Ifem.ch02

.
2
The Direct
Stiﬀness Method:
Breakdown
2–1

Chapter 2: THE DIRECT STIFFNESS METHOD: BREAKDOWN 2–2
TABLE OF CONTENTS
Page
§2.1. WHY A PLANE TRUSS? 2–3
§2.2. TRUSS STRUCTURES 2–3
§2.3. IDEALIZATION 2–5
§2.4. JOINT FORCES AND DISPLACEMENTS 2–5
§2.5. THE MASTER STIFFNESS EQUATIONS 2–7
§2.6. BREAKDOWN 2–8
§2.6.1. Disconnection . . . . . . . . . . . . . . . . . 2–8
§2.6.2. Localization . . . . . . . . . . . . . . . . . 2–8
§2.6.3. Computation of Member Stiffness Equations . . . . . . . 2–8
EXERCISES . . . . . . . . . . . . . . . . . . . . . . 2–11
2–2

2–3 §2.2 TRUSS STRUCTURES
This Chapter begins the exposition of the Direct Stiffness Method (DSM) of structural analysis.
The DSM is by far the most common implementation of the Finite Element Method (FEM). In
particular, all major commercial FEM codes are based on the DSM.
The exposition is done by following the DSM steps applied to a simple plane truss structure.
§2.1. WHY A PLANE TRUSS?
Thesimpleststructuralfiniteelementisthebar(alsocalledlinearspring)element, whichisillustrated
in Figure 2.1(a). Perhaps the most complicated finite element (at least as regards number of degrees
of freedom) is the curved, three-dimensional “brick” element depicted in Figure 2.1(b).
(a) (b)
Figure 2.1. From the simplest to a highly complex structural finite element:
(a) 2-node bar element for trusses, (b) 64-node tricubic,
curved “brick” element for three-dimensional solid analysis.
Yet the remarkable fact is that, in the DSM, the simplest and most complex elements are treated
alike! To illustrate the basic steps of this democratic method, it makes educational sense to keep
it simple and use a structure composed of bar elements. A simple yet nontrivial structure is the
pin-jointed plane truss.1
Using a plane truss to teach the stiffness method offers two additional advantages:
(a) Computations can be entirely done by hand as long as the structure contains just a few elements.
This allows various steps of the solution procedure to be carefully examined and understood
before passing to the computer implementation. Doing hand computations on more complex
finite element systems rapidly becomes impossible.
(b) The computer implementation on any programming language is relatively simple and can be
assigned as preparatory computer homework.
§2.2. TRUSS STRUCTURES
Plane trusses, such as the one depicted in Figure 2.2, are often used in construction, particularly
for roofing of residential and commercial buildings, and in short-span bridges. Trusses, whether
two or three dimensional, belong to the class of skeletal structures. These structures consist of
elongated structural components called members, connected at joints. Another important subclass
1 A one dimensional bar assembly would be even simpler. That kind of structure would not adequately illustrate some of
the DSM steps, however, notably the back-and-forth transformations from global to local coordinates.
2–3

joint
support
member
Figure 2.2. An actual plane truss structure. That shown is typical of a roof
truss used in residential building construction.
of skeletal structures are frame structures or frameworks, which are common in reinforced concrete
construction of building and bridges.
Skeletal structures can be analyzed by a variety of hand-oriented methods of structural analysis
taught in beginning Mechanics of Materials courses: the Displacement and Force methods. They
can also be analyzed by the computer-oriented FEM. That versatility makes those structures a 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.
In this and the following Chapter we will go over the basic steps of the DSM in a “hand-computer”
calculation mode. This means that although the steps are done by hand, whenever there is a
procedural choice we shall either adopt the way which is better suited towards the computer im-
plementation, or explain the difference between hand and computer computations. The actual
computer implementation using a high-level programming language is presented in Chapter 5.
Figure 2.3. The example plane truss structure, called “example truss”
in the sequel. It has three members and three joints.
To keep hand computations manageable in detail we use just about the simplest structure that can be
called a plane truss, namely the three-member truss illustrated in Figure 2.3. The idealized model
of the example truss as a pin-jointed assemblage of bars is shown in Figure 2.4(a), which also gives
its geometric and material properties. In this idealization truss members carry only axial loads,
have no bending resistance, and are connected by frictionless pins. Figure 2.4(b) displays support
conditions as well as the applied forces applied to the truss joints.
2–4

2–5 §2.4 JOINT FORCES AND DISPLACEMENTS
E(1)
A(1)
= 100
E(2)
A(2)
= 50
E(3)
A(3)
= 200
√
2
1
2
3
L(1)
= 10
L(2)
= 10
L(3)
= 10
√
2
fx1, ux1
fy1, uy1
fx2, ux2
fy2, uy2
fx3, ux3
fy3, uy3
x
y
;;;
;;;
fx3 = 2
fy3 = 1
1 2
3
(1)
(2)(3)
(a) (b)
Figure 2.4. Pin-jointed idealization of example truss: (a) geometric and
elastic properties, (b) support conditions and applied loads.
It should be noted that as a practical structure the example truss is not particularly useful — the
one depicted in Figure 2.2 is far more common in construction. But with the example truss we can
go over the basic DSM steps without getting mired into too many members, joints and degrees of
freedom.
§2.3. IDEALIZATION
Although the pin-jointed assemblage of bars (as depicted in Figure 2.4) is sometimes presented as
a real problem, it actually represents an idealization of a true truss structure. The axially-carrying
members and frictionless pins of this structure are only an approximation of a real truss. 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; see Figure 2.2. Consequently
members will carry some bending as well as direct axial loading.
Experience has shown, however, that stresses and deformations calculated for the simple idealized
problem will often be satisfactory for overall-design purposes; for example to select the cross section
of the members. Hence the engineer turns to the pin-jointed assemblage of axial force elements
and uses it to carry out the structural analysis.
This replacement of true by idealized is at the core of the physical interpretation of the finite element
method discussed in §1.4.
§2.4. JOINT FORCES AND DISPLACEMENTS
The example truss shown in Figure 2.3 has three joints, which are labeled 1, 2 and 3, and three
members, which are labeled (1), (2) and (3). These 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)
. Both E and A are assumed to
be constant along each member.
Members are generically identified by index e (because of their close relation to finite elements, see
below), which is usually enclosed in parentheses to avoid confusion with exponents. For example,
2–5

the cross-section area of a generic member is A(e)
. Joints are generically identified by indices
such as i, j or n. In the general FEM, the name “joint” and “member” is replaced by node and
element, respectively. The dual nomenclature is used in the initial Chapters to stress the physical
interpretation of the FEM.
The geometry of the structure is referred to a common Cartesian coordinate system {x, y}, which
is called the global coordinate system. Other names for it in the literature are structure coordinate
system and overall coordinate system.
The key ingredients of the stiffness method of analysis are the forces and displacements at the joints.
In a idealized pin-jointed truss, externally applied forces 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,
which we call fx and fy, respectively. The components at joint i will be denoted as fxi and fyi ,
respectively. The set of all joint forces can be arranged as a 6-component column vector:
f =







fx1
fy1
fx2
fy2
fx3
fy3







. (2.1)
The other key ingredient is the displacement field. Classical structural mechanics tells us that the
displacements of the truss are completely defined by the displacements of the joints. This statement
is a particular case of the more general finite element theory.
The x and y displacement components will be denoted by ux and uy, respectively. The values of
ux and uy at joint i will be called uxi and uyi and, like the joint forces, they are arranged into a
6-component vector:
u =







ux1
uy1
ux2
uy2
ux3
uy3







. (2.2)
In the DSM these six displacements are the primary unknowns. They are also called the degrees of
freedom or state variables of the system.2
How about the displacement boundary conditions, popularly called support conditions? This data
will tell us which components of f and u are true unknowns and which ones are known a priori. In
structural analysis procedures of the pre-computer era such information was used immediately by
the analyst to discard unnecessary variables and thus reduce the amount of bookkeeping that had
to be carried along by hand.
2 Primary unknowns is the correct mathematical term whereas degrees of freedom has a mechanics flavor. The term state
variables is used more often in nonlinear analysis.
2–6

2–7 §2.5 THE MASTER STIFFNESS EQUATIONS
The computer oriented philosophy is radically different: boundary conditions can wait until the
last moment. This may seem strange, but on the computer the sheer volume of data may not be so
important as the efficiency with which the data is organized, accessed and processed. The strategy
“save the boundary conditions for last” will be followed here for the hand computations.
§2.5. THE MASTER STIFFNESS EQUATIONS
The master stiffness equations relate the joint forces f of the complete structure to the joint dis-
placements u of the complete structure before specification of support conditions.
Because the assumed behavior of the truss is linear, these equations must be linear relations that
connect the components of the two vectors. Furthermore it will be assumed that if all displacements
vanish, so do the forces.3
If both assumptions hold the relation must be homogeneous and be expressable in component form
as follows:







fx1
fy1
fx2
fy2
fx3
fy3







=







Kx1x1 Kx1y1 Kx1x2 Kx1y2 Kx1x3 Kx1y3
Ky1x1 Ky1y1 Ky1x2 Ky1y2 Ky1x3 Ky1y3














ux1
uy1
ux2
uy2
ux3
uy3







. (2.3)
In matrix notation:
f = Ku. (2.4)
Here K is the master stiffness matrix, also called global stiffness matrix, assembled stiffness matrix,
or overall stiffness matrix. It is a 6 × 6 square matrix that happens to be symmetric, although this
attribute has not been emphasized in the written-out form (2.3). The entries of the stiffness matrix
are often called stiffness coefficients and have a physical interpretation discussed below.
The qualifiers (“master”, “global”, “assembled” and “overall”) convey the impression that there
is another level of stiffness equations lurking underneath. And indeed there is a member level or
element level, into which we plunge in the Breakdown section.
REMARK 2.1
Interpretation of Stiffness Coefficients. The following interpretation of the entries of K is highly valuable for
visualization and checking. Choose a displacement vector u such that all components are zero except the
ith
one, which is one. Then f is simply the ith
column of K. For instance if in (2.3) we choose ux2 as unit
displacement,
u =






0
0
1
0
0
0






, f =






Kx1x2
Ky1x2
Kx2x2
Ky2x2
Kx3x2
Ky3x2






. (2.5)
3 This assumption implies that the so-called initial strain effects, also known as prestress or initial stress effects, are
neglected. Such effects are produced by actions such as temperature changes or lack-of-fit fabrication, and are studied
in Chapter 4.
2–7

1
2
3
(1)
(2)(3)
y
x ¯x(1)
¯y(1)
¯x(2)
¯y(2)
¯x(3)
¯y(3)
Figure 2.5. Breakdown of example truss into individual members (1), (2) and (3),
and selection of local coordinate systems.
Thus Ky1x2, say, represents the y-force at joint 1 that would arise on prescribing a unit x-displacement at joint
2, while all other displacements vanish.
In structural mechanics the property just noted is called interpretation of stiffness coefficients as displacement
influence coefficients, and extends unchanged to the general finite element method.
§2.6. BREAKDOWN
The first three DSM steps are: (1) disconnection, (2) localization, and (3) computation of member
stiffness equations. These are collectively called breakdown steps and are described below.
§2.6.1. Disconnection
To carry out the first step of the DSM we proceed to disconnect or disassemble the structure into
its components, namely the three truss members. This step is illustrated in Figure 2.5.
To each member e = 1, 2, 3 is assigned a Cartesian system {¯x(e)
, ¯y(e)
}. Axis ¯x(e)
is aligned along
the axis of the eth
member. See Figure 2.5. Actually ¯x(e)
runs along the member longitudinal axis;
it is shown offset in that Figure for clarity. By convention the positive direction of ¯x(e)
runs from
joint i to joint j, where i < j. The angle formed by ¯x(e)
and x is called ϕ(e)
. The axes origin is
arbitrary and may be placed at the member midpoint or at one of the end joints for convenience.
These systems are called local coordinate systems or member-attached coordinate systems. In the
general finite element method they receive the name element coordinate systems.
§2.6.2. Localization
Next, we drop the member identifier (e) so that we are effectively dealing with a generic truss
member as illustrated in Figure 2.6. The local coordinate system is {¯x, ¯y}. The two end joints are
called i and j.
As shown in Figure 2.6, a generic truss member has four joint force components and four joint
displacement components (the member degrees of freedom). The member properties include the
length L, elastic modulus E and cross-section area A.
2–8

2–9 §2.6 BREAKDOWN
§2.6.3. Computation of Member Stiffness Equations
The force and displacement components of Figure 2.7(a) are linked by the member stiffness relations
¯f = K ¯u, (2.6)
which written out in full is



¯fxi
¯fyi
¯fx j
¯fyj


 =




¯Kxixi
¯Kxiyi
¯Kxix j
¯Kxiyj
¯Kyixi
¯Kyiyi
¯Kyix j
¯Kyiyj
¯Kx jxi
¯Kx jyi
¯Kx jx j
¯Kx jyj
¯Kyjxi
¯Kyjyi
¯Kyjx j
¯Kyjyj







¯uxi
¯uyi
¯ux j
¯uyj


 . (2.7)
Vectors ¯f and ¯u are called the member joint forces and member joint displacements, respectively,
whereas ¯K is the member stiffness matrix or local stiffness matrix. When these relations are
interpreted from the standpoint of the FEM, “member” is replaced by “element” and “joint” by
”node.”
There are several ways to construct the stiffness matrix ¯K in terms of the element properties L, E
and A. The most straightforward technique relies on the Mechanics of Materials approach covered
in undergraduate courses. Think of the truss member in Figure 2.6(a) as a linear spring of equivalent
stiffness ks, an interpretation depicted in Figure 2.7(b). If the member properties are uniform along
its length, Mechanics of Materials bar theory tells us that4
ks =
E A
L
, (2.8)
Consequently the force-displacement equation is
F = ksd =
E A
L
d, (2.9)
where F is the internal axial force and d the relative axial displacement, which physically is the
bar elongation.
The axial force and elongation can be immediately expressed in terms of the joint forces and
displacements as
F = ¯fx j = − ¯fxi , d = ¯ux j − ¯uxi , (2.10)
which express force equilibrium5
and kinematic compatibility, respectively.
Combining (2.9) and (2.10) we obtain the matrix relation6
¯f =



¯fxi
¯fyi
¯fx j
¯fyj


 =
E A
L



1 0 −1 0
0 0 0 0
−1 0 1 0
0 0 0 0






¯uxi
¯uyi
¯ux j
¯uyj


 = K ¯u, (2.11)
4 See for example, Chapter 2 of F. P. Beer and E. R. Johnston, Mechanics of Materials, McGraw-Hill, 2nd ed. 1992.
5 Equations F = ¯fx j = − ¯fxi follow by considering the free body diagram (FBD) of each joint. For example, take joint i
as a FBD. Equilibrium along x requires −F − ¯fxi = 0 whence F = − ¯fxi . Doing this on joint j yields F = ¯fx j .
6 The detailed derivation of (2.11) is the subject of Exercise 2.3.
2–9

i
i
i
j
j
j
(e)
d
(b)
(a)
L
¯x
¯y
fxi , uxi
fyj , uyj
ks = E A/L
fyi , uyi
fx j , ux j
F−F
_ _
_ _ _ _
_ _
Figure 2.6. Generic truss member referred to its local coordinate system {¯x, ¯y}:
(a) idealization as bar element, (b) interpretation as equivalent spring.
Hence
K =
E A
L



1 0 −1 0
0 0 0 0
−1 0 1 0
0 0 0 0


 . (2.12)
This is the truss stiffness matrix in local coordinates.
Two other methods for obtaining the local force-displacement relation (2.9) are covered in Exercises
2.6 and 2.7.
In the following Chapter we will complete the main DSM steps by putting the truss back together
and solving for the unknown forces and displacements.
2–10

2–11 Exercises
Homework Exercises for Chapter 2
The Direct Stiffness Method: Breakdown
EXERCISE 2.1
[D:5] Explain why arbitrarily oriented mechanical loads on an idealized pin-jointed truss structure must be
applied at the joints. [Hint: idealized truss members have no bending resistance.] How about actual trusses:
can they take loads applied between joints?
EXERCISE 2.2
[A:15] Show that the sum of the entries of each row of the master stiffness matrix K of any plane truss, before
application of any support conditions, must be zero. [Hint: think of translational rigid body modes.] Does the
property hold also for the columns of that matrix?
EXERCISE 2.3
[A:15] Using matrix algebra derive (2.11) from (2.9) and (2.10).
EXERCISE 2.4
[A:15] By direct matrix multiplication verify that for the generic truss member ¯f
T
¯u = F d. Can you interpret
this result physically? (Interpretation hint: look at (E2.3) below]
EXERCISE 2.5
[A:20] The transformation equations between the 1-DOF spring and the 4-DOF generic truss member may
be written in compact matrix form as
d = Td ¯u, ¯f = F Tf , (E2.1)
where Td is 1 × 4 and Tf is 4 × 1. Starting from the identity ¯f
T
¯u = F d proven in the previous exercise, and
using compact matrix notation, show that Tf = TT
d . Or in words: the displacement transformation matrix
and the force transformation matrix are the transpose of each other. (This is a general result.)
EXERCISE 2.6
[A:20] Derive the equivalent spring formula F = (E A/L) d of (2.9) by the Theory of Elasticity relations
e = d ¯u(¯x)/d ¯x (strain-displacement equation), σ = Ee (Hooke’s law) and F = Aσ (axial force deﬁnition).
Here e is the axial strain (independent of ¯x) and σ the axial stress (also independent of ¯x). Finally, ¯u(¯x)
denotes the axial displacement of the cross section at a distance ¯x from node i, which is linearly interpolated
as
¯u(¯x) = ¯uxi 1 −
¯x
L
+ ¯ux j
¯x
L
(E2.2)
Justify that (E2.2) is correct since the bar differential equilibrium equation: d[A(dσ/d ¯x)]/d ¯x = 0, is veriﬁed
for all ¯x if A is constant along the bar.
EXERCISE 2.7
[A:20] Derive the equivalent spring formula F = (E A/L) d of (2.9) by the principle of Minimum Potential
Energy (MPE). In Mechanics of Materials it is shown that the total potential energy of the axially loaded bar
is
= 1
2
L
0
A σ e d ¯x − Fd (E2.3)
where symbols have the same meaning as the previous Exercise. Use the displacement interpolation (E2.2),
the strain-displacement equation e = d ¯u/d ¯x and Hooke’s law σ = Ee to express as a function (d) of the
relative displacement d only. Then apply MPE by requiring that ∂ /∂d = 0.
2–11

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