# How to apply Pinned end condition C channel column to find the critical buckling load?

• 198 Views
• Last Post 28 February 2019
• Topic Is Solved
usmanz posted this 05 February 2019

How to apply Pinned end condition C channel column to find the critical buckling load?

EDIT: rwoolhou - moving to Structural

Order By: Standard | Newest | Votes
peteroznewman posted this 06 February 2019

When people say Pinned, some are describing a single axis that has one rotational degree of freedom while other people mean a spherical joint that has multiple rotational axes. What do you mean by Pinned?  Is it pinned at both ends?

If you mean one axis of rotation, how is the C channel oriented to that axis?

In either case, what is the dimension between the center of the axis/point and the center of the C channel?  The dimension between the center of the C and the axis/point in two directions has a large effect on the buckling load. The center of the C is already off the neutral axis so has a lower buckling load than when the point/axis is on the neutral axis.

How straight is the column?  A small bow reduces the critical buckling load compared with a perfectly straight column.

• Liked by
usmanz posted this 08 February 2019

It's 1000mm in length and straight channel column with both ends pinned end and force of 1N. I was wondering how you can you apply these condtions in fea ansys and also how to find the theoretical values for the buckling load in both cases. I would like to know fixed to fixed ends, and pinned to pinned end. All I'm interested is finding the critical buckling load for the channel. Width 200mm, sides = 69mm both, and thickness is 1.5mm.

peteroznewman posted this 08 February 2019

Does pinned mean a rotational axis of freedom?  If so where is that axis?

In the figure below are five choices places you could put the axis of the pin.

Each location will result in a different buckling load.  Where do you want the pin?

Note that some people use the word pinned for what is basically a spherical joint. Is this what you mean?

If so, where is the pivot point of the sphere in the figure above?

• Liked by
usmanz posted this 08 February 2019

Hi, thanks for all this. Basically that's the channel column I intend to use and find critical buckling. I'm trying to apply similar condtions as the paper ("local-flexural interactive buckling of standard and optimised cold-formed steel columns ") which is online. The carried out expermential test to find the critical buckling and then used fea to to conduct the same test but I tired many things but I just can't seem to get the right pinned end condtions on the top and bottom end of the channel which will be d in the diagram you have drawn and also I just use 1N force for simplicity. Can you please help me and show me how to get same results as the authors in the paper using fea ansys. Thank you very much, I really appreciate it.

peteroznewman posted this 08 February 2019

In my figure above, I showed axis C going through the centroid of the cross sectional area, which is the neutral axis of the column. I thought you would choose that axis.

I looked at the paper and they show a FEA model with a shell model of the channel, which means a surface model of the geometry. The channel makes contact with an end plate, the endplate makes contact with a roller and the roller is fixed.

Can you construct the five parts shown above in either DesignModeler or SpaceClaim?

• Liked by
usmanz posted this 08 February 2019

yes I have made the five parts so the top block and the bottom block I used inventor to constrain them with the C channel column and then used the cylinders and constrain them to the blocks but they can rotate around but only fixed in position whereas I constrain the blocks fully to the channel column so its fixed in position. I used inventor and used the option to convert the file to step file and uploaded that into ansys. Is this the correct procedure Sir?

peteroznewman posted this 09 February 2019

You can shortcut the approach used in the paper and suppress the bottom cylindrical roller body. You will instead use the cylindrical face on the bottom plate.

However, on the top cylinder, you need to split the face using a horizontal plane through the center of the cylinder.

In Mechanical, create a Joint, it will be Body-Ground and the type is Revolute. Select the cylindrical face on the underside of the bottom plate. That is your first pinned connection. You also have to create Frictional Contact between the bottom of the C channel and the top of the bottom plate.

The top needs two joints. Create a Revolute Joint between the bottom face of the cylindrical rod and the cylindrical face in the top plate. Next select the face at the top half of the cylinder and create a joint between body and ground. Set the joint type to Translational. You will have to click the + sign to access the coordinate system that was created by the joint and edit that coordinate system to make the X-axis point along the column length. Insert a Joint Load and set the type to Displacement and set the number of mm you want. You also have to create Frictional Contact between the top of the C channel and the bottom of the top plate.

This is a system that you can solve, but it will be best to insert a Contact Tool under the Connections folder and Generate Initial Contact Information. Check to see that the two frictional contacts are closed.  Next, under Analysis Settings, set Auto TIme Stepping to On and set the Initial Substeps to 100, Minimum Substeps to 100 and make the Maximum Substeps to 200.  Then set Large Deformation to On and solve.

• Liked by
usmanz posted this 09 February 2019

Hi Sir, thank you so much for your help, is it possible if you could do a video on ansys?.

usmanz posted this 09 February 2019

Using ansys, I applied the force to both faces of the top cylinder and then also used displacement and selected the 2 faces again for the cylinder and set the y=constant and x and z are free, for the bottom roller I used remote displacement or something and I set all the x,y,z components to constant and then for the y rotation I used constant so it's fixed in position and leaving the x and z in free using the bottom 2 faces of the cylinder and that gave me 19kn as the critical buckling which I'm not it's not right as the paper has 32 or 42kn can't remember exactly. Do you know where I might be going wrong? Sir

peteroznewman posted this 09 February 2019

Did you put the roller axis at the half-way point of the short dimension of the C channel or did you put it at the centroid of the cross-sectional area?  If you did not use the centroid, that could explain why you would get less than the value in the paper.

• Liked by
usmanz posted this 09 February 2019

I'm not sure what you mean by placing the roller axis half-way point or at centriod, can you please show me example. Thank u

peteroznewman posted this 09 February 2019

In the figure below, there is a cross-sectional area shaded in light green. The centroid of that area is at the point at the intersection of lines A-C. The centroid is the neutral axis for beam bending.   The line D is half-way between the top and bottom of the C channel and is not on the neutral axis but is eccentric from the neutral axis.

If you study column buckling, you will learn that as the load is applied with larger values of eccentricity (the distance from the neutral axis to the loading axis), the buckling load will reduce.  The highest buckling load is with zero eccentricity, or the load passing through the neutral axis, which is at the centroid of the cross-sectional area.

usmanz posted this 11 February 2019

What is meant by fixed ended or pinned ended in channel columns, does it mean that both the ends are fixed or just one end and also for pinned is it both ends pinned or just one because so far in ansys work bench I tried to use fixed and pinned on both ends and it does not give me any results. How do I apply the force and the fixed or pinned conditions for normal c channel.

peteroznewman posted this 11 February 2019

[EDIT: CHANGED TO MATCH MODEL BELOW THAT HAS Z AXIS COLUMN]

When the paper says Fixed-Fixed end conditions, they don't mean that literally for both ends. You can use a Fixed Support for the bottom of the bottom block. On the Top block they just mean that the block moves without any rotation or sideways displacement. It moves in a straight line down.  You can do this by picking the top face of the top block and inserting a Remote Displacement. Then you set to zero five of the six degrees of freedom: x, y and three rotations. Then for the Z coordinate, leave it free. Promote the Remote Displacement to a Remote Point. Apply a Remote Force in the Z direction to the Remote Point for the Eigenvalue Bucking Solution.

For the pinned ends, I described how to support each end above. There is a simpler way to achieve the same outcome and that is with Remote Displacements. You pick the cylindrical face of the bottom block, and set zero on five of the six degrees of freedom, leaving Rotation about X Free.  On the top you do a similar thing but instead of zero for the Y displacement, you leave that Free. Promote the Remote Displacement to a Remote Point. Apply a Remote Force in the Z direction to the Remote Point for the Eigenvalue Bucking Solution.In this way, you don't need the rods at all, you just have blocks.

• Liked by
usmanz posted this 11 February 2019

I followed your fix to fix advice and used it on a shape without any of the blocks as this is a different paper. I have attached the pictures for the fix, remote, force , deformation and critical buckling. Sir is this correct way to carry out fix to fix ends buckling in ansys and also you will see that there is small arrow to where each thing was applied.

peteroznewman posted this 11 February 2019

I edited my post above, please reread it. Now it applies a force so you can do the Eigenvalue Buckling solution.  Applying a displacement is more useful to find the nonlinear buckling solution. I think you should keep the blocks at each end from the first paper. You will have to use Bonded Contact to do an Eigenvalue Buckling solution, I don't think it will let you have frictional contact.

• Liked by
usmanz posted this 11 February 2019

at the moment I'm doing linear buckling Sir, what is a Bonded Contact and how does it work Sir. Also does the block dimension matter or can I just make up the dimensions for the block.... will different dimension block affect the result? what is frictional contact and what do I need it for?. I am sort of new to ansys.

peteroznewman posted this 28 February 2019

When you marked the post above with Is Solution, the topic was labeled as Solved and is not looked at by most people because the Solved tag implies no further help is needed. You can ask the questions above in a New Discussion if you still need answers.