CHAPTER 4:
PRINCIPLES OF STIFFNESS METHOD FOR BEAMS AND PLANE
FRAMES
4.1 INTRODUCTION:
In chapter 3 the analysis of trusses using stiffness method was
discussed. In this chapter application of stiffness method will be
extended to beams and plane frames. The procedure for application of
this method is the same as that of the trusses but the difference is only
in member stiffness matrix and deformation transformation matrix,
which will be developed in the subsequent section.
4.2 MEMBER OR ELEMENT STIFFNESS MATRIX (FLEXURAL
ELEMENT):
As a frame element is subjected not only to axial forces but also to
shear forces and bending moments, therefore three degrees of freedom
per joint of a frame element are present. A degree of freedom is an
independent deformation of a joint or a node. These are:
i
Axial deformation.
ii) End rotations.
iii) Normal translations.
Out of these three, axial deformation is normally neglected, so element or
member stiffness matrix for an element subjected to shear force and
bending moment will only be developed at this stage.
Consider a member/element shown in figure 4.1. The x-y coordinate
system shown is local coordinate system. Origin is always at the near
end. There are two forces (shear force w3 and a moment w1) acting at
near end of the joint and correspondingly there are two deformations
(vertical translations 3, and rotation 1).Similarly there are two forces
(shear force w4 and a moment w2) acting at the far end of the joint and
correspondingly two deformations (vertical translation 4 and rotation
2).
4.2.1
SIGN CONVENTION
Moments (w1, w2) and rotations (1 and 2) are positive when
clockwise and negative when counter clockwise.
Translation 3, 4 are positive when upward and negative when
downward. The local x-axis runs along the member from the first joint
to the second joint
4.2.2 DERIVATION
The load-stiffness-deformation relationship for this element is the same as
that for a truss element as expressed in equation 3.17.
[w]m = [k]m []m---------------------------------- (3.17)
As there are four forces and four corresponding deformations then the
equation 3.17 can be expanded in the following form:
 w1   k11
 w  k
 2    21
 w3  k 31
  
 w4  k 41
k12
k13
k 22
k 32
k 23
k 33
k 42
k 43
k14   1 
k 24   2  ---------------------------------- (4.1)
k 34   3 
 
k 44   4 
Where each element of the stiffness matrix is called stiffness
coefficient as discussed in chapter 2. It represents the place occupied
by it with respect to row and columns. Any stiffness coefficient may be
represented by kij; where i and j are number of rows and columns.
The above mentioned element stiffness matrix [k]m is formed by
applying a unit value of each end deformation in turn and the
corresponding column of the matrix of equation 4.1 gives the various
end forces developed at the member ends while other deformations are
restrained. This procedure is as follows:
Apply unit positive deformation (clockwise rotation) 1 = 1 and
equating all other deformations to zero (2 = 3 = 4 = 0). The element
would be deformed as shown in figure 4.2.a. From the definition of
stiffness, the forces induced at both ends due to unit clockwise rotation
of near end are as under.
w1 = k11 = Moment produced at ‘1’ due to unit clockwise rotation at 1.
w2 = k21 = Moment produced at ‘2’ due to unit clockwise rotation at 1.
w3 = k31 = Vertical reaction produced at ‘3’ due to unit clockwise rotation
at 1.
w4 = k41 = Vertical reaction produced at ‘4’ due to unit clockwise rotation
at 1.
The values of k11, k21, k31 and k41 can be obtained by using the
moment area theorems.




k11
k 21
A
B
(a)

1
k 31
k11
k 41
(+)
(b)
(-)
(c)
k 21
k11
k 21
B
A
(d)
k31
k41
L
Figure 4.2
As according to moment area theorem no.1 change in slope between two
points on an elastic curve is equal to area of the M/EI diagram between
these two points. Looking at figure 4.2’a’ change in slope between two
ends is equal to unity. Adding the areas of figure 4.2 (b) and figure 4.2
(c).
k11 .L k 21 .L

 1.0
2 EI
2 EI
------------------------------- (.4.2)
According to theorem no.2 of moment area method tangential deviation
of a certain point with respect to the tangent at another point is equal to
the moment of M/EI diagram between the two points calculated about
the point where the deviation is to be determined. From the above
definition the moment of M/EI diagram (figures 4.2 (b) and 4.2 (c) )
about the left of the member is equal to zero.
k11.L  L  k21.L  2 L 
 
 0
2EI  3  2EI  3 
----------------------- (4.3)
Following values of k11 and k21 are obtained by solving equations 4.2 and 4.3.
4 EI
,
L
2 EI
k 21 
L
k11 
---------------------------------- (4.4)
Reaction k41 and k31 can be obtained using equation of equilibrium.
Summation of moments about the right end is equal to zero (see figure
4.2 (d).
SMB=0
k31 L - k11 - k22 = 0---------------------------------- (4.5)
k 31
k11  k 21
L
--------------------- (4.6)
4 EI / L  2 EI / L
k 31 
L
---------------------------------- (4.7)
6 EI
k31 =
L2
Applying force equation of equilibrium to figure 4.2 (d)
Fy = 0
k41 - k31 = 0
k41 = k31
k41 =
6 EI
L2
------------------------------ (4.8)
----------------------------- (4.9)
These are the forces and moments as shown in figure 4.2 a, b, c, d. On
comparison with figure 4.1 the correct signs are obtained and these are
defined by the following equations.
k11 =
k31 =
4 EI
L
6 EI
 2
L
k21 =
k41 =
2EI
L
2EI
L2
------------(4.10)
This gives the first column of the element stiffness matrix. As this
matrix is symmetric so it also provides the first row. To obtain 2nd
column of stiffness matrix deformation (rotation) 2 = 1 is imposed on
the far end equating all other deformations to zero 1 = 3 = 4 = 0.
The element would be deformed as shown in figure 4.3 (a). From the
definition of stiffness as mentioned in chapter 2, the forces induced at
both ends due to unit rotation at far end can be defined as
w1 = k12 = Moment produced at
‘1’ due to unit clockwise
rotation at 2.
w2 = k22 = Moment produced at
‘2’ due to unit clockwise
rotation at 2.
w3 = k32 = Vertical reaction
produced at ‘3’ due to unit
clockwise rotation at 2.
w4 = k42 = Vertical reaction
produced at ‘4’ due to unit
clockwise rotation at 2.
k32
2
k12
A
k 42
B
k22
(a)
L
k12
(+)
(b)
(c)
The values of k12, k22, k32 and
k42 can be obtained by using
the moment area theorems.
(-)
k 22
k12
k 22
B
A
(d)
k32
k42
L
Figure 4.3




Applying moment area theorem no. 1 and using bending moment diagram
of figure 4.3 (b,c). For this case change in slope between both ends is equal
to unity so
k 22 L k 12 L

 1 ------------------------------ (4.11)
2 EI
2 EI
However according to moment area theorem no.2 the moment of M/EI
diagram about the right end of the member is equal to zero.
k 22 .L  L  k12 .L  2 L 
 
 0
2 EI  3  2 EI  3 
------------------------ (4.12)
Solving equations 4.11 and 4.12
k12
2 EI k  4 EI

22
،
L
L
--------------------------- (4.13)
Reaction k42 and k32 can be obtained using equation of equilibrium. Summation
of moments about the left end is equal to zero (see figure 4.3 (d)).
k13
k33





k23
(a)
B
A
k43
L
BENDING MOMENT DIAGRAMS FOR THE ELEMENT
(b)
k23
k13
(c)
k13
k23
k33
Figure 4.4
k43
(d)
Applying equation of force equilibrium to figure 4.5(d)
Fy = 0
k24




A
k14
k44
B

(a)
k34
L
(b)
k24
BENDING MOMENT DIAGRAM FOR THE ELEMENT
k14
(c)
k14
k24
(d)
k44
k34
L
Figure 4.5
k 34 
12 EI
--------------------------- (4.34)
L3
Correct signs can be obtained by comparing these values with figure
4.1. These are defined as
k 14 
6 EI
2
L
k 24 
6EI
2
L
k 34 
12 EI
3
L
k 44 
12 EI
3
--------- (4.35)
L
On combining the calculation given in equation 4.10, 4.17, 4.27 and 4.35,
following element stiffness matrix is obtained.
k m
 4 EI
 L
 2 EI

 L
  6 EI
 L2
 6 EI
 2
 L
2 EI
L
4 EI
L
 6 EI
L2
6 EI
L2
 6 EI
L2
 6 EI
L2
12EI
L3
 12EI
L3
6 EI 
L2 
6 EI 

2
L  --------------------------- (4.36)
12EI
 3 
L
12EI 

3
L 
And the force, stiffness and deformation relationship is as under:2 EI
 6 EI
6 EI 
 4 EI
2
2
 L

L
L
L
 w1  

2
EI
4
EI

6
EI
6
EI
w  

2
2
 2   L
L
L
L 
 w    6 EI  6 EI 12EI
12EI 
3

   2
2
3
3 
L
L
L
L
w
 4   6 EI  6 EI  12EI 12EI 
 2

2
3
3
L
L
L 
 L
 1 
 
 2
  --------------------------- (4.37)
 3
 4 
4.3 MEMBER OR ELEMENT STIFFNESS MATRIX FOR A BEAM /
FRAME ELEMENT SUBJECTED TO MOMENTS ONLY:
The element stiffness matrix for members whose ends cannot
translate but can rotate is obtained by removing third and fourth row
corresponding to 3, 4 and third and fourth columns corresponding
to w3 and w4 from the above mentioned element stiffness matrix of
equation 4.36 as shown below:1
k m
 4 EI
 L
 2 EI

 L
  6 EI
 L2
 6 EI
 2
 L
2
2 EI
L
4 EI
L
 6 EI
L2
 6 EI
L2
3
 6 EI
L2
 6 EI
L2
12EI
L3
 12EI
L3
4
6 EI 
L2  1
6 EI 

L2  2
12EI
 3  3
L
4
12EI 

L3 
The resulting stiffness matrix for a beam element is as follows:
k m  EI 
4 2
----------------------------(4.38)
L 2 4
And the force stiffness deformation relationship for a beam element is as
fallows: w1  EI 4 2  1 
w  

 
 2  L 2 4  2 
---------------------------
- (4.39)
4.4 MEMBER OR ELEMENT STIFFNESS MATRIX OF A FRAME
ELEMENT SUBJECTED TO AXIAL LOADING IN ADDITION TO
SHEARING FORCES AND BENDING MOMENTS.
The final element that is capable of axial loading in addition
to shear forces and bending moments is now considered.
The stiffness matrix for this element can be formed by
superposition of the element stiffness matrix of truss element and frame
element formed as given in the equations 2.16 and 4.20. Figure 4.6 defines
the positive forces (axial forces, shear forces and bending moments) and
deformations (axial deformations, vertical translations and rotations) for
the element.
x
, 6
2
,
w2
w6
y
1
,
1
w

, 5
w5
3
,
w3
4
,
w
4
,A
EI
L
Figure 4.6
By superposition of equation 2.16 and 4.36.The following element
stiffness matrix for six element forces and deformations is obtained.
k m
 4 EI
 L
 2 EI

 L
 6 EI
 2
 L
 6 EI
 L2

 0

 0

2 EI
L
4 EI
L
6 EI
2
L
6 EI
6 EI
6 EI
L2
6 EI
L2
6 EI
2
L
12 EI
3
L
12 EI
2
L
12 EI
3
L
12 EI
2
L
3
3
L
L
0
0
0
0
0
0
0
0
0
0
AE
L
 AE
L

0 

0 


0 

0 

 AE 

L 
AE 
L 
--------------- (4.40)
And the force, stiffness and deformation relationship is as under:
 4 EI
 w1  
   L
   2 EI
 w2  
   L
  
 w3    6 EI
   L2
 
 w   6 EI
 4   L2
  
w   0
 5 
  
  
 w6   0

2 EI
L
 6 EI
L2
6 EI
L2
0
4 EI
L
 6 EI
L2
6 EI
L2
0
 6 EI
L2
12EI
L3
 12EI
L3
0
6 EI
L2
 12EI
L3
 12EI
L3
0
0
0
0
AE
L
0
0
0
 AE
L

0 


0 


0 


0 

 AE 

L 

AE 
L 
 1 
 
 
 2 
 
 
 3 
 
 
 
 4
 
 
 5
 
 
 6 
-------- (4.41)
4.5 DEFORMATION TRANSFORMATION MATRIX:
As discussed in the case of trusses, deformation transformation matrix
is used to transform the element deformations from local coordinates
to structure deformations in global co-ordinates. Using this
deformation transformation matrix structure stiffness matrix is
obtained as given in the following equation:
[K]m = [T]Tm [k]m [T]m --------------------- (3.3)
4.5.1 DEFORMATION TRANSFORMATION MATRIX OF A FRAME
ELEMENT SUBJECTED TO AXIAL FORCE, SHEAR FORCE AND
BENDING MOMENT.
A deformation transformation matrix will also be developed to carry out
the transformation from element to global coordinates.
Consider the frame member shown in figure 4.7. Member axis, xaxis of member coordinate system makes an angle x with x-axis of the
structure coordinate system as shown in figure 4.7 (b), similarly
member axis, x-axis of member coordinate system makes angle y with
y-axis of the structure coordinate system. The cosines of these angles
are given below
l= Cos x
m = Cos y
x
,
w2
2
w
,
6
6
y
,
w1

,
w5
w
EI
1
,
4
4
,A
L
5
,
w3
3
Element forces and deformations
x
W4,
W2 , 2
y
W3 , 3
W5 ,
W1 ,
1
4
W6 ,
6
5
Structure forces and deformations
Figure 4.7
Again consider the frame member shown in figure 4.7, 1 and 2 are the
element deformations (rotation) whereas 1 and 2 are the structure
deformations (rotations). As Z-axes for both element and structure
coincides, therefore both element and structure rotations will be the same
when structure deformation 1 = 1, then the element deformation
 1

 2
 3

 4
 5

 6
 1
 0
 0

 0
 0

 0
-------------------------------- (4.43)
and when structure deformation 2 = 1, then the element deformations
 1

 2
 3

 4
 5

 6
 0
 1
 0

 0
 0

 0
-------------------------------- (4.44)
Relationship between structure and element deformations can also be
obtained in the similar manner for a unit vertical translation of the left
end i.e. 3 = 1 (see figure 4.8)
3 = 3.Cos x, 5 = 3.Cos y
3 = 1. Cos x, 5 = 1. Cos y
3=1.  = ,  5=1.m=m
FIGURE 4.8
5=3Cosy=1.m=m
3=3Cosx=1.l=l
1=2=4=6=0
=
cos
y
Y-axis
3
5
x
=
x



 0

 1.1  1 

 0

 1.m  m

 0

0

3
 1

 2
 3

 4
 5

 6

cos
3
3
y
x
 (.5)
X-axis
Y-axis
y
4

4
X-axis

Figure 4.9
4=4Cosx=1.l=l
6=4Cosy=1.m=m
5
x
For a unit vertical translation of the right end i.e. 4 = 1 (Fig.4.9) element
deformations are as under:
 1

 2
 3

 4
 5

 6

0

0
 0
  4 Cos x1.l  l
 0
  4 Cosy  1.m  m









-------------------------------- (4.46)
where 3 and 4 are vertical translations of the element.
for a unit horizontal translation (along x-axis) of the left
end.i-e; 5=1 (4.47)
(Fig.4.10)
-------------------------------element deformations are as under: 1

 2
 3

 4
 5

 6
 0 
 0 

  m

 0 
 l 

 0 
 (.7)
For a unit horizontal translation the right end I-e;6 = 1, element
deformations are as under:-
 1

 2
 3

 4
 5

 6
 0 
 0 

 0 

  m
 0 

 l 
-------------------------------- (4.48)
Y-axis
Figure 4.10
5 5 cosx=
 5sinx=-m

y
3
x
X-axis
5
5
Y-axis

y
Figure 4.11
 sinx -m
 cosx 
5
6


6=1
x
X-axis
Writing the above six equations in matrix form.
 1  1
  0
 2 
 3  0
 
 4  0
 5  0
  
 6  0
0  1 
1 0 0
0
0  2 
0 l 0  m 0   3 
 
0 0 l
0  m  4 
0 m 0
l
0   5 
 
0 0 m 0
l  6 
0
0
0
0
-------------------------------- (4.49)
 = T ---------------------------------- (3.2)
Comparing equation (4.49) with equation (3.2) deformation
transformation matrix (T) is obtained which is as follows:
1
0

0
T 
0
0

0
0 
1 0 0
0
0 
0 l 0 m 0 

0 0 l
0  m
0 m 0
l
0 

0 0 m 0
l 
0
0
0
0
-------------------- (4.50)
This is the deformation transformation matrix of a frame element subjected
to shear force, bending moment and axial forces.
To obtain the deformation transformation matrix of a beam/frame element
subjected only to shear force and bending moment, the axial deformations are
ignored. Therefore by deleting last two rows and last two columns of the
matrix in equation (4.50) following deformation transformation matrix is
obtained:
1
0
T
0

0
0
1
0
0
0
0
l
0
0
0

0

l
-------------------------------- (4.51)
Similarly to obtain the deformation transformation matrix of a beam/frame
element subjected only to bending moment, the axial and shear deformations
are ignored Therefore by deleting last two rows and last two columns of the
matrix in equation (4.51) following deformation transformation matrix is
obtained:
1 0
T

0
1


------------------------------- (4.52)
4.6 STRUCTURE STIFFNESS MATRICES
4.6.1 For a beam/frame element subjected to bending moment only:
Using the element stiffness matrix and deformation transformation matrix
structure stiffness matrix is formed. Following equation is used for this
purpose:
[K]m = [T]Tm [k]m [T]m --------------------------- (4.53)
Where,
[T]m = Deformation Transformation matrix
[k]m = Element stiffness matrix
In this case:
T  m
And
1 0


0
1


-------------------------------- (4.52)
EI
 km 
L
4 2
2 4


T  m
T
and
-------------------------------- (4.38)
1 0


0
1


Substituting these values in equation (4.53) and solving, following structure
stiffness matrix is obtained:
EI
 K m 
L
4 2
2 4


-------------------------------- (4.54)
4.6.2 Beam/Frame element subjected to Shear Force and Bending Moment only:
In this case:
1
0
T  
0

0
 k m
0
1
0
0
 4 EI
 L
 2 EI

 L
 6 EI
 L2
 6 EI
 2
 L
0
0
l
0
0
0

0

l
2 EI
L
4 EI
L
6 EI
L2
6 EI
L2
-------------------------------- (4.51)
6 EI
L2
6 EI
L2
12 EI
L3
12 EI
L3
6 EI 
L2 
6 EI 

L2 
12 EI 
L3 
12 EI 

L3 
-------------------------------- (4.36)
1
0
T
Tm  
0

0
0 0 0
1 0 0

0 l 0

0 0 l
Substituting these values in equation (4.53) the following structure stiffness
matrix is obtained:
K m
2 EI
 4 EI
 L
L

4 EI
 2 EI
 L
L

  6 EI . l  6 EI . l
 L2
L2
 6 EI
6 EI

.
l
2
2 .l
L
 L
 6 EI
6 EI

.
l
.
l

L2
L2

 6 EI
6 EI

.
l
.
l

L2
L2

12 EI 2  12 EI 2 
.l
.l

L3
L3
 12 EI 2 12 EI 2 
.l
3
3 .l
L
L

 (.55)
4.6.3 Frame element subjected to Shear Force/ Bending Moment/Axial
Forces:
Here
T  m
 k m
1
0

0

0
0

0
 4 EI
 L
 2 EI

 L
  6 EI
 2
 L
 6 EI
 L2

 0


 0
0 0 0 0
0 
1 0 0 0
0 

0 l 0 m 0 

0 0 l
0  m
0 m 0
l
0 

0 0 m 0
l 
2 EI
L
4 EI
L
 6 EI
L2
6 EI
L2
 6 EI 6 EI
L2
L2
 6 EI 6 EI
L2
L2
12 EI  12 EI
L3
L3
 12 EI 12 EI
L3
L3
0
0
0
0
0
0
-------------------------------- (4.50)

0 

0
0 


0
0 

0
0 

AE  AE 

L
L 
 AE AE 
L
L 
0
-------------------------------- (4.40)
and
T  m
T
1
0

0

0
0

0
0 0 0 0
0 

1 0 0 0
0

0 l 0 m 0 

0 0 l
0  m
0 m 0
l
0 

0 0 m 0
l 
Substituting these values in equation (4.53) and multiplying, structure stiffness
matrix is given on the next page, is obtained.
 4 EI
 L
 2 EI

 L
  6 EI l
 L2
K m   6 EI
 2 l
 L
 6 EI
 L2 m

  6 EI m
 L2
2 EI
L
4 EI
L
 6 EI
l
2
L
6 EI
l
2
L
6 EI
m
2
L
 6 EI
m
2
L
 6 EI
l
2
L
 6 EI
l
2
L
12EI 2 AE 2
l 
m
3
L
L
6 EI
l
2
L
6 EI
l
2
L
6 EI
m
2
L
6 EI
m
2
L
 12EI 2 AE 2   12EI AE 
l 
m

lm
3
3


L
L
L 
 L
 12EI 2 AE 2 12EI 2 AE 2
l 
m
l 
m
3
3
L
L
L
L
  12EI AE 
 L3  L lm
12EI AE 
 L3  L  lm
12EI AE 
 L3  L lm
  12EI AE 
 L3  L  lm
12EI
 L3

AE 
L
lm
12EI 2 AE 2
m 
l
3
L
L
 12EI 2 AE 2
m 
l
3
L
L







  12EI AE  
 L3  L lm 
 12EI 2 AE 2 
m 
l 
3
L
L

12EI 2 AE 2
m 
l 
3

L
L
 6 EI
m
2
L
 6 EI
m
2
L
12EI AE 
 L3  L  lm