6. Derivation of the finite element equations - the Galerkin method
The rod and beam elements have properties defined by classical theory, and we can thus obtain a
local matrix describing the element properties directly. The local matrix for the 3-node constant
strain triangular element for stress analysis can also be derived by direct consideration of the
deformation of a piece of material. These are called direct methods.
The 'classical' method of deriving finite element equations for continuum problems is from a
variational principle. These are integral formulations that can be difficult to interpret for some
problems, though in others they are equivalent to eg the principle of minimum potential energy,
or of virtual work etc. An integral formulation is essential to the finite element method, but it is
possible to start with a more familiar differential equation and modify it to obtain an integral
equation. This has the advantage that no integral principle need be known in advance; indeed
there is no known variational principle for some equations. We shall describe the Galerkin
method, which is itself a classical method, but which has been applied with vastly increased
usefulness to finite elements. To simplify the algebra and visualisation, we shall consider first a
1D problem, although this does not provide a very convincing motivation for finite elements,
since the solution can be obtained readily. Here, this will allow us to compare results easily.
A one-dimensional heat-conduction problem
The problem and its exact solution
A uniform metal bar, length 2m, is insulated at one end, whilst the temperature is fixed at the
other. A uniform heat source, Q per unit length, is applied to the whole bar (eg and electric
heating coil), and heat conducts along the bar. The thermal conductivity is k. Find the temperature
everywhere in the bar.
2m
k
100
Q
x
The equation of heat conduction is in this case
k
d 2T
 Q  0,
dx 2
for x in ( 0, 2 )
which is equivalent to Poisson's equation in one dimension.
Exercise
(i) Sketch a rough graph of the temperature within the bar (consider the direction in
which heat flows, and the amount flowing for different values of x as a guide to the
temperature gradient within the bar).
(ii) Integrate the equation twice to obtain the exact solution in terms of k and Q, using the
boundary conditions T  100 at x  0, and kdT dx  0 at x  2 , where the second is
a statement of zero heat flux at that boundary. Also find the heat flux at x  0 .
24
Mesh and shape functions
We'll model this problem domain as two linear elements of length 1. For each element the shape
functions will be the same:
(1)
(2)
1
2
N
1
N2
3
N2
N3
1
1
2
3
An alternative way of looking at the shape functions is to define them over the whole problem
domain, so that they are 1 at their 'own' node, decrease to 0 at surrounding nodes, and are
identically zero over all elements beyond those adjacent to the node. Thus the three shape
functions here can be seen as
N1
N
N
2
3
These can be represented mathematically as
0 x 1
0 x 1
1  x 0  x  1
 x
 0
N1  
, N2  
, N3  
1 x  2
 0
2  x 1  x  2
x  1 1  x  2
Then our overall representation of the temperature becomes simply
T  N1T1  N2 T2  N3 T3
(*)
over the whole domain, rather than a separate representation within each element. This makes it
easier to write down the integral expressions for the whole problem. It is also clear that what we
are using is a piecewise continuous function to represent the solution. Because of the
discontinuities of slope, such a function can cause problems in a differential formulation, but this
is perfectly acceptable in an integral form.
However, since any integral over a domain can be written as the sum of integrals over subdomains, when it comes to the evaluation of the integrals, it is convenient to integrate over each
element separately, and we return to our original idea of shape functions, defined locally for a
single element. This is how we come to create parts of equations, which are summed (assembled)
to form the global matrix of coefficients.
Weighted residual and Galerkin methods
We return now to the differential equation, and how we can modify it to produce an integral
equation, where we will utilise our representation (*).
The differential equation is in the form (...) = 0. If we have an approximate solution T, we can
d 2T
expect that we do not obtain zero everywhere, but instead have a residual equal to k 2  Q .
dx
This will be 0, >0 or <0 in different regions. We now require that the residual is zero on average
25
in some sense. In fact, we require that a weighted average of the residual is zero. This is called a
weighted residual method.
We assume a weighting function w(x), and thus obtain
2
k  w( x )
0
2
d 2T
dx  Q w( x )dx  0 , since k , Q are constant .
2
0
dx
Next observe that if we simply put our representation () into this equation, the second derivative
will be identically zero! To reduce the order of differentiation, and also to obtain other
advantages, we next integrate the first term by parts to obtain:
2
2 dw dT
2
 dT 
k w   k 
dx  Q wdx  0
0 dx dx
0
 dx  0
There are three gains in using this modified weak formulation:



We can use lower order shape functions, since the order of derivatives is lower.
A boundary term has appeared, allowing us to bring in the boundary conditions of the
problem.
This weak formulation is often more realistic physically than the original differential
equation, since discontinuities in geometry and materials are handled without problem within
integrals.
The boundary term will be zero if either w = 0 or dT/dx = 0 at each of x = 0 and x = 2. Otherwise
it must be evaluated to impose a boundary condition of non-zero flux, or it may be an unknown
flux to be found as part of the solution (which is the case at x = 0 for this problem).
We next consider the choice of weighting function w(x). The Galerkin method applies the same
form for w(x) as for the unknown variable. That is, we use one or more of the shape functions
N1 to N3 as w. This has the following results:




By using all of the available shape functions, we obtain as many equations as we have degrees
of freedom.
Because the derivative of N1 to N3 appear in both factors within the integrand of the second
term, we obtain a symmetric matrix.
There is also an effective squaring of errors, which, like the least-squares method for fitting
data, avoids cancellation of errors, and acts to minimise the largest errors.
Since the shape functions are greatest at and around the nodes, the weighting of the residual is
greatest there, and hence the errors tend to be smallest at and near the nodes.
Proceeding for the present pragmatically, rather than systematically, we observe that at x = 0,
N 2 and N3 are both zero, while at x = 2, dT/dx = 0. Therefore, if we use N 2 and N3 for our
weighting function w(x), the boundary term will be zero in both cases. Moreover, we will obtain
two equations, which is exactly what we need to solve for our two primary unknowns, T2 and T3 .
dN
dN
dw
dT dN1
is present, and also

T1  2 T2  3 T3 , so we need the
dx
dx
dx
dx
dx
derivatives of N1 to N3 , which are
In the modified integral
dN1 1

dx
0
dN 3 0

dx
1
dN 2  1

dx
1
26
for
0  x  1

1  x  2
Therefore,
dT   T1  T2

dx  T2  T3
and we can proceed to evaluate the integrals.
First, using N2 for w, we get
1
2
2
 k  (1)( T1  T2 )dx   ( 1)( T2  T3 )dx   Q N 2dx  0
 0

1
0
which gives us our first equation
 k  T1  2 T2  T3  Q  0.
(1)
Secondly, using N 3 for w, we get
1
2
2
 k  (0)( T1  T2 )dx   (1)( T2  T3 )dx   Q N 3dx  0
 0

1
0
which gives another equation
 k  T2  T3  12 Q  0.
(2)
Exercise
Solve the two equations (1) and (2) for the unknowns T2 , T3 in terms of k,Q and using the
known value of T1 . Compare with the exact solution already obtained.
Assume some values for k,Q, and plot graphs of the exact solution and the F.E. solution
on the same diagram (you know, since we assumed it, how the F.E. solution behaves
between the nodes). Comment.
We have now solved our original problem for the primary variable, T. However, we also wished
to find the heat flux at x = 0. Also, if we changed the boundary conditions, eg fixing T at x = 2
instead of dT/dx but not fixing T1 , we would need another equation. In practice, and to be
systematic rather than expedient, we derive all the finite element equations before considering the
boundary conditions at all. That is, we use all three of N1 to N3 for w(x) and obtain three
equations relating T1 , T2 and T3 to the known and unknown fluxes attributed to the nodes (in 2D
or 3D this may be an integrated flux for the region around the node, as described in Section 5).
Exercise
Derive the equation using N1 instead of N 2 and N3 and following exactly the same
procedure. However, note that kdT/dx at x = 0 must be kept in the equation as an
unknown. Put this term on the right-hand side.
Our final set of equations becomes
 21 
 1 1 0   T1 
 
 k 1 2 1 T2   Q1 
 21 
 0 1 1  T3 
 ?
0
 
0
Note the symmetric, banded matrix, characteristic of finite-element problems.
Exercise
27
Solve for the unknown flux kdT/dx at x = 0 using the first equation (which we derived
last!). Compare it with the exact value (which should be obvious from the original
problem) and comment. Observe whether it is an inward or outward flux and relate this to
its sign.
7. Two-dimensional problems
7.1 Two-dimensional heat conduction
Here we shall give only a summary, to show that the theory is exactly the same, but with
integration etc in two variables, not one.
The differential equation is
 kT   Q  0 ,
where k,Q may be functions of x. The finite element method permits different values of k,Q to be
used in each element. We represent the unknowns as T  N j Tj , where Tj are the nodal values.
The weighted residual gives
 N  kT dA   N QdA  0 ,
i
i
domain
domain
which is integrated by parts to give

kNi
boundary
T
ds   kN . TdA   Ni QdA  0 .
n
domain
domain
Since the integrals can be evaluated over each element and summed, we obtain


T
ds    N i QdA
   kN i . N j dA Tj    kN i

n
elements
elements each elem


each elem
boundary
The first bracket gives the matrix of coefficients, multiplying the unknown values of T.
Exercise
(i) (optional) Derive the shape functions for a general 3-node triangle with nodes at
( x1 , y1 ) etc. You should obtain the expressions
N i  ( ai  bi x  ci y ) / 2 A, where ai  x j y k  x k y j , bi  y j  y k , and ci  x k  x j ,
and A is the area of the element , given by the determinant
1 x1
y1
2 A  1 x2
y2
1 x3
y3
and where i,j,k are (1,2,3) or (2,3,1) or (3,1,2) respectively.
(ii) For the four element heat conduction problem studied before (see section 3(ii)), derive
the shape functions for each of the elements, either directly or using the general
formula from part (i). Here, nodes 1,2 and 3 refer to whichever are the 1st, 2nd and
28
3rd, numbering anticlockwise around the element. For element 3, for example, these
are 2,5,3 respectively, and you should get
N2  x  y, N5  1  x  y, N3  2  2 x .
(iii) Construct the 2x3 matrix B of derivatives of shape functions, where
 
 
B   x   N1
 
 y 
N2
N3 
for each of the four elements in turn (note that all the entries are 0, 1 or 2 for this
example). Then construct the 3x3 matrix BTB, and integrate by simply multiplying by
the area of the element and by kt (since all entries are constants). For element 3 you
should obtain the 3x3 matrix given before. If you work with the general form of the
shape functions instead, you will get the general expression for B, which you can then
use to construct the 'stiffness matrix' K. We obtain
Ki , j 


kt
bib j  ci c j .
4A
(iv) Note that each local matrix represents part of three of the equations for the system. For
element 3, we have coefficients of the three nodal variables T2 , T5 and T3 in the 2nd,
5th and 3rd equations respectively. Assemble the global matrix for the problem (the
5x5 matrix previously seen) by simply adding together the contributions from these
part-equations. Thus equation 2 gets the contribution kt( 0. 5T2  0 T5  0. 5T3 ) from
element 3, and kt( 0 T1  0. 5T2  0. 5T3 ) from element 1 (which has shape functions
N1  1  x  y, N2  x  y, N3  2 y ). The assembled equation 2 thus has left-hand
side: kt( 0 T1  1T2  1T3  0 T4  0 T5 ) .
29
7.2 Plane stress
(In the following, we use the equilibrium equations for stress components in two dimensions, and
the matrix D giving the relationship between components of stress and strain. These are derived
in Appendices 1 and 2.)
There are two degrees of freedom, u,v, and we consider three components of strain (the
component perpendicular to the plate can be calculated subsequently):
x  
u / x

  

v / y
y   

 xy  u / y  v / x 
 
or   Lu
0 
 / x
 0
 u

/

y

 v 
 / y  / x   
and three of stress
 x 
E
 
 y   1   2
 xy 
 


0 x 
1 
 
 1
0 y 

1  
0 0
  xy
2  

or   D
The equilibrium equations can be written
 y  xy
 x  xy

 0,

0
x
y
y
x
which is the same as
 
0
 y   x 
 x
 y   0 , or LT  0
 0


 y  x 

 xy 
 
Now we make use of the shape functions, and, assuming we have a 3-node triangle, we get
u  N1
v    0
  
0
N1
N2
0
0
N2
N3
0
 u1 
v 
 1
0  u2 
 
N 3  v2 
u3 
 
 v3 
or u  Nu
where u are the displacements at nodes.
We now have
  D  DLu  DLNu 
and we let B = LN, so B is the matrix of derivatives of shape functions.
30
We return to the equilibrium equations. Using the shape functions (recorded in matrix form) as
weighting functions, we obtain as our weighted residual:
 N
T
( LT )dA  0, or
 N
T
( LT DLNu )dA  0
Integrating by parts, we get (remembering that L is a first-order differential operator and that the
nodal values u are constants):
  LN  D LN dA u
T

 integrated stresses, ie forces.
If we also sum over elements, and use the definition B = LN, we obtain
  B DBdA u

T
F .
The bracketed expression gives us the global stiffness matrix for the problem.
Exercise
(i) For the original bracket problem, derive the three shape functions N1 , N2 , N3 , either
from first principles, or using the general formula derived in Exercise (i) in the
previous section. Write these as entries of the matrix N above. Then apply the matrix
of derivatives L to construct B (note that the entries of B are all constants, and keep a
common factor of 5 outside).
(ii) Calculate the matrix DB (a 3x6 matrix) and then BTDB (6x6), keeping the factors
25 E / (1   2 ) outside. Now, since all entries are constants, the integral needed to
produce the matrix of coefficients is simply this multiplied by A, the area, and t, the
thickness of the element (since the forces we consider are really forces per unit
thickness). You should now have the matrix given in section 3(i).
7.3 Related problems
Having gone through the derivation of the finite element equations for plane stress, we have
covered all the essential theory for a wider range of stress problems. For plane strain, the only
change is to the elasticity matrix D. For axisymmetric problems, a fourth component of strain and
stress ('hoop') is introduced, but the element definitions are the same and the final global matrix is
of the same size. For 3D problems, there are six components of strain and stress, and D is also
6x6. But in all cases, the general vector/matrix form above applies in exactly the same way.
31
8. Numerical integration and stress sampling
The equations relating the nodal variables take the form of integral equations, where the
unknowns are the coefficients of known shape functions, appearing within integrals. To evaluate
these integrals analytically would be complex even for regular shapes such as rectangles, and
despite using element mapping to enable the use of standard functions and domains.
The integrals can be evaluated most conveniently by numerical integration, which can be made
exact for functions of a given order over a regular domain, and a good approximation even for
mapped element domains (such as curved-sided quadrilaterals).
Moreover, the same theory can be used to obtain optimal results for stresses after the nodal
displacements are calculated for stress analysis problems.
8.1 Gaussian Quadrature (or Gauss-Legendre integration)
The formulae that are used for a two-dimensional element (which can be mapped to more
complex shapes) are of the form:
 f ( x, y)dA  w f ( x , y )  w
1
1
1
2
f ( x2 , y2 )  ......
For the standard quadrilateral element, the limits of integration are (-1,1) in both x and y.
8.1.1 Integration in a single variable
The basic idea is a formula for integration in a single variable:

1
1
f ( x)dx  w1 f ( x1 )  w2 f ( x2 ), with w1  w2  1 and - x1  x2  1 / 3 ,
(1)
which is exact if f(x) is a cubic function of x, and may be a good approximation for other
functions.
[This can be derived by writing the form of what is needed, and, recognising that there are four
parameters, requiring the result to be exact for a general cubic function, namely:

1
1
ax3  bx 2  cx  d dx  w1 (ax13  bx12  cx1  d )  w2 (ax23  bx22  cx2  d ) .
Since this is an identity for any choice of the four coefficients a,b,c,d, we can evaluate the lefthand side, then write down four equations for the coefficients and derive the four values given in
the formula (1).]
Similar formulae can be derived that are exact for higher order functions. For example,

1
1
f ( x)dx  w1 f ( x1 )  w2 f ( x2 )  w3 f ( x3 ), w1  w3  95 , w 2  89 , - x1  x3 
3
5
, x2  0 , (2)
which is exact for fifth-order polynomials (quintics).
8.1.2 Integration in two dimensions
The formulae (1) and (2) can be applied in two or three dimensions by repeating the integration
process. Thus (1) becomes, for a square with corners (-1,-1) to (1,1):
 f ( x, y) dxdy 
f (r ,r )  f (r , r )  f (r ,r )  f (r , r ), with r  1
32
3
where we refer to the four ‘Gauss points’, being the four points used to evaluate the function.
This formula is exact for functions that are cubic in both variables.
Similarly, (2) becomes :
25
81
 f ( x, y) dxdy 
 f (r ,r )  f (r , r )  f (r ,r )  f (r , r )  4081  f (r ,0)  f (r ,0)  f (0,r )  f (0, r )  6481  f (0,0),
with r 
3
5
, giving a ‘nine point’ Gauss formula, exact for functions that are fifth-order in both
variables (the four-point and nine-point schemes are illustrated).
Formulae for integration in three dimensions follow the same pattern. Similarly, formulae can be
derived for integration over triangular regions (the three-point and seven-point schemes are
illustrated), and for tetrahedra and wedge shapes in three dimensions.
8.2 Finite Element integrals using Gauss points
When we use the incomplete cubic shape functions for the eight-node quadrilateral, the first
derivatives present in N i give functions quadratic in x and y. Products of these give functions
that are quartic (fourth-order) in x and y. Therefore, even for square or rectangular elements,
exact integrals would require the nine-point Gauss formula, though this would be only
approximate for an element with curved sides.
However, if the element is not too distorted, it was shown by Barlow (1966) that better results
could be obtained by using the simpler, four-point, scheme. This is called ‘reduced integration’,
and the improved accuracy results from a cancellation of errors. At the same time, it corresponds
to the use of elements that do not match up, in displacements, at all points on the element sides.
Such elements are called ‘non-conforming’. It may seem paradoxical that a cruder integration
scheme, and elements that don’t match up so well, produces better results, but it is often so, and
finite element programs usually take account of this result. FINEL uses reduced integration by
default, but the command *full can be used to enforce the more complex scheme. More
sophisticated packages select one integration scheme or other depending on element geometry.
33
8.2.1 Stress sampling
For a stress analysis problem, the unknown displacements at the nodes are calculated first, then
the reaction forces at constrained nodes. After that, we usually want to calculate values of stress
either at selected points, or with the aim of producing contours of stress across the domain.
The stresses are evaluated from the expression:
  D  DLu  DLNu * (see section 7.2)
and in principle they could be evaluated everywhere that is needed. However, Barlow also
showed that the values thus produced can be seriously in error, an order of magnitude worse than
results for the displacements.
To illustrate what happens, consider the simple cantilever, modelled with four 8-node elements
and with only an end-load, so that a constant shear stress applies.

exact
The stresses, when calculated everywhere, show a wide variation, but note that at two values of x
in each element, the values are actually exact. It can be shown that the best two points to choose
for good stress results are precisely the Gauss points  1 / 3 if the interval (-1,1) is considered.
The best results for other points throughout the element are obtained by ‘sampling’ the stresses at
the Gauss points (using the four-point scheme), and then extrapolating from these points to
elsewhere in the element (e.g. to the nodes and the centroid, which is what FINEL does).
Sampling at these four points is optimal even when nine-point integration has been used.
A consequence of sampling the stresses within each element and extrapolating to the element
nodes (and edges generally), is that stress values do not match up between elements; this is a
further consequence of non-conforming elements, mentioned above. FINEL shows stress
contours with this feature. For crude meshes, there is a marked mismatch between the contours at
element boundaries. As the mesh is refined, there is a closer correspondence. The discontinuities
in the contours are one indication of the convergence of the results to the true values.
However, most finite element programs smooth their output, producing ‘nicer’ contour maps, but
concealing the limitations of the current mesh resolution, and hiding what may be useful
information for the engineer. FINEL offers ‘Smooth’ as an option, which gives a limited level of
contour smoothing, but this is not recommended.
34
9. Frameworks - rod and beam elements
Historically, rod and beam elements preceded continuum elements, in that they were natural
extensions of traditional matrix methods in structural engineering. Most texts, and many courses,
also start with these types of elements for the analysis of frameworks. In this course we have
concentrated on continuum problems, but we now consider briefly the principal features of
framework problems.
9.1 Rod elements
Rod elements, pin-jointed at their ends, are elastic elements that can only stretch or compress
longitudinally. Therefore, their only important properties are length, cross-sectional area and
Young's modulus. If gravity loading is required, their density is also relevant. FINEL also
calculates the Euler buckling load if the second moment of area is supplied, but this forms no part
of the primary analysis. Rod elements can be specified in FINEL by giving a problem type *2rod
or *3rod. Two node numbers are given for each element, as well as the information above.
Example data files are truss1 and beanpole.
v2
2
v1
1
u2
u1
In 2D, the variables are two displacements at each end as shown, and three for 3D problems. The
theory is very simple; Hooke's law is applied to calculate the elongation in 1D, and this is then
transformed to 2D or 3D coordinates. The assembly of the structure is, in effect, a statement of
equilibrium at each joint. Gravity loads are weights, acting in the -y direction, and shared equally
between the two nodes of any element.
In 1D, the displacements are simply u1 and u2 , and the forces on the ends X1 and X2 . With
Young's modulus E, area A and length L, we can write two (identical) equations
EA  1 1  u1   X 1 

    
L  1 1   u2   X 2 
relating displacements to forces. To transform to the general displacements above, we write these
again in terms of displacements u',v' and forces X',Y' for a rod oriented along the x-axis:
1

EA  0
L  1

0
0 1 0  u1   X 1 

0 0 0  v1   Y1 
  

0 1 0  u    X  
2
 2
0 0 0  v    Y  
 2  2 
Now, if the rod is actually at angle  to the global x-axis as shown
35
()
y,v,Y
x',u',X'
y',v',Y'

x,u,X
we get u2 '  u2 cos   v2 sin , v2 '  u2 sin   v2 cos , etc. If we define the matrices
 cos
t
  sin 
sin  
 t 0
 , and T  
,
cos 
0 t
we get u'  Tu . Note also that for this rotation matrix T 1  T T . We write the equations () as
K u   F  , and this gives us our 4x4 stiffness matrix for a rod in general orientation:
K Tu  TF , so ( T T K T ) u  F and hence Ku  F where K  T T K T .
We can write this explicitly, using c for cos , and s for sin  as
 c2
cs c2 cs


s2 cs  s2 
EA  cs
K
L  c2 cs c2
cs 


s2 
 cs  s2 cs
Note that if we impose a rigid-body motion, u1  u2 , v1  v2 , the corresponding forces are zero.
Exercise
(i) For the two rod assembly shown, where each rod has properties E,A,L, derive the 4x4
matrix for each rod (note
1   / 4    2 , so c  1 / 2 for both , while s  1 / 2 respectively ):
2
rod 1
E,A,L
rod 2
y
x
1
3
(ii) The matrix for rod 1 produces the whole of the coefficients for equations 1 and 2, and
part of those for equations 3 and 4, whilst rod 2 produces the other part of equations 3
and 4, and all of equations 5 and 6. Add together the corresponding entries to
assemble the global equations
36
 1 1 1 1 0 0   u1   X 1 

   
 1 1 1 1 0 0   v1   Y1 
EA  1 1 2 0 1 1   u2   X 2 

    
2 L  1 1 0 2 1 1  v2   Y2 
 0 0 1 1 1 1  u   X 

  3  3
 0 0 1 1 1 1   v3   Y3 
If you wish, set suitable boundary conditions at nodes 1 and 3, and apply a vertical
force Y2 to solve this problem. You can also construct a data file to run this example
with FINEL. For example, if u1  v1  u3  v3  0 , and Y2   F, X2  0 , we get
u2  0, v2   FL / EA, and X1  Y1   X3  Y3  F / 2
which also satisfies equilibrium, and ensures that the reactions act along the rods as
expected.
The derivation of the stiffness matrix in 3D follows the same pattern, but with two angles to
consider, and results in a 6x6 matrix for each element, since there are three degrees of freedom at
each end.
37
9.2 Beam elements
Beam elements have three degrees of freedom at each end (in 2D) or six (in 3D), including
bending and twisting as well as elongation. Elongation is as for rod elements, while bending is
normal thin beam theory, requring the second moment of area I to be specified for each element,
or two second moments in 3D. Consideration of torsion requires the polar second moment J also.
To derive the bending relationships in 2D, consider a beam with vertical forces and displacements
only, and with small rotations only, so that eg tan    etc.
2
v
1
Y1, v
M
2
,Y2
2
1
M
1
2D beam element showing applied forces and moments and degrees of freedom v, 
(displacements along the beam u are considered separately)
The equation of beam bending, if the curvature is small, is
d2y M

dx 2 EI
and here the moment M is linearly interpolated between the applied values M1 , M2 , so that
M  (1  x / L) M1  ( x / L) M2
Integrating twice, and using the relationship y    for small  with the conditions
y  v1 , y   1 at x  0, y  v2 , y   2 at x  L
we get the displacement as
y

 x3 
1   x2 x3 
M

  
 1   M 2   EI 1 x  v1 .
EI   2 6 L 
 6L

To relate the displacements and angles at the two ends to the forces and moments, we find y(L)y(0) and y'(L)-y'(0), ie v2  v1 and 2  1 , and finally rewrite, using
F1   F2  dM / dx  ( M1  M2 ) / L
to obtain
38
 12 EI / L3

2
 6 EI / L
 12 EI / L3

 6 EI / L2
6 EI / L2
12 EI / L3
4 EI / L
6 EI / L2
6 EI / L2
12 EI / L3
2 EI / L
6 EI / L2
6 EI / L2   v1   Y1 
   
2 EI / L    1   M1 

6 EI / L2   v2   Y2 
   
4 EI / L   2   M 2 
This (generalised) stiffness matrix is symmetric as before. Also note that if a general rigid body
motion v1  a, 1  2  b, v2  a  bL is imposed, no forces or moments are induced.
Transformation to a more general orientation is much as for rod elements, except that the angles
1 , 2 are unaffected. We write the matrix we have just obtained as rows and columns 2,3,5,6 of a
6x6 matrix, and enter the original rod matrix (not rotated) as rows and columns 1,4. Then we
define matrices t and T almost as before, namely
 cos

t    sin 

 0
sin 
cos
0
0
 t 0

0 , and T  
,
0 t

1
and obtain
K Tu  TF , so ( T T K T ) u  F and hence Ku  F where K  T T K T .
Derivation of the stiffness matrix for a beam in 3D follows the same initial stages for elongation
and bending of a bar aligned with the x-axis, but bending in two directions, with two second
moments of area I y and I z . We must also add the torsion about the axis of the bar. This works
exactly like Hooke's law, but with torque T, shear modulus G, angle of twist  and torsion
constant J. For a circular section, J is the polar second moment of area a 4 / 2 . For thin walled
sections made of rectangles, each rectangle of thickness t and breadth b  t contributes
approximately bt 3 / 3 . In general determination of J is difficult, and warping may occur. The
equations we need are
GJ  1 1   1   T1 

    
L  1 1    2   T2 
Once the elongation, two axes of bending, and torsion have been written into a 12x12 matrix, a
transformation of axes to a general orientation produces the stiffness matrix we require.
z'

y'
1
2
x'

3D beam (when not vertical) showing local axes and angle of orientation
- degrees of freedom are u , v , w ,  x ,  y ,  z at each node
39
Since the principal axes of the beam in 3D (eg is the web of an I-beam vertical or not?) may not
be along the global axes, the angle of these may need to be given also. Beam elements can be
specified using FINEL by giving a problem type *2beam or *3beam. Examples can be seen in
files truss2, pylon, conveyor.
For short, thick beams (length:breadth < 5), shear effects may be significant. FINEL incorporates
shear effects if suitable shear coefficients are specified; the theory is complex.
10. Bibliography
Most books that I have seen start with a very mathematical approach, using variational principles
to derive the finite element equations. Not many make the method really accessible. Here are a
few that have good features:
Zienkiewicz OC 1977 The Finite Element Method (McGraw-Hill) The standard reference
work, now in two volumes. Difficult to read, but a starting point and full of references to
original papers.
Huebner KH & Thornton EA
readable general text.
1982
The Finite Element Method for Engineers
Owen DRJ & Hinton E 1980 A Simple Guide to Finite Elements
introduction to the Galerkin method for heat conduction that I have seen.
A fairly
The best simple
Hinton E & Owen DRJ 1977 & 1979 Finite Element Programming (Academic Press) and
An Introduction to Finite Element Computations Two texts on programming methods.
There are many more, including recent ones I haven't looked at, and books aimed at certain
disciplines, such as fluid mechanics, structural engineering etc.
40