2023年全國(guó)碩士研究生考試考研英語(yǔ)一試題真題(含答案詳解+作文范文)_第1頁(yè)
已閱讀1頁(yè),還剩26頁(yè)未讀, 繼續(xù)免費(fèi)閱讀

下載本文檔

版權(quán)說(shuō)明:本文檔由用戶提供并上傳,收益歸屬內(nèi)容提供方,若內(nèi)容存在侵權(quán),請(qǐng)進(jìn)行舉報(bào)或認(rèn)領(lǐng)

文檔簡(jiǎn)介

1、<p><b>  英文原稿</b></p><p>  Application of Stress-based Finite Element Method to a Flexible Slider Crank Mechanism</p><p> ?。╕.L.Kuo University of Toronto W.L.Cleghorn University

2、 of Canada)</p><p>  Abstract—This paper presents a new procedure to apply the stress-based finite element method on Euler-Bernoulli beams.An approximated bending stress distribution is selected,and then the

3、 approximated transverse displacement is determined by integration.The proposed approach is applied to solve a flexible slider crank mechanism.The formulation is based on the Euler-Lagrange equation,for which the Lagran

4、gian includes the components related to the kinetic energy,the strain energy,and the work done</p><p>  Keywords:stress-based finite element method;slider</p><p>  crank mechanism;Euler-Lagrange

5、 equation.</p><p>  1.Introduction</p><p>  The displacement-based finite element method employs complementary energy by imposing assumed displacements.This method may yield the discontinuities

6、of stress fields on the inter-element boundary while employing low-order elements,and the boundary conditions associated with stress could not be satisfied.Hence,an alternative approach was developed and called the stres

7、s-based finite element method,which utilizes assumed stress functions.Veubeke and Zienkiewicz[1,2]were the first researchers intro</p><p>  The operation of high-speed mechanisms introduces vibration,acousti

8、c radiation,wearing of joints,and inaccurate positioning due to deflections of elastic links.Thus,it is necessary to perform an analysis of flexible elasto-dynamics of this class of problems rather than the analysis of r

9、igid body dynamics.Flexible mechanisms are continuous dynamic systems with an infinite number of degrees of freedom,and their governing equations of motion are modeled bynonlinear partial differential equations,bu</p&

10、gt;<p>  This paper presents a new approach for the implementation of the stress-based finite element method on the Euler-Bernoulli beams.The developed approach first selects an assumed stress function.Then,the ap

11、proximated transverse displacement function is obtained by integrating the assumed stress function.Thus,this approach can satisfy the stress boundary conditions without imposing a constraint.We apply this approach to sol

12、ve a flexible slider crank mechanism.In order to show the accuracy enhanceme</p><p>  II.Stress-based Method for Euler-Bernoulli Beams</p><p>  The bending stress of Euler-Bernoulli beams is ass

13、ociated with the second derivative of the transverse displacement,namely curvature,which can be approximated as the product of shape functions and nodal variables:</p><p>  Where is a row vector of shape fu

14、nctions for the ith element; is a column vector of nodal curvatures,y is the lateral position with respect to the neutral line of the beam,E is the Young’s modulus,and is the transverse displacement,which is a function

15、 of axial position x.</p><p>  Integrating Eq.(1)leads to the expressions of the rotation and the transverse displacement as Rotation: </p><p>  Transverse displacement: </p><p>  W

16、here and are two integration constants for the ith element,which can be determined by satisfying the compatibility.</p><p>  Substituting Eqs.(2)and(3)into(1),the finite element displacement,rotation and c

17、urvature can be</p><p>  expressed as: </p><p>  where the subscripts(C),(R)and(D)refer to curvature,rotation and displacement,respectively.By applying the variational principle,the element and

18、global equations can be obtained[11-13].</p><p>  Table 1:Comparison of the displacement-and the stress-based finite element methods for an</p><p>  Euler-Bernoulli beam element</p><p

19、>  III.Comparisons of the Displacement-and Stress-based Finite Element Methods</p><p>  The major disadvantage of the displacement-based finite element method is that the stress fields at the inter-elemen

20、t nodes are discontinuous while employing low-degree shape functions.This discontinuity yields one of the major concerns behind the discretization errors.In addition,it might use excessive nodal variables while formulati

21、ng stiffness matrices.</p><p>  The stress-based method has several advantages over the displacement-based finite method.First of all,the stress-based method produces fewer nodal variables (Table 1).Secondly

22、,when employing the stress-basedfinite method,the boundary conditions of bending stress can be satisfied,and the stress is continuous at theinter-element nodes.Finally,the stress is calculated directly from the solution

23、of the global system equations.However,the only disadvantage of the stress-based finite method is that th</p><p>  IV.Generation of Governing Equation</p><p>  The slider crank mechanism shown i

24、n Fig.1 is operated with a prescribed rigid body motion of the crank,and the governing equations are derived using a finite element formulation.The derivation procedure of the finite element equations involves:(1)derivin

25、g the kinematics of a rigid body slider crank mechanism;(2) constructing a translating and rotating beam element based on the rigid body motion of the mechanism;(3)defining a set of global variables to describe the motio

26、n of a flexible slider cra</p><p>  A.Element equation of a translating and rotating beam</p><p>  Consider a flexible beam element subjected to prescribed rigid body translations and rotations.

27、Superimposed on the rigid body trajectory,a finite number of deflection variables in the longitudinal and transverse directions is allowed.The Euler-Lagrange equation is used to derive the governing differential equation

28、s for an arbitrarily translating and rotating flexible member.Since elastic deflections are considered small,and there is a finite number of degrees of freedom,the governing equations a</p><p>  In view of h

29、igh axial stiffness of a beam,it is reasonable to consider the beam as being rigid in its longitudinal direction.Hence,the longitudinal deflection is given as </p><p>  where u1 is a nodal variable,which is

30、constant with respect to the x direction shown in Fig.2.The transverse deflection can be represented as</p><p>  The velocity of an arbitrary point on the beam element with a translating and rotating motion

31、is given as</p><p>  where is the absolute velocity of point O of the beam element shown in Fig.2;θ?is the angular velocity of the beam element; are the longitudinal and transverse displacements of an arbitr

32、ary point on the beam element,respectively;x is a longitudinal position on the beam element shown in Fig.</p><p><b>  2.</b></p><p>  If we letρbe the mass per unit volume of element

33、 material;A,the element cross-sectional area,and L the element length,then the kinetic energy of an element is expressed as</p><p>  The flexural strain energy of uniform axially rigid element with the Young

34、’s modulus,E,and second moment of area,I,is given as</p><p>  The work done by a tensile longitudinal load,(i)P,in an element that undergoes an elastic transverse deflection is given by[14]</p><p&

35、gt;  Longitudinal loads in a moving mechanism element are not constant,and depend both on the position in the element and on time.With the longitudinal elastic motions neglected,the longitudinal loads may be derived from

36、 the rigid body inertia forces,and can be expressed as</p><p>  where PR is an external longitudinal load acting at theright hand end of an element,andox</p><p>  (i )ais the absolute eaccelerat

37、ion of the point O in the x direction shown in Fig.2.</p><p>  The Lagrangian takes the form</p><p>  Substituting Eqs.(5-10)into(12),and employing the Euler-Lagrange equations,the governing equ

38、ations of motion for a rotating and translating elastic beam can be expressed in the following matrix form:</p><p>  where[Me],[Ce]and[Ke]are mass,equivalent damping,and equivalent stiffness matrices of a el

39、ement,respectively;{Fe}is a load vector of an element.When formulating the mass matrix of the coupler,the mass of the slider should be taken into account.</p><p>  B.Global equations of slider crank mechanis

40、m</p><p>  For the proposed approach to solve a flexible slider crank mechanism,the global variables are the curvatures on the nodes.For assembling all elements,it is necessary to consider the boundary condi

41、tions applied to the mechanism.Since a prescribed motion applied to the base of the crank,there is a bending moment at point O shown in Fig.1,i.e.,the curvature at point O exists.For points A and B shown in Fig.1,we pres

42、ume that both points refer to pin joints.Thus,the bendingmoments and the curvatures </p><p>  Since Eq.(13)is a matrix-form expression in terms of the vector of global variables{φ},the global equations can b

43、e obtained by directly summing up all of element equations,which can be expressed as</p><p>  where[M],[C],[K]are global mass,damping and stiffness matrices,respectively;{F}is a global load vector.</p>

44、<p>  V.Numerical simulation based on steady state</p><p>  The rotating speed of the crank is operating at 150rad/s(1432 rpm),and the system parameters of a flexible slider crank are as follows:</

45、p><p>  R2=0.15(m),R3=0.30(m),ρA=0.225(kg/m),EI=12.72(N-m2),mB=0.03375(kg)</p><p>  where R2 and R3 are the lengths of the crank and coupler,respectively;mB is the mass of the slider.</p>&l

46、t;p>  The analytical results of this paper are presented by plotting steady state transverse displacements and bending strains of midpoints on crank and coupler throughout a cycle of motion.The steady state can be obt

47、ained by adding a physical damping matrix,namely Rayleigh damping </p><p>  whereαandβare two constants,which can be determined from two given damping ratio that correspond to two unequal frequencies of vibr

48、ation[15]. In this paper,the values ofαandβare determined based on the first two natural frequencies.</p><p>  By adding physical damping to the equations of motion,the analytical solution is obtained by per

49、forming the constant time-step Newmark method over twenty cycles of motion.The initial conditions are set to zeros when performing numerical time integration.</p><p>  The error indicator is defined as </

50、p><p>  where QFE and QRef are two quantities based on a finite element solution and a reference solution,respectively.Generally,they are functions of time,and they can be arbitrarily selected,such as energy,di

51、splacement,bending strain,etc.t1 and t2 refer to the interval of timeintegration,which are usually one cycle after steady-state condition has been reached.Since an exact solution is not available,a reference solution is

52、obtained by the displacement-based finite element method based on twenty eleme</p><p>  Fig.3.Time responses of the total energy,mensionless midpoint deflection of the coupler,and</p><p>  he mi

53、dpoint strain of the coupler at the steady state condition</p><p>  VI.Numerical Simulations</p><p>  In the section,we consider the mechanism with a rigid crank.The coupler is the only flexible

54、 link.Based on the beam element constructed in Section IV.,the beam element has a rigid axial motion,but it has a transverse deflection.</p><p>  When we implement the stress-based finite elementmethod propo

55、sed in Section III.,it is necessary to consider the boundary conditions of the modeled links and the approximated degree of shape functions.In this example,we select a linear function along the axial axis to approximate

56、the strain distribution of the coupler,and the boundary conditions of the coupler are considered without zero bending moment.Thus,it is impossible to model the coupler with one element.</p><p>  In the examp

57、le,we consider the coupler discretized as two,three,four,and five elements,and its curvature distribution is approximated by a linear function as</p><p>  And then,the time responses and the errors of the to

58、tal energy,the midpoint deflection of the coupler,the midpoint strain of the coupler is obtained by the stress-based finite element method.Also,the first natural frequency is evaluated.</p><p>  The rotating

59、 speed of the crank is operating at 150rad/s(1432 rpm),and the system parameters of a flexible slider-crank are as follow[16]:R2=0.15(m),R3=0.30(m),ρA=0.225(kg/m),EI=12.72(N-m 2),mB=0.03375(kg)where R2 and R3 are the len

60、gths of the crank and coupler,respectively;mB is the mass of the slider.</p><p>  In order to compare the errors obtained by the displacement-based finite element method,we also use it to solve the mechanism

61、,and its results are based on Ref.[17].</p><p>  Table 2.Errors of the first natural frequency by both finite element methods</p><p>  Fig.3.shows the time responses of the total energy,the dime

62、nsionless midpoint deflection of coupler,and the midpoint strain of the coupler on the steady state condition.Tables 2 to 5 show the error comparisons of the first natural frequency,the total energy,the midpoint deflecti

63、on of the coupler,and the midpoint strain of the coupler by the stress-and the displacement-based finite element methods.The error calculation is based on Eq. (16).The results show that the errors from the stress-based f

64、i</p><p>  stress-based finite element method is much smaller than the errors from the displacement-based finite element method.Also,we notice that except for the errors of the first natural frequency,the er

65、rors from the stress-based finite element method are smaller than the errors from the displacement-based finite element method under the same number of elements.It illustrates that the stress-based finite element method

66、can provide much accurate approximated solutions for kineto-elasto-dynamic problems.</p><p>  VII.Conclusions</p><p>  This paper proposed a new approach to implement the stress-based finite ele

67、ment method to Euler-Bernoulli beam problems.Especially,this approach can be applied to kineto-elasto-dynamic problems.The proposed approach is to approximate the curvature of a beam. Then,we can obtain the transverse de

68、flection and the stress distribution by integrating the approximate curvature distribution.During the integration procedure, it is necessary to make the boundary conditions of a beam element satisfied,whic</p><

69、;p>  References</p><p>  [1]B.Fraeijs de Veubeke,“Displacement and equilibrium models in the finite element method”,Stress Analysis,edited by O.C.Zienkiewicz,Wiley,New York,1965.</p><p>  [2]

70、B.Fraeijs de Veubekd and O.C.Zienkiewicz,“Strain-energy bounds in finite-element analysis by slab analogy”,J.Strain Analysis,Vol.2,pp.265-271,1967.</p><p>  [3]Z.Wieckowski,S.K.Youn,and B.S.Moon,“Stressed-ba

71、sed finite element analysis of plane plasticity problems”,Int.J.Numer.Meth.Engng.,Vol.44,pp.1505-1525,1999.</p><p>  [4] H.Chanda and K.K.Tamma,“Developments encompassing stress based finite element formulat

72、ions for materially nonlinear static dynamic problems”,Comp.Struct.,Vol.59,</p><p>  No.3,pp.583-592,1996.</p><p>  [5]M.Kaminski,“Stochastic second-order perturbation approach to the stress-bas

73、ed finite element method”,Int.J.Solids and Struct.,Vol.38,No.21,pp.3831-3852,2001.[6]O.C.Zienkiewicz and R.L.Taylor,The Finite Element Method,McGraw-Hill,London,2000.</p><p>  [7]R.H.Gallagher,Finite Element

74、 Fundamentals,Prentice-Hall,Englewood Cliffs,1975.</p><p>  [8]W.L.Cleghorn,1980,Analysis and design of high-speed</p><p>  flexible mechanism,Ph.D.Thesis,University of Toronto.</p><p

75、>  [9]W.L.Cleghorn,R.G.Fenton,and B.Tabarrok,1981,“Finite element analysis of high-speed flexible mechanisms”,Mechanism and Machine Theory,16(4),407-424.</p><p>  [10]W.L.Cleghorn,R.G.Fenton,and B.Tabarro

76、k,1984,“Steady-state vibrational response of high-speed flexible mechanisms”,Mechanism and Machine Theory,19(4/5)</p><p>  [11]Y.L.Kuo,W.L.Cleghorn and K.Behdinan,“Stress-based Finite Element Method for Eule

77、r-Bernoulli Beams”,Transactions of the Canadian Society for Mechanical Engineering,Vol.30(1),pp.1-6,2006.</p><p>  [12]Y.L.Kuo,W.L.Cleghorn,and K.Behdinan“Applications of Stress-based Finite Element Method o

78、n Euler-Bernoulli Beams”,Proceedings of the 20th Canadian Congress of</p><p>  Applied Mechanics,Montreal,Quebec,Canada,May 30-Jun2,2005.</p><p>  [13]Y.L.Kuo,Applications of the h-,p-,and r-ref

79、inements of the Finite Element Method on Elasto-dynamic Problems,Ph.D.Thesis,University of Toronto,2005.</p><p>  [14]L.Meirovitch,1967,Analytical Methods in Vibrations</p><p>  Macmillan,New Yo

80、rk,436-463.</p><p>  [15]K.J.Bathe,1996,Finite Element Procedures,Prentice Hall Englewood Cliffs,NJ,USA.</p><p>  [16]A.L.Schwab and J.P.Meijaard,2002,“Small vibrations superimposed on prescribe

81、d rigid body motion”,Multibody System Dynamics,8,29-49.</p><p>  [17]Y.L.Kuo and W.L.Cleghorn,“The h-p-r-refinement FiniteElement Analysis of a High-speed Flexible Slider Crank Mechanism”,Journal of Sound an

82、d Vibration,in press.</p><p><b>  英文翻譯</b></p><p>  應(yīng)力為基礎(chǔ)的有限元方法應(yīng)用于靈活的曲柄滑塊機(jī)構(gòu)</p><p> ?。ǘ鄠惗啻髮W(xué):Y.L. Kuo .L. Cleghorn加拿大)</p><p>  摘要:本文在歐拉一伯努利梁基礎(chǔ)上提出了一種新的適用于以應(yīng)

83、力為基礎(chǔ)的有限元方法的程序。先選擇一個(gè)近似彎曲應(yīng)力的分布,然后通過(guò)一體化確定近似橫位移。該方法適用于解決靈活滑塊曲柄機(jī)構(gòu)問(wèn)題,制定的依據(jù)是歐拉-拉格朗日方程,而拉格朗日包括與動(dòng)能,應(yīng)變能有關(guān)的組件,并通過(guò)彈性橫向撓度構(gòu)成的軸向負(fù)荷的鏈接來(lái)工作。梁元模型以翻轉(zhuǎn)運(yùn)動(dòng)為基礎(chǔ),結(jié)果表明以應(yīng)力和位移為基礎(chǔ)的有限元方法。</p><p>  關(guān)鍵詞:應(yīng)力為基礎(chǔ)的有限元方法,曲柄滑塊機(jī)構(gòu),拉格-朗日方程</p>

84、<p><b>  1.前言</b></p><p>  以位移為基礎(chǔ)的有限元方法通過(guò)實(shí)行假定位移補(bǔ)充能量。這種方法可能由內(nèi)部因素產(chǎn)生不連續(xù)應(yīng)力場(chǎng),同時(shí)由于采用了低階元素,邊界條件與壓力不能得到滿足。因此,另一種被成為以應(yīng)力為基礎(chǔ)采用假定應(yīng)力的有限元方法得到了應(yīng)用和發(fā)展。Veubeke和Zienkiewicz[1-2]首先對(duì)應(yīng)力有限元素進(jìn)行了研究。之后,這種方法被廣泛用于解決應(yīng)用程

85、序中的問(wèn)題[3-5]。此外,還有各種書(shū)籍提供更加詳細(xì)的方法[6.7]。</p><p>  這一高速運(yùn)作機(jī)制采用振動(dòng),聲輻射,協(xié)同聯(lián)結(jié),和撓度彈性鏈接的準(zhǔn)確定位。因此,有必要分析靈活的彈塑性動(dòng)力學(xué)這一類的問(wèn)題,而不是分析剛體動(dòng)力學(xué)。 靈活的機(jī)制是一個(gè)由無(wú)限多個(gè)自由度組成的連續(xù)動(dòng)力學(xué)系統(tǒng),其運(yùn)動(dòng)方程是由非線性偏微分方程建立的模型,但得不到分析解決方案。Cleghorn et al[8-10] 闡述了橫向振動(dòng)上的軸向

86、荷載對(duì)靈活四桿機(jī)構(gòu)的影響。并且通過(guò)能有效預(yù)測(cè)橫向振動(dòng)和彎曲應(yīng)力的五次多項(xiàng)式建立了一個(gè)翻轉(zhuǎn)梁?jiǎn)卧?lt;/p><p>  本文提出了一種新的方法來(lái)執(zhí)行建立在歐拉一伯努利基礎(chǔ)上的以應(yīng)力為基礎(chǔ)的有限元方法。改進(jìn)后的方法首先選定了假定應(yīng)力函數(shù)。然后通過(guò)整合假定應(yīng)力函數(shù)得到橫向位移函數(shù)。當(dāng)然,這種方法能解決沒(méi)有強(qiáng)制制約因素的應(yīng)力集中問(wèn)題。我們可以通過(guò)這種方法解決靈活曲柄滑塊機(jī)構(gòu)體系中存在的問(wèn)題。目的是通過(guò)這種方法提高準(zhǔn)確性,

87、該系統(tǒng)存在的問(wèn)題也可以通過(guò)取代基有限元方法來(lái)解決。結(jié)果可以證明偏差比較。</p><p>  2.以應(yīng)力為基礎(chǔ)的歐拉一伯努利梁</p><p>  歐拉一伯努利梁的彎曲應(yīng)力與橫向位移的二階導(dǎo)數(shù)相關(guān),也就是曲率,可以近似的看做是形函數(shù)和交點(diǎn)變量:</p><p>  這里[(i)N(c)]是連續(xù)載體的形函數(shù);{(i)Øe} 是列向量的交點(diǎn)函數(shù),y是關(guān)于中性線

88、的橫向定位,E是楊氏模量,(i)v是橫向位移,x軸向定位函數(shù)。</p><p>  由方程(1)可以推導(dǎo)出橫向位移轉(zhuǎn)換方程:</p><p><b>  橫向位移:</b></p><p>  這里 (i)C1和(i)C2是兩個(gè)一體化常數(shù),可以通過(guò)滿足兼容性來(lái)確定。</p><p>  將方程(2)和(3)代入(1),可

89、以得到有限元位移和回轉(zhuǎn)曲率,如下所示:</p><p>  這里下標(biāo)(C),(R)和(D)分別代表曲率,自轉(zhuǎn)和位移。運(yùn)用變分原理,可以得到這些方程[11-13]。</p><p>  表1 分別比較以位移和應(yīng)力為基礎(chǔ)的有限元方法的歐拉-伯努利梁元素</p><p>  3.以位移和應(yīng)力為基礎(chǔ)的有限元方法的比較</p><p>  主要區(qū)別在于

90、以位移為基礎(chǔ)的有限元方法的應(yīng)力場(chǎng)存在不連續(xù)的內(nèi)部因素,同時(shí)具有低階形函數(shù)。主要是因?yàn)椴贿B續(xù)量的產(chǎn)生以及間離散分布。再者,它可能由于使用過(guò)多交點(diǎn)變量而產(chǎn)生剛度矩陣。</p><p>  以應(yīng)力為基礎(chǔ)的方法與以位移為基礎(chǔ)的方法比較具有很多優(yōu)點(diǎn)。首先,以應(yīng)力為基礎(chǔ)的方法產(chǎn)生的交點(diǎn)變量較少(如表1)。第二,使用以應(yīng)力為基礎(chǔ)的方法時(shí),彎曲應(yīng)力的邊界條件可以得到滿足。最后,應(yīng)力由體系方程直接計(jì)算得到。</p>

91、<p><b>  4.方程推導(dǎo)</b></p><p>  曲柄滑塊機(jī)構(gòu)如圖1所示,由做剛體運(yùn)動(dòng)的曲柄來(lái)運(yùn)作,該方程由有限元公式推導(dǎo)而得。有限元方程的推導(dǎo)過(guò)程如下:(1)建立剛體運(yùn)動(dòng)學(xué)曲柄滑塊機(jī)構(gòu);(2)構(gòu)建基于剛體運(yùn)動(dòng)學(xué)機(jī)構(gòu)的翻轉(zhuǎn)梁?jiǎn)卧唬?)確定一套變量用來(lái)描述靈活曲柄滑塊機(jī)構(gòu)的運(yùn)動(dòng);(4)裝配所有梁?jiǎn)卧W詈?,就可以得到有限元方程,同時(shí)該靈活曲柄滑塊機(jī)構(gòu)的時(shí)間響應(yīng)可以通過(guò)時(shí)

92、間一體化確定。</p><p>  圖1 靈活曲柄滑塊機(jī)構(gòu)</p><p><b>  A.翻轉(zhuǎn)梁的元方程</b></p><p>  考慮靈活的梁?jiǎn)卧艿絼傮w翻轉(zhuǎn)和回轉(zhuǎn)運(yùn)動(dòng)。疊加在剛體運(yùn)動(dòng)軌跡時(shí),縱向和橫向方向上允許一些撓度變量。通過(guò)拉格-朗日方程可以得到任意靈活翻轉(zhuǎn)的組件的微分方程。由于彈性變形認(rèn)為是很小的,而且自由度是有限的,這個(gè)方程是線

93、性的并且很容易畫(huà)出來(lái)。推導(dǎo)公式的元素也被很明確的列出來(lái)[8-10],并且做了簡(jiǎn)要的介紹。</p><p>  鑒于在軸向有很強(qiáng)的剛度,因此很有必要在縱向方向上合理考慮為剛性梁。所以,縱向方向如一下所示:</p><p>  (5)這里u1是交點(diǎn)變量,是關(guān)于x軸方向的常數(shù),如圖2所示。橫向可以表示為:</p><p>  翻轉(zhuǎn)梁?jiǎn)卧先我恻c(diǎn)的速度可以表示如下:<

94、;/p><p>  這里((i)Vax(i)Vay)是梁?jiǎn)卧贠點(diǎn)的絕對(duì)速度,如圖2所示; 是梁?jiǎn)卧慕撬俣?;((i)u(i)v)分別是梁?jiǎn)卧先我恻c(diǎn)縱向和橫向的位移,x是梁?jiǎn)卧v向的定位,如圖2所示。</p><p><b>  圖2 旋轉(zhuǎn)梁</b></p><p>  如果我們把 當(dāng)作組件材料的單位體積;A是組件的橫截面積,L是組件的長(zhǎng)度,組件

95、的動(dòng)能可以表示如下:</p><p>  均勻剛性組件的軸向彎曲應(yīng)變能量與楊氏模量E有關(guān),得到二階矩陣I,如下所示:</p><p>  由縱向拉伸負(fù)荷工作,(i)P,組件的橫向撓度表示如下:</p><p>  運(yùn)功機(jī)制的縱向負(fù)荷不是一成不變的,與位置和時(shí)間有關(guān)。在忽略縱向彈性形變的前提下,縱向負(fù)荷可能來(lái)自于剛性慣性力,可以表示如下:</p><

96、;p>  這里PR是元件右側(cè)的外部縱向負(fù)載, 是x軸方向上O點(diǎn)的絕對(duì)加速度。如圖2所示。 拉格-朗日形式表示如下:</p><p>  將公式(5-100)代入(12),并且運(yùn)用歐拉-拉格朗日方程,旋轉(zhuǎn)梁的運(yùn)動(dòng)方程可以表示為一下形式: </p><p>  這里[Me]、[Ce]和[Ke]分別是元件的質(zhì)量、等效阻尼和等效剛度矩陣;{Fe}是元件的載荷向量。當(dāng)建立質(zhì)量耦合矩陣時(shí),應(yīng)

97、主要考慮滑塊機(jī)構(gòu)。</p><p>  B.曲柄滑塊機(jī)構(gòu)方程</p><p>  提出解決曲柄滑塊機(jī)構(gòu)問(wèn)題的方法,變量是曲率的節(jié)點(diǎn)。裝配所有元件時(shí),考慮機(jī)構(gòu)的邊界條件是很有必要的。因?yàn)樵搫?dòng)力適用于基礎(chǔ)曲柄結(jié)構(gòu),在O點(diǎn)存在彎矩,如圖1所示,在O點(diǎn)也存在曲率。如圖1所示的A點(diǎn)和B點(diǎn),我們假定它們是很小的點(diǎn)。然而,實(shí)際上,彎矩和曲率在這兩個(gè)點(diǎn)上都為零。</p><p> 

98、 因?yàn)楣剑?3)是變量的矩陣表示方式{Ø} ,這個(gè)公式可以通過(guò)總結(jié)所有的方程來(lái)得到,可以表示如下:</p><p>  這里[M]、[C]、[K]分別是質(zhì)量、阻尼和剛度矩陣,{F}是負(fù)載向量。</p><p>  5.穩(wěn)定狀態(tài)基礎(chǔ)上的數(shù)值模擬</p><p>  曲柄的轉(zhuǎn)速是150rad/s (1432rpm),該靈活曲柄滑塊機(jī)構(gòu)的各項(xiàng)數(shù)值表示如下:R2

99、=0.15(m),R3=0.30(m), =0.225(kg/m), EI=12.72(N-m2), mB=0.03375(kg)。</p><p>  這里R2 和R3分別是曲柄和耦合器的長(zhǎng)度,mB是滑塊的質(zhì)量。</p><p>  通過(guò)曲柄和耦合器的一個(gè)運(yùn)動(dòng)周期,可以看出穩(wěn)態(tài)橫向位移和中點(diǎn)彎曲應(yīng)力的變化情況,以及分析本課題的結(jié)果??梢酝ㄟ^(guò)增加物理阻尼矩陣提高穩(wěn)定性,被稱作瑞利阻尼:&l

100、t;/p><p>  這里α和β是兩個(gè)常數(shù),可以從[15]中對(duì)應(yīng)于兩個(gè)不同頻率的振動(dòng)的阻尼比得到。本文中α和β的值取決于自然頻率。</p><p>  通過(guò)在運(yùn)動(dòng)方程中增加物理阻尼,也可以通過(guò)Newmark時(shí)間步驟觀測(cè)超過(guò)20個(gè)周期的運(yùn)動(dòng),從而得到分析結(jié)果。當(dāng)采用數(shù)值時(shí)間積分是出示條件從零開(kāi)始。</p><p><b>  誤差可以表示為:</b>

101、</p><p>  這里QFEQRef 和分別表示以有限元方法和參考方法為基礎(chǔ)的兩個(gè)值,總的來(lái)說(shuō),可以建立時(shí)間方程,而且很容易被接受,比如能量、位移、彎矩等等。t1 和t2指的是時(shí)間積分的間隔,通常指的是穩(wěn)態(tài)條件下的以個(gè)周期。因?yàn)闆](méi)有一個(gè)合適的準(zhǔn)確的方法,在本文中可以通過(guò)一個(gè)五次多項(xiàng)式表示20個(gè)元件鏈接為基礎(chǔ)的位移有限元方法得到參考值。</p><p>  Fig. 3. Time re

溫馨提示

  • 1. 本站所有資源如無(wú)特殊說(shuō)明,都需要本地電腦安裝OFFICE2007和PDF閱讀器。圖紙軟件為CAD,CAXA,PROE,UG,SolidWorks等.壓縮文件請(qǐng)下載最新的WinRAR軟件解壓。
  • 2. 本站的文檔不包含任何第三方提供的附件圖紙等,如果需要附件,請(qǐng)聯(lián)系上傳者。文件的所有權(quán)益歸上傳用戶所有。
  • 3. 本站RAR壓縮包中若帶圖紙,網(wǎng)頁(yè)內(nèi)容里面會(huì)有圖紙預(yù)覽,若沒(méi)有圖紙預(yù)覽就沒(méi)有圖紙。
  • 4. 未經(jīng)權(quán)益所有人同意不得將文件中的內(nèi)容挪作商業(yè)或盈利用途。
  • 5. 眾賞文庫(kù)僅提供信息存儲(chǔ)空間,僅對(duì)用戶上傳內(nèi)容的表現(xiàn)方式做保護(hù)處理,對(duì)用戶上傳分享的文檔內(nèi)容本身不做任何修改或編輯,并不能對(duì)任何下載內(nèi)容負(fù)責(zé)。
  • 6. 下載文件中如有侵權(quán)或不適當(dāng)內(nèi)容,請(qǐng)與我們聯(lián)系,我們立即糾正。
  • 7. 本站不保證下載資源的準(zhǔn)確性、安全性和完整性, 同時(shí)也不承擔(dān)用戶因使用這些下載資源對(duì)自己和他人造成任何形式的傷害或損失。

最新文檔

評(píng)論

0/150

提交評(píng)論