Please use this identifier to cite or link to this item: http://hdl.handle.net/11527/16425
Title: Plak sonlu elemanlarda kayma şekil değiştirmelerinin gözönüne alınması ve iki parametreli zemine oturan plakların hesabı için bir yöntem
Authors: Saygın, Ahmet Işın
Çelik, Mecit
55974
Yapı Mühendisliği
Structural Engineering
Keywords: Levhalar
Plates
Issue Date: 1996
Publisher: Fen Bilimleri Enstitüsü
Institute of Science and Technology
Abstract: Bu çalışmada, plak sonlu elemanlarda kayma şekildeğiştirmelerinin gözönüne alınmasına yönelik bir yöntem geliştirilmiş, iki parametreli zeminlerde zemin yüzey parametresi ile zemine ait karakteristik büyüklüklerin bir ardışık yaklaşım yöntemiyle hesaplanıp temeli ve çevre zeminini sonlu elemanlarla idealleştirerek yüzeysel temellerin hesabı yapılmıştır. Altı bölüm olarak sunulan bu çalışmanın birinci bölümünde konunun tanıtılması, konu ile ilgili çalışmaların gözden geçirilmesi, çalışmanın amacı ve kapsamı yer almaktadır. İkinci bölümde, kalın plaklarda şekildeğiştirmelerin eğilme ve kayma şekildeğiştirmelerinin toplamı olarak düşünülmesi halinde gerilme şekildeğiştirme bağıntıları elde edilmiştir. Daha sonra literatürde mevcut olan 16 ve 12 serbestlik dereceli ince plak sonlu elemanların eğilme ve kayma şekildeğiştirme yüzeyleri tanımlanıp, virtüel iş teoremi yardımı ile eğilme ve kayma şekildeğiştirmelerinin birlikte düşünülmesiyle oluşan rijitlik matrisleri, ve deformasyon matrisileri elde edilip tablolar halinde verilmiştir. Ayrıca 16 serbestlik dereceli halka sektörü plak sonlu elemanda kaymalı eğilme elemanının rijitlik ve deformasyon matrislerinin sayısal integrasyonla elemanda kalınlığın değişken olması hali de gözönüne alınarak elde ediliş yöntemi açıklanmıştır. Üçüncü bölümde iki parametreli zeminde zemin karakteristiklerinin tanımı yapılarak, iki parametreli zemine oturan plakların altında ve temel plağı dışında kalan noktalarda diferansiyel denklem elde edilmiş ve virtüel iş teoremi yardımı ile zemine ait karakteristik büyüklüklerin ardışık yaklaşım yöntemi ile elde edilebileceği gösterilmiştir. Bu bölümde 16 ve 12 serbestlik dereceli dikdörtgen plak sonlu elemanlara ait elastik yataklarıma ve kayma parametresi matrisleri tablolar halinde çıkartılmıştır. 16 serbestlik dereceli halka sektörü plak sonlu elemanın elastik yataklarıma ve kayma parametresi matrislerinin sayısal integrasyonla elde edilme yöntemi gösterilmiştir. Bu bölümde son olarak temel dış ortamının zemin sonlu elemanlar yardımı ile idealizasyonu yapılarak zemin yüzey parametresinin nasıl elde edilebileceği gösterilmiştir. Dördüncü bölümde sayısal uygulamaların yapılabilmesi amacıyla bu çalışma için gerekli değişikliklerin yapıldığı FORTRAN77 dilinde kodlanmış bilgisayar programının çalışma düzeni ile program ve programı oluşturan alt programların akış diyagramları verilmiştir. Beşinci bölümde yöntemin sayısal uygulamaları beş değişik örnek üzerinde yapılmış bazılarında bu çalışmaya dayanarak bulunan sonuçlarla, daha önce yapılan çözümler karşılaştırılmıştır. Altıncı bölümde bu çalışmada elde edilen sonuçlar topluca verilmiştir.
In this study, finite element formulations are developed in which shear deformations are included to bending behaviour, and the effect of subgrade reaction is considered as a combination of elastic bedding and shear deformation of the soil. Also a succesive approximation method is proposed in which elastic bedding and shear deformation characteristics of the underlying soil is determined depending on elastic properties of the soil and thickness of the compressible layer. Total deflection of the plate is taken as sum of the pure bending deflection and that of the pure shear deflection. w = wb+ws In the view of this assumption, general expressions for shear strains can be written as follows. âu aw. aw, Y = + + /jB âz âx âv âwh â x âw" Yvz = ?+. - + - where wh âz ây ' ây and ws being deflection due to the pure bending and to the pure shear. The following expressions can be written as âu âw. = 0 âz â x âv âw. --+-±=0 âz oy hence, expressions which are valid for thin plate can be applied directly. Thus internal moments are obtained as follows. M xyj Eh3 12(1- v2) 1 v v 1 0 0 0 0 1-v âx2 â2wh -2 ây2 â2w" âxây xvı Shear deformations are Txz=~âx~ r yz dy and shear forces can be obtained from aw, v7,Vy, Gh ax By using the method which is explained below the deflections w b and ws will be obtained depending on nodal displacements of the finite element without changing the degrees of freedom of the standard finite element. The stiffness matrix thus obtained for the plate element includes the effect of shear deformations and can be transformed pure bending stiffness matrix in the limit. Also stability problem called shear locking can be eliminated. The stiffness matrix including shear effect and corresponding deformation matrix can be obtained for plate finite element of various degrees of freedom. The bending deflections wb due to pure bending deformation of the full compatible rectangular plate bending finite element which is shown in Figure 2.1 can be expressed as follows. 16 W, -z wbi d, /=1 The shape functions wbj which correspond unit values of displacement dl are formed as products of cubic functions of variables x and y. The bending moments obtained by using these functions vary linearly within the element whereas shear forces are constant. For this reason, it can be seen that the field of shear deformations, ws may be choosen by products of linearly varying functions in both directions. The shape and unknows parameter of the shear plate reduced from the above argument are shown in Figure 2.3. The shear part of the deflection function ws can be given in terms of nodal displacements (dj) as below. Ts = Wsl d\ + Ws2 d2 + Ws3 d3 + Ws4 d4 where wsi are products of linearly varying auxilary functions; /,(jc), /,(>"). Thus the total deflection field of the element yields as 16 Wf = Z ;=1 i=l It is known in [50] that the stiffness matrix [£&Ji6,6 which depends only on pure bending deformation can be obtained by using the unit displacement theorem. Also shear stiffness matrix [^,]4 4 which depends only on shear deformation can be obtained as the same way, so that the term of this matrix will be xvii «j=jGhJI< âw, âwj dwi S Wj 6""*"' [ âx âx ây ây j and while the total nodal displacements in an element due to the any nodal forces are the sum of the nodal displacements depending on bending and shear deformation, the stiffness matrix included shear deformation [kT\ will be obtained as follows by using both bending and shear stiffness matrices [kb] and [*J. The shear flexibilty matrix will be obtained by fixing the displacement of the fourth node of the element The relative displacement depending on the shear deformations of the nodes 1,2,3 according to forth node due to forces Pl, P5, P9 can be obtained as given below. If the relative nodal displacements due to shear deformation for the case D,=\ are defined as the nodal displacements determining the bending deformations will be as follows dt - D, and d< = -X, 4 = -Xu d» = ~XXi The nodal forces which belong to bending effects are obtained by multiplying the bending stiffness matrix by the nodal displacement vector. k k L b9,\ "?»1,5 *"*5,5 kft9,5 If this nodal forces are multiplied by the shear flexibility matrix the nodal displacement due to the shear deformation can be obtained as and rearranging this formulation yields: xvni [[/.][*.] +w] The solution of these equations gives the part of nodal displacements due to shear effects for any £> = 1. The solution of these equations for all of the nodal displacements Z). =1 16 is obtained in the following matrix form, xu LXV. and, the stiffness matrix which includes both bending and shearing effects can be obtained by using this [X] matrix as follows. *nj = Kij ~ hs,\xu ~ kbj,5X2,i - hj.9Xv (i=l,16 j=l,16) When the system is solved for the nodal displacements the curvatures at the nodes can be obtained for each element. Thus the bending and twisting moments are: [Md] = [D][Bd][d]. The elements of [Bd] matrix which includes shear effect can be expressed by using standart pure bending matrix [Bd] as: B TiJ B bi,j *», A " B^X2j ~ BbK9XXJ (i=l,12, j=l,16) The matrices [X],[£r] and [^j^ are given in this text which belong to the rectangular plate element, have 16 degrees of freedom. The stiffness and deformation matrices which include shear deformation effects of the element developed by Adini, Clough, and Melosh having 12 degrees of freedom are obtained in the same way and the matrices [.Ar],[£7.] and [^J are shown in this text. The sector annular plate which has vertical displacement, two slopes and twisting curvature at each node is shown in Figure 2.5 The deflection wb due to the pure bending deformation in 16 degrees of freedom sector annular plate bending finite element as follows. 16 Vf, = IV/ (=1 Displacement functions wbi for unit values of displacement d, are formed by multiplying two functions which depend on radial and angular coordinates separately. If the initial point of variable x is choosen at the middle point of the plate in the radial direction xix and a = Rx-R2, auxilary cubic functions used for rectangular plate element can be used directly for the radial directions function. On the other hand, the angular direction functions have to contain trigonometric functions such as sin 0, cos 0 in order to satistify rijid body motion requirement (see 2.91). The field of shear deformation ws is a product of linear function of radial coordinate and equivalent linear function of angular coordinate Xt{0). The shape of the shear plate element and degrees of freedom of this element are shown in Figure 2.6 The displacement field due to shear effect is defined in (2.95), (2.96).Thus the total displacement field of the element will be as follows. 16 4 1=1 i=l The stiffness matrix due to bending and shearing effects seperately can be obtained numerically by using Gauss Quadrature.. Then the same procedure for obtaining stiffness and deformation matrices such as rectangular plate element which include shear effect can be used. For the plate on elastic two parameter foundation, the vertical displacement at any point in compressible soil below the plate can be expressed as follows. wz =w(x,y)0(z)- Where w(x,y) belongs to the surface displacement of the soil, 0 belongs to the displacement which is varying with depth. The boundary condition for 0(z) are z = 0-> 0(z) = 1 z = H -> 0(z) = 0 The effect of subgrade reaction can be obtained as follows qz = Cw{x,y)-2CT\-^-r- + -j^r- where C = E.{\-v,) Jp0(z)V dz (l+vs)(l-vs)ziX dz H 2Cr = GsJ02(z)dz z=0 where Es, vs are material properties of the soil. The differential equation of the plate on two parameter foundation which takes the reaction of the soil and the external loading q into account, can be written as follows xx D A A w - 2CT A w + Cw = q At the points out of the plate domain the following differential equation will be valid -2CrAw + Cw = 0 provided that there is no external loading. In these expressions C corresponds to the Winkler coefficient, CT to the shear parameter which depends on the shear deformation of the soil. C and CT depend on the material properties, the thickness of compressible layer of the soil and the function of 0(z). The function of 0(r), which satisfy the boundary condition at z=0 and z=H, Shy and y is a parameter of soil surface which can be obtained as in chapter 3.1 and [44] as follows 2 H\\-2vs) r JJ V C a v dy) \dxdy 2(1- v.) J J w2dxdy If the function 0(z) is substituted into the expression for C, and CT it yields Es(l-vs) y (Sh2y + 2y) (l+vs)(l-2vs)H 4Sh2y ır r H(Sh2r-2r) r~% *Sh2y As seen in these expressions C and CT depend on the material properties, the thickness of compressible layer of the soil, the coefficient of /..whereas y depends on the dimension, stiffness, and the external loads distributed on the foundation. The parameter y can be evaluated after determining w(x,y) which satisfies differantial equation below and around the plate, by integrating the numerator and denominator in expression of y2 for below and around of the foundation domain. Then it is obvious that for computing the parameter y the successive approximation method have to be used. In finite element two parameter soil effects can be expressed by adding the matrices [C] and [CT] to the stiffness matrices. The terms of these matrices can be computed as follows. xxi Cu^cWwfWjdxdy rr I dw, aw TiJ TJJ [ ax ax dy dy J 7 It is obvious that the function wt having 16 degrees of freedom rectangular finite element is given as follows Wi = Wbi + XU K,l - Wb,l ) + X2,i K,2 ' Wb,5 ) + *%, K.3 - Wb,9 ). where wi has been computed as sum of the bending and shearing fields for the deformation of Dt = 1. The reaction matrices can be computed as the matrices [^46] [^4rf] and [X]iJ are known, but these terms depend on the dimension, the ratio of bending and shearing stiffness of the element then contain long and complex expressions. For this reason the assumption w, s wbi can be made and the reaction matrices of the soil for both 16 and 12 degrees of freedom elements can be computed. Cij = c\)wbtwbJdxdy dw.t dwu The matrices [C] and [Cr] for 16 degrees of freedom sector annular plate finite element can be computed numerically by following expressions. N M aar i rt=\ m=\ " r -or VVff u aarlaWjdwj dw, dWj 'T,J Ti^J^JU n m 4 y dr dr rgQrdQ\ where n is the number of the Gaussian integral points in radial direction, m is the number of the Gaussian integral point in circumferantel direction. While the deflection shape is computed numerically, assumption is not necessary for this element. The differential equation which is valid at the points out of the plate region where no external load is considered can be given as: â2w(x,y) â2w(x,y) Cw(x,y)-2Cj = 0 âxz dy1 the changes of deflections around the soil can be determined by finite element idealization in soil domain. The wideness of the soil region around the plate that is divided into finite elements has to be as much as the computed deflection on this boundary is close to zero. It appears from evaluated examples that the extent of this region can be chosen as the thickness of the compressible layer of soil. xxn By assuming the differential equation, the outside soil region of the plate has the same behavior of the shear plate having shear rigidity 2CT = Gh'. By using this analogy the displacement field of the finite element in the soil can be determined. If [d] shows the nodal vertical displacement of the soil element then * = 2>,4=k],M and the nodal forces of the elements depend on the displacements as follows [c][J\+lcT][J\ = [p]. The displacement field of the rectangular finite element will be expressed depending on vertical displacement of the nodes can be written as Ud]s=[l2(x)l2(x) lx{x)l2(x) l{{y)l2{x) /,(*)/, OO] The matrices of the soil can be obtained by integrating the function as follows. CtJ = Cİ\wtWjdA nv^cjjj- dw. dw. dw, dw, a,... =2CJJ1 ', 7+ J J\dA ox ox o y d y J The displacement field of the sector type soil finite element having four degrees of freedom can be obtained by multiplying in similar way the linear function in radial direction and the equivalent to linear function Xt{0) in angular direction. [Ad)s=[X2{9)l2(x) Xx{0)l2{x) X2{0)l,(x) A,(0)1^)] Then the matrices of the soil can be obtained by the use of the function for the case D, = 1 as follows. 
Description: Tez (Doktora) -- İstanbul Teknik Üniversitesi, Fen Bilimleri Enstitüsü, 1996
Thesis (Ph.D.) -- İstanbul Technical University, Institute of Science and Technology, 1996
URI: http://hdl.handle.net/11527/16425
Appears in Collections:Yapı Mühendisliği Lisansüstü Programı - Doktora

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