139 Element and Global Stiffness and Mass Matrices 10.1 APPLICATION OF THE PRINCIPLE OF VIRTUAL WORK Elements of variational calculus were discussed in Chapter 3, and the Principle of Virtual Work was introduced in Chapter 5. Under static conditions, the principle is repeated here as (10.1) As before, δ represents the variational operator. We assume for our purposes that the displacement, the strain, and the stress satisfy representations of the form (10.2) in which E and S are written as one-dimensional arrays in accordance with traditional finite-element notation. For use in the Principle of Virtual Work, we need D ′ , which introduces the factor 2 into the entries corresponding to shear. We suppose that the boundary is decomposed into four segments: S = S I + S II + S III + S IV . On S I , u is prescribed, in which event δ u vanishes. On S II , the traction ττ ττ is prescribed as ττ ττ 0 . On S III , there is an elastic foundation described by ττ ττ = ττ ττ 0 − A ( x ) u , in which A ( x ) is a known matrix function of x . On S IV , there are inertial boundary conditions, by virtue of which ττ ττ = ττ ττ 0 − Bü . The term on the right now becomes (10.3) 1 0 δδρδτ ESdV u udV u dS ij ij i i i i ∫∫∫ += ˙˙ . ux ) x T (()() ()(),t t , t , ,===ϕϕΦΦγγββΦΦγγESE T xD δτ δ δ δ udS dS dS t dS t ii ∫∫ ∫ ∫ = − − ++ γγΦΦϕϕττ γγΦΦϕϕϕϕΦΦγγ γγΦΦϕϕϕϕΦΦγγ TT 0 TT T TT T x xA x xB x () () () () () () ˙˙ (). SS S S S II III IV III IV 0749_Frame_C10 Page 139 Wednesday, February 19, 2003 6:04 PM © 2003 by CRC CRC Press LLC 140 Finite Element Analysis: Thermomechanics of Solids The term on the left in Equation 10.1 becomes (10.4) in which K is called the stiffness matrix and M is called the mass matrix. Canceling the arbitrary variation and bringing terms with unknowns to the left side furnishes the equation as follows: (10.5) Clearly, elastic supports on S III furnish a boundary contribution to the stiffness matrix, while mass on the boundary segment S IV furnishes a contribution to the mass matrix. Sample Problem 1: One element rod Consider a rod with modulus E, mass density ρ , area A , and length L . It is built in at x = 0. At x = L , there is a concentrated mass m to which is attached a spring of stiffness k , as illustrated in Figure 10.1. The stiffness and mass matrices, from the domain, reduce to the scalar values K → EA / L , M → ρ AL / 3, M S → m , K S → k . The governing equation is . Sample Problem 2: Beam element Consider a one-element model of a cantilevered beam to which a solid disk is welded at x = L . Attached at L is a linear spring and a torsional spring, the latter having the property that the moment developed is proportional to the slope of the beam. FIGURE 10.1 Rod with inertial and compliant boundary conditions. δ δρ ESdV udV ij ij ii ∫∫ ∫∫ == ′ == δδγγγγΦΦββββΦΦ δδγγγγρρΦΦϕϕϕϕΦΦ TTT TTT KKxDx MM xx (), ( ) ( ) ˙˙ ˙˙ (), ( ) ( ) , t dV ut dV ()()( ) ˙˙ KK MM f+++ = SS ttγγγγ() fx KxAx MxBx T 0 TT TT = = = + ∫ ∫ ∫ ΦΦϕϕττ ΦΦϕϕϕϕΦΦ ΦΦϕϕϕϕΦΦ () () () () () . dS dS dS S S SS S S II III III IV ()( ) ˙˙ EA L AL kmf++ + = γγ ρ 3 E,A,L,ρ P m k 0749_Frame_C10 Page 140 Wednesday, February 19, 2003 6:04 PM © 2003 by CRC CRC Press LLC Element and Global Stiffness and Mass Matrices 141 The shear force V 0 and the moment M 0 act at L . The interpolation model, incorpo- rating the constraints w (0, t ) = − w ′ (0, t ) = 0 a priori , is (10.6) The stiffness and mass matrices, due to the domain, are readily shown to be (10.7) The stiffness and mass contributions from the boundary conditions are (10.8) The governing equation is now (10.9) 10.2 THERMAL COUNTERPART OF THE PRINCIPLE OF VIRTUAL WORK For our purposes, we focus on the equation of conductive heat transfer as . (10.10) FIGURE 10.2 Beam with translational and rotational inertial and compliant boundary conditions. z y E,I,A,L,ρ M 0 k T k x V 0 m r wxt xx LL LL wLt wLt (,) ( ) (,) (,) .= −− − ′ − 23 23 2 1 23 KM= = EI L L LL AL L LL 3 2 13 35 11 210 11 210 1 105 2 12 6 64 ,. ρ KM S T S mr k k m = = 0 0 0 0 2 2 , . () ˙˙ (,) ˙˙ (,) () (,) (,) .MM KK+ − ′ ++ − ′ = SS wLt wLt wLt wLt V M 0 0 kc t e ∇= ∂ ∂ 2 T T ρ 0749_Frame_C10 Page 141 Wednesday, February 19, 2003 6:04 PM © 2003 by CRC CRC Press LLC 142 Finite Element Analysis: Thermomechanics of Solids Multiplying by the variation of T − T 0 , integrating by parts, and applying the divergence theorem furnishes (10.11) Now suppose that the interpolation models for temperature in the current element furnish a relation of the form . (10.12) The terms on the left in Equation 10.11 can now be written as (10.13) K T and M T can be called the thermal stiffness (or conductance) matrix and thermal mass (or capacitance) matrix, respectively. Suppose that the boundary S has four zones: S = S I + S II + S III + S IV . On S I , the temperature is prescribed as T 1 , from which we conclude that δ T = 0. On S II , the heat flux is prescribed as n T q 1 . On S III , the heat flux satisfies n T q = n T q 1 − h 1 (T − T 0 ), while on S IV , n T q = n T q 1 − h 2 d T/dt. The governing finite-element equation is now (10.14) 10.3 ASSEMBLAGE AND IMPOSITION OF CONSTRAINTS 10.3.1 R ODS Consider the assemblage consisting of two rod elements, denoted as e and e + 1 [see Figure 10.3(a)]. There are three nodes, numbered n, n + 1, and n + 2. We first consider assemblage of the stiffness matrices, based on two principles: (a) the forces at the nodes are in equilibrium, and (b) the displacements at the nodes are continuous. Principle (a) implies that, in the absence of forces applied externally to the node, at node n + 1, the force of element e + 1 on element e is equal to and opposite the force of element e on element e + 1. It is helpful to carefully define global (assemblage δδρδ ∇∇ + ∂ ∂ = ∫∫ ∫ TT nqTT T T TkdV c t dV dS e . TT T −= ∇= =− 0 ϕϕΦΦθθββΦΦθθΦΦθθ TT TT T T ttkt TTT xxqx() (), () (), () () δδ δρ δ ρ ∇∇ → = ∂ ∂ →= ∫∫ ∫∫ T TTTTTT eTTeTTTT kdV t t k dV c t dV t t c dV TT T T θθθθΦΦββββΦΦ θθθθΦΦϕϕϕϕΦΦ TTT TTT KK MM () (), () ˙ (), . [] ˙ () [ ]() ()MM KK f TTS TTS T ttt+++=θθθθ MK fnq TT TS T T TT T TTSTTTT TT T T T hdS hdS tdS == ==++ ∫∫ ∫ ΦΦϕϕϕϕΦΦΦΦϕϕϕϕΦΦ ΦΦϕϕ 21 1 SS II III IV IV III S S S , () , Ω Ω 0749_Frame_C10 Page 142 Wednesday, February 19, 2003 6:04 PM © 2003 by CRC CRC Press LLC Element and Global Stiffness and Mass Matrices 143 level) and local (element level) systems of notation. The global system of forces is shown in (a), while the local system is shown in (b). At the center node, (10.15) since no external load is applied. Clearly, . The elements satisfy (10.16) and in this case, k (e) = k (e+1) = EA/L. These relations can be written as four separate equations: (10.17) FIGURE 10.3 Assembly of rod elements. P n P n+2 P 1 (e+1) P 2 (e+1) P 1 (e) P 2 (e) n n+1 e+1e e e+1 a. forces in global system b. forces in local system PP ee 12 1 0 () ( ) −= + PP P P e n e n21 1 2 () ( ) == + + and k u u P P k u u P P e n n e e e n n e e () () () () () () 11 11 11 11 1 2 1 1 1 2 2 1 1 1 − − = − − − = − + + + + + + ku ku P ku ku P ku ku P ku ku P e n e n e e n e n e e n e n e e n e n e () () () () () () () () () () () () ( ( ( −=− −+ = −=− −+ = + + + + + + + + + + + + 12 11 1 1 1 22 1 1 1 1 21 1 i) ii) iii) iv)( 0749_Frame_C10 Page 143 Wednesday, February 19, 2003 6:04 PM © 2003 by CRC CRC Press LLC 144 Finite Element Analysis: Thermomechanics of Solids Add (ii) and (iii) and apply Equation 10.15 to obtain (10.18) and in matrix form (10.19) The assembled stiffness matrix shown in Equation 10.19 can be visualized as an overlay of two element stiffness matrices, referred to global indices, in which there is an intersection of the overlay. The intersection contains the sum of the lowest entry on the right side of the upper matrix and the highest entry on the left side of the lowest matrix. The overlay structure is depicted in Figure 10.4. FIGURE 10.4 Assembled beam stiffness matrix. ku ku P ku k k u k u ku ku P e n e n e e n ee n e n e n e n e () () () () () ( ) ( ) () () () [] ( ( ( −=− −++ − = −−= + + + + + + + + + + + 12 1 1 1 2 1 1 1 21 1 0 i) ii iii) iv) kk kkk k kk u u u P P ee eee e ee n n n n n () () () () ( ) ( ) () () − −+− − = − ++ ++ + ++ 0 0 0 11 11 1 21 K (1) K (2) K (3) K (4) k 22 +k 11 K (5) K (N–2) K (N–1) K (N) (4) (5) 0749_Frame_C10 Page 144 Wednesday, February 19, 2003 6:04 PM © 2003 by CRC CRC Press LLC Element and Global Stiffness and Mass Matrices 145 Now, the equations of the individual elements are written in the global system as (10.20) The global stiffness matrix (the assembled stiffness matrix K (g) of the two-element member) is the direct sum of the element stiffness matrices: . Generally, K (g) = ∑ e . In this notation, the strain energy in the two elements can be written in the form (10.21) The total strain energy of the two elements is Finally, notice that K (g) is singular: the sum of the rows is the zero vector, as is the sum of the columns. In this form, an attempt to solve the system will give rise to “rigid-body motion.” To illustrate this reasoning, suppose, for simplicity’s sake, that k (e) = k (e+1) , in which case equilibrium requires that P n = P n+2 . If computations were performed with perfect accuracy, the equation would pose no difficulty. How- ever, in performing computations, errors arise. For example, P n is computed as and . Computationally, there is now an unbalanced force, . In the absence of mass, this, in principle, implies infinite accelerations. In the finite-element method, the problem of rigid-body motion can be detected if the output exhibits large deformation. The problem is easily suppressed using constraints. In particular, symmetry implies that u n+1 = 0. Recalling Equation 10.19, we now have , (10.22) in which R is a reaction force that arises to enforce physical symmetry in the presence of numerically generated asymmetry. The equation corresponding to the second equation is useless in predicting the unknowns u n and u n+2 since it introduces the KKK K KK eee e ee () () ( ) ( ) () () ( ) ( ) ˜ , ˜ ˜ , ˜ . →→ = − − =− − ++ ++ 11 11 110 110 000 00 0 01 1 011 kk ee KKK gee() () ( ) ˜˜ =+ +1 ˜ () K e VK VK eeee eeee e T () () () () () () ()() () ˜ ˜ (). = = = ++ ++ 1 2 1 2 11 12 γγγγ γγγγ γγ ΤΤ ΤΤ uu u nn n 1 2 γγγγ Tg K () . ˆ PP nnn =+ ε ˆ ,PP PPP nnnnn++++ =+ == 1111 ε εε n n + − 1 EA L u u P R P n n n n 110 12 1 011 0 21 − −− − = − ++ 0749_Frame_C10 Page 145 Wednesday, February 19, 2003 6:04 PM © 2003 by CRC CRC Press LLC 146 Finite Element Analysis: Thermomechanics of Solids new unknown R. The first and third equations are now rewritten as , (10.23) with the solution . (10.24) To preserve symmetry, it is necessary for u n + u n+1 = 0. However, the sum is computed as (10.25) The reaction force is given by R = −[ ε n + ε n+1 ], and, in this case, it can be considered as a measure of computational error. Note that Equation 10.23 can be obtained from Equation 10.22 by simply “striking out” the second row of the matrices and vectors and the second column of the matrix. The same assembly arguments apply to the inertial forces as to the elastic forces. Omitting the details, the kinetic energies T of the two elements are (10.26) 10.3.2 BEAMS A similar argument applies for beams. The potential energy and stiffness matrix of the e th element can be written as (10.27) EA L u u P P n n n n 10 01 11 = −+ + ++ ε ε uP A L uP A L nn n n =− + = + ++ [] [] εε E , E 21 uu nn nn A L += + () + + 1 1 εε E . T =+ + 1 2 1 ˙ [ ˜˜ ˙ () () ( ) () γγ]]γγ ee ee T MM ˜ / / ˙ { ˙˙ ˙ } () () () () () M ee e e e T = = = ++ m mAl uu u ee e 112 12 1 1 3 12 ρ γγ V www w bb b eee e () () () () () () () {} ee e 11 e 12 e 21 eT 22 e T K K KK KK = = =− ′ − ′ ++ 1 2 11 γγγγ γγ T 0749_Frame_C10 Page 146 Wednesday, February 19, 2003 6:04 PM © 2003 by CRC CRC Press LLC Element and Global Stiffness and Mass Matrices 147 In a two-element beam model analogous to the previous rod model, V (g) = V (e) + V (e+1) , implying that . (10.28) Generally, in a global coordinate system, K (g) = ∑∑ ∑∑ e K (e) . 10.3.3 TWO-DIMENSIONAL ELEMENTS We next consider assembly in 2-D. Consider the model depicted in Figure 10.5 consisting of four rectangular elements, denoted as element e, e + 1, e + 2, and e + 3. The nodes are also numbered in the global system. Locally, the nodes in an element are numbered in a counterclockwise fashion. Suppose there is one degree of freedom per node (e.g., x-displacement) and one corresponding force. FIGURE 10.5 2-D assembly process. K KK 0 KKK K 0K K g 11 e 12 e 21 eT 22 e 11 e 12 e 21 eT 22 e () () () () () ( ) ( ) () () =+ ++ ++ 11 11 e+3,4 e+3,1 e+3,2 e+2,1 e+2,2 e+1,3 e+1,2e+1,1e,2e,1 e,4 e,3 e+2,4 e+1,4 e+2,3 e+3 e+3 e+2 e+2 e e e+1 e+1 e+3,3 789 123 6 5 4 0749_Frame_C10 Page 147 Wednesday, February 19, 2003 6:04 PM © 2003 by CRC CRC Press LLC 148 Finite Element Analysis: Thermomechanics of Solids In the local systems, the force on the center node induces displacements according to (10.29) Globally, (10.30) Adding the forces of the elements on the center node gives (10.31) Taking advantage of the symmetry of the stiffness matrix, this implies that the fifth row of the stiffness matrix is (10.32) Finally, for later use, we consider damping, which generates a stress proportional to the strain rate. In linear problems, it leads to a vector-matrix equation of the form . (10.33) At the element level, the counterpart of the kinetic energy and the strain energy is the Rayleigh Damping Function, D (e) , given by , and the con- sistent damping force on the e th element is (10.34) fkukukuku fkukukuku e e e e e e e e e e e e e e e e e , () () () () () () () () 3311322333 344 14 41 1 11 4 2 1 12 43 1 13 44 1 =+++ =+++ + + + + + + + + ,, ,, ,, ,, ,, ,, ,, ,,ee e, e e e e e e e e e e e e e fkukukuku fkukuk + + + + + + + + + + + + + + + =+++ =++ 14 21 11 2 21 12 2 22 13 2 23 14 2 24 32 21 3 31 2 2 3 32 2 , ,,,,,,,, ,, ,, , () () () () () () ,, , , ,3 3 33 24 3 34 () ()e e e e uku + + + + + uu uu uu uu uuuuuuuu uuuuuuu u uuu eeee eeee eeee ee ,,,, ,,,, ,,,, ,, 11 2 2 3 5 4 6 11 2 12 3 13 4 14 5 21 5 22 4 23 9 24 8 31 6 32 →→→→ →→→→ →→→→ →→ ++++ ++++ ++ uuu uu u ee 5 33 8 34 7++ →→ ,, fkukkukuk ku kk k kuk eee e ee ee e e 5 31 1 32 41 1 242 1 343 1 12 2 4 33 44 1 11 2 22 3 5 3 =++ [] +++ [] ++ + + [] + ++++ ++ + , () , () , () , () , () , () , () , () , () , () ,, () , () , () , () , () , () 421 3 6 24 3 714 2 23 3 813 2 9 ee eeee ku kuk k uku + [] +++ [] + + ++++ κ 5 31 32 41 1 42 1 43 1 12 2 33 44 1 11 2 22 3 T =+ [] + [] { +++ [] } ++++ ++ + kkkkkk kk kk symmetry eeeeee ee e e , () , () , () , () , () , () , () , () , () , () KK MDKf ˙˙ ˙ ()γγγγγγ++=t D () () () () ˙˙ ee T ee D= 1 2 γγγγ f d t () () () () () ˙ ˙ . eeee D= ∂ ∂ = γγ γγD 0749_Frame_C10 Page 148 Wednesday, February 19, 2003 6:04 PM © 2003 by CRC CRC Press LLC [...]... γ 3 Write down the assembled mass and stiffness matrices of the following three -element configuration (using rod elements) The elastic modulus is E, the mass density is ρ, and the cross-sectional area is A © 2003 by CRC CRC Press LLC 0749_Frame_C10 Page 150 Wednesday, February 19, 2003 6:04 PM 150 Finite Element Analysis: Thermomechanics of Solids y L L/2 L 4 Assemble the stiffness coefficients associated... node depends on the displacements (velocities and accelerations) of the nodes of the elements connected at the given node, thus determining the bandwidth 10. 3 EXERCISES 1 The equation of static equilibrium in the presence of body forces, such as gravity, is expressed by ∂Sij ∂X j = − bi Without the body forces, the dynamic Principle of Virtual Work is derived as ∫ ∫ δ Eij Sij dV + δ ui ρ ∂ 2ui dV...0749_Frame_C10 Page 149 Wednesday, February 19, 2003 6:04 PM Element and Global Stiffness and Mass Matrices 149 The Rayleigh Damping Function is additive over the elements Accordingly, if th ˆ D ( e ) is the damping matrix of the e element referred to the global system, the assembled damping matrix is given by D( g ) = ˆ ∑D (e) (10. 35) e It should be evident that the global... following figure, assuming plane-stress elements The modulus is E, and (1) (2) (3) the Poisson’s ratio is ν K , K , and K denote the stiffness matrices of the elements K (3) K (1) n K (2) 5 Suppose that a rod satisfies δ Ψ = 0, in which Ψ is given by Ψ= ∫ L 0 2 1 du Pu( L) dx − E 2 dx Use the interpolation model u( x ) = {1 u( x e , t ) x}Φ u( xe+1, t ) For an element xe < x < xe+1, find... are such that −1 = axe + b, +1 = axe+1 + b 7 Next, regard the nodal-displacement vector as a function of t Find the matrix Me such that •• u( x e , t ) u( xe , t ) fe + Ke Me = , u( xe+1, t ) u( xe+1, t ) fe+1 © 2003 by CRC CRC Press LLC x 0749_Frame_C10 Page 151 Wednesday, February 19, 2003 6:04 PM Element and Global Stiffness and Mass Matrices 151 in which ρ is the... and natural coordinates 8 Show that, for the rod under gravity, a two -element model gives the exact answer at x = 1, as well as a much better approximation to the exact displacement distribution E A ρ g L 9 Apply the method of the previous exercise to consider a stepped rod, as shown in the figure, with each segment modeled as one element Is the displacement at x = 2L still exact? E1 A1 L1 ρ1 g E2 A2... derived as ∫ ∫ δ Eij Sij dV + δ ui ρ ∂ 2ui dV = δ uiτ i dS ∂t 2 ∫ How should the second equation be modified to include body forces? 2 Consider a one-dimensional system described by a sixth-order differential equation: Q d 6q = 0, dx 6 Q a constant Consider an element from xe to xe+1 Using the natural coordinate ξ = −1 when x = xe′ = +1 when x = xe+1, for an interpolation model with the minimum order that . 0749_Frame_C10 Page 139 Wednesday, February 19, 2003 6:04 PM © 2003 by CRC CRC Press LLC 140 Finite Element Analysis: Thermomechanics of Solids The term on the left in Equation 10. 1 becomes (10. 4) in. finite -element equation is now (10. 14) 10. 3 ASSEMBLAGE AND IMPOSITION OF CONSTRAINTS 10. 3.1 R ODS Consider the assemblage consisting of two rod elements, denoted as e and e + 1 [see Figure 10. 3(a)] iv)( 0749_Frame_C10 Page 143 Wednesday, February 19, 2003 6:04 PM © 2003 by CRC CRC Press LLC 144 Finite Element Analysis: Thermomechanics of Solids Add (ii) and (iii) and apply Equation 10. 15 to obtain (10. 18) and