購(gòu)買設(shè)計(jì)請(qǐng)充值后下載，，資源目錄下的文件所見(jiàn)即所得，都可以點(diǎn)開(kāi)預(yù)覽，，資料完整，充值下載可得到資源目錄里的所有文件。。?！咀ⅰ浚篸wg后綴為CAD圖紙，doc,docx為WORD文檔，原稿無(wú)水印，可編輯。。。具體請(qǐng)見(jiàn)文件預(yù)覽，有不明白之處，可咨詢QQ:12401814

輸送帶的二維動(dòng)態(tài)特性 3.1.1非線性梁架（構(gòu)架）元 如果只有帶的縱向變形是主要素，那么梁架元就可用于模型的皮帶彈性反應(yīng)。梁架元組成部分有如圖2所示的兩個(gè)結(jié)點(diǎn)， P和Q ，四個(gè)位移參數(shù)確定部分載體X： xT = [up vp uq vq] ????????????(1) 對(duì)平面運(yùn)動(dòng)的梁架元有三個(gè)獨(dú)立的剛體運(yùn)動(dòng)，因此（這公式）仍然是描述一個(gè)變形的參數(shù)。 圖2 ：梁架元的精確位移 梁架元軸的長(zhǎng)度變化， [ 7 ] ： ε1 = D1(x) = ∫1 o ds2 - ds2o dξ ???????? ????(2) 2ds2o DSO是限元未變形的長(zhǎng)度，DS是限元變形的長(zhǎng)度，ξ是沿著有限元軸的無(wú)量綱長(zhǎng)度。 圖3 ：張帶的靜態(tài)凹陷 雖然帶呈彎曲狀態(tài)，但梁架元并沒(méi)有變形，這可能考慮到帶小數(shù)值凹陷的靜態(tài)影 響。靜態(tài)帶凹陷的比率是有定義的（見(jiàn)圖3 ） ： K1 = δ/1 = q1/8T ??????????(3) 其中q是暴露在外面帶和散裝物料的重量在豎直方向上分布的荷載， 1是帶輪間距，而T是帶的張力。，帶凹陷的縱向變形影響取決于[ 7 ] ： εs = 8/3 K2s ?????????????(4) 產(chǎn)生了非線性梁架元總的縱向變形。 3.1.2梁架元 圖4 ：節(jié)點(diǎn)的精確位移和旋轉(zhuǎn)的梁架元。 如果帶的橫向位移是主要因素，那么梁架元就可以用來(lái)模擬皮帶。同樣對(duì)于擁有六個(gè)位移參數(shù)的梁架元的平面運(yùn)動(dòng)來(lái)說(shuō)，相當(dāng)于三個(gè)獨(dú)立的剛體運(yùn)動(dòng)。因此就剩下三個(gè)變形參數(shù)是：縱向變形參數(shù)ε1 ，兩個(gè)彎曲變形參數(shù)ε2和ε3 。 圖5 ：梁架元的彎曲變形的 梁架元彎曲變形的參數(shù)可以定義為梁架元的組成載體（見(jiàn)圖4 ） ： xT = [up vp μp uq vq μq] ???????? (5) 和如圖5的變形結(jié)構(gòu) ε2 = D2(x) = e2p1pq ?????? ??(6) 1o ε3 = D3(x) = -eq21pq 1o 3.2繞過(guò)托輥及帶輪的帶運(yùn)動(dòng) 當(dāng)繞過(guò)托輥或帶輪的時(shí)候，帶運(yùn)動(dòng)是受到約束的。為了說(shuō)明（弄清楚）這些制約因素，影響制約因素（邊界）的條件都必須添加到用來(lái)代模擬帶的有限元中來(lái)。這可以通過(guò)使用多體動(dòng)力學(xué)進(jìn)行描述。多體機(jī)置動(dòng)力學(xué)的經(jīng)典描述，建立起由若干約束條件連接起來(lái)的剛體或剛性鏈接。在（變形）輸送帶的有限元描述里，帶被分離成多個(gè)有限元，有限元之間的聯(lián)系是可變形的。有限元是由節(jié)點(diǎn)連接的，因此分配了位移參數(shù)。要確定帶的運(yùn)動(dòng)，排除了剛體模型的變形模式。如果一個(gè)帶繞過(guò)托輥，，決定托輥上帶的位置（如見(jiàn)圖6）的帶長(zhǎng)度為ξ，被添加到組件矢量，如：式（6） ，因此產(chǎn)生了7個(gè)位移矢量參數(shù)。 圖6 ：由托輥支撐的帶 梁架元有兩個(gè)獨(dú)立的剛體運(yùn)動(dòng)，因此依然有五個(gè)變形參數(shù)存在。其中已經(jīng)在3.1中給出了ε1 , ε2和ε3 ，確定了帶的變形。剩下ε4和ε5 ，確定帶和托輥之間的相互作用，見(jiàn)圖7 。 圖7 ：兩個(gè)約束條件的梁架元有限元。 這些變形參數(shù)可以假設(shè)成無(wú)限剛度的彈性。這意味著： ε4 = D4(x) = (rξ + u ξ)e2 - rid.e2 = 0 ε5 = D5(x) = (r ξ + uξ)e1 - rid.e1 = 0 ?????????????? (7) 如果模擬的是ε4 > 0的時(shí)候，那么帶將脫離托輥，而描述帶的有限元上的約束條件也將去除。 3.3滾動(dòng)阻力 為了使一種模型能應(yīng)用于帶式輸送機(jī)有限元模型的滾動(dòng)阻力，已經(jīng)制定了一種計(jì)算滾動(dòng)阻力的近似公式， [ 8 ] 。帶運(yùn)動(dòng)中，暴露在帶外面的總滾動(dòng)阻力的組成部分，這三部分是耗能的主要部分，可以區(qū)分為包括：壓痕滾動(dòng)阻力，托輥的慣性（加速滾動(dòng)阻力）和軸承滾動(dòng)阻力（軸承阻力） 。確定滾動(dòng)阻力因素的參數(shù)包括直徑和托輥的材料，以及各種帶參數(shù)，如速度，寬度，材料，緊張狀態(tài)，環(huán)境溫度，帶橫向負(fù)荷，托輥間距和槽角。總滾動(dòng)阻力的因素，可以表示成總滾動(dòng)阻力和帶垂直負(fù)荷之間的比例，定義為： ft = fi + fa + fb ??? ?????????????(8) Fi是壓痕滾動(dòng)阻力的系數(shù)，F(xiàn)A是加速阻力系數(shù)，而FB是軸承阻力系數(shù)。這些組成系數(shù)由下面的[9]確定： Fi = CFznzh nhD-nD VbnvK-nk NTnT ???????(9) fa = Mred ?2u ?Fzb ???t2 fb = ????Mf???? ??Fzbri FZ是帶垂直方向上分布的負(fù)載和散裝物料的負(fù)載的總和， H是帶的覆蓋厚度，D是托輥的直徑，Vb是帶速，KN是帶負(fù)荷的名義百分之比，T是環(huán)境溫度，Mred是托輥的折算質(zhì)量，B是帶的寬度， U是帶的縱向位移，MF是總的軸承阻力矩和RI是軸承內(nèi)部半徑。在計(jì)算滾動(dòng)阻力中，皮帶的動(dòng)力性能及機(jī)械性能和皮帶上覆蓋的材料發(fā)揮著重要作用。這使得帶的選擇和帶上覆蓋材料，盡量減少由動(dòng)力阻力引起的能源消耗。 3.4帶驅(qū)動(dòng)系統(tǒng) 在穩(wěn)定性的帶運(yùn)動(dòng)情況下，為了能夠測(cè)定帶式輸送機(jī)驅(qū)動(dòng)系統(tǒng)的旋轉(zhuǎn)組件的影響，這個(gè)帶式輸送機(jī)的總模型必須是含有驅(qū)動(dòng)系統(tǒng)模型。驅(qū)動(dòng)系統(tǒng)的旋轉(zhuǎn)元件，就像一個(gè)減速箱，參照了3.2節(jié)中所述的約束條件。帶有減速比的減速箱，可以用帶兩個(gè)位移參數(shù)的減速元件來(lái)代替， μp和μq ，像一個(gè)剛體的（旋轉(zhuǎn)）運(yùn)動(dòng)，因此就剩下一個(gè)變形參數(shù)： εred = Dred(x) = iμp + μq = 0 ??????????????(10) 要確定電式扭矩感應(yīng)式電機(jī)，是否適應(yīng)所謂的兩軸式電動(dòng)機(jī)。該相電壓的矢量v可從（11）獲得： v = Ri + ωsGi + L ?i/?t ???? ????????????? (11) 在（11）式中I是相電流矢量，R是模型的相電阻， c是模型的相電感抗，L是模型的相感系數(shù)而ωs是電機(jī)轉(zhuǎn)子的角速度。電磁轉(zhuǎn)矩等于： Tc = iTGi ??????????????(12) 電機(jī)模型和驅(qū)動(dòng)系統(tǒng)機(jī)械組件是由驅(qū)動(dòng)系統(tǒng)的運(yùn)動(dòng)方程聯(lián)系著的： Ti = Iij ?2?j + Cik ??k ??Kil? ???? ????????????(13) ?t2 ?t 其中T是扭矩矢量，I是模型的慣量，C是模型的阻尼，K是矩陣剛度和?是電機(jī)旋轉(zhuǎn)軸的角速度。 模擬啟動(dòng)或停止程序控制反饋的程序可以添加到帶式驅(qū)動(dòng)系統(tǒng)模型中，用來(lái)控制驅(qū)動(dòng)扭矩。 3.5運(yùn)動(dòng)方程 整個(gè)帶式輸送機(jī)模型的運(yùn)動(dòng)方程可以得出潛在功率的原則， [ 7 ] ： fk - Mkl ?2x1 / ?t2 = σ1Dik ???????????????(14) 其中F是阻力矢量，M是模型的質(zhì)量而σ是拉格朗日乘數(shù)的矢量，可能解釋為雙重壓力矢量to張力矢量ε 。為了解決帶有X這一組方程，方程一體化是必要的。但是一體化的結(jié)果，必須確保滿足約束條件。如果(8)式中應(yīng)變?yōu)榱悖敲幢仨毤m正一體化結(jié)果，如見(jiàn)[ 7 ] ?？梢允褂媚Ｐ偷姆答佭x擇，例如限制提升物質(zhì)垂直方向上的運(yùn)動(dòng)。這種違逆動(dòng)力學(xué)的問(wèn)題可以用下面公式表示。鑒于帶模型及其驅(qū)動(dòng)系統(tǒng)的提升運(yùn)動(dòng)眾所周知，根據(jù)系統(tǒng)自由度和它的比例（速度）可以確定其他元件的運(yùn)動(dòng)。它超出了本文所討論關(guān)于此項(xiàng)的所有細(xì)節(jié)范圍。 3.6實(shí)例 為了在長(zhǎng)距離帶式輸送機(jī)系統(tǒng)設(shè)計(jì)階段能夠正確設(shè)計(jì)，應(yīng)用了有限元法。例如帶強(qiáng)度的選擇，可以減少的盡量減少，使用模型模擬的結(jié)果確定傳送帶的最大張力。以有限元模型的功能作為例子，應(yīng)該考慮到在兩個(gè)托輥位置范圍之間穩(wěn)定移動(dòng)帶的橫向振動(dòng)。在運(yùn)輸機(jī)的設(shè)計(jì)階段這必須被確定，才得以確?？諑У墓舱?。 對(duì)于皮帶輸送機(jī)的設(shè)計(jì)來(lái)說(shuō)，托輥和移動(dòng)帶間相互作用影響是很重要的。托輥的及帶輪的幾何不完善性，導(dǎo)致帶脫離托輥和帶輪能支撐的位置，在帶和支撐帶輪之間產(chǎn)生一種橫向振動(dòng)。這對(duì)帶施加了一部分的交互軸向應(yīng)力。如果這部分力是比皮帶的預(yù)應(yīng)力小，那么帶將在它的固有頻率中振動(dòng)，否則帶將被迫振動(dòng)。皮帶是會(huì)受迫振動(dòng)的，例如受托輥的偏心率影響。在輸送帶返程中，這種振動(dòng)特別值得注意。由于受迫振動(dòng)的頻率取決于帶輪和托輥的角速度，因此對(duì)于帶的速度，確定在帶輪和托輥之間，帶在自然頻率狀況下，橫向振動(dòng)中帶速影響，這個(gè)是很重要的。如果受迫振動(dòng)的頻率接近于皮帶橫向振動(dòng)的固有頻率，將發(fā)生共振現(xiàn)象。 有限元模型的模擬結(jié)果可用于確定穩(wěn)定移動(dòng)的帶的橫向振動(dòng)頻率范圍。該頻率是利用快速傅立葉技術(shù)從時(shí)域范圍到頻域范圍，帶橫向位移變換后得到的結(jié)果。除了使用有限元模型外也可以運(yùn)用近似分析法。 皮帶可以模擬成一個(gè)預(yù)應(yīng)力梁。如果皮帶的彎曲硬度可以被忽略，橫向位移比托輥間距還小，Ks << 1 ，并且?guī)г黾拥拈L(zhǎng)度相對(duì)于橫向位移的原始長(zhǎng)度來(lái)說(shuō)是微不足道，帶的橫向振動(dòng)可近似為下列線性微分方程，如見(jiàn)圖15 ： ?2v = (c22 - C2b) ?2v - 2Vb ?2v ?????????????? ???(15) ?t2 ?x2 ?x?t 其中V是皮帶的橫向位移和C2是橫向波的波速度，由（16）式定義： c2 = √g1/8Ks??????????????????(16) 首先，圖5中帶的橫向固有頻率范圍可從公式（16）獲得，如果假定v(O,t)=v(l,t)=0： fb = ??1?? c2 (1 - ?2) ??????? ???????????(17) 21 ?是無(wú)量綱的速比，由（18）式確定： ? = Vb / c2??????????????????(18) FB是不同帶的各自獨(dú)立的頻率范圍，由于輸送帶長(zhǎng)度方向上帶張力變化。托輥的受迫振動(dòng)頻率，使托輥產(chǎn)生了一個(gè)偏心率等于： fi = Vb / πD ??????????????????(19) 其中D是托輥的直徑。為了設(shè)計(jì)一個(gè)在托輥間距中無(wú)支撐的共振，這受到以下條件限制： L ≠ πD (1-?2) ????????? ??????????(20) 2? 由線性微分方程（16）所取得的成果不過(guò)是只適用于小數(shù)值的速比?。對(duì)于大數(shù)值的速比?來(lái)說(shuō)，如高速運(yùn)輸機(jī)或低的帶張力，在（16）式中所有非線性條件就顯得重要的。因此，數(shù)值模擬的運(yùn)用，有限元模型的開(kāi)發(fā)，都是為了確定帶橫向振動(dòng)線性和非線性頻率之間的比例范圍。這些關(guān)系已被確定適合不同的數(shù)值的?，例如說(shuō)一個(gè)功能凹陷的比率Ks。 使用快速傅里葉技術(shù)將橫向位移結(jié)果的轉(zhuǎn)化為頻譜。從這些頻譜中獲得的頻率與公式（18）獲得的頻率相比，其產(chǎn)生了圖8所顯示的曲線。從這一數(shù)字可見(jiàn)，對(duì)小于0.3的?來(lái)說(shuō)，計(jì)算誤差很小。對(duì)于大數(shù)值的?來(lái)說(shuō)，運(yùn)用線性近似值法產(chǎn)生的計(jì)算誤差達(dá)到10 ％以上。運(yùn)用了皮帶采用非線性梁架元的有限元模型，因此可以準(zhǔn)確地確定大數(shù)值?的橫向振動(dòng)。 對(duì)于小數(shù)值?的橫向振動(dòng)的頻率也可以用公式（18）準(zhǔn)確地預(yù)測(cè)。然而，它不能分析，例如帶凹陷和縱向波的傳播之間的相互作用，或者同樣可以看成有限元模型的脫離托輥的皮帶。 這決定帶應(yīng)力和橫向振動(dòng)頻率之間的關(guān)系可以用于皮帶張力監(jiān)測(cè)系統(tǒng)。 圖8 ：由兩個(gè)托輥支撐的帶的橫向振動(dòng)線性和非線性頻率之間的比例。 4 實(shí)驗(yàn)驗(yàn)證 為了使模擬的結(jié)果能夠得到驗(yàn)證，實(shí)驗(yàn)中使用了動(dòng)態(tài)試驗(yàn)設(shè)備，如圖9所示。 圖9 ：動(dòng)態(tài)試驗(yàn)設(shè)施 使用這試驗(yàn)設(shè)施能夠確定的兩個(gè)托輥的間距和卸荷扁帶的橫向振動(dòng)，例如返程部分的。聲音裝置是用來(lái)測(cè)量皮帶的位移。此外，還有在試驗(yàn)中為我們所知的張緊力，帶速，電機(jī)轉(zhuǎn)矩，托輥轉(zhuǎn)子與托輥的距離。 5 為例 由于最具有成本效益帶式輸送機(jī)的操作條件中出現(xiàn)了寬度范圍為0.6m- 1.2m[ 2 ] 的各種皮帶 ，可通過(guò)變換不同的帶速改變帶的輸送能力，。然而在帶速度被改變之前，應(yīng)確定帶和托輥之間的相互作用，以確保無(wú)支撐的帶的共振。為了說(shuō)明穩(wěn)定移動(dòng)的帶的橫向位移這一點(diǎn)，測(cè)量了兩個(gè)托輥的間隔。帶的總長(zhǎng)度L是52.7m，托輥間距I是3.66m，靜態(tài)凹陷的比例常數(shù)是2.1 ％ ，?為0.24而帶速Vb為 3.57m/ s。 這個(gè)信號(hào)的后期轉(zhuǎn)化由如圖5所示的快速傅里葉技術(shù)頻譜獲得。在圖5中 出 圖10 ：帶穩(wěn)定移動(dòng)時(shí)橫向振動(dòng)頻率 現(xiàn)了3個(gè)頻率。第一頻率是由帶結(jié)合處所引起的： fs = Vb/L = 0.067 Hz 第二個(gè)頻率，出現(xiàn)在1.94赫茲，是由皮帶的橫向振動(dòng)所造成的。 第三個(gè)頻率出現(xiàn)在10.5Hz，是由托輥的旋轉(zhuǎn)所造成的，從圖11所示的數(shù)值模擬獲得。 圖11 ：計(jì)算共振區(qū)的不同托輥的直徑D. 貫穿實(shí)驗(yàn)表明皮帶速度和托輥間距。 圖11顯示的是拖過(guò)帶與托輥互動(dòng)引起的共振區(qū)可以預(yù)測(cè)三個(gè)托輥的直徑。該帶式輸送機(jī)的托輥直徑為0.108M，從而可以預(yù)測(cè)皮帶速度鄰近0.64M/S的共振現(xiàn)象。為了驗(yàn)證結(jié)果，在啟動(dòng)運(yùn)輸機(jī)的時(shí)候測(cè)量了帶的最大橫向位移跨度。 圖12 ：測(cè)量橫向振動(dòng)和帶靜態(tài)凹陷幅度的標(biāo)準(zhǔn)差的比例。 在圖12中，可以看出橫向振動(dòng)的最大振幅發(fā)生在帶速為0.64M/S處，正如有限元模型模擬預(yù)測(cè)的結(jié)果一樣。因此，帶速度不應(yīng)選擇臨近0.64米/ s的。雖然是用扁帶進(jìn)行實(shí)驗(yàn)和理論的驗(yàn)證的，但是這種應(yīng)用技術(shù)也可運(yùn)用于槽型帶中。 6．結(jié)論 帶式輸送機(jī)有限元模型中梁架元的應(yīng)用，帶橫向位移的模擬，從而使能夠設(shè)計(jì)出帶無(wú)支撐的共振。對(duì)于小數(shù)值的?來(lái)說(shuō)，采用梁架元代替線性微分方程預(yù)測(cè)共振現(xiàn)象的優(yōu)勢(shì)是同樣可以預(yù)測(cè)到皮帶縱向和橫向位移的之間的相互作用以及從模擬中預(yù)見(jiàn)皮帶脫離托輥。 The Two-Dimensional Dynamic Behavior of Conveyor Belts 3.1.1 NON LINEAR TRUSS ELEMENT If only the longitudinal deformation of the belt is of interest then a truss element can be used to model the elastic response of the belt. A truss element as shown in Figure 2 has two nodal points, p and q, and four displacement parameters which determine the component vector x: xT = [up vp uq vq] ????????????(1) For the in-plane motion of the truss element there are three independent rigid body motions therefore one deformation parameter remains which describes Figure 2: Definition of the displacements of a truss element the change of length of the axis of the truss element [7]: ε1 = D1(x) = ∫1 o ds2 - ds2o dξ ???????? ????(2) 2ds2o where dso is the length of the undeformed element, ds the length of the deformed element and ξ a dimensionless length coordinate along the axis of the element. Figure 3: Static sag of a tensioned belt Although bending, deformations are not included in the truss element, it is possible to take the static influence of small values of the belt sag into account. The static belt sag ratio is defined by (see Figure 3): K1 = δ/1 = q1/8T ??????????(3) where q is the distributed vertical load exerted on the belt by the weight of the belt and the bulk material, 1 the idler space and T the belt tension. The effect of the belt sag on the longitudinal deformation is determined by [7]: εs = 8/3 K2s ?????????????(4) which yields the total longitudinal deformation of the non linear truss element: 3.1.2 BEAM ELEMENT Figure 4: Definition of the nodal point displacements and rotations of a beam element. If the transverse displacement of the belt is being of interest then the belt can be modelled by a beam element. Also for the in-plane motion of a beam element, which has six displacement parameters, there are three independent rigid body motions. Therefore three deformation parameters remain: the longitudinal deformation parameter, ε1, and two bending deformation parameters, ε2 and ε3. Figure 5: The bending deformations of a beam element The bending deformation parameters of the beam element can be defined with the component vector of the beam element (see Figure 4): xT = [up vp μp uq vq μq] ???????? (5) and the deformed configuration as shown in Figure 5: ε2 = D2(x) = e2p1pq ?????? ??(6) 1o ε3 = D3(x) = -eq21pq 1o 3.2 THE MOVEMENT OF THE BELT OVER IDLERS AND PULLEYS The movement of a belt is constrained when it moves over an idler or a pulley. In order to account for these constraints, constraint (boundary) conditions have to be added to the finite element description of the belt. This can be done by using multi-body dynamics. The classic description of the dynamics of multi-body mechanisms is developed for rigid bodies or rigid links which are connected by several constraint conditions. In a finite element description of a (deformable) conveyor belt, where the belt is discretised in a number of finite elements, the links between the elements are deformable. The finite elements are connected by nodal points and therefore share displacement parameters. To determine the movement of the belt, the rigid body modes are eliminated from the deformation modes. If a belt moves over an idler then the length coordinate ξ, which determines the position of the belt on the idler, see Figure 6, is added to the component vector, e.g. (6), thus resulting in a vector of seven displacement parameters. Figure 6: Belt supported by an idler. There are two independent rigid body motions for an in-plane supported beam element therefore five deformation parameters remain. Three of them, ε1, ε2 and ε3, determine the deformation of the belt and are already given in 3.1. The remaining two, ε4 and ε5, determine the interaction between the belt and the idler, see Figure 7. Figure 7: FEM beam element with two constraint conditions. These deformation parameters can be imagined as springs of infinite stiffness. This implies that: ε4 = D4(x) = (rξ + u ξ)e2 - rid.e2 = 0 ε5 = D5(x) = (r ξ + uξ)e1 - rid.e1 = 0 ?????????????? (7) If during simulation ε4 > 0 then the belt is lifted off the idler and the constraint conditions are removed from the finite element description of the belt. 3.3 THE ROLLING RESISTANCE In order to enable application of a model for the rolling resistance in the finite element model of the belt conveyor an approximate formulation for this resistance has been developed, [8]. Components of the total rolling resistance which is exerted on a belt during motion three parts that account for the major part of the dissipated energy, can be distinguished including: the indentation rolling resistance, the inertia of the idlers (acceleration rolling resistance) and the resistance of the bearings to rotation (bearing resistance). Parameters which determine the rolling resistance factor include the diameter and material of the idlers, belt parameters such as speed, width, material, tension, the ambient temperature, lateral belt load, the idler spacing and trough angle. The total rolling resistance factor that expresses the ratio between the total rolling resistance and the vertical belt load can be defined by: ft = fi + fa + fb ??? ?????????????(8) where fi is the indentation rolling resistance factor, fa the acceleration resistance factor and fb the bearings resistance factor. These components are defined by: Fi = CFznzh nhD-nD VbnvK-nk NTnT ???????(9) fa = Mred ?2u ?Fzb ???t2 fb = ????Mf???? ??Fzbri where Fz is distributed vertical belt and bulk material load, h the thickness of the belt cover, D the idler diameter, Vb the belt speed, KN the nominal percent belt load, T the ambient temperature, mred the reduced mass of an idler, b the belt width, u the longitudinal displacement of the belt, Mf the total bearing resistance moment and ri the internal bearing radius. The dynamic and mechanic properties of the belt and belt cover material play an important role in the calculation of the rolling resistance. This enables the selection of belt and belt cover material which minimise the energy dissipated by the rolling resistance. 3.4 THE BELT'S DRIVE SYSTEM To enable the determination of the influence of the rotation of the components of the drive system of a belt conveyor, on the stability of motion of the belt, a model of the drive system is included in the total model of the belt conveyor. The transition elements of the drive system, as for example the reduction box, are modelled with constraint conditions as described in section 3.2. A reduction box with reduction ratio i can be modelled by a reduction box element with two displacement parameters, μp and μq, one rigid body motion (rotation) and therefore one deformation parameter: εred = Dred(x) = iμp + μq = 0 ??????????????(10) To determine the electrical torque of an induction machine, the so-called two axis representation of an electrical machine is adapted. The vector of phase voltages v can be obtained from: v = Ri + ωsGi + L ?i/?t ???? ????????????? (11) In eq. (11) i is the vector of phase currents, R the matrix of phase resistance's, C the matrix of inductive phase resistance's, L the matrix of phase inductance's and ωs the electrical angular velocity of the rotor. The electromagnetic torque is equal to: Tc = iTGi ??????????????(12) The connection of the motor model and the mechanical components of the drive system is given by the equations of motion of the drive system: Ti = Iij ?2?j + Cik ??k ??Kil? ???? ????????????(13) ?t2 ?t where T is the torque vector, I the inertia matrix, C the damping matrix, K the stiffness matrix and ? the angle of rotation of the drive component axis's. To simulate a controlled start or stop procedure a feedback routine can be added to the model of the belt's drive system in order to control the drive torque. 3.5 THE EQUATIONS OF MOTION The equations of motion of the total belt conveyor model can be derived with the principle of virtual power which leads to [7]: fk - Mkl ?2x1 / ?t2 = σ1Dik ???????????????(14) where f is the vector of resistance forces, M the mass matrix and σ the vector of multipliers of Lagrange which may be interpret as the vector of stresses dual to the vector of strains ε. To arrive at the solution for x from this set of equations, integration is necessary. However the results of the integration have to satisfy the constraint conditions. If the zero prescribed strain components of for example e.g. (8) have a residual value then the results of the integration have to be corrected, also see [7]. It is possible to use the feedback option of the model for example to restrict the vertical movement of the take-up mass. This inverse dynamic problem can be formulated as follows. Given the model of the belt and its drive system, the motion of the take-up system known, determine the motion of the remaining elements in terms of the degrees of freedom of the system and its rates. It is beyond the scope of this paper to discuss all the details of this option. 3.6 EXAMPLE Application of the FEM in the desian stage of long belt conveyor systems enables its proper design. The selected belt strength, for example, can be minimised by minimising, the maximum belt tension using the simulation results of the model. As an example of the features of the finite element model, the transverse vibration of a span of a stationary moving belt between two idler stations will be considered. This should be determined in the design stage of the conveyor in order to ensure resonance free belt support. The effect of the interaction between idlers and a moving belt is important in belt-conveyor design. Geometric imperfections of idlers and pulleys cause the belt on top of these supports to be displaced, yielding a transverse vibration of the belt between the supports. This imposes an alternating axial stress component in the belt. If this component is small compared to the prestress of the belt then the belt will vibrate in it's natural frequency, otherwise the belt's vibration will follow the imposed excitation. The belt can for example be excitated by an eccentricity of the idlers. This kind of vibrations is particularly noticeable on belt conveyor returns. Since the frequency of the imposed excitation depends on the angular speed of the pulleys and idlers, and thus on the belt speed, it is important to determine the influence of the belt speed on the natural frequency of the transverse vibration of the belt between two supports. If the frequency of the imposed excitation approaches the natural frequency of transverse vibration of the belt, resonance phenomena occur. The results of simulation with the finite element model can be used to determine the frequency of transverse vibration of a stationary moving belt span. This frequency is obtained after transformation of the results of the transverse displacement of the belt span from the time domain to the frequency domain using the fast fourier technique. Besides using the finite element model also an analytical approach can be used. The belt can be modelled as a prestressed beam. If the bending stiffness of the belt is neglected, the transverse displacements are small compared to the idler space, Ks << 1, and the increase of the belt length due to the transverse displacement is negligible compared to its initial length, the transverse vibration of the belt can be approximated by the following linear differential equation, also see Figure 5: ?2v = (c22 - C2b) ?2v - 2Vb ?2v ?????????????? ???(15) ?t2 ?x2 ?x?t where v is the transverse displacement of the belt and c2 the wave speed of the transverse waves defined by, [1]: c2 = √g1/8Ks??????????????????(16) The first natural transverse frequency of the belt span of Figure 5 can be obtained from eq. (16) if it is assumed that v(O,t)=v(l,t)=0: fb = ??1?? c2 (1 - ?2) ??????? ???????????(17) 21 where ? is the dimensionless speed ratio defined by: ? = Vb / c2??????????????????(18) The frequency fb is different for each individual belt span since the belt tension varies over the length of the conveyor. The excitation frequency of an idler which has a single eccentricity is equal to: fi = Vb / πD ??????????????????(19) where D is the diameter of the idler. In order to design a resonance free belt support the idler space is subjected to the following condition: L ≠ πD (1-?2) ????????? ??????????(20) 2? The results obtained with the linear differential equation (16) however are valid only for low values of the ratio ?. For higher values of ?, as is the case for high-speed conveyors or low belt tensions, the non-linear terms in the full form of e.g. (16) become significant. Therefore numerical simulations using, the FEM model have been made in order to determine the ratio between the linear and the non-linear frequency of transverse vibration of a belt span. These relations have been determined for different values of ? as a function of the sag ratio Ks. The results for the transverse displacements were transformed to a frequency spectrum using a fast-fourier technique. The frequencies obtained from these spectra were compared to the frequencies obtained from e.g. (18) which yielded the curves as shown in Figure 8. From this figure it follows that for ? smaller that 0.3 the calculation errors are small. For higher values of ? the calculation error made by a linear approximation is more than 10 %. Application of a finite element model of the belt which uses non-linear beam elements therefore enables an accurate determination of the transverse vibrations for high values of ?. For lower values of ? the frequencies of transverse vibration can also be predicted accurate by e.g. (18). However it is not possible to analyse, for example, the interaction between the belt sag and the propagation of longitudinal waves or the lifting of the belt off the idlers as can be done with the finite element model. The determined relation between the belt stress and the frequency of transverse vibrations can also be used in belt tension monitoring systems. Figure 8: Ratio between the linear and the non-linear frequency of transverse vibration of a belt span supported by two idlers. 4. EXPERIMENTAL VERIFICATION In order to be able to verificate the results