Lecture D Problems using CST APL705 Finite Element Method Two-dimensional Problems using Constant Strain Triangles To formulate D problems we will follow similar steps as in the case of D FE modeling. Here the displacements, traclon, distributed body force are funclons of the posilon given by (x,y) The displacement vector is given by u=[u v Where u and v are x and y components of u Stresses and strains are σ = [σ x,σ y,τ xy The body force and traclon vectors are f = [ f x, f y ε = [ε x,ε y,γ xy T = [T x,t y
FE Modeling of D Problem For elemental volume calculalon we have V = tda Where t is the thickness normal to element (along z direclon) Units of bod force force per unit volume and that of traclon is force per unit area. The strain displacement relalon is ε = u x, v y, u y + v x The stress strain relalon is matrix σ = Dε T where D is the material D FE Modeling Two dimensional problems discussed here we have two degrees of freedom at each node as follows The global displacement vector is Q = [ Q T, Q, Q... Q N ] where N is the total number of dof The given D domain is triangulated first and the triangulalon data is stored as nodal coordinates and conneclvity informalon. The coordinates are stored in an NNx matrix where NN is total number of nodes in the triangulalon. Q i i y x Q i-
Triangulated D domains Here we show two examples triangular meshes. First one straight edges which is tria -ngulated evenly. The other domain with curved boundary D Element ConnecLvity Considering an element, the conneclvity of a typical element in the triangular mesh is shown here q 6 The element displacement vector is shown here as q = [q, q, q, q 4, q 5, q 6 A typical conneclvity table is given as Element no./ local nodes N 6 6 8 (x,y) u q q q 4 The element displacement vector q can be extracted from global Q using conneclvity table v q 5 q
The Local and Global Connect The nodal coordinates shown here as (x,y ),(x,y ) and (x,y ) have a global correspondence through the conneclvity table The local representalon of nodal coordinates and degrees of freedom is a way of clearly represenlng the element characterislcs. k = L T k 'L Constant Strain Triangle Recall the shape funclons used to interpolate the nodal displacements in D problems. Here in D problems we determine the displacements inside an element from the nodal displacements using linear shape funclons. By inspeclon we see that N +N +N =. This means that they are not linearly independent. 4
D Shape funclons The plot of third funclon is given here Note that the funclon N is at node and 0 at nodes and. Similarly N and N are also have their values as at nodes and respeclvely and 0 at other nodes. RepresenLng the independent shape funclons by ξ and η, we have N = ξ N = η and N = ξ η Here ξ and η are called the natural coordinates. D Shape funclons Now we look at the analogy of D and D shape funclons In D case, x-coordinate mapped onto ξ coordinate and shape funclon is a funclon of ξ. Here in D problems, x, y coordinates map onto ξ, η and the shape funclon is a funclon of ξ and η. The shape funclons are now represented by area coordinates. A point P(x,y) divides the triangle into three areas A, A and A as shown. Now express the shape funclons as area ralos. N + N + N = A A + A A + A A = For every point inside element N +N +N = 5
Isoparametric RepresentaLon Now displacements inside the element are expressed in terms of the nodal values of the unknown displacement field. u = N (ξ,η)q + N (ξ,η)q + N (ξ,η)q 5 v = N (ξ,η)q + N (ξ,η)q 4 + N (ξ,η)q 6 Now using the definilons of shape funclons N = ξ N = η and N = ξ η We write the displacements u,v as u = (q q 5 )ξ + (q q 5 )η + q 5 v = (q q 6 )ξ + (q 4 q 6 )η + q 6 As in D we represent N as a matrix now. Isoparametric RepresentaLon As in D we represent N as a matrix now. N 0 N 0 N 0 N = 0 N 0 N 0 N The displacements in matrix form { u} = N We know that for an isoparametric representalon the coordinates inside the triangular element also can be represented in terms of the nodal coordinates as x = N (ξ,η)x + N (ξ,η)x + N (ξ,η)x y = N (ξ,η)y + N (ξ,η)y + N (ξ,η)y [ ]{ q} OR x = (x x )ξ + (x x )η + x y = (y y )ξ + (y y )η + y 6
Isoparametric RepresentaLon The (x,y) coordinates inside the elements are wrihen as x = (x x )ξ + (x x )η + x y = (y y )ξ + (y y )η + y x ij = x i x j) By changing the notalon as x = x ξ + x η + x y = y ξ + y η + y ( ) 7