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Chapter 16
Frame Analysis
Using The Stiffness Method
Preliminary Definitions & Concepts
• Global & member coordinates. Similar to truss global system will be identified using x, y axes and local
or member coordinate will be identified using x’,y’ axes.
If we consider the axial forces and the effect of both bending & shear,
then each node on a frame can have three degree of freedom, namely, a
horizontal displacement, a vertical displacement & a rotation.
Lowest code numbers will be used to identify the unknown
displacements.
Stiffness Matrix
• Positive Sign Convention
Stiffness Matrix
• Member stiffness matrix (Shear & Bending)
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
– Resultant forces caused by both displacements are
'N N
AEq d
L 'F N
AEq d
L
''N F
AEq d
L ''F F
AEq d
L
N N F
AE AEq d d
L L
F N F
AE AEq d d
L L
Stiffness Matrix
• Member stiffness matrix
' ' ' ' ' '
3 2 3 2
2 2
3 2 3 2
2 2
0 0 0 0
12 6 12 60 0
6 4 6 20 0
'
0 0 0 0
12 6 12 60 0
6 2 6 40 0
x y z x y zN N N F F F
EA EA
L L
EI EI EI EI
L L L L
EI EI EI EI
L L L Lk
EA EA
L L
EI EI EI EI
L L L L
EI EI EI EI
L L L L
q k d
Load-displacement Relationship
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 Nx y
N Ny x
N N
F x y F
y xF F
F F
d D
d D
d D
d D
d D
d D
d TD
'
'
'
'
'
'
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 Nx y
N Ny x
N N
F x y F
y xF F
F F
Q q
Q q
Q q
Q q
Q q
Q q
TQ T q
• Force Transformation
Transformation Matrices
Member Global Stiffness Matrix
• Stiffness matrix
– We will determine the stiffness matrix for a member which relates
the member’s global force components Q to its global
displacements D.
T T
q k d and d TD q k TD
Q T q Q T k TD
'Tk T k T
Q k D
Member Global Stiffness Matrix
'Tk 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
Application for Beam Analysis
The beam can be related to the displacements using the structure stiffness
equation.
k 11 12 u
u 21 22 k
=Q K K D
Q K K D
Q KD
11 12k u kQ K D K D
21 22u u kQ K D K D
• Member Forces
Where: q0 is the fixed end reactions if the beam subjected to intermediate loading.
0
0
'
'
q k d q
k TD q
Application for Beam Analysis
The beam can be related to the displacements using the structure stiffness
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
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 60 0
6 4 6 20 0
0 0 0 0
12 6 12 60 0
6 2 6 40 0
x
y
z
x
y
z
Nx
N
N
F
F
F
EA EA
L L
EI EI EI EIq
L L L Lq
EI EI EI EIq L L L L
q EA EA
L LqEI EI EI EI
qL 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
zz
xx
y
y
z
z
N
Ny
NN
y x
NN
x y F F
y x F
F
F
F
qD
qD
qD
D q
Dq
Dq
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.
Example 1
Example 1Member stiffness matrix
Member 1
Example 1
Beam stiffness matrix Member 2
Example 1Q KD
k 11 12 u
u 21 22 k
=Q K K D
Q K K D
Example 1
Example 1
Example 1
Internal Loading: Member 1
Example 1
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.
Example 2Member stiffness matrix Member 1
Example 2Member stiffness matrix
Member 1
Example 2Member stiffness matrix
Member 2
Example 2Member stiffness matrix
Member 2
Example 2Q KD
k 11 12 u
u 21 22 k
=Q K K D
Q K K D
Example 2
Example 2
Internal Loading: Member 1
'
1 1 1q k T D
Example 2
