9. Test Yourself – Worked Solution #2

Indeterminate Structures & The Moment Distribution
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Questions and discussion for this lecture live here. Fire away by hitting Reply below :fire:

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I have been wondering for a while now. What is the main reason we don’t work from shear and use the integral to generate the formula for the moment when we want to find the point at where moment becomes zero? We already know what C, the constant of the integral, is given that we have done moment distribution. Is there a deeper reason for doing moment formulas first and then work back by differentiating to get the shear formula?

Personally, I find it easier to work out the formula pertaining to shear and integrate that with respect to x to find the formula for the moment with respect to x (and C has always been the original moment on the left of the member segment). This approach is working so far, so I wanted to make sure I was not hamstringing myself down the road by gravitating towards that approach. Figured it might be good to ask the expert!

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Hi

First, there is no issue with your approach - if you’re comfortable integrating v(x) to get m(x), carry on. There are no deeper or strategic reasons for the workflow I favour. My approach when determining shear and moment diagrams is generally to:

  1. Work out the shear diagram by following the loads across the structure, calculus is usually overkill here.
  2. I’ll then usually evaluate M(x) by making a cut
  3. To identify the location for local max/min in the moment diagram I’ll typically just identify the point of zero shear and directly evaluate the moment at that location.

Having knowledge of the relationship between loading, shear and moment will allow you to seamlessly ‘jump’ from one diagram to the next. With this knowledge, it’s purely a matter of personal preference how you build your diagrams.

I hope that’s helpful.

S

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Sean,

Awesome. I think I will keep on with the approach of integrating v(x) to find m(x). For some reason, whenever I try to build a formula directly for m(x) I will occasionally forget to multiply the lever arm component and have to back-track after noticing that the units don’t make sense.

We all think a bit differently. :slight_smile:

Thank you very much Sean.

1 Like
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Forgetting the lever arm happens to us all every now and then!
S

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Hi Sean,
A new member here, and thank you for all your contributions to the community.

I have a question about the BMD.
On joint B, about 09:47 timestamp into the video, why did you consider -17.42 (BC) but not 17.42 (BA) when drawing the BMD at this joint.
I realize you used the span moment between member AB instead.

This is quite confusing because in video 7, (Multi-iteration distribution), you drew the BMD considering BA and BC separately (-50, +50)

I would appreciate if you can shed more light on this grey area.
Johnson