A mechanical component is subjected to a normally distributed force with a mean of 1000 N and a standard deviation of 200 N. The designer has used a factor of safety of 1.5 based on mean values. However due to variations in dimensions, the strength of the component is normally distributed with a

Question

A mechanical component is subjected to a normally distributed force with a mean of 1000 N and a standard deviation of 200 N. The designer has used a factor of safety of 1.5 based on mean values. However due to variations in dimensions, the strength of the component is normally distributed with a mean of 1500 N and a standard deviation of 150 N.
(i) What percentage of failure would be expected?
(ii) It is required to reduce the standard deviation of part strength by better quality control in order to achieve a failure rate of only 1%. What should be the standard deviation of the strength assuming other factors without any change?
(iii) It is required to improve the mean strength of the component by using better material in order to achieve a failure rate of 1%. What should be the mean strength assuming other factors unchanged?

Accepted Answer

[latex] ext { Given } \mu_{S}=1500 N \quad \hat{\sigma}_{S}=150 N [/latex].

[latex] (f s)=1.5 \quad \mu_{F}=1000 N \quad \hat{\sigma}_{F}=200 N [/latex].

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

[latex] \mu_{S}=1500 N ext { and } \hat{\sigma}_{S}=150 N [/latex].

Step II Population of force (F)
F denotes the population of force.

[latex] \mu_{F}=1000 N \quad ext { and } \quad \hat{\sigma}_{F}=200 N [/latex].

Step III Population of margin of safety (m)
A third population of margin of safety is denoted by m. It is obtained by subtracting the force population from the population of strength. Therefore,

[latex] \mu_{m}=\mu_{S}-\mu_{F}=1500-1000=500 N [/latex].

[latex] \hat{\sigma}_{m}=\sqrt{\left(\hat{\sigma}_{S} ight)^{2}+\left(\hat{\sigma}_{F} ight)^{2}}=\sqrt{(150)^{2}+(200)^{2}} [/latex].

[latex] =250 N / mm ^{2} [/latex].

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

[latex] Z_{0}=\frac{m-\mu_{m}}{\hat{\sigma}_{m}}=\frac{0-500}{250}=-2 [/latex].

As shown in Fig. 24.25(a), the area under the normal curve from [latex] Z=-\infty ext { to } Z=Z_{0} [/latex] or –2 indicates the probability of failure of the component. From Table 24.6, the area below the normal curve from Z = 0 to Z = 2 is 0.4772. The shaded area below the normal curve from Z = –2 to Z = –∞ is (0.5 – 0.4772) or 0.0228. Therefore, the probability of failure of the components is 0.0228 or 2.28%. (i)

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

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 [latex] Z=Z_{0} ext { to } Z=+\infty [/latex] 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 [latex] Z=Z_{0} [/latex] should be (0.99 – 0.5) or 0.49. From Table 24.6, the corresponding value of [latex] Z_{0} [/latex] is approximately 2.33.

[latex] Z_{0}=\frac{m-\mu_{m}}{\hat{\sigma}_{m}} \quad ext { or } \quad-2.33=\frac{0-500}{\hat{\sigma}_{m}} [/latex].

[latex] herefore \hat{\sigma}_{m}=214.59 N [/latex].

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

[latex] ext { or }(214.59)^{2}=\left(\hat{\sigma}_{S} ight)^{2}+(200)^{2} \quad herefore \hat{\sigma}_{S}=77.77 N [/latex].

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 [latex] Z_{0} [/latex] is approximately –2.33.

[latex] Z_{0}=\frac{m-\mu_{m}}{\hat{\sigma}_{m}} \quad ext { or } \quad-2.33=\frac{0-\mu_{m}}{250} [/latex].

[latex] herefore \quad \mu_{m}=2.33(250)=582.5 N [/latex].

[latex] \mu_{m}=\mu_{S}-\mu_{F} \quad ext { or } \quad 582.5=\mu_{S}-1000 [/latex].

[latex] herefore \quad \mu_{S}=1582.5 N [/latex].

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).

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