# Dynamic Analysis Cantilever Beam (Staad Pro vs Analysis)

1. Jul 6, 2012

Hello Phys Forum
I’m trying to solve a dynamic problem that involves a cantilever beam fixed onto a support that experiences a time history load on the free end of the beam as shown below:

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Cross section Beam is rectangular beam with 1m width and height.

The objective is to obtain the natural frequency of the beam. I believe this is the equation to use:
\omega_n = \frac{\alpha_n}{L^2} \sqrt{\frac{EI}{m}}

Please correct me if my approach is incorrect.

The paramenters are the following:
\alpha_n = 3.57
L (Length) = 5m
E = 205 KN/mm^2
I = 1m^2
m = 2kg (0.204N)

So the natural frequency \omega_n should be

\omega_n = 45.72 Hz

Now, when I run it through Staad Pro (haven’t tried AnSys yet I heard its powerful tool). With the same parameters where I use a Time History Load of -0.204 N onto the end node of the beam and applied the same parameters plus the self loads of 1 throughout the system, my results indicate that its 22.684 Hz for the 1st Mode. Could you guys tell me where I am going wrong all this time is it the analytical solution my Staad Pro or both?!

The script of the Staad Pro is defined below.

START JOB INFORMATION
ENGINEER DATE 06-Jul-12
END JOB INFORMATION
INPUT WIDTH 79
UNIT METER NEWTON
JOINT COORDINATES
1 0 0 0; 2 5 0 0;
MEMBER INCIDENCES
1 1 2;
DEFINE MATERIAL START
ISOTROPIC STEEL
E 2.05e+011
POISSON 0.3
DENSITY 76819.5
ALPHA 1.2e-005
DAMP 0.03
TYPE STEEL
STRENGTH FY 2.532e+008 FU 4.078e+008 RY 1.5 RT 1.2
END DEFINE MATERIAL
MEMBER PROPERTY
1 PRIS YD 1 ZD 1
CONSTANTS
MATERIAL STEEL ALL
SUPPORTS
1 FIXED
DEFINE TIME HISTORY DT 0.0013888
TYPE 1 FORCE
1 -0.204 2 0
ARRIVAL TIME
1 2
DAMPING 0
2 FY 1 1 1.000000
CALCULATE RAYLEIGH FREQUENCY
SELFWEIGHT X 1 LIST ALL
SELFWEIGHT Y 1 LIST ALL
SELFWEIGHT Z 1 LIST ALL
PERFORM ANALYSIS PRINT ALL
FINISH

2. Jul 6, 2012

### Mech_Engineer

For a start your area moment of inertia (a.k.a. Second Moment of Area) is wrong, both in value and units. Calculation of the area moment of inertia for a rectangular cross-section is achieved through this equation:

http://en.wikipedia.org/wiki/List_of_area_moments_of_inertia

So in the case of your beam I = 1/12 m^4. BUT your beam is 1m x 1m x 5m, which if it's made out of steel (probably based on your modulus of elasticity) means it weighs almost 40,000kg! Are you taking into account this beam's self-weight?! Who cares about a 2kg weight at the end of it?

3. Jul 6, 2012

Sorry, I'm bit new to this area.

OK just realised that cross sect area is corrected (thanks!)
Indeed the beam is Steel, how did you work out the beams mass though 40000kg.

Did you use the relation E = Stress/Strain where
Strain = Disp/Orig Length
Stress = Force/Area

And the beams mass....have you tried it on Staad Pro? I mean would you have it as a distributed load on the beam or a selfload of -40000kg in the Y dir?

For the free 2kg I thought thats needed to work out the natural freq.

4. Jul 8, 2012

### Mech_Engineer

"I" is not the cross-sectional area, it's the beam's section modulus. It's important to note its units are (length)^4

You stated the beam is rectangular, 1m x 1m x 5m. This means it is 5 m^3 in volume, and since steel's density is around 7850 kg/m^3 the beam is (do the math here) around 39,250 kg. Are you sure you want to analyze a beam of this size? It seems pretty ridiculous when you're hanging a mere 2kg off its end...

I don't have Staad pro, all I've done is look at the numbers you've provided so far. I don't think the analytical equation you've provided takes into account the beam's self-weight.

What's the "SELFWEIGHT X 1 LIST ALL" command? That sounds like self-weight of the beam to me...

Sure when you've got a beam where the weight of the beam is small compared to the weight being hung off it... but when the beam weighs 20,000 times more than the "load" you're putting on it, all you really care about is the beam's reaction to its own self weight!

You have to look at the overall context of what you're trying to analyze: you're looking at a 1m square SOLID STEEL bar, with a 2kg weight on it's end. It just doesn't seem like a useful analysis.

5. Aug 9, 2012

Hi thanks for the heads up mech_engineer. I looked up the problem again and it turns out that its a cantilever beam with a force applied to the end of the beam. Hence for my analysis is the following:

$I = \frac{bh^{3}}{12}$
$K = \frac{3EI}{L^{3}}$
$M = \frac{F}{g}$
[itex]\omega = \sqrt{\frac{K}{M}}[\itex]

Then it turnsout it matches the staad pro solution so thats one problem sorted the next one now is similar but a distributed load of 20N along a 10m cantilever beam. Tried making the loading force F = 200N and use the same idea but I'm not getting the natural freq as required on the Staad Solution.

0.576 cycles/sec

Parameters are the following:

b = 0.05m
h = 0.1m
E = 2.05*10^{11}
F/m = 20N per meter
L = 10m

Any idea where I might be going wrong?
Thanks again