General elastic beam with an elastic foundation General elastic beam with an elastic foundation Figure 1 shows a
Figure 1 shows a beam-column on an elastic foundation. beam-column on an elastic foundation. The beam is connected to The beam is connected to a continuousa continuous series of foundation springs.
series of foundation springs. The other end of The other end of the foundation spring has a the foundation spring has a known displacement,known displacement,
,, which is almostwhich is almost always zero. always zero. The differential equation of The differential equation of equilibrium of an initially straight beam, ofequilibrium of an initially straight beam, of flexural stiffness
flexural stiffness EI EI , resting on a Winkler foundation, of stiffness, resting on a Winkler foundation, of stiffness k k per unit length, with a transverse per unit length, with a transverse load per unit length of
load per unit length of w w , subjected to a tensile axial load, subjected to a tensile axial load N N acting along the x-axis is: acting along the x-axis is:
(())
.
.
(1)
(1)
where
where v(x)v(x) is the transverse displacement of the beam. is the transverse displacement of the beam. If the axial force,
If the axial force, N N , is present, then this is usually called a Beam-Column model. If the axial load is , 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 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.
by solving a linear algebraic system. Beams on elastic
Beams on elastic foundations are fairly foundations are fairly common (like pipelines, rail common (like pipelines, rail lines or drill lines or drill strings). strings). However,However, most introductions to beams assume no axial load,
most introductions to beams assume no axial load, nor any foundation support. nor any foundation support. Then, the aboveThen, the above general beam equation reduces to that covered in an introduction to solid mechanics:
general beam equation reduces to that covered in an introduction to solid mechanics:
()
()
..
(2)(2) The homogeneous solution ofThe homogeneous solution of this equation is this equation is simply a cubic simply a cubic polynomial with four constants. polynomial with four constants. EitherEither beam equation is
beam equation is a fourth-order ordinary differential equation. a fourth-order ordinary differential equation. Therefore it will Therefore it will generally need fourgenerally need four boundary conditions.
boundary conditions.
Figure 1 A beam-column on an elastic foundation Figure 1 A beam-column on an elastic foundation Related physical quantities are the slope,
Related physical quantities are the slope,
(())
(())
, the bending moment,, the bending moment,()
()
()()
, and, and the transverse shear force,the transverse shear force,
()
()
()()
. . The sign The sign conventions are that conventions are that the position,the position, x x , is positive, is positive to the right, the deflection,to the right, the deflection, v v , point forces, P , point forces, P , and the line load, w , and the line load, w , are positive in the, are positive in the y y -direction-direction (upward), and the
(upward), and the slopes and slopes and moments are positive in moments are positive in the counter-clockwise direction. the counter-clockwise direction. Since theSince the foundation springs are restrained, there can be no
foundation springs are restrained, there can be no rigid body motions of the rigid body motions of the beam. beam. In other words,In other words, the assembled equations should never be singular and it is acceptable to just have non-essential the assembled equations should never be singular and it is acceptable to just have non-essential boundary conditions applied
boundary conditions applied to the to the beam. beam. A simple A simple Winkler foundation model like this Winkler foundation model like this one can one can pushpush or pull on the beam as needed and no gaps can occur.
The distributed load per unit length can include point transverse shear loads
The distributed load per unit length can include point transverse shear loads , V , V , by using the Dirac, by using the Dirac Delta distribution.
Delta distribution. Likewise, employing Likewise, employing a doublet distribution a doublet distribution in definingin defining w(x)w(x) allows for the inclusion allows for the inclusion of point couples, or moments,
of point couples, or moments, M M . . Engineers designing beams Engineers designing beams are usually are usually interested in the localinterested in the local moment and transvers shear force since they define the stress levels and the material failure criteria. moment and transvers shear force since they define the stress levels and the material failure criteria. The exact solution of the homogeneous (
The exact solution of the homogeneous ( w==0 w==0 ) general form given above is) general form given above is given in terms ofgiven in terms of
hyperbolic sines and cosines. Based on Tong’s Theorem, exact solutions at the nodes are obtained if hyperbolic sines and cosines. Based on Tong’s Theorem, exact solutions at the nodes are obtained if such functions are
such functions are used as the used as the interpolation functions in a interpolation functions in a finite element model. finite element model. Advanced elementsAdvanced elements of that type have been applied with excellent results.
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 To apply the Galerkin weak form the governing ODE is multiplied by v(x) and the integral over the length of the beam is
length of the beam is set to zero. set to zero. The highest derivative term is always The highest derivative term is always present, and needs to bepresent, and needs to be integrated twice by
integrated twice by parts to reduce parts to reduce the inter-element continuity requirement. the inter-element continuity requirement. That integral becomesThat integral becomes
(())()()
(())(())
(())(())
(())()()
which brings the
which brings the non-essential boundary conditions into the integral non-essential boundary conditions into the integral form. form. The first term is The first term is thethe product of the displacement and the point shear force at both ends, while the second 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
product of the slope and the point moment at both ends. the point moment at both ends. Thus, the contributions from the highestThus, the contributions from the highest derivative term in the ODE are
derivative term in the ODE are
()
()
()
()
(())()()
Therefore, the integral form contains second derivatives (
Therefore, the integral form contains second derivatives (
⁄⁄
))
as its highest derivative term. as its highest derivative term. The presence of second derivatives in the integral form means that calculus requires the elements to 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 (have inter-element continuity of the deflection and the slope ( v(x)v(x) and and
(())
(())
). ). Such Such elementselements are said to have an inter-element continuity ofare said to have an inter-element continuity of
. . Therefore, each shared beam Therefore, each shared beam node must have node must have twotwo degrees of freedom (dof), at least.degrees of freedom (dof), at least.
The presence of the second derivatives in the integral form also means that the essential boundary The presence of the second derivatives in the integral form also means that the essential boundary conditions (EBC) will specify
conditions (EBC) will specify
or or
or b or both at a oth at a point. point. The non-essential boundary conditions (NBC)The non-essential boundary conditions (NBC) are that the transverse shear force,are that the transverse shear force,
, and/or the moment,, and/or the moment,
, is given at a point., is given at a point. When the deflection is given, thenWhen the deflection is given, then the shear force is a the shear force is a reaction quantity. reaction quantity. When the slope is specified,When the slope is specified, then the moment is a reaction.
then the moment is a reaction.
Calculus continuity requires the elements share
Calculus continuity requires the elements share their slope and their slope and deflection values. deflection values. Thus, classic beamThus, classic beam elements approximated by at least a cubic polynomial having
elements approximated by at least a cubic polynomial having
continuity can give nodally exact continuity can give nodally exact deflections and slopes.based on Hermite interpolation polynomials.
based on Hermite interpolation polynomials. The classic beam element is oThe classic beam element is often said to be ften said to be a Hermitea Hermite cubic.
cubic. Some typical Some typical Hermite polynomials are Hermite polynomials are given in given in the Appendix.the Appendix. A line el
A line element witement with only twh only two nodes o nodes will provwill provide the fouide the four constants nr constants needed to eeded to define a define a cubiccubic polynomial. Such elements are referred
polynomial. Such elements are referred to here as to here as classic beam elements. classic beam elements. But since designers But since designers areare interested in the second and third derivatives of the solution, classic beam elements give very poor 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.
design results unless a large number of them are used. For a typical beam (
For a typical beam ( k = 0 k = 0 ) with a cross-sectional moment of inertia,) with a cross-sectional moment of inertia, I I , length,, length, LL, a depth of, a depth of hh, and, and material with an elastic modulus of
material with an elastic modulus of E E and a coefficient of thermal expansion of and a coefficient of thermal expansion of α α , the corresponding, the corresponding
matrices for the equilibrium of a single classic beam element, are matrices for the equilibrium of a single classic beam element, are
, (3), (3) wherewhere
and and
are the load per unit length at the first (left) and second end, respectively, and are the load per unit length at the first (left) and second end, respectively, and
is is the temperature increase, overthe temperature increase, over hh, from the top to the bottom of the beam. Also, the, from the top to the bottom of the beam. Also, the
and and
terms terms are point shear forces and point moments applied to the nodes as external loadings and/or unknown are point shear forces and point moments applied to the nodes as external loadings and/or unknown reactions.reactions. In matrix In matrix symbol notation this symbol notation this equation is wequation is written asritten as
..
(4)(4)The classic beam element has a deflection given by (for
The classic beam element has a deflection given by (for
⁄⁄
):):(())
((
))
((
))
((
))
((
))
(5)(5) A slope oA slope off
(())
((
))
((
))
(
(
)
)
((
)
)
(6)(6) A linear bA linear bending moending moment ofment of
(())
((
))
((
))
((
))
((
))
(7)(7) and a constant transverse shear force ofand a constant transverse shear force of
(())
(())
(())
(())
(())
. . (8)(8)Many exact solutions for beams not on an elastic foundation are tabulated in typical structural design Many exact solutions for beams not on an elastic foundation are tabulated in typical structural design manuals.
manuals. They show They show that for tthat for the most he most common load and common load and support conditions the support conditions the exact solution is exact solution is aa third, fourth or fifth degree polynomial.
third, fourth or fifth degree polynomial. Thus, a fifth degree polynomial will generally give Thus, a fifth degree polynomial will generally give a verya very accurate or exact beam solution with only
accurate or exact beam solution with only a few elements, perhaps with a few elements, perhaps with only one element. only one element. However,However, if a foundation 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 The Hermite family of interpolation polynomials can be increased in order by increasing the number of nodes and/or by
nodes and/or by increasing the continuity level at increasing the continuity level at each node. each node. Here, a Here, a more accurate three-nodemore accurate three-node
beam element will be presented.beam element will be presented. This will be created by This will be created by adding a third (middle) node to adding a third (middle) node to the element.the element. That adds two more degrees of freedom and raises the polynomial to a fifth degree (a quintic
That adds two more degrees of freedom and raises the polynomial to a fifth degree (a quintic polynomial).
polynomial). For simplicity, all of For simplicity, all of the coefficients in Eq. (1) the coefficients in Eq. (1) are taken as are taken as constants evaluated at constants evaluated at thethe center of the
center of the element length. element length. That allows the That allows the corresponding finite element matrices of the corresponding finite element matrices of the fifth-dfifth-degreeegree element to be expressed in closed form in terms of the element deflection and slope at each of its element to be expressed in closed form in terms of the element deflection and slope at each of its three nodes [
three nodes [
The element The element interpolation relations and the interpolation relations and the corresponding slope,corresponding slope, moment and shear distributions are:moment and shear distributions are:
(())
((
((
))
((
))
((
))
))
((
))
((
))
(9)(9)(())
(())
((
))
((
))
((
))
((
))
((
))
((
))
(10)(10)()
()
()
()
((
))
((
))
((
))
((
))
((
))
((
))
(11)(11)()
()
()
()
((
))
((
))
((
))
((
))
((
))
((
))
(12)(12)The 6 by 6 element matrix equations of equilibrium of a single element are: The 6 by 6 element matrix equations of equilibrium of a single element are:
((
))
(13)(13)where {
where {
}}TT = [ = [
] are the generalized nodal displacements and rotations, and the] are the generalized nodal displacements and rotations, and the
are are the generalized resultant nodal loading the generalized resultant nodal loading forces and moments. forces and moments. TheThe
terms are stiffnessterms are stiffness matrices arising from thematrices arising from the first three terms in the first three terms in the differential equation. differential equation. Usually, only the Usually, only the first is non-first is non-zero.
zero. This system is singular until enough This system is singular until enough boundary conditions are applied to prevent rigid bodyboundary conditions are applied to prevent rigid body translations and rotations.
translations and rotations.
Since this element has three nodes, the loading per unit length can be input at each of those nodes. 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,
The resultant element external loading vectors due to point forces,
, and couples,, and couples,
, and the, and the quadratic distributed load values,quadratic distributed load values,
, at each node (or constant, at each node (or constant
, or linear, or linear
∫∫
(14)(14) where thewhere the
are the six Hermite beam interpolations, and the are the six Hermite beam interpolations, and the
are the three quadratic Lagrangian are the three quadratic Lagrangian interpolationinterpolations for s for the distributed loads. the distributed loads. The integral of their The integral of their product forms a product forms a rectangular six by rectangular six by threethree array to convert
array to convert distributed loads to concentrated shears and distributed loads to concentrated shears and moments at the moments at the beam nodes. beam nodes. Likewise,Likewise, it can be shown that a temperature difference through the depth of a beam causes only end moment it can be shown that a temperature difference through the depth of a beam causes only end moment loadings given by
loadings given by
⁄⁄
.. A constaA constant transvnt transverse loaerse load per unit ld per unit length redength reduces the uces the resultanresultant load ant load and moment d moment vector tovector to
,,
(15)(15) The symmetric flexural stiffness matrix for the three noded quintic element isThe symmetric flexural stiffness matrix for the three noded quintic element is
(16)(16)The symmetric beam-column (or geometric stiffness) matrix due to any axial load is The symmetric beam-column (or geometric stiffness) matrix due to any axial load is
(17)
(17)
The element foundation stiffness is The element foundation stiffness is
(18)
(18)
Since it is common for beams to have piecewise constant properties, the classic beam element and 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 beam analysis problems can be solved in Using either the classic or quintic beam element some beam analysis problems can be solved in closed antalytic form.
closed antalytic form. As an example of As an example of the three-node beam, consider a fixed-fixed beam with athe three-node beam, consider a fixed-fixed beam with a constant line
constant line load. load. The essential The essential boundary conditions boundary conditions areare
. The. The external point force and moment at center node 2 are zero (external point force and moment at center node 2 are zero (
). ). The The middle two middle two rowsrows define the remaining unknown center point generalized displacements:define the remaining unknown center point generalized displacements:
. . (19)(19) Multiplying by the inverse of the square matrix gives the middle node solutions:Multiplying by the inverse of the square matrix gives the middle node solutions:
⁄⁄
⁄⁄
(20)(20) which are exact.which are exact. The slope was The slope was expected to expected to be zero due be zero due to symmetry. to symmetry. It could have It could have been used been used as aas a boundary condition to compute
boundary condition to compute
from one equation. from one equation.Here, the reactions on the left are found from the first two rows of the equilibrium equations (since all Here, the reactions on the left are found from the first two rows of the equilibrium equations (since all the displacements are now known):
the displacements are now known):
(21)(21)
⁄⁄
..Likewise, at the right end, utilizing the last two rows of the equilibrium equations Likewise, at the right end, utilizing the last two rows of the equilibrium equations
⁄⁄
..
Note that the net resultant force is
Note that the net resultant force is – –wLwL, which is equal and opposite to the applied transverse load., which is equal and opposite to the applied transverse load. The net external
The net external moment is zero. moment is zero. Usually static equilibrium is taught Usually static equilibrium is taught in undergraduate classes in undergraduate classes withwith Newton’s Laws.
Newton’s Laws. You can use them to You can use them to verify that the computed reactverify that the computed reactions do indeed satisfy tions do indeed satisfy that thehat the sum of the forces is zero, and that the sum of the moments, taken at any reference point, is zero. 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
Since a 5-th degree element was used this problem is element was used this problem is exact everywhere. exact everywhere. That includes the momentThat includes the moment and shear diagrams.
and shear diagrams.
The exact shear diagrams of
The exact shear diagrams of
can have a discontinuity (due to point loads can have a discontinuity (due to point loads P P ) and the exact) and the exact moment diagram ofmoment diagram of
can have a discontinuity (due to point momentscan have a discontinuity (due to point moments M M ). ). The The mesh mesh should should bebe constricted such that any point source occursconstricted such that any point source occurs at an interface node at an interface node between elements. between elements. Those itemsThose items can be discontinuous at element interfaces because the calculus requirement for splitting integrals on can be discontinuous at element interfaces because the calculus requirement for splitting integrals on says that
says that vv andand v’ v’ must be must be continuous there. continuous there. That explains why That explains why a two nodea two node
Hermite interpolation Hermite interpolationwas not chosen
was not chosen to create a quintic beam. to create a quintic beam. That choice could be That choice could be made only if the made only if the beam was notbeam was not allowed to have point s
allowed to have point sources. ources. Inside any element the solution is continuous and Inside any element the solution is continuous and thus isthus is
at any at any non-interface node. In other words, inside any standard element you can calculate an infinite number non-interface node. In other words, inside any standard element you can calculate an infinite numberof continuous derivatives (but the vast majority is z
of continuous derivatives (but the vast majority is zero). ero). If you were If you were to apply a point source to apply a point source at anat an interior node then the second and third derivatives would still be continuous and the shear and interior node then the second and third derivatives would still be continuous and the shear and moment diagrams in that element would be wrong.
moment diagrams in that element would be wrong.
Next consider the involvement of the foundation in the post-processing needed to recover the Next consider the involvement of the foundation in the post-processing needed to recover the reactions on each beam element.
reactions on each beam element. Before the deflections of a Before the deflections of a beam on an elastic foundation (BOEF)beam on an elastic foundation (BOEF) are computed, a typical element free body
are computed, a typical element free body diagram would be similar to Figure 1. diagram would be similar to Figure 1. As before, anyAs before, any external point load,
external point load, P P , and/or any point couple,, and/or any point couple, M M , should be placed at an element interface node and, should be placed at an element interface node and they go directly into the
they go directly into the first or second equilibrium row of the node first or second equilibrium row of the node at that point. at that point. A distributed lineA distributed line load is integrated to become a c
load is integrated to become a consistent load vector. onsistent load vector. Those two effects are detailed in Those two effects are detailed in Eq. 14. BothEq. 14. Both the beam stiffness and the foundation stiffness resist the external applied loads. The element
the beam stiffness and the foundation stiffness resist the external applied loads. The element equilibrium equation (for no axial force) is
equilibrium equation (for no axial force) is
((
))
(22)(22)The assembled equations have a similar form and are solved for the system displacements, The assembled equations have a similar form and are solved for the system displacements, v v ..
Returning to the element to get the beam member reactions, the now known element degrees of Returning to the element to get the beam member reactions, the now known element degrees of freedom,
freedom,
,, are gathered. are gathered. The free body The free body diagram of the element now diagram of the element now has a significant changehas a significant change because the foundation is now applying a known pressure that opposes the displacement of the because the foundation is now applying a known pressure that opposes the displacement of the beam.beam. That That pressure,pressure,
(())
()()
, is a polynomial of the same degree as the assumed, is a polynomial of the same degree as the assumed displacements.displacements. That state is That state is shown in Figure shown in Figure 2, which defines 2, which defines the sought element reaction the sought element reaction vector,vector,
..Now, there is a new consistent load vector to evaluate during post processing to find the element Now, there is a new consistent load vector to evaluate during post processing to find the element reactions.
reactions. It comes from the It comes from the foundation pressure and is foundation pressure and is proportional to the deproportional to the deflectionflection::
∫∫
(())
∫∫
.. (23)(23) The foundation stiffness caused the new load, but that load and others must be resisted by only the The foundation stiffness caused the new load, but that load and others must be resisted by only the beam stiffness.beam stiffness. Now the Now the beam member equilibrium equations define beam member equilibrium equations define the reactions on the reactions on the nodes the nodes ofof the beam, the beam,
::
(24)(24) or or
..
(25)(25) Now that there is more than one dof per node the scatter/gather operations previously used for axial Now that there is more than one dof per node the scatter/gather operations previously used for axial models have become more generalized.models have become more generalized. The usual convention for numbering the The usual convention for numbering the unknowns is tounknowns is to count all of the unknowns
count all of the unknowns at a node at a node before moving on to another before moving on to another node. node. Therefore, the displacementsTherefore, the displacements are associated with the odd number of rows and columns in a matrix, while the slopes contributions are associated with the odd number of rows and columns in a matrix, while the slopes contributions
Copyright J.E. Akin. All rights reserved. Page8 of11
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 graphically illustrates the sub-matrices scatter operations for a line element with two unknowns per node.
Figure 2 Beam element reaction vector, including foundation pressure
The general numbering scheme that relates local element equation numbers to the system equation numbers refer to Figure 4. As a detailed example of the change in scatter operations, consider a long pipeline with hundreds of three node beam element that are numbered in a random fashion. Figure 5 shows typical element connectivity data for the hypothetical example. Since there are two dof per node there are six local equations associated with each quintic beam element. They have to be scattered to a much larger set of system equilibrium equations. Figure 6 illustrates how all of the local generalized force vector terms are scattered, as well as showing the same information for six typical terms selected for the element local stiffness matrix.
Figure 3 Scattering sub-matrices for line elements with two dof per node Figure 3 Scattering sub-matrices for line elements with two dof per node
Figure 4 Calculating equation numbers for general scatter operations Figure 4 Calculating equation numbers for general scatter operations
Figure 5 Example data for gather/scatter of element number 21 Figure 5 Example data for gather/scatter of element number 21
Figure 6 Scattering typical matrix terms from example element number 21 Figure 6 Scattering typical matrix terms from example element number 21 Appendix
Appendix
Figure 8 The first three Hermite interpolation options for elements with two nodes Figure 8 The first three Hermite interpolation options for elements with two nodes
