機(jī)械裝備壽命的可靠構(gòu)建與優(yōu)化設(shè)計(jì)
摘要：維護(hù)設(shè)計(jì)對于機(jī)電產(chǎn)品或系統(tǒng)的壽命周期來講，是一種重要的設(shè)計(jì)方法?；跈C(jī)構(gòu)作用可能會出現(xiàn)失敗的幾率，機(jī)械系統(tǒng)的可靠性建模被發(fā)展了起來。基于部件可能會出現(xiàn)失敗的情況，機(jī)械系統(tǒng)的可靠性建模就被發(fā)展了起來而系統(tǒng)最小的可靠性和最穩(wěn)定的可靠性系數(shù)被定義為機(jī)械系統(tǒng)在壽命周期內(nèi)大致的可靠性。其次，提出維護(hù)的一個(gè)基于可靠性的設(shè)計(jì)最優(yōu)化模型，總生活周期消耗被考慮作為設(shè)計(jì)目標(biāo)和系統(tǒng)可靠性。最終，維護(hù)的基于可靠性的設(shè)計(jì)最優(yōu)化方法通過組分設(shè)計(jì)示范被說明。
關(guān)鍵詞:維護(hù); 可靠性; 模擬; 優(yōu)化設(shè)計(jì)
1. 介紹
在一個(gè)機(jī)械產(chǎn)品期間的生命周期，維護(hù)是非常重要的，可以保持產(chǎn)品可利用時(shí)間和延長它的壽命。關(guān)于維護(hù)的研究機(jī)械產(chǎn)品的大致被分類為以下三種：(1)如何擬定維護(hù)政策或(和)如何優(yōu)選考慮系統(tǒng)可靠性和維修費(fèi)用的維護(hù)期間[1?4]。(2)發(fā)展維護(hù)方法和工具保證系統(tǒng)維護(hù)到低成本和短的修理時(shí)間，例如發(fā)展特別的維護(hù)工具[5?9]。(3)在設(shè)計(jì)程序期間，為維護(hù)設(shè)計(jì)，系統(tǒng)可維護(hù)性地被評估和被改進(jìn)[10?12]。維護(hù)在設(shè)計(jì)之初就開始了。明顯地，維護(hù)的設(shè)計(jì)方法論，是其中一個(gè)在產(chǎn)品的生命周期的最佳的有效的維護(hù)手段，吸引許多研究者的興趣。然而，維護(hù)設(shè)計(jì)的研究主要在于兩個(gè)領(lǐng)域。一個(gè)是在產(chǎn)品設(shè)計(jì)選擇的可維護(hù)性評估; 另一個(gè)是為方便維護(hù)設(shè)計(jì)的特殊結(jié)構(gòu)的零件。在本文中，根據(jù)時(shí)間對失敗零件的密度函數(shù)，要調(diào)查接受維護(hù)的一個(gè)機(jī)械系統(tǒng)的零件的服務(wù)壽命。然后機(jī)械系統(tǒng)的可靠性模型被重建并且被仿真。最終，維護(hù)的新穎的設(shè)計(jì)最優(yōu)化方法通過一個(gè)鏈?zhǔn)皆O(shè)計(jì)被發(fā)展并說明。
2. 可靠性塑造維護(hù)的機(jī)械系統(tǒng)
2.1模型假定
在一個(gè)機(jī)械系統(tǒng)運(yùn)行一段時(shí)間之后，由于失敗的被替換分開，主要可靠模型是不適用于改變系統(tǒng)，因而應(yīng)重建可靠性模型。在本文談?wù)摰臋C(jī)械系統(tǒng)有以下特征：①一個(gè)系統(tǒng)包括很多同樣的零件，零件的數(shù)量在系統(tǒng)期間的一生周期是恒定的。②所有零件的時(shí)間對失敗密度分布函數(shù)是相同的，并且替換件也和原始的零件一樣有失敗分布函數(shù)。③每個(gè)部分的失敗是一個(gè)任意的獨(dú)立事件，也就是說一部分的失敗不影響其他部分在系統(tǒng)的中的失敗。
2.2為維護(hù)的可靠性建模
一個(gè)機(jī)械系統(tǒng)的可靠性取決于它的零件，可靠性和失敗的可能性取決于他們的工作壽命。在這里，根據(jù)時(shí)間的密度分布函數(shù)對零件的失效，應(yīng)計(jì)算好部件在機(jī)械系統(tǒng)中工作壽命，然后開發(fā)機(jī)械系統(tǒng)的可靠性模型。在一個(gè)機(jī)械系統(tǒng)的服務(wù)期間，發(fā)生故障的有些零件要求及時(shí)替換，因此機(jī)械系統(tǒng)的部分的壽命分布被改變了。推測在機(jī)械系統(tǒng)運(yùn)行一段時(shí)間后tn = nτ，其中τ是維護(hù)活動之間的時(shí)間，即維護(hù)間隔時(shí)間之后， τ單位可以是幾小時(shí)、幾天、幾個(gè)月或者幾年。 如果pi (pt)代表零件的年齡比例在n用年齡iτ，因而部分的年齡分布在時(shí)間上表示為矩陣{p0 (tn)， p1 (tn)， L、pi (tn)， L， pn (tn)}。零件的失敗密度函數(shù)和部分的壽命分布在系統(tǒng)的確定壽命分布在下次或者在下一段時(shí)間區(qū)內(nèi)留下來的部分目錄。壽命分布取決于每一段時(shí)間內(nèi)每一部分部件在下一段時(shí)間區(qū)內(nèi)失效的幾率。要發(fā)現(xiàn)零件的失效可能性失敗密度函數(shù)是從零開始的。存留下來的數(shù)量在下一個(gè)時(shí)間段得到提升，失效的部分被新的部件替換被重新返回第一個(gè)盒子里。
最初的，所有零件是新的病在第一個(gè)盒子里。 即在t0=0，在第一個(gè)箱子的部分是
P0(t0)=1 (1)
在t1=τ，第一個(gè)箱子的年齡分?jǐn)?shù)和第二個(gè)箱子代表：
P1(t1)=p0(t0)[1-∫f(x)dx]
P0(t1)=p0(t0) ∫f(x)dx (2)
兩種壽命盒子內(nèi)的部分，生存并且到下個(gè)壽命箱子，而不合格部件的部分被新的零件替換，從這兩個(gè)箱子到第一個(gè)箱子。
在T2 = 2τ，計(jì)算前三個(gè)箱子的比例
P2(t2)=p1(t1)[1-∫f(x)dx]
P1(t2)=p0(t1)[1-∫f(x)dx]
P0(t2)= p1(t1) ∫f(x)dx+ p0(t1) ∫f(x)dx (3)
……
因此，通過使用以下等式，在tn = nτ時(shí)，在每個(gè)箱內(nèi)的部分部件將被進(jìn)行如下計(jì)算：
當(dāng)P0(tn)是壽命在tn的部件總量的一小部分時(shí)，代表了部件剛剛被投入使用。這就意味著P0(tn)是這部分的失效率，或者說是小部件的代替。換種說法就是說，這些在第一個(gè)盒子里的一小部分部件在t0,t1,L,tn是用來取代失效部件的新部件。
一系列的系統(tǒng)包含了N個(gè)有相同失效概率分布的部件，每一個(gè)部分只是一系列的單元，每個(gè)單元是相對獨(dú)立的。在同一系列系統(tǒng)里任何一個(gè)單元的失效表現(xiàn)為一個(gè)系統(tǒng)的失效，按照可能增長的原則，一系列系統(tǒng)的可靠性就是：
由于組成系統(tǒng)的部件數(shù)量是恒定的，在此，機(jī)械系統(tǒng)維護(hù)的系統(tǒng)穩(wěn)定性被定義為：
3. 維護(hù)可靠性的模擬仿真
模擬仿真的結(jié)果顯示了系統(tǒng)的穩(wěn)定性在工作期間是不斷變化著的。一個(gè)系統(tǒng)的可靠性經(jīng)歷了幾次波動，有時(shí)是最大值而有時(shí)是最小值，最終達(dá)到一個(gè)穩(wěn)定的值。系統(tǒng)穩(wěn)定性的震動會周期性的衰減，這段時(shí)間是部件μ的預(yù)期壽命（根據(jù)韋伯分布，參數(shù)β近似于大α的預(yù)期壽命）。對于機(jī)械系統(tǒng)的設(shè)計(jì)和維護(hù)，系統(tǒng)穩(wěn)定性的最小值跟穩(wěn)定值是最重要的。系統(tǒng)的最低穩(wěn)定性出現(xiàn)在初始階段，但系統(tǒng)可靠性的穩(wěn)定值出現(xiàn)在運(yùn)行一段很長時(shí)間后。在此，為了后面方便討論，系統(tǒng)維護(hù)的最低可靠性和穩(wěn)定可靠性被定義為基于如圖6所示的系統(tǒng)穩(wěn)定性的仿真結(jié)果中。
由于它發(fā)生在初始階段，系統(tǒng)最小可靠性會在從t=0到t=2μ的仿真結(jié)果的不相關(guān)聯(lián)的可靠性值中找到。最小可靠性被定義為：
假設(shè)仿真時(shí)間是T0和Rmax、Rmin分別代替了在t∈[T ,T + 2μ]的最大值和最小值。一旦當(dāng)最大可靠性值和最小可靠性值的比值Rmin/Rmax>ε滿足，系統(tǒng)可靠性被認(rèn)為在T0時(shí)達(dá)到一個(gè)穩(wěn)定的值。因此，系統(tǒng)穩(wěn)定性或者說穩(wěn)定可靠性被定義為：
ε ≤ 1是穩(wěn)定的要求，通常是98%。如果T0不存在，系統(tǒng)穩(wěn)定性是不穩(wěn)定的。
4. 可靠性的優(yōu)化設(shè)計(jì)模型庫
對于維護(hù)的一個(gè)基于可靠性優(yōu)化設(shè)計(jì)模型被用來跟耗費(fèi)維護(hù)的系統(tǒng)可靠性和壽命周期消費(fèi)來代替，上述模型對于計(jì)算系統(tǒng)的部件替換率，最小可靠性和系統(tǒng)可靠性有幫助。在這個(gè)模型里，壽命周期的消耗被認(rèn)為是一個(gè)設(shè)計(jì)目標(biāo)，而系統(tǒng)的可靠性被認(rèn)為是設(shè)計(jì)約束條件。我們的工作目標(biāo)就是要去找到一個(gè)最小消耗的設(shè)計(jì)方法并同時(shí)滿足這個(gè)系統(tǒng)規(guī)定參數(shù)。
4.1 模型的壽命周期損耗
機(jī)械系統(tǒng)的壽命周期損耗包含著產(chǎn)品成本和維護(hù)成本。系統(tǒng)維護(hù)成本是來源于以下所列的項(xiàng)目：（1）替代部件的成本；（2）操作損耗包括替換部件時(shí)的資源損耗（比如： 勞動、裝備）；（3）替換部件時(shí)的生產(chǎn)間隔造成的間接成本；（4）替換部件的準(zhǔn)備工作成本。在前面的三個(gè)項(xiàng)目參與了每次維護(hù)時(shí)替代部件的數(shù)量。替換越多的部件就會耗費(fèi)越多的資源，占用越多的生產(chǎn)時(shí)間，因而帶來巨大的損失并增長了維護(hù)成本。最后一項(xiàng)沒有參與替換部件的數(shù)量上但參與了每次維護(hù)跟替換上。結(jié)果，機(jī)械系統(tǒng)的維護(hù)成本被保密為替換部件數(shù)量上的成本考慮和維護(hù)次數(shù)上的成本考慮。在這種方法下，對于一個(gè)包含固定數(shù)量N部件的機(jī)械系統(tǒng)，在它運(yùn)行了一段時(shí)間M，它的壽命周期損耗模型包含了生產(chǎn)成本和維護(hù)成本，表示為：
在式子9，C是系統(tǒng)內(nèi)每一部分部件總的壽命周期損耗。c0, c1, c2分別表示部件生產(chǎn)系數(shù)，更換成本系數(shù)和準(zhǔn)備成本系數(shù)，這些數(shù)據(jù)是統(tǒng)計(jì)分析領(lǐng)域的數(shù)據(jù)。m = M /τ，M代表著系統(tǒng)壽命。式子10等號右邊的首項(xiàng)代表系統(tǒng)的生產(chǎn)成本，式子9右邊的第二項(xiàng)表示系統(tǒng)的維護(hù)成本。在式子9里，由于部件的替換成本包含著不僅僅是替換失效部件的部件生產(chǎn)成本，而且有用于資源的成本和用于替換的間接成本。顯然，式子10里表示的不是絕對成本，而是相對成本。式子9也可以表示為：
4.2 基于可靠性的設(shè)計(jì)與優(yōu)化
假設(shè)系統(tǒng)的一類部件有n種設(shè)計(jì)方案。X= (x1,x2,L,xn), 它們的失效密度函數(shù)被表示為每一種方案，X= (x1,x2,L,xn) 它們的失效密度函數(shù)被表示為F=(f1(t),f2(t),L,fn(t)) 作為每一個(gè)方案。
對于一個(gè)維護(hù)的固定間隔τ0，它的基于可靠性優(yōu)化設(shè)計(jì)的模型I的維護(hù)被表示為：
顯然的，最小壽命周期損耗和可靠性取自上述模型的一段特定的時(shí)間段。對于任何一個(gè)的n種設(shè)計(jì)方案，它的成本和可靠性取決于維護(hù)間隔τ。最小的成品成本可以取自于最優(yōu)化的維護(hù)間隔。所謂的最佳的維護(hù)間隔，顧名思義地，就是將維護(hù)間隔優(yōu)化到最小的壽命周期成本，因此基于可靠性設(shè)計(jì)和優(yōu)化的模型Ⅱ的維護(hù)克表示為：
在式子11和式子12里，C是取自式子9或式子10，Rm、Rs分別表示系統(tǒng)的最小可靠性和穩(wěn)定的可靠性。Rm0、Rs0是系統(tǒng)允許的可靠性值。通常來講，Rm0= （0.75~0.95）Rs0，這也就意味著系統(tǒng)穩(wěn)定性在整個(gè)壽命周期內(nèi)允許在某一定的程度上變化，但變化范圍不會超過穩(wěn)定可靠性值的5%~25%。
4.3 根據(jù)系統(tǒng)可靠性模仿的設(shè)計(jì)最優(yōu)化
顯然，系統(tǒng)平穩(wěn)的可靠性、極小的可靠性和部分在設(shè)計(jì)模型的替換率可以從可靠性模仿而獲得。所以，維護(hù)的設(shè)計(jì)最優(yōu)化是基于模擬的設(shè)計(jì)方法。在設(shè)計(jì)模型，可靠性模仿的輸入的情況是時(shí)間對失敗密度系統(tǒng)部分F，系統(tǒng)服務(wù)生活M的分布函數(shù)，并且生活周期消耗系數(shù)是c0，c1，c2。
為固定的間隔時(shí)間維護(hù)，輸入的情況在固定的維護(hù)間隔時(shí)間τ 0增加。維護(hù)的時(shí)間與M/τ 0明顯地是相等的在一生周期期間。 至于維護(hù)間隔時(shí)間需要被優(yōu)選的情況，維護(hù)的時(shí)間是獲得的被環(huán)繞的M/τ在另外維護(hù)間隔時(shí)間。另外，系統(tǒng)的設(shè)計(jì)選擇必須滿足系統(tǒng)可靠性的要求，因而Rm, Rs得出來了。 終于，一個(gè)優(yōu)選設(shè)計(jì)選擇和它極小的信度、平穩(wěn)的可靠性和生活周期費(fèi)用被得出了。 設(shè)計(jì)最優(yōu)化流程圖維護(hù)的顯示作為式子二，設(shè)計(jì)最優(yōu)化二個(gè)模型維護(hù)的是聯(lián)合。最可能，一個(gè)模型的解答通常是與另一個(gè)模型不同。
5. 設(shè)計(jì)示范
有鏈?zhǔn)絺魉蜋C(jī)鏈接圓環(huán)的三個(gè)設(shè)計(jì)選擇，產(chǎn)品使用期限M等于100個(gè)月。時(shí)間的密度分布函數(shù)對圓環(huán)的失敗的是Weibull的作用，并且他們的發(fā)行參量和費(fèi)用系數(shù)生命周期在表1被列出如下。
假設(shè)極小的可靠性和平穩(wěn)的可靠性的要求是R0 = 0.85， R0 = 0.75。 考慮系統(tǒng)維護(hù)間隔時(shí)間從一系列的等效區(qū)別價(jià)值被挑選，離散最優(yōu)化方法被采取。兩個(gè)設(shè)計(jì)模型的模仿結(jié)果維護(hù)在表2.圖8到圖11被列出說明系統(tǒng)可靠性和總生活周期費(fèi)用隨系統(tǒng)的工作次數(shù)變化。
注： τ 0是間隔時(shí)間固定周期維護(hù)的設(shè)計(jì)模型Eq (11)，和最宜的間隔時(shí)間優(yōu)選周期維護(hù)的設(shè)計(jì)模型Eq(12).
當(dāng)系統(tǒng)維護(hù)間隔固定，最宜的設(shè)計(jì)選擇如顯示從模仿結(jié)果在表2列出了，Eq(11)是選擇τ 0 = 1 2 x。其中1x不滿足系統(tǒng)可靠性壓抑，并且共計(jì)選擇2 x的壽命消耗低于選擇3 x。 從這個(gè)例子，了解到不可能有將遇見系統(tǒng)可靠性為不適當(dāng)?shù)墓潭ǖ木S護(hù)間隔時(shí)間壓抑的設(shè)計(jì)選擇。當(dāng)系統(tǒng)維護(hù)間隔時(shí)間被優(yōu)選時(shí)，最宜的設(shè)計(jì)選擇被獲得了。Eq(12)是選擇3 x。 在這個(gè)例子中，所有設(shè)計(jì)選擇符合系統(tǒng)可靠性的要求，并且共計(jì)選擇3 x的壽命消耗是最低的，相應(yīng)地系統(tǒng)維護(hù)間隔時(shí)間τ *1.8選擇了3 x。 顯示易變的維護(hù)周期警察導(dǎo)致設(shè)計(jì)選擇另外選擇，并且共計(jì)生活費(fèi)用可以是通過優(yōu)選維護(hù)間隔時(shí)間減少。
幾個(gè)有趣的結(jié)果能從圖3到圖6中被找到。.
（1）當(dāng)固定的間隔時(shí)間(τ0 = 1)是堅(jiān)定的，選擇的系統(tǒng)可靠性τ0 = x2不僅滿足所有設(shè)計(jì)要求，而且接近對要求價(jià)值。選擇x1的可靠性滿足平穩(wěn)的可靠性的要求，但是不滿足極小的可靠性的要求竟管它最便宜。 雖然選擇3 x滿足系統(tǒng)可靠性的要求，平穩(wěn)的可靠性或極小值可靠性，它有最高的總壽命周期消耗。
圖3.設(shè)計(jì)選擇的可靠性模仿固定的維護(hù)間隔時(shí)間
圖4. 設(shè)計(jì)選擇的生活周期費(fèi)用模仿固定的維護(hù)間隔時(shí)間
（2）當(dāng)維護(hù)間隔時(shí)間被優(yōu)選時(shí)，最宜的間隔時(shí)間的選擇根據(jù)系統(tǒng)可靠性的令人滿意要求前提。 至于選擇1 x，為了符合系統(tǒng)可靠性的要求，維護(hù)間隔時(shí)間減退，但是它的總生活費(fèi)用增加有些。 為選擇τ * = 0.8x2，維護(hù)間隔時(shí)間在優(yōu)化以后保留常數(shù)，也，因此意味著間隔時(shí)間τ =1是這個(gè)選擇的最宜的間隔時(shí)間。 為選擇3 x，由于優(yōu)化、維護(hù)間隔時(shí)間增量、τ =1.8和在系統(tǒng)可靠性和設(shè)計(jì)要求之間的區(qū)別減少，因而它有更低的總生活周期費(fèi)用。 其外，三個(gè)設(shè)計(jì)選擇被優(yōu)選，系統(tǒng)可靠性和總生活周期費(fèi)用他們的曲線趨向?qū)谢蛨?jiān)固性，并且費(fèi)用區(qū)別在三個(gè)選擇之中的減少。
圖5.設(shè)計(jì)選擇的可靠性模仿最宜的維護(hù)間隔時(shí)間
圖6.設(shè)計(jì)選擇的生活周期費(fèi)用模仿最宜的維護(hù)間隔時(shí)間的
(3) 當(dāng)系統(tǒng)要求高的可靠性時(shí)，相應(yīng)地，維護(hù)間隔時(shí)間將減少，并且維修費(fèi)用將上升。 相反，當(dāng)系統(tǒng)要求低可靠性，相應(yīng)地，維護(hù)間隔時(shí)間將延遲，因此維修費(fèi)用將減少，系統(tǒng)維護(hù)費(fèi)用減退受系統(tǒng)可靠性要求支配。系統(tǒng)可靠性平穩(wěn)的價(jià)值和最小值隨著維護(hù)間隔時(shí)間的增加單調(diào)減少總生活周期費(fèi)用隨著維護(hù)間隔時(shí)間的增加而減少。結(jié)果，穩(wěn)定的極小的間隔時(shí)間系統(tǒng)可靠性價(jià)值和最小值滿足設(shè)計(jì)要求將得到設(shè)計(jì)選擇的極小的總生活周期費(fèi)用。必須指出設(shè)計(jì)選擇的系統(tǒng)可靠性比要求價(jià)值不是相等與，而是少許更多由于離散最優(yōu)化的采用。
(4) 當(dāng)系統(tǒng)的設(shè)計(jì)選擇決定時(shí)，設(shè)計(jì)選擇最宜的選擇取決于不僅維護(hù)系統(tǒng)可靠性和系統(tǒng)服務(wù)生活的間隔時(shí)間而且還有要求。 例如，當(dāng)間隔時(shí)間被固定時(shí)(τ 0 = 1)，并且需要的系統(tǒng)可靠性減少從對，從Eq.(11)獲得的最宜的設(shè)計(jì)選擇是選擇0.75 m R = 0.70 m R = x1而不是選擇2 x。當(dāng)系統(tǒng)服務(wù)生活轉(zhuǎn)換從M = 100到50時(shí)，最宜的設(shè)計(jì)選擇被獲得了。 Eq(12)是替換3 x而選擇1 x顯示作為圖6。這就意味著，因?yàn)楦哔|(zhì)量材料做的零件有長的產(chǎn)品使用期限，設(shè)計(jì)選擇得到更低的總壽命周期成本，竟管他們會有更高的生產(chǎn)成本。
6. 結(jié)論
在產(chǎn)品期間的生命周期維護(hù)是其中一項(xiàng)重要任務(wù)。零件的替換將導(dǎo)致系統(tǒng)可靠性和生活周期費(fèi)用的變動?；诹慵臅r(shí)間失效密度函數(shù)，平穩(wěn)的可靠性、極小的可靠性和生活周期費(fèi)用可以通過可靠性模型的系統(tǒng)可靠性的重建和模仿得到。本文開發(fā)維護(hù)的基于可靠性的設(shè)計(jì)最優(yōu)化方法學(xué)，總生活周期費(fèi)用被看待，當(dāng)作為設(shè)計(jì)的設(shè)計(jì)對象和系統(tǒng)可靠性壓抑。它提供一種新的方法做在機(jī)械系統(tǒng)之間的可靠性和總生活周期費(fèi)用的一種交易在設(shè)計(jì)最優(yōu)化的維護(hù)。
鳴謝
這項(xiàng)工作得到了湖南科學(xué)技術(shù)大學(xué)的劉博士的大力支持。 筆者相當(dāng)感激能得到其參考資料的注釋，極大地改進(jìn)了目前這項(xiàng)工作。
參考文獻(xiàn)
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Reliability Modeling and Design Optimization for Mechanical Equipment Undergoing Maintenance Liangpei HUANG Key Lab of Health Maintenance for Mechanical Equipment Hunan University of Science and Technology Xiangtan, China E-mail: huanglp413@ Wenhui YUE Key Lab of Health Maintenance for Mechanical Equipment Hunan University of Science and Technology Xiangtan, China Abstract—Design for maintenance is an important design methodology for the life cycle design of electromechanical products or systems. Based on the time-to-failure density function of the part, the reliability model of the mechanical system are developed and the system minimum reliability and steady reliability are defined for maintenance based on reliability simulation during the life cycle of the mechanical system. Secondly, a reliability-based design optimization model for maintenance is presented, in which total life cycle cost is considered as design objective and system reliability as design constrain. Finally, the reliability-based design optimization method for maintenance is illustrated by means of component design demonstrations. Keywords- Maintenance; Reliability; Simulation; Design optimization I. INTRODUCTION During the life cycle of a mechanical product, maintenance is very important to keep the product available and prolong its life. Studies on maintenance for mechanical products are roughly classified into the following three catalogs. (1) How to formulate maintenance policy or (and) how to optimize maintenance periods considering system reliability and maintenance cost [1?4] . (2) To develop maintenance methods and tools to ensure system maintenance to both low cost and short repair time, such as special maintenance toolboxes developed [5?9] . (3) To design for maintenance, during design procedure, system maintainability is evaluated and is improved [10?12] . Maintenance starts at design. Obviously, design methodology for maintenance, which is one of best effective maintenance means in the life-cycle of a product, attracts many researchers’ interests. However, research on design for maintenance is mainly centralized on two fields. One is maintainability evaluation on product design alternatives, the other is some peculiar structures of parts designed for convenient maintenance. In this paper, based on the time-to- failure density function of the part, distributions of service age of parts for a mechanical system that undergoes maintenance are investigated. Then the reliability model of the mechanical system is reconstructed and simulated. Finally, a novel design optimization methodology for maintenance is developed and illustrated by means of design of a link ring for the chain conveyor. II. RELIABILITY MODELING OF MECHANICAL SYSTEM FOR MAINTENANCE A. Model assumptions After a mechanical system runs some time, due to replacement of fail parts, primary reliability model is inapplicable to changed system, thus the reliability model should be reconstructed. The mechanical system discussed in this paper has following characteristics: ① System consists of a large number of same type parts, in which the number of parts is constant during the whole life cycle of the system. ② The time-to-failure density distribution functions of all parts are same, also, replacement parts have the same failure distribution functions as the original parts. ③ Failure of each part is a random independent event, i.e., failure of one part does not affect failure of other parts in the system. B. Reliability modeling for maintenance Reliability of a mechanical system depends on its parts, yet reliability and failure probability of which rest on their service ages. Herein, according to the density distribution function of time to failure of the part, part service age distribution of the mechanical system is calculated, then reliability model of the mechanical system for maintenance is developed. During the service of a mechanical system, some parts that fail require to be replaced in time, hence age distribution of parts of the mechanical system undergoing maintenance has been changed. Supposed that after the mechanical system runs some time n tnτ= , where τ is time between maintenance activities, i.e., maintenance interval, the unit of τ can be hours, days, months, or years. If () in p t represents age proportion of parts at with age n t iτ , thus age distribution of parts at time denotes matrix n t { } 01 (), (), , (), , () nn in nn p tptptptLL .The failure density function of parts and current age distribution of parts in the system determine age distribution at next time, or the portion of the contents of each bin that survive to the next time step. An age distribution obtained at each time step for each part population determines failure rate for the following time step. To find failure probability of parts the failure density function is integrated from zero to . The portion of the population that survives advances to the next age box, and the n t This project is supported National Natural Science Foundation of China [50875082] 978-1-4244-4905-7/09/$25.00?2009 IEEE 1029 portion that fail is replaced by new parts to become zero age to reenter the first box. Initially, all parts are new and zero age in the first box. That is, at , the portion in the first box is 0 0t = 00 () 1pt = (1) At 1 t τ= , age fractions of the first box and the second box are represented as (2) 11 0 0 0 01 00 0 () ()1 ()d () () ()d pt pt fx x pt pt fx x τ τ ? ?? =? ?? ?? ? ? = ? ∫ ∫ ? Portions of both age boxes survive and advance to the next age box, and portions of failed parts from both boxes replaced by new parts appear in the first box. At 2 2t τ= , the proportions of the first three boxes are calculated as 2 22 11 0 12 01 0 2 02 11 01 00 () ()1 ()d () ()1 ()d () () () () () pt pt fx x pt pt fx x p t p t f xdx p t f xdx τ τ ττ ? ?? =? ??? ?? ? ? ?? =? ? ?? ?? ? ? =+ ? ? ∫ ∫ ∫∫ (3) …… So, at n tnτ= , portions of parts in each box are calculated by using the following equations. 11 0 (1) 121 0 (2) 231 0 2 211 0 101 0 01 () ( )1 ()d () ( )1 ()d () ( )1 ()d () ( )1 ()d () ( )1 ()d () ( ) n nn n n n nn n n n nn nn nn nn nin pt p t fx x p tpt fxx p tpt fxx pt pt fx x pt pt fx x pt pt f τ τ τ τ τ ?? ? ??? ? ??? ? ? ? ?? =? ?? ?? ? =? ? ? ? =? ? ? ?? =? ?? ?? ?? =? ?? ?? = ∫ ∫ ∫ ∫ ∫ LL 1 (1) 0 0 ()d n i i xx τ ? + = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ∑ ∫ ? ? ? ? ? ? (4) Where 0 () n p t is the fraction of population of parts with age 0 at , representing parts that have just been put into service. It means that n t 0 () n p t is failure rate of parts, or replacement rate of failed parts. In other word, the fractions of parts in the first box at are new parts that replace these failed parts. 01 ,, , n tt tL A series system consists of N parts that have the same failure density distribution, each part is just a series unit, and each unit is relatively independent. In series system the failure of any one unit results in system failure, in according to the principle of probability multiplication, the reliability of series systems becomes () 0 0 () 1 () in pt Nn i n i Rt fxdx τ = ?? ′ =? ?? ?? ∏ ∫ (5) Since the number of parts that comprise the system is constant, here, the system reliability of the mechanical system for maintenance is defined as () 0 0 () 0 0 () () 1()d 1()d in in N nn pt Nn i N i ptn i i Rt R t fx x fx x τ τ = = ′== ?? ?= ?? ?? ?? ? ?? ?? ∏ ∫ ∏ ∫ (6) III. RELIABILITY SIMULATION FOR MAINTENANCE Simulation results show that system reliability varies during service. The reliability of a system experiences several oscillations, sometimes is maximum value and then minimum value, finally reaches steady value. Oscillations of system reliability periodically decay, and the period is about the expected life time of parts μ (for Weibull distribution, the parameter β approximates expected life at bigα ). For design and maintenance of mechanical systems, minimum value and steady value of system reliability are of importance. Minimum reliability of the system appears at beginning stage, but steady reliability value of the system appears after running a long time. Here, to conveniently discuss later, minimum reliability and steady reliability of the system for maintenance are defined based on simulation results of system reliability shown as in Fig.6. t () Rt 0 T max R min R max R min R 0 T 0 T 1 T 2 T Figure 1. System reliability parameters definition As it appears at initial phase, minimum reliability of the system can be found in discrete reliability values of simulation 1030 results from to 0t = 2t μ= . Minimum reliability is defined as ()min ( ) , 0,1, ,R Rt i n==L (7) Supposed that simulation time is , and 0 T max min ,R R represent maximum value and minimum value of 00 [, 2]tTT μ∈+ respectively. Once when ratio of maximum reliability value and minimum reliability value min max /RR ε> is satisfied, system reliability is regarded as arriving at steady value at time . Thus system reliability, or called as steady reliability, is defined as 0 T max min ( s RR R=+)/2 (8) 1≤ε is the stabilization requirement, which could usually be 98%. If does not exist, system reliability will be unsteady. 0 T IV. OPTIMIZATION DESIGN MODEL BASED ON RELIABILITY A reliability-based design optimization model for maintenance is presented to make a trade-off between the system reliability and life-cycle cost of parts that includes maintenance cost, in which the above models are helpful to calculate part replacement rate of the system, minimum reliability and system reliability. In the model, the cost of life cycle is considered as a design objective, and the reliability of the system is considered as design constraint. The task is to find a design having the minimum cost and satisfying the constraints. A. Model of life cycle cost Life cycle costs of mechanical systems include production costs and maintenance costs. System maintenance costs are from items listed as follows. (1) cost of parts’ replacement, (2) operation cost including cost of resources spent (i.e. labor, equipment) for replacing parts, (3) indirect cost resulting from production interrupt caused by replacing parts, (4) preparation work cost for replacing parts.The foregoing three items are concerned with the number of replacing parts every time of maintenance. The more parts replaced will consume more resource, occupy more production time, thus bring tremendous loss and increase maintenance cost. The last item is not concerned with the number of replacing parts but times of maintenance or replacement. As a result, maintenance costs of mechanical systems are classified as cost considering part replacement number and cost considering maintenance times. In this way, for a mechanical system with a constant number of parts , after it runs for time N M , its life cycle cost model, including production cost and maintenance cost, is represent as [] 010 1 () m i i Cc cpt c = =+ + ∑ 2 (9) In Eq. (9), C is total cost of life cycle of the system for per part in the system. denote coefficient of part production cost, coefficient of replacement cost and coefficient of preparation cost respectively, and these coefficients can be confirmed by statistical analysis of field datum. 210 c,c,c /mMτ= , where M represents life of the system. The first term of right-hand side of Eq.(10) represents production cost of the system, the second term of right-hand side of Eq. (9) represents maintenance cost of the system. In Eq. (9), , because part replacement cost includes not only production cost of the part that replaces the failed part, but also costs that are spent for resources, and indirect cost caused by replacement. Obviously, the cost that Eq.(10) denotes is not absolute cost, but relative cost. Eq.(9) is also represented as 1 cc≥ 0 + ∑0210 1 () m i i Cc mc c pt = =+ (10) B. Model of reliability-based design and optimization Supposed that a type part of the system has n design alternatives, 12 (, , , ) n X xx x= L , their failure density functions are expressed as corresponding to each alternative. () 12 (), (), ,L () n Fftft ft= For a fixed maintenance interval 0 τ , its reliability-based design optimization model I for maintenance is represented as: 0 0 min ( ), s.t. mm ss Cx x X RR RR ∈ ≥ ≥ (11) Apparently, the minimum life cycle cost and reliability obtained from the above model is responding to the fixed period. For any one of n design alternatives, its cost and reliability depend on the maintenance interval τ . The achievable minimum cost could be obtained from the optimization of the maintenance interval. For the optimal maintenance interval, namely, maintenance interval is optimized to minimized the life cycle cost, thus reliability- based design and optimization model Ⅱ for maintenance is expressed as 0 0 min ( , ), s.t. mm ss Cx x X RR RR τ ∈ ≥ ≥ (12) In Eq.(11) and Eq.(12), C is obtained from Eq.(9) or Eq.(10). , ms R R denotes minimum reliability and steady reliability of the system respectively. 0 , ms 0 R R is allowable reliability values of the system. In general, 00 (0.75 ~ 0.95) ms R R= ,which implies system reliability allows to vary in some certain degree during whole life cycle, but variation scope is not over 5%~25% of steady reliability. 1031 C. Design optimization based on system reliability simulation Obviously, system steady reliability, minimum reliability and part replacement rate in design models can be derived from reliability simulation. Therefore, design optimization for maintenance is a design methodology based on simulation. In design models, input conditions of reliability simulation are the time to failure density distribution functions of the system part F , system service life M and coefficients of life cycle cost are . For maintenance of fixed interval, input conditions add in fixed maintenance interval 210 c,c,c 0 τ . Times of maintenance are clearly equal to 0 /M τ during whole life cycle. As to the situation that maintenance interval needs to be optimized, times of maintenance are rounded /M τ to obtain at different maintenance interval. In addition, Design alternatives for the system must satisfy requirement of system reliability, thus 0 , ms 0 R R are given. Finally, an optimal design alternative and its minimum reliability, steady reliability and life cycle cost are outputted. The flow chart of design optimization for maintenance is shown as Fig.2, in which two models of design optimization for maintenance are integrated. Most possibly, the solution of one model is usually different from another model. F 012 ,,ccc , ii ms R R i c i x , ii ms R R i c i x Figure 2. Chart of design optimization for maintenance V. DESIGN DEMONSTRATION There are three design alternatives for link rings of chain conveyors, the service life M of which is equal to 100 months. The density distribution function of the time to failure of rings is the Weibull distribution, and their distribution parameters and cost coefficients of life cycle are listed in Table 1 as below. TABLE I. DESIGN ALTERNATIVES AND THEIR PARAMETERS Scheme x α β 0 c /(yuan) 1 c /(yuan) 2 c /(yuan) 1 5 10 3 7 0.5 2 4 12 5 10 0.5 3 4 20 10 18 0.5 Suppose that the requirement of minimum reliability and steady reliability is . Considering that system maintenance interval is selected from a series of equivalent difference values, discrete optimization method is adopted. Simulation results of two design models for maintenance are listed in Table 2. Fig. 8 to Fig. 11 illustrates that system reliability and total life cycle costs vary with service time of the system. 00 0.85, 0.75RR== sm TABLE II. SIMULATION RESULTS OF DESIGN ALTERNATIVES Scheme x 0 τ /(month) m R s R C /(RMB) 1 1 0.7266 0.8531 131.1384 2 1 0.7917 0.8677 157.6288 Identical cycle 3 1 0.8654 0.9252 180.0370 Scheme x 0 τ /(month) m R s R C /(RMB) 1 0.8 0.7734 0.8807 146.8879 2 1 0.7917 0.8677 157.6288 Optimization cycle 3 1.8 0.7820 0.8588 141.5058 Notes: 0 τ is interval of design model for fixed cycle maintenance Eq. (11), and is optimum interval of design model for optimizing cycle maintenance Eq. (12). * τ As shown from simulation results listed in Table 2, when system maintenance interval is fixed , optimum design alternative derived from Eq. (11) is alternative 0 1τ = 2 x . Alternative 1 x does not satisfy system reliability constrains, and total life costs of alternative 2 x is lower than that of alternative 3 x . From the example, it could be understood that there could not be a design alternative that would meet system reliability constrains for an inappropriate fixed maintenance interval. When system maintenance interval is optimized, optimum design alternative derived from Eq. (12) is alternative 3 x . In the case, all design alternatives meet the requirements of system reliability, and total life costs of alternative 3 x is the lowest, correspondingly system maintenance interval * τ is 1.8 in the alternative 3 x . It is shown that variable maintenance cycle police leads to different choice of design alternatives, and total life costs can be reduce by optimizing maintenance interval. Several interesting results could be found from Fig. 3-Fig. 6. (1) When a fixed interval ( ) is determined, system reliability of alternative 0 1τ = 2 x not only satisfies all design requirements but also approaches to requirement value. Reliability of alternative 1 x satisfies the requirement of steady 1032 reliability, but does not satisfy the requirement of minimum reliability in spite of its lowest cost. Though alternative 3 x satisfies the requirement of system reliability, either steady reliability or minimum reliability, it has highest total life cycle costs. 1 x 2 x 3 x () Rt t Figure 3. Reliability simulation of design alternatives for a fixed maintenance interval 1 x 2 x 3 x t C /yuan Figure 4. Life cycle costs simulation of design alternatives for a fixed maintenance interval (2) When maintenance interval is optimized, selection of optimum interval is based on the premise of satisfying requirements of system reliability. As to alternative 1 x , in order to meet requirements of system reliability, maintenance interval decreases, , but its total life cost increases somewhat. For alternative * 0.8τ = 2 x , maintenance interval keeps constant after optimization, also, which means that interval 1τ = is an optimum interval for this alternative. For alternative 3 x , due to optimization, maintenance interval increases, 1.8τ = , and the difference between system reliability and design requirements reduces, thus it has lower total life cycle costs. Besides, three design alternatives are optimized, their curves of system reliability and total life cycle costs trend to centralization and consistence, and difference of costs among three alternatives reduces. 1 x 2 x 3 x () Rt t Figure 5. Reliability simulation of design alternatives for an optimum maintenance interval 1 x 2 x 3 x t C Figure 6. Life cycle costs simulation of design alternatives for an optimum maintenance interval (3) When the system requires high reliability, correspondingly, maintenance interval will reduce and maintenance costs will rise. On the contrary, when system requires low reliability, correspondingly, maintenance interval will delay, so maintenance costs will reduce, as the decrease of system maintenance costs is subject to system reliability requirement. The steady value and minimum value of system reliability monotonously reduce with the increase of maintenance interval, also, total life cycle costs reduce with the increase of maintenance interval. As a result, minimum interval that steady value and minimum value of system reliability satisfy design requirements will obtain minimum total life cycle costs for the design alternative. It must be pointed out that system reliability of the design alternative is not equal to but l