\text { Given } \mu_{d}=50 mm \quad \hat{\sigma}_{d}=0.125 mm .
\mu_{M}=1750 \times 10^{3} N – mm \quad \hat{\sigma}_{M}=150 \times 10^{3} N – mm .
Step I Population of diameter (d)
d denotes the population of diameters. For this population
\mu_{d}=50 mm \quad \text { and } \quad \hat{\sigma}_{d}=0.125 mm .
Step II Population of bending moment (M)
M denotes the population of values of bending moment. For this population,
\mu_{M}=1750 \times 10^{3} N – mm .
\text { and } \hat{\sigma}_{M}=150 \times 10^{3} N – mm .
\text { Step III Population of }\left(\pi d^{3} / 32\right)( Z )
Z denotes a third population. It is obtained by using the expression,
Z=\frac{\pi}{32} d^{3} .
In the above expression (π/32) is constant and using the equations in Table 24.7
Table 24.7 Mean and standard deviations of the function Z
| \text { Standard deviation } \hat{\sigma}_{Z} |
\text { Mean } \mu_{Z} |
Function |
| 0 |
a |
Z = a |
| a \hat{\sigma}_{X} |
a \mu_{X} |
Z = aX |
| \hat{\sigma}_{X} |
\mu_{X} \pm a |
Z = X ± a |
| \sqrt{\left(\hat{\sigma}_{X}\right)^{2}+\left(\hat{\sigma}_{Y}\right)^{2}} |
\mu_{X} \pm \mu_{Y} |
Z = X ± Y |
| \sqrt{\mu_{X}^{2}\left(\hat{\sigma}_{Y}\right)^{2}+\mu_{Y}^{2}\left(\hat{\sigma}_{X}\right)^{2}+\left(\hat{\sigma}_{X}\right)^{2}\left(\hat{\sigma}_{Y}\right)^{2}} |
\mu_{X} \mu_{Y} |
Z = X Y |
| \frac{1}{\mu_{Y}}\left[\frac{\mu_{X}^{2}\left(\hat{\sigma}_{Y}\right)^{2}+\mu_{Y}^{2}\left(\hat{\sigma}_{X}\right)^{2}}{\mu_{Y}^{2}+\left(\hat{\sigma}_{Y}\right)^{2}}\right]^{1 / 2} |
\mu_{X} / \mu_{Y} |
Z = X/Y |
| \frac{1}{2}\left(\frac{\hat{\sigma}_{X}}{\mu_{X}}\right)\left[4 \mu_{X}^{2}+\left(\hat{\sigma}_{X}\right)^{2}\right] |
\mu_{x}^{2}+\hat{\sigma}_{x}^{2} |
Z = X² |
| 3 \mu_{X}^{2}\left(\hat{\sigma}_{X}\right)+3\left(\hat{\sigma}_{X}\right)^{3} |
\mu_{X}^{3}+3 \mu_{X}\left(\hat{\sigma}_{X}\right)^{2} |
Z = X³ |
| \frac{\hat{\sigma}_{X}}{\mu_{X}^{2}}\left[1+\left(\frac{\hat{\sigma}_{X}}{\mu_{X}}\right)^{2}\right] |
\frac{1}{\mu_{X}}\left[1+\left(\frac{\hat{\sigma}_{X}}{\mu_{X}}\right)^{2}\right] |
Z = 1/X |
\mu_{Z}=\frac{\pi}{32}\left[\mu_{d}^{3}+3 \mu_{d}\left(\hat{\sigma}_{d}\right)^{2}\right] .
=\frac{\pi}{32}\left[50^{3}+3(50)(0.125)^{2}\right] .
= 12 272.08 mm³.
\hat{\sigma}_{Z}=\frac{\pi}{32}\left[3 \mu_{d}^{2} \hat{\sigma}_{d}+3\left(\hat{\sigma}_{d}\right)^{3}\right] .
=\frac{\pi}{32}\left[3(50)^{2}(0.125)+3(0.125)^{3}\right] .
= 92.04 mm³.
Step IV Population of bending stress (σ)
A fourth population of bending stress is denoted by σ. It is obtained by dividing the population of values of bending moment M by the population Z.
From Table 24.7,
\mu_{\sigma}=\frac{\mu_{M}}{\mu_{Z}}=\frac{1750 \times 10^{3}}{12272.08}=142.6 N / mm ^{2} .
\hat{\sigma}_{\sigma}=\frac{1}{\mu_{Z}}\left[\frac{\mu_{M}^{2} \hat{\sigma}_{Z}^{2}+\mu_{Z}^{2} \hat{\sigma}_{M}^{2}}{\mu_{Z}^{2}+\hat{\sigma}_{Z}^{2}}\right]^{1 / 2} .
=\frac{1}{12272.08}\left[\frac{\left(1750 \times 10^{3}\right)^{2} \times(92.04)^{2}+(12272.08)^{2} \times\left(150 \times 10^{3}\right)^{2}}{(12272.08)^{2}+(92.04)^{2}}\right]^{1 / 2} .
\hat{\sigma}_{\sigma}=12.27 N / mm ^{3} .
Step V Comments on analysis
The bending stress population has a mean of 142.6 N/mm² and standard deviation of 12.27 N/mm².
We can predict the mean and standard deviation of this population. However, the population is not a normally distributed random variable.