Weibull Distribution-Yield Strength

Let s consider sample problem 2 again. This time the ten experimental data will be analysed using the Weibull distribution. As first we list the ten results x, (j. — 1, 2, 

10) in ascending order. Second, we assign a probability P to each result by considering its frequency of occurrence according to the mean rank Eqs. 4.40 or 4.41. 

Then we plot lnln[l/(l - P)] versus the logarithm of the yield strength, as shown in Fig. 4.24a. The Weibull exponent is equal to 40.87 with the shape factor Go = 236.4 MPa. If we want to know the yield strength of the material with 99 % probability of survival we may write 

Against 218.5 MPa calculated with the normal distribution. We may also want to assess the volume effect and ask what the yield strength would be using a specimen of 20 mm diameter and 200 mm length. The volume ratio is 

We need to infer the scale factor 0 20 mm for the 20 mm diameter samples. From Eq. 4.45 

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Figure 8.29. Analysis of Steel Yield Strength by Weibull distribution function.

Ultimate strength Yielding Toughness Weibull distribution Stress concentration Trousers tear Simple shear test Energy to cause rupture Impact tester Izod test Charpy test... 

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