Question 24.19: A mechanical component is subjected to a normally distribute...

Step V Standard deviation to achieve 1% failure rate When the manufacturing process is kept under better quality control, the failure is reduced to 1%. As shown in Fig. 24.25(b), the area below the normal curve from Z=Z_{0} \text { to } Z=+\infty should be 0.99 for a failure of 1%. The area below the normal curve from Z = 0 to Z = + ∞ is 0.5. Therefore, the area below the normal curve from Z = 0 to Z=Z_{0} should be (0.99 – 0.5) or 0.49. From Table 24.6, the corresponding value of Z_{0} is approximately 2.33.

Z_{0}=\frac{m-\mu_{m}}{\hat{\sigma}_{m}} \quad \text { or } \quad-2.33=\frac{0-500}{\hat{\sigma}_{m}} .

\therefore \hat{\sigma}_{m}=214.59 N .

\left(\hat{\sigma}_{m}\right)^{2}=\left(\hat{\sigma}_{S}\right)^{2}+\left(\hat{\sigma}_{F}\right)^{2} .

\text { or }(214.59)^{2}=\left(\hat{\sigma}_{S}\right)^{2}+(200)^{2} \quad \therefore \hat{\sigma}_{S}=77.77 N .

Therefore, it is necessary to reduce the standard deviation of strength from 150 to 77.77 N to reduce the failure of the components to 1%.          (ii)

Step VI Mean strength to achieve 1% failure rate
When the mean strength of the component is improved by using a better material, the failure is reduced to 1%. As discussed in part (ii), the corresponding value of Z_{0} is approximately –2.33.

Z_{0}=\frac{m-\mu_{m}}{\hat{\sigma}_{m}} \quad \text { or } \quad-2.33=\frac{0-\mu_{m}}{250} .

\therefore \quad \mu_{m}=2.33(250)=582.5 N .

\mu_{m}=\mu_{S}-\mu_{F} \quad \text { or } \quad 582.5=\mu_{S}-1000 .

\therefore \quad \mu_{S}=1582.5 N .

Therefore, it is necessary to improve the mean strength of the material from 1500 to 1582.5 N in order to reduce the failure of the components to 1%.               (iii).