# formula for M(x) - Shear Force and Bending Moment Diagrams

I've watched the video "Understanding Shear Force and Bending Moment Diagrams" from "The efficient engineer" and i got a doubt in this exercise

For $$4 I arrive to $$V=-6x+42$$ and by doing sum of the moments equal to zero I got $$M-21-18x-6(x-4)(x-4)/2 = 0$$ which simplifying is equal to $$M=3x^2-6x-51$$ which is different from the formula of the solution from the video

I already understand that the formula for $$V(x)$$ that I arrive is right but I can't understand if the formula that I arrive for $$M(x)$$ is right or not..

could someone explain it to me? My diagramm was: Let's clear the question first: The shear and moment diagrams indicate this is a cantilever beam supported at the left end, and the system is in equilibrium under the applied loads (in green) and the support reactions (in yellow). The range 4<x<7 indicates the internal reactions at a location below the uniform load are in question ("x" is measured from the fixed support towards the free end).

Now let's cutoff the unloaded beam at the free end (L = 1 m), and let's denote the cutoff location as point "A". Essentially, this becomes a question on the internal reactions at the location measured a distance "a" from the end point A", and "a" is less than, or equal to 3 m.

So the solutions should be: V(a) = wa = 6a, and M(a) = wa^2/2= 6a^2/2. Now let's check a = 3 m; V(a=3)= 18, and M(a=3) = 27 (note the moment is clockwise). Hope this helps.

I am not sure which end you are writing the equation from (support at the left end or the right/cut)

However I think:

• I think you are forgetting the 120 kNm

• Also the uniform load creates a counterclockwise rotation about the cut/section, so it has the same sign with 120 kNm, which is the opposite of the 21 and the 18*(4+x)

• also note that x is defined from the right hand side, while i suspect you have defined x staffing from the left. That accounts for the difference in your V(x) equation, and the one given by the video. It probably has to do with the signs you have arrived at.

I'll try to update it with a full solution and some images later on.

UPDATE: I'll use the subscript $$y$$ for your solution, and $$v$$ for the video solution. So $$M_y(x)$$ is the bending moment you calculate, and $$x_y$$ is definition of x.

if you write your equation it results in : $$M_y(x_y) = 21 + 18*x_y - 120 - 6*\frac{(x_y - 4)^2}{2}$$

which when expanded results in $$M_y(x_y) = -147 + 42 x_y - 3 x_y^2$$

You correctly calculated that the video solution expands to

$$M_v(x_v) = -27 + 18 x_v - 3 x_v^2$$

Seemingly there is a discrepancy until you notice that:

$$x_y = 4+x_v$$

So when you substitute $$x_y = 4+x_v$$ in your solution:

$$M_y = -147 + 42 (4+x_v) - 3(4+x_v)^2$$

magically it transforms (after expansion) to:

$$M_y = -27 + 18 x_v - 3 x_v^2$$

So you see the two are equivalent.

# Final thought

In my humble opinion, your way of taking the x consistently from the left end of the beam is less confusing, and you should stick to it.

Pedagogically, I don't know why they opted to change the system of reference (for me the only benefit is simpler calculations). However, eventually it all becomes quite complicated (logistically) when you try to draw the diagram because you need to account every time for the different system.

• You probably already noticed but if you b expand you don't get the ssme result. The reason is that x is defined differently in the two equations. E.g. take your x=4 where the uniform load starts. It's x=3 for the video solution. Further to the right increases your x while reduces the video's. If you account for that then the result should be alright. Dec 26 '20 at 22:09
• I finished the update. I hope it is clear. Your diagram was very helpful, in trying to explain, because I needed to understand the system of coordinates. Let me know how it goes. Dec 26 '20 at 22:43
• I'm glad this time, we solved it with minimal back and forth. :-) Dec 27 '20 at 12:32