Demiryolu üstyapısı hesaplarının sonlu elemanlar yöntemiyle analizi
Başlık çevirisi mevcut değil.
- Tez No: 75185
- Danışmanlar: PROF. DR. İNAL SEÇKİN
- Tez Türü: Yüksek Lisans
- Konular: İnşaat Mühendisliği, Civil Engineering
- Anahtar Kelimeler: Belirtilmemiş.
- Yıl: 1998
- Dil: Türkçe
- Üniversite: İstanbul Teknik Üniversitesi
- Enstitü: Fen Bilimleri Enstitüsü
- Ana Bilim Dalı: İnşaat Mühendisliği Ana Bilim Dalı
- Bilim Dalı: Ulaştırma Bilim Dalı
- Sayfa Sayısı: 123
Özet
Demiryolu üstyapısı hesaplan klasik yöntemde elastik yatağa oturan kiriş olarak çözülebilmektedir. Demiryolu üstyapısı hesaplan sonlu elemanlar yöntemiyle de modellenebilir ve çözülebilir. Bu çalışmada, balast tabakası üzerine oturan ray ve traverslerden oluşan yapı, elastik yatağa oturan kiriş olarak sonlu elemanlar yöntemiyle modellendirilerek çözülmüştür. Sonlu elemanlar yöntemiyle modellenen Winkler elastik yatak modeli klasik yöntem ile de çözülerek karşılaştırmıştır. Sonlu elemanlar yöntemiyle boyuna ve enine doğrultuda bulunan tesirler klasik yöntemle yaklaşık aynı sonuçları vermiştir. Sonlu elemanlarla kurulan modelde balast tabakası yay ve düzlem eleman olarak modellendirilmiştir. Balast tabakasının yay olarak çözümünde yatak katsayısı, düzlem eleman olarak çözümünde ise elastisite modülü kullanılmıştır. Balast tabakası altında sonlu elemanlar yöntemiyle bulunan gerilmeler, Boussinesq yöntemiyle de çözülerek karşılaştırmıştır. Bunun yanında balast tabakası, travers ve raylardan oluşan üstyapı, sonlu elemanlar yönteminde iki boyutlu ızgara kiriş olarak modellenip çözülmüştür. Sonlu elemanlarla yapılan analizlerde SAP90 paket programı kullanılmıştır.
Özet (Çeviri)
It is possible to solve displacements of beams which cause forces on elastic foundation with classic methods. Models which are formed with finite element method are more realistic than classic method due to advantage of finite element methods. Rail, sleeper and ballast layer which form truck structure can be considered and can be determined all deformations which each other's having separately. A powerful program and sensible using are necessary to solve problem realistically. While a beam which have a width of b restraints on a Winkler elastic foundation, bending curve differential equation and moment differential equation depend on each other. From both of equations depending on lateral earth pressure as, the forth degree differential equation is equal displacement of y. Figure 1 Beam on elastic foundation. A beam which is on Winkler foundation is presented in Figure 1. While acting of single force on a elastic beam, forth root of constant term of equation is known as elastic length L. L = i I4EI cb (1) If the beam's length is 1 and 1/L > 2n, the beam is accepted as long beam. Table 1 gives the closed-form solution of the basic differential equations for several loading utilising the Winkler concept. Sometimes one structure is supported by another, but stress analysis is required for only the first of the two. Then it suffices to model the effect of the second structure on the first. We need not model the second structure in such detail that stresses within it can be determined. Examples include a rail on a roadbed or pavement slab on soil. The rail or the slab must be analysed; the supporting effect of the roadbed or soil must be modelled. Elastic support can be represented by a foundation stiffness matrix, fks] for a foundation element and [Ksl^Z[ks] for the entire foundationstructure. If [k] is the stiffness matrix of the supported structure, then [k]+[Ks] is the net stiffness matrix of the supported structure on its elastic foundation. Table 1 Closed-Form Solutions Of Infinite Beam On Elastic Foundation The A, B, C, D and k coefficients are * = i cb_ 4EI (2) A = e~AX(Costoc + SinXx) (3) B = e'^Sinkx C = e^(CosXx-SinXx) (4) (5) D = e_AXCoskx (6) Analysing of railway truck structure longitudinally, sleepers and rails are considered together as a beam on elastic foundation and used long beam's equations (Figure 2). EbnB Sleepers Rail uA Figure 2 Rail and sleepers on elastic foundation. Consider a uniform beam that deforms in the plane of the paper and undergoes no axial deformation. This element has four d.o.f: a lateral displacement co and rotation 0 at each end (Fig. 1). Nodal forces F; and Fj correspond to nodaldisplacements (O; and coj. Nodal moments M; and Mj correspond to nodal rotations 0; and 0j. Positive directions of these quantities are shown in Fig. 3. The element equation is expressed as: [k]{d} = {P} (7)) Where [k] is a 4 by 4 stiffness matrix for the present beam element, {d} is the element nodal displacement vector. Vector {P} contains nodal forces and moments applied to the element to maintain the deformation state {d}. The element stiffness equation is written out as: L21 k41 (8) We must express each stiffness coefficient in [k] in terms of element geometry and elastic modulus. To determine the stiffness coefficients in single column of [k], we set the corresponding d.o.f. to utility while keeping all other d.o.f. zero, and calculate the values of Fj, Mj, Fj and Mj needed in order to produce this deformation state. Thus, the first column of [k] is determined by activating only the first d.o.f Specifically, for the case Oi=l and 6i=©j=0j=O, we see kn=F;, k2i=Mj, k3i=Fj and k4i=Mj. As a result, stiffness matrix of a beam is written as: EI w-o 12 6L -12 6L 6L 4L2 -6L 2L2 -12 -6L 12 -6L 6L 2L2 -6L 4L2 (9) V^ Cûi K- (a) P # (b) -y «M Figure 3 (a) A uniform beam element and its four nodal d.o.f (b) Associated nodal forces and moments System's width which consists of sleepers and rail is convert the real width using below equation:b3b' (10) af A simple elastic beam's finite element model is used in railway truck structure. The beam which consists of rail-sleepers is resided on springs. Modules of subgrade reaction of elastic foundation is used in determining of spring coefficients as follows: K;=cb (L.+Lj) (11) The long beam which is on the springs displaces vertical direction about to acting from the nodal forces and moments. Springs which are under beam act a force about wk the opposite direction (Fig. 4). K; I Mid\ KjCûi (a) L (b) M;.A KjOJj Figure 4 (a) A simple elastic beam on springs, (b) Forces and moments acting on a simple elastic beam's nodes. If the springs reactions considered in Eqs. 8, then the characteristic matrix of a simple elastic beam on springs determine as below: (12) If the stress analysis is required throughout the ballast layer, it is necessary to model the ballast layer. The ballast layer is modelled as asolid (plane) element in finite element method. The layer can be divided into finite number of quadrangular or triangular elements. Modulus of subgrade reaction is not used in stress analysis. It is converted to stress-strain modules E, as below:c = (13) b(l-|i2) Width of beam, Poisson's ratio and modulus of subgrade reaction are expressed as b, H and ks in turn in order. In fig. 5, there is a plane rectangular bilinear element. The displacement function of the element in each corner is expressed as below: (J) = a, + a2x + a3y + a4xy (14) There are shape function as much as displacement parameters. These are expressed as below: (15) (16) (17) (18) 4ab The plane rectangular bilinear element displaces x and y directions. This four nodes element has eight d.o.f. This element is based on the bilinear displacement field: u = a, + a2x + a3y + a4xy v = a5 + a6x + a7y + a8xy In terms of nodal d.o.f, the displacement field: {U} = [N]{d} (19) (20) (21) The strain displacement matrix is [Bk8 = dldx 0 0 d/dy d/8y d/dx [N] (22)As a result, the element stiffness matrix depending on the strain-displacement matrix [B], stress-strain modules Es and thickness of element t is presented as: b a [k]Sx« = /J[B]ltx.,[E],x.,[B]3Xlttdxdy -b-a (23) y#v |«- a - »e - a - »| U4 U| v4 fV3 Vl JtL + ~TfC "2 *-*¦ V2 Figure 5 Plane rectangular bilinear element (eight d.o.f).
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