Vibrational loads

The stress-life and strain-life approaches covered in Introduction to fatigue assume that you know the load signal in the time domain. In some cases, such as for loads on wind turbines or oil platforms, you do not know the load signal because measurement would take years. Or you need to simulate the test on a shaker table where the load is given as sine sweeps or the spectrum is described as a power spectral density (PSD).

The analysis of random acoustic loads leads to a description of the local stresses as PSDs. Hence, there are still two cases to distinguish:

• Random excitation (described by power spectrum densities and cross correlations)

• Deterministic (sine or harmonic) excitation

PSDs, cross correlations, and moments

First, we need some definitions for random excitations. We will denote by X(t) a random process. (Formally, one should describe it by X(t,p) where for fixed T, X(T,p) is a random variable and for a fixed P, X(t,P) is a deterministic function in time. For simpler reading, we omit the probabilistic dependency in the following.)

When we talk about probabilistic processes, it makes sense to look at expectation values. For further simplicity, we consider only ergodic processes, where these expectation values coincide with the averages in time:

Mean function:

{m_x}(t) \equiv \bar X(t) \equiv E\left{ {X(t)} \right} = \mathop {\lim }\limits_{T \to \infty } \frac{1}{{2T}}\int\limits_{ - T}^T {X(t)dt}

This also means that in those cases, the mean function is constant. Ergodic processes are always stationary.

Variance function:

\sigma _x^2\left( t \right) \equiv E\left{ {{{\left[ {X\left( t \right) - \bar X\left( t \right)} \right]}^2}} \right}

Auto-correlation function:

{R_X}(\tau ) = E[X(t)X(t + \tau )]\mathop { = \lim }\limits_{T \to \infty } \frac{1}{{2T}}\int\limits_{ - T}^T {X(t)X(t + \tau )dt}

The auto-correlation function defines how a signal is correlated with itself, with a time separation τ.

Cross-correlation function:

{R_{XY}}(\tau ) = E[X(t)Y(t + \tau )]\mathop { = \lim }\limits_{T \to \infty } \frac{1}{{2T}}\int\limits_{ - T}^T {X(t)Y(t + \tau )dt}

The cross-correlation functions show the similarity and dependency between two random functions. Thinking about the random functions as load functions, this describes the dependencies between loads. For example, the output load on a rear wheel is very similar to the load on a front wheel, just shifted by a (velocity-dependent) time interval.

The PSD functions can be defined as the Fourier transform of the auto-correlation function:

{S_X}(\omega ) = FT({R_X}(\tau )) = \int\limits_{ - \infty }^\infty {{R_X}(\tau ){e^{2\pi i\omega \tau }}d\tau }

It can be shown that this corresponds to the density of the power, for example, generally the square of the modulus of the Fourier transform – which often is also used as the definition of a PSD.

{S_X}(\omega ) = \mathop { = \lim }\limits_{T \to \infty } \frac{1}{{2T}}E\left( {{{\left| {FT({X_{2T}}(t))} \right|}^2}} \right)

where {X_{2T}}(t) = X(t) if \left| t \right| < T or else 0.

The definition by the Fourier transform allows the extension to cross power spectra:

{S_{XY}}(\omega ) = FT({R_{XY}}(\tau )) = \int\limits_{ - \infty }^\infty {{R_{XY}}(\tau ){e^{2\pi i\omega \tau }}d\tau }

Considering n random functions X1 to Xn, you can combine these into a matrix, called the cross power matrix. This matrix is Hermitian:

{S_{XY}}(\omega ),, = \overline {{S_{YX}}(\omega )}

And if X(t) real valued:

{S_{XY}}(\omega ),, = \overline {{S_{XY}}( - \omega )}

From loads to stresses—transfer functions

When loads are applied to a structure, they induce local stresses and strains. If you apply a given deterministic load to a structure, you can calculate the stresses, for example with finite element methods. For a large range of loads, this behavior is (mostly) linear. If you apply twice the load, the local stresses and strains also double. In the time domain, this means that if you know the load over time and the stresses for one load, such as load = 1, you can calculate the stresses for all points in time by multiplying the stress for load = 1 (c(x)) with the actual load.

\sigma \left( {t,x} \right) = L\left( t \right)*c\left( x \right)

Very similarly, you can obtain displacements and stresses in the frequency domain. Here you use transfer functions: the response per unit input at each frequency of interest. Also, multiplied with the load at the frequency, you get the stress.

\sigma \left( {t,x} \right) = L\left( f \right)*T\left( {f,x} \right)

In the frequency case, this means you have a transfer function at each point. To handle this data more efficiently, you can use the assumption of linearity and a modal representation of the transfer functions.

T\left( {f,x} \right) = \sum {m{c_i}\left( f \right)*M{\sigma _i}\left( x \right)}

where

M{\sigma _i} represents a modal basis of stresses and

m{c_i}\left( f \right) are the modal coordinates of the transfer function.

The modal basis is typically calculated by a finite element analysis (for example, Simcenter Nastran SOL103).

The modal coordinates for transfer functions for a unit force or an enforced unit motion can be calculated by a modal forced response analysis using the fact that if you write the equations of motion in the modal basis, you end up with a decoupled system of differential equations for the modal coordinates in the time domain and a linear system of equations in the frequency domain. The only inputs needed are the modes, modal damping, modal masses, and the point (and direction) of excitation.

Using the transfer function, you can also calculate the local stress PSDs from a load PSD.

PS{D_\sigma }\left( x \right) = T{\left( {f,x} \right)^2}*PS{D_{load}}

When applying a cross power matrix, you have to multiply the cross power matrix accordingly from the left and the right with the vector of transfers. Keep in mind that in this case, you do not get scalar stress PSDs but a tensor of PSDs in the coordinate directions and cross powers between the directions. To get to a stress PSD, like in the multiaxial case of time-based fatigue, you have to use a projection (critical plane approach) or an equivalent stress value, such as von Mises.

Random excitations and stress cycles

If the excitations are random and given by a cross power matrix, you must use probabilistic approaches to get to real stress cycles. First, you assume that you locally know a uni-axial stress PSD.

The PSD provides information on a random process of the local stresses. You can calculate one very useful characteristic directly from the PSD. The root mean square (rms) value is defined as the square root of the area under the PSD curve.

Some basic analysis tells you that the moments give some basic statistical information:

• Expected zero-crossings

• Peaks

• Irregularity factor

Two of the most important statistical parameters are the number of zero crossings and the number of peaks in the signal. The figure below shows an interval cut out from a wide-band signal.

The red points indicate the upward zero crossings and the blue points the upper turning points. The more upper turning points exist for which no zero crossings exist, the more irregular the signal is. Basic analysis shows that you can estimate the number of zero crossings (E(Z)) and the number of upper turning points (E(P)) from the PSD.

E(Z) represents the number of (upward) zero crossings, or the mean level crossings for a signal with a non-zero mean. E(P) represents the number of peaks in the same sample. These are both specified for a typical unit interval (1s) sample. The irregularity factor (γ) is defined as the number of upward zero crossings divided by the number of peaks.

In this case, the number is 2, and the number of peaks is 3, so the irregularity factor is equal to 0.66. This number can theoretically only fall in the range 0 to 1. For a value of 1, the samples must get close to a harmonic loading, meaning the process must be narrow band. The broader the process is (that is, the more different frequencies are found in the samples), then the value for the irregularity factor tends towards 0.

To calculate these expectation values from a given PSD, you must look at the moments from a PSD.

{m_n} = \int\limits_0^\infty {{f^n}S(f)df} is called n-th moment of S.

With this you can calculate the values mentioned above:

rms(S) = \sqrt {{m_0}}

E(Z) = \sqrt {\frac{{{m_2}}}{{{m_0}}}}

E(P) = \sqrt {\frac{{{m_4}}}{{{m_2}}}}

\gamma = \frac{{E(Z)}}{{E(P)}} = \frac{{{m_2}}}{{\sqrt {{m_0}{m_4}} }}

The moments and expectations values give overall information about the process; however, for a fatigue analysis, you also need the distribution of cycles. The acoustic loads are random processes and therefore you can expect to get only the probability distribution of cycles, defined as:

Probability({\sigma _{a,\min }} < {\sigma _a} < {\sigma {a,\max }}) = \int\limits{{\sigma _{a,\min }}}^{{\sigma _{a,\max }}} {p({\sigma _a}),d{\sigma _a}}

This means that the number of cycles to be expected in a rainflow count during a test time T is:

E(RFM({\sigma {a,\min }}{,{a,\max }}),T) = T \cdot E(P)\int\limits_{{\sigma _{a,\min }}}^{{\sigma _{a,\max }}} {p({\sigma _a}),d{\sigma _a}}

The step to find p({\sigma _a}) for a given random process defined by its PSD, called amplitude estimator in the following section, is not unique and separates different methods.

Narrow-band solution

Bendat 1964 presented the theoretical basis for the first of these of these amplitude estimators, the narrow-band solution:

{p_{NB}}({\sigma _a}) = \frac{{2{\sigma _a}}}{{{m_0}}}{e^{ - \frac{{2\sigma _a^2}}{{{m_0}}}}}

As the name indicates, this solution is suitable only for a specific class, namely a narrow band of response conditions. On the other hand, investigations show that the narrow-band solution is always more conservative than assumptions of broader band conditions. Therefore, if you know that you have a narrow-band condition, you should also use this parameter.

Dirlik

Because real-life load spectra for durability problems are often far from narrow band, people developed several correction methods. Most were developed with reference to offshore platform design where interest in the techniques has existed for many years. Dirlik produced an empirical closed-form expression for the probability distribution of rainflow ranges, which was obtained using extensive computer simulations to model the signals using the Monte Carlo technique.

The complete formula reads:

{p_D}({\sigma _a}) = \frac{{\frac{{{D_1}}}{Q}{e^{\frac{{ - \zeta }}{Q}}} + \frac{{{D_2}\zeta }}{{{R^2}}}{e^{\frac{{ - {\zeta ^2}}}{{2{R^2}}}}} + {D_3}\zeta {e^{\frac{{ - {\zeta ^2}}}{2}}}}}{{2\sqrt {{m_0}} }}

where:

\zeta = \frac{{{\sigma _a}}}{{\sqrt {{m_0}} }}

{D_1} = \frac{{2{{({\xi m} - {\gamma ^2})}{}}}}{{1 + {\gamma ^2}}}

{D_2} = \frac{{1 - \gamma - {D_1} + D_1^2}}{{1 - R}}

{D_3} = 1 - {D_1} - {D_2}

Q = \frac{{5/4\left( {\gamma - {D_3} - {D_2}R} \right)}}{{{D_1}}}

R = \frac{{\gamma - {\xi _m} - D_1^2}}{{1 - {\gamma _m} - {D_1} + D_1^2}}

{\xi _m} = \frac{{{m_1}}}{{{m_0}}}\sqrt {\frac{{{m_2}}}{{{m_4}}}}

Dirlik's empirical amplitude estimator has been shown to be far superior, in terms of accuracy, than the previously available correction factors. Further on, Bishop provided theoretical verification of the method.

Equivalent sine wave

The equivalent sine wave method can be used for a very fast overview. It simply estimates a sine wave showing the same {m_0} and {m_2}.

Learn more

Quasi-static superposition, modal superposition, and transient analysis

Multiaxial fatigue

Rainflow projector

Temperature-dependent materials for Durability calculations

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