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<article article-type="research-article" dtd-version="1.1" specific-use="sps-1.9" xml:lang="en" xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink">
	<front>
		<journal-meta>
			<journal-id journal-id-type="publisher-id">dyna</journal-id>
			<journal-title-group>
				<journal-title>DYNA</journal-title>
				<abbrev-journal-title abbrev-type="publisher">Dyna rev.fac.nac.minas</abbrev-journal-title>
			</journal-title-group>
			<issn pub-type="ppub">0012-7353</issn>
			<issn pub-type="epub">2346-2183</issn>
			<publisher>
				<publisher-name>Universidad Nacional de Colombia</publisher-name>
			</publisher>
		</journal-meta>
		<article-meta>
			<article-id pub-id-type="doi">10.15446/dyna.v87n215.84909</article-id>
			<article-categories>
				<subj-group subj-group-type="heading">
					<subject>Artículos</subject>
				</subj-group>
			</article-categories>
			<title-group>
				<article-title>Stress-strength Weibull analysis with different shape parameter β and probabilistic safety factor</article-title>
				<trans-title-group xml:lang="es">
					<trans-title>Análisis estrés-resistencia Weibull con diferente parámetro de forma β y su factor de seguridad probabilístico</trans-title>
				</trans-title-group>
			</title-group>
			<contrib-group>
				<contrib contrib-type="author">
					<name>
						<surname>Baro-Tijerina</surname>
						<given-names>Manuel</given-names>
					</name>
					<xref ref-type="aff" rid="aff1"><sup>
 <italic>a</italic>
</sup></xref>
				</contrib>
				<contrib contrib-type="author">
					<name>
						<surname>Piña-Monárrez</surname>
						<given-names>Manuel Román</given-names>
					</name>
					<xref ref-type="aff" rid="aff2"><sup>
 <italic>b</italic>
</sup></xref>
				</contrib>
				<contrib contrib-type="author">
					<name>
						<surname>Villa-Covarrubias</surname>
						<given-names>Baldomero</given-names>
					</name>
					<xref ref-type="aff" rid="aff3"><sup>
 <italic>c</italic>
</sup></xref>
				</contrib>
			</contrib-group>
			<aff id="aff1">
				<label>a</label>
				<institution content-type="original"> Industrial and Manufacturing Department at IIT Institute, Universidad Autónoma de Ciudad Juárez, Ciudad Juárez, México. &amp; Industrial and Technology Department, Instituto Tecnológico Superior de Nuevo Casas Grandes, Casas Grandes, México. al164467@alumnos.uacj.mx</institution>
				<institution content-type="normalized">Universidad Autónoma de Ciudad Juárez</institution>
				<institution content-type="orgname">Universidad Autónoma de Ciudad Juárez</institution>
				<addr-line>
					<city>Juárez</city>
				</addr-line>
				<country country="MX">Mexico</country>
				<email>al164467@alumnos.uacj.mx</email>
			</aff>
			<aff id="aff2">
				<label>b</label>
				<institution content-type="original"> Industrial and Manufacturing Department at IIT Institute, Universidad Autónoma de Ciudad Juárez, Ciudad Juárez, México. manuel.pina@uacj.mx</institution>
				<institution content-type="normalized">Universidad Autónoma de Ciudad Juárez</institution>
				<institution content-type="orgname">Universidad Autónoma de Ciudad Juárez</institution>
				<addr-line>
					<city>Juárez</city>
				</addr-line>
				<country country="MX">Mexico</country>
				<email>manuel.pina@uacj.mx</email>
			</aff>
			<aff id="aff3">
				<label>c</label>
				<institution content-type="original"> IIT Institute, Universidad Autónoma de Ciudad Juárez, Ciudad Juárez, México. baldomero.villa@uacj.mx</institution>
				<institution content-type="normalized">Universidad Autónoma de Ciudad Juárez</institution>
				<institution content-type="orgname">Universidad Autónoma de Ciudad Juárez</institution>
				<addr-line>
					<city>Juárez</city>
				</addr-line>
				<country country="MX">Mexico</country>
				<email>baldomero.villa@uacj.mx</email>
			</aff>
			<pub-date pub-type="epub" publication-format="electronic">


				<day>04</day>
				<month>01</month>
				<year>2021</year>
			</pub-date>
			<pub-date date-type="collection" publication-format="electronic">
				<season>Oct-Dec</season>
				<year>2020</year>
			</pub-date>
			<volume>87</volume>
			<issue>215</issue>
			<fpage>28</fpage>
			<lpage>33</lpage>
			<history>
				<date date-type="received">
					<day>03</day>
					<month>02</month>
					<year>2020</year>
				</date>
				<date date-type="rev-recd">
					<day>03</day>
					<month>07</month>
					<year>2020</year>
				</date>
				<date date-type="accepted">
					<day>27</day>
					<month>07</month>
					<year>2020</year>
				</date>
			</history>
			<permissions>
				<license license-type="open-access" xlink:href="https://creativecommons.org/licenses/by-nc-nd/4.0" xml:lang="en">
					<license-p>The author; licensee Universidad Nacional de Colombia</license-p>
				</license>
			</permissions>
			<abstract>
				<title>Abstract</title>
				<p>Since products are subjected to a random variable stress-strength, their reliability must be determined using the stress-strength analysis. Unfortunately, when both, stress and strength, follow a Weibull distribution with different shape parameters, the reliability stress-strength has not a close solution. Therefore, in this paper, the formulation to perform the analysis stress-strength Weibull with different shape parameters is derived. Furthermore, the formulation to determine the safety factor that corresponds to the designed reliability is also given. And because the relationship between the derived safety factor and the designed reliability is unique, then because reliability is random, the derived safety factor is random.</p>
			</abstract>
			<trans-abstract xml:lang="es">
				<title>Resumen</title>
				<p>Cuando los productos son sometidos a estrés variables, su confiabilidad debe ser estimada a partir de la resistencia que éstos tienen ante el estrés al que son sometidos, para lo que usamos el análisis estrés-resistencia. Desafortunadamente, cuando ambos, el estrés y la resistencia siguen un comportamiento Weibull con diferentes parámetros de forma, no existe una solución cerrada en la estimación de la confiabilidad. Por lo tanto, en este artículo, se presenta la formulación para poder realizar el análisis estrés-resistencia Weibull con diferente parámetro de forma. Además, la formulación para determinar el factor de seguridad probabilístico correspondiente al producto es dado. Y como la relación entre el factor de seguridad y la confiabilidad son únicas, entonces, la confiabilidad es aleatoria y el factor de seguridad también.</p>
			</trans-abstract>
			<kwd-group xml:lang="en">
				<title><bold>
 <italic>Keywords</italic>:</bold></title>
				<kwd>probabilistic safety factor</kwd>
				<kwd>Weibull distribution</kwd>
				<kwd>stress-strength analysis</kwd>
				<kwd>common Weibull shape parameter</kwd>
				<kwd>variable stress-strength</kwd>
				<kwd>β estimation directly</kwd>
			</kwd-group>
			<kwd-group xml:lang="es">
				<title>Palabras clave:</title>
				<kwd>factor de seguridad probabilístico</kwd>
				<kwd>distribución Weibull</kwd>
				<kwd>análisis estrés-resistencia</kwd>
				<kwd>parámetro de forma común Weibull</kwd>
				<kwd>estrés-resistencia variables</kwd>
				<kwd>estimación directa de β</kwd>
			</kwd-group>
			<counts>
				<fig-count count="1"/>
				<table-count count="2"/>
				<equation-count count="28"/>
				<ref-count count="20"/>
				<page-count count="6"/>
			</counts>
		</article-meta>
	</front>
	<body>
		<sec sec-type="intro">
			<title>1. Introduction</title>
			<p>In general, the reliability of a component is given by its capability to withstand the applied stress. Therefore, due to both the stress and the strength variables are random, then in the reliability field, the analysis is performed by using a probability density functions <bold>
 <italic>(pdf)</italic>
</bold> to model the stress variable and a pdf to model the strength variable [<xref ref-type="bibr" rid="B1">1</xref>]. Moreover, because in general two types of failure modes exist, to know the fatigue and shock failures modes. And since fatigue occurs due to the repeated application of the load (cyclical load behavior), and the shock occurs when the applied stress (s) is higher than the product’s strength (S) (s&gt;S). Then in both cases, the reliability of the analyzed component can be accessed by applying the stress-strength reliability analysis [<xref ref-type="bibr" rid="B2">2</xref>]. Unfortunately, if the stress and the strength variables follow both a Weibull distribution <bold>
 <italic>W~(β,η)</italic>
</bold> with different shape parameter <bold>
 <italic>β</italic>
</bold> , then a close solution of the Weibull/Weibull stress-strength function does not exist [<xref ref-type="bibr" rid="B3">3</xref>]. Therefore, in response to this problems researchers have being proposing some solutions as [<xref ref-type="bibr" rid="B4">4</xref>-<xref ref-type="bibr" rid="B12">12</xref>]. Unfortunately, their approaches are not efficient to determine the reliability of the analyzed product. This because of the solution for different shape parameter <bold>
 <italic>(β</italic>
</bold> 
 <sub>
 <italic>S</italic>
</sub> 
 <bold>
 <italic>≠β</italic>
</bold> 
 <sub>
 <italic>s</italic>
</sub> 
 <italic>)</italic> has been possible only in the case of (<bold>
 <italic>2β</italic>
</bold> 
 <sub>
 <italic>S</italic>
</sub> 
 <bold>
 <italic>=β</italic>
</bold> 
 <sub>
 <italic>s</italic>
</sub> ), (<bold>
 <italic>β</italic>
</bold> 
 <sub>
 <italic>S</italic>
</sub> 
 <bold>
 <italic>=2βs</italic>
</bold> ). But a general solution has not been found yet. Even [<xref ref-type="bibr" rid="B7">7</xref>] mention that it is important to continue studying the Weibull/Weibull stress-strength behavior to generate a general solution for different shape parameter combinations.</p>
			<p>Therefore, in this paper, based on the desired reliability index and the value of the estimated shape parameters η a common shape parameter β<sub>
 <italic>c</italic>
</sub> is derived. And then it is used to determine the reliability of the designed component. Moreover, based on the stress-strength analysis and the estimated Weibull parameters the safety factor which corresponds to the found reliability is also determined. And because its formulation is based only on the desired (or observed) R(t) index, given by the stress/strength Weibull analysis (see section 4.3), then the derived SF index could be considered as a probabilistic index also. In particular, it is highlighted that because the formulated SF<sub>i</sub> index is based on the Weibull stress-strength analysis performed by using the addressed common shape parameter, the SF index also represents the corresponding stress-strength reliability R(t) index. </p>
			<p>The structure of the paper is as follows: in section 2, the generalities of the Weibull distribution are given. Section 3 presents the stress strength of Weibull formulation. In section 4 the estimation of β<sub>
 <italic>S</italic>
</sub> =β<sub>
 <italic>s</italic>
</sub> , and β<sub>
 <italic>S</italic>
</sub> ≠β<sub>
 <italic>s</italic>
</sub> are both given. In section 5 an application of the methodology is sowed. Finally, section 6 presents the conclusion and references. </p>
		</sec>
		<sec>
			<title>2. Generalities of the Weibull distribution</title>
			<p>Due to its flexibility to model the design, production, and wear out or aging phases of a product, the Weibull distribution is widely used in reliability and lifetime analysis [<xref ref-type="bibr" rid="B2">2</xref>]. And since all products are subjected to a random (variant) environmental stress, and they have a strength to withstand the applied stress, which means that both the stress and the strength are modeling by a probability density function, then their reliability analysis is performed by using the stress-strength analysis given in the next section [<xref ref-type="bibr" rid="B13">13</xref>]. The compound Reliability Weibull/Weibull stress/strength probability density function is based on the Weibull distribution [<xref ref-type="bibr" rid="B14">14</xref>] given by:</p>
			<p>
				<disp-formula id="e1">
					<graphic xlink:href="2346-2183-dyna-87-215-28-e1.png"/>
				</disp-formula>
			</p>
			<p>with cumulative failure function and reliability function given by:</p>
			<p>
				<disp-formula id="e2">
					<graphic xlink:href="2346-2183-dyna-87-215-28-e2.png"/>
				</disp-formula>
			</p>
			<p>
				<disp-formula id="e3">
					<graphic xlink:href="2346-2183-dyna-87-215-28-e3.png"/>
				</disp-formula>
			</p>
			<p>where the estimation of the β and η parameters is performed by using the linear form of <xref ref-type="disp-formula" rid="e2">eq. (2)</xref> as</p>
			<p>
				<disp-formula id="e4">
					<graphic xlink:href="2346-2183-dyna-87-215-28-e4.png"/>
				</disp-formula>
			</p>
			<p>which represents the linear model given by:</p>
			<p>
				<disp-formula id="e5">
					<graphic xlink:href="2346-2183-dyna-87-215-28-e5.png"/>
				</disp-formula>
			</p>
			<p>with Y<sub>
 <italic>i</italic>
</sub> =ln(-ln(1-F(t<sub>
 <italic>i</italic>
</sub> )),b<sub>
 <italic>0</italic>
</sub> =-β ln(η), and x<sub>
 <italic>i</italic>
</sub> =ln(t<sub>
 <italic>i</italic>
</sub> ). In practice F(t<sub>
 <italic>i</italic>
</sub> )in <xref ref-type="disp-formula" rid="e4">Eq. (4)</xref> is estimated by the median rank approach to estimate the corresponding reliability of a data set and is given by [<xref ref-type="bibr" rid="B16">16</xref>]:</p>
			<p>
				<disp-formula id="e6">
					<graphic xlink:href="2346-2183-dyna-87-215-28-e6.png"/>
				</disp-formula>
			</p>
			<p>In <xref ref-type="disp-formula" rid="e7">eq. (7)</xref>, form [<xref ref-type="bibr" rid="B2">2</xref>], the sample size n is determined to ensure the desired reliability index as:</p>
			<p>
				<disp-formula id="e7">
					<graphic xlink:href="2346-2183-dyna-87-215-28-e7.png"/>
				</disp-formula>
			</p>
			<p>Now based on the formulation from the above, let determine the corresponding Weibull/Weibull reliability strength function.</p>
		</sec>
		<sec>
			<title>3. Stress-strength Weibull reliability formulation</title>
			<p>The stress/strength function is the one we must use to determine the reliability of a product which is subjected to variant stress [<xref ref-type="bibr" rid="B8">8</xref>]. And since the Weibull/Weibull stress-strength function does not have the closure property [<xref ref-type="bibr" rid="B16">16</xref>,<xref ref-type="bibr" rid="B17">17</xref>], implying that it has not a close solution for β<sub>
 <italic>s</italic>
</sub> ≠β<sub>
 <italic>S</italic>
</sub> , then the purpose of this section is to present the formulation of the stress-strength Weibull/Weibull reliability function when (β<sub>
 <italic>s</italic>
</sub> =β<sub>
 <italic>S</italic>
</sub> ), as well as when (β<sub>
 <italic>s</italic>
</sub> ≠β<sub>
 <italic>S</italic>
</sub> ). In general, the stress/strength reliability function, as shown in <xref ref-type="fig" rid="f1">Fig. 1</xref>, is given by the probability that the strength (S=Y) will be higher than the applied stress (s=X) R(t)=(P(Y&gt;X)) [<xref ref-type="bibr" rid="B13">13</xref>]. Therefore, the inference region represents the corresponding cumulative failure function [<xref ref-type="bibr" rid="B18">18</xref>].</p>
			<p>
				<fig id="f1">
					<label>Figure 1</label>
					<caption>
						<title>Stress-Strength Graphical Representation.</title>
					</caption>
					<graphic xlink:href="2346-2183-dyna-87-215-28-gf1.jpg"/>
					<attrib>Source: The Authors.</attrib>
				</fig>
			</p>
			<p>Thus, based on the Weibull distribution defined in <xref ref-type="disp-formula" rid="e1">Eq. (1)</xref>, the Weibull/Weibull stress/strength formulation for β<sub>
 <italic>s</italic>
</sub> =β<sub>
 <italic>S</italic>
</sub> is as follows.</p>
			<sec>
				<title>3.1. Stress-strength Weibull function for ( 𝛃 𝟏 = 𝛃 𝟐 )</title>
				<p>When the shape Weibull parameters are equal to β<sub>
 <italic>s</italic>
</sub> =β<sub>
 <italic>S</italic>
</sub> [<xref ref-type="bibr" rid="B19">19</xref>], the Weibull/Weibull stress/strength function is given as:</p>
				<p>
					<disp-formula id="e8">
						<graphic xlink:href="2346-2183-dyna-87-215-28-e8.png"/>
					</disp-formula>
				</p>
				<p>Unfortunately, because the stress behavior is determined by the environment on which the product is performing, and the product’s strength depends on the manufacturing process and material properties, then the stress and the product’s strength variables are independent each other, and as a consequence the probability that β<sub>
 <italic>s</italic>
</sub> =β<sub>
 <italic>S</italic>
</sub> is very low. Thus, now let present the stress-strength formulation when β<sub>S</sub>≠β<sub>s</sub>.</p>
				<p>The formulation is as follows.</p>
			</sec>
			<sec>
				<title>3.2. Stress-Strength Weibull Formulation for (β<sub>1</sub> ≠β<sub>2</sub>)</title>
				<p>Let first mentioning that for β<sub>S</sub>≠β<sub>s</sub> the Weibull distribution has not the closure property [<xref ref-type="bibr" rid="B19">19</xref>] given by:</p>
				<p>
					<disp-formula id="e9">
						<graphic xlink:href="2346-2183-dyna-87-215-28-e9.png"/>
					</disp-formula>
				</p>
				<p>It is to say, the closure property defined in <xref ref-type="disp-formula" rid="e9">Eq. (9)</xref> for the stress variable X<sub>1</sub>=X and the strength variable X<sub>2</sub>=Y holds only for β<sub>
 <italic>s</italic>
</sub> =β<sub>
 <italic>S</italic>
</sub> Therefore, the fact that the stress/strength Weibull/Weibull reliability function, for β<sub>
 <italic>S</italic>
</sub> ≠β<sub>
 <italic>s</italic>
</sub> has not a close solution is given as</p>
				<p>
					<disp-formula id="e10">
						<graphic xlink:href="2346-2183-dyna-87-215-28-e10.png"/>
					</disp-formula>
				</p>
				<p>Thus, in order to determine the reliability of a product when β<sub>
 <italic>S</italic>
</sub> ≠β<sub>
 <italic>s</italic>
</sub> , first a common β<sub>
 <italic>c</italic>
</sub> value must be determined and second, it has to be used in <xref ref-type="disp-formula" rid="e8">eq. (8)</xref> to determine the corresponding reliability. The common β<sub>
 <italic>c</italic>
</sub> value is determined as follows.</p>
			</sec>
		</sec>
		<sec>
			<title>4. Common Beta parameter estimation (β<sub>S</sub>≠β<sub>S</sub>)=βc</title>
			<p>The common β<sub>
 <italic>c</italic> 
</sub> value which can be used to determine the reliability given by <xref ref-type="disp-formula" rid="e8">eq. (8)</xref>, based on the desired known reliability R(tw), is given by:</p>
			<p>
				<disp-formula id="e11">
					<graphic xlink:href="2346-2183-dyna-87-215-28-e11.png"/>
				</disp-formula>
			</p>
			<p>In <xref ref-type="disp-formula" rid="e11">eq. (11)</xref> η<sub>
 <italic>S</italic>
</sub> and η<sub>
 <italic>S</italic> 
</sub> are the Weibull strength and stress parameters respectively. Now, because the common approach consists of using a safety factor to deal with uncertainties, let derive the corresponding formula to relate the used safety factor with the desired reliability. </p>
			<sec>
				<title>4.1. Probabilistic Weibull safety factor by using the common β value</title>
				<p>The Weibull safety factor is based on the strength and stress scale parameters (η<sub>S</sub>,η<sub>s</sub>), of the two parameters Weibull distribution is given by:</p>
				<p>
					<disp-formula id="e12">
						<graphic xlink:href="2346-2183-dyna-87-215-28-e12.png"/>
					</disp-formula>
				</p>
				<p>Thus, the SF<sub>
 <italic>w</italic>
</sub> index for β<sub>
 <italic>S</italic>
</sub> =β<sub>
 <italic>s</italic>
</sub> is directly given by </p>
				<p>
					<disp-formula id="e13">
						<graphic xlink:href="2346-2183-dyna-87-215-28-e13.png"/>
					</disp-formula>
				</p>
				<p>And for β<sub>
 <italic>S</italic>
</sub> ≠β<sub>
 <italic>s</italic>
</sub> , because by using <xref ref-type="disp-formula" rid="e10">Eq. (10)</xref> in terms of <xref ref-type="disp-formula" rid="e13">eq. (13)</xref> with the common βc value</p>
				<p>
					<disp-formula id="e14">
						<graphic xlink:href="2346-2183-dyna-87-215-28-e14.png"/>
					</disp-formula>
				</p>
				<p>Then, the SF<sub>
 <italic>w</italic>
</sub> index for any desired R(w) index is given by</p>
				<p>
					<disp-formula id="e15">
						<graphic xlink:href="2346-2183-dyna-87-215-28-e15.png"/>
					</disp-formula>
				</p>
				<p>Or equivalently R(tw) in terms of known SF<sub>
 <italic>w</italic> 
</sub> index is given by </p>
				<p>
					<disp-formula id="e16">
						<graphic xlink:href="2346-2183-dyna-87-215-28-e16.png"/>
					</disp-formula>
				</p>
				<p>Also, notice that from <xref ref-type="disp-formula" rid="e11">Eq. (11)</xref>, η<sub>
 <italic>S</italic>
</sub> in terms of SF<sub>
 <italic>w</italic> 
</sub> is given by</p>
				<p>
					<disp-formula id="e17">
						<graphic xlink:href="2346-2183-dyna-87-215-28-e17.png"/>
					</disp-formula>
				</p>
				<p>Finally, the corresponding mean <inline-formula id="e18">
						<inline-graphic xlink:href="2346-2183-dyna-87-215-28-ie18.png"/>
					</inline-formula> and standard deviation σ<sub>wS</sub> of the strength variable (or the mean <inline-formula id="e19">
						<inline-graphic xlink:href="2346-2183-dyna-87-215-28-ie18.png"/>
					</inline-formula>, and stress standard deviation σ<sub>wS</sub> of the stress, variable), by using the desired η<sub>
 <italic>S</italic>
</sub> (or η<sub>
 <italic>S</italic>
</sub> ) value is determined as follows.</p>
				<p>The mean <inline-formula id="e20">
						<inline-graphic xlink:href="2346-2183-dyna-87-215-28-ie18.png"/>
					</inline-formula> (or <inline-formula id="e21">
						<inline-graphic xlink:href="2346-2183-dyna-87-215-28-ie21.png"/>
					</inline-formula>)is given by:</p>
				<p>
					<disp-formula id="e22">
						<graphic xlink:href="2346-2183-dyna-87-215-28-e22.png"/>
					</disp-formula>
				</p>
				<p>And the standard deviation σ<sub>wS</sub> (or σ<sub>wS</sub>) is given by </p>
				<p>
					<disp-formula id="e23">
						<graphic xlink:href="2346-2183-dyna-87-215-28-e23.png"/>
					</disp-formula>
				</p>
				<p>Now let present the steps to determine the common β<sub>
 <italic>c</italic>
</sub> value.</p>
			</sec>
			<sec>
				<title>4.3. Steps to determine common β<sub>
 <italic>C</italic>
</sub> , SF<sub>
 <italic>w</italic>
</sub> and the stress-strength analysis for (β<sub>
 <italic>1</italic>
</sub> ≠β<sub>
 <italic>2</italic>
</sub> )</title>
				<p>In this section the steps of the proposed methodology to estimate the reliability index R(t<sub>
 <italic>w</italic>
</sub> ) and the corresponding safety factor SF<sub>
 <italic>w</italic>
</sub> are given. However, it is important to notice that 1) the derived safety factor is unique to the used reliability index R(t<sub>
 <italic>w</italic>
</sub> ) 2) by using the common β<sub>
 <italic>c</italic>
</sub> value, the stress-strength analysis for (β<sub>
 <italic>1</italic>
</sub> ≠β<sub>
 <italic>2</italic>
</sub> ) can easily be performed. The steps are as follows: </p>
				<p>Step 1. Determine the desired reliability R(t<sub>
 <italic>w</italic>
</sub> ) index to perform the analysis.</p>
				<p>Step 2. By using the R(t<sub>
 <italic>w</italic>
</sub> ) index of step 1, estimate the corresponding cumulative failure probability as: </p>
				<p>
					<disp-formula id="e24">
						<graphic xlink:href="2346-2183-dyna-87-215-28-e24.png"/>
					</disp-formula>
				</p>
				<p>Step 3. By using the R(t<sub>
 <italic>w</italic>
</sub> ) index of step 1 in <xref ref-type="disp-formula" rid="e7">Eq. (7)</xref>, determine the corresponding n value. </p>
				<p>Step 4. Collect the set of n stress data and the set of n strength data.</p>
				<p>Step 5. Determine the average (μ) of the stress and strength collected data. </p>
				<p>Step 6. Determine the logarithm of each one of the stress and strength collected data and determine their log-average values as</p>
				<p>
					<disp-formula id="e25">
						<graphic xlink:href="2346-2183-dyna-87-215-28-e25.png"/>
					</disp-formula>
				</p>
				<p>Then determine the exponential of the average of the estimated logarithms as</p>
				<p>
					<disp-formula id="e26">
						<graphic xlink:href="2346-2183-dyna-87-215-28-e26.png"/>
					</disp-formula>
				</p>
				<p>Step 7. With the average of step 5 and step 6 calculate the corresponding Ratio as</p>
				<p>
					<disp-formula id="e27">
						<graphic xlink:href="2346-2183-dyna-87-215-28-e27.png"/>
					</disp-formula>
				</p>
				<p>Step 8. By using the result of step 7 estimate the maximum and minimum stresses σ<sub>
 <italic>max</italic>
</sub> and σ<sub>
 <italic>min</italic>
</sub> values of the stress and the strength data as</p>
				<p>
					<disp-formula id="e28">
						<graphic xlink:href="2346-2183-dyna-87-215-28-e28.png"/>
					</disp-formula>
				</p>
				<p>
					<disp-formula id="e29">
						<graphic xlink:href="2346-2183-dyna-87-215-28-e29.png"/>
					</disp-formula>
				</p>
				<p>Step 9. By using the median rank approach estimate the average of the generated n y<sub>
 <italic>i</italic>
</sub> elements as</p>
				<p>
					<disp-formula id="e30">
						<graphic xlink:href="2346-2183-dyna-87-215-28-e30.png"/>
					</disp-formula>
				</p>
				<p>Step 10. Using the maximum and minimum stress σ<sub>
 <italic>1</italic>
</sub> , σ<sub>
 <italic>2</italic>
</sub> and μ<sub>
 <italic>y</italic>
</sub> values [<xref ref-type="bibr" rid="B13">13</xref>], determine the corresponding Weibull shape β value as</p>
				<p>
					<disp-formula id="e31">
						<graphic xlink:href="2346-2183-dyna-87-215-28-e31.png"/>
					</disp-formula>
				</p>
				<p>Step 11. By using the estimated β value of step 10, determine the corresponding Weibull scale η parameters as </p>
				<p>
					<disp-formula id="e32">
						<graphic xlink:href="2346-2183-dyna-87-215-28-e32.png"/>
					</disp-formula>
				</p>
				<p>The addressed β value of step 10, and the η value of step 11, are the Weibull stress distribution W<sub>
 <italic>s</italic>
</sub> (β(<sub>
 <italic>s</italic>
</sub> ,),η<sub>
 <italic>s</italic>
</sub> ).</p>
				<p>Step 12. By using the strength σ<sub>
 <italic>1</italic>
</sub> , σ<sub>
 <italic>2</italic>
</sub> values of step 5 in Eq. (<xref ref-type="disp-formula" rid="e28">28</xref> and <xref ref-type="disp-formula" rid="e29">29</xref>) determine the Weibull strength family W<sub>
 <italic>S</italic>
</sub> (β(<sub>
 <italic>S</italic>
</sub> ,),η<sub>
 <italic>S</italic>
</sub> ).</p>
				<p>Step 13. With the Weibull family of both the stress, strength and the desired reliability estimate the common β<sub>
 <italic>c</italic>
</sub> by using <xref ref-type="disp-formula" rid="e11">eq. (11)</xref>.</p>
				<p>Step 14. By using the βc, η<sub>
 <italic>s</italic>
</sub> and η<sub>
 <italic>s</italic>
</sub> values in <xref ref-type="disp-formula" rid="e8">eq. (8)</xref>, determine the corresponding reliability index R(t<sub>
 <italic>w</italic>
</sub> ).</p>
				<p>Step 15. By using <xref ref-type="disp-formula" rid="e13">eq. (13)</xref> calculate the corresponding probability safety factor SF<sub>
 <italic>w</italic>
</sub></p>
			</sec>
		</sec>
		<sec>
			<title>5. Numerical application</title>
			<p>In this section the numerical application to determine the reliability R(t<sub>
 <italic>w</italic>
</sub> ).index and its corresponding safety factor SF<sub>
 <italic>w</italic> 
</sub> is showed. In section 5.1 the reliability index for β<sub>
 <italic>S</italic>
</sub> =β<sub>
 <italic>s</italic>
</sub> , is presented. Section 5.2 presents the reliability index estimation besides to the corresponding probabilistic safety factor in the case of β<sub>
 <italic>s</italic>
</sub> ≠β<sub>
 <italic>S</italic>
</sub></p>
			<sec>
				<title>5.1. Numerical application for β<sub>
 <italic>S</italic>
</sub> =β<sub>
 <italic>s</italic>
</sub></title>
				<p>The stress in this application is the usage mileage distribution, the strength is the miles to failure distribution of the component. <xref ref-type="table" rid="t1">Table 1</xref> gives the data for the distribution of stress and strength. </p>
				<p>
					<table-wrap id="t1">
						<label>Table 1</label>
						<caption>
							<title>Mileage distribution per year</title>
						</caption>
						<graphic xlink:href="2346-2183-dyna-87-215-28-gt1.png"/>
						<table-wrap-foot>
							<fn id="TFN1">
								<p>Source: [<xref ref-type="bibr" rid="B20">20</xref>]</p>
							</fn>
						</table-wrap-foot>
					</table-wrap>
				</p>
				<p>Since from this data the Weibull stress parameters are β<sub>s</sub>=12.2171, and η<sub>s</sub>=12791 and the Weibull strength distribution parameters are β<sub>s</sub>=12.2171 and η<sub>s</sub>=15041, then, from <xref ref-type="disp-formula" rid="e8">eq.(8)</xref> the reliability index is R(w)=0.878643, and from <xref ref-type="disp-formula" rid="e13">eq.(13)</xref> the corresponding Weibull safety factor is SF<sub>
 <italic>w</italic>
</sub> =1.1759. Now let present the stress-strength analysis for β<sub>
 <italic>S</italic>
</sub> ≠β<sub>
 <italic>s</italic>
</sub> .</p>
			</sec>
			<sec>
				<title>5.1. Numerical application for β<sub>
 <italic>S</italic>
</sub> ≠β<sub>
 <italic>s</italic>
</sub> .</title>
				<p>A shaft testing subjected to bending and torsion stress load is presented. The shaft is subjected to random variable stress, then its behavior must be represented by a probability density function. On the other hand, all elements or products possess an inherent variable strength to support the stress exposed, in such a way in <xref ref-type="table" rid="t2">Table 2</xref> the corresponding test results are presented.</p>
				<p>Step 1. The desired reliability to perform the analysis is 95%. </p>
				<p>Step 2. From <xref ref-type="disp-formula" rid="e21">eq. (21)</xref>, the failure probability is 5% i.e. F(t<sub>
 <italic>w</italic>
</sub> )=0.05.</p>
				<p>Step 3. From <xref ref-type="disp-formula" rid="e7">eq. (7)</xref>, the number of observations to be collected is n=19.49572575≅20.</p>
				<p>Step 4. The experiment data is taken from <xref ref-type="table" rid="t2">Table 2</xref>. </p>
				<p>
					<table-wrap id="t2">
						<label>Table 2</label>
						<caption>
							<title>Stress-Strength Design Shaft</title>
						</caption>
						<graphic xlink:href="2346-2183-dyna-87-215-28-gt2.png"/>
						<table-wrap-foot>
							<fn id="TFN2">
								<p>Source: The Authors</p>
							</fn>
						</table-wrap-foot>
					</table-wrap>
				</p>
				<p>Step 5. From <xref ref-type="disp-formula" rid="e21">eq. (21)</xref>, the average of the experiment data is μ<sub>
 <italic>s</italic>
</sub> =9142854.736 for the stress and μ<sub>
 <italic>S</italic>
</sub> =34847504.61 for the strength.</p>
				<p>Step 6. Since the average of the log stress data is μ<sub>
 <italic>Xs</italic>
</sub> = 15.75, and the average of the log strength data is μ<sub>
 <italic>Xs</italic>
</sub> =17.3, then from <xref ref-type="disp-formula" rid="e22">eq.(22)</xref> μ<sub>
 <italic>ys</italic>
</sub> =6989990.705 and μ<sub>yS</sub>=32633020,09.</p>
				<p>Step 7. From <xref ref-type="disp-formula" rid="e23">eq.(23)</xref>, the ratio of the stress and strength data are Ratio<sub>
 <italic>s</italic>
</sub> =5893371.078 and Ratio<sub>
 <italic>S</italic>
</sub> =12224343.63.</p>
				<p>Step 8. From <xref ref-type="disp-formula" rid="e24">eq.(24</xref> and <xref ref-type="disp-formula" rid="e24">25</xref>), the maximum and minimum values of the stress are σ<sub>1</sub>=15036225.81 and σ<sub>2</sub>=3249483.65 and for the strength they are σ<sub>1</sub>=47071848.24 and σ<sub>2</sub>=22623160.98.</p>
				<p>Step 9. From <xref ref-type="disp-formula" rid="e26">eq. (26)</xref>, μ<sub>
 <italic>yi</italic>
</sub> =-0.5444.</p>
				<p>Step 10. From <xref ref-type="disp-formula" rid="e27">eq. (27)</xref>, the strength and stress shape parameters are β<sub>s</sub>=1.44 for the stress and β<sub>S</sub>=3.00 for the strength.</p>
				<p>Step 11. From <xref ref-type="disp-formula" rid="e28">eq. (28)</xref>, the strength and stress scale parameters are η<sub>s</sub>=10201500.14 for the stress and η<sub>S</sub>=39126161.24 for the strength.</p>
				<p>Step 12. From <xref ref-type="disp-formula" rid="e11">eq. (11)</xref>, the common shape parameter is βc=2.19038.</p>
				<p>Step 13. From <xref ref-type="disp-formula" rid="e8">eq. (8)</xref> the corresponding reliability index R(t<sub>
 <italic>w</italic>
</sub> )=0,9499997.</p>
				<p>Step 14. From <xref ref-type="disp-formula" rid="e13">eq. (13)</xref> the corresponding probabilistic safety factor is S<sub>
 <italic>fw</italic>
</sub> = 3.835.</p>
			</sec>
		</sec>
		<sec sec-type="conclusions">
			<title>6. Conclusion</title>
			<p>Because all products or elements are subject to a random stress variable, and, the product or element possesses an inherent random strength to support the applied stress, then, the reliability of products must be determined by using the stress-strength reliability methodology. And because the Weibull distribution does not possess the additive property then the stress-strength analysis is not possible to perform, but by the formulated of a common shape βc parameter the stress-strength Weibull analysis can be performed. And because the βc value only depends on the desired reliability of the analysis, then for this reliability value it is unique. Since the derived SFw safety factor depends on the desired reliability, then it is a random variable, And because the SFw value is unique for the corresponding reliability index, then in practice the 𝑆𝐹𝑤 index can be used to represent the R(t) index and vice versa.</p>
		</sec>
	</body>
	<back>
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		<fn-group>
			<fn fn-type="other" id="fn1">
				<label>M. Baro-Tijerina,</label>
				<p> is a researcher-professor in the Industrial and Manufacturing Department at the Technological Institute of Nuevo Casas Grandes, Mexico, He completed his Engineering career in Industrial and System at the ITSNCG. After concluding his master’s degree in Industrial Engineering at Universidad Autónoma de Ciudad Juárez, México, getting the highest mark. By last he completed his Technology PhD in 2020 getting also the highest mark. Contact al164467@alumnos.uacj.mx; Cellphone (52) 636-115-5553. ORCID: 0000-0003-1665-8379</p>
			</fn>
			<fn fn-type="other" id="fn2">
				<label>M.R. Piña-Monárrez,</label>
				<p> is a researcher-professor in the Industrial and Manufacturing Department at the Autonomous University of Ciudad Juárez, México. He completed his PhD. in Science in Industrial Engineering in 2006 at the Instituto Tecnológico de Ciudad Juárez, México. He had conducted research on system design methods including robust design, reliability and multivariate process control. He is member of the National Research System (SNI), of the National Council of Science and Technology (CONACYT) in México. ORCID: 0000-0002-2243-340</p>
			</fn>
			<fn fn-type="other" id="fn3">
				<label>B. Villa Covarrubias,</label>
				<p> has done his B.E. (Mechanical Engineering) and is a doctoral student at the Universidad Autónoma de Ciudad Juárez. His research interest is the study of reliability of mechanical components using Weibull distribution and mechanical design.ORCID: 0000-0002-6271-8072</p>
			</fn>
			<fn fn-type="other" id="fn4">
				<label>How to cite:</label>
				<p> Baro-Tijerina, M, Piña-Monárrez, M.R. and Villa-Covarrubias, B, Stress-strength Weibull analysis with different shape parameter β and probabilistic safety factor. DYNA, 87(215), pp. 28-33, October - December, 2020.</p>
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</article>