Tekil yük altında ray eğilme momenti travers eğilme momenti ilişkisi
The relation between rail bending moment and sleepers bending moment
- Tez No: 66458
- Danışmanlar: PROF. DR. İNAL SEÇKİN
- Tez Türü: Yüksek Lisans
- Konular: İnşaat Mühendisliği, Civil Engineering
- Anahtar Kelimeler: Belirtilmemiş.
- Yıl: 1997
- Dil: Türkçe
- Üniversite: İstanbul Teknik Üniversitesi
- Enstitü: Fen Bilimleri Enstitüsü
- Ana Bilim Dalı: İnşaat Mühendisliği Ana Bilim Dalı
- Bilim Dalı: Belirtilmemiş.
- Sayfa Sayısı: 126
Özet
ÖZET Bu çalışmada, Winkler'in zemin itkisinin o noktadaki çökmeye Oranlı olduğu hipotezine dayanarak, Travers eğilme momenti, ray eğilme momenti karşılaştırılmasına gidilmiştir. Raylar sonsuz kiriş, traversler ise sonlu kiriş şeklinde gruplandırılarak Winkler hipotezi uygulanmıştır. Raylarda, sonsuz uzakta çökmenin sıfır olduğu, yük altında dönmenin sıfır olduğu, sağ taraf için kesme kuvvetinin etkiyen yükünün yarısı olduğu kabullerine dayanarak eğilme momenti denklemi elde edilmiştir. Traverslerde eğilme momentinin bulunmasında başlangıç değerleri yöntemi, süperpozisyon yöntemi, tesir çizgileri ile çözüm, Zimmermann metodu kullanılır. Başlangıç değerleri ve tesir çizgileri ile çözümde sol uç başlangıç noktası alınır ve bu noktadaki çökme ve dönme tesbit edilerek denklemler oluşturulur. Tesir çizgisinden çözümde tablodan hesap yapılır. Süperpozisyon metodunda sonlu kirişin tesbitinde sonsuz kiriş hesapları süperpoze edilerek sonuca ulaşılır. Çalışmada travers eğilme momenti için Zimmermann metodu seçilmiştir. Raylarla traverslerin eğilme momentleri yatak katsayısı, travers uzunluğu, ray cinsine bağlı olarak karşılaştırılmış ve Oran katsayıları elde edilmiştir.
Özet (Çeviri)
SUMMARY The condition of sleepers and rail is observed as the beams on elastics foundation. In the major part of this work the analysis of bending of beams on an elastic foundation is developed on the assumption that the reaction forces of the foundation are proportional at every point to the deflection of the beam at that point. This assumption was introduced first by E. Winkler in 1 867 and formed the basis of Zimmermann's classical work on the analysis of the railroad track, published in 1888. While the theory of beams on elastic foundation holds rigidly for most of the problems mentioned above, its application to soil foundations should be regarded only as a practical approximation. The physical properties of soils are obviously of a much more complicated nature than that which could be accurately represented by such a simple mathematical relationship as the one assumed by Winkler. There are, however some important points which can be brought up in supporting the application of this theory to soil foundations. First, under certain conditions the elasticity of soil is undeniable. Second, the foundations deforms only along the portion directly under loading, If we take these things into consideration, there is reason to believe that the Winkler theory, in spite of its simplicity, may often more accurately represent the actual conditions existing in soil foundations than do some of the more complicated analysis advanced. Consider a straight beam supported along its entire length by an elastic medium and subjected to vertical forces acting in the principal plane of the symmetrical cross section. Because of this action the beam will deflect, producing continuously distributed reaction forces in the supporting medium. Regarding these reaction forces we make the fundamental assumption that their intensity p at any point is proportional to the deflection of the beam yat that point p = k.y XIIIThis assumption implies the statement that the supporting medium is elastic; in other words, that its material follows Hooke's law. Its elasticity can be characterized by the force which distributed over a unit area, will cause a deflection equal to unity. This constant of the supporting medium, C kg/cm3, is called the modulus of the foundation. Assume that the beam has a uniform cross section and that b is its constant width, which is supported on the foundation. A unit deflection of this beam will cause reaction b.C in the foundation; consequently, at a point where the deflection is y the intensity of distributed reaction will be P=b.C.y. (1) Let us take an infinitely small element enclosed between two vertical cross sections a distance dx apart on the beam under consideration (Figl) T. q,dx /f* mm M t pdx 7 M +rJM 1 T*dT Figl. An infinitely small element on the beam Considering the equilibrium of the element in Figure 2, we find that the summation of the vertical forces gives T - (T+dT) + b.C.y.dx - q.dx = 0 (2) From this equation and the others below T=dM/dx, E.I.(d2y/dx2)= - M (3) XIVand the knowledge that along the unloaded parts of the beam where no distributed load is acting, q=0, the equation will take the form d4y E.I.-4 = -b.C.y (4) dx With the notation 4 E I x the general solution y of equation (4) maybe expressed in a convenient form y = e^.(kı-SİnÇ +k3.cosi; ) + e“^.(k2.sin£ + Iq.cosC ) (6) Here L includes the flexural rigidity medium, and is an important factor influencing the shape of the elasticline. For this reason the term L is frequently referred to as the characteristic length and the factor w=l/L is called the characteristic of the system and its dimension is length”1. Consequently, £ = x/L will be an absolute number. Expression 6 represents the general solution for the deflection line, but with no q loading. An additional term is necessary where is a distributed load. By differentiation and using formula 3 we get. M= -E.I. -y. (e^.( k,.sinÇ +k3.cosÇ ) + e“^.(k2.sinÇ + L,.cos£ )) (7) Usually the rails are behaving as infinite beams an elastic foundation. In the present problem., dealing with a beam of unlimited length, it is reasonable to assume that in an infinite from the application of the load the deflection of the beam must approach zero, that is, if x-> oo then y-»0. This condition can be fulfilled only if in xvthe equation above the terms connected with e^ vanish, so ki and k3 will become zero. So equation 6 will take the form y=e”^.(k2.cos£ + k4.sin{;) (8) From the condition of symmetry we know that dx (9) x=0 that is, -(k2-k4)=0, from which we find k2=k4=k. This last constant of the equation. y= k.e“^. (cos£, + sini; ) (10) can be obtained from the consideration that the sum of the reaction forces will equilibrium with the load P, that is to 2J”b.C.y.dx = P (11) 0 CO Since2.b.C.k.Je“ç.(cos^ + sin4).dx = 2,.b.C.k.L from 2,b.C.k.(L) = P 0 we get k=P/(2.b.C.L) and, substituting this in above we have y = 2.b.C.L,e_ç.(cos^ + sin^) (12) Taking the successive derivatives of y with respect to x, we obtain the expression for M on the right side of the beam P.L M= ~r~-(e ç.(cosi;-sinÇ)) (13) XVIIn rails inorder to solve bending moment problem, we use the equation (13) We observe sleepers as we come across with a beam of finite length. For a beam of finite length the correct solution is the one which besides fullfilling the differential equation of the elastic line, also satisfies the required conditions at both ends of the beam. Today, we have four method in observing sleeper bending. First one is the zimmermanns' method, second one superposition, third one the method of initial conditions, fourth one is the method of solving by the help of influence lines. In Zimmermann's bending moment equation Mt=-L,[K] where P is concentrated load, Lt is flexural rigidity and [K] = -.(u, + u,)-vp.ch£,cosÇ + up.sh!;.sinÇ -> x 1 1./2 ? lt/2 Fig 2. Zimmerman's approximation to sleepers Here |.iı=e~Çl.(Cos^-sin^) u.,= e^2.(CosÇ2 -sin£2) (24a-d) up= u.chp.cosp + v.shp.sinp xvnVp^ v.chp.cosp + w.shp.sinp lt/2/L=X, r/lt=p.x/lt=£ x,/lt=^, x2/lt=£2 - 2 + cos2A,-sin2A. + e-:* u = sin2X. + sin2A, - cos2X. + sin2^-e 2X v = sin2A. + sin2X - 2-cos2X, + sin2A.-e w= - -2k sin2A, + sin2A, By the help of this formula for any section the bending moment of a sleeper can be computed. In railway, sleepers have two critical places. Firstone a negative moment in the middle (x=0) and the other under the load (x=r). Consider an infinitely long beam subjected to a given loading, The aim will be to obtain from it a solution for the beam of finite length which is under the same loading and has free ends at A and B. The method will be to make the bending moments and shearing forces vanish at points A and B on the infinite beam. For this reason fictive P0, M0 forces are used to obtain this condition. If S=7/L, M the bending moment in infinitely long beam on section A and B and T is the shearing force in the same beam and section. Po= 4.E,.[T.(1+DS)+-.(l-ris)] = 0 where 2.M Mo = -2.Lt.Ej.[T.(l+us) + - -. (1-D.)] = 0 1 es Ei = n~^ZZ ^7. n, = e”s.(cosS + SinS) 2sınhS + sınS u.=e“s(cosS-sinS), D,=e”s.cosS. xvuiMA P Ta I P MB Jn (a) P P I i B (b) / Fİg3. Getting a finite beam from an infinite one In order to calculate the bending moment for a given section, the.given loading and fictive'Po, M“ loading are used, The calculation by these loads are made as if there is a infinitey long beam. But the sum of the results of these calculations will give the value for the finite beam under the given loading. Consider a finitely long beam subjected to symmetrical two concentrated loads. Because of the free ends, bending moment and shearing force at these ends will be zero. If the left-end is considered to be the origin, Yo will be downward deflection 80 angular deflection, Mo bending moment, To shearing force at tihs point. Bending moment and shearing force on free ends in zero. There remains two unknown quantities; y0 and 9o. From the left end to the right, when ever a concentrated load is passed, a new term is added to the section force formulas. Ul + u2 It Fig 4. initial condition application in sleepers. Bending moment and shearing feree formulas at the right end are (S=lt/Lt) xixM,r=b.C.Lt2-yo. F3(s)+b.C.L,3e0. F4(s)-Lt.P. (F2 (S - ^-) + F2 (S - ^)) = 0 T”=b.C.Lt.y". F2(s)+b.C.L,2.e0.F3(s)-P (F, (S - -1-) + F, (S - -1-)) = 0 From these two equations two unknowns can be found by the help of the condition M\t = T/t = 0 Bending moment equation between left-endand first concentrated load is £=x/L.) Mx = b.C.L,2.yo.F3(0 + b.C.Lt2.9o.F4(Ç) Bending moment equation between two concentrated loads Mx = b.C.L,2.y0.F3(Ç) + b.C.Ll3.0o.F4(^)-L,.P. F2 Ç - ^~ Bending moment equation between second concentrated load and the right end. Mx = b.C.L»2.y0.F3(O + b.C.Ll3.9o.F4(^)-Lt.P. Fı(£)=coshÇ.cosŞ, F2(4 )=0,5(cosh£.sinÇ +sinh£ xos£ ) f O + F. r u v\ \ L.J \ LJJ F3(Ç)=0,5.sinhÇ.sinÇ F4=0,25(cosh£.sin£, -sinhJ;.cosÇ ) So, bending moment of a given section can be calculated. xxIn calculation with influence lines, there are tables written for different relative stiffness values, The rows of these tables gives bending moment variation along the beam and the columns give influence lines. Taking the left end as the original point, the distance between the load and the origin divided by the whole length gives all changing for different rows. The same rule is applied to the given sections over whole length changing for different columns. An operator should calculate the distance between load and origin and the distance between the given section and the origin. After dividing these terms by the whole length, operator should choose correct rows and collumns respectively, and multiply them with P.L inorder to find the bending moment value for the given section. If the variation of bending moments along the sleeper is calculated, a diagram below will occur. t e t Fig 5. Variation of Bending Moments of a sleeper. Here, There will be two critical points, one in the midde the other under the rail. So the maximum bending moment can change for different reasons Zimmermann's method is preferred to other methods, in comparing sleeper and rail bending moment. For this reason, Pr.Lr Pt r, Mr=- -, Mt=-.Lt.[K] 4 2 As it is known that Pt = 0,6 Pr, the equation will become, Mt Lt,, - XXIIn this study the q tables is given for different types of modulus of foundation, sleeper length, different sleeper materials, group of railr and the place of sleeper bending moment (in the midde, or under rail) q is called as ratio coefficient. xxn
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