THE MEAN SQUARE FORCE DUE TO RANDOM FORCES ON A RIGID BEAM

Revision B

By Tom Irvine

December 24, 2014

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Consider a rigid beam subjected to two discrete random forces P1(t) and P2(t).

Figure 1. Rigid Beam with Two Random Forces

Each of the loads varies randomly with time about a mean value of zero. The average properties of the total loads will therefore depend upon the extent to which these loads are dependent, which is the correlation between the loads.

The combined load Pc(t) at any instant is

Pc(t) = P1 + P2 (1)

The average of this sum is zero.

The mean square value of the total load is

(2)

(3)

(4)

The correlation term is .

Now assume a homogeneous pressure field such that

(5)

By substitution,

(6)

(7)

Define a cross-correlation coefficient r.

(8)

Note that the cross-correlation coefficient may vary from -1 to +1.

(9)

Now consider a distributed pressure p(x,t) acting on the beam of length L, which p(x,t) is a random function of both space and time with a zero mean. Assume that the beam has unit width. Note that p(x,t) thus has dimensions of force/length.

The total load on the beam Q(t) at any instant is

(10)

The total value must also be a random function of time, with a zero mean value.

The mean square value can be found from the autocorrelation function as follows

(11)

where

/ is the delay
T / is the signal period

The autocorrelation at zero delay is

(12)

where

/ is the standard deviation
μ / is the mean

Again the mean value is assumed to be zero.

(13)

(14)

Let be the space-time correlation function of the distributed pressure.

(15)

(16)

Now if the pressure field is homogeneous, the mean square pressure at all points will be uniform and equal to

Furthermore, the correlation function has the same shape everywhere and depends only on .

Thus,

(17)

where is the cross-correlation coefficient.

By substitution,

(18)

Let x = x1.

Note that

(19)

Hold x constant and take the derivative.

(20)

(21)

Set the delay equal to zero to calculate the mean square value of Q(t).

(22)

Extreme Case: Nearly Correlation Coefficient

If the correlation function is constant, or nearly so, then

(23)

The mean square load is

(24)

(25)

(26)

Extreme Case: Correlation Coefficient Drops Very Rapidly

Figure 2. Rapidly Decreasing Cross-Correlation Coefficient

The curve in Figure 2 is one possible rapidly-dropping function.

In this case, the following function has little dependence on x.

(27)

Replace this function by the constant l, called the “correlation length.”

The mean square load is thus

(28)

Now let

(29)

where n is the number of correlation lengths required to fill the span.

The mean square load becomes

(30)

Note that as the correlation length becomes smaller, the mean square load does likewise.

Significance of Correlation Length

The integral weights each position with the appropriate value of the space correlation function. The result is an equivalent length over which the load is perfectly correlated.

Figure 3. Cross-correlation Distribution (left) and Equivalent Area (right)

Recall the embedded term from equation (21). Note that this term varies both with space and delay time, which will have significance for beams and plates.

Consider two extremes.

1.  The applied pressure may vary randomly in space and time. A continuity condition yields spatial correlation for closely-space points. But the pressures at a pair of point far away from one another are in no way correlated. Such as pressure field may exist in the near field of a jet.

2.  The applied pressure may be in the form of a traveling wave with uniform speed v. Assume that the wave maintains its shape as it travels. Complete correlation now occurs at any separation distance x provided the correct delay time t = x / v is used. This may occur for:

a)  An acoustic field in which the waves are effectively plane.

b)  The hydrodynamic pressures fluctuations in a boundary layer. These are associated with “bubbles” of turbulence which are convected downstream at a certain speed. Here the model only applies over short distances because the turbulence for a given bubble decays as it is convected.

Generalized Force Due to Random Plane Waves

Figure 4. Oblique Pressure Field

The fluctuations in a time at a point may be resolved into a continuous frequency spectrum. The pressure distribution may be resolved into a continuous spectrum of traveling waves of all wavelengths. The velocity of the plane waves, perpendicular to the wavefronts, is the velocity of sound c. The trace velocity in the beam is

(31)

The pressure at a point is

(32)

The pressure distribution on the beam is

(33)

Now represent the right-hand side of equation (33) in Fourier integral form.

(34)

The total load acting on the beam is

(35)

The trace wavelength lt is related to the acoustic wavelength l by

(36)

Note that

(37)

(38)

By substitution,

(39)

Note that is the Fourier spectrum of p(t).

(40)

(41)

(42)

(43)

The total load Q(t) can also be represented in terms of its Fourier spectrum .

(44)

Compare equations (44) and (43).

(45)

Define a power spectrum Sq(ω).

(46)

(47)

Equation (47) may be simplified using the derivation in Appendix A.

(48)

The power spectrum of p(t) is

(49)

Thus,

(50)

Note that is a “wavelet filter function” as shown in Figure 4. The power spectrum goes to zero at integer values of (L/). The principal part of the generalized force comes from those components which have long wavelengths compared to the length of the beam.

Again,

(51)

Figure 4. Wavelength Filter Function

Reference

1.  E. Richards & D. Mead, Noise and Acoustic Fatigue in Aeronautics, Wiley, New York, 1968.

APPENDIX A

Trigonometric Derivation

(A-1)

(A-2)

(A-3)

(A-4)

(A-5)

(A-6)

(A-7)

(A-8)

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