External moments when assembling internal force functions

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Discussion Overview

The discussion revolves around the analysis of a beam subjected to distributed loads and an external moment. Participants are examining the correct formulation of bending moment functions and the impact of external forces on these calculations. The scope includes technical reasoning and homework-related problem-solving in structural mechanics.

Discussion Character

  • Technical explanation
  • Homework-related
  • Debate/contested

Main Points Raised

  • The original poster presents a beam problem with specific loads and moments, detailing their bending moment calculations and expressing confusion over discrepancies in results.
  • Some participants suggest that the external moment at 0m must be included in the bending moment function, similar to how reaction forces are included in shear force calculations.
  • One participant emphasizes the importance of sketching shear force and bending moment diagrams to aid in constructing the functions accurately.
  • There is a discussion about whether the external moment should be considered only at x=0 or if it affects values at x=6m as well, leading to further confusion for the original poster.
  • Participants express differing views on how to approach the problem, with some insisting on the necessity of including the external moment in all calculations while others focus on the initial conditions.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to including the external moment in the bending moment calculations. There are multiple competing views on how to handle the moment and its effects on the bending moment values throughout the beam.

Contextual Notes

There are unresolved issues regarding the assumptions made in the calculations, particularly concerning the treatment of external moments and their influence on bending moment values at different points along the beam.

Bill Nye Tho
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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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