#### Transcript Document

```CHAPTER 5
THE ANALYSIS OF BEAMS & FRAMES
5.1 Introduction
For a two-dimensional frame element each node has the capability of
translating in two directions and rotating about one axis. Thus each
node of plane frame has three degrees of freedom. Similarly three
structure forces (vertical force, shear force and bending moment) act
at a node.
For a two-dimensional beam element each node has two degrees of
freedom (one rotation and one translation). Similarly two structure
forces (vertical force and a bending moment) act at a node. However in
some structures a node has one degree of freedom either rotation or
translation. Therefore they are subjected to a moment or a force as the
case may be.
The structure stiffness matrices for all these cases have been
developed in the previous chapter which can be summarized as
under:i)
ii)
iii)
Beams and frames subjected to bending moment
Beams and frames subjected to shear force and bending moment
Beams and frames subjected to shear force, bending moment and axial
forces
Following steps provide a procedure for the determination of unknown
deformation, support reactions and element forces (axial forces, shear
forces and bending moment) using the force displacement relationship
(W=K). The same procedure applies both to determinate and
indeterminate structures.
5.2 PROCEDURE TO ANALYSE BEAMS AND FRAMES USING DIRECT
STIFFNESS METHOD
5.2.1.
Identifying the components of the structural system or labeling the
Structures & Elements.
As a first step, divide the structure into some finite number of elements by
defining nodes or joints. Nodes may be points of supports, points of
concentrated loads, corners or bends or the points where the internal
forces or displacements are to be determined. Each element extends
between the nodes and is identified by arbitrary numbers (1,2,3).
a) Structure Forces and Deformations
At a node structure forces are assumed to act in their positive direction. The
positive direction of the forces is to the right and upward and positive
moments and rotations are clockwise. Start numbering the known forces
first and then the unknown forces.
Structures Forces not acting at the joints
Stiffness method is applicable to structures with structure forces acting at
nodes only. However if the structure is subjected to concentrated loads
which are not acting at the joints or nodal points or if it is subjected to
following procedure.
i) All the joints are considered to be fixed. [Figure-5(b)]
ii) Fixed End Moments (FEM’s) and Reactions are calculated using the
formulae given in the table as annex-I
iii) If more than one FEM and reactions are present then the net FEM and
Reaction is calculated. This is done by algebric summation. [Figure-5(c)]
Equivalent structure forces or loads at the joints/nodes are obtained by
reversing the signs of net FEM’s & Reactions. [Figure-5(d)]
or
Reversing the signs of Net FEM’s or reaction gives the equivalent structure
v) Equivalent element forces are calculated from these equivalent structure
loads using equation 5.2, 5.3 and 5.4 as explained in article number 5.
vi) Final element forces are obtained by the following equation
w = wE + wF
where
wE = Equivalent element forces
wF = Element forces while considering the elements to be fixed.
A
B
M FAB
M FBA
C
M FBC
(a) Actual Structure
M FCB
(b) Fixed End Moments
& Reactions (wf)
R1
R2
R3
( M FBA+ M FBC
M FAB
R1
( R2 + R3 )
( M FBA+ M FBC
M FAB
R4
M FCB
(c) Net Fixed End Moments
& Reactions
R4
M FCB
(d)
R1
( R2 + R3 )
Fig.5.1
R4
b) Element Forces
Specify the near and far end of each element. Draw free body diagram of
each member showing its local co-ordinate and element forces. Arbitrary
numbers can identify all the element forces of the structure.
5.2.2.
Calculation of Structure Stiffness Matrices of the members
Properties of each element like its length, cross-sectional area, moment of
inertia, direction cosines, and numbers identifying the structure forces
acting at its near and far ends can be systematically tabulated. Using
values of these parameters in equation 4.53 structure stiffness matrix of
each member can be formed by applying equation 4.54,4.55 and 4.56
depending upon the situation.
5.2.3.
Formation of Structure Stiffness Matrix of the Entire Structure
According to the procedure discussed in chapter 3 article 3.1.3 stiffness
matrix [K] of the entire structure is formed.
5.2.4.
Calculation of Unknown Structure Forces and Displacements
Following relation expresses the force-displacement relationship of the
structure in the global coordinate system:
[W] = [K] []
Where
[W] is the structure load vector
[K] is the structure stiffness matrix
[] is the displacement vector
Partitioning the above equation into known and unknown portions as
shown below:
Where
 
 
   K   
   K   
 Wk   K11


 Wu   K21
12
u
22
k
u = unknown deformation
k = known deformation
Expansion of the above leads to the following equation.
[Wk] = [K11] [u] + [K12] [k]
--------------------- (A)
[Wu] = [K21] [u] + K22 k
--------------------- (B)
As k = 0
So, unknown structure displacement [u] can be calculated by
solving the relation (A), which takes the following form.
[u] = [K11]-1 [Wk]
---------------------(C)
Unknown structure force i.e. reactions can be calculated by solving equation
B which takes the following form
Wu = [K21] [u]
---------------------- (D)
5.2.5.
Calculation of element forces:
Finally element forces at the end of the member are computed using the
following equation (E).
w = k
 = T
w = kT --------------- (E)
where [w] is the element force vector
[kT] is the product of [k] and [T] matrices of the element
where [w] is the element force vector
[kT] is the product of [k] and [T] matrices of the element
[] is the structure displacement vector for the element.
Following are the [kT] matrices for different elements used in the
subsequent examples.
Case-I
Beam/frame subjected to bending moment only
 4 EI
 L
kT    2EI

 L
2 EI 
L 
4 EI 

L 
Case-II
Beams subjected to Shear Forces & Bending Moment
 4 EI
 L
 2 EI

kT     6LEI

 L2
 6 EI

 L2
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 

L2 
 12EI 
L3 
12EI 

L3 
Case-IIIFor frame element subjected to axial force, shear force and bending
moment.
kT m
 4 EI
 L
 2 EI

 L
  6 EI
 L2
  6 EI

 L2


0


0


2 EI
L
4 EI
L
 6 EI
L2
6 EI
L2
0
0
 6 EI.l
L2
 6 EI.l
L2
 12EI.l 


3
 L 
 12EI.l 


3
 L 
 AE 

m
 L 
 AE 

m
 L 
6 EI.l
L2
6 EI.l
L2
 12EI.l 


3
 L 
 12EI.l 


3
 L 
 AE 

m
 L 
 AE 

m
 L 
6 EI.m
L2
6 EI.m
L2
  12EI.m 


L3


  12EI.m 


L3


 AE.l 


 L 
 AE.l 


 L 
 6 EI.m 
L2 
 6 EI.m 

L2 
  12EI.m 


L3


  12EI.m 


L3


 AE.l 


 L 
 AE.l 


 L 
Plotting bending moment and shearing force diagrams:
Bending moment and shearing force diagrams of the structure
are plotted using the element forces calculated in step-5.
Examples on the next pages have been solved using the
above-mentioned procedure.
5.3 ILLUSTRATIVE EXAMPLES
EXAMPLE 5.1 Solve the beam shown in the figure using stiffness method.
Solution
Numbering element and structure forces
10k
10k
5'
2kip/ft
5'
W1 , 1
15'
6'
6'
W3 , 3
W2 , 2
W4 , 4
w1, 1
w2, 2
1
w4 ,4
w3 , 3
2
Element forces and deformation
w5 , 5
w6 ,6
3
Calculating fixed end moments and equivalent joint loads
  12.5
 12.5


 37.5
FEF' s   

37
.
5


  39


39

W F
-12.5k'
10k
12.5k'
-39
2k/ft
-37.5k'
37.5k'
Fixed end moments
W1F   12.5
W    25

 Net Fixed End Moments  2 F   
W3 F    1.5

 

W
39

 4F  
-1.5
-25
-12.5
1
2
Net fixed end moment
39
3
10k
2k/ft
39
W K
W1E  12.5
W   25

  2E   
W3 E   1.5

 

W

39

 4E  
1.5
25
12.5
1
2
-39
3
Equivalent joint Moments
Calculating structure stiffness matrices of element
Following table lists the properties needed to form structure stiffness
matrices of elements.
Member
Length (ft)
I
J
1
10
1
2
2
15
2
3
3
12
3
4
Structure stiffness matrices are:
1
2
EI 4 2 1
K 1   
10 2 4 2
2
K 2

3
EI 4 2 2

15 2 4 3
3
1
4
2
0.4 0.2 1
EI

0
.
2
0
.
4

2
2

3
0.267 0.13  2
EI

0
.
13
0
.
267

3
3
4
 0.34 0.167 3
EI 4 2 3
K3     EI

12 2 4 4
0.167 0.34  4
Forming structure stiffness matrix of the entire structure
Using relation [K] = [K]1 + [K]2 + [K]3 structure stiffness matrix of
the entire structure is:
1
2
3
4
0 .2
0
0 1
 0.4
0.2 0.4  0.267
2
0
.
13
0

K   EI 
 0
0.13 0.267  0.34 0.167 3


0
0
0
.
167
0
.
33

4
1
2
3
4
0
0 1
 0 .4 0 .2
0.2 0.667 0.13
2
0

K   EI
0
0.13 0.597 0.167 3


0
0
0
.
167
0
.
33

4
Finding unknown deformations
Unknown deformations are obtained by using the following equation
] = [K]-1 [W]
 1 
0.4






 
0.2
 2


 1 


EI 
 3 
 0









 0
 4 
0.2
0
0.667
0.13
0.13
0.597
0
0.167
2.97
 1 







 
  0.94
 2


 1 


EI 
 3 

0.24









  0.123
 4 
18.8
 1 









 

24.85
 2



 1 




EI 
 3 

35












  132.736
 4 
0


0



0.167



0.33
 0.94
0.24
1.88
 0.49
 0.49
2.08
 0.25
 1.05
1
12.5 




 25 






 1.5 






  39
 0.123


 0.25



 1.05



3.56
12.5 




 25 






 1.5 






  39
Calculating element forces
Using relation [w]=[kT][] we get
 w1E 
0.4






w 
0.2
2
E









 w3 E 
 0




  EI 



w
 4E 
 0






w 
 0
5E









 w6 E 
 0
0.2
0
0.4
0
0.267
0.13
0.13
0.267
0
0.34
0
0.167
0
 12.5






 13.7
0 
18.8














 11.325
0  24.85



1







EI



0 
35
12
.
67














  10.27
0.167 - 132.74









0.34
  39.88
Actual forces on the structure are obtained by superimposing the fixed end
reactions on above calculated forces.
 w1   w1F   w1E    12.5   12.5   0 
 w   w   w   12.5   13.7   26.2 
 2   2F   2E  
 
 

 w3   w3 F   w3 E   37.5  11.325   26.2
 





w
w
w
37
.
5
12
.
67
50
.
1
4
4
F
4
E
  
 
 
 
 

 w5   w5 F   w5 E    39    10.27   50.1
  
 
 
 
 

 w6   w6 F   w6 E   39   39.08  0 
 w 1   w 1F   w 1E    12.5   12.5   0 
 w   w   w   12.5   13.7   26.2 
 2   2F   2E  
 
 

 w 3   w 3F   w 3E   37.5  11.325  26.2
 





w
w
w
37
.
5
12
.
67
50
.
1
 4   4F   4E  
 
 

 w 5   w 5 F   w 5 E    39    10.27   50.1
  
 
 
 
 

w
w
w
39

39
.
08
0
 
 

 6   6 F   6 E  
EXAMPLE 5.2
Analyse the frame shown in the figure using stiffness method.
Solution
2kip/ft
W2 ,2
50k'
W1 , 1
Numbering element forces and deformations
W3 , 3
10'
20'
Structure forces and deformations
w1 , 1
w2 ,2
w3 ,3
1
2
Element forces and deformations
w4 ,4
Finding Fixed end moments and equivalent structure load
 w1F   66.67
 w   66.67 

FEM ' s    2 F   
 w3 F   0 

 

w
0

 4F  
W F
W1F   66.67 
 Net fixed end moments W2 F    66.67
W3 F   0 
2kip/ft
66.67k'
Net fixed end
moment
-16.67k'
W 1E   16.67
Equivalent joint
moment
Calculating structure stiffness matrices of elements.
Following table shows the properties of the elements required to form structure
stiffness matrices of elements.
Members
Length (ft)
i
j
1
20
2
1
2
10
1
3
Structure stiffness matrices of both elements are:
2
1
2
1
0.2 0.1 2
EI 4 2 2
K 1     EI 
1
0
.
1
0
.
2
20 2 4 1


1
K 2
EI 4


10 2
3
1
2 1
0.4
  EI 
0.2
4 3
3
0.2 1

0.4 3
Forming structure stiffness matrix for the entire structure.
Structure stiffness matrix for the entire frame is obtained using relation
[K] = [K]1 + [K]2
1
2
0.6

K   EI  0.1

0.2
0.1
0.2
0
3
0.2 1

0 2

0.4 3
finding unknown deformation
Unknown deformation can be calculated by using equation
[]u = [K11]-1 [W]k
This can be done by partitioning the structure stiffness matrix into known
and unknown deformations and forces
Wk   K11
 
Wu   K 21
K12   u 
 
K 22   k 
[u] = [K11]-1[Wu]
16.67
0.6 0.1 0.2




W

EI
0
.
1
0
.
2
0
 2E 


 W3E 
0.2 0 0.4
1 
  1
 0  EI
 0 
1  1 
 27.78
1     16.667 
EI  0.6 
EI
Finding Unknown Reactions
Unknown reactions can be calculated using the following equation.
W u

K 21 u
W2 E 
 0.1

EI
W 
0.21 / EI  27.78
 
 3E 
W2 E    2.78 
W    5.556

 3E  
W=WF+WE
W2   66.67   2.78   69.45
W    0    5.556    5.56 
 
 

 3 
Calculating element forces(Moments)
Using relation
w  kT 
 w1E 
0.2
w 
 0.1
2
E

  EI 
 w3 E 
0



0
 w4 E 
0
  2.78 
0




0.2 0  
1   5.56 

- 27.78

0.4 0.2 
EI  11.11
  0 


0.2 0.4 
  5.56 
0.1
OR
 w1E 
0.2 0.1  0 
 2.78

EI
1
/
EI

w 
 0.1 0.2  27.78
 5.56





 2E 
 w3 E 
O.4 0.2  27.78
 11.11

EI
1
/
EI

w 
 0.2 0.4  0 
  5.56 





 4E 
Actual forces acting on the structure can be found by superimposing the
fixed end reactions on the forces calculated above.
w  wF  wE
 w1    2.78   66.667  69.45
 w    5.56   66.667   61.11 
 2  



 w3   11.11 
   11.11
0
  
 
 

w

5
.
56
0

5
.
56
 
 

 4 
69.45
61.11
11.11
11.11
1
Final element forces
61.11
2
69.45
5.56
BMD
5.56
EXAMPLE 5.3
Analyse the shown beam using direct stiffness method.
Beam subjected to shear and moment.
STEP-1 Numbering the forces and deformations
5kip/ft (0.41667kip/inch)
2kips
10'
15'
W1
W4
W2
W3
W5
W6
Structure forces and deformations
w2
w1
1
w3
w5
w6
2
w4
w7
Element forces and deformations
w8
Finding fixed end forces and equivalent joint loads
 w1F    30 
 w   30 
 2F  

 w3 F   1 

 

w
1

FEF" s   4 F   
w5 F
 1125

 

w
1125
 6F  

 w   37.5 
 7F  

 w8 F   37.5 
30k"
30k"
2kips
1k
1k
5kip/ft (0.41667kip/inch)
-1125k"
1125k"
37.5k
Fixed end forces
37.5k
30k"
1095k"
Net fixed end forces
[W]F=Net fixed end moments and forces =
30k"
1095k"
W1F    30 
W   1095
 2F  

W3 F   1125 



W
1
4
F

 

W5 F   38.5 

 

W6 F   37.5 
W1E   30 
W   1095 

 2E  
Calculating Structure Stiffness Matrices of Elements
Following table shows the properties of the elements required to form
structure stiffness matrices of elements
L
I
J
K
L
1
120
1
2
4
5
2
180
2
3
5
6
M
From structural elements
Member-1
1
2
4
5
0.016667 0.00041667 0.00041667 1
 0.0333


0.016667
0.033333  0.00041667 0.00041667 2

K 1  EI 
 0.00041667  0.00041667 0.00000694  0.00000694 4


 0.00041667 0.00041667  0.00000694 0.00000694 5
Member-2
2
K 2
3
0.0111
 0.0222
 0.0111
0.0222
 EI 
 0.000185  0.000185

 0.000185 0.000185
5
6
 0.000185 0.000185  2
 0.000185 0.000185  3
0.00000206 0.00000206 5

0.00000206 0.00000206 6
Forming structure stiffness matrix for the entire structure
Structure stiffness matrix for the entire beam is obtained using the relation
[K] = [K]1 + [K]2
1
2
3
0.016667
0
 0.0333
 0.016667
0.0555
0.0111


0
0.0111
0.0222
K   EI 
0
 0.00041667  0.00041667
 0.00041667 0.00023148  0.000185

0
0.00018519 0.000185

4
5
6
 0.00041667 0.00041667
0

 0.00041667 0.00023148 0.00018519
0
 0.000185
0.000185 

 0.00000694  0.000000694
0

 0.00000694 0.000009 0.00000206

0
0.0000206 0.00000206
1
2
3
4
5
6
Finding unknown deformations
Unknown deformation can be calculated using equation
[u] = [K11]-1 [Wk]
This can be done by partitioning the structure stiffness matrix into known
and unknown deformations and forces.
Wk   K11 K12   u 
W    K K   
22   k 
 u   21
1
2
3
4
0.016667
0
W1E 
 0.0333
W 
 0.016667
0.0555
0.0111
2
E



W3 E 

0
0.0111
0.0222

  EI 
0
W4 E 
 0.00041667  0.00041667
W5 E 
 0.00041667 0.00023148  0.000185



0
0.00018519 0.000185

W6 E 
5
  1 
 0.00041667 0.00023148 0.00018519  2 
0
 0.000185
0.000185    3 
 
 0.00000694  0.000000694
0
  4 
 0.00000694 0.000009 0.00000206   5 
 
0
0.0000206 0.00000206  6 
 0.00041667 0.00041667
Using Equation [u] = [K11]-1 [Wk]
 1  1  0.0333 0.016667
 


EI
 2 
0.016667 0.0555 
6
1
 30  1  10535.2946




EI
1095
 22870.5886 
Finding Unknown Reactions
Unknown reactions can be calculated using the following equation
[W]u = [K]21 []u
0
0.0111 
W3 E 

253.86






W4 E 
 0.00041667  0.00041667  10535.2946 1   5.14 

  EI 




W5 E 
 0.00041667 0.00023148  22870.5886  EI  0.91 






W6 E 

 4.235 
0
0.00018519
0
W = WE + WF
W3  253.86 1125 1378.86
  
 
 

W4    5.14   1    4.14 
 



W5   0.91   38.5   39.41 
  
 
 

W6   4.235   37.5   41.735 
Since all the deformations are known to this point we can find the element
forces in each member using the relation
[w]m = [kT]m []m
Member-1:
0.0333
0.016667  0.00041667 0.0001667   10535.2946  30 
 w1E 

w 
 0.016667
  22870.58863 586.7647
0
.
033333

0
.
00041667
0
.
00041667
2
E
   EI 
 1 


 w3 E 
 0.000416667  0.00041667 0.00000694  0.00000694 EI 
   5.1397
0
 

 
 

w
0
.
00041667
0
.
00041667

0
.
00000694
0
.
00000694
0
5
.
1397

 
 

 4E 
Superimposing the fixed end forces for member-1 on the above w’s we
get
w = wE + wF
0
 w1   30   30 

 w  586.7647  30   616.764
 2  



 w3    5.1397  1   4.1397
  
 
 

w
5
.
1397
1
6
.
1397
 
 

 4 
Member 2 :
0.0111
 w5 E 
 0.0222
w 

0.0222
 6 E   EI  0.0111
 w7 E 
 0.000185  0.000185



 0.000185 0.000185
 w8 E 
 0.000185 0.000185  22870.5886  508.2353
  254.1176
 0.000185 0.000185  1 
0


  4.23529
0.00000206 0.00000206 EI 
0
 
 

0.00000206 0.00000206 
0
  4.23529 
Superimposing the fixed end forces
w  wE  wF
 w5   508.2353  1125  616.764
 w   254.1176  1125   1379.117 
 6  



 w7   4.23529  37.5   33.2647 
  
 
 

w
4
.
23529
37
.
5
41
.
73529
 
 

 8 
OR
wE
0.01667
0
 w1E 
 0.0333
w 
 0.01667
0.0333
0
 2E 

 w3 E 
 0.00041667  0.00041667
0



w
0.00041667 0.00041667
0
  4 E   EI 
 w5 E 

0
0.0222
0.0111



0
0.0111
0.0222
 w6 E 

w 

0
 0.000185  0.000185
 7E 

0
0.000185
0.000185
 w8 E 

wE
 w1E   30 
 w  586.53
 2E  

 w3 E    5.14 

 

w
5
.
14

  4E   
 w5 E  508.42

 

w
254
.
21
 6E  

 w    4.24 
 7E  

w
 8 E   4.24 
w = wE +wF
 0.00041667 0.00041667

  10535.2946
 0.00041667 0.00041667
0



 22870.5886 
0.00000694  0.00000694
0



 1
 0.00000694 0.000000694
0
0



0
 0.000185 0.000185  
0
 EI


0
 0.000185 0.000185  
0


0
0.00000206 0.00000206 
0


0
0.00000206 0.00000206
0
 w1   30    30 
 w  586.53  30 
 2 
 

 w3    5.14   1 
  
 

w
5
.
14
1


w  4  
 w5  508.42  1125
  
 

w
 6   254.21  1125 
 w    4.24   37.5 
 7 
 

 w8   4.24   37.5 
 w1   0 
 w   616.53 
 2 

 w3    4.14 
  

w
6
.
14

w  4  
 w5   616.58
  

 w6   1379.2 
 w   33.26 
 7 

 w8   41.74 
0
616.7647k"
1
-4.139
1379.177kip"
-616.7647k"
2
-6.139
41.735
33.26
711.432
33.2647
-4.139
-
+
-
-6.139
Shear Force Diagram
-248.34
-41.735
-
+
-616.7647
-1379.117
Bending Moment Diagram
EXAMPLE 5.4
Analyse the frame by STIFFNESS METHOD
3k/ft
B
5'
C
4k
6'
5'
D
A
10'
W2,2
W5,5
W1,1
W4,4
W6,6
W3,3
W9,9
W12,12
W10,10
W7,7
W8,8
W11,11
Structure Forces
and Deformation
w9,9
w10,10
2
w11,11
w12,12
w7,7
w8,8
w4,4
w17,17
w2,2
w6,6
w13,13
w15 ,15
1
3
w16,16
w5 ,5
w14,14
w1,1
w18,18
w3,3
Element Forces
and Deformations
 w1F    60 
 w   60 
 2F  

 w3 F   0 

 

w
0
 4F  

 w5 F   2 

 

w
2
 6F  

 w   300
 7F  

w
300
 8F  

 w   15 

FEM' s   9 F   
 w10 F   15 

 

w
0
 11F  

 w12 F   0 

 

w
0
 13 F  

w   0 
 14 F  

w
0
 15 F  

w   0 
 16 F  

 w17 F   0 

 

 w18 F   0 
3k/ft
2
300k"
300k"
15k
60k"
15k
2k
4k
1
3
2k
60k"
Fixed End Moments and reactions
[W]F=
 W1F   240
 W   15 
 2F  

 W3F    2 

 

W
 4 F   300 
 W 5 F   15 

 

W6 F   0 
W    0 
 7F  

 W8 F   0 
W   0 
 9F  

W10 F    60 

 

W
0
 11F  

W12 F    2 
15k
240 K"
2k
15k
300 K"
Net fixed end moments
0
[W]E=
W1E   240 
W    15 
 2E  

W3 E   2 

 


W

300
 4E  

W5 E    15 

 

W
0
 6E  

W   0 

 7E  
15k
15k
300 K"
240 K"
2k
0
0
Finding the structure stiffness matrices of elements.
Next we will calculate the structure stiffness matrices for each element
using the properties of members tabulated in below where E = 29000 ksi
I = 100 inch4 and A = 5 inch2 (same for all members):
Member
Length
(in.)
l
m
I
J
K
L
M
N
1
120
0
1
10
1
11
2
12
3
2
120
1
0
1
4
2
5
3
6
3
72
0
-1
4
7
5
8
6
9
For member-1
we have
10
1
11
2
12
3
0
0
1208.333  1208.333 10
 96666.667 48333.333
 48333.333 96666.667
 1
0
0
1208
.
333

1208
.
333



0
0 1208.333  1208.333
0
0 11
K  1  

0
0

1208
.
333
1208
.
333
0
0

 2
 1208.333 1208.333
0
0
20.13889  20.13889 12


0
0  20.13889 20.13889 3
 1208.333  1208.333
For member-2
1
K  2
4
2
5
3
6
1208.333
0
0
 96666.667 48333.333  1208.333
 48333.333 96666.667  1208.333

1208
.
333
0
0


 1208.333  1208.333
20.13889  20.13889
0
0


1208
.
333
1208
.
333

20
.
13889
20
.
13889
0
0



0
0
0
0
1208.333  1208.333


0
0
0
0  1208.333 1208.333

1
4
2
5
3
6
For member-3
4
K  3
7
5
8
6
9
0
0  3356.481 3356.481
161111.111 80555.556
 80555.556 161111.111

0
0

3356
.
481
3356
.
481



0
0
2013.889  2013.889
0
0


0
0

2013
.
889
2013
.
889
0
0


  3356.481  3356.481
0
0
93.2356  93.2356


3356.481
0
0  93.2356 93.2356
 3356.481
Finding structure stiffness matrix for the entire frame.
Using relation [K] = [K]1+[K]2+[K]3 we get the following structure stiffness
matrix.
(See next slide)
4
7
5
8
6
9
1
2
3
4
5
1208.3
193333.3  1208.3  1208 .33 48333.3

1  1208.3 1228.47
0
 1208.3
 20.138

2
 1208.3
0
1228 .47
0
0
3 
48333.3  1208.3
0
257777.78
1208.3
4

 20.139
0
1208.3
2034.03
5  1208.3
0
 1208.3
 3356.48
6 0
K   
7 0
0
0
 80555.56
0
8 0
0
0
0
 2013.89

9
0
0
0
3356.48
0
10 
 48333.3
0
 1208.33
0
0
11 

 1208 .3
0
0
0
12  0
 1208.3
0
 20.139
0
0
6
0
0
 1208.3
 3356.48
0
1301.568
 3356.48
0
 93.2356
0
0
0
7
8
9
10
11
12
1208.3 

0
 1208.3
0 

0
0
0
 1208.3
0
 20.139
80555.56
0
3356.48
0
0
0 
0
 2013 .89
0
0
0
0 

 3356.48
0
 93.2356
0
0
0 

161111.1
0
3356.48
0
0
0 

0
2013.89
01
0
0
0 
3356.48
0
93.2356
0
0
0 
0
0
0
96666.67
0
1208.3 

0
0
0
0
1208.3
0 

0
0
0
1208.3
0
20.139 
0
0
0
0
0
0
48333.3
0
Finding unknown deformations
Unknown deformation can be calculated using equation
[u] = [K11]–1[Wk]
This can be done by partitioning the structure stiffness matrix into known
and unknown deformations and forces.
Wk  K 11
 W   K
 u   21
K 12   u 
K 22   k 
Unknown deformations can be calculated using the equation ;
[]u=[K11]-1[W]k
Solving the above equation we get ,
 1  193333.333
  
 2    1208.333
     1208.333
 3 
 4    48333.333
    1208.333
 5 
0
 6  
  
0
 7 
 1208.333  1208.333 48333.333 1208.333
1228.4718
0
 1208.333  20.13889
0
1228.722
0
0
 1208.333
0
257777.778 1208.333
 20.13889
0
1208.333
2034.0278
0
 1208.333  3356.4814
0
0
0
80555.556
0
0
0


0
0


 1208.333
0

 3356.481 80555.556 

0
0

1301.568  3356.481
 3356.481 161111.111
1
 240 
  15 


 2 


 300
  15 


 0 
 0 


 1   .00184 
  

 2    .01202
    .039859 
 3 

 4    .001527
     .007682
 5 

  6   .037023 
   .001535 

 7 
Finding Unknown reactions
Unknown reactions can be calculated using the following equation:
Wu   K 21 u 
 0.00184 


0
0
0
0
 2013.889
0
0
 W8 E  
   0.01202
W  
0
0
0
3356.481
0
 93.2356 3356.481  0.039859 
 9E  


W10 E   48333.333
  0.001527
0
 1208.333
0
0
0
0

 


0
 1208.333
0
0
0
0
0
W11E  
  0.007682
W12 E   1208.333
  0.037023 
0
 20.13889
0
0
0
0
 0.001535 


 15.4707 
 3.42501


  40.7704 


 14.5242 
 1.42062 
Now we will superimpose the fixed end reactions on the above calculated
structure forces.
[W]=[W]E+[W]F
 W8   15.47   0   15.47 
 W   3.42  0    3.42 
 9 
 
 

W10    40.77    60   19.23
  
 
 

W
14
.
52
0
14
.
52
11
  
 
 

W12   1.42    2    0.58 
Finding the unknown element forces.
Up to this point all the deformations are known to us, we can find the element forces
in each element using relation
[w]m = [kT]m[]m
For member-1
0
0
1208.333  1208.333 
0
 w1   96666.667 48333.333

 w   48333.333 96666.667


0
0
1208.333  1208.333  0.00184 
 2 
 w5   1208.333  1208.333

0
0
 20.13889 20.13889  
0
 


w
1208
.
333
1208
.
333
0
0
20
.
13889

20
.
13889

0
.
01202
6
  


 w3  



0
0
1208.3330  1208.333
0
0
0
  


0
0
 1208.333 1208.333
0
0
  0.039859
 w4  
Equation gives
w1 = 40.77 kips-inch
c.w.
w2 = 129.704 kips-inch
c.w.
w5 = –1.421 kips
Downward
w6 = 1.421 kips
Upward
w3 =14.524 kips
Rightward
w4 = –14.524 kips
Leftward
Superimposing the fixed end reactions in their actual direction we get
w1 = 41.0269 – 60 = -18.973 kips-inch
c.c.w.
w2 = 130.2173 + 60 = 190.2173 kips-inch
c.w.
w5 = 1.427036 – 2 = -3.1427036 kips
Rightward
w6 = -1.427036 – 2 =0.57296 kips
Rightward
w3 = 14.5289 kips
Upward
w4 = -14.5289 kips
Downward
For member-2
0
0
 w7   96666.667 48333.333  1208.333 1208.333
  0.001845 
 w   48333.333 96666.667  1208.333 1208.333
   0.001528
0
0
 8 


 w9   1208.333  1208.333 20.13889  20.13889
   0.012024
0
0
 


0
0
 w10   1208.333 1208.333  20.13889 20.13889
  0.0076822
 w11  
0
0
0
0
1208.333  1208.333  0.0398595
  


0
0
0
0
 1208.333 1208.333   0.0370233
 w12  
Above equation gives
w7 = 109.782 kips-inch
w8 = -53.25353 kips-inch
w9 = - 0.47048 kips
w10 = 0.47048 kips
w11 = 3.427 kips
w12 = -3.427 kips
c.w.
c.c.w.
Downward
Upward
Rightward
Leftward
Similarly for member-2 we have to superimpose the fixed end reactions
w7 = 109.782 – 300 = -190.218 kips-inch
c.c.w.
w8 = –53.25353 + 300 = 246.746 kips-inch
c.w.
w9 = –0.471071+15 = 14.5289 kips
Upward
w10 = 0.471071 +15 = 15.471 kips
Upward
w11 = 3.427 kips
Rightward
w12 = –3.427 kips
Leftward
And finally for member-3 we get
0
0
 3356.481 3356.481   0.0015278
 w13  161111.111 80555.556
 w   80555.556 161111.111
0
0
 3356.481 3356.481  0.00153523
 14  
 w15    3356.481  3356.481
0
0
93.2356  93.2356  0.00768219
 


w
3356
.
481
3356
.
481
0
0

93
.
2356
93
.
2356
0
16
  


 w17  
  0.0370233 
0
0
 2013.889 2013.889
0
0
  


0
0
2013.889  2013.889
0
0
0
 

 w18  
Solving the above equation we get
w13 = -246.74227 kips-inch
w14 = 0 kips-inch
w15 = 3.427kips
w16 = -3.427 kips
w17 = 15.4711 kips
w18 = -15.4711kips
c.c.w.
Upward
Downward
Rightward
Leftward
PLOTTING THE BENDING MOMENT AND SHEARING FORCE
DIAGRAM.
According to the forces calculated above bending moment and
shearing force diagrams are plotted below:
231.959k"
14.528k
+
-190.218k"
+
-
-
+
-15.471k
-246.747k"
-
+
15.407k"
-3.427k
-
3.427k
3.1427k
Shearing Force
Diagram
-18.973
Bending moment
diagram
EXAMPLE 5.5 To analyse the frame shown in the figure
using direct stiffness method.
3k/ft
10'
40'
10'
W 1,1
W7,7
W3 ,3
W9 ,9
W6 ,6
W2 , 2
W4 ,4
W 8,8
W5 ,5
Structure Forces and Deformation
w7, 7
w2,2
w6, 6
w 11,11
w 12,12
w9, 9
w1, 1
w5, 5
w8, 
w4, 4
w3, 3
Element Forces and Deformation
w10, 10
Fixed end moments
[W]F=Net
fixed
end
forces=
 w1F   0 
w   0 
 2F  

 w3 F   0 

 

 w4 F   0 
 w5 F   0 

 

w
0
 6F  

 w    4800
 7F  

 w8 F   4800 
 w   60 
 9F  

 w10 F   60 

 

 w11F   0 
 w12 F   0 
W1F   4800
W   60 
 2F  

W3 F   0 

 

W
0
4
F

 

W5 F    0 

 

W6 F   0 
W   4800 
 7F  

W8 F   60 
W   0 

 9F  
4800k"
0
60
4800k"
0
60
W1E  4800
W     60 
 2E  

W3 E   0 
The properties of each member are shown in the table below.
E = 29 x 103 ksi , I = 1000 inch4 and A = 10 inch2 are same for all members.
Member
Length
l
m
I
J
K
L
M
N
1
169.7056 0.707 0.707
inch
4
1
5
2
6
3
2
480 inch
1
7
2
8
3
9
1
0
Using the structure stiffness matrix for frame element in general form we get struc
stiffness matrix for member-1.
K 1
683536.666 341768.33  4272.064 4272.064 4272.064
 341768.33 683536.666  4272.064 4272.064 4272.064

  4272.064  4272.064 890.0045  890.0045 818..804

4272.064  890.0045 890.045
 818.804
 4272.064
 4272.064
4272.064
818.804
 818.804 890.0045

818.804  890.0045
  4272.064  4272.064  818.804
 4272.064 4
 4272.064 1
 818.804  5

818.804  2
 890.0045 6

890.0045  3
Similarly for member-2 we get,
K 2
0
0
241666.667 120833.33  755.21 755.21
1
 120833.33 241666.667  755.21 755.21
7
0
0


  755.21
2
 755.21
3.15
 3.15
0
0


755
.
21
755
.
21

3
.
15
3
.
15
0
0

8

0
0
0
0
604.167  604.167 3


0
0
0
0

604
.
167
604
.
167

 9
Calculating the structure stiffness matrix for the entire frame.
After getting the structure stiffness matrices for each element we can find the structure
stiffness matrix for the whole structure using following relation:
[K]1+[K]2 = [K]
1
2
1  925203.32
2  3516.854
3  4272.064

4  341768.33
K   5  4272.064

6  4272.064
7  120833.33

8  755.21
9 
0
3516.854
893.195
818.804
4272.064
 890.0045
 818.804
 755.21
 3.15
0
3
 4272.064
818.804
1494.1715
 4272.064
 818.804
 890.0045
0
0
 604.167
4
341678.33
4272.064
 4272.064
683536.66
 4272.064
4272.064
0
0
0
5
 4272.06
 890.0045
 818.804
 4272.064
890.0045
818.804
0
0
0
6
7
8
9
4272.064 120833.33 755.21
0


 818.804  755.21  3.15
0

 890.0045
0
0
 604.167

4272.064
0
0
0


818.804
0
0
0

890.0045
0
0
0


0
241666.66 755.21
0

0
755.21
3.15
0

0
0
0
604.167 
Finding unknown deformations
Unknown deformation can be calculated using equation
[u] = [K11]–1[Wk]
This can be done by partitioning the structure stiffness matrix into known and
unknown deformations and forces.
Wk  K 11 K 12   u 
 W   K
  
K
22   k 
 u   21
Unknown deformations can be calculated using the equation ;
[u] = [K11]-1 [Wk]
  1  925203.333 3516.854  4272.064
    3516.854 893.195 818.804 
 2 

  3    4272.064 818.804 1494.1715
1
4800
  60 


 0 
Solving above equation we get
1 = 0.00669
2 = - 0.223186
3 = 0.141442
c.w.
downward
Rightward
Wu   K 21  u 
W 4 E   341768.33 4272.064
W   4272.064  890.0045
 5E  
W6 E   4272.064  818.804


W7 E   120833.33  755.21
W8 E   755.21
 3.15

 
0
0
W9 E  
 728.716 
 54.2432 


 85.4417 


976
.
927


 5.75539 


 85.4546
 4272.064  0.00669 
 818.804   0.223186
 890.0045  0.141442 


0




0


 604.167  

Now we will superimpose the fixed end forces on the
above calculated equivalent forces.
W=WE+WF
W4   728.716   0 
W   54.2432   0 
 5 
 

W6   85.4417   0 
 


W
976
.
927
4800
 7 
 

W8   5.75539   60 
  
 

W9   85.4546  0 
 728.716 
 54.2432 


 85.4417 


5776
.
927


 65.75539



85
.
4546


Calculating the unknown element forces
All the deformations are known up to this point, therefore we can calculate the
forces in all elements using relation
[w] =[kT][]
For member-1 we get
[w]1=[kT]1[]1
 w1E 
w 
 2E 
 w3 E  
 w4 E 
 
 w5 E 
 w6 E 
683536.665 341768.332
341768.332 683536.665

  6041.668  6041.668

6041.668
 6041.668

0
0

0
0

 4272.063 4272.063
 4272.063 4272.063
50.34676  50.34676
 50.34676 50.34676
1208.321
 1208.321
 1208.321 1208.321
4272.063  4272.063 
0



4272.063  4272.063 0.00669 


 50.34676 50.34676  
0


50.34676  50.34676  0.22319


0
1208.321  1208.321 

  0.14144 

 1208.321 1208.321  
Solving the above equation we get following values of
w1E= 729.0578 kip-inch c.w.
w2E= 3015.829 kip-inch c..w.
w3E= –22.066 kip
Downward
w4E= 22.0669 kip
Upward
w5E= 98.773 kip
Rightward
w6E= –98.773 kip
Leftward
w = wE+wF
As wF are zero therefore w’s will be having the same values as those of wE ‘s
For member-2 we get
[w]2=[kT]2[]2
 w7 E  241666.667 120833.333  755.208
 w  120833.333 241666.667  755.208
 8E  
3.1467
 w9 E     755.208  755.208

 w10 E 
755.208
755.208
 3.1467

 
0
0
0
 w11E  

 w12 E  
0
0
0

  0.00669 


755.208
0
0
0


  0.223186
 3.1467
0
0


0
3.1467
0
0


0
604.16667  604.16667  0.141442 


0
0
 604.16667 604.16667  
755.208
0
0
solving the above equation we get the structure forces
for the equivalent structure loads provided by member-2
w7E= 1785.543 kip-inch c.w.
w8E= 977.047 kip-inch c..w.
w9E= –5.7553 kip
Downward
w10E= 5.7553 kip
Upward
w11E= 85.454 kip
Rightward
w12E= –85.454 kip
Leftward
Superimposing the fixed end forces in their actual direction we get the
w7= 1785.543 – 4800 = -3014.698 kip-inch
c.c.w.
w8= 977.047 + 4800 = 5776.927 kip-inch
c..w.
w9= –5.7553 +60 = 54.244 kip
Upward
w10= 5.7553 + 60 = 65.754 kip
Upward
w11 = 85.454 kip
Rightward
w12= – 85.454 kip
Leftward
Plotting the bending moment and shearing force diagrams.
Bending moment and the shearing force diagram can be drawn according to
the element forces calculated above.
54.244k
65.754k
-85.454k
85.454k
-3014.698k"
5776.927k"
2870.376 kip-inch
54.244 kip
+
-
263.016 "
-
+
729.0578
kip-inch
+
-
-
-65.7541 kips
-22.059kip
-3015.829
kip-inch
-5797.047 kip-inch
Shearing force Diagram
Bending Moment Diagram
P
b
a
MA
A
B
MB
MA  
P ab2
L2
MB 
P a2 b
L2
L
P
L/2
MA
L/2
A
B
MB
MA  
P ab2
L2
P a2 b
MB 
L2
L
A
B
w
MA
MB
L
wL2
MA  12
wL2
MB 
12
L/2
A
w
B
MA
MB
MA  -
5
wL2
192
MB 
11
wL2
192
L
w
A
B
MA
MB
MA  -
L
L/2
L/2
w
A
B
MA
MB
wL2
30
5wL2
MA  96
wL2
MB 
20
5wL2
MB 
96
L
a
MA
A
MA  -
b
M
B
MB
L
Mb  3a 
- 1

L L 
MB 
Ma  3b 
 - 1
L L 
The
End
```