Spring constant in an elevator

  • Thread starter serene37
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  • #1
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hi,

i am required to find the spring constant of the spring(s) below a typical elevator, you know the big metal springs locater underneath the elevator when it lands on the ground floor right. i have to find the spring constant/stiffness of the springs. anyone knows how to go about doing it w/o tearing up the lift? i know i have to use an accelerometer... thanks for ur help.

serene:bugeye:
 

Answers and Replies

  • #2
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If you're allowed to measure the compression of the spring, then use the conservation of energy principle.
 
  • #3
Q_Goest
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Hi serene,
Without tearing up the lift? So I assume you're looking at an existing elevator installation? If so, I'd suggest simply measuring the spring dimensions and calculating the spring constant. I'd assume you're referring to a coil spring, in which case you only need the overall diameter, diameter of the wire used, and number of free turns.

Note that elevators generally have a hydraulic shock absorber of some sort to absorb impact if the elevator drops accidentally. They aren't springs. If they were simply springs, the elevator would be accelerated upward after impact.
 
  • #4
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I suppose that you need to find the spring constant to put in an elevator physics problem.
You must know the total kinetic energy that the spring must transform in potential energy. That is what is the kinetic energy [tex]{1 \over 2}mV^2[/tex] that the springs must accept before shortening too much. The kinetic energy depends on the mass of the cabin and the drop height.
The potential energy of a compressed or stretched spring is [tex]{1 \over 2}kL^2[/tex] where k is the spring constant (newtons/meter) and L is the variation in length of the spring.
This gives you the constant for a single spring. If there are several springs in parallel, the constant is the result obtained divided by the number of springs.

Note that elevators generally have a hydraulic shock absorber of some sort to absorb impact if the elevator drops accidentally. They aren't springs. If they were simply springs, the elevator would be accelerated upward after impact.
All elevators I have seen have springs and not shock absorbers. Have you ever looked inside an elevator duct?
 
Last edited:
  • #5
ppl.. d problem is simple..
suppose d maximum gforce a person can bear is g(f).
then maximum spring force
kX(m) = m(l) . ( g + g(f) )
where m(l) is mass of lift including mass inside it.X is maximum compression of spring. maximum force is applied at this point according to hooks law.

and energy equation is
1/2 m(l) V(sqr) = 1/2 k X(sqr)
this gives u X. and u know kX. So now u noe K.

V is the velocity d lift falls at. It cn be calculated from d hieght at which it is falling

m(l).g.h = 1/2 m.V(sqr).
 
  • #6
springs are used in some cases ... instead of hydraulic absorbers.
 

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