The document discusses the displacement or stiffness method for analyzing structures. It begins by introducing the stiffness method, which allows the same approach to analyze both statically determinate and indeterminate structures. The key concepts are that nodal displacements are the unknowns, and equilibrium equations relating the unknown displacements to known stiffness coefficients are written and solved. It then defines stiffness, stiffness coefficients, and stiffness matrices. The stiffness of a member is the force required to produce a unit displacement. Stiffness coefficients relate the forces and displacements at nodes. Element stiffness matrices are developed and combined to form the overall structure stiffness matrix, which relates the total nodal forces and displacements.

1. BASIC CONCEPTS OF DISPLACEMENT OR STIFFNESS
METHOD:
2.1 INTRODUCTION
Displacement or stiffness method allows one to use the same method to
analyse both statically determinate and indeterminate structures, whereas
the force or the flexibility method requires a different procedure for each of
these two cases. Furthermore, it is generally easier to formulate the
necessary matrices for the computer operations using the displacement
method. Once these matrices are formulated, the computer calculations can
be performed efficiently.
As discussed previously in this method nodal displacements are the basic
unknown. However like slope deflection and moment distribution methods,
the stiffness method does not involve the ideas of redundancy and
indeterminacy. Equilibrium equations in terms of unknown nodal
displacements and known stiffness coefficients (force due a unit
displacement) are written. These equations are solved for nodal
displacements and when the nodal displacements are known the forces in the
members of the structure can be calculated from force displacement
relationship
CHAPTER 2

2. 2.2 STIFFNESS, STIFFNESS COEFFICIENT AND STIFFNESS
MATRIX:
The stiffness of a member is defined as the force which is to be applied at some
point to produce a unit displacement when all other displacement are restrained
to be zero.
If a member which behaves elastically is subjected to varying axial
tensile load (W) as shown
in fig. 2.1 and a graph is drawn of load (W) versus displacement (∆) the result
will be a straight line as shown in fig. 2.2, the slope of this line is called
stiffness.

3. FIG.2.1 FIG.2.2
(Members subjected to varying axial load ( (Graph of load verses
displacement)
Displacement
Load
∆
∆ ∆
W
2
2W
W

4. Mathematically it can be expressed as
K=W/∆ ------ 2.1
In other words Stiffness ‘K’ is the force required at a certain point to
cause a unit displacement at that point.
Equation 2.1 can be written in the following form
W = K ∆ ----- 2.2
Where,
W = Force at a particular point
K = Stiffness
∆ = Unit displacement of the particular point.
The above equation relates the force and displacement at a single point.
This can be extended for the development of a relationship between
load and displacement for more than one point on a structure.

5. 1 2
W = K
∆ = 0
∆ = 0
(a)
(b)
(c)
Fig:2.3
1
2
2
1
1
1 11
2
W = K2 21
W = K1 12
W = K2 22
1
2
∆ =1
L
∆ =1

6. Let us consider a beam of fig. 2.3 and two points (nodes) 1, and 2. If a
unit displacement is induced at point ‘1’ while point ‘2’ is restrained
from deflecting up or down (see the definition of stiffness). then the
forces “W1”and “W2” can be expressed in terms of “∆1” in equation
2.2 as:
W = K ∆ ---------- (2.2)
when ∆1 = 1
W1 = K11. ∆1 = K11
See fig. 2.3(b)
W2 = K21. ∆1 = Kwhere,
K11 = force at 1 due to unit displacement at 1
K21 = force at 2 due to unit displacement at 1
These are known as stiffness co-efficients.
If a unit displacement is induced at a point “2” while point “1” is
restrained from deflecting up or down,then the forces W1 and W2
can be expressed in terms of “ D2” in equation 2.2 as;

7. when ∆2 = 1
W1 = K12. ∆2 = K12
See fig. 2.3(c)
W2 = K22. ∆2 = K22
where,
K12 = force at 1 due to unit displacement at 2
K22 = force at 2 due to unit displacement at 2
First subscript indicates the point of force and second the point of
deformation. As forces W1, W2 are proportional to the deformations
∆1 and ∆2, the following equation for the beam of fig.2.3 can be
written as
W1 = K11 ∆1 + K12 ∆2 ---------- (2.3)
W2 = K21 ∆1 + K22 ∆2 ---------- (2.4)
Rewriting this in matrix form
W
W
K K
K K
1
2
11 12
21 22
1
2





 =












∆
∆
--------- (2.5)

8. where
is called stiffness matrix
Elements of the stiffness matrix are known as
stiffness coefficients. So stiffness coefficients can be
defined as the forces at points (nodes) caused by
introducing various
unit deformations one at a time. is called force
vector
and is called displacement or deformation
vector.
K K
K K
11 12
21 22






W
W
1
2






∆
∆
1
2







9. The expression (2.5) expresses the equilibrium at each of the node points
in terms of stiffness co-efficients and the unknown nodal
deformation and can be written as:
W = K ∆ ---------- (2.6)
The matrix K contains the stiffness co-efficients and it relates the forces
W to the deformations ∆ and is called stiffness matrix. W and ∆ are
called force and deformation vectors. The term “force” and the
symbol “W” refers to the moments as well as forces and the term
“deformation” and symbol “∆” refer to the both rotations and
deflection.
2.3 STIFFNESS OR DISPLACEMENT METHOD FOR TRUSSES
2.3.1 Element and structure stiffness matrix.
Application of stiffness method requires subdividing the structure
into series of elements. The load-deformation characteristics of a
structure are obtained from load-deformation characteristics of
elements. It means that stiffness matrix of a structure [K] is formed
from the stiffness matrices of the individual elements which make
up the structure. Therefore it is important first to develop element
stiffness matrix. The stiffness matrix for a truss element is
developed in subsequent section.

10. w, δ1 1
1 2
w ,δ22
L
2.3.2 Stiffness Matrix of an Axially loaded Element (An Individual Truss
Member)
For the development of an element stiffness matrix for a truss member, let us
consider an axially load member of length ‘L’, area ‘A’ and modulus of
elasticity ‘E’. The ends (nodes) of the member are denoted by 1 and 2 as
shown in fig. 2.4(a).
fig2,.4(a)
(a) Element forces, w1, w2 and deformations, δ1, δ2
(b) Deformation introduced at node ‘1’ with node ‘2’ restrained.

11. (b) Deformation introduced at node ‘1’ with node ‘2’ restrained
(c)Deformation introduced at node ‘2’ with node ‘1’ restrained.
δ=1 δ=0
21
k k2111
1 2
δ =1 δ =0
21
k k2111
1 2

12. 1
Fig:2.4
w2
δ1=1
w
δ2=1
)d)Member forces and deformations of the actual members.
The vectors in fig. 2.4(a) define the forces w1, w2 and the
corresponding deformation δ1 and δ2 at the ends of the
member. These also define their positive directions.
As shown in fig. 2.4(b) a positive deformation δ1, at node ‘1’ is
introduced. while node ‘2’ is assumed to be restrained by a
temporary pin support. Expressing the end forces in terms
of δ:
As
(from stress-strain relationship) --------
(2.7)
δ =
wL
AE

13. w
AE
L
=
δ
--------(2.8)
when δ = 1
when δ1 = 1
-------(2.9)
where ,
k11
is the force at 1 due to unit displacement at 1
k21
is the force at 2 due to unit displacement at 1
The first subscript denotes the location of the node at which the
force acts and second subscript indicates the location of
displacement. As forces and deformations are positive when they act
to the right, so k11
is positive while k21
is negative.
L
AE
kw ==
k
AE
L
and k
AE
L
11 21= = −

14. Similarly if end ‘1’ is restrained while end ‘2’ is deformed in the positive
direction a distance d2
= 1 from fig. 2.4(c).
k
AE
L
and k
AE
L
12 22= − = ----- (2.10)
where, k12
is the force at 1 due to unit displacement at 2
k22
is the force at 2 due to unit displacement at 2
To evaluate the resultant forces w1
and w2
in terms of displacement δ1
and δ2
w1
= k11
δ1
+ k12
δ2
---------- (2.11)
w2
= k21
δ1
+ k22
δ2
---------- (2.12)

15. Expressing in matrix form
w
w
k k
k k
1
2
11 12
21 22
1
2





 =












δ
δ ---------- (2.13)
















+−
−+
=





2
1
2
1
δ
δ
L
AE
L
AE
L
AE
L
AE
w
w












+−
−+
=





2
1
2
1
11
11
δ
δ
L
AE
w
w
---------- (2.14)
It can be written as
w = k δ ---------- (2.15)
[ ] 





+−
−+
=
11
11
L
AE
k ---------- (2.16)

16. This “k” is called element stiffness matrix. It can be observed that sum of the
elements in each column of element stiffness matrix “k” equals zero. It is due
to the reason that co-efficients in each column represent the forces produced
by a unit displacement of one end while the other end is restrained (see fig.
2.4(b)). Since the bar is in equilibrium in the x-direction the forces must be
equal to zero.
Similarly all co-efficients along the main diagonal must be positive
because these terms are associated with the forces acting at the node at which
a positive displacement is introduced into the structure and correspondingly
the force is the same (positive) as the displacement.
2.3.3 Composite stiffness matrix
Equation 2.16 gives the stiffness matrix for an element of
a truss. A great advantage of subdividing a structure into a series
of elements is that the same element stiffness matrix can be used
for all the elements of a structure. Stiffness matrix comprising of
all the element stiffness matrices is called composite stiffness
matrix. Composite stiffness matrix is a square matrix and its size
depends upon number of members. Order of the the composite
stiffness matrix is 2m × 2m, where m is the number of members.
Let us consider a truss shown in fig. 2.5.

18. This truss is subdivided into three elements. Forces and deformations are
shown in fig. 2.5(b).
Stiffness matrix of element no.1
[ ]k
AE
L
1
1 1
1 1
=
−
−






Stiffness matrix of element no.2
[ ]k
AE
L
2
6 5 6 5
6 5 6 5
=
−
−






/ /
/ /
Stiffness matrix of element no.3
[ ]k
AE
L
3
6 5 6 5
6 5 6 5
=
−
−






/ /
/ /

19. Composite stiffness matrix of all elements is given by:
[ ]K
AE
L
c =
−
−
−
−
−
−




















1 1 0 0 0 0
1 1 0 0 0 0
0 0 12 12 0 0
0 0 12 12 0 0
0 0 0 0 12 12
0 0 0 0 12 12
. .
. .
. .
. .
Relationship between forces and displacement from
equation
w = kc
δ ----------- (2.17)
















































−
−
−
−
−
−
=
























6
5
4
3
2
1
6
5
4
3
2
1
2.12.10000
2.12.10000
002.12.100
002.12.100
000011
000011
δ
δ
δ
δ
δ
δ
L
AE
w
w
w
w
w
w

20. It can be seen that the some of the elements in each column of matrix
‘kc’ is zero. It is due to the reason that the stiffness co-efficient in each
column represents the force produced by unit deformation of one end while
other is restrained. Since the member is in equilibrium the sum of the forces
must be zero.
However all co-efficients along the main diagonal must be positive
because these terms are associated with the force acting at the end at which
positive deformation is introduced. As deformation is positive so force
produced is also positive.
2.3.4 Structure stiffness matrix:
Stiffness matrix of a structure can be generated from stiffness
matrices of the elements into which a structure has been subdivided. The
composite stiffness matrix [kc] describes the force deformation relationship
of the individual elements taken one at a time, whereas structure stiffness
matrix [K] describes the load deformation characteristics of the entire
structure. In order to obtain structure stiffness matrix [K] from composite
stiffness matrix [kc] a deformation transformation matrix is used which is
described in the subsequent section.

21. 2.3.5 Deformation transformation matrix:
Deformation transformation matrix relates internal
element or member deformation to the external nodal
structure deformation. It is simply a geometric
transformation of co-ordinates representing the
compatibility of the deformations of the system.
Following is the relationship between element and
structure deformation.
δ = T ∆ ------------ (2.18)
where
δ = element deformation
∆ = structure deformation
T = deformation transformation matrix
As work done by structure forces = work done by element forces

22. [ ] [ ] [ ] [ ]
1
2
1
2
∆
T T
W w= δ ------------ (2.19)
as
[ ] [ ][ ]∆= KW ------------ (2.6)
[ ] [ ][ ]w kc= δ ------------ (2.17)
substituting values of W and w from equation (2.6) and (2.17) into equation (2.19)
[ ] [ ][ ] [ ] [ ][ ]
1
2
1
2
∆ ∆
T T
cK k= δ δ
[ ] [ ][ ] [ ] [ ][ ]δδ c
TT
kK =∆∆ ------------ (2.20)
[ ] [ ][ ]∆= Tδ -------------(2.18)
[ ] [ ][ ] [ ] [ ] [ ][ ][ ]∆ ∆ ∆ ∆
T T T
cK T k T=
[ ] [ ][ ] [ ] [ ] [ ][ ][ ]∆ ∆ ∆ ∆
T T T
cK T k T=
so
[ ] [ ] [ ][ ]K T k T
T
c= -------------(2.21)

23. Therefore structure stiffness matrix [K] can be obtained from composite
element stiffness matrix [kc] if latter is pre-multiplied by [T]T and
post-multiplied by [T].
2.3.6 Formation of deformation transformation matrix:
As we know
δ = T ∆
Let Tij represent the value of element deformation “δi”
caused by a unit structure displacement [∆j]. The total value of each
element deformation caused by all structure deformations may be
written as
δ
δ
δ
1
2
11 1 12 2 1
21 1 22 2 2
1 1 2 2
   
m
n n
n n
m m mn n
T T T
T T T
T T T












=
+ +
+ +
+ +












∆ ∆ ∆
∆ ∆ ∆
∆ ∆ ∆
where δ1
, δ2
, δ3
--- δn
represent set of element deformations and ∆1
, ∆2
---
∆n
the set of structure deformations.
In matrix form

24. 











∆
∆
∆












=












nmnmm
n
n
m TTT
TTT
TTT

2
1
21
22221
11211
2
1
δ
δ
δ
so
[ ]












=
mnmm
n
n
TTT
TTT
TTT
T




21
22221
11211
------------(2.22)
is called deformation transformation matrix. Matrix [T] is
usually a rectangular matrix. Its columns are obtained by
applying a unit values of structure deformation ∆1
through ∆n
one at a time and determining the corresponding element
deformations δ1
through δm
. Formation of the transformation
matrix of a truss, beam and a frame element is explained in the
subsequent chapters.