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Cantilever beam with non-udl on - solution provided, need help understanding

  • Thread starter louza8
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  • #1
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Problem and Solution (albeit to a beam of different length L) provided in attachments.

Hi, I have been able to follow the provided solution to get the moment equations and understand the x1 and x2 positional references etc. I also understand that integrating the moment equations provides the solutions to the slope and further integration the solution to the displacement. I also understand in order to get the constants of integration I have to use known boundary conditions.

My hang up is how the solution has gone from this part:

EI*y''=wL(3x2 - L)/12
EI*y'=wx2^2/8 + C2

Perhaps my calculus is a bit weak these days after a break from studies. Little help?
 

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Answers and Replies

  • #2
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Could someone have a look at this one please? I'm still struggling with the understanding. Much appreciated.
 
  • #3
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So nobody here can help me along with solving this problem or did you run into similar issues?
 
  • #4
AlephZero
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The solution you attached is too small to read. That's could be why you aren't getting any help. (The picture showing the question is just about readable, but not the solution)
 
  • #5
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Hi AlephZero, thanks for the tip.

I have added what I hope are more readable versions for the solution :)

To be specific I know how to get the moment equation here:
EI*y''=wL(3x2 - L)/12
But don't know how they've integrated the above to get here:
EI*y'=wx2^2/8 + C2

I thought there would have been an Lx term in there ?
 

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  • #6
nvn
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But I don't know how they've integrated the above to get here:
EI*y' = wx2^2/8 + C2

I thought there would have been an Lx term in there?
louza8: Why did you write C2 here, instead of C3? And why did you omit here the L they wrote? And why did you say Lx here? :confused:

Nonetheless, you are correct, in that their second equation currently seems wrong. Feel free to post what you think the second equation instead should be, and we can tell you whether or not it seems correct.
 
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  • #7
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hi nvn,

thanks for your reply. both were mistakes. i was trying to be clear but was not, apologies.

anyways...i managed to solve the question by simplifying the load using singularity functions then integrating 4 times. i then went and checked the above method (with the correct integrals for the slope) and got the same answer and was satisfied.
 
  • #8
nvn
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louza8: In the last paragraph of post 7, do you mean you got a different answer, or the same final answer, compared to the final answer in the attached files in post 5?
 
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  • #9
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Hi nvn. I used the method outlined in the attached file to solve the homework question (with correct integration) but the homework question is for beam length 2L not L as shown in the attached. Hence, the answers differ due to the longer beam under a greater total load deflecting more.

To check my answer I used an alternative method. I simplified the loading using singularity functions into 3 components. (1 big triangle acting down for entire length 2L, 1 rectangle of beam length L acting up between A-B and 1 triangle of length L acting up between A-B). By integrating the functions describing the loading and using known boundary conditions, I arrived at the same answer as I determined using the method outlined in the attached.
 
  • #10
nvn
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louza8: OK, so you are aware that the answers in the files you attached are wrong, right?
 
  • #11
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nvm - yeah I determined (after posting) that it would be incorrect due to the incorrect integration and the resulting on flow effects. It is quite frustrating when solutions are provided that are incorrect, particularly when I'm just trying to wrap my head around concepts/methods.
 
  • #12
nvn
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It is quite frustrating when solutions are provided that are incorrect ...
True. And you are correct regarding the resulting on-flow effects. The final answers in the attached files in posts 5 and 1 are completely wrong.
 
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