FA

1. Chapter 16
Frame Analysis
Using The Stiffness Method

2. Preliminary Definitions & Concepts
• Global & member coordinates.
 Similar to truss global system will be identified using x, y axes and local
or member coordinatewill be identified using x’,y’ axes.
 If we consider the axial forces and the effect of both bending & shear,
then each nodeon a frame can have threedegree of freedom, namely, a
horizontal displacement,a vertical displacement & a rotation.
 Lowest code numbers will be used to identify the unknown
displacements.

3. Stiffness Matrix
• Positive Sign Convention

4. Stiffness Matrix
• Member stiffness matrix (Shear & Bending)

5. Member Stiffness Matrix
• Case I
– Positive displacement dN on the near end
• Case II
– Positive displacement dF on the far end
• Case I + Case II
– Resultantforces caused by both displacements are
'N N
AE
q d
L
 'F N
AE
q d
L
 
''N F
AE
q d
L
  ''F F
AE
q d
L

N N F
AE AE
q d d
L L
 
F N F
AE AE
q d d
L L
  

6. Stiffness Matrix
• Member stiffness matrix
' ' ' ' ' '
3 2 3 2
2 2
3 2 3 2
2 2
0 0 0 0
12 6 12 6
0 0
6 4 6 2
0 0
'
0 0 0 0
12 6 12 6
0 0
6 2 6 4
0 0
x y z x y z
N N N F F F
EA EA
L L
EI EI EI EI
L L L L
EI EI EI EI
L L L L
k
EA EA
L L
EI EI EI EI
L L L L
EI EI EI EI
L L L L
 

 
 
 

 
 
 

 

 

 
 
 
  
 
 
 

 
 

7. q k d


Load-displacement Relationship

8. Transformation Matrices
• Displacement Transformation
'
'
'
'
'
'
0 0 0 0
0 0 0 0
0 0 1 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0 0 1
x x
y y
z z
x x
y y
z z
N N
x y
N N
y x
N N
F x y F
y x
F F
F F
d D
d D
d D
d D
d D
d D
 
 
 
 
   
 
   
 
   

 
   
 
   
  
   
 
   
 

   
 
   
 
   
   
d TD


9. '
'
'
'
'
'
0 0 0 0
0 0 0 0
0 0 1 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0 0 1
x x
y y
z z
x x
y y
z z
N N
x y
N N
y x
N N
F x y F
y x
F F
F F
Q q
Q q
Q q
Q q
Q q
Q q
 
 
 
 
   

 
   
 
   
 
   
 
   
  
   

 
   
 
   
 
   
 
   
   
T
Q T q

• Force Transformation
Transformation Matrices

10. Member Global Stiffness Matrix
• Stiffness matrix
– We will determine the stiffness matrix for a member which relates
the member’s global force componentsQ to its global
displacementsD.
T T
q k d and d TD q k TD
Q T q Q T k TD
 
   

  
'
T
k T k T

Q k D


11. Member Global Stiffness Matrix
'
T
k T k T

' ' ' ' ' '
2 2 2 2
3 3 2 3 3 2
2 2 2 2
3 3 2 3 3 2
12 12 6 12 12 6
12 12 6 12 12 6
x y z x y z
x y x y y x y x y y
x y y x x x y y x
N N N F F F
A I A I I A I A I I
L L L L L L L L L L
A I A I I A I A I I
L L L L L L L L L L
k E
         
         
       
      
       
       
       
    
       
       

2 2 2 2
2 2 2 2
3 3 2 3 3 2
2 2 2 2
3 3 2 3 3
6 6 4 6 6 2
12 12 6 12 12 6
12 12 6 12 12
x
y x y x
x y x y y x y x y y
x y y x x x y y x
I I I I I I
L L L L L L
A I A I I A I A I I
L L L L L L L L L L
A I A I I A I A I
L L L L L L L L L
   
         
        
 
       
    
       
       
     
     
     
     
2
2 2 2 2
6
6 6 2 6 6 4
x
y x y x
I
L
I I I I I I
L L L L L L

   
 
 
 
 
 
 
 
 
 
 
 
 
 
 

 
 
 
 
 
 
 
 

12. Application for Beam Analysis
The beam can be related to the displacements using the structurestiffness
equation.
k 11 12 u
u 21 22 k
=
Q K K D
Q K K D
     
     
     
Q KD

11 12
k u k
Q K D K D
 
21 22
u u k
Q K D K D
 
• Member Forces
Where: q0 is the fixed end reactions if the beam subjected to intermediateloading.
0
0
'
'
q k d q
k TD q
 
 

13. Application for Beam Analysis
The beam can be related to the displacements using the structurestiffness
equation.
k 11 12 u
u 21 22 k
11 12
21 22
=
k u k
u u k
Q K K D
Q K K D
Q K D K D
Q K D
Q
K D
KD
     
     
     
 
 


14. Application for Beam Analysis
• Member Force
'
0
'
0
q k d q
q k T D q
 
 
'
'
'
'
'
'
3 2 3 2
2 2
3 2 3 2
2 2
0 0 0 0
12 6 12 6
0 0
6 4 6 2
0 0
0 0 0 0
12 6 12 6
0 0
6 2 6 4
0 0
x
y
z
x
y
z
N
x
N
N
F
F
F
EA EA
L L
EI EI EI EI
q
L L L L
q
EI EI EI EI
q L L L L
q EA EA
L L
q
EI EI EI EI
q
L L L L
EI EI EI EI
L L L L
 
 

 
 
   

   
   
   

   

   
  
 
   
   
    
 
 
 
 

 
 
 
 
 
 
 
 
'
'
'
'
'
'
0
0
0
0
0
0
0 0 0 0
0 0 0 0
0 0 1 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0 0 1
x
x
y
y
z
z
x
x
y
y
z
z
N
N
y
N
N
y x
N
N
x y F F
y x F
F
F
F
q
D
q
D
q
D
D q
D
q
D
q
 
 
 
 
 
 
   
 
   
 

   
 
   
    
   
 
   
 
 
  
 
   
   
 
   
 

15. Example 1
• Determine the loading at the joints of the two-member
frame shown. Take I=500 in4, A=10 in2 and E=29(103) ksi
for both members.

16. Example 1

17. Example 1
Member stiffness matrix
Member 1

18. Example 1
Beam stiffness matrix
Member 2

19. Example 1
Q KD

k 11 12 u
u 21 22 k
=
Q K K D
Q K K D
     
     
     

20. Example 1

21. Example 1

22. Example 1
Internal Loading: Member 1

23. Example 1

24. Example 2
• Determine the loadings at the ends of each member of the
frame. Take I=600 in4, A=12 in2 and E=29(103) ksi for each
member.

25. Example 2
Member stiffness matrix
Member 1

26. Example 2
Member stiffness matrix
Member 1

27. Example 2
Member stiffness matrix
Member 2

28. Example 2
Member stiffness matrix
Member 2

29. Example 2
Q KD

k 11 12 u
u 21 22 k
=
Q K K D
Q K K D
     
     
     

30. Example 2

31. Example 2
Internal Loading: Member 1
'
1 1 1
q k T D


32. Example 2