104
CHAPTER 5. THE FINITE ELEMENT METHOD (FEM)
Figure 5.4: Various planar roof trusses.
good choice to illustrate the transition from the hand-calculation methods
taught in undergraduate courses, to the fully automated finite element
analysis procedures available in commercial programs.
5.2.2
Idealization & the example truss
The first analysis step carried out by a structural engineer is to replace the
actual physical structure by a mathematical model. This model represents
an idealization of the actual structure. For truss structures, by far the
most common idealization is the pin-jointed truss, which directly maps to
a FEM model.
The replacement of real by idealized is at the core of the physical interpretation of the finite element method. The axially-carrying-load members
and frictionless pins of the pin-jointed truss are only an approximation of
the physical one. For example, building and bridge trusses usually have
members joined to each other through the use of gusset plates, which are
attached by nails, bolts, rivets or welds. Consequently members will carry
some bending as well as direct axial loading.
Experience has shown, however, that stresses and deformations calculated using the simple idealized model will often be satisfactory for pre-
5.2. THE DIRECT STIFFNESS METHOD (DSM)
105
liminary design purposes; for example to select the cross section of the
members. Hence the engineer turns to the pin-jointed assemblage of axialforce- carrying elements and uses it to perform the structural analysis.
To keep hand computations manageable we will use just about the simplest structure that can be called a plane truss, namely the three-member
truss illustrated in Fig. 5.5(a). The idealized model of this physical truss
as a pin-jointed assemblage of bars as shown in Fig. 5.5(b), In this idealization truss members carry only axial loads, have no bending resistance,
and are connected by frictionless pins. Geometric, material and fabrication properties of the idealized truss are given in Fig. 5.5(c), while idealized
loads and support conditions are provided in Fig. 5.5(d).
(a)
3
(b)
Idealization as Pin-Jointed Truss
and FEM Discretization
1
fy3 , uy3
(c)
3(10,10)
E (3) = 100, A(3) = 2 2,
(3)
L = 10 2.
(3)
y
fx1 , ux1
1(0,0)
fy1 , uy1
45 o
x
(1)
E (1) = 50, A(1) = 2,
(1)
L = 10.
2
f y3 = 1
fx3 , ux3
(d)
3
f x3 = 2
45o
(2)
E (2) = 50, A(2) = 1,
(2)
L = 10.
(3)
y
fx2 , ux2
2(10,0)
fy2 , uy2
1
x
(1)
(2)
2
Figure 5.5: The three-member simple example truss: (a) physical structure; (b) idealization as a pin-jointed bar assemblage; (c) geometric, material and fabrication properties; (d) support conditions and applied loads.
The pin-jointed idealization of the example truss, pictured in Fig. 5.5,
has three joints, which are labeled 1, 2 and 3, and three members, which
are labeled with the numbers in parentheses: (1), (2) and (3). Those members connect joints 1-2, 2-3, and 1-3, respectively. The member lengths
are denoted by L(1) , L(2) and L(3) , their elastic moduli by E (1) , E (2) and
E (3) , and their cross-sectional areas by A(1) , A(2) and A(3) , respectively.
Both E and A are assumed for simplicity to be constant along each mem-
106
CHAPTER 5. THE FINITE ELEMENT METHOD (FEM)
ber. In the general FEM, the names ‘‘joint’’ and ‘‘member’’ are replaced by
node and element, respectively. The geometry of the structure is referred
to a common Cartesian coordinate system {x, y}, which is usually called
the global, structure or overall coordinate system. In our example case, its
origin is positioned at joint 1; see Fig. 5.5.
In an idealized pin-jointed truss, all forces, externally applied as well as
reactions, can act only at the joints. All member axial forces can be characterized by the x and y components of these forces, denoted by fx and
fy , respectively. Hence, the force components at joint i will be identified
as fxi and fyi , respectively. The set of all joint forces can be arranged as a
column vector, with 6 components in our case, which we may call f . Furthermore, classical structural mechanics dictates that the displacements
of the truss are completely defined by the the joints’ displacements. This
statement is a particular case of the more general finite element theory.
The x and y displacement components are usually denoted by ux and uy ,
respectively. Their values at joint i will be denoted as uxi and uyi and the
set of displacement values can be arranged, as in the cases of joint forces,
in a column vector denoted as u. Hence, the forces and displacements
vectors at the joints for our truss example will be:




f =



fx1
fy1
fx2
fy2
fx3
fy3




,







u=



ux1
uy1
ux2
uy2
ux3
uy3








In the DSM these displacements contained in u are the primary unknowns. In pertinent literature they are also called degrees of freedom or
state variables of the system.
The Master Stiffness Equations
The master stiffness equations relate the joint forces f to the joint displacements u before specification of any support conditions. As we have
assumed a linear behavior for the truss, the stiffness equations need to be
linear relations. Furthermore, we also assume that no prestress / initial
stress is taken into consideration, which means that zero displacements
5.2. THE DIRECT STIFFNESS METHOD (DSM)
107
imply zero forces. Hence, we may write that:








fx1
fy1
fx2
fy2
fx3
fy3


 
 
 
=
 
 
 
Kx1,x1
Ky1,x1
Kx2,x1
Ky2,x1
Kx3,x1
Ky3,x1
Kx1,y1
Ky1,y1
Kx2,y1
Ky2,y1
Kx3,y1
Ky3,y1
Kx1,x2
Ky1,x2
Kx2,x2
Ky2,x2
Kx3,x2
Ky3,x2
Kx1,y2
Ky1,y2
Kx2,y2
Ky2,y2
Kx3,y2
Ky3,y2
Kx1,x3
Ky1,x3
Kx2,x3
Ky2,x3
Kx3,x3
Ky3,x3
Kx1,y3
Ky1,y3
Kx2,y3
Ky2,y3
Kx3,y3
Ky3,y3
 
 
 
 
·
 
 
 
ux1
uy1
ux2
uy2
ux3
uy3








or simply:
f = Ku
(5.1)
Here K is the master stiffness matrix, also called global stiffness matrix,
assembled stiffness matrix, or overall stiffness matrix. The terms master,
global and overall signify the fact that there is another stiffness matrix of
local nature that corresponds to structure’s elements. In our example,
the master stiffness matrix is a 6 × 6 square matrix that happens to be
symmetric. The elements of the stiffness matrix are often called stiffness
coefficients.
5.2.3
Breakdown stage
The breakdown stage is performed using three individual steps:
• disconnection,
• localization and
• formation of the element/member stiffness equations.
Disconnection
To carry out the first breakdown step, we begin by discarding all loads and
supports (the so-called boundary conditions). Next, we disconnect or disassemble the structure into its components. For the case of the pin-jointed
truss, disconnection can be visualized as a removal of the pin connectors.
To each member e = 1, 2, 3 we assign a local Cartesian system {x̄e , ȳ e }
commonly called the element coordinate system. Axis x̄e is aligned along
the axis of the eth member as can be seen in Fig. 5.6. By convention, the
positive direction of x̄e is defined from i to j where i < j . The angle formed
by the local and global x−axis, i.e., x and x̄e , is called the orientation angle φe with counter-clockwise direction being the positive one. The local
system’s origin can be placed on any point of the element but they are
commonly positioned on the midpoint or on one of the endpoints.
108
CHAPTER 5. THE FINITE ELEMENT METHOD (FEM)
(a)
3
(b)
3
Remove loads and supports,
_
and disconnect pins
y (3)
_
x (3)
_
x (2)
(3)
y
y (2)
_
y(1)
1
1
2
(2)
_
_
x (1)
x
(1)
2
Figure 5.6: Disconnection step: (a) idealized truss, (b) removal of loads
and support.
Localization
We are examining each element e separately as shown in Fig. 5.7 where
{x̄, ȳ} is the local element coordinate system, the joints at its ends are i and
j , and a pair of forces and displacements are attached to each endpoint.
_ _
fyi, u yi
_ _
fxi, u xi
_
y
_
x
_ _
fyj, u yj
_ _
fxj, u xj
j
i
L
F
d
Figure 5.7: Idealized element considered separately with its interpretation
as a mechanical spring.
Element/Member stiffness equations
There are several ways to construct the element stiffness matrix K̄ , however, the most straightforward one is to consider the spring equivalence
assumption shown in Fig. 5.7. Under this assumption and considering
uniform material properties we know that the spring’s stiffness ks is equal
to EA
, where E is the elastic modulus, A the constant cross-sectional area
L
of the element and L its length. Hence, the axial force F will be:
F = ks d =
EA
d,
L
5.2. THE DIRECT STIFFNESS METHOD (DSM)
109
where d is the relative axial displacement. Going now back to our element
and using the notation introduced in Fig. 5.7, we have that:
 
K̄xi,xi
f¯xi
 f¯yi   K̄yi,xi
 
f̄ = K̄ ū ⇒ 
 f¯xj  =  K̄xj,xi
f¯yj
K̄yj,xi

K̄xi,yi
K̄yi,yi
K̄xj,yi
K̄yj,yi
K̄xi,xj
K̄yi,xj
K̄xj,xj
K̄yj,xj
 
K̄xi,yj
ūxi


K̄yi,yj   ūyi
·
K̄xj,yj   ūxj
K̄yj,yj
ūyj




(5.2)
Looking once again at Fig. 5.7, it is obvious that F = f¯xj = −f¯xi and
d = ūxj − ūxi and hence, if we plug these in Eq. 5.2, we get:
  EA
f¯xi
L
 f¯yi   0
 
f̄ = 
 f¯xj  =  − EA
L
f¯yj
0
0 − EA
L
0
0
0 EA
L
0
0

 
0
ūxi


0   ūyi
·
0   ūxj
ūyj
0


 = K̄ ū.

(5.3)
Hence,

1
EA 
 0
K̄ =
L  −1
0

0 −1 0
0 0 0 

0 1 0 
0 0 0
(5.4)
is the element’s stiffness matrix in local coordinates.
5.2.4
Assembly stage
The assemble stage has two main steps: globalization and merging. In
globalization we are aiming in transforming the element stiffness equations from the local (element’s) coordinate system to the system’s global
coordinate system. Then, merging deals with the collection of globalized
element equations and the assemblage of the global stiffness matrix.
Globalization
Looking at Fig. 5.8, it is easy to derive the transformations mapping displacements and forces from the local to the global coordinate system.
Specifically, we are going to have6 :
110
CHAPTER 5. THE FINITE ELEMENT METHOD (FEM)
(a) Displacement
transformation
_
uyj
uyj
(b) Force
transformation
_
_
fyj
fyj
_
_
y
uyi
_
uyi
x
_
φ
uxi
i
j
_
uxj
y
uxi
_
y
fyi
_
fyi
x
_
fxj
uxj
x
_
fxj
φ
fxi
i
j
fxi
Figure 5.8: Transformation of node (joint) displacements and force components from the local {x̄, ȳ} to the global coordinate system {x, y}.
uxi = cos(φ)ūxi + cos(90 + φ)ūyi = cos(φ)ūxi − sin(φ)ūyi ,
uyi = sin(φ)ūxi + sin(90 + φ)ūyi = sin(φ)ūxi + cos(φ)ūyi ,
uxj = cos(φ)ūxj − sin(φ)ūyj ,
uyj = sin(φ)ūxj + cos(φ)ūyj .
which can be written in matrix form as:

 
uxi
cos φ − sin φ
0
0
 uyi   sin φ cos φ
0
0

 
 uxj  =  0
0
cos φ − sin φ
uyj
0
0
sin φ cos φ
 

ūxi
  ūyi 
·

  ūxj  ,
ūyj
(5.5)
and, in more compact form for the eth element, as
ue = T ūe ,
(5.6)
where T denotes the transformation matrix above.
Looking carefully
at
R(φ)
0
=, where
0
R(φ)
R(φ) is the rotation matrix for angle φ. Since, R(φ) is an isometric matrix,
we have that RT (φ) = R−1 (φ) and hence, for the inverse transformation
matrix T it is easy to see that its structure is:
we are going to have:

cos φ sin φ
0
0
 − sin φ cos φ
0
0 
 ue ⇒
ue = 

0
0
cos φ sin φ 
0
0
− sin φ cos φ

ūe =
6
RT (φ)
0
0
RT (φ)
Remember that: cos(a+b) = cos(a) cos(b)−sin(a) sin(b) and sin(a+b) = sin(a) cos(b)+
sin(b) cos(a)
5.2. THE DIRECT STIFFNESS METHOD (DSM)
111
ūe = T T ue .
(5.7)
Similarly, for the forces in the global coordinate system we will have:
f e = T f̄ e .
(5.8)
Furthermore, we know that element’s stiffness equations in global coordinates are f e = K e ue , while the same equations in local coordinates are
f̄ e = K̄ e ūe . Hence, we can write that:
K e ue = T f̄ e = T K̄ e ūe = T K̄ e T T ue ⇒ K e = T K̄ e T T
This last equation enables us to compute the stiffness matrix in global
coordinates from the element’s stiffness matrix in local coordinates using
the transformation matrix T . Hence, if we compute the product T K̄ e T T ,
we are going get:
cos(φe )2
cos(φe ) sin(φe )
− cos(φe )2
2

cos(φ
)
sin(φ
)
sin(φ
)
−
cos(φ
e
e
e
e ) sin(φe )
K e = ks 

− cos(φe )2
− cos(φe ) sin(φe )
cos(φe )2
− cos(φe ) sin(φe )
− sin(φe )2
cos(φe ) sin(φe )


− cos(φe ) sin(φe )

− sin(φe )2
,
cos(φe ) sin(φe ) 
sin(φe )2
(5.9)
E e Ae
.
Le
where ks =
Hence, the element stiffness equations in global coordinates for the
three elements in our example are going to be (see Fig. 5.5 for the values
used):
Element (1): joints 1 − 2, E = 50, A = 2, L = 10 and φ = 0o .





(1)
fx1
(1)
fy1
(1)
fx2
(1)
fy2


1


0

 = 10 

−1

0
  (1)
ux1
0 −1 0

(1)
uy1
0 0 0 


0 1 0   u(1)
x2
(1)
0 0 0
uy2





(5.10)
Element (2): joints 2 − 3, E = 50, A = 1, L = 10 and φ = 90o .
  (2) 
ux2
0 0 0 0




 0 1 0 −1   u(2)


  y2 
(5.11)
 = 5

 0 0 0 0   u(2) 


x3 
(2)
0 −1 0 1
uy3
√
√
Element (3): joints 1 − 3, E = 100, A = 2 2, L = 10 2 and φ = 45o .
 (3) 
  (3) 

fx1
ux1
0.5
0.5 −0.5 −0.5
 (3) 
 (3) 


0.5
0.5 −0.5 −0.5   uy1 
 fy1 
(5.12)
 (3)  = 20 



−0.5
−0.5
0.5
0.5   u(3)
 fx3 
x3 
(3)
(3)
−0.5 −0.5 0.5
0.5
fy3
uy3

(2)
fx2
(2)
fy2
(2)
fx3
(2)
fy3


112
CHAPTER 5. THE FINITE ELEMENT METHOD (FEM)
Merging
The key operation of the assembly process is the ‘‘placement’’ of the contribution of each member (element) to the master stiffness equations. The
process is technically called merge of individual members or elements.
Merging is performed by following the two rules below:
1. Compatibility of displacements: The displacements of all members
that meet a joint are the same.
2. Force equilibrium: The sum of internal forces exerted by all members
that meet at a joint balances the external force applied to the joint.
Specifically, we first augment the global stiffness equations of each
element (Eqs 5.10,5.11 and 5.12) by adding zero rows and columns as
appropriate to complete the missing force and displacement vectors in each
case. Hence the augmented equations will become:
Element (1):










(1)
fx1
(1)
fy1
(1)
fx2
(1)
fy2
(1)
fx3
(1)
fy3




0 −10 0 0 0 

0 0 0 0 0 


0 10 0 0 0 


0 0 0 0 0 

0 0 0 0 0 

0 0 0 0 0
10

  0
 
  −10
=
  0
 
  0

0
(1)
ux1
(1)
uy1
(1)
ux2
(1)
uy2
(1)
ux3
(1)
uy3










Element (2):

(2)









fx1
(2)
fy1
(2)
fx2
(2)
fy2
(2)
fx3
(2)
fy3
(3)




 
 
 
=
 
 
 

0
0
0
0
0
0
0
0
0
0
0
0
0 0
0 0
0 0
0 5
0 0
0 −5


0 0


0 0 


0 0 


0 −5  

0 0 

0 5
ux1
(2)
uy1
(2)
ux2
(2)
uy2
(2)
ux3
(2)
uy3


ux1
(3)
uy1
(3)
ux2
(3)
uy2
(3)
ux3
(3)
uy3
(2)









Element (3):










fx1
(3)
fy1
(3)
fx2
(3)
fy2
(3)
fx3
(3)
fy3

10
10 0

  10
10 0
 
  0
0 0
=
  0
0 0
 
  −10 −10 0

−10 −10 0

0 −10 −10 

0 −10 −10 


0 0
0 


0 0
0 

0 10
10  

0 10
10
(3)










5.2. THE DIRECT STIFFNESS METHOD (DSM)
113
Due to the compatibility of displacements, me may drop the element index
in the displacements vector and hence get:










(1)
fx1
(1)
fy1
(1)
fx2
(1)
fy2
(1)
fx3
(1)
fy3




















(3)
fx1
(3)
fy1
(3)
fx2
(3)
fy2
(3)
fx3
(3)
fy3



 
 
 
=
 
 
 

(2)
fx1
(2)
fy1
(2)
fx2
(2)
fy2
(2)
fx3
(2)
fy3


0 −10 0 0 0
ux1
 uy1
0 0 0 0 0 


0 10 0 0 0 
  ux2

0 0 0 0 0 
  uy2
0 0 0 0 0   ux3
0 0 0 0 0
uy3
10
0
−10
0
0
0



 
 
 
=
 
 
 

0
0
0
0
0
0
0
0
0
0
0
0
0 0
0 0
0 0
0 5
0 0
0 −5

10
10 0

  10
10 0
 
  0
0 0
=
  0
0 0
 
  −10 −10 0

−10 −10 0

0 0
ux1


0 0   uy1

0 0 
  ux2

0 −5  
 uy2

 ux3
0 0
0 5
uy3

















0 −10 −10
ux1


0 −10 −10   uy1

0 0
0 
  ux2

0 0
0 
 uy2
0 10
10   ux3
0 10
10
uy3








These three last equations can be represented in compact matrix notation
as:
f (1) = K (1) u,
f (2) = K (2) u,
f (3) = K (3) u.
According now to the second rule:
f = f (1) + f (2) + f (3) = (K (1) + K (2) + K (3) )u
(5.13)
Thus, all we have to do is add the three augmented stiffness matrices and
we arrive at the master stiffness equations:








fx1
fy1
fx2
fy2
fx3
fy3


 
 
 
=
 
 
 

20
10 −10 0 −10 −10
ux1


10
10
0
0 −10 −10   uy1

−10 0
10
0
0
0 
  ux2

0
0
0
5
0
−5 
  uy2
−10 −10 0
0
10
10   ux3
−10 −10 0 −5 10
15
uy3








(5.14)
114
CHAPTER 5. THE FINITE ELEMENT METHOD (FEM)
This approach to the assembly process is conceptually the easiest to follow
and understand and is also suitable for hand computations. However, the
approach followed in computer programs differs for reducing the memory
requirements required by this approach.
5.2.5
Solution stage
Having formed the master stiffness equations we can now proceed to the
solution phase. If one attempted to solve the linear system in Eq. 5.14,
he/she would find out that the matrix is singular. The physical interpretation of this singularity is that we have not posed the boundary conditions
and thus our structure can freely move in space.
To eliminate rigid body motions and render the system non-singular
we must apply the physical support conditions as displacement boundary
conditions. Specifically, looking at Fig. 5.5 we see that:
ux1 = uy1 = uy2 = 0
and
fy3 = 1, fx3 = 2, fx2 = 0.
Hence, we can eliminate rows and columns 1,2 and 4 from the stiffness
matrix (due to the fact that the corresponding displacements are 0) and
hence get:


  
10 0 0
ux2
0
 0 10 10   ux3  =  2  ,
0 10 15
uy3
1
which, if solved by e.g., Gauss elimination, will give as:


ux2 = 0
 ux3 = 0.4 
uy3 = −0.2
and therefore, the overall solution (using the known displacements also)
will be:


ux1 = 0
 uy1 = 0

 ux2 = 0

 uy2 = 0

 ux3 = 0.4
uy3 = −0.2



.



5.2. THE DIRECT STIFFNESS METHOD (DSM)
5.2.6
115
Results interpretation (Post-processing) stage
In the previous stage of the DSM we computed the joint displacements
but usually the engineers need additional information on other mechanical quantities; such as the reaction forces at the supports, or the internal
member forces. Such quantities are said to be derived because they are
recovered from the displacement solution. The recovery of derived quantities is part of the so-called post-processing steps of the DSM. Two such
steps, for our example, are described below.
Recovery of reaction forces
If we multiply K with u, we will be able to compute the complete force
vector:




f = Ku = 




20
10 −10 0 −10 −10
0
 0
10
10
0
0 −10 −10 


−10 0
10
0
0
0 
 0

0
0
0
5
0
−5 
 0

 0.4
−10 −10 0
0
10
10
−10 −10 0 −5 10
15
−0.2


 
 
 
=
 
 
 
−2
−2
0
1
2
1








In this way we recover the already known applied forces (as expected) and
additionally we get fx1 = −2,fy1 = −2 and fy2 = 1 which are the reactions
at the support points.
Recovery of internal forces and stresses
Often the structural engineer is not so much interested in displacements
as in internal forces and stresses. In the case of pin-jointed trusses, we
examine here, the only internal forces are the axial member forces, which
we’ll denote as F (1) , F (2) and F (3) for the three elements in our example.
Furthermore, the axial stress σ e will obviously be computed by dividing
the corresponding axial force F e with cross-sectional area (Ae ) of the member/element. The computation of axial forces begins with the calculation
of local displacements at each element using Eq. 5.7. Therefore, the elongation de will be given by de = ūexj − ūexi and, obviously, the internal axial
force will then be:
Fe =
E e Ae e
d.
Le
To make this more clear, let’s follow the process for the 2nd element in
our example: The displacements required from the global solution are:
116
CHAPTER 5. THE FINITE ELEMENT METHOD (FEM)
u(2) = [ux2 , uy2 , ux3 , uy3 ]T = [0, 0, 0.4, −0.2]T . We have that ū(2) = T T u(2)
and hence:

 
ūx2
cos(90)
 ūy2   − sin(90)

 
 ūx3  = 
0
ūy3
0

 
ūx2
0
 ūy2   −1

 
 ūx3  =  0
ūy3
0
sin(90)
0
cos(90)
0
0
cos(90)
0
− sin(90)

1 0 0
0


0 0 0  0
0 0 1   0.4
0 −1 0
−.02

0
ux2


0
  uy2
sin(90)   ux3
cos(90)
uy3
 

0
  0 
=

  −0.2  .
−.4
Hence, d(2) = ūx3 − ūx2 = −0.2 − 0 = −0.2 and consequently:
F (2) = −0.2 50·1
= −1.
10


⇒
