Fem by Logan

CIVL 7117 Finite Elements Methods in Structural Mechanics Page 1

A First Course in Finite Elements

Introduction

The finite element method has become a powerful tool for the numerical so-lution of a wide range of engineering problems. Applications range from deforma-tion and stress analysis of automotive, aircraft, building, and bridge structures to field analysis of heat flux, fluid flow, magnetic flux, seepage, and other flow prob-lems.

With the advances in computer technology and CAD systems, complex prob-lems can be modeled with relative ease. Several alternative configurations can be tried out on a computer before the first prototype is built. All of this suggests that we need to keep pace with these developments by understanding the basic theory, modeling techniques, and computational aspects of the finite element method.

In this method of analysis, a complex region defining a continuum is discre-tized into simple geometric shapes called finite elements. The material proper-ties and the governing relationships are considered over these elements and ex-pressed in terms of unknown values at element corners. An assembly process, duly considering the loading and constraints, results in a set of equations. Solu-tion of these equations gives us the approximate behavior of the continuum.

Historical Background

Basic ideas of the finite element method originated from advances in aircraft structural analysis. In 1941, Hrenikoff presented a solution of elasticity problems using the “frame work method.” Courant’s paper, which used piecewise polyno-mial interpolation over triangular subregions to model torsion problems, appeared in 1943. Turner et al. derived stiffness matrices for truss, beam, and other ele-ments and presented their findings in 1956. The term finite element was first coined and used by Clough in 1960.

In the early 1960s, engineers used the method for approximate solution of problems in stress analysis, fluid flow, heat transfer, and other areas. A book by


Argyris in 1955 on energy theorems and matrix methods laid a foundation for fur-ther developments in finite element studies. The first book on finite elements by Zienkiewicz and Chung was published in 1967. In the late 1960s and early 1970s, finite element analysis was applied to nonlinear problems and large de-formations. Oden’s book on nonlinear continua appeared in 1972.

Mathematical foundations were laid in the 1970s. New element development, convergence studies, and other related areas fall in this category.

Today, developments in mainframe computers and availability of powerful mi-crocomputers have brought this method within reach of students and engineers working in small industries.

Stress and Equilibrium

Consider a three-dimensional body of volume V having a surface S:

A point in the body is located by x, y, and z coordinates. On part of the

boundary, a distributed force per unit area T, also called traction is applied. Un-


der the force, the body deforms. The deformation of a point x (x, y, z) is given by the three components of its displacement vector:

[ ]Twvu ,,=u The distributed force per unit volume is given by force vector:

[ ]Tzyx FFF ,,=F

The traction T is given by its components at points along the surface:

[ ]Tzyx TTT ,,=T

A load Pi acting at a point i is given by its three components:

[ ] Tizyx PPP ,,=iP

The stresses acting on the element volume dV are:

When the volume dV shrinks to a point, the stresses may be represented by

placing its components in a (3 x 3) symmetric matrix. Stress can be represented by the six independent components:


[ ]Τ= yzxzxyzyx τττσσσσ

where σx, σy, and σz are called normal stress and τxy, τxz, and τyz are called shear stress. Consider the equilibrium of the element volume dV. Forces are developed by multiplying the stresses by the corresponding areas. Writing the equations of equilibrium, recognizing the dV = dx dy dz:

0=+∂

∂+

∂

∂+

∂∂

=∑ xxzxyx

x fzyx

F ττσ

0=+∂

∂+

∂∂

+∂

∂=∑ y

yzxxyy f

zyxF

τστ

0=+∂

∂+

∂

∂+

∂∂

=∑ zzyzxz

z fzyx

F σττ

Boundary Conditions

There are displacement boundary conditions and surface loading conditions.

u = c on Su

where c is a given displacement


Consider the equilibrium if an elemental tetrahedron ABCD:

where DA, DB, and DC are parallel to the x, y, and z axes, respectively, and the area ABC, denoted by dA, lies on the surface. If the unit vector normal to the surface dA is given as:

[ ]Tzyx nnn ,,=n

then the areas:

dAnx=BDC dAny=ADC dAnz=ADB

Consider equilibrium in each direction:

xzxzyxyxx Tnnn =++ ττσ

yzyzyyxxy Tnnn =++ τστ

yzzyxyxxz Tnnn =++ σττ


These conditions must be satisfied on the boundary, ST, where the tractions are applied. Point loads must be treated as loads distributed over small but finite areas.

Strain-Displacement Relations

The strains in vector form that corresponds to the stress are:

where εx, εy, and εz are normal strains and γxy, γxz, and γxz, are the engineering shears strains.

We can approximate the shear strains by considering a small deformation of the dx-dy face of the unit volume dV:

, , , , ,T

u v w u v u w u vx y z y x z x z y

ε ⎡ ⎤∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂= + + +⎢ ⎥∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎣ ⎦

[ ]Tyzxzxyzyx γγγεεεε ,,,,,=


Stress-Strain Relations

For linear elastic materials, the stress-strain relations come from the general-ized Hooke’s Law. For isotropic materials, the two material properties are Young’s modulus (or the modulus of elasticity) E and Poisson’s ratio ν. Consider-ing an elemental volume inside the body, Hooke’s Law gives:

EEEzyx

xσν

σνσε −−=

EEEzyx

yσν

σσνε −+−=

EEEzyx

zσσ

νσνε +−−=

where the shear modulus (or modulus of rigidity), G, is given by:

)1(2 ν+=

EG

From Hooke’s Law, strain and stress are related by:

( )zyxzyx Eσσσνεεε ++

−=++

)21(

By substituting the above relationships into Hooke’s Law we get an inverse rela-tionship:

εσ D=

Gxz

xzτγ =

Gxy

xyτ

γ =Gyz

yzτ

γ =


where D is a symmetric (6 x 6) material matrix:

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−−

−−

−+=

νν

νν

νν

νν

ν

νν

ν

νν

5.000000

05.00000

00

5.0000

000

1

000

1

000

1

)21)(1(ED

Special Cases

One-Dimension. In one-dimension, the normal stress s along x and the cor-responding normal strain e. The stress-strain relationship is simply:

εσ E=

Two-Dimension. A thin planar body subjected to in-plane loading on its edge surface is said to be plane stress. For example, consider the thin plates shown below.

Here the stresses σz, τxz, and τyz are assumed to be zero. Generally, members that are thin (those with a small z dimension compared to the in-plane x and y dimensions) and shoes loads only in the x-y plane can be consider under plane stress.


For plane stress, Hooke’s Law reduces to:

EEyx

xσ

νσε −=

EEyx

yσσνε +−=

( )yxz Eσσνε +−=

xyxy Eτνγ )1(2 +

=

The inverse relationship σ = Dε reduces to:

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

xy

y

x

xy

y

x E

γεε

ν

νν

ντσσ

2)1(

00

01

0

1

1 2

If the body of uniform cross section is subjected to a transverse loading along

its length, a small thickness in the loaded area can be approximated by plane strain. For example:

Here the strains εz, γxz, and γyz are assumed to be zero.


The inverse relationship σ = Dε reduces to:

( )( ) ⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−

−+=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

xy

y

x

xy

y

x E

γεε

νν

νν

ν

νντσσ

21

00

01

0

1

211

Temperature Effects

If there is a temperature change ∆T(x, y, z) with respect to the original state, then an additional deformation can be estimated. For isotropic material, the tem-perature rise ∆T results in a uniform strain; this depends on the coefficient of lin-ear expansion of the material.

The temperature strain dose not cause any stresses when the body is free to deform. The temperature strain is represented as an initial strain:

[ ]TTTT 0,0,0,,,0 ∆∆∆= αααε

The stress-strain relationship becomes:

)( 0εεσ −= D

In plane stress, we get:

[ ]TTT 0,,0 ∆∆= ααε

In plane strain, we get:

[ ]TTT 0,,)1(0 ∆∆+= αανε

For plane stress and plane strain σ, ε, and D are defined by the corresponding equations given above.


Potential Energy and Equilibrium; the Rayleigh- Ritz Method

In mechanics of solids, our problem is to determine the displacement u of the body, satisfying the equilibrium equations.

Note that stresses are related to strains, which, in turn, are related to dis-

placements. This leads to requiring the solution of set of second-order partial dif-ferential equations. Solution of these equations is generally referred to as an ex-act solution. Such exact solutions are available for simple geometries and load-ing conditions. For problems of complex geometries and general boundary and loading conditions, obtaining exact solutions is an almost impossible task. Ap-proximate solution methods usually employ potential energy or variational methods, which place less stringent conditions on the functions.

Potential Energy, Π

The total potential energy Π of an elastic body is defined as the sum of the to-tal strain energy, U, and the work potential, WP:

Π = Strain Energy + Work Potential (U) (WP)


For linear elastic materials, the strain energy per unit volume is:

εσ T

21

The total strain energy U is given as:

∫=V

T dVεσ21U

The potential work WP is given as:

∑∫∫ −−−= iT

iS

T

V

T dSdV PuTufuWP

The total potential energy for a general elastic body is:

∑∫∫∫ −−−=Π iT

iS

T

V

T

V

T dSdVdV PuTufuεσ21

This is a conservative system, where the work potential is independent of the path taken. In other words, if the system is displaced from a given configuration and brought back to its original state, the forces do zero work regardless of the path.

Kinematically admissible displacements are those that satisfy the single-valued nature of displacements (compatibility) and the boundary conditions.

Principle of Minimum Potential Energy – For conservative sys-tems, of all the kinematically admissible displacement fields, those corresponding to equilibrium extermize the total potential energy. If the extremum condition is a minimum, the equilibrium state is stable.


Example

Consider a discrete connected system. The figure below shows a system of springs.

The total potential energy of the system is:

( ) 33112

442

332

222

1121 qFqFkkkk −−+++=Π δδδδ

where δ1, δ 2, δ 3, and δ 4 are the extensions of the four springs.

Therefore, total potential energy of the system is:

[ ] 33112

342

2332

222

211 )()(21 qFqFqkqqkqkqqk −−+−++−=Π

where q1, q2, and q3 are the displacements of nodes 1, 2, and 3, respectively.

22 q=δ

233 qq −=δ 34 q−=δ

211 qq −=δ


For equilibrium of this three degree-of-freedom system, we need to minimize Π with respect to the displacements q1, q2, and q3.

0 1, 2, 3i

iq

∂Π= =

∂

Therefore, the three equations are:

0)( 12111

=−−=∂

Π∂ Fqqkq

0)()( 233222112

=−−+−−=∂

Π∂ qqkqkqqkq

0)( 3342333

=−+−=∂

Π∂ Fqkqqkq

These equations can be written in matrix form as:

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+−

−++

−−

3

1

3

2

1

43

3

3

321

1

1

1

00

0 F

F

qqq

kkk

kkkk

kk

k

Alternately, we could write the equations of equilibrium for each node separately.


111 Fk =δ

0331122 =−− δδδ kkk

34433 Fkk =− δδ

Notice the equations for the displacements were obtained in a routine manner using the potential energy approach, without any reference to free body dia-grams. This feature makes the potential energy approach attractive for large and complex problems.

Rayleigh-Ritz Method

For continua, the total potential energy, Π, can be used for finding an ap-proximate solution. The Rayleigh-Ritz method involves the construction of an as-sumed displacement field [u, v, w]:

ltoizyxau ii 1),,( == ∑ φ

mtoljzyxav jj 1),,( +== ∑ φ

lmnntomkzyxaw kk >>+== ∑ 1),,(φ

The functions φi are usually taken as polynomials. Displacements u, v, and w must be kinematically admissible (that is u, v, and w must satisfy boundary conditions). Introducing stress-strain and strain-displacement relationships gives:

),...,,( 21 naaaΠ=Π where n is the number of independent unknowns. The extremum with respect to ai, (i = 1 to n) gives a set of n equations:

niai

,,2,10 ==∂

Π∂


Example

Consider the linear elastic one-dimensional rod with a body force shown below:

The potential energy of this system is:

10

2

221 udx

dxduEA

L

−⎟⎠⎞

⎜⎝⎛=Π ∫

where u1 = u(x=1). Consider the polynomial function:

2321 xaxaau ++=

The kinematically admissible function u must satisfy the boundary conditions u = 0 at both (x = 0) and (x = 2). Therefore:

0420 321 =+= aaa


Hence:

32 2aa −=

23 1 3( 2 ) ( 1)u a x x u x u a= − + = = = −

)1(23 −= xadxdu

The potential energy of this system using the function u is:

)(2)1(421

3

2

0

223 adxxaEA +−=Π ∫

22 23 3

0

2 (1 2 ) 2EAa x x dx a= − + +∫

23 3

4 23

a a= +

Applying the Rayleigh-Ritz method gives:

0238

33

=+=∂

Π∂ aa

Solving for a3 gives:

( )23 1 30.75 0.75 ( ) 0.75 2a u a u x x x= − = − = = −


The stress in the bar is given by:

)1(5.1 xdxduE −==σ

Notice that an exact solution is obtained if a piecewise polynomial interpolation is used in the construction of u.

Galerkin’s Method

Galerkin’s method uses the set of governing equations in the development of an integral form. It is usually presented as one of the weighted residual methods. For our discussion, let us consider a general representation of a governing equa-tion on a region V:

PLu = For the one-dimensional rod considered in the pervious example, the govern-

ing equation is:

0=⎟⎠⎞

⎜⎝⎛

dxduEA

dxd


If we consider L as the following operator:

()dxdEA

dxd

operating on u, the exact solution needs to satisfy L at every point x. If we seek an approximate solution, u , it introduces an error e(x), called the residual:

PuLxe −= ˆ)(

Approximate methods revolve around setting the residual relative to a weight-ing function Wi, to zero.

ntoidVPLuWv

i 10)( ==−∫

The choice of the weighting function, Wi, leads to various approximation

methods. In the Galerkin method, the weighting functions, Wi, are chosen from the basis functions used for constructing u . Let u be represented by:

∑=

=n

iiiGQu

1

ˆ

where Gi, i = 1 to n, are basis functions (usually polynomials of x, y, z). Here we choose the weighting function to a linear combination of the basis functions Gi. Consider an arbitrary function φ given by:

∑=

=n

iiiG

1

φφ

where the coefficients φi are arbitrary, except for requiring that φ satisfy homoge-neous (zero) boundary conditions where u is prescribed.


Galerkin’s Method in Elasticity

Consider the equations of equilibrium we developed earlier. Galerkin’s method requires:

yyyzyxy

Vxx

xzxyx fzyx

fzyx

φτστ

φττσ⎟⎟⎠

⎞⎜⎜⎝

⎛+

∂∂

+∂

∂+

∂∂

+⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+

∂∂

+∂

∂+

∂∂

∫

0=⎥⎦

⎤⎟⎟⎠

⎞⎜⎜⎝

⎛+

∂∂

+∂

∂+

∂∂

+ dVfzyx zz

zyzxz φσττ

where

, ,T

x y zφ φ φ φ⎡ ⎤= ⎣ ⎦

is an arbitrary displacement consistent with the displacements, u. Consider inte-gration by parts using the following formula:

xV V S

dV dV n dSx xα θθ α αθ∂ ∂

= − +∂ ∂∫ ∫ ∫

where α and θ are functions of (x, y, z). For multi-dimensional problems the above equation is referred to as Green-Gauss theorem or the divergence theo-rem.

Galerkin’s Method – Chose basis functions Gi. Determine the coefficients φi such that

where φi are arbitrary except for requiring that φ satisfy homogeneous boundary conditions. The solution of the resulting equations for Gi then yields the approximate solution u .

∑∫=

==−n

iii

V

GdVPuL1

where0)ˆ( φφφ


xzxzyxyxx Tnnn =++ ττσ

( ) xzxzyxyxx PdSnnn =++ ττσ

Using the Green-Gauss theorem on the equations of equilibrium yields:

∫∫ +−V

T

V

T dVfdV φφεσ )(

( )x x y xy z xz xS

n n n dSσ τ τ φ+ + +∫

( )x xy y y z yz yS

n n n dSτ σ τ φ+ + +∫

( )x xz y yz z z zS

n n n dSτ τ σ φ+ + +∫

where ε(φ) is the strain field corresponding to the arbitrary displacement field φ.

T

yxzxzyzyx

xyxzyzzyx ⎥⎦

⎤⎢⎣

⎡∂∂

+∂∂

∂∂

+∂∂

∂∂

+∂∂

∂∂

∂∂

∂∂

=φφφφφφφφφφε ,,,,,)(

On the boundary we have: At a point loads: These are the natural boundary conditions in the problem. Therefore the Galerkin “weak form” or “variational form” for three-dimensional stress analysis is:

yzyzyyxxy Tnnn =++ τστ

yzzyxyxxz Tnnn =++ σττ

( ) yzyzyyxxy PdSnnn =++ τστ

( ) yzzyxyxxz PdSnnn =++ σττ


0)( =−−− ∑∫∫∫ PTf T

S

T

V

T

V

T dSdVdV φφφφεσ

where φ is an arbitrary displacement consistent with the boundary conditions. For problems of linear elasticity, the above equation is precisely the principle of vir-tual work. The function φ is the admissible virtual displacement. The principle of virtual work may be stated as follows:

Example

Let consider the pervious problem and solve it by Galerkin’s approach. The equilibrium equation is:

2at00at00 ===== xuxudxduEA

dxd

Multiplying the differential equation above by φ and integrating by parts gives:

1 22 2

0 10 0

0d du du d du duEA dx EA dx EA EAdx dx dx dx dx dx

φφ φ φ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠∫ ∫

where φ is zero at (x = 0) and (x = 2) and EA(du/dx) is the tension in the rod, which make a jump of magnitude of 2 at (x = 1). Therefore:

Principle of Virtual Work – A body is in equilibrium if the internal work equals the external virtual work for every kinematically admissible displace-ment field.


02 1

2

0

=+− ∫ φφ dxdxd

dxduEA

If we use the same polynomial function for u and φ and If u1 and φ1 are the values at (x = 1), we get:

( ) ( ) 12

12 22 φφ xxuxxu −=−=

Substituting these and E = A = 1 in the above integral:

02)22(2

0

211 =⎥

⎦

⎤⎢⎣

⎡+−− ∫ dxxuφ

0238

11 =⎟⎠⎞

⎜⎝⎛ +− uφ

This is to be satisfied for every φ1. We get:

75.01 =u


The stress in the bar is given by:

)1(5.1 xdxduE −==σ

Problems:

1. Obtain the D matrix given below using the generalized Hook’s law relation-ships.

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−−

−−

−+=

νν

νν

νν

νν

ν

νν

ν

νν

5.000000

05.00000

00

5.0000

000

1

000

1

000

1

)21)(1(ED


2. Determine the displacement of nodes of the spring system shown below using both equilibrium and principle of minimum potential energy method.

3. Use the Rayleigh-Ritz method to find the displacement of the midpoint of the rod shown below (assume ρg = 1 and u = a1 + a2x + a3x2):

4. Use Galerkin’s method to find the displacement of the midpoint of the rod in Problem 3.


Role of Computers in Finite Element Methods

Until the early 1950s, matrix methods and the associated finite element method were not readily adaptable for solving complicated problems because of the large number of algebraic equations that resulted. Hence, even though the finite element method was being used to describe complicated structures, the re-sulting large number of equations associated with the finite element method of structural analysis made the method extremely difficult and impractical to use.

With the advent of the computer, the solution of thousands of equations in a matter of minutes became possible. The development of the computer resulted in computational program development. Numerous special-purpose and general-purpose programs have been written to handle various complicated structural (and non-structural) problems. To use the computer, the analyst, having defined the finite element model, inputs the information into the computer. This formation may include the position of the element nodal coordinates, the manner in which elements are connected together, the material properties of the elements, the applied loads, boundary conditions, or constraints, and the kind of analysis to be performed. The computer then uses this information to generate and solve the equations necessary to carry out the analysis.

General Steps of the Finite Element Method

The following section presents the general steps for applying the finite ele-ment method to obtain solutions of structural engineering problem. Typically, for the structural stress-analysis problem, the engineer seeks to determine dis-placements and stresses throughout the structure, which is in equilibrium and is subjected to applied loads. For many structures, it is difficult to determine the dis-tribution of deformation using conventional methods, and thus the finite element method is necessarily used.

There are two general approaches associated with the finite element method. One approach, called the force, or flexibility method, uses internal forces as the unknowns of the problem. To obtain the governing equations, first the equilib-rium equations are used. Then necessary additional equations are found by in-troducing compatibility equations. The result is a set of algebraic equations for determining the redundant or unknown forces.


The second approach, called the displacement, or stiffness method, as-sumes the displacements of the nodes as the unknowns of the problem. The governing equations are expressed in terms of nodal displacements using the equations of equilibrium and an applicable law relating forces to displacements.

These two approaches result in different unknowns (forces or displacements) in the analysis and different matrices associated with their formulations (flexibil-ities or stiffnesses). It has been shown that, for computational purposes, the dis-placement (or stiffness) method is more desirable because its formulation is sim-pler for most structural analysis problems. Consequently, only the displacement method will be used throughout this text.

The finite element method involves modeling the structure using small inter-connected elements called finite elements. A displacement function is associ-ated with each finite element. Every interconnected element is linked, directly or indirectly, to every other element through common (or shared) interfaces, includ-ing nodes and/or boundary lines and/or surfaces. The total set of equations de-scribing the behavior of each node results in a series of algebraic equations best expressed in matrix notation.

Step 1 - Discretize and Select Element Types

Step 1 involves dividing the body into an equivalent system of finite elements with associated nodes and choosing the most appropriate element type. The total number of elements used and their variation in size and type within a given body are primarily matters of engineering judgment. The elements must be made small enough to give usable results and yet large enough to reduce computational ef-fort. Small elements (and possibly higher-order elements) are generally desirable where the results are changing rapidly, such as where changes in geometry oc-cur, whereas large elements can be used where results are relatively constant.

The primary line elements, consist of bar (or truss) and beam elements. They have a cross-sectional area but are usually represented by line segments. In general, the cross-sectional area within the element can vary, but it will be considered to be constant throughout this text.


These elements are often used to model trusses and frame structures. The

simplest line element (called a linear element) has two nodes, one at each end, although higher-order elements having three nodes or more (called quadratic, cubic, etc. elements) also exist. The line elements are the simplest of elements to consider and will be used to illustrate many of the basic concepts of the finite element method.

The basic two-dimensional (or plane) elements are loaded by forces in their

own plane (plane stress or plane strain conditions). They are triangular or quadri-lateral elements.


The simplest two-dimensional elements have corner nodes only (linear ele-

ments) with straight sides or boundaries although there are also higher-order elements, typically with mid-side nodes (called quadratic elements) and curved sides. The elements can have variable thicknesses throughout or be constant. They are often used to model a wide range of engineering problems.

The most common three-dimensional elements are tetrahedral and hexahe-dral (or brick) elements; they are used when it becomes necessary to perform a three-dimensional stress analysis. The basic three dimensional elements have corner nodes only and straight sides, whereas higher-order elements with mid-edge nodes (and possible mid-face nodes) have curved surfaces for their sides.



The axisymmetric element is developed by rotating a triangle or quadrilateral about a fixed axis located in the plane of the element through 360°. This element can be used when the geometry and loading of the problem are axisymmetric

Step 2 - Select a Displacement Function

Step 2 involves choosing a displacement function within each element. The function is defined within the element using the nodal values of the element. Lin-ear, quadratic, and cubic polynomials are frequently used functions because they are simple to work with in finite element formulation. The functions are expressed in terms of the nodal unknowns (in the two-dimensional problem, in terms of an x and a y component). Hence, the finite element method is one in which a continu-ous quantity, such as the displacement throughout the body, is approximated by a discrete model composed of a set of piecewise-continuous functions defined within each finite domain or finite element.


Step 3 - Define the Strain/Displacement and Stress/Strain Relationships

Strain/displacement and stress-strain relationships are necessary for deriving the equations for each finite element. For one-dimensional small strain deforma-tion, say, in the x direction, we have strain εx, related to displacement u by:

dxdu

x =ε

In addition, the stresses must be related to the strains through the stress-strain law (generally called the constitutive law). The ability to define the mate-rial behavior accurately is most important in obtaining acceptable results. The simplest of stress-strain laws, Hooke’s law, often used in stress analysis, is given by:

xx Eεσ =

Step 4 - Derive the Element Stiffness Matrix and Equations

Initially, the development of element stiffness matrices and element equations was based on the concept of stiffness influence coefficients, which presupposes a background in structural analysis. We now present alternative methods used in this text that do not require this special background.

Direct Equilibrium Method - According to this method, the stiffness matrix and element equations relating nodal forces to nodal displacements are obtained using force equilibrium conditions for a basic element, along with force-deformation relationships. This method is most easily adaptable to line or one-dimensional elements (spring, bar, and beam elements).

Work or Energy Methods - To develop the stiffness matrix and equations for two- and three-dimensional elements, it is much easier to apply a work or energy method. The principle of virtual work (using virtual displacements), the princi-ple of minimum potential energy, and Castigliano’s theorem are methods fre-quently used for the purpose of derivation of element equations. We will present the principle of minimum potential energy (probably the most well known of the three energy methods mentioned here).


Methods of Weighted Residuals - The methods of weighted residuals are useful for developing the element equations (particularly popular is Galerkin’s method). These methods yield the same results as the energy methods, wher-ever the energy methods are applicable. They are particularly useful when a functional such as potential energy is not readily available. The weighted resid-ual methods allow the finite element method to be applied directly to any differen-tial equation.

Step 5 - Assemble the Element Equations and Introduce Boundary Condi-

tions

The individual element equations generated in Step 4 can now be added to-gether using a method of superposition (called the direct stiffness method) whose basis is nodal force equilibrium (to obtain the global equations for the whole structure). Implicit in the direct stiffness method is the concept of continu-ity, or compatibility, which requires that the structure remain together and that no tears occur anywhere in the structure. The final assembled or global equation written in matrix form is:

[ ] dKF = where F is the vector of global nodal forces, [K] is the structure global or total stiffness matrix, and d is now the vector of known and unknown structure nodal degrees of freedom or generalized displacements.


Step 6 - Solve for the Unknown Degrees of Freedom

(or Generalized Displacements)

Once the element equations are assembled and modified to account for the boundary conditions, a set of simultaneous algebraic equations that can be writ-ten in expanded matrix form as:

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⋅⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⋅⋅⋅⋅⋅

⋅⋅

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⋅

nnnnn

n

n

n d

dd

KKK

KKKKKK

F

FF

2

1

21

22221

11211

2

1

where n is the structure total number of unknown nodal degrees of freedom. These equations can be solved for the d’s by using an elimination method (such as Gauss’s method) or an iterative method (such as Gauss Seidel’s method).

Step 7 - Solve for the Element Strains and Stresses

For the structural stress-analysis problem, important secondary quantities of strain and stress (or moment and shear force) can be obtained in terms of the displacements determined in Step 6.

Step 8 - Interpret the Results

The final goal is to interpret and analyze the results for use in the de-sign/analysis process. Determination of locations in the structure where large de-formations and large stresses occur is generally important in making de-sign/analysis decisions. Post-processor computer programs help the user to in-terpret the results by displaying them in graphical.


Applications of the Finite Element Method

The following applications will illustrate the variety, size, and complexity of problems that can be solved using the finite element method and the typical dis-cretization process and kinds of elements used.

The first example is of a control tower for a railroad. The tower is a three-dimensional frame comprising a series of beam-type elements. The 48 elements are labeled by the circled numbers, whereas the 28 nodes are indicated by the encircled numbers. Each node has three rotation and three displacement com-ponents associated with it. The rotations and displacements are called the de-grees of freedom.

The next example is the determination of displacements and stresses in an underground box culvert subjected to ground shock loading from a bomb explo-sion. The discretized model that included a total of 369 nodes, 40 one-dimensional bar or truss elements used to model the steel reinforcement in the box culvert, and 333 plane strain two-dimensional triangular and rectangular elements used to model the surrounding soil and concrete box culvert. With an assumption of symmetry, only half of the box culvert must be analyzed. This problem requires the solution of nearly 700 unknown nodal displacements.

Another two-dimensional problem is that of a hydraulic cylinder rod end. It was modeled by 120 nodes and 297 plane strain triangular elements. Symmetry was also applied to the whole rod end so that only half of the rod end had to be ana-lyzed, as shown. The purpose of this analysis was to locate areas of high stress concentration in the rod end.

The next example shows a chimney stack section that is four form heights high (or a total of 32 ft high). The engineer used 584 beams to model the vertical and horizontal stiffeners making up the formwork, whereas 252 flat-plate ele-ments were used to model the inner wooden form and the concrete shell.

The next example shows the finite element discretized model of a proposed steel die used in a plastic film-making process. Two hundred forty axisymmetric elements were used to model the three-dimensional die.







The next example illustrates the use of a three-dimensional solid element to model a swing casting for a backhoe frame. The three-dimensional hexahedral elements are necessary to model the irregularly shaped three-dimensional cast-ing. Two-dimensional models certainly would not yield accurate engineering solu-tions to this problem.


Advantages of the Finite Element Method

The finite element method has been applied to numerous problems, both structural and non-structural. This method has a number of advantages that have made it very popular.

1. Model irregularly shaped bodies quite easily 2. Handle general load conditions without difficulty 3. Model bodies composed of several different materials because the ele-

ment equations are evaluated individually 4. Handle unlimited numbers and kinds of boundary conditions 5. Vary the size of the elements to make it possible to use small elements

where necessary 6. Alter the finite element model relatively easily and cheaply 7. Include dynamic effects 8. Handle nonlinear behavior existing with large deformations and nonlin-

ear materials The finite element method of structural analysis enables the designer to detect

stress, vibration, and thermal problems during the design process and to evalu-ate design changes before the construction of a possible prototype.


Introduction to the Stiffness (Displacement) Method

Introduction

This section introduces some of the basic concepts on which the direct stiff-ness method is based. The linear spring is simple and an instructive tool to illus-trate the basic concepts. The steps to develop a finite element model for a linear spring follow our general 8 step procedure.

1. Discretize and Select Element Types - Linear spring elements 2. Select a Displacement Function - Assume a variation of the dis-

placements over each element. 3. Define the Strain/Displacement and Stress/Strain Relationships -

We next illustrate how to assemble the total stiffness matrix for a struc-ture comprising an assemblage of spring elements by using elementary concepts of equilibrium and compatibility.

4. Derive the Element Stiffness Matrix and Equations - Define the stiff-ness matrix for an element and then consider the derivation of the stiff-ness matrix for a linear-elastic spring element.

5. Assemble the Element Equations to Obtain the Global or Total Equations and Introduce Boundary Conditions - We then show how the total stiffness matrix for the problem can be obtained by superimpos-ing the stiffness matrices of the individual elements in a direct manner. The term direct stiffness method evolved in reference to this method. After establishing the total structure stiffness matrix, impose boundary conditions (both homogeneous and nonhomogeneous).

6. Solve for the Unknown Degrees of Freedom (or Generalized Dis-placements) - Solve for the nodal displacements.

7. Solve for the Element Strains and Stresses - The reactions and inter-nal forces association with the bar element.

8. Interpret the Results


Step 1 - Select Element Type

Consider the linear spring shown below. The spring is of length L and is sub-jected to a nodal tensile force, T directed along the x axis.


A displacement function u is assumed.

xaau ˆˆ 21 += In general, the number of coefficients in the displacement function is equal to the total number of degrees of freedom associated with the element. We can write the displacement function in matrix forms as:

[ ]⎭⎬⎫

⎩⎨⎧

=2

1ˆ1âa

xu

We can express u as a function of the nodal displacements d by evaluating u at each node and solving for a1 and a2.

11ˆ)0ˆ(ˆ adxu x ===

122ˆ)ˆ(ˆ aLadLxu x +===


Solving for a2:

Ldda xx 12

2

ˆˆ −=

Substituting a1 and a2 into u gives:

xxx dx

Lddu 1

12 ˆˆˆˆ

ˆ +⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

In matrix form:

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −=

x

x

dd

Lx

Lxu

2

1

ˆˆˆˆ

1ˆ

Or in another form:

[ ]⎭⎬⎫

⎩⎨⎧

=x

x

ddNNu

2

121 ˆ

ˆˆ

where N1 and N2 are defined as:

The functions Ni are called interpolation functions because they describe how the assumed displacement function varies over the domain of the element. In this case the interpolation functions are linear.


Tensile forces produce a total elongation (deformation) δ of the spring. For lin-ear springs, the force T and the displacement δ are related by Hooke’s law:

δkT =

LxNˆ

11 −=LxNˆ

2 =


where deformation of the spring δ is given as:

)0(ˆ)(ˆ uLu −=δ

xx dd 12ˆˆ −=δ


We can now derive the spring element stiffness matrix as follows:

Rewrite the forces in terms of the nodal displacements:

( )xxx ddkfT 121ˆˆˆ −=−=

( )xxx ddkfT 122ˆˆˆ −==

We can write the last two force-displacement relationships in matrix form as:

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡ −−

=⎭⎬⎫

⎩⎨⎧

x

x

x

x

dd

kk

kk

ff

2

1

2

1

ˆˆ

ˆˆ

This formulation is valid as long as the spring deforms along the x axis. The co-efficient matrix of the above equation is called the local stiffness matrix k :

⎥⎦

⎤⎢⎣

⎡ −−

=kk

kk

k

Tf x −=1 Tf x =2


Step 5 - Assemble the Element Equations

and Introduce Boundary Conditions

The global stiffness matrix and the global force vector are assembled using the nodal force equilibrium equations, and force/deformation and compatibility equations.

[ ] ∑=

==N

e

ekK1

)(ˆK

∑=

==N

e

efF1

)(ˆF

where k and f are the element stiffness and force matrices expressed in global coordinates.

Step 6 - Solve for the Nodal Displacements

Solve the displacements by imposing the boundary conditions and solving the following set of equations:

KdF =

Step 7 - Solve for the Element Forces

Once the displacements are found, the forces in each element may be calcu-lated from:

δkT =


Example 1 - Spring Problem

Consider the following two-spring system shown below:

where the element axis x coincides with the global axis x. For element 1:

1 11 1

3 31 1

x x

x x

f dk kf dk k

−⎧ ⎫ ⎧ ⎫⎡ ⎤=⎨ ⎬ ⎨ ⎬⎢ ⎥−⎣ ⎦⎩ ⎭ ⎩ ⎭

For element 2:

3 2 2 3

2 2 2 2

x x

x x

f k k df k k d

−⎧ ⎫ ⎡ ⎤ ⎧ ⎫=⎨ ⎬ ⎨ ⎬⎢ ⎥−⎩ ⎭ ⎣ ⎦ ⎩ ⎭

Both continuity and compatibility require that both elements remain connected at node 3.

(1) (2)3 3x xd d=

We can write the nodal equilibrium equation at each node as:

(1)1 1x xF f=

(2)2 2x xF f=

(1) (2)3 3 3x x xF f f= +


Therefore the force-displacement equations for this spring system are:

1 1 1 1 3x x xF k d k d= −

2 2 3 2 2x x xF k d k d= − +

3 1 1 1 3 2 3 2 2x x x x xF k d k d k d k d= − + + − In matrix form the above equations are:

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+−−−−

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

x

x

x

x

x

x

ddd

kkkkkkkk

FFF

3

2

1

2121

22

11

3

2

1

00

or

KdF = where F is the global nodal force vector, d is called the global nodal dis-placement vector, and K is called the global stiffness matrix.

Assembling the Total Stiffness Matrix by Superposition

Consider the spring system defined in the last example:

The elemental stiffness matrices may be written for each element. For element 1:


⎥⎦

⎤⎢⎣

⎡ −−

=1

1

1

1)1(

31

kk

kk

k

dd xx

For element 2:

⎥⎦

⎤⎢⎣

⎡ −−

=2

2

2

2)2(

23

kk

kk

k

dd xx

Write the stiffness matrix in global format for element 1 as follows:

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−

)1(3

)1(2

)1(1

)1(3

)1(2

)1(1

1

101000101

x

x

x

x

x

x

fff

ddd

k

For element 2:

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

)2(3

)2(2

)2(1

)2(3

)2(2

)2(1

2

110110

000

x

x

x

x

x

x

fff

ddd

k

Apply the force equilibrium equations at each node.

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧+

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

x

x

x

x

x

x

x

FFF

ff

f

f

3

2

1

)2(3

)2(2

)1(3

)1(1 00

The above equations give:

1 1 1 1

2 2 2 2

1 2 1 2 3 3

00

x x

x x

x x

k k d Fk k d F

k k k k d F

−⎡ ⎤ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥− =⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥− − +⎣ ⎦ ⎩ ⎭ ⎩ ⎭


To avoid the expansion of the each elemental stiffness matrix, we can use a more direct, shortcut form of the stiffness matrix. For element 1:

⎥⎦

⎤⎢⎣

⎡ −−

=1

1

1

1)1(

31

kk

kk

k

dd xx

For element 2:

⎥⎦

⎤⎢⎣

⎡ −−

=2

2

2

2)2(

23

kk

kk

k

dd xx

The global stiffness matrix may be constructed by directly adding terms associ-ated with the degrees of freedom in k(1) and k(2) into their corresponding locations in the K as follows:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+−−

−−=

21

2

1

2

2

1

1

321

00

kkkk

kk

k

kddd xxx

K

Boundary Conditions

In order to solve the equations defined by the global stiffness matrix, we must apply some form of constraints or supports or the structure will be free to move as a rigid body.

Boundary conditions are of two general types: homogeneous boundary conditions (the most common) occur at locations that are completely prevented from movement; nonhomogeneous boundary conditions occur where finite


non-zero values of displacement are specified, such as the settlement of a sup-port.

Consider the equations we developed for the two-spring system. We will con-sider node 1 to be fixed d1x = 0. The equations describing the elongation of the spring system become:

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+−−−−

x

x

x

x

x

FFF

dd

kkkkkkkk

3

2

1

3

2

2121

22

11 00

0

Expanding the matrix equations gives:

xx dkF 311 −=

xxx dkdkF 22322 +−=

( ) xxx dkkdkF 321223 ++−= The second and third equation may be written in matrix form as:

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡+−

− x

x

x

x

FF

dd

kkk

kk

3

2

3

2

21

2

2

2

Once we have solved the above equations for the unknown nodal displacements, we can use the first equation in the original matrix to find the support reaction.

xx dkF 311 −= For homogeneous boundary conditions, we can delete the row and column cor-responding to the zero-displacement degrees-of-freedom.

Let’s again look at the equations we developed for the two-spring system. However, this time we will consider a nonhomogeneous boundary condition at node 1 d1x = δ. The equations describing the elongation of the spring system be-come:


⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+−−−−

x

x

x

x

x

FFF

dd

kkkkkkkk

3

2

1

3

2

2121

22

11

00 δ

Expanding the matrix equations gives:

xx dkkF 3111 −= δ

xxx dkdkF 32222 −=

( ) xxx dkkdkkF 3212213 ++−−= δ By considering the second and third equations because they have known nodal forces we get:

xxx dkdkF 32222 −=

( ) xxx dkkdkkF 3212213 ++−=+ δ Writing the above equations in matrix form gives:

⎭⎬⎫

⎩⎨⎧

+=

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡+−

− δ13

2

3

2

21

2

2

2

kFF

dd

kkk

kk

x

x

x

x

For nonhomogeneous boundary conditions, we must transfer the terms from the stiffness matrix to the right-hand-side force vector before solving for the unknown displacements.

Once we have solved the above equations for the unknown nodal displace-ments, we can use the first equation in the original matrix to find the support re-action.

1 1 1 3x xF k k dδ= −


Example 2 -Spring Problem

Consider the following three-spring system:

The elemental stiffness matrices for each element are:

⎥⎦

⎤⎢⎣

⎡ −−

=11

11

1000)1(

31

k ⎥⎦

⎤⎢⎣

⎡ −−

=11

11

2000)2(

43

k

⎥⎦

⎤⎢⎣

⎡ −−

=11

11

3000)3(

24

k

Using the concept of superposition (the direct stiffness method), the global stiff-ness matrix is:

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−

−

−−

=

5000200030000

20003000

01000

30000

30000

010000

1000

K

The global force-displacement equations are:

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−

−

−−

x

x

x

x

x

x

x

x

FFFF

dddd

4

3

2

1

4

3

2

1

5000200030000

20003000

01000

30000

30000

010000

1000


We have homogeneous boundary conditions at nodes 1 and 2 (d1x = 0 and d2x = 0). Deleting the first two rows and the first two columns and substituting for the known force at node 4 (F4x = 5000 lb) gives:

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

− 50000

50002000

20003000

4

3

x

x

dd

Solving for d3x and d4x gives:

.1115.

1110

43 indind xx ==

To obtain the global forces, substitute the displacement in the force-displacement equations.

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−

−

−−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

1115

1110

4

3

2

1

00

5000200030000

20003000

01000

30000

30000

010000

1000

x

x

x

x

FFFF

Solving for the forces gives:

lbFlbF xx 11000,45

11000,10

21 −=−=

lbFF xx 11000,550 43 ==

Next, use the local element equations to obtain the force in each spring. For element 1:

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡ −−

=⎭⎬⎫

⎩⎨⎧

1110

3

1 010001000

10001000

ˆˆ

x

x

ff

The local forces are:


lbflbf xx 11000,10ˆ

11000,10ˆ

31 =−=

A free-body diagram of the spring element 1 is shown below.

For element 2:

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡ −−

=⎭⎬⎫

⎩⎨⎧

1115

1110

4

3

20002000

20002000

ˆˆ

x

x

ff



11000,10ˆ

43 =−=


For element 3:

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡ −−

=⎭⎬⎫

⎩⎨⎧

030003000

30003000

ˆˆ 11

15

2

4

x

x

ff



11000,45ˆ

24 −==




Consider the following four-spring system:

The spring constant k = 200 kN/m and the displacement δ = 20 mm. Therefore, the elemental stiffness matrices are:

mkNkkkk /11

11

200)4()3()2()1(⎥⎦

⎤⎢⎣

⎡ −−

====

Using superposition (the direct stiffness method), the global stiffness matrix is:

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−−

−−

−−

=

200200

000

200400200

00

0200400200

0

00

200400200

000

200200

K



⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−−

−−

−−

x

x

x

x

x

x

x

x

x

x

FFFFF

ddddd

5

4

3

2

1

5

4

3

2

1

200200

000

200400200

00

0200400200

0

00

200400200

000

200200

Applying the boundary conditions (d1x = 0 and d5x = 20 mm) and the known forces (F2x, F3x, and F4x equal to zero) give:

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−

−−−

x

x

x

x

Fddd

5

4

3

2

000

02.0200200002004002000020040020000200400

Rearranging the first three equations gives:

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

−

−−

400

400200

0

200400200

0200400

4

3

2

x

x

x

ddd

Solving for d2x, d3x, and d4x gives:

mdmdmd xxx 015.001.0005.0 432 === Solving for the forces F1x and F5x gives:

kNF x 0.1)005.0(2001 −=−=

kNF x 0.1)02.0(200)015.0(2005 =+−=


Next, use the local element equations to obtain the force in each spring. For element 1:

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡ −−

=⎭⎬⎫

⎩⎨⎧

005.00

200200

200200

ˆˆ

2

1

x

x

ff


kNfkNf xx 0.1ˆ0.1ˆ21 =−=

For element 2:

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡ −−

=⎭⎬⎫

⎩⎨⎧

01.0005.0

200200

200200

ˆˆ

3

2

x

x

ff



For element 3:

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡ −−

=⎭⎬⎫

⎩⎨⎧

015.001.0

200200

200200

ˆˆ

4

3

x

x

ff



For element 4:

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡ −−

=⎭⎬⎫

⎩⎨⎧

02.0015.0

200200

200200

ˆˆ

5

4

x

x

ff





Consider the following spring system:

The boundary conditions are:

0431 === xxx ddd The compatibility condition at node 2 is:

xxxx dddd 2)3(

2)2(

2)1(

2 === Using the direct stiffness method: the elemental stiffness matrices for each ele-ment are:

⎥⎦

⎤⎢⎣

⎡ −−

=1

1

1

1)1(

21

kk

kk

k ⎥⎦

⎤⎢⎣

⎡ −−

=2

2

2

2)2(

32

kk

kk

k

⎥⎦

⎤⎢⎣

⎡ −−

=3

3

3

3)3(

42

kk

kk

k


Using the concept of superposition (the direct stiffness method), the global stiff-ness matrix is:

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−−

++−

−=

3

3

2

2

3

2

321

1

1

1

0

0

0

0

00

4321

k

kkk

kk

kkkk

kk

K

Applying the boundary conditions (d1x = d3x = d4x = 0) and the known forces (F2x

= P) gives:

( ) xdkkkP 2321 ++= Solving the equation gives:

3212 kkk

Pd x ++=

Solving for the forces gives:

xxxxxx dkFdkFdkF 234223211 −=−=−=

Potential Energy Approach to Derive Spring Element Equations

One of the alternative methods often used to derive the element equations and the stiffness matrix for an element is based on the principle of minimum potential energy. This method has the advantage of being more general than the methods involving nodal and element equilibrium equations, along with the stress/strain law for the element.

Thus, the principle of minimum potential energy is more adaptable for the de-termination of element equations for complicated elements (those with large numbers of degrees of freedom) such as the plane stress/strain element, the axi-


symmetric stress element, the plate bending element, and the three-dimensional solid stress element.

Total Potential Energy

The total potential energy is defined as the sum of the internal strain energy U and the potential energy of the external forces Ω:

Ω+=Upπ

Strain energy is the capacity of the internal forces (or stresses) to do work through deformations (strains) in the structure; Ω is the capacity of forces such as body forces, surface traction forces, and applied nodal forces to do work through the deformation of the structure.

Recall the force-displacement relationship for a linear spring:

kxF = The differential internal work (or strain energy) dU in the spring is the internal force multiplied by the change in displacement which the force moves through:

( )dxkxFdxdU ==

The total strain energy is:

( ) 2

0 21 kxdxkxdUU

x

L

=== ∫∫

The strain energy is the area under the force-displacement curve. The potential energy of the external forces is the work done by the external forces:


Fx−=Ω Therefore, the total potential energy is:

Fxkxp −= 2

21π

The concept of a stationary value of a function G is shown below:

The function G is expressed in terms of x. To find a value of x yielding a station-ary value of G(x), we use differential calculus to differentiate G with respect to x and set the expression equal to zero.

0=dxdG

We can replace G with the total potential energy πp and the coordinate x with a discrete value di. To minimize πp we first take the variation of πp (we will not cover the details of variational calculus):

nn

pppp d

dd

dd

dδ

πδ

πδ

πδπ

∂∂

++∂∂

+∂∂

= ...22

11

The principle states that equilibrium exist when the di define a structure state such that δπp = 0 for arbitrary admissible variations δdi from the equilibrium state. An admissible variation is one in which the displacement field still satisfies the boundary conditions and interelement continuity.


To satisfy δπp = 0, all coefficients associated with δdi must be zero independently, therefore:

0,,2,10 =∂∂

==∂∂

dorni

dp

i

p ππ


Consider the following linear-elastic spring system subjected to a force of 1,000 lb. Evaluate the potential energy for various displacement values and show that the minimum potential energy also corresponds to the equilibrium position of the spring.


The total potential energy is defined as the sum of the internal strain energy U and the potential energy of the external forces Ω:

The variation of πp with respect to x is:

0=∂∂

= xx

pp δ

πδπ

Since δx is arbitrary and might not be zero, then:

0=∂∂

xpπ

Using our express for πp, we get:

( ) xlbxinlbFxkxp 1000)/(50021

21 22 −=−=π

.0.210005000 inxxx

p =−==∂∂π

If we had plotted the total potential energy function πp for various values of de-formation we would get:

Fx−=Ω2

21 kxU =

Ω+=Upπ


Let’s derive the spring element equations and stiffness matrix using the principal of minimum potential energy. Consider the linear spring subjected to nodal forces shown below:

The total potential energy πp is:

( ) xxxxxxp dfdfddk 2211

2

12ˆˆˆˆˆˆ

21

−−−=π

Expanding the above express gives:

( ) xxxxxxxxp dfdfddddk 22112

1212

2ˆˆˆˆˆˆˆ2ˆ

21

−−−−=π

Minimizing the total potential energy πp:

210 toidi

p ==∂∂π


( ) xxxp fddk

d 1121

ˆˆ2ˆ2210 −+−==

∂∂π

( ) xxxp fddk

d 2122

ˆˆ2ˆ2210 −−==

∂∂π

In matrix form the above equations are:

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡ −−

=⎭⎬⎫

⎩⎨⎧

x

x

x

x

dd

kk

kk

ff

2

1

2

1

ˆˆ

ˆˆ

This result is identical to stiffness matrix we obtained using equilibrium.


Obtain the total potential energy of the spring system shown below and find its minimum value.

The potential energy πp for element 1 is:


131)1( ˆ

21

−−−=π



342)2(

21

−−−=π




423)3(

21

−−−=π

The total potential energy πp for the spring system is:

∑=

=3

1

)(

e

epp ππ

Minimizing the total potential energy πp:

)1(11131

1

0 xxxx

p fdkdkd

−+−==∂∂π

)3(24323

2

0 xxxx

p fdkdkd

−−==∂∂π

)2(3

)1(332421131

3

0 xxxxxxx

p ffdkdkdkdkd

−−+−−==∂∂π

)3(4

)2(443233242

4

0 xxxxxxx

p ffdkdkdkdkd

−−+−−==∂∂π

In matrix form:

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

++

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

+−−

−+

−−

)3(

4

)2(

4

)2(

3

)1(

3

2

1

4

3

2

1

32

2

3

2

21

1

3

3

1

1 00

0

0

0

0

xx

xx

x

x

x

x

x

x

ffff

ff

dddd

kkkk

kkk

k

k

kk

k

Using the following force equilibrium equations:

xx Ff 1

)1(1 = xx Ff 2

)3(2 =



⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−

−

−−

x

x

x

x

x

x

x

x

FFFF

dddd

4

3

2

1

4

3

2

1

5000200030000

20003000

01000

30000

30000

010000

1000

The above equations are identical to those we obtained through the direct stiff-ness method.

Problems:

5. Do problems 2.2, 2.9, and 2.17 on pages 59 - 62 in your textbook “A First Course in the Finite Element Method” by D. Logan.

xxx Fff 4)3(

4)2(

4 =+xxx Fff 3

)2(

3

)1(

3 =+

CIVL 7117 Finite Element Methods in Structural Mechanics Page 70

Development of Truss Equations

Introduction

Having set forth the foundation on which the direct stiffness method is based, we will now derive the stiffness matrix for a linear-elastic bar (or truss) element using the general steps outlined in Chapter 1. We will include the introduction of both a local coordinate system, chosen with the element in mind, and a global or reference coordinate system, chosen to be convenient (for numerical purposes) with respect to the overall structure.

We will also discuss the transformation of a vector from the local coordinate system to the global coordinate system, using the concept of transformation ma-trices to express the stiffness matrix of an arbitrarily oriented bar element in terms of the global system. We will solve example plane truss problems to illus-trate the procedure of establishing the total stiffness matrix and equations for so-lution of a structure.

Next we will describe how to handle inclined, or skewed, supports. We will then extend the stiffness method to include space trusses. We will develop the transformation matrix in three-dimensional space and analyze a space truss.

We will then use the principle of minimum potential energy and apply it to the bar element equations. Finally, we will introduce Galerkin’s residual method and then apply it to derive the bar element equations.

Derivation of the Stiffness Matrix for a Bar Element

Consider the derivation of the stiffness matrix for the linear-elastic, constant cross-sectional area (prismatic) bar element show below. This application is di-rectly applicable to the solution of pin-connected truss problems. Consider the idealized element as:


where T is the tensile force directed along the axis at nodes 1 and 2, x is the lo-cal coordinate system directed along the length of the bar, and x and y are the global coordinate system. The bar element has a constant cross-section A, an initial length L, and modulus of elasticity E. The nodal degrees of freedom are the local axial displacements xd1

ˆ and xd2ˆ at the ends of the bar.

The strain-displacement relationship is:

xdudEˆˆ

== εεσ

From equilibrium of forces, assuming no distributed loads acting on the bar, we get:

constant== TA xσ Combining the above equations gives:

constantˆˆ

== TxdudAE


taking the derivative of the above equation with respect to the local coordinate x gives:

0ˆˆ

ˆ=⎟

⎠⎞

⎜⎝⎛

xdudAE

xdd

where u is the axial displacement function in the x direction and A and E can be written as functions of x .

The following assumptions are considered in deriving the bar element stiffness matrix:

1. The bar cannot sustain shear force: 0ˆˆ21 == yy ff .

2. Any effect of transverse displacement is ignored. 3. Hooke’s law applies: stress is related to strain: xx Eεσ =


We will consider the linear bar element shown below.



A linear displacement function u is assumed.

xaau ˆˆ 21 += The number of coefficients in the displacement function, ai, is equal to the total number of degrees of freedom associated with the element. Applying the bound-ary conditions and solving for the unknown coefficients gives:

xxx dx

Lddu 1

12 ˆˆˆˆ

ˆ +⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

In matrix form:

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −=

x

x

dd

Lx

Lxu

2

1

ˆˆˆˆ

1ˆ

Or in another form:

[ ]⎭⎬⎫

⎩⎨⎧

=x

x

ddNNu

2

121 ˆ

ˆˆ

where N1 and N2 are the interpolation functions gives as:

LxN

LxN

ˆˆ1 21 =−=

The linear displacement function u plotted over the length of the bar element is shown below.



The stress-displacement relationship is:

Ldd

xdud xx

x12

ˆˆˆˆ −

==ε


We can now derive the element stiffness matrix as follows:

xAT σ=

Substituting the stress-displacement relationship into the above equation gives:

⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

LddAET xx 12ˆˆ

The nodal force sign convention, defined in element figure, is:


TfTf xx =−= 21ˆˆ

therefore,

⎟⎟⎠

⎞⎜⎜⎝

⎛ −=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −=

LddAEf

LddAEf xx

xxx

x12

221

1

ˆˆˆˆˆˆ

Writing the above equations in matrix form gives:

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−

=⎭⎬⎫

⎩⎨⎧

x

x

x

x

dd

LAE

ff

2

1

2

1

ˆˆ

11

11

ˆˆ

Notice that AE/L for a bar element is analogous to the spring constant k for a spring element.

Step 5 - Assemble the Element Equations and

Introduce Boundary Conditions

The global stiffness matrix and the global force vector are assembled using the nodal force equilibrium equations, and force/deformation and compatibility equations.

[ ] ∑∑==

====N

e

eN

e

e fFkK1

)(

1

)( ˆˆ FK

where k and f are the element stiffness and force matrices expressed in global coordinates.


Solve the displacements by imposing the boundary conditions and solving the following set of equations:

KdF =


Step 7 - Solve for the Element Forces

Once the displacements are found, the stress and strain in each element may be calculated from:

xxxx

x EL

ddxdud εσε =

−== 12

ˆˆˆˆ


Consider the following three-bar system shown below. Assume for elements 1 and 2: A = 1 in2 and E = 30 (106) psi and for element 3: A = 2 in2 and E = 15 (106) psi.

Determine (a) the global stiffness matrix, (b) the displacement of nodes 2 and 3, and (c) the reactions at nodes 1 and 4. For elements 1 and 2:

( )( )in

lbin

lbkk ⎥⎦

⎤⎢⎣

⎡ −−

=⎥⎦

⎤⎢⎣

⎡ −−

×==

11

11

1011

11

3010301 6

6)2()1(

2elementfornumbersnode32



For element 3:

( )( )in

lbin

lbk ⎥⎦

⎤⎢⎣

⎡ −−

=⎥⎦

⎤⎢⎣

⎡ −−

×=

11

11

1011

11

3010152 6

6)3(


where, as before, the numbers above the matrices indicate the displacements associated with the matrix. Assembling the global stiffness matrix by the direct stiffness methods gives:

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−

−−

=

1100

1210

0121

0011

106

4321

K

xxxx dddd

Relating global nodal forces related to global nodal displacements gives:

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−

−−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

x

x

x

x

x

x

x

x

dddd

FFFF

4

3

2

1

6

4

3

2

1

1100

1210

0121

0011

10


041 == xx dd Applying the boundary conditions and the known forces (F2x = 3000 lb.) gives:

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−

=⎭⎬⎫

⎩⎨⎧

x

x

dd

3

26

21

12

100

3000

Solving for d2x and d3x gives:


.001.0.002.0 32 indind xx == The global nodal forces are calculated as:

lbs

FFFF

x

x

x

x

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

−

−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−

−−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

10000

30002000

0001.0002.00

1100

1210

0121

0011

106

4

3

2

1

Selecting Approximation Functions for Displacements

Consider the following guidelines, as they relate to the one-dimensional bar element, when selecting a displacement function.

1. Common approximation functions are usually polynomials, where the function is expressed in terms of the shape functions.

2. The approximation function should be continuous within the bar element. 3. The approximating function should provide interelement continuity for all

degrees of freedom at each node for discrete line elements, and along common boundary lines and surfaces for two- and three-dimensional elements.

For the bar element, we must ensure that nodes common to two or more elements remain common to these elements upon deformation and thus prevent overlaps or voids between elements.

The linear function is then called a conforming (or compatible) function

for the bar element because it ensures both the satisfaction of continuity between adjacent elements and of continuity within the element.


4. The approximation function should allow for rigid-body displacement and for a state of constant strain within the element.

The interpolation function must allow for a rigid-body displacement, that

means the function must be capable of yielding a constant value. Consider the follow situation;

xx ddaau 2111ˆˆˆ ===

therefore,

( ) 1212211ˆˆˆ aNNdNdNu xx +=+=

since u = a1 then:

( ) 1211ˆ aNNau +== this means that:

121 =+ NN The displacement interpolation function must add to unity at every point within the element so the u will yield a constant value when a rigid-body displacement occurs.

Transformation of Vectors in Two Dimensions

In many problems it is convenient to introduce both local and global (or refer-ence) coordinates. Local coordinates are always chosen to conveniently repre-sent the individual element. Global coordinates are chosen to be convenient for the whole structure.

Given the nodal displacement of an element, represented by the vector d in the figure below, we want to relate the components of this vector in one coordi-nate system to components in another.


Let’s consider that d does not coincident with either the local or global axes. In

this case, we want to relate global displacement components to local ones. In so doing, we will develop a transformation matrix that will subsequently be used to develop the global stiffness matrix for a bar element. We define the angle θ to be positive when measured counterclockwise from x to x . We can express vector displacement d in both global and local coordinates by:

jdiddd yxyxˆˆˆˆ +=+= jid

where i and j are unit vectors in the x and y directions, and i and j are unit vec-tor in the x and y direction.


Consider the following diagram:

Using vector addition:

iba =+ Using the law of cosines, we get:

θθ cos||cos|||| == aia

Similarly:

θθ sin||sin|||| == bib The vector a is in the i direction and b is in the j direction, therefore:

( ) ( ) ( )( )jjbbiia ˆsinˆ||ˆcosˆ|| −=−=== θθa The vector i can be rewritten as:

ji ˆsinˆcos θθ −=i


In a similar manner, we could get an expression for j

jij ˆcosˆsin θθ += If we use the above express to describe the displacement vector we get:

( ) ( ) jdidjidjidjdid yxyxyxˆˆˆˆˆcosˆsinˆsinˆcos +=++−=+ θθθθ

If we combine like coefficients of i and j we get:

xyx ddd ˆsincos =+ θθ

yyx ddd ˆcossin =+− θθ


θθ

sincos

ˆˆ

==

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

=⎭⎬⎫

⎩⎨⎧

SC

dd

CSSC

dd

y

x

y

x

The above equation relates the global displacement d to the local displacements d . The matrix above is called the transformation matrix.

⎥⎦

⎤⎢⎣

⎡− C

SSC

The figure below shows xd expressed in terms of the global coordinates x and y.


The magnitude of xd is:

yxx SdCdd +=ˆ

Example 2 - Bar Element Problem

The global nodal displacement at node 2 is d2x = 0.1 in. and d2y = 0.2 in. for the bar element shown below. Determine the local x displacement.

Using the following expression we just derived, we get:

yxx SdCdd +=ˆ


.223.0)2.0(60sin)01.0(60cos2 ind oox =+=

Global Stiffness Matrix

We will now use the transformation relationship developed above to obtain the global stiffness matrix for a bar element. We known that for a bar element in local coordinates we have:

dkfdd

LAE

ff

x

x

x

x ˆˆˆˆˆ

11

11

ˆˆ

2

1

2

1 =⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−

=⎭⎬⎫

⎩⎨⎧

We want to relate the global element forces f to the global displacements d for a bar element with an arbitrary orientation.

kdf

dddd

k

ffff

y

x

y

x

y

x

y

x

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

2

2

1

1

2

2

1

1

Using the relationship between local and global components, we can develop the global stiffness matrix. We already know the transformation relationships:

xyx ddd ˆsincos =+ θθ

Combining both expression xd1

ˆ and xd2ˆ , in matrix form, we get:

dTd

dddd

SCSC

dd

y

x

y

x

x

x *

2

2

1

1

2

1 ˆ0000ˆ

ˆ=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎦

⎤⎢⎣

⎡=

⎭⎬⎫

⎩⎨⎧

where

⎥⎦

⎤⎢⎣

⎡=

SCSC

T00

00*


A similar expression for the force transformation can be developed.

fTf

ffff

SCSC

ff

y

x

y

x

x

x *

2

2

1

1

2

1 ˆ0000ˆ

ˆ=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎦

⎤⎢⎣

⎡=

⎭⎬⎫

⎩⎨⎧

Substituting the global force expression into element force equation gives:

dkfTdkf ˆˆˆˆˆ * =⇒= Substituting the transformation between local and global dispalcements gives:

dTkfT ** ˆ= The matrix T* is not a square matrix so we cannot invert it. Let’s expand our rela-tionship between local and global displacement.

Tdd

dddd

CS

SC

CS

SC

dddd

y

x

y

x

y

x

y

x

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

ˆ00

00

00

00

ˆˆˆˆ

2

2

1

1

2

2

1

1

where T is:

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−=

CS

SC

CS

SC

T00

00

00

00

We can write a similar expression for the relationship between local and global forces.


Tff

ffff

CS

SC

CS

SC

ffff

y

x

y

x

y

x

y

x

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

ˆ00

00

00

00

ˆˆˆˆ

2

2

1

1

2

2

1

1

Therefore our original local coordinate force-displacement expression

dkfdd

LAE

ff

x

x

x

x ˆˆˆˆˆ

11

11

ˆˆ

2

1

2

1 =⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−

=⎭⎬⎫

⎩⎨⎧

May be expanded

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡ −

−=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

y

x

y

x

y

x

y

x

dddd

LAE

ffff

2

2

1

1

2

2

1

1

ˆˆˆˆ

0000

0101

0000

01

01

ˆˆˆˆ


TdkTfdkf ˆˆˆˆ =⇒= Multiply both side by T -1 we get:

TdkTf ˆ1−= where T-1 is the inverse of T. It can be shown that:

TTT =−1 The global force-displacement equations become:

TdkTf T ˆ=


Where the global stiffness matrix k is:

TkTk T ˆ= Expanding the above transformation gives:

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡−−

−−

−−

−−

=

2

2

2

2

2

2

2

2

SCSSCS

CSCCSC

SCSSCS

CSC

CSC

LAEk

We can assemble the total stiffness matrix by using the above element stiffness matrix and the direct stiffness method.

[ ] ∑∑==

====N

e

eN

e

e fFkK1

)(

1

)( FK

dKF = Local forces can be computed as:

1 1 1

1 1 1

222

222

ˆ ˆ1 0 1 0 1 0 1 0 0 0ˆ ˆ0 0 0 0 0 0 0 0 0 0ˆ ˆ1 0 1 0 1 0 1 0 0 0

0 0 0 0 0 0 0 0 0 0ˆˆ

x x x

y y y

xxx

yyy

f d dC Sf d dS CAE AE

dC SL LdfdS Cdf

⎧ ⎫ ⎧ ⎫ ⎧ ⎫− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎪ ⎪ −⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥= =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪−⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎩ ⎭⎩ ⎭⎩ ⎭

1 1 1 2 2

1

1 1 2 22

2

ˆ

ˆ 0ˆ

0ˆ

x x y x y

y

x y x yx

y

f Cd Sd Cd Sdf AE

Cd Sd Cd SdLf

f

⎧ ⎫ + − −⎡ ⎤⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎢ ⎥=⎨ ⎬ ⎢ ⎥− − + +⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎣ ⎦⎩ ⎭



For the bar element shown below, evaluate the global stiffness matrix. As-sume the cross-sectional area is 2 in.2, the length is 60 in., and the E is 30 x 106 psi.

Therefore:

2130sin

2330cos30 ===== ooo SCθ

The global elemental stiffness matrix is:

( )in

lbk

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−−

−−

−−

×=

41

43

41

43

43

43

43

43

41

43

41

43

43

43

43

43

601030)2( 6

Simplifying the global elemental stiffness matrix is:

inlbk

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡−−

−−

−−

−−

=

250.0433.0250.0433.0

433.0750.0433.0750.0

250.0433.0

250.0433.0

433.0750.0

433.0750.0

106

Computation of Stress for a Bar in the x-y Plane

For a bar element the local forces are related to the local displacements by:


⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−

=⎭⎬⎫

⎩⎨⎧

x

x

x

x

dd

LAE

ff

2

1

2

1

ˆˆ

11

11

ˆˆ

The force-displacement equation for xf2 is:

[ ]⎭⎬⎫

⎩⎨⎧

−=x

xx d

dL

AEf2

12 ˆ

ˆ11ˆ

Therefore the stress is:

[ ] 12

2

ˆˆ1 1

ˆxx

x

df EA L d

σ⎧ ⎫⎪ ⎪= = − ⎨ ⎬⎪ ⎪⎩ ⎭

The stress in terms of global displacement is:

[ ]1

11 1 2 2

2

2

0 01 1

0 0

x

yx y x y

x

y

ddC SE E Cd Sd Cd SddC SL Ld

σ

⎧ ⎫⎪ ⎪

⎡ ⎤ ⎪ ⎪ ⎡ ⎤= − = − − + +⎨ ⎬⎢ ⎥ ⎣ ⎦⎣ ⎦ ⎪ ⎪

⎪ ⎪⎩ ⎭


For the bar element shown below, determine the axial stress. Assume the cross-sectional area is 4 x 10-4 m2, the length is 2 m, and the E is 210 GPa. The global displacements are known as d1x = 0.25 mm, d1y = 0, d2x = 0.5 mm, and d2y = 0.75 mm.


mKN⎥

⎦

⎤⎢⎣

⎡++−−

×= )75.0(

43)5.0(

21)0(

43)25.0(

21

210210 6

σ

MPamkN 32.81/1032.81 3 =×=σ

Solution of a Plane Truss

We will now illustrate the use of equations developed above along with the di-rect stiffness method to solve the following plane truss example problems. A plane truss is a structure composed of bar elements all lying in a common plane that are connected together by frictionless pins. The plane truss also must have loads acting only in the common plane.

Example 5 - Plane Truss Problem

The plane truss shown below is composed of three bars subjected to a down-ward force of 10 kips at node 1. Assume the cross-sectional area A = 2 in.2 and E is 30 x 106 psi for all elements. Determine the x and y displacement at node 1 and stresses in each element.

Element Node 1 Node 2 θ C S 1 1 2 90o 0 1 2 1 3 45o 0.707 0.707 3 1 4 0o 1 0


The global elemental stiffness matrix for element 1 is:

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡−

−

×=

101

0

0000

1010

0000

120)1030)(2( 6

)1(

2211

k

yxyxdddd


⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡−−

−−

−−

−−

×=

1111

1111

11

11

11

11

2240)1030)(2( 6

)2(

3311

k

yxyxdddd


⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡ −

−×

=

0000

0101

0000

01

01

120)1030)(2( 6

)3(

4411

k

yxyxdddd


The total global stiffness matrix is:

inlbK

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

×=

−−−

−−

−

−−

−

−−−

00000000

01000001

00354.0354.000354.0354.0

00354.0354.000354.0354.0

0000101

0

00000000

00354.0354.01

0354.1354.0

01354.0354.0

00354.0354.1

5105

The total global force-displacement equations are:

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

=

−

0000001

1

4

4

3

3

2

2

000,100

ydxd

yFxFyFxFyFxF

K

The above equations reduce to:

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡×=

⎭⎬⎫

⎩⎨⎧

− y

x

dd

1

15

354.1354.0

354.0354.1

105000,10

0

Solving the equations gives:

.1059.1.10414.0 21

21 indind yx

−− ×−=×=

The stress in an element is:

[ ]yxyx SdCdSdCdLE

2211 ++−−=σ


Element Node 1 Node 2 θ C S

1 1 2 90o 0 1 2 1 3 45o 0.707 0.707 3 1 4 0o 1 0

The stress in element 1 is:

[ ] psid y 965,3120

10301

6)1( =−

×=σ


[ ] psidd yx 471,1)707.0()707.0(120

103011

6)2( =+

×−=σ


[ ] psid x 035,1120

10301

6)3( −=−

×=σ

Let’s check equilibrium at node 1: 0)2)(035,1()707.0)(2)(471,1( 22 =−−=∑ inpsiinpsiFx

0000,10)707.0)(2)(471,1()2)(965,3( 22 =−+=∑ inpsiinpsiFy


Example 6 - Plane Truss Problem

Develop the element stiffness matrices and system equations for the plane truss below. Assume the stiffness of each element is constant. Use the number-ing scheme indicated. Solve the equations for the displacements and compute the member forces. All elements have a constant value of AE/L

45

P1

P2

1

3

2

The node numbering is given in the diagram above (Note that this is the optimum numbering configuration). Develop the element information:

Member Node 1 Node 2 Elemental Stiffness

θ

1 1 2 k 0 2 2 3 k 3π/4 3 1 3 k π/2

Compute the elemental stiffness matrix for each element. The general form of the matrix is:

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡−

−−−

−−

−−

=

2

2

2

2

2

2

2

2

SCSS

CS

CSCCSC

SCSSCS

CSC

CSC

LAEk


For element 1:

For element 2:

For element 3:

Assemble the global stiffness matrix by superimposing the elemental global ma-trices.

y

x

y

x

dddd

kk

yxyxdddd

2

2

1

1

)1(

0000

0101

0000

0101

2211

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡ −

−=

y

x

y

x

dddd

kk

yxyxdddd

3

3

2

2

)2(

1111

1111

1111

1111

2

3322

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−

−

−

−

−−

=

y

x

y

x

dddd

kk

yxyxdddd

3

3

1

1

)3(

1010

0000

1010

0000

3311

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡−

−

=

y

x

y

x

y

x

dddddd

kK

yxyxyx dddddd

3

3

2

2

1

1

311120

111100

111100

111302

200020

000202

2

332211

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−

−

−

−

−

−−

−

−

−=


The unconstrained (no boundary conditions satisfied) equations are

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

−=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−

−

−

−

−

−−

−

−

−

y

x

y

x

y

x

y

x

y

x

FFP

PFF

dddddd

k

3

3

2

1

1

1

3

3

2

2

1

1

311120

111100

111100

111302

200020

000202

2

The displacement at nodes 1 and 3 are zero in both directions. Applying these conditions to the system equations gives

⎭⎬⎫

⎩⎨⎧−

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −− 2

1

2

2

11

13

2 PP

ddk

y

x

Solving this set of equations is fairly easy. The solution is

kPPd

kPPd yx

212

212

3−=

−=

Using the force-displacement relationship the force in each member may be computed. Member (element) 1:

011221

1 =−=⎟⎠⎞

⎜⎝⎛ −−= yx fPP

kPPkf

1 22 1 2 2 0x y

P Pf k P P fk−⎛ ⎞= = − =⎜ ⎟

⎝ ⎠

Member (element) 2:

1 2 1 22 2

31 1 22 2x

P P P Pf k Pk k

⎡ ⎤− −⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= + − = −⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎣ ⎦


1 2 1 23 2

31 1 22 2x

P P P Pf k Pk k

⎡ ⎤− −⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= − + =⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎣ ⎦

00 32 == yy ff

Member (element) 3:

0000

33

11

==

==

yx

yx

ffff

The solution to this simple problem can be readily checked by using simple static equilibrium equations.

Example 7 - Plane Truss Problem Consider the two bar truss shown below. Determine the displacement in the y di-rection of node 1 and the axial force in each element. Assume E = 210 GPa and A = 6 x 10 -4 m 2



8.054sin6.0

53cos )1()1( ==== θθ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡−−

−−

−−

−−

××=

−

64.048.064.048.0

48.036.048.036.0

64.048.0

64.048.0

48.036.0

48.036.0

5)106(10210 46

)1(k

Simplifying the above expression gives:

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡−−

−−

−−

−−

=

64.048.064.048.0

48.036.048.036.0

64.048.0

64.048.0

48.036.0

48.036.0

200,25)1(

2211

k

yxyxdddd


1sin0cos )2()2( == θθ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡−−

−

××=

−

1010

0000

1010

0000

4)106)(10210( 46

)2(k

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡−

−

=

25.1025.1

0

0000

25.1025.10

0000

200,25)2(

3311

k

yxyxdddd


The total global equations are:

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡−−

−−−

−

−−

−−

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

y

x

y

x

y

x

y

x

y

x

y

x

dddddd

FFFFFF

3

3

2

2

1

1

3

3

2

2

1

1

25.1000

25.10

000000

00

64.048.064.048.0

00

48.036.048.036.0

25.10

64.048.089.148.0

00

48.036.048.036.0

200,25

The displacement boundary conditions are:

033221 ===== yxyxx ddddd δ

By applying the boundary conditions the force-displacement equations reduce to:

)89.148.0(200,25 1ydP += δ

Solving the equation gives:

δ25.0)101.2( 51 −×= − Pd y

By substituting P = 1000 kN and δ = -0.05 m in the above equation gives:

md y 0337.01 =

The local element forces for element 1 are:

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧=

−=

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡ −−

=⎭⎬⎫

⎩⎨⎧

y

x

y

x

x

x

dd

dd

ff

2

2

1

1

2

1 0337.005.0

8.00

6.00

08.0

06.0

11

11

200,25

The element forces are:


kNfkNf xx 7.766.76 21 =−= The local element forces for element 2 are:

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧=

−=

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡ −−

=⎭⎬⎫

⎩⎨⎧

y

x

y

x

x

x

dd

dd

ff

3

3

1

1

3

1 0337.005.0

10

00

01

00

11

11

500,31

The element forces are:

kNfkNf xx 10611061 31 −==

Transformation Matrix and Stiffness Matrix for a Bar in Three-Dimensional Space

Let’s derive the transformation matrix for the stiffness matrix for a bar element in three-dimensional space as shown below:

The coordinates at node 1 are x1, y1, and z1, and the coordinates of node 2 are x2, y2, and z2. Also, let θx, θy, and θz be the angles measured from the global x, y,


and z axes, respectively, to the local x axis. The three-dimensional vector repre-senting the bar element is gives as:

kdjdidddd zyxzyxˆˆˆˆˆˆ ++=++= kjid

where i, j, and k are unit vectors in the x, y, and z directions, and i , j , and k are unit vector in the x , y , and z direction. Taking the dot product of the above

equation with i gives:

xzyx dikdijdiid ˆ)ˆ()ˆ()ˆ( =•+•+•

By the definition of the dot product we get:

2 1 2 1 2 1ˆ ˆ ˆx y z

x x y y z zi i C j i C k i CL L L− − −

• = = • = = • = =

where

212

212

212 )()()( zzyyxxL −+−+−=

and

zzyyxx CCC θθθ coscoscos ===

where Cx, Cy, and Cz are projections of i on to i, j, and k, respectively.

zzyyxxx dCdCdCd ++=ˆ


The transformation between local and global displacements is:

dTd *ˆ =

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⎥⎦

⎤⎢⎣

⎡=

⎭⎬⎫

⎩⎨⎧

z

y

x

z

y

x

zyx

zyx

x

x

dddddd

CCCCCC

dd

2

2

2

1

1

1

2

1000

000ˆˆ

where

⎥⎦

⎤⎢⎣

⎡=

zyx

zyx

CCCCCC

T000

000*

The transformation from the local to the global stiffness matrix is:

TkTk T ˆ=

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡ −−

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=zyx

zyx

z

y

x

z

y

x

CCCCCC

LAE

CCC

CCC

k000

00011

110

00

000



⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−

−−

−

−−−

−−−

−−

−

−−−

=

2

2

2

2

2

2

2

2

2

2

2

2

z

zy

zx

z

zy

zx

zy

y

yx

zy

y

yx

zx

yx

x

zx

yx

x

z

zy

zx

z

zy

zx

zy

y

yx

zy

y

yx

zx

yx

x

zx

yx

x

CCCCCCCCCC

CCCCCCC

CCC

CCCC

CCCCC

C

CCCCC

CCCCC

CCCCC

CCCCC

CCCC

CCCCC

C

LAEk

The global stiffness matrix can be written in a more convenient form as:

⎥⎦

⎤⎢⎣

⎡ −−

=λλ

λλ

LAEk

where

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=2

2

2

z

zy

zx

zy

y

yx

zx

yx

x

CCCCC

CCC

CC

CCCC

Cλ

Example 8 - Space Truss

Consider the space truss shown below. The modulus of elasticity, E, is 1.2 x 106 psi for all elements. Node 1 is constrained from movement in the y direction.

To simplify the stiffness matrices for the three elements, we will express each element in the following form:

⎥⎦

⎤⎢⎣

⎡ −−

=λλ

λλ

LAEk


Consider element 3:

)3(14

)3(14

)3(14

LzzC

LyyC

LxxC zyx

−=

−=

−=

where

214

214

214

)3( )()()( zzyyxxL −+−+−=

.5.86)48()72( 22)3( inL =−+−= then

550.05.86

480833.05.86

72−=

−==−=

−= zyx CCC


Therefore, the matrix λ is:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

30.0046.0

000

46.0069.0

λ

The global element stiffness matrix is:

⎥⎦

⎤⎢⎣

⎡ −−

×=

λλ

λλ

5.86102.1)(187.0( )6

444111

k

zyxzyxdddddd

Consider element 1:

.5.80)36()72( 22)1( inL =+−= then

045.05.80

3689.05.80

72===−=

−= zyx CCC


⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ −−=

000

020.040.0

040.0

79.0λ


⎥⎦

⎤⎢⎣

⎡ −−

×=

λλ

λλ

5.80102.1)(302.0( )6

222111

k

zyxzyxdddddd


Consider element 2:

.108)72()36()72( 222)2( inL =++−= then

667.01087233.0

10836667.0

10872

====−=−

= zyx CCC


⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ −−

−−=

45.045.045.0

45.011.022.0

45.022.0

45.0λ


⎥⎦

⎤⎢⎣

⎡ −−

×=

λλ

λλ

108102.1)(729.0( )6

333111

k

zyxzyxdddddd


000 444333222 ========= zyxzyxzyx ddddddddd

01 =yd

Canceling the rows and the columns associated with the boundary conditions re-duces the global stiffness matrix to:

⎥⎦

⎤⎢⎣

⎡ −−

=44502450

24509000

11

K

zxdd



⎭⎬⎫

⎩⎨⎧

−=

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −− 1000

044502450

24509000

1

1

z

x

dd


.264.0.072.0 11 indind zx −=−= The local forces in an element are:

ˆ

ˆ

ix

iy

ix x y z x y z iz

x y z x y z jxjx

jy

jz

dd

f C C C C C C dAEC C C C C C dLf

dd

⎧ ⎫⎪ ⎪⎪ ⎪

⎧ ⎫ ⎪ ⎪− − −⎡ ⎤⎪ ⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬⎢ ⎥− − −⎣ ⎦⎪ ⎪ ⎪ ⎪⎩ ⎭⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

The stress in an element is:

[ ]

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

−−−=

jz

jy

jx

iz

iy

ix

zyxzyx

dddddd

CCCCCCLEσ



[ ]

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−

−

−×

=

000264.00072.0

045.089.0045.089.05.80102.1 6

)1(σ

psi955)1( −=σ


[ ]

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−

−

−−−×

=

000264.00072.0

667.033.0667.0667.033.0667.0108

102.1 6)2(σ

psi423,1)2( =σ The stress in element 3 is:

[ ]

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−

−

−−×

=

000264.00072.0

55.0083.055.0083.05.86102.1 6

)3(σ

psi843,2)3( =σ


Inclined, or Skewed, Supports

If a support is inclined, or skewed, at some angle α for the global x axis, as shown below, the boundary conditions on the displacements are not in the global x-y directions but in the x’-y’ directions.

We must transform the local boundary condition of d’3y = 0 (in local coordinates) into the global x-y system. Therefore, the relationship between of the compo-nents of the displacement in the local and the global coordinate systems at node 3 is:

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

=⎭⎬⎫

⎩⎨⎧

y

x

y

x

dd

dd

3

3

3

3

cossin

sincos

''

αα

αα

We can rewrite the above expression as:

[ ] ⎥⎦

⎤⎢⎣

⎡−

==αα

αα

cossin

sincos

][' 3333 tdtd


We can apply this sort of transformation to the entire displacement vector as:

'][][' 11 dTdordTd T== where the matrix [T1]T is:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

][]0[]0[

]0[][]0[

]0[]0[][

][

3

1

tI

IT T

Both the identity matrix [I] and the matrix [t3] are 2 x 2 matrices.

The force vector can be transformed by using the same transformation.

fTf ][' 1= In global coordinates, the force-displacement equations are:

dKf ][= Applying the skewed support transformation to both sides of the equation gives:

dKTfT ]][[][ 11 = By using the relationship between the local and the global displacements, the force-displacement equations become:

1 1' [ ][ ][ ] 'Tf T K T d=


Therefore the global equations become:

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

y

x

y

x

y

x

T

y

x

y

x

y

x

dddddd

TKT

FFFFFF

3

3

2

2

1

1

11

3

3

2

2

1

1

''

]][][[

''

Example 9 - Space Truss

Consider the plane truss shown below. Assume E = 210 GPa, A = 6 x 10-4 m2 for element 1 and 2, and A = 6 2 x 10-4 m2 for element 3.

Determine the stiffness matrix for each element. The global elemental stiffness matrix for element 1 is:

1sin0cos )1()1( == θθ


1 1 2 2

6 4(1)

0 0 0 00 1 0 1(210 10 )(6 10 )0 0 0 010 1 0 1

x y x yd d d d

k−

⎡ ⎤⎢ ⎥−× × ⎢ ⎥=⎢ ⎥⎢ ⎥−⎣ ⎦


0sin1cos )2()2( == θθ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡ −

−××

=−

0000

0101

0000

01

01

1)106)(10210( 46

)2(

3322

k

yxyx dddd


22sin

22cos )3()3( == θθ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡−−

−−

−−

−−

××=

−

1111

1111

1111

11

11

22)1026)(10210( 46

)3(

3311

k

yxyxdddd


Using the direct stiffness method, the global stiffness matrix is:

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡−−

−−−

−

−−−

−

−−

×=

5.05.0005.05.0

5.05.1015.05.0

001010

010100

5.05.0105.15.0

5.05.0005.05.0

10260,1 5

mNK

We must transform the global displacements into local coordinates. Therefore the transformation [T1] is:

1

3 2 22 2

2 22 2

1 0 0 0 0 00 1 0 0 0 0

[ ] [0] [0]0 0 1 0 0 0

[ ] [0] [ ] [0]0 0 0 1 0 0

[0] [0] [ ]0 0 0 00 0 0 0

IT I

t

−

⎡ ⎤⎢ ⎥⎢ ⎥⎡ ⎤⎢ ⎥⎢ ⎥= = ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥⎢ ⎥⎣ ⎦

The first step in the matrix transformation to find the product of [T1][K].

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

×=

−−

−

−−−

−

−−−

−0

707.0005.05.0

70.0414.1

015.05.0

001010

707.0707.0

0100

0707.0

105.15.0

0707.0

005.05.0

5101260]][[ mNKT1


The next step in the matrix transformation to find the product of [T1][K][T1]T.

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

×=

−−

−−−

−

−−−

−5.05.00

707.000

5.05.10

707.0707.0707.0

001010

707.0707.0

0100

0707.0

105.15.0

0707.0

005.05.0

5101260]][][[ mNT

11 TKT

The displacement boundary conditions are:

0'3211 ==== yyyx dddd

By applying the boundary conditions the global force-displacement equations are:

⎭⎬⎫

⎩⎨⎧

==

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−

×0'000,1

'5.1707.0

707.01

1012603

2

3

25

x

x

x

x

FkNF

dd


mmdmmd xx 61.5'91.11 32 == The global nodal forces are calculated as:

⎪⎪

⎭

⎪⎪

⎬

⎫

⎪⎪

⎩

⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−−

−−−

−

−−−061.5091.11

00

5.05.00

707.000

5.05.10

707.0707.0707.0

001010

707.0707.0

0100

0707.0

105.15.0

0707.0

005.05.0

1

3

3

2

2

1

1

''

B

FFFFFF

y

x

y

x

y

x

mmNB 2

1 10260,1 ×=


Therefore:

kNFkNF yx 500500 11 −=−=

kNFF yy 707'0 32 ==

Potential Energy Approach to Derive Bar Element Equations Let’s derive the equations for a bar element using the principle of minimum

potential energy. The total potential energy, πp, is defined as the sum of the inter-nal strain energy U and the potential energy of the external forces Ω:

Ω+= Upπ

The differential internal work (strain energy) dU in a one-dimensional bar element is:

xx dxzydU εσ ))()(( ∆∆∆=

If we let the volume of the element approach zero, then:

dVddU xx εσ= Summing the differential energy over the whole bar gives:

∫ ∫⎭⎬⎫

⎩⎨⎧=

Vxx dVdU

xε

εσ0


For a linear-elastic material (Hooke’s law) as shown below:

The internal strain energy statement becomes

∫=V

xx dVU εσ21

The potential energy of the external forces is:

∑∫∫=

−−−=ΩM

1iixixxb dSdV dfuTuX

SV

ˆˆˆˆˆˆ

where bX is the body force (force per unit volume), xT is the traction (force per

unit area), and ixf is the nodal concentrated force. All of these forces are consid-ered to act in the local x direction.

Apply the following steps when using the principle of minimum potential en-ergy to derive the finite element equations.

1. Formulate an expression for the total potential energy 2. Assume a displacement pattern 3. Obtain a set of simultaneous equations minimizing the total potential en-

ergy with respect to the displacement parameters.

xx Eεσ =


Consider the following bar element, as shown below:

We can approximate the axial displacement as:

[ ]⎭⎬⎫

⎩⎨⎧

=x

x

ddNNu

2

121 ˆ

ˆˆ


LxN

LxN

ˆˆ1 21 =−=

Using the stress-strain relationships, the axial strain is:

⎭⎬⎫

⎩⎨⎧

⎥⎦⎤

⎢⎣⎡==

x

xx d

dxd

dNxd

dNxdud

2

121

ˆˆ

ˆˆˆˆ

ε


∫∫

∫

−−

−−=

SV

xx1x1x

L

oxxp

uTuX

dfdfxd2A

dSdV xb ˆˆˆˆ

ˆˆˆˆˆ 22εσπ


][ˆˆ11

2

1 dBdd

LL x

x

xx =

⎭⎬⎫

⎩⎨⎧

⎥⎦⎤

⎢⎣⎡−= εε

where [B] are:

⎥⎦⎤

⎢⎣⎡−=

LLB 11

The axial stress-strain relationship is:

xx D εσ ][= where [D] = [E] for the one-dimensional stress-strain relationship and E is the modulus of elasticity. Therefore, stress can be related to nodal displacements as:

dBDxˆ]][[=σ

The total potential energy expressed in matrix form is:

∫∫∫ −−−=S

T

V

TTL

oxxp TuXuPdxd

2A dSdV xb

ˆˆˆˆˆˆεσπ

where P represented the concentrated nodal loads. If we substitute the rela-tionship between u and d into the energy equations we get:

[ ] [ ] [ ]

[ ] [ ] ∫∫

∫

−−

−=

S

TT

V

TT

TL

o

TTT

p

TNdXNd

PdxddBDBd2A

dSdV xbˆˆˆˆ

ˆˆˆˆπ

In the above expression for potential energy πp is a function of the d , that is: πp = πp )ˆ,ˆ( 21 xx dd . However, [B] and [D] and the nodal displacements d are not a function of x . Integration the energy expression with respect to x gives:


fddBDBdAL TTTT

pˆˆˆ][][][ˆ

2−=π

where

∫∫ ++=SV

dSXNdVXNPf bT

bT ][][ˆ

We can define the surface tractions and body-force matrices as:

∫=S

x dSTNf ˆ][ˆ Ts

∫=V

dVXNf bT

bˆ][ˆ

Minimization of πp with respect to each nodal displacement requires that:

0ˆ0ˆ21

=∂∂

=∂∂

x

p

x

p

ddππ

For convenience, let’s define the following

dBDBdU TTT ˆ][][][ˆ* = or

[ ]⎭⎬⎫

⎩⎨⎧

⎥⎦⎤

⎢⎣⎡−

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧−=

x

xxx d

dLL

E

L

LddU2

121

*

ˆˆ11][1

1ˆˆ


( )2221

212

* ˆˆˆ2ˆxxxx dddd

LEU +−=


The loading on a bar element is given as:

xxxx

Tfdfdfd 2211ˆˆˆˆˆˆ +=

Therefore, the minimum potential energy is:

( ) 0ˆˆ2ˆ22ˆ 121

1

=−−=∂∂

xxxx

p fddL

AEdπ

( ) 0ˆˆ2ˆ22ˆ 221

2

=−+−=∂∂

xxxx

p fddL

AEdπ

The above equations can be written in matrix form as:

( ) 0ˆˆ

ˆˆ

11

11

ˆ2

1

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛⎥⎦

⎤⎢⎣

⎡ −−

=∂∂

x

x

x

xp

ff

dd

LAE

dπ

The stiffness matrix for a bar element is:

[ ] ⎥⎦

⎤⎢⎣

⎡ −−

=11

11ˆ

LAEk

This form of the stiffness matrix obtained from the principle of minimum potential energy is identical to the stiffness matrix derived from the equilibrium equations.

Example 10 - Bar Problem

Consider the bar shown below:


The energy equivalent nodal forces due to the distributed load are:

∫=S

x dSTNf ˆ][ˆ Ts

∫⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧ −=

⎭⎬⎫

⎩⎨⎧

=L

xdxC

Lx

Lx1

fff

0

ˆˆˆ

ˆ

ˆˆˆ2x

1x

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧ −=

⎭⎬⎫

⎩⎨⎧

3CL6

CL

3LxC

3LxC

2xC

ff

2

2L

3

32

0

ˆ

ˆˆ

ˆˆ

2x

1x

The total load is the area under the distributed load curve, or:

2))((

21 2CLCLLF ==

The equivalent nodal forces for a linearly varying load are:

loadtotaltheof31

31 ==Ff x

loadtotaltheof32

32

2 ==Ff x


Example 11 - Bar Problem

Consider the axially loaded bar shown below. Determine the axial displace-ment and axial stress. Let E = 30 x 10 6 psi, A = 2 in.2, and L = 60 in. Use (a) one and (b) two elements in the finite element solutions.

The one-element solution:

The distributed load can be converted into equivalent nodal forces using:

∫=S

x dSTNF ˆ][ T

The above expression can be expressed as:

∫ −⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧ −=

⎭⎬⎫

⎩⎨⎧

=L

2x

1x dx10x

Lx

Lx1

FF

F0

or


⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−

−=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−

+−=

⎭⎬⎫

⎩⎨⎧

310L

610L

310L

310L

210L

FF

2

2

2

22

2x

1x

or

⎭⎬⎫

⎩⎨⎧

−−

=⎭⎬⎫

⎩⎨⎧

lb12,000lb6,000

2x

1x

FF

The stiffness matrix for element 1 is:

⎥⎦

⎤⎢⎣

⎡ −−

=11

11

106)1(k

The element equations are:

⎭⎬⎫

⎩⎨⎧

−−

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −− 000,12

000,601

111

102

16

x

x

Rd


.006.01 ind x −= The axial stress-strain relationship is:

][ xx D εσ =

dBEx ][ =σ

⎟⎠⎞

⎜⎝⎛ −

=⎭⎬⎫

⎩⎨⎧

⎥⎦⎤

⎢⎣⎡−=

LddE

dd

LLE xx

x

x 12

2

111

)(000,360

006.001030 6 Tpsi=⎟⎠⎞

⎜⎝⎛ +

×=


The two-element solution:

The distributed load can be converted into equivalent nodal forces. For element 1, the total force of the triangular-shaped distributed load is:

lbinlbin 500,4.)/300.)(30(21

−=

Based on equations developed for the equivalent nodal force of a triangular dis-tributed load, develop in the one-element problem, the nodal forces are:

⎭⎬⎫

⎩⎨⎧

−−

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−

−=

⎭⎬⎫

⎩⎨⎧

lblb

ff

x

x

000,3500,1

)500,4(32

)500,4(31

)1(2

)1(1

For element 2, the applied force is in two parts: a triangular-shaped distributed load and a uniform load. The uniform load is:

lbinlbin 000,9.)/300.)(30( −= The nodal forces for element 2 are:

⎭⎬⎫

⎩⎨⎧−−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎦⎤

⎢⎣⎡ +−

⎥⎦⎤

⎢⎣⎡ +−

=⎭⎬⎫

⎩⎨⎧

lblb

ff

x

x

500,7000,6

)500,4(32)000,9(

21

)500,4(31)000,9(

21

)2(3

)2(2


The final nodal force vector is:

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−−−

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

500,7000,9500,1

33

2

1

xx

x

x

RFFF

The element stiffness matrices are:

⎥⎦

⎤⎢⎣

⎡ −−

==11

112)2()1(

32

21

LAEkk

The assembled global stiffness matrix is:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

−

−−×=

110

121

011

102 6K

The assembled global force-displacement equations are:

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−−−

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

−

−−×

500,7000,9500,1

0110

121

011

102

3

2

1

6

x

x

x

Rdd

After the eliminating the row and column associated with d3x, we get:

⎭⎬⎫

⎩⎨⎧−−

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡ −−

×000,9500,1

21

11

1022

16

x

x

dd


.00525.0.006.0 21 indind xx −=−=


The axial stress-strain relationship is:

)(75000525.0

006.0301

30111

2

1)1( TpsiEdd

LLE

x

xx =

⎭⎬⎫

⎩⎨⎧

−−

⎥⎦⎤

⎢⎣⎡−=

⎭⎬⎫

⎩⎨⎧

⎥⎦⎤

⎢⎣⎡−=σ

)(250,50

00525.0301

301)2( TpsiEx =

⎭⎬⎫

⎩⎨⎧−

⎥⎦⎤

⎢⎣⎡−=σ

Comparison of Finite Element Solution to Exact Solution

In order to be able to judge the accuracy of our finite element models, we will develop an exact solution for the bar element problem. The exact solution for the displacement may be obtained by:

∫=x

dxxPAE 0

)(1δ

where the force P is shown on the following free-body diagram.

Therefore:

1

32

3551 CAExdxx

AE

x

o+== ∫δ

Applying the boundary conditions:

AELCC

AExL

35

350)(

3

11

3

−=⇒+==δ

[ ] 25)10(21)( xxxxP ==


The exact solution for axial displacement is:

( )33

35)( LxAE

L −=δ

The exact solution for axial stress is:

Ax

AxPx

25)()( ==σ

A plot of the exact solution for displacement as compared to several different finite element solutions is shown below.

A plot of the exact solution for axial stress as compared to several different fi-

nite element solutions is shown below.


Galerkin’s Residual Method and Its Application to a One-Dimensional Bar

There are a number of weighted residual methods. However, the Galerkin’s method is more well-known and will be the only weighted residual method dis-cussed in this course.

In weighted residual methods, a trial or approximate function is chosen to ap-proximate the independent variable (in our case, displacement) in a problem de-fined by a differential equation. The trial function will not, in general, satisfy the governing differential equation. Therefore, the substitution of the trial function in the differential equation will create a residual over the entire domain of the prob-lem.

minimum=∫V

RdV


In the residual methods, we require that a weighted value of the residual be a minimum over the entire domain of the problem. The weighting function allows the weighted integral of the residuals to go to zero.

0=∫V

dVWR

Using Galerkin’s weighted residual method, we require the weighting functions to be the interpolation functions or the shape functions, Ni. Therefore:

nidVRNV

i ,,2,10 ==∫

Example 12 - Bar Element Formulation

Let’s derive the bar element formulation using Galerkin’s method. The govern-ing differential equation is:

0ˆˆ

ˆ=⎟

⎠⎞

⎜⎝⎛

xdudAE

xdd

Applying Galerkin’s method we get:

nixdNxdudAE

xddL

i ,,2,10ˆˆˆ

ˆ0

==⎟⎠⎞

⎜⎝⎛∫

We now apply integration by parts using the following general formula:

∫∫ −= vduuvudv

where u and v are variable in the general equation. If we assume the following:

xdxd

dNduNu ii ˆ

ˆ==

xdudAEvxd

xdudAE

xdddv

ˆˆˆ

ˆˆ

ˆ=⎟

⎠⎞

⎜⎝⎛=


then integration by parts gives:

0ˆˆˆ

ˆˆˆ

00

=−⎟⎠⎞

⎜⎝⎛ ∫

Li

L

i xdxd

dNxdudAE

xdudAEN

Recall that:

xx dxd

dNdxd

dNxdud

22

11 ˆ

ˆˆ

ˆˆˆ

+=

or

⎭⎬⎫

⎩⎨⎧

⎥⎦⎤

⎢⎣⎡−=

x

x

dd

LLxdud

2

1

ˆˆ11

ˆˆ

Our original weighted residual expression, with the approximation for u be-comes:

L

o

i

x

xL

i

xdudAEN

ddxd

LLxddNAE ⎟

⎠⎞

⎜⎝⎛=

⎭⎬⎫

⎩⎨⎧

⎥⎦⎤

⎢⎣⎡−∫ ˆ

ˆˆˆ

ˆ11ˆ

2

1

0

Substituting N1 for the weighting function Ni gives:

L

x

xL

xdudAEN

ddxd

LLxddNAE

01

2

1

0

1

ˆˆ

ˆˆ

ˆ11ˆ ⎟

⎠⎞

⎜⎝⎛=

⎭⎬⎫

⎩⎨⎧

⎥⎦⎤

⎢⎣⎡−∫

01

02

1

0 ˆˆˆ

ˆˆ

ˆˆ

ˆ111==

==⎭⎬⎫

⎩⎨⎧

⎥⎦⎤

⎢⎣⎡−⎥⎦

⎤⎢⎣⎡−∫

x

x

xx

xL

xdudAEf

xdudAE

ddxd

LLLAE

( ) xxx fddL

AE121ˆˆˆ =−


Substituting N2 for the weighting function Ni gives:

L

ox

xL

xdudAEN

ddxd

LLxddNAE ⎟

⎠⎞

⎜⎝⎛=

⎭⎬⎫

⎩⎨⎧

⎥⎦⎤

⎢⎣⎡−∫ ˆ

ˆˆˆ

ˆ11ˆ 2

2

1

0

2

Lx

x

Lxx

xL

xdudAEf

xdudAE

ddxd

LLLAE

==

==⎭⎬⎫

⎩⎨⎧

⎥⎦⎤

⎢⎣⎡−⎥⎦

⎤⎢⎣⎡

∫ ˆˆˆ

ˆˆ

ˆˆ

ˆ1112

2

1

0

( ) xxx fddL

AE212ˆˆˆ =−

In matrix form the equations may be written as:

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎥⎦

⎤⎢⎣

⎡ −−

x

x

x

x

ff

dd

LAE

2

1

2

1

ˆˆ

ˆˆ

11

11

Problems:

6. Verify the global stiffness matrix for a three-dimensional bar. Hint: First, expand T* to a 6 x 6 square matrix, then expand k to 6 x 6 square matrix by adding the appropriate rows and columns of zeros, and finally, perform the matrix triple product k = TT k T.

7. Do problems 3.3, 3.7, 3.12, 3.16, 3.18, 3.22, 3.36, 3.43, 3.47, 3.49, and 3.55 on pages 120 - 136 in your textbook “A First Course in the Finite Ele-ment Method” by D. Logan.


Symmetry and Bandwidth

Introduction

In this chapter, we will introduce the concepts of symmetry to reduce the size of a problem and of banded-symmetric matrices and bandwidth. In many in-stances, we can use symmetry to facilitate the solution of a problem. Symmetry means correspondence in size, shape, and position of loads; material properties; and boundary conditions that are mirrored about a dividing line or plane. Use of symmetry allows us to consider a reduced problem instead of the actual problem. Thus, the order of the total stiffness matrix and total set of stiffness equations can be reduced.

Example 1 - Truss Symmetry Problem

Solve the plane truss problem shown below. The truss is composed of eight elements and five nodes.

A vertical load of 2P is applied at node 4. Nodes 1 and 5 are pin supports. Bar elements 1, 2, 7, and 8 have axial stiffnesses of 2 AE and bars 3, 4, 5, and 6 have axial stiffnesses of AE.


In this problem, we will use a plane of symmetry. The vertical plane perpen-dicular to the plane truss passing through nodes 2, 4, and 3 is the plane of sym-metry because identical geometry, material, loading, and boundary conditions occur at the corresponding locations on opposite sides of this plane. For loads such as 2P, occurring in the plane of symmetry, one-half of the total load must be applied to the reduced structure. For elements occurring in the plane of symme-try, one-half of the cross-sectional area must be used in the reduced structure.

The truss information may be summarized in the following table.

Element θ C S C2 S2 CS 1 45° 0.707 0.707 0.5 0.5 0.5

2 315° 0.707 -0.707 0.5 0.5 -0.5

3 0° 1 0 1 0 0

4 270° 0 -1 0 1 0

5 90° 0 1 0 1 0


1 1 2 2

(1)

1 1 1 11 1 1 11 1 1 121 1 1 1

x y x yd d d d

AEkL

− −⎡ ⎤⎢ ⎥− −⎢ ⎥=⎢ ⎥− −⎢ ⎥− −⎣ ⎦


1 1 3 3

(2)

1 1 1 11 1 1 11 1 1 121 1 1 1

x y x yd d d d

AEkL

− −⎡ ⎤⎢ ⎥− −⎢ ⎥=⎢ ⎥− −⎢ ⎥− −⎣ ⎦



1 1 4 4

(3)

1 0 1 00 0 0 01 0 1 00 0 0 0

x y x yd d d d

AEkL

−⎡ ⎤⎢ ⎥−⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦


2 2 4 4

(4)

0 0 0 00 1 0 10 0 0 020 1 0 1

x y x yd d d d

AEkL

⎡ ⎤⎢ ⎥−⎢ ⎥=⎢ ⎥⎢ ⎥−⎣ ⎦


3 3 4 4

(5)

0 0 0 00 1 0 10 0 0 020 1 0 1

x y x yd d d d

AEkL

⎡ ⎤⎢ ⎥−⎢ ⎥=⎢ ⎥⎢ ⎥−⎣ ⎦

Since elements 4 and 5 lie in the plane of symmetry, one half of their original ar-eas have been used in developing the stiffness matrices. The displacement boundary conditions are:

1 1 2 3 4 0x y x x xd d d d d= = = = =


By applying the boundary conditions the force-displacement equations reduce to:

Solution by Partitioning the Matrix

We can solve the above equations by separating the matrices in submatrices (indicated by the dashed lines). Consider a general set of equations shown be-low:

or

0212111 =+ dKdK

FdKdK =+ 222121 Solving the first equation for d1 gives:

2121

111 dKKd −−= Substituting the above equation in the second matrix equation gives:

( ) FdKdKKK =+− −

2222121

1121 Simplifying this expression gives:

( ) FdKKKK =− −

212

1

112122 or

Fdkc =2 where

2 2

3 3

4 4

2 0 1 00 2 1 0

21 1 2

x x

x x

x x

F dAEF d

LF d P

−⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥= − =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥− −⎩ ⎭ ⎣ ⎦ ⎩ ⎭ ⎩ ⎭

11 12 1

21 22 2

0K K dK K d F

⎡ ⎤ ⎧ ⎫ ⎧ ⎫=⎨ ⎬ ⎨ ⎬⎢ ⎥

⎩ ⎭⎣ ⎦ ⎩ ⎭


121

112122 KKKKkc−−=

Therefore, the displacements d2 are:

( ) Fkd c1

2−=

If we apply this solution technique to our example global stiffness equations we get:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−

−⎥⎦

⎤⎢⎣

⎡⎥⎦⎤

⎢⎣⎡ −−−=

−

2121

10

01

21

21]1[

1

LAEkc

Simplifying:

⎥⎦⎤

⎢⎣⎡=

⎭⎬⎫

⎩⎨⎧

⎥⎦⎤

⎢⎣⎡−=

21

21]1[

LAE

LAEkc

Therefore:

( )AE

Lkc

21 =−

Therefore, the displacements d2 are:

AEPLdd y

242 −==

The remaining displacements can be found by substituting the result for d4y in the global force-displacement equations.

⎥⎦⎤

⎢⎣⎡−

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−

−⎥⎦

⎤⎢⎣

⎡−=

⎭⎬⎫

⎩⎨⎧

AEPL

dd

y

y 2

2121

10

01

3

2


Expanding the above equations gives the values for the displacements.

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−

−=

⎭⎬⎫

⎩⎨⎧

AEPLAEPL

dd

y

y

3

2

Banded-Symmetric Matrices and Bandwidth

The coefficient matrix (stiffness matrix) for the linear equations that occur in structural analysis is always symmetric and banded. Because a meaningful analysis generally requires the use of a large number of variables, the implemen-tation of compressed storage of the stiffness matrix is desirable both from the viewpoint of fitting into memory (immediate access portion of the computer) and computational efficiency.

Another method, based on the concept of the skyline of the stiffness matrix, is often used to improve the efficiency in solving the equations. The skyline is an envelope that begins with the first nonzero coefficient in each column of the stiff-ness matrix (see the following figure). In skyline, only the coefficients between the main diagonal and the skyline are stored. In general, this procedure takes even less storage space in the computer and is more efficient in terms of equa-tion solving than the conventional banded format.

A matrix is banded if the nonzero terms of the matrix are gathered about the main diagonal. To illustrate this concept, consider the plane truss shown on be-low. We can see that element 2 connects nodes 1 and 4. Therefore, the 2 x 2 submatrices at positions 1-1, 1-4, 4-1, and 4-4 will have nonzero coefficients. The total stiffness matrix of the plane truss, shown in the figure below, denotes non-zero coefficients with X’s. The nonzero terms are within the some band. Using a banded storage format, only the main diagonal and the nonzero upper codiago-nals need be stored.


We now define the semibandwidth nb as nb = nd(m + 1), where nd is the num-ber of degrees of freedom per node and m is the maximum difference in node numbers determined by calculating the difference in node numbers for each ele-ment of a finite element model. In the example for the plane truss shown above, m = 4 - 1 = 3 and nd = 2, so that nb = 2(3 + 1) = 8.

Execution time (primarily, equation-solving time) is a function of the number of equations to be solved. Without using banded storage of global stiffness matrix K, the execution time is proportional to (1/3)n3, where n is the number of equa-tions to be solved. Using banded storage of K, the execution time is proportional to n(nb)2. The ratio of time of execution without banded storage to that using banded storage is then (1/3)(n/nb)2. For the plane truss example, this ratio is (1/3)(24/8)2 = 3. Therefore, it takes about three times as long to execute the solu-tion of the example truss if banded storage is not used.


Several automatic node renumbering schemes have been computerized. This option is available in most general-purpose computer programs. Alternatively, the wavefront or frontal method is becoming popular for optimizing equation solution time. In the wavefront method, elements, instead of nodes, are automatically renumbered.

In the wavefront method the assembly of the equations alternates with their solution by Gauss elimination. The sequence in which the equations are proc-essed is determined by element numbering rather than by node numbering. The first equations eliminated are those associated with element 1 only. Next the con-tributions to stiffness coefficients from the adjacent element, element 2, are eliminated. If any additional degrees of freedom are contributed by elements 1 and 2 only these equations are eliminated (condensed) from the system of equa-tions. As one or more additional elements make their contributions to the system of equations and additional degrees of freedom are contributed only by these elements, those degrees of freedom are eliminated from the solution. This repeti-tive alternation between assembly and solution was initially seen as a wavefront that sweeps over the structure in a pattern determined by the element number-ing.

The wavefront method, although somewhat more difficult to understand and to program than the banded-symmetric method, is computationally more efficient. A banded solver stores and processes any blocks of zeros created in assembling the stiffness matrix. These blocks of zero coefficients are not stored or processed using the wavefront method. Many large-scale computer programs are now using the wavefront method to solve the system of equations.

Problems

8. Do problem B.9 on page 651 in your textbook “A First Course in the Finite Element Method” by D. Logan.

9. Develop a proposal for a truss analysis problem to be handed in with your take-home mid-term exam.


10. Solve the following truss problems. You may use SAP2000 to do truss analysis.

a) For the plane truss shown below, determine the nodal displacements and element stresses. Nodes 1 and 2 are pin joints. Let E = 107 psi and the A = 2.0 in2 for all elements.

1

3

5

7

2

4

6

85 k

10 k

10 k

10 ft

10 ft

10 ft

10 ft


b) For the 25-bar truss shown below, determine the displacements and ele-mental stresses. Nodes 7, 8, 9, and 10 are pin connections. Let E = 107 psi and the A = 2.0 in2 for the first story and A = 1.0 in2 for the top story. Table 1 lists the coordinates for each node. Table 2 lists the values and directions of the two loads cases applied to the 25-bar space truss.

Table 1. Coordinates for the 25-Bar Truss

Node x (in) y (in) z (in)

1 -37.5 0.0 200.0 2 37.5 0.0 200.0 3 -37.5 37.5 100.0 4 37.5 37.5 100.0 5 37.5 -37.5 100.0 6 -37.5 -37.5 100.0 7 -100.0 100.0 0.0 8 100.0 100.0 0.0 9 100.0 -100.0 0.0

10 -100.0 -100.0 0.0 Note: 1 in = 2.54 cm

Table 2. Multiple Loading Conditions for the 25-Bar Truss

Case Node Fx (kip) Fy (kip) Fz (kip)

1 1.0 10.0 -5.0 2 0.0 10.0 -5.0 3 0.5 0.0 0.0

1

6 0.5 0.0 0.0 1 0.0 20.0 -5.0 2 2 0.0 -20.0 -5.0

Note: 1 kip = 4.45 kN

1 2

3 4

5 6

7 9

8

10


c) For the 72-bar truss shown below, determine the displacements and ele-mental stresses. Nodes 1, 2, 3, and 4 are pin connections. Let E = 107 psi and the A = 1.0 in2 for the first two stories and A = 0.5 in2 for the top two stories. Table 3 lists the values and directions of the two loads cases ap-plied to the 72-bar space truss.

Table 3. Multiple Loading Conditions for the 72-Bar Truss

Case Node Fx (kip) Fy (kip) Fz (kip) 17 0.0 0.0 -5.0 18 0.0 0.0 -5.0 19 0.0 0.0 -5.0

1

20 0.0 0.0 -5.0 2 17 5.0 5.0 -5.0

Note: 1 kip = 4.45 kN.

120 in

60 in

60 in

60 in

60 in

1

5

9

13

17

2

6

10

14

18

1

2

3

4

5 6

9

7

8 10

11

12

13

14 15

16 17

18

(a)

(b)


Development of Beam Equations

Introduction

In this section, we will develop the stiffness matrix for a beam element, the most common of all structural elements. The beam element is considered to be straight and to have constant cross-sectional area. We will derive the beam ele-ment stiffness matrix by using the principles of simple beam theory.

The degrees of freedom associated with a node of a beam element are a transverse displacement and a rotation. We will discuss procedures for handling distributed loading and concentrated nodal loading. We will include the nodal shear forces and bending moments and the resulting shear force and bending moment diagrams as part of the total solution.

To further acquaint you with the potential energy approach for developing stiffness matrices and equations, we will again develop the beam bending ele-ment equations using this approach. Finally, the Galerkin residual method is ap-plied to derive the beam element equations.

Beam Stiffness

A beam is a long, slender structural member generally subjected to transverse loading that produces significant bending effects as opposed to twisting or axial effects. The bending deformation is measured as a transverse displacement and a rotation. The degrees of freedom at each node of a beam element will be a transverse displacement and a rotation (as opposed to only an axial displace-ment for the bar element).

Consider the beam element shown below. The beam is of length L with axial local coordinate x and transverse local coordinate y . The local transverse nodal

displacements are given by iyd and the rotations by iφ . The local nodal forces

are given by iyf and the bending moments by im . We initially neglect all axial ef-

fects.


At all nodes, the following sign conventions are used:

1. Moments are positive in the counterclockwise direction. 2. Rotations are positive in the counterclockwise direction. 3. Forces are positive in the positive y direction.

4. Displacements are positive in the positive y direction.

The differential equation governing simple linear-elastic beam behavior can be

derived as follows. Consider the beam shown below.

Write the equations of equilibrium for the differential element:


0ˆ2ˆˆ)ˆ(ˆ)(0 2 ≈⎟⎠⎞

⎜⎝⎛+−−+==∑ − xdxdxdxwxVdMdMMM sideright

dxxwdVVVFy )ˆ()(0 −+−==∑ From force and moment equilibrium of a differential beam element, we get:

xddMVdMxVdM sideright ˆ

or0ˆ0 ==+−⇒=∑ −

⎟⎠⎞

⎜⎝⎛−=⇒−==−−⇒=∑ xddM

xddw

xddVwdVxwdFy ˆˆˆ

or0ˆ0

The curvature k of the beam is related to the moment by:

EIMk ==

ρ1

where ρ is the radius of the deflected curve, v is the transverse displacement function in the y direction, E is the modulus of elasticity, and I is the principle moment of inertia about y direction, as shown below.

The curvature for small slopes xdvd ˆ/ˆ=θ is given as:

2

2

ˆˆ

xdvdk =

Therefore:

2

2

2

2

ˆˆ

ˆˆ

xdvdEIM

EIM

xdvd

=⇒=


Substituting the moment expression into the moment-load equations gives:

( )xwxdvdEI

xdd ˆ

ˆˆ

ˆ 2

2

2

2

−=⎟⎠

⎞⎜⎝

⎛

For constant values of EI, the above equation reduces to:

( )xwxdvdEI ˆ

ˆˆ4

4

−=⎟⎠

⎞⎜⎝

⎛


We will consider the linear-elastic beam element shown below.


Assume the transverse displacement function v is:

432

23

1 ˆˆˆ axaxaxav +++= The number of coefficients in the displacement function, ai, is equal to the total number of degrees of freedom associated with the element (displacement and rotation at each node). The boundary conditions are:

yy dLxvdxv 21ˆ)ˆ(ˆˆ)0ˆ(ˆ ====

21ˆ

ˆ)ˆ(ˆˆ

ˆ)0ˆ(ˆ φφ =

==

=xd

Lxvdxd

xvd


Applying the boundary conditions and solving for the unknown coefficients gives:

41ˆ)0(ˆ adv y ==

432

23

12ˆ)(ˆ aLaLaLadLv y +++==

31ˆ)0(ˆ a

xdvd

== φ

322

12 23ˆ)(ˆ aLaLadx

Lvd++== φ

Solving these equations for a1, a2, a3, and a4 gives:

( ) ( ) ( ) ( ) yyyyy dxxL

ddL

xL

ddL

v 112

212123

212213ˆˆˆˆˆˆ21ˆˆ3ˆˆˆ1ˆˆ2ˆ ++⎥⎦

⎤⎢⎣⎡ +−−−+⎥⎦

⎤⎢⎣⎡ −+−= φφφφφ


dNv ˆ][ˆ = where

[ ]4321

2

2

1

1

][ˆ NNNNNd

d

dy

y

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

=

φ

φ

and

( ) ( )322332

32331 ˆˆ2ˆ1ˆ3ˆ21 LxLxLx

LNLLxx

LN +−=+−=

( ) ( )22334

2333 ˆˆ1ˆ3ˆ21 LxLx

LNLxx

LN −=+−=


N1, N2, N3, and N4 are called the shape functions for a beam element.

N1

-0.200

0.000

0.200

0.400

0.600

0.800

1.000

0.00 1.00

N2

-0.200

0.000

0.200

0.400

0.600

0.800

1.000

0.00 1.00

N3

-0.200

0.000

0.200

0.400

0.600

0.800

1.000

0.00 1.00

N4

-0.200

0.000

0.200

0.400

0.600

0.800

1.000

0.00 1.00


The stress-displacement relationship is:

( )xdudyxx ˆˆˆ,ˆ =ε

where u is the axial displacement function. We can relate the axial displacement to the transverse displacement by considering the beam element shown below:


xdvdyuˆˆˆˆ −=

One of the basic assumptions in simple beam theory is that planes remain planar after deformation, therefore:

( ) ⎟⎠

⎞⎜⎝

⎛−= 2

2

ˆˆˆˆ,ˆ

xdvdyyxxε

Moments and shears are related to the transverse displacement as:

( ) ( ) ⎟⎠

⎞⎜⎝

⎛=⎟

⎠

⎞⎜⎝

⎛= 3

3

2

2

ˆˆˆ

ˆˆˆˆ

xdvdEIxV

xdvdEIxm


Using beam theory sign convention for shear force and bending moment we obtain the following equations:

( )221133

3

1ˆ6ˆ12ˆ6ˆ12

ˆ)0(ˆˆˆ φφ LdLd

LEI

xdvdEIVf yyy +−+===

( )221133

3

2ˆ6ˆ12ˆ6ˆ12

ˆ)(ˆˆˆ φφ LdLd

LEI

xdLvdEIVf yyy −+−−==−=

( )22

212

132

2

1ˆ2ˆ6ˆ4ˆ6

ˆ)0(ˆˆˆ φφ LdLLdL

LEI

xdvdEImm yy +−+=−=−=

( )22

212

132

2

2ˆ4ˆ6ˆ2ˆ6

ˆ)(ˆˆˆ φφ LdLLdL

LEI

xdLvdEImm yy +−+===

xdvdˆˆ

≈θxdds ˆ≈



⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−

−−=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

2

2

1

1

2

2

2

2

3

2

2

1

1

ˆˆˆˆ

46

26

612612

2646

612612

ˆ

ˆˆ

ˆ

φ

φ

y

y

y

y

d

d

LL

LL

L

L

LLLL

L

LLEI

mfmf

where the stiffness matrix is:

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−

−−=

2

2

2

2

3

46

26

612612

26

46

612612

LL

LL

L

L

LL

LL

L

LLEIk

Step 5 - Assemble the Element Equations and Introduce Boundary Conditions

Consider a beam modeled by two beam elements, shown below:

Assume the EI to be constant throughout the beam. A force of 1,000 lb and mo-ment of 1,000 lb-ft are applied to the mid-point of the beam. The beam element stiffness matrices are:


⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−

−−=

2

2

2

2

3)1(

46

26

612612

26

46

612612

2211

LL

LL

L

L

LL

LL

L

LLEIk

ydyd φφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−

−−=

2

2

2

2

3)2(

46

26

612612

26

46

612612

3322

LL

LL

L

L

LL

LL

L

LLEIk

ydyd φφ

In this example, the local coordinates coincide with the global coordinates of the whole beam (therefore there is no transformation required for this problem). The total stiffness matrix can be assembled as:


1 1 3 0y yd dφ= = =

By applying the boundary conditions the beam equations reduce to:

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧⋅

−

3

2

2

22

223

4262806024

0000,1

000,1

φφ

yd

LLLLLL

LEIftlb

lb

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−++−−+−+−−

−−

=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

3

3

2

2

1

1

22

2222

22

3

3

3

2

2

1

1

46260061261200

2644662661266121261200264600612612

φ

φ

φ

y

y

y

y

y

y

d

d

d

LLLLLL

LLLLLLLLLLLL

LLLLLL

LEI

MFMFMF


Solving the above equations gives:

⎪⎪⎪⎪⎪

⎭

⎪⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪⎪

⎩

⎪⎪⎪⎪⎪

⎨

⎧

−−

−−

−−

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

radEI

LL

radEI

LL

lbEI

LL

d y

1252125

46252125

12

23753875

3

2

2

φφ

Example 1 - Beam Problem

Consider the beam shown below. Assume that EI is constant and the length is 2L.

The beam element stiffness matrices are:

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−

−−=

2

2

2

2

3)1(

46

26

612612

26

46

612612

2211

LL

LL

L

L

LL

LL

L

LLEIk

ydyd φφ


⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−

−−=

2

2

2

2

3)2(

3

46

26

612612

26

46

612612

322

LL

LL

L

L

LL

LL

L

LLEIk

ydyd φφ

The local coordinates coincide with the global coordinates of the whole beam (therefore there is no transformation required for this problem). The total stiffness matrix can be assembled as:

The governing beam equations are:

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−−−

−−

=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

3

3

2

2

1

1

3

3

3

2

2

1

1

2462260061261200

2262802266120246120022624600612612

φ

φ

φ

y

y

y

y

y

y

d

d

d

LLLLLL

LLLLLLL

LLLLLL

LEI

MFMFMF


0332 === φyy dd

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−−−

−−

=

2462260061261200

2262802266120246120022624600612612

3

LLLLLL

LLLLLLL

LLLLLL

LEIK


By applying the boundary conditions the beam equations reduce to:

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧−

2

1

1

2

2

2

23

826

246

6612

00

φφ

yd

LLL

LLL

LL

LEI

P


⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧−

1

34

37

2

2

1

1

L

EIPL

d y

φφ

The positive signs for the rotations indicate that both are in the counterclockwise direction. The negative sign on the displacement indicates a deformation in the

y− direction.

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−−−

−−

=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧ −

00103

2462260061261200

2262802266120246120022624600612612

4

37

3

3

2

2

1

1L

y

y

y

LLLLLL

LLLLLLL

LLLLLL

LP

MFMFMF

The local nodal forces for element 1:

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

−

−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧−

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−

−−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

PLP

P

LLLLLL

LLLLLL

LP

mfmf L

y

y

0

001

4626612612

2646612612

4ˆ

ˆˆ

ˆ 37

22

22

2

2

1

1



⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

−=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−

−−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

PLP

PLP

LLLLLL

LLLLLL

LP

mfmf

y

y

5.05.1

5.1

0010

4626612612

2646612612

4ˆ

ˆˆ

ˆ

22

22

3

3

2

2

The free-body diagrams for the each element are shown below.

Combining the elements gives the forces and moments for the original beam.

Therefore, the shear force and bending moment diagrams are:



Consider the beam shown below. Assume E = 30 x 106 psi and I = 500 in.4 are constant throughout the beam. Use four elements of equal length to model the beam.


⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−

−−=

++

2

2

2

2

3)(

46

26

612612

26

46

612612

1)1(

LL

LL

L

L

LL

LL

L

LLEIk i

iyidiiyd φφ

Using the direct stiffness method, the four beam element stiffness matrices are superimposed to produce the global stiffness matrix. As shown on the next page. The boundary conditions for this problem are:

1 1 3 5 5 0y y yd d dφ φ= = = = =


Ele

men

t 1

Ele

men

t 2

Ele

men

t 3

Ele

men

t 4


After applying the boundary conditions the global beam equations reduce to:

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−

−

=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

0000,1000000,10

80200024600

2682600280006024

4

4

3

2

2

22

222

22

3

lb

lb

d

d

LLL

LLLLLLLL

LEI

y

y

φ

φφ

Substituting L = 120 in., E = 30 x 106 psi, and I = 500 in.4 into the above equa-tions and solving for the unknowns gives:

0048.0 43242 ===−== φφφindd yy The global forces and moments can be determined as:

ftkipsMMMM

ftkipsM

kipsFkipsFkipsFkipsF

kipsF

y

y

y

y

y

·25000

·25

51010105

5

4

3

2

1

5

4

3

2

1

−=====

=====


⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

−=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

−⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−

−−=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

ftkkips

ftkkips

LL

LL

L

L

LL

LL

L

LLEI

mfmf

y

y

·255

·255

0048.000

46

26

612612

26

46

612612

ˆ

ˆˆ

ˆ

2

2

2

2

3

2

2

1

1


⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

−

−−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧−

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−

−−=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

ftkkips

ftkkips

LL

LL

L

L

LL

LL

L

LLEI

mfmf

y

y

·255

·255

000048.0

46

26

612612

26

46

612612

ˆ

ˆˆ

ˆ

2

2

2

2

3

3

3

2

2



⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

−=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

−⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−

−−=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

ftkkips

ftkkips

LLLL

L

L

LLLL

L

LLEI

mfmf

y

y

·255

·255

0048.000

4626

612612

2646

612612

ˆ

ˆˆ

ˆ

2

2

2

2

3

4

4

3

3


⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

−

−−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧−

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−

−−=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

ftkkips

ftkkips

LLLL

L

L

LLLL

L

LLEI

mfmf

y

y

·255

·255

000048.0

4626

612612

2646

612612

ˆ

ˆˆ

ˆ

2

2

2

2

3

5

5

4

4

Note: Due to symmetry about the vertical plane at node 3, we could have worked just half the beam, as shown below.



Consider the beam shown below. Assume E = 210 GPa and I = 2 x 10-4 m4 are constant throughout the beam and the spring constant k = 200 kN/m. Use two beam elements of equal length and one spring element to model the struc-ture.


⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−

−−=

2

2

2

2

3)1(

46

26

612612

26

46

612612

2211

LL

LL

L

L

LL

LL

L

LLEIk

ydyd φφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−

−−=

2

2

2

2

3)2(

3

46

26

612612

26

46

612612

322

LL

LL

L

L

LL

LL

L

LLEIk

ydyd φφ


The spring element stiffness matrix is:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−=⇒⎥

⎦

⎤⎢⎣

⎡ −−

=kk

kkk

kk

kk

k

dddd yyyy

0000

0)3()3(

43343 φ

Using the direct stiffness method and superposition gives the global beam equa-tions.

The boundary conditions for this problem are:

1 1 2 4 0y y yd d dφ= = = =


⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧−=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−+−

−=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

0

0

4626'126

268

3

3

2

22

22

3

3

3

2

PdLLLLkL

LLL

LEI

MFM

yy

φ

φ


EIkLk

kEIPL

kEIPL

kEIPL

d y

3

2

3

2

3

3

2

'

'71219

'71217

'71213

=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⎟⎠⎞

⎜⎝⎛

+−

⎟⎠⎞

⎜⎝⎛

+−

⎟⎠⎞

⎜⎝⎛

+−

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

φ

φ

EIkLk

d

d

d

d

kkLLLL

kLkLLLLLLLL

LLLLLL

LEI

FMFMFMF

y

y

y

y

y

y

y

y

3

4

3

3

2

2

1

1

22

222

22

3

4

3

3

2

2

1

1

'

'0'00000462600

'6'1261200026802606120246120002646000612612

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−−+−−−−−−

−−

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

φ

φ

φ


Substituting L = 3 m, E = 210 GPa, I = 2 x 10-4 m4, and k = 200 kN/m in the above equations gives:

radradmd y

00747.000249.00174.0

3

2

3

−=−=−=

φφ

Substituting the solution back into the global equations gives:

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

−

⋅−−

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

kN

kN

kNmkN

kN

FMFMFMF

y

y

y

y

5.30

5004.116

7.699.69

4

3

3

2

2

1

1

A free-body diagram, including forces and moments acting on the beam is shown below.


Distributed Loadings

Beam members can support distributed loading as well as concentrated nodal loading. Therefore, we must be able to account for distributed loading. Consider the fixed-fixed beam subjected to a uniformly distributed loading w shown the fig-ure below. The reactions, determined from structural analysis theory, are called fixed-end reactions. In general, fixed-end reactions are those reactions at the ends of an element if the ends of the element are assumed to be fixed (dis-placements and rotations are zero). Therefore, guided by the results from struc-tural analysis for the case of a uniformly distributed load, we replace the load by concentrated nodal forces and moments tending to have the same effect on the beam as the actual distributed load.

The figure below illustrates the idea of equivalent nodal loads for a general beam. We can replace the effects of a uniform load by a set of nodal forces and moments.


Work Equivalence Method

This method is based on the concept that the work done by the distributed load is equal to the work done by the discrete nodal loads. The work done by the distributed load is:

( ) ( )∫=0

ˆˆˆˆ xdxvxwW ddistribute

where ( )xv ˆˆ is the transverse displacement. The work done by the discrete nodal forces is:

yyyynodes dfdfmmW 22112211ˆˆˆˆˆˆˆˆ +++= φφ

The nodal forces can be determined by setting Wdistributed = Wnodes for arbitrary displacements and rotations.

nodesddistribute WW =

Example 4 - Load Replacement

Consider the beam, shown below, and determine the equivalent nodal forces for the given distributed load.

Using the work equivalence method or:

nodesddistribute WW = we get:

( ) ( ) yyyy dfdfmmxdxvxwL

221122110

ˆˆˆˆˆˆˆˆˆˆˆˆ +++=∫ φφ


Evaluating the left-hand-side of the above expression using ( )xw ˆ = -w and ( )xv ˆˆ equal to:

( ) ( ) ( ) ( ) ( ) yyyyy dxxL

ddL

xL

ddL

xv 112

212123

212213ˆˆˆˆˆˆ21ˆˆ3ˆˆˆ1ˆˆ2ˆˆ ++⎥⎦

⎤⎢⎣⎡ +−−−+⎥⎦

⎤⎢⎣⎡ ++−= φφφφφ

gives:

( ) ( ) ( ) ( ) ( ) yyyyy dwLwLwLddLwwLddLwxdxvwL

11

2

21

2

1221

2

210

ˆˆ2

ˆˆ23

ˆˆˆˆ4

ˆˆ2

ˆˆˆ −−++−−+−−=∫ φφφφφ

Using a set of arbitrary nodal displacements, such as:

10 1221 ==== φφyy dd The resulting nodal equivalent force or moment is:

12232

4)1(ˆ

222

2

1

wLwLwLwLm −=⎟⎠

⎞⎜⎝

⎛+−−=

Using another set of arbitrary nodal displacements, such as:

10 2121 ==== φφyy dd The resulting nodal equivalent force or moment is:

1234)1(ˆ

222

2

wLwLwLm =⎟⎠

⎞⎜⎝

⎛−−=

Setting the nodal rotations equal zero except for the yd1

ˆ and yd2ˆ gives:

22)1(1

LwLwLwLWf y −=−+−=

22)1(2

LwLwLWf y −=−=


General Formulation

We can account for the distributed loads or concentrated loads acting on a beam elements by considering the following formulation for a general structure:

0FKdF −= where F0 are the equivalent nodal forces, expressed in terms of the global-coordinate components. These force would yield the same displacements as the original distributed load. If we assume that the global nodal forces are not pre-sent (F = 0) then:

KdF =0 We now solve for the displacements, d, given the nodal forces F0. Next, substi-tute the displacements and the equivalent nodal forces F0 back into the original expression and solve for the global nodal forces.

Example 5 - Load Replacement

Consider the beam shown below, determine the equivalent nodal forces for the given distributed load.

The work equivalent nodal forces are shown above. Using the beam stiffness equations, with the boundary conditions applied, we can solve for the displace-ments

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡ −−

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧−

2

2232 ˆ

ˆ

46

612

12

2φ

ydLL

LLEI

wL

wL


Therefore:

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−

−=

⎭⎬⎫

⎩⎨⎧

EIwL

EIwL

d y

6

8ˆˆ

3

4

2

2

φ

In this case, the method of equivalent nodal forces gives the exact solution for the displacements and rotations. To obtain the global nodal forces, we will first define the product of Kd to be Fe, where Fe is called the effective global nodal forces. Therefore:

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

−−

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−

−−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

EIwL

EIwL

e

ye

e

ye

LLLLLL

LLLLLL

LEI

MFMF

6

8

22

22

3

1

2

1

1

3

4

00

4626612612

2646612612


⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

−=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

12

2

125

2

2

2

2

2

1

1

wL

wL

wL

wL

MFMF

e

ye

e

ye

Using the above expression and the fix-end moments in:

0FKdF −= gives the correct global nodal forces as:


⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

−

−

−

−

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

−=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

0

02

12

2

12

2

12

2

125

22

2

2

2

2

2

2

1

1 wL

wL

wL

wL

wL

wL

wL

wL

wL

wL

MFMF

y

y

Example 6 - Cantilever Beam

Consider the beam, shown below, determine the vertical displacement and ro-tation at the free-end and the nodal forces, including reactions. Assume EI is constant throughout the beam.

We will use one element and replace the concentrated load with the appropriate nodal forces. The beam stiffness equations become:

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡ −−

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧−

2

223 ˆ

ˆ

46

612

8

2φ

ydLL

LLEI

PL

P

Therefore:

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−

−=

⎭⎬⎫

⎩⎨⎧

EIPL

EIPL

d y

8

485

ˆˆ

2

3

2

2

φ

To obtain the global nodal forces, we begin by evaluating the effective nodal

forces.


⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

−−

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−

−−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

EIPL

EIPL

e

ye

e

ye

LLLLLL

LLLLLL

LEI

MFMF

8

485

22

22

3

1

2

1

1

2

3

00

4626612612

2646612612


⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

−=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

8

2

83

2

1

2

1

1

PL

P

PL

P

MFMF

e

ye

e

ye

Using the above expression in the following equation, gives:

0FKdF −= The correct global nodal forces as:

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

−

−

−

−

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

−=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

0

02

8

2

8

2

8

2

83

2

2

2

1

1 PL

P

PL

P

PL

P

PL

P

PL

P

MFMF

y

y

In general, for any structure in which an equivalent nodal force replacement is made, the actual nodal forces acting on the structure are determined by first evaluating the effective nodal forces Fe for the structure and then subtracting off the equivalent nodal forces F0 for the structure. Similarly, for any element of a structure in which equivalent nodal force replacement is made, the actual local nodal forces acting on the element are determined by first evaluating the effec-


tive local nodal forces )(ˆ ef for the element and then subtracting off the equivalent local nodal forces 0f associated only with the element.

Beam Element with Nodal Hinge

Consider the beam, shown below, with an internal hinge. An internal hinge causes a discontinuity in the slope of the deflection curve at the hinge and the bending moment is zero at the hinge.

For a beam with a hinge on the right end, the moment 2m is zero and we can partition the matrix to eliminate the degree of freedom associated with 2φ .

We can condense out the degree of freedom by using the partitioning method discussed earlier. Recall, the form of kc

][]][[][ 211

221211 KKKKkc−−=

[ ]LLLL

LLL

LEIL

LLL

LLEIkc 626

41

626

12612

646

12612

22

23

23

−⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−−

−−=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−

−−=

2

2

2

2

3

46

26

612612

26

46

612612

ˆ

LL

LL

L

L

LL

LL

L

LLEIk


Therefore, the condensed stiffness matrix is:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−

=11

113 2

3

LLLL

L

LEIkc

The element force-displacement equations are:

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

y

y

y

y

d

d

LLLL

L

LEI

fmf

2

1

12

3

2

1

1

ˆˆˆ

11

113

ˆˆ

ˆ

φ

Expanding the element force-displacement equations and maintaining 2m = 0 gives:

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

2

2

1

12

3

2

2

1

1

ˆˆˆˆ

00000110011

3

ˆ

ˆˆ

ˆ

φ

φ

y

y

y

y

d

d

LLLL

L

LEI

mfmf

The element force-displacement equations maintaining 1m = 0 gives:

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−

−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

2

2

1

1

2

3

2

2

1

1

ˆˆˆˆ

0101

0000101

3

ˆ

ˆˆ

ˆ

φ

φ

y

y

y

y

d

d

LLLL

L

LEI

mfmf


Example 7 - Beam With Hinge

In the following beam, shown below, determine the vertical displacement and rotation at node 2 and the element forces for the uniform beam with an internal hinge at node 2. Assume EI is constant throughout the beam.

The stiffness matrix for element 1 (with hinge) is:

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−

=

2

2

1

12

3)1(

ˆˆˆˆ

00000110011

3

2211

φ

φ

φφ

y

y

d

d

aaaa

a

aEIk

ydy

d

The stiffness matrix for element 2 (without hinge) is:

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−

−−=

2

2

2

2

3)2(

46

26

612612

26

46

612612

3322

bb

bb

b

b

bb

bb

b

bbEIk

ydy

d φφ


03131 ==== φφyy dd



3 3 22

22

3 12 6

6 4 0φ

⎡ ⎤+⎢ ⎥ −⎧ ⎫ ⎧ ⎫=⎢ ⎥ ⎨ ⎬ ⎨ ⎬⎩ ⎭⎩ ⎭⎢ ⎥

⎢ ⎥⎣ ⎦

yd Pa b bEI

b b


( )

( )

3 3

3 32

3 32

3 3

3

2φ

⎧ ⎫−⎪ ⎪+⎧ ⎫ ⎪ ⎪=⎨ ⎬ ⎨ ⎬

⎩ ⎭ ⎪ ⎪⎪ ⎪+⎩ ⎭

y

a b Pb a EIda b P

b a EI

The element force-displacement equations for element 1 are:

( ) ⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

+−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

EIabPbaa

aaaa

aEI

fmf

y

y

33

33

23

2

1

1

3

0

0

11

113

ˆˆ

ˆ

Therefore:

3

3 3

1 3

1 3 3

32

3 3

ˆ

ˆˆ

⎧ ⎫⎪ ⎪+⎧ ⎫ ⎪ ⎪

⎪ ⎪ ⎪ ⎪⎪ ⎪ =⎨ ⎬ ⎨ ⎬+⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎩ ⎭

−⎪ ⎪+⎩ ⎭

y

y

b Pb afab Pm

b af b P

b a


The element force-displacement equations for element 2 are:

( )

( )

3 3

3 32

2 2 3 22

3 3 33

2 23

ˆ 312 6 12 6ˆ 6 4 6 2ˆ 12 6 12 6 2

6 2 6 4ˆ 00

⎧ ⎫−⎪ ⎪+⎧ ⎫ −⎡ ⎤ ⎪ ⎪

⎪ ⎪ ⎢ ⎥ ⎪ ⎪−⎪ ⎪ ⎪ ⎪⎢ ⎥= −⎨ ⎬ ⎨ ⎬⎢ ⎥− − − +⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪−⎣ ⎦⎩ ⎭ ⎪ ⎪⎪ ⎪⎩ ⎭

y

y

a b Pb a EIf b b

m b b b bEI a b Pb b b b a EIf

b b b bm

Therefore:

3

3 3

2

23

33 3

33

3 3

ˆ0ˆ

ˆ

ˆ

⎧ ⎫−⎪ ⎪+⎪ ⎪⎧ ⎫ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬

⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪+⎩ ⎭ ⎪ ⎪

⎪ ⎪⎪ ⎪+⎩ ⎭

y

y

a Pb a

fm

a Pfb amba P

b a

Potential Energy Approach to Derive Beam Element Equations

Let’s derive the equations for a beam element using the principle of minimum potential energy. The procedure for applying the principle of minimum potential energy is similar to that used for the bar element. The total potential energy, πp, is defined as the sum of the internal strain energy U and the potential energy of the external forces Ω:

Ω+=Upπ The differential internal work (strain energy) dU in a one-dimensional beam ele-ment is:

∫=V

xx dVU εσ21


For a single beam element, shown below, subjected to both distributed and con-centrated nodal forces, the potential energy due to forces (or the work done these forces) is:

∑∑∫==

−−−=Ω2

1

2

1

ˆˆˆˆˆî

iii

iyiyS

y mdPdSvT φ

If the beam element has a constant cross-sectional area A, then the differential volume of the beam is given as:

xddAdV ˆ= and the differential element where the surface loading acts is given as:

xdbdS ˆ= where b is the width of the beam element. Therefore the total potential energy is:

( )∑∫∫ ∫=

−−−=2

1ˆˆ21 ˆˆˆˆˆˆˆˆ

iiiiyiy

xy

x Axxp mdPxdbvTxddA φεσπ

The strain-displacement relationship is:

2

2

ˆˆˆ

xdvdyx −=ε


We can express the strain in terms of nodal displacements and rotations as:

dL

LxLL

LxL

LxLL

Lxyxˆ2661246612

3

2

33

2

3 ⎥⎦

⎤⎢⎣

⎡ −+−−−−=ε

or: dByx

ˆ][ˆ−=ε where [B] is:

⎥⎦

⎤⎢⎣

⎡ −+−−−= 3

2

33

2

3

2661246612][L

LxLL

LxL

LxLL

LxB

The stress-strain relationship in one-dimension is:

xx E εσ ][= where E is the modulus of elasticity. Therefore:

dBEyxˆ]][[ˆ−=σ

The total potential energy can be written in matrix form as:

PdxdvTbxddAT

x

Ty

x Vx

Txp

ˆˆˆˆˆˆˆˆ

21 −−= ∫∫ ∫ εσπ

If we define, yTbw ˆ= as a line load (load per unit length) in the y direction and the substitute the definitions of σx and εx the total potential energy can be written in matrix form as:

PdxdNdwxdAddBBdyE TLTT

x A

TT

pˆˆˆ][ˆˆˆ][][ˆ

2 0ˆ

2 −−= ∫∫ ∫π

Use the following definition for moment of inertia:

∫=A

dAyI 2


Then the total potential energy expression becomes:

PdxdNdwxddBBdEI TLTT

x

TT

pˆˆˆ][ˆˆˆ][][ˆ

2 0ˆ

−−= ∫∫π

Differentiating the total potential energy with respect to the displacement and ro-tations ( 2121

ˆ,ˆ,ˆ,ˆ ffdd yy ) and equating each term to zero gives:

0ˆˆ][ˆˆ][][0ˆ

=−− ∫∫ PxdNwdxdBBEIL

T

x

T

The nodal forces vector is:

0ˆˆ][ˆ0

=−− ∫ PxdNwfL

T

The elemental stiffness matrix is:

[ ] ∫=x

T xdBBEIkˆ

ˆ][][ˆ

Integrating the above matrix expression gives:

[ ]⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−

−−=

2

2

2

2

3

46

26

612612

26

46

612612

ˆ

LL

LL

L

L

LL

LL

L

LLEIk

Galerkin’s Method to Derive Beam Element Equations

The governing differential equation for a one-dimensional beam is:

4

4

ˆ0

ˆd vEI wdx

⎛ ⎞+ =⎜ ⎟

⎝ ⎠

We can define the residual R as:


4

40

ˆ ˆ 0 1, 2, 3, and 4ˆ

L

id vEI w N dx idx

⎛ ⎞⎛ ⎞+ = =⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠∫

If we apply integration by parts twice to the first term we get:

( ) ( )( ) ( ) ( )( )[ ]Lxxxixxxi

L

xxixx

L

ixxxx vNvNEIxdNvEIxdNvEI0ˆˆˆ,ˆˆˆ

0ˆˆ,ˆˆ

0ˆˆˆˆ ˆˆˆˆˆˆ −+= ∫∫

where the subscript x indicates a derivative with respect to x . Since dNv ˆ][ˆ = , then the second derivative of v with respect to x is:

dL

LLxL

LxL

LLxL

Lxv xxˆ2ˆ66ˆ124ˆ66ˆ12ˆ

3

2

33

2

3ˆˆ ⎥⎦

⎤⎢⎣

⎡ −+−−−=

or dBv xxˆ][ˆˆ =

Therefore the integration by parts becomes:

( ) ( ) ˆ ˆ ˆ, ,00 0

ˆ ˆˆˆ ˆ[ ] 0 1, 2, 3, and 4L L L

i ii xx i xN EI B dx d N w dx NV N m d i⎡ ⎤+ + − = =⎣ ⎦∫ ∫ The above expression is really four equations (one for each Ni) and can be writ-ten in matrix form as:

ˆ 00 0

ˆ ˆˆˆ ˆ[ ] [ ] [ ] [ ] [ ]L L LT T T T

xB EI B dx d N w dx N m N V= − + −∫ ∫

The interpolation function in the last two terms can be evaluated:

( ) ( ) ]1000[ˆ][]0010[0ˆ][ ˆˆ ==== LxNxN xx

( ) ( ) ]0100[ˆ][]0001[0ˆ][ ==== LxNxN


Therefore, the last two terms of the matrix form of the Galerkin formulation be-come (see the figure below):

)0(ˆ2)0(ˆ1 miVi ⇒=⇒=

)(ˆ4)(ˆ3 LmiLVi ⇒=⇒=

When beam elements are assembled, as shown below, two shear forces and two moments form adjoining elements contribute to the concentrated force and the concentrated moment at the node common to both elements.


Problems:

11. Verify the four beam element equations are contained in the following ma-trix expression.

0ˆˆ][ˆˆ][][0

=−− ∫∫ PxdNwdxdBBEIL

T

A

T

12. Do problems 4.6, 4.11, 4.19, 4.23, 4.29 and 4.34 on pages 181 - 187 in your textbook “A First Course in the Finite Element Method” by D. Logan.

13. Work problem 4.21 (on page 185 in your textbook “A First Course in the Fi-nite Element Method” by D. Logan) using the SAP2000 analysis system (you may used any available structural analysis software system). Start with one element and increase the number of beam elements until you get good agreement with the exact solution. Plot you results for each discreti-zation to the exact solution. How many elements are enough to accurately model this problem?


Plane Frame and Grid Equations

Introduction

Many structures, such as buildings and bridges, are composed of frames and/or grids. This chapter develops the equations and methods for solution of plane frames and grids. First, we will develop the stiffness matrix for a beam element arbitrarily oriented in a plane. We will then include the axial nodal displacement degree of freedom in the local beam element stiffness matrix. Then we will combine these results to develop the stiffness matrix, including axial deformation effects, for an arbitrarily oriented beam element. We will also consider frames with inclined or skewed supports.

Two-Dimensional Arbitrarily Oriented Beam Element

We can derive the stiffness matrix for an arbitrarily oriented beam element, shown in the figure below, in a manner similar to that used for the bar element. The local axes x and y are located along the beam element and transverse to the beam element, respectively, and the global axes x and y are located to be convenient for the total structure.

The transformation from local displacements to global displacements is given in matrix form as:


θθ

sincos

ˆˆ

==

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡−

=⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

SC

dd

CSSC

dd

y

x

y

x

Using the second equation for the beam element, we can relate local nodal degrees of freedom to global degree of freedom:

1

1 1

1 1

22

22

2

ˆ 0 0 0 0ˆ 0 0 1 0 0 0 ˆ

ˆ 0 0 0 00 0 0 0 0 1ˆ

φ φ

φφ

⎧ ⎫⎪ ⎪⎧ ⎫ −⎡ ⎤ ⎪ ⎪⎪ ⎪ ⎢ ⎥ ⎪ ⎪⎪ ⎪ ⎢ ⎥= = − +⎨ ⎬ ⎨ ⎬⎢ ⎥−⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎣ ⎦⎩ ⎭ ⎪ ⎪⎩ ⎭

X

y y

y x yXy

y

dd dS C

d Sd CddS Cdd

For a beam we will define the following as the transformation matrix:

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−

=

10000000000001000000

CS

CS

T

Notice that the rotations are not affected by the orientation of the beam. Substituting the above transformation into the general form of the stiffness matrix

TkTk T ˆ= gives:

Let’s know consider the effects of an axial force in the general beam transformation.

2 2

2 2

2 2

3 2 2

2 2

2 2

12 12 6 12 12 612 12 6 12 12 66 6 4 6 6 212 12 6 12 12 6

12 12 6 12 12 66 6 2 6 6 4

⎡ ⎤− − − −⎢ ⎥− −⎢ ⎥⎢ ⎥− −

= ⎢ ⎥− −⎢ ⎥

⎢ ⎥− − − −⎢ ⎥− −⎢ ⎥⎣ ⎦

S SC LS S SC LSSC C LC SC C LCLS LC L LS LC LEIk

L S SC LS S SC LSSC C LC SC C LCLS LC L LS LC L


Recall the simple axial deformation, define in the spring element:

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡ −−

=⎭⎬⎫

⎩⎨⎧

x

x

x

x

dd

LAE

ff

2

1

2

1

ˆˆ

11

11

ˆˆ

Combining the axial effects with the shear force and bending moment effects, in local coordinates, gives:

where

321 LEIC

LAEC ==

11 1 1

11 2 2 2 22 2

1 2 2 2 2 1

1 1 22

2 2 2 222 2 2

2 2 2 22 2

ˆˆ 0 0 0 0ˆˆ 0 12 6 0 12 6ˆˆ 0 6 4 0 6 2

ˆ ˆ0 0 0 00 12 6 0 12 6 ˆˆ0 6 2 0 6 4ˆ ˆ

φ

φ

⎧ ⎫⎧ ⎫ −⎡ ⎤ ⎪ ⎪⎪ ⎪ ⎢ ⎥ ⎪ ⎪−⎪ ⎪ ⎢ ⎥ ⎪⎪ ⎪ ⎢ ⎥−⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬⎢ ⎥−⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪⎢ ⎥− − −⎪ ⎪ ⎪⎢ ⎥

−⎣ ⎦⎪ ⎪ ⎪⎩ ⎭ ⎩

xx

xy

xx

yy

df C Cdf C LC C LC

m LC C L LC C LC C df

C LC C LC dfLC C L LC C Lm

⎪⎪

⎪⎪⎪⎪⎭


Therefore:

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−−

−

=

222

222

2222

11

222

222

2222

11

460260612061200000

26046061206120

0000

ˆ

LCLCLCLCLCCLCC

CCLCLCLCLC

LCCLCCCC

k

The above stiffness matrix include the effects of axial force in the x direction, shear force in the y , and bending moment about the z axis. The local degrees of freedom may be related to the global degrees of freedom by:

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

2

2

2

1

1

1

2

2

2

1

1

1

1000000000000000010000000000

ˆˆˆˆˆˆ

φ

φ

φ

φ

y

x

x

x

y

x

x

x

dd

dd

CSSC

CSSC

dd

dd

where the transformation matrix, including axial effects is:

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−

=

1000000000000000010000000000

CSSC

CSSC

T


Substituting the above transformation into the general form of the stiffness matrix TkTk T ˆ= gives:

The analysis of a rigid plane frame can be undertaken by applying stiffness matrix. A rigid plane frame is defined here as a series of beam elements rigidly connected to each other; that is, the original angles made between elements at their joints remain unchanged after the deformation. Furthermore, moments are transmitted from one element to another at the joints. Hence, moment continuity exists at the rigid joints. In addition, the element centroids, as well as the applied loads, lie in a common plane. We observe that the element stiffnesses of a frame are functions of E, A, L, I, and the angle of orientation θ of the element with respect to the global-coordinate axes.

EkL

=


Rigid Plane Frame Example

Consider the frame shown in the figure below.

The frame is fixed at nodes 1 and 4 and subjected to a positive horizontal force of 10,000 lb applied at node 2 and to a positive moment of 5,000 lb-in. applied at node 3. Let E = 30 x 106 psi and A = 10 in.2 for all elements, and let I = 200 in.4 for elements 1 and 3, and I = 100 in.4 for element 2. Element 1: The angle between x and x is 90°

10 == SC where

( )32

22 0.10120

)200(66167.0120

)200(1212 inLIin

LI

====

3

6

/000,250120

1030 inlbLE

=×

=


Therefore, for element 1:

(1)

1 1 1 2 2 2

0.167 0 10 0.167 0 100 10 0 0 10 010 0 800 10 0 400

250,0000.167 0 10 0.167 0 10

0 10 0 0 10 010 0 400 10 0 800

φ φ

− − −⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥−

= ⎢ ⎥−⎢ ⎥⎢ ⎥−⎢ ⎥

−⎣ ⎦

d d d dx y x y

lbk in

Element 2: The angle between x and x is 0°

01 == SC

( )32

22 0.5120

)100(660835.0120

)100(1212 inLIin

LI

====


inlbk

yd

xd

yd

xd

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−

−−

−

=

400502005050835.0050835.00

00100010200504005050835.0050835.0000100010

000,250)2(

333222φφ


10 −== SC

( )32

22 0.10120

)200(66167.0120

)200(1212 inLIin

LI

====

3

6

/000,250120

1030 inlbLE

=×

=



inlbk

yd

xd

yd

xd

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−−−−

−−

=

80001040001001000100100167.0100167.0

40001080001001000100

100167.0100167.0

000,250)3(

444333φφ


1 1 1 4 4 4 0x y x yd d d dφ φ= = = = = =


⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−

−−

×=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

3

3

3

2

2

2

5

12005102005050835.10050835.00

100167.10001020050120051050835.0050835.1000010100167.10

105.2

000,50000000,10

φ

φ

y

x

y

x

dd

dd


2

2

2

3

3

3

0.2110.00148

0.001530.209

0.001480.00149

φ

φ

⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪−

=⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪

−⎩ ⎭⎩ ⎭

x

y

x

y

d ind in

radd ind in

rad


Element 1: The element force-displacement equations can be obtained using dTkf ˆˆ = . Therefore, dT is:

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−−

=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−===

===

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−

=

radinin

radind

ind

dd

dT

y

x

y

x

00153.0211.0

00148.0000

00153.000148.0

211.0000

100000001000010000000100000001000010

2

2

2

1

1

1

φ

φ

Recall the elemental stiffness matrix is:

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−−

−

=

222

222

2222

11

222

222

2222

11

460260612061200000

26046061206120

0000

ˆ

LCLCLCLCLCCLCC

CCLCLCLCLC

LCCLCCCC

k

Therefore, the local force-displacement equations are:

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−−

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−

−−

×==

radinin

dTkf

00153.0211.0

00148.0000

80010040010010167.0010167.00

10010001040010080010010167.0010167.0000100010

105.2ˆˆ 5)1(

Simplifying the above equations gives:

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⋅−

⋅

−

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

inklblbinklblb

mffmff

y

x

y

x

223990,4700,3

376990,4700,3

ˆ

ˆˆˆ

ˆˆ

2

2

2

1

1

1


Element 2: The element force-displacement equations are:

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−−

−

−

=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−=−==

−===

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

radinin

radinin

radind

indradind

ind

dT

y

x

y

x

00149.000148.0

209.000153.0

00148.0211.0

00149.000148.0209.0

00153.000148.0

211.0

100000010000001000000100000010000001

3

3

3

2

2

2

φ

φ


⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−−

−

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−

−−

×==

radinin

radinin

dTkf

00149.000148.0

209.000153.0

00148.0211.0

400502005050833.0050833.00

00100010200504005050833.0050833.0000100010

105.2ˆˆ 5)2(


⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⋅−

−⋅−

−

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

inklblbinklblb

mffmff

y

x

y

x

221700,3010,5

223700,3010,5

ˆ

ˆˆˆ

ˆˆ

3

3

3

2

2

2


⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

===

−=−==

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−

=

000

00149.0209.0

00148.0

000

00149.000148.0209.0

100000001000010000000100000001000010

4

4

4

3

3

3

radin

in

dd

radind

ind

dT

y

x

y

x

φ

φ



⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−

−−

×==

000

00149.0209.0

00148.0

80010040010010167.0010167.00

10010001040010080010010167.0010167.0000100010

105.2ˆˆ 5)3( radin

in

dTkf


⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⋅−−

⋅=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

inklblbinklblb

mffmff

y

x

y

x

375010,5700,3

226010,5700,3

ˆ

ˆˆˆ

ˆˆ

4

4

4

3

3

3



The frame is fixed at nodes 1 and 3 and subjected to a positive distributed load of 1,000 lb/ft applied along element 2. Let E = 30 x 106 psi and A = 100 in.2 for all elements, and let I = 1,000 in.4 for all elements.


First we need to replace the distributed load with a set of equivalent nodal forces and moments acting at nodes 2 and 3. For a beam with both end fixed, subjected to a uniform distributed load, w, the nodal forces and moments are:

2 3(1,000 / )40 20

2 2y ywL lb ft ftf f k= = − = − = −

2 2

2 3(1,000 / )(40 ) 133,333 1,600

12 12wL lb ft ftm m lb ft k in= − = − = − = − ⋅ = ⋅

If we consider only the parts of the stiffness matrix associated with the three degrees of freedom at node 2, we get: Element 1: The angle between x and x is 45º

707.0707.0 == SC where

( )6

3 222

3

30 10 12 12(1,000)58.93 / 0.0463509 12 30 2

6 6(1,000) 11.7855112 30 2

E Ik in inL L

I inL

×= = = =

×

= =×


2 2 2

(1)

50.02 49.98 8.3358.93 49.98 50.02 8.33

8.33 8.33 4000

x yd d

kk in

φ

⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥−⎣ ⎦

Simplifying the above equation:

2 2 2

(1)

2,948 2,945 4912,945 2,948 491491 491 235,700

x yd d

kk in

φ

⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥−⎣ ⎦



01 == SC where

( )

63 2

22

3

30 10 12 12(1,000)62.5 / 0.0521480 12 40

6 6(1,000) 12.512 40

E Ik in inL L

I inL

×= = = =

×

= =×

Therefore, for element 2: 2 2 2

(2)

100 0 062.50 0 0.052 12.5

0 12.5 4,000

x yd d

kk in

φ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

Simplifying the above equation:

2 2 2

(2)

6,250 0 00 3.25 781.250 781.25 250,000

x yd d

kk in

φ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

The global beam equations reduce to:

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⋅−−

2

2

2

700,485290491290951,2945,2491945,2198,9

600,1200

φy

x

dd

inkk


⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−−=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

radin

indd

y

x

0033.00097.0

0033.0

2

2

2

φ



⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−−−

=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−−

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−

=

radinin

radin

indT

0033.00092.0

00452.0000

0033.00097.0

0033.0000

1000000707.0707.00000707.0707.00000001000000707.0707.00000707.0707.0

Recall the elemental stiffness matrix is a function of values C1, C2, and L

( ) ink

ink

LEIC

LAEC 2273.0

23012

)000,1(1030893,5230121030)100(

3

6

32

6

1 =×

×===

××

==


(1)

5,893 0 10 5,893 0 0 00 2.730 694.8 0 2.730 694.8 0

10 694.8 117,900 0 694.8 117,000 0ˆ ˆ5,893 0 0 5,983 0 0 0.00452

0 2.730 694.8 0 2.730 694.8 0.00920 694.8 117,000 0 694.8 235,800 0.0033

−⎡ ⎤ ⎧⎢ ⎥ ⎪−⎢ ⎥ ⎪⎢ ⎥ ⎪−

= = ⎨⎢ ⎥− −⎢ ⎥⎢ ⎥− − − −⎢ ⎥

− −⎣ ⎦

f kTdin

inrad

⎫⎪⎪⎪⎬

⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭


1

1

1

2

2

2

ˆ 26.64ˆ 2.268ˆ 389.1ˆ 26.64

2.268ˆ778.2ˆ

⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪− ⋅⎪ ⎪ =⎨ ⎬ ⎨ ⎬−⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪

− ⋅⎩ ⎭⎪ ⎪⎩ ⎭

x

y

x

y

f kf km k in

kfkf

k inm



⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−−

=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−−

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

000

0033.00097.0

0033.0

000

0033.00097.0

0033.0

100000010000001000000100000010000001

radin

in

radin

in

dT

Recall the elemental stiffness matrix is a function of values C1, C2, and L

( ) ink

ink

LEIC

LAEC 2713.0

4012)000,1(1030250,6

40121030)100(

3

6

32

6

1 =×

×===

××

==


(2)

6,250 0 0 6,250 0 0 0.00330 3.25 781.1 0 3.25 781.1 0.00970 781.1 250,000 0 781.1 125,000 0.0033ˆ ˆ

6,250 0 0 6,250 0 0 00 3.25 781.1 0 3.25 781.1 00 781.1 125,000 0 781.1 250,00 0

− −⎡ ⎤ ⎧ ⎫⎢ ⎥ ⎪ ⎪− −⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎪− −

= = ⎨ ⎬⎢ ⎥− ⎪⎢ ⎥⎪⎢ ⎥− − −⎪⎢ ⎥

−⎣ ⎦ ⎩

inin

radf kTd

⎪

⎪⎪⎪⎭


⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⋅−

−⋅−

−

=

inkkkinkkk

dk

50.41258.263.20

57.83258.263.20

ˆˆ


To obtain the actual element local forces, we must subtract the equivalent nodal forces.

2

2

2

3

3

3

ˆ 20.63 0 20.63ˆ 2.58 20 17.42ˆ 832.57 1600 767.4ˆ 20.63 0 20.63

2.58 20 22.58ˆ412.50 1600 2,013ˆ

⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪− −⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪− ⋅ − ⋅ ⋅⎪ ⎪ = − =⎨ ⎬ ⎨ ⎬ ⎨ ⎬− −⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪ ⎪ ⎪

− ⋅ ⋅ − ⋅⎩ ⎭ ⎩ ⎭⎪ ⎪⎩ ⎭

x

y

x

y

f k kf k k km k in k in k in

k kfk k kfk in k in k inm

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭


Consider the frame shown in the figure below. In this example will illustrate the equivalent joint force replacement method for a frame subjected to a load acting on an element instead of at one of the joints of the structure. Since no distributed loads are present, the point of application of the concentrated load could be treated as an extra joint in the analysis.

This approach has the disadvantage of increasing the total number of joints, as well as the size of the total structure stiffness matrix K. For small structures solved by computer, this does not pose a problem. However, for very large structures, this might reduce the maximum size of the structure that could be analyzed.

The frame is fixed at nodes 1, 2, and 3 and subjected to a concentrated load of 15 k applied at mid-length of element 1. Let E = 30 x 106 psi, A = 8 in2, and let I = 800 in4 for all elements.


Solution Procedure 1. Express the applied load in the element 1 local coordinate system (here

x is directed from node 1 to node 4).

2. Next, determine the equivalent joint forces at each end of element 1,

using the table in Appendix D (see figure below).


3. Then transform the equivalent joint forces from the local coordinate

system forces into the global coordinate system forces, using the equation fTf T ˆ= . These global joint forces are shown below.

4. Then we analyze the structure, using the equivalent joint forces (plus

actual joint forces, if any) in the usual manner.


5. The final internal forces developed at the ends of each element may be obtained by subtracting Step 2 joint forces from Step 4 joint forces.

Element 1: The angle between x and x is 63.43°

895.0447.0 == SC where

( )2 3

2212 12(800) 6 6(800)0.0334 8.95

44.7 1244.7 12= = = =

××

I Iin inL L

6

330 10 55.9 /44.7 12

E k inL

×= =

×


(1)

4 4 4

90.0 178 448178 359 244448 244 179,000

kin

d dx y

k

φ

⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥−⎣ ⎦

Element 2: The angle between x and x is 116.57°

895.0447.0 =−= SC where

( )2 3

2212 12(800) 6 6(800)0.0334 8.95

44.7 1244.7 12= = = =

××

I Iin inL L

6

330 10 55.9 /44.7 12

E k inL

×= =

×


(2)

4 4 4

90.0 178 448178 359 244448 244 179,000

kin

d dx y

k

φ

−⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥⎣ ⎦


Element 3: The angle between x and x is 0° (The author of your textbook directed the element from node 4 to 3. In general, as we have discussed in class, we usually number the element numerically or from 3 to 4. In this case the angle between x and x is 180°)

6330 101 0 50 /

50 12EC S k inL

×= = = =

×

( )2 3

2212 12(800) 6 6(800)0.0267 8.0

50 1250 12= = = =

××

I Iin inL L


(2)

4 4 4

400 0 00 1.334 4000 400 160,000

kin

d dx y

k

φ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

The global beam equations reduce to:

4

4

4

7.5 582 0 8960 0 719 400

900 896 400 518,000 φ

−⎧ ⎫ ⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥− ⋅⎩ ⎭ ⎣ ⎦ ⎩ ⎭

x

y

k dd

k in


4

4

4

0.01030.0009560.00172φ

−⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪−⎩ ⎭ ⎩ ⎭

x

y

d ind in

rad


Element 1: The element force-displacement equations can be obtained using ˆ ˆf kTd= . Therefore, dT is:

895.0447.0

1000000000000000010000000000

==

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−

= SC

CSSC

CSSC

T

0.447 0.895 0 0 0 0 0 00.895 0.447 0 0 0 0 0 00 0 1 0 0 0 0 00 0 0 0.447 0.895 0 0.0103 0.003740 0 0 0.895 0.447 0 0.000956 0.009630 0 0 0 0 1 0.00172 0.00172

⎡ ⎤ ⎧ ⎫ ⎧ ⎫⎢ ⎥ ⎪ ⎪ ⎪ ⎪−⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪

= =⎨ ⎬ ⎨ ⎬⎢ ⎥ − −⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥−⎪ ⎪ ⎪ ⎪⎢ ⎥− −⎣ ⎦ ⎩ ⎭ ⎩ ⎭

Tdin inin in

rad rad

Recall the elemental stiffness matrix is:

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−−

−

=

222

222

2222

11

222

222

2222

11

460260612061200000

26046061206120

0000

ˆ

LCLCLCLCLCCLCC

CCLCLCLCLC

LCCLCCCC

k

( ) ink

ink

LEIC

LAEC 155.0

72.4412)800(10302.447

72.44121030)8(

3

6

32

6

1 =××

===×

×==



⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−

−

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−−

−

==

radinin

dTkf

00172.000963.000374.0

000

000,1795.5000490,895.50005.500868.105.500868.10

0044700447490,895.5000000,1795.5000

5.500868.105.500868.100044700447

ˆˆ)1(


(1)

1.670.88

158ˆ ˆ ˆ1.670.88

311

⎧ ⎫⎪ ⎪−⎪ ⎪⎪ ⎪− ⋅

= = ⎨ ⎬−⎪ ⎪⎪ ⎪⎪ ⎪− ⋅⎩ ⎭

kk

k inf kd

kk

k in

To obtain the actual element local forces, we must subtract the equivalent nodal forces.

1

1

1

4

4

4

ˆ 1.67 3.36 5.03ˆ 0.88 6.71 7.59ˆ 158 900 1,058ˆ 1.67 3.36 1.68

0.88 6.71 5.83ˆ311 900 589ˆ

⎧ ⎫ −⎧ ⎫ ⎧ ⎫ ⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪− −⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪− ⋅ ⋅ − ⋅⎪ ⎪ = − =⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨− −⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ −⎪ ⎪ ⎪ ⎪ ⎪ ⎪

− ⋅ − ⋅ ⋅⎩ ⎭ ⎩ ⎭⎪ ⎪⎩ ⎭

x

y

x

y

f k k kf k k km k in k in k in

k k kfk k kf

k in k in k inm

⎫⎪

⎪ ⎪⎪ ⎪

⎬⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭



895.0447.0

1000000000000000010000000000

=−=

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−

= SC

CSSC

CSSC

T

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−

=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−

−

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−

−−−

=

radinin

radinin

dT

00172.000879.000546.0

000

00172.0000956.0

0103.0000

1000000447.0895.00000895.0447.00000001000000447.0895.00000895.0447.0


( ) ink

ink

LEIC

LAEC 155.0

72.4412)800(10302.447

72.44121030)8(

3

6

32

6

1 =××

===×

×==

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−−

−

==

radinin

dTkf

00172.000879.000546.0

000

000,1795.5000490,895.50005.500868.105.500868.10

0044700447490,895.5000000,1795.5000

5.500868.105.500868.100044700447

ˆˆ)2(



(2)

2.440.877

158ˆ ˆ ˆ2.440.877

312

⎧ ⎫⎪ ⎪−⎪ ⎪⎪ ⎪− ⋅

= = ⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪− ⋅⎩ ⎭

kk

k inf kd

kk

k in

Since there are no applied loads on element 2, there are no equivalent nodal forces to account for. Therefore, the above equations are the final local nodal forces

Element 3: The element force-displacement equations can be obtained using ˆ ˆf kTd= . Therefore, dT is:

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−

−

=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−

−

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

000

00172.0000956.0

0103.0

000

00172.0000956.0

0103.0

100000010000001000000100000010000001

radinin

radinin

dT



( ) ink

ink

LEIC

LAEC 111.0

5012)800(1030400

50121030)8(

3

6

32

6

1 =×

×===

××

==

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−

−

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−−

−

==

000

00172.0000956.0

0103.0

000,1604000000,804000400335.10400335.100040000400000,804000000,1604000

400335.10400335.100040000400

ˆˆ)3(

radinin

dTkf


(3)

4.120.687

275ˆ ˆ ˆ4.120.687

137

−⎧ ⎫⎪ ⎪−⎪ ⎪⎪ ⎪− ⋅

= = ⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪− ⋅⎩ ⎭

kk

k inf kd

kk

k in

Since there are no applied loads on element 3, there are no equivalent nodal forces to account for. Therefore, the above equations are the final local nodal forces. The free-body diagrams are shown below.



The frame shown on the right is fixed at nodes 2 and 3 and subjected to a concentrated load of 500 kN applied at node 1. For the bar, A = 1 x 10-3 m2, for the beam, A = 2 x 10-3 m2, I = 5 x 10-5 m4, and L = 3 m. Let E = 210 GPa for both elements.


Beam Element 1: The angle between x and x is 0°

01 == SC where

345

252

5

2 103

)105(661067.6)3(

)105(1212 mLIm

LI −

−−

−

=×

=×=×

=

36

6

/10703

10210 mkNLE

×=×

=


mkNk

yx dd

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡×=

20.010.0010.0067.00002

1070 3)1(

111 φ

Bar Element 2: The angle between x and x is 45°

707.0707.0 == SC

where

( )m

kNm

mkNmk

yx dd

⎥⎦

⎤⎢⎣

⎡×=

−

5.05.05.05.0

24.4/1021010 2623

)2(

11

mkNk

yx dd

⎥⎦

⎤⎢⎣

⎡×=

354.0354.0354.0354.0

1070 3)2(

11

Assembling the elemental stiffness matrices we obtain the global stiffness matrix

mkNK⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡×=

20.010.0010.0421.0354.00354.0354.2

1070 3


The global equations are:

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡×=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧−

1

1

13

20.010.0010.0421.0354.00354.0354.2

10700

5000

φy

x

dd

mkNkN


⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧−=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

radmm

dd

y

x

0113.00225.0

00388.0

1

1

1

φ

Bar Element: The bar element force-displacement equations can be obtained using ˆ ˆf kTd= .

Therefore, the forces in the bar element are:

( ) kNSdCdL

AEf yxx 670ˆ111 −=+=

( ) kNSdCdL

AEf yxx 670ˆ113 =+−=

Beam Element: The beam element force-displacement equations can be obtained using ˆ ˆ ˆf kd= . Since the local axis coincides with the global coordinate system, and the displacements at node 2 are zero. Therefore, the local force-displacement equations are:

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−=

⎭⎬⎫

⎩⎨⎧

y

x

y

x

x

x

dddd

SCSC

LAE

ff

3

3

1

1

3

1

0000

1111

ˆˆ


32

1

222

222

2222

11

222

222

2222

11

460260612061200000

26046061206120

0000

ˆ

LEIC

LAEC

LCLCLCLCLCCLCC

CCLCLCLCLC

LCCLCCCC

k

=

=

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−−

−

=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⋅−

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−−

−

×==

000

0113.00225.0

00388.0

20.010.0010.010.0010.0067.0010.0067.00

00200210.010.0020.010.0010.0067.0010.0067.00002002

1070ˆˆˆ 3)1(

mkNmm

dkf

Substituting numerical values into the above equations gives:

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⋅−

−

−

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

mkNkNkN

kNkN

mffmff

y

x

y

x

3.785.26

4730.05.26

473

ˆ

ˆˆˆ

ˆˆ

2

2

2

1

1

1




The frame is fixed at nodes 1 and 3 and subjected to a moment of 20 kN-m applied at node 2. Assume A = 2 x 10-2 m2, I = 2 x 10-4 m4, and E = 210 GPa for all elements.


10 == SC where

344

242

4

2 1034

)102(66105.1)4(

)102(1212 mLIm

LI −

−−

−

×=×

=×=×

=

37

6

/1025.54

10210 mkNLE

×=×

=

Therefore, the stiffness matrix for element 1, considering only the parts associated with node 2, is:

mkNk

yx dd

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡×=

08.0003.002003.00015.0

1025.5 5)1(

222 φ



01 == SC where

344

252

4

2 104.25

)102(66106.9)5(

)102(1212 mLIm

LI −

−−

−

×=×

=×=×

=

37

6

/102.45

10210 mkNLE

×=×

=

Therefore, the stiffness matrix for element 2, considering only the parts associated with node 2, is:

mkNk

yx dd

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡×=

08.0024.00024.00096.00002

102.4 5)2(

222 φ

Assembling the elemental stiffness matrices we obtain the global stiffness matrix:

mkNK⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

0756.00101.00158.00101.00500.100158.008480.0

106


⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⋅ 2

2

26

0756.00101.00158.00101.00500.100158.008480.0

1020

00

φy

x

dd

mkN



62

62

42

4.95 102.56 102.66 10φ

−

−

−

⎧ ⎫− ×⎧ ⎫⎪ ⎪⎪ ⎪ = − ×⎨ ⎬ ⎨ ⎬

⎪ ⎪ ⎪ ⎪×⎩ ⎭ ⎩ ⎭

x

y

d md m

rad

Element 1: The beam element force-displacement equations can be obtained using ˆ ˆf kTd= .

6 6

6 6

4 4

0 1 0 0 0 0 0 01 0 0 0 0 0 0 0

0 0 1 0 0 0 0 00 0 0 0 1 0 4.95 10 2.56 100 0 0 1 0 0 2.56 10 4.95 100 0 0 0 0 1 2.66 10 2.66 10

− −

− −

− −

⎡ ⎤ ⎧ ⎫ ⎧ ⎫⎢ ⎥ ⎪ ⎪ ⎪ ⎪−⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪

= =⎨ ⎬ ⎨ ⎬⎢ ⎥ − × − ×⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥− − × ×⎪ ⎪ ⎪ ⎪⎢ ⎥

× ×⎣ ⎦ ⎩ ⎭ ⎩ ⎭

Tdm mm m

rad rad


⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−−

−

=

222

222

2222

11

222

222

2222

11

460260612061200000

26046061206120

0000

ˆ

LCLCLCLCLCCLCC

CCLCLCLCLC

LCCLCCCC

k

( )

2 66

1

6 4

2 33

(2 10 )210 10 1.05 104

210 10 (2 10 ) 656.254

kNm

kNm

AECL

EICL

−

−

× ×= = = ×

× ×= = =


⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

×××−

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−−

−

×==

−

−

−

radmm

dTkf

4

6

63

)1(

1066.21095.41056.2

000

83043035.1035.10

002000020043083035.1035.100020000200

1025.5ˆˆ

Solving for the forces and moments gives:

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⋅−

−⋅

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

mkNkNkNmkN

kNkN

mffmff

y

x

y

x

17.112.4

69.259.5

2.469.2

ˆ

ˆˆˆ

ˆˆ

2

2

2

1

1

1

Element 2: The beam element force-displacement equations can be obtained using ˆ ˆf kTd= .

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

××−×−

=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

××−×−

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

−

−

−

−

−

−

radmm

radmm

dT

4

6

6

4

6

6

1066.21056.21095.4

000

1066.21056.21095.4

000

100000010000001000000100000010000001


( )

2 66

1

6 4

2 33

(2 10 )210 10 0.84 105

210 10 (2 10 ) 3365

kNm

kNm

AECL

EICL

−

−

× ×= = = ×

× ×= = =


⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

××−×−

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−−

−

×==−

−

−

0001066.2

1056.21095.4

840.20440.2040.296.0040.296.00

0020000200440.20840.2040.296.0040.296.000020000200

102.4ˆˆ4

6

6

3)2(

radmm

dTkf


2

2

2

3

3

3

ˆ 4.16ˆ 2.69ˆ 8.92ˆ 4.16

2.69ˆ4.47ˆ

⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⋅⎪ ⎪ =⎨ ⎬ ⎨ ⎬−⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪

⋅⎩ ⎭⎪ ⎪⎩ ⎭

x

y

x

y

f kNf kNm kN m

kNfkNfkN mm

Inclined or Skewed Supports

If a support is inclined, or skewed, at some angle α for the global x axis, as shown below, the boundary conditions on the displacements are not in the global x-y directions but in the x’-y’ directions.


We must transform the local boundary condition of d’3y = 0 (in local coordinates) into the global x-y system. Therefore, the relationship between of the components of the displacement in the local and the global coordinate systems at node 3 is:

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

3

3

3

3

3

3

1000cossin0sincos

'''

φαααα

φy

x

y

x

dd

dd

We can rewrite the above expression as:

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−==

1000cossin0sincos

][' 3333 αααα

tdtd

We can apply this sort of transformation to the entire displacement vector as:

'][or][' dTddTd Tii ==

where the matrix [Ti] is:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

][]0[]0[

]0[][]0[

]0[]0[][

][

3tI

ITi

Both the identity matrix [I] and the matrix [t3] are 3 x 3 matrices.

The force vector can be transformed by using the same transformation.

fTf i ][' = In global coordinates, the force-displacement equations are:

dKf ][=


Applying the skewed support transformation to both sides of the force-displacement equation gives:

dKTfT ii ]][[][ = By using the relationship between the local and the global displacements, the force-displacement equations become:

']][][['']][][[][ dTKTfdTKTfT Tii

Tiii =⇒=

Therefore the global equations become:

⎪⎪⎪⎪⎪⎪

⎭

⎪⎪⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪⎪⎪

⎩

⎪⎪⎪⎪⎪⎪

⎨

⎧

=

⎪⎪⎪⎪⎪⎪

⎭

⎪⎪⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪⎪⎪

⎩

⎪⎪⎪⎪⎪⎪

⎨

⎧

1

3

3

2

2

2

1

1

1

3

3

3

2

2

2

1

1

1

''

]][][[

''

φ

φ

φ

y

x

y

x

y

x

Tii

y

x

y

x

y

x

dd

dd

dd

TKT

MFFMFFMFF

Grid Equations

A grid is a structure on which the loads are applied perpendicular to the plane of the structure, as opposed to a plane frame where loads are applied in the plane of the structure. Both torsional and bending moment continuity are maintained at each node in a grid element. Examples of a grid structure are floors and bridge deck systems. A typical grid structure is shown in the figure below.


A representation of the grid element is shown below:

The degrees of freedom for a grid element are: a vertical displacement iyd (normal to the grid), a torsional rotation ixφ about the x axis, and a bending rotation izφ about the z axis. The nodal forces are: a transverse force iyf a torsional moment ixm about the x axis, and a bending moment izm about the z axis.

Let’s derive the torsional rotation components of the element stiffness matrix. Consider the sign convention for nodal torque and angle of twist shown the figure below.

A linear displacement function φ is assumed.

xaa ˆ21 +=φ Applying the boundary conditions and solving for the unknown coefficients gives:

xxx x

L 112 ˆˆˆˆ

φφφφ +⎟⎟⎠

⎞⎜⎜⎝

⎛ −=


Or in matrix form:

[ ]⎭⎬⎫

⎩⎨⎧

==x

xNN2

121 ˆ

ˆˆφφ

φ


LxN

LxN =−= 21

ˆ1

To obtain the relationship between the shear strain γ and the angle of twist φ consider the torsional deformation of the bar as shown below.

If we assume that all radial lines, such as OA, remain straight during twisting or torsional deformation, then the arc length AB is:

φγ ˆˆmax RdxdAB == Therefore;

xdRd

ˆˆ

max

φγ =

At any radial position, r, we have, from similar triangles OAB and OCD:


( )xxLr

xddr 12

ˆˆˆˆ

φφφγ −==

The relationship between shear stress and shear strain is:

τ γ=G where G is the shear modulus of the material. From elementary mechanics of materials, we get:

ˆ τ=x

JmR

where J is the polar moment of inertia for a circular cross section or the torsional constant for non-circular cross sections. Rewriting the above equation we get:

( )2 1ˆ ˆˆ φ φ= −x x x

GJmL

The nodal torque sign convention gives:

xxxx mmmm ˆˆˆˆ 21 =−= Therefore;

( ) ( )1 1 2 2 2 1ˆ ˆ ˆ ˆˆ ˆφ φ φ φ= − = −x x x x x x

GJ GJm mL L


⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡−

−=

⎭⎬⎫

⎩⎨⎧

x

x

x

x

LGJ

mm

2

1

2

1

ˆˆ

1111

ˆˆ

φφ


Combining the torsional effects with shear and bending effects, we obtain the local stiffness matrix equations for a grid element.

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−−−−

−−

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

z

x

y

z

x

y

LEI

LEI

LEI

LEI

LGJ

LGJ

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LGJ

LGJ

LEI

LEI

LEI

LEI

z

x

y

z

x

y

d

d

mmf

mmf

2

2

2

1

1

1

4626

612612

2646

612612

2

2

2

1

1

1

ˆˆˆˆˆˆ

000000

0000

000000

ˆˆ

ˆˆˆ

ˆ

22

2323

22

2323

φφ

φφ

The transformation matrix relating local to global degrees of freedom for a grid is:

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−=

CSSC

CSSC

TG

00000000

00100000000000000001

where θ is now positive taken counterclockwise from x to x in the x-z plane: therefore;

cos sinj i j ix x z zC S

L Lθ θ

− −= = = =

The global stiffness matrix for a grid element arbitrary oriented in the x-z plane is given by:

GGT

GG TkTk ˆ=


Grid Example Consider the frame shown in the figure below.

The frame is fixed at nodes 2, 3, and 4, and is subjected to a load of 100 kips applied at node 1. Assume I = 400 in.4, J = 110 in.4, G = 12 x 10 3 ksi, and E = 30 x 10 3 ksi for all elements.

To facilitate a timely solution, the boundary conditions at nodes 2, 3, and 4 are applied to the local stiffness matrices at the beginning of the solution.

000

444

333

222

=========

zxy

zxy

zxy

ddd

φφφφφφ

Beam Element 1:

447.036.221020sin894.0

36.22200cos )1(

12)1(

12 =−

=−

==−=−

=−

==L

zzSL

xxC θθ

where

3 3

3 3 2 212 12(30 10 )(400) 6 6(30 10 )(400)7.45 1,000

(22.36 12) (22.36 12)× ×

= = = =× ×

EI EIk kinL L


3 34 4(30 10 )(400) (12 10 )(110)179,000 4,920(22.36 12) (22.36 12)

× ×= = ⋅ = = ⋅

× ×EI GJk in k inL L

The global stiffness matrix for element 1, considering only the parts associated with node 1, and the following relationship:

GGT

GG TkTk ˆ=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−=

894.0447.00447.0894.00001

894.0447.00447.0894.00001

TGG TT

inkk

zxyd

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

000,1790000,10920,40

000,1045.7ˆ )1(

111 φφ

Therefore, the global stiffness matrix is

inkk

zxyd

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−−=

000,144600,69894600,69700,39447

89444745.7)1(

111 φφ

Beam Element 2:

447.036.22100sin894.0

36.22200cos )2(

13)2(

13 −=−

=−

==−=−

=−

==L

zzSL

xxC θθ

where

3 3

3 3 2 212 12(30 10 )(400) 6 6(30 10 )(400)7.45 1,000

(22.36 12) (22.36 12)× ×

= = = =× ×

EI EIk kinL L


3 34 4(30 10 )(400) (12 10 )(110)179,000 4,920

(22.36 12) (22.36 12)× ×

= = ⋅ = = ⋅× ×

EI GJk in k inL L


GGT

GG TkTk ˆ=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−=

894.0447.00447.0894.00001

000,1790000,10920,40

000,1045.7

894.0447.00447.0894.00001

)2(k


inkk

zxyd

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−

=000,144600,69894600,69700,39447

89444745.7)2(

111 φφ

Beam Element 3:

110

100sin010

2020cos )3(14

)3(14 −=

−=

−===

−=

−==

LzzS

LxxC θθ

where

kLEIink

LEI 000,5

)1210()400)(1030(66/3.83

)1210()400)(1030(1212

2

3

23

3

3=

××

==×

×=

inkL

GJinkLEI

⋅=×

×=⋅=

××

= 000,11)1210(

)110)(1012(000,400)1210(

)400)(1030(44 33


GGT

GG TkTk ˆ=


⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=

010100

001

000,4000000,50000,110000,503.83

010100001

)3(k


⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

000,11000000,400000,50000,53.83

)3(

111

k

zxyd φφ

Superimposing the three elemental stiffness matrices gives:

1 1 1

98.2 5,000 1,7905,000 479,000 01,790 0 299,000

φ φ

−⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥−⎣ ⎦

y x zd

K


1 1

1 1

1 1

100 98.2 5,000 1,7900 5,000 479,000 00 1,790 0 299,000

φφ

= − −⎧ ⎫ ⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥= =⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥= −⎩ ⎭ ⎣ ⎦ ⎩ ⎭

y y

x x

z z

F k dMM


⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−

−=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

radrad

ind

z

x

y

0169.00295.0

83.2

1

1

1

φφ


Element 1: The grid element force-displacement equations can be obtained using dTkf GG

ˆˆ = .

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧−−

=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−

−

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−

−−−

=

000

00192.00339.0

83.2

000

0169.00295.0

83.2

894.0447.00000447.0894.00000001000000894.0447.00000447.0894.00000001

radradin

radradin

dTG


⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧−−

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−−−−

−−

==

000

00192.00339.0

83.2

000,1790000,1500,890000,10920,400920,40000,1045.7000,1045.7500,890000,1000,1790000,10920,400920,40

000,1045.7000,1045.7

ˆˆ)1(

radradin

dTkf


1

1

1

2

2

2

ˆ 19.2ˆ 167ˆ 2,480ˆ 19.2

167ˆ2,260ˆ

⎧ ⎫ −⎧ ⎫⎪ ⎪ ⎪ ⎪− ⋅⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪− ⋅⎪ ⎪ =⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⋅⎪ ⎪ ⎪ ⎪

− ⋅⎩ ⎭⎪ ⎪⎩ ⎭

y

x

z

y

x

z

f km k inm k in

kfk inmk inm


Element 2: The grid element force-displacement equations can be obtained using ˆ ˆ

G Gf k T d= .

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧−−

=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−

−

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−

−−−

=

000

0283.00188.0

83.2

000

0169.00295.0

83.2

894.0447.00000447.0894.00000001000000894.0447.00000447.0894.00000001

radradin

radradin

dTG


⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧−−

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−−−−

−−

==

000

0283.00188.0

83.2

000,1790000,1500,890000,10920,400920,40000,1045.7000,1045.7500,890000,1000,1790000,10920,400920,40

000,1045.7000,1045.7

ˆˆ)2(

radradin

dTkf


1

1

1

3

3

3

ˆ 7.23ˆ 92.5ˆ 2,240ˆ 7.23

92.5ˆ295ˆ

⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪− ⋅⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪− ⋅⎪ ⎪ =⎨ ⎬ ⎨ ⎬

−⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⋅⎪ ⎪ ⎪ ⎪

− ⋅⎩ ⎭⎪ ⎪⎩ ⎭

y

x

z

y

x

z

f km k inm k in

kfk inm

k inm



G Gf k T d= .

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧ −

=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−

−

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−

=

000

0295.00169.0

83.2

000

0169.00295.0

83.2

010000100000

001000000010000100000001

radradin

radradin

dTG


⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧ −

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−−−−

−−

==

000

0295.00169.0

83.2

000,4000000,5000,2000000,50000,1100000,110000,503.83000,503.83000,2000000,5000,4000000,5

0000,1100000,110000,503.83000,503.83

ˆˆ)3(

radradin

dTkf


1

1

1

4

4

4

ˆ 88.1ˆ 186ˆ 2,340ˆ 88.1

186ˆ8,240ˆ

⎧ ⎫ −⎧ ⎫⎪ ⎪ ⎪ ⎪⋅⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪− ⋅⎪ ⎪ =⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪− ⋅⎪ ⎪ ⎪ ⎪

− ⋅⎩ ⎭⎪ ⎪⎩ ⎭

y

x

z

y

x

z

f km k inm k in

kfk inmk inm


To check the equilibrium of node 1 the local forces and moments for each element need to be transformed to global coordinates. Recall, that:

1ˆˆ −==⇒= TTfTfTff TT Since we are only checking the forces and moments at node 1, we need only the upper-left-hand portion of the transformation matrix TG. Therefore; for Element 1:

1

1

1

1 0 0 19.2 19.20 0.894 0.447 167 1,2600 0.447 0.894 2,480 2,150

− −⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥= − − − ⋅ = ⋅⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥− − ⋅ ⋅⎩ ⎭ ⎣ ⎦ ⎩ ⎭ ⎩ ⎭

y

x

z

f k km k in k inm k in k in


Therefore; for Element 2:

1

1

1

1 0 0 7.23 7.230 0.894 0.447 92.5 1,0800 0.447 0.894 2,240 1,960

−⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥= − − ⋅ = ⋅⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥− − − ⋅ − ⋅⎩ ⎭ ⎣ ⎦ ⎩ ⎭ ⎩ ⎭

y

x

z


Therefore; for Element 3:

1

1

1

1 0 0 88.1 88.10 0 1 2,340 2,3400 1 0 186 186

− −⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥= − ⋅ = − ⋅⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥− − ⋅ − ⋅⎩ ⎭ ⎣ ⎦ ⎩ ⎭ ⎩ ⎭

y

x

z


The forces and moments that are applied to node 1 by each element are equal in magnitude and opposite direction. Therefore the sum of the forces and moments acting on node 1 are:

The forces and moments accurately satisfy equilibrium considering the amount of truncation error inherent in results of the calculations presented in this example.

kF y 07.01.882.1923.71001 =++−−=∑

inkM x ⋅=+−−=∑ 0.0340,2080,1260,11

inkM z ⋅−=++−=∑ 0.4186060,1150,21


Grid Example Consider the frame shown in the figure below.

The frame is fixed at nodes 1 and 3, and is subjected to a load of 22 kN applied at node 2. Assume I = 16.6 x 10-5 m4, J = 4.6 x 10-5 m4, G = 84 GPa, and E = 210 GPa for all elements.

To facilitate a timely solution, the boundary conditions at nodes 1 and 3 are applied to the local stiffness matrices at the beginning of the solution.

00

333

111

======

zxy

zxy

dd

φφφφ

Beam Element 1: the local x axis coincides with the global x axis

030sin1

33cos )1(

12)1(

12 ==−

====−

==L

zzSL

xxC θθ

where

mkNLEI /1055.1

)3()106.16)(10210(1212 4

3

56

3 ×=××

=−

kNLEI 4

2

56

2 1032.2)3(

)106.16)(10210(66×=

××=

−

mkNLEI ·1065.4

3)106.16)(10210(44 4

56

×=××

=−

mkNL

GJ ·10128.03

)106.4)(1084( 456

×=××

=−


The global stiffness matrix for element 1, considering only the parts associated with node 2, may be obtained from the following relationship:

GGT

GG TkTk ˆ=

mkNk⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

100010001

65.4032.20128.0032.2055.1

100010001

104)1(


mkNk

zxyd

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−=

65.4032.20128.0032.2055.1

104)1(

222 φφ

Beam Element 2: the local x axis is located from node 2 to node 3

3 2 3 2(2) (1)

0 3cos 0 sin 13 3

x x z zC SL L

θ θ− − −= = = = = = = = −

The global stiffness matrix for element 2, considering only the parts associated with node 2, may be obtained using:

GGT

GG TkTk ˆ=

mkNk⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=

010100

001

65.4032.20128.0032.2055.1

010100001

104)2(


mkNk

zxyd

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

128.000065.432.2032.255.1

104)2(

222 φφ


Superimposing the two elemental stiffness matrices gives:

mkNK

zxyd

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−=

78.4032.2078.432.2

32.232.210.310 4

222 φφ


⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

==

−=

x

x

y

z

x

y d

MM

kNF

2

2

24

2

2

2

78.4032.2078.432.232.232.210.3

1000

22

φφ


2

2

2

0.002590.001260.00126

φφ

−⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪−⎩ ⎭ ⎩ ⎭

y

x

z

d mradrad


G Gf k T d= .

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−

−=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−

−

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

radrad

m

radrad

mdTG

00126.000126.0

00259.0000

00126.000126.0

00259.0000

100000010000001000000100000010000001



⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−

−

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−−−−

−−

==

radrad

mdTkf

00126.000126.0

00259.0000

65.4032.233.2032.20128.000128.0032.2055.132.2055.1

33.2032.265.4032.20128.000128.0032.2055.132.2055.1

10ˆˆ 4)1(


1

1

1

2

2

2

ˆ 11.0ˆ 1.50ˆ 31.0ˆ 11.0

1.50ˆ1.50ˆ

⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪− ⋅⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⋅⎪ ⎪ =⎨ ⎬ ⎨ ⎬

−⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⋅⎪ ⎪ ⎪ ⎪

⋅⎩ ⎭⎪ ⎪⎩ ⎭

y

x

z

y

x

z

f kNm kN mm kN m

kNfkN mmkN mm


G Gf k T d= .

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧−

=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

−

−

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−

=

000

00126.000126.0

00259.0

000

00126.000126.0

00259.0

010000100000

001000000010000100000001

radrad

m

radrad

m

dTG



⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧−

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−−−−

−−

==

000

00126.000126.0

00259.0

65.4032.233.2032.20128.000128.0032.2055.132.2055.1

33.2032.265.4032.20128.000128.0032.2055.132.2055.1

10ˆˆ 4)2(

radrad

m

dTkf


2

2

2

3

3

3

ˆ 11.0ˆ 1.50ˆ 1.50ˆ 11.0

1.50ˆ31.0ˆ

⎧ ⎫ −⎧ ⎫⎪ ⎪ ⎪ ⎪⋅⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪− ⋅⎪ ⎪ =⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪− ⋅⎪ ⎪ ⎪ ⎪

− ⋅⎩ ⎭⎪ ⎪⎩ ⎭

y

x

z

y

x

z

f kNm kN mm kN m

kNfkN mm

kN mm

The resulting free-body diagrams:


Beam Element Arbitrarily Oriented in Space In this section,we will develop a beam element that is arbitrarily oriented in

three-dimensions. This element can be used to analyze three-dimensional frames. Let consider bending about axes, as shown below.

The y axis is the principle axis for which the moment of inertia is minimum, Iy. The right-hand rule is used to establish the z axis and the maximum moment of inertia, Iz. Bending in the zx ˆˆ − plane: The bending in the zx − plane is defined by ym . The stiffness matrix for bending the in the x-z plane is:

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−

−−=

3

2

3

2

2

2

3

2

3

2

2

2

4

4626

612612

2646

612612

ˆ

LLLL

LL

LL

LLLL

LL

LL

LEI

k yY

where Iy is the moment of inertia about the y axis (the weak axis). Bending in the yx ˆˆ − plane: The bending in the yx − plane is defined by ˆ zm . The stiffness matrix for bending the in the yx − plane is:

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−

−−=

3

2

3

2

2

2

3

2

3

2

2

2

4

4626

612612

2646

612612

ˆ

LLLL

LL

LL

LLLL

LL

LL

LEIk z

z

where Iz is the moment of inertia about the z axis (the strong axis).


Direct superposition of the bending stiffness matrices with the effects of axial forces and torsional rotation give:

The global stiffness matrix may be obtained using:

TkTk T ˆ= where

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

33

33

33

33

x

x

x

x

T

λλ

λλ

where

zyxzyxzyxzyx dddddd 222222111111ˆˆˆˆˆˆˆˆˆˆˆˆ φφφφφφ


⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

zzzyzx

yzyyyx

xzxyxx

x

CCCCCCCCC

ˆˆˆ

ˆˆˆ

ˆˆˆ

33λ

where the direction cosines,

jiC ˆ , are defined as shown below

The direction cosines of the x axis are:

kjix xzxyxx ˆˆˆ coscoscosˆ θθθ ++= where

nL

zzmL

yylL

xxxzxyxx =

−==

−==

−= 12

ˆ12

ˆ12

ˆ coscoscos θθθ

The y axis is selected to be perpendicular to the x and the z axes is such a way that the cross product of global z with x results in the y axis as shown in the figure below.

yxz ˆˆ =×


3 3 0x

l m n

m lD Dln mn DD D

λ

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥= −⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦

jDli

Dm

nml

kji

Dyxz +−===× 1001ˆˆ

where 22 mlD +=

The z axis is determined by the condition that yxz ˆˆˆ ×=

kDjD

mniDnl

lmnmlkji

Dyxz +−−=

−=×=

0

1ˆˆˆ

Therefore, the transformation matrix becomes:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

zzzyzx

yzyyyx

xzxyxx

x

CCCCCCCCC

ˆˆˆ

ˆˆˆ

ˆˆˆ

33λ

There are two exceptions that arise when using the above expressions for mapping the local coordinates to the global system: (1) when the positive x coincides with z; and (2) when the positive x is in the opposite direction as z. For the first case, it is assumed that y is y.

0 0 10 0 01 0 0

λ⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥−⎣ ⎦

In case two, it is assumed that y is y.

0 0 10 0 01 0 0

λ−⎡ ⎤

⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


If the effects of axial force, both shear forces, twisting moment, and both bending moments are considered, the stiffness matrix for a frame element is:


In this case the symbol φ are:

2 2

12 12y zy z

s s

EI EIGA L GA L

φ φ= =

where As is the effective beam cross-section in shear. Recall the shear modulus of elasticity or the modulus of rigidity, G, is related to the modulus of elasticity and the Poisson’s ratio, ν as:

( )2 1

EGν

=+

If φy and φz are set to zero, the stiffness matrix reduces to that shown previously on page 235. This is the form of the stiffness matrix used by SAP2000 for its frame element.

zyxzyxzyxzyx dddddd 222222111111ˆˆˆˆˆˆˆˆˆˆˆˆ φφφφφφ


Example Frame Application

A bus subjected to a static roof-crush analysis. In this model 599 frame elements and 357 nodes are used.

Concept of Substructure Analysis

Sometimes structures are too large to be analyzed as a single system or treated as a whole; that is, the final stiffness matrix and equations for solution exceed the memory capacity of the computer. A procedure to overcome this problem is to separate the whole structure into smaller units called substructures. For example, the space frame of an airplane, as shown below, may require thousands of nodes and elements to completely model and describe the response of the whole structure. If we separate the aircraft into substructures,


such as parts of the fuselage or body, wing sections, etc., as shown below, then we can solve the problem more readily and on computers with limited memory.


Problems

14. Do problems 5.3, 5.8, 5.13, 5.28, 5.41, and 5.43 on pages 240 - 263 in your textbook “A First Course in the Finite Element Method” by D. Logan.

15. Do problems 5.23, 5.25, 5.35, 5.39, and 5.55 on pages 240 - 263 in your textbook “A First Course in the Finite Element Method” by D. Logan. You may use the SAP2000 to do frame analysis.


Development of the Plane Stress and Plane Strain Stiffness Equations

Introduction

In Chapters 2 through 5, we considered only line elements. Line elements are connected only at common nodes, forming framed or articulated structures such as trusses, frames, and grids. Line elements have geometric properties such as cross-sectional area and moment of inertia associated with their cross sections. However, only one local coordinate along the length of the element is required to describe a position along the element (hence, they are called line elements). Nodal compatibility is then enforced during the formulation of the nodal equilib-rium equations for a line element.

This chapter considers the two-dimensional finite element. Two-dimensional (planar) elements are thin-plate elements such that two coordinates define a po-sition on the element surface.

The elements are connected at common nodes and/or along common edges

to form continuous structures. Nodal compatibility is then enforced during the formulation of the nodal equilibrium equations for two-dimensional elements. If proper displacement functions are chosen, compatibility along common edges is also obtained.


The two-dimensional element is extremely important for: (1) Plane stress analysis, which includes problems such as plates with

holes, fillets, or other changes in geometry that are loaded in their plane resulting in local stress concentrations; and

(2) Plane strain analysis, which includes problems such as a long under-ground box culvert subjected to a uniform load acting constantly over its length or a long cylindrical control rod subjected to a load that remains constant over the rod length (or depth).


Plane Stress Problems

Plane Strain Problems We begin this chapter with the development of the stiffness matrix for a basic

two-dimensional or plane finite element, called the constant-strain triangular element. The constant-strain triangle (CST) stiffness matrix derivation is the simplest among the available two-dimensional elements.

We will derive the CST stiffness matrix by using the principle of minimum po-tential energy because the energy formulation is the most feasible for the devel-opment of the equations for both two- and three-dimensional finite elements.


General Steps in the Formulation of the Plane Triangular Element Equations

We will now follow the steps described in Chapter 1 to formulate the governing equations for a plane stress/plane strain triangular element. First, we will de-scribe the concepts of plane stress and plane strain. Then we will provide a brief description of the steps and basic equations pertaining to a plane triangular ele-ment.

Plane Stress Plane stress is defined to be a state of stress in which the normal stress and

the shear stresses directed perpendicular to the plane are assumed to be zero. That is, the normal stress σz and the shear stresses τxz and τyz are assumed to be zero. Generally, members that are thin (those with a small z dimension com-pared to the in-plane x and y dimensions) and whose loads act only in the x-y plane can be considered to be under plane stress.

Plane Strain Plane strain is defined to be a state of strain in which the strain normal to the

x-y plane εz and the shear strains γxz and γyz are assumed to be zero. The as-sumptions of plane strain are realistic for long bodies (say, in the z direction) with constant cross-sectional area subjected to loads that act only in the x and/or y di-rections and do not vary in the z direction.

Two-Dimensional State of Stress and Strain The concept of two-dimensional state of stress and strain and the stress/strain

relationships for plane stress and plane strain are necessary to understand fully the development and applicability of the stiffness matrix for the plane stress/plane strain triangular element.

A two-dimensional state of stress is shown in the figure below. The infinitesi-mal element with sides dx and dy has normal stresses σz and σy acting in the x and y directions (here on the vertical and horizontal faces), respectively. The shear stress τxy acts on the x edge (vertical face) in the y direction. The shear stress τyx acts on the y edge (horizontal face) in the x direction.


Since τxy equals τyx, three independent stress exist:

Tx y xyσ σ σ τ⎡ ⎤= ⎣ ⎦

Recall, the relationships for principal stresses in two-dimensions are:

σ σ σ σσ τ σ

+ −⎛ ⎞= + + =⎜ ⎟

⎝ ⎠

22

1 max2 2x y x y

xy

min2

2

2 22στ

σσσσσ =+⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−

+= xy

yxyx

Also, θp is the principal angle which defines the normal whose direction is per-pendicular to the plane on which the maximum or minimum principle stress acts.

2tan2 xy

px y

τθ

σ σ=

−


The general two-dimensional state of strain at a point is show below.

The general definitions of normal and shear strains are:

xv

yu

xv

xu

xyyx ∂∂

+∂∂

=∂∂

=∂∂

= γεε

The strain may be written in matrix form as:

Tx y xyε ε ε γ⎡ ⎤= ⎣ ⎦


For plane stress, the stresses σz, τxz, and τyz are assumed to be zero. The stress-strain relationship is:

( ) ⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

xy

y

x

xy

y

x E

γεε

νν

ν

ντσσ

15.0000101

1 2

where

( )⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

=ν

νν

ν15.0000101

1][

2

ED

is called the stress-strain matrix (or the constitutive matrix), E is the modulus of elasticity, and ν is Poisson’s ratio.

For plane strain, the strains εz, γxz, and γyz are assumed to be zero. The stress-strain relationship is:

( )( ) ⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−

−+=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

xy

y

x

xy

y

x E

γεε

ννν

νν

νντσσ

5.0000101

211

where

( )( )⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−

−+=

ννν

νν

νν5.0000101

211][ ED

The partial differential equations for plane stress are:

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂−

∂∂+

=∂∂

+∂∂

yxv

yu

yu

xu 2

2

2

2

2

2

2

21 ν

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂−

∂∂+

=∂∂

+∂∂

yxu

yv

yv

xv 2

2

2

2

2

2

2

21 ν


Steps in the Formulation of the Element Stiffness Equations

Consider the problem of a thin plate subjected to a tensile load as shown in the figure below:

Step 1 - Discretize and Select Element Types Discretize the thin plate into a set of triangular elements. Each element is de-

fine by nodes i, j, and m.

We use triangular elements because boundaries of irregularly shaped bodies can be closely approximated, and because the expressions related to the triangular element are comparatively simple. This discretization is called a coarse-mesh generation if few large elements are used. Each node has two degrees of free-dom: displacements in the x and y directions. We will let ui and vi represent the node i displacement components in the x and y directions, respectively. The nodal displacements for an element with nodes i, j, and m are:


⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

m

j

i

ddd

d

where the nodes are ordered counterclockwise around the element, and

⎭⎬⎫

⎩⎨⎧

=i

ii v

ud

Therefore:

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

=

m

m

j

j

i

i

vuvuvu

d

Step 2 - Select Displacement Functions The general displacement function is:

⎭⎬⎫

⎩⎨⎧

=Ψ),(),(

yxvyxu

i

The functions u(x, y) and v(x, y) must be compatible with the element type.

Step 3 - Define the Strain-Displacement and Stress-Strain Relationships

The general definitions of normal and shear strains are:

xv

yu

xv

xu

xyyx ∂∂

+∂∂

=∂∂

=∂∂

= γεε

For plane stress, the stresses σz, τxz, and τyz are assumed to be zero. The stress-strain relationship is:

( ) ⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

xy

y

x

xy

y

x E

γεε

νν

ν

ντσσ

15.0000101

1 2


For plane strain, the strains εz, γxz, and γyz are assumed to be zero. The stress-strain relationship is:

( )( ) ⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−

−+=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

xy

y

x

xy

y

x E

γεε

ννν

νν

νντσσ

5.0000101

211

Step 4 - Derive the Element Stiffness Matrix and Equations Using the principle of minimum potential energy, we can derive the element

stiffness matrix.

dkf ][= This approach is better than the direct methods used for one-dimensional ele-ments.


The final assembled or global equation written in matrix form is:

dKF ][= where F is the equivalent global nodal loads obtained by lumping distributed edge loads and element body forces at the nodes and [K] is the global structure stiffness matrix.

Step 6 - Solve for the Nodal Displacements Once the element equations are assembled and modified to account for the

boundary conditions, a set of simultaneous algebraic equations that can be writ-ten in expanded matrix form as:

Step 7 - Solve for the Element Forces (Stresses) For the structural stress-analysis problem, important secondary quantities of

strain and stress (or moment and shear force) can be obtained in terms of the displacements determined in Step 6.


Derivation of the Constant-Strain Triangular Element Stiffness Matrix and Equations

Consider the problem of a thin plate subjected to a tensile load as shown in the figure below:

Step 1 - Discretize and Select Element Types The basic triangular element, shown below, is define by nodes i, j, and m la-

beled in a counterclockwise manner.

The nodal displacement matrix is:

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

=

m

m

j

j

i

i

vuvuvu

d

Step 2 - Select Displacement Functions We will select a linear displacement function for each triangular element, de-

fined as:

⎭⎬⎫

⎩⎨⎧

++++

=⎭⎬⎫

⎩⎨⎧

=Ψyaxaayaxaa

yxvyxu

i654

321

),(),(


A linear function ensures that the displacements along each edge of the element and the nodes shared by adjacent elements are equal.

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⎥⎦

⎤⎢⎣

⎡=

⎭⎬⎫

⎩⎨⎧

++++

=Ψ

6

5

4

3

2

1

654

321

10000001

aaaaaa

yxyx

yaxaayaxaa

i

To obtain the values for the a’s substitute the coordinated of the nodal points into the above equations:

= + + = + +1 2 3 4 5 6i i i i i iu a a x a y v a a x a y jjjjjj yaxaavyaxaau 654321 ++=++=

mmmmmm yaxaavyaxaau 654321 ++=++= Solving for the a’s and writing the results in matrix forms gives:

[ ] uxaaaa

yxyxyx

uuu

mm

jj

ii

m

j

i1

3

2

1

111

−=⇒⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

(xj, yj)

(xi, yi) (xm, ym)

uiuj

um

u

y

x

Linear representation of u(x, y)


The inverse of the [x] matrix is:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=−

mji

mji

mji

Ax

γγγβββααα

21][ 1

where

mm

jj

ii

yxyxyx

A111

2 =

is the determinant of [x].

( ) ( ) ( )jimimjmji yyxyyxyyxA −+−+−=2 where A is the area of the triangle and

jmimjimjmji xxyyxyyx −=−=−= γβα

mijimjmimij xxyyxyyx −=−=−= γβα

ijmjimjijim xxyyxyyx −=−=−= γβα The values of a may be written matrix form as:

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

m

j

i

mji

mji

mji

uuu

Aaaa

γγγβββααα

21

3

2

1

and

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

m

j

i

mji

mji

mji

vvv

Aaaa

γγγβββααα

21

6

5

4


We will now derive the displacement function in terms of the coordinates x and y.

[ ]⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

3

2

1

1aaa

yxu

Substituting the values for a into the above equation gives:

[ ]1 12

i j m i

i j m j

i j m m

uu x y u

Au

α α αβ β βγ γ γ

⎡ ⎤ ⎧ ⎫⎪ ⎪⎢ ⎥= ⎨ ⎬⎢ ⎥⎪ ⎪⎢ ⎥ ⎩ ⎭⎣ ⎦

Expanding the above equations

[ ]1 12

i i j j m m

i i j j m m

i i j j m m

u u uu x y u u u

Au u u

α α αβ β βγ γ γ

⎡ ⎤+ +⎢ ⎥= + +⎢ ⎥⎢ ⎥+ +⎣ ⎦

Multiplying the matrices in the above equations gives

( ) ( ) ( ) 1( , )2 i i i i j j j j m m m mu x y x y u x y u x y u

Aα β γ α β γ α β γ= + + + + + + + +

A similar expression can be obtained for the y displacement

( ) ( ) ( ) 1( , )2 i i i i j j j j m m m mv x y x y v x y v x y v

Aα β γ α β γ α β γ= + + + + + + + +

The displacements can be written in a more convenience form as:

mmjjiimmjjii vNvNvNyxvuNuNuNyxu ++=++= ),(),( where

( ) ( ) ( )yxA

NyxA

NyxA

N mmmmjjjjiiii γβαγβαγβα ++=++=++=21

21

21

The elemental displacements can be summarized as:

⎭⎬⎫

⎩⎨⎧

++++

=⎭⎬⎫

⎩⎨⎧

=Ψmmjjii

mmjjiii vNvNvN

uNuNuNyxvyxu

),(),(


In another form the above equations are:

0 0 0 [ ]

0 0 0

i

i

i j m j

i j m j

m

m

uv

N N N uN d

N N N vuv

ψ

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎡ ⎤

Ψ = =⎨ ⎬⎢ ⎥⎣ ⎦ ⎪ ⎪

⎪ ⎪⎪ ⎪⎩ ⎭

where

[ ] ⎥⎦

⎤⎢⎣

⎡=

mji

mji

NNNNNN

N000

000

The linear triangular shape functions are illustrated below

Step 3 - Define the Strain-Displacement and Stress-Strain Relationships Elemental Strains: The strains over a two-dimensional element are:

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

∂∂

+∂∂∂∂∂∂

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

xv

yu

yvxu

xy

y

x

γεε

ε

j

i m

1

Ni

y

x

j

i m 1

Nj

y

x

j

i m

1

Nm

y

x


Substituting our approximation for the displacement gives:

( )mmjjiix uNuNuNx

uxu

++∂∂

==∂∂

,

mxmjxjixix uNuNuNu ,,,, ++= where the comma indicates differentiation with respect to that variable. The de-rivatives of the interpolation functions are:

( )A

NA

NA

yxxA

N mxm

jxj

iiiixi 2222

1,,,

βββγβα ===++∂∂

=

Therefore:

( )mmjjii uuuAx

u βββ ++=∂∂

21

In a similar manner, the remaining strain terms are approximated as:

( )12 i i j j m m

v v v vy A

γ γ γ∂= + +

∂

( )mmmmjjjjiiii vuvuvuAx

vyu γβγβγβ +++++=

∂∂

+∂∂

21

We can write the strains in matrix form as:

or

[ ]⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

m

j

i

mji

ddd

BBBε

0 0 01 0 0 0

2

i

ix i j m

jy i j m

jxy i i j j m m

m

m

uuvxuvvy Auu v

y x v

ε β β βε ε γ γ γ

γ γ β γ β γ β

⎧ ⎫⎧ ⎫∂ ⎪ ⎪⎪ ⎪∂ ⎪ ⎪⎪ ⎪⎧ ⎫ ⎡ ⎤

⎪ ⎪⎪ ⎪∂⎪ ⎪ ⎢ ⎥== = =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥∂⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥⎩ ⎭ ⎣ ⎦⎪ ⎪ ⎪ ⎪∂ ∂+⎪ ⎪ ⎪ ⎪∂ ∂⎩ ⎭ ⎩ ⎭

0 0 01 0 0 0

2

i

ix i j m

jy i j m

jxy i i j j m m

m

m

uuvxuvvy Auu v

y x v

ε β β βε ε γ γ γ

γ γ β γ β γ β

⎧ ⎫⎧ ⎫∂ ⎪ ⎪⎪ ⎪∂ ⎪ ⎪⎪ ⎪⎧ ⎫ ⎡ ⎤

⎪ ⎪⎪ ⎪∂⎪ ⎪ ⎢ ⎥== = =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥∂⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥⎩ ⎭ ⎣ ⎦⎪ ⎪ ⎪ ⎪∂ ∂+⎪ ⎪ ⎪ ⎪∂ ∂⎩ ⎭ ⎩ ⎭


where

[ ] [ ] [ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

mm

m

m

m

jj

j

j

j

ii

i

i

i AB

AB

AB

βγγ

β

βγγ

β

βγγ

β0

0

210

0

210

0

21

These equations can be written in matrix form as:

][ dB=ε Stress-Strain Relationship: The in-plane stress-strain relationship is:

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

xy

y

x

xy

y

x

Dγεε

τσσ

][

where [D] for plane stress is:

( )⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

=ν

νν

ν15.0000101

1][

2

ED

and [D] for plane strain is:

( )( )⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−

−+=

ννν

νν

νν5.0000101

211][ ED

In-plane stress can be related to displacements by:

]][[ dBD=σ


Step 4 - Derive the Element Stiffness Matrix and Equations The total potential energy is defined as the sum of the internal strain energy U

and the potential energy of the external forces Ω:

spbp U Ω+Ω+Ω+=π where the strain energy is:

∫=V

T dVU 21 σε

or

∫=V

T dVDU ][21 εε

The potential energy of the body force term is:

∫ Ψ−=ΩV

Tb dVX

where Ψ is the general displacement function, and X is the body weight per unit volume.

The potential energy of the concentrated forces is:

Pd Tp −=Ω

where P are the concentrated forces, and d are the nodal displacements.

The potential energy of the distributed loads is:

∫ Ψ−=ΩS

Ts dST

where Ψ is the general displacement function, and T are the surface tractions. Then the total potential energy expression becomes:

1 [ ] [ ][ ] [ ] [ ] 2

T T T TT T Tp

V V S

d B D B d dV d N X dV d P d N T dSπ = − − −∫ ∫ ∫


The nodal displacements d are independent of the general x-y coordinates, therefore

1 [ ] [ ][ ] [ ] [ ] 2

T T T TT T Tp

V V S

d B D B dV d d N X dV d P d N T dSπ = − − −∫ ∫ ∫

We can define the last three terms as:

∫∫ ++=S

T

V

T dSTNPdVXNf ][][

Therefore:

1 [ ] [ ][ ]2

T TTp

V

d B D B dV d d fπ = −∫

Minimization of πp with respect to each nodal displacement requires that:

[ ] [ ][ ] 0p T

V

B D B dV d fdπ∂

= − =∂ ∫

The above relationship requires:

[ ] [ ][ ]T

V

B D B dV d f=∫

The stiffness matrix can be defined as:

[ ] [ ] [ ][ ]T

V

k B D B dV= ∫

For an element of constant thickness, t, the above integral becomes:

∫=A

T dydxBDBtk ]][[][][

The integrand in the above equation is not a function of x or y (global coordi-nates); therefore, the integration reduces to:

∫=A

T dydxBDBtk ]][[][][


or ]][[][][ BDBtAk T=

where A is the area of the triangular element. Expanding the stiffness relationship gives:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

][][][][][][][][][

][

mmmjmi

jmjjji

imijii

kkkkkkkkk

k

where each [kii] is a 2 x 2 matrix define as:

tABDBktABDBktABDBk mT

iimjT

iijiT

iii ]][[][][]][[][][]][[][][ === Recall:

[ ] [ ] [ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

mm

m

m

m

jj

j

j

j

ii

i

i

i AB

AB

AB

βγγ

β

βγγ

β

βγγ

β0

0

210

0

210

0

21

Step 5 - Assemble the Element Equations to Obtain the Global Equations and Introduce the Boundary Conditions

The global stiffness matrix can be found by the direct stiffness method.

∑=

=N

e

ekK1

)( ][][

The global equivalent nodal load vector is obtained by lumping body forces

and distributed loads at the appropriate nodes as well as including any concen-trated loads.

∑=

=N

e

efF1

)(

The resulting global equations are:


][ dKF = where d is the total structural displacement vector.

In the above formulation of the element stiffness matrix, the matrix has been derived for a general orientation in global coordinates. Therefore, no transforma-tion form local to global coordinates is necessary. However, for completeness, we will now describe the method to use if the local axes for the constant-strain triangular element are not parallel to the global axes for the whole structure.

To relate the local to global displacements, force, and stiffness matrices we

will use:

TkTkTffTdd T ˆˆˆ === The transformation matrix T for the triangular element is:

where C = cos θ and S = sin θ, and θ is shown in the figure above.


0 0 0 00 0 0 0

0 0 0 00 0 0 00 0 0 00 0 0 0

C SS C

C ST

S CC SS C

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥

= ⎢ ⎥−⎢ ⎥⎢ ⎥⎢ ⎥

−⎣ ⎦


Step 7 - Solve for Element Forces and Stress

Having solved for the nodal displacements, we can obtain strains and stresses in x and y directions in the elements by using:

[ ] [ ][ ] B d D B dε σ= =

Plane Stress Example

Consider the structure shown in the figure below.

Assume plane stress conditions. All coordinates are shown on the figure. Let E = 30 x 106 psi, ν = 0.25, and t = 1 in. Assume the element nodal displacements have been determined to be u1 = 0.0, v1 = 0.0025 in., u2 = 0.0012 in., v2 = 0.0, u3 = 0.0, and v3 = 0.0025 in. Determine the element stiffness matrix and the element stresses. First, we calculate the element β’s and γ’s as:

220110 −=−=−=−=−=−= jmimji xxyy γβ

0002)1(0 =−=−==−−=−= mijimj xxyy γβ

202101 =−=−=−=−−=−= ijmjim xxyy γβ Therefore, the [B] matrix is:


[ ]0 0 0 1 0 2 0 1 0

1 10 0 0 0 2 0 0 0 22 2(2)

2 1 0 2 2 1

i j m

i j m

i i j j m m

BA

β β βγ γ γ

γ β γ β γ β

⎡ ⎤ − −⎡ ⎤⎢ ⎥ ⎢ ⎥= = −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− − −⎣ ⎦⎣ ⎦

For plane stress conditions, the [D] matrix is:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−×

=375.0000125.0025.01

)25.0(11030][ 2

6

D

Substitute the above expressions for [D] and [B] into the general equations for the stiffness matrix:

]][[][][ BDBtAk T=

6

1 0 20 2 1

1 0.25 0 1 0 2 0 1 02 0 1(2)30 10 10.25 1 0 0 2 0 0 0 2

4(0.9375) 2(2)2 0 20 0 0.375 2 1 0 2 2 1

1 0 20 2 1

k

− −⎡ ⎤⎢ ⎥− −⎢ ⎥ − −⎡ ⎤ ⎡ ⎤⎢ ⎥× ⎢ ⎥ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦⎢ ⎥−⎢ ⎥

−⎣ ⎦ Performing the matrix triple product gives:

inlbk

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−−−−−−−

−−−−−−−

−−−

×=

375.425.175.01625.325.025.15.25.1225.05.075.05.15.1075.05.1

120412625.325.075.01375.425.125.05.05.1225.15.2

104 6

The in-plane stress can be related to displacements by:

]][[ dBD=σ


6

0.00.0025

1 0.25 0 1 0 2 0 1 00.001230 10 10.25 1 0 0 2 0 0 0 2

0.9375 2(2) 0.00 0 0.375 2 1 0 2 2 1

0.00.0025

x

y

xy

inin

in

σστ

⎧ ⎫⎪ ⎪⎪ ⎪⎧ ⎫ − −⎡ ⎤ ⎡ ⎤⎪ ⎪×⎪ ⎪ ⎢ ⎥ ⎢ ⎥= −⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥

⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦⎩ ⎭ ⎪ ⎪⎪ ⎪⎩ ⎭

The stresses are:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

psipsipsi

xy

y

x

000,15800,4200,19

τσσ

Recall, the relationships for principal stresses and principal angle in two-dimensions are:

max2

2

1 22στ

σσσσσ =+⎟⎟

⎠

⎞⎜⎜⎝

⎛ −+

+= xy

yxyx

min2

2

2 22στ

σσσσσ =+⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−

+= xy

yxyx

⎥⎦

⎤⎢⎣

⎡

−= −

yx

xyp σσ

τθ

2tan

21 1

Therefore:

( ) psi639,28000,152

800,4200,192

800,4200,19 22

1 =−+⎟⎠⎞

⎜⎝⎛ −

++

=σ

( ) psi639,4000,152

800,4200,192

800,4200,19 22

1 −=−+⎟⎠⎞

⎜⎝⎛ −

−+

=σ

op 3.32

800,4200,19)000,15(2tan

21 1 −=⎥⎦

⎤⎢⎣⎡

−−

= −θ


Treatment of Body and Surface Forces

The general force vector is defined as:

∫∫ ++=S

T

V

T dSTNPdVXNf ][][

Body Force

Let’s consider the first term of the above equation.

[ ] = ∫ Tb

V

f N X dV

where

⎧ ⎫= ⎨ ⎬⎩ ⎭

b

b

XX

Y

where Xb and Yb are the weight densities in the x and y directions, respectively. The force may reflect the effects of gravity, angular velocities, or dynamic inertial forces.

The integration of the fb is simplified if the origin of the coordinate system is chosen at the centroid of the element, as shown in the figure below. With the ori-gin placed at the centroid, we can use the definition of a centroid.

For a given thickness, t, the body force term becomes: [ ] [ ] T T

bV A

f N X dV t N X dA= =∫ ∫

0=∫A

x dA

0=∫A

y dA


Recall the interpolation functions for a place stress/strain triangle:

( ) ( ) ( )yxA

NyxA

NyxA

N mmmmjjjjiiii γβαγβαγβα ++=++=++=21

21

21

Therefore the terms in the integrand are:

0β γ= =∫ ∫i iA A

x dA y dA

and 23i j mAα α α= = =

The body force at node i is given as:

3⎧ ⎫

= ⎨ ⎬⎩ ⎭

bbi

b

XtAfY

The general body force vector is:

3

bix b

biy b

bjx bb

bjy b

bmx b

bmy b

f Xf Yf XtAff Yf Xf Y

⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪= =⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎩ ⎭⎩ ⎭

Surface Force

The third term in the general force vector is defined as:

[ ] = ∫ Ts

S

f N T dS


Let’s consider the example of a uniform stress p acting between nodes 1 and 3 on the edge of element 1 as shown in figure below.

The surface traction becomes:

0

⎧ ⎫ ⎧ ⎫= =⎨ ⎬ ⎨ ⎬

⎩ ⎭⎩ ⎭

x

y

p pT

p

and [N]T is:

1

1

2

2

3

3

00

0[ ]

00

0

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

T

NN

NN

NN

N

evaluated at x=a

Therefore, the traction force vector is:

1

1

2

20 0

3

3

00

00 0

00

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥ ⎧ ⎫

= ⎢ ⎥ ⎨ ⎬⎩ ⎭⎢ ⎥

⎢ ⎥⎢ ⎥⎣ ⎦

∫ ∫t L

s

NN

N pf dy dz

NN

N

evaluated at x=a


After some simplification, the traction force vector is:

1

2

0

3

0

0

0

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

∫L

s

N p

N pf t dy

N p

The interpolation function for i = 1 is:

( )12

α β γ= + +i i i iN x yA

For convenience, let’s choose the coordinate system shown in the figure below.

Recall:

α = −i j m j mx y y x

with i = 1, j = 2, and m = 3, we get

1 2 3 2 3α = −x y y x If we substitute the coordinates of the triangle show above in the above equation we get:


1 0α = Similarly, we can find:

1 10β γ= = a Therefore, the interpolation function, N1 is:

1 2=

ayNA

The remaining interpolation function, N2 and N2 are:

2 3( )2 2− −

= =L a x Lx ayN N

A A

Substituting the interpolation function in the traction force vector expression gives:

1

1

2

2

3

3

100

2 010

⎧ ⎫ ⎡ ⎤⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎢ ⎥⎪ ⎪= =⎨ ⎬ ⎢ ⎥⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎣ ⎦⎩ ⎭

s x

s y

s xs

s y

s x

s y

fff pLtffff

Explicit Expression for the Constant-Strain Triangle Stiffness Matrix

Usually the stiffness matrix is computed internally by computer programs, but since we are not computers, we need to explicitly evaluate the stiffness matrix. For a constant-strain triangular element, considering the plane strain case, recall that:

[ ] [ ] [ ][ ]= Tk tA B D B where [D] for plane strain is:


( )( )⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−

−+=

ννν

νν

νν5.0000101

211][ ED

Substituting the appropriate definition into the above triple product gives:

00

1 00

[ ] 1 004 (1 )(1 2 )

0 0 0.50

0

β γγ β

ν νβ γ

ν νγ βν ν

νβ γ

γ β

⎡ ⎤⎢ ⎥⎢ ⎥ −⎡ ⎤⎢ ⎥ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥+ − ⎢ ⎥ ⎢ ⎥−⎣ ⎦⎢ ⎥⎢ ⎥⎣ ⎦

i i

i i

j j

j j

m m

m m

tEkA

0 0 0

0 0 0β β β

γ γ γγ β γ β γ β

⎡ ⎤⎢ ⎥×⎢ ⎥⎢ ⎥⎣ ⎦

i j m

i j m

i i j j m m

Therefore the global stiffness matrix is:


The stiffness matrix is a function of the global coordinates x and y, the material properties, and the thickness and area of the element.

Finite Element Solution of a Plane Stress Problem

Consider the thin plate subjected to the surface traction shown in the figure below.

Assume plane stress conditions. Let E = 30 x 106 psi, ν = 0.30, and t = 1 in. De-termine the nodal displacements and the element stresses.

Discretization Let’s discretize the plate into two elements as shown below:

This level of discretization will probably not yield practical results for displace-ment and stresses: however; it is useful example for a longhand solution.


The tensile traction forces can be converted into nodal forces as follows:

1

1

2

2

3

3

1 1 5,0000 0 00 0 01,000 (1 )10

2 20 0 01 1 5,0000 0 0

s x

s y

s xs

s y

s x

s y

f lbff pLt psi in infff lbf

⎧ ⎫ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎪ ⎪= = = =⎨ ⎬ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎩ ⎭

The governing global matrix equations are:

[ ] =F K d Expanding the above matrices gives:

1 11

1 11

2 22

2 22

3 3 3

3 3 3

4 4 4

4 4 4

0000

[ ] [ ]5,000

05,000

0

⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪= = =⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎭⎩ ⎭ ⎩ ⎭ ⎩ ⎭

x xx

y yy

x xx

y yy

x x x

y y y

x x x

y y y

F dRF dRF dRF dR

K KF d dlbF d dF d dlbF d d

⎪⎪⎪⎪⎪

where [K] is an 8 x 8 matrix before deleting the rows and columns accounting for the boundary supports.

Assemblage of the Stiffness Matrix The global stiffness matrix is assembled by superposition of the individual ele-ment stiffness matrices. The element stiffness matrix is:

[ ] [ ] [ ][ ]= Tk tA B D B


For element 1: the coordinates are xi = 0, yi = 0, xj = 20, yj = 10, xm = 0, and ym = 10. The area of the triangle is:

The matrix [B] is:

0 0 01[ ] 0 0 0

2

β β βγ γ γ

γ β γ β γ β

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

i j m

i j m

i i j j m m

BA

We need to calculate the element β’s and γ’s as:

10 10 0 0 20 20β γ= − = − = = − = − = −i j m i m jy y x x

1 10 0 10 0 0 0β γ= − = − = = − = − =j m j i my y x x

0 10 10 20 0 20β γ= − = − = − = − = − =m i j m i jy y x x

Therefore, the [B] matrix is:

0 0 10 0 10 01 1[ ] 0 20 0 0 0 20

20020 0 0 10 20 10

−⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥− −⎣ ⎦

B in


61 0.3 0

30 10[ ] 0.3 1 00.91

0 0 0.35

⎡ ⎤× ⎢ ⎥= ⎢ ⎥

⎢ ⎥⎣ ⎦

D psi

2=

bhA

2(20)(10) 100 .2

= =A in


Therefore:

6

0 0 200 20 0

1 0.3 010 0 030(10 )[ ] [ ] 0.3 1 0

200(0.91) 0 0 100 0 0.35

10 0 200 20 10

−⎡ ⎤⎢ ⎥−⎢ ⎥ ⎡ ⎤⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥−⎢ ⎥

−⎣ ⎦

TB D


6

0 0 76 20 0

10 3 030(10 )[ ] [ ]200(0.91) 0 0 3.5

10 3 76 20 3.5

TB D

−⎡ ⎤⎢ ⎥− −⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥− −⎢ ⎥

−⎣ ⎦

The element stiffness matrix is:

[ ] [ ] [ ][ ]= Tk tA B D B therefore:

6

0 0 76 20 0

10 3 0(0.15)(10 )[ ] [ ][ ] 1(100)0.91 0 0 3.5

10 3 76 20 3.5

TtA B D B

−⎡ ⎤⎢ ⎥− −⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥− −⎢ ⎥

−⎣ ⎦

0 0 10 0 10 0

1 0 20 0 0 0 20200

20 0 0 10 20 10

−⎡ ⎤⎢ ⎥× −⎢ ⎥⎢ ⎥− −⎣ ⎦



1 1 3 3 2 2

140 0 0 70 140 700 400 60 0 60 4000 60 100 0 100 6075,000[ ]

0.91 70 0 0 35 70 35140 60 100 70 240 130

70 400 60 35 130 435

u v u v u v

k

− −⎡ ⎤⎢ ⎥− − −⎢ ⎥⎢ ⎥− −

= ⎢ ⎥− −⎢ ⎥⎢ ⎥− − −⎢ ⎥

− − −⎣ ⎦

For element 2: the coordinates are xi = 0, yi = 0, xj = 20, yj = 0, xm = 20, and ym = 10. The area of the triangle is:

We need to calculate the element β’s and γ’s as:

0 10 10 20 20 0β γ= − = − = − = − = − =i j m i m jy y x x

1 10 0 10 0 20 20β γ= − = − = = − = − = −j m j i my y x x

0 0 0 20 0 20β γ= − = − = = − = − =m i j m i jy y x x

Therefore, the [B] matrix is:

10 0 10 0 0 01 1[ ] 0 0 0 20 0 20

2000 10 20 10 20 0

−⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥− −⎣ ⎦

B in

2(20)(10) 100 .2

= =A in



61 0.3 0

30 10[ ] 0.3 1 00.91

0 0 0.35

⎡ ⎤× ⎢ ⎥= ⎢ ⎥

⎢ ⎥⎣ ⎦

D psi

Therefore:

6

10 0 00 0 10

1 0.3 010 0 2030(10 )[ ] [ ] 0.3 1 0

200(0.91) 0 20 100 0 0.35

0 0 200 20 0

−⎡ ⎤⎢ ⎥−⎢ ⎥ ⎡ ⎤⎢ ⎥− ⎢ ⎥= ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥⎢ ⎥⎣ ⎦

TB D


6

10 3 00 0 3.5

10 3 730(10 )[ ] [ ]200(0.91) 0 20 3.5

6 0 76 20 0

− −⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥−

= ⎢ ⎥−⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦

TB D

The element stiffness matrix is:

[ ] [ ] [ ][ ]= Tk tA B D B therefore:

6

10 3 00 0 3.5

10 3 7(0.15)(10 )[ ] [ ][ ] 1(100)0.91 0 20 3.5

6 0 76 20 0

− −⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥−

= ⎢ ⎥−⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦

TtA B D B


10 0 10 0 0 0

1 0 0 0 20 0 20200

0 10 20 10 20 0

−⎡ ⎤⎢ ⎥× −⎢ ⎥⎢ ⎥− −⎣ ⎦


1 1 4 4 3 3

100 0 100 60 0 600 35 70 35 70 0100 70 240 130 140 6075,000[ ]

0.91 60 35 130 435 70 4000 70 140 70 140 060 0 60 400 0 400

u v u v u v

k

− −⎡ ⎤⎢ ⎥− −⎢ ⎥⎢ ⎥− − −

= ⎢ ⎥− − −⎢ ⎥⎢ ⎥− −⎢ ⎥− −⎣ ⎦

Element 1:

Element 2:

1 1 2 2 3 3 4 4

28 0 28 14 0 14 0 0

0 80 12 80 12 0 0 0

28 12 48 26 20 14 0 0

14 80 26 87 12 7 0 0375,000[ ]

0.91 0 12 20 12 20 0 0 0

14 0 14 7 0 7 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

− −

− −

− − −

− − −=

− −

− −

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

u v u v u v u v

k

1 1 2 2 3 3 4 4

20 0 0 0 0 12 20 12

0 7 0 0 14 0 14 7

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0375,000[ ]

0.91 0 14 0 0 28 0 28 14

12 0 0 0 0 80 12 80

20 14 0 0 28 12 48 26

12 7 0 0 14 80 26 87

− −

− −

=− −

− −

− − −

− − −

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

u v u v u v u v

k


Using the superposition, the total global stiffness matrix is:

1 1 2 2 3 3 4 4

48 0 28 14 0 26 20 12

0 87 12 80 26 0 14 7

28 12 48 26 20 14 0 0

14 80 26 87 12 7 0 0375,000[ ]

0.91 0 26 20 12 48 0 28 14

26 0 14 7 0 87 12 80

20 14 0 0 28 12 48 26

12 7 0 0 14 80 26 87

− − −

− − −

− − −

− −=

− − −

− − −

− − −

− − −

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

u v u v u v u v

k

The governing global matrix equations are:

[ ] =F K d

1

1

2

2

48 0 28 14 0 26 20 12

0 87 12 80 26 0 14 7

28 12 48 26 20 14 0 0

14 80 26 87 12 7 0 0

0 26 20 12 48 0 28 14

26 0 14 7 0 87 12 80

20 14 0 0 28 12 48 26

12 7 0 0 14 80 26 87

375,0000.915,000

0500

0

− − −

− − −

− − −

− −

− − −

− − −

− − −

− − −

⎧ ⎫ ⎡⎪ ⎪ ⎢⎪ ⎪ ⎢⎪ ⎪ ⎢⎪ ⎪⎪ ⎪ =⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪ ⎣⎩ ⎭

x

y

x

y

RRRR

lb

lb

1

1

2

2

3

3

4

4

⎧ ⎫⎤⎪ ⎪⎥⎪ ⎪⎥⎪ ⎪⎥⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥ ⎨ ⎬⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎪ ⎪⎦ ⎩ ⎭

x

y

x

y

x

y

x

y

dddddddd

Applying the boundary conditions:

1 1 2 2 0= = = =x y x yd d d d

The governing equations are:

3

3

4

4

5,000 48 0 28 140 0 87 12 80375,000

0.915,000 28 12 48 260 14 80 26 87

x

y

x

y

dlbddlbd

⎧ ⎫−⎧ ⎫ ⎡ ⎤⎪ ⎪⎪ ⎪ ⎢ ⎥−⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥− −⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪− −⎩ ⎭ ⎣ ⎦ ⎩ ⎭


Solving the equations gives:

( )3

3 6

4

4

609.64.2

10663.7104.1

−

⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎩ ⎭⎩ ⎭

x

y

x

y

dd

indd

The exact solution for the displacement at the free end of the one-dimensional bar subjected to a tensile force is:

66

(10,000)20 670 1010(30 10 )

δ −= = = ××

PL inAE

The in-plane stress can be related to displacements by:

[ ][ ] =s D B d

( )2

1 0 0 0 0 1 0 0 0 0

2 (1 )0 0 0.5 1

ν β β βσ ν γ γ γ

νν γ β γ β γ β

⎧ ⎫⎪ ⎪⎪ ⎪⎡ ⎤⎡ ⎤⎪ ⎪⎪ ⎪⎢ ⎥⎢ ⎥= ⎨ ⎬⎢ ⎥⎢ ⎥− ⎪ ⎪⎢ ⎥⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

ix

iyi j m

jxi j m

jyi i j j m m

mx

my

dddEdAdd

Element 1:

( )

1

11 3 2

31 3 22

31 1 3 3 2 2

2

2

1 0 0 0 0 1 0 0 0 0

2 (1 )0 0 0.5 1



⎧ ⎫⎪ ⎪⎪ ⎪⎡ ⎤ ⎡ ⎤ ⎪ ⎪⎪ ⎪⎢ ⎥ ⎢ ⎥= ⎨ ⎬⎢ ⎥ ⎢ ⎥− ⎪ ⎪⎢ ⎥ ⎢ ⎥− ⎣ ⎦⎣ ⎦ ⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

x

y

x

y

x

y

dddEdAdd


6 6

0.00.0

1 0.3 0 0 0 10 0 10 0609.630(10 )(10 ) 0.3 1 0 0 20 0 0 0 20

0.96(200) 4.20 0 0.35 20 0 0 10 20 10

0.00.0

σστ

−

⎧ ⎫⎪ ⎪⎪ ⎪⎧ ⎫ −⎡ ⎤ ⎡ ⎤⎪ ⎪⎪ ⎪ ⎢ ⎥ ⎢ ⎥= −⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥

⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦⎩ ⎭ ⎪ ⎪⎪ ⎪⎩ ⎭

x

y

xy

The stresses are:

1,0053012.4

σστ

⎧ ⎫ ⎡ ⎤⎪ ⎪ ⎢ ⎥=⎨ ⎬ ⎢ ⎥⎪ ⎪ ⎢ ⎥⎣ ⎦⎩ ⎭

x

y

xy

psipsipsi

Element 2:

( )

1

11 4 3

41 4 32

41 1 4 4 3 3

3

3

1 0 0 0 0 1 0 0 0 0

2 (1 )0 0 0.5 1



⎧ ⎫⎪ ⎪⎪ ⎪⎡ ⎤ ⎡ ⎤ ⎪ ⎪⎪ ⎪⎢ ⎥ ⎢ ⎥= ⎨ ⎬⎢ ⎥ ⎢ ⎥− ⎪ ⎪⎢ ⎥ ⎢ ⎥− ⎣ ⎦⎣ ⎦ ⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

x

y

x

y

x

y

dddEdAdd

6 6

0.00.0

1 0.3 0 10 0 10 0 0 0663.730(10 )(10 ) 0.3 1 0 0 0 0 20 0 20

0.96(200) 104.10 0 0.35 0 10 20 10 20 0

609.64.2

σστ

−

⎧ ⎫⎪ ⎪⎪ ⎪⎧ ⎫ ⎡ ⎤ ⎡ ⎤⎪ ⎪⎪ ⎪ ⎢ ⎥ ⎢ ⎥= −⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥

⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦⎩ ⎭ ⎪ ⎪⎪ ⎪⎩ ⎭

x

y

xy

The stresses are:

9951.22.4

σστ

⎧ ⎫ ⎡ ⎤⎪ ⎪ ⎢ ⎥= −⎨ ⎬ ⎢ ⎥⎪ ⎪ ⎢ ⎥−⎣ ⎦⎩ ⎭

x

y

xy

psipsipsi


The principal stresses and principal angle are:

22

1995 1.2 995 1.2 ( 2.4) 995

2 2− +⎛ ⎞= + + − =⎜ ⎟

⎝ ⎠s psi

22

2995 1.2 995 1.2 ( 2.4) 1.1

2 2σ − +⎛ ⎞= − + − = −⎜ ⎟

⎝ ⎠psi

11 2( 2.4) 02 995 1.2

θ − −⎡ ⎤= ≈⎢ ⎥+⎣ ⎦o

p tan


WinFElt FEM Program To demonstrate a simple FEM solution for CST we will use the FElt (Finite ELe-menT) program. FElt is a free system for introductory level finite element analy-sis. It is primarily intended as a teaching tool for introductory type courses in finite elements - probably in the mechanical/structural/civil fields. In a command line environment, FElt uses an intuitive, straightforward input syntax to describe prob-lems. It also includes a graphical user interface for workstations that allows the user to set-up, solve and post-process the problem in a single CAD-like environ-ment. The Windows interface, WinFElt, consists of an editor and an encapsula-tor for the command line applications. It has some graphical post-processing ca-pabilities. The current version of FElt knows how to solve linear static and dynamic struc-tural and thermal analysis problems; it can also do modal and spectral analysis for dynamic problems. FElt's element library currently contains fourteen ele-ments. FElt uses an intuitive, ASCII based syntax for problem definition. This powerful syntax allows you to substitute analytic functions in place of numeric values (sin(60) instead of 0.866025) and even more importantly allows for time-dependent forcing and boundary conditions to be specified as analytic functions of time or in the more traditional fashion as a series of discrete time, magnitude pairs. This feature makes it quite easy to specify a wide range of functions. We can generate a color plot of stress contours for the above wrench problem with one simple menu selection after we have solved the problem in velvet. Dis-placement contours, and two- and three-dimensional wire frame plots of the dis-placed shape are also available post-processing options.


WinFElt Manual Version 1.1.3 Problem description

The problem description section is used to define the problem title and the number of nodes and ele-ments in the problem. The problem description section is the only section which you cannot repeat within a given input file. The format for a problem-description is given below. problem description [ title = string ] [ nodes = integer ] [ elements = integer ] [ analysis = static | transient | modal | static-thermal | transient-thermal | spectral ]

nodes = integer elements = integer

These numbers will be used for error checking so the specifications given here must match the actual number of nodes and elements given in the definition sections. Note that the definitions for nodes and elements do not have to be given in numerical order, as long as nodes 1 ... m and elements 1 ... n (where m is the number of nodes and n is the number of elements) all get defined in one of the element and node definition sections in the file.

analysis = problem type

Defines the type of problem that you wish to solve. Currently it can either be static, transient, static-substitution, modal, static-thermal, transient-thermal, or spectral. If you do not specify anything, static analysis will be assumed.

Analysis parameters

The analysis parameters section is required only if you are doing some type of transient, modal, or spectral analysis (e.g., analysis=transient, analysis=spectral in the problem description section). For modal analysis it is simply used to set the type of element mass matrices that will be formed, but for tran-sient and spectral analyses it contains information that further defines the problem and the parameters for the numerical integration in time. analysis parameters [ alpha = expression ] [ beta = expression ] [ gamma = expression ] [ step = expression ] [ stop = expression ] [ Rm = expression ] [ Rk = expression ] [ nodes = [ node-list] ] [ dofs = [ dof-list] ] [ mass-mode = limped | consistent ] The alpha, beta, and gamma parameters are used in numerical integration schemes (transient and tran-sient-thermal analysis). start, stop, and step define the range of time or frequency interest for transient or spectral analyses. In transient analyses, start is meaningless and duration and dt can be used as ali-ases for stop and step, respectively. Rk and Rm are global Rayleigh (stiffness and mass) damping pro-portionality constants. The node-list is a comma or white space separated list of node numbers that are of interest in the analysis. Similarly, the dof-list is a list of the degrees of freedom (Tx, Ty , Tz, Rx, Ry, and Rz) that are of interest.


An object-definition section defines objects of a specified type. Objects include nodes, elements, materi-als, constraints, forces, and distributed loads. Each of these types of objects is discussed below. Multiple object-definition sections are allowed and the sections may occur in any order.

Nodes

Nodes are points in Cartesian space to which elements are attached. A node must have a constraint and may have an optional force. A node is identified by a natural number. The syntax is as follows: nodes node-definitions where a node-definition takes the following form: node-number [ x = expression ] [ y = expression ] [ z = expression ] [ constraint = constraint-name ] [ force = force-name ] [ mass = expression ] The node-number starts the definition. Each node must have a unique number. If a Cartesian coordinate is not given then the coordinate of the previous node is used. Similarly, if no constraint is given then the constraint applied to the previous node is used. As above, the assignments can appear in any order and any number of times. As indicated above, some objects are identified by their name and some by their number. Elements and nodes have numbers while materials, forces, loads, and constraints have names.

Elements

Elements are linear, planar, or solid objects which are attached to nodes. Each element must have a ma-terial and may have optional loads. Furthermore, each element has a type, or definition. Like nodes, ele-ments are identified by a unique natural number. Elements of specific type are defined with the following syntax: element-type elements element-definition where an element-type is one of the following: spring truss beam beam3d CSTPlaneStrain CSTPlaneStress iso2d-PlaneStrain iso2d-PlaneStress quad-PlaneStrain quad-PlaneStress timoshenko htk brick ctg rod and an element-definition has the following form: element-number nodes = [ node-list ] [ material = material-name ] [ load = load-name-list ] The element-number starts the definition. Each element must have a unique number. If no material is given then the material applied to the previous element is used. The load-name-Iist is a list of up to three loads to apply to the element. The node-list is a comma or white space separated list of node numbers.


Each type of element requires a certain number of nodes and in some cases a special "null node" which is numbered zero may be used to indicate a gap or filler in the list.

Materials

Elements are made of a type of material. Each material has a name and certain physical properties not all of which may be used by anyone element. The syntax for defining materials is as follows: material properties material-definitions where material-definition has the following form: material-name [ E = expression ] # Young's modulus [ Ix = expression ] # moment of inertia about x-x axis [ Iy = expression ] # moment of inertia about y-y axis [ Iz = expression ] # moment of inertia about z-z axis [ A = expression ] # cross-sectional area [ J = expression] # polar moment of inertia [ G = expression ] # bulk (shear) modulus [ t = expression ] # thickness [ rho = expression ] # density [ nu = expression ] # Poisson's ratio [ kappa = expression ] # shear force correction [ Rk = expression ] # Rayleigh damping coefficient (K) [ Rm = expression ] # Rayleigh damping coefficient (M) [ Kx = expression ] # thermal conductivity in the x-direction [ Ky = expression ] # thermal conductivity in the y-direction [ Ky = expression ] # thermal conductivity in the z-direction [ c = expression ] # heat capacitance The material-name starts the definition. If an attribute of a material is not specified then that attribute is zero. The assignments may occur in any order.

Constraints

Constraints are applied to nodes to indicate about which axes a node can move. The syntax for defining a constraint is as follows: constraints constraint -definitions where constraint-definition has the following form: constraint-name [ tx = c I u I expression ] # boundary translation along x axis [ ty = c I u I expression ] # boundary translation along y axis [ tz = c I u I expression ] # boundary translation along z axis [ rx = c I u I expression I h ] # boundary rotation about x axis [ ry = c I u I expression I h ] # boundary rotation about y axis [ rz = c I u I expression I h ] # boundary rotation about z axis [ itx = expression ] # initial displacement along x axis [ ity = expression ] # initial displacement along y axis [ itz = expression ] # initial displacement along z axis [ irx = expression ] # initial rotation about x axis [ iry = expression ] # initial rotation about y axis [ irz = expression ] # initial rotation about z axis


[ vx = expression ] # initial velocity along x axis [ vy = expression ] # initial velocity along y axis [ vz = expression ] # initial velocity along z axis [ ax = expression ] # initial acceleration. along x axis [ ay = expression ] # initial acceleration. along y axis [ az = expression ] # initial acceleration. along z axis The constraint-name starts the definition. A value of c for a boundary condition indicates that the axis is constrained; a value of u indicates that the axis is unconstrained. An expression indicates a displacement (non-zero) boundary condition and may contain the t variable for time varying boundary conditions in tra-nsient analysis problems. The initial displacement, velocity and acceleration specifications are only used in transient problems. A value of h for a rotational boundary condition indicates a hinge. By default, all axes are unconstrained.

Forces

Forces, or point loads, may be applied to nodes. The syntax for a force definition is as follows: forces force-definitions where a force-definition has the following form: force-name [ Fx = expression ] # force along x axis [ Fy = expression ] # force along y axis [ Fz = expression ] # force along z axis [ Mx = expression ] # moment about x axis [ My = expression ] # moment about y axis [ Mz = expression ] # moment about z axis [ Sfx = expression ] # frequency-domain spectra of force along x axis [ Sfy = expression ] # frequency-domain spectra of force along y axis [ Sfz = expression ] # frequency-domain spectra of force along z axis [ Smx = expression ] # frequency-domain spectra of moment about x axis [ Smy = expression ] # frequency-domain spectra of moment about y axis [ Smz = expression ] # frequency-domain spectra of moment about z axis The force-name starts the definition. If the force or moment is not specified then it is assumed to be zero. The expressions for forces may be time-varying. Time-varying expressions include the single variable t to represent the current time in the solution of a dynamic problem or consist of a list of discrete (time, value) pairs. Frequency varying expressions for spectra can also use w to represent the independent variable (radial frequency).

Loads

Distributed loads, or loads for short, are applied to elements. The syntax for a defining a distributed load is as follows: distributed loads load-definitions where a load-definition has the following form: load-name [ direction = dir ] # direction [ values = pair-list ] # local nodes and magnitudes The load-name starts the definition. The dir is one of LocalX, LocalY, LocalZ (local coordinate system), GlobalX, GlobalY, GlobalZ (global coordinate system), parallel, or perpendicular. The pair-list is a se-


quence of pairs. A pair is a node number and an expression enclosed in parentheses. The node number refers to the position within the element rather than referring to an actual node.

Expressions

An expression can be either constant or time-varying. As discussed above, time-varying expressions con-tain the variable t or consist of a list of discrete (time, value) pairs. If a time-varying expression is given where a constant expression is expected, the expression is evaluated at time zero. An expression has one of the following forms, where all operators have the precedences and associativities given to them in the C programming language. expression ? expression: expression # in-line conditional expression II expression # logical or expression && expression # logical and expression I expression # integer inclusive or expression êxpression # integer exclusive or expression & expression # integer and expression == expression # equality expression != expression # inequality expression < expression # less than expression > expression # greater than expression <= expression # less than or equal expression >= expression # greater than or equal expression << expression # integer shift left expression >> expression # integer shift right expression + expression # addition expression -expression # subtraction expression * expression # multiplication expression / expression # division expression % expression # integer remainder -expression # arithmetic negation ! expression # logical negation ~ expression # integer bitwise negation ( expression ) # enforce precedence sin ( expression ) # sine cos ( expression ) # cosine tan ( expression ) # tangent pow ( expression, expression ) # power (exponentiation) exp ( expression ) # exponential log ( expression ) # natural logarithm log10 ( expression ) # base-IO logarithm sqrt ( expression ) # square root hypot ( expression, expression ) # Euclidean distance floor ( expression ) # floor ceil ( expression ) # ceiling Cmod ( expression, expression ) # floating point remainder Cabs ( expression ) # absolute value number # literal value t # current time Finally, a discretely valued expression has the following syntax, where the optional + indicates that the list represents one cycle of an infinite waveform. ( expression ',' expression ) ...[ + ]


Below are a few examples of loading expressions

5 10 15 20 t(s)

500

F (lbs)

F (0,0.0) (4.99,500.0)= +

F 100 * fmod( ,5.0)t=

1 2 t(s)

1500

F (lbs)

F (0,0.0) (0.5,1500.0)(1,0)= +

F (fmod( ,1.0) 0.5 ? 3000 * fmod( ,0.5) :3000 * (0.5 - fmod( ,0.5)))

t tt

= <

2 4 6 8 t(s)

2000

F (lbs)

F (fmod( ,4.0) 2.0 ? 2000 : 0)t= <

F (0,2000) (1.99,2000) (2,0) (3.99,0)= +

12 t(s)

1500

F (lbs)

F (0,0.0) (3,650) (9,650) (12,0) (13,0)=

F 3 ? 650/3 * :( 9 ? 650 : ( 12 ? 650/3 * (12 ) : 0))

t tt t t

= << <= −

10 9 3


WinFElt Example 1: Use WinFElt to determine the nodal displacements and the element stresses.

Assume plane stress conditions. Let E = 30 x 106 psi, ν = 0.30, and t = 1 in. Con-sider the following discretization of two plane stress CST.

WinFElt input file: problem description title="CST Sample Problem (Logan 6.2, p.291)" nodes=4 elements=2 nodes 1 x=0 y=0.0 constraint=pin 2 x=0 y=10.0 constraint=pin 3 x=20.0 y=10.0 constraint=free force=point 4 x=20.0 y=0.0 constraint=free force=point CSTPlaneStress elements 1 nodes=[1,3,2] material=steel 2 nodes=[1,4,3] material=steel material properties steel e=30e06 nu=0.30 t=1.0 forces point fx=5000 constraints pin tx=c ty=c free tx=u ty=u end


WinFElt output file: ** CST Sample Problem (Logan 6.2, p.291) ** Nodal Displacements ----------------------------------------------------------------------------- Node # DOF 1 DOF 2 DOF 3 DOF 4 DOF 5 DOF 6 ----------------------------------------------------------------------------- 1 0 0 0 0 0 0 2 0 0 0 0 0 0 3 0.00060958 4.1633e-06 0 0 0 0 4 0.0006637 0.00010408 0 0 0 0 Element Stresses ------------------------------------------------------------------------------- 1: 1004.8 301.44 2.4019 1004.8 301.43 0.19566 2: 995.2 -1.201 -2.4019 995.2 -1.2068 -0.13812 Reaction Forces ----------------------------------- Node # DOF Reaction Force ----------------------------------- 1 Tx -5000 1 Ty -3002.4 2 Tx -5000 2 Ty 3002.4

The results are similar to those presented in the textbook on page 298. In addi-tion, WinFElt has some graphical visualization capabilities. By selecting the ap-propriate options on the Controls menu, a color contour of the displacements can be plotted.


If more elements are utilized it may be possible to visualize the smoothed stress contours.

Note that the stress contours are not correct and most likely due to the fact the CST elements do not provide continuous inter-element stress values. WinFElt Example 2: Use WinFElt to determine the nodal displacements and the element stresses for the thin plate shown below.

Assume plane stress conditions. Let E = 210 GPa, ν = 0.30, and t = 5 mm. Con-sider the following discretization of four plane stress CST.


WinFElt input file: problem description title="CST Sample Problem (Logan 6.11, p.305)" nodes=5 elements=4 nodes 1 x= 0.0 y= 0.0 constraint=free 2 x=400.0 y= 0.0 constraint=free 3 x=400.0 y=400.0 constraint=pin 4 x= 0.0 y=400.0 constraint=pin 5 x=200.0 y=200.0 constraint=free force=point CSTPlaneStress elements 1 nodes=[1,2,5] material=steel 2 nodes=[2,3,5] 3 nodes=[3,4,5] 4 nodes=[1,5,4] material=steel material properties steel e=210 nu=0.30 t=5.0 forces point fy=-30 constraints pin tx=c ty=c free tx=u ty=u end

WinFElt output file: ** CST Sample Problem (Logan 6.11, p.305) ** Nodal Displacements ----------------------------------------------------------------------------- Node # DOF 1 DOF 2 DOF 3 DOF 4 DOF 5 DOF 6 ----------------------------------------------------------------------------- 1 -0.0016515 -0.012505 0 0 0 0 2 0.0016515 -0.012505 0 0 0 0 3 0 0 0 0 0 0 4 0 0 0 0 0 0 5 4.9049e-19 -0.016279 0 0 0 0 Element Stresses ------------------------------------------------------------------------------- 1: 0.00059891 -0.003784 -1.0875e-19 0.00059891 -0.003784 -1.4217e-15 2: 0.0031171 0.0075 0.003716 0.0096226 0.00099449 -29.735 3: 0.0056352 0.018784 -1.9808e-19 0.018784 0.0056352 0 4: 0.0031171 0.0075 -0.003716 0.0096226 0.00099449 29.735 Reaction Forces ----------------------------------- Node # DOF Reaction Force ----------------------------------- 3 Tx 6.2341 3 Ty 15 4 Tx -6.2341 4 Ty 15


WinFElt Example 3: Use WinFElt to determine the nodal displacements and ro-tations, element forces, and the reactions for the rigid frame shown below (as-sume E = 3 x 106 psi, A = 15 in2, and I = 250 in4).

WinFElt input file: problem description title="Simple Frame Problem (Logan 5.8 p.243)" nodes=4 elements=3 nodes 1 x= 0.0 y= 0.0 constraint=fixed 2 x= 0.0 y=240.0 constraint=free 3 x=240.0 y=240.0 constraint=free 4 x=240.0 y= 0.0 constraint=fixed beam elements 1 nodes=[1,2] material=steel load=uni 2 nodes=[2,3] 3 nodes=[3,4] material=steel material properties steel E=30e06 A=15 Ix=250 distributed loads uni direction=perpendicular values=(1,-20.833) (2,-20.833) constraints fixed tx=c ty=c rz=c free tx=u ty=u rz=u end

WinFElt output file: ** Simple Frame Problem (Logan 5.8 p.243) ** Nodal Displacements ----------------------------------------------------------------------------- Node # DOF 1 DOF 2 DOF 3 DOF 4 DOF 5 DOF 6 ----------------------------------------------------------------------------- 1 0 0 0 0 0 0 2 0.24746 0.00038057 0 0 0 -0.00019435 3 0.2469 -0.00038057 0 0 0 -0.00072536 4 0 0 0 0 0 0 Element Stresses ------------------------------------------------------------------------------- 1: -713.57 3959.2 2.8118e+05 713.57 1040.7 69034 2: 1040.7 -713.57 -69034 -1040.7 713.57 -1.0222e+05 3: 713.57 1040.7 1.0222e+05 -713.57 -1040.7 1.4756e+05


Reaction Forces ----------------------------------- Node # DOF Reaction Force ----------------------------------- 1 Tx -3959.2 1 Ty -713.57 1 Rz 2.8118e+05 4 Tx -1040.7 4 Ty 713.57 4 Rz 1.4756e+05


Problems

16. Do problems 6.5, 6.6, 6.9, 6.10, and 6.13 on pages 301 - 306 in your text-book “A First Course in the Finite Element Method” by D. Logan.

17. Rework the plane stress problem given on page 291 in your textbook “A First Course in the Finite Element Method” by D. Logan using WinFElt to do analysis. Start with the simple two element model. Continuously refine your discretization by a factor of two each time until your FEM solution is in agreement with the exact solution for both displacements and stress. How many elements did you need?


Practical Considerations in Modeling; Interpreting Results

Introduction

In this section we will discuss some modeling considerations and guidelines, including mesh size, natural subdivisions, and the use of symmetry and associ-ated boundary conditions,

We will also introduce the concept of static condensation, which enables us to apply the basis of the CST stiffness matrix to a quadrilateral element.

Finite Element Modeling

Finite element modeling is partly an art guided by visualizing physical interac-tions taking place within a body, In modeling the user is confronted with the diffi-cult task of understanding physical behavior taking place and understanding the physical behavior of various elements available for use. Matching the appropriate finite element to the physical behavior being modeling is one of many decisions that must be made by the modeler. Understanding the boundary conditions can be one of the most difficult tasks a modeler must face in construction a useable finite element model.

Aspect Ratio and Element Shape The aspect ratio is define as the ratio of the longest dimension to the shortest

dimension of a quadrilateral element, In general, as the aspect ratio increases, the inaccuracy of the finite element solution increases,

Consider the five different finite element model shown in the figure below. A plot of the resulting error in the displacement at point A of the beam verse aspect ratio is given. In addition, the numerical answers are given in the following table.



Case Aspect Ratio

Number of Nodes

Number of Elements

Point A Point B % Error

1 1.1 84 60 -1.093 -0.346 5.2 2 1.5 85 64 -1.078 -0.339 6.4 3 3.6 77 60 -1.014 -0.238 11.9 4 6.0 81 64 -0.886 -0.280 23.0 5 24.0 85 64 -0.500 -0.158 56.0

Exact Solution -1.152 -0.360

In general, elements that yield the best results are compact and regular in

shape will: (1) aspect ratios near one; and (2) corner angles of quadrilaterals near 90°.


Use of Symmetry The use of symmetry will often expedite the modeling of a problem. Symme-

try allows us to consider a reduced problem instead of the actual problem. This will allow us to use a finer discretization of element with less computational cost.

Natural Subdivisions at Discontinuities There are a variety of natural subdivisions for finite element discretizations.

For example, natural locations of nodes occur at concentrated loads or disconti-nuities in loading, other types of boundary conditions, and abrupt changes in ge-ometry of materials.

Sizing of Elements and Mesh Refinement A discretization depends on the geometry of the structure, the loading, and the

boundary conditions. For example, areas of high, rapidity changing stresses re-quire a finer mesh then regions where the stress is constant.


Here the use of symmetry is applied to a soil mass subjected to a foundation

loading (66 nodes and 50 elements). Note that at the place of symmetry the dis-placements in the direction perpendicular to the plane must be zero. This is mod-eled by rollers at nodes 2 - 6.


The figure above illustrates the use of triangular elements for transitions from smaller quadrilaterals to larger quadrilaterals. The transitions are required since CST elements do not have immediate nodes along their edges. If an element had an intermediate node, the resulting equations would be inconsistent with the en-ergy formulation for the CST equations.




Infinite Medium A typical example of infinite medium is a soil foundation problem. The guide-

line for the finite element model is that enough material must be included such that the displacements at nodes and stresses within the elements become negli-gibly small at locations far from the foundation load. The level of discretization can be determined by a trail-and-error procedure in which the horizontal and ver-tical distances from the load are varied and the resulting effects on the displace-ments and stresses are observed. For a homogeneous soil mass, experience has shown the influence of a footing becomes insignificant if the horizontal dis-tance of the model is taken as approximately four and six times the width of the footing and the vertical distance is taken as approximately four to ten times the width of the footing.

Checking the Model The discretized finite element model should be checked carefully before re-

sults are computed. Ideally, a model should be checked by an analyst not in-volved in the preparation of the model, who is then more likely to be objective. Preprocessors with their detailed graphical display capabilities now make it com-paratively easy to find errors, particularly with a misplaced node or missing ele-ment or a misplaced load or boundary condition. Preprocessors include the abil-ity to color, shrink, rotate, and section a model mesh.

Checking the Results and Typical Postprocessor Results An analyst should probability spend as much time processing, checking, and

analyzing results as spent in data preparation.


Plate with a hole in the center.

Deformed shape of plate superimposed over an unreformed shape


The wrench in this example is modeled by 307 constraint strain triangular ele-ments (plane stress assumption). Below is a plot of the deformed shape of the wrench over the original mesh.


On the left is a plot of the stress in the wrench and the plot on the right is a

plot of the defected shape. This analysis was preformed using WinFElt finite element structural analysis system.


Equilibrium and Compatibility of Finite Element Results

An approximate solution for a stress analysis problem using the finite element method based on assumed displacement fields does not generally satisfy all the requirements for equilibrium and compatibility that an exact theory-of-elasticity solution satisfies. However, remember that relatively few exact solutions exist. Hence, the finite element method is a very practical one for obtaining reasonable, but approximate, numerical solutions.

We now describe some of the approximations generally inherent with finite element solutions.

1. Equilibrium of nodal forces and moments is satisfied. This is true be-cause the global equation F = Kd is a nodal equilibrium equation whose solution for d is such that the sums of all forces and moments applied to each node are zero. Equilibrium of the whole structure is also satisfied because the structure reactions are included in the global forces, and hence, in the nodal equilibrium equations.

2. Equilibrium within an element is not always satisfied. However, for the constant-strain bar and the constant-strain triangle, element equilibrium is satisfied. Also the cubic displacement function is shown to satisfy the basic beam equilibrium differential equation, and hence, to satisfy ele-ment force and moment equilibrium.

3. Equilibrium is not usually satisfied between elements. A differential ele-ment including parts of two adjacent finite elements is usually not in equilibrium (see the figure below). For line elements, such as used for truss and frame analysis, interelement equilibrium is satisfied. However, for two- and three-dimensional elements, interelement equilibrium is not usually satisfied. Also, the coarseness of the mesh causes this lack of interelement equilibrium to be even more pronounced.


4. Compatibility is satisfied within an element as long as the element dis-

placement field is continuous; hence, individual elements do not tear apart.

5. In the formulation of the element equations, compatibility is invoked at the nodes. Hence, elements remain connected at their common nodes. Similarly, the structure remains connected to its support nodes because boundary conditions are invoked at these nodes.

6. Compatibility may or may not be satisfied along interelement bounda-ries. For line elements such as bars and beams, interelement bounda-ries are merely nodes. The constant-strain triangle remain straight sided when deformed and therefore, interelement compatibility exists for these elements. Incompatible elements, those that allow gaps or overlaps be-tween elements, can be acceptable and even desirable.


Convergence of Solution

When the mesh size is reduced - that is the number of elements is increased - we are ensured of monotonic convergence of the solution when compatible and complete displacement functions are sued.

Case Number of Elements Number of Nodes Aspect Ratio

Point A

1 12 21 2 -0.740 2 24 39 1 -0.980 3 32 45 3 -0.875 4 64 85 1.5 -1.078 5 80 105 1.2 -1.100

Exact Solution -1.152


Interpretation of Stresses

In the stiffness or displacement formulation of the finite element method, used in this course, the primary quantities determined are the interelement nodal dis-placements of the assemblage. Secondary quantities, such as stress and strain, are computed based on these nodal displacements. In the case of the bar and constant-strain triangles, stresses are constant over the element. For these ele-ments, it is common practice to assign the stress to the centroid of the element with acceptable results.

An alternative procedure sometimes is to use an average (possibly weighted) value of the stresses evaluated at each node of the element. This averaging method is often based interpolating the element nodal values using the element shape functions. The averaging method is called smoothing. While the results from smoothing may be pleasing to the eye, they may not indicate potential prob-lems with the model and the results. You should always view the unsmoothed contour plots as well. Highly discontinuous contours between elements in a re-gion of an unsmoothed plot indicate modeling problems and typically require ad-ditional refinement of the element mesh in the suspect region.

Static Condensation

Let’s consider the concept of static condensation and used it to develop the stiffness matrix of a quadrilateral element. Consider a general quadrilateral ele-ment as shown below.


An imaginary node 5 is temporary introduced at the intersection of the diagonals of the quadrilateral to create four triangles. We can superimpose the stiffness matrices of the four triangles to create the stiffness matrix of the quadrilateral element, where the internal imaginary node 5 degrees of freedom are said to be condensed out so that they never enter into the final equations.

Let’s start by partitioning the equilibrium equations:

where di is the vector of displacements corresponding to the imaginary internal node, Fi is the vector of loads at the internal node, and de and Fe are the actual displacements and loads, respectively. Rewriting the above equations we gives:

+ =11 12e i eK d K d F

21 22e i iK d K d F+ = Solving for di gives:

1 122 21 22i e id K K d K F− −= − +

Substituting the above equation, we obtain the condensed equilibrium equation:

=c c ck d F where

−= − 111 12 22 21ck K K K K

−= − 112 22c e iF F K K F

where kc and Fc are called the condensed stiffness matrix and the condensed load vector, respectively. An advantage of the four-CST quadrilaterals is that the solution becomes less dependent on the skew of the subdivision mesh. The skew means a directional stiffness bias that is built into a model through certain discretization patterns.

11 12

21 22

e e

i i

K K d FK K d F⎡ ⎤ ⎧ ⎫ ⎧ ⎫

=⎨ ⎬ ⎨ ⎬⎢ ⎥⎣ ⎦ ⎩ ⎭ ⎩ ⎭


The stiffness matrix of a typical triangular element, call it element 1, labeled with nodes 1, 2, and 5 is given as:

(1) (1) (1)11 12 15

(1) (1) (1) (1)21 22 25(1) (1) (1)51 52 55

k k kk k k k

k k k

⎡ ⎤⎢ ⎥⎡ ⎤ = ⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦

where kij

(1) is a 2 x 2 matrix. The assembled stiffness matrix for the quadrilateral is:


Example Problem

Consider the quadrilateral with internal node 5 and dimensions as shown be-low. Apply the static condensation technique.

Using the CST stiffness matrix for plain strain, we get:

(1) (3)

1 2 5

3 4 5

1.5 1.0 0.1 0.2 1.6 1.21.0 3.0 0.2 2.6 0.8 5.60.1 0.2 1.5 1.0 1.6 1.2

[ ] [ ]4.16 0.2 2.6 1.0 3.0 0.8 5.6

1.6 0.8 1.6 0.8 3.2 0.01.2 5.6 1.2 5.6 0.0 11.2

Ek k

− −⎡ ⎤⎢ ⎥− − −⎢ ⎥⎢ ⎥− − −

= = ⎢ ⎥− −⎢ ⎥⎢ ⎥− − −⎢ ⎥− − −⎣ ⎦

(2) (4)

4 1 5

2 3 5

1.5 1.0 0.1 0.2 1.6 1.21.0 3.0 0.2 2.6 0.8 5.60.1 0.2 1.5 1.0 1.6 1.2

[ ] [ ]4.16 0.2 2.6 1.0 3.0 0.8 5.6

1.6 0.8 1.6 0.8 3.2 0.01.2 5.6 1.2 5.6 0.0 11.2

Ek k

− −⎡ ⎤⎢ ⎥− − −⎢ ⎥⎢ ⎥− − −

= = ⎢ ⎥− −⎢ ⎥⎢ ⎥− − −⎢ ⎥− − −⎣ ⎦


The resulting assembled matrix before static condensation is:

After partitioning, the condensed stiffness matrix is:


Flowchart for the Solution of Place Stress/Strain Problems The following flowchart is typical for a finite element process used for the

analysis of plane stress and plane strain problems.


Problems


19. Work problems 7.7, 7.10, and 7.15 on pages 332 - 343 in your textbook “A First Course in the Finite Element Method” by D. Logan using WinFElt.


Development of the Linear-Strain Triangle Equations

Introduction

In this section we will develop a higher-order triangular element, called the linear-strain triangle (LST). This element has many advantages over the con-stant-strain (CST). The LST element has six nodes and twelve displacement de-grees of freedom. The displacement function for the triangle is quadratic.

Derivation of the Linear-Strain Triangular Elemental Stiffness Matrix and Equations

The procedure to derive the LST element stiffness matrix and element equa-tions is identical to that used for the CST element.

Step 1 - Discretize and Select Element Types Consider the triangular element shown in the figure below:

Each node has two degrees of freedom: displacements in the x and y directions. We will let ui and vi represent the node i displacement components in the x and y directions, respectively.


The nodal displacements for an LST element are:

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪

⎧ ⎫ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪= =⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎪ ⎪

⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

1

1

2

1 2

2 3

3 3

4 4

5 4

6 5

5

6

6

uvu

d vd ud v

dd ud vd u

vuv

where the nodes are ordered counterclockwise around the element, and

Step 2 - Select Displacement Functions Consider a straight-sided triangular element shown below

The variation of the displacements over the element may be expressed as

= + + + + +2 21 2 3 4 5 6( , )u x y a a x a y a x a xy a y

= + + + + +2 27 8 9 10 11 12( , )v x y a a x a y a x a xy a y

The displacement compatibility among adjoining elements is satisfied because the three nodes defining adjacent sides define a unique a parabola. The CST and LST triangles are variations of the Pascal triangles as show below.

1

3

2

y

x

6

4

5


Terms in Pascal Triangle Polynomial Degree

# of Terms

1 0 (constant) 1

x y

1 (linear) 3

x2 xy y2

2 (quadratic) 6

x3 x2y xy2 y3

3 (cubic) 10

The general element displacement functions are:

1

22 2

2 2

11

12

·1 0 0 0 0 0 0·0 0 0 0 0 0 1

aa

x y x xy yx y x xy y

aa

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎡ ⎤

Ψ = ⎨ ⎬⎢ ⎥⎣ ⎦ ⎪ ⎪

⎪ ⎪⎪ ⎪⎩ ⎭

or [ ] *M aΨ =

To obtain the values for the a’s substitute the coordinated of the nodal points into the above equations:


Solving for the a’s and writing the results in matrix forms gives:

or

[ ] 1a x u−=

where [x] is the 12 x 12 matrix on the right-hand-side of the above equation. The “best” way to invert [x] is to use a computer. Note that only the 6 x 6 part of [x] really need be inverted. The general displacement expressions in terms of shape functions and the nodal degrees of freedom are:

[ ] N dΨ =

where [ ] [ ][ ] 1*N M x −

=


Step 3 - Define the Strain-Displacement and Stress-Strain Relationships Elemental Strains: The strains over a two-dimensional element are:

εε ε

γ

⎧ ⎫∂⎪ ⎪

∂⎪ ⎪⎧ ⎫⎪ ⎪∂⎪ ⎪= =⎨ ⎬ ⎨ ⎬∂⎪ ⎪ ⎪ ⎪

⎩ ⎭ ⎪ ⎪∂ ∂+⎪ ⎪∂ ∂⎩ ⎭

x

y

xy

uxvy

u vy x

or

1

2

11

12

0 1 0 2 0 0 0 0 0 0 0·

0 0 0 0 0 0 0 1 0 0 2·

0 0 1 0 2 0 1 0 2 0

aa

x yx y

x y x yaa

ε

⎧ ⎫⎪ ⎪⎪ ⎪⎡ ⎤⎪ ⎪⎢ ⎥= ⎨ ⎬⎢ ⎥⎪ ⎪⎢ ⎥⎣ ⎦ ⎪ ⎪⎪ ⎪⎩ ⎭

Observe that the strains are linear over the triangular element; therefore, the element is called a linear-strain triangle (LST). The above equation may be written in matrix form as:

[ ] 'M aε =

where [M ’] is based on derivatives of [M*]. If we substitute the values of a’s into the above equation gives:

[ ] B dε =

where [B] is a function of the nodal coordinates (x1, y1) through (x6, y6).

[ ] [ ][ ] 1'B M x −=


The stresses are given as:

[ ]x x

y y

xy xy

Dσ εσ ετ γ

⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭

For plane stress, [D] is:

( )⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

=ν

νν

ν15.0000101

1][

2

ED

For plane strain, [D] is:

( )( )⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−

−+=

ννν

νν

νν5.0000101

211][ ED

Step 4 - Derive the Element Stiffness Matrix and Equations The stiffness matrix can be defined as:

[ ] [ ] [ ][ ]T

V

k B D B dV= ∫

However, [B] is now a function of x and y; therefore, we must integrate the above expression to develop the element stiffness matrix. The [B] matrix is:

[ ]1 2 3 4 5 6

1 2 3 4 5 6

1 1 2 2 3 3 4 4 5 5 6 6

0 0 0 0 0 01 0 0 0 0 0 0

2B

A

β β β β β βγ γ γ γ γ γ

γ β γ β γ β γ β γ β γ β

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

where the β’s and the γ’s are functions of x and y as well as the nodal coordi-nates. The stiffness matrix is a 12 x 12 matrix and is very cumbersome to com-pute in explicit form. However, if the origin of the coordinates is the centroid of the element, the integrations become more amenable. Typically, the integration are computed numerically.


The element body forces and surface forces should net be automatically lumped at the nodes. The following integration should be computed.

[ ] Tb

V

f N X dV= ∫

[ ] Ts

S

f N T dS= ∫

The element equations are:

[ ] [ ] [ ]

1 11 12 13 1,12 1

1 21 22 23 2,12 1

2 31 32 33 3,12 2

6 12,1 12,2 12,3 12,12 6

12 1 12 12 12 1

x

y

x

y

f k k k k uf k k k k vf k k k k u

f k k k k v

⎧ ⎫ ⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎢ ⎥ ⎪ ⎪⎪ ⎪ ⎢ ⎥ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬

⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎩ ⎭⎣ ⎦⎩ ⎭

× × ×

Step 5, 6, and 7 Assembling the global stiffness matrix, determining the global displacements,

and calculating the stresses, are identical to the procedures used for CST ele-ments.

Example LST Stiffness Determination

Consider a straight-sided triangular element shown below:


The triangle has a base dimension of b and a height h, with mid-side nodes. We can calculate the coefficients a1 through a6 by evaluating the displacement u at each node.

1 1(0,0)u u a= =

22 1 2 4( ,0)u u b a a b a b= = + +

23 1 3 6(0, )u u h a a h a h= = + +

2 2

4 1 2 3 4 5 6,2 2 2 2 2 4 2b h b h b bh hu u a a a a a a⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = + + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2

5 1 3 60,2 2 2h h hu u a a a⎛ ⎞ ⎛ ⎞= = + +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

2

6 1 2 4,02 2 2b b bu u a a a⎛ ⎞ ⎛ ⎞= = + +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Solving the above equations simultaneously for the a’s gives:

6 1 2 5 1 31 1 2 3

4 3 4 3u u u u u ua u a ab h

− − − −= = =

( ) ( )2 6 1 1 4 5 64 52

2 2 4u u u u u u ua a

b bh− + + − −

= =

( )3 5 16 2

2 2u u ua

h− +

=

The u displacement equation is:

( )2 6 1 26 1 2 5 1 31 2

2 24 3 4 3( , )u u uu u u u u uu x y u x y x

b h b⎡ ⎤− +− − − −⎡ ⎤ ⎡ ⎤= + + + ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

( ) ( )1 4 5 6 3 5 1 22

4 2 2u u u u u u uxy y

bh h⎡ ⎤ ⎡ ⎤+ − − − +

+ +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦


The v displacement equation can be determined in a manner identical to that used for the u displacement:

( )2 6 1 26 1 2 5 1 31 2

2 24 3 4 3( , )v v vv v v v v vv x y v x y x

b h b⎡ ⎤− +− − − −⎡ ⎤ ⎡ ⎤= + + + ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

( ) ( )1 4 5 6 3 5 1 22

4 2 2v v v v v v vxy y

bh h⎡ ⎤ ⎡ ⎤+ − − − +

+ +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

The general form of the displacement expressions in terms of the shape func-tions is given as:

1

11 2 3 4 5 6

1 2 3 4 5 66

6

0 0 0 0 0 00 0 0 0 0 0

uv

N N N N N NuN N N N N Nv

uu

⎧ ⎫⎪ ⎪⎪ ⎪⎡ ⎤⎧ ⎫ ⎪ ⎪=⎨ ⎬ ⎨ ⎬⎢ ⎥

⎩ ⎭ ⎣ ⎦ ⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

where the shape functions are:

2 2 2 2

1 2 32 2 2 23 3 2 4 2 2 21 x y x xy y x x y yN N Nb h b bh h b b h h

= − − + + + = − + = − +

2 2

4 5 62 24 4 4 4 4 4 4xy y xy y x xy xN N Nbh h bh h b bh b

= = − − = − −

The element interpolation functions N have two basic shapes. The behavior of the functions N1, N2, and N3 is similar except referenced at different nodes. The shape function N1 is shown below:

2

1

3 1

N1

y

x

6 4

5


The second type of shape function is valid for functions N4, N5, and N6. The func-tion N5 is shown below:

The element strain is gives as:

[ ] e B d=

where the [B] matrix is:

[ ]1 2 3 4 5 6

1 2 3 4 5 6

1 1 2 2 3 3 4 4 5 5 6 6

0 0 0 0 0 01 0 0 0 0 0 0

2B

A

β β β β β βγ γ γ γ γ γ

γ β γ β γ β γ β γ β γ β

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

where the β’s and the γ’s are:

1 2 34 43 4 0hx hxh y hb b

β β β= − + + = − + =

4 5 684 4 4 4hxy y h y

bβ β β= = − = − −

1 2 34 43 4 0by byb x b

h hγ γ γ= − + + = = − +

4 5 684 4 4 4byx b x x

hγ γ γ= = − − = −

2

1

3 1

N5

y

x

6 4

5


The stiffness matrix for a constant thickness element can be obtained by substi-tuting the β’s and the γ’s into the [B] and then substituting [B] into the following expression and evaluating the integral numerically.

[ ] [ ] [ ][ ]T

V

k B D B dV= ∫

Comparison of Elements

For a given number of nodes, a better representation of true stress and dis-placement is generally obtained using LST elements than is obtained using the same number of nodes a finer subdivision of CST elements. For example, a sin-gle LST element gives better results than four CST elements.

Consider the following cantilever beam with E = 30 x 106 psi, ν = 0.25, and t = 1 in.

Table 1 lists the series of tests run to compare results using the CST and LST elements. Table 2 shows comparisons of free-end (tip) deflection and stress for each element type used to model the cantilever beam.


Table 1. Comparison of CST and LST results

Series of Test Runs Number of Nodes

Degrees of Freedom, nd

Number of Elements

A-1 4 x 16 85 160 128 CST A-2 8 x 32 297 576 512 CST B-1 2 x 8 85 160 32 LST B-2 4 x 16 297 576 128 LST

Table 2. Comparison of CST and LST results

Runs nd Bandwidth, nd Tip Deflection

(in) σx

(ksi) Location

(x,y) A-1 160 14 -0.29555 67.236 (2.250,11.250) A-2 576 22 -0.33850 81.302 (1.125,11.630) B-1 160 18 -0.33470 58.885 (4.500,10.500) B-2 576 22 -0.35159 69.956 (2.250,11.250)

Exact Solution -0.36133 80.000 (0,12)

From Table 2, we can observe that: • The larger the number of degrees of freedom for a given type of triangular

element, the closer the solution converges to the exact one (compare run A-l to run A-2, and B-l to B-2).

• For a given number of nodes, the LST analysis yields somewhat better results than the CST analysis (compare run A-l to run B-l).

• Although the CST element is rather poor in modeling bending, we observe from Table 2 that the element can be used to model a beam in bending if suf-ficient number of elements is used through the depth of the beam.

• In general, both the LST and CST analyses yield sufficient results for most plane stress/strain problems provided a sufficient number of elements are used.


Most commercial programs incorporate the use of CST and/or LST elements for plane stress/strain problems although these elements are used primarily as transition elements (usually during mesh generation). Also, recall that finite ele-ment displacements will always be less than the exact ones, because finite ele-ment models are always predicted to be stiffer than the actual structures when using the displacement formulation of the finite element method.

A comparison of CST and LST models of a plate subjected to parabolically distributed edge loads is shown in the figure below. The LST model converges to the exact solution for horizontal displacement at point A faster than does the CST model. However, the CST model is quite acceptable even for modest numbers of degrees of freedom. For example, a CST model with 100 nodes (200 degrees of freedom) often yields nearly as accurate a solution as does an LST model with the same number of degrees of freedom.


The results of Table 2 indicate: • That the LST model might be preferred over the CST model for plane stress

applications when a relatively small number of nodes is used.

• That the use of triangular elements of higher order, such as the LST, is not visibly more advantageous when large numbers of nodes are used, particu-larly when the cost of formation of the element stiffnesses, equation band-width, and overall complexities involved in the computer modeling are consid-ered.

Problems



Development of the Plate Bending Element

Introduction

In this section we will begin by describing elementary concepts of plate bend-ing behavior and theory. The plate element is one of the more important struc-tural elements and is used to model and analyze such structures as pressure vessels, chimney stacks, and automobile parts. A large number of plate bending element formulations exist that would require lengthy chapter to cover. The pur-pose in this chapter is to present the derivation of the stiffness matrix for one of the most common plate bending finite elements and then to compare solutions to some classical problems for a variety of bending elements in the literature.

Basic Concepts of Plate Bending

A plate can be considered the two-dimensional extension of a beam in simple bending. Both plates and beams support loads transverse or perpendicular to their plane and through bending action. A plate is a flat (if it were curved, it would be a shell). A beam has a single bending moment resistance, while a plate re-sists bending about two axes and has a twisting moment. We will consider the classical thin-plate theory or Kirchhoff plate theory.

Basic Behavior of Geometry and Deformation Consider the thin plate in the x-y plane of thickness t measured in the z direc-

tion shown in the figure below:


The plate surfaces are at z = ±t/2, and its midsurface is at z = 0. The basic ge-ometry of the plate is as follows:

1. The plate thickness is much smaller than its inplane dimensions b and c (that is, ort b c ). If t is more than about one-tenth the span of the plate, then transverse shear deformation must be accounted for and the plate is then said to be thick.

2. The deflection w is much less than the thickness t (than is, / 1w t ).

Kirchhoff Assumptions Consider the differential slice cut from the plate by planes perpendicular to the x axis as show in the figure below:

Loading q causes the plate to deform laterally or upward in the z direction and, the defection w of point P is assumed to be a function of x and y only; that is w=w(x, y) and the plate does not stretch in the z direction. The line a-b drawn perpendicular to the plate surface before loading remains perpendicular to the surface after loading. These conditions are consistent with the Kirchhoff assump-tions:

1. Normals remain normal. This implies that transverse shears strains γyz = 0 and γxz = 0. However γxy does not equal to zero. Right angles in the plane of the plate may not remain right angles after loading. The plate may twist in the plane.

2. Thickness changes can be neglected and normals undergo no extension. This means that εz = 0.


3. Normal stress σz has no effect on in-plane strains εx and εy in the stress-strain equations and is considered negligible.

4. Membrane or in-plane forces are neglected here, and the plane stress re-sistance can be superimposed later (that is, the constant-strain triangle be-havior of Chapter 6 can be superimposed with the basic plate bending ele-ment resistance). Therefore, the in-plane deflections in the x and y direc-tions at the midsurface, t = 0, are assumed to be zero; u(x, y, 0) = 0 and v(x, y, 0) = 0.

Based on Kirchhoff assumptions, at any point P the displacement in the x direc-tion due to a small rotation α is:

wu z zx

α ∂ = − = − ∂

At the same point, the displacement in the y direction is:

wv z zy

α ∂= − = − ∂

The curvatures of the plate are then given as the rate of change of the angular displacements of the normals and defined as:

2 2 2

2 22

x y xyw w wx y x y

κ κ κ∂ ∂ ∂= − = − = −

∂ ∂ ∂ ∂

Using the definitions for in-plane strains, along with the curvature relationships, the in-plane strain/displacement equations are:

2 2 2

2 2 2x y xyw w wz z zx y x y

ε ε γ∂ ∂ ∂= − = − = −

∂ ∂ ∂ ∂

The first of the above equations is used in beam theory. The remaining two equa-tions are new to plate theory.

Stress/Strain Relationship Based on the third Kirchhoff assumption, the plane stress equations that relate in-plane stresses to in-plane strains for an isotropic material are:


( ) ( )2 21 1x x y y y x xy xyE E Gσ ε νε σ ε νε τ γν ν

= + = + =− −

The in-plane normal stresses and shear stress are shown acting on the edges of the plate shown in figure below:

Similar to the stress variation in a beam, the stresses vary linearly in the z direction from the midsurface of the plate. The transverse shear stresses τyz and τxz are also present, even though transverse shear deformation is neglected. These stresses vary quadratically through the plate thickness. The bending moments acting along the edge of the plate can be related to the stresses by:

/ 2 / 2 / 2

/ 2 / 2 / 2

t t t

x x y y xy xyt t t

M z dz M z dz M z dzσ σ τ− − −

= = =∫ ∫ ∫

Substituting strains for stresses gives:

( ) ( )/ 2 / 2 / 2

2 2/ 2 / 2 / 21 1

t t t

x x y y y x xy xyt t t

E EM z dz M z dz M zG dzε νε ε νε γν ν− − −

= + = + = − − ∫ ∫ ∫ Using the strain/curvature relationships, the moment expression become:

( ) ( ) (1 )2x x y y y x xy xy

DM D M D M νκ νκ κ νκ κ−= + = + =

where D = Et3/[12(1 - ν2)] is called the bending rigidity of the plate.


The maximum magnitude of the normal stress on each edge of the plate are lo-cated at the top or bottom at z = t/2.

26 x

xMt

σ =

The equilibrium equations for plate bending are important in selecting the ele-ment displacement fields. The governing differential equations are:

0yx QQ qx y

∂∂+ + =

∂ ∂

0xyxx

MM Qx y

∂∂+ − =

∂ ∂

0y xyy

M MQ

y x∂ ∂

+ − =∂ ∂

where q is the transverse distributed loading and Qx and Qy are the transverse shear line loads as shown below.

Substituting the moment/curvature expressions in the last two differential equa-tions list above, solving for Qx and Qy, and substituting the results into the first equation listed above, the governing partial differential equation for isotropic, thin-plate bending may be derived as:

4 4 4

4 2 2 42w w wD q

x x y y ∂ ∂ ∂

+ + = ∂ ∂ ∂ ∂

where the solution to the thin-plate bending is a function of the transverse dis-


placement w. If we neglect the differentiation with respect to the y direction, the above equation simplifies to the equation for a beam and the flexural rigidity D of the plate reduces to the EI of the beam when the Poisson effect is set to zero. Potential Energy of a Plate The total potential energy of a plate is given as:

( )12 x x y y xy xy

V

U dVσ ε σ ε τ γ= + +∫

The potential energy can be expressed in terms of moments and curvatures as:

( )12 x x y y xy xy

A

U M M M dAκ κ κ= + +∫

Derivation of a Plate Bending Element Stiffness Matrix and Equations Numerous finite elements for plates bending have been developed on the years, references cite 88 different elements. In this section, we will introduce the basic 12-degree-of-freedom rectangular element shown below.

The formulation will be developed consistently with the stiffness matrix and equa-tions for the bar, beam, plane stress/strain elements of previous chapters.

Step 1 - Discretize and Select Element Types Consider the 12-degree-of-freedom plate element shown in the figure above.

Each node has 3 degrees of freedom – a transverse displacement w in the z di-rection, a rotation θx about the x axis, and a rotation θy about the y axis.


The nodal displacements at node i are:

i

xi

yi

wd θ

θ

=

where the rotations are related to the transverse displacements by:

x yw wy x

θ θ∂ ∂= = −∂ ∂

The negative sign on θy is due to the fact that a negative displacement w is re-quired to produce a positive rotation about the y axis.

The total element displacement matrix is:

i

j

m

n

dd

ddd

=

Step 2 - Select Displacement Functions Since the plate element has 12 degrees of freedom, we select a 12-term poly-

nomial in x and y as:

2 2 3 2 21 2 3 4 5 6 7 8 9

3 3 310 11 12

( , )w x y a a x a y a x a xy a y a x a x y a xya y a x y a xy

= + + + + + + + +

+ + +

The function given above is an incomplete quartic polynomial; however, it is complete up to the third order (first ten terms), and the choice of the two more terms from the remaining five terms of the complete quartic must be made. The choice of x3y and y3x ensure that we will have continuity in the displacement among the interelement boundaries. The terms x4 and y4 would yield discontinui-ties along the interelement boundaries. The final term x2y2 cannot be paired with any other term so it is also rejected. The displacement function approximation also satisfies the basic differential equation over the unloaded part of the plate. In addition, the function accounts for rigid-body motion and constant strain in the


plate. However, interelement slope discontinuities along common boundaries of elements are not ensured.

To observe this discontinuities in slope, evaluate the polynomial and its slopes along a side or edge. For example, consider side i-j, the function gives:

2 31 2 4 7( , )w x y a a x a x a x= + + +

22 4 72 3w a a x a x

x∂

= + +∂

2 33 5 8 11

w a y a x a x a xy

∂= + + +

∂

The displacement w is cubic while the slope /w x∂ ∂ is the same as in beam bending. Based on the beam element, recall that the four constants a1, a2, a4, and a7 can be defined by invoking the endpoint conditions of wi, wj, θi, and θi. Therefore, w and /w x∂ ∂ are completely define along this edge. The normal slope

/w y∂ ∂ is cubic in x: however; only two degrees of freedom remain for definition of this slope while four constant exist a3, a5, a8, and a11. This slope is not uniquely defined and a slope discontinuity occurs. The solution obtained form the finite element analysis using this element will not be a minimum potential energy solu-tion. However, this element has proven to give acceptable results.

The constant a1 through a12 can be determined by expressing the 12 simulta-neous equation linking the values of w and its slope at the nodes when the coor-dinates take their appropriate values.

12 2 3 2 2 3 3 3

22 2 3 2

32 2 2 3

12

10 0 1 0 2 0 2 3 30 1 0 2 0 2 2 0 2

awx y x xy y x x y xy y x y xy a

w x y x xy y x xy ay

x y x xy y x y yw

ax

∂ = ∂ − − − − − − − ∂− ∂

or in matrix form as:

[ ] P aψ =


where [P] is the 3 x 12 first matrix on the right-hand side of the above equation.

Next, evaluate the matrix at each node point

2 2 3 2 2 3 3 3

2 2 3 2

2 2 2 3

2 2 3 2 2 3 3 3

1

0 0 1 0 2 0 2 3 3

0 1 0 2 0 2 2 0 2

1

0 1 0 2 0

i i i i i i i i i i i i i i i i

i i i i i i i i i

i i i i i i i i i

j j j j j j j j j j j j j j

n n

i

xi

yi

j

yn

x y x x y y x x y x y y x y x y

x y x x y y x x y

x y x x y y x y y

x y x x y y x x y x y y x y x y

x y

w

w

θθ

θ

− − − − − − − −

− − −

=

2 2 3

1

2

3

4

122 2 0 2n n n n n n nx x y y x y y

aaaa

a− − − − −

In compact matrix form the above equations are:

[ ] d C a= Therefore, the constants a can be solved for by: [ ] 1a C d−

= Substituting the above expression into the general form of the matrix gives: [ ] [ ] [ ] 1 orP C d N dψ ψ−

= = where [N] = [P][C]-1 is the shape function matrix.

Step 3 - Define the Strain (Curvature)/Displacement and Stress (Moment)/Curvature Relationships

Recall the general form of the curvatures:

2 2 2

2 22

x y xyw w wx y x y

κ κ κ∂ ∂ ∂= − = − = −

∂ ∂ ∂ ∂


The curvature matrix can be written as:

4 7 8 11

6 9 10 122 2

5 8 9 11 12

2 6 2 62 2 6 6

2 4 4 6 6

x

y

xy

a a x a y a xya a x a y a xy

a a x a y a x a y

κκκ

− − − − = − − − − − − − − −

or in matrix form as: [ ] Q aκ =

where [Q] is the coefficient matrix multiplied by the a’s in the curvature matrix equations.

1

2

32 2

12

0 0 0 2 0 0 6 2 0 0 6 00 0 0 0 0 2 0 0 2 6 0 60 0 0 0 2 0 0 4 4 0 6 6

ax y xy a

Q x y xy ax y x y

a

− − − − = − − − − − − − − −

Therefore: [ ] [ ] [ ] [ ] 1 orQ a Q C d B dκ κ κ−

= ⇒ = =

where [ ] [ ] [ ] 1B Q C −

= The moment/curvature matrix for a plate is given by:

[ ] [ ] x x

y y

xy xy

MM M D D B d

M

κκκ

= = =

where the [D] matrix for isotropic materials is:

( ) ( )

3

2

1 0[ ] 1 0

12 1 0 0 0.5 1

EtDν

νν

ν

= − −


Step 4 - Derive the Element Stiffness Matrix and Equations The stiffness matrix is given by the usually form of the stiffness matrix as:

[ ] [ ] [ ][ ]Tk B D B dxdy= ∫∫

The stiffness matrix for the four-node rectangular element is of a 12 x 12.

The surface force due to distributed loading q acting per unit area in the z di-rection is:

[ ] [ ]Ts sF N q dxdy= ∫∫

for a uniform load q acting over the surface of an element of dimensions 2b x 2c the forces and moments at node i are:

3

3

wi

xi

yi

fqcbf c

f bθ

θ

= −

with similar expression at nodes j, m, and n.

The element equations are given by:

11 12 1,12

21 22 2,12

31 32 3,12

41 42 4,12

12,1 12,2 12,12

wi i

xi xi

yi yi

w j j

yn yn

k k k

k k k

k k k

k k k

k k k

f wfff w

f

θ

θ

θ

θθ

θ

=

The remaining steps of assembling the global equations, applying boundary con-ditions, and solving the equations for nodal displacements and slopes follow the standard procedures introduced in previous chapters.


The figure below shows a number of plate element formulations results for a square plate simply supported all around and subjected to a concentrated vertical load applied at the center of the plate. The results show the upper and lower bound solutions behavior and demonstrate convergence of solution for various plate elements. Included in these results is the 12-term polynomial plate element introduced in this chapter.


The figure below shows comparisons of triangular plate formulations for the same centrally loaded simply supported plate. From both figures, we can observe a number of different formulations with results that converge for above and be-low. Some of these elements produce better results than others.


The figure below shows results for some selected Mindlin plate theory elements. Mindlin plate elements account for bending deformations and for transverse shear deformation.


Computer Solution for a Plate Bending Problem

Consider the clamped plate show below subjected to a 100 lb load applied at the center (let E = 30 x 106 psi and ν = 0.3).

The exact solution for the displacement at the center of the plate is w = 0.0056PL2/D. Substituting the values for the variables gives a numerical value of w = 0.0815 in. The table below shows the results of modeling this plate structure using SAP2000 (the educational version allows only 100 nodes) compares to the exact solution.

Number of square elements

Displacement at the center (in) % error

4 0.09100 11.6

16 0.09334 14.5

36 0.08819 8.2

64 0.08584 5.3

256 0.08300 1.8

1,024 0.08209 0.7

4,096 0.08182 0.3 Exact Solution 0.08154 --


The figures below show non-node-averaged contour plot for the normal stress σx and σy.

The next set of plots shows the non-node-averaged moments Mx and My.

Both sets of plots seem to indicate good interelement continuity between normal stresses in the x and y direction and bending moments about the x and y axes.


The next set of plots shows the shear stress τxy and the node-average shear stress τxy.

The next set of plots shows the twisting moment Mxy and the node-average twist-ing moment Mxy.

Both sets of plots indicate interelement discontinuities for shear stress and twist-ing moment. However, if the node-average plots are viewed, the discontinuities are smoothed out and not visible.


Problems

21. Do problems 12.1 and 12.6 on pages 455 - 457 in your textbook “A First Course in the Finite Element Method” by D. Logan using SAP2000.


Structural Dynamics

Introduction

This chapter provides an elementary introduction to time-dependent problems. We will introduce the basic concepts using the single-degree-of-freedom spring-mass system. We will include discussion of the stress analysis of the one-dimensional bar, beam, truss, and plane frame.

We will provide the basic equations necessary for structural dynamic analysis and develop both the lumped- and the consistent-mass matrices involved in the analyses of the bar, beam, truss, and plane frame. We will describe the assembly of the global mass matrix for truss and plane frame analysis and then present numerical integration methods for handling the time derivative.

We will provide longhand solutions for the determination of the natural fre-quencies for bars and beams, and then illustrate the time-step integration proc-ess involved with the stress analysis of a bar subjected to a time dependent forc-ing function.

Dynamics of a Spring-Mass System

In this section we will discuss the motion of a single-degree-of-freedom spring-mass system as a introduction to the dynamic behavior of bars, truss, frames. Consider the single-degree-of-freedom spring-mass system subjected to a time-dependent force F(t) as shown in the figure below. The term k is the stiff-ness of the spring and m is the mass of the system.

The free-body diagram of the mass is shown below. The spring force T = kx and the applied force F(t) act on the mass, and the mass-times-acceleration term is


shown separately.

Applying Newton’s second law of motion, f = ma, to the mass, we obtain the equation of motion in the x direction:

( )F t kx mx− = where a dot over a variable indicates differentiation with respect to time; (·) () /d dt= . The standard form of the equation is:

( )mx kx F t+ = The above equation is a second-order linear differential equation whose solution for the displacement consists of a homogeneous solution and a particular solu-tion. The homogeneous solution is the solution obtained when the right-hand-side is set equal to zero. A number of useful concepts regarding vibrations are available when considering the free vibration of a mass; that is when F(t) = 0.

Let’s define the following term:

2 km

ω =

The equation of motion becomes:

2 0x xω+ = where ω is called the natural circular frequency of the free vibration of the mass (radians per second). Note that the natural frequency depends on the spring stiffness k and the mass m of the body. The motion describe by the homogeneous equation of motion is called simple harmonic motion. A typical displacement/time curve is shown below.


where xm denotes the maximum displacement (or amplitude of the vibration). The time interval required for the mass to complete one full cycle of motion is called the period of the vibration τ (in seconds) and is defined as:

2πτω

=

The frequency in hertz (Hz = 1/s) is f = 1/τ = ω /(2π).

Direct Derivation of the Bar Element

Let’s derive the finite element equations for a time-dependent (dynamic) stress analysis of a one-dimensional bar.

Step 1 - Select Element Type We will consider the linear bar element shown below.

where the bar is of length L, cross-sectional area A, and mass density ρ (with typical units of lb-s2/in4), with nodes 1 and 2 subjected to external time-dependent loads, ˆ ( )e

xf t .


Step 2 - Select a Displacement Function A linear displacement function is assumed in the x direction.

1 2ˆ û a a x= + The number of coefficients in the displacement function, ai, is equal to the total number of degrees of freedom associated with the element. We can express the displacement function in terms of the shape functions:

[ ] 11 2

2

ˆˆ

ˆx

x

du N N

d

= =


1 2

ˆ ˆ1 x xN N

L L= − =

Step 3 - Define the Strain/Displacement and Stress/Strain Relationships The stress-displacement relationship is:

ˆ ˆ[ ]xdu B ddx

ε = =

where

1

2

ˆ1 1 ˆ[ ]ˆ

x

x

dB d

L L d

= − =

The stress-strain relationship is given as:

ˆ[ ] [ ][ ]x xD D B dσ ε= =

Step 4 - Derive the Element Stiffness Matrix and Equations The bar element is typically not in equilibrium under a time-dependent force;

hence, f1x ≠ f2x. We must apply Newton’s second law of motion, f = ma, to each node. Write the law of motion as the external force fxe minus the internal force equal to the nodal mass times acceleration. Therefore:


2 21 2

1 1 1 2 2 22 2

ˆ ˆˆ ˆ ˆ ê ex xx x x x

d df f m f f mt t

∂ ∂= + = +

∂ ∂

where m1 and m2 are obtained by lumping the total mass of the bar equally at the two nodes such that:

1 22 2AL ALm mρ ρ

= =

In matrix form, the above equations are:

212

1 1 1

222 2 2

2

ˆˆ ˆ 0ˆ ˆ ˆ0

xe

x x

ex x x

df f m t

mf f dt

∂ ∂= +

∂ ∂

If we replace f with [ ] k d we get:

ˆ ˆ ˆ ˆˆ( )ef t k d m d = +

where the elemental stiffness matrix is:

2

2

ˆ1 1ˆ ˆ1 1

dAEk dL t

∂− = = ∂−

and the lumped-mass matrix is:

1 0ˆ2 0 1ALm ρ

=

Let’s derive the consistent-mass matrix for a bar element. The typical

method for deriving the consistent-mass matrix is the principle of virtual work; however, an even simpler approach is to use D’Alembert’s principle. The effec-tive body force is:

êX uρ= −

The nodal forces associated with Xe are found by using the following:


[ ] Tb

V

f N X dV= ∫

Substituting Xe for X gives:

ˆ[ ]TbV

f N u dVρ= −∫

The second derivative of the u with respect to time is:

ˆ ˆˆ ˆ[ ] [ ]u N d u N d= =

where u and u are the nodal velocities and accelerations, respectively.

[ ] [ ] ˆ ˆˆTb

V

f N N d dV m dρ= − = − ∫

where

[ ] [ ]ˆ T

V

m N N dVρ= ∫

The mass matrix is called the consistent mass matrix because it is derived using the same shape functions use to obtain the stiffness matrix. Substituting the shape functions in the above mass matrix equations gives:

ˆ1 ˆ ˆˆ 1

ˆV

xx xLm dVL Lx

L

ρ

− = −

∫

or

0

ˆ1 ˆ ˆˆ ˆ1

ˆ

Lx

x xLm A dxL Lx

L

ρ

− = −

∫

or


2

20

ˆ ˆ ˆ1 1

ˆˆ ˆ ˆ

1

L

x x xL L L

m A dxx x xL L L

ρ

− − = −

∫

Evaluating the above integral gives:

2 1ˆ6 1 2ALm ρ

=


The global stiffness matrix and the global force vector are assembled using the nodal force equilibrium equations, and force/deformation and compatibility equa-tions.

[ ] [ ] ( ) F t K d M d= +

where

[ ] [ ] ( ) ( ) ( )

1 1 1

N N Ne e e

e e eK k M m F f

= = =

= = = ∑ ∑ ∑

Numerical Integration in Time

We now introduce procedures for the discretization of the equations of motion with respect to time. These procedures will allow the nodal displacements to be determined at different time increments for a given dynamic system. The general method used is called direct integration. There are two classifications of direct integration: explicit and implicit. We will formulate the equations for two direct in-tegration methods. The first, and simplest, is an explicit method known as the central difference method. The second, more complicated but more versatile than the central difference method, is an implicit method known as the New-mark-Beta (or Newmark’s) method. The versatility of Newmark’s method is evi-denced by its adaptation in many commercially available computer programs.


Central Difference Method The central difference method is based on finite difference expressions for the

derivatives in the equation of motion. For example, consider the velocity and the acceleration at time t:

1 1 1 1

2( ) 2( )i i i i

i id d d dd d

t t+ − + −− −

= =∆ ∆

where the subscripts indicate the time step for a given time increment of ∆t. The acceleration can be expressed in terms of the displacements (using a Taylor se-ries expansion) as:

1 12

2( )

i i ii

d d ddt

+ −− +=

∆

We generally want to evaluate the nodal displacements; therefore, we rewrite the above equation as:

21 12 ( )i i i id d d d t+ −= − + ∆

The acceleration can be expressed as:

( )1i i id d−= −M F K

To develop an expression of di+1, first multiply the nodal displacement equation by M and substitute the above equation for id into this equation.


( )( )21 12i i i i id d d d t+ −= − + − ∆M M M F K

Combining terms in the above equations gives:

( ) ( )2 21 12i i i id t t d d+ −

= ∆ + − ∆ − M F M K M

To start the computation to determine 1 1 1, , andi i id d d+ + + we need the displacement at time step i-1. Using the central difference equations for the velocity and accel-eration and solving for di-1

2

1( )( )

2i i i itd d t d d−

∆= − ∆ +

Procedure for solution: 1. Given: d0, 0d , and Fi(t)

2. If the acceleration is not given, solve for 0d at t = 0.

( )10 0 0d d−= −M F K

3. Solve for d-1 at t = -∆t

2

1 0 0 0( )( )

2td d t d d−

∆= − ∆ +

4. Solve for d1 at t = ∆t using the value of d-1 from Step 3

( ) ( ) 2 211 0 0 12d t t d d−

− = ∆ + − ∆ − M F M K M

5. With d0 given and d1 determined in Step 4 solve for d2

( ) ( ) 2 212 1 1 02d t t d d− = ∆ + − ∆ − M F M K M

6. Solve for 1d

( )11 1 1d d−= −M F K

7. Solve for 1d using the central difference equation

2 01 2( )

d ddt

−=

∆


8. Repeat Steps 5, 6, and 7 to obtain the displacement, acceleration, and velocity for other time steps.


Example Problem

Determine the displacement, acceleration, and velocity at 0.05 second time in-tervals for up to 0.2 seconds for the one-dimensional spring-mass system shown in the figure below.

The time-dependent forcing function is given as:

Consider the above spring-mass system as a single degree of freedom problem represented by the displacement d.

Procedure for solution: 1. At time t = 0:

0 00 0d d= = 2. The initial acceleration at t = 0:

( )120 0 0

2,000 100(0) 62.8331.83

ind d s− −

= − = =M F K



2

1 0 0 0( )( )

2td d t d d−

∆= − ∆ +

2

1(0.05)0 (0.05)0 (62.83) 0.0785

2d in− = − + =

4. Solve for d1 at t = ∆t (0.05 seconds) using the value of d-1 from Step 3:

( ) ( ) 2 211 0 0 12d t t d d−

− = ∆ + − ∆ − M F M K M

( ) ( ) ( ) ( ) ( )( ) 2 21

1 0.05 2,000 2 31.83 0.05 100 0 31.83 0.078531.82

0.0785

d

in

= + − −

=

5. Solve for d2 at t = 0.10 seconds:

( ) ( ) 2 212 1 1 02d t t d d− = ∆ + − ∆ − M F M K M

( ) ( ) ( ) ( ) ( ) ( )( ) 2 22

1 0.05 1,500 2 31.83 0.05 100 0.0785 31.83 031.82

0.274

d

in

= + − −

=

6. Solve for the acceleration 1d at time t = 0.05:

( ) ( )121 1 1

1 1,500 100 0.0785 46.8831.83

ind d s− = − = − = M F K


( )2 0

10.274 0 2.74

2( ) 2 0.05d d ind st

− −= = =

∆


8. Repeat Steps 5, 6, and 7 to obtain the displacement, acceleration, and velocity for the next time step. Repeating Step 5:

( ) ( ) 2 213 2 2 12d t t d d− = ∆ + − ∆ − M F M K M

( ) ( ) ( ) ( ) ( ) ( )( ) 2 23

1 0.05 1,000 2 31.83 0.05 100 0.274 31.83 0.078531.82

0.546

d

in

= + − −

=

Repeating Step 6:

( ) ( )122 2 2

1 1,000 100 0.274 30.5631.83

ind d s− = − = − = M F K

Repeating Step 7:

( )3 1

20.546 0.0785 4.68

2( ) 2 0.05d d ind st

− −= = =

∆

The following table summarizes the results for the remaining time steps as com-pared with the exact solution.

t (s) F(t) (lb) id (in/s2) id (in/s) id (in) id (exact)

0.00 2,000 62.83 0.00 0.000 0.0000 0.05 1,500 46.88 2.74 0.0785 0.0718 0.10 1,000 30.56 4.68 0.274 0.2603 0.15 500 13.99 5.79 0.546 0.5252 0.20 0 -2.68 6.07 0.854 0.8250 0.25 0 -3.63 5.91 1.154 1.132


Newmark’s Method Newmark’s equations are given as:

1 1( ) (1 )i i i id d t d dγ γ+ + = + ∆ − +

( )2 11 12( ) ( )i i i i id d t d t d dβ β+ +

= + ∆ + ∆ − +

where β and γ are parameters. The parameter β is typically between 0 and ¼, and γ is often taken to be ½. For example, if β = 0 and γ = ½ the above equation reduce to the central difference method. To find di+1 first multiply the above equation by the mass matrix M and substi-tute into this the expression for acceleration. Recall the acceleration is:

( )10 0 0d d−= −M F K

The expression Mdi+1 is:

( ) [ ]2 211 1 12( ) ( ) ( )i i i i i id d t d t d t dβ β+ + += + ∆ + ∆ − + ∆ −M M M M F K

Combining terms gives:

( ) ( )2 2 2 11 1 2( ) ( ) ( ) ( )i i i i it d t d t d t dβ β β+ ++ ∆ = ∆ + + ∆ + ∆ −M K F M M M

Dividing the above equation by β (∆t)2 gives:

1 1' 'i id + +=K F where

21'

( )tβ= +

∆K K M

( ) 211 1 22' ( ) ( )

( )i i i i id t d t dt

ββ+ +

= + + ∆ + − ∆ ∆MF F

The advantages of using Newmark’s method over the central difference method are that Newmark’s method can be made unconditionally stable (if β = ¼ and γ = ½) and that larger time steps can be used with better results.


Procedure for solution of Newmark’s Method:

1. Given: d0, 0d , and Fi(t)

2. If the acceleration is not given, solve for 0d at t = 0.

( )10 0 0d d−= −M F K

3. Solve the displacement d1 at time t = ∆t

1 1' 'd =K F


4. Solve for 1d (original Newmark equation for 1id + rewritten for 1id + )

( )2 11 1 0 0 022

1 ( ) ( )( )

d d d t d t dt

ββ

= − − ∆ − ∆ − ∆

5. Solve for 1d

1 0 0 1( ) (1 )d d t d dγ γ = + ∆ − +

6. Repeat Steps 3, 4, and 5 to obtain the displacement, acceleration, and velocity for the next time step.

Example Problem

Determine the displacement, acceleration, and velocity at 0.1 second time in-tervals for up to 0.5 seconds for the one-dimensional spring-mass system shown in the figure below.

The time-dependent forcing function is given as:

Consider the above spring-mass system as a single degree of freedom problem represented by the displacement d. Use Newmark’s method with β = 1/6 and γ = ½.


Procedure for solution: 1. At time t = 0: 0 00 0d d= =

2. If the acceleration is not given, solve for 0d at t = 0:

( )1 20 0 0

100 70(0) 56.5 /1.77

d d in s− −= − = =M F K

3. Solve the displacement d1 at time t = 0.1 seconds:

1 1' 'd =K F

2 216

1 1' 70 (1.77) 1,132 /( ) (0.1)

lb intβ

= + = + =∆

K K M

( ) 211 1 0 0 022' ( ) ( )

( )d t d t d

tβ

β = + + ∆ + − ∆ ∆

MF F

( ) ( )21 11 2 621

6

1.77' 80 0 (0.1)0 (0.1) 56.5 280(0.1)

lb = + + + − = F

11

' 280 0.248' 1,132

d in= = =FK

4. Solve for 1d at time t = 0.1 seconds:

( )2 11 1 0 0 022

1 ( ) ( )( )

d d d t d t dt

ββ

= − − ∆ − ∆ − ∆

( )2 1 121 2 621

6

1 0.248 0 (0.1)0 (0.1) 56.5 35.4(0.1)

ind s = − − − − =

5. Solve for 1d

1 0 0 1( ) (1 )d d t d dγ γ = + ∆ − +

( )1 11 2 20 (0.1) (1 )56.5 35.4 4.59 ind s = + − + =

6. Repeat Steps 3, 4, and 5 to obtain the displacement, acceleration, and velocity for the next time step (t = 0.2 s).


Repeating Step 3:

( ) 212 2 1 1 122' ( ) ( )

( )d t d t d

tβ

β = + + ∆ + − ∆ ∆

MF F

( ) ( )21 12 2 621

6

1.77' 60 0.248 (0.1)4.59 (0.1) 35.4 934(0.1)

lb = + + + − = F

11

' 934 0.825' 1,132

d in= = =FK

Repeating Step 4:

( )2 12 2 1 1 122

1 ( ) ( )( )

d d d t d t dt

ββ

= − − ∆ − ∆ − ∆

( )2 1 122 2 621

6

1 0.825 0.248 (0.1)4.59 (0.1) 35.4 1.27(0.1)

ind s = − − − − =

Repeating Step 5:

2 1 1 2( ) (1 )d d t d dγ γ = + ∆ − +

( )1 12 2 24.59 (0.1) (1 )35.4 1.27 6.42 ind s = + − + =

The following table summarizes the results for the time steps through t = 0.5 sec-onds.

t (s) F(t) lb id (in/s2) id (in/s) di (in)

0 100 56.6 0 0 0.1 80 35.4 4.59 0.248 0.2 60 1.27 6.42 0.825 0.3 48.6 -26.2 5.17 1.36 0.4 45.7 -42.2 1.75 1.72 0.5 42.9 -42.2 -2.45 1.68


Natural Frequencies of a One-Dimensional Bar

Before solving the structural stress dynamic analysis problem, let’s consider how to determine the natural frequencies of continuous elements. Natural fre-quencies are necessary in vibration analysis and important when choosing a proper time step for a structural dynamics analysis.

Natural frequencies are obtained by solving the following equation:

0d d+ =M K The standard solution for d is given as ( ) ' i td t d e ω= where 'd is the part of the nodal displacement matrix called natural modes that is assumed to independent of time, i is the standard imaginary number, and ω is a natural frequency. Differentiating the above equation twice with respect to time gives:

( )2' i td d e ωω= −

Substituting the above expressions for d and d into the equation of motion gives:

2 ' ' 0i t i td e d eω ωω− + =M K Combining terms gives:

( )2 ' 0i te dω ω− =K M

Since eiωt is not zero, then:

( )2 ' 0dω− =K M

The above equations are a set of linear homogeneous equations in terms of dis-placement mode 'd . There exist a non-trivial solution if and only if the determi-nant of the coefficient matrix of 'd is zero.

2 0ω− =K M


Example Problem

Determine the first two natural frequencies for the bar shown in the figure be-low. Assume the bar has a length 2L, modulus of elasticity E, mass density ρ, and cross-sectional area A.

Let’s discretize the bar into two elements each of length L as shown below. We need to develop the stiffness matrix and the mass matrix (either the lumped- mass of the consistent-mass matrix). In general, the consistent-mass matrix has resulted in solutions that compare more closely to available analytical and ex-perimental results than those found using the lumped-mass matrix. However, when performing a long hand solution, the consistent-mass matrix is more diffi-cult and tedious to compute; therefore, we will use the lumped-mass matrix.

The elemental stiffness matrices are:

(1) (2)

1 2 2 3

1 1 1 1ˆ ˆ1 1 1 1

AE AEk kL L

− − = = − −

The global stiffness matrix is:

[ ]1 1 01 2 1

0 1 1

AEKL

− = − − −


The lumped-mass matrices are:

(1) (2)

1 2 2 3

1 0 1 0ˆ ˆ2 20 1 0 1AL ALm mρ ρ

= =

The global lumped-mass matrix is:

[ ]1 0 00 2 0

20 0 1

ALM ρ =

Substituting the above stiffness and lumped-mass matrices into the natural fre-quency equation

( )2 ' 0dω− =K M and applying the boundary condition d1x = 0 (or 1' 0d = ) gives:

22

3

'2 1 2 0 0'21 1 0 1 0

dAE ALdL

ρω −

− = −

Set the determinant of the coefficient matrix equal to zero as:

2 1 2 00

21 1 0 1AE ALL

ρλ−

− = −

where λ = ω2. Dividing the above equation by ρAL and letting 2E

Lµρ

= gives:

20

2

µ λ µλµ µ

− −=

− −

Evaluating the determinant of the above equations gives:

2 2λ µ µ= ± or

1 20.60 3.41λ µ λ µ= =


For comparison, the exact solution gives λ = 0.616µ, whereas the consistent-mass approach yields λ = 0.648 µ. Therefore, for bar elements, the lumped-mass approach can yield results as good as, or even better than, the results from the consistent-mass approach. However, the consistent-mass approach can be mathematically proven to yield an upper bound on the frequencies, whereas the lumped-mass approach has no mathematical proof of boundedness. The first and second natural frequencies are given as:

1 1 2 20.77 1.85ω λ µ ω λ µ= = = = The term µ may be computed as:

66 2

2 230 10 4.12 10

(0.00073)(100)E sL

µρ

−×= = = ×

Therefore, first and second natural frequencies are:

3 31 21.56 10 / 3.76 10 /rad s rad sω ω= × = ×

In general, an n-degree-of-freedom discrete system has n natural modes and frequencies. A continuous system actually has an infinite number of natural modes and frequencies. The lowest modes and frequencies are approximated most often; the higher frequencies are damped out more rapidly and are usually less important. Substituting λ1 into the following equation

22

3

'2 1 2 0 0'21 1 0 1 0

dAE ALdL

ρω −

− = −

gives:

(1) (1) (1) (1)2 3 2 31.4 ' ' 0 ' 0.7 ' 0d d d dµ µ µ µ− = − + =

where the superscripts indicate the natural frequency. It is customary to specify the value of one of the natural modes 'd for a given µ i or ωi and solve for the remaining values. For example, if (1)

3' 1d = than the solution for (1)2' 0.7d = . Simi-

larly, if we substitute λ2 and let (2)3' 1d = the solution of the above equations gives


(2)2' 0.7d = − . The modal response for the first and second natural frequencies are

shown in the figure below.

The first mode means that the bar is completely in tension or compression, de-pending on the excitation direction. The second mode means that bar is in com-pression and tension or in tension and compression.

Time-Dependent One-Dimensional Bar Example

Consider the one-dimensional bar system shown in the figure below.

Assume the boundary condition d1x = 0 and the initial conditions d0 = 0 and 0d = 0. Let ρ = 0.00073 lb-s2/in.4, A = a in.2, E = 30 x 106 psi, and L = 100 in. The bar will be discretized into two elements as shown below.


The elemental stiffness matrices are:

(1) (2)

1 2 2 3

1 1 1 1ˆ ˆ1 1 1 1

AE AEk kL L

− − = = − −

The global stiffness matrix is:

[ ]1 1 01 2 1

0 1 1

AEKL

− = − − −

The lumped-mass matrices are:

(1) (2)

1 2 2 3

1 0 1 0ˆ ˆ2 20 1 0 1AL ALm mρ ρ

= =


[ ]1 0 00 2 0

20 0 1

ALM ρ =

Substitute the global stiffness and mass matrices into the global dynamic equa-tions gives:

1 1 1

2 2

3 3 3

1 1 0 1 0 01 2 1 0 2 0 0

20 1 1 0 0 1 ( )

x x

x x

x x

d d RAE ALd dL

d d F t

ρ − − − + =

−

where R1 denotes the unknown reaction at node 1.

For this example, we will used the central difference method, because it is easier to apply, for the numerical time integration. It has been mathematically shown that the time step ∆t must be less than or equal to two divided by the highest natural frequency.


2

max

tω

∆ ≤

For practical results, we should use a time step defined by:

3 24 max

tω

∆ ≤

An alternative guide (used only for a bar) for choosing the approximate time step is:

x

Ltc

∆ =

where L is the element length, and xx

Ec ρ= is called the longitudinal wave

velocity. Evaluating the time step estimates gives:

33

3 2 1.5 0.40 104 3.76 10max

t sω

− ∆ = = = × ×

3

6

100 0.48 1030 10

0.00073x

Lt sc

−∆ = = = ××

Guided by these estimates for time step, we will select ∆t = 0.25 x 10-3 s.

Procedure for solution: 1. Given: d1x = 0 (fixed end), all nodal displacements, velocities are zero at time t = 0, d0 = 0 and 0d = 0, also 1xd = 0 at all times.

2. Solve for 0d at t = 0.

( )10 0 0d d−= −M F K

122

03 0

0 0 2 1 020 1 1,000 1 1 0

x

x t

d AEdAL Ld ρ

=

− = = − −

Applying the boundary conditions d1x = 0 and 1xd = 0 and simplifying gives:


220

3 0

0 020001 27,400

x

x

d ind sALd ρ = = =


2

1 0 0 0( )( )

2td d t d d−

∆= − ∆ +

Applying the initial conditions for d0 and 0d and 1xd from Step 2 gives:

3 22 3

33 1

0 0(0.25 10 )0 (0.25 10 )(0)2 27,400 0.856 10

x

x

din

d

−−

−−

×= − × + = ×

4. Solve for d1 at t = ∆t using the value of d-1 from Step 3

( ) ( ) 2 211 0 0 12d t t d d−

− = ∆ + − ∆ − M F M K M

( )1 222 3

3 1

0 0 2 02 2(0.073)0.25 100.073 20 1 1,000 0 1

x

x

dd

− = × +

( ) ( )23 43

2 1 0 2 0 00.0730.25 10 30 1021 1 0 0 1 0.856 10

−−

− − × × − − ×

Simplifying the above equation gives:

122

3 33 1

0 0 020.073 0 1 0.0625 10 0.0312 10

x

x

dd − −

= − × ×

The nodal displacements at t = 0.25 x 10-3 are:

23

3 1

00.858 10

x

x

din

d −

= ×

5. With d0 given and d1 determined in Step 4 solve for d2

( ) ( ) 2 212 1 1 02d t t d d− = ∆ + − ∆ − M F M K M


( )1 222 3

3 2

0 0 2 02 2(0.073)0.25 100.073 20 1 1000 0 1

x

x

dd

− = × +

( ) ( )23 43

2 1 0 2 0 00.0730.25 10 30 1021 1 0.858 10 0 1 0

−−

− − × × − − ×


−

− −

× = − × ×

3122

3 33 2

0 0 0.0161 1020.073 0 1 0.0625 10 0.0466 10

x

x

dd


32

33 2

0.221 102.99 10

x

x

din

d

−

−

×=

×

6. Solve for 1d

( )11 1 1d d−= −M F K

12 42

33 1

0 0 2 1 02 (30 10 )0.073 0 1 1000 1 1 0.858 10

x

x

dd −

− = − × − ×


22

3 1

3,52620,345

x

x

d insd

=


2 01 2( )

d ddt

−=

∆


( )

3

32

1 33 1

00.221 1002.99 10 0.442

5.982 0.25 10x

x

d ind sd

−

−

−

× −

× = ⇒ = ×

8. Repeat Steps 5, 6, and 7 to obtain the displacement, acceleration, and velocity for other time steps. Repeating Step 5:

( ) ( ) − = ∆ + − ∆ − 2 21

3 2 2 12d t t d dM F M K M

( )1 222 3

3 3

0 0 2 02 2(0.073)0.25 100.073 20 1 1000 0 1

x

x

dd

− = × +

( ) ( )−

−−−

− × − × × − − ××

323 433

2 1 2 0 00.221 10 0.0730.25 10 30 1021 1 0 1 0.858 102.99 10


3122

3 33 3

0 0 0.080 1020.073 0 1 0.0625 10 0.135 10

x

x

dd

−

− −

× = + × ×


32

33 3

1.096 105.397 10

x

x

din

d

−

−

×=

×

Repeating Step 6:

( )12 2 2d d−= −M F K

312 42

33 2

0 0 2 1 0.221 102 (30 10 )0.073 0 1 1000 1 1 2.99 10

x

x

dd

−

−

− × = − × − ×


Simplifying the above equation gives: =

22

3 2

10,5004,600

x

x

d insd

Repeating Step 7:

−=

∆3 1

2 2( )d dd

t

( )

3

332

2 33 2

01.096 100.858 105.397 10 2.192

9.0782 0.25 10x

x

d ind sd

−

−−

−

× − × × = ⇒ = ×

Beam Element Mass Matrices and Natural Frequencies

We will develop the lumped- and consistent-mass matrices for time-dependent beam analysis. Consider the beam element shown in the figure below.

The basic equations of motion are:

[ ] [ ] ( ) F t K d M d= +

where the stiffness matrix is:

2 2

3

2 2

1 1 2 2

12 6 12 66 4 6 2ˆ12 6 12 66 2 6 4

d dy y

L LL L L LEIk

L L LL L L L

φ φ

− − = − − − −

and the lumped-mass matrix is:


1 1 2 2

1 0 0 00 0 0 0ˆ

2 0 0 1 00 0 0 0

d dy y

ALm

φ φ

ρ =

The mass in lumped equally into each transitional degree of freedom; however, the inertial effects associated with any possible rotational degrees of freedom is assumed to be zero. A value for these rotational degrees of freedom could be assigned by calculating the mass moment of inertia about each end node using basic dynamics as:

3

24ALI ρ

=

The consistent-mass matrix can be obtained by applying

[ ] [ ]ˆ T

V

m N N dVρ= ∫

[ ]1

21 2 3 4

30

4

ˆ ˆL

A

NN

m N N N N dA dxNN

ρ

=

∫ ∫

where

( ) ( )3 2 3 3 2 2 31 23 3

1 1ˆ ˆ ˆ ˆ2 3 2N x x L L N x L x L xLL L

= − + = − +

( ) ( )3 2 3 2 23 43 3

1 1ˆ ˆ ˆ ˆ2 3N x x L N x L x LL L

= − + = −

Substituting the shape functions into the above mass expression and integrating gives:


2 2

2 2

156 22 54 1322 4 13 3ˆ[ ]

420 54 13 156 2213 3 22 4

L LL L L LALm

L LL L L L

ρ−

− = − − − −

Example Problem

Determine the first natural frequency for the beam shown in the figure below. Assume the bar has a length 2L, modulus of elasticity E, mass density ρ, and cross-sectional area A.

Let’s discretize the beam into two elements each of length L. We will use the lumped-mass matrix. We can obtained the natural frequencies by using the fol-lowing equation.

2K M 0ω− =

The boundary conditions are d1x = d3x = 0 and φ1 = φ3 = 0. Therefore the global stiffness matrix is:

2 2

3 2

24 0K

0 8

yd

EIL L

φ

=


2 0M

2 0 0ALρ

=

Substituting the global stiffness and mass matrices into the global dynamic equa-tions gives:

23 2

24 0 1 00

0 8 0 0EI ALL L

ω ρ

− =


Dividing by ρAL and simplify

24

24EIAL

ωρ

=

or

24.90 EIL A

ωρ

=

The exact solution for the first natural frequency is:

25.59 EIL A

ωρ

=

Example Problem

Determine the first natural frequency for the beam shown in the figure below. Assume the bar has a length L = 30 in, modulus of elasticity E = 3 x 107 psi, mass density ρ = 0.00073 lb-s2/in, and cross-sectional area A 1 in2, moment of inertia I = 0.0833 in4, and Poisson’s ratio ν = 0.3.

Let’s discretize the beam into two elements each of length L = 15 in. We will use the lumped-mass matrix. We can obtained the natural frequencies by using the following equation.

2K M 0ω− =

The problem is similar to the previous problem. The solution for the first natural frequency is:

23.148 EI

L Aω

ρ=

The exact solution for the first natural frequency is:


23.516 EI

L Aω

ρ=

According to vibration theory for a clamped-free beam, the higher natural fre-quencies to the first natural frequency is given as:

32

1 1

6.2669 17.5475ωωω ω

= =

The figure below shows the first, second, and third mode shapes corresponding to the first three natural frequencies for the cantilever beam.


The table below shows various finite element solutions compared to the exact so-lution.

ω1, (rad/s) ω2, (rad/s) Exact Solution 228 1,434

Finite Element Solution Using 2 elements 205 1,286 Using 6 elements 226 1,372

Using 10 elements 227.5 1,410 Using 30 elements 228.5 1,430 Using 60 elements 228.5 1,432

Truss and Plane Frame Analysis

The dynamics of trusses and plane frames are preformed by extending the concepts of bar and beam element. The truss element requires the same trans-formation of the mass matrix from local to global coordinates as that used for the stiffness matrix given as:

ˆT TTm m=

Truss Elements Since the motion of the element is now in two- or three-dimension, the bar ele-ment mass matrix must be reformulated to account for the axial and transverse inertial properties in the x and y directions.


Considering two-dimensional motion, the axial and the transverse displacement are given as:

1

1

2

2

ˆ

ˆˆ ˆ ˆ0 01ˆˆ ˆ ˆ0 0ˆ

x

y

x

y

d

du L x xLv L x x d

d

−

= −

The shape functions for the matrix are:

[ ]ˆ ˆ0 01

ˆ ˆ0 0L x x

NL L x x

− = −

The consistent-mass matrix can be obtained by applying:

ˆ[ ] [ ] [ ]T

V

m N N dVρ= ∫

2 0 1 00 2 0 1ˆ

6 1 0 2 00 1 0 2

ALm ρ =

The lumped-mass matrix for two-dimensional motion is obtained by simply lump-ing mass at each node and remembering that mass is the same in both the x and y directions, The lumped-mass matrix is:

1 0 0 00 1 0 0ˆ

2 0 0 1 00 0 0 1

ALm ρ =

Frame Elements The plane frame element requires combining the bar and beam elements to

obtain the local mass matrix. There are six degrees of freedom associated with a plane frame element.


The plane frame analysis requires first expanding and then combining the bar

and beam mass matrices to obtain the local mass matrix. The bar and beam mass matrices are expanded to a 6 x 6 and superimposed. Combining the local axis consistent-mass matrices for the bar and beam elements gives:

22

2 2

2 16 6

156 54 1322420 420 420 420

13 322 4420 420 420 420

1 26 6

54 13 156 22420 420 420 420

13 3 22 4420 420 420 420

0 0 0 00 00 0ˆ[ ]

0 0 0 00 00 0

LL

L LL L

L L

L L L L

m ALρ

−

−

−

− − −

=

The resulting lumped-mass matrix for a plane frame element is give as:

ˆ ˆ ˆ ˆˆ1 1 2 2 21

1 0 0 0 0 00 1 0 0 0 00 0 0 0 0 0ˆ[ ]

2 0 0 0 1 0 00 0 0 0 1 00 0 0 0 0 0

d d d dx y x y

ALm

φ φ

ρ

=

The global mass matrix for the plane frame element arbitrarily oriented in x-y co-ordinates is transformed by:


[ ]0 0 0

0 0 0i j m

i j m

N N NN

N N N

=

ˆT TTm m= where the transformation matrix is given as:

−

−

=

1000000000000000010000000000

CSSC

CSSC

T

Long-hand solution to the truss and frame problem are quite tedious and lengthy; therefore, we will use a computer problem to generate approximation for the mo-tion of truss and frame structures.

Plane Stress/Strain Elements The plane stress/strain constant-strain triangle consistent-mass matrix is ob-tained using the shape functions given below as:

The consistent-mass matrix can be obtained by applying: ˆ[ ] [ ] [ ]T

V

m N N dVρ= ∫

where dV = tdA The CST global consistent-mass Matrix is:


2 0 1 0 1 00 2 0 1 0 11 0 2 0 1 0

[ ]12 0 1 0 2 0 1

1 0 1 0 2 00 1 0 1 0 2

tAm ρ

=

Example Problem

Determine the motion of the frame structure shown below.

Assume the modulus of elasticity E = 3 x 107 psi. The mass densities ρ are ob-tained by dividing the total mass of each floor by the cross-sectional area and length the element. For example, consider the element 6:

( )( )( )2

266

104 30 15121

386.4 ins

psf ft ftW lb sM ing⋅= = =

22

46 2121 0.0136

(24.7 )(360 )

lb sin lb s

inin inρ

⋅⋅= =


Use Newmark’s method with β = ¼ and γ = ½. The following is the input file for WinFElt.

problem description title=îdynamic frame analysisî nodes=8 elements=9 analysis=transient

analysis parameters beta=0.25 gamma=0.5 alpha=0.0 duration=0.8 dt=0.05 nodes=[8,6,3] dofs=[Tx] mass-mode=lumped

nodes 1 x=0 y=0 constraint=fixed 2 x=360 y=0 3 x=0 y=180 constraint=free force=f1 4 x=360 5 x=0 y=300 force=f2 6 x=360 7 x=0 y=420 force=f3 8 x=360

beam elements 1 nodes=[1,3] material=wall_bottom 2 nodes=[3,5] material=wall_top 3 nodes=[5,7] 4 nodes=[7,8] material=floor_top load=top_wt 5 nodes=[5,6] material=floor_bottom load=bottom_wt 6 nodes=[3,4] load=bottom_wt 7 nodes=[8,6] material=wall_top 8 nodes=[6,4] 9 nodes=[4,2] material=wall_bottom

material properties wall_bottom A=13.2 Ix=249 E=30e6 rho=0.0049 wall_top A=6.2 Ix=107 E=30e6 rho=0.0104 floor_top A=12.3 Ix=133 E=30e6 rho=0.01315 floor_bottom A=24.7 Ix=237 E=30e6 rho=0.0136

distributed loads top_wt direction=perpendicular values=(1,-62.5) (2,-62.5) bottom_wt direction=perpendicular values=(1,-130) (2,-130)

forces f1 Fx=1000*(t < 0.2 ? 25*t : 5) f2 Fx=800*(t < 0.2 ? 25*t : 5) f3 Fx=500*(t < 0.2 ? 25*t : 5)

constraints fixed Tx=c Ty=c Rz=c free Tx=u Ty=u Rz=u

end


The following is the WinFElt output

------------------------------------------------------------------ time Tx(8) Tx(6) Tx(4) ------------------------------------------------------------------ 0 0 0 0 0.05 0.0054775 0.0046834 0.0047332 0.1 0.032795 0.028894 0.026946 0.15 0.10231 0.092341 0.078059 0.2 0.23314 0.21232 0.16267 0.25 0.43808 0.39636 0.27818 0.3 0.71526 0.63623 0.41441 0.35 1.0528 0.91484 0.56253 0.4 1.4335 1.2132 0.71742 0.45 1.8341 1.5124 0.87255 0.5 2.2269 1.7954 1.0185 0.55 2.5809 2.0464 1.1461 0.6 2.8674 2.2528 1.2504 0.65 3.0641 2.4041 1.3297 0.7 3.1589 2.4924 1.3824 0.75 3.1498 2.511 1.4031 0.8 3.0433 2.4556 1.3844


Example Problem

Determine the motion of the frame structure shown below. This problem is the same as the previous example, except for the loading function F(t) and the time duration.


The following is the WinFElt output

------------------------------------------------------------------ time Tx(8) Tx(6) Tx(4) ------------------------------------------------------------------ 0 0 0 0 0.05 0.0054775 0.0046834 0.0047332 0.1 0.032795 0.028894 0.026946 0.15 0.10231 0.092341 0.078059 0.2 0.21123 0.19359 0.14374 0.25 0.32881 0.29952 0.18933 0.3 0.43722 0.38244 0.20996 0.35 0.52949 0.43491 0.22408 0.4 0.59182 0.45828 0.23644 0.45 0.61612 0.45612 0.23879 0.5 0.59851 0.42714 0.22158 0.55 0.53509 0.37319 0.1881 0.6 0.42314 0.30076 0.1489 0.65 0.27506 0.21612 0.11208 0.7 0.11427 0.11995 0.07504 0.75 -0.041956 0.012366 0.028051 0.8 -0.18758 -0.10174 -0.034383 0.85 -0.31656 -0.21548 -0.10737 0.9 -0.41638 -0.31995 -0.17587 0.95 -0.48337 -0.40479 -0.22728 1 -0.52583 -0.45861 -0.25632 1.05 -0.55228 -0.47809 -0.26688 1.1 -0.55737 -0.46626 -0.26286 1.15 -0.5321 -0.42613 -0.24354 1.2 -0.4741 -0.3581 -0.20324 1.25 -0.38481 -0.26715 -0.13927 1.3 -0.26441 -0.16487 -0.062339 1.35 -0.11615 -0.058684 0.0067415 1.4 0.046209 0.05035 0.051067 1.45 0.20774 0.16045 0.076603 1.5 0.35642 0.26056 0.10478 1.55 0.48124 0.34101 0.15095 1.6 0.56682 0.40072 0.20559 1.65 0.60441 0.44251 0.24564 1.7 0.59778 0.46276 0.25798 1.75 0.55525 0.45409 0.24836 1.8 0.48211 0.41509 0.22947 1.85 0.38454 0.3486 0.20213 1.9 0.27609 0.25766 0.15744 1.95 0.16472 0.14464 0.088264 2 0.046915 0.019038 0.0033795


Problems

22. Do problems 16.5 and 16.11 on pages 611-613 in your textbook “A First Course in the Finite Element Method” by D. Logan.

23. Do problems 16.14 and 16.16 on pages 613 - 614 in your textbook “A First Course in the Finite Element Method” by D. Logan using WinFElt.

Documents

Fem by Logan