Dears,

i am a beginner in ANSYS and need help in simulating Stub column test in Workbech. How to provide a axial load to column that is vertically pinned at both ends?

Thanks in advace

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- Last Post 30 April 2018

vineesh karthikeyan
posted this
20 February 2018
- Last edited 20 February 2018

Dears,

i am a beginner in ANSYS and need help in simulating Stub column test in Workbech. How to provide a axial load to column that is vertically pinned at both ends?

Thanks in advace

peteroznewman
posted this
20 February 2018
- Last edited 20 February 2018

First draw the column geometry in SpaceClaim or import a solid model.

The pinned end is easily done in Mechanical with a Remote Displacement.

If Y is up, then set x,y,z and Ry to zero at the bottom point.

Set x and z to zero at the top point but set y to -10 mm or some value to load the column.

You have to turn on Large Deformation under Analysis Settings.

Are you expecting the failure to be a buckling type failure?

vineesh karthikeyan
posted this
27 February 2018

Thank you sir. im expecting a global buckling failure similar to the second picture. The result i get is shown in picture

peteroznewman
posted this
27 February 2018
- Last edited 28 February 2018

How did you constrain the top of the column?

Did you introduce any imperfections into the geometry to induce the buckling?

vineesh karthikeyan
posted this
01 March 2018

Sir, support conditions i provided are free at top and fixed at bottom and length of the specimen is half of original length so that it is equivalent to pined at both ends according to Euler's buckling load theory. The geometrical imperfections is corresponding the the lowest energy mode ( here it is mode1) from the eigen value buckling analysis. The above result is mode1 from linear analysis. Actually im doing a non linear analysis for this specimen but im not getting an expected buckling shape ( i.e buckling at the middle hieght ). so, i cant proceed to non-linear analysis.

Thanks in advace

vineesh karthikeyan
posted this
01 March 2018

If large deflection is turned ON, I get the message - linear buckling analysis cannot be done

peteroznewman
posted this
01 March 2018
- Last edited 01 March 2018

You can simulate half the original length, but there is a Symmetry BC that you have to apply to do that correctly. Leaving the edges free is incorrect. You can use the Symmety feature in DesignModeler to create those Symmetry BCs automatically for you, or you can manually add them. For a shell model, assuming the column axis is along the Y axis, add two Supports: a Displacement, Y=0 and Rotation, Rx=0 and Rz=0.

Fixing the bottom edge is not simulating a pinned end, it is simulating a fixed end. To simulate a pinned end, use a Remote Displacement on those bottom edges. Set the behavior to Rigid. Set X,Y,Z to zero and set Ry = 0. Leaving Rx and Rz free creates the pinned end.

Yes, you cannot perform a linear buckling analysis with Large Deflection on. But with that on, you can perform a Static Structural analysis and track the force displacement curve to the point of zero slope, which is the nonlinear critical buckling load.

peteroznewman
posted this
04 March 2018

You should take a look at this discussion that has some useful information.

vineesh karthikeyan
posted this
11 April 2018

Thanks sir.

vineesh karthikeyan
posted this
11 April 2018
- Last edited 11 April 2018

Sir, i have done axial compression test in stainless steel (grade 304) rectangular hollow section of dimension 75 x 25mm, thickness 1.02mm and length of specimen 640mm. The axial compression was was applied in Universal testing machine (UTM). I haven't given any particular constraints in the ends where i just kept on the specimen vertically in the UTM and axial compression was applied. There was a local buckling at the ends and the failure occurred in the local buckling at mid-height. so in order to simulate this nonlinear buckling what type of end condition should i provide? And which mode should to actually transferred using finite element modeler so that same buckled shaped can be obtained in nonlinear analysis.

Thanks in advance

peteroznewman
posted this
11 April 2018

It sounds like the flat ends of the column were placed on a flat plate at the top and bottom of the UTM. A Fixed Support on the bottom and a Remote Displacement with Behavior set to Rigid, (not Deformable). You can set all the degrees of freedom to zero except for the compression distance. This should prevent the local buckling at the end.

You can modify the geometry with a very small deviation at the location where you want the local buckling failure to occur mid-span. Tweak Face is a useful tool in SpaceClaim to create this local buckled shape if you are using shell elements built on a surface model. Make sure you have sufficiently small elements to allow local buckling to occur. You can add a single long curve (Mode 1 in a Fixed/Fixed Modal Analysis) instead of the Tweak Face.

vineesh karthikeyan
posted this
12 April 2018

Thanks sir, how can i transfer the 2nd buckling mode to new static structural analysis using finite element modeler in ANSYS.

vineesh karthikeyan
posted this
12 April 2018

Sir, the material model i need to use for the analysis is based on two stage Ramberg-Osgood equation. The stress-strain equation is based on 2 stage Ramberg-osgood equation. So, how can i modify the stress-strain curve in the Engineering data to 2stage Ramberg-Osgood equation? Also I need to incorporate two parameters proof stress (e) and strain-hardening exponent (n) into the engineering data.

peteroznewman
posted this
12 April 2018
- Last edited 12 April 2018

Explicit Dynamics supports this material:

Is this what you mean? Materials with a strain-rate parameter will exhibit that behavior in a dynamic analysis. You are performing a static analysis of a nonlinear buckling model, so the strain rate would be zero. Is that correct?

For Static Structural models, I have used a plasticity model that has kinematic hardening. Either Bilinear or Multilinear.

Or there is this...

Go to the Online Help (if you are on 18.2) and read the Mechanical APDL, Material Reference.

vineesh karthikeyan
posted this
23 April 2018

Thank you sir

vineesh karthikeyan
posted this
23 April 2018

sir, how to obtain the failure load in the non-linear buckling analysis i.e by the large deflection is set to ON. The specimen is stainless steel hollow section (rectangular, square &circular cross-section). i have obtained a load-deflection curve in the static analysis with large deflection ON even though it didnt converge. i have applied 100kN and at 69.15kN the solution stopped to converge and the bi-linear isotropic hardening was set ton in engineering data.

In the second case, i have ran another non-linear analysis with large deflection set ON for the same specimen with same boundary condition such that the bi-linear is set to OFF and the safety factor tool (theory- Mohr-coloumb stress) is used. The safety factor min is plotted against force reaction and the load corresponding to safety factor =1 have been taken as the failure load. So, is this method correct?

maurya
posted this
23 April 2018
- Last edited 23 April 2018

Hello Peter sir,

Vineesh told "Sir, support conditions i provided are free at top and fixed at bottom and length of the specimen is half of original length so that it is equivalent to pined at both ends according to Euler's buckling load theory. The geometrical imperfections is corresponding the the lowest energy mode ( here it is mode1) from the eigen value buckling analysis."

But i have doubt that we cannot modeled it as symmetry because boundary conditions are different.

i am attaching picture with explanation as well according to formula fixed- free length is 2times to pin pin just for comparing the columns.

"Fixing the bottom edge is not simulating a pinned end, it is simulating a fixed end. To simulate a pinned end, use a Remote Displacement on those bottom edges. * Set the behavior to Rigid.* Set X,Y,Z to zero and set Ry = 0. Leaving Rx and Rz free creates the pinned end."

other thing i want to know the **Set the behavior to Rigid **

**which condition decides when to use rigid or deformable.**

thankyou

maurya
posted this
23 April 2018
- Last edited 23 April 2018

hello Vineesh

instead of applying force(mentioned above) you can apply one newton force and then does static structure solution and then import the data to eigenvalue buckling analyses. in buckling analyses in analyses setting set the no. of modes.

Solve it > after solution find total deformation with that you will get a table which describe the load multiplier (it is value which you can apply without failure).

i am attaching one ansys 18.1 file

vineesh karthikeyan
posted this
25 April 2018

Thank you sir, but the theoretical load obtained from the eigen value analysis is at the bifurcation point ryt? so the load obtained from the eigenvalue analysis should be greater than the failure load obtained from the non-linear analysis. please correct me if im wrong. Thanks inadvance!

peteroznewman
posted this
25 April 2018
- Last edited 25 April 2018

@maurya, the symmetry plane is halfway along the column where a Mode 1 buckling shape would have symmetry below and above the plane that slices the model into an upper and lower half. Do you agree?

In general, I would not recommend using symmetry in a buckling problem unless you are sure you only want a symmetric solution and you are on a Student license and can't solve the problem without using symmetry.

peteroznewman
posted this
25 April 2018

I usually want rigid behavior on the entities selected for a remote displacement.

An example of when I want deformable behavior is when a hollow section like a pipe has an end load that deforms the shape of the cross-section in the middle. If I cut the pipe in half and use a remote force on the cut face (or edge for shell elements) of the pipe, I would use deformable because a rigid behavior would impose an unnatural cross-section at the cut boundary.

maurya
posted this
25 April 2018
- Last edited 25 April 2018

hello sir

YES , that's correct point i actually think that symmetry was because of its equivalent length is twice of pinned thats why i asked.

thankyou

maurya
posted this
30 April 2018
- Last edited 30 April 2018

hello VIneesh

yes critical load obtained is at bifurcation point.

linear buckling is for:

1) Relatively inexpensive analyses.

2) Mode shape can be used for initial geometric imperfection for non linear.

According to me in linear analyses the buckling load is greater but it is the point of instability so difference is due to type of imperfection so it is not very much.

Deepak

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