X Force Keygen Inventor Professional 2014 Keygen
Edelira Longinotti
Having some problems with my Dynamic Simulation. I'm in a prestudy phase of an project and trying to quickly check how big cylinder needed to lift an lever arm. I'm trying to find the unknown force in the cylinder.
I'm designing a hinged cover for something right now that will use a pair of gas extension springs for assisted opening. I'm currently working to figure out how much force it will take to open the cover using different values of spring - it's available in 50, 100, 150, 200, etc., lb versions.
I've tried running simulations using a force applied to the moving end of the gas spring, and I've trued running simulations using a spring joint applied between the fixed and and the moving end of the gas spring, and I'm getting dramatically different results between the two. For instance, using a 200 lb force shows me that I need about 25 extra lbs of force applied to the handle of the cover to open it, where when I use that same 200 lbs on a spring joint, I need 50-75 lbs to keep the cover closed, because it will open on its own.
I have been using unknown force, actually. I've been setting up all the joints and / or forces, then using unknown force to find the the force needed to move the cover at any given point on a 90 degree rotation of the cover. What I'm trying to do is figure out what force to use on the gas springs to bring the force needed to start the cover opening to right around 15-20 lbs.
I didn't include the unknown force stuff - I didn't actually set up any simulations on the assembly that I attached earlier. Gravity would, of course, be defined as "down" relative to the model. I used one of the frame corner edges to define it.
This is how I set up the unknown force operation. I'm looking to find out how much force it will take on the pick hole (in this case, in the real part it's actually a handle) to open the cover, when using gas springs that have a certain pre-set force.
Actually you don't need to have either of these in the assembly to calculate how much force would be required to open the cover unless the weigh enough that their weight needs to be considered in the calculations.
Now, I've tried to set up the unknown force a couple different ways. I've used pre-set external load forces on the hydraulic parts along the long axis of the gas spring, with the unknown force on the pick hole, acting on the revolution joint between the cover and frame. I've set force joints, both jack and spring, between the pivot points of the (deleted) gas spring, with the same unknown force on the pick hole. I've set a constant force on the pick hole, with the unknown force being a jack between the spring pivot points.
As far as the different results, I just ran through all the possibe rigs again, and I'm getting roughly the same results using a force between the two halves of the gas spring as I do using a force joint of, of equal power, set up as a jack between the two pivot points. It appears to be when I use a spring (rather than a jack) that the issues happen. Since I'm getting almost identical results between the other two methods, I'm guessing those are the right answers, and the spring joint is causing issues because spring power is expressed in lbs/inch, rather than lbforce.
This puts the preload on the spring. The amount of force will be the (free length - minimum length)*the stiffness lbs/inch rating for the spring force. So: if Stiffness is 10lb/inch and the change of spring length (compression) is 6in the force is 60lbf for the spring.
I have a simple shaft with a radial load applied to one end and a fixed constraint on the opposite end. When I run the analysis on it, I am confused about what it reports for the "Reaction Force Magnitude". Since I only have one radial load (500 lbs in the Y-direction), I would expect the reaction force to be equal and opposite to this, but it is not even close.
I think you need to specify the exact direction of the load. I got the same result as yours if no direction is specify. Though it looks like there is only one direction, it is actually around the cylinder, so the upper portion shrink in size. If I specify the direction of the force, the result is similar to NASTRAN. Please share the file, if the result is still different.
I was trying to see if my spring would react as expected if I were to pull that other part, and I figured out that adding a force to said part was the thing to do. So I went to "dynamic simulation" but the "force" button is grayed out. Can someone help me with this ? Sorry if this is kind of a dumb question, I'm a bit new to Inventor.
This is an intro to Autodesk Inventors stress analysis feature. When making parts and assemblies it is extremely important to know where the critical stress and fail points are in your design. The stress analysis feature removes the guesswork and over engineering in your design.
The two things needed to start this process is a Inventor Part or Assembly file, and some estimated or actual calculated input forces for your part.
Let's start by opening your Inventor part file.
To start a new stress analysis go over to ENVIRONMENTS tab on your ribbon, click on it and on the left side of your screen you will see the stress analysis feature (rainbow colored cube). Click on the icon and then click on create simulation. That will bring up a screen of initial settings. You can choose a static analysis or modal analysis.
The simple definition of static would be your main input force which is not affected by time/temp/atmospheric pressure. Modal is dynamic forces (vibration) that will have secondary effects on your part.
For this trial we are going to use a static stress analysis. Select it and click OK.
Defining what type of material your part is determines the amount and types of forces it can handle before it fails. This should be the first step before you continue on with your test. Under your ribbon you will have a materials section with an icon that says ASSIGN, click on this. It will bring up a pop up window displaying your part material. If you already defined what the material was when you made your part, that material will be displayed. If you haven't yet defined it, click on materials and select the desired one. When you select a material you can also double click on it to see the preset settings for that type if you need to verify or change them.
Once you are happy with your material and settings click OK and then we will be ready to define our constraints and input forces.
Defining constraints for your part is important in order for your analysis to perform correctly. If I have input forces pushing up on my part, but nothing constrained or held in place to hold the part down, there would be no stress to display. In this trial I am going to used a fixed constraint the center of my part, which is the middle red cylinder (simulation of steering stem shaft). This means that this area of my part will remain in place while the input forces are affecting my part. Use the constraints tools in the ribbon to define them on your part.
Now that this is complete we can move on to input forces.
The input forces are what are going to actually be exerted on your part. These are defined using the LOADS tools in your ribbon. I am going to add two vertical forces on each of the outer red cylinders (simulated fork tubes). These loads will be both perpendicular to the top and bottom faces of the clamp for this simulation. Once I have defined where the load will be exerted the next step is to define the amount of force (magnitude) in Newtons. For this I am going to use 4448 newtons which is approximately 1000 lbs of force.
Now that our constraints and input loads are set, we can run our simulation.
Click on the simulate tab in your ribbon (rainbow colored cube) and the simulation window will pop up. If you have any errors they will be displayed here. If there are no errors, click on RUN.
Inventor will run the simulation and show you the different stress types and amounts using the color coded visualization chart. Anything in RED is bad and changed to your design should be implemented.
This instructable was just an intro to the stress analysis feature, but there are almost limitless settings allowing you to get the most accurate data possible. You can import forces from the dynamic simulation feature, see the stress loads in different factors, as well as get a full report under the report icon in your ribbon. Take some time and familiarize yourself with the aspects of this feature and ensure that your part will do the job intended for it.
Come check out the Autodesk software at Techshop and start your stress analysis today!
Here is a video of how to do that. Remember though that FORCE is a magnitude and values will always be positive and without direction. You'll have to look at the component forces to calculate the direction yourself.
I am not sure I understand your question about applying a force on the cylinder. Currently you have defined a forced displacement, if you want a applied force instead of an applied displacement, then:
2. As you can see, in order to close this door under the external force 30 kN and self-weight, this cylinder's force is more than 244 kN (driving force) -> it means that with 75kN, this cylinder cannot drive the door under this external force --> this statement is correct or not?
It is a bit confusing, because, at the first step, this cylinder force is at least 244kN to close the door, but now, with 75kN this cylinder still operate this door? How to make this point more clear?
2) Well, you see, the issue with the applied motion and then the force required is that the load initially is in the direction of rotation. So the load is assisting the motion and the applied displacement is actually countering it (the results output is a magnitude and not directional). That is why when you use the force instead it moves very quickly.
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