121 Element Fields in Linear Problems This chapter presents interpolation models in physical coordinates for the most part, for the sake of simplicity and brevity. However, in finite-element codes, the physical coordinates are replaced by natural coordinates using relations similar to interpola- tion models. Natural coordinates allow use of Gaussian quadrature for integration and, to some extent, reduce the sensitivity of the elements to geometric details in the physical mesh. Several examples of the use of natural coordinates are given. 9.1 INTERPOLATION MODELS 9.1.1 O NE -D IMENSIONAL M EMBERS 9.1.1.1 Rods The governing equation for the displacements in rods (also bars, tendons, and shafts) is (9.1) in which u ( x , t ) denotes the radial displacement, E, Α and ρ are constants, x is the spatial coordinate, and t denotes time. Since the displacement is governed by a second-order differential equation, in the spatial domain, it requires two (time- dependent) constants of integration. Applied to an element, the two constants can be supplied implicitly using two nodal displacements as functions of time. We now approximate u ( x , t ) using its values at x e and x e + 1 , as shown in Figure 9.1. The lowest-order interpolation model consistent with two integration constants is linear, in the form (9.2) We seek to identify ΦΦ ΦΦ m1 in terms of the nodal values of u . Letting u e = u ( x e ) and u e + 1 = u ( x e + 1 ), furnishes (9.3) 9 EA u x A u t ∂ ∂ = ∂ ∂ 2 2 2 2 ρ , ux t x t t ut ut xx m T mm m e e m T (,) () (), () () () () ( ).== = + ϕϕΦΦγγ γγ ϕϕ 111 1 1 1 1, ut x t u t x t e e mm e e mm () ( ) (), () ().== () ++ 11 11 1 1 11 ΦΦγγ ΦΦγγ 0749_Frame_C09 Page 121 Wednesday, February 19, 2003 5:09 PM © 2003 by CRC CRC Press LLC 122 Finite Element Analysis: Thermomechanics of Solids However, from the meaning of γγ γγ m 1 ( t ), we conclude that (9.4) 9.1.1.2 Beams The equation for a beam, following Euler-Bernoulli theory, is: (9.5) in which w ( x , t ) denotes the transverse displacement of the beam’s neutral axis, and I is a constant. In the spatial domain, there are four constants of integration. In an element, they can be supplied implicitly by the values of w and w ′ = ∂ w /∂ x at each of the two element nodes. Referring to Figure 9.2, we introduce the interpolation model for w ( x , t ): (9.6) FIGURE 9.1 Rod element. FIGURE 9.2 Beam element. u e x e x e +1 u e +1 x w e t w e w e t +1 w e+1 x e x e+1 x ΦΦ m e e e ee ee e x x l xx lx x 1 1 1 1 1 1 1 1 11 = = − − =− + − + + ,. EI w x A u t ∂ ∂ + ∂ ∂ = 4 4 2 2 0 ρ , wx t x t x x x x t w w w w bbb b m e e e e (,) () (), () , () .== () = − ′ − ′ + + ϕϕΦΦγγϕϕγγ ΤΤΤΤ 111 1 23 1 1 1 1 0749_Frame_C09 Page 122 Wednesday, February 19, 2003 5:09 PM © 2003 by CRC CRC Press LLC Element Fields in Linear Problems 123 Enforcing this model at x e and at x e + 1 furnishes (9.7) 9.1.1.3 Beam Columns Beam columns are of interest, among other reasons, in predicting buckling according to the Euler criterion. The z –displacement w of the neutral axis is assumed to depend only on x and the x –displacement. Also, u is modeled as (9.8) in which u 0 ( x ) represents the stretching of the neutral axis. It is necessary to know u 0 ( x ), w ( x ) and at x e and x e + 1 . The interpolation model is now (9.9) and 9.1.1.4 Temperature Model: One Dimension The temperature variable to be determined is T − T 0 , in which T 0 is a reference temperature assumed to be independent of x . The governing equation for a one- dimensional conductor is (9.10) ΦΦ b1 23 2 11 2 1 3 11 2 1 1 01 2 3 1 012 3 = −− − −− − ++ + ++ − xx x xx xx x xx ee e ee ee e ee . ux z u x z wx x (, ) () () ,=− ∂ ∂ 0 ∂ ∂ wx x () ux zt x z x x x mm bb (, ,) ( ) ( ) ,=−11 11 23 11 ΦΦγγΦΦγγ γγγγ m e e b e e e e ut ut w w w w ll , = = − ′ − ′ + + + () () , 1 1 1 ΦΦΦΦ m b 1 1 1 23 2 11 2 1 3 11 2 1 1 11 1 01 2 3 1 012 3 = − − = −− − −− − + ++ + ++ − l xx xx x xx xx x xx e ee ee e ee ee e ee ,. kA x Ac t e ∂ ∂ = ∂ ∂ 2 2 TT ρ . 0749_Frame_C09 Page 123 Wednesday, February 19, 2003 5:09 PM © 2003 by CRC CRC Press LLC 124 Finite Element Analysis: Thermomechanics of Solids This equation is formally the same as for a rod equation (see Equation 9.1), furnishing the interpolation model for the element as (9.11) 9.1.2 I NTERPOLATION M ODELS : T WO D IMENSIONS 9.1.2.1 Membrane Plate Now suppose that the displacements u ( x , y , t ) and v ( x , y , t ) are modeled on the triangular plate element in Figure 9.3, using the values u e ( t ), v e ( t ), u e + 1 (t), v e+1 (t), u e+2 (t), and v e+2 (t). This element arises in plane stress and plane strain, and is called a membrane plate element. A linear model suffices for each quantity because it provides three coefficients to match three nodal values. The interpolation model now is (9.12) . 9.1.2.2 Plate with Bending Stresses In a plate element experiencing bending only, the in-plane displacements, u and v, are expressed by (9.13) FIGURE 9.3 Triangular plate element. z Y middle surface X e+1 ,Y e+1 X e+2 ,Y e+2 X e ,Y e X T T TT TT (, ) () (), () () () .xt x t t t t mm e e −= = − − + 01 11 1 0 10 ϕϕΦΦθθθθ T ux y t vx y t t t m m m m u v (, , ) (, , ) () () = ϕϕ ϕϕ ΦΦ ΦΦ γγ γγ ΤΤ ΤΤ ΤΤ ΤΤ 2 2 2 2 0 0 0 0 T T T T γγγγϕϕΦΦ ΤΤ uvmm () () () () ,() () () () ,,t ut ut ut t vt vt vt x y xy xy e e e e e e ee ee = = = = + + + + ++1 2 1 2 2211 11 1 11 22 1 xy ee++ − ux y z t z w x vx y z t z w y (, , ,) , (, , ,) ,=− ∂ ∂ =− ∂ ∂ 0749_Frame_C09 Page 124 Wednesday, February 19, 2003 5:09 PM © 2003 by CRC CRC Press LLC Element Fields in Linear Problems 125 in which z = 0 at the middle plane. The out-of-plane displacement, w, is assumed to be a function of x and y only. An example of an interpolation model is introduced as follows to express w(x, y) throughout the element in terms of the nodal values of : (9.14) . It follows that (9.15) 9.1.2.3 Plate with Stretching and Bending Finally, for a plate experiencing both stretching and bending, the displacements are assumed to satisfy (9.16) w w x w y , and ∂ ∂ ∂ ∂ wxyt xy t bbb (,,) (, ) ()=ϕϕΦΦγγ 222 T ϕϕ b xy x y x xy y x xy yx y 2 223223 1 1 2 T (,) ( )=+ γγ b e e e e e e e e e w x w y w x w y w x w y w 2 1 1 1 2 2 2 T () tw w=− () − − () − − () − ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ + + + + + + ΦΦ b ee e ee e e eeee e ee e eee ee eee e ee e xy x xy y x xyxy y xy x xyy xx xxy y xy x x 2 1 223 1 2 3 2 1 2 2 1 2 22 11 1 2 1 010 2 0 3 0 00 1 0 2 0 3 1 − ++ + = + −−− −−+ −−− −+− () () () eee e e ee e e e ee e eee ee ee yy x xyxy y xy x xyy xy xx ++ + + ++ + + + ++ + +++ ++ ++ + −−− −−+ −−− −+ 11 1 2 1 3 1 2 11 1 1 1 3 11 1 2 11 1 2 1 2 11 1 2 1 2 010 2 0 3 0 00 1 0 2 0 () () ( 111 1 2 22 2 2 22 2 2 2 3 1 2 22 2 2 2 3 22 2 2 22 1 2 2 2 3 1 010 2 0 3 0 00 yy xy x xy y x xyxy y xy x xyy ee eeeeee e eeee e ee e eee ++ ++++++ + ++++ + ++ + +++ − + −−− −−+ ) () () −−−− −+− ++ +++ + 10 2 0 3 22 1 2 2 2 22 2 2 xy xxy y ee eee e () u(x, y, z, t) v(x, y, z, t) w(x, y, z, t) z zt b b b bb x y = − − ∂ ∂ ∂ ∂ ϕϕ ϕϕ ΤΤ ΤΤ ΤΤ ϕϕ ΦΦγγ 2 2 2 22 (). ux y zt u x y z t z w x vx y z t v x y zt z w y (,,,) (,,,) , (,,,) (,,,)=− ∂ ∂ =− ∂ ∂ 00 0749_Frame_C09 Page 125 Wednesday, February 19, 2003 5:09 PM © 2003 by CRC CRC Press LLC 126 Finite Element Analysis: Thermomechanics of Solids and w is a function only of x, y, and t(not z). Here, z = 0 at the middle surface, while u 0 and v 0 represent the in-plane displacements. Using the nodal values of u 0 , v 0 , and w 0 , a combined interpolation model is obtained as (9.17) 9.1.2.4 Temperature Field in Two Dimensions In the two-dimensional, triangular element illustrated in Figure 9.3, the linear inter- polation model for the temperature is (9.18) 9.1.2.5 Axisymmetric Elements An axisymmetric element is displayed in Figure 9.4. It is applicable to bodies that are axisymmetric and are submitted to axisymmetric loads, such as all-around pressure. The radial displacement is denoted by u, and the axial displacement is denoted by w. The tangential displacement v vanishes, while radial and axial dis- placements are independent of θ . Now u and w depend on r, z, and t. There are two distinct situations that require distinct interpolation models. In the first case, none of the nodes are on the axis of revolution (r = 0), while in the second case, one or two nodes are, in fact, on the axis. In the first case, the linear FIGURE 9.4 Axisymmetric element. ux yzt vx yzt wx yzt uxyt vxyt wxyt z z b b b x y (, , ,) (, , ,) (, , ,) (, ,) (, ,) (, ,) = + − − ∂ ∂ ∂ ∂ 0 0 0 2 2 2 ϕϕ ϕϕ ΤΤ ΤΤ ΤΤ ϕϕ = − − ∂ ∂ ∂ ∂ ΦΦγγ ϕϕ ϕϕ ϕϕ ΦΦ ΦΦ ΦΦ γγ γγ γγ ΤΤ ΤΤ ΤΤ ΤΤ ΤΤ ϕϕ ϕϕ bb m b m b b m m b u v w t z z t t t x y 22 2 2 2 2 2 2 2 2 () () () () 0 0 00 00 00 00 T T TT . T T T T T T T T−= = − − − ++0222 2 0 10 20 ϕϕΦΦθθθθ ΤΤΤΤ mm e e e ,( ). r e+2 e e+1 θ 0749_Frame_C09 Page 126 Wednesday, February 19, 2003 5:09 PM © 2003 by CRC CRC Press LLC Element Fields in Linear Problems 127 interpolation model is given by (9.19) Now suppose that there are nodes on the axis, and note that the radial displace- ments are constrained to vanish on the axis. For reasons shown later, it is necessary to enforce the symmetry constraints a priori in the formulation of the displacement model. In particular, suppose that node e is on the axis, with nodes e + 1 and e + 2 defined counterclockwise at the other vertices. The linear interpolation model is now (9.20) A similar formulation can be used if two nodes are on the axis of symmetry so that the u displacement in the element is modeled using only one nodal displacement, with a coefficient vanishing at each of the nodes on the axis of revolution. 9.1.3 INTERPOLATION MODELS: THREE DIMENSIONS We next consider the tetrahedron illustrated in Figure 9.5. A linear interpolation model for the temperature can be expressed as (9.21) ur z t wr z t t t a a a a ua wa (, , ) (, , ) () () . = ϕϕ ϕϕ ΦΦ ΦΦ γγ γγ 1 1 1 1 1 1 TT TT TT TT 0 0 0 0 ϕϕΦΦγγγγ aa ee ee ee ua e e e wa e e e rz rz rz rz u u u w w w 1111 22 1 11 2 11 2 1 1 1 1 T == = = ++ ++ − + + + + (), , , ur zt wr z t t t a a a a ua wa (, , ) (, , ) () () = ϕϕ00 00ϕϕ ΦΦ00 00ΦΦ γγ γγ 2 2 2 2 2 2 TT TT TT TT ϕϕΦΦ aea eee eee rzz rzz rzz 22 11 22 1 T =− = − − ++ ++ − (), . TT−= 0333 ϕϕΦΦθθ TT T ϕϕ ΦΦ 3 3 111 222 333 1 3123 1 1 1 1 1 T T T T TT T T T T T T T = = =− − − − +++ +++ +++ − ++ + () {} xyz xyz xyz xyz xyz eee eee eee eee eo e o e o e o θ 0749_Frame_C09 Page 127 Wednesday, February 19, 2003 5:09 PM © 2003 by CRC CRC Press LLC 128 Finite Element Analysis: Thermomechanics of Solids For elasticity with displacements u, v, and w, the corresponding interpolation model is (9.22) 9.2 STRAIN-DISPLACEMENT RELATIONS AND THERMAL ANALOGS 9.2.1 STRAIN-DISPLACEMENT RELATIONS: ONE DIMENSION For the rod, the strain is given by . An estimate for ε implied by the interpolation model in Equation 9.3 has the form (9.23) FIGURE 9.5 Tetrahedral element. x e+1 e+2 e+3 y e z 0 ux y zt vx y z t wx y zt t t t u v w (, , ,) (, , ,) (, , ,) () () () = ϕϕ0000 00ϕϕ00 0000ϕϕ ΦΦ0000 00ΦΦ00 0000ΦΦ γγ γγ γγ 3 3 3 3 3 3 TTT TTT TTT γγγγγγ u e e e e v e e e e w e e t ut ut ut ut t vt vt vt vt t wt wt () () () () () ,() () () () () ,() () ( = = = + + + + + + +1 2 3 1 2 3 1 )) () () wt wt e e + + 2 3 ε == ∂ ∂ E u x 11 ε (,) () (),xt x=ββΦΦγγ mmm t 111 T 0749_Frame_C09 Page 128 Wednesday, February 19, 2003 5:09 PM © 2003 by CRC CRC Press LLC Element Fields in Linear Problems 129 For the beam, the corresponding relation is (9.24) from which the consistent approximation is obtained: (9.25) For the beam column, the strain is given by (9.26) 9.2.2 STRAIN-DISPLACEMENT RELATIONS: TWO DIMENSIONS In two dimensions, the (linear) strain tensor is given by (9.27) We will see later the two important cases of plane stress and plane strain. In the latter case, E zz vanishes and s zz is not needed to achieve a solution. In the former case, s zz vanishes and E zz is not needed for solution. In traditional finite-element notation, we obtain [cf. (Zienkiewicz and Taylor, 1989)] (9.28) ββ ϕϕ m m d dx 1 1 01 T T ==() ε (, , ) ,xzt z w x =− ∂ ∂ ε (,,) () (),xzt z x bbb =− ββΦΦγγ 111 T t ββ ϕϕ b b d dx xx 1 1 2 012 3 T T ==() ε (,,) () (),xzt z x bbb =− ββΦΦγγ mmm t 111 T ββββββ ΦΦ00 00ΦΦ γγ γγ mb m b m mb m b z t t 1 1 1 1 1 1 1 TT T =− () () () E() .xy EE EE xx xy xy yy u x u y v x u y v x v y , = = () () ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ 1 2 1 2 ′ = = εεββΦΦ γγ γγ E E E xx yy xy m T m u v 22 2 2 ˆ 0749_Frame_C09 Page 129 Wednesday, February 19, 2003 5:09 PM © 2003 by CRC CRC Press LLC 130 Finite Element Analysis: Thermomechanics of Solids The prime in e′ is introduced temporarily to call attention to the fact that it does not equal VEC(E L ). Hereafter, the prime will not be displayed. For a plate with bending stresses only, (9.29) from which . (9.30) For a plate experiencing both membrane and bending stresses, the relations can be combined to furnish (9.31) 9.2.3 AXISYMMETRIC ELEMENT ON AXIS OF REVOLUTION For the toroidal element with a triangular cross section, it is necessary to consider two cases. If there are no nodes on the axis of revolution, then (9.32) ββ 00 00ΦΦ ΦΦ00 00ΦΦ ΤΤ ΤΤ ΤΤ ΤΤ ΤΤ ΤΤΤΤ ϕϕ ϕϕ ϕϕϕϕ m m m mm m m m x y yx 2 2 2 22 2 2 2 1 2 1 2 = = ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ , ˆ ′ =− ∂ ∂ ∂ ∂ ∂ ∂∂ e (, , , ) ,xyzt z w x w y w xy 2 2 2 2 2 ′ =− = ∂ ∂ ∂ ∂ ∂ ∂∂ e (, , , ) () (),xyzt z x t x y xy b T bb b b T b T b T ββΦΦγγββ ϕϕ ϕϕ ϕϕ 222 2 2 2 2 2 2 2 2 2 ′ =e (, , ,) (, , ) ()xyzt xyz t mb mb mb ββΦΦγγ 222 T ββββββΦΦ ΦΦ00 00ΦΦ γγ γγ γγ mb m bmb m b mb m b t t t 2 2 22 2 2 2 2 2 TT T =− () = = z ,,() () () e(, , ) () () rzt t t u r u r w z u z w r a a a ua wa = () = ∂ ∂ ∂ ∂ ∂ ∂ + ∂ ∂ 1 2 1 1 1 1 1 ββ ΦΦ ΦΦ γγ γγ T 0 0 0749_Frame_C09 Page 130 Wednesday, February 19, 2003 5:09 PM © 2003 by CRC CRC Press LLC [...]... (9. 35) 07 49_ Frame_C 09 Page 132 Wednesday, February 19, 2003 5: 09 PM 132 Finite Element Analysis: Thermomechanics of Solids 9. 2.6 THERMAL ANALOG IN THREE DIMENSIONS Again referring to the tetrahedral element, the relation for the temperature gradient is T ∇T = β 3T Φ 3T θ 3 , T β 3T 0 = 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 (9. 36) 0 9. 3 STRESS-STRAIN-TEMPERATURE... axisymmetric triangular element, find the strain-displacement T matrix β in the cases in which two of the nodes are located on the axis of revolution Note that for axisymmetric problems, ur = 0 at r = 0 © 2003 by CRC CRC Press LLC 07 49_ Frame_C 09 Page 138 Wednesday, February 19, 2003 5: 09 PM 138 Finite Element Analysis: Thermomechanics of Solids 7 The stress-strain relations in two-dimensional plane stress... −1 (9. 51) 07 49_ Frame_C 09 Page 136 Wednesday, February 19, 2003 5: 09 PM 136 Finite Element Analysis: Thermomechanics of Solids For use in the Principle of Virtual Work, Da is modified to furnish Da , given by ′ 1 0 Da = E ′ 0 0 0 0 1 0 0 1 0 0 −ν −ν 1 −ν −ν 1 0 0 1 0 −ν 0 −ν 2 0 0 0 0 0 1+ν −1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 (9. 52)... xy (9. 42) In traditional finite -element notation, this can be written as Sxx Exx Syy = Dm 21 Eyy S E xy xy (9. 43) in which Dm21 1 = E −ν 0 © 2003 by CRC CRC Press LLC −ν 1 0 0 0 1+ ν −1 1 E ν = 1−ν2 0 ν 1 0 0 0 1+ ν (9. 44) 07 49_ Frame_C 09 Page 134 Wednesday, February 19, 2003 5: 09 PM 134 Finite Element Analysis: Thermomechanics. .. setting γ m1 or γ b1 equal to zero vectors, respectively © 2003 by CRC CRC Press LLC 07 49_ Frame_C 09 Page 133 Wednesday, February 19, 2003 5: 09 PM Element Fields in Linear Problems 133 9. 3.3 TWO-DIMENSIONAL ELEMENTS 9. 3.3.1 Membrane Response In two-dimensional elements, several cases can be distinguished We first consider elements in plane stress It is convenient to use Hooke’s Law in the form Exx = 1 [ S... ) (9. 38) Letting s = VEC(S) and e = VEC(EL), the stress-strain relations are written using Kronecker product operators as s = De, D = 2 µI 9 + λii T , (9. 39) and D is the tangent-modulus tensor introduced in the previous chapters 9. 3.2 ONE-DIMENSIONAL MEMBERS For a beam column, recalling the strain-displacement model, S11 = σ( x, z, t ) = E ε = − z Eβ T 1 ( x )Φmb1 γ mb1 (t ) mb (9. 40) The cases of. .. xy xy (9. 49) 9. 3.4 ELEMENT FOR PLATE WITH MEMBRANE AND BENDING RESPONSE Plane stress is also applicable, consequently: Sxx Exx ˆ Syy = Dm 21 Eyy = Dm 21βT 2 ( x, y, z )Φmb 2 γ mb 2 (t ) mb S E xy xy (9. 50) 9. 3.5 AXISYMMETRIC ELEMENT It is sufficient to consider the case in which none of the nodes of the element are located on the axis of revolution... 1+ν 0 0 0 0 0 0 0 0 0 1+ν (9. 53) −1 and for the Principle of Virtual Work, 1 −ν −ν D3 = EJ 6 ′ 0 0 0 © 2003 by CRC CRC Press LLC −1 0 0 0 J6 0 0 1+ν (9. 54) 07 49_ Frame_C 09 Page 137 Wednesday, February 19, 2003 5: 09 PM Element Fields in Linear Problems 1 0 0 J6 = 0 0 0 9. 3.7 ELEMENTS FOR 137 0 0 0 1 0 0 0 0 1 0 0 0 0 2 0 0...07 49_ Frame_C 09 Page 131 Wednesday, February 19, 2003 5: 09 PM Element Fields in Linear Problems 0 1 2 β T1 = a 0 0 131 1 0 0 1 z r 0 0 0 0 0 0 0 1 2 0 0 0 1 0 0 1 2 in which the prime is no longer displayed If element e is now located on the axis of revolution, we obtain Φa 2 T e(r, z, t ) = ba 2 0 β T2 a 9. 2.4 THERMAL ANALOG 1 1 = 0... 2 plane stress (9. 48) plane strain 07 49_ Frame_C 09 Page 135 Wednesday, February 19, 2003 5: 09 PM Element Fields in Linear Problems 135 9. 3.3.2 Two-Dimensional Members: Bending Response Thin plates experiencing bending only are assumed to be in a state of plane stress The tangent-modulus matrix is given in Equation 9. 47, and an approximation for the strain is obtained as Sxx Exx ˆ . 11 1 07 49_ Frame_C 09 Page 137 Wednesday, February 19, 2003 5: 09 PM © 2003 by CRC CRC Press LLC 138 Finite Element Analysis: Thermomechanics of Solids 7. The stress-strain relations in two-dimensional. (,,,)=− ∂ ∂ =− ∂ ∂ 00 07 49_ Frame_C 09 Page 125 Wednesday, February 19, 2003 5: 09 PM © 2003 by CRC CRC Press LLC 126 Finite Element Analysis: Thermomechanics of Solids and w is a function only of x, y, and. = = () () ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ 1 2 1 2 ′ = = εεββΦΦ γγ γγ E E E xx yy xy m T m u v 22 2 2 ˆ 07 49_ Frame_C 09 Page 1 29 Wednesday, February 19, 2003 5: 09 PM © 2003 by CRC CRC Press LLC 130 Finite Element Analysis: Thermomechanics of Solids The prime in e′ is introduced