To calculate a hoist as in the attached image, how do you define the crane attachment point at the top and the motor attachment point at the bottom.
In Simulate you can not leave two parts free.
Blocking for example the point of the crane will give a false representation of the deformation.
Simulate can leave two parts free. (some degrees of freedom (d.o.f.) free)
IF you have applied all loads from a Free Body Diagram (FBD) and determinate Static case.
This means you are in static equilibrium.
Then turn on inertia relief check box at static Analysis Definition. (this resolves any numerical rounding errors that make the part move and not be static)
Probably best load type is BEARING Load. for hoist pin and mass pin (motor)
Another way may be cut in half and use planar symmetry since the hoist rig appears symmetric from top to bottom. Again inertia relief.
Another way instead of inertia relief for any case is to hold some area where stress is insignificant in the missing DOF. If FBD and loads are accurate then the stress will not be very much. There is a specific technique from PTC on constraining parts in static equilibrium. I could not find it in their knowledge base so it is attached here.
I found a related article at PTC support.
Article - CS59477
And here is the static equilibrium technique. It was redone for a thermal study but it is the same method.
Thank you for your model.
But in the case of my Cé, the fixation is in a circle, how can I put points there to fix the degrees of freedom as on your bridge?
I thank you for your very detailed explanations and your models.
On the other hand I did not understand the utility of the spring in the model test_d_new.
(Can not read the video)
I put a force measurement on the spring by choosing spring.
But for simulation the measurement does not appear in User defined?
Now it's my fault, I didn't orient the axes correctly for the spring.
That's why it didn't appear in Measure.
To have the Measurements in the calculations it is not necessary to use the Measure function to load them?.
It seems to me that the Measurements are taken automatically in the calculations?.
Thank you for this new model and explanations.
I'll look into it right now.
( I'm surprised, I no longer receive notifications of new messages by e-mail? but I haven't changed anything.
Every day I have to go see the discussions. )
Oops, I had the explanation right in front of me.
But unfortunately I don't read English. (neither speaks nor writes it).
Thanks for the highlight.
The principle of "spring" is still very obscure for me.
In your different examples it is placed in different places.
Can we find a document that explains the procedure to follow for using the "spring"?
In this case use a spring because constraint on the weighted link dependent node is not allowed.
The general spring replicates the desired constraint because it has stiffness settings for all 6 d.o.f.
Some of the ways springs are used:
To constrain a model so it is a static case. (eliminate d.o.f.)
To model some component stiffness as a simplification.
To model bearing/joint stiffness less than rigid. (skunks example is held at a bearing area)
To stabilize the solving for a component held only by contact friction. (use a low stiffness spring)
The option inertia relief also constrains any free d.o.f. And in the case of the unbalanced load mass of weight at the bottom joint vs mass of weight plus mass of lifting rig at the top joint, the inertia relief applies gravity to the model until the load is balanced and all d.o.f are held = static case. That is why I did not have to include gravity in the unbalanced model because inertia relief is a type of gravity load. Inertia relief is why no spring is needed for adding constraint on free d.o.f.
Inertia relief is a powerful tool because it allows the engineer to use a free body diagram with only loads applied and no constraints. Of course the free body diagram must be force and moment balanced as usual but there are small numerical errors that the inertia relief can "fix". In the case of any free body diagram with a gravity load, the gravity load is omitted when doing the FEA with inertia relief because the inertia relief will apply the gravity automatically to re-balance the forces.
Hello Skunks, Sweetpeahub.
Great, I thank you for all these explanations and drawings as well as for the time spent on the subject.
I will study all this information.