Undergrad Calculating force when landing on a beam

[some bloke]

TL;DR

I need to work out the equations to determine the load capacity needed for a beam to catch a falling weight.

I am designing a machine at the moment, and I am struggling to get my head around impact forces. I have established that the force applied is mgh/d, but I am struggling to work out the value for d.

Taking it in isolation, assuming that the falling weight does not "give" in any significant way when it lands on the end of the beam, then the value for "d" is going to be the amount of flex that the beam experiences, but this in turn is related to the force, which is related to the distance, and I'm going in circles.

I want, ideally, as little "d" as possible - a solid landing, as it were, akin to the weight landing on a concrete pavement. But, if I plug "d" in as 0, we get infinite force, so I'm coming up stumped again.

Current values I'm using to try and work this out are:

m=125kg
g=9.81ms-2
h=1m
Beam can be treated as cylindrical, 0.5m long and made of steel, fixed rigidly at one end and the weight is caught a the other, after falling 1m.

Beam diameter is going to depend on the force (which depends on the distance, which depends on the flex, which depends on the beam diameter ow my head)

The premise is similar to someone jumping on a diving board.

Please can someone help me to tie this together!

 

[Chestermiller]

Are you interested in determining the force as a. function of time?

 

[jrmichler]

A diagram is always helpful. This shows what I THINK you are asking:

If this diagram is correct, the procedure is as follows:

1) Calculate the spring constant of the cantilever beam from the beam deflection equation. The spring constant is force divided by deflection, with units of lbs/inch or N/m.

2) Calculate the kinetic energy of the falling mass at the point where it first contacts the beam. The kinetic energy is equal to the potential energy at its starting point.

3) The beam is a spring. The potential energy in a spring is 0.5Kx^2, where K is the spring constant and x is the deflection. When the mass contacts the spring, kinetic energy is converted to potential energy in the spring. When the velocity of the mass reaches zero, it has zero kinetic energy, and the spring has maximum potential energy. The spring potential energy is now equal to the peak mass kinetic energy.

4) Calculate the spring deflection at the point of maximum potential energy from the spring potential energy equation.

5) From the spring deflection and spring constant, calculate the force.

6) From the force, calculate spring stress. If the stress is above the yield point, the calculations get much more complex because they involve plastic deformation.

7) Stand well back because that spring will now convert its potential energy into mass kinetic energy and fling the mass back up with the velocity at which it came down. That's if the spring did not yield.

8) If this spring will be impacted more than a few times, do a fatigue life calculation.

 

[sophiecentaur]

The above is pretty well a comprehensive description of how to do it. I could add that the maximum force will depend on the 'spring constant' of the beam. A stiff beam will 'carry' the same energy as a flexible beam but the force on the stiff beam will be more (could be a lot more) than the force on the flexible beam. This is pretty obvious, of course but, if you are not yet committed to a particular beam, you could choose an optimum, to protect the falling object from too much force. There are tables which will give you all the information about strength and flexing for many different cross sections of beam and materials. I used that sort of information, years ago, when planning the loads on steel joists for a loft extension. I remember that searching was not a hard problem.

 

[jbriggs444]

I would add to this that you could treat the collision with the beam as inelastic. Instead of needing to store the entire kinetic energy of the faller as potential energy associated with the beam deflection, you only need to store the kinetic energy of the beam tip/faller assembly as determined by momentum conservation.

 

[sophiecentaur]

A damper would surely be a good idea, to avoid the energy going elsewhere and doing damage. The OP could look into car suspension systems, perhaps.

 

[jbriggs444]

assuming that the falling weight does not "give" in any significant way

I missed this part on the first pass. It sounds as if there is an assumption that if there is no "give" that there is no dissipation of energy. However, such is not the case. Energy dissipation (elasticity or coefficient of restitution) is a separate parameter from "give" (stiffness). You can idealize a collision to infinite stiffness without idealizing it to perfect elasticity.