Example

A simply supported beam of span L = 9.144 m is loaded by a uniformly distributed load w in kN/m and a concentrated load P (in kN) applied at the midspan. The maximum deflection of the beam at the midspan can be calculated as

where E is the Young's modulus and I is the moment of inertia of the cross-section of the beam. A beam with EI = 182,262 kN m2 is selected to carry the load. Suppose w is a normal random variable with a mean of 35.03 kN/m and a standard deviation of 5.25 kN/m and P is a lognormal random variable with mean of 111.2 kN and a standard deviation of 11.12 kN. Further assume w and P are statistically independent. The allowable deflection, da, is considered to be 38.1 mm. The task is to calculate the probability of failure of the beam in deflection satisfying the allowable deflection criterion.

Solution

This example is a problem of estimating the reliability index and the corresponding probability of failure when the limit state function contains independent nonnormal variables. This type of problem is very common in practice. The procedure is explained using the 8 tasks identified earlier.

Task 1 — Define the limit state function. Substituting deterministic values of all the parameters in Equation 12.20, the limit state function for the problem can be defined as g() = 0.0381 — (4.99444 x 10—4w + 8.73919 x 10—5P) (12.21)

where w is a normal random variable with specified mean and standard deviation values, that is, w~ N(35.03 kN/m, 5.25 kN/m) and P is a lognormal random variable with parameters 1P and ZP, that is, P ~ ln(lP, ZP). These parameters can be calculated as [14]

dP =-= 0.1 ffi ZP and 1P = ln 111.2 --x 0.12 = 4.706

The probability of failure or the reliability index of the beam in deflection corresponding to Equation 12.21 is evaluated using the eight tasks identified earlier. The results are summarized in Table 12.1. For further clarification, the detail calculations for the third and the final interations are given below.

Third iteration

Task 2. As shown in Table 12.1, an initial value of b of 3.0 is assumed to start the iteration. Task 3. Using Equation 12.17, the coordinates of the new checking point for w and P can be shown to be w* = 35.03 + 0.926 x 3 x 5.25 = 49.615 p* = 110.02 + 0.377 x 3 x 12.22 = 123.841

Task 4. Since P is a lognormal random variable, the equivalent normal mean and standard deviation at the checking point are calculated using Equations 12.15 and 12.16

0 0

Post a comment