Statics Springs Design Problem

In summary, the problem involves determining the stiffness and unstretched length of a spring used in a plate storage system for a restaurant's self-serve salad bar. The plates are always kept protruding above the tabletop by 60mm and have a mass of 0.509 kg each. The support A also has a mass of 0.509 kg. The spring can be compressed by a maximum of 40% of its initial unstretched length before its coils begin to touch. The question also asks for the number of plates that can be stored. Using the equation for spring force and summing forces in the y direction, the stiffness is found to be 0.33 N/mm and the maximum number of plates that can be stored
  • #1
jklops686
40
0
Statics Springs "Design Problem"

Homework Statement



Here's a problem I've been very determined to figure out, but I can't. I attached the picture. This single problem is 5% of my grade, so I'd appreciate any help!

A plate storage system for a self-serve salad bar in a restaurant is shown. As plates are added to or withdrawn from the stack, the spring force and stiffness are such that the plates always protrude above the tabletop by about 60 mm. If each plate has 0.509 kg mass, and if the support A also has 0.509 kg mass, determine the stiffness k and unstretched length L0 of the spring. Assume the spring can be compressed by a maximum of 40% of its initial unstretched length before its coils begin to touch. Also specify the number of plates that can be stored. Assume the system has guides or other mechanisms so the support A is always horizontal.

Homework Equations



Spring force=stiffness*(L-Lo) L=final elongation and Lo=initial elongation

The Attempt at a Solution


There's 3 parts to the spring equation and I can only get one by a random number of plates (since you can't tell how many exactly their are in the picture) and taking their weight force. By summing the forces in the y direction i get Force of spring-mg=0, so Fs=mg. That's what I tried and got me...nowhere. I'm guessing I should find the stiffness first since it's listed first in the problem statement, but I'm not sure how to, or what to use for elongation.
 

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  • #2
I haven't solved it but I'd try writing two equations based on the one you cited:

1) An equation with just one plate.
2) An equation with nPlates where n is unknown but causes the spring to be compressed by 40%

See if that's enough equations to solve for the unknowns. If not then pick another number of plates and write another equation?

Something like that anyway :-)

Basically for each plate you add you know the change in the displacement and the change in mass. You might be able plot a graph and extrapolate to zero plates?
 
  • #3
CWatters said:
I haven't solved it but I'd try writing two equations based on the one you cited:

1) An equation with just one plate.
2) An equation with nPlates where n is unknown but causes the spring to be compressed by 40%

See if that's enough equations to solve for the unknowns. If not then pick another number of plates and write another equation?

Something like that anyway :-)

Basically for each plate you add you know the change in the displacement and the change in mass. You might be able plot a graph and extrapolate to zero plates?

Thanks I know what you mean. So the spring goes down 15mm every time a plate is added. But do you think the unstretched length of the spring is 660mm (so then when there's 4 plates on top protruding 60mm it will be level with the table top)? Because, that's what I assumed, so i took 40% of 660mm and got like 396mm. So that would me the max compressed. then i took 396/15 and got max of 24 plates. So i added up the weight of the plates to get spring force and solved for stiffness 247=-k(396-660) so k≈0.4 N/mm.

Could this be right!? I really hope so. I think it's similar to what you told me to do with 2 equations, but i just used one with the max spring lengths and trays assuming the spring unstretched is about 660mm.

Thanks
 
  • #4
jklops686 said:
So the spring goes down 15mm every time a plate is added.
Yes, and you know the weight of each plate, so that tells you k.
when there's 4 plates on top protruding 60mm it will be level with the table top)?
Not quite. You're overlooking that the bottom plate effectively occupies 30mm height. So what does that make the platform height with no plates?
 
  • #5
haruspex said:
Yes, and you know the weight of each plate, so that tells you k.

Not quite. You're overlooking that the bottom plate effectively occupies 30mm height. So what does that make the platform height with no plates?

Okay k stiffness changes after each plate is added then?

And okay would I add 15mm to compensate for the first plate height making it 375mm unstretched length?
 
  • #6
jklops686 said:
Okay k stiffness changes after each plate is added then?
No, the stiffness (modulus) is a constant.
And okay would I add 15mm to compensate for the first plate height making it 375mm unstretched length?
The OP is wrong in referring to 'unstretched' length. The fully relaxed state of the spring would be its uncompressed length, and that must exceed 600mm. And that's not simply the length with no plates - the platform A also weighs as much as a plate. So it's like the length with -1 plates.
 
  • #7
haruspex said:
No, the stiffness (modulus) is a constant.

The OP is wrong in referring to 'unstretched' length. The fully relaxed state of the spring would be its uncompressed length, and that must exceed 600mm. And that's not simply the length with no plates - the platform A also weighs as much as a plate. So it's like the length with -1 plates.

Okay because when I was adding one plate on and subtracting 15mm for length and then solving for k several times, I was getting different answers. This must be because I had the wrong uncompressed length?
 
  • #8
jklops686 said:
Okay because when I was adding one plate on and subtracting 15mm for length and then solving for k several times, I was getting different answers. This must be because I had the wrong uncompressed length?
Possibly, but getting k is much simpler than that - just take the weight of one plate and divide by the change in length of the spring its addition causes: ΔF/Δx.
 
  • #9
haruspex said:
Possibly, but getting k is much simpler than that - just take the weight of one plate and divide by the change in length of the spring its addition causes: ΔF/Δx.

k=0.33 Everything makes sense now. Thanks a lot
 

Related to Statics Springs Design Problem

1. How do you determine the stiffness of a spring?

The stiffness of a spring is determined by its spring constant, which is the force required to stretch the spring by a certain distance. This constant can be calculated by dividing the applied force by the displacement of the spring.

2. What factors affect the design of a static spring?

The design of a static spring is affected by several factors including the desired load capacity, the material used, the spring's geometry, and the environment in which it will be used. These factors must be carefully considered in order to design an effective and safe spring.

3. How do you choose the right material for a static spring?

The right material for a static spring depends on the application it will be used for. Factors to consider include the required strength and stiffness, the operating temperature, and the corrosive environment. Common materials for static springs include steel, titanium, and various alloys.

4. Can static springs be designed for non-linear behavior?

Yes, static springs can be designed to exhibit non-linear behavior, meaning the force required to displace the spring changes as it is stretched. Non-linear springs are often used in applications where a constant force is required, such as in a valve or brake system.

5. How do you test the performance of a static spring?

The performance of a static spring can be tested by subjecting it to a known load and measuring the displacement and the applied force. This data can then be compared to the predicted behavior based on the design calculations. The spring should also be tested under various environmental conditions to ensure it can withstand the intended use.

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