Reference no: EM13576899

Question: Rivets used to secure the stainless steel sheet metal of aircraft wings are designed to withstand certain shearing forces. It is known that because of random variations in location the load on a rivet is Weibull distributed with a scale parameter of 750 pounds and a shape parameter of 3.0. The strength of the rivets in pounds is also random because of variations in the material characteristics and the dimensional tolerances. It has been found that the strength is also Weibull distributed with the same shape parameter but with a scale parameter of 1000 pounds.

Let x denote the load on a given rivet and let y denote the strength of that rivet. The rivet will not fail if x is less than y. The Reliability may therefore be expressed as P[X<Y]. Use @Risk to estimate the reliability. One approach is to set up a cell to yield random values of X and another cell to yield random values of Y. Then compute X-Y in a third cell and make it an output cell. From the simulated distribution of X-Y determine the probability that X-Y is less than 0.

A second approach is to compute a binary variable having the value 1 if X<Y and 0 otherwise. The average value of the binary variable will be an estimate of the reliability. Whichever method you choose use 1000 iterations. This problem may be solved analytically but for other distributions this may not be possible. The simulation will be valid for any choices of the distributions of load and strength. The theoretical value for the present case is 70.3%. If you do it correctly your answer should be close to this.