Chapter 14
Truss Analysis
Using The Stiffness Method
Stiffness Method
• Fundamentals of the stiffness method
– There are essentially two ways in which structures can be analyzed using
matrix methods.
• Flexibility method
• Stiffness method
– The stiffness method can be used to analyzed both statically determinate
& indeterminate structures, whereas the flexibility method required a
different procedure for each of these cases.
– The stiffness method yields the displacements & forces directly, whereas
with the flexibility method the displacements are not obtained directly.
– Application of this method requires subdividing the structure into a series
of discrete finite elements & identifying their end points as nodes.
– The force-displacement properties of each element are determined & then
related to one another using the force equilibrium equations written at the
nodes
– These relationships, for the entire structure, are then grouped together into
what is called the structure stiffness matrix K.
– Once the K is established, the unknown displacements of the nodes can
then be determined for any given loading on the structure.
– When the displacements are known, the external & internal forces in the
structure can be calculated using the force-displacement relations for each
member.
Stiffness Method
Preliminary Definitions & Concepts
• Member & node identifications– We will specify each member by a number enclosed within a square,
& use a number enclosed within a circle to identify the nodes.
– Also the ‘near’ & ‘far’ ends of the member must be identified, this will
be done using an arrow written along the member with the head of the
arrow directed toward the far end.
• Global & member coordinates. – We will use two different type of coordinate systems, global or
structure coordinate system and local or member coordinate system.
– Global system x, y, used to specify the sense of each of the external
force & displacement components at the nodes.
– Local system x’,y’ used to specify the sense of direction of members
displacements & internal loadings.
Preliminary Definitions & Concepts
• Degrees of freedom – The unconstrained truss has two degree of freedom or two possible
displacements for each joint (node).
– Each degree of freedom will be specified on the truss using a code
number, shown at the joint or node, & referenced to its positive global
coordinate direction using an associated arrow.
• For example– The truss has eight degree of freedom or eight possible displacements.
– 1 through 5 represent unknown or
unconstrained degree of freedom.
– 6 through 8 represent constrained
degree of freedom
Lowest code numbers will always be used
to identify the unknown displacements.
Highest code numbers will be used to
identify the known displacements
Preliminary Definitions & Concepts
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
– These load-displacement equations written in matrix form
1 1
1 1
N N
F F
q dAE
q dL
'q k d
1 1'
1 1
AEk
L
or
where
The matrix, k` is called the member stiffness matrix.
It is of the same form for each member of the truss.
Member Stiffness Matrix
Transformation Matrices• Displacement & force transformation
– We will now develop a method for
transforming the member forces q and
displacements d defined in local coordinate
to global coordinates
cos F Nx x
x x
L
cos F Ny y
y y
L
2 2
cos F Nx x
F N F N
x x
x x y y
2 2
cos F Ny y
F N F N
y y
x x y y
Transformation Matrices• Displacement transformation matrix
– In global coordinate each end of the member can have two independent
displacements.
– Joint N has displacements DNx & DNy in global coordinate
cos cosN Nx x Ny yd D D
N Nx x Ny yd D D
or
– Also joint F has displacements DFx & DFy
– Displacements at N & F
cos cosF Fx x Fy yd D D
F Fx x Fy yd D D
or
N Nx x Ny yd D D
F Fx x Fy yd D D
0. 0.
0. 0.
Nx
x y NyN
x y FxF
Fy
D
Dd
Dd
D
d TDor
0. 0.
0. 0.
x y
x y
T
where
Transformation Matrices
• Force transformation matrix
– Force qN applied to the near end of the member
– Force qF applied to the far end of the member
– Rewrite in a matrix form:
cosNx N xQ q or
cosNy N yQ q
Nx N xQ q Ny N yQ q
cosFx F xQ q or
cosFy F yQ q
Fx F xQ q Fy F yQ q
0.
0.
0.
0.
Nx x
Ny y N
Fx x F
Fy y
Q
Q q
Q q
Q
TQ T qor where
0.
0.
0.
0.
x
yT
x
y
T
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. 'q k d d TDand 'q k T D
TQ T q 'TQ T k TD Q kD
'Tk T k T0.
0. 0. 0.1 1
0. 0. 0.1 1
0.
x
y x y
x x y
y
AEk
L
Member Global Stiffness Matrix
2 2
2 2
2 2
2 2
xx x y x x y
yy x y y x y
xx x y x x y
yy x y y x y
N
NAEk
FL
F
x y x yN N F F
'Tk T k T
Example 1 Determine the structure stiffness matrix for the two-member truss
shown. AE is constant
Solution:
Establish the x, y global system
Identify each joint & member numerically.
Member 1:
Determine x & y, where L = 3ft
3 01
3x
0 00
3y
2 2
2 2
2 2
2 2
xx x y x x y
yy x y y x y
xx x y x x y
yy x y y x y
N
NAEk
FL
F
x y x yN N F F
1
0.333 0. 0.333 0. 1
0. 0. 0. 0. 2
0.333 0. 0.333 0. 3
0. 0. 0. 0. 4
k AE
1 2 3 4
Dividing each element by L = 3ft
22 1
0.3333
x
L
Member 2:Determine x & y, where L = 5ft
3 00.6
5x
4 00.8
5y
2 2
2 2
2 2
2 2
xx x y x x y
yy x y y x y
xx x y x x y
yy x y y x y
N
NAEk
FL
F
x y x yN N F F
2
0.072 0.096 0.072 0.096 1
0.096 0.128 0.096 0.128 2
0.072 0.096 0.072 0.096 5
0.096 0.128 0.096 0.128 6
k AE
1 2 5 6
Dividing each element by L = 5ft 0.8 0.6
0.0965
Structure stiffness matrix
K = k1 + k2
0.333 0 0.333 0 0 0 1 0.072 0.096 0 0 0.072 0.096
0. 0 0. 0 0 0 2 0.096 0.128 0 0 0.096 0.128
0.333 0 0.333 0 0 0 3 0. 0. 0 0 0. 0.
0. 0 0. 0 0 0 4 0. 0. 0 0 0. 0.
0. 0 0. 0 0 0 5 0.072 0.096 0 0 0.072 0.096
0. 0 0. 0 0 0 6 0.096 0.128 0 0 0
K AE AE
1
2
3
4
5
.096 0.128 6
1 2 3 4 5 6 1 2 3 4 5 6
0.405 0.096 0.333 0. 0.072 0.096
0.096 0.128 0. 0. 0.096 0.128
0.333 0. 0.333 0. 0. 0.
0. 0. 0. 0. 0. 0.
0.072 0.096 0. 0. 0.072 0.096
0.096 0.128 0. 0. 0.096 0.128
K AE
Example 2
Determine the structure stiffness matrix for the truss shown. AE is constant
Member 1:
Determine x & y, where L = 10ft
10 01
10x
0 00
10y
2 2
2 2
2 2
2 2
xx x y x x y
yy x y y x y
xx x y x x y
yy x y y x y
N
NAEk
FL
F
x y x yN N F F
1
0.1 0 0.1 0 1
0 0 0 0 2
0.1 0 0.1 0 6
0 0 0 0 5
k AE
1 2 6 5
Dividing each element by L = 10ft
2
10.1
10
Member 2:
Determine x & y, where L = 14.14ft10 0
0.70714.14
x
10 0
0.70714.14
y
2 2
2 2
2 2
2 2
x x y x x y
y x y y x y
x x y x x y
y x y y x y
AEk
L
2
0.035 0.035 0.035 0.035 1
0.035 0.035 0.035 0.035 2
0.035 0.035 0.035 0.035 7
0.035 0.035 0.035 0.035 8
k AE
1 2 7 8
Member 3:
Determine x & y, where L = 10ft0 0
0.10
x
10 0
110
y
3
0 0 0 0 1
0 0.1 0 0.1 2
0 0 0 0 3
0 0.1 0 0.1 4
k AE
1 2 3 4
Member 4:Determine x & y, where L = 10ft
10 01
10x
10 100.
10y
2 2
2 2
2 2
2 2
x x y x x y
y x y y x y
x x y x x y
y x y y x y
AEk
L
4
0.1 0 0.1 0 3
0 0 0 0 4
0.1 0 0.1 0 7
0 0 0 0 8
k AE
3 4 7 8
Member 5:
Determine x & y, where L = 14.14ft10 0
0.70714.14
x
0 10
0.70714.14
y
5
0.035 0.035 0.035 0.035 3
0.035 0.035 0.035 0.035 4
0.035 0.035 0.035 0.035 6
0.035 0.035 0.035 0.035 5
k AE
3 4 6 5
Member 6:Determine x & y, where L = 10ft
10 100.
10x
10 01
10y
2 2
2 2
2 2
2 2
x x y x x y
y x y y x y
x x y x x y
y x y y x y
AEk
L
6
0 0 0 0 6
0 0.1 0 0.1 5
0 0 0 0 7
0 0.1 0 0.1 8
k AE
6 5 7 8
Structure stiffness matrix
K = k1 + k2 + k3 + k4 + k5 + k6
0.1 0 0 0 0 0.1 0 0 1
0 0 0 0 0 0 0 0 2
0 0 0 0 0 0 0 0 3
0 0 0 0 0 0 0 0 4
0 0 0 0 0 0 0 0 5
0.1 0 0 0 0 0.1 0 0 6
0 0 0 0 0 0 0 0 7
0 0 0 0 0 0 0 0 8
K AE
1 2 3 4 5 6 7 8
0.035 0.035 0 0 0 0 0.035 0.035 1
0.035 0.035 0 0 0 0 0.035 0.035 2
0 0 0 0 0 0 0 0 3
0 0 0 0 0 0 0 0 4
0 0 0 0 0 0 0 0 5
0 0 0 0 0 0 0 0 6
0.035 0.035 0 0 0 0 0.035 0.035 7
0.035 0.035 0 0 0 0 0.035 0.035 8
AE
1 2 3 4 5 6 7 8
0 0 0 0 0 0 0 0 1
0 0.1 0 0.1 0 0 0 0 2
0 0 0 0 0 0 0 0 3
0 0.1 0 0.1 0 0 0 0 4
0 0 0 0 0 0 0 0 5
0 0 0 0 0 0 0 0 6
0 0 0 0 0 0 0 0 7
0 0 0 0 0 0 0 0 8
AE
0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 2
0 0 0.1 0 0 0 0.1 0 3
0 0 0 0 0 0 0 0 4
0 0 0 0 0 0 0 0 5
0 0 0 0 0 0 0 0 6
0 0 0.1 0 0 0 0.1 0 7
0 0 0 0 0 0 0 0 8
AE
0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 2
0 0 0.035 0.035 0.035 0.035 0 0 3
0 0 0.035 0.035 0.035 0.035 0 0 4
0 0 0.035 0.035 0.035 0.035 0 0 5
0 0 0.035 0.035 0.035 0.035 0 0 6
0 0 0 0 0 0 0 0 7
0 0 0 0 0 0 0 0 8
AE
0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 2
0 0 0 0 0 0 0 0 3
0 0 0 0 0 0 0 0 4
0 0 0 0 0.1 0 0 0.1 5
0 0 0 0 0 0 0 0 6
0 0 0 0 0 0 0 0 7
0 0 0 0 0.1 0 0 0.1 8
AE
1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8
1 2 3 4 5 6 7 81 2 3 4 5 6 7 8
0.135 0.035 0 0 0 0.1 0.035 0.035
0.035 0.135 0 0.1 0 0 0.035 0.035
0 0 0.135 0.035 0.035 0.035 0.1 0
0 0.1 0.035 0.135 0.035 0.035 0 0
0 0 0.035 0.035 0.135 0.035 0 0.1
0.1 0 0.035 0.035 0.035 0.135 0 0
0.035 0.035 0.1 0 0 0 0.1
K AE
1
2
3
4
5
6
35 0.035 7
0.035 0.035 0 0 0.1 0 0.035 0.135 8
1 2 3 4 5 6 7 8
0. 0. 0.1 0. 0.035 0. 0.135
0. 0. 0. 0. 0.035 0. 0.035
Application of Stiffness Method• Truss analysis
– The global force components Q acting on the truss can be
related to its global displacements D using
Qk, Dk = know external loads & displacements. The
loads here exist on the truss as part of the
problem
Qu, Du = unknown loads & displacements. The loads
here represent the unknown support reactions
K = structure stiffness matrix
11 12
21 22
k u k
u u k
Q K D K D
Q K D K D
k 11 12 u
u 21 22 k
=
Q KD
Q K K D
Q K K D
Application of Stiffness Method
– The member forces can be determined using
0. 0.1 1
0. 0.1 1
Nx
x y NyN
x y FxF
Fy
D
Dq AE
Dq L
D
'q k TD
Nx
Ny
F x y x y
Fx
Fy
D
DAEq
DL
D
Example 3
• Determine the force in each member of the two member truss
shown. AE is constant.
Structure stiffness matrix: from previous example
0.405 0.096 0.333 0. 0.072 0.096
0.096 0.128 0. 0. 0.096 0.128
0.333 0. 0.333 0. 0. 0.
0. 0. 0. 0. 0. 0.
0.072 0.096 0. 0. 0.072 0.096
0.096 0.128 0. 0. 0.096 0.128
K AE
Displacements and loads
Q KD
1 1
2 2
3 3
4 4
5 5
6
0.405 0.096 0.333 0. 0.072 0.096
0.096 0.128 0. 0. 0.096 0.128
0.333 0. 0.333 0. 0. 0.
0. 0. 0. 0. 0. 0.
0.072 0.096 0. 0. 0.072 0.096
0.096 0.128 0. 0. 0.096 0.128
Q D
Q D
Q DAE
Q D
Q D
Q
6D
The known external displacements are D3 = D4 = D5 = D6 = 0.
Determine the unknown displacements by
The known external loads are Q1 = 0., Q2 = -2k
Rewrite the matrix
1
2
3
4
5
6
0 0.405 0.096 0.333 0. 0.072 0.096
2 0.096 0.128 0. 0. 0.096 0.128
0.333 0. 0.333 0. 0. 0. 0
0. 0. 0. 0. 0. 0. 0
0.072 0.096 0. 0. 0.072 0.096 0
0.096 0.128 0. 0. 0.096 0.128 0
D
D
QAE
Q
Q
Q
1
2
0 0.405 0.096 0
2 0.096 0.128 0
DAE
D
1 20 0.405 0.096AE D D
1 22 0.096 0.128AE D D 1
4.505D
AE
2
19.003D
AE
The support reactions are now obtained by
The force in each member obtained by
3
4
5
6
0.333 0 04.05
0 0 0
0.072 0.096 19.003 0
0.096 0.128 0
Q
Q AEAE
Q
AEQ
3 0.333(4.505) 1.5Q K
4 0Q
5 0.072(4.505) 0.096( 19.003) 1.5Q K
6 0.096(4.505) 0.128( 19.003) 2.0Q K
Nx
Ny
F x y x y
Fx
Fy
D
DAEq
DL
D
Member 1:
1
4.505
19.0031 0 1 0 1.5
3
0
0
AE
AEq k
AE
1x 0y 3L ft
Member 2:
2
4.505
19.0030.6 0.8 0.6 0.8 2.5
5
0
0
AE
AEq k
AE
0.6x 0.8y 5L ft
Example 4
• Determine the support reactions and the force in member 2 of
the truss shown in figure. AE is constant
Example 4
• The Stiffness matrix has been determine in Example 2 using the
same notation as shown.
Example 5
Determine the force in member 2 of the assembly if the support at joint 1
settles downward 25mm. AE = 8103 kN
Member 1:
1
0, 1, 3
0
0.002580000 1 0 1 8.333
0.005563
0.021875
x y L m
q kN
Member 2:
2
0.8, 0.6, 5
0.00556
0.02187580000.8 0.6 0.8 0.6 13.9
05
0
x y L m
q kN
Member 3:
3
1, 0, 4
0
080001 0 1 0 11.11
0.005564
0.021875
x y L m
q kN
Members Forces
Nodal Coordinate
To solve this problem
A set of nodal coordinate system x’’, y’’ located at the inclined
support will be used
If the support is an inclined roller
The zero deflection cannot be defined using single horizontal and
vertical global coordinate system
Nodal Coordinate
'' ''
'' ''
0.
0.
0.
0.
Nx x
Ny y N
Fx x F
Fy y
Q
Q q
Q q
Q
The Nodal Forces
'' '' ''
''
0. 0.
0. 0.
Nx
x y NyN
x y FxF
Fy
D
Dd
Dd
D
The Nodal Displacements
Nodal Coordinate
''
''
2
'' ''
2
'' ''
2
'' '' '' '' ''
2
'' '' '' '' ''
xx x y x x x y
yy x y y x y y
x x y x x x y x
x y y y x y y y
N
NAEk
FL
F
'' ''x y x yN N F F
'Tk T k T
The Stiffness Matrix
'' '' ''
''
0.
0. 0. 0.1 1
0. 0. 0.1 1
0.
x
y x y
x x y
y
AEk
L
– The member forces can be determined using
'' '' ''
''
0. 0.1 1
0. 0.1 1
Nx
x y NyN
F x y Fx
Fy
D
Dq AE
Dq L
D
'q k TD
''
''
'' ''
Nx
Ny
F x y x y
Fx
Fy
D
DAEq
DL
D
Nodal Coordinate
The Member Forces
Example 6
• Determine the support reaction for the truss shown
Member 1:
'' ''
1
1, 0, 0.707, 0.707, 4
0
011 0 0.707 0.707 22.5
127.34
0
x y x yL m
EAq kN
EA
Member 2:
'' ''
2
0, 1, 0.707, 0.707, 3
352.5
157.510 1 0.707 0.707 22.5
127.33
0
x y x yL m
EAq kN
EA
Member 3:
3
0.8, 0.6, 5
0
010.8 0.6 0.8 0.6 37.5
352.55
157.5
x y L m
EAq kN
EA
Members Forces
Thermal Changes and Fabrication Errors
Thermal Effects
If a truss member of length L is subjected to a temperature
increase T, The member undergo an increase in length of
L TL
0
0
N
F
q AE T
q AE T
Then
Transforming into global coordinate
0
0
0
0
0.
0. 1
0. 1
0.
Nxx x
Ny y y
x xFx
y yFy
Q
QAE T AE T
Q
Q
Thermal Changes and Fabrication Errors
Fabrication Errors
If a truss member is made too long by an amount L, then the force
q0 needed to keep the member at its design length L
0
0
N
F
AE Lq
L
AE Lq
L
In global coordinates
0
0
0
0
Nxx
Ny y
xFx
yFy
Q
Q AE L
LQ
Q
Thermal Changes and Fabrication Errors
0Q KD Q
Q0 : is the column matrix for the entire truss of the
initial fixed-end force caused by the temperaturechanges and fabrication errors of the members
defined in previous equations
kk 11 12 u 0
u 21 22 k u 0
=QQ K K D
Q K K D Q
Matrix Analysis
Thermal Changes and Fabrication Errors
The Member forces
'
0q k TD q
0
Nx
Ny
F x y x y F
Fx
Fy
DAE T
DAEq q
DLLAED L
Example 7
• Determine the force in member 1 and 2 of the pin-connected
assembly if the member 2 was made 0.01m too short before it
was fitted into place. Take AE=8(103)
Example 8
Member 2 of the truss shown is subjected to an increase in temperature
of 150 F. Determine the force developed in member 2. E=29(106)lb/in2.
Each member has across sectional area of A=0.75 in2
Springs Structures
1 1'
1 1
Where:
:is the spring stiffness
s
s
k k
k
'q k dks
ks q2q1
d2d1
Example 9
Given: For the spring system shown,
K1 = 100 N/mm, K2 = 200 N/mm,
K3 = 300 N/mm
P = 500 N.
Find:
(a) The global stiffness matrix
(b) Displacements of nodes 1 and 2
(c) The reaction forces at nodes 3 and 4
(d) The force in the spring 2
1 23 41 2 3
Solution
(a) The global stiffness matrix
1
100 0 100 0 1
100 100 3 0 0 0 0 2
100 100 1 100 0 100 0 3
0 0 0 0 4
k
2
200 200 0 0 1
200 200 1 200 200 0 0 2
200 200 2 0 0 0 0 3
0 0 0 0 4
k
3
0 0 0 0 1
300 300 2 0 300 0 300 2
300 300 4 0 0 0 0 3
0 300 0 300 4
k
Solution
The global stiffness matrix
100 0 100 0 1 200 200 0 0 1 0 0 0 0 1
0 0 0 0 2 200 200 0 0 2 0 300 0 300 2
100 0 100 0 3 0 0 0 0 3 0 0 0 0 3
0 0 0 0 4 0 0 0 0 4 0 300 0 300 4
K
K = k1 + k2 + k3
300 200 100 0 1
200 500 0 300 2
100 0 100 0 3
0 300 0 300 4
K
1 23 41 2 3
Application of Stiffness Method– The global force components Q acting on the Structure can
be related to its global displacements D using
k 11 12 u
u 21 22 k
=
Q KD
Q K K D
Q K K D
Qk, Dk = know external loads & displacements. The loads
here exist on the truss as part of the problem
Qu, Du = unknown loads & displacements. The loads here
represent the unknown support reactions
K = structure stiffness matrix
11 12
21 22
k u k
u u k
Q K D K D
Q K D K D
(b) Displacements of nodes 1 and 2
D3 = D4 = 0., Q1 = 0 and Q2 = P = 500
1 1
2 2
10
300 200 0. 11
200 500 500 15
11
D D
D D
1 1
2 2
3 3
4 4
300 200 100 0
200 500 0 300
100 0 100 0
0 300 0 300
Q KD
Q D
Q D
Q D
Q D
1
2
3
4
0 300 200 100 0
500 200 500 0 300
100 0 100 0 0
0 300 0 300 0
D
D
Q
Q
3
3 4
4
10
0300 200 100 0 11 1000
500200 500 0 300 15 11
11100 0 100 0 4500
0 110 300 0 300
0
Q
Q Q
Q
(c) The reaction forces at nodes 3 and 4
(d) The force in the spring 2
22 2 1 1
22 2 2 2
2 2
1 1
2 2
2 2
10 1000
200 200 11 11
200 200 15 1000
11 11
k k d q
k k d q
q q
q q
Space Truss Analysis
2 2 2
2 2 2
2 2 2
cos
cos
cos
F N F Nx x
F N F N F N
F N F Ny y
F N F N F N
F N F Nz z
F N F N F N
x x x x
L x x y y z z
y y y y
L x x y y z z
z z z z
L x x y y z z
0 0 0
0 0 0
x y z
x y z
T
Stiffness Matrix
2 2
2 2
2 2
2 2
2 2
2 2
x y z x y z
x x y x z x x y x z x
y x y y z y x y y z y
x z y z z x z y z z z
x y x x z x x y x z x
x y y y z x y y y z y
x z y z z x z y z z z
N N N F F F
N
N
NAEk
FL
F
F
0
0
0 0 00 1 1
0 0 00 1 1
0
0
x
y
x y zz
x y zx
y
z
AEk
L
Stiffness Matrix
2
2
2
xyz xyz
xyz xyz
x x y x z
xyz x y y y z
x z y z z
C Ck
C C
AEC
L