[ 𝐼 ]− + − ( )=0. (1) - rice university per unit length, 𝜌 , replaced with the stiffness...

Quintic beam closed form matrices (revised 3/24/15)

Copyright J.E. Akin. All rights reserved. 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



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



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.



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


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


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.



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



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



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)



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



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):



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



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



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:



|(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.



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



Figure 5 Scattering sub-matrices for beam elements with two dof per node



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.

[ 𝐼 ]− + − ( )=0. (1) - rice university per unit length, 𝜌 , replaced with the stiffness...

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