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Quintic beam closed form matrices (revised 3/24/15)
Copyright J.E. Akin. All rights reserved. Page 1 of 16
General elastic beam with an elastic foundation
Introduction
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, 𝑣𝑓,
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:
𝑑2
𝑑𝑥2[𝐸𝐼
𝑑2𝑣
𝑑𝑥2] − 𝑁
𝑑2𝑣
𝑑𝑥2+ 𝑘𝑣 − 𝑤(𝑥) = 0. (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:
𝑑2
𝑑𝑥2[𝐸𝐼
𝑑2𝑣
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
Quintic beam closed form matrices (revised 3/24/15)
Copyright J.E. Akin. All rights reserved. Page 2 of 16
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.
Typical exact analytic solutions for statically determinate and statically indeterminate beams are
shown in Figures 2 and 3. A classic cubic beam element will give analytically exact node
displacements and slopes, but not give accurate interior moments and gives poor interior shear force
values. Using several cubic elements will give reasonable estimates for the moment and shear. One
quintic beam element will yield the exact answers everywhere, including the moment and shear, for
the four cases included in those figures.
Figure 2 AISC sample indeterminate beam tables
Quintic beam closed form matrices (revised 3/24/15)
Copyright J.E. Akin. All rights reserved. Page 3 of 16
Figure 3 AISC sample determinate and indeterminate beam tables
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.
Quintic beam closed form matrices (revised 3/24/15)
Copyright J.E. Akin. All rights reserved. Page 4 of 16
The exact solution of the homogeneous (w==0) 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.
Galerkin weak form
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
𝐼𝐸 = ∫ 𝑣𝑑2
𝑑𝑥2
𝐿
0
(𝐸(𝑥)𝐼(𝑥)𝑑2𝑣
𝑑𝑥2)𝑑𝑥
𝐼𝐸 = [𝑣𝑑
𝑑𝑥(𝐸(𝑥)𝐼(𝑥)
𝑑2𝑣
𝑑𝑥2)]0
𝐿
− [𝑑𝑣
𝑑𝑥(𝐸(𝑥)𝐼(𝑥)
𝑑2𝑣
𝑑𝑥2)]0
𝐿
+∫𝑑2𝑣
𝑑𝑥2
𝐿
0
(𝐸(𝑥)𝐼(𝑥)𝑑2𝑣
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
𝐼𝐸 = [𝑣 𝑉(𝑥)]0𝐿 − [
𝑑𝑣
𝑑𝑥 𝑀(𝑥)]
0
𝐿
+∫𝑑2𝑣
𝑑𝑥2
𝐿
0
(𝐸(𝑥)𝐼(𝑥)𝑑2𝑣
𝑑𝑥2)𝑑𝑥
Therefore, the integral form contains second derivatives (𝑑2𝑣 𝑑𝑥2) ⁄ 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 C1. 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 C1 continuity can give nodally exact
deflections and slopes. One-dimensional polynomial interpolation with C1 or C2 continuity usually is
based on Hermite interpolation polynomials. The classic beam element is often said to be a Hermite
cubic.
Quintic beam closed form matrices (revised 3/24/15)
Copyright J.E. Akin. All rights reserved. Page 5 of 16
From the Galerkin weak form, the elastic stiffness matrix is
𝑲𝑬𝒆 = ∫ 𝑩𝒆𝑻
𝐿𝑒𝐸𝐼(𝑥) 𝑩𝒆 𝑑𝑥, 𝑩𝒆 =
𝑑2𝑯(𝑥)
𝑑𝑥2
If the axial load is present then the beam geometric stiffness is
𝑲𝑵𝒆 = ∫
𝑑𝑯(𝑥)
𝑑𝑥
𝑇
𝐿𝑒𝑁𝑒
𝑑𝑯(𝑥)
𝑑𝑥𝑑𝑥
If the elastic foundation is present it has a stiffness matrix similar to the mass matrix, with the mass
per unit length, 𝜌𝑒𝐴𝑒, replaced with the stiffness per unit length, 𝑘𝑒:
𝑲𝒌𝒆 = ∫ 𝑯𝒆𝑻
𝐿𝑒𝑘𝑒 𝑯𝒆𝑑𝐿
For dynamic response or natural frequency calculations it is necessary to include the physical mass matrix defined for any Hermite beam element as
𝒎𝒆 = ∫ 𝑯𝒆𝑻
𝐿𝑒𝜌𝑒𝐴𝑒 𝑯𝒆𝑑𝐿
The resultant of the load per unit, w(x), is
𝑭𝒘𝒆 = ∫ 𝑯𝒆𝑻
𝐿𝑒𝑤(𝑥) 𝑑𝑥 = ∫ 𝑯𝒆𝑻
𝐿𝑒 𝒉𝑒(𝑥) 𝑑𝑥 𝒘𝒆 ≡ 𝑹𝒆 𝒘𝒆
when the load is interpolated in the Lagrange form, 𝑤(𝑥) = 𝒉𝑒(𝑥) 𝒘𝒆, which requires the line load to
be input at the nodes of the element, like a property of the element.
When the top and bottom of the beam are at different temperatures that causes a transverse
temperature gradient of ∆𝑇 ℎ⁄ , where h is the depth of the beam. That introduces a curvature to the
beam, or a moment if it is restrained. The resultant of the initial transverse thermal strain is
𝑭𝜶𝒆 = ∫ 𝑩𝒆
𝑻
𝐿𝑒𝐸𝐼𝑒(𝑥)
𝛼𝑒∆𝑇(𝑥)
ℎ(𝑥)
𝑒
𝑑𝑥
that for constant properties reduces to
𝑭𝜶𝒆 =
𝛼𝑒∆𝑇𝑒𝐸𝐼𝑒
ℎ𝒆∫ 𝑩𝒆
𝑻
𝐿𝑒(𝑥)𝑑𝑥
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
Quintic beam closed form matrices (revised 3/24/15)
Copyright J.E. Akin. All rights reserved. Page 6 of 16
says that v and v’ must be continuous there. That explains why a two node 𝐶2 Hermite interpolation
was not chosen to create a quintic beam (below). 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.
Cubic beam element
A line element with only two nodes with two dof per node will provide the four constants needed to
define a cubic polynomial element (L2_C1). 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.
For a typical beam (k = 0) 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
matrices for the equilibrium of a single classic beam element, are
𝐸𝐼
𝐿3[
12 6𝐿 6𝐿 4𝐿2
−12 6𝐿−6𝐿 2𝐿2
−12 −6𝐿 6𝐿 2𝐿2
12 −6𝐿 −6𝐿 4𝐿2
] {
𝑣1𝜃1𝑣2𝜃2
} = {
𝑉1𝑀1𝑉2𝑀2
} +𝐿
60[
21 93𝐿 2𝐿9 21−2𝐿 −3𝐿
] {𝑤1𝑤2} +
𝛼∆𝑇𝐸𝐼
ℎ{ 0 1 0−1
}, (3)
where 𝑤1 and 𝑤2 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 𝑉𝑗 and 𝑀𝑗 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
𝑲𝒆𝒗𝒆 = 𝑭𝑷𝒆 + 𝑭𝒘
𝒆 + 𝑭𝜶𝒆 . (4)
The classic beam element has a deflection given by (for r = x L⁄ ):
𝑣(𝑥) = 𝑣1(1 − 3𝑟2 + 2𝑟3) + 𝜃1(𝑟 − 2𝑟
2 + 𝑟3)𝐿 + 𝑣2(3𝑟2 − 2𝑟3) + 𝜃2(𝑟
3 − 𝑟2)𝐿 (5)
A slope of
𝜃(𝑥) = 𝑣1(6𝑟2 − 6𝑟)/𝐿 + 𝜃1(1 − 4𝑟 + 3𝑟
2) + 𝑣2(6𝑟 − 6𝑟2)/𝐿 + 𝜃2(3𝑟
2 − 2𝑟) (6)
A linear bending moment of
𝑀(𝑥) = [𝑣1(12𝑟 − 6) + 𝜃1(6𝑟 − 4)𝐿 + 𝑣2(6 − 12𝑟) + 𝜃2(6𝑟 − 2)𝐿]/𝐿2 (7)
and a constant transverse shear force of
𝑉(𝑥) = [𝑣1(12) + 𝜃1(6)𝐿 + 𝑣2(−12) + 𝜃2(6)𝐿]/𝐿3. (8)
For the classic cubic beam the mass matrix becomes
Quintic beam closed form matrices (revised 3/24/15)
Copyright J.E. Akin. All rights reserved. Page 7 of 16
𝒎𝒆 =𝜌𝑒𝐴𝑒𝐿𝑒
420[
156 22𝐿𝑒
22𝐿𝑒 4𝐿𝑒2
54 −13𝐿𝑒
13𝐿𝑒 −3𝐿𝑒2
54 13𝐿𝑒
−13𝐿𝑒 −3𝐿𝑒2156 −22𝐿𝑒
−22𝐿𝑒 4𝐿𝑒2
] (9)
Many exact solutions for beams not on an elastic foundation are tabulated in typical structural design
manuals. Four cases from the American Institute for Steel Construction (AISC) tables are given in
Figure 2. 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. If the elastic foundation is present it has a stiffness matrix similar to the mass matrix, with the
mass per unit length, 𝜌𝑒𝐴𝑒, replaced with the stiffness per unit length, 𝑘𝑒. If the axial load is present
then the cubic beam geometric stiffness is
𝑲𝑵𝒆 =
𝑁𝑒
30 𝐿𝑒[
36 3𝐿𝑒
3𝐿𝑒 4𝐿𝑒2−36 3𝐿𝑒
−3𝐿𝑒 −𝐿𝑒2
−36 −3𝐿𝑒
3𝐿𝑒 −𝐿𝑒236 −3𝐿𝑒
−3𝐿𝑒 4𝐿𝑒2
]
Both of these stiffness matrices would be added to the square matrix in Equation 3.
Quintic beam element
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 C1
beam element will be presented, L3_C1. 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 [𝑣1 𝜃1 𝑣2 𝜃2 𝑣3 𝜃3]. The element interpolation relations and the corresponding
slope, moment and shear distributions are:
𝑣(𝑟) = [𝑣1(1 − 23𝑟2 + 66𝑟3 − 68𝑟4 + 24𝑟5) + 𝜃1(𝑟 − 6𝑟
2 + 13𝑟3 − 12𝑟4 + 4𝑟5)𝐿 +𝑣2(16𝑟
2 − 32𝑟3 + 16𝑟4) + 𝜃2(−8𝑟2 + 32𝑟3 − 40𝑟4 + 16𝑟5)𝐿
+𝑣3(7𝑟2 − 34𝑟3 + 52𝑟4 − 24𝑟5) + 𝜃3(−𝑟
2 + 5𝑟3 − 8𝑟4 + 4𝑟5)𝐿] (11)
𝜃(𝑟) = 𝑣′(𝑟) = [𝑣1(−46𝑟 + 198𝑟2 − 272𝑟3 + 120𝑟4)/𝐿 + 𝜃1(1 − 12𝑟
+ 39𝑟2 − 48𝑟3 + 20𝑟4)] +𝑣2(32𝑟 − 96𝑟
2 + 64𝑟3)/𝐿 + 𝜃2(−16𝑟 + 96𝑟2 − 160𝑟3 + 80𝑟4)
+𝑣3(14𝑟 − 102𝑟2 + 208𝑟3 − 120𝑟4)/𝐿 + 𝜃3(−2𝑟 + 15𝑟
2 − 32𝑟3 + 20𝑟4)] (12)
𝑀(𝑟) = 𝐸𝐼𝑣′′(𝑟) = 𝐸𝐼[𝑣1(−46 + 396𝑟 − 816𝑟2 + 480𝑟3)/𝐿 + 𝜃1(12 + 78𝑟 − 144𝑟
2 + 80𝑟3) +𝑣2(32 − 192𝑟 + 192𝑟
2)/𝐿 + 𝜃2(−16 + 192𝑟 − 480𝑟2 + 320𝑟3)
+𝑣3(14 − 204𝑟 + 624𝑟2 − 480𝑟3)/𝐿 + 𝜃3(−2 + 30𝑟 − 96𝑟
2 + 80𝑟3)]/𝐿 (13)
Quintic beam closed form matrices (revised 3/24/15)
Copyright J.E. Akin. All rights reserved. Page 8 of 16
𝑉(𝑟) = 𝐸𝐼𝑣′′′(𝑟) = 𝐸𝐼[𝑣1(396 − 1,632𝑟 + 1,440𝑟2)/𝐿 + 𝜃1(78 − 288𝑟 + 240𝑟
2)
+𝑣2(−192 + 384𝑟)/𝐿 + 𝜃2(192 − 960𝑟 + 960𝑟2)
+𝑣3 (−204 + 1,248𝑟 − 1,350𝑟2)/𝐿 + 𝜃3(30 − 192𝑟 + 240𝑟
2)]/𝐿2 (14)
The 6 by 6 element matrix equations of equilibrium of a single element are:
(𝑲𝑬𝒆 +𝑲𝑵
𝒆 +𝑲𝒌𝒆){𝒗𝒆} = {𝑭𝒆} (15)
where {𝒗𝒆}T = [𝑣1, 𝜃1 , 𝑣2, 𝜃2, 𝑣3, 𝜃3] are the generalized nodal displacements and rotations, and the
{𝑭𝒆} are the generalized resultant nodal loading forces and moments. The 𝑲𝒆 terms are stiffness
matrices arising from the first three terms in the differential equation. Usually, only the first is non-
zero. 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, 𝑉𝑘, and couples, 𝑀𝑘, and the
quadratic distributed load values, 𝑤𝑘, at each node (or constant 𝑤1 = 𝑤2 = 𝑤3, or linear 𝑤2 =
(𝑤1 + 𝑤3) 2⁄ values) are
𝑭𝑷𝒆 =
{
𝑉1𝑀1𝑉2𝑀2
𝑉3𝑀3}
, 𝑭𝒘𝒆 = ∫ 𝑯𝒆𝑻𝒉𝒘𝒘𝒆
𝐿𝑒𝑑𝑙 =
𝐿
14,700
[ 1,995 1,540 −105105𝐿 140𝐿 0560 6,720 560
−280𝐿 0 280𝐿−105 1,540 1,9950 −140𝐿 −105𝐿]
{
𝑤1𝑤2𝑤3} (16)
where the 𝑯𝒆 are the six Hermite beam interpolations, and the 𝒉𝒘 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 𝑭𝜶𝒆 = 𝛼 ∇𝑇 𝐸𝐼 ℎ⁄ [0 1 0 0 0 − 1].
A constant transverse load per unit length reduces the resultant load and moment vector to
𝑭𝑤𝑇 =
𝑤𝐿
14,700[3,430 245𝐿 7,840 0 3,430 −245𝐿], 𝑓𝑜𝑟 𝑤1 = 𝑤2 = 𝑤3 = 𝑤 (17)
The symmetric flexural stiffness matrix for the three noded quintic element is
𝑲𝑬 =𝐸𝐼
35𝐿3
[ 5,092 1,138𝐿 −3,584
1,138𝐿 332𝐿2 −896𝐿−3,584 −896𝐿 7,168
1,920𝐿 −1,508 242𝐿
320𝐿2 −242𝐿 38𝐿2
0 −3,584 896𝐿
1,920𝐿 320𝐿2 0−1,508 −242𝐿 −3,584
242𝐿 38𝐿2 896𝐿
1,280𝐿2 −1,920𝐿 320𝐿2
−1,920𝐿 5,092 −1,138𝐿
320𝐿2 −1,138𝐿 332𝐿2 ]
(18)
The symmetric beam-column (or geometric stiffness) matrix due to any axial load is
Quintic beam closed form matrices (revised 3/24/15)
Copyright J.E. Akin. All rights reserved. Page 9 of 16
𝑲𝑵 =𝑁
5,670 𝐿
[ 15,012 351 𝐿
351 𝐿 252 𝐿2−13,824 2,160 𝐿
−432 𝐿 −72 𝐿2−1,188 −81 𝐿
81𝐿 −45 𝐿2
−13,824 −432 𝐿
2,160 𝐿 −72 𝐿2 27,648 0
0 2,304 𝐿2−13,824 252 𝐿
−2,160 𝐿 −72 𝐿2
−1,188 81𝐿
−81 𝐿 −45 𝐿2−13,824 −2,160 𝐿
252 𝐿 −72 𝐿2 15,012 −351 𝐿
−351 𝐿 252 𝐿2]
(19)
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
𝑲𝒌 =𝑘 𝐿
13,860
[ 2,092 114 𝐿
114 𝐿 8 𝐿2880 −160 𝐿88 𝐿 −12 𝐿2
262 −29 𝐿29 𝐿 −3 𝐿2
880 88 𝐿−160 𝐿 −12 𝐿2
5,632 0
0 128 𝐿2880 −88 𝐿160 𝐿 −12 𝐿2
262 29 𝐿−29 𝐿 −3 𝐿2
880 160 𝐿−88 𝐿 −12 𝐿2
2,092 −114 𝐿
−114 𝐿 8 𝐿2 ]
(20)
If the mass matrix, 𝐦𝐞, 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 cross-
sectional area of the beam (so ρA is the mass-per unit length).
Example solutions
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 closed analytic form.
As an example of the three-node beam, consider a fixed-fixed beam with a constant line load. That is
case 15 in Figure 2. The essential boundary conditions are 𝑣1 = 0, 𝜃1 = 0, 𝑣3 = 0, 𝜃3 = 0. The
external point force and moment at center node 2 are zero (𝑉2 = 0,𝑀2 = 0). The middle two rows
define the remaining unknown center point generalized displacements:
𝐸𝐼
35𝐿3[7,168 0
0 1,280𝐿2] {𝑣2𝜃2} = {
00} +
𝑤𝐿
14,700{7,8400
} + {00}. (21)
Multiplying by the inverse of the square matrix gives the middle node solutions:
{𝑣2𝜃2} =
35𝑤𝐿4
14,700𝐸𝐼[1 7,168⁄ 0
0 1 1,280𝐿2⁄] {7,8400
} =𝑤𝐿4
384𝐸𝐼{10} (22)
which are exact. The slope was expected to be zero due to symmetry. It could have been used as a
boundary condition to compute 𝑣2 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):
Quintic beam closed form matrices (revised 3/24/15)
Copyright J.E. Akin. All rights reserved. Page 10 of 16
𝐸𝐼
35𝐿3[−3,584 1,920𝐿
−896𝐿 320𝐿2]𝑤𝐿4
384𝐸𝐼{10} = {
𝑉1𝑀1} +
𝑤𝐿
14,700{3,430245𝐿
} (23)
{𝑉1𝑀1} =
−𝑤𝐿
2{1𝐿 6⁄
}.
Likewise, at the right end, utilizing the last two rows of the equilibrium equations
{𝑉3𝑀3} =
−𝑤𝐿
2{1
−𝐿 6⁄}.
Both of the reactions are exact. Note that the net 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. You can use them 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.
The interpolated deflection is
𝑣(𝑥) = 0 + 0 + 𝑣2(16𝑟2 − 32𝑟3 + 16𝑟4) + 0 + 0 + 0
Substituting that
𝑣2 =𝑤𝐿4
384𝐸𝐼
and that here r = x/L the deflection estimate is
𝑣(𝑥) =𝑤
24𝐸𝐼[𝑥2𝐿2 − 2𝑥3𝐿 + 𝑥4] =
𝑤 𝑥2
24𝐸𝐼(𝐿 − 𝑥)2
which is exact. The fifth degree polynomial estimate captured the exact fourth degree solution for the
deflections. Since the finite element deflections are exact the shear and moment are also exact.
Since a 5-th degree element was used this problem is exact everywhere.
To model this problem with the cubic element requires using two elements joined at the mid-span, or
one element with the symmetry condition that the slope is zero at the plane of symmetry. Then,
𝐿𝑒 = 𝐿 2⁄ for the single element symmetry model. Applying the above EBC and the requirement from
symmetry that the middle slope is zero gives one equation with one unknown:
𝐸𝐼
(𝐿 2⁄ )3[12]𝑣2 = 0 +
𝑤(𝐿 2⁄ )
60{30} + 0
which again is the exact nodal displacement
𝑣2 =𝑤𝐿4
384𝐸𝐼
but when substituted into the deflection equation for the left half of the span the result is
Quintic beam closed form matrices (revised 3/24/15)
Copyright J.E. Akin. All rights reserved. Page 11 of 16
𝑣(𝑥) = 𝑣2(3𝑟2 − 2𝑟3) =
𝑤𝐿4
384𝐸𝐼[3 (
𝑥
𝐿 2⁄)2
− 2(𝑥
𝐿 2⁄)3
]
𝑣(𝑥) =𝑤
24𝐸𝐼(3
4𝑥2𝐿2− 𝑥3𝐿)
which, of course is missing the fourth degree term in the exact solution.
The cubic approximation of the moment (on the left half of the span) is
𝑀(𝑥) = 𝐸𝐼(𝑥)𝑑2𝑣(𝑥)
𝑑𝑥2=𝑤
12(3
4𝐿2 − 3𝑥𝐿)
which is a straight line, while the exact moment (and the quintic moment) is
𝑀𝑒𝑥𝑎𝑐𝑡(𝑥) =𝑤
12(6𝐿𝑥 − 𝐿2 − 6𝑥2)
Likewise, the cubic shear approximation is just a constant
𝑉(𝑥) = 𝐸𝐼(𝑥)𝑑3𝑣(𝑥)
𝑑𝑥3=−𝑤𝐿
4
while the exact shear (and the quintic shear), with a different sign convention, is
𝑉𝑒𝑥𝑎𝑐𝑡(𝑥) = −𝑤 (𝐿
2− 𝑥).
In other words, the cubic shear estimate is just the average shear, for this problem.
Element reactions
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)
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
(𝑲𝑬𝒆 +𝑲𝒌
𝒆){𝒗𝒆} = {𝑭𝑷𝒆 } + {𝑭𝒘
𝒆 } (24)
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, 𝒗𝒆, 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) = −𝑘𝑓𝑣(𝑥), is a polynomial of the same degree as the assumed
Quintic beam closed form matrices (revised 3/24/15)
Copyright J.E. Akin. All rights reserved. Page 12 of 16
displacements. That state is shown in Figure 4, which defines the sought element reaction vector,
{𝑭𝑹𝒆 }.
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:
𝑭𝒑𝒆 = ∫ 𝑯𝒆𝑻
𝑳𝒆𝑝(𝑥)𝑑𝑥 = −∫ 𝑯𝒆𝑻
𝑳𝒆𝑘𝑓𝑯
𝒆𝑑𝑥 𝒗𝒆 = −𝑲𝒇𝒆𝒗𝒆. (25)
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, {𝑭𝑹𝒆 }:
𝑲𝑬𝒆 {𝒗𝒆} = {𝑭𝑹
𝒆 } + {𝑭𝒑𝒆} + {𝑭𝑷
𝒆 } + {𝑭𝒘𝒆 } (26)
so the reactions are obtained as
𝑲𝑬𝒆 {𝒗𝒆} − {𝑭𝒑
𝒆} − {𝑭𝑷𝒆 } − {𝑭𝒘
𝒆 } = {𝑭𝑹𝒆 }. (27)
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 (ng) 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.
Natural frequencies
To determine the natural frequencies of a beam-column on an elastic foundation the eigen-problem is
|(𝑲𝑬 +𝑲𝑵
+𝑲𝒌 ) − 𝝎𝟐𝑴| = 0 (28)
To determine the natural frequencies of a beam-column the eigen-problem is
|(𝑲𝑬 +𝑲𝑵
) − 𝝎𝟐𝑴| = 0 (29)
Axial load effects
When the axial load is an constant, say 𝑁𝐵, then it factors out of each element matrix and becomes a
global constant:
Quintic beam closed form matrices (revised 3/24/15)
Copyright J.E. Akin. All rights reserved. Page 13 of 16
|(𝑲𝑬 +𝑁𝐵𝑲𝒏
) − 𝝎𝟐𝑴| = 0 (30)
where 𝑲𝒏 is formed with a unit positive load. This shows that any axial load, 𝑁𝐵, 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.
Beam buckling
When the global constant, 𝑁𝐵, is an unknown constant 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 𝑭𝒓𝒆𝒇. In general, a linear static analysis is completed and the stresses are
determined:
(𝑲𝑬 +𝑲𝒌
){𝒗 } = 𝑭𝒓𝒆𝒇 (31)
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 𝑭𝒓𝒆𝒇 , 𝑲𝑵
→ 𝐵𝐹𝑆𝑲𝒓𝒆𝒇
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:
|𝑲𝑬 +𝑲𝒌
+ 𝐵𝐹𝑆 𝑲𝒓𝒆𝒇 | = 0 (32)
This is solved for the value of the BFS. Then the load that would theoretically cause buckling is
𝐵𝐹𝑆 𝑭𝒓𝒆𝒇
. The solution of the eigen-problem also yields the relative buckling mode shape (eigen-
vector), 𝒗𝑩𝑭𝑺. The magnitude of the mode shape displacements are arbitrary and most commercial
software normalizes them to range from 0 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 are exactly equal to the theoretical critical loads.
0↔1 Buckling predicted The applied loads exceed the theoretical critical loads.
-1↔0 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.
Quintic beam closed form matrices (revised 3/24/15)
Copyright J.E. Akin. All rights reserved. Page 14 of 16
The buckling estimate depends heavily on the geometry of the model. If the part has small
irregularities that have been omitted, they can drastically reduce the actual buckling load.
Figure 4 Beam element reaction vector, including foundation pressure
Quintic beam closed form matrices (revised 3/24/15)
Copyright J.E. Akin. All rights reserved. Page 15 of 16
Figure 5 Scattering sub-matrices for beam elements with two dof per node
Quintic beam closed form matrices (revised 3/24/15)
Copyright J.E. Akin. All rights reserved. Page 16 of 16
Exercises
1. Solve case 12 of Figure 2 using one cubic element to show it gives the exact end slope and
the exact reactions. Calculate the element shear and compare it to the exact shear in Figure
2.
2. Solve case 18 of Figure 2 using one cubic element to show it gives the exact end deflection
and slope and the exact reactions. Calculate the element shear and compare it to the exact
shear in Figure 2.
3. Solve case 20 of Figure 2 using one cubic element to show it gives the exact end deflections
and the exact reactions. Calculate the element shear and compare it to the exact shear in
Figure 2.
4. Modify case 15 to have a triangular load of zero at the left and w at the right. Use one quintic
element to solve for the center deflection and slope. Recover the four reactions. Hint, this is
like the example solution but with a new element force (and moment) resultant given by
Equation 16.