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Lecture 8: Flexibility Method - Frames
Washkewicz College of Engineering
Analysis of Plane Frames
Plane frames are two-dimensional structures
constructed with straight elements connected
together by rigid and/or hinged connections.
Frames are subjected to loads and reactions
that lie in the plane of the structure.
Under the action of external loads, the
elements of a plane frame are subjected to
axial forces similar to truss members as well as
bending moments and shears one would see in
a beam. Hence the analysis of plane frame
members can be conveniently conducted by
treating the frame as a composite structure
with beam elements that can be subject to axial
loads. These elements are usually rigidly
connected or semi-rigidly connected
depending on the amount of rotational restraint
designed into the connection.
Moment
connections
throughout
3D Frame
Lecture 8: Flexibility Method - Frames
Washkewicz College of Engineering
Equivalent Joint Loads
The calculations of displacements in larger more extensive structures by the means of the
matrix methods derived later requires that the structure be subject to loads applied only at the
joints. Thus in general, loads are categorized into those applied at joints, and those that are
not. Loads that are not applied to joints must be replaced with statically equivalent loads.
Consider the statically indeterminate frame with a distributed load between joints B and C:
2
Lecture 8: Flexibility Method - Frames
Washkewicz College of Engineering
The frame on the previous page is statically equivalent to the following two structure:
The frame to the left is statically
equivalent to the original frame in
the sense that the reactions would
be the same. However the internal
bending moment in the cross beam
would not be the same. The frame
would not be kinematically
consistent with the original frame.
However, the reactions at the
foundations and the joint actions
would be equivalent. Keep in mind
we are trying to find these unknown
actions (forces and moments).
Lecture 8: Flexibility Method - Frames
Washkewicz College of Engineering
Example 8.1
Redundants
are forces
and moments
Zero and nonzero
displacements in
the released
structure due to
external loads are
shown
The plane frame shown at the left has fixed
supports at A and C. The frame is acted upon
by the vertical load P as shown. In the analysis
account for both flexural and axial
deformations. The flexural rigidity EI is
constant. The axial rigidity EA is also constant.
Joint B is a rigid connection and we will
endeavor to preserve equilibrium at Joint B and
throughout the structure.
The structure is statically indeterminate to the
third degree. A released structure is obtained by
cutting the frame at joint B and the released
actions Q1, Q2 and Q3 are redundants. Find the
magnitude and direction of these redundants.
Lecture 8: Flexibility Method - Frames
Washkewicz College of Engineering
The displacements in the released structure caused by P and corresponding to Q1, Q2
and Q3 are depicted in the previous figures. These displacements in the released
structure caused by the external load are designated DQL1, DQL2 and DQL3 respectively.
The displacement DQL1 consists of the sum of two translations which are found by
analyzing the released structure as a set of two cantilever beams AB and CB.
First analyze the cantilever beam AB. The load P will cause a downward translation at
B and a clockwise rotation at B. There is no axial displacement thus
D 
QL1 AB

0
The displacements DQL2 and DQL3 in the released structure AB consist of a vertical
displacement and a rotations, i.e.,
D 
QL 2 AB
5PL3
 
48EI
D 
QL 3 AB
PL2
 
8EI
But these displacements in the released structure AB have corresponding displacements
in the released structure CB.
Lecture 8: Flexibility Method - Frames
Washkewicz College of Engineering
However, since there is no load on member CB in the released structure, there will be no
displacement at end B and
D 
QL1 CB

D 
QL 2 CB

D 
QL 3 CB
 0
Even though the displacements at B in CB are zero, the total displacement of joint B
would be the summation of the two DQL components from AB and CB. Thus the DQL
matrix is
DQL
 0  0 




 5PL3

 
 0 
 48EI

2
  PL  0 
 8EI



2
PL
48EI
 0 



5
L


 6 


We need to assemble the flexibility matrix F. Consider the released structure with
Q1
 1
Q2
 Q3
 0
Lecture 8: Flexibility Method - Frames
Washkewicz College of Engineering
The displacements corresponding to unit values of Q1, Q2
and Q3 are shown as flexibility coefficients F11, F21 and F31.
If both axial and flexural deformations are considered, the
displacements at end B of member AB are
F11 AB

L
AE
F21 AB
 0
F31 AB
The displacements at end B of member BC are
F11 CB

H3
3EI
F21 CB
 0
F31 CB
H2
 
2 EI
The flexibility coefficients take on the following values
F11 
L
H3

AE 3EI
F21  0
F31
H2
 
2 EI
 0
Lecture 8: Flexibility Method - Frames
Washkewicz College of Engineering
The same analysis must be made with
Q2
 1
Q1
 Q3
 0
For frame sections AB and CB
AB :
F12 AB
 0
F22 AB

L3
3EI
CB :
F12 CB
 0
F22 CB

H
AE
F32 AB

F32 CB
 0
Which leads to flexibility coefficients
F21  0
F22

L3
H

3EI AE
L2
2 EI
F32

L2
2 EI
Lecture 8: Flexibility Method - Frames
Washkewicz College of Engineering
Once again from frame sections AB and CB (Q3 = 1,
Q1 = Q2 = 0)
AB :
CB :
L2
2 EI
F13 AB
 0
F23 AB

F13 CB
H2
 
2 EI
F23 CB
 0
F33 AB

L
EI
F33 CB

H
EI
leading to flexibility coefficients
F13
H2
 
2 EI
F23

L2
2 EI
F33

L
H

EI EI
Lecture 8: Flexibility Method - Frames
Washkewicz College of Engineering
Assembly of the flexibility matrix leads to
F
 L
H3


 EA 3EI
 
0


H2
 
2 EI

0
L3
H

3EI EA
L2
2 EI
H2 


2 EI 
L2 
2 EI 
L H
 
EI EI 
Now let
P  10 K
L  H  12 ft
 144 inches
E
 30,000 ksi
I
 200 in 4
A  10 in 2
Lecture 8: Flexibility Method - Frames
Washkewicz College of Engineering
When these numerical values
DQL
and
L
EA
H3
3EI


H
EA

0




   0.5184 
 0.00432


144
 0.00048 in / kip
30,00010
1443
330,000200
 0.165888 in / kip
The axial compliance (flexibility) of each component is quite small relative to the
flexural compliance (flexibility). We will ignore the axial compliance of the beam
and the column when assembling the flexibility matrix. The inverse of compliance is
stiffness. This is equivalent to stating that the axial stiffness of the beam and column
is so large relative to flexural stiffness of the beam and column that axial
displacements are negligible.
Lecture 8: Flexibility Method - Frames
Washkewicz College of Engineering
Omitting axial deformations leads to the following flexibility matrix
F 
 0.1659
 
0.0
 0.001728
0.0
0.1659
0.001728
 0.001728 
0.001728 
0.000048 
When the flexibility matrix is inverted we obtain
F 1
 0.0015  0.0009 0.0868 
 110 4  0.0009 0.0015  0.0868
 0.0868  0.0868 8.3315 
Lecture 8: Flexibility Method - Frames
Washkewicz College of Engineering
With F-1 and DQL we can compute the unknown redundants utilizing the matrix
equation
Q

F 1  DQ 

D  
QL
i.e.,
Q
 0.0015  0.0009 0.0868 
 110 4  0.0009 0.0015  0.0868
 0.0868  0.0868 8.3315 
0.0
 0.0 
 


0
.
0

0
.
5189
 


0.0
0.00432





 0.9414 kips




 
4.0692 kips

 90.3818 kip  inches 


Note that the displacements associated with the redundants in the original structural,
represented by the matrix {DQ} are zero because joint B is a rigid connection. One can
rationalize the rotation DQ3 is zero from this assumption.
Lecture 8: Flexibility Method - Frames
Washkewicz College of Engineering
Now examine the structure by omitting axial deformations. The flexibility matrix
F 
 0.1664
 
0.0
 0.001728
0.0
0.1664
0.001728
0.001728 
0.001728 
0.000048 
Finding the inverse of this matrix (homework assignment) and substitution into
Q

F 1  DQ 

D  
QL
leads to
Q
 0.916272 kips 


 
4.03466 kips

88.2336 kip  inches 


which is less then 3% different from the computations where axial deformations are
included. This frequently happens in the analysis of typical frames and bolsters the
assumption that DQ1 and DQ2 (axial deformations associated with axial force redundants)
are zero. This allows the seasoned engineer to make judgments about considering only
bending in frame analyses.
Lecture 8: Flexibility Method - Frames
Washkewicz College of Engineering
Example 8.2
Lecture 8: Flexibility Method - Frames
Washkewicz College of Engineering
Example 8.3