External moments when assembling internal force functions

In summary, the problem involves a 6 meter long beam with a roller support at 0m and a pin support at 6m. There are two triangular distributed loads, one from 0m to 3m with a maximum load of 2kN/m and the other from 3m to 6m with a maximum load of 2kN/m. There is also a counter-clockwise moment of 18 kNm at 0m. The vertical reactions at A and C are 6kN and 0kN respectively. The shear functions are fine, but there is a problem with the bending functions. The bending function from 0m to 3m is f(x) = -(2/
  • #1
Bill Nye Tho
48
0

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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  • #2
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.
 
  • #3
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.
 
  • #4
Thanks SteamKing. When I include the couple at x = 0 m, it checks out but at x = 6 m i still get -18.
 
  • #5
Why don't you post your work? That way, everything you've done should be clear.
 
  • #6
What part should I post? My functions are the same on paper as my original post.
 
  • #7
That's your problem. They shouldn't be.

When x = 0, M = 18 kN-m
 
  • #8
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.
 
  • #9
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.
 

1. What is an external moment?

An external moment is a force that acts on an object at a distance from its center of mass, causing it to rotate. It is typically represented as a vector with a magnitude and direction.

2. How do external moments affect internal force functions?

External moments can cause changes in the internal force functions of an object. For example, if an object experiences an external moment that causes it to rotate, the internal force functions such as stress and strain may change due to the new orientation of the object.

3. Can external moments be controlled?

External moments can be controlled to some extent by adjusting the forces acting on an object. However, external moments can also be unpredictable, especially in real-world situations where there may be multiple external forces acting on an object.

4. How do external moments impact the structural integrity of an object?

External moments can have a significant impact on the structural integrity of an object. If the external moment exceeds the object's ability to withstand it, it can lead to structural failure, such as bending or breaking.

5. How are external moments calculated?

External moments are calculated by multiplying the force acting on an object by the distance from the object's center of mass. The direction of the vector is also taken into account when calculating the external moment.

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