Chapter 14 Truss Analysis Using
the Stiffness Method
Structural Mechanics 2
Dept of Arch Eng, Ajou Univ
Outline
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Fundamentals of the stiffness method
Member stiffness matrix
Displacement and force transformation matrix
Member global stiffness matrix
Truss stiffness matrix
Application of the stiffness method for truss analysis
Nodal coordinates
Trusses having thermal changes & fabrication errors
Space-truss analysis
14-1 Fundamentals of the
Stiffness Method
• Subdivides the structure into a series of
discrete finite elements represented by
members and joints
• The force-displacement relations
–
–
–
–
–
Equilibrium condition at each node
Stiffness matrix
Unknown displacement vector
Given loading
Unknown external and internal forces
14-2 Member Stiffness Matrix
+
=
14-2 Member Stiffness Matrix
• For +ve displacement, dN in local x’ and
y’ coordinates, the forces developed at
the ends of the members are
AE
q'N 
dN
L
AE
q'F  
dN
L
14-2 Member Stiffness Matrix
• For +ve displacement, dF at the far end,
keeping the near end pinned, results in
member forces
AE
dF
q' ' N  
L
AE
dF
q' 'F 
L
14-2 Member Stiffness Matrix
• By superposition, the resultant forces
caused by both displacements are
AE
AE
qN 
dN 
dF
L
L
AE
AE
qF 
dF 
dN
L
L
14-2 Member Stiffness Matrix
• These load-displacement equations may
be written in matrix form as
q N  AE  1  1 d N 
 
 


qF  L  1 1 d F 
q  k 'd 
AE  1  1
k '  
: stiffness matrix

L  1 1
14-3 Displacement and Force
Transformation Matrices
• Transformation of member forces q and
displacement d defined in local
coordinates to global coordinates
• Global coordinates convention: +ve x to
the right and +ve y upward
14-3 Displacement & Force
Transformation Matrices
• direction cosines
xF  x N
x  cos  x 
L

xF  x N
( xF  x N ) 2  ( y F  y N ) 2
yF  yN
 y  cos  y 
L

yF  yN
( xF  x N ) 2  ( y F  y N ) 2
14-3 Displacement & Force
Transformation Matrices
• Displacement Transformation Matrix
d N  DN x cos  x  DN y cos  y
14-3 Displacement & Force
Transformation Matrices
• Displacement Transformation Matrix
d F  DFx cos  x  DFy cos  y
14-3 Displacement & Force
Transformation Matrices
• Displacement Transformation Matrix
Let x  cos  x ;  y  cos  y
d N  DN x x  DN y  y ; d F  DFx x  DFy  y
In matrix form,
 DN x 
D 


0
0


d
 N
x
y
 Ny 

 D 
  
 d F  0 0 x  y   Fx 
 DFy 


d   T D
※ [T] transforms the 4 global displacement into 2 local x’ displacement
14-3 Displacement & Force
Transformation Matrices
• Force Transformation Matrix
QN x  q N cos  x  q N x
QFx  qF cos  x  q F x
QN y  q N cos  y  q N  y
QFy  qF cos  y  q F  y
14-3 Displacement & Force
Transformation Matrices
• Force Transformation Matrix
– In matrix form
QN x  x
Q  
 Ny   y
Q   
 Fx  0
 QFy  0

 
Q  T  q
0 

0  q N 


x  qF 

 y 
T
※ [T]T transforms the 2 local forces q into 4 global force components Q
14-4 Member Global Stiffness Matrix
Since q  k d  and d   T D,
q  k T D
Also since Q  T  q,
T
Force - Displacement is obtained as
Q  T  k T D  k D
T
where k   T  k T 
T
14-4 Member Global Stiffness Matrix
• Performing the matrix operation yields
14-5 Truss stiffness matrix
• Stiffness matrix [K] for entire truss can be
obtained by assembling all member
stiffness matrices [k] in global coordinates
• The 4 code numbers to identify the 2
global degrees of freedom at each end of
a member
• Appropriate for analysis by computer
programming
Example 14.1 Determine the structure
stiffness matrix. Use constant AE.
(F)
(Constrained)
Two unknown
displacement
(N)
(Constrained)
Stiffness matrix:
Member 1
Direction cosines:
x 
30
 1;
3
y 
00
0
3
(F)
Example 14.1 Determine the structure
stiffness matrix. Use constant AE.
(F)
(Constrained)
Two unknown
displacement
(N)
(Constrained)
Stiffness matrix:
Member 2
Direction cosines:
x 
30
 0.6;
5
y 
40
 0.8
5
(F)
Example 14.1 Determine the structure
stiffness matrix. Use constant AE.
• Algebraically added to form structure
stiffness matrix
14-6 Application of the Stiffness
Method for Truss Analysis
• The relationship between global force
components Q acting on a truss and its
global displacements D
Q  K D
• The structure stiffness equation
14-6 Application of the Stiffness
Method for Truss Analysis
• Expanding the structure stiffness
equation
Qk  K11 Du  K12 Dk
Qu  K 21 Du  K 22 Dk
• Often Dk = 0 since the supports are not
displaced, then
Qk  K11 Du
• Solving unknown displacement
Du   K11  Qk 
1
• Then unknown reactions at supports
Qu   K 21 Du 
14-6 Application of the Stiffness
Method for Truss Analysis
• The member forces can be determined
using q  k T D
• Expanding this expression
• Since with qN = -qF for equilibrium
Example 14.3 Determine the force in
each member. AE is constant.
6
y
①
[2]
2
②
1
5
Since D3=D4=D5=D6=0 and Q1=0,
Q2=-2kN,
0 3
0
 4
4
Dk    
[1] ③
0 5
x
3
0 6
Structure stiffness equation:
0  1
Qk    
 2  2
Example 14.3 - solution
• Separating and solving unknown
displacements
0 
0.405 0.096   D1  0
  
   AE 

 2 
0.096 0.128   D2  0
4.505
19.003
D1 
; D2 
AE
AE
• Substituting for unknown reactions
Example 14.3 - solution
• Then reactions at supports are
Q3  1.5kN ; Q4  0kN ; Q5  1.5kN ; Q6  2.0kN
• The force in each member
For member 1,
x  1,  y  0, L  3m  q1  1.5kN
For member 2,
x  0.6,  y  0.8, L  5m  q2  2.5kN
14-7 Nodal Coordinates
• A truss supported by a roller placed on an
incline
• The condition of zero displacement at node 1 is
defined only along the y” axis, while the
displacement along the x” axis will have
displacement components along both global
coordinates axes x and y
14-7 Nodal Coordinates
• Consider truss member 1 having a global
coordinate system x, y at the near node
and a nodal coordinate system x”, y” at
the far node
14-7 Nodal Coordinates
d N  D N x cos  x  D N y cos  y
d F  D F x " cos  x "  D F y " cos  y "
 DN x 


d N   x  y 0 0   DN y 

 

 d F  0 0 x  y   DFx " 
D 
 Fy " 
14-7 Nodal Coordinates
QN x  q N cos  x
QN y  q N cos  y
QFx "  q F cos  x"
QFy "  qF cos  y"
QN x  x

 
QN y   y

 
QFx "  0
Q  0
 Fy "  
0 

0  q N 
 

 x " q F 

 y" 
14-7 Nodal Coordinates
• Member stiffness matrix
T
k  T k 'T
Example 14.6 Determine the
support reactions
Example 14.6 Determine the
support reactions
Member 1
 x  1,  y  0, x"  0.707,  y "  0.707
Example 14.6 Determine the
support reactions
Member 2
  x  0,  y  1, x"  0.707,  y "  0.707
Example 14.6 - solution
• Member 3
 x  0.8,  y  0.6
Example 14.6 - solution
• Assembling the matrices to determine the
structure stiffness matrix
 157.5
 127.3
352.5
D1 
; D2 
; D3 
AE
AE
AE
Q4  31.8kN ; Q5  7.5kN ; Q6  22.5kN
14-8 Trusses having thermal
changes and fabrication errors
• Superposition method
– Calculate fixed end forces to prevent
movement of the nodes by temperature or
fabrication
– Calculate the displacement due to equal but
opposite forces placed on the truss at the
nodes
– Determine the actual forces in the members
and the reactions by superposing the results
14-8 Trusses having thermal
changes and fabrication errors
• For statically indeterminate truss,
increment in length of a member due to
T is L =  TL
• A decrease in length due to a
compressive force q0 is L' = q0L/AE
• Equating the two gives q0 = AE T
14-8 Trusses having thermal
changes and fabrication errors
• This force will hold the nodes of the
member fixed
(q N ) 0  AET
(q F ) 0   AET
14-8 Trusses having thermal
changes and fabrication errors
• If a truss member is made too long by an
amount L before it is fitted into a truss,
the force q0 needed to keep the member
at its design length L is q0 = AEL /L
AEL
(q N ) 0 
L
AEL
(q F ) 0  
L
14-8 Trusses having thermal
changes and fabrication errors
• If the member is too short, then L
becomes negative and these forces will
reverse
• In global coordinates, these forces are
14-8 Trusses having thermal
changes and fabrication errors
• The initial force-displacement relationship
due to temperature changes and
fabrication errors
Q  K D Q0 
• Where Qo is the initial fixed-end forces
caused by temperature changes and
fabrication errors
14-8 Trusses having thermal
changes and fabrication errors
• Carrying out the multiplication on the right
hand side,
Qk   K11 Du  K12 Dk  Qk 0
Qu   K 21 Du  K 22 Dk  Qk 0
• Using the superposition procedure, the
unknown displacements are determined
from the first equation by subtracting
K12Dk and (Qk)0 from both sides and then
solving for Du
14-8 Trusses having thermal
changes and fabrication errors
• The member forces are determined by
superposition
q  k T D q0 
• The force at the far end of the member
Example 14.7 Determine the force in
member 1 and 2 if member 2 was
made 0.01 m too short before it was
fitted into place. Use AE = 8103kN.
Since L = -0.01m,
 x  0.8  y  0.6
3m
Forces due to shortage in
global coordinates
4m
Example 14.7 Determine the force in
member 1 and 2 if member 2 was
made 0.01 m too short before it was
fitted into place. Use AE = 8103kN.
Structure stiffness matrix:
Example 14.7 Determine the force in
member 1 and 2 if member 2 was
made 0.01 m too short before it was
fitted into place. Use AE = 8103kN.
Partitioning the matrices to obtain the equation for the
unknown displacement
D1  0.003704m
D2  0.002084m
Example 14.7 Determine the force in
member 1 and 2 if member 2 was
made 0.01 m too short before it was
fitted into place. Use AE = 8103kN.
• Member 1
x  0,  y  1, L  3m, AE  8 10 kN
3
q1  5.56kN
• Member 2
x  0.8,  y  0.6, L  5m, AE  8 10 kN
3
q2  9.26kN
14-9 Space-truss analysis
• To account for the 3-D aspects of the
problem, additional elements must be
included in the transformation matrix [T]
14-9 Space-truss analysis
• The direction cosines
x  cos  x 

xF  x N
L
xF  x N
( xF  x N ) 2  ( y F  y N ) 2  ( z F  z N ) 2
yF  yN
 y  cos  y 
L
yF  yN

( xF  x N ) 2  ( y F  y N ) 2  ( z F  z N ) 2
z  cos  z 

zF  zN
L
zF  zN
( xF  x N ) 2  ( y F  y N ) 2  ( z F  z N ) 2
14-9 Space-truss analysis
• The transformation matrix in 3D
• Member stiffness matrix
14-9 Space-truss analysis
• Member stiffness matrix in global
coordinates