土木工程外文資料翻譯---timoshenko和剪切模型梁的動(dòng)力學(xué)研究（英文）

1、On the Comparison of Timoshenko and Shear Models in Beam DynamicsNoël Challamel1Abstract: The classical Timoshenko beam model and the shear beam model are often used to model shear building behavior both for stabili

2、ty or dynamic analysis. This technical note questions the theoretical relationship between both models for large values of bending to shear stiffness parameter. The simply supported beam is analytically studied for both

3、models. Asymptotic solutions are obtained for large values of bending to shear stiffness parameter. In the general case, it is proven that the shear beam model cannot be deduced from the Timoshenko model, by considering

4、large values of bending to shear stiffness parameter. This is only achieved for specific geometrical parameter in the present example. As a conclusion, the capability of the shear model to approximate Timoshenko model fo

5、r large values of bending to shear stiffness parameter is firmly dependent on the material and geometrical characteristics of the beam section and on the boundary conditions.DOI: 10.1061/?ASCE?0733-9399?2006?132:10?1141?

6、CE Database subject headings: Shear waves; Structural dynamics; Dynamic models; Beams; Comparative studies.IntroductionThe classical Timoshenko beam model ?Timoshenko 1921, 1922? and the shear beam model ?see for instanc

7、e Kausel 2002? are often used to model shear building behavior both for stability or dynamic analysis. This technical note questions the theoretical relationship between both models for large values of bending to shear s

8、tiffness parameter. Aristizabal-Ochoa ?2004? compares these two models and suggests a relationship between these two models by considering large values of a dimensionless parameter s2, a bending to shear stiffness parame

9、ter. This technical note shows on a simple example that this parameter may be not suffi- cient to link both theories.Equations of Motion—Timoshenko ModelThe governing equations of the Timoshenko model arem?2y?t2 ? ASG?2y

10、?x2 + ASG???x = 0?1?mr2?2??t2 ? ASG? ?y?x ? ?? ? EI?2??x2 = 0The beam is made of a homogeneous linear elastic material with moduli E ?Young modulus? and G ?transverse shear modulus?.Its transverse cross section is doubly

11、 symmetric with an effective shear area denoted by AS and a principal moment of inertia I=Ar2 ?r?radius of gyration of the cross section and A?total area?. The effective area AS can also be denoted by ?A. ?, the so-calle

12、d shear correction coefficient, is a dimensionless factor which gives the ratio of the average strain on a section to the shear strain at the centroid. Its value is dependent on the shape of cross section, but also on th

13、e material’s Poisson ratio ? ?see, for instance, Cowper 1966?. The uniform mass per unit of length is denoted by m; y?average displacement; and ??average slope, both function of the time t and the spatial coordinate x. T

14、he rotation angle ? can be evaluated from the first of these equilibrium relationships Eq. ?1? and substituted into the second, leaving the transverse displacement y as the only variableEI?4y?x4 + m?2y?t2 ??mr2 + m EIASG

15、? ?4y?x2 ? t2 + m2 r2ASG?4y?t4 = 0 ?2?One would obtain the same differential operator for the cross- sectional rotation angle ?Cheng 1970?EI?4??x4 + m?2??t2 ??mr2 + m EIASG? ?4??x2 ? t2 + m2 r2ASG?4??t4 = 0 ?3?The simply

16、 supported beam of length L is studied. The boundary conditions are then reduced to???x?0,t? = ???x?L,t? = y?0,t? = y?L,t? = 0 ?4?Of course, other types of boundary conditions can be treated ?Huang 1961, for instance? bu

17、t a comparison with other beam models will be facilitated by the available closed-form solutions of this simply supported beam. The solution of Eq. ?2? is sought of the formy?x,t? = ??x?sin ?nt ?5?1Associate Professor, L

18、aboratoire de Génie Civil et Génie Mécanique ?LGCGM?, INSA de Rennes, 20 av. des Buttes de Coësmes, 35043 Rennes cedex, France. E-mail: noel.challamel@insa-rennes.fr Note. Associate Editor: Hayder A.

19、Rasheed. Discussion open until March 1, 2007. Separate discussions must be submitted for individual papers. To extend the closing date by one month, a written request must be filed with the ASCE Managing Editor. The manu

20、script for this techni- cal note was submitted for review and possible publication on March 17, 2005; approved on January 25, 2006. This technical note is part of the Journal of Engineering Mechanics, Vol. 132, No. 10, O

21、ctober 1, 2006. ©ASCE, ISSN 0733-9399/2006/10-1141–1145/$25.00.JOURNAL OF ENGINEERING MECHANICS © ASCE / OCTOBER 2006 / 1141J. Eng. Mech. 2006.132:1141-1145.Downloaded from ascelibrary.org by University of Live

22、rpool on 04/19/15. Copyright ASCE. For personal use only; all rights reserved.The function of the lateral deflection ? is now expressed as follows:??x? = C1 cos??nx? + C2 sin??nx? + C3 cos??nx? + C4 sin??nx??17?Note the

23、difference in the eigenfunctions between Cases 1 and 2. The natural frequencies are simply obtained using the boundary conditions in Eq. ?8??n = n? and ?n = n? ?18?It can be remarked that this second case is often omitte

24、d in the literature.Asymptotic AnalysisEigenfunctions depend on the frequency-dependent Eq. ?12?. In any case, it is not difficult to show that in both cases, the natural frequency is given by the following equation:bn 4

25、R2s2 ? bn 2?n2?2?R2 + s2? + 1? + n4?4 = 0 ?19?The Euler–Bernoulli solution is found by settingR = s = 0 Þ bn = n2?2 ?20?It can be moreover stated thats2 ? 1 ?21?This assumption is often formulated in order to justif

26、y the as- sumption of the shear model ?see for instance Aristizabal-Ochoa 2004?. In this case, Eq. ?19? is reduced tobn 4R2s2 ? bn 2n2?2?R2 + s2??1 + ?? + n4?4 = 0where? = 1n2?2?R2 + s2? ?22?In case of Eq. ?21?, ??small

27、parameter ???1/?n?s?2?. The fol- lowing perturbation expansion for the solution of Eq. ?22? is obtained from the asymptotic sequence ?see for instance Bush 1992?bn 2 = bn 2?0? + ?bn 2?1? + ?2bn 2?2? + ¯ ?23?Substitu

28、ting the assumed expansion into Eq. ?22? yields the equationbn 2 =?n2?2R2 ?1 + ?R2 + s2s2 ? R2? + ¯orn2?2s2 ?1 + ?R2 + s2R2 ? s2? + ¯??24?The case R=s is a special case for which another perturbation expansion

29、has to be consideredbn 2 = bn 2?0? + ??bn 2?1? + ?bn 2?2? + ¯ ?25?The expansion leads in this case tobn 2 =?n2?2s2 ?1 + ?2?? + ¯orn2?2s2 ?1 ? ?2?? + ¯?26?However, the first term shows that the two roots ar

30、e merged for this specific case ?double frequency phenomenon—see Abbas and Thomas 1977 or more recently Geist and McLaughlin 1997?. By considering only the constant term in the asymptotic ex- pansion, the natural frequen

31、cies are finally expressed in both cases bybn =?n?s + ¯orn?R + ¯?27?Another asymptotic case can be studied for the more artificial caseR = 0; s ? 1 Þ bn 2 = n2?2s2 ?1 ? ?? + ¯ ?28?Equations of Motion—

32、Shear ModelOn the other hand, for the shear beam considered in the technical note ? ???x = 0?equation of motion is simply reduced to the classical uncoupled shear-wave equationm?2y?t2 ? ASG?2y?x2 = 0 with y?0,t? = y?L,t?

33、 = 0 ?29?Using Eq. ?5?, the following differential equation is obtained:ASG?? + m?n 2? = 0 ?30?The dimensionless parameters previously introduced are used againbn 2s2 = m?n 2L2GAS ?31?The frequency equation givesbn = n?s

34、 ?32?Comparison of Both ModelsEqs. ?27? and ?32? are not equivalent. These differences show that the shear model is not obtained asymptotically from the Timoshenko model for large values of s. These equations can also be

35、 summarized asJOURNAL OF ENGINEERING MECHANICS © ASCE / OCTOBER 2006 / 1143J. Eng. Mech. 2006.132:1141-1145.Downloaded from ascelibrary.org by University of Liverpool on 04/19/15. Copyright ASCE. For personal use on