moments do not depend on the distance that they act from a particular point when it comes to summing the moments of a system around said point, the moment on the arm at point A does need to be included when performing the calculations to find the moment around point A.

Do bending moments always start and end at zero on the diagram?

Hi Eunice. I think for the problems we were given the bending moment just so happened to end at zero, but this is not always the case. Example 9.B.1 in the lecture book is a good example of a problem where the bending moments start and end at values other than zero. Also, double check your starting value for this problem because I think you'll find that it doesn't start at zero.

Don't forget both moments in your equilibrium equations. I at first thought you didn't include one of the moments, but both are needed to get the correct calculation.

I have a bit of a conceptual question on problems involving the shear force V like this problem. Sometimes when I solve, assuming that V points downward, I will get a negative number for V. In a normal force diagram setting, this would indicate that the force is actually towards the other direction (pointing upward in this case). However, it seems through the practice problems we did in class that this is not the case. Would anyone be able to explain the reason for the difference in methods? Thanks!

The moments are drawn in the negative direction but M is given as a positive number. Can we assume that the M given is just a magnitude.

Yes, I believe this is correct. You will still need to manipulate the signs when calculating your moment equations as the moments are clockwise like you mentioned.

In the diagrams, since there are not required calculations at C, does the M(x) line directly from M(0) to M(2L/3)?

Yes, you do not need to make a cut between AB. The segments that you should have after making the cuts are 0<x<6 and 6<x<9 since L is 9ft. It is crucial to note the direction of the shear forces and the moments provided in the diagram when setting up your equations for each segment.

Always recall the sign conventions for the internal resultants of V(x) and M(x) to solve correctly and to include the two moments within your calculation process.

When I did my sketch for my M(x), I am getting a jump (not continuous) at point B (6ft). Is this correct? I thought the moment equation would never make a jump.

The bending moment diagram shouldnt have any jumps in it. At point B you should get the same value for M(x) for both your moment equation for segment AB and segment BD. I would double check your signs and equations.

I got a jump as well, and I think this happens whenever a moment is applied at that point on the overall FBD. I found that the change/jump in y value at x=6 will be equal to the given couple of 2M at that point on the beam.

When I sum the moments about A, do I include the moment M at A in my calculation?

Yes, you should include all pure moments regardless of your chosen point.

When you do segment AB, do you include the moment that is present at A?

Yes

If you are having difficulties with your numbers adding up right in this problem, double-check to make sure you are including moments in your equation! The moments in this equation act in the opposite direction to the moment about the cut, so make sure your signs match accordingly! Hopefully this saves somebody a lot of trouble because it took me a long time to figure out that I was making this minor mistake.

Do we not consider the force P at point D during the calculations for the section between B and D?

Yes, I think we don't have to consider the force P at point D for calculating V and M for section BD, but make sure you do consider moments at points A and B.

Correct, if you think about slowly uncovering the diagram from left to right, you only take the forces/moments into account once they are "uncovered".

Is the moment at A positive?

Note. it would make sense for your bending moment diagram to be discontinuous due to the added force couple at point B