Lecture 2 Direct Stiffness matrix method Introduction.pptx

Matrix Analysis of
Structures
Prepared by
Engr. Ehsan Ullah Khan
Structural Analysis
Lecture by
Engr. Sana Gul

Introduction
• Two methods are available to analyze indeterminate
structures, depending on whether we satisfy force
equilibrium or displacement compatibility conditions -
They are: Force method and Displacement Method
• Force Method satisfies displacement compatibility and
force-displacement relationships; it treats the forces as
unknowns - Two methods which we will be studying are
Method of Consistent Deformation and (Iterative
Method of) Moment Distribution
• Displacement Method satisfies force equilibrium and
force-displacement relationships; it treats the
displacements as unknowns - Two available methods are
Slope Deflection Method and Stiffness (Matrix)
method

BASIC CONCEPTS OF DISPLACEMENT OR
STIFFNESS ( MATRIX) METHOD:
2.1 INTRODUCTION
Displacement or stiffness method allows one to use
the same method to analyze 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.

BASIC CONCEPTS OF DISPLACEMENT OR STIFFNESS
METHOD:
• In the stiffness (Matrix) 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.

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.

FIG.2.1 FIG.2.2
(Members subjected to varying axial load ( (Graph of load Vs Displacement)
Displacement
Load

 
W
2
2W
W

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.

1 2
W = K
= 0

(a)
(b)
(c)
Fig:2.3
1
2
2
1
1
1 11
2
W = K
2 21
W = K
1 12
W = K
2 22
1
2

L

𝑊1 =𝐾 11
𝑊1 =𝐾 12
𝑊 2=𝐾 22
𝑊 2=𝐾 21
∆1=1
∆2=0
∆2=1
∆1=0

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 = K21
where,
K11 = force at 1 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;

when 2 = 1
W1 = K12. 2 = K12
See fig. 2.3(c)
W2 = K22. 2 = K22
where,
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)

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







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.

w


1 1
1 2
w
2
2
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, 1, 2
(b) Deformation introduced at node ‘1’ with node ‘2’ restrained.

(b) Deformation introduced at node ‘1’ with node ‘2’
restrained
(c)Deformation introduced at node ‘2’ with node ‘1’
restrained.


 


2
1
k k21
11
1 2


 


2
1
k k21
11
1 2

)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 1 and 2 at the ends of the
member. These also define their positive directions.
As shown in fig. 2.4(b) a positive deformation 1, 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
:
As
(from stress-strain relationship) --------(2.7)
 
wL
AE

w
AE
L


--------(2.8)
when d = 1
when 1 = 1
-------(2.9)
where ,
k11
is the force at 1 due to unit displacement at 1
k21
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
k
w 

k
AE
L
and k
AE
L
11 21
 

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
k22
To evaluate the resultant forces w1
and w2
in terms of
displacement 1
and 2
w1
= k11
1
+ k12
2
---------- (2.11)
w2
= k21
1
+ k22
2
---------- (2.12)

Expressing in matrix form
w
w
k k
k k
1
2
11 12
21 22
1
2





 













 ---------- (2.13)



























2
1
2
1


L
AE
L
AE
L
AE
L
AE
w
w























2
1
2
1
1
1
1
1


L
AE
w
w ---------- (2.14)
It can be written as
w = k  ---------- (2.15)
  










1
1
1
1
L
AE
k
---------- (2.16)

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.
  










1
1
1
1
L
AE
k

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.

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









 
k
AE
L
2
6 5 6 5
6 5 6 5









/ /
/ /
 
k
AE
L
3
6 5 6 5
6 5 6 5









/ /
/ /

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
 ----------- (2.17)















































































6
5
4
3
2
1
6
5
4
3
2
1
2
.
1
2
.
1
0
0
0
0
2
.
1
2
.
1
0
0
0
0
0
0
2
.
1
2
.
1
0
0
0
0
2
.
1
2
.
1
0
0
0
0
0
0
1
1
0
0
0
0
1
1






L
AE
w
w
w
w
w
w

 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.

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.
 = T  ------------ (2.18)
where
 = element deformation
 = structure deformation
T = deformation transformation matrix
As work done by structure forces = work done by element forces

       
1
2
1
2

T T
W w
  ------------ (2.19)
as
    

 K
W ------------ (2.6)
    
w kc
  ------------ (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
c
K k
  
         

 c
T
T
k
K 

 ------------ (2.20)
    

 T

 -------------(2.18)
            
   
T T T
c
K T k T

            
   
T T T
c
K T k T

so
      
K T k T
T
c
 -------------(2.21)

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
 = T 
Let Tij represent the value of element deformation “i”
caused by a unit structure displacement [j]. The total value of each
element deformation caused by all structure deformations may be
written as



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 1
, 2
, 3
--- n
represent set of element deformations and 1
,
2
--- n
the set of structure deformations.
In matrix form









































n
mn
m
m
n
n
m T
T
T
T
T
T
T
T
T





2
1
2
1
2
22
21
1
12
11
2
1



so
 













mn
m
m
n
n
T
T
T
T
T
T
T
T
T
T






2
1
2
22
21
1
12
11
------------(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 1
through m
. Formation of the
transformation matrix of a truss, beam and a frame element is explained in
the subsequent chapters.

Lecture 2 Direct Stiffness matrix method Introduction.pptx

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