The document discusses the direct stiffness method for analyzing trusses. [1] The direct stiffness method treats each individual truss element as a structure and calculates the element stiffness matrix using the element's local coordinate system. The total structure stiffness matrix is obtained by superimposing the stiffness matrices of all elements. [2] To transform between the local and global coordinate systems, a deformation transformation matrix [T] is derived. This matrix relates the element deformations in the local system to the structure deformations in the global system. [3] The structure stiffness matrix of each element [K]m is obtained by transforming the element stiffness matrix [k]m from the local to the global system using [

1. Direct Stiffness
Method - Trusses
Engr. Sana Gul
Lectures prepared by
Engr. Ehsan Ullah Khan
Structural Analysis

2. 2
DIRECT STIFFNESS METHOD FOR TRUSSES:
3.1 INTRODUCTION
In the previous Lectures the procedure for obtaining
the structure stiffness matrix was discussed. The structure
stiffness matrix was established by the following equation:
[K] = [T]T [kc] [T] -----------(2.21)
However if a large and complicated structure is to be
analyzed and if more force components are to be included
for an element then the size of composite stiffness matrix
[kc] and deformation transformation matrix [T] will be
increased.

3. 3
In this chapter an alternative procedure called the
“direct stiffness method” is introduced. This procedure
provides the basis for most computer programs to analyze
structures. In this method each individual element is
treated as a structure and structure stiffness matrix is
obtained for this element using the relationship:
[K]m = [T]T
m[k]m [T]m -----------(3.1)
Where
[K]m=Structure stiffness matrix of an individual element.
[T]m= Deformation Transformation matrix of an individual
element.
[k]m = Member Stiffness matrix of an individual element.
Total structure stiffness matrix can be obtained by
superimposing the structure stiffness matrices of the
individual elements.

4. 4
As all members of a truss are not in the same
direction i.e. inclination of the longitudinal axes of the
elements varies, therefore stiffness matrices are to be
transformed from element coordinate system to structure
or global coordinate system.
When the matrices for all the truss elements have
been formed then adding or combining together the
stiffness matrices of the individual elements can generate
the structure stiffness matrix [K] for the entire structure,
because of these considerations two systems of coordinates
are required.
i) Local or member or element coordinate system:
In this coordinate system x-axis is collinear with the
longitudinal axis of the element or member. Element
stiffness is calculated with respect to this axis. This system
is illustrated in figure 3.1

5. 5
Element 'B'
Element 'A'
A B
Structure or
globel axes
Local or member
or element axes for
element B
Local or member
or element axes
for element 'A'
Figure 3.1

6. 6
ii) Structure or global coordinate system:
A single coordinate system for the entire structure is chosen, with
respect to which stiffness of all elements must be written.
3.2 PROCEDURE FOR THE FORMATION OF TOTAL STRUCTURE
STIFFNESS MATRIX FOR AN ELEMENT USING DIRECT STIFFNESS
METHOD:
Following is the procedure for the formation of structure
stiffness matrix:
i) Formation of the element stiffness matrix using equation 2.16.
 













1
1
1
1
L
AE
k m
----------- (2.16)
ii) Formation of the deformation transformation matrix [T] for
a single element:
[]m = [T]m []m -----------(3.2)

7. 7
where
[]m= Element or member deformation matrix.
[]m= Structure deformation matrix of an element or member
[T]m= Element or member deformation transformation matrix.
(iii)Formation of a structure stiffness matrix [K] or an element using
the relation
[K]m = [T]T
m[k]m [T]m -----------(3.1)
The above mentioned procedure is discussed in detail in the subsequent
discussion.
3.2.1 The formation of element stiffness matrix in local co-
ordinates:
It has already been discussed in the previous chapter. However it
is to be noted that for horizontal members the structure stiffness matrix
and element stiffness matrix are identical because both member
coordinate systems and structure coordinate systems are identical.
But for inclined members deformation transformation matrices are to be
used because member coordinate system and structure coordinate
system are different; therefore their structure stiffness matrix and
element stiffness matrix will also be different.

8. 8
3.2.2 The formation of deformation transformation matrix:
As the main difference between the previously discussed
method and direct stiffness method is the formation of the
deformation transformation matrix. In this article
deformation transformation matrix for a single element
will be derived.
Before the formation of deformation transformation matrix
following conventions are to be established in order to
identify joints, members, element and structure
deformations.
1) The member is assigned a direction. An arrow is written
along the member, with its head directed to the far end of
the member.
2) i, j, k, and l, are the x and y structure deformations at
near and far (tail & head) ends of the member as shown in
figure 3.2. These are positive in the right and upward
direction.

9. The formation of deformation transformation
matrix:
3) r and s are the element deformations at near and far
(tail & head) ends of the member as shown in figure 3.2.
4)The member axis (x-axis of member coordinate system)
makes an angle x, with the x-axis of the structure
coordinate system as shown in figure 3.2.
5)The member axis (x-axis of member coordinate system)
makes an angle y, with the y-axis of the structure co-
ordinate system as can be seen from figure 3.2.
6) The cosines of these angles are used in subsequent
discussion Letters l and m represent these respectively.
9

10. 10
Figure 3.2
Near
end
Far
end
Member
direction

x
y
 (b)
r
s


i
j
k

Near
end
Far
end
Member
direction





(a)
x
y
l
i
j
k
r
s

11. 11
   2
1
2
2
1
2
1
2
1
2
Y
Y
X
X
X
X
L
X
X






 = Cos x and m = Cos y
 = Cos x =
m = Cos y =
   2
1
2
2
1
2
1
2
1
2
Y
Y
X
X
Y
Y
L
Y
Y






The algebraic signs of θ‘s will be automatically accounted for
the members which are oriented in other quadrants of X-Y
plan.
In order to form deformation transformation matrix, once
again consider the member of a truss shown in figure 3.1.

12. 12
s=0
r
j
i
s=0
j

r
=1
(b)
(a) =1
j
rjcosy=m
r
cosxl
x
i
i
i
y
x

13. 13

r=0


r=0

y
x
Figure 3.3
(d)
(c)
=1
k
=1



s



1

c
o
s

y
=
m
1
l
k
k
skcosx=
1
s
s
x
y

14. 14
Following four cases are considered to form deformation transformation matrix.
Case-1: Introduction of horizontal deformation to the structure i = 1, while far
end of the member is hinged (restrained against movement). From the geometry
of figure 3.3(a)
r = i Cos x = 1. Cos x = Cos x = l
r =l ---------- (3.3)
s = 0 ---------- (3.4)
Case-2: Introduction of vertical deformation to the structure j = 1, while near
end of the member is hinged (restrained against movement). From the geometry of
figure 3.3 (b)
r = j Sin x = j Cos y = 1. Cos y = m
r = m ---------- (3.5)
s = 0 ---------- (3.6)
Case-3: Introduction of horizontal deformation to the structure k = 1, while
near end of the member is hinged (restrained against movement).From the
geometry of figure 3.3(c)
s = k Cos x = 1. Cos x = Cos x = l
r = 0 ---------- (3.7)
s = l ---------- (3.8)

15. 15
Case-4: Introduction of vertical deformation to the structure L = 1, while
near end is hinged
From the geometry of figure 3.3(d)
L = 1
r = 0 ---------- (3.9)
s =  L
Cos y = m ---------- (3.10)
Effect of four structure deformations and two member deformations can be
written as
r = i Cos x + j Cos y + k .0 +  L .0
s = i .0 + j .0 + k Cos x +  L Cos y
r = l.i + m.j + 0.k + 0.  L ----------- (3.11)
s = 0.i + 0.j + l.k + m.  l ----------- (3.12)
Arranging equations 3.11 and 3.12 in matrix from





























L
k
j
i
s
r
m
m


0
0
0
0

 ----------- (3.13)

16. 16
Comparing this equation with the following equation
[]m = [T]m []m ---------------(3.2)
After comparing equation (3.14) and (3.2) following deformation
transformation matrix is obtained.
  






m
m
T m 

0
0
0
0
- ---------- (3.14)
This matrix [T]m transforms four structure deformation (i, j, k, L )
into two element deformation (r and s).
3.2.3 Formation of structure stiffness matrix:
Structure Stiffness matrix of an individual member is first to be
transformed from member to structure coordinates. This can be done by
using equation 3.1.
[K] = [T]T
m [k]m [T]m --------(3.1)
  






m
l
m
l
m
T
0
0
0
0
----- (3.15)

17. 17
  








1
1
1
1
L
AE
k m
------(2.16)
 
L
k
j
i
m
m
m
m
L
k
j
i
K m







































0
0
0
0
1
1
1
1
0
0
0
0
 
L
k
j
i
m
m
m
m
L
AE
m
m
L
k
j
i
K m



































0
0
0
0
 
L
k
j
i
L
k
j
i
m
m
m
m
m
m
m
m
m
m
m
m
L
AE
K m

























2
2
2
2
2
2
2
2












--------------------------------(3.16)

18. 18
3.2.4 ASSEMBLING OF THE INDIVIDUAL STRUCTURE ELEMENT
STIFFNESS MATRICES TO FORM TOTAL STRUCTURE
STIFFNESS MATRIX
Combining the stiffness matrix of the individual members can
generate the stiffness matrix of a structure However the
combining process should be carried out by identifying the truss
joint so that matrix elements associated at particular member
stiffness matrices are combined. The procedure of formation of
structure stiffness matrix is as follows:
Step 1: Form the individual element stiffness matrices for each
member.
Step 2: Form a square matrix, whose order should be equal to that
of structure deformations.
Step 3: Place the elements of each individual element stiffness
matrix framed into structure in the corresponding rows and
columns of structure stiffness matrix of step-2.
Step 4: If more than one element are to be placed in the same
location of the structure stiffness matrix then those elements
will be added.

19. 19
1
2 3
2
3
2
1
Structure
stiffness
matrix for
member-2
Structure
stiffness
matrix for
member-1
Structure
stiffness matrix
Following figure shows the above mentioned process.

20. 20
 
2
1
2
1
k
k
k
k
K
1
1
1
1
1 









  
3
2
3
2
k
k
k
k
K
2
2
2
2
2 










     2
1 K
K
K 

 
3
2
1
k
k
k
k
k
k
k
k
K
3
2
1
















2
2
2
2
1
1
1
1
0
0
In the above matrix the element in the second
row and second column is k1+k2 where k1 is for
member 1 and k2 is for member 2.This is
because both k1 and k2 have structure
coordinates 2-2.

21. 21
Illustrative Examples Regarding the Formation of [K] Matrix:
L/2
(x1,y1) = (0,0)
(x1,y1) = (L/2,2L/3)
(x2,y2) = (L/2,2L/3)
(x2,y2) = (L,0)
(x1,y1) = (L,0)
(x2,y2) = (0,0)
(a)
(b)
(c)
(Structure forces and deformations)
(Free body diagram indicating structure forces
and coordinates)
2L/3
1 1
2 2
6 4
5 3
6 4
5 3
1
5
6
3
4
2
12
11.5
1
1
2
2
3
3
L/2
Example 3.1: Form the structure stiffness matrix for the following
truss by direct stiffness method.

22. 22
Member Length Near
End
X1 Y1
Far End
X2, Y2
i j k L l=Cosx =
X2-X1/
Length
m=Cosy
Y2-Y1/
Length
1 5L/6 0 0 L/2 2L/3 5 6 1 2 0.6 0.8
2 5L/6 L/2 2L/
3
L 0 1 2 3 4 0.6 -0.8
3 L L 0 0 0 3 4 5 6 -1 0
 
l
k
j
i
m
lm
m
lm
lm
l
lm
l
m
lm
m
lm
lm
l
lm
l
l
k
j
i
L
AE
K m

















2
2
2
2
2
2
2
2
According to eq. (3.16) we get

23. 23
 
2
1
6
5
768
.
0
576
.
0
768
.
0
576
.
0
576
.
0
432
.
0
576
.
0
432
.
0
768
.
0
576
.
0
768
.
0
576
.
0
576
.
0
432
.
0
576
.
0
432
.
0
2
1
6
5
1 














L
AE
K
 
4
3
2
1
768
.
0
576
.
0
768
.
0
576
.
0
576
.
0
432
.
0
576
.
0
432
.
0
768
.
0
576
.
0
768
.
0
576
.
0
576
.
0
432
.
0
576
.
0
432
.
0
4
3
2
1
2 














L
AE
K
 
6
5
4
3
0
0
0
0
0
00
.
1
0
00
.
1
0
0
0
0
0
00
.
1
0
00
.
1
6
5
4
3











L
AE
3
K

24. 24
 
6
5
4
3
2
1
768
.
0
576
.
0
000
.
0
000
.
0
768
.
0
576
.
0
576
.
0
432
.
1
000
.
0
000
.
1
576
.
0
432
.
0
000
.
0
000
.
0
768
.
0
576
.
0
768
.
0
576
.
0
000
.
0
000
.
1
576
.
0
432
.
1
576
.
0
432
.
0
768
.
0
576
.
0
768
.
0
576
.
0
536
.
1
000
.
0
576
.
0
432
.
0
576
.
0
432
.
0
000
.
0
864
.
0
6
5
4
3
2
1





























L
AE
K

25. 25
If
wr is positive then member is in compression.
ws is negative then member is in compression.
wr is negative then member is in tension.
ws is positive then member is in tension.
3.5 ANALYSIS OF TRUSSES USING DIRECT STIFFNESS METHOD:
Basing on the derivations in the preceding sections of this chapter
a truss can be completely analyzed. The analysis comprises of
determining.
i) Joint deformations.
ii) Support reactions.
iii) Internal member forces.
As the first step in the analysis is the determination of unknown joint
deformation. Using the equations can do this.
[W] = [K] []
The matrices [W], [K] and [] can be divided in submatrices in the
following form




















k
u
u
k
K
K
K
K
W
W
22
21
12
11
----------- (3.20)

26. 26
Where
Wk = Known values of loads at joints.
Wu = Unknown support reaction.
u = Unknown joint deformation.
k = Known deformations, generally zero due to support
conditions.
K11, K12, K21, K22 are the sub-matrices of [K]
Expanding equation 3.20
Wk = K11 u + K12 k ----------- (3.21)
Wu = K21 u + K22 k ----------- (3.22)
If the supports do not move, then k = 0 therefore equation 3.21
& 3.22 can be written as
Wk = K11 u ----------- (3.23)
Wu = K21 u ----------- (3.24)
By pre-multiplying equation 3.23 by [K11]-1 following equation is
obtained
[K11]-1 [WK]=[K11]-1 [K11][U]
[u] = [K11]-1 [Wk] ----------- (3.25)
Substituting value u from equation 3.25 into equation 3.24
[Wu] = [K21] [K11]-1 [Wk] ----------- (3.26)

27. 27
Using equation 3.25, 3.26 and 3.19, joint deformations, support reactions
and internal member forces can be determined respectively.
As this method does not depend upon degree of indeterminacy so it can
be used for determinate as well as indeterminate structures.
Using the basic concepts as discussed in the previous pages, following
are the necessary steps for the analysis of the truss using stiffness
method.
1-Identify the separate elements of the structure numerically and specify
near end and far end of the member by directing an arrow along the
length of the member with head directed to the far end as shown in the
fig. 3.2
2-Establish the x,y structure co-ordinate system. Origin be located at one
of the joints. Identify all nodal co-ordinates by numbers and specify
two different numbers for each joint (one for x and one for y). First
number the joints with unknown displacements.
3-Form structure stiffness matrix for each element using equation 3.16.
4-Form the total structure stiffness matrix by superposition of the element
stiffness matrices.
5-Get values of unknown displacements using equation 3.25.
6-Determine support reaction using equation 3.26.
7-Compute element or member forces using equation 3.19.

28. 28
3.6- Illustrative Examples Regarding Complete Analysis of Trusses:
Example 3.3: Solve truss in example 3.1 to find member forces.
L/2
(x1,y1) = (0,0)
(x1,y1) = (L/2,2L/3)
(x2,y2) = (L/2,2L/3)
(x2,y2) = (L,0)
(x1,y1) = (L,0)
(x2,y2) = (0,0)
(a)
(b)
(c)
(Structure forces and deformations)
(Free body diagram indicating structure forces and
coordinates)
2L/3
1 1
2 2
6 4
5 3
6 4
5 3
1
5
6
3
4
2
12
11.5
1
1
2
2
3
3

29. 29
The [K] matrix for the truss as formed in example 3.1 is:
 
6
5
4
3
2
1
768
.
0
576
.
0
000
.
0
000
.
0
768
.
0
576
.
0
576
.
0
432
.
1
000
.
0
000
.
1
576
.
0
432
.
0
000
.
0
000
.
0
768
.
0
576
.
0
768
.
0
576
.
0
000
.
0
000
.
1
576
.
0
432
.
1
576
.
0
432
.
0
768
.
0
576
.
0
768
.
0
576
.
0
536
.
1
000
.
0
576
.
0
432
.
0
576
.
0
432
.
0
000
.
0
864
.
0
6
5
4
3
2
1





























L
AE
K
Using the relation [W] = [K] [] we get





















k
u
u
k
K
K
K
K
W
W
22
21
12
11

30. 30













































































6
5
4
3
2
1
6
5
4
3
2
1
6
5
4
3
2
1
768
.
0
576
.
0
000
.
0
000
.
0
768
.
0
576
.
0
576
.
0
432
.
1
000
.
0
000
.
1
576
.
0
432
.
0
000
.
0
000
.
0
768
.
0
576
.
0
768
.
0
576
.
0
000
.
0
000
.
1
576
.
0
432
.
1
576
.
0
432
.
0
768
.
0
576
.
0
768
.
0
576
.
0
536
.
1
000
.
0
576
.
0
432
.
0
576
.
0
432
.
0
000
.
0
864
.
0
L
AE
W
W
W
W
W
W
From equation (3.25) [u] = [K11]-1 [Wk]






















2
1
536
.
1
000
.
0
000
.
0
864
.
0
2
1
1
W
W
AE
L
































12
5
.
11
651
.
0
000
.
0
000
.
0
157
.
1
2
1
AE
L
















812
.
7
31
.
13
2
1
AE
L

31. 31
Using the relation:
[Wu] = [K21] [u]








































812
.
7
310
.
13
768
.
0
576
.
0
576
.
0
432
.
0
768
.
0
576
.
0
576
.
0
432
.
0
6
5
4
3
W
W
W
W






























1.66
1.25
13.66
10.25
6
W
5
W
4
W
3
W
Now the member forces can be found using the relation:

32. 32
   



















L
k
j
i
m
m
L
AE
wr 

    09
.
2
812
.
7
31
.
13
00
.
0
00
.
0
8
.
0
6
.
0
8
.
0
6
.
0
5
6
1 

















AE
L
L
AE
w
kips
    09
.
17
00
.
0
00
.
0
812
.
7
31
.
13
8
.
0
6
.
0
8
.
0
6
.
0
5
6
2 

















AE
L
L
AE
w kips
    00
.
0
00
.
0
00
.
0
00
.
0
00
.
0
0
.
0
0
.
1
0
.
0
0
.
1
3 














AE
L
L
AE
w kips

33. 33
11.5k
12k
1.25k 10.25k
13.66k
1
7
.
0
9
k
i
p
2
.
0
9
k
i
p
s
1.66k
0 kips
Final Results & sketch

34. 34
Thank You

35. 35
Example 3.4:
L L
P
1
2
3
4
5
6
7
8
(L,L)
(0,0)
(L,0)
(2L,0)
(L,L)
(L,L)
y
x
1
2
3
2 2
1
1
3
4
6
5
7
8
1
2
(Structure forces and deformations)
(Structure to be analysed)
L
1
2
3
(Free body diagram showing structure forces and
coordinates)
Solve the truss in example 3.2 to find member forces.

36. 36
The [K] matrix for the truss as formed in example 3.2 is as follows:
 
8
7
6
5
4
3
2
1
5
.
5
.
0
0
0
0
5
.
5
.
5
.
5
.
0
0
0
0
5
.
5
.
0
0
414
.
1
0
0
0
414
.
1
0
0
0
0
0
0
0
0
0
0
0
0
0
5
.
5
.
5
.
5
.
0
0
0
0
5
.
5
.
5
.
5
.
5
.
5
.
414
.
1
0
5
.
5
.
414
.
2
0
5
.
5
.
0
0
5
.
0
5
.
0
1
8
7
6
5
4
3
2
1
2



































L
AE
K
Now as W = k

37. 37





























































































8
7
6
5
4
3
2
1
8
7
6
5
4
3
2
1
5
.
5
.
0
0
0
0
5
.
5
.
5
.
5
.
0
0
0
0
5
.
5
.
0
0
414
.
1
0
0
0
414
.
1
0
0
0
0
0
0
0
0
0
0
0
0
0
5
.
5
.
5
.
5
.
0
0
0
0
5
.
5
.
5
.
5
.
5
.
5
.
414
.
1
0
5
.
5
.
414
.
2
0
5
.
5
.
0
0
5
.
0
5
.
0
1
8
7
6
5
4
3
2
1
2L
AE
W
W
W
W
W
W
W
W





















k
u
u
k
k
k
k
k
W
W
22
21
12
11
As k = 0 or 3, 4, 5, 6, 7,8, all are zero due to supports and using
[Wk] = [k11] [u] we get
W
W
AE
L
1
2
1
2
2
1 0
0 2 414





 












.


Which gives

38. 38
W1 = 0
W2 = P
Also using [Wu] = [k21] [u] we have
0
2
1 0
0 2 414
1
2
P
AE
L





 












.


Or






















P
AE
L 0
414
.
2
0
0
1
2
1
2
1













2 1
2 414
2 414 0
0 1
0
L
AE P
.
.
.
Or

39. 39


1
2
2
2 414
0





 






L
AE P
.
Or
2
2
2 414

LP
AE
.
Or
2 05858
 .
PL
AE
Now Wu = [K21] [u]
W
W
W
W
W
W
AE
L
3
4
5
6
7
8
1
2
2
5 5
5 5
0 0
0 1414
5 5
5 5
























 
 


























. .
. .
.
. .
. .


W
W
W
W
W
W
P
3
4
5
6
7
8
1
2 414
5 5
5 5
0 0
0 1414
5 5
5 5
0
























 
 


























.
. .
. .
.
. .
. .

40. 40
W
W
W
W
W
W
P
P
P
P
P
3
4
5
6
7
8
1
2 414
05
05
0
1414
05
05













































.
.
.
.
.
.
W3 = 0.207 P
W4 = -0.207 P
W5 = 0
W6 = -0.5858 P
W7 = -0.207 P
W8 = -0.207 P
Now as

































l
k
j
i
s
r
m
l
m
l
m
l
m
l
L
AE
w
w
So
 
2
1
8
7



















l
k
j
s
i
m
l
m
l
L
AE
w

41. 41
So, for member # 1:
 
























AE
L
P
L
AE
w
s
w
5858
.
0
0
0
0
707
.
0
707
.
0
707
.
0
707
.
0
1
Or
ws= 0.707 x 0.5858 P
w1= 0.414 P
Similarly, for member # 2:
 
6
5
2
1
0
0
5858
.
0
1
0
1
0
2

















 AE
PL
L
AE
w
ws
Or w2 = 1 x 0.5858 P
Or w2 = 0.5858 P

42. 42
For member # 3:
 
4
3
2
1



















l
k
j
s
i
m
l
m
l
L
AE
W
 

















0
0
5858
.
0
707
.
707
.
707
.
707
.
3 AE
PL
L
AE
w
Or
w3= 0.707 x 0.5858 P
w3= 0.414 P

43. 43
P
0.207P 0.59P 0.207P
0.207P
0.207P
0.414P 0.414P
0.586P
w1 = 0.414 P
w2 = 0.586 P
w3 = 0.414 P

44. 44
Example 3.5 Analyze the truss shown in the figure.
5k
2k
W1,1
W7,7
W8,8
W5,5
W6,6
W3,3
W4,4
W2,2
1
2
3
4
5
8' 6'
8'
Truss to be analysed
Structure forces and
displacements
A and E are constant for each member.

45. 45
Member Length l m i j k l
1 10 -0.6 -0.8 7 8 1 2
2 11.34 -0.7071 0.7071 1 2 3 4
3 8 1 0 3 4 5 6
4 6 1 0 5 6 7 8
5 8 0 1 1 2 5 6
Using the properties given in the above table we can find the structures
stiffness matrices for each element as follows.

46. 46
Element forces and displacements
w , 5
w , 
4
5
4
w , 
w , 
2
3 3
w , 
9 9
6
w , 
3
6
w , 
5
w , 
10
7
10
7
1
2 2
w , 
w , 
4
8
1
8
1

47. 47
6
5
4
3
0
0
0
0
0
125
.
0
0
125
.
0
0
0
0
0
0
125
.
0
0
125
.
0
k
6
5
4
3
4
3
2
1
044
.
0
044
.
0
044
.
0
044
.
0
044
.
0
044
.
0
044
.
0
044
.
0
044
.
0
044
.
0
044
.
0
044
.
0
044
.
0
044
.
0
044
.
0
044
.
0
k
4
3
2
1
2
1
8
7
064
.
0
048
.
0
064
.
0
048
.
0
048
.
0
036
.
0
048
.
0
036
.
0
064
.
0
048
.
0
064
.
0
048
.
0
048
.
0
036
.
0
048
.
0
036
.
0
k
2
1
8
7
3
2
1

























































48. 48
8
7
4
3
125
.
0
0
125
.
0
0
0
0
0
0
125
.
0
0
125
.
0
0
0
0
0
0
k
8
7
4
3
8
7
6
5
0
0
0
0
0
167
.
0
0
167
.
0
0
0
0
0
0
167
.
0
0
167
.
0
k
8
7
6
5
5
4






























Using equation [K] = [K1]+[K2]+[K3]+[K4]+[K5] we get the structure
stiffness matrix of (8x8) dimensions. Partitioning this matrix with respect
to known and unknown deformations we get [K11] and [K12] portions as
follows.

49. 49































665
.
9
983
.
62
2
5
233
.
0
004
.
0
004
.
0
08
.
0
1
2
1
 
2
1
233
.
0
004
.
0
004
.
0
08
.
0
2
1
11 






K
 
8
7
6
5
4
3
064
.
0
048
.
0
048
.
0
036
.
0
125
.
0
0
0
0
044
.
0
044
.
0
044
.
0
004
.
0
2
1
12




























K
Using equation [Du]=[K11]–1[Wk] we get the
Putting the above value in equation [Wu]=[K21][Du]

50. 50















































































405
.
2
803
.
1
208
.
1
0
197
.
3
197
.
3
665
.
9
983
.
62
064
.
0
048
.
0
048
.
0
036
.
0
125
.
0
0
0
0
044
.
0
044
.
0
044
.
0
004
.
0
6
5
4
3
2
1
W
W
W
W
W
W
Element forces can be calculated using equation as follows.

51. 51
 
 
 
 
  kip
w
kip
w
kip
w
kip
w
kip
w
208
.
1
0
0
665
.
9
983
.
62
1
0
1
0
8
1
0
0
0
0
0
0
1
0
1
6
1
0
0
0
0
0
0
1
0
1
8
1
540
.
4
0
0
665
.
9
983
.
62
707
.
0
707
.
0
707
.
0
707
.
0
314
.
11
1
006
.
3
665
.
9
983
.
62
0
0
8
.
0
6
.
0
8
.
0
6
.
0
10
1
5
4
3
2
1


















































































52. 52
4.54 kips
3.197 K
3.197 K
1.208
kips
-
3
.
0
0
6
k
i
p
s
2 K
5 K
1.803 K
2.405 K
0 K
1.208 K