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DRAFT –1
Chapter 3 Truss Element
By trigonometric relations, we have,
δ
1
= u
1
cos
θ
+ u
2
sin
θ
= c u
1
+ s u
2
δ
2
= u
3
cos
θ
+ u
4
sin
θ
= c u
3
+ s u
4
where, cos
θ
= c, and sin
θ
= s
Writing the above equations in a matrix form, we get,
u
1
δ
1
c s 0 0 u
2
=
u
3
(3.2)
δ
1
0 0 c s u
4
Or, in short form,
δ
= T u
Where T is called Transformation matrix.
Along with equation (3.2), we also need an equation that relates the local and global
forces.
3.3.5 Relationship Between Local and Global Forces
By using trigonometric relations similar to the previous section, we can derive the desired
relationship between local and global forces. However, it will be easier to use the work-
energy concept for this purpose. The forces in local coordinates are: R
1
and R
2
, and in
global coordinates: f
1
, f
2
, f
3
, and f
4
, see Figure 3.6 for their directions.
Since work done is independent of a coordinate system, it will be the same whether we
use a local coordinate system or a global one. Thus, work done in the two systems is
equal and given as,
W =
δ
T
R = u
T
f, or in an expanded form,
R
1
f
1
W =
δ
1
δ
2
= u
1
u
1
u
1
u
1
f
2
R
2
f
3
f
4
= {
δ
}
T
{R} = {u}
T
{f}
ME 273 Lecture Notes © by R. B. Agarwal
3-6
DRAFT –1
Chapter 3 Truss Element
Substituting
δ
= T u in the above equation, we get,
[[T] {u}]
T
{R} = {u}
T
{f}, or
{u}
T
[T]
T
{R} = {u}
T
{f}, dividing by {u}
T
on both sides, we get,
[T]
T
{R}
=
{f}
(3.3)
Equation (3.3) can be used to convert local forces into global forces and vice versa.
F
4
, u
4
R
2
,
δ
2
f
2
, u
2
2
f
3
u
3
1
θ
R
1
,
δ
1
f
1
, u
1
Figure 3.6 Local and Global Forces
3.3.6 Finite Element Equation in Local Coordinate System
Now we will derive the finite element equation in local coordinate system. This equation
will be converted to global coordinate system, which can be used to generate a global
structural equation for the given structure. Note that, we can not use the element
equations in their local coordinate form, they must be converted to a common coordinate
system, the global coordinate system.
Consider the element shown below, with nodes 1 and 2, spring constant k, deflections
δ
1
,
and
δ
2
, and forces R
1
and R
2
. As established earlier, the finite element equation in local
coordinates is given as,
R
1
k -k
δ
1
1
k
2
=
δ
1
,
R
1
R
2
-k
k
δ
1
δ
2
, R
2
Figure 3.7 A Truss Element
Recall that, for a truss element, k = AE/L
ME 273 Lecture Notes © by R. B. Agarwal
3-7
DRAFT –1
Chapter 3 Truss Element
Let k
e
= stiffness matrix in local coordinates, then,
AE/L
-AE/L
k
e
=
Stiffness matrix in local coordinates
-AE/L
AE/L
3.3.7 Finite Element Equation in Global Coordinates
Using the relationships between local and global deflections and forces, we can convert
an element equation from a local coordinate system to a global system.
Let k
g
= Stiffness matrix in global coordinates.
In local system, the equation is:
R = [k
e
]{
δ
}
(A)
We want a similar equation, but in global coordinates. We can replace the local force R
with the global force f derived earlier and given by the relation:
{f} = [T
T
]{R}
Replacing R by using equation (A), we get,
{f} = [T
T
] [[k
e
]{
δ
}],
and
δ
can be replaced by u, using the relation
δ
= [T]{u}, therefore,
{f} = [T
T
] [k
e
] [T]{u}
{f} = [k
g
] { u}
Where, [k
g
] = [T
T
] [k
e
] [T]
Substituting the values of [T]
T
, [T], and [k
e
], we get,
c
0
[k
g
] = s
0
AE/L -AE/L
c
s
0
0
0
c
-AE/L AE/L
0
0
c
s
0
s
ME 273 Lecture Notes © by R. B. Agarwal
3-8
DRAFT –1
Chapter 3 Truss Element
Simplifying the above equation, we get,
c
2
cs -c
2
-cs
cs
s
2
-cs -s
2
[k
g
] =
-c
2
-cs
c
2
cs (AE/L)
-cs
-s
2
cs s
2
This is the global stiffness matrix of a truss element. This matrix has several noteworthy
characteristics:
The matrix is symmetric
Since there are 4 unknown deflections (DOF), the matrix size is a 4 x 4.
The matrix represents the stiffness of a single element.
The terms c and s represent the sine and cosine values of the orientation of
element with the horizontal plane, rotated in a counter clockwise direction
(positive direction).
The following example will illustrate its application.
(1)
1
260 AL
Examples
3
For the truss structure shown: 150 AL
(3)
(2)
ST
1. Find displacements of joints 2 and 3
300
0.4 kN
2. Find stress, strain, & internal
forces
in each member.
2
A
AL
= 200 mm
2
, A
ST
= 100 mm
2
All other dimensions are in mm.
Solution
Let the following node pairs form the elements:
Element Node
Pair
(1)
1-3
(2)
1-2
(3)
2-3
Using Shigley’s Machine Design book for yield strength values, we have,
ME 273 Lecture Notes © by R. B. Agarwal
3-9
DRAFT –1
Chapter 3 Truss Element
S
y
(AL)
= 0.0375kN/mm
2
(375 Mpa)
S
y
(ST)
= 0.0586kN/mm
2
(586 Mpa)
E
(AL)
= 69kN/mm
2
, E
(ST)
= 207kN/mm
2
A
(1)
= A
(2)
= 200mm
2
, A
(3)
=100mm
2
Find the stiffness matrix for each element
u
2
u
6
Element (1)
(1)
L
(1 )
= 260 mm,
u
1
u
5
E
(1)
= 69kN/mm
2
, 1
260 mm
3
A
(1)
= 200mm
2
θ
= 0
c = cos
θ
= 1,
c
2
= 1
s = sin
θ
= 0, s
2
= 0
cs = 0
2
2
2
2
2
2
(1)
2
2
2
2
(1)
1
69
/
200
53.1
/
260
[
]
1 1
0
1 0
2 0
1
0
0
[
]
(53.1)
5
1 0
1
0
6 0
0
0
0
g
g
EA
kN mm
mm
kN mm
L
mm
c
cs
c
cs
cs
s
cs
s
AE
K
L
c
cs
c
cs
cs
s
cs
s
kN
K
mm
=
×
×
=
−
−
−
−
=
−
−
−
−
−
=
−
ME 273 Lecture Notes © by R. B. Agarwal
3-10