A mechanical component is subjected to a mean stress of 100 N/mm^2 and a standard deviation of 10 N/mm^2. The material of the component has a mean strength of 130 N/mm^2 and a standard deviation of 15 N/mm^2. (i) Find the probability of failure for the component. (ii) If better manufacturing

Question

A mechanical component is subjected to a mean stress of 100 N/mm² and a standard deviation of 10 N/mm². The material of the component has a mean strength of 130 N/mm² and a standard deviation of 15 N/mm².

(i) Find the probability of failure for the component.

(ii) If better manufacturing control reduces the
standard deviation of material strength to 10 N/mm², find the probability of failure.

(iii) If we consider only mean values of the data in design, fi nd out the factor of safety.

Accepted Answer

[latex] ext { Given } \mu_{S}=130 N / mm ^{2} \quad \hat{\sigma}_{S}=15 N / mm ^{2} [/latex].

[latex] \mu_{\sigma}=100 N / mm ^{2} \quad \hat{\sigma}_{\sigma}=10 N / mm ^{2} [/latex].

Step I Population of strength (S)
S denotes the population of strength. For this population,

[latex] \mu_{S}=130 N / mm ^{2} ext { and } \hat{\sigma}_{S}=15 N / mm ^{2} [/latex].

Step II Population of stress (σ)
The population of stress is denoted by σ.

[latex] \mu_{\sigma}=100 N / mm ^{2} \quad ext { and } \quad \hat{\sigma}_{\sigma}=10 N / mm ^{2} [/latex].

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,

[latex] \mu_{m}=\mu_{S}-\mu_{\sigma}=130-100=30 N / mm ^{2} [/latex].

[latex] \hat{\sigma}_{m}=\sqrt{\left(\hat{\sigma}_{S} ight)^{2}+\left(\hat{\sigma}_{\sigma} ight)^{2}}=\sqrt{(15)^{2}+(10)^{2}} [/latex].

[latex] =18.03 N / mm ^{2} [/latex].

Step IV Probability of failure for component
When m = 0, the standard variable [latex] Z_{0} [/latex] is given by

[latex] Z_{0}=\frac{m-\mu_{m}}{\hat{\sigma}_{m}}=\frac{0-30}{18.03}=-1.664 [/latex].

As shown in Fig. 24.23(a), the area under the normal curve from [latex] Z=-\infty ext { to } Z=Z_{0} [/latex] 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,

[latex] ext { Area }=0.4515+\frac{(0.4525-0.4515)}{(1.67-1.66)}(1.664-1.66) [/latex].

= 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

[latex] \left(\hat{\sigma}_{S}=10 N / mm ^{2} ight) [/latex].

When the manufacturing process is kept under better control,

[latex] \hat{\sigma}_{S}=10 N / mm ^{2} [/latex].

[latex] \hat{\sigma}_{m}=\sqrt{\left(\hat{\sigma}_{S} ight)^{2}+\left(\hat{\sigma}_{\sigma} ight)^{2}}=\sqrt{(10)^{2}+(10)^{2}} [/latex].

= 14.14 N/mm².

When m = 0, the standard variable [latex] Z_{0} [/latex] is given by

[latex] Z_{0}=\frac{m-\mu_{m}}{\hat{\sigma}_{m}}=\frac{0-30}{14.14}=-2.122 [/latex].

As shown in Fig. 24.23(b), the area under the normal curve from [latex] Z=-\infty ext { to } Z=Z_{0} [/latex] 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,

[latex] ext { Area }=0.4830+\frac{(0.4834-0.4830)}{(2.13-2.12)}(2.122-2.12) [/latex].

= 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

[latex] ext { Factor of safety }=\frac{\mu_{S}}{\mu_{\sigma}}=\frac{130}{100}=1.3 [/latex] (iii).

Question 24.17: A mechanical component is subjected to a mean stress of 100 ...

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