NPTEL STRUCTURAL RELIABILITY

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1 NPTEL Course On STRUCTURAL RELIABILITY Module # 0 Lecture Course Format: eb Instructor: Dr. Arunasis Chakraborty Department of Civil Engineering Indian Institute of Technology Guwahati

2 . Lecture 0: System Reliability Till now, the discussions are concentrated on failure probability or reliability analysis at the elemental level of the structures. The examples discussed in the previous chapters are on design of any beam or column of the civil structures. Some idea about load model and code calibration is also given in the preceding lectures. A civil structure is consisted number of such elements. For the whole structure, its strength depends on the strength of this elements and its probabilistic behaviour is dependent on the of the individual component s performance, which is random in nature. If civil engineering structures are compared with a system then to evaluate its performance under certain condition, reliability of its element must be known to us. The reliability of component is the performance under the condition for which it is made. For a system, failure can occurred in various ways and depends on the combination of its component failure. Combining this failure modes structure reliability can be obtained. In a broad way system is classified in three classes: () series system, () parallel or redundant system and () mixed system. Series System Figure.. Series system In this system, as like in a chain, if a single component s performance is below the satisfactory level i.e. failure the system undergoes failure. So series system failure depends on its weakest element. A statistically determined structure is an example of series system because failure any one of its member will be reason of whole structure failure. As shown in Figure.., if it is assumed that only columns will undergo failure then it will represent a series system. To illustrate the concept, a block diagram is shown in Figure.., where block is indicated as weakest. So failure of will be cause of failure of whole system.

A civil structure is consisted number of such elements.

3 Lecture 0: System Reliability N Figure.. Block diagram for series system For reliability of the system, every component should function satisfactory p ss = P(S S S n ).. In Eq..., S is the event that component i will work satisfactory, p ss is the probability of survival of the system. If the events S i are independent, then p ss = P S P S P S n n = p fi i=.. where, p fi is the probability of failure of the component. This type of system is also known as weakest link model. In civil engineering field, the value of p fi is very small. So, Eq... can be further modified as p ss n p fi i= as p fi.. and p fs n p fi i=.. here, p fs is probability of failure of system. Parallel or Redundant System If the elements in a structure oriented such a way that failure of one or two components does not influence the failure of the whole structure then this structural system will be classed in parallel system. In Figure.. shows a system in which failure considered only in columns. So failure of one or two column will not be reason for failure of the structure. In Figure.., a block diagram is shown for a parallel system.

. where, p fi is the probability of failure of the component. This type of system is also known as weakest link model. In civil engineering field, the value of p fi is very small. So, Eq.

4 Lecture 0: System Reliability Figure.. Parallel system The reliability of this system would be N Figure.. Block diagram for parallel system p ss = p fs = P S c S c S n c.. here, S i c is the event that component i does not function satisfactory. If the events S i c are independent then Eq... can be written as

. Block diagram for parallel system p ss = p fs = P S c S c S n c.

5 p ss = P S c P S c P S n c n = p fi i= Lecture 0: System Reliability.. Mixed System This class of system is combination of series and parallel system. A real structure is combination of the series and parallel system. To perform the objective, both series and parallel system is required in action together. In Figure.., the block diagram model is shown. To evaluate mixed system reliability, it can be visualized as consisting two subsystems A and A. Here, A is A A Figure.. Block diagram for mixed system parallel system and A is series system. Reliability of the system is given by p ss = P E E..7 where, E i is the event that subsystem A i function satisfactory. Now for individual subsystem the system reliability can be computed as discussed above. So, mixed system reliability would be p ss = P E P E = p fs p fs..8 here, p fsi belongs to each subsystem. To get the above expression it is assumed that events A i are statically independent. Example Ex#0: For a series system, given failure probability of its four components are given as

., the block diagram model is shown. To evaluate mixed system reliability, it can be visualized as consisting two subsystems A and A. Here, A is A A Figure.

6 p f = 0.; p f = 0.; p f = 0.; p f = 0. Lecture 0: System Reliability (i) compute the system reliability. (ii) compute the system reliability assuming system as parallel and (iii) considering mixed system shown in Figure.., compare the system reliability. Solution: (i) For series system, system reliability would be (Eq...) p ss = 0. ( 0.)( 0.)( 0.) = 0.0 (ii) For parallel system, system reliability would be (Eq...) (ii) For mixed system Reliability of subsystem A is Reliability of subsystem A is p ss = (0.)(0.)(0.)(0.) = P(E ) = (0.)(0.) = 0.98 P(E ) = ( 0.)( 0.) = 0. Hence, the reliability of the mixed system would be (Eq...8) p ss = P E P E = 0.98 (0.) = 0. In this example it is observe that with same component different system reliability can be achieved by orienting the component in number of combination. Ex#0: A rigid steel frame is shown in Figure.. and its properties are given as

Solution: (i) For series system, system reliability would be (Eq...) p ss = 0. ( 0.)( 0.)( 0.) = 0.0 (ii) For parallel system, system reliability would be (Eq.

7 Lecture 0: System Reliability μ M = μ M = μ M = μ M7 = 7 knm σ M = σ M = σ M = σ M7 = 7. knm μ M = μ M = μ M = 7 knm σ M = σ M = σ M = 0. knm μ = 80 knm σ = 7 knm 7 7 (a) (b) (c) (d) (e) Figure.. (a) Rigid Frame, failure modes (b), (c), (d), (e) and block diagram of modes where, M i is plastic moment capacity of section i. Here, all M i, are independent and all variables are normally distributed. Evaluate the system reliability.

knm μ = 80 knm σ = 7 knm 7 7 (a) (b) (c) 7 7 7 (d) (e) Figure.

8 Lecture 0: System Reliability Solution: The frame will face failure if hinge form locations are, and or, and or, and or, and as shown in Figure.. (b)-(e). From virtual work theory in plastic analysis, for safety of frame of failure mode M θ + M θ + M θ > θ where, θ is the virtual rotation of section. So, safety margin is G = M + M + M So, the survival probability of the frame under any mode is p s = P(G 0). To evaluate the parameter of G is μ G = μ M + μ M + μ M μ = = 80 knm σ G = =.7 knm Here, all variables are independent, normally distributed and as G depends on them linearly. So, G will be also a normal variable. The probability of failure of the frame under the mode is p f = P G < 0 = Φ μ G σ G = Φ 80.7 = Φ(.7) =.98 0 Similarly for other modes shown in Figure.. (c)-(e), failure probability is calculated below Table.. Failure probability calculation Mode No G i μ Gi σ Gi p fi β M + M + M M + M + M M + M + M M + M + M From block model shown in Figure.. it is observed that this system is a series system. It is intentionally assume that all failure modes are statistically independent (though it is not case). So system reliability can be evaluated from Eq... 7

So, safety margin is G = M + M + M So, the survival probability of the frame under any mode is p s = P(G 0). To evaluate the parameter of G is μ G = μ M + μ M + μ M μ = 7 + 7 + 7 80 = 80 knm σ G = 7.

p ss = p fi The probability of failure of the system is i= =.98 0 0.8 0 0.78 0 0.

9 p ss = p fi The probability of failure of the system is i= = = Lecture 0: System Reliability p fs = p ss = =

Module 2. Analysis of Statically Indeterminate Structures by the Matrix Force Method. Version 2 CE IIT, Kharagpur

Module 2. Analysis of Statically Indeterminate Structures by the Matrix Force Method. Version 2 CE IIT, Kharagpur Module Analysis of Statically Indeterminate Structures by the Matrix Force Method esson 11 The Force Method of Analysis: Frames Instructional Objectives After reading this chapter the student will be able