-
Chapter 4
Plane Trusses
-
4.1 What is Truss Structure?
A truss is a static structure consisting of straight slender
members, inter-
connected at joints, to form triangular units.
There are two types of trusses:
1. The pitched truss or common truss - characterized by its
triangular
shape. It is most often used for roof construction.
2. The parallel chord or flat truss – characterized by its
parallel top and bottom chords. It is often used for floor
construction.
-
Examples of pitched truss structures.
-
Pitched trusses used in bridge construction.
-
A complex truss structure used in bridge construction.
-
Truss structures in an International Space Station (ISS).
-
4-2 Vierendeel Truss
A special truss structure, characterized by its rigid upper and
lower
beams, connected by vertical beams. The joints are also
rigid.
Used in construction of some bridges and in the frame of the
“late” Twin
Towers of the World Trade Centre.
A Vierendeel bridge
The “late” twin towers of the world trade centre.
-
4-3 Plane Trusses A typical two-dimensional plane truss is
shown. It comprises of two-force
members, connected by frictionless joints. All loads and
reaction forces are
applied at the joints only.
Note: There are two displacement components at a given node j,
denoted by Q2j-1 and Q2j.
-
4-4 Local & Global Coordinate Systems
Local and global coordinate systems are
shown at left.
In local coordinate (x’), every node has one
degree of freedom, while in global coordinate
(x, y), every node has two degrees of
freedom.
The nodal displacements, in the local
coordinate system is,
In the global coordinate system,
Local and global coordinate systems.
{ }' ' '1 2T
q q q =
{ } [ ]1 2 3 4T
q q q q q=
-
4-5 Relation Between Coordinate Systems Consider a deformed
truss member as shown. We can now establish a
relationship between {q’} and {q} as follows:
θθ
θθ
sincos
sincos
43
'
2
21
'
1
qqq
qqq
+=
+=
-
4-6 Direction Cosines
θcos=l and θsin=m
The relation between {q’} and {q} can now be written as
To eliminate the θ terms from previous equations, we define
direction
cosines, such that
which can be written in matrix form as
)()(
)()(
43
'
2
21
'
1
mqlqq
mqlqq
+=
+=
=
4
3
2
1
'
2
'
1
00
00
q
q
q
q
ml
ml
q
q
-
where [L] is a rectangular matrix called the transformation
matrix,
given by,
Or, in a condensed matrix form
{ } [ ]{ }qLq ='
[ ]
=
ml
mlL
00
00
-
4-7 Formula for Evaluating l and m
Using a trigonometry relation, we observe
Note: Coordinates (xi, yi) are based on local coordinate
system.
2 1
2 1
cos
sin
e
e
x xl
l
y ym
l
θ
θ
−= =
−= =
( ) ( )212
2
12yyxxle −+−=
-
4-8 Element Stiffness Matrix A truss element is a
one-dimensional (bar) element, when it is viewed in
local coordinate system.
Thus, element stiffness matrix for a truss element in local
coordinate,
The internal strain energy in the truss element, in local
coordinate system is,
Substituting
[ ] [ ]' 1 1
1 1truss bar
e
AEk k
l
− = = −
{ } [ ] { }''''2
1qkqU truss
T
e =
{ } [ ]{ }qLq =' we get,
[ ]{ }( ) [ ] [ ]{ }( )
{ } [ ] [ ] [ ]( ){ }qLkLqU
qLkqLU
truss
TT
e
truss
T
e
''
''
2
1
2
1
=
=
We need the expression for [k] when viewed in
global coordinate…
-
In the global coordinates system,
Since internal strain energy is independent of coordinate
system, Ue = U’e.
Therefore,
Simplifying,
{ } [ ] { }qkqU trussT
e2
1=
[ ] [ ] [ ] [ ]
[ ]
'
0
0 1 1 0 0
0 1 1 0 0
0
T
truss truss
truss
e
k L k L
l
m l mAEk
l l ml
m
=
− = ⋅ −
[ ] [ ]
2 2
2 2
2 2
2 2
truss
e
l lm l lm
lm m lm mAEk k
l l lm l lm
lm m lm m
− −
− − = = − − − −
-
EXERCISE 4.1
Write the element stiffness matrix for each element .
Use: E = 180 GPa; d = 15 mm for all members.
P = 50 kN
0.8 m
0.6 m
1
2
3
1
2
3
-
4-9 System of Linear Equations The system of linear equations
for a single plane truss element in local
coordinate system can be expressed as
Where {q} is nodal displacement vector and {f} is nodal force
vector, in the global
coordinate direction. Substituting, we get
Note: To assemble the global stiffness matrix, a local-global
nodal connectivity
will be required.
[ ]{ } { }fqk =
2 2
1 1
2 2
2 2
2 2
3 3
2 2
4 4
e
q fl lm l lm
q flm m lm mAE
q fl l lm l lm
q flm m lm m
− −
− − ⋅ = − − − −
-
Exercise 4-2
Reconsider Exercise 4-1. a) Assemble global system of linear
equations for
the structure; b) Apply the boundary conditions, and c) Write
the reduced
system of linear equations.
P = 50 kN
0.8 m
0.6 m
A
B
C
Use: E = 180 GPa; d = 15 mm for all members.
Boundary conditions:
Q1 = Q2 = Q6 = 0
(homogeneous type)
-
4-10 Stress Calculations Normal stress in a plane truss element,
in local coordinate system is,
In the global coordinate system, since
Expanding the [B] and [L] matrices,
Or
[ ]{ }'qBE=σ
{ } [ ]{ }qLq ='
[ ] [ ] { }E B L qσ = ⋅ ⋅
[ ]
−−=
4
3
2
1
q
q
q
q
mlmll
E
e
σ
[ ]
1
2
3
4
0 011 1
0 0
q
ql mE
ql ml
q
σ
= − ⋅
e
-
Exercise 4-3
Reconsider Exercise 4-2. a) Determine the unknown nodal
displacements
at B and C; b) Compute the stresses in the member AC and BC.
P = 50 kN
0.8 m
0.6 m
A
B
C
Use: E = 180 GPa; d = 15 mm for all members.
