Rubber compression strength question

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Discussion Overview

The discussion revolves around determining the impact force strength of a urethane rubber door stop when subjected to the force of a swinging metal door. Participants explore the calculations involved in estimating the impact force based on the door's weight and speed, as well as the rubber's compression characteristics.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant describes the setup involving a 50t press and the force required to compress the rubber, raising a question about the impact force the rubber can withstand when a door swings open.
  • Another participant suggests that additional information, such as the speed of the door, is necessary for accurate calculations.
  • A participant calculates the distance the door travels when swung open and proposes using speeds of 10 ft/sec and 20 ft/sec for further analysis.
  • One participant emphasizes the need to treat the door as a rotating body, introducing concepts of moment of inertia and energy stored in the door, and provides a formula to equate the energy dissipated in the rubber stop with the energy of the door.
  • A follow-up question asks for the impact force in pounds produced by the door at the specified speeds and whether the rubber can handle this force without fully compressing.

Areas of Agreement / Disagreement

Participants express varying views on the calculations and considerations needed to assess the impact force. There is no consensus on the exact impact force values or the adequacy of the rubber's compression resistance against the door's force.

Contextual Notes

Participants highlight the importance of speed and the door's rotational dynamics in the calculations, indicating that assumptions about the door's motion and the rubber's properties may affect the results.

Who May Find This Useful

Individuals interested in material science, mechanical engineering, or those involved in product design and testing of impact-resistant materials may find this discussion relevant.

cliffd
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Need help in determining the impact force strength of a door stop rubber. It is a urethane rubber formed in a triangular shape. I used a 50t press with a 2" round shaft to compress rubber product and it required 100 lbs of force to compress. I believe this worked out to 18.75 ft lbs to compress rubber. My question is : if a 36" x 80" metal door weighing 200lbs is swung open, what type of impact force can this rubber take? (contact surface of the rubber is 2" x 3/4")
 
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I think more info might be required, such as how fast the door could be moving.
 
If my calculations are correct, the circumference of a 6 ft circle would be 28.26 ft. A 3 ft door would basically swing 90 degrees from the closed to the open postion where it would contact the door stop. If we divide the total circumference of a 6 ft circle by 4 we should get the distance traveled by the door from its closed to its open position. If it took 1 second to open the door, that would be approximately 7 ft/sec. It takes less than one second to open a door.Let's use 10 ft/sec and 20 ft/sec for calculating.
 
Because it's a hinged door you should really treat it as a rotating body so it has a moment of inertia and then energy stored in the door will be proportional to it's angular velocity rather than it's linear velocity... but you could approximate and do it the way you have..

Energy stored in the door

= 0.5 x mass x velocity2

Energy dissipated in the stop

= Force x distance

so equate these two..

Force x distance = 0.5 x mass x velocity2

Force = 0.5 x mass x velocity2/distance

"distance" is the amount the stop is compressed.

Probably need to add some sort of safety margin? Factor of two?
 
So what type of force in lbs would this door produce traveling at 10ft/sec or at 20ft/sec?
How would this compare with the compression resistance of the door stop rubber? Would the door stop rubber be able to handle the impact force created by the door at either speed without completely compressing together (front portion of rubber stop would need to travel approximately 1 1/4" to contact rear portion portion of rubber stop)
 

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