*Part A*

Determine *β* and the MPPF in the *X* space and the *U* space. FORM is used to solve part A. The computation continues until the solution converges with a tolerance error of 0.0001. <sup>j</sup>*βi*�*βi*�1<sup>j</sup> *βi*�<sup>1</sup> <sup>&</sup>lt;0*:*<sup>0001</sup> *:*

*Assume the initial MPPF as the given mean of each variable. Transform the LSF Z*ð Þ *X to Z*ð Þ *U :*

$$Z(\mathbf{U}) = 24(\mathbf{0}.2u\_1 + \mathbf{2})^2 + \mathbf{13}(\mathbf{0}.32u\_2 + \mathbf{2})^2 + \left(\mathbf{0}.4u\_3 + \mathbf{4}\right)^2 - \mathbf{10}\mathbf{0}$$

*Compute the initial values of the design point in U space:*

$$\mathbf{x}\_1 = \mu\_{\mathbf{x}\_1} = \mathbf{2}$$

$$\mathbf{x}\_2 = \mu\_{\mathbf{x}\_2} = \mathbf{2}$$

*Perspective Chapter: Probabilistic Modeling of Failure – Nonlinear Approximation DOI: http://dx.doi.org/10.5772/intechopen.109266*

$$\mathbf{x}\_{3} = \boldsymbol{\mu}\_{\mathbf{x}\_{3}} = 4$$

$$\boldsymbol{\mu}\_{1} = \frac{\mathbf{x}\_{1} - \boldsymbol{\mu}\_{\mathbf{x}\_{1}}}{\sigma\_{\mathbf{x}1}} = \frac{2 - 2}{0.2} = \mathbf{0}$$

$$\boldsymbol{\mu}\_{2} = \frac{\mathbf{x}\_{2} - \boldsymbol{\mu}\_{\mathbf{x}\_{2}}}{\sigma\_{\mathbf{x}2}} = \frac{2 - 2}{0.32} = \mathbf{0}$$

$$\boldsymbol{\mu}\_{3} = \frac{\mathbf{x}\_{3} - \boldsymbol{\mu}\_{\mathbf{x}\_{3}}}{\sigma\_{\mathbf{x}3}} = \frac{4 - 4}{0.4} = \mathbf{0}$$

*Compute the initial estimate of Z U*ð Þ:

$$Z(0,0,0) = 64$$

*Compute the partial derivatives of the LSF:*

$$\frac{\partial Z(U^\*)}{\partial u\_1} = 1.92u\_1 + 19.2 = 19.2$$

$$\frac{\partial Z(U^\*)}{\partial u\_2} = 2.6624u\_2 + 16.64 = 16.64$$

$$\frac{\partial Z(U^\*)}{\partial u\_3} = 0.32u\_3 + 3.2 = 3.2$$

*Compute the standard deviation of the LSF*:

$$
\sigma\_{\overline{x}} = \sqrt{\sum\_{i=1}^{n} \left(\frac{\partial Z(U^\*)}{\partial u\_i}\right)^2} = \sqrt{\left(19.2\right)^2 + \left(16.64\right)^2 + \left(3.2\right)^2} = 25.6080
$$

*Compute the initial reliability index β:*

$$
\beta = \frac{\mu\_x}{\sigma\_x} = \frac{64}{25.6080} = 2.4992
$$

*Compute the directional cosines αi*:

$$\alpha\_1 = -\frac{\left(\frac{\partial Z(U^\*)}{\partial u\_1}\right)}{\sqrt{\sum\_{i=1}^n \left(\frac{\partial Z(U^\*)}{\partial u\_i}\right)^2}} = -\frac{19.2}{25.6080} = -0.74981$$

$$\alpha\_2 = -\frac{\left(\frac{\partial Z(U^\*)}{\partial u\_2}\right)}{\sqrt{\sum\_{i=1}^n \left(\frac{\partial Z(U^\*)}{\partial u\_i}\right)^2}} = -\frac{16.64}{25.6080} = -0.64981$$

$$\alpha\_3 = -\frac{\left(\frac{\partial Z(U^\*)}{\partial u\_3}\right)}{\sqrt{\sum\_{i=1}^n \left(\frac{\partial Z(U^\*)}{\partial u\_i}\right)^2}} = -\frac{3.2}{25.6080} = -0.1250$$

*The initial calculation of the probability of failure is:*

$$P\_f = \Phi(-\beta) = \mathbf{1} - \Phi(\beta) = \mathbf{1} - \Phi(2.4992) = 0.006222$$

#### *Iteration 1*

*Determine the new MPPF/design point in U space and X space:*

$$u\_1 = \beta a\_1 = 2.4992 \ast (-0.7498) = -1.8738$$

$$u\_2 = \beta a\_2 = 2.4992 \ast (-0.64980) = -1.6240$$

$$u\_3 = \beta a\_3 = 2.4992 \ast (-0.1250) = -0.3123$$

$$\alpha\_1 = \beta a\_1 \sigma\_{\mathbf{x}\_1} + \mu\_{\mathbf{x}\_1} = u\_1 \sigma\_{\mathbf{x}\_1} + \mu\_{\mathbf{x}\_1} = -1.8738 \ast 0.2 + 2 = 1.6252$$

$$\alpha\_2 = \beta a\_2 \sigma\_{\mathbf{x}\_2} + \mu\_{\mathbf{x}\_2} = u\_2 \sigma\_{\mathbf{x}\_2} + \mu\_{\mathbf{x}\_2} = -1.6240 \ast 0.32 + 2 = 1.4803$$

$$\alpha\_3 = \beta a\_3 \sigma\_{\mathbf{x}\_3} + \mu\_{\mathbf{x}\_3} = u\_3 \sigma\_{\mathbf{x}\_3} + \mu\_{\mathbf{x}\_3} = -0.3123 \ast 0.4 + 4 = 3.8751$$

*Compute the LSF in terms of the new design points:*

$$Z(U^\*) = Z\ (-1.8738, -1.624, -0.3123) = 6.8972$$

*Compute the partial derivatives and the standard deviation of the LSF at the new design points:*

*The partial derivatives:*

$$\frac{\partial Z(U^\*)}{\partial u\_1} = 1.92u\_1 + 19.2 = 15.6022$$

$$\frac{\partial Z(U^\*)}{\partial u\_2} = 2.6624u\_2 + 16.64 = 12.3163$$

$$\frac{\partial Z(U^\*)}{\partial u\_3} = 0.32u\_3 + 3.2 = 3.1001$$

*The standard deviation of Z:*

$$
\sigma\_{\overline{x}} = \sqrt{(15.6022)^2 + (12.3163)^2 + (3.1001)^2} = 20.1179
$$

*Compute the new β in terms of the new design points:*

$$\beta = \frac{Z(\mathbf{U}^\*) - \sum\_{i=1}^n \frac{\partial Z(U)}{\partial u\_i} \left( u\_i^\* \right)}{\sqrt{\sum\_{i=1}^n \left( \frac{\partial Z(U^\*)}{\partial u\_i} \right)^2}}$$

$$= \frac{6.8972 - (15.6022 \ast (-1.8738) + 12.3163 \ast (-1.6240) + 3.1001 \ast (-0.3123))}{20.1179}$$

$$= 2.8384$$

We examine the tolerance:

*Perspective Chapter: Probabilistic Modeling of Failure – Nonlinear Approximation DOI: http://dx.doi.org/10.5772/intechopen.109266*

$$\frac{|\beta\_i - \beta\_{i-1}|}{\beta\_{i-1}} = \frac{|2.8384 - 2.4992|}{2.4992} = 0.1357$$

$$0.1357 > 0.0001$$

Since the error is greater than the established tolerance (0.0001), the computation should continue until convergence is obtained.

*The next step is to compute the directional cosines:*

$$a\_1 = -\frac{(78.011 \ast 0.2)}{20.1179} = -0.7755$$

$$a\_2 = -\frac{(38.488 \ast 0.32)}{20.1179} = -0.6122$$

$$a\_3 = -\frac{(7.7502 \ast 0.4)}{20.1179} = -0.1541$$

*The computation continues until β converges following the steps mentioned in Appendix A. Once convergence is obtained the probability of failure is calculated.*

**Table 5** shows the results for the next iterations. As the table indicates, the convergence occurred at the third iteration with an error less than the tolerance.


#### **Table 5.** *Summary of the results – Example 2.0, Part A.*

$$\frac{|\beta\_i - \beta\_{i-1}|}{\beta\_{i-1}} = \frac{|2.84463 - 2.84461|}{2.84461} = 0.000007031$$

$$0.000007031 \ll 0.0001$$

*β* is calculated to be 2.84463 and the MPPF is located at *U*(�2.1911, �1.7580, �0.4478) and *X*(1.5618, 1.4375, 3.8209).

The probability of failure is computed in terms of the final value of *β* :

$$P\_f = \Phi(-\beta) = \mathbf{1} - \Phi(\beta) = \mathbf{1} - \Phi(2.84463) = 0.002223$$

#### *Part B*

Determine the probability of failure using SORM by applying *Breitung, Hohenbichler*, and *Tvedt* methods.

*Determine the length of the gradient vector at the MPPF:* The partial derivatives at the third iteration are:

$$\frac{\partial Z(U^\*)}{\partial u\_1} = 14.9931$$

$$\frac{\partial Z(U^\*)}{\partial u\_2} = 11.9596$$

$$\frac{\partial Z(U^\*)}{\partial u\_3} = 3.0567$$

$$|\nabla Z(U^\*)| = \sqrt{\sum\_{i=1}^n \left(\frac{\partial Z(U^\*)}{\partial u\_i}\right)^2} = \sqrt{(14.9931)^2 + (11.9569)^2 + (3.0567)^2} = 19.4208$$

*Construct the matrix* **R01** *using Eq. (11):*

$$\mathbf{R\_{01}} = \begin{bmatrix} \mathbf{1} & \mathbf{0} & \mathbf{0} \\ \mathbf{0} & \mathbf{1} & \mathbf{0} \\ -\frac{\mathrm{d}Z(U^{\prime})\_{\mathrm{du}\_{1}}}{\sqrt{\sum\_{i=1}^{n}\left(\frac{\mathrm{d}Z(U^{\prime})}{\mathrm{du}\_{i}}\right)^{2}}} & -\frac{\mathrm{d}Z(U^{\prime})\_{\mathrm{du}\_{1}}}{\sqrt{\sum\_{i=1}^{n}\left(\frac{\mathrm{d}Z(U^{\prime})}{\mathrm{du}\_{i}}\right)^{2}}} & -\frac{\mathrm{d}Z(U^{\prime})\_{\mathrm{du}\_{3}}}{\sqrt{\sum\_{i=1}^{n}\left(\frac{\mathrm{d}Z(U^{\prime})}{\mathrm{du}\_{i}}\right)^{2}}} \\ \mathbf{R\_{01}} = \begin{bmatrix} \mathbf{1} & \mathbf{0} & \mathbf{0} \\ \mathbf{0} & \mathbf{1} & \mathbf{0} \\ -0.7720\mathbf{1} & -0.6158\mathbf{1} & -0.1573\mathbf{9} \end{bmatrix} \end{bmatrix}$$

The last row is the directional cosines of the reliability index, *β*, at the MPPF.

*Perform Gram-Schmidt orthogonalization for the matrix R***<sup>01</sup>** *using Eqs.* (12) and (13)*, and perform orthonormalization of each row vector to come up with the matrix R:*

$$r\_{\mathfrak{n}} = r\_{0\mathfrak{n}} = r\_{0\mathfrak{z}}$$

$$r\_i = r\_{0i} - \sum\_{j=i+1}^{n} \frac{r\_j r\_{0i}^t}{r\_j r\_j^t} r\_j$$

*Perspective Chapter: Probabilistic Modeling of Failure – Nonlinear Approximation DOI: http://dx.doi.org/10.5772/intechopen.109266*

*i=3*

$$r\_3 = r\_{03} = \begin{bmatrix} -0.77201 & -0.61581 & -0.15739 \end{bmatrix}$$

The remaining elements of the matrix are calculated as: *i=2*

$$r\_2 = r\_{02} - \sum\_{j=2}^{n} \frac{r\_j r\_{02}^t}{r\_j r\_j^t} r\_j = \begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix} - \frac{r\_3 r\_{01}^t}{r\_3 r\_3^t} r\_3$$

$$= \begin{bmatrix} -0.77201 & -0.61581 & -0.15799 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$$

$$= \begin{bmatrix} 0 & 1 & 0 \end{bmatrix} - \frac{\begin{bmatrix} -0.77201 & -0.61581 & -0.15799 \end{bmatrix} \begin{bmatrix} -0.77201 & -0.61581 & -0.15799 \end{bmatrix}}{\begin{bmatrix} -0.61581 & -0.61581 \\ -0.61581 \end{bmatrix}} \begin{bmatrix} -0.61581 & -0.61581 & -0.61573 \end{bmatrix}$$

$$r\_2 = \begin{bmatrix} -0.47541 & 0.62077 & -0.096925 \end{bmatrix}$$

Normalizing the elements of the row vector, *r*<sup>2</sup> becomes:

$$r\_2 = \frac{[-0.47541 \quad 0.62077 \quad -0.096925]}{\sqrt{(-0.47541)^2 + 0.62077^2 + (-0.096925)^2}}$$

$$r\_2 = [-0.6034 \ 0.78789 - 0.12302]$$

Following the same procedures row vector *r*<sup>1</sup> becomes:

$$r\_1 = [0.19977\ 0 - 0.97984]$$

The matrix *R* becomes:

$$\mathbf{R} = \begin{bmatrix} 0.19977 & 0 & -0.97984 \\ -0.6034 & 0.78789 & -0.12302 \\ -0.77201 & -0.61581 & -0.15739 \end{bmatrix}.$$

*Compute the second-order derivative of the LSF at the design point, U* <sup>∗</sup> *, using Equation (20) to obtain the H matrix:*

$$H = \frac{1}{|\nabla Z(U^\*)|} \begin{bmatrix} \frac{\partial^2 Z(U^\*)}{\partial u\_1^2} & \frac{\partial^2 Z(U^\*)}{\partial u\_1 \partial u\_2} & \frac{\partial^2 Z(U^\*)}{\partial u\_1 \partial u\_3} \\\\ \frac{\partial^2 Z(U^\*)}{\partial u\_2 \partial u\_1} & \frac{\partial^2 Z(U^\*)}{\partial u\_2^2} & \frac{\partial^2 Z(U^\*)}{\partial u\_2 \partial u\_3} \\\\ \frac{\partial^2 Z(U^\*)}{\partial u\_3 \partial u\_1} & \frac{\partial^2 Z(U^\*)}{\partial u\_3 \partial u} & \frac{\partial^2 Z(U^\*)}{\partial u\_3^2} \end{bmatrix}$$

$$\eta = \frac{1}{19.4208} \begin{bmatrix} 1.9200 & 0 & 0 \\ 0 & 2.6624 & 0 \\ 0 & 0 & 0.3200 \end{bmatrix} = \begin{bmatrix} 0.0989 & 0 & 0 \\ 0 & 0.1371 & 0 \\ 0 & 0 & 0.0165 \end{bmatrix}$$

**159**

Compute *RHRt* matrix:

$$\mathbf{RHR}^t = \begin{bmatrix} 0.0197650 & -0.0099306 & -0.0127056 \\ -0.0099306 & 0.1213464 & -0.0201427 \\ -0.0127056 & -0.0201427 & 0.1113186 \end{bmatrix}$$

*Compute the eigenvalues of the matrix B to obtain principal curvatures (ki's):*

The principal curvatures, *ki*'s, are computed by solving the eigenvalues of *RHR<sup>t</sup>* . To do that, the last column and the last row are dropped from the matrix. The above matrix becomes:

$$\mathbf{RHR}^t = \begin{bmatrix} 0.0197650 & -0.0099306 \\ -0.0099306 & 0.1213464 \end{bmatrix}$$

Applying the eigenvalue method to obtain the principal curvatures:

*k* ¼ ½ � 0*:*018803 0*:*122308

*Calculate the probability of failure Pf using Eqs. (22)–(24):*

*Breitung method*

$$P\_{f\_{Brediting}} = \Phi(-\beta) \prod\_{i=1}^{n-1} \frac{1}{\sqrt{(1+\beta k\_i)}}$$

$$= \Phi(-2.8446\ ) \frac{1}{\sqrt{(1+2.8446(0.018803))(1+2.8446(0.122308))}}$$

$$P\_{f\_{Brediting}} = 0.0018656$$

*Hohenbichler method*

$$P\_{f\_{Holonbubble}} = \Phi(-\beta) \prod\_{i=1}^{n-1} \frac{1}{\sqrt{\left(1 + \frac{\phi(-\beta)}{\Phi(-\beta)} k\_i\right)}}$$

$$= \frac{\Phi(-2.8446\ )}{\sqrt{\left(1 + \left(\frac{\phi(-2.8446\ )}{\Phi(-2.8446\ )}\right)(0.018803)\right)\left(1 + \left(\frac{\phi(-2.8446\ )}{\Phi(-2.8446\ )}\right)(0.122308)\right)}}$$

$$= 0.0018363$$

*Tvedt method*

$$f\mathbf{1} = \prod\_{i=1}^{n-1} \frac{\mathbf{1}}{\sqrt{(\mathbf{1} + \beta k\_i)}} = 0.83918$$

$$f\mathbf{2} = (\beta \Phi(-\beta) - \phi(-\beta)) \left( f\mathbf{1} - \left( \prod\_{i=1}^{n-1} \frac{\mathbf{1}}{\sqrt{((\mathbf{1} + (\mathbf{1} + \beta)k\_i)}} \right) \right)^2$$

**160**

*Perspective Chapter: Probabilistic Modeling of Failure – Nonlinear Approximation DOI: http://dx.doi.org/10.5772/intechopen.109266*

$$= ((2.8446)(0.0022231) - 0.0069786)(0.83918 - (0.9657)(0.82472))$$

$$f2 = -2.7977541E - 05$$

$$f3 = (1 + \beta)(\beta \Phi(-\beta) - \phi(-\beta))\left(f1 - \text{Real}\left(\prod\_{i=1}^{n-1} \frac{1}{\sqrt{(1 + (j + \beta)k\_i)}}\right)\right)$$

$$= (1 + 2.8446)(2.8446(\Phi(-2.8446)) - \phi(-2.8446))\left(0.83918$$

$$- \text{Real}\left(\frac{1}{\sqrt{(1 + (j + 2.8446)(0.018803))(1 + (j + 2.8446)(0.122308))}\right)\right)$$

$$f3 = -7.583962E - 06$$

$$P\_{T\_{\text{Total}}} = \Phi(-\beta f1 + f2 + f3$$

$$= (0.0022231)(0.83918) + (-2.79775E - 05) + (-7.58396E - 06)$$

$$P\_{f\_{Tuedt}} = 0.0018300$$

See **Table 6**.

#### *Part C*

Compare the results of parts *A* and *B* with results obtained by Monte Carlo simulation:


**Table 6.**

*Analytical methods results – Example 2.0, Part B.*

The simulation was conducted using 2e5 and 1e6 simulation cycles as shown in **Table 4**. The results obtained by *Breitung*,*Tvedt,* and *Hohenbichler* methods are close to the Monte Carlo simulation results when using 1e6 simulation cycles.

#### **3.3 Example 3**

Determine the second-order reliability index (*βSORM*) for the probability of failure calculated in example 2.

#### **Solution**

The reliability index for the second-order LSF can be calculated by taking the inverse of the cumulative distribution of the probability of failure as:

$$
\boldsymbol{\beta}\_{\text{SORM}} = \boldsymbol{\Phi}^{-1} \left[ \mathbf{1} - P\_f \right]
$$

$$
\beta\_{Breditung} = \Phi^{-1}[1 - 0.0018656] = 2.90004$$

$$
\beta\_{Teedt} = \Phi^{-1}[1 - 0.0018300] = 2.90606$$

$$
\beta\_{Huberichler} = \Phi^{-1}[1 - 0.0018363] = 2.90499$$

$$
\beta\_{MC} = \Phi^{-1}[1 - 0.0018360] = 2.90505$$

The absolute difference between Monte Carlo method and the other methods is calculated below:

> Breitung method ¼ j2*:*90505 � 2*:*90004j ¼ 0*:*00500 Tvedt method ¼ j2*:*90505 � 2*:*90606j ¼ 0*:*00101 Hohenbichler method ¼ j2*:*90505 � 2*:*90499j ¼ 0*:*00006

In this example, the reliability index obtained by *Hohenbichler* is the closest to the reliability index obtained by *Monte Carlo* method.

## **4. Summary and concluding remarks**

This chapter presented commonly used analytical probabilistic methods for determining the probability of failure for nonlinear and higher-order systems. They included Breitung, Hohenbichler, and Tvedt methods. Furthermore, Monte Carlo Simulation was used to estimate the probability of failure to validate the results obtained by the methods mentioned above. It is concluded that the results obtained by Breitung, Tvedt, and Hohenbichler methods are closer to the results obtained by Monte Carlo Simulation when using 1e6 simulation cycles. Future work should include the analysis of multi-dimensional and higher-order failure functions having correlated variables. Moreover, it is recommended that future work apply finite element method to gain more insight into how the methods compare with each other.

## **A. An overview of FORM**

## **A.1 Failure modeling**

The performance function, also called the limit state function (LSF), is formulated in terms of a system's load and capacity. Both the load, *L,* and the capacity, *C*, have an impact on the performance of the system. The system fails when the load exceeds the capacity or when the LSF becomes less than zero. The LSF can be expressed as:

$$Z = \mathbb{C} - L \tag{25}$$

The capacity *C* and the load *L* are formulated in terms of random variables, *x*1, *x*2, … *xn*:

$$Z(\mathbf{X}) = Z(\mathbf{x}\_1, \mathbf{x}\_2, \dots, \dots, \dots, \mathbf{x}\_n) \tag{26}$$

*Perspective Chapter: Probabilistic Modeling of Failure – Nonlinear Approximation DOI: http://dx.doi.org/10.5772/intechopen.109266*

These variables are assumed to be statistically independent random variables having a normal distribution. The probability of failure can be expressed as:

$$P\_f = P(Z < 0) \tag{27}$$

This is the probability that the LSF becomes less than zero. In **Figure A1**, this is the region that exists above the straight line for linear LSF and above the curve for nonlinear LSF. The safe region or nonfailure region is below the curve, and this is the region where the LSF becomes greater than zero. The LSF becomes at the limit state when the LSF becomes equal to zero as **Figure A1** indicates.

Also, the figure shows the reliability index, *β*, which is the shortest distance from the origin to the surface. It is calculated as the ratio of the mean, *μZ*, to the standard deviation, *σZ*, of the LSF.

$$
\beta = \frac{\mu\_Z}{\sigma\_Z} \tag{28}
$$

$$
\mu\_Z \approx Z(\mu\_{X\_1}, \mu\_{X\_2}, \dots, \dots, \dots, \dots, \mu\_{X\_n}) \tag{29}
$$

$$\sigma\_{\mathbf{z}} = \sqrt{\sum\_{i=1}^{n} \left(\frac{\partial Z(X^\*)}{\partial \mathbf{x}\_i}\right)^2 \sigma\_{\mathbf{x}\_i}^2} \tag{30}$$

Then, the probability of failure is expressed in terms of *β* as:

$$P\_f = \phi(-\beta) = \mathbf{1} - \Phi(\beta) \tag{31}$$

**Figure A1.** *Representation of linear and nonlinear LSF – original coordinates (X domain).*

## **A.2 First Order Reliability Method (FORM)**

This section will use the Hasofer and Lind method to calculate the probability of failure by FORM [9]. The first step is to transform the normal random variables into standard normal variables as:

$$
\mu\_i = \frac{\varkappa\_i - \mu\_{\varkappa\_i}}{\sigma\_{\varkappa\_i}}, \quad i = 1, 2, \ldots, n \tag{32}
$$

$$\omega\_{i} = \mu\_{i} \* \sigma\_{\mathbf{x}\_{i}} + \mu\_{\mathbf{x}\_{i}}, \quad i = 1, 2, \ldots, n \tag{33}$$

where *μxi* and *σxi* are the mean and the standard deviation of the random variable *xi* respectively, and *ui* is the transformed standard normal variable. The LSF is then formulated in terms of the standard normal variables as:

$$Z(U) = Z(\mu\_1 \* \sigma\_{\ge 1} + \mu\_{\ge 1}, \mu\_2 \* \sigma\_{\ge 1} + \mu\_{\ge 1}, \dots, \mu\_n \* \sigma\_{\ge 1} + \mu\_{\ge n}) = \mathbf{0} \tag{34}$$

**Figure A2** shows the representation of linear and nonlinear LSF and the most probable point of failure, *MPPF* : *u*<sup>∗</sup> 1 , *u*<sup>∗</sup> <sup>2</sup> Þ, or sometimes referred to as the design point. The next subsection outlines the steps for calculating the probability of failure using FORM.

**Figure A2.** *Representation of linear and nonlinear LSF – transformed coordinates (U domain).*
