Dunkeley Formula

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Transverse Vibration of Beam Lab Report

Introduction
This experiment aimed to study the forced damped transverse vibration of a beam and verify Dunkerley’s empirical formula.

According to Dunkerley’s empirical formula,

1f2=1fb2+1fω2
where f referring to the natural frequency of heavy beam with central load fb referring to the natural frequency of heavy beam only
fω referring to the natural frequency of light beam with central load M By using the results of light damping, Dunkerley’s empirical formula would be studied and verified by comparing the calculated resonant frequency f of the whole system and the experimental value.

Apparatus
1. Beam vibration apparatus which consists of :
(1) A rectangular bar-length
Cross section
Mass
Young’s modulus E
Motor and rotor assembly – mass
Mass of one plate
2. DC power supply
3. B & K accelerometer 4369 with magnetic support and cable holder – mass = 66.7gm 4. B & K charge amplifier 2635
5. Oscilloscope
Procedure
1. The charge amplifier was set to the following
(1) Transducer Sensitivity range switch 1-11
(2) Transducer Sensitivity conditioning knobs (221)
(3) mV/unit output switch 100
(4) Upper frequency limit 0.1kHz
(5) Accel-Vel-displ switch 1mm (1Hz)
2. The oscilloscope was set to:
(1) Sweep Time/cm (20~50) ms
(2) Gain (5~50) mV/cm
3. The output was connected from the charge amplifier to channel 1 or 2 of the oscilloscope. 4. The accelerometer was attached by means of magnetic support to the steel plate of the motor assembly. 5. With light damping, the motor by was started increasing the voltage supply of the DC power supply from 20 volt to 40 volt in step of 1 volt. 6. Te frequency was recorded and amplitude of vibration from the trace on the oscilloscope, (allow the motor speed to settle before making each recording). 7. The entire procedure was repeated with heavy damping.

Theory
A. Dunkerley’s formula for beam vibration is
1f2=1fb2+1fω2
Where f= natural frequency of heavy beam with central load
fb= natural frequency of heavy beam only
fω= natural frequency of light beam with central load M
Derivation of fb

Let y= deflection of beam
X= distance along beam
m= mass of beam per unit length
E= Young’s modulus
I= Second moment of area of beam section
Mb= Bending moment
F= shearing force

Considering the equilibrium of a small element of the vibration beam Moment equilibrium
∂Mb∂X=F
Force equilibrium
∂Mb∂X=-m∂2y∂X2
Theory of deflection of beams gives
Mb=EI∂2y∂X2
Elimination M and F from these equations gives the following equation of motion. EI∂2y∂X2+m∂2y∂X2=0 ---------- (1)
Assume vibrating motion to be sinusoidaly=y0sinωt, where y0 is a function of x only. Equation (1) becomes d4ydX4-α4y0=0 ---------- (2)
where α4=mω2EI ---------- (3)
General solution of Equation (2) is
y0=Acosαx+Bsinαx+Ccoshαx+Dsinhαx
For a beam simply supported at both ends.
At x=0, y0=0 and d2ydx2=0
∴A+C=0
α2-A+C=0
∴A=0 and C=0 for α≠0
At
x=L, y0=0 and d2ydx2=0
∴BsinαL+DsinhαL=0
-BsinαL+DsinhαL=0
i.e. BsinαL=0 and DsinhαL=0
∴D=0 for α≠0
Also, either B=0 or sinαL=0
If B=0, y0=0 and therefore the only condition under which vibration can occur is that sinαL=0 αL=nπ, n = 1, 2, 3, ……
From equation (3)
ω=α2EIm
=n2π2L2EIm , n=1,2,3
For fundamental mode,n=1
fb=π2L2EIm Hz
Derivation of fω
y=static deflection of beam due to mas M
Strain energy at extreme position due to mass M
=work done in deflecting mass M at a distance u
=12Mguyu
Kinetic energy at equilibrium position due to M=12M(ωu)2
Equating S.E to K.E
12Mguyu=12M(ωu)2
ω=gy
fω=12πgy Hz
For a light beam carrying a concentrated mass M at centre and dimply supported at...
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