\text { Given } \mu_{S}=130 N / mm ^{2} \quad \hat{\sigma}_{S}=15 N / mm ^{2} .
\mu_{\sigma}=100 N / mm ^{2} \quad \hat{\sigma}_{\sigma}=10 N / mm ^{2} .
Step I Population of strength (S)
S denotes the population of strength. For this population,
\mu_{S}=130 N / mm ^{2} \text { and } \hat{\sigma}_{S}=15 N / mm ^{2} .
Step II Population of stress (σ)
The population of stress is denoted by σ.
\mu_{\sigma}=100 N / mm ^{2} \quad \text { and } \quad \hat{\sigma}_{\sigma}=10 N / mm ^{2} .
Step III Population of margin of safety (m)
The population of margin of safety is denoted by m.
It is obtained by subtracting the stress population from the population of strength. Therefore,
\mu_{m}=\mu_{S}-\mu_{\sigma}=130-100=30 N / mm ^{2} .
\hat{\sigma}_{m}=\sqrt{\left(\hat{\sigma}_{S}\right)^{2}+\left(\hat{\sigma}_{\sigma}\right)^{2}}=\sqrt{(15)^{2}+(10)^{2}} .
=18.03 N / mm ^{2} .
Step IV Probability of failure for component
When m = 0, the standard variable Z_{0} is given by
Z_{0}=\frac{m-\mu_{m}}{\hat{\sigma}_{m}}=\frac{0-30}{18.03}=-1.664 .
As shown in Fig. 24.23(a), the area under the normal curve from Z=-\infty \text { to } Z=Z_{0} or –1.664 indicates the probability of failure of the component. From Table 24.6, the area below the normal curve from Z = 0 to Z = 1.664 is given by,
\text { Area }=0.4515+\frac{(0.4525-0.4515)}{(1.67-1.66)}(1.664-1.66) .
= 0.4519.
Table 24.6 Areas under normal curve from 0 to Z
9 |
8 |
7 |
6 |
5 |
4 |
3 |
2 |
1 |
0 |
Z |
.0359 |
.0319 |
.0279 |
.0239 |
.0199 |
.0160 |
.0120 |
.0080 |
.0040 |
.0000 |
0.0 |
.0754 |
.0714 |
.0675 |
.0636 |
.0596 |
.0557 |
.0517 |
.0478 |
.0438 |
.0398 |
0.1 |
.1141 |
.1103 |
.1064 |
.1026 |
.0987 |
.0948 |
.0910 |
.0871 |
.0832 |
.0793 |
0.2 |
.1517 |
.1480 |
.1443 |
.1406 |
.1368 |
.1331 |
.1293 |
.1255 |
.1217 |
.1179 |
0.3 |
.1879 |
.1844 |
.1808 |
.1772 |
.1736 |
.1700 |
.1664 |
.1628 |
.1591 |
.1554 |
0.4 |
.2224 |
.2190 |
.2157 |
.2123 |
.2088 |
.2054 |
.2019 |
.1985 |
.1950 |
.1915 |
0.5 |
.2549 |
.2518 |
.2486 |
.2454 |
.2422 |
.2389 |
.2357 |
.2324 |
.2291 |
.2258 |
0.6 |
.2852 |
.2823 |
.2794 |
.2764 |
.2734 |
.2704 |
.2673 |
.2642 |
.2612 |
.2580 |
0.7 |
.3133 |
.3106 |
.3078 |
.3051 |
.3023 |
.2996 |
.2967 |
.2939 |
.2910 |
.2881 |
0.8 |
.3389 |
.3365 |
.3340 |
.3315 |
.3289 |
.3264 |
.3238 |
.3212 |
.3186 |
.3159 |
0.9 |
.3621 |
.3599 |
.3577 |
.3554 |
.3531 |
.3508 |
.3485 |
.3461 |
.3438 |
.3413 |
1.0 |
.3830 |
.3810 |
.3790 |
.3770 |
.3749 |
.3729 |
.3708 |
.3686 |
.3665 |
.3643 |
1.1 |
.4015 |
.3997 |
.3980 |
.3962 |
.3944 |
.3925 |
.3907 |
.3888 |
.3869 |
.3849 |
1.2 |
.4177 |
.4162 |
.4147 |
.4131 |
.4115 |
.4099 |
.4082 |
.4066 |
.4049 |
.4032 |
1.3 |
.4319 |
.4306 |
.4292 |
.4279 |
.4265 |
.4251 |
.4236 |
.4222 |
.4207 |
.4192 |
1.4 |
.4441 |
.4429 |
.4418 |
.4406 |
.4394 |
.4382 |
.4370 |
.4357 |
.4345 |
.4332 |
1.5 |
.4545 |
.4535 |
.4525 |
.4515 |
.4505 |
.4495 |
.4484 |
.4474 |
.4463 |
.4452 |
1.6 |
.4633 |
.4625 |
.4616 |
.4608 |
.4599 |
.4591 |
.4582 |
.4573 |
.4564 |
.4554 |
1.7 |
.4706 |
.4699 |
.4693 |
.4686 |
.4678 |
.4671 |
.4664 |
.4656 |
.4649 |
.4641 |
1.8 |
.4767 |
.4761 |
.4756 |
.4750 |
.4744 |
.4738 |
.4732 |
.4726 |
.4719 |
.4713 |
1.9 |
.4817 |
.4812 |
.4808 |
.4803 |
.4798 |
.4793 |
.4788 |
.4783 |
.4778 |
.4772 |
2.0 |
.4857 |
.4854 |
.4850 |
.4846 |
.4842 |
.4838 |
.4834 |
.4830 |
.4826 |
.4821 |
2.1 |
.4890 |
.4887 |
.4884 |
.4881 |
.4878 |
.4875 |
.4871 |
.4868 |
.4864 |
.4861 |
2.2 |
.4916 |
.4913 |
.4911 |
.4909 |
.4906 |
.4904 |
.4901 |
.4898 |
.4896 |
.4893 |
2.3 |
.4936 |
.4934 |
.4932 |
.4931 |
.4929 |
.4927 |
.4925 |
.4922 |
.4920 |
.4918 |
2.4 |
.4952 |
.4951 |
.4949 |
.4948 |
.4946 |
.4945 |
.4943 |
.4941 |
.4940 |
.4938 |
2.5 |
.4964 |
.4963 |
.4962 |
.4961 |
.4960 |
.4959 |
.4957 |
.4956 |
.4955 |
.4953 |
2.6 |
.4974 |
.4973 |
.4972 |
.4971 |
.4970 |
.4969 |
.4968 |
.4967 |
.4966 |
.4965 |
2.7 |
.4981 |
.4980 |
.4979 |
.4979 |
.4978 |
.4977 |
.4977 |
.4976 |
.4975 |
.4974 |
2.8 |
.4986 |
.4986 |
.4985 |
.4985 |
.4984 |
.4984 |
.4983 |
.4982 |
.4982 |
.4981 |
2.9 |
.4990 |
.4990 |
.4989 |
.4989 |
.4989 |
.4988 |
.4988 |
.4987 |
.4987 |
.4987 |
3.0 |
.4993 |
.4993 |
.4992 |
.4992 |
.4992 |
.4992 |
.4991 |
.4991 |
.4991 |
.4990 |
3.1 |
.4995 |
.4995 |
.4995 |
.4994 |
.4994 |
.4994 |
.4994 |
.4994 |
.4993 |
.4993 |
3.2 |
.4997 |
.4996 |
.4996 |
.4996 |
.4996 |
.4996 |
.4996 |
.4995 |
.4995 |
.4995 |
3.3 |
.4998 |
.4997 |
.4997 |
.4997 |
.4997 |
.4997 |
.4997 |
.4997 |
.4997 |
.4997 |
3.4 |
.4998 |
.4998 |
.4998 |
.4998 |
.4998 |
.4998 |
.4998 |
.4998 |
.4998 |
.4998 |
3.5 |
.4999 |
.4999 |
.4999 |
.4999 |
.4999 |
.4999 |
.4999 |
.4999 |
.4998 |
.4998 |
3.6 |
.4999 |
.4999 |
.4999 |
.4999 |
.4999 |
.4999 |
.4999 |
.4999 |
.4999 |
.4999 |
3.7 |
.4999 |
.4999 |
.4999 |
.4999 |
.4999 |
.4999 |
.4999 |
.4999 |
.4999 |
.4999 |
3.8 |
.5000 |
.5000 |
.5000 |
.5000 |
.5000 |
.5000 |
.5000 |
.5000 |
.5000 |
.5000 |
3.9 |
The shaded area below the normal curve from Z = –1.664 to Z = –∞ is (0.5 – 0.4519) or 0.0481. Therefore, the probability of failure of the components is 0.0481 or 4.81%. (i)
Step V Probability of failure for component
\left(\hat{\sigma}_{S}=10 N / mm ^{2}\right) .
When the manufacturing process is kept under better control,
\hat{\sigma}_{S}=10 N / mm ^{2} .
\hat{\sigma}_{m}=\sqrt{\left(\hat{\sigma}_{S}\right)^{2}+\left(\hat{\sigma}_{\sigma}\right)^{2}}=\sqrt{(10)^{2}+(10)^{2}} .
= 14.14 N/mm².
When m = 0, the standard variable Z_{0} is given by
Z_{0}=\frac{m-\mu_{m}}{\hat{\sigma}_{m}}=\frac{0-30}{14.14}=-2.122 .
As shown in Fig. 24.23(b), the area under the normal curve from Z=-\infty \text { to } Z=Z_{0} or – 2.122 indicates the probability of failure of the component. From Table 24.6, the area below the normal curve from Z = 0 to Z = 2.122 is given by,
\text { Area }=0.4830+\frac{(0.4834-0.4830)}{(2.13-2.12)}(2.122-2.12) .
= 0.4831.
The shaded area below the normal curve from Z = –2.122 to Z = –∞ is (0.5 – 0.4831) or 0.0169. Therefore, the probability of failure of the components is 0.0169 or 1.69%. (ii)
Step VI Factor of safety
When only mean values are considered
\text { Factor of safety }=\frac{\mu_{S}}{\mu_{\sigma}}=\frac{130}{100}=1.3 (iii).