General elastic beam with an elastic foundation Figure 1 shows a beam-column on an elastic foundation. The beam is connected to a continuous series of foundation springs. The other end of the foundation spring has a known displacement, v f, which is almost always zero. The differential equation of equilibrium of an initially straight beam, of flexural stiffness EI, resting on a Winkler foundation, of stiffness k per unit length, with a transverse load per unit length of w, subjected to a tensile axial load N acting along the x-axis is: d 2 [EI d2 v dx 2 dx 2] N d2 v dx2 + kv w(x) =. (1) where v(x) is the transverse displacement of the beam. If the axial force, N, is present, then this is usually called a Beam-Column model. If the axial load is an unknown constant, then it is a buckling load that has to be found by an eigen-solution rather than by solving a linear algebraic system. Beams on elastic foundations are fairly common (like pipelines, rail lines or drill strings). However, most introductions to beams assume no axial load, nor any foundation support. Then, the above general beam equation reduces to that covered in an introduction to solid mechanics: d 2 [EI d2 v dx 2 dx2] = w(x). (2) The homogeneous solution of this equation is simply a cubic polynomial with four constants. Either beam equation is a fourth-order ordinary differential equation. Therefore it will generally need four boundary conditions. Figure 1 A beam-column on an elastic foundation Related physical quantities are the slope, θ(x) = v (x), the bending moment, M(x) = EIv (x), and the transverse shear force, V(x) = EIv (x). The sign conventions are that the position, x, is positive to the right, the deflection, v, point forces, P, and the line load, w, are positive in the y-direction (upward), and the slopes and moments are positive in the counter-clockwise direction. Since the foundation springs are restrained, there can be no rigid body motions of the beam. In other words, the assembled equations should never be singular and it is acceptable to just have non-essential boundary conditions applied to the beam. A simple Winkler foundation model like this one can push or pull on the beam as needed and no gaps can occur. Copyright J.E. Akin. All rights reserved. Page 1 of 11
The distributed load per unit length can include point transverse shear loads, V, by using the Dirac Delta distribution. Likewise, employing a doublet distribution in defining w(x) allows for the inclusion of point couples, or moments, M. Engineers designing beams are usually interested in the local moment and transvers shear force since they define the stress levels and the material failure criteria. The exact solution of the homogeneous (w==) general form given above is given in terms of hyperbolic sines and cosines. Based on Tong s Theorem, exact solutions at the nodes are obtained if such functions are used as the interpolation functions in a finite element model. Advanced elements of that type have been applied with excellent results. To apply the Galerkin weak form the governing ODE is multiplied by v(x) and the integral over the length of the beam is set to zero. The highest derivative term is always present, and needs to be integrated twice by parts to reduce the inter-element continuity requirement. That integral becomes L I E = v d2 dx 2 (E(x)I(x) d2 v dx2) dx I E = [v d dx (E(x)I(x) d2 L v dx 2)] [ dv dx (E(x)I(x) d2 L v dx 2)] + d2 v dx 2 Copyright J.E. Akin. All rights reserved. Page 2 of 11 L (E(x)I(x) d2 v dx2) dx which brings the non-essential boundary conditions into the integral form. The first term is the product of the displacement and the point shear force at both ends, while the second term is the product of the slope and the point moment at both ends. Thus, the contributions from the highest derivative term in the ODE are I E = [v V(x)] L [ dv L dx M(x)] L + d2 v dx 2 (E(x)I(x) d2 v dx2) dx Therefore, the integral form contains second derivatives (d 2 v dx 2 ) as its highest derivative term. The presence of second derivatives in the integral form means that calculus requires the elements to have inter-element continuity of the deflection and the slope (v(x) and θ(x) = v (x)). Such elements are said to have an inter-element continuity of C 1. Therefore, each shared beam node must have two degrees of freedom (dof), at least. The presence of the second derivatives in the integral form also means that the essential boundary conditions (EBC) will specify v or θ or both at a point. The non-essential boundary conditions (NBC) are that the transverse shear force, V = EIv, and/or the moment, M = EIv, is given at a point. When the deflection is given, then the shear force is a reaction quantity. When the slope is specified, then the moment is a reaction. Calculus continuity requires the elements share their slope and deflection values. Thus, classic beam elements approximated by at least a cubic polynomial having C 1 continuity can give nodally exact deflections and slopes. One-dimensional polynomial interpolation with C 1 or C 2 continuity usually is
based on Hermite interpolation polynomials. The classic beam element is often said to be a Hermite cubic. Some typical Hermite polynomials are given in the Appendix. A line element with only two nodes will provide the four constants needed to define a cubic polynomial. Such elements are referred to here as classic beam elements. But since designers are interested in the second and third derivatives of the solution, classic beam elements give very poor design results unless a large number of them are used. All of the matrices for a cubic beam are well known and are just summarized here for comparison to the three-node quintic beam. For a typical beam (k = ) with a cross-sectional moment of inertia, I, length, L, a depth of h, and material with an elastic modulus of E and a coefficient of thermal expansion of α, the corresponding flexural stiffness transverse load and thermal load matrices for the equilibrium of a single classic beam element, are EI [ L 3 12 6L 12 6L v 1 V 1 6L 4L 2 6L 2L 2 θ ] { 1 M 12 6L 12 6L v } = { 1 } + 2 V 2 6L 2L 2 6L 4L 2 θ 2 M 2 21 9 L 3L 6 9 2L 21 2L 3L Copyright J.E. Akin. All rights reserved. Page 3 of 11 ] { w 1 w } + α TEI { 2 h 1 }, (3) where w 1 and w 2 are the load per unit length at the first (left) and second end, respectively, and T is the temperature increase, over h, from the top to the bottom of the beam. Also, the V j and M j terms are point shear forces and point moments applied to the nodes as external loadings and/or unknown reactions. In matrix symbol notation this equation is written as K e v e = F P e + F w e + F α e. (4) The classic beam element has a deflection given by (for r = x L): A slope of v(x) = v 1 (1 3r 2 + 2r 3 ) + θ 1 (r 2r 2 + r 3 )L + v 2 (3r 2 2r 3 ) + θ 2 (r 3 r 2 )L (5) θ(x) = v 1 (6r 2 6r)/L + θ 1 (1 4r + 3r 2 ) + v 2 (6r 6r 2 )/L + θ 2 (3r 2 2r) (6) A linear bending moment of M(x) = [v 1 (12r 6) + θ 1 (6r 4)L + v 2 (6 12r) + θ 2 (6r 2)L]/L 2 (7) and a constant transverse shear force of V(x) = [v 1 (12) + θ 1 (6)L + v 2 ( 12) + θ 2 (6)L]/L 3. (8) If a Winkler foundation is present then the foundation stiffness is For the classic cubic beam this becomes 1 K e f = k e H e dl (9) L e H e T
156 22L e K e f = ρe A e L e 22L e 42 54 4L e 2 13L e 13L e 3L e 2 54 13L e 13L e 156 3L e 2 22L e ] (1) 22L e 4L e 2 If an axial load is present, the geometric stiffness matrix after integration by parts is K e = H e T L e N e d2 e H dx 2 dl K e N = Ne 3L e [ 36 3L e 3L e 4L e 2 36 3L e 3L e L e 2 36 3L e 3L e 36 L e 2 3L e ]. (11) 3L e 4L e 2 For dynamic response or natural frequency calculations it is necessary to include the physical mass matrix defined as For the classic cubic beam this becomes m e = H e T L e ρ e A e H e dl 156 22L e m e = ρe A e L e 22L e 42 54 4L e 2 13L e 13L e 3L e 2 54 13L e 13L e 156 3L e 2 22L e ]. (12) 22L e 4L e 2 Many exact solutions for beams not on an elastic foundation are tabulated in typical structural design manuals. They show that for the most common load and support conditions the exact solution is a third, fourth or fifth degree polynomial. Thus, a fifth degree polynomial will generally give a very accurate or exact beam solution with only a few elements, perhaps with only one element. However, if a foundation or an axial load is present polynomial approximations will no longer be exact at the nodes. The Hermite family of interpolation polynomials can be increased in order by increasing the number of nodes and/or by increasing the continuity level at each node. Here, a more accurate three-node C 1 beam element will be presented. This will be created by adding a third (middle) node to the element. That adds two more degrees of freedom and raises the polynomial to a fifth degree (a quintic polynomial). For simplicity, all of the coefficients in Eq. (1) are taken as constants evaluated at the center of the element length. That allows the corresponding finite element matrices of the fifth-degree element to be expressed in closed form in terms of the element deflection and slope at each of its three nodes [v 1 θ 1 v 2 θ 2 v 3 θ 3 ]. The element interpolation relations and the corresponding slope, moment and shear distributions are: Copyright J.E. Akin. All rights reserved. Page 4 of 11
v(r) = [v 1 (1 23r 2 + 66r 3 68r 4 + 24r 5 ) + θ 1 (r 6r 2 + 13r 3 12r 4 + 4r 5 )L +v 2 (16r 2 32r 3 + 16r 4 ) + θ 2 ( 8r 2 + 32r 3 4r 4 + 16r 5 )L +v 3 (7r 2 34r 3 + 52r 4 24r 5 ) + θ 3 ( r 2 + 5r 3 8r 4 + 4r 5 )L] (13) θ(r) = v (r) = [v 1 ( 46r + 198r 2 272r 3 + 12r 4 )/L + θ 1 (1 12r + 39r 2 48r 3 + 2r 4 )] +v 2 (32r 96r 2 + 64r 3 )/L + θ 2 ( 16r + 96r 2 16r 3 + 8r 4 ) +v 3 (14r 12r 2 + 28r 3 12r 4 )/L + θ 3 ( 2r + 15r 2 32r 3 + 2r 4 )] (14) M(r) = EIv (r) = EI[v 1 ( 46 + 396r 816r 2 + 48r 3 )/L + θ 1 (12 + 78r 144r 2 + 8r 3 ) +v 2 (32 192r + 192r 2 )/L + θ 2 ( 16 + 192r 48r 2 + 32r 3 ) +v 3 (14 24r + 624r 2 48r 3 )/L + θ 3 ( 2 + 3r 96r 2 + 8r 3 )]/L (15) V(r) = EIv (r) = EI[v 1 (396 1,632r + 1,44r 2 )/L + θ 1 (78 288r + 24r 2 ) +v 2 ( 192 + 384r)/L + θ 2 (192 96r + 96r 2 ) +v 3 ( 24 + 1,248r 1,35r 2 )/L + θ 3 (3 192r + 24r 2 )]/L 2 (16) The 6 by 6 element matrix equations of equilibrium of a single element are: (K E e + K N e + K k e ){v e } = {F e } (17) where {v e } T = [v 1, θ 1, v 2, θ 2, v 3, θ 3 ] are the generalized nodal displacements and rotations, and the {F e } are the generalized resultant nodal loading forces and moments. The K e terms are stiffness matrices arising from the first three terms in the differential equation. Usually, only the first is nonzero. This system is singular until enough boundary conditions are applied to prevent rigid body translations and rotations. Since this element has three nodes, the loading per unit length can be input at each of those nodes. The resultant element external loading vectors due to point forces, V k, and couples, M k, and the quadratic distributed load values, w k, at each node (or constant w 1 = w 2 = w 3, or linear w 2 = (w 1 + w 3 ) 2 values) are F P e = V 1 M 1 V 2 M 2 V 3 { M 3 }, F w e = H et H w w e L e dl = L 14,7 [ 1,995 1,54 15 15L 14L 56 6,72 56 28L 28L 15 1,54 1,995 14L 15L] w 1 { w 2 } (18) w 3 where the H e are the six Hermite beam interpolations, and the H w are the three quadratic Lagrangian interpolations for the distributed loads. The integral of their product forms a rectangular six by three array to convert distributed loads to concentrated shears and moments at the beam nodes. Likewise, it can be shown that a temperature difference through the depth of a beam causes only end moment loadings given by F e α = α T EI h [ 1 1]. A constant transverse load per unit length reduces the resultant load and moment vector to Copyright J.E. Akin. All rights reserved. Page 5 of 11
F w T = wl 14,7 [3,43 245L 7,84 3,43 245L], for w 1 = w 2 = w 3 = w (19) The symmetric flexural stiffness matrix for the three noded quintic element is K E = EI 35L 3 5,92 1,138L 3,584 1,138L 332L 2 896L 3,584 896L 7,168 1,92L 32L 2 1,58 242L 3,584 [ 242L 38L 2 896L 1,92L 1,58 242L 32L 2 242L 38L 2 3,584 896L 1,28L 2 1,92L 32L 2 1,92L 5,92 1,138L 32L 2 1,138L 332L 2 ] (2) The symmetric beam-column (or geometric stiffness) matrix due to any axial load is K N = N 63 L 1,668 39 L 1,536 24 L 132 9 L 39 L 28 L 2 48 L 8 L 2 9 L 5 L 2 1,536 48 L 3,72 1,536 48 L 24 L 8 L 2 256 L 2 24 L 8 L 2 132 9L 1,536 24 L 1,668 39 L [ 9 L 5 L 2 48 L 8 L 2 39 L 28 L 2 ] (21) The geometric stiffness matrix always depends on the existing stresses in the element. Here that is just a uniform axial stress, N / A. There is a general equation for the geometric stiffness matrix, involving all the stresses at a point, which comes from a non-linear analysis. It will not be given here. The element foundation stiffness is K k = k L 13,86 2,92 114 L 88 16 L 262 29 L 114 L 8 L 2 88 L 12 L 2 29 L 3 L 2 88 88 L 5,632 88 88 L 16 L 12 L 2 128 L 2 16 L 12 L 2 262 29 L 88 16 L 2,92 114 L [ 29 L 3 L 2 88 L 12 L 2 114 L 8 L 2 ] (22) If the mass matrix, m e, is included for dynamic response or vibration analysis then the above matrix is the element mass matrix when k is replaced by ρa where ρ is the mass density and A is the crosssectional area of the beam (so ρa is the mass-per unit length). Since it is common for beams to have piecewise constant properties, the classic beam element and the more accurate quintic beam element can be programmed in closed form. Using either the classic or quintic beam element some small beam analysis problems can also be solved in closed analytic form. As an example of the three-node beam, consider a fixed-fixed beam with a constant line load. The essential boundary conditions are v 1 =, θ 1 =, v 3 =, θ 3 =. The external point force and moment at center node 2 are zero (V 2 =, M 2 = ). The middle two rows define the remaining unknown center point deflection and slope: Copyright J.E. Akin. All rights reserved. Page 6 of 11
EI [7,168 35L 3 1,28L 2] {v 2 θ } = { 2 } + wl {7,84 14,7 } + { }. (23) Multiplying by the inverse of the square matrix gives the middle node solutions: { v 2 θ 2 } = 35wL4 14,7EI 7,168 [1 1 1,28L 2 ] { 7,84 } = wl4 {1 384EI } (24) which are exact. The slope, θ 2, was expected to be zero due to symmetry. It could have been used as a boundary condition to compute v 2 from one equation. Here, the reactions on the left are found from the first two rows of the equilibrium equations (since all the displacements are now known): EI [ 3,584 35L 3 896L 1,92L 32L 2 ] wl4 {1 384EI } = { V 1 } + wl M {3,43 1 14,7 245L } (25) { V 1 } = wl M { 1 1 2 L 6 }. Likewise, at the right end, utilizing the last two rows of the equilibrium equations { V 3 } = wl M { 1 }. (26) 3 2 L 6 Note that the total resultant force is wl, which is equal and opposite to the applied transverse load. The net external moment is zero. Usually static equilibrium is taught in undergraduate classes with Newton s Laws. They can used to verify that the computed reactions do indeed satisfy that the sum of the forces is zero, and that the sum of the moments, taken at any reference point, is zero. Since a 5-th degree element was used this problem is exact everywhere. That includes the moment and shear diagrams. The exact shear diagrams of EIv can have a discontinuity (due to point loads P) and the exact moment diagram of EIv can have a discontinuity (due to point moments M). The mesh should be constructed such that any point source occurs at an interface node between elements. Those items can be discontinuous at element interfaces because the calculus requirement for splitting integrals on says that v and v must be continuous there. That explains why a two node C 2 Hermite interpolation was not chosen to create a quintic beam. That choice could be made only if the beam was not allowed to have point sources. Inside any element the solution is continuous and thus is C at any non-interface node. In other words, inside any standard element you can calculate an infinite number of continuous derivatives (but the vast majority is zero). If you were to apply a point source at an interior node then the second and third derivatives would still be continuous and the shear and moment diagrams in that element would be wrong. Next consider the involvement of the foundation in the post-processing needed to recover the reactions on each beam element. Before the deflections of a beam on an elastic foundation (BOEF) Copyright J.E. Akin. All rights reserved. Page 7 of 11
are computed, a typical element free body diagram would be similar to Figure 1. As before, any external point load, P, and/or any point couple, M, should be placed at an element interface node and they go directly into the first or second equilibrium row of the node at that point. A distributed line load is integrated to become a consistent load vector. Those two effects are detailed in Eq. 14. Both the beam stiffness and the foundation stiffness resist the external applied loads. The element equilibrium equation (for no axial force) is (K E e + K k e ){v e } = {F P e } + {F w e } (27) The assembled equations have a similar form and are solved for the system displacements, v. Returning to the element to get the beam member reactions, the now known element degrees of freedom, v e, are gathered. The free body diagram of the element now has a significant change because the foundation is now applying a known pressure that opposes the displacement of the beam. That pressure, p(x) = k f v(x), is a polynomial of the same degree as the assumed displacements. That state is shown in Figure 2, which defines the sought element reaction vector, {F R e }. Now, there is a new consistent load vector to evaluate during post processing to find the element reactions. It comes from the foundation pressure and is proportional to the deflection: F e p = H e T p(x)dx = L e H e T k L e f H e dx v e = K e f v e. (28) The foundation stiffness caused the new load, but that load and others must be resisted by only the beam stiffness. Now the beam member equilibrium equations define the reactions on the nodes of the beam, {F R e }: or K E e {v e } = {F R e } + {F p e } + {F P e } + {F w e } (29) K E e {v e } {F p e } {F P e } {F w e } = {F R e }. (3) Now that there is more than one dof per node the scatter/gather operations previously used for axial models have become more generalized. The usual convention for numbering the unknowns is to count all of the unknowns at a node before moving on to another node. Therefore, the displacements are associated with the odd number of rows and columns in a matrix, while the slopes contributions are occurring in the even ones. For previous one-dimensional scalar unknowns scatter simply meant adding a scalar term at any node into the system equilibrium matrix and source vector. Now, that changes to adding a two by two sub-matrix from each element node into the system square matrix and a two by one column matrix into the system source vector. The prior algorithms for doing the scatter and gather still work because they had the number of degrees of freedom per node (n g ) as an input argument. It is just for hand solutions that you need to clearly understand this change. Figure 3 Copyright J.E. Akin. All rights reserved. Page 8 of 11
graphically illustrates the sub-matrices scatter operations for a line element with two unknowns per node. To determine the natural frequencies of a beam-column on an elastic foundation the eigen-problem is (K E + K N + K k ) ω 2 M = (31) To determine the natural frequencies of a beam-column the eigen-problem is (K E + K N ) ω 2 M = (32) When the axial load is an unknown constant, say N B, then it factors out of each element matrix and becomes a global constant: where K n is formed with a unit positive load. (K E + N B K n ) ω 2 M = (33) This shows that any axial load, N B, has an influence on the natural frequency, ω. In general, a tensile force (+) increases the natural frequency while a compression force (-) lowers the natural frequency. When the global constant, N B, is unknown it is called the buckling factor of safety (BFS) and the equations lead to a buckling eigen-problem to determine the BFS. The BFS is the factor of safety against buckling, or the ratio of the buckling loads to the applied loads. The following table gives the interpretation of possible values for the buckling factor of safety: Since the geometric stiffness matrix depends on the stress that means that it depends on all of the assembled loads, say F ref. In general, a linear static analysis is completed and the stresses are determined: (K E + K k ){v } = F ref (34) Then the geometric stiffness matrix is calculated from those stresses. The stress, and thus the geometric stiffness matrix, is proportion to the resultant load, so as the loads are scaled by the buckling factor the geometric stiffness increases by the same amount F BFS F ref, K N BFSK ref The scaling value that renders the combined stiffness to have a zero determinant (that is to become unstable) is calculated from a buckling eigen-problem as: K E + K k + BFS K ref = (35) This is solved for the value of the BFS. Then the load that would theoretically cause buckling is BFS F ref. The solution of the eigen-problem also yields the relative buckling mode shape (eigen- Copyright J.E. Akin. All rights reserved. Page 9 of 11
vector), v BFS. The magnitude of the mode shape displacements are arbitrary and most commercial software normalizes them to range from to 1. BFS Status Note >1 Buckling not predicted The applied loads are less than the theoretical critical loads. 1 Buckling predicted The applied loads exceed the theoretical critical loads. 1 Buckling predicted The applied loads are exactly equal to the theoretical critical loads. -1 Buckling not predicted The buckling occurs when the directions of the applied loads are all reversed. -1 Buckling not predicted The buckling occurs when the directions of the applied loads are all reversed. < -1 Buckling not predicted Buckling is not predicted even if you reverse all loads. Figure 2 Beam element reaction vector, including foundation pressure Copyright J.E. Akin. All rights reserved. Page 1 of 11
Figure 3 Scattering sub-matrices for beam elements with two dof per node Copyright J.E. Akin. All rights reserved. Page 11 of 11