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"but that has no practical value in your question" ...

203.12.195.242

In Reply to: RE: Calling all mechanical engineers - lever problem ... :-)) posted by Inmate51 on October 05, 2014 at 14:02:40

I'm not an engineer ... but my first degree was in applied mathematics, Inmate51.

As such, my approach is to move from the simplest case ... to more complex cases. You asked what is my situation:

* the simplest case (of my situation) is ... I'm interested in the apportioning of the weight of a bearing & platter which is located somewhere inside a right-angled triangle, on 3 springs which are fixed at the apexes of said triangle.

The "3-pointed star" comes from there being 3 (imaginary) arms of the 'star' which connects the central bearing location to the 3 springs.

The desired result is to get the weight of the bearing & platter apportioned equally, so that the 3 springs are equally compressed.

Hence my statement to ignore the weight of the (imaginary) 'star'.

* the more complex case is then to account for the weights of 2 arms which are located just off-board of 2 of the springs.

Regards,

Andy

Follow Ups:

RE: "but that has no practical value in your question" ... - Inmate51 10:19:50 10/06/14 (0)
I sent your initial post to my son, who is a fifth year mechanical engineering student. Haven't got a solution yet - he's probably ciphering the problem. Or, with any luck, he's spending his time on classes.

:)
Right angled triangle - Bill the K 19:57:18 10/05/14 (6)
What are the dimensions of this triangle?

Call this triangle ABC.

Tell us the distances of the load from A, B and C.

3 equations will solve the problem.

Cheers
Bill
- Thank you Bill ... - andyr 00:22:38 10/06/14 (5)
  Here is a sketch of the setup:
  
  I thought it easier to draw the 3 lines connecting the weight to the position of the 3 springs (than the triangle). The weight (8.44 lbs) is located at the central cross.
  
  Regards,
  
  Andy
  - I have the answer - Bill the K 06:23:31 10/06/14 (4)
    andyr
    
    Let us call the support 7.5" away as A, 8" away as B and 10" away as C. The triangle ABC is right angled at B and has side AB 15.1", BC 8 and AC 16.9" in length.
    
    From my calculations I find Reaction at A as 4.3 lbs, at B as 2.61 lbs and at C as 1.53 lbs with the total load being 8.44 lbs.
    
    Had to refer back to old textbooks like Timoshenko's Plates and Shells and the two other books he wrote with Young and Goodyear. I had to do some derivation to get the values for the triangular shape. It all turned into a simple equation of the ratio of the distance of the load to the opposite side of the triangle to the distance of the support to the same side. Thus it is a matter of 3 equations. Interestingly the sum of these 3 ratios add up to one in any triangle.
    
    Thanks for the opportunity to exercise the old grey cells with some structural engineering.
    
    Regards
    Bill
    - Brilliant, Bill - thank you, thank you! :-)) ... - andyr 13:41:39 10/06/14 (0)
      I intuited that the answer lay in the concept you said ( the ratio of the distance of the load to the opposite side of the triangle to the distance of the support to the same side ) but I didn't know how to express it, mathematically.
      
      OK, so:
      * 4.30 @ A
      * 2.61 @ B, and
      * 1.53 @ C.
      
      That was the simple case ! :-)) For the real-life situation, we need to get more complicated ... and allow for the added affect of 2 arm weights near B and C!! :-))
      
      See my modified sketch:
      
      It seems to me that if the arms were placed at apex B (arm weight = 0.99 lb) and apex C (arm weight = 1.65 lb) then the weight distribution would simply become:
      * 4.30 @ A
      * 3.60 @ B, and
      * 3.18 @ C.
      
      But because the arms are cantilevered out from B & C, there is a negative action on the apex which is at the other end of the particular triangle side?
      
      So:
      * the 0.99 lb arm outboard from B reduces the weight at C, and
      * the 1.65 lb arm outboard from C reduces the weight at A?
      
      But how to calculate the resulting apex totals, with these cantilevered weights? :-))
      
      FYI, here is a pic of the 1st version of my 'SkeletaLinn' subchassis - which I completed in July, last year. This has the bearing positioned closer to A, causing an excessive compression of the spring at that apex (so I had to get a custom-made spring made up, which was 30% stiffer than the Linn springs).
      
      Hence the weight analysis for v2! :-))
      
      Thanks, Bill,
      
      Andy
    - No you don't have the answer - I do. - mgbpuff 08:28:23 10/06/14 (2)
      A + B + C MUST = 9 By definition I got 4.62# for A(7.5"); 2.81# for B(8"); and 1.57# for C (10")
      - Actually - mgbpuff 08:37:12 10/06/14 (1)
        
        In Reply to: RE: No you don't have the answer - I do. posted by mgbpuff on October 06, 2014 at 08:28:23
        
        The problem I solved is the one framed for Krieger below and I thought the weight was 9# but I see he changed that to 8.44 #. Therefore, if I ratio my solution from 9# to 8.44#, I get the same results you did. You're correct - my bad!
        
        Edits: 10/06/14 10/06/14 10/06/14
        
        RE: Actually - Bill the K 09:45:03 10/06/14 (0)
        
        In Reply to: RE: Actually posted by mgbpuff on October 06, 2014 at 08:37:12
        
        Hey thats wonderful. The answers match which means they are correct. Great.
        
        Cheers
        Bill

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