Solving Deflection Using Double Integration: Simplify or Integrate?

AI Thread Summary
The discussion focuses on the use of the double integration method for calculating deflection in a cantilever beam, specifically whether to simplify the integral before integrating. The participant experiences discrepancies in the constant of integration, resulting in different values when simplifying versus integrating directly. It is emphasized that integration should be performed over the entire beam, and matching conditions at the cut point are crucial for consistency. The confusion arises from the differing outcomes based on the approach taken, despite the understanding that slopes should be equal at the cut point. The conversation highlights the importance of method consistency in obtaining accurate results in beam deflection calculations.
coolguy16
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When using the double integration method to solve for deflection, do you simplify the integral before integrating or do you just integrate it? When i tried it with simplifying and without simplifying, I get a different value for the constant which results in different answers. For example in the question attached, between L<x<2L, i got EIv''=-3PL+2Px-P(x-L). When I integrated this and solved for the constants I got C3=PL^2. When I simplified the moment equation to EIv''=Px-2PL, i got C3=PL^2/2. I am kinda confused here. Which way is correct? Thanks
 

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coolguy16 said:
When using the double integration method to solve for deflection, do you simplify the integral before integrating or do you just integrate it? When i tried it with simplifying and without simplifying, I get a different value for the constant which results in different answers. For example in the question attached, between L<x<2L, i got EIv''=-3PL+2Px-P(x-L). When I integrated this and solved for the constants I got C3=PL^2. When I simplified the moment equation to EIv''=Px-2PL, i got C3=PL^2/2. I am kinda confused here. Which way is correct? Thanks
I'm not sure what you are doing here.

AFAIK, you can't use double integration over just part of the beam, and then determine the constant of integration at each step. You've got to integrate over the entire beam. In any event, since this is a cantilever beam, you know the slope and deflection are both zero only at x = 0, where the beam is fixed, not at x = L.
 
SteamKing said:
I'm not sure what you are doing here.

AFAIK, you can't use double integration over just part of the beam, and then determine the constant of integration at each step. You've got to integrate over the entire beam. In any event, since this is a cantilever beam, you know the slope and deflection are both zero only at x = 0, where the beam is fixed, not at x = L.

Thanks for replying. Yes I know you have to integrate over the entire beam. I used matching conditions since if I make a cut at x=L, I know the deflections will be similar at that point for each side. Between 0 and L, I solved for the constants and got zero since its at a fixed end. For C3 and C4 between L and 2L however I equated two equations together in order to solve for it. However, I'm getting different values for the constant depending on if I simplify the integral then solve or I just solve for it directly. Please let me know if you want me to demonstrate the steps I took if I'm not clear. Thanks!
 
coolguy16 said:
Thanks for replying. Yes I know you have to integrate over the entire beam. I used matching conditions since if I make a cut at x=L, I know the deflections will be similar at that point for each side. Between 0 and L, I solved for the constants and got zero since its at a fixed end. For C3 and C4 between L and 2L however I equated two equations together in order to solve for it. However, I'm getting different values for the constant depending on if I simplify the integral then solve or I just solve for it directly. Please let me know if you want me to demonstrate the steps I took if I'm not clear. Thanks!
I think you should integrate over the entire length of the beam. You might have to employ singularity functions to do this, but I think you will obtain the correct result.

You can split this beam in two at the step, and there will be a total of four constants of integration to determine for the deflections. Simplifying the integrals should not affect the determination of these constants.
 
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SteamKing said:
I think you should integrate over the entire length of the beam. You might have to employ singularity functions to do this, but I think you will obtain the correct result.

While the deflections at the step should be equal on either side, the slopes are not.
We have not learned singularity functions and the professor said only double integration should be used if these questions appeared on a exam. Also, the slopes should be equal at that very point since its identical on both sides. I'm just unclear as of why I'm getting completely different answers if I simplify the integral before integrating it or integrating it as it is.
 
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