Conclusion, Stiffness Method, Trusses
For the truss we are analyzing, there are 8 displacements, in global coordinates, of the
pins connecting the various members. Those 8 displacements were numbered so that
displacements 1 through 5 were unknown, and 6 through 8 were known.
There are also 8 forces, in global coordinates, acting at the pins. The external loads,
forces 1 through 5, four of which are zero, are known. The reaction loads, forces 6 through 8,
are unknown. We could, of course, determine the reaction forces from statics. However, the
whole purpose of this method is to analyze statically indeterminate trusses, so we will pretend
this option is not available to us.
MAE3501 - 22 - 1
The matrix equation relating the global forces to the global displacements.
Q  KD
(22.1)
If all the displacements were known, all the forces could be determined from equation
(22.1). But some of the displacements are not known.
Pre-multiply equation (22.1) by K  1 .
K 1 Q  K 1 K D  D
(22.2)
If all the forces were known, all the displacements could be determined from equation
(22.2). But some of the forces are not known.
The solution is to partition the matrices in the following manner. The subscripts U and K
denote unknown and known.
DU 
D    
 D K 
(22.3)
DU
 D1 
   
 D 5 
(22.4)
DK
D 6 
   
 D 8 
(22.5)
Q K 
Q    
 Q U 
QK
(22.6)
Q1 
   
Q 5 
(22.7)
MAE3501 - 22 - 2
QU
Q 6 
   
Q 8 
K 11
K 
K 21


(22.8)
K 12

K 22
(22.9)
K 11 is a 5 by 5 matrix.
K 12 is a 5 by 3 matrix.
K 21 is a 3 by 5 matrix.
K 22 is a 3 by 3 matrix.
Substitute the partitioned matrices into equation (22.1).
Q K 



 Q U 
 K 11


K 21


K 12 
 
K 22 
DU 



 D K 
(22.10)
Expand equation (22.10), yielding the following two matrix equations.
Q K  K 11 D U  K 12 DK
(22.11)
Q U  K 21 D U  K 22 DK
(22.12)
Rearrange equation (22.11).
K 11 D U  Q K  K 12 DK
(22.13)
Pre-multiply equation (22.13) by K 11 1 .
K 11 1 K 11 D U  D U  K 11 1  Q K  K 12 DK 
(22.14)
All the quantities on the right side of equation (22.14) are known or can be determined.
It is now possible to determine D U , the unknown-displacement matrix.
MAE3501 - 22 - 3
After the value of the unknown-displacement matrix has been determined, all the
quantities on the right side of equation (22.12) are known. It is now possible to determine Q U ,
the unknown-force matrix.
Note that the partitioning will always work, because the known forces will always occur
at the unknown displacements, and vice versa.
All that is left is to use the results to determine the forces in the members of the truss.
The mathematics necessary for this calculation has been performed, but was buried in the
derivations.
Consider the displacements, in both local and global coordinates, of a member of a truss.
Note that an additional angle has been determined.
D NY
Y
dN
X
D NX
N
The displacement of the near end, in the local coordinate system of the member, is
denoted by d N .
d N   X D NX   Y D NY
(22.15)
The displacement of the far end, in the local coordinate system of the member, is denoted
by d F .
d F   X DFX   Y DFY
(22.16)
The change in length of the member, positive for elongation, is determined by subtracting
equation (22.15) from equation (22.16).
L  dF  d N
 L    X DNX   Y DNY   X DFX   Y DFY
(22.17)
Define a row matrix, denoted by  .
    X ,   Y ,  X ,  Y 
MAE3501 - 22 - 4
(22.18)
Recall the definition of D for a single member.
 D NX 
D 
D   NY 
 D FX 


 D FY 
(22.19)
For example, for member 1,
 D1 
D 
D   2
D 3 
 
D4 
(22.20)
The subscripts in equation (22.20) refer to the entries in the global-displacement D
column matrix.
Express equation (22.17) in matrix form.
L   D
(22.21)
The member force is denoted by q .
q
AE L
L
(22.22)
MAE3501 - 22 - 5
Homework
Complete the analysis of the truss. Determine all the unknown displacements, the
“unknown” reaction forces, and the “unknown” forces in all the members. You may check your
work using the method of joints.
The correct K, the global stiffness matrix.
 0.354  0.354
  0.354 0.854

  0.354 0.354

 0.354
6  0.354
K  0.125 (10)
 0
0

0
 0
 0
0

 0.5
 0
 0.354
0.354
1.061
 0.354
 0.354
0.354
 0.354
 0.354
0.354
0
0
0
0 
 0.354
0
0
0
 0.5 
 0.354  0.354 0.354  0.354  0.354

1.061
0.354  0.354  0.354  0.354
0.354
0.854  0.354  0.5
0 

 0.354  0.354 0.354
0
0 
 0.354  0.5
0
0.854
0.354 

 0.354
0
0
0.354
0.854 
MAE3501 - 22 - 6