The CR Formulation: BE Plane Beam

6 The CR Formulation: BE Plane Beam 6

Chapter 6: THE CR FORMUATION: BE PANE BEAM TABE OF CONTENTS Page 6. Introduction..................... 6 4 6.2 CR Beam Kinematics................. 6 4 6.2. Coordinate Systems............... 6 4 6.2.2 Degrees of Freedom............... 6 4 6.2.3 Partial Derivatives................ 6 6 6.2.4 Arbitrary Base Configuration............ 6 8 6.2.5 Stress Resultants................ 6 9 6.3 The Deformational Strain Energy............. 6 9 6.4 Internal Force Vector and Tangent Stiffness Matrix....... 6 9 6.4. Internal Force Vector.............. 6 0 6.4.2 Material Stiffness Matrix............. 6 0 6.4.3 Geometric Stiffness Matrix............ 6 6. Notes and Bibliography................. 6 6. Exercises...................... 6 3 6 2

TABE OF CONTENTS Page 6. Introduction..................... 6 4 6.2 CR Beam Kinematics................. 6 4 6.2. Coordinate Systems............... 6 4 6.2.2 Degrees of Freedom............... 6 4 6.2.3 Partial Derivatives................ 6 6 6.2.4 Arbitrary Base Configuration............ 6 8 6.2.5 Stress Resultants................ 6 9 6.3 The Deformational Strain Energy............. 6 9 6.4 Internal Force Vector and Tangent Stiffness Matrix....... 6 9 6.4. Internal Force Vector.............. 6 0 6.4.2 Material Stiffness Matrix............. 6 0 6.4.3 Geometric Stiffness Matrix............ 6 6. Notes and Bibliography................. 6 6. Exercises...................... 6 3 6 3

Chapter 6: THE CR FORMUATION: BE PANE BEAM 6.. Introduction This Chapter uses the CR kinematic description to construct a geometrically nonlinear, 2-node, Bernoulli-Euler plane beam. Unlike the T beam element presented in Chapters 0-, we can do a C (Hermitian) beam from scratch, without need of MacNeal s RBF trick to finish the job. The reason is that with the CR approach it is as easy to do C (Bernoulli-Euler) or C 0 (Timoshenko) beams, and the former has a better performing geometric stiffness matrix. 6.2. CR Beam Kinematics The CR formulation of the beam motion is quite similar to that of the bar element in many respects, and much of the development can be reused. Only the major differences will be noted here. 6.2.. Coordinate Systems As in Chapter 0, we consider a plane, straight, prismatic beam element with two end nodes. The element is initially aligned with the global X axis in the initial configuration C 0, with the origin O 0 located at the element midpoint. This configuration is assumed to be straight and undeformed although it may be under initial uniform axial stress with resultant N 0. The bar properties include the elastic modules, E, the cross section area A 0 and the moment of inertia I 0 about the neutral axis. The reference length is 0. The motion on the {X, Y } plane carries it to the current configuration C. The corotated configuration C R is selected as depicted in Figure 6... The longitudinal axis passes through the current position of the end nodes. This defines the local axis x e. The origin of {x e, y e } is placed halfway between the nodes. This forms an angle ψ with X. 2. The C R nodes are placed at an equal distance from the C nodes. Hence the corotated axes {xr e, ye R }, including origin, coincide with {x e, y e }. The new ingredient is the rotation angle θ about Z or z. With C R chosen as indicated, the deformation part of these rotations is easily extracted: θ = θ ψ. Other choices for C R are possible. The foregoing one has the advantages of simplicity, and of being compatible with that of the bar element discussed in the previous Chapter. 6.2.2. Degrees of Freedom The CR plane beam element has six degrees of freedom, which are placed in the vectors u X ū e X u Y ū e 2 d Y 0 θ u =, ū e θ e θ = u X2 ū e = X2 u Y 2 ū e 2 d. (6.) Y 2 0 θ 2 θ 2 e θ 2 See Figure 6.2 for a picture of the global displacements and Figure 6.3 for the deformational displacements. 6 4

Current configuration // Y _ C // Y _ θ 2 6.2 CR BEAM KINEMATICS φ = ψ+ϕ ψ // X _ ϕ 2(x 2,y 2) // X u Y2 Y, y Base or initial configuration C 0 X, x Y _ γ (X,Y ) θ (x,y ) u X u Y Corotated configuration X _ 2(X 2,Y 2) ϕ u X2 C R Reference length 0, cross section area A 0, second moment of inertia I 0, and elastic modulus E Figure 6.. Kinematics of corotational C plane beam element. Deformations in current configuration grossly exaggerated for visibility. 2 θ 2 //X, x C u X θ //X, x u Y2 u Y Y, y X, x C 0 0 20 u X2 Figure 6.2. Global displacements of CR plane beam element initially aligned with X axis. y e d/2 θ _ C R d/2 2 _ θ 2 x e 0 C Figure 6.3. Deformational displacements of CR plane beam element in its local (CR) system. Deformations grossly exaggerated for visibility. 6 5

Chapter 6: THE CR FORMUATION: BE PANE BEAM Proceeding as in the general formulation specialized to the 2D case, we can obtain the following relation: ū e x c ψ s ψ 0 0 0 0 u X u X0 2 0( c ψ ) ū e y s ψ c ψ 0 0 0 0 u Y u Y 0 2 ū e θ e 0 0 0 0 0 θ 0s ψ ψ = ū e = + x2 0 0 0 c ψ s ψ 0 u X2 u X0 2 ū e 0 0 0 s y2 ψ c ψ 0 u Y 2 u Y 0 0(c ψ ) (6.2) 2 0s ψ θ 2 e 0 0 0 0 0 θ 2 ψ Here c ψ and s ψ and the angle ψ are implicitly defined by the displacements through the trigonometric relations s ψ = sin ψ = y, c ψ = cos ψ = x, ψ = arctan y (6.3) x where x = 0 + u X2 u X, y = u Y 2 u Y, and = 2 x + 2 y (6.4) is the bar length in the current configuration, ignoring the bending deformation. We note the following relations = = c ψ, u X2 u X c ψ u X2 = c ψ u X = s2 ψ, = = s ψ, u Y 2 u Y c ψ u Y 2 = c ψ u Y = s ψc ψ, c ψ θ s ψ = s ψ = s ψc ψ u X2 u X, s ψ = s ψ = c2 ψ u Y 2 u Y, ψ = ψ = s ψ u X2 u X, ψ = ψ = c ψ u Y 2 u Y, which are useful in the calculations that follow. 6.2.3. Partial Derivatives = = 0, θ θ 2 = c ψ θ 2 = 0, s ψ = s ψ = 0, θ θ 2 ψ = ψ = 0. θ θ 2 (6.5) The first and second partial derivatives of the deformations d, θ and θ 2 with respect to the node displacements are necessary for the computations of internal forces and stiffness matrices. Using (6.5) and Mathematica, one obtains for the first derivatives: δū e x δū e 2 c ψ 2 s ψ 0 2 c ψ 2 s ψ 0 y s ψ c ψ 0 / cψ 2 δ θ e 0/ 0 s ψ c ψ 0 / cψ 2 0/ 0 s ψ / c ψ / s ψ / c ψ / 0 δū e = x2 δū e 2 c ψ 2 s ψ 0 2 c ψ 2 s ψ 0 y2 s ψ c ψ 0 / cψ 2 0/ 0 s ψ c ψ 0 / cψ 2 0/ 0 s ψ / c ψ / 0 s ψ / c ψ / δ θ e 2 δu X δu Y δθ δu X2 δu Y 2 δθ 2 (6.6) 6 6

6.2 CR BEAM KINEMATICS θ 2 // Y _ φ = ψ+ϕ ψ // X _ ϕ 2(x 2,y 2) // X // Y _ C u Y 2 θ (x,y ) _ X Y, y X, x _ Y C 0 u Y 2(X,Y ) 2 2 u X2 ϕ u X (X,Y ) Figure 6.4. Beam element with arbitrarily oriented base configuration C 0, forming an angle ϕ with X. Corotated configuration not shown to reduce clutter. Deformations in current configuration grossly exaggerated for visibility. Since u e x = ue x2 = 2 d, θ e = θ, and θ e 2 = θ 2 we get δd c ψ s ψ 0 c ψ s ψ 0 δ θ = s ψ / c ψ / s ψ / c ψ / 0 δ θ 2 s ψ / c ψ / 0 s ψ / c ψ / δu X δu Y δθ δu X2 δu Y 2 δθ 2 The second derivatives of deformation variables are sψ 2 s ψ c ψ 0 s 2 ψ s ψ c ψ 0 2 d u u = s ψ c ψ cψ 2 0 s ψ c ψ cψ 2 0 sψ 2 s ψ c ψ 0 sψ 2 s ψ c ψ 0 s ψ c ψ cψ 2 0 s ψ c ψ cψ 2 0 2s ψ c ψ cψ 2 s2 ψ 0 2s ψ c ψ sψ 2 c2 ψ 0 2 θ u u = cψ 2 s2 ψ 2s ψ c ψ 0 sψ 2 c2 ψ 2s ψ c ψ 0 2 2s ψ c ψ sψ 2 c2 ψ 0 2s ψ c ψ cψ 2 s2 ψ 0 sψ 2 c2 ψ 2s ψ c ψ 0 cψ 2 s2 ψ 2s ψ c ψ 0 (6.7) (6.8) (6.9) 6 7

Chapter 6: THE CR FORMUATION: BE PANE BEAM 2 θ 2 u u = 2 6.2.4. Arbitrary Base Configuration 2s ψ c ψ cψ 2 s2 ψ 0 2s ψ c ψ sψ 2 c2 ψ 0 cψ 2 s2 ψ 2s ψ c ψ 0 sψ 2 c2 ψ 2s ψ c ψ 0 2s ψ c ψ sψ 2 c2 ψ 0 2s ψ c ψ cψ 2 s2 ψ 0 sψ 2 c2 ψ 2s ψ c ψ 0 cψ 2 s2 ψ 2s ψ c ψ 0 (6.0) The foregoing relations can be generalized to the case of a base (initial) configuration C 0 not aligned with the X axis as shown in Figure 6.4. Given the node coordinates and displacements shown in that figure, it is easily shown (cf. Chapter 0) that cos ϕ = X 2 / 0, sin ϕ = Y 2 / 0, cos φ = cos(ψ + ϕ) = x 2 /, sin φ = sin(ψ + ϕ) = y 2 /, cos ψ = (X 2 x 2 + Y 2 y 2 )/( 0 ) and sin ψ = (X 2 y 2 Y 2 x 2 )/( 0 ). The preceding transformation rules remain correct if ψ is replaced by φ = ϕ + ψ, except for the deformation angle computation, which remain θ = θ ψ and θ 2 = θ 2 ψ because the θs are measured from X. The relation between deformational and global displacements become d = 0 = u X2 c φ + u Y 2 s φ + 0 ( c φ ) θ = θ ψ (6.) θ 2 = θ 2 ψ The first derivatives of deformation variables, expressed as variations to get a neast matrix form, are δu X δd c φ s φ 0 c φ s φ 0 δu Y δ θ = s φ / c φ / s φ / c φ / 0 δθ (6.2) δu δ θ 2 s φ / c φ / 0 s φ / c φ / X2 δu Y 2 δθ 2 The second derivatives of deformation variables are sφ 2 s φ c φ 0 s 2 φ s φ c φ 0 2 d u u = s φ c φ cφ 2 0 s φ c φ cφ 2 0 sφ 2 s φ c φ 0 sφ 2 s φ c φ 0 (6.3) s φ c φ cφ 2 0 s φ c φ cφ 2 0 2s φ c φ cφ 2 s2 φ 0 2s φ c φ sφ 2 c2 φ 0 2 θ u u = cφ 2 s2 φ 2s φ c φ 0 sφ 2 c2 φ 2s φ c φ 0 2 2s φ c φ sφ 2 c2 φ 0 2s φ c φ cφ 2 s2 φ 0 (6.4) sφ 2 c2 φ 2s φ c φ 0 cφ 2 s2 φ 2s φ c φ 0 6 8

6.4 INTERNA FORCE VECTOR ANDTANGENT STIFFNESS MATRIX 2 θ 2 u u = 2 6.2.5. Stress Resultants 2s φ c φ cφ 2 s2 φ 0 2s φ c φ sφ 2 c2 φ 0 cφ 2 s2 φ 2s φ c φ 0 sφ 2 c2 φ 2s φ c φ 0 2s φ c φ sφ 2 c2 φ 0 2s φ c φ cφ 2 s2 φ 0 sφ 2 c2 φ 2s φ c φ 0 cφ 2 s2 φ 2s φ c φ 0 (6.5) The stress resultants in the reference configuration (either C 0 or C R ) are N 0, M 0 and M0 2. The initial shear force is V 0 = (M 0 M0 2 )/ 0. See Figure 6.5 for sign conventions. Denote by N, V and M the stress resultants in the current configuration. Whereas N and V are constant along the element, M = M(x e ) varies linearly along the length because this is a Hermitian or model, which relies on cubic transverse displacements. Consequently we will define its variation by the two node values M and M 2. The shear V is recovered from equilibrium as V = (M M 2 )/, which is also constant. The stress resultants can be obtained from the deformations as N = N 0 + EA 0 0 d, M = M 0 2EI 0 0 (2 θ + θ 2 ), M 2 = M2 0 + 2EI 0 ( θ + 2 θ 2 ), V = M M 2 0 = V 0 0 + 2EI 0 ( θ θ 2 ). (6.6) 6.3. The Deformational Strain Energy The next step in the CR formulation is to work out the deformational strain energy of the beam. The basic choices are:. A linear beam 2. A nonlinear T beam In the interest of resuing the linear element, we select the first choice. The strain energy of the beam for small strains can be written U = U a + U b + U g (6.7) where U a, U b and U g are the energy taken by axial (bar) deformation, bending deformation, and initial-stress geometric effects, respectively. We adopt the following energy expressions: U a = N 0 d + 2 (N N 0 )d 2 = N 0 0 e + 2 EA 0 0 e 2, [ ] T [ ][ ] θ U b = M2 0 θ 2 M 0 θ + EI 0 4 2 θ, 2 θ 2 2 4 θ 2 U g = 2 [ θ θ 2 ] T N 0 0 30 0 [ ][ ] 4 θ. 4 θ 2 (6.8) The 2 2 matrices appearing in U b and U g may be derived from those given in Chapters 5 and 5, respectively, of Przemieniecki s book. This book, howevr, omits the initial stress terms. J. S. Przemieniecki, Theory of Matrix Structural Analysis, Dover, New York, 985. 6 9

Chapter 6: THE CR FORMUATION: BE PANE BEAM M 2 V N C M V N C 0 V 0 M 0 2 N 0 N 0 V 0 M 0 Figure 6.5. Plane beam stress resultants displaying positive sign conventions. Axial forces N and transverse shear forces V are constant along the length, but bending moments M vary linearly. Hence two nodal values of M are required. 6.4. Internal Force Vector and Tangent Stiffness Matrix The internal force vector and tangent stiffness matrix of the corrotational element are then obtained by the usual formulas: p = U u, K = p u = K M + K G (6.9) To develop these quantities it is necessary to find the first and second partial derivatives of d, θ and θ 2 in terms of the node displacements. 6.4.. Internal Force Vector Using the partial derivatives compiled above and Mathematica, one obtains the following expression for the internal forces. p = p a + p b + p g (6.20) where p a = U a u = N [ c φ s φ 0 c φ s φ 0 ] T p b = U b u = [ Vs φ Vc φ M Vs φ Vc φ M 2 ] T p g = U g u = N 0 0 30 [ 3s φ( θ + θ 2 )/ 3c φ ( θ + θ 2 )/ 4 θ θ 2 3s φ ( θ + θ 2 )/ 3c φ ( θ + θ 2 )/ 4 θ 2 θ ] T (6.2) 6 0

6. Notes and Bibliography 6.4.2. Material Stiffness Matrix Carrying out the computations one obtains the following compact expression for the material stiffness: K M = T T K M0 T (6.22) where EA 0 0 EA 0 0 0 2EI 6EI 2 0 2EI 6EI 2 K M0 = 0 6EI 4EI 0 6EI 2EI, (6.23) EA 0 0 EA 0 0 0 2EI 2 6EI 0 2EI 2 6EI 0 6EI 2EI 0 6EI 4EI is the stiffness matrix of the linear beam-column element, and T is the transformation matrix c φ s φ 0 0 0 0 s φ c φ 0 0 0 0 0 0 0 0 0 T =, (6.24) 0 0 0 c φ s φ 0 0 0 0 s φ c φ 0 0 0 0 0 0 which introduces the effect of finite rigid body motions. 6.4.3. Geometric Stiffness Matrix The expression for the geometric stiffness is a bit more complicated. It can be presented in a compact form as follows: K G = T T KG N T + KV G (6.25) where T is the transformation matrix (6.24), KG N is the well known geometric stiffness for a Hermitian beam element under axial force: 0 36 3 0 36 3 KG N = N 0 3 4 2 0 3 2 (6.26) 30 0 36 3 0 36 3 0 3 2 0 3 4 2 and the remaining term introduces the effect of varying moments through the transverse shear force in C: sin 2φ cos 2φ 0 sin 2φ cos 2φ 0 cos 2φ sin 2φ 0 cos 2φ sin 2φ 0 KG V = V (6.27) sin 2φ cos 2φ 0 sin 2φ cos 2φ 0 cos 2φ sin 2φ 0 cos 2φ sin 2φ 0 in which sin 2φ = 2s φ c φ and cos 2φ = cφ 2 s2 φ. Note that K G is symmetric. 6

Chapter 6: THE CR FORMUATION: BE PANE BEAM Notes and Bibliography Most of the foregoing expressions were obtained in the writer s 966 Berkeley thesis [208], but not using the CR approach (which was not fully developed until the mid 980s). One thesis result omitted here is the more detailed expression of the contribution of the varying bending moment to the geometric stiffness. For most practical problems these refinements are not important. Similar results can be found in the 99 book by Crisfield [72, Ch. 7]. 6 2

Exercises Homework Exercises for Chapter 6 The Corotational Description: 2D C Beam EXERCISE 6. Complete the derivation of p for the 2-node C beam element and implement in Mathematica, using the same inputs as in Chapter 9 Addendum. (Implemented and posted on Web) EXERCISE 6.2 Complete the derivation of K for the 2-node C beam element and implement in Mathematica, using the same inputs as in Chapter 9 Addendum. (Implemented and posted on Web) EXERCISE 6.3 A plane 2-node C beam element has properties 0 = 6, E = 3000, A 0 = 2, I 0 = 2, N 0 = 5 in the initial state C 0 along X, with node at (0,0) and node 2 at ( 0, 0). The beam rotates by 45 about the origin so that at the current configuration C node stays at {0, 0} while node 2 moves to {( 0 + d)/ 2,( 0 + d)/ 2}, where d = 0 /000. The rotational freedoms at C are θ = θ 2 = 45 = π/4 radians. Compute p, K M and K G at the current configuration, and compare those quantities with those of the C 0 beam element presented in Chapter 9, using RBF for the latter. Note: AMathematica implementation of this C element has been posted on the Web as a Mathematica 4. Notebook PlaneBeamC.nb. The element checks out when moving about the reference configuration C 0. It gives excellent buckling values for the problem of Exercise 9.3. More tests are needed, however, for an arbitrary configuration to make sure the internal force vector and the tangent stiffness are consistent. EXERCISE 6.4 Confirm the previous statement by repeating the buckling calculations of Exercise 9.3 using the CR beam element provided in the Mathematica Notebook mentioned above (extract the material and stiffness matrices, ignore the rest). Compare the speed of convergence of the CR and T element for the cantilever buckling problem. 6 3