Plane Trusses - Universiti Teknologi Malaysiataminmn/SME 3033 Finite Element...
Transcript of Plane Trusses - Universiti Teknologi Malaysiataminmn/SME 3033 Finite Element...
PLANE TRUSS ELEMENTS
MNTamin, CSMLab
SME 3033 – FINITE ELEMENT METHOD
Plane Trusses (Initial notes are designed by Dr. Nazri Kamsah)
PLANE TRUSS ELEMENTS
MNTamin, CSMLab
SME 3033 – FINITE ELEMENT METHOD 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.
PLANE TRUSS ELEMENTS
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SME 3033 – FINITE ELEMENT METHOD
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 2
T
q q q
1 2 3 4
Tq q q q q
PLANE TRUSS ELEMENTS
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SME 3033 – FINITE ELEMENT METHOD
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:
sincos
sincos
43
'
2
21
'
1
qqq
qqq
PLANE TRUSS ELEMENTS
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SME 3033 – FINITE ELEMENT METHOD
4-6 Direction Cosines
cosl and sinm
The relation between {q’} and {q} can now be written as
To eliminate the 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
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SME 3033 – FINITE ELEMENT METHOD
where [L] is a rectangular matrix called the transformation matrix,
given by,
Or, in a condensed matrix form
qLq '
ml
mlL
00
00
PLANE TRUSS ELEMENTS
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SME 3033 – FINITE ELEMENT METHOD
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
2 2
2 1 2 1
cos
sin
( ) ( )
e
e
e
x xl
l
y ym
l
l x x y y
PLANE TRUSS ELEMENTS
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SME 3033 – FINITE ELEMENT METHOD 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 bare
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…
PLANE TRUSS ELEMENTS
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SME 3033 – FINITE ELEMENT METHOD
In the global coordinates system,
Since internal strain energy is independent of coordinate system, Ue = U’e.
Therefore,
Simplifying,
qkqU truss
T
e2
1
'
0
0 1 1 0 0
0 1 1 0 0
0
T
truss truss
trusse
k L k L
l
m l mAEk
l l ml
m
2 2
2 2
2 2
2 2
trusse
l lm l lm
lm m lm mAEk k
l l lm l lm
lm m lm m
PLANE TRUSS ELEMENTS
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SME 3033 – FINITE ELEMENT METHOD Exercise 4-1
For each element of a plane truss structure shown,
a) Determine the transformation matrix, [L]e;
b) Write the element stiffness matrix, [k]e;
c) Assemble the global stiffness matrix, [K].
P = 50 kN
0.8 m
0.6 m
A
B
C
Use: E = 180 GPa; d = 15 mm for all members.
PLANE TRUSS ELEMENTS
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SME 3033 – FINITE ELEMENT METHOD
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 21 1
2 22 2
2 23 3
2 24 4
e
q fl lm l lm
q flm m lm mAE
q fl l lm l lm
q flm m lm m
PLANE TRUSS ELEMENTS
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SME 3033 – FINITE ELEMENT METHOD
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 = Q4 = 0
(homogeneous type)
PLANE TRUSS ELEMENTS
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SME 3033 – FINITE ELEMENT METHOD 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
qLq '
E B L q
1
2
3
4
0 011 1
0 0
q
ql mE
ql ml
q
4
3
2
1
q
q
q
q
mlmll
E
PLANE TRUSS ELEMENTS
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SME 3033 – FINITE ELEMENT METHOD
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.
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SME 3033 – FINITE ELEMENT METHOD
Homework 4-1
For the truss structure shown, rod (1)
has modulus E = 29 x 103 ksi and a
yield point Y = 36 ksi. Rod (2) has
modulus E = 10 x 103 ksi and a yield
point Y = 60 ksi. If the horizontal
force P = 10 kips, determine the
cross-sectional areas of the two rods
so that the yield stress in each rod is
not exceeded.
Note: This is a trial and error
problem, so you may use Nastran
simulation to help you solve the
problem. Work in group.
PLANE TRUSS ELEMENTS
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SME 3033 – FINITE ELEMENT METHOD Example 4-1
For a plane truss shown, determine: a) displacements at nodes 2 and 3; b)
stresses in each element; c) reaction forces at the supports.
Use: Ee = 29.5 x 106 psi and Ae = 1 in2 for all elements.
PLANE TRUSS ELEMENTS
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SME 3033 – FINITE ELEMENT METHOD Solution
1. Establish nodal coordinate data and element connectivity
information
Node x y
1 0 0
2 40 0
3 40 30
4 0 30
Element connectivity
Element Node 1 Node 2
1 1 2
2 3 2
3 1 3
4 4 3
Nodal coordinate
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SME 3033 – FINITE ELEMENT METHOD
2. Compute the direction cosines
Element le l m
1 40 1 0
2 30 0 -1
3 50 0.8 0.6
4 40 1 0
3. Write the stiffness matrix for each element
0000
0101
0000
0101
40
1105.29 6)1(
k
Element 1
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SME 3033 – FINITE ELEMENT METHOD
For elements 2, 3, and 4
1010
0000
1010
0000
30
1105.29 6)2(
k
36.048.036.048.0
48.064.048.064.0
36.048.036.048.0
48.064.048.064.0
50
1105.29 6)3(
k
0000
0101
0000
0101
40
1105.29 6)4(
k
PLANE TRUSS ELEMENTS
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SME 3033 – FINITE ELEMENT METHOD
4. Assemble the global stiffness matrix [K] for the entire truss structure
6
22.68 5.76 15.0 0 7.68 5.76 0 0
5.76 4.32 0 0 5.76 4.32 0 0
15.0 0 15.0 0 0 0 0 0
0 0 0 20.0 0 20.0 0 029.5 10
7.68 5.76 0 0 22.68 5.76 15.0 0600
5.76 4.32 0 20.0 5.76 24.32 0 0
0 0 0 0 15.0 0 15.0 0
0 0 0 0 0 0 0 0
K
PLANE TRUSS ELEMENTS
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SME 3033 – FINITE ELEMENT METHOD
5. Assemble the global system of linear equations.
1
2
3
6
Q22.68 5.76 15.0 0 7.68 5.76 0 0
Q5.76 4.32 0 0 5.76 4.32 0 0
Q15.0 0 15.0 0 0 0 0 0
Q0 0 0 20.0 0 20.0 0 029.5 10
7.68 5.76 0 0 22.68 5.76 15.0 0600
5.76 4.32 0 20.0 5.76 24.32 0 0
0 0 0 0 15.0 0 15.0 0
0 0 0 0 0 0 0 0
4
5
6
7
8
0
0
20000
0
Q 0
Q 25000
Q 0
Q 0
PLANE TRUSS ELEMENTS
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SME 3033 – FINITE ELEMENT METHOD
6. Impose boundary conditions & write the reduced system of
linear equations
25000
0
20000
32.2476.50
76.568.220
0015
600
105.29
6
5
36
Q
Q
Q
7. Solve the reduced equations using the Gaussian elimination method,
we get
in
1025.22
1065.5
1012.27
3
3
3
6
5
3
Q
Q
Q
We have; Q1 = Q2 = Q4 = Q7 = Q8 = 0.
Using elimination method, the above equations reduced to
PLANE TRUSS ELEMENTS
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SME 3033 – FINITE ELEMENT METHOD
8. Compute the stress in each element
Element 1
psi20000
0
1012.27
0
0
010140
105.293
6
1
Element 2
psi21880
0
1012.27
1025.22
1065.5
101030
105.293
3
3
6
2
psi4167
psi5208
4
3
In the same manner,