External moments when assembling internal force functions

AI Thread Summary
The discussion revolves around calculating bending moment functions for a beam subjected to distributed loads and a counter-clockwise moment. The user correctly identifies the vertical reactions but struggles with the bending moment calculations, particularly at the endpoints and the midpoint. It is emphasized that the applied moment at 0m must be included in all bending moment calculations along the beam. The confusion arises from the user not consistently applying the external moment in their calculations, leading to discrepancies in the results. Constructing shear force and bending moment diagrams is recommended to clarify the relationships and ensure accurate calculations.
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Homework Statement



I have a 6 meter long beam that is supported by a roller at 0m and a pin at 6m.

There is a triangular distributed load that begins at 0m to 3m which is 2kN/m. (Max load is at 0m)

There is an additional triangular distributed load that begin at 3m which mirrors the first triangular load (2kN/m, max load is at 6m)

And there is a counter-clockwise moment acting at 0m of 18 kNm

Homework Equations



ƩFx = 0
ƩFy = 0
ƩMr = 0

Vertical Reaction @ A (0m) = 6kN
Vertical Reaction @ C (6m) = 0kN

The Attempt at a Solution



So my shear functions are fine but my bending functions always run into a similar problem.

From 0m to 3m my bending function is f(x) = -(2/3)^2(x)^3(.5) + 6(x)
From 3m to 6m my bending function is f(x) = -(.5)(2)(3)(x-1) + 6(x) - ((.5)(2/3)(x-3)^3)/3

My method for drawing internal force diagrams involves checking shared values of each function. So in this case, I like to use 3m in each function to see if I get the same thing, since I know that the bending function is continuous.

When I plug 3 into my first equation I get: -12 kNm
When I plug 3 into my second equation I get: -12 kNm

Everything checks out!

Now, if I calculate my bending forces at 0 m, I get 0 kNm; If I calculate my bending forces at 6 m I get -18kNm.

So basically I now have a bending force diagram that begins at 0, hits -12 kNm at 3m, finally hits -18 kNm at 6m.

I'm looking at the solution and for some reason it's reversed. So I decided to see what I did wrong by including the external force into my functions and while I get the correct answer at 0m and 6m, my 3m answer is completely off.

Is my solution correct? Should it be reversed? Why/Why not? (I solve left to right)
 
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You've got to include the applied couple of 18 kN-m @ x = 0 m in your bending moment function, just like you include the reaction forces at x = 0 m and at x = 6 m in your shear force functions.
 
With these problems, make a sketch of the shear force and bending moment curves. It's much easier to construct your shear force and bending moment functions if you have a sketch.
 
Thanks SteamKing. When I include the couple at x = 0 m, it checks out but at x = 6 m i still get -18.
 
Why don't you post your work? That way, everything you've done should be clear.
 
What part should I post? My functions are the same on paper as my original post.
 
That's your problem. They shouldn't be.

When x = 0, M = 18 kN-m
 
SteamKing said:
That's your problem. They shouldn't be.

When x = 0, M = 18 kN-m

This is where my confusion lies.

I include this couple in calculation only at x=0? That I can accept but then at 6m I have to also include it to get a bending moment of 0.
 
Look, it will be easier to construct the shear force and bending moment functions if you construct the shear force and bending moment diagrams to check yourself. The applied couple at x = 0 affects ALL of the subsequent bending moment values, just like the reaction force RA affects ALL subsequent shear force values.
 

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