TL Bar Elements: Formulation

Transcription

1 9 TL Bar Elements: Formulation 9 1

2 Chapter 9: TL BAR ELEMENTS: FORMULATION TABLE OF CONTENTS Page 9.1 Introduction The TL Plane Bar Element Plane Bar Kinematics Plane Bar GL Strain Plane Bar Strain Derivatives Plane Bar PK Axial Stress Plane Bar Potential Energy Plane Bar Residual and Internal Force Vectors Plane Bar Stiffness Matrix Plane Bar Implementation *Plane Bar Verification Using Finite Differences *The TL Space Bar Element Space Bar Kinematics Space Bar Strains and Strain Derivatives Space Bar Potential Energy Space Bar Residual And Internal Force Vectors Space Bar Stiffness Matrix Space Bar Implementation *Mallett-Marcal Stiffness Decomposition Notes and Bibliography Exercises

3 9. THE TL PLANE BAR ELEMENT 9.1. Introduction The basic concepts of nonlinear continuum mechanics reviewed in Chapter 7 are applied to the development of finite element equations of a two-node bar element moving in D (plane bar) and 3D (space bar). Their chief use is modeling of truss structures, taut cables, and certain types of edge reinforcements. Both elements are based on the Total Lagrangian (TL) kinematic description. There are two ways to construct TL finite elements: 1. The Standard Formulation (SF). The Core Congruential Formulation (CCF). The standard formulation is easier to describe and is that followed in this Chapter. The second one is more flexible and powerful but more difficult to teach because it proceeds in stages. The CCF will not be covered in this course, but older Chapters explaining it are posted on the web site. 9.. The TL Plane Bar Element This is a prismatic bar element that can be used to model pin-jointed plane truss structures of the type sketched in Figure 9.1. Such structures may undergo large displacements and rotations but their strains are assumed to remain small so that the material stays in the linear elastic range. These assumptions allows us to consider only geometric nonlinear effects. 1 Motion Reference configuration (same as base in TL description) Current configuration Figure 9.1. A plane truss structure undergoing large displacements while the material stays in the linear elastic range. Figure 9. shows a two-node bar element appropriate to model members of such truss structures. The element moves in the (X, Y ) plane. In the reference (base) configuration the element has cross section area A (constant along the element) and length. In the current configuration the cross section area and length become A and L, respectively. The material has an elastic modulus E that links the axial-stress and axial-strain measures defined below. The element has four node 1 An important application in Aerospace is the deployment of space trusses. In Civil Engineering it would be the deployment of geodesic domes. For those applications, however, a three-dimensional version, namely the space bar element covered in 9.3, is more useful. 9 3

4 Chapter 9: TL BAR ELEMENTS: FORMULATION (a) Reference (base) configuration C X (b) (X,Y ) Elastic modulus E, x-sec area A, length, GL strain, PK stress s FEM idealization C 1 (X 1,Y 1) u Y Current configuration C Elastic modulus E, x-sec area A, length L, GL strain e, PK stress s x Y, y u Y1 u X1 f Y1 1 (x 1,y 1) X, x Global system X,Y,Z : material frame x,y,z : spatial frame f X1 C u X f Y (x,y ) f X Figure 9.. TL plane bar lement: (a) bar segment moving in D; (b) FEM idealization. displacements and associated node forces. These quantities are collected in the vectors u X1 x 1 X 1 f X1 u u = Y 1 y = 1 Y 1 f, f = Y 1, (9.1) u X x X f X u Y y Y f Y Because this Chapter deals primarily with the formulation of individual elements, the identification superscript e will be omitted to reduce clutter until assemblies are considered in the next one Plane Bar Kinematics According to bar theory, the element remains straight in any configuration. To describe its plane motion it is sufficient to consider a generic point P of material coordinates X located on the longitudinal axis of the reference configuration C. That point maps to point P of spatial coordinates x in the current configuration C. See Figure 9.3. The C coordinates can be parametrically interpolated from the end nodes as X(ξ) = [ X (ξ) Y (ξ) ] [ 1 = (1 ξ)x (1 + ξ)x ] [ N1 (ξ)x 1 (1 ξ)y (1 + ξ)y = 1 + N (ξ)x N 1 (ξ)y 1 + N (ξ)y ], (9.) with a similar formula for the C coordinates: [ ] [ 1 x(ξ) x(ξ) = = (1 ξ)x (1 + ξ)x ] [ ] N1 (ξ)x y(ξ) 1 (1 ξ)y (1 + ξ)y = 1 + N (ξ)x. (9.3) N 1 (ξ)y 1 + N (ξ)y Here ξ is the usual isoparametric coordinate that varies from 1atnode1to+1 at node, whereas N 1 (ξ) = 1 (1 ξ) and N (ξ) = 1 (1 + ξ) are the well known linear shape functions in terms of ξ. The displacement field is obtained by subtraction: [ ] [ ] u u(ξ) = x(ξ) X(ξ) = X (ξ) N1 (ξ) u = X1 + N (ξ) u X, (9.4) u Y (ξ) N 1 (ξ) u Y 1 + N (ξ) u Y 9 4

5 9. THE TL PLANE BAR ELEMENT Bar longitudinal axis C (X,Y ) X Y, y 1 (X 1,Y 1) P (X) X(ξ) = X(ξ) Y(ξ) = N 1 (ξ) X 1 + N (ξ) X N 1 (ξ) Y 1 + N (ξ) Y X, x u(ξ) = x(ξ) X(ξ) = u (ξ) N (ξ) u + N (ξ) u X 1 X1 X = u Y (ξ) N 1 (ξ) u Y1 + N (ξ) uy 1 (x 1,y 1) x(ξ) = x(ξ) y(ξ) = N 1 (ξ) x 1 + N (ξ) x N 1 (ξ) y 1 + N (ξ) y P(x) C (x,y ) x Figure 9.3. Displacement field interpolation for the plane TL bar element. Equation (9.4) may be expressed in matrix form as u(ξ) = [ ] [ ] u X1 u X (ξ) N1 (ξ) N = (ξ) u Y 1 = N(ξ) u. (9.5) u Y (ξ) N 1 (ξ) N (ξ) u X u Y Here N(ξ) is the 4 shape function matrix. The element kinematics defined by these equations is pictured in Figure Plane Bar GL Strain As discussed in Chapter 8, in the Total Lagrangian (TL) description the Green-Lagrange (GL) strains and the second Piola-Kirchhoff (PK) stresses are the standard conjugate measures in the formulation of the internal energy. The only GL strain that appears in the internal (strain) energy expression is the GL axial strain e 1 e, which is most expediciously defined using the length change as e = L L L, (9.6) rather than through displacement gradients (see next Remark). Because of the linear displacement assumptions (9.5) the strain e is constant over the element. Expression (9.6) can be maneuvered into a quadratic matrix function of the node displacements. Lateral strains due to Poisson s ratio ν do appear, but do not contribute to the strain energy because the associated PK stresses are zero in classical bar theory. 9 5

6 Chapter 9: TL BAR ELEMENTS: FORMULATION X 1 (X,Y ) X _ ψ Y 1 C X X 1 c = = cos ψ Y 1 c Y = = sin ψ Y, y 1 (X 1,Y 1) x 1 L c x = = cos ψ y L c 1 y = = sin ψ X, x 1 (x = X +u, y =Y +u ) 1 1 X1 1 1 Y1 x 1= X 1+ ux1 C L ψ y =Y + u 1 1 Y1 (x = X +u, y =Y + u ) X x _ Y Figure 9.4. Geometric interpretation of quantities used in the study of element kinematics. To expedite the derivation it is convenient to introduce the following auxiliary variables: X 1 = X X 1, Y 1 = Y Y 1, u X1 = u X u X1, u Y 1 = u Y u Y 1, c X = X 1 = cos ψ, c Y = Y 1 = sin ψ, c x = c X + u X1 = x 1 = L cos ψ, c y = c Y + u Y 1 = y 1 = L sin ψ. (9.7) These quantities can be geometrically interpreted as illustrated in Figure 9.4. In particular, c X and c Y are the {X, Y } direction cosines of the bar longitudinal axis in the reference configuration. On the other hand, c x and c y are the scaled (by L/ ) direction cosines of the current configuration. The squared bar length expressions in e in terms of nodal displacements are L = (X 1 + u X1 ) + (Y 1 + u Y 1 ), L L = X 1 u X1 + Y 1 u Y 1 + X 1 + Y 1, (9.8) whence e = L L L = 1 (c X u X1 + c Y u Y 1 ) + 1 ( u L X1 + u ) Y 1. (9.9) This may be compactly expressed in matrix form as e = B u + 1 ut Hu, (9.1) in which B = 1 [ c X c Y c X c Y ], H = 1 L 9 6 P = 1 L (9.11)

7 9. THE TL PLANE BAR ELEMENT The expression (9.1) shows that the GL strain splits naturally as e = e L + e N, in which e L = B u, where B is the 1 4 rectangular matrix given by (9.11), depends linearly on the node displacements u if C is viewed as frozen. This is called the linear part of the GL strain. e N = 1 ut Hu, where H is the 4 4 symmetric matrix given by (9.11), depends quadratically on the node displacements if C is viewed as frozen. This is called the nonlinear part of the GL strain. For use below it is convenient to introduce the matrix B = 1 [ c x c y c x c y ] = B + u T H, (9.1) which is similar to B but with direction cosines evaluated in the current configuration. 3 Remark 9.1. The GL strain measure (9.6) can be obtained, though more labouriosly, by using the gradientbased formulation of Chapter 8. The procedure is worked out in Examples 7.7 and 8.3 of previous Chapters. The result (9.6) emerges in (8.8). Caution must be exercised, however, in how stretching and rigid rotation are applied in sequence. If done in the wrong order, the GL axial strain becomes coupled to lateral stretches and rotation angle, which is physically absurd since it violates the principle of observer invariance. Remark 9.. The symmetric numerical matrix P = L H of (9.11) satisfies P = P, P 3 = 4P, etc. Its eigenvalues are {,,, }. If the 1 s are replaced by 1, P becomes a projection matrix: ( 1 P) = 1 P. The latter property is useful in the corotational kinematic description of this element Plane Bar Strain Derivatives For development of the internal force vector and tangent stiffness matrix, the first and second derivatives of e with respect to the state vector u (the nodal displacements) will be needed. 4 The first derivative is e u = (B u + 1 ut Hu) u = B T + HT u = B T + Hu= BT, (9.13) in which use is made of (9.1). The second derivative of e with respect to u is e u u = e u u = (BT + Hu) u = BT u = H. (9.14) Remark 9.3. The foregoing formulas may be also be obtained using the chain rule with L as intermediate variable. For example, e u = e L L u = L L L [ cos ψ sin ψ cos ψ sin ψ ]T = 1 [ c x c y c x c y ] T = B T, (9.15) and likewise for the second derivative. This technique, which can be readily generalized to other strain measures e = f (L, ), is elaborated in Chapter The relation B B = u T H may be checked by direct computation. Like B, B isa1 4 row matrix, not a vector. 4 If you are rusty in matrix calculus as regards taking derivatives with respect to a vector, please skim

8 Chapter 9: TL BAR ELEMENTS: FORMULATION Plane Bar PK Axial Stress The stress measure conjugate to GL strains is the second Piola-Kirchhoff (PK) stress tensor. The only component that appears in the internal energy is the axial stress s. Assuming that the strain e in the reference configuration vanishes, s is related to e through the constitutive equation s = s + Ee, (9.16) where s is the PK axial stress in C, which is constant over the element, and E is the elastic modulus. The axial force based on this stress is F = A s. (9.17) Note that this is not the true axial force in the current configuration C, which would be F true = A σ, (9.18) in which σ denotes the true or Cauchy stress in C and A is the actual cross-section area there. The connection between σ and s is worked out in Exercise 9.3 for a bar of isotropic material. Since s and E are fixed, the stress derivatives with respect to the state vector are s u = E e u = E BT, s u u = E e u u = E H. (9.19) Plane Bar Potential Energy We consider only element node forces f that are conservative and proportional. Consequently f = λq. (9.) Here q denotes the incremental load vector. The Total Potential Energy (TPE) of the element in the current configuration, expressed in terms of GL strains and PK stresses, is = U W = (s e + 1 Ee ) dv f T u = A (s e + 1 Ee ) d X λq T u, (9.1) V in which X is directed along the bar longitudinal axis in C, as shown in Figure 9.4. In accordance with the TL description, all integrals are carried out over the reference (base) configuration C. Since the integrands of (9.1) are constant over the element, we get = U W, U = V (s e + 1 Ee ), W = λ q T u. (9.) Here V = A is the bar volume in C. This expression is separable because the internal energy U depends only on u through e, and not on λ. 9 8

9 9. THE TL PLANE BAR ELEMENT f = F c X1 x 1 f = F c Y1 C y F f = F c Y y f = F c X x Figure 9.5. Geometrical interpretation of the internal force vector (9.3). The axial force F = A s would be positive as shown Plane Bar Residual and Internal Force Vectors The residual force vector is r = / u. Since is separable, r = p f, where f = W/ u = λq is the external force. The internal force can be expanded as p = U ( ) u = V e s u + Ee e = V s e u u. (9.3) Using (9.13) we arrive at p = V s B T = A s B T = F B T = F [ c x c y c x c y ] T. (9.4) Here F = A s is the current configuration axial force measured per unit area of the reference configuration. 5 The interpretation of c x and c y is depicted in Figure 9.4 and that of F in Figure 9.5. The relation between F = A s and the true axial force F true = A σ is worked out in an Exercise Plane Bar Stiffness Matrix Since p does not depend on λ, the tangent stiffness matrix is obtained simply by differentiating the internal force (9.3) with respect to the state vector: K = p u = ( V s e ) = V E e u u u e u + V s e u u = K M + K G, (9.5) in which denotes the dyadic (outer) product of two vectors. This indicates that K splits naturally into two parts: K M and K G, which are called the material stiffness matrix and geometric stiffness matrix, respectively, in the FEM literature. For the material stiffness we get K M = V E e u e u = EA B T B = EA c x c x c y c x c x c y c x c y c y c x c y c y c x c x c y c x c x c y c x c y c y c x c y c y (9.6) Here we have replaced e/ u = B T from the first of (9.19), and applied the dyadic product formula recalled in (7.46). The matrix expression in (9.6) looks formally similar to the stiffness matrix of. 5 This is the PK axial force; cf. (9.17). 9 9

10 Chapter 9: TL BAR ELEMENTS: FORMULATION a linear bar element, 6 except that B is now a function of u. The dependence of K M on material properties (here just the elastic modulus E) explains the name material stiffness, which has become standard. For the geometric stiffness, use (9.14) to get 1 1 K G = V s H = F H = F P = F 1 1. (9.7) This component of K depends only on the PK stress state in the current configuration, since F = A s. No material properties appear. Thus the name geometric stiffness applied to K G. 7 Remark 9.4. Assuming that E, A and are nonzero, the rank of K M is obviously one since B is a 1 4 matrix. On the other hand, the rank of the numerical matrix P is since its eigenvalues are,,,. Consequently K G has rank if s is nonzero and otherwise. The rank of K = K M + K G is 1 if the current configuration is unstressed and otherwise, as shown in Exercise 9.4. That is, the rank deficiency of K is 3 and, respectively. Implications in the context of stability analysis are considered in later Chapters. Remark 9.5. The addition of K G increases the bar stiffness if the current configuration is in tension (s > ), but it reduces it if the current configuration is in compression (s < ). This is in accord with physical intuition. The main effect of this stiffness is on the rotational rigid-body motions of the bar about the Z axis Plane Bar Implementation The Mathematica modules TLPlaneBarIntForce and TLPlaneBarTanStiff form the internal force vector and the tangent stiffness matrix, respectively, of a TL plane bar element. They are listed in Figure 9.6. Both modules are invoked by the same calling sequence: pe=tlplanebarintforce[enodxy,em,a,s,edisxy,numer]; (9.8) { KeM,KeG }=TLPlaneBarTanStiff[enodXY,Em,A,s,edisXY,numer]; (9.9) The arguments are enodxy List of X, Y coordinates of element end nodes in reference configuration, configured as {{X 1,Y 1 },{ X,Y }} Em Bar elastic modulus E. Assumed to stay constant during motion. A Bar cross section area A in reference configuration. s PK axial stress s in reference configuration. edisxy List of X, Y displacements of element end nodes from reference to current configuration, configured as {{u X1, u Y 1 },{ u X, u Y }}, numer A logical flag: True to request numerical floating point work. Else set to False. The module TLPlaneBarIntForce returns one function value 6 To which it reduces if u =. In the linear case c x and c y become the cosine and sine, respective, of the orientation angle ψ in the reference configuration. That angle is shown in in Figure As described under Notes and Bibliography, the confusing name initial stress stiffness was often used for K G in pre-197 FEM publications. (Confusing because K G depends on the current stress, rather than the initial stress, unless it is evaluated at C.) It is pointed out in that K M was presented as a two-matrix combination by some authors. 9 1

11 9. THE TL PLANE BAR ELEMENT TLPlaneBarIntForce[enodXY_,Em_,A_,s_,edisXY_,numer_]:= Module[{X1,Y1,X,Y,x1,y1,x,y,X1,Y1, x1,y1,ll,l,ll,l,e,f=s*a,f,pe}, {{X1,Y1},{X,Y}}=enodXY; {X1,Y1}={X-X1,Y-Y1}; LL=X1^+Y1^; L=Sqrt[LL]; {{x1,y1},{x,y}}=enodxy+edisxy; {x1,y1}={x-x1,y-y1}; LL=x1^+y1^; L=Sqrt[LL]; e=(ll-ll)/(*ll); If [!numer, {L,L}=PowerExpand[{L,L}]]; F=F+Em*A*e; pe=f*{-x1,-y1,x1,y1}/l; If [numer,pe=n[pe],pe=simplify[pe]]; Return[pe]]; TLPlaneBarTanStiff[enodXY_,Em_,A_,s_,edisXY_,numer_]:= Module[{X1,Y1,X,Y,x1,y1,x,y,X1,Y1,x1,y1, LL,L,LL,L,e,B,BB,P,F=s*A,F,KeM,KeG}, {{X1,Y1},{X,Y}}=enodXY; {X1,Y1}={X-X1,Y-Y1}; LL=X1^+Y1^; L=Sqrt[LL]; {{x1,y1},{x,y}}=enodxy+edisxy; {x1,y1}={x-x1,y-y1}; LL=x1^+y1^; L=Sqrt[LL]; If [!numer, {L,L}=PowerExpand[{L,L}]]; B={-x1,-y1,x1,y1}; BB=Outer[Times,B,B]/LL; P={{1,,-1,},{,1,,-1},{-1,,1,},{,-1,,1}}; e=(ll-ll)/(*ll); F=F+Em*A*e; KeM=(Em*A/L)*BB; KeG=(F/L)*P; If [numer,{kem,keg}=n[{kem,keg}],{kem,keg}=simplify[{kem,keg}]]; Return[{KeM,KeG}]]; Figure 9.6. Implementation of the internal force and tangent stiffness of the TL plane bar element. pe Element internal forces returned as the 4-vector { p X1, p Y 1, p X, p Y }. The module TLPlaneBarTanStiff returns the material and geometric components as two separate matrices: { KeM,KeG } KeM is the 4 4 element material stiffness matrix whereas KeG is the 4 4 element geometric stiffness matrix. The tangent stiffness is the sum of those two. The separate return of K M and K G is useful in stability analysis. Example 9.1. This example pertains to the plane bar element pictured in Figure 9.7. The unstretched bar is taken as reference configuration geometry as defined by the position coordinates (X 1 =, Y 1 = ) and (X = 6, Y = ) of the end nodes. (The rotated frame { X, Ȳ } shown in red is used in the next example.) The reference bar length is = (6 ) + ( ) = 4. Other data includes: elastic modulus E = 16, cross-section area A = 6, reference (initial) PK stress s = 5, and reference GL strain e =. The node displacements are (u X1, u Y 1 ) = (, 1) and (u X, u Y ) = (1, 5). As a result, the end nodes in the current configuration C are located at (x 1 = + = 4, y 1 = 1 = 1) and (x = = 7, y = 5 = 3). Compute the internal force and tangent stiffness matrix in C. The current length is L = (7 4) + ( 3 1) = 5. The current GL strain is e = (L L )/(L ) = (5 16)/3 = 9/3 = 8.15%, whence the PK stress is s = s + Ee= 7. The scaled direction cosines in C are c x = x 1 / = 3/4 and c y = y 1 / = 4/4 = 1. The matrices required for the internal force and tangent stiffness computations are B = 1 4 [ 3/4 1 3/4 1 ], H = 1 16 P = (9.3) 1 1

12 Chapter 9: TL BAR ELEMENTS: FORMULATION Unstretched configuration with PK stress s = 5, GL strain e =. This is C in Example 9.1. Not used in Examples X-sec area A = 6, length L = 4, elastic modulus E = 16 Y, y 1 (,) Y, y X, x φ= X, x u = 1 Y1 u = X1 1 (4,1) (1.6,3.8) (6,) u = 5 Y X-sec area A, length L =5. For isochoric material, A=4/5=4.8 Stretched configuration with PK stress s=7, GL strain e=9/3. This is C in Example 9.1, C in Examples u =1 X (7, 3) (6.6,3.8) Figure 9.7. TL plane bar element example, with data as shown. ClearAll[Em,A,s,numer]; enodxy={{1,},{5,}}; Em=16; A=6; s=5; edisxy={{,-1},{1,-5}}; numer=false; pe= TLPlaneBarIntForce[enodXY,Em,A,s,edisXY,numer]; {KeM,KeG}=TLPlaneBarTanStiff[enodXY,Em,A,s,edisXY,numer]; Ke=KeM+KeG; Print["pe=",pe//MatrixForm, " KeM=",KeM//MatrixForm, "\nkeg=",keg//matrixform, " Ke=",Ke//MatrixForm, "\neigenvalues of KeM=",Chop[Eigenvalues[N[KeM]]], "\neigenvalues of KeG=",Chop[Eigenvalues[N[KeG]]], "\neigenvalues of Ke=", Chop[Eigenvalues[N[Ke]]]]; pe= KeG= KeM= Ke= Eigenvalues of KeM={75.,,, } Eigenvalues of KeG={1., 1.,, } Eigenvalues of Ke= {96., 1.,, } Figure 9.8. Verification of Example 9.1 using the Mathematica modules of Figure 9.6. Replacing these in equations (9.3) through (9.7) yields p = 315, K M = , K G = 15 15, K = K M +K G = (9.31) 9 1

13 9. THE TL PLANE BAR ELEMENT The foregoing results are verified with the script of Figure 9.8, which calls the Mathematica modules of Figure 9.6. The script computes the eigenvalues of K M, K G and K, which shows their ranks to be 1, and, respectively. Note that the eigenvalues of K are the sum of those of K M and K G. Cf. Exercise 9.4. Example 9.. To facilitate linkage to the next Example, the stretched bar in Figure 9.7 is now anointed as the reference configuration C. (Current-to-reference resettings of this kind are commonly used in the Updated Lagrangian (UL) description.) Find the tangent stiffness and internal force vector assuming zero displacements there. The reference length is now = 5; the section area is reset to A = 4/5 = 4.8, which assumes a constant bar volume (isochoric material), and the PK stress is s = 7. Calling the modules with enodxy={{4,1 },{ 7,-3 }}, edisxy={{, },{, }}, Em=16, A=4/5, and s=7 gives p = 1.6, K M = , K G = , K = K M +K G = (9.3) Comparing to (9.31) a reduction of about half in the tangent stiffness can be observed. The reason is primarily geometric: the squared lengths ratio is L /L = 16/5 = 64% whereas the area ratio is 4.8/6 = 8%. Dramatic overstiffness in extension is typical of GL-strain-based formulations that undergo large deformations. Example 9.3. Consider again the example of Figure 9.7. The global axes are rotated by φ = For this angle, tan φ = 4/3, sin φ = 4/5 and cos φ = 3/5, whence X x becomes parallel to the stretched bar. The red-colored rotated axes are X x and Ȳ ȳ. Coordinates transform as X = (3X 4Y )/5 and Ȳ = (4X + 3Y )/5. Transformed node coordinates of the stretched bar are shown in red. As in Example 9., this is declared the reference configuration with zero displacements. Calling the modules with enodxy={{8/5,19/5 },{ 33/5,19/5 }}, edisxy={{, },{, }}, Em=16, A=4/5, and s=7 gives p = 336, K M = , K G = , K = K M + K G = 8 8, (9.33) in which matrix sparseness reflects the X-to-element alignment. Next, introduce the node displacement transformation matrix and apply it to (9.33) as cos φ sin φ sin φ cos φ T = cos φ sin φ sin φ cos φ = , (9.34) p = T T p, K M = T T K M T, K M = T T K M T, K = T T KT. (9.35) 9 13

14 Chapter 9: TL BAR ELEMENTS: FORMULATION TLPlaneBarIntEnergy[enodXY_,Em_,A_,s_,edisXY_,numer_]:= Module[{X1,Y1,X,Y,x1,y1,x,y,X1,Y1,x1,y1,LL,L,LL,L,e, F=s*A,Ue}, {{X1,Y1},{X,Y}}=enodXY; {X1,Y1}={X-X1,Y-Y1}; LL=X1^+Y1^; L=Sqrt[LL]; {{x1,y1},{x,y}}=enodxy+edisxy; {x1,y1}={x-x1,y-y1}; LL=x1^+y1^; L=Sqrt[LL]; e=(ll-ll)/(*ll); Ue=A*L*(s*e+Em*e^/); If [!numer, Ue=Simplify[Ue]]; Return[Ue]]; enodxy={{1,},{5,}}; Em=16; A=6; s=5; edisxy={{,-1},{1,-5}}; (* Check internal force *) pe=tlplanebarintforce[enodxy,em,a,s,edisxy,true]; pecfd=table[,{4}]; delta=.1; For [i=1,i<=4,i++, dv={,,,}; dv[[i]]=delta; dv=partition[dv,]; edisxyp=edisxy+dv; edisxym=edisxy-dv; Uep=TLPlaneBarIntEnergy[enodXY,Em,A,s,edisXYp,True]; Uem=TLPlaneBarIntEnergy[enodXY,Em,A,s,edisXYm,True]; pecfd[[i]]=(uep-uem)/(*delta) ]; Print["pe=",pe//MatrixForm," pe by CFD=",peCFD//MatrixForm]; (* Check tangent stiffness *) {KeM,KeG}=TLPlaneBarTanStiff[enodXY,Em,A,s,edisXY,True]; Ke=KeM+KeG; KeCFD=Table[,{4},{4}]; delta=.1; For [i=1,i<=4,i++, dv={,,,}; dv[[i]]=delta; dv=partition[dv,]; edisxyp=edisxy+dv; edisxym=edisxy-dv; pep=tlplanebarintforce[enodxy,em,a,s,edisxyp,true]; pem=tlplanebarintforce[enodxy,em,a,s,edisxym,true]; KeCFD[[i]]=(pep-pem)/(*delta) ]; Print["Ke=",Ke//MatrixForm," Ke by CFD=",KeCFD//MatrixForm]; pe= pe by CFD= Ke= Ke by CFD= Figure 9.9. Numerical verification of plane bar internal force vector and tangent stiffness matrix modules using central finite differences (CFD). This reproduces exactly the results (9.3). Note that K G is invariant. To pass from (9.3) to (9.33) use the inverse transformation by replacing T by T T and T T by T. It is reassuring that some rules learned in linear FEM can be reused in nonlinear analysis. This is especially useful for D elements moving in 3D *Plane Bar Verification Using Finite Differences For a conservative and separable system, the internal force vector can be derived from the internal energy by differentiation. Likewise the tangent stiffness matrix can be obtained by differentiating the internal force vector. Replacing analytical derivatives by finite differences provides a simple way to gain confidence on element implementation. The procedure is illustrated in Figure 9.9 for checking the modules of Figure 9.6. This verification uses central finite difference (CFD) approximations of analytical derivatives. To test the internal force vector an internal energy module called TLPlaneBarIntEnergy is implemented. It is shown at the top of Figure 9.9. The script of Figure 9.9 systematically perturbs the nodal displacements supplied in edisxy by adding and subtracting a small number δ to each component in turn. This results in two calls to TLPlaneBarIntEnergy 9 14

15 9.3 *THE TL SPACE BAR ELEMENT for each internal force vector component, and two calls to TLPlaneBarIntForce for each stiffness matrix row. The difference between these plus-minus incremented values is divided by δ, stored in a CFD vector or matrix, and compared to the analytical output. The input data of Example 9.1 is used in the script. The output is shown at the bottom of Figure 9.9. It can be observed that an increment of δ =.1 provides satisfactory numerical agreement. (In practice one can play with this value until the comparison is acceptable.) The process can be repeated by setting up other input data sets. It is also feasible to keep certain inputs, such as Em and A in this case, as symbolic variables, but that may clutter the output. This kind of verification is easy to implement. As in all tests of this nature, it can detect the presence of coding errors by taking potshots at the input data domain. But it cannot certify their absence. 8 Such certification would require a symbolic test and passage to the limit. However, this would typically result in highly unwieldy expressions, even for simple elements like this one. So in practice it is only applied using numerical data *The TL Space Bar Element Extension of previous section results to three dimensions to formulate a space bar element is straightforward. Consequently the ensuing formulation can proceed fairly quickly the main changes being the inclusion of state and force components in the third dimension Space Bar Kinematics A bar segment moves in 3D space as pictured in Figure 9.1(a). The FEM idealization as two node line element is shown in Figure 9.1(b). Element configurations are referred to a RCC global frame. Following the usual TL convention, the material coordinates {X, Y, Z} and the spatial coordinates {x, y, z} of coincide in the global frame. Element nodes have coordinates (X i, Y i, Z i ), i = 1,, in the reference configuration C and (x i, y i, z i ), i = 1,, in the current configuration C. The element has six node displacements and associated node forces. These quantities are collected in the 6-vectors u = u X1 u Y 1 u Z1 u X u Y u Z = x 1 X 1 y 1 Y 1 z 1 Z 1 x X y Y z Z, f = f X1 f Y 1 f Z1 f X f Y f Z The squared bar lengths in the reference and current configurations are (9.36) L = (X X 1 ) + (Y Y 1 ) + (Z Z 1 ) = X1 + Y 1 + Z 1, L = (x x 1 ) + (y y 1 ) + (z z 1 ) = x1 + y 1 + z 1 = (X 1 + u X1 ) + (Y 1 + u Y 1 ) + (Z 1 + u Z1 ). (9.37) Here coordinate differences are abbreviated as usual as X 1 = X X 1, x 1 = x x 1, etc., while displacement differences are abbreviated as u X1 = u X u X1, u x1 = u x u x1, etc. As in the 8 Testing can only show the presence of errors, but not their absence. (E. W. Dijkstra). 9 15

16 Chapter 9: TL BAR ELEMENTS: FORMULATION (a) Current configuration C x (b) C u X u Z (x,y,z ) Elastic modulus E, x-sec area A, length L, GL strain e, PK stress s FEM idealization u X1 u Z1 1 (x 1,y 1,z 1) u Y Elastic modulus E, x-sec area A, length, GL strain, PK stress s Y, y u Y1 1 (X 1,Y 1,Z 1) Reference (base) configuration C X C X, x Z, z Global system X,Y,Z : material frame x,y,z : spatial frame (X,Y,Z ) Figure 9.1. TL space bar element: (a) bar segment moving in 3D space; (b) FEM idealization. Unlike Figure 9., node forces are not pictured to reduce visual clutter. case of the plane bar, it is convenient to introduce c X = X 1, c Y = Y 1, c Y = Z 1 c x = c X + u X1 = x 1, c y = c Y + u Y 1 = y 1, c z = c Z + u Z1 = z 1. (9.38) The geometric interpretation is similar to that for the plane bar. c X, c Y and c Z are the direction cosines of the reference configuration bar axis X with respect to X, Y, and Z, respectively. On the other hand, (L/ ) c x, (L/ ) c y and (L/ ) c z are direction cosines of the current configuration bar axis x with respect to x, y and z, respectively, which happen to coincide with X, Y, and Z. The main difference with (9.7) is that these coefficients are no longer expressible in terms of a single angle, since three orientation angles (for example, the Euler angles) are needed in 3D Space Bar Strains and Strain Derivatives The GL axial strain is still defined by (9.6) and is constant along the element. The lateral strains, while generally nonzero, do not contribute to the internal energy. The squared-bar-length expressions appearing in e in terms of nodal displacements are L = X 1 + Y 1 + Z 1, L = (X 1 + u X1 ) + (Y 1 + u Y 1 ) + (Z 1 + u Z1 ), L L = X 1 u X1 + Y 1 u Y 1 + Z 1 u Z1. (9.39) Using (9.38) the GL strain (9.6) can be expressed as e = 1 (c X u X1 + c Y u Y 1 + c Z u Z1 ) + 1 ( u L X1 + u Y 1 + ) u Z1. (9.4) 9 16

17 9.3 *THE TL SPACE BAR ELEMENT This can be placed in matrix form as e = B u + 1 ut Hu, (9.41) in which B = 1 [ c X c Y c Z c X c Y c Z ], H = 1 L P = 1 (9.4) 1 1 L The numerical matrix P = L H has properties similar to those of the eponymous matrix for the 1 plane bar, which are noted in Remark 9.. In particular, P is a projector. For use below we introduce B = 1 [ c L x c y c z c x c y c z ] = B + u T H, (9.43) This is analogous to B but with direction cosines evaluated in the current configuration. The matrix forms of the first- and second-order derivatives of e with respect to the state vector are given by (9.13) and (9.14), which are repeated for convenience: e u = (B u + 1 ut Hu) u e u u = e u u = (BT + Hu) u = B T + HT u = B T + Hu= BT, = BT u = H. (9.44) Space Bar Potential Energy The PK axial stress s is related to e through the constitutive equation (9.16). The axial force associated with this stress is F = A s. The stress derivatives with respect to the state vector are given by (9.19). We assume that the element is subjected to node forces f that are conservative and proportional. Consequently the external force is f = λ q, in which q denotes the (constant) incremental load vector. The associated work is W = f T u = λ q T u. The Total Potential Energy (TPE) of the element in the current configuration, expressed in terms of GL strains and PK stresses, is = U W = (s e + 1 Ee ) dv f T u = A (s e + 1 Ee ) d X λ q T u. (9.45) V All integrals are carried out over the reference (=base) configuration C. Since the integrand is constant over the element, we get = U W, U = V (s e + 1 Ee ), W = λ q T u. (9.46) in which V = A is the bar volume in C. This energy expression is separable because the internal energy U depends only on u through e, and not on λ. 9 17

18 Chapter 9: TL BAR ELEMENTS: FORMULATION Space Bar Residual And Internal Force Vectors The finite element residual equations are obtained by taking the gradient of (9.46) with respect to u. Since is separable, r = p f, in which f = W/ u = λq is the external force. The internal force can be expanded as p = U ( ) u = V e s u + Ee e = V s e u u. (9.47) Using (9.44) we arrive at the matrix expression p = V s B T = F B T = F [ c x c y c x c y ] T. (9.48) Here F = A s is the axial force in the current configuration measured per unit area of the reference configuration Space Bar Stiffness Matrix Because the residual equations are separable, the tangent stiffness matrix is obtained simply by differentiating the internal force p given by (9.47) with respect to the state vector: u: K = ( V s e ) = V E e u u u e u + V s e u u = K M + K G, (9.49) As for the plane bar, K splits naturally into the material and geometric stiffness matricesk M and K G, respectively. For the material stiffness, expanding the dyadic product we get K M = EA B T B = EA c x c x c y c x c z c x c x c y c x c z c x c y c y c y c z c x c y c y c y c z c x c z c y c z c z c x c z c y c z c z c x c x c y c x c z c x c x c y c x c z c x c y c y c y c z c x c y c y c y c z c x c z c y c z cz c x c z c y c z cz. (9.5) This differs from (9.5) in the dimension of the matrix of direction cosine products. The rank is still 1. For the geometric stiffness, use (9.14) to get K G = V s H = F H = F (9.51) This differs from the plane bar K G given in (9.14) in the numerical matrix P = L H, which is now 6 6 and has rank

19 9.3 *THE TL SPACE BAR ELEMENT TLSpaceBarIntForce[enodXYZ_,Em_,A_,s_,edisXYZ_,numer_]:= Module[{X1,Y1,Z1,X,Y,Z,x1,y1,z1,x,y,z,X1,Y1,Z1, x1,y1,z1,ll,l,ll,l,e,f=s*a,f,pe}, {{X1,Y1,Z1},{X,Y,Z}}=enodXYZ; {X1,Y1,Z1}={X-X1,Y-Y1,Z-Z1}; LL=X1^+Y1^+Z1^; L=Sqrt[LL]; {{x1,y1,z1},{x,y,z}}=enodxyz+edisxyz; {x1,y1,z1}={x-x1,y-y1,z-z1}; LL=x1^+y1^+z1^; L=Sqrt[LL]; If [!numer, {L,L}=PowerExpand[{L,L}]]; e=(ll-ll)/(*ll); F=F+Em*A*e; pe=f*{-x1,-y1,-z1,x1,y1,z1}/l; If [numer,pe=n[pe],pe=simplify[pe]]; Return[pe]]; TLSpaceBarTanStiff[enodXYZ_,Em_,A_,s_,edisXYZ_,numer_]:= Module[{X1,Y1,Z1,X,Y,Z,x1,y1,z1,x,y,z,X1,Y1,Z1, x1,y1,z1,ll,l,ll,l,e,b,bb,p,f=s*a,f,kem,keg}, {{X1,Y1,Z1},{X,Y,Z}}=enodXYZ; {X1,Y1,Z1}={X-X1,Y-Y1,Z-Z1}; LL=X1^+Y1^+Z1^; L=Sqrt[LL]; {{x1,y1,z1},{x,y,z}}=enodxyz+edisxyz; {x1,y1,z1}={x-x1,y-y1,z-z1}; LL=x1^+y1^+z1^; L=Sqrt[LL]; If [!numer, {L,L}=PowerExpand[{L,L}]]; B={-x1,-y1,-z1,x1,y1,z1}; BB=Outer[Times,B,B]/LL; P={{1,,,-1,,},{,1,,,-1,},{,,1,,,-1}, {-1,,,1,,},{,-1,,,1,},{,,-1,,,1}}; e=(ll-ll)/(*ll); F=F+Em*A*e; KeM=(Em*A/L)*BB; KeG=(F/L)*P; If [numer,{kem,keg}=n[{kem,keg}],{kem,keg}=simplify[{kem,keg}]]; Return[{KeM,KeG}]]; Figure Implementation of the internal force and tangent stiffness of the TL space bar element Space Bar Implementation The Mathematica modules TLSpaceBarIntForce and TLSpaceBarTanStiff form the internal force vector and the tangent stiffness matrix, respectively, of a TL space bar element. They are listed in Figure Both modules are invoked by the same calling sequence: pe=tlspacebarintforce[enodxyz,em,a,s,edisxyz,numer]; (9.5) { KeM,KeG }=TLSpaceBarTanStiff[enodXYZ,Em,A,s,edisXYZ,numer]; (9.53) The arguments are enodxyz List of X, Y, Z coordinates of element end nodes in reference configuration, configured as {{X 1,Y 1,Z 1 },{ X,Y,Z }} Em Bar elastic modulus E. Assumed to stay constant during motion. A Bar cross section area A in reference configuration. s PK axial stress s in reference configuration. edisxyz List of X, Y displacements of element end nodes from reference to current configuration, configured as {{u X1, u Y 1, u Z1 },{ u X, u Y, u Z }}, numer A logical flag: True to request numerical floating point work. Else set to False. The module TLSpaceBarIntForce returns one function value pe Element internal forces returned as the 6-vector { p X1, p Y 1, p Z1, p X, p Y, p Z }. 9 19

20 Chapter 9: TL BAR ELEMENTS: FORMULATION The module TLSpaceBarTanStiff returns the material and geometric components as two separate matrices: { KeM,KeG } KeM is the 6 6 element material stiffness matrix whereas KeG is the 6 6 element geometric stiffness matrix. The tangent stiffness is the sum of those two. The separate return of K M and K G is useful in stability analysis. ClearAll[Em,A,s,numer]; enodxyz={{3,,-6},{5,7,8}}; Em=45; A=15; s=-8; edisxyz={{3,1,4},{3,4,6}}; numer=false; pe= TLSpaceBarIntForce[enodXYZ,Em,A,s,edisXYZ,numer]; {KeM,KeG}=TLSpaceBarTanStiff[enodXYZ,Em,A,s,edisXYZ,numer]; Ke=KeM+KeG; Print["pe=",pe//MatrixForm, " KeM=",KeM//MatrixForm, "\nkeg=",keg//matrixform, " Ke=",Ke//MatrixForm, "\neigenvalues of KeM=",Chop[Eigenvalues[N[KeM]]], "\neigenvalues of KeG=",Chop[Eigenvalues[N[KeG]]], "\neigenvalues of Ke= ",Chop[Eigenvalues[N[Ke]]]]; pe= KeG= KeM= Ke= Eigenvalues of KeM={196.,,,,, } Eigenvalues of KeG={38., 38., 38.,,, } Eigenvalues of Ke= {1334., 38., 38.,,, } Figure 9.1. Verification of Example 9.4 using the Mathematica modules of Figure Example 9.4. This numerical example pertains to a space bar element with reference configuration geometry defined by the position coordinates (X 1 = 3, Y 1 =, Z 1 = 6) and (X = 5, Y = 7, Z = 8) of the end nodes. The reference bar length is = = 15. Other inputs include: elastic modulus E = 45, cross-section area A = 15, reference (initial) PK stress s = 8, and reference GL strain e =. The node displacements are u X1 = 3, u Y 1 = 1, u Z1 = 4, u X = 3, u Y = 4, u Z = 6. As a result, the end nodes in the current configuration are located at (x 1 = 6, y 1 = 3, z 1 = ) and (x = 8, y = 11, z = 14). Compute the internal force and tangent stiffness matrix in the current configuration. The current length is L = = 18. The current GL strain is e = (L L )/(L ) = 11/5 = %, whence the PK stress is s = s + Ee= = 19. The scaled direction cosines in C are c x = x 1 / = /15, c y = y 1 / = 8/15, and c z = z 1 / = 16/15. Check: cx +c y +c z = L/ = 18/15. The matrices required for the internal force and tangent stiffness computations are B = [ /15 8/15 16/15 /15 8/15 16/15 ], H = P. (9.54) 5 9

21 9.3 *THE TL SPACE BAR ELEMENT Replacing we get p = 34 38, K M = , K G = , K = K M +K G = , (9.55) These results are verified with the script of Figure 9.1, which calls the Mathematica modules of Figure The script computes the eigenvalues of K M, K G and K = K M + K G ; which shows their ranks to be 1, 3 and 3, respectively. Again the eigenvalues of K are the sum of those of K M and K G *Mallett-Marcal Stiffness Decomposition In 1968 Mallett and Marcal [488] introduced a three-way split of the tangent stiffness matrix. This is briefly covered as it has historical interest and may help the reader understand the literature of the time. In addition, solution methods based on further rearrangements of that split survive in some commercial FEM programs. The material stiffness matrix K M is presented in [488] as the combination K M = K L + K D, in which The linear stiffness matrix K L is that evaluated at the reference configuration, assuming zero initial stresses. This may be obtained by replacing (for the space bar) c x, c y, and c z by c X, c Y and c Z, respectively. The initial displacement matrix K D = K M K L accounts for the material stiffness change from the reference to the current configuration. The end result is that the tangent stiffness matrix appears as a combination of three matrices: linear, initial displacement, and geometric: K = K L + K D + K G, (9.56) whence (4.7) becomes (K L + K D + K G ) u = q λ. Looks like a needless complication: three matrices on the left instead of just two! Quid est nisi miserabilis insania? But there is method in the madness. The idea is to selectively transfer some terms to the right hand side. For example, Bathe and Wilson [5,53] do that with the initial displacement term: (K L + K G ) u = q λ K D u = q λ f D. (9.57) When this ODE is discretized in pseudo-time to set up an incremental marching scheme, the force vector f D = K D u can be efficiently evaluated, using the previous solution(s), in element-by-element fashion. On the left hand side, only K G is changing. This solution procedure is still used in several commercial FEM codes. The split (9.57) has computational advantages when the deviation from the reference configuration is small, as in small-deformation material nonlinearity. It also accommodates the Linearized Prebuckling (LPB) model of stability analysis described in Chapter 9 because LPB assumes that prebuckling displacements can be neglected, whence K D. Except for those scenarios, K L + K G is not the tangent stiffness, which precludes the use of a true Newton corrector and may hamper finding critical points by the SST (see 9.3.). A more drastic rearrangement moves all nonlinear terms to the right: K L u = q λ (K D + K G ) u = q λ f D f G. (9.58) 9 1

22 Chapter 9: TL BAR ELEMENTS: FORMULATION This leads to the initial stiffness method, also known as the pseudo force method, as well as a few other names. All geometric nonlinearities are accounted for through corrective force vectors, avoiding matrix reassembly. This was popular in early FEM days as a quick and dirty way to introduce nonlinear capabilities into an existing linear code. Only one factorization (that of K L ) is needed during the analysis procedure. Aside from that, (9.58) has few redeeming features. Notes and Bibliography Work in DSM-based finite elements for geometrically nonlinear analysis started at Boeing in the late 195s. (See 1.6 for a summary.) Much of it appeared in internal reports and memoranda difficult to access. The first systematic account of this early period was provided by Martin 9 at the 1965 Wright-Patterson conference [493]. His paper starts Development and application of the direct stiffness method to geometrically nonlinear problems has been underway since Only a portion of that work has appeared in the general technical literature. Furthermore, the derivations which have been given for stiffness matrices required for the large deflection problem have often been confusing and, in some cases, have led to incorrect conclusions. Most of this work, as well as the confusion alluded to by Martin, pertains to what is now called the geometric stiffness matrix. The early name for it was initial stress matrix. A bar geometric stiffness was first presented in [8]. If this is compared to (9.7) one may observe that half of the nonzero entries are missing. Consequently that inaugural K G is not invariant with respect to the choice of coordinate frames, and has the wrong rank. For beams and plates, a veritable geometric stiffness matrix zoo appeared in print during the 196s. 1 Such discrepancies were clarified in review articles by Martin [495] and Carey [17], who showed it to be the result of using different order theories in the strain energy to build approximations. Another contributor to zoo crowding was use of different shape functions for different energy terms. Further clarification on matrix derivations was provided by Rajasekaran and Murray [636]. This was the departing point for the CCF formulation [33,37]. From current perpective those past controversies are useless. The present derivation is consistent with continuum mechanics, and leaves no room for ambiguities. A three-way decomposition of the tangent stiffness was popular during the 197s and is still used in some nonlinear commercial codes. See above for more details. For a generalization that includes a unified kinematic description that embodies TL as special case, as well as hyperelastic material behavior, see Chapter Harold C. Martin was Professor of Aeronautics and Astronautics at the University of Washington, Seattle, WA. He was an active consultant to Boeing and JPL. His work with Jon Turner and Ray Clough over resulted in the seminal paper [798], widely considered as the start of the FEM as used today. 1 Some of the published results were plainly wrong, for example those in [4]. Incorrect conclusions were unintended consequences of using purely intuitive geometric arguments. 9

23 Exercises Homework Exercises for Chapter 9 TL Bar Elements: Formulation λ will be negative for compression E, A f = λ P (1) u Y u X () (a) E, A (b) Y, y S H 1 α (3) Y, y S X, x E, A 3 L (1) Bar element models weightless string in reference configuration E, A g gravity field 4 ;; : FEM node (e) : element number Point mass node 1 u Y 1 u X X, x Figure E9.1. Problems for the first two exercises: (a) a 3-element TL-plane-bar FEM model for Exercise 9.1; (b) a 1-element TL-plane-bar model of a classical pendulum for Exercise 9.. EXERCISE 9.1 [A/C:5] You go to work as a nonlinear-fem engineer for a car company. Your boss assigns you the job of designing a component of a wheel suspension system that can be modeled by the 3-bar structure depicted in Figure E9.1(a), which is a slight modification of the Mises truss. The model has 3 TL-plane-bar elements, with the the dimensions and properties shown in the figure. It is only subjected to vertical load f Y = λp, λ<, at node. The length S, bar section areas A and elastic modulus E are known, and the reference load P may be taken as unity. The rise angle α>is the only design variable. Find the largest α for which bifurcation, which your boss says is bad for the wheels, cannot occur. (For the -bar arch example structure a.k.a. the Mises truss that maximum α was shown to be defined by tan α /.) To study the occurrence of bifurcation, it is enough to set up the tangent stiffness matrix assuming that the X (horizontal) displacement u X is zero. The following tangent stiffness matrix is obtained for S =, E = 1, A = 1 and u X = : K = [ ] K XX K XY = K XY K YY u Y 4 + u Y Hu Y + u Y (1 + H ) u Y 4 + 6u Y 16 + H + 6Hu Y + 3u Y (1 + H ) 3 (E9.1) Bifurcation occurs if K XX, which is quadratic in u Y : a + bu Y + cu Y, vanishes. Physical bifurcation points occur if the discriminant b 4ac (real roots). Study when this happens as a function of H, and deduce 9 3

24 Chapter 9: TL BAR ELEMENTS: FORMULATION the largest α for which bifurcation cannot occur. [Note that the null eigenvector z cr for bifurcation failure is [ 1 ] T, whereas for limit point failure it is [ 1] T. Those can be used to verify the critical point type.] EXERCISE 9. [A:C:] Although this course focuses on statics, this exercise deals with the effect of the geometric stiffness on vibrations. (Don t be scared, the problem is exceedingly simple.) Consider the well known pendulum configuration idealized in Figure E9.1(b). A lumped mass m is suspended by a weightless elastic string. The pendulum is in static equilibrium at the vertical position shown. That position is taken as the (initially stressed) reference configuration C. The gravity acceleration is g. The arm is modeled by a -node TL-plane-bar element, which at C is under a tensile prestress s = mg/a. The tangent stiffness matrix for the cable element in the reference configuration is K = K M + K G, which is upon removing the degrees of freedom at the fixed node. Because of the prestress the geometric stiffness does not vanish. The order- vibration eigenproblem is K φ i = ω i M φ i, i = 1,, (E9.) in which i is the mode index, ω i is the i th circular frequency in radians per second, φ i the associated eigenvector (vibration mode shape) that includes the horizontal and vertical displacements of node 1, and the mass matrix is M = [ m m ]. (E9.3) Compute the two frequencies ω 1 and ω. One of them, say ω 1, pertains to pendulum motions while the other one pertains to a bar mode associated with axial arm oscillations. Discuss (1) what happens to ω 1 and ω if E, which characterizes the inextensional arm limit, and () whether the classical pendulum small-oscillations frequency ω P = g/l is obtained. (The latter result is impossible to get with linear FEM since then K G is missing.) EXERCISE 9.3 [A:15] Consider an individual TL plane-bar-element under axial load P. The material is linearly isotropic, with elastic modulus E and Poisson s ratio ν. The reference area under P = isa, whereas the current area under P is A. Find the relation between the true (Cauchy) axial stress σ = σ xx in the bar and the PK axial stress s = s XX. Hint: study the change in cross section area as function of ν. Compare your solution with that of Example 8.4 in Chapter 8 if ν = 1. EXERCISE 9.4 (Requires knowledge of eigensystem analysis.) [A:] Show that the material stiffness and geometric stiffness matrices for the plane bar element have ranks 1 and, respectively. The tangent stiffness matrix, which is their sum, has rank and not 1 + = 3. Explain the reason. Hint: look at the eigenvectors. 9 4