TWO-DIMENSIONAL
MATRIX STIFFNESS
ANALYSIS
Analytical Model
As you are aware, matrix methods
do not involve any new fundamental principles. The fundamental
relationships of equilibrium,
compatibility and constitutive
(member force-displacement)
relationships are simply expressed
in the form of matrix equations so
that the numerical computations
can be efficiently performed on a
computer.
1
Again, in matrix stiffness analysis,
the structure is modeled as an
assemblage of straight members
connected at their ends to joints or
nodes. Kassimali (1999) defines a
member (or element) as: part of the
structure for which the member
force-displacement relations are
valid. A joint is defined to be a
point used to idealize a structural
connection. A node can be a joint
or simply a mathematical break in
the member in order to model a 2
change in geometry, loading or
simply to provide intermediate
displacement and force calculated
results.
Prior to performing an analysis of
the structure, an analytical or
mathematical model of the structure
must be defined. The model is
represented by a line diagram of the
structure on which all nodes and
elements are identified by numbers.
NOTE: Element force-displacement
relations herein are exact for
piecewise prismatic members only.3
Figure 17.1 – Frame Structure
(Kassimali, 1999)
4
1
Overall geometry and behavior of a
structure is referenced to a global or
structural Cartesian coordinate
system XYZ – plane of the structure
lying in the XY plane. However, the
basic force-displacement relationships are derived in terms of forces
and displacements aligned with the
element (member), which is referred
to as the local coordinate system
xyz. Furthermore, design forces are
defined in terms of the local
5
coordinate system.
Displacement degrees of freedom
are numbered starting at the lowest
node number and proceeding
sequentially to the highest node
number. Order of the nodal
degrees of freedom is: X-axis, Yaxis and rotation about Z-axis. If a
DOF does not exist, e.g., continuous beam model that neglects
axial deformation or truss structure
in which the joint rotations are not
defined since truss joints are
presumed to be frictionless hinges,
then that DOF is eliminated from
7
the sequence.
x
Y
Global and Local Coordinate
Systems
y
frame
element
X
z
Z
(a) Global Coordinate
System
(b) Local Coordinate
System
Figure A – 2D Stiffness Analysis
Coordinate Systems
In Fig. A:
x-axis  aligned with the element
centroidal axis
y-axis  normal to the element x-axis
and consistent with the right hand
rule positive coordinate sign
convention
z-axis  out of plane local axis parallel
and in the same direction as the
global Z-axis; y-axis is normal to
6
the x-z plane using the right hand
rule positive sign convention
Furthermore, if a displacement is
known, then that DOF is also not
numbered.
Figure 17.2(a) shows a continuous
beam subjected to transverse
(lateral) loading only. For such a
beam, the axial deformation is zero
and the corresponding analytic
model is shown in Fig. 17.2(b).
8
Figure 17.2 Continuous Beam Structure
2
Similarly, Fig. 17.3 shows a truss
structure and Fig. 17.1 shows an
example frame structure. NOTE:
The frame structure element
includes axial and bending
deformation, which results in three
displacement degrees of3 freedom
per node.
(a) Truss
Figure 17.3 Truss Structure
(17.1)
10
(a) Frame Member: Global and Local Coordinate Systems
(17.2a)
Q 2  k 21u1  k 22 u 2  k 23u 3  k 24 u 4
(17.2b)
 k 25u 5  k 26 u 6  Qf 2
Q3  k 31u1  k 32 u 2  k 33u 3  k 34 u 4
(17.2c)
 k 35u 5  k 36 u 6  Qf 3
Q 4  k 41u1  k 42 u 2  k 43u 3  k 44 u 4
 k 45u 5  k 46 u 6  Qf 4
[S] {d} = {P} – {Pf}
9
Considering a plane frame element
with three nodal degrees of
freedom ( NNDF) and six element
degrees of freedom ( NEDF) as
shown in Fig. C, the element
stiffness equations are
 k15u 5  k16 u 6  Qf1
As stated previously, the nodal
displacement vector {d} for the
analytical model of the structure is
obtained by solving a system of
simultaneous equations
The structure stiffness matrix [S] is
obtained by assembling the
stiffness matrices for the individual
elements of the structure.
(b) Analytical Model
Q1  k11u1  k12 u 2  k13u 3  k14 u 4
Member Stiffness Equations in
Local Coordinates
(17.2d)
(b) Frame Member End Displacements and Forces
Figure C. Frame Element Terminology and
Deformation Behavior
12
3
Q5  k 51u1  k 52 u 2  k 53u 3  k 54 u 4
(17.2e)
 k 55u 5  k 56 u 6  Qf 5
Q6  k 61u1  k 62 u 2  k 63u 3  k 64 u 4
(17.2f)
 k 65u 5  k 66 u 6  Qf 6
Recall that kij represents the force
at the location and in the direction
of Qi required, along with other end
forces, to cause a unit value of the
displacement uj while all other end
displacements are zero. Such
forces per unit displacement are
defined as stiffness coefficients –
first subscript identifies the force
and the second subscript identifies
the displacement. Equations (17.2)
can be expressed in matrix form as
 Q1 
Q 
 2
 Q3 
 
Q 4 
 Q5 
 
Q6 
 k11
k
 21
 k 31
k
 41
 k 51

 k 61
k12
k 22
k13
k 23
k14
k 24
k15
k 25
k 32
k 33
k 34
k 35
k 42
k 43
k 44
k 45
k 52
k 53
k 54
k 55
k 62
k 63
k 64
k 65
  Qf1 Qf 2
Qf 3
Qf 4
k16 
k 26 

k 36 
k 46 
k 56 

k 66 
Qf 5
 u1 
u 
 2
u3 
 
u 4 
u5 
 
u 6 
Qf 6 T
(17.3)
13
The element stiffness matrix of
(17.3) expresses the end forces in
terms of the element material and
geometric properties (stiffness
coefficients) times the element
displacements ( u i , i 1,2,,6 ) plus
the contribution of the element
loads in the kinematically
determinate state (fixed-end forces
Qfi , i 1,2,,6 ). As shown in Fig.
C, the element end displacements
are measured relative to the
undeformed position of the element
in the local coordinate system.
15
14
Translational forces and displacements are positive when they are in
the positive x or y coordinate
directions and counterclockwise
moments and rotations are positive.
A more succinct representation of
(17.3) is
(17.4)
{Q}  [k]{u}  {Qf }
{Q}  local coordinate element
force vector
{Qf }  local coordinate element
fixed-end force vector
{u}  local coordinate element
displacement vector
[k] 
local coordinate element
stiffness matrix
4
Stiffness Coefficient Derivations
The frame element stiffness
equations of (17.3) can be rewritten
as
{Qa }2x1  [k aa ]2x2 [k ab ]2x4  {u a }2x1 
{Q}6x1  



{Qb }4x1  [kba ]4x2 [kbb ]4x4  {ub}4x1 
{Q } 
  fa 2 x1 
{Qfb }4 x1 
(1)
Where {Q a }   Q1 Q 4 T = axial
member end force vector; {Qb } =
 Q2 Q3 Q5 Q6 T = bending
member end force vector; [kaa] =
axial member stiffness matrix;
17
Equation (1) simply rearranges the
degrees of freedom listed in (17.3)
into axial and bending components
and clarifies that the two deformation modes are uncoupled ([kab] =
[kba]T = [0]2x4) for straight members
in the local xy coordinate system.
Since the frame element is straight,
the axial and bending deformation
modes are uncoupled. Thus, the
derivation of the stiffness
coefficients can be carried out
separately.
19
 k11 k14  [kbb] = beam member
k
;
 41 k 44  stiffness matrix =
 k 22
k
 32
 k 52
k
 62
k 23
k 33
k 53
k 25
k 35
k 55
k 63
k 65
k 26  [k ] = [k ]T =
ab
ba
k 36  axial-beam
;
k 56  coupling
k 66  stiffness matrix
= [0]2x4; {ua} = axial member
displacement vector = <u1 u4>T; {ub}
= beam member displacement
vector = <u2 u3 u5 u6>T; subscript f ≡
fixed-end member force vector; and
subscripts 1, 2, …, 6 coincide with
the degrees of freedom shown in
18
Fig. 17.4 and Eq. (17.3).
Axial Stiffness Coefficients
Consider the frame member of Fig.
C subjected to u1 = 1 with all other
displacements equal to zero as
shown in Fig. A.1 below.
Figure A.1
 u1 1; u 2  u 3  u 4  u 5  u 6  0 
The force Q1 (= k11) to cause u1 = 1
with all others equal to zero can be
obtained using Hooke’s law (see
your Mechanics of Deformable
Solids textbook):
20
5
x  E x
(a)
k 41  Q 4 u 1
1
Substituting
x 
Fx
A

Remember
Q1
;
A
x 
L
L

u1
L

1
L
leads to
Q1   x A  E x A 
EA
L
Since u1 = 1 with all others equal to
zero no bending deformation is
introduced as shown in Fig. A.1:
k 21  k 31  k 51  k 61  0
EA
k11  Q1 u 1 
1
L
Similarly, imposing u4 = as shown in
Fig. A.2 leads to:
Stiffness coefficient k41 at end e is
obtained from equilibrium:
 Fx  0  k11  k 41
 k 41   k11  
EA
L
Figure A.2  u 4 1; u1  u 2  u 3  u 5  u 6  0 
21
EA
k 44  Q4 u 1 
4
L
 Fx  0  k14  k 44
 k14   k 44  
EA
 Q1 u 1
4
L
k 24  k 34  k 54  k 64  0
22
Beam Stiffness Coefficients
You may recall from your Mechanics of Deformable Solids course
that the differential equation of a
beam composed of linearly elastic,
homogenous material and loaded
in the plane of symmetry of it crosssection can be expressed as
d 2u y
dx
2

M
EI
(2)
in which uy = deflection of the
beam’s centroidal axis (which
coincides with the neutral axis) in
23
24
6
the y direction at a distance x from
the origin of the x-y coordinate
system; and M = internal bending
moment at the same location x and
follows the usual beam sign
convention.
To calculate the stiffness coefficients in the first column of [kbb],
subject the member to u2 = 1, u3 =
u5 =u6 = 0 as shown in Fig. B.1.
(NOTE: Axial displacements are
not considered since they are
uncoupled from the bending
displacements.) Recall that kij =
force at dof i due to a unit
displacement at dof j with all other
displacements equal to zero.
Taking a cut at an arbitrary distance
x in Fig. 1 results in
M   k 32  k 22 x
(3)
Substituting (3) into (2):
d 2u y
Figure B.1 – First Column of [kbb] = {kbb}1
dx
1
x2 

  k 32 x  k 22
  C1
dx
EI 
2 
uy 

1
(k 32  k 22 x)
EI
25
Integrating (4) twice with respect to
x leads to
du y
2
(5a)
1
x2
x3 
 k 22
  k 32
  C1 x  C2 (5b)
EI 
2
6 
where C1 and C2 are constants of
integration. A total of four unknowns
are included in (5a, b) – the two
constants of integration and the two
stiffness coefficients. Referring to
Fig. B.1 and imposing the
displacements provides four
boundary conditions, i.e.
27
(4)
26
End b, x  0 :
(x  0)  u 3  0  C1
u y (x  0)  u 2  1  C2
(6a)
End e, x  L :

L2 
(x  L)  u 6  0    k 32 L  k 22
  C1

2 
(6b)

2
3
1
L
L 
u y (x  L)  u 5  0    k 32
 k 22 

EI 
2
6 
 C1 L  C2
Equations (6a) show that C1 = 0
and C2 = 1 for Fig. B.1. Substituting
these results into (6b) leads to
28
7
k 22 
12EI
k 32 
6EI
(7a)
L3
(7b)
L2
The stiffness coefficients at end e
can be calculated directly from
equilibrium of the beam element,
i.e.


Fy  0  k 52   k 22  
12EI
(7c)
L3
M e  0  k 62   k 32  k 22 L 
6EI
L2
(7d)
All four beam bending cases are in
your class notes. Pick one of the
29
remaining three for me to solve.
Combining the derived axial and
beam stiffness coefficients, the
element stiffness matrix coefficients
of (17.2) can be expressed in
matrix form as
A
0
0 A
0
0 
L
L

6I
12I 6I 
 0 12I3
0

L
L2
L3
L2 

6I
4I
0

0  6I2 2I
L
L 
L2
L

[k]  E
 A 0
A
0
0
0 
L
 L

 0 12I  6I 0 12I  6I 

L3
L2
L3
L2 


6I
2I
0  6I2 4I
2
 0

L
L
L
L
(17.5)
31
Combining (7), (12), (17) and (22)
from your class notes leads to
 k 22
k
[k bb ]   32
 k 52

 k 62
 12
 3
 L
 6
 L2
 EI 
  12
 L3

 6
 L2
k 23
k 25
k 33
k 53
k 63
k 35
k 55
k 65
6
2
L
4
L
6

L2
2
L


12
3
L
6
L2
12
L3
6

L2
k 26 
k 36 

k 56 

k 66 
6 

L2 
2 
L 

6

L2 

4 
L 
(17.6)
which is obviously symmetric, i.e.,
kij = kji . Equation (17.6) is the
30
beam element stiffness matrix.
Derivation of Frame
Member Fixed-End Forces
The fixed-end force vector of (1)
has contributions due to axial
({Qfa}) and bending ({Qfb})
member loads. Consider the beam
shown in Fig. B.5, where px is the
distributed axial load and py is a
uniform transverse load.
Figure B.5 – Frame Element Subjected to
Uniform Axial and Transverse Loadings
32
8
Fixed-End Axial Forces
Fixed-end axial forces are obtained
by solving the following differential
equation
du x Q x

dx
EA
(23)
x
du x ()
d 
d
x
Q x (x)   Qf1 
 p x d
0
where ux = x-axis (axial) displacement; and Qx = axial force (or axial
stress resultant). The solution of
(23) is expressed as

where Qa(x) depends on the
distribution of the axial for local xaxis loading. For the uniformly
distributed axial loading of Fig. B.5:

Q x ( )
d
EA
(24)
  Qf1  p x x
(25)
Substituting (25) into (24) and
integrating over the element length
L:
u x (x  L)  u x (x  0)
x
Qf1 L p x L2

EA
2EA

33
(26)
34
where ux(x=0) = u1; and ux(x=L) = u4.
Imposing the boundary conditions
u1 = u4 = 0 on (26) and solving for
Qf1:
p L
(27)
Qf1  x
2
Fixed-End Bending Forces
Fixed-end bending forces are
obtained by solving the following
differential equation
Fixed-end force Qf4 is obtained
from equilibrium:
The solution of (29) is expressed
as
 Fx  0  Qf1  Qf 4  px L
p L
 Qf 4  x
2
(28)
d2u y
dx

2

M
EI
d 2 u y ()
x
d
2
d 
(29)

M()
d
EI
x
 d 2 u (x)



M() 
y 

d  dx  
d dx


 EI

dx 2


x  x
x  x

35
(30a)

(30b)
36
9
where M() depends on the
distribution of the bending or y-axis
loading. For example the uniformly
distributed loading py shown in Fig.
B.5 results in the following moment
expression
M(x)  Qf 2 x  Qf 3  p y
x2
2
(31)
Substituting (31) into (30a) and
integrating results in
(x  L)  (x  0)

3
L
L
(Qf 2 L  2Qf 3 ) 
py
2EI
6EI
(32a)
p y L2
12
and Qf 2 
pyL
2
u y (x  L)  u y (x  0)

2
(32b)
4
pyL
L
(Qf 2 L  3Qf 3 ) 
6EI
24EI
where uy(x=0) = u2; and uy(x=L) =
u5. Imposing the boundary
conditions u3 = u6 = 0 on (32a) and
u2 = u5 = 0 on (32b) leads to
p y L3
L
(Qf 2 L  2Qf 3 ) 
2EI
6EI
(33a)
p y L4
L2
(Qf 2 L  3Qf 3 ) 
6EI
24EI
(33b)
Fixed-End Forces:
Thermal Loading
Solving (33a, b) for Qf3 and Qf2
gives
Qf 3 
du
where   y ; (x=0) = u3; and
dx
(x=L) = u6. Similarly, substituting
(31) into (30b) and integrating:
(34)
Axial Deformation
Fixed-end forces Qf5 and Qf6 are
obtained from equilibrium:
pyL
(35)
Fy  0  Qf 5 
2


M e  0   Qf 2 L  Qf 3  Qf 6  p y
 Qf 6  Qf 2 L  Qf 5 
L2
2
p y L2
2
2

pyL
12
(36)
39
Figure B.6 – Axial Force Member
Subjected to Uniform
Temperature Loading
40
10
The prismatic bar of Fig. B.6(a) is
subjected to a constant temperature load of Ta. For this case
x 
L
L
 Ta
 L  LTa
if permitted to freely expand as
shown in Fig. B.6(a) and  = linear
coefficient of thermal expansion. If
the expansion is prevented, as
shown in Fig. B.6(b), the force
required to prevent the free
expansion is
Q  EA
L  EATa
L
Thus, the fixed-end force at end e
of the member is
Qf 4   Q   EATa
and the fixed-end force at the
beginning of the member (b) is
obtained from x-axis equilibrium to
give
Qf1   Qf 4  EATa
(37b)
These axial fixed-end forces are
independent of the length of the
member.
41
Bending Deformation
(37a)
42
In the beam bending case, the
beam of Fig. B.7 is presumed to be
doubly symmetric (bisymmetrical)
and prismatic. The entire bottom
surface of the beam is heated to Tl
and the top surface Tu, which
produces a linear temperature
gradient Tb = (Tl – Tu) through the
depth of the beam. If free to
expand, the axis of the beam would
elongate
L = L(Tl + Tu)/2
Figure B.7 – Thermal Gradient
43
Through Beam
44
11
The axial fixed-end force can be
calculated in this case by
substituting (Tl + Tu)/2 for Ta in
(37a, b).
Neglecting second-order effects,
which is the case for linear analysis
1
d (Tl  Tu )
(38)
dx


dx
h

For the bending case, the beam
will curve into a circular arc as
shown in Fig. B.7(b). Figure B.7(c)
is an enlarged view of Fig. B.7(b)
from which it can be seen that
where  = radius of curvature and
 = curvature. Thus
(Tl  Tu )
d 
dx
h
d (Tl  Tu ) Tb



dx
h
h

2
d d u y M


dx
EI
dx 2
Substituting (38) into (39) and
solving for M leads to
Qf 3   Qf 6  M 
45
(39)
EI(Tl  Tu ) (40)
h
46
12