Hi all
My project is to build a motorized wine cellar hatch.
However, simulation experts will be able to tell me, is it possible with solidworks to calculate what will be the effort needed to lift the hatch?
The information is:
Hatch width: 900mm
Hatch weight: 130kg including glazing
The cylinder fixing pins are not yet in place and are therefore free.
The hinges are on the length of the hatch, so the opening to be lifted is 900mm
Hello
There is no need to do a simulation in the RDM sense to solve this problem.
The approach consists of first simulating on a kinematic and static diagram (with Meca3D) as can be proposed @m.blt
Not having Méca3D, we can use the attached excel sheet which allows you to test several scenarios, perhaps in a more interactive way. I used it for an inspection hatch for an in-ground pool room.
I don't know if it struck you during your design but there is a trap in proceeding with a jack under the door that will probably not be a gas spring given the efforts. Indeed, if the jack breaks down, you don't have access to the cellar anymore and if you are in it and alone it's more annoying.
In this case I recommend that the cylinder rod is not attached to the chassis but that it fits into a shoe. Thus, in the event of a problem with the cylinder and by lifting the hatch slightly, the cylinder releases itself under its own weight.
Kind regards
Motion (part of solidworks premium) allows you to do this kind of calculation
Let's use the most suitable tools...
Hello@benoit_3
If it is a question of making a static study on a simple mechanism such as the hatch lifting system, our old memories of graphic statics can apply. Three forces acting on the trapdoor, they are concurrent and their sum is zero.
A simple (!) sketch in SolidWorks, and you're done. With the possibility of "animating" the mechanism to know the forces in the successive positions of the opening, by moving with the mouse the end axis of the "Hatch angle" dimension.
The results are the forces of the cylinder and the bearing on the hatch, as well as the length of the cylinder.
The only difficulty is to set up the skeleton sketch so that the diagram does not fall apart.
A criterion for the choice of the arrangement of the connections: to make a compact and "internal" assembly to the frame of the hatch. Disadvantages: the forces are greater than for other configurations, the cylinder(s) work by pulling, etc...
And let's keep Motion or Meca3D for more complex cases.
The SolidWorks (2018) file is attached, with no guarantee from its author...
M.Blt
PS: problem following the verification of the operation...
When loading the assembly, SolidWorks loses the coincidence relationship of the origin of the "weight" vector with the center of gravity of the hatch.
To re-establish the link, simply edit Sketch1 of the Dynamic part, and reconstruct the geometry (
or Ctrl-B). This makes the sketch leave and re-establish the lost coincidence.
A little hassle that deserves a question about Lynkoa...
Hello
In the example I give, the power required is lower as well as the sizing of the parts which is less important.
However, it all depends on the space you have under the chassis.
If no space then the solution proposed by @ m.blt is better suited.
If there is a staircase and there is room on the side for one or two jacks then the solution I propose does not require modifying to strengthen the model as proposed by @benoit_3 in his first image and adapting it to the larger forces.
It should be noted that the good system proposed by @M.Blt does not allow the safety problem to be taken into account in the event of a power failure. A low-power electric actuator costs less. Just like the small lever takes up space on the back of the chassis which means making a housing in the wall if there is a wall ;-)
But finally @ m.blt answers the initial question (with solidworks to calculate what will be the effort needed to lift the hatch).
Kind regards
Thank you for your answers.
I tested both solutions, the first, the one from zozo mp.
I arrive at a result of 1400N
For the second one, from M blt, I like the force return system of the cylinder, but since I am going to work with a central cylinder, I have to leave room for the descent on the stairs so it is not applicable in my situation
In conclusion, I ordered a 200kg cylinder so that he wouldn't panic at each effort.
I am always impressed by your responsiveness. Thank you all
Your two answers solved my problem but I can only choose one.
So I'm going to choose which one I'm going to implement.
Have a good day and thank you again