Rubber compression strength question

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SUMMARY

The discussion centers on calculating the impact force that a urethane rubber door stop can withstand when a 200 lbs metal door swings open. Using a 50-ton press, it was determined that 100 lbs of force compresses the rubber, equating to 18.75 ft-lbs of energy. The calculations suggest that at speeds of 10 ft/sec and 20 ft/sec, the door generates significant energy upon impact, necessitating a comparison with the rubber's compression resistance. The analysis emphasizes the importance of considering the door's moment of inertia and angular velocity in determining the impact force.

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  • Understanding of basic physics principles, particularly energy conservation and momentum.
  • Familiarity with the properties of urethane rubber and its compression characteristics.
  • Knowledge of moment of inertia and angular velocity concepts.
  • Experience with force and energy calculations in mechanical systems.
NEXT STEPS
  • Research "Energy conservation in rotating bodies" to understand the dynamics of hinged doors.
  • Study "Urethane rubber compression testing" to gain insights into material properties.
  • Explore "Calculating impact forces in mechanical systems" for practical applications.
  • Investigate "Safety factors in mechanical design" to determine appropriate margins for impact resistance.
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Mechanical engineers, product designers, and anyone involved in the design and testing of impact-resistant materials, particularly in door stop applications.

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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