I have a small problem with the sizing of the structure and I would like your enlightened opinions before making a return. I have to make a " table " (basically a reinforced plate with 4 legs) to support the fall of a mass of 1.1T from a height of 10 meters. My problem is that I can get a value of the Kinetic Energy at the moment of impact, in joules, but Solidworks asks me for a value in Newtons. I understood that it took a real test to detect the deformation and bring out the value in N, but this is impossible for me to achieve. Is there a " sleight of hand " that already frames an approximate value of the force to be applied? Thank you TABLE V2.SLDPRT (283.1 KB)
Hello Yes, that's what I had seen, but if I isolate the N's, I still have to find the distance m, which must be the deformed for me (J/m=N). What I didn't have... or indeed I am going to look too far...
It's a problem of depreciation. The distance m depends precisely on the flexibility of your table, and therefore on the deformation you will choose to accept.
It's a fast dynamic calculation. There is specialized software for this (like those used for car crashtests). The impact behaviour of materials can be quite different from that of a quasi-static tensile test.
Hello At the end of a 10m fall, the kinetic energy of the mass is of the order of 108e+3 J, energy that must be dissipated between the moment of contact and the immobilization of the whole. The question is: What are the elements that contribute to this dissipation?
We can imagine that the " table " is a steel structure, with relatively stiff elastic behavior.
But nothing is specified about the mass of the fall. If it is a ball of pizza dough (a large one, I grant you), it will have the ability to absorb most of the kinetic energy through its very important elasto-visco-plastic deformation. It is a " soft " shock. If it is a Twingo (or a Fiat 500) dropped from the 3rd floor of a building, the deformation of the car will probably be less significant than that of the pizza dough, and the table will be more heavily used. If it is a steel ball, it is close to a " hard " shock. The top and the ball will have to share the dissipation, with much higher forces than in the previous cases, because of their respective significant stiffness.
A simple dynamic model can be developed, based on a contact interface consisting of springs and dampers. The difficulty lies in giving values to their stiffness and damping factors.
We would have to know more about the nature and structure of the table and the moving mass, to try to estimate these coefficients... Kind regards.
It seems that it is possible to perform a simulation of your problem directly on solidworks. Only available from a professional simulation license at least.
Hello, we have the Premium and pro simulation, but the only thing they offer is a simulation of the object falling itself on a floor that is not analyzed. What interests me is the fall of an object on my object of analysis.
Here it's about seeing how my structure reacts to the impact of a one-ton pebble, not how my structure reacts when it falls. As usual, I'm asked to check a structure that has already been built in a hurry... I doubt that it doesn't work at all but I still want to be able to provide a reliable result, and even know how to do it for my professional culture... I put the TQC part in PJ in my first post
Hello I put the coin in PJ in my 1st post. Regarding the mass that impacts, it is iron ore. It can be considered that it is sufficiently agglomerated not to deform on impact...
In itself, it is just the relative movement that is reversed. I don't know how to set up this kind of study on SW, but to make the table fall with a sufficient speed for it to have an equivalent kinetic energy?
Hello Do you have more information about the form... of your object that will collide with your already calculated table? I made a simplified shape to have the mass of one ton, I have a section of 0.26 m², the height compatible I get it with the density. S235 base for your table and for the projectile.
The values you provided: a mass of 1.1 tons (1100 kg) and a fall height of 10 meters.
First, let's calculate the velocity at the moment of impact using the formula for the conservation of mechanical energy:
m * g * h = (1/2) * m * v^2
where m is the mass (1100 kg), g is the acceleration due to gravity (9.8 m/s^2) and h is the fall height (10 meters).
1100 kg * 9.8 m/s^2 * 10 m = (1/2) * 1100 kg * v^2
107800 J = 550 v^2
Solving this equation, we can find the value of the velocity v:
v^2 = 107800 J / 550 kg v^2 ≈ 196.18 m^2/s^2 v ≈ √196.18 m/s v ≈ 14 m/s
I agree with the joules of the previous answer. And at first glance, it's the feet that are the first problem...
thank you
