Elastik zemin üzerine oturan kirişler
Başlık çevirisi mevcut değil.
- Tez No: 39200
- Danışmanlar: PROF.DR. HASAN ENGİN
- Tez Türü: Yüksek Lisans
- Konular: İnşaat Mühendisliği, Civil Engineering
- Anahtar Kelimeler: Belirtilmemiş.
- Yıl: 1993
- Dil: Türkçe
- Üniversite: İstanbul Teknik Üniversitesi
- Enstitü: Fen Bilimleri Enstitüsü
- Ana Bilim Dalı: Belirtilmemiş.
- Bilim Dalı: Belirtilmemiş.
- Sayfa Sayısı: 68
Özet
Bu çalışmada, elastik zemine oturan ağırlıksız kiriş lerin statik ve dinamik tekil yükler altındaki elastik davranışları incelenmiştir. Zeminin basınç ve çekmede farklı davranış gösterdiği kabul edilmiştir. Çalışmanın birinci bölümünde konu ile ilgili genel ta nımlar ve açıklamalar yapılmıştır. Problem ilgili çalış malar özetlenmiş, bu çalışmalar hakkında çok kısa açıkla malar yapılmış ve Winkler zeminine oturan, kesiti sabit olan bir kirişin yönetici denklemi çıkarılmıştır. ikinci bölümde, problem ilk olarak, zeminin çekme ve basınçta yatak katsayıları eşit alınarak, ortasından sta tik bir tekil yükle yüklenmiş sonlu ve sonsuz kiriş için çözüm yapılarak, elastik eğri ifadesi elde edilmiş tir. Daha sonra ayni problem harmonik tekil yük hali için ele alınmıştır, değişik yatak katsayısı oranları ve yüklemeler için kirişe ait yer değiştirmeye ait sayısal dedegerler elde edilmiştir. Yine zeminin çekme ve basınç ta yatak katsayıları eşit alınarak, ortasından dinamik tekil yükle yüklenmiş, sonsuz ve sonlu kiriş için çözüm yapılmıştır. Zeminin çekme ve basınçta farklı davranış göstermesi halinde zorlama frekansı değerlerine göre tes- bit edilmiş zemin parametrileri ve kiriş geometrisi için ortaya çıkan beş değişik durum göz önüne alınarak matema tiksel çözümler yapılmış, sayısal sonuçlar elde edilmiş tir. Sonuçlar bölümünde, yükleme frekansına, kiriş rijit- ligine, kiriş boyuna, yatak katsayıları oranına ve yükün şiddetine bağlı olarak, nümerik sonuçlara ait grafik ve tablolar verilmiştir.
Özet (Çeviri)
In this work, the behavior of weightless beam of fi nite/infinite length subjected to a static or dynamic concentrated load and supported by an elastic foundation is investigated. It is assumed that the foundation behaves differently in compression and tension. The problem of beams on elastic foundation has been examined firstly by E. Winkler in 1867. In“Winkler”foundation it is assumed that the stress is proportional to the deflection and the foundation has been composed of independent springs. The foundation is characterized by the foundation modulus which is generally denoted by k. In engineering applications, there are some important problems which can be handled successfully by means of Winkler hypothesis : the frames of ships, space vehicles grid systems at plates and bridges, continuous founda tions in one or two directions, rotationary shell and perpendicular piles under the effect of horizontal load. Mathematical solution of problems of beams on elastic foundation is highly tiresome and difficult, but todays widely using of computers has greatly eliminated this difficulty. The hypothesis that a foundation has such an elastic behavior may seem strange at first, but experiments done on railway tracks show that the foundation behaves elas- tically. If one observes a train moving on a railroad track, one notes that everytime a wheel passes over a crosstie, the crosstie moves down and afterwards comes up again. However, Winkler's hypothesis that the effect produced by a concentrated force on the foundation appli es only at the point of application is not exact since near-by points of foundation are also effected. 2. FUNDAMENTAL EQUATIONS Consider a weightless beam resting on a Winkler foundation and subjected to a concentrated load at the center. For unit breadth of the beam, the reaction of founda tion for some small length of the beam is proportional to the deflection y. The proportionately coefficient has been shown by k called as foundation modulus.The foundation modulus is different in compression and tension region. So the differential equation of the def lection of the beam resting on this foundation is also different. The foundation modulus are kt and ka., in compression and tension regions respectively. I 1 Figure 1 Beam on elastic foundation The differential equation which governs the response of the beam in figure. 1 is as follows EI d*Vi + k, y* (1) If the beam is loaded with a concentrated load P at x =? 0, its deflection is shown in figure. 2. The posi tive deflection y of the beam and loading is assumed to be positive downward. The length of the beam is 2L. The system is symmetric according to loading. At the both sides the deflections of the beam, in the same distance from the loading, are equal. So the problem is. solved for right half part of the beam. I Figure 2 The deflection of the beam I- Figure 3 Right half part of the beam Tİ2.1 Statically Loaded Beam Before going into the solution let us define the fol- loving quantities. Denote 34*=4EI/ki and introduce the dimensionless quantities. cci - xt / B. PP=P/'43k»), r=L/3 r4=Xi/0, e* - ka / ki, i>i (2) Now the governing equation of the elastic curve will become, d*yt + 4 * yi = 0 in the compression (3) da*“* region d*y, + 49-* * y± = 0 in the tension (4) dcci1* region The deflection of the beam in compression reagion the foundation modulus of intensity ki directed upward and in tension region the foundation modulus of intensity ka directed downward. Where EI is the flexural rigidity of the beam. kt, k2 are the subgrade modulus in compression and tension re gions respectively, y is the deflection of the beam (taken positive downward) and ai=rt is the location (unknown) where the deflection of the beam is zero. Clearly y(r±)=0. We observe that (3) represents an even operator y. Using some dimensionless magnitudes the solution can be written in the compression and tension regions respectively. y* = AiSha±sinai+BiChaiCosai+ CıchaiSina4+DıSha±cosaı y±= AiSheaisin8ai+Bich6o:iCos0cu + C ± ch6a i s i n6a ı +D ıshöaı cbs6a' * or y±= A** e”*1 * since* + B4* ea,x * cosa* + Çı* e-0,i * sina4 + D±* e-oci * cosai y*= At* e*»“*1 * sinea* + Bi* e0a,i * cosGai + Ci* e-eo,i * sinecu + D** e-0-1 * cosOai (5) iiWhere A*, B±. Ct. D4, At, Bı, Ç».. D4 are unknown coef ficients of integration for the compression and tension regions and I\ parameters are also unknown, and 9 is, 8* = ksj/k* The boundary conditions are; yi' =0. -EI.yin,| =0 at=0 jai=0 At the crossing points deflection is zero and slope, bending moment and shear force must- be continuous. yi I =0 |ai=ri y**i| =0 y”* J = y' **i| ai=r± ai+i=0..II I...II I y i i - y ±-x i \f 'I * I _ trn I I y 1 1 y *.-*?*? i Ja»=ri Jai+i=0 The end of the beam is free. -EI*y“”**| =0 I CCn-t- 1 =r'r»-»- 1 -EI*y,“r,*l| =0 i CCn-«- 1 =rn+ 1 n Where rrwi-r-Z T± (6) i-1 In addition to this, for the single foundation modulus the finite or infinite beams subjected to concentrated load are solved. In case of infinite beam the boundary conditions are y =0 cc=0 -EI.y'”| - P/2 cc=0 Tillyi -o (7) |cc=°o If the beam is finite and pressed against an elastic foundation by a concentrated load P at the origin as shown in figure. 2. The boundary conditions are ; y'l =0. -EI.y“'J =-P/2 | oc=0 J a=0 -EI*y”| =0, -EI*y'"| =0 (8) 2.2 Dynamically Loaded Beam If beam is subjected to a harmonic concentrated load Po.cos£t, the governing equation in compression and tension region respectively with dimensionless quantities will be d*y. 6i~=l-G3/wa (9) e*=|©i| l - 0. e24=k2/ki-22/w2 (10) dcc±* 6a-|ea| The governing equations change depending on the ratio of (Qa/w2). Q is frequency of the concentrated load, w is subjected frequency of the beam. If the beam length is finite and ki and ka are different, five different cases occure as follows Case.l : If the signs of the 0i and 0=> are positive the solution of the governing equation in compression and tension region respectively will be y*= AiSh9ıoCiSin0ıo:±+Bıch0ıo:ıcos6ıo:ı + CiCh0ıCCiSin0ıacı+D±sh0ıa:iCos0ıo:± yi= AiSh02aiSin02ai+BiCh02aiCos©2Cti + Ctch0aaiSin0=aı+Dısh0zaiCos02o:t (11) Case. 2 : If the sign of the 0* is positive and 02 is equal to zero, the solutions of the governing equations are ixyt=» Aısh6ıcCiSineıact+Bıch6»cctcos6ıat + CiCheiaisineiai+DtSheiaicosoxai yt= Aıaı3+Bıaıa!+Cıa1+Dt (12) Case. 3 : If the sign Q% is positive and the sign of the 0a is negative, the solution of the governing equa tions are y*= AiSh6ıaiSin6ıat+BtCh6ıa4cos6ıat+ Cıch6ıaıSineıaı+DiSh03La±cos©xai yt= Aishözaı+Bicheaaı+Ctsinezaı+Dtcoseaa* v (13) Case. 4 : If 6* is equal to zero and the sign of the 6_a is negative, the solution of the governing equation in compression and tension region respectively will be yı= AiCC^+BıOCi^+CiCCi+Di yt= Aısh92ai+BiCheaai+ Cisineacu+Dicoseacu (14) Case. 5 : If the signs of the 9» and 6=» are negative, the solution of the governing equation in compression and tension region respectively will be y±= AiSh6ıCCi+BiCh9ıa:ı+ CiSİn9ıat+Di.cos9ıCU y±= AiSh92ai+BiCh92ai+ CiSin03sCCi+DiCos02o:i (15) To find the unknown values of the equations for elas tic beam in each case, the boundary conditions can be written as in static solution.
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