Argument: stiffness of divided spring + angle of evaluation

1. Mar 18, 2012

springsvt

This is not a HW question, but a question that I need help solving. I am having an argument with a co-worker about the solution.

There is a 4ft sq sheet of plywood with an 18" wide rubber band stretched across it. To measure the spring constant (stiffness) of the rubber band a 6" diameter circular metal plate is slipped under the center of the rubber band and pull out taking force readings at 1", 2", 3", 4" and so on from the face of the plywood.

An assumption is that the rubber band is fixed at the edges of the plywood and does not slip off the plate during the evaluation. Additionally, the rubber band is being evaluated within its linear range, before the yield stress (just go with it)

1) Does the spring constant change when (essentially) dividing the spring in half as described above? (i don’t think so because when you cut a metal spring in half the spring constant of the sides does not change)

2) In calculating the stiffness of the rubber band, does the angle of evaluation matter? i.e., does the measured force require a vector adjustment depending on the angle between the rubber band and the plywood? (i don’t think so) he says that:
where
F= force
S = stiffness
D= distance pulled
A= angle of evaluation
2 = number of springs
F=2*s*d*sin (a) meaning that the s changes depending on the distance pulled from the plywood
I say that
F=2*s*d meaning that the s is constant and does not change when the distance pulled from the plywood (good ole’ Hooke)

3) does the fact that the plate is round/does not cover the entire width of the rubber band have anything to do with the evaluated stiffness? (I don’t think it does, as long as the volume of material evaluated does not change during the test)

Your answers will help sway him but if I can get a text references I would really win, so references/key search words would be great.

2. Mar 18, 2012

tal444

For 1), I know for sure that spring constant DOES change if you cut a spring in half. K will be doubled, in other words, you need twice the amount of force to stretch the spring the same distance.

3. Mar 18, 2012

springsvt

yes i understand that. if you cut a spring in half it will take 2x the amount of force to stretch the same distance. that is a simple manipulation of Hooke. However it does not really help me in the situation described above.

and if it does, i don't understand how.

4. Mar 19, 2012

Staff: Mentor

You should be able to analyze the situation if you start with an appropriate drawing. Here's one that might help:

I think you'll find that the effective spring constant (dF/dD) of the setup will change as D is increased.

Why don't you have a go at writing an equation for the force F in the diagram? You'll probably want to take into account the initial length (Lo) and tension (FT = Fo) in the elastic due to its being stretched across the plywood.

Attached Files:

• Fig2.gif
File size:
3 KB
Views:
178
5. Mar 20, 2012

springsvt

from your reply i am guessing that you agrees with my co-worker that

F=2*s*ΔL*sin (θ)

where L' can be found using the law of cos. and the θ can be found using trig (S.A.S.)

i have found a couple of others that agree with this statement and understand the problem now from a statics view point (i think).

however, in the case of the problem at hand, the numbers do not work if the sin θ is used in the equation and work just fine if it is not included. I am looking for a reason why that would be the case. when a spring is evaluated at a angle do the universal coordinate systems shift? meaning that x,y become x',y' where x' lies along the plane of the spring?

6. Mar 20, 2012

Staff: Mentor

I don't agree entirely I think that the force equation will end up being slightly more complicated than either of your choices.
Fortunately, physics results are not subject to popular vote The scenario can be analyzed to find a definitive result.
Well certainly hooks law is going to apply, by definition, along the length of the elastics, with ΔL being the change in length of the "spring" element. The geometry of the setup will dictate how the resulting force gets translated into the horizontal direction for any given distance D and thus angle θ.

7. Mar 20, 2012

springsvt

ok, so going back to your original drawing/post it looks like i may have been using the wrong θ. If i use the θ between F & Ft the results come more in line. But i still don't understand how the eq. gets more complicated.

unless... you are talking about how the rubber band applies different forces to the plate depending on the deformation of the rubber band during the evaluation process. if that is the case i would require a FEA analysis and we are not going there. i am looking to keep this problem simple.

thanks again

8. Mar 20, 2012

Staff: Mentor

Understanding comes from doing Have you attempted to analyze the setup mathematically? It's not particularly difficult.
No, I assume ideal components. But I make sure to include the "realistic" assumption that the band was stretched to fit on the plywood so that it has an initial tension even when lying flat against the plywood.

9. Mar 20, 2012

springsvt

well, to be honest, what little math background i have is failing me. i don't understand what you mean when you say "Have you attempted to analyze the setup mathematically?" i mistakenly thought that is what we were discussing.

Can you at give me a starting point? and maybe i can figure it out from there....

thanks again.

10. Mar 20, 2012

Staff: Mentor

You need to start by finding an expression for the tension in the elastic segments with respect to the current length of the segment. Using Hook's law that would be

$F = F_o + k_s(L - L_o)$

with Fo being the initial tension (elastic lying flat against the plywood), Lo being the initial length of the segment, and ks being the spring constant.

After that, relate L to the parameters D and/or θ as you wish via geometry. Finally find the horizontal component of F.

11. Mar 20, 2012

springsvt

right, in my original post i did not step that far back into the linear form and i prob should have. Yes, i understand how that works.

And by using the alternate θ you solved the problem that i was having.

Thank you for your help and may you enjoy a wonderful motorcycle ride today. i wishing i was playing with my 76 xs650 instead of sitting here doing this.