Vibration Measurements PDF

Free un-damped oscillations of a single degree of freedom system Free length (no load) position For dynamic analysis start here For static deflection due to mass m load (weight = mg) we have Free un-damped oscillations of a single deg...

Free un-damped oscillations of a single degree of freedom system Free length (no load) position For dynamic analysis start here For static deflection due to mass m load (weight = mg) we have Free un-damped oscillations of a single degree of freedom system Pulling the mass m from the static equilibrium position by an initial disturbance force and then releasing it will cause the mass to oscillate. The oscillation, if un-damped, can be controlled by the following dynamic equation: The solution to the second order differential equation is of the form: Where C1 and C2 are constants to be evaluated by knowing the initial conditions (x and dx/dt at time t=0)  Free un-damped oscillations of a single degree of freedom system Natural or normal angular frequency in rad/sec. Linear frequency in Hz Period of oscillation in seconds Free un-damped oscillations of a single degree of freedom system For initial displacement xo and initial velocity vo the solution is of the form: general case For initial displacement xo and zero initial velocity vo = 0 the solution is of the form: special case  P = F/A Piezoelectric Transducer Overlapping area, A E=gtP t Crystal When a force is applied to the plates, a stress will be produced in the crystal and a corresponding deformation. This deformation will produce a potential difference, proportional to the impressed force, at the surface of the crystal, and the effect is called the piezoelectric effect. Used in pressure transducers for dynamic measurements. Good for high frequency (rapid dynamic changes) measurements. Piezoelectric Transducers A class of piezoelectric material when mechanically strained in a preferred manner generate an open circuit EMF (displaced electric charge) F t + + + + + + + t Vo - - - - - - - Metallic electrodes All piezoelectric materials have a Curie Vapor deposited on temperature above which piezoelectric top and bottom activity ceases to exist. surfaces Piezoelectric Transducers κε0 A transudcer capacitance = C X = F t κ = dielectric constant of material stress F/A Young's modulus Y = = strain δt/t net charge displacement = Q = d × F C d = charge sensitivity of material(C/m 2 )/(N/m 2 ) A capacitor is a passive element that stores energy in the form of an electric field (electric charge). DC does not flow through a capacitor, rather, charges are displaced from one side of the capacitor through the circuit to the other side, establishing the electric field. Capacitor’s voltage-current relationship is defined as: t 1 q(t) V(t) = C 0 I(τ)d(τ) = C , dV or, I(t) = C dt where : q(t) = accumulated charge in coulumbs C = capacitance in farads (F = coulumbs/volts) I(t) = displacement current Typical capacitor range (1pF to 1000  F) Piezoelectric Transducers Open circuit voltage across transducer is : Q dF F Eo = = = g  t V C  o A / t  A where, g is the voltage sensitivity of thetransducer in (V / m) / ( N / m 2 ) Mechanical input frequency should be well below the natural (resonance) frequency of the piezoelectric transducer. High resonance frequency ➔ can measure frequencies in the KHz range Variable Capacitive sensor: A change in capacitance due to temperature, humidity or other physical parameter can be converted to an analog output voltage by and ac bridge circuit. It can measure force, acceleration, depth, etc. by relating it to the displacement of one or both capacitive electrodes κεo A capacitance = C = d where, κ = dielectric constant εo = permittivity of free space (8.85  10 − 12 F / m) Differential pressure sensor Seismic Sensor (vibration) housing damper Seismic Displacement mass m transducer x1 = xo cos 1t spring wrokpiece x1 = xo cos 1t x2 (t ) Seismic Sensor A voltage- (vibration) divider potentiometer is used for sensing A piezoelectric relative transducer is displacement. used to measure vibration. A strain gage is used for sensing relative displacement.  c Damper Seismic Displacement mass m m x measurement output k Spring Static m equilibrium Workpiece xm xs = xs o cos t k ( xm − x s ) c( xm − xs ) m = seismic element mass k = spring deflection constant c = viscous damping constant xm = absolute displacement of mass M xs = displacement of support ( workpiece ) = xso cos  t xr = xm − xs = relative displacement t = time variable  = support exciting frequency n = undamped natural frquency of seismic system = k / m  = phase angle Applying Newton’s second law to the free body of mass M, we find the differential equation for the motion of the mass as: d 2 xm dxr m 2 +c + k xr = 0 dt dt In this form, each term represents a force [ inertia force + damping force + spring force ]. Substitute Sm = Ss + Sr ➔ d 2 xr dxr d 2 xs m 2 +c + k xr = − m dt dt dt 2 and , xs = xso cos t Therefore, d 2 xr dxr m 2 + c + k xr = m x so  2 cos t dt dt  = c / cc = damping factor where cc = 2 m n = critical damping constant n = k / m , → k = m  2 n c  c cc  2 m n   = = = 2 k cc k mn 2 n Linear differential equation of the sec ond order. Solution : xr = e − t /  A cos 1 −  2 nt + B sin 1 −  2 nt    + ( m / k ) xso  2 cos ( t −  ) , 2 2 1 − (  / n )  +  2 (  / n )  2     Steady state solution (after several time constants ) xso (  / n ) 2 xr = 2 1 − (  / n )  +  2 (  / n )  2 2     2 3 3 Newtonian Accelerometer M From Newton's 2nd law : 0 = Mx2 + B(x2 - x1 )+ k s (x2 - x1 ) Newtonian Accelerometer From Newton's 2nd law : 0 = Mx2 + B(x2 - x1 )+ k s (x2 - x1 ) Laplace transfomr this ODE and put in transfer function form : X 2 (s) (sB/k s )+1 = 2 (1) X 1 (s) (s M/k s )+(sB/k s )+1 LVDT output; Vo = ko ( X 1 − X 2 ) (2) Substituting (1) in (2); Vo (s) ko (s 2 M/k s ) = 2 X 1 (s) (s M/k s )+(sB/k s )+1 Servo Accelerometer Design Force V1=KD(x1-x2) Phase relationships among displacement, velocity, and acceleration. (6)  x = X sin(2 f )t = X sin t v = 2 f X cos(2 f )t  peak velocity = 2 f X =  X a = − 4 2 f 2 X sin(2 f )t  peak acceleration = 4 2 f 2 X =  2 X X pk − pk = 2 X It can be seen that low-frequency motion is likely to exhibit low- amplitude accelerations even though displacement may be large. It can also be seen that high-frequency motion is likely to exhibit low- amplitude displacements, even though acceleration is large. Examples: At 1 Hz, 1 in. pk-pk displacement is only ~0.05 g acceleration; 10 in. is ~0.5 g At 1000 Hz, 1g acceleration is only ~0.00002 in pk-pk. displacement; 100 g is ~0.002 in pk-pk.