Section 1 Load Resistance Fundamentals Structural Reliability Analysis

Section 1: Load & Resistance (Fundamentals) Structural Reliability Analysis – Basics A structural component fails when the component encounters an extreme load, or the component is subjected to a combination of loads that attain sufficient magnitude to exhaust the capacity/resistance of the component. In order to quantify the reliability of a component, the failure state must be clearly defined. Typically structural failure is defined in either two ways: • Complete loss of load carrying capacity • Loss in serviceability Regardless of which type of failure state is adopted, carefully constructed mathematical descriptions are a necessity in defining either.

Section 1: Load & Resistance (Fundamentals) The mathematics and the design process requires the design engineer to: • Predict the magnitude of extreme (maximum) load events, and/or • Predict the strength (or capacity) of the material used to fabricate the component Thus it becomes necessary to synthesize probabilistic models for both load and capacity. Initially we will consider components that are subjected to a single load random variable, and a material that can be characterized by a single resistance random variable. Under these conditions we will characterize the probabilistic response of a component by the following methods: 1. Probability statements, e. g. , 2. Integral definitions (and interpretations) of reliability and probability of failure. 3. Approximate methods for calculating component reliability. 4. Develop bounds on component reliability. In all the methods above probability density functions will be assume for the resistance and load random variables. Some situations will yield expressions for component reliability that are exact. Most of the time we are not so lucky.

Section 1: Load & Resistance (Fundamentals) Example 1. 1 If the resistance random variable (R) and the load random variable (S) are normally distributed, establish an expression for the probability of failure (Pf). Solution: Here Pf = Pr ( R – S < 0) = Pr (M < 0) where M=R–S = Safety Margin Thus m. M = E [M] = E [R – S] = E [R] – E[S]

Section 1: Load & Resistance (Fundamentals) m. M = m. R - m. S Similarly s. M 2 = VAR [ M ] = E [ (M – E [M])2 ] = E [( R – S – m. R + m. S )2 ] = E [ (R – m. R )2 + (S – m. S )2 – 2(R – m. R )(S – m. S )] =E [(R – m. R )2 + E[(S – m. S)2] – E[2(R – m. R )(S – m. S )] = s. R 2 + s. S 2 – COV[R , S] However, under the assumption that R and S are independent random variables COV [R , S] = 0 Since R and S are normal, and since M is a linear combination of two normally distributed random variables, then M is normally distributed. If R and S are not normally distributed, then M is only approximately normal (see Central Limit Theorem) and the results here are only approximations. This can either be good or bad.

Section 1: Load & Resistance (Fundamentals) Making use of the Standard Normal probability density function with and

Section 1: Load & Resistance (Fundamentals) Thus We are now in a position to formally define the reliability index (b) as

Section 1: Load & Resistance (Fundamentals) The reliability index is depicted in the figure below.

Section 1: Load & Resistance (Fundamentals) The previous example demonstrated that under limiting conditions a closed form expression can be evaluated for the probability of failure is defined by a safety margin (R – S). Keep in mind that both the load and resistance random variables were assumed to be normally distributed. There is one other case that can be somewhat easily generated when the factor of safety is used to define failure instead of a safety margin. In order to make this example somewhat straightforward, one must assume that the load and resistance random variable are both lognormally distributed. Under these strong assumptions one can demonstrate that Where (R/S) is the traditional definition of a factor of safety and

Section 1: Load & Resistance (Fundamentals)