III. Reliability in Design and Development
A. Reliability design techniques
2. Stress-strength analysis (Evaluate)
Apply stress-strength analysis method of computing probability of failure, and interpret the results.
When the stress is larger than the specific units strength; that is a problem.
Additional References
Stress Strength Normal Assumption (article)
The Stress – Strength Concept in Practice (article)
SOR 059 How to Set Environmental Specifications for Testing (podcast)
Quick Quiz
1-81. Normal stress and strength distributions are constructed for four different parts as described in the following:
For part 1, the average and standard deviation of the stress and strength distributions are the same.
For part 2, the average and standard deviation of the stress distribution are both higher than for the strength curve.
For part 3, both distributions have the same standard deviation but the strength average is higher than the stress average.
For part 4, Both distributions have the same variance but the stress average is slightly higher than the strength average.
Identify the part with the highest reliability.
(A) part 1
(B) part 2
(C) part 3
(D) part 4
(C) part 3
This takes some careful reading and knowledge that when the strength curve is higher (stronger) than the stress curve for most scales of stress/strength application, then the part is more reliable. Parts 2 and 4 have the mean stress higher than the mean strength suggesting a lower reliability then with part 1, which still isn’t a great situation. Part 3 has a higher mean strength than stress, which will result in a higher reliability than other the other parts.
1-87. An oxygen cylinder is rated at 2200 psi, with a standard deviation of 190 psi. If the expected stress will be 1750 psi with a standard deviation of 300 psi, what is the probability of failure of the cylinder?
(A) 0.0345
(B) 0.1025
(C) 0.1817
(D) 0.8975
(B) 0.1025
This is a normal based stress-strength calculation assuming both distributions are normal and independent. Use the following formula to determine the z-value representing the probability of failure.
$$ Z=\frac{{{\mu }_{x}}-{{\mu }_{y}}}{\sqrt{\sigma _{x}^{2}+\sigma _{y}^{2}}}$$
where the x subscripts are for the strength parameters and y is for the stress parameters. Not it uses variance, not the standard deviations. Inserting the values and calculation we fine z is
$$ Z=\frac{2200-1750}{\sqrt{190_{{}}^{2}+300_{{}}^{2}}}=0.1025$$
Then it is off to the z-table, to determine the probability of failure. In this case you should find the probability value, in the body of the table is between z-values of 1.26 and 1.27, with probabilities of 0.1038 and 0.1020, respectively. Use intropolation to determine the value for a z-value of 1.27.
Interpolation uses the idea equivalent ratios
$$ \frac{{{z}_{between}}-{{z}_{lower}}}{{{z}_{higher}}-{{z}_{lower}}}=\frac{{{P}_{between}}-{{P}_{lower}}}{{{P}_{higher}}-{{P}_{lower}}}$$
In this case, insert all the known values and solve for the unknown probability corresponding to a z-value of 1.27.
$$ \begin{array}{l}\frac{1.267-1.26}{1.27-1.26}=\frac{{{P}_{between}}-0.1038}{0.1020-0.1038}\\{{P}_{between}}=\left( \frac{1.267-1.26}{1.27-1.26} \right)\left( 0.1020-0.1038 \right)+0.1038\\{{P}_{between}}=0.1025\end{array}$$
1-103. A metal component has a strength of 7,000 psi and a standard deviation of 800 psi. The component must withstand a load with a mean value of 4,500 psi and a standard deviation of 400 psi. Assuming that both strength and load are normally distributed, calculate the probability of failure for the component.
(A) 0.0026
(B) 0.0186
(C) 0.9814
(D) 0.9974
(A) 0.0026
This is a normal based stress-strength calculation assuming both distributions are normal and independent. Use the following formula to determine the z-value representing the probability of failure.
$$ Z=\frac{{{\mu }_{x}}-{{\mu }_{y}}}{\sqrt{\sigma _{x}^{2}+\sigma _{y}^{2}}}$$
where the x subscripts are for the strength parameters and y is for the stress parameters. Not it uses variance, not the standard deviations. Inserting the values and calculation we fine z is
$$ Z=\frac{7000-4500}{\sqrt{800_{{}}^{2}+400_{{}}^{2}}}=2.795 $$
Then it is off to the z-table, to determine the probability of failure. In this case you should find the probability value, in the body of the table is between z-values of 2.79 and 2.80, with probabilities of 0.0026 and 0.0026, respectively. There is no need to use interpolation to determine the value for a z-value of 2.795, as it is the same for the bounding values thus the resulting probability of failure is 0.0026.
Hi Fred,
Check the area of the tail you looked up in the example. I think you picked up the wrong value for z=1.39.
You are right Raye, I used the value for 1.49, not 1.39… an honest mistake misreading the table. Should have used a ruler or paper edge to track across the row of numbers accurately.
Cheers,
Fred