Questions and discussion for this lecture live here. Fire away by hitting Reply below 
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!
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:
- Work out the shear diagram by following the loads across the structure, calculus is usually overkill here.
- I’ll then usually evaluate
M(x)by making a cut - 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
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. 
Thank you very much Sean.
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Forgetting the lever arm happens to us all every now and then!
S
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
Based on your question, I’m assuming that you’re wondering why is it that in the final BMD for the structure, at joint B, we show a hogging moment of -17.42kNm? I’m guessing that what’s confusing you is that this seems to be obtained by looking only at sub-structure BC and ignoring sub-structure BA? If this is not what’s confusing you please follow up, but I’ll assume this is where the confusion is.
The moment shown for member BA (on the left side of joint B) is +17.42 kNm, while the moment shown for member BC (right side of joint B) is -17.42kNm. These are not actually in conflict with each other - remember, we must have moment equilibrium at the joint. Both of these moments indicate tension on the top side of the beam and this is what’s reflected on the final BMD.
A simple way to visualise the influence of an internal bending moment (revealed by a cut) is to imagine it applied to the end of a cantilever; so, if the +17.42 kNm that we see applied to BA was applied to a cantilever (fixed at A), it would induce tension on the top side of the cantilever. Similarly, if the moment -17.42 kNm we see applied to BC was applied to a cantilever, fixed at C, it would also generate tension on the top side of that cantilever. So, you see, two moments with equal magnitude and different signs on either side of a joint, are in fact indicating the same thing - tension on the top of the beam, which is what we see in the final BMD at this location.
I talk a little more about sign conventions for bending moments in this lecture:
I hope that helps.
S