NPTEL

Course On

STRUCTURAL

RELIABILITY Module # 05

Lecture 2

Course Format: Web

Instructor:

Dr. Arunasis Chakraborty

Department of Civil Engineering

Indian Institute of Technology Guwahati

Course Instructor: Dr. Arunasis Chakraborty

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2. Lecture 02: Example

Example

Ex # 01.

A cantilever steel beam of span = 2 is subjected to a point load = 100 at the free end. For this section, resisting moment capacity is 12. For collapse prevention condition limit state defined as below

Table 5.2.1 Random variables

S. No. Random Variable

()

1 1(kN/m2) 250 103 25 103

2 2(m3) 1.2 103 6.0 105

Solution. PDF for both the random variable is taken as Gaussian distribution. Figure 5.2.2 shows

the distribution of 1 and similarly Figure 5.2.3 shows the distribution of 2. In this problem response surface () is obtained by 3 iterations. For each iteration, value is calculated as shown in Table 5.2.2. In Figure 5.2.4, the original limit surface and limit surfaces of successive

iteration are shown. From this figure it can be clear that after 3 iteration () is enough

() = 12

Lecture 02: Example

Course Instructor: Dr. Arunasis Chakraborty

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approximated to replace the original (). That can be concluded from Table 5.2.2. Here, value converge after 3 iterations.

Table 5.2.2 value in each iteration

Iteration

1 2.9814

2 3.1468

3 3.1441

Figure 5.2.2 PDF of variable

Lecture 02: Example

Course Instructor: Dr. Arunasis Chakraborty

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Figure 5.2.3 PDF of variable

Figure 5.2.4 Response Surfaces in each iteration

Lecture 02: Example

Course Instructor: Dr. Arunasis Chakraborty

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Ex # 02.

Case 1: A cantilever steel beam of span = is subjected to two randomly varying point load

at the middle of the span and at the free end. Flexural rigidity of the beam is . kN-

m2. For the serviceability criteria, limit state would be as below

Table 5.2.3 Random variables

S. No. Random Variable

()

1 1(kN) 5 0.75

2 2(kN) 55 11

Correlation matrix is given

= . .

. .

Solution: The joint PDF of the variables are shown in Figure 5.2.5. In this case also random

variables are considered with Gaussian distribution. Figure 5.2.6 shows, by 2 successive

iterations () is close enough to replace the original limit surface (). Also from Table 5.2.4

it can be observed that value has been converged.

Table 5.2.4 value in each iteration

Iteration

1 5.0764

2 5.0764

2 1

1

=

325

5P1L3

48

P2L3

3

2

Lecture 02: Example

Course Instructor: Dr. Arunasis Chakraborty

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Figure 5.2.5 Joint PDF of and

Figure 5.2.6 Response surface in each iteration

Lecture 02: Example

Course Instructor: Dr. Arunasis Chakraborty

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Case 2: For the same problem, resisting moment capacity of the section is . For collapse

prevention condition limit state is defined as below

Table 5.2.5 Random variables

S. No. Random Variable

()

1 1 (kN/m2) 250 103 25 103

2 2 (m3) 2.2 103 1.1 104

3 1 (kN) 100 10

4 2 (kN) 100 10

Correlation matrix is given

=

. . . .

. . . .

. . . .

. . . .

Solution. In this case, joint PDF is shown in Figure 5.2.7 and considered distribution for all the

random variables are Gaussian. From Table 5.2.6, it is observable that after 3 iterations value of

is converged.

2 1

1

= 12 P1L1 P2L2

2

Lecture 02: Example

Course Instructor: Dr. Arunasis Chakraborty

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Table 5.2.6 value in each iteration

Iteration

1 3.7180

2 3.9049

3 3.9035

Figure 5.2.7 Joint PDF of and