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forced vibration calculator 3
- Thread starter vik
- Start date
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3
- #2
I doubt if you will find any shareware that will solve this problem. It would take a respectable <br>
finite element program to give you results that you could trust. These usually run in the several<br>
thousands of dollars range. However, if you only need the natural frequency, the calculations<br>
are straight forward enough to do with a calculator for a rectangular plate loaded in the center <br>
with fixed edges. fn=[sqrt(k/m)]<br>
<br>
where fn=natural frequency (first fundamental mode or eigenvalue) (Hz)<br>
k=spring rate (lb/in)<br>
m=mass of plate (lbm)<br>
<br>
You would first have to calculate the spring rate of the plate. This could be done by calculating<br>
the deflection in the center of the plate, inverting it and multiplying the result by the load used<br>
to calculate the deflection lb/in. The deflection formula for a rectangular plate with a load in the<br>
center from Roark is: y = (u * W * b^2)/(E * t^3)<br>
where y=deflection (in.)<br>
W=load applied to center of plate (lb)<br>
a=long length of plate (in)<br>
b=short length of plate (in)<br>
E=modulus of elasticity (10.6x10^6) (psi)<br>
t=thickness of plate (in)<br>
u=empirical data from the table:<br>
a/b u<br>
1.0 .0611<br>
1.2 .0706<br>
1.4 .0754<br>
1.6 .0777<br>
1.8 .0786<br>
2.0 .0788<br>
>2 .0791<br>
<br>
example: Given a rectangular plate 20" x 15" x .25" with a load in the center of 50 lbs<br>
a=20 in<br>
b=15 in<br>
W=50 lbs<br>
t=.25 in<br>
a/b=1.33<br>
u=.0738 (using linear interpolation for 1.33 between 1.2 and 1.4)<br>
<br>
y=[(.0738)(50)(15)^2]/[(10.6E6)(.25)^3] = .00501 in<br>
<br>
k=50/.005 = 9974 lb/in<br>
m=Vol*density of aluminum = (20*15*.25)(.0955) = 7.16 lbm<br>
<br>
fn=(sqrt(9974/7.16)) = 37.3 Hz
finite element program to give you results that you could trust. These usually run in the several<br>
thousands of dollars range. However, if you only need the natural frequency, the calculations<br>
are straight forward enough to do with a calculator for a rectangular plate loaded in the center <br>
with fixed edges. fn=[sqrt(k/m)]<br>
<br>
where fn=natural frequency (first fundamental mode or eigenvalue) (Hz)<br>
k=spring rate (lb/in)<br>
m=mass of plate (lbm)<br>
<br>
You would first have to calculate the spring rate of the plate. This could be done by calculating<br>
the deflection in the center of the plate, inverting it and multiplying the result by the load used<br>
to calculate the deflection lb/in. The deflection formula for a rectangular plate with a load in the<br>
center from Roark is: y = (u * W * b^2)/(E * t^3)<br>
where y=deflection (in.)<br>
W=load applied to center of plate (lb)<br>
a=long length of plate (in)<br>
b=short length of plate (in)<br>
E=modulus of elasticity (10.6x10^6) (psi)<br>
t=thickness of plate (in)<br>
u=empirical data from the table:<br>
a/b u<br>
1.0 .0611<br>
1.2 .0706<br>
1.4 .0754<br>
1.6 .0777<br>
1.8 .0786<br>
2.0 .0788<br>
>2 .0791<br>
<br>
example: Given a rectangular plate 20" x 15" x .25" with a load in the center of 50 lbs<br>
a=20 in<br>
b=15 in<br>
W=50 lbs<br>
t=.25 in<br>
a/b=1.33<br>
u=.0738 (using linear interpolation for 1.33 between 1.2 and 1.4)<br>
<br>
y=[(.0738)(50)(15)^2]/[(10.6E6)(.25)^3] = .00501 in<br>
<br>
k=50/.005 = 9974 lb/in<br>
m=Vol*density of aluminum = (20*15*.25)(.0955) = 7.16 lbm<br>
<br>
fn=(sqrt(9974/7.16)) = 37.3 Hz
The method suggested has a couple of problems. They are as follows:<br>
<br>
1.) It was assumed that the entire mass of the plate is lumped at the center. This is not correct, and will give a natural frequency that is too low, because the effective mass is too large. To get a more realistic result, Rayleigh's method (see any good vibrations text for details) should be used. Although the exact mode shape is unknown, the approprate number of half sine waves in the x and y direction to match the boundary conditions and midspan displacement should give a closer result.<br>
<br>
2.) It was assumed that the force-deflection characteristics of the plate are linear. This is true only if the maximum deflection is much smaller than the plate thickness, and the plate span is much larger than the plate thickness. Larger deflections will lead to membrane stresses in addition to the bending stresses in the plate, which causes the plate to effectively become stiffer as deflections get larger. This will cause the natural frequency of vibration to be amplitude dependant. This also has benefits, as it limits the maximum displacement at resonance.
<br>
1.) It was assumed that the entire mass of the plate is lumped at the center. This is not correct, and will give a natural frequency that is too low, because the effective mass is too large. To get a more realistic result, Rayleigh's method (see any good vibrations text for details) should be used. Although the exact mode shape is unknown, the approprate number of half sine waves in the x and y direction to match the boundary conditions and midspan displacement should give a closer result.<br>
<br>
2.) It was assumed that the force-deflection characteristics of the plate are linear. This is true only if the maximum deflection is much smaller than the plate thickness, and the plate span is much larger than the plate thickness. Larger deflections will lead to membrane stresses in addition to the bending stresses in the plate, which causes the plate to effectively become stiffer as deflections get larger. This will cause the natural frequency of vibration to be amplitude dependant. This also has benefits, as it limits the maximum displacement at resonance.
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