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Product Design - Real World Physics Problem

  1. Dec 9, 2009 #1
    Hi. I am hoping that someone can explain, in the language of physics, the solution to a problem encountered during a real-world product design. I have already solved the problem. I am simply looking for why the solution worked in terms of the language of physics.

    Please see the attached PDF to view the assembled product.

    I have a product that has 3 components -- (1) a bracket that is vertically affixed to a wall such that it is perfectly level (2) a 3/8 diameter solid steel rod that is bent at each end with the end that is inserted into the bracket bent downward at a 93 degree angle and the opposite end bent upwards at a 90 degree angle and (3) a container that is placed atop the up-turned end of the arm that weighs approx. 8 pounds (when it is filled). The rod can simply be lifted out of the wall bracket and the container can simply be lifted off of the up-turned end of the arm (i.e. there are no additional parts or mechanisms securing either the rod or container).

    Previously, the down-turned end of the rod (which fits into the bracket) was bent at a 90 degree angle, but we had to hyper-extend that angle to 93 degrees in order to keep the rod level (parallel to the floor) when the 8 pound container was placed atop it. In other words, when both bends were 90 degrees, the weight of the container caused the rod to slope downward.

    Can someone explain in terms of physics (additional vertical force needed to counter-balance the container, etc.) why this is the case. Can someone actually give me an equation that shows why 93 degrees seemed to work? Or can you show me that it really should have been 92 or 94 degrees, etc.?

    As a math geek (but in statistics instead of physics), I sincerely want to understand the physics behind this. I am sure this is elementary for all of you, but I truly appreciate it. If additional measurements of any of the components are needed, I can provide them.
     

    Attached Files:

  2. jcsd
  3. Dec 10, 2009 #2
    This is an engineering problem. No solid is perfectly rigid, so loads cause distortions. I'd look up some of the easy formulas, but I wasted all my patience getting registered. J.E. Gordon has written some of the best books ever written on any subject to painlessly explain structures.
     
  4. Dec 10, 2009 #3

    minger

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    Check out this page:
    http://www.advancepipeliner.com/Resources/Others/Beams/Beam_Deflection_Formulae.pdf

    Basically you have a cantilevered beam. It will deflect according to:
    [tex]
    \begin{equation}
    \begin{split}
    \delta_{max} &= \frac{Pl^3}{3EI} \\
    \theta_{max} &= \frac{Pl^2}{2EI}
    \end{split}
    \end{equation}
    [/tex]
    Where delta is the maximum deflection and theta is the maximum angle at the free end. In this case, P is the applied single point load at the free end, l is the total length, E is the Modulus of Elasticity (property of the material), and I which is the area moment of inertia.
     
  5. Dec 10, 2009 #4
    well when the rod was in the wall was it seated so that there was no spaces between the rod and the fixture. if there was minor spaces in that fixture i could see how it would change angles once a weight is applied.
     
  6. Dec 10, 2009 #5

    FredGarvin

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    One of the things we need to know is the length of the rod.

    Just using very rough cantilever beam equations, here is a quick chart for estimating the required load to deflect the end of the beam the required amount to bring it level with the wall end, i.e. L*Sin(3°).

    This assumes a lot. I assumed a steel rod and constant cross section as well as neglecting shear terms as well as angular affects on the load at the end of the rod and, most likely, some large deflection issues with yielding. Hopefully your rod is about 24" long.
     

    Attached Files:

  7. Dec 10, 2009 #6
    Thank you for your reply. The rod is 11.5 inches long (including up and down-turned ends). The down-turned end is 1.88 inches long and the up-turned end is 1.37 inches long.
     
  8. Dec 10, 2009 #7
    There is a very small amount of space between the rod and the bracket (i.e. it is not held completely rigid by the bracket). This is so that the rod may be swiveled from side to side while inside the bracket (as indicated on the drawing).
     
  9. Dec 10, 2009 #8

    Mech_Engineer

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    Fred, how did you calculate your second moment of area? Yours is almost a factor of 20 lower than what I got (I calculated I = 0.016in^4).
     
  10. Dec 10, 2009 #9

    minger

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    You should also consider a "loose" fit of the bar into the bracket. How tight is it exaclty, and how much does it deflect simply due to the fit? It may be very low if the tolerances are small enough, but it may be the driving factor.
     
  11. Dec 10, 2009 #10

    Mech_Engineer

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    To achieve a 3 degree deflection over the legnth of the bar, it looks to me like your box weighs over 550 pounds. [STRIKE]The stresses are high for a normal steel as well, it's likely you're yielding the material near the bend radius at the mounting point for the bar.[/STRIKE] [STRIKE]This is a large deflection beam bending problem, but for a quick approximation the beam bending formulas can at least give you an idea of what you're up against.[/STRIKE]

    EDIT- I see that you say in your orignial post the box only weighs in at 8 pounds. This tells me that the 3 degrees of deflection you are seeing is due to a loose fit in the mount at the wall, NOT due to beam bending. An 8 pound load on the bar you're describing should only deflect it .139" at it's end, or about .7 degrees; on the other hand if the hole the bar is mounted in is .1" larger, that would account for the 3 degrees.
     
    Last edited: Dec 10, 2009
  12. Dec 10, 2009 #11

    FredGarvin

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    [tex]I=\frac{\pi d^4}{64}[/tex] I thought I saw a rod of 3/8". Let me go back and double check.
     
  13. Dec 10, 2009 #12
    ha loose fit thats exactly what i was talking about.
     
  14. Dec 10, 2009 #13

    Mech_Engineer

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    For a circular cross-section:

    [tex]I_{xy}=\frac{\pi R^{4}}{4}[/tex]

    EDIT- Gah, that damn factor of 2 and radius vs. diameter! It'll get you every time! I was off by 2^4=16.
     
    Last edited: Dec 10, 2009
  15. Dec 10, 2009 #14
    Thank you for taking the time to address this. Given the measurements above (and the .1" loose fit), is there a simple equation that would point to the 3 degree adjustment?
     
  16. Dec 10, 2009 #15

    Mech_Engineer

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    What size hole is the bar being mounted in?

    1.88in*sin(3deg)=.0983in
     
  17. Dec 13, 2009 #16
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