This document outlines the procedure for analyzing beams and frames using the direct stiffness method. It discusses: 1) Dividing the structure into finite elements and defining nodes with degrees of freedom. Forces and deformations are defined at nodes. 2) Calculating the structure stiffness matrix of each element based on its properties. 3) Assembling the overall structure stiffness matrix. Equations are set up relating known/unknown loads and displacements to solve for unknowns. 4) Computing element forces from the determined displacements using appropriate transformation matrices for each element type. Bending moment and shear force diagrams can then be plotted.

1. THE ANALYSIS OF BEAMS & FRAMES
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) Beams and frames subjected to bending moment
ii) Beams and frames subjected to shear force and bending moment
iii) Beams and frames subjected to shear force, bending moment and axial
forces

2. 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.
2. PROCEDURE TO ANALYSE BEAMS AND FRAMES USING DIRECT
STIFFNESS METHOD
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.

3. 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
distributed loads then equivalent joint loads are calculated using the
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
loads on the joints/nodes.
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

4. where
wE = Equivalent element forces
wF = Element forces while considering the elements to be fixed.
FAB
FAB
M
M
Equivalent Joint Loads
Fixed End Moments
& Reactions
Fixed End Moments
& Reactions
Actual Structure
Fig.5.1
FBA
( + )
R
( + )
1
R
R1
( M
R
2 3
FBC
+
R2
M
R3
R4
FCB
M
4
R
(d)
A B C
FBA
FBA
R
R1
( M
FAB M
M
+
2
FBC
3
R
FBC
M
FCB
M
R4
(c)
(b)
FCB
M
(a)
Net
(wf)
(WE)
M

5. 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.
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.
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.
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

6. Partitioning the above equation into known and unknown portions as
shown below:
 
 
   
   
 
 
W
W
K K
K K
k
u
u
k

























11 12
21 22


Where
Wk = known loads
Wu = unknown loads
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)

7. 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
 











L
EI
L
EI
L
EI
L
EI
kT
4
2
2
4

8. Case-II Beams subjected to Shear Forces & Bending Moment
 



























3
3
2
2
3
3
2
2
2
2
2
2
12
12
6
6
12
12
6
6
6
6
4
2
6
6
2
4
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
kT
Case-IIIFor frame element subjected to axial force, shear force and bending
moment.
 

























































































 





 



















 






 




















L
l
AE
L
l
AE
m
L
AE
m
L
AE
L
l
AE
L
l
AE
m
L
AE
m
L
AE
L
m
EI
L
m
EI
L
l
EI
L
l
EI
L
EI
L
EI
L
m
EI
L
m
EI
L
l
EI
L
l
EI
L
EI
L
EI
L
m
EI
L
m
EI
L
l
EI
L
l
EI
L
EI
L
EI
L
m
EI
L
m
EI
L
l
EI
L
l
EI
L
EI
L
EI
kT m
.
.
0
0
.
.
0
0
.
12
.
12
.
12
.
12
6
6
.
12
.
12
.
12
.
12
6
6
.
6
.
6
.
6
.
6
4
2
.
6
.
6
.
6
.
6
2
4
3
3
3
3
2
2
3
3
3
3
2
2
2
2
2
2
2
2
2
2

9. 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.

10. 3. ILLUSTRATIVE EXAMPLES
EXAMPLE 1.1 Solve the beam shown in the figure using stiffness method.
Solution
6'
w ,
W , W , W ,
Structure loads and deformations
w ,
w ,
1 1
1
2 2
1
1
w , w ,
4 4
Element forces and deformation
3 3
2
5
2 2 3 3
5'
10k
5'
2kip/ft
15' 6'
10k
W ,
w ,
3
5 6 6
4 4
Numbering element and structure forces

11.  
























39
39
5
.
37
5
.
37
5
.
12
5
.
12
's
FEF
10k
2k/ft 2k/ft
10k
Fixed end moments
-12.5k'
-37.5k'
-39
12.5k'
37.5k'
39
Calculating fixed end moments and equivalent joint loads
 






























39
5
.
1
25
5
.
12
Moments
End
Fixed
Net
W
4
3
2
1
F
F
F
F
W
W
W
W
F
39
Net fixed end moment
-12.5 -25 -1.5
1 3
2

12.  



























39
5
.
1
25
5
.
12
4
3
2
1
E
E
E
E
W
W
W
W
W K
12.5 25 1.5
1 2 3
Equivalent joint Moments
-39
Equivalent joint loads are :
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

13. Structure stiffness matrices are:
 
2
1
2
1
1
2
1
2
1
4
.
0
2
.
0
2
.
0
4
.
0
EI
4
2
2
4
10 











 
EI
K
 
3
2
3
2
2
3
2
3
2
267
.
0
13
.
0
13
.
0
267
.
0
EI
4
2
2
4
15 











 
EI
K
 
4
3
4
3
3
4
3
4
3
34
.
0
167
.
0
167
.
0
34
.
0
EI
4
2
2
4
12
EI
K 













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:

14.  
4
3
2
1
33
.
0
167
.
0
0
0
167
.
0
34
.
0
267
.
0
13
.
0
0
0
13
.
0
267
.
0
4
.
0
2
.
0
0
0
2
.
0
4
.
0
4
3
2
1














 EI
K
 
4
3
2
1
33
.
0
167
.
0
0
0
167
.
0
597
.
0
13
.
0
0
0
13
.
0
667
.
0
2
.
0
0
0
2
.
0
4
.
0
EI
K
4
3
2
1













Finding unknown deformations
Unknown deformations are obtained by using the following equation
] = [K]-1 [W]

15. 




























































































































































































































736
.
132
35
85
.
24
8
.
18
1
39
5
.
1
25
5
.
12
56
.
3
05
.
1
25
.
0
123
.
0
05
.
1
08
.
2
49
.
0
24
.
0
25
.
0
49
.
0
88
.
1
94
.
0
123
.
0
24
.
0
94
.
0
97
.
2
1
39
5
.
1
25
5
.
12
33
.
0
167
.
0
0
0
167
.
0
597
.
0
13
.
0
0
0
13
.
0
667
.
0
2
.
0
0
0
2
.
0
4
.
0
1
4
3
2
1
4
3
2
1
1
4
3
2
1
EI
EI
EI

16. Calculating element forces
Using relation [w]=[kT][] we get




















































































































































88
.
39
27
.
10
67
.
12
325
.
11
7
.
13
5
.
12
1
132.74
-
35
24.85
18.8
34
.
0
167
.
0
0
0
167
.
0
34
.
0
0
0
0
267
.
0
13
.
0
0
0
13
.
0
267
.
0
0
0
0
4
.
0
2
.
0
0
0
2
.
0
4
.
0
6
5
4
3
2
1
EI
EI
w
w
w
w
w
w
E
E
E
E
E
E

17. Actual forces on the structure are obtained by superimposing the fixed end
reactions on above calculated forces.




































































































































0
1
.
50
1
.
50
2
.
26
2
.
26
0
08
.
39
27
.
10
67
.
12
325
.
11
7
.
13
5
.
12
39
39
5
.
37
5
.
37
5
.
12
5
.
12
6
5
4
3
2
1
6
5
4
3
2
1
6
5
4
3
2
1
E
E
E
E
E
E
F
F
F
F
F
F
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w




































































































































0
1
.
50
1
.
50
2
.
26
2
.
26
0
08
.
39
27
.
10
67
.
12
325
.
11
7
.
13
5
.
12
39
39
5
.
37
5
.
37
5
.
12
5
.
12
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
E
6
E
5
E
4
E
3
E
2
E
1
F
6
F
5
F
4
F
3
F
2
F
1
6
5
4
3
2
1

19. EXAMPLE 1.2
Analyse the frame shown in the figure using stiffness method.
Solution
2kip/ft 50k'
10'
W ,
W ,
W ,
1 1
2 2
3 3
w ,
w ,
1 1
2 2 w ,
3 3
w ,
4 4
1
2
Structure forces and deformations
Element forces and deformations
20'
Numbering element forces and deformations

20.  


























0
0
67
.
66
67
.
66
'
4
3
2
1
F
F
F
F
w
w
w
w
s
FEM
-16.67k'
Equivalent joint
moment
66.67k'
Net fixed end
moment
2kip/ft
Finding Fixed end moments and equivalent structure load
 
























0
67
.
66
67
.
66
moments
end
fixed
Net
3
2
1
F
F
F
F
W
W
W
W
Equivalent Joint Loads
   
67
.
16
1 

E
W

21. 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:
 
1
2
2
.
0
1
.
0
1
.
0
2
.
0
1
2
4
2
2
4
20
1 












 EI
EI
K
 
3
1
4
.
0
2
.
0
2
.
0
4
.
0
3
1
4
2
2
4
10
2

















 EI
EI
K
2 1 2 1
1 3 1 3

22. Forming structure stiffness matrix for the entire structure.
Structure stiffness matrix for the entire frame is obtained using relation
[K] = [K]1 + [K]2
 
3
2
1
4
.
0
0
2
.
0
0
2
.
0
1
.
0
2
.
0
1
.
0
6
.
0












 EI
K
finding unknown deformation
Unknown deformation can be calculated by using equation
[]u = [K11]-1 [W]k
1 2 3
This can be done by partitioning the structure stiffness matrix into known
and unknown deformations and forces

23. 


























k
u
u
k
K
K
K
K
W
W
22
21
12
11
EI
EI
W
W
E
E
1
0
0
4
.
0
0
2
.
0
0
2
.
0
1
.
0
2
.
0
1
.
0
6
.
0
67
.
16 1
3
2































   
EI
EI
78
.
27
667
.
16
6
.
0
1
1
1











[u] = [K11]-1[Wu]
Finding Unknown Reactions
Unknown reactions can be calculated using the following equation.
  
u
W    u
K 
21

24.  





























556
.
5
78
.
2
78
.
27
/
1
2
.
0
1
.
0
3
2
3
2
E
E
E
E
W
W
EI
EI
W
W
W=WF+WE































56
.
5
45
.
69
556
.
5
78
.
2
0
67
.
66
3
2
W
W
Calculating element forces(Moments)
Using relation
    

 kT
w




















































56
.
5
11
.
11
56
.
5
78
.
2
1
0
27.78
-
0
4
.
0
2
.
0
0
2
.
0
4
.
0
0
0
2
.
0
1
.
0
0
1
.
0
2
.
0
4
3
2
1
EI
EI
w
w
w
w
E
E
E
E

25. OR

























































56
.
5
11
.
11
/
1
0
78
.
27
4
.
0
2
.
0
2
.
0
4
.
56
.
5
78
.
2
/
1
78
.
27
0
2
.
0
1
.
0
1
.
0
2
.
0
4
3
2
1
EI
O
EI
w
w
EI
EI
w
w
E
E
E
E
Actual forces acting on the structure can be found by superimposing the
fixed end reactions on the forces calculated above.
E
F w
w
w 



























































56
.
5
11
.
11
11
.
61
45
.
69
0
0
667
.
66
667
.
66
56
.
5
11
.
11
56
.
5
78
.
2
4
3
2
1
w
w
w
w

26. 1
2
69.45 61.11 11.11
5.56
69.45
61.11
11.11
5.56
BMD
Final element forces

27. EXAMPLE 1.3
Analyse the shown beam using direct stiffness method.
Beam subjected to shear and moment.
15'
2kips
10'
5kip/ft (0.41667kip/inch)
W1 W2 W3
w1
W5
W4
w7
w5
w8
w6
W6
w4
w3
w2
1 2
Structure forces and deformations
Element forces and deformations
STEP-1 Numbering the forces and deformations

28.  
























































5
.
37
5
.
37
1125
1125
1
1
30
30
"
8
7
6
5
4
3
2
1
F
F
F
F
F
F
F
F
w
w
w
w
w
w
w
w
s
FEF
37.5k
Fixed end forces
30k"
37.5k
5kip/ft (0.41667kip/inch)
1k 1k
1125k"
2kips 30k"
-1125k"
Finding fixed end forces and equivalent joint loads

29. 30k" 1095k"
Net fixed end forces
[W]F=Net fixed end moments and forces =











































5
.
37
5
.
38
1
1125
1095
30
6
5
4
3
2
1
F
F
F
F
F
F
W
W
W
W
W
W
Equivalent joint loads
30k" 1095k"

30. Equivalent joint loads = 












1095
30
2
1
E
E
W
W
Calculating Structure Stiffness Matrices of Elements
Following table shows the properties of the elements required to form
structure stiffness matrices of elements
M
L I J K L
1 120 1 2 4 5
2 180 2 3 5 6

31. From structural elements
Member-1
1 2 4 5
 
5
4
2
1
00000694
.
0
00000694
.
0
00041667
.
0
00041667
.
0
00000694
.
0
00000694
.
0
00041667
.
0
00041667
.
0
00041667
.
0
00041667
.
0
033333
.
0
016667
.
0
00041667
.
0
00041667
.
0
016667
.
0
0333
.
0
1




















 EI
K
Member-2
2 3 5 6
 
6
5
3
2
00000206
.
0
00000206
.
0
000185
.
0
000185
.
0
00000206
.
0
00000206
.
0
000185
.
0
000185
.
0
000185
.
0
000185
.
0
0222
.
0
0111
.
0
000185
.
0
000185
.
0
0111
.
0
0222
.
0
2
















 EI
K

32. 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
 
6
5
4
3
2
1
00000206
.
0
0000206
.
0
0
000185
.
0
00018519
.
0
0
00000206
.
0
000009
.
0
00000694
.
0
000185
.
0
00023148
.
0
00041667
.
0
0
000000694
.
0
00000694
.
0
0
00041667
.
0
00041667
.
0
000185
.
0
000185
.
0
0
0222
.
0
0111
.
0
0
00018519
.
0
00023148
.
0
00041667
.
0
0111
.
0
0555
.
0
016667
.
0
0
00041667
.
0
00041667
.
0
0
016667
.
0
0333
.
0
6
5
4
3
2
1





























 EI
K
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.





















k
u
u
k
K
K
K
K
W
W
22
21
12
11

33. 











































































6
5
4
3
2
1
6
5
4
3
2
1
00000206
.
0
0000206
.
0
0
000185
.
0
00018519
.
0
0
00000206
.
0
000009
.
0
00000694
.
0
000185
.
0
00023148
.
0
00041667
.
0
0
000000694
.
0
00000694
.
0
0
00041667
.
0
00041667
.
0
000185
.
0
000185
.
0
0
0222
.
0
0111
.
0
0
00018519
.
0
00023148
.
0
00041667
.
0
0111
.
0
0555
.
0
016667
.
0
0
00041667
.
0
00041667
.
0
0
016667
.
0
0333
.
0
6
5
4
3
2
1
EI
W
W
W
W
W
W
E
E
E
E
E
E
Using Equation [u] = [K11]-1 [Wk]





































5886
.
22870
2946
.
10535
1
1095
30
0555
.
0
016667
.
0
016667
.
0
0333
.
0
1
1
2
1
EI
EI
Finding Unknown Reactions
Unknown reactions can be calculated using the following equation
[W]u = [K]21 []u





























































235
.
4
91
.
0
14
.
5
86
.
253
1
5886
.
22870
2946
.
10535
00018519
.
0
0
00023148
.
0
00041667
.
0
00041667
.
0
00041667
.
0
0111
.
0
0
6
5
4
3
EI
EI
W
W
W
W
E
E
E
E

34. W = WE + WF





































































735
.
41
41
.
39
14
.
4
86
.
1378
5
.
37
5
.
38
1
1125
235
.
4
91
.
0
14
.
5
86
.
253
6
5
4
3
W
W
W
W
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:

























































1397
.
5
1397
.
5
7647
.
586
30
0
0
58863
.
22870
2946
.
10535
1
00000694
.
0
00000694
.
0
00041667
.
0
00041667
.
0
00000694
.
0
00000694
.
0
00041667
.
0
000416667
.
0
00041667
.
0
00041667
.
0
033333
.
0
016667
.
0
0001667
.
0
00041667
.
0
016667
.
0
0333
.
0
4
3
2
1
EI
EI
w
w
w
w
E
E
E
E
Superimposing the fixed end forces for member-1 on the above w’s we
get
w = wE + wF

35. 




















































1397
.
6
1397
.
4
764
.
616
0
1
1
30
30
1397
.
5
1397
.
5
7647
.
586
30
4
3
2
1
w
w
w
w




















































73529
.
41
2647
.
33
117
.
1379
764
.
616
5
.
37
5
.
37
1125
1125
23529
.
4
23529
.
4
1176
.
254
2353
.
508
8
7
6
5
w
w
w
w























































23529
.
4
23529
.
4
1176
.
254
2353
.
508
0
0
0
5886
.
22870
1
00000206
.
0
00000206
.
0
000185
.
0
000185
.
0
00000206
.
0
00000206
.
0
000185
.
0
000185
.
0
000185
.
0
000185
.
0
0222
.
0
0111
.
0
000185
.
0
000185
.
0
0111
.
0
0222
.
0
8
7
6
5
EI
EI
w
w
w
w
E
E
E
E
F
E w
w
w 

Member 2 :
Superimposing the fixed end forces

36. OR
 
EI
EI
w
w
w
w
w
w
w
w
w
E
E
E
E
E
E
E
E
E
1
0
0
0
0
5886
.
22870
2946
.
10535
00000206
.
0
00000206
.
0
0
000185
.
0
000185
.
0
0
00000206
.
0
00000206
.
0
0
000185
.
0
000185
.
0
0
000185
.
0
000185
.
0
0
0222
.
0
0111
.
0
0
000185
.
0
000185
.
0
0
0111
.
0
0222
.
0
0
0
000000694
.
0
00000694
.
0
0
00041667
.
0
00041667
.
0
0
00000694
.
0
00000694
.
0
0
00041667
.
0
00041667
.
0
0
00041667
.
0
00041667
.
0
0
0333
.
0
01667
.
0
0
00041667
.
0
00041667
.
0
0
01667
.
0
0333
.
0
8
7
6
5
4
3
2
1




















































































 
























































24
.
4
24
.
4
21
.
254
42
.
508
14
.
5
14
.
5
53
.
586
30
8
7
6
5
4
3
2
1
E
E
E
E
E
E
E
E
E
w
w
w
w
w
w
w
w
w
w = wE +wF

37. 




















































































5
.
37
5
.
37
1125
1125
1
1
30
30
24
.
4
24
.
4
21
.
254
42
.
508
14
.
5
14
.
5
53
.
586
30
8
7
6
5
4
3
2
1
w
w
w
w
w
w
w
w
w
























































74
.
41
26
.
33
2
.
1379
58
.
616
14
.
6
14
.
4
53
.
616
0
8
7
6
5
4
3
2
1
w
w
w
w
w
w
w
w
w

38. Shear Force Diagram Bending Moment Diagram
-4.139
-6.139
33.2647
-41.735
-248.34
-616.7647
711.432
-1379.117
1 2
-4.139
0 616.7647k" 1379.177kip"
-616.7647k"
-6.139 33.26 41.735
- - - -
+
+

39. 10'
5'
5'
3k/ft
4k
A
B
C
D
6'
EXAMPLE 1.4
Analyse the frame by STIFFNESS METHOD

40. W11,11
W10,10
W12,12
W3,3
W2,2 W5,5
W6,6
W1,1 W4,4
W7,7
W9,9
W8,8
StructureForces
and Deformation

41. w11,11
w9,9 w10,10
w8,8
w7,7
w12,12
w3,3
w1,1
w5,5
w6,6
w4,4
w2,2
w17,17
w15,15
w13,13
w14,14
w16,16
w18,18
Element Forces
and Deformations
1
3
2

42. 3k/ft
4k
2k
2k
60k"
60k"
300k" 300k"
15k 15k
1
2
3
 




















































































































0
0
0
0
0
0
0
0
15
15
300
300
2
2
0
0
60
60
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
w
s
FEM'
Fixed End Moments and reactions

43. 
















































































2
0
60
0
0
0
0
15
300
2
15
240
12
11
10
9
8
7
6
5
4
3
2
1
F
F
F
F
F
F
F
F
F
F
F
F
W
W
W
W
W
W
W
W
W
W
W
W
2k
240 K"
15k
0
15k
300 K"
Net fixed end moments
[W]F=

44. 2k
240 K"
15k
0
15k
300 K"
0
Equivalent joint loads
[W]E=
















































0
0
15
300
2
15
240
7
6
5
4
3
2
1
E
E
E
E
E
E
E
W
W
W
W
W
W
W

45. 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

46.  
3
12
2
11
1
10
13889
.
20
13889
.
20
0
0
333
.
1208
333
.
1208
13889
.
20
13889
.
20
0
0
333
.
1208
333
.
1208
0
0
333
.
1208
333
.
1208
0
0
0
0
333
.
1208
333
.
1208
0
0
333
.
1208
333
.
1208
0
0
667
.
96666
333
.
48333
333
.
1208
333
.
1208
0
0
333
.
48333
667
.
96666
1





























K
For member-1
we have
10 1 11 2 12 3

47. For member-2
 
6
3
5
2
4
1
333
.
1208
333
.
1208
0
0
0
0
333
.
1208
333
.
1208
0
0
0
0
0
0
13889
.
20
13889
.
20
333
.
1208
333
.
1208
0
0
13889
.
20
13889
.
20
333
.
1208
333
.
1208
0
0
333
.
1208
333
.
1208
667
.
96666
333
.
48333
0
0
333
.
1208
333
.
1208
333
.
48333
667
.
96666
2





























K
1 4 2 5 3 6

48. For member-3
4 7 5 8 6 9
 
9
6
8
5
7
4
2356
.
93
2356
.
93
0
0
481
.
3356
481
.
3356
2356
.
93
2356
.
93
0
0
481
.
3356
481
.
3356
0
0
889
.
2013
889
.
2013
0
0
0
0
889
.
2013
889
.
2013
0
0
481
.
3356
481
.
3356
0
0
111
.
161111
556
.
80555
481
.
3356
481
.
3356
0
0
556
.
80555
111
.
161111
3





























K
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)

49.  






































































139
.
20
3
.
1208
139
.
20
3
.
1208
3
.
1208
3
.
1208
67
.
96666
33
.
1208
3
.
48333
0
48
.
3356
2356
.
93
48
.
3356
01
89
.
2013
0
0
89
.
2013
0
48
.
3356
1
.
161111
48
.
3356
0
56
.
80555
2356
.
93
48
.
3356
568
.
1301
48
.
3356
3
.
1208
0
0
0
03
.
2034
3
.
1208
139
.
20
3
.
1208
48
.
3356
0
56
.
80555
48
.
3356
3
.
1208
78
.
257777
3
.
1208
3
.
48333
139
.
20
3
.
1208
3
.
1208
0
0
3
.
1208
3
.
1208
0
138
.
20
3
.
1208
47
.
1228
3
.
1208
3
.
1208
3
.
48333
3
.
1208
3
.
48333
3
.
1208
3
.
193333
12
11
10
9
8
7
6
5
4
3
2
1
12
11
10
9
8
7
6
5
4
3
2
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
3
.
1208
0
0
0
0
0
0
0
0
0
0
0
0
2356
.
93
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
89
.
2013
0
0
0
0
0
0
0
0
0
0
47
.
1228
0
0
0
0
0
0
0
0
0
0
0
33
.
1208
K

50. 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.
Unknown deformations can be calculated using the equation ;
[]u=[K11]-1[W]k
Solving the above equation we get ,





















k
u
u
k
22
21
12
11
K
K
K
K
W
W


























































































 
0
0
15
300
2
15
240
111
.
161111
481
.
3356
0
556
.
80555
0
0
0
481
.
3356
568
.
1301
0
4814
.
3356
333
.
1208
0
0
0
0
0278
.
2034
333
.
1208
0
13889
.
20
333
.
1208
556
.
80555
481
.
3356
333
.
1208
778
.
257777
0
333
.
1208
333
.
48333
0
333
.
1208
0
0
722
.
1228
0
333
.
1208
0
0
13889
.
20
333
.
1208
0
4718
.
1228
333
.
1208
0
0
333
.
1208
333
.
48333
333
.
1208
333
.
1208
333
.
193333
7
6
5
4
3
2
1
1

51. Finding Unknown reactions
Unknown reactions can be calculated using the following equation:























































001535
.
037023
.
007682
.
001527
.
039859
.
01202
.
00184
.
7
6
5
4
3
2
1
    



















































































42062
.
1
5242
.
14
7704
.
40
42501
.
3
4707
.
15
001535
.
0
037023
.
0
007682
.
0
001527
.
0
039859
.
0
01202
.
0
00184
.
0
0
0
0
0
13889
.
20
0
333
.
1208
0
0
0
0
0
333
.
1208
0
0
0
0
0
333
.
1208
0
333
.
48333
481
.
3356
2356
.
93
0
481
.
3356
0
0
0
0
0
889
.
2013
0
0
0
0
12
11
10
9
8
21
E
E
E
E
E
u
u
W
W
W
W
W
K
W

52. Now we will superimpose the fixed end reactions on the above calculated
structure forces.
[W]=[W]E+[W]F









































































58
.
0
52
.
14
23
.
19
42
.
3
47
.
15
2
0
60
0
0
42
.
1
52
.
14
77
.
40
42
.
3
47
.
15
12
11
10
9
8
W
W
W
W
W
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






































































039859
.
0
0
01202
.
0
0
00184
.
0
0
0
0
333
.
1208
333
.
1208
0
0
0
0
333
.
1208
3330
.
1208
0
0
13889
.
20
13889
.
20
0
0
333
.
1208
333
.
1208
13889
.
20
13889
.
20
0
0
333
.
1208
333
.
1208
333
.
1208
333
.
1208
0
0
667
.
96666
333
.
48333
333
.
1208
333
.
1208
0
0
333
.
48333
667
.
96666
4
3
6
5
2
1
w
w
w
w
w
w

53. 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

54. For member-2








































































0370233
.
0
0398595
.
0
0076822
.
0
012024
.
0
001528
.
0
001845
.
0
333
.
1208
333
.
1208
0
0
0
0
333
.
1208
333
.
1208
0
0
0
0
0
0
13889
.
20
13889
.
20
333
.
1208
333
.
1208
0
0
13889
.
20
13889
.
20
333
.
1208
333
.
1208
0
0
333
.
1208
333
.
1208
667
.
96666
333
.
48333
0
0
333
.
1208
333
.
1208
333
.
48333
667
.
96666
12
11
10
9
8
7
w
w
w
w
w
w
Above equation gives
w7 = 109.782 kips-inch c.w.
w8 = -53.25353 kips-inch c.c.w.
w9 = - 0.47048 kips Downward
w10 = 0.47048 kips Upward
w11 = 3.427 kips Rightward
w12 = -3.427 kips Leftward

55. 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
0370233
.
0
0
00768219
.
0
00153523
.
0
0015278
.
0
0
0
889
.
2013
889
.
2013
0
0
0
0
889
.
2013
889
.
2013
0
0
2356
.
93
2356
.
93
0
0
481
.
3356
481
.
3356
2356
.
93
2356
.
93
0
0
481
.
3356
481
.
3356
481
.
3356
481
.
3356
0
0
111
.
161111
556
.
80555
481
.
3356
481
.
3356
0
0
556
.
80555
111
.
161111
18
17
16
15
14
13
w
w
w
w
w
w

56. Solving the above equation we get
w13 = -246.74227 kips-inch c.c.w.
w14 = 0 kips-inch
w15 = 3.427kips Upward
w16 = -3.427 kips Downward
w17 = 15.4711 kips Rightward
w18 = -15.4711kips Leftward
PLOTTING THE BENDING MOMENT AND SHEARING FORCE
DIAGRAM.
According to the forces calculated above bending moment and
shearing force diagrams are plotted below:

57. -3.427k
14.528k
-15.471k
3.427k
+
+
+ -
Shearing Force
Diagram
231.959k"
-246.747k"
-190.218k"
-18.973
15.407k"
Bending moment
diagram
3.1427k
+
- -
-

58. EXAMPLE 1.5 To analyse the frame shown in the figure
using direct stiffness method.
40'
10'
10'
3k/ft

59. W1,1
W3,3
W2,2
W5,5
W6,6
W8,8
W9,9
W7,7
W4,4
Structure Forces and Deformation
w1,1
w2,2
w5,5
w6,6
w3,3
w4,4
Element Forces and Deformation
w7,7
w11,11
w9,9
w8,
w12,12
w10,10

60. 













































































0
0
60
60
4800
4800
0
0
0
0
0
0
12
11
10
9
8
7
6
5
4
3
2
1
F
F
F
F
F
F
F
F
F
F
F
F
w
w
w
w
w
w
w
w
w
w
w
w
Fixed end moments

























































0
60
4800
0
0
0
0
60
4800
9
8
7
6
5
4
3
2
1
F
F
F
F
F
F
F
F
F
W
W
W
W
W
W
W
W
W
4800k"
0
60
4800k"
0
60
[W]F=Net
fixed end
forces=

61. Equivalent joint loads=






















0
60
4800
3
2
1
E
E
E
W
W
W
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
inch
0.707 0.707 4 1 5 2 6 3
2 480 inch 1 0 1 7 2 8 3 9

62. Using the structure stiffness matrix for frame element in general form we get struc
stiffness matrix for member-1.
 
3
6
2
5
1
4
0045
.
890
0045
.
890
804
.
818
804
.
818
064
.
4272
064
.
4272
0045
.
890
0045
.
890
804
.
818
804
.
818
064
.
4272
064
.
4272
804
.
818
804
.
818
045
.
890
0045
.
890
064
.
4272
064
.
4272
804
.
818
804
..
818
0045
.
890
0045
.
890
064
.
4272
064
.
4272
064
.
4272
064
.
4272
064
.
4272
064
.
4272
666
.
683536
33
.
341768
064
.
4272
064
.
4272
064
.
4272
064
.
4272
33
.
341768
666
.
683536
1





































K
Similarly for member-2 we get,
 
9
3
8
2
7
1
167
.
604
167
.
604
0
0
0
0
167
.
604
167
.
604
0
0
0
0
0
0
15
.
3
15
.
3
21
.
755
21
.
755
0
0
15
.
3
15
.
3
21
.
755
21
.
755
0
0
21
.
755
21
.
755
667
.
241666
33
.
120833
0
0
21
.
755
21
.
755
33
.
120833
667
.
241666
2





























K

63. 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 3 4 5 6 7 8 9
 



















































167
.
604
0
0
0
0
0
167
.
604
0
0
0
15
.
3
21
.
755
0
0
0
0
15
.
3
21
.
755
0
21
.
755
66
.
241666
0
0
0
0
21
.
755
33
.
120833
0
0
0
0045
.
890
804
.
818
064
.
4272
0045
.
890
804
.
818
064
.
4272
0
0
0
804
.
818
0045
.
890
064
.
4272
804
.
818
0045
.
890
064
.
4272
0
0
0
064
.
4272
064
.
4272
66
.
683536
064
.
4272
064
.
4272
33
.
341768
167
.
604
0
0
0045
.
890
804
.
818
064
.
4272
1715
.
1494
804
.
818
064
.
4272
0
15
.
3
21
.
755
804
.
818
0045
.
890
064
.
4272
804
.
818
195
.
893
854
.
3516
0
21
.
755
33
.
120833
064
.
4272
06
.
4272
33
.
341678
064
.
4272
854
.
3516
32
.
925203
9
8
7
6
5
4
3
2
1
K

64. 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.





















k
u
u
k
22
21
12
11
K
K
K
K
W
W
Unknown deformations can be calculated using the equation ;
[u] = [K11]-1 [Wk]






































0
60
4800
1715
.
1494
804
.
818
064
.
4272
804
.
818
195
.
893
854
.
3516
064
.
4272
854
.
3516
333
.
925203
1
3
2
1
Solving above equation we get
1 = 0.00669 c.w.
2 = - 0.223186 downward
3 = 0.141442 Rightward

65.     































































































4546
.
85
75539
.
5
927
.
976
4417
.
85
2432
.
54
716
.
728
141442
.
0
223186
.
0
00669
.
0
167
.
604
0
0
0
15
.
3
21
.
755
0
21
.
755
33
.
120833
0045
.
890
804
.
818
064
.
4272
804
.
818
0045
.
890
064
.
4272
064
.
4272
064
.
4272
33
.
341768
9
8
7
6
5
4
21
E
E
E
E
E
E
u
u
W
W
W
W
W
W
K
W

66. Now we will superimpose the fixed end forces on the
above calculated equivalent forces.
W=WE+WF





















































































4546
.
85
75539
.
65
927
.
5776
4417
.
85
2432
.
54
716
.
728
0
60
4800
0
0
0
4546
.
85
75539
.
5
927
.
976
4417
.
85
2432
.
54
716
.
728
9
8
7
6
5
4
W
W
W
W
W
W

67. 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








































































14144
.
0
0
22319
.
0
0
321
.
1208
321
.
1208
321
.
1208
321
.
1208
0
0
321
.
1208
321
.
1208
321
.
1208
321
.
1208
0
0
34676
.
50
34676
.
50
34676
.
50
34676
.
50
668
.
6041
668
.
6041
34676
.
50
34676
.
50
34676
.
50
34676
.
50
668
.
6041
668
.
6041
063
.
4272
063
.
4272
063
.
4272
063
.
4272
665
.
683536
332
.
341768
063
.
4272
063
.
4272
063
.
4272
063
.
4272
332
.
341768
665
.
683536
0.00669
0
6
5
4
3
2
1
E
E
E
E
E
E
w
w
w
w
w
w
Solving the above equation we get following values of
w’s for the equivalent loading condition provided by member-1
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

68. 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


































































0
141442
.
0
0
223186
.
0
0
00669
.
0
16667
.
604
16667
.
604
0
0
0
0
16667
.
604
16667
.
604
0
0
0
0
0
0
1467
.
3
1467
.
3
208
.
755
208
.
755
0
0
1467
.
3
1467
.
3
208
.
755
208
.
755
0
0
208
.
755
208
.
755
667
.
241666
333
.
120833
0
0
208
.
755
208
.
755
333
.
120833
667
.
241666
12
11
10
9
8
7
E
E
E
E
E
E
w
w
w
w
w
w
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

69. Superimposing the fixed end forces in their actual direction we get the
element forces for the actual loading condition.
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.

70. -3014.698k" 5776.927k"
54.244k 65.754k
-85.454k
85.454k

71. Shearing force Diagram Bending Moment Diagram
-22.059kip
54.244 kip
-65.7541 kips
263.016 " 729.0578
kip-inch
-3015.829
kip-inch
2870.376 kip-inch
-5797.047 kip-inch
- - -
-
+
+
+

72. a
L
L/2
L
L
A
A
b
M
B
L/2
w
B
A B
L
L
L
A
A
A
L/2
w
w B
B
w B
a
L/2
L
MA
A
b
L/2
P
P
B
MB
MA
MB
MA
MB
MA
MB
MA
MB
MA
MB
MA
MB
2
2
B
2
2
A
L
b
Pa
M
L
Pab
M 


2
2
B
2
2
A
L
b
Pa
M
L
Pab
M 


12
wL
-
M
2
A 
12
wL
M
2
B 
2
A wL
192
5
-
M  2
B wL
192
11
M 
30
wL
-
M
2
A 
20
wL
M
2
B 
96
5wL
-
M
2
A 
96
5wL
M
2
B 






 1
-
L
3a
L
Mb
-
MA






 1
-
L
3b
L
Ma
MB

73. The
End