July 4, 2019 at 9:55 amshantashreejena97Subscriber
I have a pneumatic artificial muscle set up in which a vertical chain and sprocket assembly attached in a shaft. Both the end of the chain assembly is attached with the hooks. Can anyone let me know about the boundary conditions of the system that I should take to perform the modal analysis?
July 4, 2019 at 10:28 ampeteroznewmanSubscriber
Please insert some images so we can better advise you.
July 4, 2019 at 11:51 am
July 4, 2019 at 3:04 pmpeteroznewmanSubscriber
Thank you for the image. I would like to know what the hooks are attached to above the top edge of the image. If you can show more, that would be helpful.
Is there a chain tensioning mechanism? If so please describe it. I believe the lowest frequencies in this assembly are going to be the motion of the chain links. The tension in the chain will have a large effect on the frequencies. If there is a chain tensioning mechanism, then you must have a Static Structural analyisis to Pre-tension the chain, and you link the Solution cell of Static Structural to the Setup cell of the Modal analysis in Workbench.
Modal analysis is all about stiffness and mass. It looks like the structure has a base that is fixed to ground, perhaps at four bolt holes that are not visible in the image. Use a Fixed Support at the bolt holes. If there is a large mass at the other end of the structure or the hooks, that mass must be represented in the model. Again, what is above the hooks? A significant amount of mass I would imagine.
Create a Revolute Joint between the shaft and the bearing to leave the rotational degree of freedom of the shaft free.
Delete the chain solid body and replace it with a Line body in DesignModeler, or a Beam in SpaceClaim. Now you can mesh the Line body/Beam with Link elements.
Each line body connects to the sprocket at the tangent point with a spherical joint and to a hook using a spherical joint. What supports the hook, I don't know.
If you don't care about the vibration of the chain, just mesh the line bodies with one element each by using a Mesh Size control to force that to happen.
July 5, 2019 at 7:02 amshantashreejena97Subscriber
Thank you for your reply.
Below is the figure of my whole set up. One end of the hook is connected to the spring(here I have taken cylindrical structure instead of the spring which is shown in the left-hand side) and the other end is connected to the muscle with GI Wire hook at its two sides(shown in the right-hand side). A weight of 500g is attached to the shaft with the help of a rod. I want to know at certain compressive force(acted on both the faces) in the muscle how much angle will be the weight lifted? Can you help me regarding the boundary conditions to solve this?
July 5, 2019 at 7:07 am
July 5, 2019 at 7:40 amshantashreejena97Subscriber
As you said above I have deleted the chain assemblies. As I have never Sketched a line body from the solid body can you let me know the entire procedure to draw that line body by showing a demo of it? Also, I have doubt on how could I mate that line body with the teeth of the sprocket?
Below is the figure of my setup after the chains are deleted.
July 5, 2019 at 11:32 ampeteroznewmanSubscriber
Thank you for the detailed explanation and additional images, that is very helpful.
A simpler approach to replacing the chain is to add two springs, one on each side of the sprocket tangent point, the other end to each hook. A spring can be scoped to a cylindrical face on a sprocket tooth, and the other end to a face of the hook. You can move around the coordinates after scoping to get the chain to be in a straight line parallel to the Y axis if you want. The chain has an axial spring rate of some kN/mm per meter of length. You might need to request that data from the chain manufacturer, or measure it in the lab. The downside of using a spring instead of link elements to represent the chain is the spring has no mass, which affects a Modal analysis slightly.
Now the real mechanism has a low rate spring (compared with the chain), on the other side of the hook. Replace the cylindrical structure that represents the real spring with a Spring connection, and insert the proper spring rate. This is a spring that might have a pretension applied that pulls on the muscle through the chain and sprocket. You can apply a pretension to this spring.
Modal analysis doesn't include loads, so applying forces is not allowed in a Modal analysis. You would use a Static Structural model to apply loads if you are interested in the deformation of deformable parts. Or you might use a Rigid Dynamics model to apply loads if you are interested in the motion of rigid parts and the forces acting through them. Let's say you are in one of those.
Instead of applying a force to the weight, insert a Gravity load into your model. But make sure that the properties of the weight show that it has the correct mass. If not, you can add a point mass at the center of the face to make up the difference if it is too light, or you can create a custom material and adjust the density until the mass is correct. The second technique must be used if the initial mass is too heavy.
If you want to plot muscle tension vs. shaft angle, that is most easily done in a Rigid Dynamics model. Let's make a very simple model that you can get some similar results from easily. Make a copy of the file you are working in so you can go back to the original file later. In Workbench, RMB on cell 1 Modal and Replace with... Rigid Dynamics. Open the geometry in SpaceClaim and create a new component and call it Arm. Move all the solids such as the weight, the rod, the disk, the shaft and the sprocket into that component called Arm. That is one rigid body. You can pick all the other solids in the Structure and set to Ignore for Physics so that they don't transfer.
Now open the Model cell. In Mechanical, insert a Revolute joint to ground on the shaft body. Insert a Standard Earth Gravity load and check that it is pointing in the correct direction. Add a Joint load to the Revolute joint: type: Displacement and Magnitude, Function, =10*time. Under Analysis settings, if you want to move 180 degrees, enter 18 seconds for the End time. Solve the model. Into the Solution, put a Total Deformation result and you can animate to see the Arm rotate. Add a Probe, Joint, Total Moment to the Solution branch. Now you can plot Torque vs time. Add a Probe, Joint, Relative Rotation. Now you can plot Torque vs Angle. Of course you could have obtained the same plot using Excel and a few simple formulas.
I understand this is not exactly what you want in the end, but you have to get some experience with successfully building, solving and obtaining results from a model first. Let me know if you have success building this simple Rigid Dynamics model. You can get the same plot from Static Structural, it would just take a lot more time to solve the model.
July 5, 2019 at 1:06 pmshantashreejena97Subscriber
Thanks for the detailed explanation, Peter. I will definitely go through it. But I have another doubt. To the bolted joints should I give bolt pretension load? and if yes then for the tightening torque of 12Nm how much will be the preload for a bolt length of 3.5cm(head length.5cm and shank 3cm) for a standard M5 bolt?
July 5, 2019 at 2:38 pmpeteroznewmanSubscriber
If you are doing Rigid Dynamics, a Bolt Pretension load is irrelevant as you are using rigid bodies.
In Static Structural, you might represent the bolt by a simple beam that is scoped to an imprinted face where the head would have touched on one end, and the cylindrical face where the threads would hold on the other end. You can pretension this beam if you want. You must also put frictional contact between the parts.
For an M5 bolt you must also say what the material is. Only a Class 12.9 material can take a 12 Nm torque, which is very high at 100% of Proof Stress. The threads have to be in a strong material like 1018 steel and have 15 mm of thread engagement. A more reasonable torque is 8.9 Nm which is at 70% of Proof Strength of a Class 12.9 M5 and requires only 10.6 mm of thread engagement in a 1018 steel. The clamp force is 2,166 lbf or 9,635 N for an 8.9 Nm torque.
July 5, 2019 at 3:33 pmshantashreejena97Subscriber
The bolt, I am using, is an M5 Allen bolt made up of AISI 316L Stainless steel. Is the Tightening torque and preload force same as that of 1018 steel?
July 5, 2019 at 5:58 pmpeteroznewmanSubscriber
An M5 bolt made of AISI 316L steel reaches 100% of Proof Strength with only 4.5 Nm of torque.
To tighten to 70% of Proof Strength, use a tightening torque of 3.1 +/- 0.2 Nm of torque.
There are two failures: (1) breaking the bolt by exceeding its proof strength, and (2) stripping the thread out of the other part that had the tapped hole. If the other part is Aluminum, then a low torque will strip the threads. If the other part is 1018 steel, it will take a much higher torque to strip the threads. This is a separate consideration from the bolt material.
July 5, 2019 at 6:09 pmshantashreejena97SubscriberActually all the bolts in the setup is attached to the beams made up of aluminium alloy. Can you tell me what exact value of preload should I take for the bolt pretension?
July 5, 2019 at 6:15 pmpeteroznewmanSubscriber
- Aluminum comes is a wide variety of strengths.
- M5 316SS Class 70 Fastener Torque for Aluminum holes:
- 6061-T6 is 3.1 Nm and the minimum thread engagement is 3 mm but for
- 5052-H32, it is the same torque but the minimum thread engagement is 7.3 mm.
July 5, 2019 at 6:26 pmshantashreejena97SubscriberThe aluminium alloy I have used is of 6063 T5 type. Now what will be its preload for the M5 Allen bolt of 3.5cm total length(shank+head)?
July 5, 2019 at 6:36 pmpeteroznewmanSubscriber
The torque is still 3.1 Nm The different material shear strengths change the minimum thread engagement as you can see for the two materials I showed above. My torque spreadsheet doesn't have 6063 T5. You can look up the shear strengths for the three materials and interpolate from the data I gave you.
July 5, 2019 at 6:40 pmshantashreejena97SubscriberActually I don't know how to calculate preload force from the tightening torque. Can you tell me how to calculate the preload force from torque or the value of force from the above data?
July 5, 2019 at 6:56 pmpeteroznewmanSubscriber
T = K x D x F where T = torque, K = nut factor, D = bolt diameter, F = bolt tension.
We are getting off-topic from the Title of this discussion and it is getting rather long to scroll down to the bottom. How about you close this discussion by marking one of my posts above with Is Solution? Start a New Discussion to ask more questions about your model. You can open that new discussion with a link to this first discussion. You can also open a separate New Discussion to ask more questions about fasteners.
July 5, 2019 at 7:30 pmshantashreejena97SubscriberThank you so much for all your help.
July 7, 2019 at 7:53 amshantashreejena97Subscriber
According to you I have given revolute joint between the shaft and bearing, replace spring in place of chain and real spring and assign gravity, added fixed joint to the bolt hole of the base plate, gave pretension to all the bolts and give two equal and opposite force to both side face of muscle and also force(i.e load of weight) at the weight hung below. but its showing error(shown below in figure). I could not recognize my fault. Can you help me to solve it?
July 7, 2019 at 6:30 pmpeteroznewmanSubscriber
Suppress all the Bolts and change the Frictional contact where the plates were being squeezed by the bolts to Bonded contact now that the Bolts are gone. Try to solve that problem first.
If that doesn't work, attach your .wbpz file to a New Discussion and I will take a look.
July 8, 2019 at 3:38 amshantashreejena97SubscriberActually all my contact regions are bonded type except in between shaft and bearing region. Should I change any other parts to any other type contact?
July 8, 2019 at 10:40 ampeteroznewmanSubscriber
The idea of bolts and pretension is to squeeze together a frictional contact pair. If that contact pair is bonded, then there is no need for the bolts.
Do what the Error message says to do. Under the Solution Branch of the Outline is the Solution Information folder. Click on that. The default result is that the Solution Output text listing appears where the graphics window was before. Click in that text and hit Ctrl-F to open the find dialog and type error. Read what it says. Post the few lines of text that describe the error in your reply.
On the Details window for the Solution Information folder is a line that say Newton-Raphson Residual and the default is 0, type a 5 there
Independent of that, under Analysis Settings, make sure that Large Deflection is turned on, Auto Time Stepping is turned on, set the Initial Substeps to 10.
Now try to solve again.
July 8, 2019 at 3:07 pmshantashreejena97Subscriber
Some kind of error occurred in the shaft region. I have attached the error dialogue box below. Kindly tell me what exactly I have to do to eliminate this. I have also attached analysis settings details and force table(which acts on both the top and bottom surface of the muscle and is compressive in nature) for your further reference.
July 9, 2019 at 10:46 ampeteroznewmanSubscriber
At what time between 0 - 2500 s did that error occur?
If it occurred late, you might need smaller and better shaped elements. Click on the Solution Information folder and on the line that says Newton-Raphson Residuals, type 5 instead of the default 0. Then click Solve. You will get 5 plots under the Solution Information folder that shows where the elements are having difficulty obtaining equilibrium. That is where you need better/smaller elements.
If the error occurred early, perhaps your constraints are in conflict and need to be reviewed for compatibility.
I'm concerned by the "force table (which acts on both the top and bottom surface of the muscle and is compressive in nature)". Please show in detail where those forces are applied. Are you familiar with a Free Body Diagram? Take the Muscle Body out of the system and apply the force it generates to the rest of the system. There is an upward force on the chain. That moves the sprocket which lifts the weight on the end of the rod. That is similar to the model I suggested above that you build to calculate the torque on the shaft to lift the weight. Did you build that model?
There is a downward force on the frame. Do you care about deflection of the frame?
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