QTZ40塔式起重機-塔身的優(yōu)化設(shè)計

資源目錄里展示的全都有，所見即所得。下載后全都有,請放心下載。原稿可自行編輯修改=【QQ：401339828 或11970985 有疑問可加】 摘要 本次設(shè)計在參照同類塔式起重機基礎(chǔ)上，對QTZ40型塔式起重機進行總體設(shè)計及塔身分析設(shè)計。在塔身設(shè)計工程中，采用了有限元法對其進行分析計算，采用了ANSYS10.0軟件進行分析。 按照整機主要性能參數(shù)，確定各機構(gòu)類型及鋼結(jié)構(gòu)型式，主要確定了塔身的結(jié)構(gòu)參數(shù)，并按照風載荷平行于起重臂方向，風載荷與起重臂方向呈45度角時兩種工況分析。通過對塔身作適當?shù)暮喕?，?yīng)用ANSYS10.0軟件建立塔身的有限元模型，施加各工況載荷，進行求解，進而可得各工況下各節(jié)點受力情況及各單元所受軸向力、軸向應(yīng)力大小及各工況下塔身的變形撓度大小，并能演示塔身加載過程的動畫，清晰的展現(xiàn)了各工況下塔身的受力性能。 本次設(shè)計分析通過修改模型參數(shù)共準備了三種不同方案，進而對不同模型方案進行分析比較。由比較不同模型在相同工況下的受力狀況及剛度狀況，綜合分析強度和剛度條件，可得出受力最為合理的一組模型參數(shù)，通過對此組參數(shù)下模型進行強度及剛度校核，進而獲得塔身的最終參數(shù)結(jié)果。 關(guān)鍵詞：QTZ40型塔式起重機 塔身 有限元分析 ANSYS10.0 畢業(yè)設(shè)計（論文）任務(wù)書 課題 名稱 QTZ40塔式起重機——塔身的優(yōu)化設(shè)計 系： 機械工程系 專業(yè)： 機械設(shè)計制造及其制動化 班級： 姓名： 學號： 起迄日期： 年3月25日~ 2013年 6月21日 設(shè)計（論文）地點: 指導教師： 輔導教師： 發(fā)任務(wù)書日期： 年3月 5 日 1、畢業(yè)設(shè)計（論文）目的： 本次畢業(yè)設(shè)計是對機械專業(yè)學生在畢業(yè)前的一次全面訓練，目的在于鞏固和擴大學生在校所學的基礎(chǔ)知識和專業(yè)知識，訓練學生綜合運用所學知識分析和解決問題的能力。是培養(yǎng)、鍛煉學生獨立工作能力和創(chuàng)新精神的最佳手段。畢業(yè)設(shè)計要求每個學生在工作過程中，要獨立思考，刻苦鉆研，有所創(chuàng)新、解決相關(guān)技術(shù)問題。通過畢業(yè)設(shè)計，使學生掌握塔式起重機的總體設(shè)計、塔身的設(shè)計、整體穩(wěn)定性計算等內(nèi)容，為今后步入社會、走上工作崗位打下良好的基礎(chǔ)。 2、畢業(yè)設(shè)計（論文）任務(wù)內(nèi)容和要求（包括原始數(shù)據(jù)、技術(shù)要求、工作要求等）： （1） 設(shè)計任務(wù)： ① 總體參數(shù)的選擇（QTZ40級別） ② 結(jié)構(gòu)形式 （2） 總體設(shè)計 ① 主要技術(shù)參數(shù)性能 ② 設(shè)計原則 ③ 平衡重的計算 ④ 塔機的風力計算 ⑤ 整機傾翻穩(wěn)定性的計算 （3） 塔身的設(shè)計和計算 ① 塔身的形式及尺寸 ② 設(shè)計塔身的強度、穩(wěn)定性及剛度驗算 ③ 聯(lián)接套焊縫強度的計算 ④ 高強度螺栓強度的計算 ⑤ 塔身腹肝的計算 （4） 要求 ① 主要任務(wù)：學生應(yīng)在指導教師指導下獨立完成一項給定的設(shè)計任務(wù)，編寫符合要求的設(shè)計說明書，并正確繪制機械與電氣工程圖紙，獨立撰寫一份畢業(yè)論文，并繪制有關(guān)圖表。 ② 知識要求：學生在畢業(yè)設(shè)計工作中，應(yīng)綜合運用多學科的理論、知識與技能，分析與解決工程問題。通過學習、鉆研與實踐，深化理論認識、擴展知識領(lǐng)域、延伸專業(yè)技能。 ③ 能力培養(yǎng)要求：學生應(yīng)學會依據(jù)技術(shù)課題任務(wù)，完成資料的調(diào)研、收集、加工與整理，正確使用工具書；培養(yǎng)學生掌握有關(guān)工程設(shè)計的程序、方法與技術(shù)規(guī)范，提高工程設(shè)計計算、圖紙繪制、編寫技術(shù)文件的能力；培養(yǎng)學生掌握實驗、測試等科學研究的基本方法；鍛煉學生分析與解決工程實際問題的能力。 ④ 綜合素質(zhì)要求：通過畢業(yè)設(shè)計，學生應(yīng)掌握正確的設(shè)計思想；培養(yǎng)學生嚴肅認真的科學態(tài)度和嚴謹求實的工作作風；在工程設(shè)計中，應(yīng)能樹立正確的生產(chǎn)觀、經(jīng)濟觀與全局觀。 ⑤ 設(shè)計成果要求： 1） 凡給定的設(shè)計內(nèi)容，包括說明書、計算書、圖紙等必須完整，不得有未完的部分，不應(yīng)出現(xiàn)缺頁、少圖紙現(xiàn)象。 2） 對設(shè)計的全部內(nèi)容，包括設(shè)計計算、機械構(gòu)造、工作原理、整機布置等，均有清晰的了解。對設(shè)計過程、計算步驟有明確的概念，能用圖紙完整的表達機械結(jié)構(gòu)與工藝要求，有比較熟練的認識圖紙能力。對運輸、安裝、使用等亦有一般了解。 3） 說明書、計算書內(nèi)容要精練，表述要清楚，取材合理，取值合適，設(shè)計計算步驟正確，數(shù)學計算準確，各項說明要有依據(jù)，插圖、表格及字跡均應(yīng)工整、清楚、不得隨意涂改。制圖要符合機械機械制圖標準，且清潔整齊。 4） 對國內(nèi)外塔式起重機情況有一般的了解，對各種塔式起重機有一定的分析、比較能力。 其他各項應(yīng)符合本資料有關(guān)部分提出的要求。 3、畢業(yè)設(shè)計（論文）成果要求（包括圖表、實物等硬件要求）： ① 計算說明書一份 內(nèi)容包括：設(shè)計任務(wù)要求的選型、設(shè)計計算內(nèi)容、畢業(yè)實習報告等。作到內(nèi)容完整，論證充分（包括經(jīng)濟性論證），字跡清楚，插圖和表格正規(guī)（分別進行統(tǒng)一編號）、批準，字數(shù)要求不少于2萬字；撰寫中英文摘要；提倡學生應(yīng)用計算機進行設(shè)計、計算與繪圖。 ② 圖紙一套 不少于四張零號圖紙量。 4、主要參考文獻： [1] 哈爾濱建筑工程學院主編.工程起重機.北京：中國建筑工業(yè)出版社 [2] 董剛、李建功主編.機械設(shè)計.機械工業(yè)出版社 [3] 機械設(shè)計手冊.化學工業(yè)出版社（5冊） [4] GB/T9462—1999 塔式起重機技術(shù)條件 [5] GB/T13752—1992 塔式起重機設(shè)計規(guī)范 [6] GB5144—1994 塔式起重機安全規(guī)程 [7] 劉品主編《互換性與測量技術(shù)基礎(chǔ)》（第2版）哈爾濱：哈爾濱工業(yè)大學出版社　2001 [8] 徐灝主編《機械設(shè)計手冊》（第2版）　北京：機械工業(yè)出版社　2000 [9] 曹雙寅主編《工程結(jié)構(gòu)設(shè)計原理》　南京：東南大學出版社　2002 [10]劉鴻文主編《材料力學 Ⅰ》（第4版） 高等教育出版社 [11]《QTZ50塔式起重機使用說明書》 [12]張青 張瑞軍 編著《工程起重機結(jié)構(gòu)與設(shè)計》北京：化學工業(yè)出版社，2008.7 [13]中華人民共和國國家標準GB/T　13752-92　《塔式起重機設(shè)計規(guī)范》　北京：中國標準出版社　1993 [14]范俊祥主編《塔式起重機》 中國建材工業(yè)出版社 [15]許鎮(zhèn)宇、邱宣懷主編：《機械零件》 人民教育出版社 [16]劉佩衡主編《塔式起重機使用手冊》 北京：機械工業(yè)出版社，2002 [17]中國建設(shè)部《鋼結(jié)構(gòu)設(shè)計規(guī)范》 2003.12 [18]黃靖遠 龔劍霞 賈延林 《機械設(shè)計學》北京工業(yè)出版社，2002 [19]GB/T　75144-1994 《塔式起重機安全規(guī)程》　 [20]孫在魯 著《塔式起重機應(yīng)用技術(shù)》北京：中國建材工業(yè)出版社，2003.5 [21]邢靜忠編著《ANSYS應(yīng)用實例與分析》北京：科學出版社，2006 [22]張方瑞主編《ANSYS應(yīng)用基礎(chǔ)與實例教程》北京：電子工業(yè)出版社，2006.9 [23]陳精一編著《ANSYS工程分析實例教程》北京：中國鐵道出版社，2007.5 5、本畢業(yè)設(shè)計（論文）課題工作進度計劃： 起 迄 日 期 工 作 內(nèi) 容 2013.3.25-2013.3.28 2013.3.29-2013.4.13 2013.4.14-2013.4.20 2013.4.21-2013.5.15 2013.5.16-2013.6.5 2013.6.6-2013.6.19 2013.6.20-2013.6.21 熟悉整理資料 方案選擇及總體設(shè)計 繪制總圖 塔身設(shè)計 繪制塔身裝配及結(jié)構(gòu)圖紙 繪制零件圖紙 準備論文及答辯 教研室審查意見： 教研室主任簽字： 年 月 日 系審查意見： 系主任簽字： 年 月 日 畢業(yè)實習報告 系 別 機械工程系 專 業(yè) 機械設(shè)計制造及其自動化 班 級 姓 名 學 號 指導教師 實習成績 畢業(yè)實習報告 1．實習內(nèi)容 大四下學期的4月11日至18日我在張家口神力起重設(shè)備有限公司進行為期一周的畢業(yè)實習。首先簡單介紹一下張家口神力起重設(shè)備有限公司。張家口神力起重設(shè)備有限公司，位于張家口市陵園北街3號，占地面積15000M3，建筑面積5100 M3。，現(xiàn)有員工150人(固定)，其中高級工程師3人，中級職稱10人，有證焊工18人，經(jīng)省級培訓專業(yè)技術(shù)人員33人。公司技術(shù)先進設(shè)備齊全，是起重行業(yè)比較早的中小企業(yè)。經(jīng)營范圍：少先吊、單梁橋式起重機、雙梁橋式起重機、懸掛式起重機、門式起重機、提升機、BR1-2類壓力容器制造，Ｂ級起重機安裝、改造。公司的前身為張家口市起重設(shè)備總廠，始建于1956年。從1970年由一般機械產(chǎn)品轉(zhuǎn)為起重機械產(chǎn)品。產(chǎn)品由原來的外加工逐步發(fā)展到擁有起重機械、壓力容器、起重機安裝、改造等國家頒發(fā)有資質(zhì)證的制造企業(yè)。現(xiàn)在有效資產(chǎn)約1500萬元，資產(chǎn)負債率為零。多年來，生產(chǎn)的各種吊車及壓力容器已銷往全國各個省市，獲得客戶的一致好評，由其是2002年后經(jīng)公司工程技術(shù)人員和廣大職工不解的努力和研究，通過不斷的改進，成功的制造出散粒物裝集裝箱用的專用起重機QE10+10、1-5型、經(jīng)北京、河北、東北三省、廣東、福建、等地區(qū)廣大用戶使用后反映此設(shè)備目前是裝散粒物最經(jīng)濟速度最快的一種設(shè)備。 本次實習是在完成課程之后進行最后的綜合實習，是貫徹理論聯(lián)系實際的一次實踐，在公司里我了解了起重機的概況。起重機械是指用于垂直升降或者垂直升降并水平移動重物的機電設(shè)備，其范圍規(guī)定為額定起重量大于或者等于0．5t的升降機額定起重量大于或者等于1t，且提升高度大于或者等于2m的起重機和承重形式固定的電動葫蘆等。多數(shù)起重機械在吊具取料之后即開始垂直或垂直兼有水平的工作行程，到達目的地后卸載，再空行程到取料地點，完成一個工作循環(huán)，然后再進行第二次吊運。一般來說，起重機械工作時,取料、運移和卸載是依次進行的,各相應(yīng)機構(gòu)的工作是間歇性的。起重機械主要用于搬運成件物品，配備抓斗后可搬運煤炭、礦石、糧食之類的散狀物料，配備盛桶后可吊運鋼水等液態(tài)物料。有些起重機械如電梯也可用來載人。在某些使用場合，起重設(shè)備還是主要的作業(yè)機械，例如在港口和車站裝卸物料的起重機就是主要的作業(yè)機械，通常起重機械由起升機構(gòu)（使物品上下運動）、運行機構(gòu)（使起重機械移動）、變幅機構(gòu)和回轉(zhuǎn)機構(gòu)（使物品作水平移動），再加上金屬機構(gòu)，動力裝置，操縱控制及必要的輔助裝置而成。 簡單介紹一下起重機的類型，在建筑工程中所用的起重機械，根據(jù)其構(gòu)造和性能的不同，起重機械按結(jié)構(gòu)不同可分為輕小型起重設(shè)備、升降機、起重機和架空單軌系統(tǒng)等幾類。輕小型起重設(shè)備主要包括起重小車、吊具、千斤頂、手動葫蘆、電動葫蘆和普通絞車，大多體積小、重量輕、使用方便。除電動葫蘆和絞車外，絕大多數(shù)用人力驅(qū)動，適用于工作不繁重的場合。它們可以單獨使用，有的也可作為起重機的起升機構(gòu)。有些輕小型起重設(shè)備的起重能力很大，如液壓千斤頂?shù)钠鹬亓恳堰_ 750噸。升降機主要作垂直或近于垂直的升降運動，具有固定的升降路線，包括電梯、升降機、礦井提升機和料斗升降機等。起重機是在一定范圍內(nèi)垂直提升并水平搬運重物的多動作起重機械。架空單軌系統(tǒng)具有剛性吊掛軌道所形成的線路，能把物料運輸?shù)綇S房各部分，也可擴展到廠房的外部，在一定范圍內(nèi)垂直提升和水平搬運重物的多動作起重機械，又稱吊車。屬于物料搬運機械。起重機的工作特點是間歇性運動，即在一個工作循環(huán)中取料、運移、卸載等動作的相應(yīng)機構(gòu)是交替工作的。 在實習崗位上我了解到了起重機的結(jié)構(gòu)及按結(jié)構(gòu)分類起重機運行機構(gòu)一般只用四個主動和從動車輪，如果起重機很大，常用增大車輪的辦法來降低輪壓。當車輪超過四個時，必須采用鉸接均衡車架裝置，使起重機的載荷均勻地分布在各車輪上。 橋架的金屬結(jié)構(gòu)由主梁和端梁組成，分為單主梁橋架和雙梁橋架兩類。單主梁由單根主梁和位于跨度兩邊的端梁組成，雙梁橋架又兩根主梁和端梁組成。主梁與端梁剛性連接，端梁兩端裝有車輪，用以支承橋架在高架上運行。主梁上焊有軌道，供其中小車運行。橋式主梁的結(jié)構(gòu)類型較多比較典型的有箱型結(jié)構(gòu)和空腹桁架結(jié)構(gòu)。箱型結(jié)構(gòu)又可分為正軌箱型雙梁、偏軌箱型雙梁、偏軌箱型單主梁等幾種。正軌箱型雙梁是廣泛采用的一種基本形式，主梁由上、下翼緣板和兩側(cè)的垂直腹板組成，小車鋼軌布置在上翼緣板的中心線上，它的結(jié)構(gòu)簡單，制造方便，適于成批生產(chǎn)，但自重較大。偏軌箱型雙梁和偏軌箱型單主梁量的截面都是由上、下翼緣板和不等厚的主副腹板組成，小車鋼軌布置在主腹板上方，箱體內(nèi)的短加筋板可以省去，其中偏軌箱型單主梁是由一根寬翼緣箱型主梁代替兩根主梁，自重較輕，單制造較復雜，而對于四桁架式結(jié)構(gòu)是在上水平桁架表面一般鋪有走臺板，自重輕，剛度大，但與其他結(jié)構(gòu)相比，外形尺寸大，制造較復雜，疲勞強度較低，已較少生產(chǎn)。 普通橋式起重機主要采用電力驅(qū)動，一般是在司機室內(nèi)操縱，也有遠距離控制的，起重量可達500噸，跨度可達60米，簡易梁式起重機又稱梁式起重機，其結(jié)構(gòu)組成與普通橋式起重機類似，起重量、跨度和工作速度均較小，橋架主梁是由工字鋼或其他型鋼和板鋼組成的簡單截面梁，用手拉葫蘆或者電動葫蘆配上簡易小車作為起重小車，小車一般在工字鋼的下翼緣上運行，橋架可以沿高架上的軌道運行，也可以沿懸吊在高架下面的軌道運行，這種起重機稱為懸掛梁式起重機。 冶金專用橋式起重機在鋼鐵生產(chǎn)過程中可參與特定的工藝操作，其基本結(jié)構(gòu)與普通橋式起重機相似，但在起重小車上還裝有特殊的工作機構(gòu)或裝置。這種起重機的工作特點是使用頻率、條件惡劣、工作級別高。 在工人師傅的細心指導下我懂得了起重機的維護方法，即起重機的潤滑是保證機器正常運轉(zhuǎn)，延長使用壽命，提高效率及保證安全的重要措施之一，凡是有軸承和孔動配合的部位以及有摩擦面的機械部分，都要定期進行潤滑。在操作起重機的時候安全是最重要的。牢固的安全觀念：安全工作是任何時候時候都是安全工作的頭等大事，在工作中，時時刻刻牢記安全落實，到到位，徹底做到安全意識在腦海中扎根。 2．實習結(jié)果 綜合上述，經(jīng)過這段時間的努力奮斗，我的理論知識和實際操作有了一定的提高和進步 ，希望我通過這次學習到的理論與自己的工作實踐緊密聯(lián)系起來，立足于本職工作崗位，勤懇踏實，不斷提高自己的理論與操作水平，努力提高自身素質(zhì)，使自己能適應(yīng)社會經(jīng)濟發(fā)展的客觀要求，做一名合格的當代工人。 3．實習總結(jié) 轉(zhuǎn)眼間，短短一周的畢業(yè)實習已悄然過去了。在這次實訓中，我們都真正把課堂所學的理論知識運用到實際業(yè)務(wù)中，掌握了許多平時生活中很少接觸的實踐技能。這無疑對于我們機械設(shè)計與制造專業(yè)的同學來說意義重大！雖然是短暫的一周畢業(yè)實習，但不僅讓我們對起重機的基本情況、基本知識等有了更加深刻的了解，并且為我們畢業(yè)設(shè)計提供了豐富的素材并且為我們以后走上工作崗位打下良好的基礎(chǔ)。與此同時，此次認知實習還讓我深刻感覺到自己的專業(yè)知識還不夠扎實，對起重機的設(shè)計規(guī)范以及注意事項還不夠熟練?？偟膩碚f，在此次認知實習中，我無論是專業(yè)知識還是操作技能上得到了很多的進步。 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: Wenhui YUE Key Lab of Health Maintenance for Mechanical Equipment Hunan University of Science and Technology Xiangtan, China AbstractDesign 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 14. (2) To develop maintenance methods and tools to ensure system maintenance to both low cost and short repair time, such as special maintenance toolboxes developed59. (3) To design for maintenance, during design procedure, system maintainability is evaluated and is improved1012. 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: 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. 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. 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 timentn=, where is time between maintenance activities, i.e., maintenance interval, the unit of can be hours, days, months, or years. If ( )inp t represents age proportion of parts at with age nti, thus age distribution of parts at time denotes matrix nt01( ),( ),( ),( )nninnnp tp tp tp tLL.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 ntThis project is supported National Natural Science Foundation of China 50875082 978-1-4244-4905-7/09/$25.002009 IEEE 1029portion 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 00t =00( )1p t= (1) At1t=, age fractions of the first box and the second box are represented as (2) 1100001000( )( ) 1( )d( )( )( )dp tp tf xxp tp tf xx=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. At22t=, the proportions of the first three boxes are calculated as 22211012010202110100( )( ) 1( )d( )( ) 1( )d( )( )( )( )( )p tp tf xxp tp tf xxp tp tf x dxp tf x dx=+ (3) So, at ntn=, portions of parts in each box are calculated by using the following equations. 110(1)1210(2)231022110101001( )() 1( )d( )() 1( )d( )() 1( )d( )() 1( )d( )() 1( )d( )()nnnnnnnnnnnnnnnnnnnninp tptf xxptptf xxptptf xxp tp tf xxp tp tf xxp tp tf=L L1(1)00( )dniixx+= (4) Where 0( )np tis the fraction of population of parts with age 0 at , representing parts that have just been put into service. It means that nt0( )np 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, ,nt ttLA 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 ()00( )1( )inp tNniniR tf x dx= (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 ()00()00( )( )1( )d1( )dininNnnp tNniNip tniiR tR tf xxf xx= (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 (for Weibull distribution, the parameter approximates expected life at big). 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( )R t0TmaxRminRmaxRminR0T0T1T2T 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 1030results from to 0t =2t=. Minimum reliability is defined as ()min( ) ,0,1,miRR tin=L (7) Supposed that simulation time is, and 0Tmaxmin,RR represent maximum value and minimum value of 00,2 tT T+ respectively. Once when ratio of maximum reliability value and minimum reliability value minmax/RR is satisfied, system reliability is regarded as arriving at steady value at time . Thus system reliability, or called as steady reliability, is defined as 0Tmaxmin(sRRR=+)/2 (8) 1 is the stabilization requirement, which could usually be 98%. If does not exist, system reliability will be unsteady. 0TIV. 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 NM, its life cycle cost model, including production cost and maintenance cost, is represent as 0101( )miiCcc p tc=+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.210c ,c ,c/mM=, 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 1cc0+02101( )miiCcmccp t=+ (10) B. Model of reliability-based design and optimization Supposed that a type part of the system has n design alternatives,12( ,)nXx xx=L, their failure density functions are expressed as corresponding to each alternative. ()12( ),( ),L( )nFf tf tf t=For a fixed maintenance interval0, its reliability-based design optimization model I for maintenance is represented as: 00min( ),s.t.mmssC x xXRRRR (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. 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 for maintenance is expressed as 00min( , ),s.t.mmssC xxXRRRR (12) In Eq.(11) and Eq.(12), C is obtained from Eq.(9) or Eq.(10). ,msRRdenotes minimum reliability and steady reliability of the system respectively. 0,ms0RRis allowable reliability values of the system. In general, 00(0.75 0.95)msRR=,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. 1031C. 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 partF, system service life M and coefficients of life cycle cost are . For maintenance of fixed interval, input conditions add in fixed maintenance interval210c ,c ,c0. Times of maintenance are clearly equal to0/Mduring whole life cycle. As to the situation that maintenance interval needs to be optimized, times of maintenance are rounded /M to obtain at different maintenance interval. In addition, Design alternatives for the system must satisfy requirement of system reliability, thus0,ms0RR 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. F012,c c c,iimsRRicix,iimsRRicix 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 Schemex 0c/(yuan) 1c/(yuan)2c/(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. 000.85,0.75RR=smTABLE II. SIMULATION RESULTS OF DESIGN ALTERNATIVES Schemex 0/(month) mRsR 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 Schemex 0/(month) mRsR C/(RMB)1 0.8 0.7734 0.8807 146.8879 2 1 0.7917 0.8677 157.6288 Optimizationcycle 3 1.8 0.7820 0.8588 141.5058 Notes: 0is interval of design model for fixed cycle maintenance Eq. (11), andis optimum interval of design model for optimizing cycle maintenance Eq. (12). *As shown from simulation results listed in Table 2, when system maintenance interval is fixed , optimum design alternative derived from Eq. (11) is alternative 01=2x. Alternative 1x does not satisfy system reliability constrains, and total life costs of alternative2x is lower than that of alternative3x. 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 3x. In the case, all design alternatives meet the requirements of system reliability, and total life costs of alternative 3x is the lowest, correspondingly system maintenance interval * is 1.8 in the alternative3x. 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 01=2x not only satisfies all design requirements but also approaches to requirement value. Reliability of alternative 1x satisfies the requirement of steady 1032reliability, but does not satisfy the requirement of minimum reliability in spite of its lowest cost. Though alternative3xsatisfies the requirement of system reliability, either steady reliability or minimum reliability, it has highest total life cycle costs. 1x2x3x( )R tt Figure 3. Reliability simulation of design alternatives for a fixed maintenance interval 1x2x3xtC/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 1x, in order to meet requirements of system reliability, maintenance interval decreases, , but its total life cost increases somewhat. For alternative*0.8=2x, maintenance interval keeps constant after optimization, also, which means that interval 1= is an optimum interval for this alternative. For alternative 3x, due to optimization, maintenance interval increases, 1.8=, 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. 1x2x3x( )R tt Figure 5. Reliability simulation of design alternatives for an optimum maintenance interval 1x2x3xtC 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 little more than the requirement values due to adoption of discrete optimization. (4) When design alternatives of the system are decided, the optimum choice of design alternatives depends on not only maintenance interval but also requirement of system reliability and system service life. For example, when interval is fixed (01=), and system reliability required reduces from to, the optimum design alternative derived from Eq.(11) is alternative00.75mR =00.70mR =1x instead of alternative2x. When system service life switches from 100=M to 50, the optimum design alternative obtained from Eq. (12) is alternative 1x replacing alternative3x shown as Fig. 6. It 1033means that, since parts made by high quality materials have long service life, the design alternative obtains lower total life cycle costs in spite of their higher production costs. VI. CONCLUSION Maintenance is one of critical tasks during life cycle of the product. Replacement of parts will cause the chang