**Abstract**

The failure of systems to meet the specified requirements may have adverse effects on their integrity and reliability. The systems could be mechanical, electrical, structural, telecommunications, or electronic that are designed and built to satisfy certain technical specifications and operational requirements. Failure does not necessarily mean the occurrence of a disaster or damage to the system, but also the degraded performance of such systems is considered a failure. One of the essential indicators of the performance and reliability of a system is the probability of failure which is computed by probabilistic methods. One of these methods is the first-order reliability method (FORM). Using FORM to estimate the probability of failure of systems having a nonlinear or a higher-order performance function may provide inaccurate results that may lead to misleading conclusions. To resolve this issue, the second-order reliability method (SORM) is recommended to estimate the probability of failure. This chapter presents commonly used probabilistic approximation methods to estimate the probability of failure for nonlinear performance functions. Illustrative examples to demonstrate the application of these methods are provided at the end of the chapter.

**Keywords:** second-order approximation, failure analysis, probabilistic methods in engineering, second-order reliability, reliability engineering

## **1. Introduction**

Probability of failure can be calculated using analytical as well as simulation methods. Simulation methods such as the Monte Carlo technique can be used to calculate the probability of failure. It is an essential step in the calculation process to validate the analytical results. One of the analytical methods to compute the probability of failure is the first-order reliability method (FORM) which is based on the first-order expansion of the Taylor series. This method may provide inaccurate results when approximating nonlinear or higher-order performance functions. It should be noted that the performance function can be referred to as the limit state function (LSF).

The second-order reliability method, abbreviated as SORM, is used to resolve the nonlinearity issues of the performance function or the LSF. It uses the second-order expansion of the Taylor series to include the curvature of the LSF in the calculation to achieve better accuracy of the results.

**Figure 1.** *Representation of the linear and nonlinear LSF.*

**Figure 1** shows the failure regions for linear and nonlinear LSFs. It also shows the most probable point of failure (MPPF) as the tangent point on the limit-state surface, *Z U* <sup>∗</sup> ð Þ¼ 0, and the reliability index, *<sup>β</sup>*, is the shortest distance from the origin to the limit-state surface.

Breitung [1], Hohenbichler and Rackwitz [2], Tvedt [3, 4], and many other researchers came up with methods to calculate the probability of failure for nonlinear second-order LSFs. These methods use *β* obtained by FORM and the principal curvature, *ki*, of the limit state surface in the calculation. Essentially, *ki* estimates how much the curve diverges from the straight line, and it is an essential part of the calculation process. It is added to the calculation to provide a more accurate estimate of the probability of failure obtained by FORM. This chapter presents commonly used techniques that provide solutions to nonlinear second-order LSFs. The focus of this chapter is on SORM with the assumption that the random variables of the LSF are uncorrelated. A brief discussion about SORM and the commonly used methods for calculating the probability of failure are provided in the next section. Numerical examples are provided in Section 3 to demonstrate the calculation steps and provide in-depth understanding of these methods. Appendix A provides an overview of FORM and the steps involved in its calculation.

## **2. Second Order Reliability Method (SORM)**

The LSF may exhibit nonlinearity when random variables have a non-normal distribution or a nonlinear relationship. Moreover, transforming uncorrelated variables to correlated variables may present nonlinearity in the LSF. SORM is used to deal with the nonlinearity issues to provide more accurate results. It uses the second-order Taylor series expansion of the LSF at the most probable point of failure (MPPF). The vectorized form of the second-order Taylor Series expansion is used to express the LSF around the MPPF as:

$$\mathbf{Z}(\mathbf{U}) \approx \mathbf{Z}(\mathbf{U}^\*) + \nabla \mathbf{Z}(\mathbf{U}^\*)^t (\mathbf{U} - \mathbf{U}^\*) + \mathbf{0}.\\ \mathbf{5}(\mathbf{U} - \mathbf{U}^\*)^t \nabla^2 \mathbf{Z}(\mathbf{U}^\*)(\mathbf{U} - \mathbf{U}^\*) \tag{1}$$

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

where:


Since *<sup>U</sup>* <sup>∗</sup> is located on the limit-state surface, *Z U* <sup>∗</sup> ð Þ equals zero. *Z U*ð Þ is normalized by dividing Eq. (1) by <sup>j</sup>∇*Z U* <sup>∗</sup> ð Þj which yields:

$$Z\_{norm}(U) \approx \frac{\nabla Z(U^\*)^t (U - U^\*)}{|\nabla Z(U^\*)|} + \frac{\mathbf{0.5}(U - U^\*)^t \nabla^2 Z(U^\*)(U - U^\*)}{|\nabla Z(U^\*)|}\tag{2}$$

where <sup>j</sup>∇*Z U* <sup>∗</sup> ð Þj is the length of the gradient vector of the LSF at the MPPF and is equal to:

$$|\nabla Z(U^\*)| = \sqrt{\sum\_{i=1}^n \left(\frac{\partial Z(U^\*)}{\partial u\_i}\right)^2} \tag{3}$$

The approximation of LSF at the MPPF is expressed as:

$$Z\_{norm}(U) \approx \frac{\nabla Z(U^\*)^t (U - U^\*)}{|\nabla Z(U^\*)|} + \frac{\mathbf{0.5} (U - U^\*)^t \nabla^2 Z(U^\*) (U - U^\*)}{|\nabla Z(U^\*)|} = \mathbf{0} \tag{4}$$

From the first-order reliability approximation (FORM approximation) we have:

$$\frac{\nabla Z(U^\*)^t U^\*}{|\nabla Z(U^\*)|} = -\beta \tag{5}$$

$$\frac{\nabla Z(U^\*)}{|\nabla Z(U^\*)|} = a \tag{6}$$

*α: the directional cosines, α*1, … , *α<sup>n</sup> the components of the unit gradient vector α: These are the cosines of the angles between the vector β and the axes.*

Denoting <sup>∇</sup>2*Z U* <sup>∗</sup> ð Þ <sup>j</sup>∇*Z U* <sup>∗</sup> ð Þj as *H:*

$$\mathbf{H} = \frac{\nabla^2 \mathbf{Z}(\mathbf{U}^\*)}{|\nabla \mathbf{Z}(\mathbf{U}^\*)|} \tag{7}$$

**Figure 2.** *Second-order failure surface shown in the rotated coordinates.*

Using Eqs. (5) and (7), Eq. (4) yields:

$$Z\_{norm}(U) \approx \frac{\nabla Z(U^\*)^t U}{|\nabla Z(U^\*)|} + \beta + 0.5(U - U^\*)^t \mathbf{H}(U - U^\*) = \mathbf{0} \tag{8}$$

To go further with the solution of SORM, the *U* variables are transformed into a new set of standard normal random variables denoted as *U*<sup>0</sup> *<sup>i</sup>* where the axis of the last variable coincides with the *β* vector as indicated in **Figure 2** [5–7].

This is an orthogonal transformation where:

$$\mathbf{U}' = \mathbf{R}\mathbf{U} \tag{9}$$

*R*: orthogonal *n* by *n* rotation matrix, where *n* is the number of variables. Since *R* is an orthogonal rotation matrix, *<sup>R</sup>*�**<sup>1</sup>** <sup>¼</sup> *<sup>R</sup><sup>t</sup>*

$$\mathbf{U} = \mathbf{R}^t \mathbf{U}'\tag{10}$$

The matrix *R* is constructed in two steps. First, define an initial matrix, *R***<sup>01</sup>** that consists of rows representing the unit vectors of the axes of the input variables. Since, we want to make *β* vector coincide with the axis of the last variable, we can simply substitute the last row of matrix *R***<sup>01</sup>** with the directional cosines of *β*. The following row vectors are the components of the *R***<sup>01</sup>** matrix:

*r*<sup>01</sup> ¼ ½ � 100 *:::::::::* 0 *r*<sup>02</sup> ¼ ½ � 010 *:::::::::* 0 *: : :* (11)

$$r\_{0\mathfrak{n}} = \left[ -\frac{\partial Z(U^\*)\_{\rangle\_{\partial U\_1}}}{|\nabla Z(U^\*)|} \quad . \quad . \quad . \quad . -\frac{\partial Z(U^\*)\_{\rangle\_{\partial U\_\bullet}}}{|\nabla Z(U^\*)|} \right]$$

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

The Gram-Schmidt process is applied to orthogonalize the matrix [8]. The next step is to orthonormalize the matrix by dividing each element of the matrix by the length of its corresponding row vector.

Perform the Gram-Schmidt process for the matrix **R01** using Eqs. (12) and (13), in reverse order:

$$r\_n = r\_{0n} \tag{12}$$

$$r\_i = r\_{0i} - \sum\_{j=i+1}^{n} \frac{r\_j r\_{0i}^t}{r\_j r\_j^t} r\_j \quad i = 1, 2, \dots, n-1 \tag{13}$$

Then orthonormalization is performed for each vector to produce the matrix *R*. Using Eq. (10), Eq. (8) yields:

$$Z\_{norm}(U') \approx -U\_n' + \beta + 0.5(U' - U'^\*)^t \mathbf{RHR}^t (U' - U'^\*) = \mathbf{0} \tag{14}$$

The *H* matrix is the second-order derivative of the LSF at the design point *U* <sup>∗</sup> divided by the length of the gradient vector, <sup>j</sup>∇*Z U* <sup>∗</sup> ð Þj, and is formulated as:

$$H = \frac{1}{|\nabla Z(\boldsymbol{U}^{\prime\prime})|} \begin{bmatrix} \frac{\partial^2 Z(\boldsymbol{U}^{\prime\prime})}{\partial \boldsymbol{u}\_1^2} & \frac{\partial^2 Z(\boldsymbol{U}^{\prime\prime})}{\partial \boldsymbol{u}\_1 \partial \boldsymbol{u}\_2} & \cdots & \frac{\partial^2 Z(\boldsymbol{U}^{\prime\prime})}{\partial \boldsymbol{u}\_1 \partial \boldsymbol{u}\_n} \\\\ \frac{\partial^2 Z(\boldsymbol{U}^{\prime\prime})}{\partial \boldsymbol{u}\_2 \partial \boldsymbol{u}\_1} & \ddots & \ddots & \ddots & \vdots \\\\ \vdots & \ddots & \ddots & \ddots & \ddots \\\\ \frac{\partial^2 Z(\boldsymbol{U}^{\prime\prime})}{\partial \boldsymbol{u}\_n \partial \boldsymbol{u}\_1} & \cdots & \cdots & \frac{\partial^2 Z(\boldsymbol{U}^{\prime\prime})}{\partial \boldsymbol{u}\_n^2} \end{bmatrix} \tag{15}$$

Denoting *RHR<sup>t</sup>* as *B*, Eq. (14) becomes:

$$Z\_{norm}(U') \approx -U\_n' + \beta + \mathbf{0.5(U'^t \mathbf{B} U')} = \mathbf{0} \tag{16}$$

Then the solution of the second-order approximation is given as:

$$U\_n' = \beta + 0.5(\mathbf{U}^t \mathbf{B} \mathbf{U}') \tag{17}$$

*B* is an *(n* � *1)* by *(n* � *1)* matrix and *U*<sup>0</sup> is a *1* by *(n* � *1)* vector. The eigenvalues of matrix *B* are computed to obtain the main curvatures, *ki* 0 *s* , of the LSF at the MPPF. These *ki* 0 *s* are used in the calculation of the probability of failure for nonlinear LSF as indicated in the next subsection [5–7]. The 0*:*5 *U*0*<sup>t</sup> BU*<sup>0</sup> � � term is simplified to *n*

0*:*5 P�1 *i*¼1 *kiU*0<sup>2</sup> *<sup>i</sup>* . Thus, Eq. (17) is expressed in terms of the curvatures, *ki* 0 *s*, as:

$$U\_n' = \beta + 0.5 \sum\_{i=1}^{n-1} k\_i U\_i^2 \tag{18}$$

If *ki* >0, the LSF will have a convex failure region, and if *ki* < 0, it will have a concave failure region as shown in **Figure 3**. The figure shows the failure surfaces for an arbitrary LSF with three random variables.

#### **Figure 3**.

*Convex and concave failure regions for an arbitrary LSF with three random variables.*

The calculation of <sup>∇</sup>*Z U* <sup>∗</sup> ð Þ and <sup>∇</sup><sup>2</sup> *Z U* <sup>∗</sup> ð Þ can be performed in the X domain using the following equations:

$$\nabla Z\left(U^{\prime \*}\right) = \frac{\partial Z\left(U^{\prime \*}\right)}{\partial u\_i} = \frac{\partial Z(X^\*)}{\partial \varkappa\_i} \sigma\_{xi} \tag{19}$$

$$\nabla^2 Z(\mathbf{U}^{\prime \*}) = \frac{\partial^2 Z(\mathbf{U}^{\prime \*})}{\partial u\_i^2} = \frac{\partial^2 Z(\mathbf{X}^\*)}{\partial \mathbf{x}\_i^2} \sigma\_{\mathbf{x}\_i}^2 \tag{20}$$

where *σxi* is the standard deviation of the random variable *xi*.

To summarize, the main objective of using SORM is to improve the estimate of the probability of failure that was obtained by FORM. The calculation includes three key steps. These are, the calculation of the reliability index and its associated design points either in the U domain or X domain using FORM, determining the principal curvatures, and determining the probability of failure using one of the commonly used methods for nonlinear LSFs.

#### **2.1 Probability of failure**

Breitung, Hohenbichler, and Tvedt methods use a correction factor which is expressed in terms of *ki* and *β* to adjust the probability of failure obtained by FORM. The probability of failure for nonlinear LSF is then formulated as:

$$P\_f = \Phi(-\beta)(\mathcal{CF}) \tag{21}$$

The first term is the cumulative distribution function (cdf) of beta, *CF* is the SORM correction factor and *β* is the reliability index obtained by FROM.

#### *2.1.1 Breitung formulation*

Breitung formulation is expressed as:

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

$$P\_{f\_{Brewing}} = \Phi(-\beta) \prod\_{i=1}^{n-1} \frac{1}{\sqrt{(1+\beta k\_i)}} \tag{22}$$

#### *2.1.2 Hohenbichler formulation*

Hohenbichler formulation is expressed in terms of the cdf of beta (the first term), probability density function of beta (pdf) which is the upper term in the denominator and the cdf of beta which is the lower term in the denominator as:

$$P\_{f\_{Holenbichler}} = \Phi(-\beta) \prod\_{i=1}^{n-1} \frac{1}{\sqrt{\left(\mathbf{1} + \frac{\phi(-\beta)}{\Phi(-\beta)} k\_i\right)}} \tag{23}$$

#### *2.1.3 Tvedt formulation*

Tvedt formulation has three parts as indicated below:

$$P\_{f\_{Tend}} = \Phi(-\beta)f1 + f2 + f3 \tag{24}$$

$$f1 = \prod\_{i=1}^{n-1} \frac{1}{\sqrt{(1+\beta k\_i)}}$$

$$f2 = (\beta \Phi(-\beta) - \phi(-\beta)) \left( f1 - \left(\prod\_{i=1}^{n-1} \frac{1}{\sqrt{(1+(1+\beta)k\_i)}}\right)\right)$$

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

#### **2.2 Calculation steps**


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


## **3. Examples**

#### **3.1 Example 1**

The limit state function for a system has been formulated as:

*Z X*ð Þ¼ <sup>2</sup>*:*<sup>5</sup> *<sup>x</sup>*<sup>2</sup> <sup>1</sup> � *<sup>x</sup>*<sup>2</sup> <sup>2</sup> � 0*:*167 where *x*<sup>1</sup> � *N*ð Þ 2, 0*:*2 , *x*<sup>2</sup> � *N*ð Þ 2, 0*:*32 *:*



**Table 1.** *MCS results – Example 1.0, Part C.*

#### **Solution**

#### *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 :*

$$Z(\mathbf{U}) = 2.5(0.2\mu\_1 + 2)^2 - (0.32\mu\_2 + 2)^2 - 0.167$$

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

$$\begin{aligned} \mathbf{x}\_1 &= \mu\_{\mathbf{x}\_1} = 2\\ \mathbf{x}\_2 &= \mu\_{\mathbf{x}\_2} = 2 \end{aligned}$$

$$\begin{aligned} \mu\_1 &= \frac{\mathbf{x}\_1 - \mu\_{\mathbf{x}\_1}}{\sigma\_{\mathbf{x}1}} = \frac{2 - 2}{0.2} = \mathbf{0} \\ \mu\_2 &= \frac{\mathbf{x}\_2 - \mu\_{\mathbf{x}\_2}}{\sigma\_{\mathbf{x}2}} = \frac{2 - 2}{0.32} = \mathbf{0} \end{aligned}$$

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

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

$$Z(0,0) = 5.83300$$

*Compute the partial derivatives of the LSF:*

$$\frac{\partial Z(U^\*)}{\partial u\_1} = 0.2u\_1 + 2 = 2$$

$$\frac{\partial Z(U^\*)}{\partial u\_2} = -0.2048u\_2 - 1.28 = -1.28$$

*Compute the standard deviation of the LSF*:

$$\sigma\_{\overline{z}} = \sqrt{\sum\_{i=1}^{n} \left(\frac{\partial Z(U^\*)}{\partial u\_i}\right)^2} = \sqrt{(2)^2 + (-1.28)^2} = 2.3745$$

*Compute the initial reliability index β:*

$$
\beta = \frac{\mu\_x}{\sigma\_x} = \frac{5.8330}{2.3745} = 2.4565
$$

*Compute the directional cosines αi*:

$$a\_1 = -\frac{\left(\frac{\partial Z(U^\*)}{\partial u\_1}\right)}{\sqrt{\sum\_{i=1}^n \left(\frac{\partial Z(U^\*)}{\partial u\_i}\right)^2}} = -\frac{2}{2.3745} = -0.8423$$

$$a\_2 = -\frac{\left(\frac{\partial Z(U^\*)}{\partial u\_2}\right)}{\sqrt{\sum\_{i=1}^n \left(\frac{\partial Z(U^\*)}{\partial u\_i}\right)^2}} = -\frac{-1.28}{2.3745} = 0.5391$$

#### *Iteration 1*

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

$$u\_1 = \beta a\_1 = 2.4565 \ast (-0.8423) = -2.0690$$

$$u\_2 = \beta a\_2 = 2.4565 \ast (0.5391) = 1.3242$$

$$\alpha\_1 = \beta a\_1 \sigma\_{\ge 1} + \mu\_{\ge 1} = u\_1 \sigma\_{\ge 1} + \mu\_{\ge 1} = -2.0690 \ast 0.2 + 2 = 1.5862$$

$$\alpha\_2 = \beta a\_2 \sigma\_{\ge 1} + \mu\_{\ge 1} = u\_2 \sigma\_{\ge 1} + \mu\_{\ge 1} = 1.3242 \ast 0.32 + 2 = 2.4237$$

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

$$Z(U^\*) = Z(-2.0690, 1.3242\ ) = 0.2485$$

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

*The partial derivatives:*

$$\frac{\partial Z(U^\*)}{\partial u\_1} = 1.5862$$

*Failure Analysis – Structural Health Monitoring of Structure and Infrastructure Components*

$$\frac{\partial Z(U^\*)}{\partial u\_2} = -1.5512$$

*The standard deviation of Z:*

$$
\sigma\_x = \sqrt{(1.5862)^2 + (-1.5512)^2} = 2.2186
$$

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

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

$$= \frac{0.2485 - (1.5862 \ast (-2.0690) + (-1.5512 \ast 1.3242))}{2.2186}$$

$$\beta = 2.5171$$

We examine the tolerance:

$$\frac{|\beta\_i - \beta\_{i-1}|}{\beta\_{i-1}} = \frac{|2.5171 - 2.45650|}{2.45650} = 0.02466$$

$$0.02466 > 0.0001$$

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:*

$$a\_1 = -\frac{1.5862}{2.2186} = -0.7150$$

$$a\_2 = -\frac{-1.5512}{2.2186} = 0.6992$$

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

**Table 2** shows the results for the next iterations. As the table indicates, convergence occurred at the third iteration with an error less than the tolerance.

$$\frac{|\beta\_i - \beta\_{i-1}|}{\beta\_{i-1}} = \frac{|2.51171 - 2.51170|}{2.51170} = 0.000003981$$

$$0.000003981 \ll 0.0001$$

*β* is calculated to be *2.51171* and the MPPF is located at *U(-1.7758, 1.7762)* and *X(1.6448, 2.5684)*.

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

$$P\_f = \Phi(-\beta) = \mathbf{1} - \Phi(\beta) = \mathbf{1} - \Phi(2.51171) = 0.006007$$


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

**Table 2.**

*Summary of results – Example 1.0, Part A.*

### *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:*

$$\frac{\partial Z(U^\*)}{\partial u\_1} = 1.6448$$

$$\frac{\partial Z(U^\*)}{\partial u\_2} = -1.6438$$

$$|\nabla Z(U^\*)| = \sqrt{\sum\_{i=1}^n \left(\frac{\partial Z(U^\*)}{\partial u\_i}\right)^2} = \sqrt{\left(1.6448\right)^2 + \left(-1.6438\right)^2} = 2.3254$$

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

$$\mathbf{R\_{01}} = \begin{bmatrix} 1 & 0\\ -\frac{dZ(U^\*)\_{\acute{\beta}\mathbf{u}\_1}}{|\nabla Z(U^\*)|} & -\frac{dZ(U^\*)\_{\acute{\beta}\mathbf{u}\_2}}{|\nabla Z(U^\*)|} \end{bmatrix} = \begin{bmatrix} \mathbf{1} & \mathbf{0} \\ -\mathbf{0.7073} & \mathbf{0.7069} \end{bmatrix}.$$

The second 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:*

$$r\_n = r\_{02}$$

The last element of the matrix is calculated as:

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

*i=2*

$$r\_2 = r\_{02} = [-0.7073 \ 0.7069]$$

*i=1*

$$r\_1 = r\_{01} - \sum\_{j=i+1}^{n} \frac{r\_j r\_{02}^t}{r\_j r\_j^t} r\_j = \begin{bmatrix} 1 & 0 \end{bmatrix} - \frac{r\_1 r\_{02}^t}{r\_1 r\_1^t} r\_1$$

$$= \begin{bmatrix} 1 & 0 \end{bmatrix} - \frac{[-0.7073 & 0.7069] \begin{bmatrix} 1 \\ 0 \end{bmatrix}}{[-0.7073 & 0.7069] \begin{bmatrix} -0.7073 \\ 0.7069 \end{bmatrix}} [-0.7073 & 0.7069]$$

$$r\_1 = [0.4997 & 0.5000]$$

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

$$r\_2 = \frac{[0.4997 \, 0.50000]}{\sqrt{0.4997^2 + 0.5000^2}} = [0.7073 \, 0.7069]$$

The matrix *R* becomes:

$$\mathbf{R} = \begin{bmatrix} \mathbf{0.7069} & \mathbf{0.7073} \\ -\mathbf{0.7073} & \mathbf{0.7069} \end{bmatrix}$$

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

$$H = \frac{1}{|\nabla Z(U^\*)|} \begin{bmatrix} \frac{\partial^2 Z(U^\*)}{\partial u\_1^2} & \frac{\partial^2 Z(U^\*)}{\partial u\_1 \partial u\_2} \\ \frac{\partial^2 Z(U^\*)}{\partial u\_2 \partial u\_1} & \frac{\partial^2 Z(U^\*)}{\partial u\_2^2} \end{bmatrix} = \frac{1}{2.3254} \begin{bmatrix} 0.2 & 0 \\ 0 & -0.2048 \end{bmatrix} = \begin{bmatrix} 0.0860 & 0 \\ 0 & -0.0881 \end{bmatrix}$$

Compute *RHRt* matrix: *RHR<sup>t</sup>* **<sup>=</sup>** �0*:*<sup>0011</sup> �0*:*<sup>0870</sup> �0*:*<sup>0870</sup> �0*:*<sup>0010</sup> � �

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

The principal curvatures, *ki* 0 *s*, are computed by solving the eigenvalues of *RHR<sup>t</sup>* . The last column and last row are dropped from the above matrix, then the eigenvalues are obtained to determine the *ki* values. From the above matrix, there is only one *k* at the MPPF, with a value of *(*�*0.0011)*.

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

*Breitung method*

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

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

$$=\Phi(-2.5117)\frac{1}{\sqrt{(1+2.5117)(-0.0011)}}$$

$$P\_{f\_{Breatung}} = 0.00601565$$

*Hohenbichler method*

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

$$= \Phi(-2.5117) \frac{1}{\sqrt{\left(1 + \frac{\phi(-2.5117)}{\Phi(-2.5117)} (-0.0011)\right)}}$$

$$P\_{f\_{Huberible}} = 0.00601671$$

*Tvedt method*

$$f1 = \prod\_{i=1}^{n-1} \frac{1}{\sqrt{(1+\beta k\_i)}} = 1.00137$$

$$f2 = (\beta \Phi(-\beta) - \phi(-\beta)) \left(f1 - \left(\prod\_{i=1}^{n-1} \frac{1}{\sqrt{(1+(1+\beta)k\_i)}}\right)\right)$$

$$= (-0.00193)(1.00137 - (1.00192)) = 1.05670E - 06$$

$$f2 = 1.05670E - 06$$

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

$$= (1+2.5117)(2.5117(\Phi(-2.5117)) - \phi(-2.5117))$$

$$\left(1.00137 - \text{Real}\left(\frac{1}{\sqrt{(1+(j+2.5117))(-0.0011)}}\right)\right)$$

$$f3 = -3.03430E - 9$$

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

$$= (6.00743E - 3)(1.00137) + (1.05670E - 06) + (-3.03430E - 9)$$

$$P\_{f\_{T\text{mid}}} = 6.01671E - 03$$

See **Table 3**. *Part C*

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

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


#### **Table 3.**

*Summary of SORM results – Example 1.0 Part B.*
