Learning Forum

[avikr1525]

I am having difficulty adding constraints and choosing the type of contact in Static Structural.

My problem is that I have to do a crash simulation of Helicopter. The Helicopter weights 4000 Kg and at the skid, I want to place a crash test device. The Skid is 3m long as 20 cm wide. The SKid material is Steel and the crash device is Aluminium Alloy.(See attached diagram files)

The crash device is bolted between the two skids. When the Helicopter falls from a height of 80mm, I need to calculate the fatigue life.

1. How should I calculate the force? Should I consider here impact load.

2. Which boundary conditions need to be applied.

3. WHich contact type to be used?

[peteroznewman]

Please use the Insert Image button to put the image into the post as well as attaching the image files because ANSYS staff are not permitted to open attachments.

If you have a mass hanging from a thread, suspended 80 mm above a surface and you cut the thread, gravity will accelerate the mass. The velocity of the mass when it reaches the surface will be sqrt[2*g*h] or 1.25 m/s.

A helicopter is not generally in free fall, but it is reasonable to set an impact velocity limit for the device and your numbers suggest that limit is 1.25 m/s.

Most helicopters I have seen have two parallel skids, one on each side, so you are suggesting that this crash device is on each side of the helicopter.

Helicopters can take on a whole range of orientations while landing and it is reasonable to expect that it would touch-down on one end of one skid first. That first contact point then creates a reaction force that causes the helicopter to rotate about that first point until either the rest of that skid touches or to rotate about a different axis until the other skid touches and finally the two skids are on the landing pad. Therefore, another specification is the angle of tilt of the helicopter relative to the landing pad. Let's assume that the pilot will have the skid level within 200 mm, so over 3 m that means an angle of about 4 degrees.

It looks like you have used shell elements to mesh this geometry, which is a good way to go.

Start with a simple model, and add more detail as each simple model solves and gives you some information.

You should use a Transient Structural model.  Bring this geometry into SpaceClaim. Duplicate the crash device and make a parallel copy for the other side of the helicopter, maybe 2 m away to represent the width of the helicopter. Create a component called Pad. In that component, draw a rectangular surface that is a bit larger than 3 m x 2 m and is tangent to the bottom rail. Use the Move Tool on that component/surface. Reposition the triad of the Move tool to the vertex at the front outer corner of one bottom skid. Now rotate the surface 3 degrees about X and 3 degrees about Z (assuming Y is up). That is the landing pad and the first contact point.

In Mechanical, create a remote point scoped to the two top skid surfaces. It will show up in the middle. Edit the Coordinates of that remote point and raise it up by 2 m to where you estimate the center of mass of the helicopter is located. Make sure the behavior of that remote point is Rigid. Later, you might replace that with the actual frame of the landing gear and the helicopter frame.

Assign a Point Mass to the Remote Point. That is your 4000 kg. Later you could add some moments of inertia.

Create a Coordinate system called Ynorm and align the Y axis with the normal of the Pad surface. You can do that with Hit Point Normal.

Make the Pad a Rigid Surface. Create a Fixed Joint to hold that Pad surface to Ground.

Create Frictional Contact between the Pad surface and the two bottom skids. The Pad surface must be on the Target side because it is rigid. Set the Pinball Radius to 0.5 m so that the contact on the other skid is included. Insert a Contact Tool under the Connections folder and check that the contact status is Closed.

Under the Initial Velocity folder, assign the 1.25 m/s Initial Velocity to all the bodies except the Pad.

Add an Acceleration Load of -9.8 m/s^2 in the Y coordinate using the Coordinate System Ynorm. This is gravity pulling the helicopter to the Pad.

Set the End Time to 1 s, the Initial time increment to 1e-3  s, the Minimum time increment to 1e-5 and the maximum time increment to 1e-2.

You might need more than 1 s to see both skids land on the ground.  You might find the peak stress is in the first 0.1 s and there is no need to simulate the full landing.

You will probably get some convergence problems along the way. You can reduce the contact Normal Stiffness to help that to converge.

Good luck!

[avikr1525]

Thanks for your insights. I want to add on a few things which I got recently while discussing with my course coordinator. He told to do this problem only in Static Structural. Moreover, he gave a hint to use the Energy dissipated of the falling Helicopter as Input parameter. I think the Energy dissipated will be the Potential Energy(mgh). So, how to make use of Energy here. I am very new to this topic so I don't have much idea. I have successfully modeled it and now I am remaining to give input constraints.n

[peteroznewman]

It's very simple if you assume the helicopter is going to touch down perfectly flat so the entire length of both skids touches at the same time.nThe Potential Energy PE = mgh as you said.nThe Kinetic Energy KE = 1/2 mv^2 and if you equate that with the PE you get the Impact Velocity I gave above.nThe Strain Energy SE = 1/2 kx^2 so equate that with the PE to compute the peak deformation x after impact.nTo do that, you have to do a quick Static Structural model to compute k. Compress the skids by an arbitrary value of x, such as 2 mm, and request the reaction force F then compute k = F/x for the arbitrary value of x.nFinally, the peak deformation after impact is x = sqrt(2mgh/k).nReuse the Static Structural model from above, but change the value of x to the peak deformation you just calculated with the last formula and the solution will show the peak stress in the skids upon landing. That is the value you can use for Fatigue calculations.nn

[avikr1525]

Thanks a lot..I will try like this and update youn

[avikr1525]

Any specific reason, why are you using spring force equations for Force and Strain Energy here.nnAlso, I applied displacement in -y dirn and through Force probe got the values for Forces is x,y and z dirn. So, I used the resultant value here instead of y-dirn Force value. That is correct??nnn

[peteroznewman]

At a high level, a structure is a spring. If you push in the X direction with a force F, you will get a displacement x and so the linear elastic constant ratio of force to displacement is called k as in F = kx.nIf you apply a displacement in -y, take the y component of force to compute the spring constant k.n

[avikr1525]

Many Thanks for your reply!nI tried to find some more articles related to When and Why at high level a structure behaves as a spring but I couldn't found. Could you explain me or give some reference as I want to know more about it.n

[avikr1525]

I tried to find the value of k using F=kx. I gave displacement of 40mm in negative y dirn. My k value came as = 2801.8/0.04, which is 70045. After that I found the actual value of x using the strain energy equation x = sqrt(2mgh/k). Here I took m = 4000/2(because the helicopter has two skids so weight will be evenly distributed on each one)nin another try I gave displacement as 20mm in negative y dirn but in that case my F_y value was -2039.6N so k came as 101980.nWhy I am getting different k values?nI don't want the upper part of skid to penetrate inside the Cylindrical cone as shown as figure. The skid is welded to the cone so, I want the cone to be compressed by the skid. How can I do that?n

[peteroznewman]

The equation F=kx is only useful for linear elastic deformations. 40 mm is too large for linear elastic deformation. Try 4 mm instead.nWhen you use 4 mm, the cone will not penetrate the skid.n

[avikr1525]

Okay so, I tried with 4mm and then I also tried for 5 mm. For 4mm, k= 48000 and for 5mm its 40300. For elastic simulations k should be same, right?nShould I attache here the Archive file, so you could have a quick look.n

[peteroznewman]

Yes, the value of k is for the linear elastic part of the deformation.nWhen you create the workbench .wbpz project archive file, rename the extension to .7z and you can use the paperclip icon to attach it to your reply.nThe file size limit is 50 MB so you might need to clear the mesh, then save and archive to keep the file size small.n

[avikr1525]

Thanks again, I really appreciate your help. I am looking to complete this in next few days.nI have attached here the .7 file. When you open it you will find two Analysis, select the MID - Surface Analysis file name.nAlso, I have attached the Homework description. I will help to know about the material am using and also a few details which I have missed earlier.n

[peteroznewman]

UPDATE: you can now attach .wbpz files directly to replies, no need to put them inside a .7z file.nDESIGN TOUCH DOWNnA simple estimate is to do a purely linear model and obtain the spring rate k for one cone. To do this, change the material 6061-O to AL, which is a purely linear elastic material. Suppress the skids and use a Remote Displacement (Rigid) on the top of the cone and a Fixed Support on the flange of the cone. Set the Displacement to -1 mm. The reaction force for HALF a cone is 47,783 N so ONE FULL cone has a stiffness of 95,566 N/mm.nAt 1 mm deformation of the top, the stress was 4,184 MPa. To have infinite life, the stress must stay below 80 MPa, so the allowable deformation is 80/4184 = 0.019 mm. That is a ridiculously small value, but this is a homework exercise and not real life. In real life you need a much lower spring rate to get much lower stresses with much larger displacements. By the way, if you remesh with smaller elements, the stress goes up and you get an even smaller allowable, but let's not go there.nNow that you have the allowable peak deformation on a normal touchdown of 0.019 mm, and you have the potential energy of a 2500 kg mass dropping 30 mm, you can calculate the spring rate needed for the system to limit the peak deformation to 0.019 mm. Divide the calculated spring rate for the system by the spring rate for one cone and you have the number of cones needed to achieve infinite life.nEMERGENCY LANDINGnIn a separate analysis system, use the 6061-O material to include the work done plastically deforming one cone to crush it down to a small percentage of its original height. Divide the kinetic energy of the impact by the work done to crush one cone and you can compute the number of cones needed to absorb all the kinetic energy. There is one more requirement and that is to limit the peak deceleration to 10g. Plot the Force vs Displacement data for one cone during the crushing operation. The area under that curve is Work done on the cone. That work is subtracted from the kinetic energy of the helicopter mass, so you can then estimate the velocity vs displacement, which will change as the number of cones change. You can then differentiate the velocity to estimate the acceleration and compare that with the 10g requirement. I would let a solver do all that for me in an Explicit Dynamics model.n

[avikr1525]

Thank You !! I will try to do what you have said.nCould you send me the .wbpz file in which you worked? The Remote displacement thing scares me a lot. I am getting stress of 2084 MPa. Just want a check if you applied Remote Displacement(Rigid) and displacement on the top cap end of the Cone or also on the the adjacent curve.nSorry, I don't have that much experience with Ansys so maybe sometimes I ask for silly thingsnThis is pretty cool tool to do fatigue analysis, Prof. told to use this.n

[peteroznewman]

EXPLICIT DYNAMICS for EMERGENCY LANDINGnHere is 100 kg being crashed into a fixed top plate at 2.426 m/sn

[peteroznewman]

If you differentiate the velocity plot, you get the acceleration-time history. Notice that the peak deceleration exceeds the 10g specification.nI expect that if more mass was used on this one cone, then the peak deceleration would be reduced in magnitude. Once you have a solution that meets the 10g spec, divide the total mass of the helicopter by that mass and you will know the total number of cones needed.nThe data in this and the previous post took 66 minutes to solve on my 4-core laptop.nANSYS 2020 R1 archive attached.nn

[avikr1525]

Thanks!!nnFor the design touch down, I was calculating the number of Cones, I am getting a pretty weird value..its in order of 10000. The length of skid is only 2500 mm. So, practically the value should be 15-20 or maybe less than that. Have u calculated that?n

[peteroznewman]

I got the same order of magnitude for the number of cones needed to achieve infinite life. You can input my numbers to see if you calculate the same number I got. nIt's not a weird value, it simply tells you that this initial design does not meet the requirements.nEngineers often go through several concepts before they find one they want to optimize, then iterate on the parameters of the better concept until it meets all the requirements. As I said above, this cone design is a long way away from meeting the requirements.n