Introduction to
Stiffness Analysis
The stiffness method of analysis is
the basis of all commercial
structural analysis programs.
Focus of this chapter will be
development of stiffness equations
that only take into account bending
deformations, i.e., ignore axial
member, a.k.a. slope-deflection
method.
In the stiffness method of analysis,
we write equilibrium equations in
1
terms of unknown joint (node)
Definitions and Terminology
Positive Sign Convention:
Counterclockwise moments and
rotations along with transverse
forces and displacements in the
positive y-axis direction.
Fixed-End Forces: Forces at the
“fixed” supports of the kinematically determinate structure.
Member-End Forces: Calculated
forces at the end of each element/
member resulting from the
applied loading and deformation
of the structure.
3
displacements. The number of
unknowns in the stiffness method of
analysis is known as the degree of
kinematic indeterminacy, which
refers to the number of node/joint
displacements that are unknown
and are needed to describe the
displaced shape of the structure.
One major advantage of the
stiffness method of analysis is that
the kinematic degrees of freedom
are well-defined.
2
Stiffness Analysis
Procedure
The steps to be followed in
performing a stiffness analysis can
be summarized as:
1. Determine the needed displacement unknowns at the nodes/
joints and label them d1, d2, …,
dn in sequence where n = the
number of displacement
unknowns or degrees of
freedom.
4
1
2. Modify the structure such that it
is kinematically determinate or
restrained, i.e., the identified
displacements in step 1 all
equal zero.
3. Calculate the member fixed-end
forces in this kinematically
restrained state at the nodes/
joints of the restrained structure
due to the member applied
loads. Tables of member fixedend forces due to member loads
for the kinematically restrained
members are available later in
5
these notes. The member
fixed-end forces are vectorially
added at the nodes/joints to
produce the equivalent fixed-end
structure forces, which are
labeled Pfi for i = 1, 2, …, n later
in the notes.
4. Introduce a unit displacement at
each displacement degree of
freedom identified in step 1 one
at a time with all others equal to
zero and without any loading on
the structure, i.e., di = 1 with d1,
…, di-1, di+1, …, dn = 0 for i = 1,
2, …, n. Sketch the displaced
structure for each of these cases.
Determine the member-end forces
introduced as result of each unit
displacement for the kinematically
restrained structure. These
member-end forces define the
member-end stiffness coefficients, i.e., forces per unit
displacement.
5. Eliminate the error introduced in
step 3 to permit the displacement at the nodes/joints. This is
accomplished by applying the
negative of the forces calculated
in step 3 and defines the kinematically released structure.
The member-end stiffness
coefficients are vectorially added
at the nodes/joints to produce the
structure stiffness coefficients,
which are labeled Sij for i = 1, 2, 7
…, n and j = 1, 2, …, n.
6
6. Calculate the unknown node/
joint displacements.
7. Calculate the member-end
forces.
8
2
Illustration
To illustrate the stiffness method of
analysis, we will first consider
continuous beam structures. Start
off by considering the two-span
beam shown in Figure 1.
1: Determine the degree of
kinematic indeterminacy. The only
unknown node/joint displacement
occurs at node B and it is a
rotational displacement. Thus, the
rotation at node B is labeled d1.
2: Kinematically restrain the
structure such that the displacements identified in step 1 equal
zero. See Figure 2.
Fig. 1 – Two-Span Continuous Beam
Figure 2 – Kinematically Restrained Two-Span
Beam of Figure 1
9
The heavy vertical line drawn
through the horizontal roller support
at B signifies that node B is “fixed”
against displacement. Thus, the
rotational displacement d1 = 0 for
the kinematically restrained
structure of Figure 2.
3: Calculate the element/member
fixed-end forces for the kinematically restrained structure and
vectorially add to obtain the fixedend forces for the structure.
11
10
Figure 3 – “Fixed-End” Forces for the
Kinematically Restrained Two-Span Beam of
Figure 1
Since span element (member) A-B
is not loaded, it will not produce any
fixed-end forces. However,
element (member) B-C is loaded
and the “fixed-end” forces are
labeled in Figure 3. They are
simply the support reactions for the
“fixed-fixed” beam.
12
3
Calculate the fixed-end forces for
the structure by vectorially adding
the member-end fixed-end forces.
Figure 4 – Joint Equilibrium at
the Kinematic Degree of
Freedom for the Restrained
Two-Span Beam of Figure 1
4: Impose a unit displacement at
each kinematic degree of freedom
(DOF) to establish the structure
stiffness equations.
Figure 4 shows that
 M  0  Pf1 
B
qL2
12
Pf1 is drawn counterclockwise in
Figure 4 since our sign
convention is counterclockwise
moments are positive.
13
Figure 5 shows the displaced shape
of the two-span beam for d1 = 1 as
well as the displaced shapes and
member-end stiffness coefficients
for the two elements comprising the
two-span beam of Figure 1.
Member-end stiffness coefficients
are defined as the member-end
forces resulting from the imposition
of the single unit displacement for
the structure as shown in Figure 5.
Derivation of the member-end
stiffness coefficients (forces) shown
in Figure 5 and others will be
15
covered later in the notes.
Figure 5 – Kinematically Restrained Two-Span
Beam of Figure 1 Subjected to a Unit
Displacement d1 = 1
14
The structure stiffness equations
are expressed as
[S] {d} = {P} – {Pf}
where [S] is the structure stiffness
matrix; {d} is the structure
displacement vector; {P} is the
applied structure concentrated
force vector; and {Pf} is the
structure fixed-end force vector
calculated in step 3. The applied
structure concentrated force vector
{P} lists the point forces for each
structure displacement DOF. It
contains nonzero entries only at 16
4
the displacement DOF where a
point force or moment is applied at
the corresponding displacement
DOF.
The structure stiffness matrix
coefficients are obtained by
performing equilibrium at the nodes
for each structure DOF using the
member-end stiffness coefficients.
These structure stiffness matrix
coefficients are designated as Sij
and i = 1, 2, …, n and j = 1, 2, …, n.
Sij  force at displacement DOF i
due to a unit displacement at
DOF j (i.e., dj = 1) with all other
displacement DOF equal to
zero (i.e., di = 0 for i = 1, …, j-1,
j+1, …, n). Stiffness coefficients have units of force/
displacement (or moment/
rotation).
The structure stiffness coefficients
are obtained by performing
equilibrium calculations at the
structure displacement degrees of
freedom.
17
For example structure:
{d} = {d1} = d1  unknown
{P} = {0} = 0
5: Eliminate the error introduced in
the kinematically restrained
structure:
[S] {d} = {P} – {Pf}
{Pf} = {Pf1} = qL2/12
Figure 6 – Equilibrium
at Kinematic DOF 1 for
the Two-Span Beam of
Figure 1
6: Calculate the unknown structure
displacements
{d} = [S]-1 ({P} – {Pf})
For the example structure:
Performing node equilibrium at
displacement DOF 1 gives (see
Figure 6) gives
S11 = (4EI/L)AB + (4EI/L)BC
= 8EI/L
18
d1 = L/8EI (-qL2/12)
= -qL3/96EI
19
7: Calculate the member-end
forces.
20
5
 Vfb 
M 


{Qfb }   fb 
V
fe


 M fe 
The member fixed-end forces are
defined as
The beam member stiffness
equations can be written as
{Qb }4x1  [k bb ]4x4 {u b }4x1  {Qfb }4x1
 Vb 
M 


{Qb }   b 
V
e


 M e 
= Member Fixed-End
Force Vector
= Member-End Force
Vector
Qfi = Qi
in the kinematically determinate
state due to member loading.
 vb 
 
 
{u b }   b 
 ve 
 e 
= Member-End
Displacement Vector
22
21
Simplified Member End
Force Calculations
6L 12 6L 
 12

2
2
EI  6L 4L 6L 2L 
[k bb ]  3 
L 12 6L 12 6L 


 6L 2L2 6L 4L2 
Vb  12EI
(v b  ve )  6EI
(b  e )  Vfb
L3
L2
(internal shear at beginning
node b)
= Member Bending Stiffness
Matrix
The ij member stiffness coefficient
can be expressed mathematically
as
k ij  Qi u 1
j
M b  6EI
(v b  ve )  2EI
(2b  e )  M fb
L
L2
(internal moment at beginning
node b)
Ve   12EI
(v b  ve )  6EI
(b  e )  Vfe
3
2
L
L
(internal shear at end node e)
all other u k  0 (k  j)
M e  6EI
(v b  ve )  2EI
(b  2e )  M fe
L
L2
(internal moment at end node e)
23
24
6
In order to apply equation (1) to the
calculation of element end forces,
compatibility between element
displacements ui (i=1,2,3,4) and the
structure displacements dj (j=1,2,
…, n) must be established. For the
example beam:
eAB  d1 ; bBC  d1 ; all others  0
{Qfb }AB  {0}
{Qfb }BC  qL / 2  1 L / 6 1  L / 6 T
Since only one displacement is
nonzero for each member, the
member end forces are
 6EI 
 2 
AB  L 
V
 b
 2EI 
M 
 L 


{Q}AB   b   

  6EI 
 Ve 
 L2 
 M e 


 4EI 
 L 
 6EI 
 2 
BC  L 
 Vb 
 4EI 


 L 
M b 
BC
{Q}  

 
  6EI 
 Ve 
 L2 
 M e 


 2EI 
 L 

21qL
48
qL2
24
27qL
48
 qL 
  16 


 qL2 

3
 qL   48 



 96EI   qL 
 16 

2
 qL 
 24 
 1 
 qL3  qL  L / 6 




 96EI  2  1 
 L / 6 
5qL2

48
T
25
The shear force and bending
moment diagrams are given below
for the example structure.
26
Mathematical Expression of
Stiffness Superposition
As stated in step 5, the structure
stiffness equations are expressed
as
[S] {d} + {Pf} = {P}
[S]  structure stiffness matrix
{d}  structure displacement
vector
{Pf}  structure fixed-end
force vector
{P}  structure node/joint
force vector
27
28
7
Referencing the previous figure on
stiffness superposition, the first
column of the figure represents the
kinematically determinate state,
i.e.
{d}  0  {Pf }  {P}
The second column of the figure
represents the kinematically
released state, i.e.
[S] {d} = {P} - {Pf}
from which we can calculate the
displacements
{d} = [S]-1 ({P} - {Pf})
29
30
The last column of the figure is the
addition of the first two columns.
Example Continuous Beam: 2
DOF (EI = constant)
31
32
8
Modified Stiffness Calculations
– Zero End Moment
33
Simplification of the stiffness
analysis is possible if we take into
account the fact that the bending
moment at an end simple support
is zero (node C in the previous
example). This leads to a reduction of one rotational degree of
freedom for each zero moment
location. Inclusion of this modifycation results in a reduction of the
number of member displacement
and force degrees of freedom from
34
4 to 3.
Commensurate with this reduction
is a change in the member
stiffness coefficients as well as the
member fixed-end forces.
35
36
9
Modified Bending
Stiffness Matrices
Modified Member End
Force Calculations
For the propped cantilever beam
of (a – Simple-Fixed (S-F) Beam):
 1 1 L 

3EI 
[kS-F
]


1
1

L
bb


L3 
2
 L L L 
 {QfMb }3x1
For the propped cantilever beam
of (b – Fixed-Simple (F-S) Beam):
 1 L 1 

3EI 
F-S
[k bb
]  3  L L2  L 
L 
1  L 1 


M
{QbM }3x1  [k bb
]3x3 {u bM }3x1
{QbM } 
modified memberend force vector
{QfMb } 
modified member
fixed-end force
vector
Superscript M  S-F for the
“simple-fixed” beam
37
Simple-Fixed Beam
S F
VbS F  3EI
(v b  ve )  3EI
e  Vfb
L3
L2
Superscript M  F-S for the
“fixed-simple” beam 38
Example Continuous Beam –
EI = constant
Modified Equations
q
A
S F
VeS F   3EI
(v b  ve )  3EI
  Vfe
3
2 e
C
B
L
L
L
S F
3EI
3EI
M e  2 (v b  ve ) 
e  MSfe F
L
L
L
d1 = 1
S11 = (4EI/L)AB + (3EI/L)BC = 7EI/L
Fixed-Simple Beam
F S
VbFS  3EI
(v b  ve )  3EI
  Vfb
3
2 b
L
L
FS
F S
M b  3EI
(v b  ve )  3EI
b  M fb
2
L
L
F S
F S
Ve
  3EI
(v b  ve )  3EI
  Vfe
3
2 b
L
L
39
d1 = (S11)-1 (-Pf1) = L/(7EI) (-qL2/8)
= -qL3/56EI
Compatibility:
eAB  d1  BC
b
40
10
Stiffness Coefficient Summary
Member AB End Forces
3qL
Vb  6EI
d  28
2 1
L
In stiffness analysis of structures
that ignore axial deformation,
there are only four sets of
member – displacement stiffness
results that need to be applied for
prismatic members (elements).
qL2
M b  2EI
d   28
L 1
3qL
Ve   6EI
d1   28
L2
qL2
M e  4EI
d   14
L 1
Member BC End Forces
4qL
FS
Vb  3EI
d  Vfb
 7
2 1
L
qL2
FS
M b  3EI
d  M fb
 14
L 1
(a) “Fixed-Fixed” Beam Subjected to a
Unit Transverse Displacement
FS
Ve   3EI
d  Vfe
 7
2 1
L
3qL
2
F S
F S
F S
Vfb
 8 ; M fb
 8 ; Vfe
 8
5qL
qL
3qL
(b) “Fixed-Fixed” Beam Subjected to a
Unit Rotational Displacement
F1   F3 
6EI
L2
; F2  2F4 
F1   F3 
12EI
6EI
; F2  F4 
L3
L2
42
(c) “Fixed-Simple” Beam Subjected to
a Unit Transverse Displacement
4EI
L
Calculations for displacements at
end e are similar to those shown
for end b for the “fixed-fixed”
beam.
43
F1   F3 
3EI
L3
; F2 
3EI
L2
What would the forces (stiffness
coefficients) be a unit
displacement at end e?
44
11
Braced Frame Example
(d) “Fixed-Simple” Beam Subjected to
a Unit Rotational Displacement
F1   F3 
3EI
L2
; F2 
3EI
L
Calculations for the “simple-fixed”
displacements at end e are
similar to the “fixed-simple” beam.
Results from figures (a) – (d) can
be applied directly to frames
provided axial deformation is
45
ignored.
Unbraced Frame Example
46
Unbraced Frame Example - Model
40 kN
12 kN
12 kN
3 kN/m
24 m-kN
24 m-kN
LAB = LDC = 12 m
LBC = 20 m
IAB = IDC = I
IBC = 2I
Equivalent Loaded
Structure
d3
d1
d2
Kinematic DOF
47
48
12