Given a piece of 3" x 4" x 1/4 angle iron (3" leg vertical), 30" long and continuously welded throughout the cross-sectional perifery at each end, can anyone help me with an approximate natural frequency or with the formula to do so myself? (Give a man a fish....teach a man to fish....)

Thanks,

Danny

I will look in Mark's handbook when I get home.

One thing that strikes me is that the side of the angle that is bending along it's 1/4" dimension will contribute only weight but negligible stiffness. So the problem will amount to Looking at a 1/4" plate bending along it's 3" or 4" dimension with added weight from the other half.

According to my textbook (Mechanical Vibrations by Rao)Chapter 8.5 - Lateral Vibration of Beams:

w (rad/s) = (B*L)^2 * sqrt [ (E*I) / (rho * A * L^4) ]

(B*L) is read from Figure 8.15. The first mode for a fixed-fixed beam gives a (B*L) value of 4.730041

Hope this helps.

Steve

Pete,
I think you and I are thinking along the same lines. With the use of angle the front edge of the horizontal 4" leg is like a piece of plate while the back edge with the 3" leg mounted vertically underneath is much stiffer. It certainly appears to be behaving like you would expect given the difference in front to back stiffness.

Steve,

Thanks for the info. I'm on the road for the rest of the week, but maybe I can try my luck this weekend.

In Den Hartog page 154 I find an equation which ends up being identical to the one provided by Steve.

w = 22.4/L^2 * sqrt(E*I/mu)
22.4 is 4.730041^2
1/L^2 outside the root is equivalent to 1/L^4 inside the root.
mu is Den Hartog's symbol for mass per foot which is A*rho

E is an easy one. 30E6 psi for steel.
There will be a lot of units to convert which is not too tough.

The part I can't figure is the I. For simple rectangular beam we use b*h^3/12. And you can break up your L-shape into 2 rectangles. But I think you have to use to parallel axis theorem to see the effect of those I's from the location you want. That's the part I'm a little fuzzy on... where to we want to look at the I's from. Somewhere near the center but where? Maybe someone out there can help with figuring the correct I to use? (Steve? William?)

I've got a nifty little program for those kinds of things. 'I' depends on how you bend it. Take a look at the screenshot.

That's neat Patrick. I am wondering how did they pick the location for the x axis among many possible parallel axes we could have picked. Is it the centroid of the area?

Using Patrick's Ixx, I computed 770hz (calculations shown in attached pdf). That seems way too high to me. Maybe I made some errors? I wouldn't be surprised.

Assuming the result is right, you could calculate that the frequency bending about the Y axis would be 770hz * sqrt(Iyy/Ixx). Likewise for that other bending mode 770hz *sqrt(Izz/Ixx)

This message has been edited. Last edited by: electricpete,

Lshaped.pdf (62 Kb, 34 downloads)

It's through the centroid. The y, x and y dimensions above note the location.

Thanks Patrick.

With the axis on the centroid it seems like the effect of both halves of the L has big effect on I (different than what I thought initially). Maybe that is the philosophy of I-beams. It gets strength from the top and bottom horizontal pieces for bending in vertical direction even though the horizontal piece by itself would have very little strength for bending in that direction?

I see one error in my calculation already. For area, I used A=11.75*0.25 inch^2. It should have been A=6.75*0.25 inch^2. Since area appears in the denominator of the root (part of mu), the corrected answer would be 770 * sqrt(11.75/6.75) which is even higher than before. That sounds much higher than I would have expected.

This message has been edited. Last edited by: electricpete,

I revised my previous calculation to correct the error in area. (see file Lshaped2.pdf below). I added some explanation of the variables. I also added a calculation of centroids and Ixx and Izz about centroids.

I set the orientation as shown in Patrick's figure assuming 4" plate is vertical and 3" horizontal - I think that is opposite the original post.

I got Ixx and Iyy close to predicted by Patrick. I tried several combinations of addressing exactly what happens to the 1/4" dimension at the intersection of the two plates (1/4"x1/4" overlap, 1/4"x1/4" hole, 3" plate butts 4" plate, 4" butts 3"). I couldn't get an exact match but pretty close. Izz looks too difficult - I didn't even try.

No change in the corrected results above. Frequencies of 1015hz, 710hz, 515hz, depending on the orientation.

Once again can anyone else double-check this or else do an independent calculation?

Lshaped2.pdf (93 Kb, 15 downloads)

Pete some more information about Ixx etc. can be found in Vector Mechanics for Engineers Statics, from Ferdinand P. Beer and E. Russel Johnston. Didn't check the formulas yet, bought the book somewhere in 1989 and didn't look in it for the past 10 years.

Pete - I got 717 Hz using the Y-Y axis (Iyy) and some slightly different numbers for properties and sizes.

BL = 4.730041
E = 30 * 10^6 psi
Iyy = 1.36 in^4
rho = 0.284 lbm / in^3
A = 1.69 in^2
L = 30 in

I looked up Iyy, rho, and A in Mechanics of Materials (2nd Edition) by Beer & Johnston.

Danny - Are you going to test it and let us know?

quote:

they got it wrong... they thought like I thought before that the vertical part of I-beam is the only significant role for supporting vertical loads but in fact the horizontal part would play a much bigger role assuming you had an I-beam with square dimensions.... the reason is because the horizontal portions are farthest from the centroil and therefore their area contributes the most to I.

I'm going to correct myself. "Supporting load" was an ambiguous phrase. If an I-beam were loaded vertically:
1 - The horizontal members contribute the most to resisting bending (due to contribution to moment of inertia as discussed above)
2 - The vertical members are subjected to the most shear force.

This message has been edited. Last edited by: electricpete,

The bending about Izz identified in Patrick's post may be of most interest since it is the lowest frequency. But I would have never thought to look for that mode - after all for rotating machinery we just look at horizontal and vertical bending (neglecting rocking modes that pivot about the center which don't seem relevant to this mounting, and higher order bending modes which generate higher frequencies). So what is different about this situation?

Does Izz represent a mode of vibration?
How do we know how many modes of vibration need to be looked at? Is the axis of Izz at 45-degrees to the x,y coordinate axes or is it parallel to the hypoteneuse of the 3-4-5 right triangle?

Any other comments to help understand Izz?
Thanks.

This message has been edited. Last edited by: electricpete,

I appreciate all the help here.

Everything here suggests a real stretch for resonance since all the calculations point to somewhere around 700-750 hz and my vibration is at around 20.

The rocking mode sounds like what is happening in my case and it appears so visually.

To answer Steve's question about am I going to test it, that depends on whether my client is willing to take some downtime and actually pay me to di it. So far they apparently think that I enjoy this work so much that I do it for free.
I would ceratinly like to know the real world answer to the question, but not bad enough to do charity work for a large property management company.

Let me also add, that as most noticed, this probably would not be a problem it they had used 4" channel rather than save maybe two pounds of steel and used angle. The question of rocking would be eliminated by the added stiffness in all directions due to the extra vertical leg.

Thanks again,

Danny

This message has been edited. Last edited by: Danny Harvey,

I think the calculations were mathematically correct for the problem we solved... especially since Steve and I got similar results for the bending mode associated with Iyy. I am still a little bit confused about Izz as expressed above (can anyone answer those questions?)

Thinking about the original problem and why we got such a high resonant frequency - we modeled a structural support brace (high stiffness) with no supported mass other than the brace itself (low mass). It makes sense we get a high natural frequency for that setup.

I am suspecting maybe in the real world problem there is some more vibrating mass associated with what is attached to the brace. By the way, what is attached to the brace on each end (or anywhere else for that matter)? Just curious.

I talked with the client yesterday and he would prefer not to take the unit out of service this summer (at least not long enough to do any substantial testing), but he is going to add a stiffener to the unsupported horizontal leg of the angle.

If that doesn't help, he will be forced to make corrective actions during off-peak hours and they will include checking the torque tube and couplings for runout, bends, worn bushings, etc.; getting phase measurements, and alignment.

To answer Pete's questions, this angle is welded on each end to 8" channel and has the motor bolted directly to the 4" leg of the angle and centered between the channels. The bolt holes for the motor hold-down bolts are drilled about 1" from the unsupported edge-in probably the worst possible place you could have put them.

I'm told now that the gearbox was replaced and everything was bolted back together in the same position as before, so the chances for misalignment are great.

The funny thing is that the other cell has a high amplitude at 1785 rpm (high speed) and probably has worse misalignment than the one in question, but has had stiffeners added so that there isn't the signal amplification caused by resonance. At about 1.8 ips @1785 cpm, they are not too concerned about that unit, just the one at 2.5 ips at 1190 cpm.

I think Pete and Steve did arrive at the correct answer for Iyy. Maybe repeating the calculations treating the horizontal leg as a plate would be more true to the situation. Probably somewhere in between the two is the actual answer. I'm not even sure that I would be able to get the correct answer by performing bump tests either.

Once again, thanks for your help on this. Just taking the time to collectively think through the problem is a big help even if our calculations don't come up with the answer I had expected.

Good Luck,

The math problem we solved was a piece of angle-iron which is fixed to an immovable wall at each end with no mass attached anywhere.

The resonant frequency of your real-world system will be decreased by the mass of the motor attached to the center of the angle and perhaps by the flexibility of the 8" channel to which the angle-iron is attached at its end. More details on motor mass and channel configuration would be required to come up with a better estimate of resonant frequency.

Even though we misunderstood the real-world setup, it was an interesting excercize for me.

Pete,

I'm sorry that I didn't make it clear that the motor was bolted to the angle. Other than that, the example is essentially correct. There is certainly some flexibility in the channel frame, especially since it is attached to the wooden structure of the cooling tower.

You guys have been kind enough to give me the tools to finish an estimate myself, so I will give it a try on my own.

I truly appreciate you and all the others taking time to help me out with this.

Regards,

Danny

I had a hard no understanding about equation presented for you :
wn=22,4/(L*L)*sqr(E*I/mu)
=(22,4/L)*sqr((E*I)/(L*L*mu))
.
I could to see that 22,4 is near(3pi/2)*(3pi/2)althought there isny any relation, after I discover that this is a general convercion factor
.
Reviewing my books i found:
wn=(n/(2L))*sqr(F/mu)
. where:
F is restoration elastic force, like a strain force in a cord.
n is the number of waves length (lambda) between two nodes, nodes are the ends of the 30 inch rod.
.
In the next exercise has justification use n=1 because a mass-spring system has only one natural frequency.
.
Making equal equation of my book and your equation, I could to see that F=(E*I)/(L*L), the restoration elastic force wich produce wn
.
Sustituting F in my equation :
wn=(n/(2L))*sqr((E*I)/(L*L*mu)) wich can be writted
wn=n*sqr((E*I)/(4*L*L*L*L*mu))
.
Using this equation, in order to compare results obtained by Pete, for example in x-x axes, n=1 and mu=9,8651lbm/ft
.
wn=sqr((30E6 lbf/in*in*2,68in*in*in*in)/-
-/(4*30in*30in*30in*30in*9,8651lbm7ft))
.
Applying right convercion, like:
1lbf=0,454kgf , 1lbm=0,454kgm
1in=0,0254m , 1kgf=9,81N= 9,81kgm*m/(sec*sec)
.
We will obtain :
.
in x-x:wn=107,97rad/sec=17,18cicle/sec=1031cpm
near 1015cpm obtained by Pete
.
In the same way:
.
in y-y:wn=715,34cpm using Iyy=1,29in*in*in*in
near 710,4cpm obtained by Pete
.
in z-z:wn=523,93cpm using Izz=0,692in*in*in*in
near 515,7cpm obtained by Pete
.
Quetions:
1.-What is the correctInertia wich we have to use in the presented problem?
2.- What could be the usefull of F=(E*I)/(L*L)in the solution of this problem or another kind of problem?
.
Regards, MikeK