Consider the fatigue life of a structural joint for which the S/N curve has been found to be Nf = K (Sr)^-4. In actual use, the joint will be subjected to a narrowband stochastic stress { X(t) } that is mean-zero and has standard deviation σX. This narrowband stress can be written as

Question

Consider the fatigue life of a structural joint for which the S/N curve has been found to be $N_{f}=K(S_{r})^{-4}$ . In actual use, the joint will be subjected to a narrowband stochastic stress $\left\{X(t)\right\}$ that is mean-zero and has standard deviation $\sigma _{X}$ . This narrowband stress can be written as $X(t)=A(t)\cos [\omega _{c}t+\theta (t)]$ , with $A(t)$ and $\theta (t)$ being independent and $\theta (t)$ uniformly distributed on the set of possible values. Rather than being Rayleigh distributed, it is believed that $A(t)$ has the one-sided Gaussian distribution

p_{A(t)}(u)=\frac{1}{\pi^{1/2}\sigma _{X}}\exp \left(\frac{-u^{2}}{4\sigma ^{2}_{X}} \right)  U(u)

Compare the fatigue life predicted by the Rayleigh approximation, the use of $S_{r}=2A(t)$ and $p_{A(t)}(u)$ in Eq. 11.49 $\frac{1}{E(N(T))} \equiv E(\Delta D)=K^{-1}E(S_{r}^{m})$ , and the use of Eq. 11.66

$GR=\left(1+\frac{m(m-1)(kurtosis-3)}{24} \right) ^{-1}$ for $kurtosis \gt 3$

$GR=1+\frac{m(m-1)(3- kurtosis)}{24} ^{-1}$ for $kurtosis \lt 3$

[latex] p_{A(t)}(u)=\frac{1}{\pi^{1/2}\sigma _{X}}\exp \left(\frac{-u^{2}}{4\sigma ^{2}_{X}} ight) U(u) [/latex]

Compare the fatigue life predicted by the Rayleigh approximation, the use of [latex] S_{r}=2A(t) [/latex] and [latex] p_{A(t)}(u) [/latex] in Eq. 11.49 [latex]\frac{1}{E(N(T))} \equiv E(\Delta D)=K^{-1}E(S_{r}^{m}) [/latex], and the use of Eq. 11.66

[latex]GR=\left(1+\frac{m(m-1)(kurtosis-3)}{24} ight) ^{-1} [/latex] for [latex] kurtosis \gt 3[/latex]

[latex]GR=1+\frac{m(m-1)(3- kurtosis)}{24} ^{-1} [/latex] for [latex] kurtosis \lt 3[/latex]

Accepted Answer

Before starting on the fatigue computation, it is appropriate to check for consistency of the model. In particular, we find that

[latex] E[A^{2}(t)]=\int_{0}^{\infty }{u^{2}} p_{A(t)}(u)du=2\sigma ^{2}_{X} [/latex]

so the narrowband process gives [latex] E[X^{2}(t)]=E[A^{2}(t)]/2=\sigma ^{2}_{X} [/latex] (see Example 4.3). Thus, the model is consistent with [latex] \sigma _{X} [/latex] being the standard deviation of the stress. Now we note that the Rayleigh approximation from Eqs. 11.50 [latex]E(\Delta D)=K^{-1}(2)^{3m/2}\lambda _{0}^{m/2}\Gamma \left(1+\frac{m}{2} ight) [/latex] and 11.51 [latex] E(T)= \frac{2\pi \sigma _{X}}{E(\Delta D)\sigma _{X}} =\frac{2\pi }{E(\Delta D)}\left(\frac{\lambda _{0}}{\lambda _{2}} ight) ^{1/2}[/latex] is [latex] E(T_{Ray})=\pi K/[64(\lambda _{0})^{3/2}(\lambda _{2})^{1/2}] [/latex] , as in Example 11.9. For this narrowband process we can say that [latex] \lambda _{2}=\omega ^{2}_{c}\lambda _{0} [/latex], and we also know that [latex] \lambda _{0}=\sigma ^{2}_{X} [/latex] . This allows the Rayleigh approximation to be rewritten as

[latex] E(T_{Ray})=\pi K/(64\omega _{c}\sigma ^{4}_{X}) [/latex]

Using [latex] S_{r}=2A(t) [/latex] in Eq. 11.49 [latex]\frac{1}{E(N(T))} \equiv E(\Delta D)=K^{-1}E(S_{r}^{m}) [/latex] gives

[latex] E(\Delta D)=K^{-1}E(S^{4}_{r})=2^{4}K^{-1}\int_{0}^{\infty }{u^{4}} p_{A(t)}(u)du=192K^{-1}\sigma ^{4}_{X} [/latex]

and Eq. 11.51 then gives the expected fatigue life as [latex] E(T)=\pi K/[96\omega _{c}(\sigma _{X})^{4}] [/latex] . Thus, this gives a prediction of [latex] E(T) [/latex] that is 33% smaller than the Rayleigh approximation.

In order to use Eq. 11.66

[latex]GR=\left(1+\frac{m(m-1)(kurtosis-3)}{24} ight) ^{-1} [/latex] for [latex] kurtosis \gt 3[/latex]

[latex]GR=1+\frac{m(m-1)(3- kurtosis)}{24} ^{-1} [/latex] for [latex] kurtosis \lt 3[/latex]

we must find the kurtosis of the stress process. To do this, we first integrate over the possible values for [latex] heta (t) [/latex] to find that

[latex] E[X^{4}(t)]=E[A^{4}(t)]\int_{0}^{2\pi }{\frac{\cos ^{4}(\psi )}{2\pi} } d\psi =\frac{3}{8} E[A^{4}(t)] [/latex]

From the integral used in the previous step, we find that [latex] E[A^{4}(t)]=12\sigma ^{4}_{X} [/latex] so [latex] E[X^{4}(t)]=4.5\sigma ^{4}_{X} [/latex] . Thus this process has [latex] kurtosis=4.5 [/latex] , and Eq. 11.66 then gives [latex] GR=24/[24+12(1.5)]\approx 0.571 [/latex] , which predicts a 43% reduction in fatigue life as compared with the Rayleigh approximation (which would be appropriate for a Gaussian process). This value is relatively consistent with the 33% reduction calculated in the previous step. To neglect the non-Gaussian effect would be a more serious error, because it would result in a fatigue life prediction that was significantly too large.

Question 11.12: Consider the fatigue life of a structural joint for which th...

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