×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Are you an
Engineering professional?
Join Eng-Tips Forums!
• Talk With Other Members
• Be Notified Of Responses
• Keyword Search
Favorite Forums
• Automated Signatures
• Best Of All, It's Free!

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

#### Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

# Natural frequency of clamped-hinged column

## Natural frequency of clamped-hinged column

(OP)
I have a column clamped at the bottom and for the top, I use hinged BC but free at the Y axis. I need to calculate its natural frequency at the 1st, 2nd, and 3rd mode shapes. I did calculate it with the FEA program but I need to calculate it manually too. I found in a paper the natural freq. equation as fn=k^2/(2*pi()*L^2)*sqrt(EI/m). It said that k is a dependent parameter on the boundary condition. The paper doesn't have the same BC as my problem.
I tried to use the k value for clamped-clamped (assuming this is the closes to my problem's k value ). So, I calculate it and I got another problem. Another problem is the paper uses kNm2, m, and kg/mm units. When I use the same unit it has a very big gap with the FEA results, like 95 Hz (FEA) and 0.003 Hz (manually). BUT, when I use Nmm2, mm, kg/mm unit the gap is still big but it's become 95 Hz (FEA) and 76 Hz (manually). This is understandable because I used the clamped-clamped k value.
So my questions are:
1. Does someone have the list of k parameter values for 1st, 2nd, & 3rd mode shapes with clamped-hinged BC for me to use for that equation?
2. Does anyone have another natural frequency of clamped-hinged column equation for me to use? Because I'm not confident with that equation (the units problem).

### RE: Natural frequency of clamped-hinged column

If you calculate the clamped-clamped condition you should get a higher natural frequency than the FEA result and not a lower one.

There are a lot of ways to calculate the natural frequency. Here's one based on the maximum deflection because of it's own weight.
n = k*946*SQRT(1/f)
with n = natural frequency in rpm
f = maximal deflection because of it's own weight using correct boundary conditions but the beam in a horizontal position.
I don't know the k value for your boundary conditions but it's 1 if both ends are simply supported and 1.3 if it's clamped on both sides. The correct k value should be between these 2.

### RE: Natural frequency of clamped-hinged column

Just out of curiosity, are you able to supply the length, section properties, weight and material modulus? People may be able to compare results using different methods, and suggest approaches.

### RE: Natural frequency of clamped-hinged column

(OP)
I have figured out the unit error. I made an Excel, I attached it below here. Maybe someone is curious and it could help someone in the future.

I found the source of the equation. The equation is by Stokey. The parameter k value I asked for was not effective length value (k). It was kl value from Stokey. I called it k before because it said so in the paper I read. I've found a pdf that contains the source of the equation and list for the kl value. This is the link to Stokey's document:
https://perso.univ-rennes1.fr/lalaonirina.rakotoma...
The kl value I was looking for is in Table 7.3 column (E).

Actually, I was comparing my natural frequency FEA results and the result from the equation (Stokey eq.). When I calculate natural freq. with clamped-hinged kl value, it doesn't match with the FEA results. So, I don't think clamped-hinged is correct when describing my model. I got 24% errors for every mode shapes when comparing it with my FEA results.

I think the problem here is the boundary condition of my FEA model.

Let's talk about how I modeled my column's BC in FEA. My FEA model has ENCASTRE (clamped) BC at the bottom column and YASYMM at the top. This makes my column is not clamped-clamped nor clamped-hinged but clamped YASYMM. YASYMM is a boundary with U1=U3=UR2=0. Why did I choose YASYMM for the top BC? Because I want to make my column can deflect only on the y-axis. U2 is the Y-axis. So when U1=U3=UR2=0, U2 is free and my column will only deflect on Y-axis. Am I right? Please correct me if I'm wrong. Also, I don't really get the UR1, UR2, & UR3 tbh because when I check the box it or not it doesn't make any difference with the natural frequency. Why do I need to make my column only deflects on the y-axis? Because I will give my column variation of axial force so the force will have any impact on the natural frequency.

Okay, enough of that. Let's go back to the problem. So, as you can see in Table 7.3, Stokey doesn't have kl value for clamped-YASYMM. I can't calculate with clamped-hinged because as I said before it got 24% errors.

I tried using U1=U2=U3=0 on my FEA model and analyzed it. And when I compared the FEA results with my calculation using Stokey eq. with clamped-hinged kl value, it has errors but the second mode shape has 0,4% errors. Maybe it really is because of the BC.

Oh one more things, I modeled my FEA model with clamped or hinged BC for top BC and analyze its natural freq. Both got same results. Even though different BC, clamped-clamped and clamped-hinged. It has same natural freq. I said it before the UR1, UR2, UR3 (what do you call this? rotation BC? haha) didn't do anything.

Anyway, thank you Steven for the advice but I can't calculate the deflection when I don't know what my column's BC is. Hmm, is U1=U3=0 and U2=free BC called roller BC? Also, is there a way to calculate the deflection of a column with clamped-roller BC? I'm so confused, sorry guys.. Thank you for reading!

That's it guys I'm so tired I hope someone read this 3-3

### RE: Natural frequency of clamped-hinged column

I wonder if the section would "feel" Imin which would be at 45deg (with the square section becoming a diamond section) ?

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

### RE: Natural frequency of clamped-hinged column

For my manual calculations I use this site, maybe it can help you to.
Since its a different formula, well at least written different.

https://www.roymech.co.uk/Useful_Tables/Vibrations...

### RE: Natural frequency of clamped-hinged column

Just for information, using the info in the spreadsheet, I calculated f1, f2 and f3 as 23.56 Hz, 69.63 Hz and 138.73 Hz.

### RE: Natural frequency of clamped-hinged column

(OP)
>Just for information, using the info in the spreadsheet, I calculated f1, f2 and f3 as 23.56 Hz, 69.63 Hz and 138.73 Hz.

Which equation did you use, sir?

### RE: Natural frequency of clamped-hinged column

I derived them from 1st principles. I obtained the displaced shape functions generated under the first three compression buckling loads and then used the displacement functions in the energy method, where you equate bending strain energy to kinetic energy.

### RE: Natural frequency of clamped-hinged column

@dyahaw._

You say you used the BC that resembles a roller to take the external load into account but the external loads shouldn't have an influence on the natural frequency.

Furthermore, a beam is loaded in pure compression only theoretically, in reality the force will be off-center from the neutral axis of the beam. This is because of manufacturing variations (beams are slightly bend, etc.) thus the load not being in it's exact center along it's full length. This means that a beam that is free on 1 end and loaded by compression will also have a bending moment so in reality it will not only deflection in the U2-direction but in the U1 and/or U3 direction as well. (This is why beams will fail by buckling rather than only deforming in the U2-direction which would be pure compression.)

So you should figure out what the appropriate BC is for your application. I assume it's not a beam that is free on 1 end because that would make this really easy .

### RE: Natural frequency of clamped-hinged column

Another approach which is fun is Rayleigh Ritz. Assume a couple of plausible mode shapes with an arbitrary weighting on one of them, differentiate the frequency with respect to that, and find the minimum. Rinse and repeat.

Cheers

Greg Locock

New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?

#### Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

#### Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Close Box

# Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

• Talk To Other Members
• Notification Of Responses To Questions
• Favorite Forums One Click Access
• Keyword Search Of All Posts, And More...

Register now while it's still free!