1 1 BAR & TRUSS FINITE ELEMENT Direct Stiffness Method FINITE ELEMENT ANALYSIS AND APPLICATIONS 2 INTRODUCTION TO FINITE ELEMENT METHOD • What is the finite element method (FEM)? – A technique for obtaining approximate solutions of differential equations. – Partition of the domain into a set of simple shapes (element) – Approximate the solution using piecewise polynomials within the element F Structure Element u x Piecewise-Linear Approximation 0 0 xy xx x xy yy y b x y b x y 3 • How to discretize the domain? – Using simple shapes (element) – All elements are connected using “nodes”. – Solution at Element 1 is described using the values at Nodes 1, 2, 6, and 5 (Interpolation). – Elements 1 and 2 share the solution at Nodes 2 and 6. INTRODUCTION TO FEM cont. 1 2 3 1 2 3 4 5 6 7 8 Elements Nodes
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1
BAR & TRUSS FINITE ELEMENT
Direct Stiffness Method
FINITE ELEMENT ANALYSIS AND APPLICATIONS
2
INTRODUCTION TO FINITE ELEMENT METHOD
• What is the finite element method (FEM)?
– A technique for obtaining approximate solutions of differential
equations.
– Partition of the domain into a set of simple shapes (element)
– Approximate the solution using piecewise polynomials within the
element
F
Structure
Element
u
x
Piecewise-Linear Approximation
0
0
xyxxx
xy yy
y
bx y
bx y
3
• How to discretize the domain?
– Using simple shapes (element)
– All elements are connected using “nodes”.
– Solution at Element 1 is described using the values at Nodes 1, 2, 6,
and 5 (Interpolation).
– Elements 1 and 2 share the solution at Nodes 2 and 6.
INTRODUCTION TO FEM cont.
1 2 3
1 2 3 4
5 6 7 8
Elements
Nodes
2
4
INTRODUCTION TO FEM cont.
• Methods
– Direct method: Easy to understand, limited to 1D problems
– Variational method
– Weighted residual method
• Objectives
– Determine displacements, forces, and supporting reactions
– Will consider only static problem
5
1-D SYSTEM OF SPRINGS
• Bodies move only in horizontal direction
• External forces, F2, F3, and F4, are applied
• No need to discretize the system (it is already discretized!)
• Rigid body (including walls) NODE
• Spring ELEMENT
F2
F3
F41 2
3
4
5
6
1
2
3
4 5
6
SPRING ELEMENT
• Element e
– Consist of Nodes i and j
– Spring constant k(e)
– Force applied to the nodes:
– Displacement ui and uj
– Elongation:
– Force in the spring:
– Relation b/w spring force and nodal forces:
– Equilibrium:
e
i j
( ), ei iu f ( ), e
j ju f
( )e
j iu u
e e e e
j iP k k u u
e e
jf P
0
e e e e
i j i jf f or f f
,
e e
i jf f
3
7
SPRING ELEMENT cont.
• Spring Element e
– Relation between nodal forces and displacements
– Matrix notation:
– k: stiffness matrix
– q: vector of DOFs
– f: vector of element forces
e e
i i j
e e
j i j
f k u u
f k u u
( )
( )
e e ei i
ee ej j
u fk k
u fk k
( )
( )
( )
( ) ( ) ( )[ ]
ei ie
ej j
e e e
u f
u f
k
k q f
k q f
8
SPRING ELEMENT cont.
• Stiffness matrix
– It is square as it relates to the same number of forces as the
displacements.
– It is symmetric.
– It is singular, i.e., determinant is equal to zero and it cannot be
inverted.
– It is positive semi-definite
• Observation
– For given nodal displacements, nodal forces can be calculated by
– For given nodal forces, nodal displacements cannot be determined
uniquely
( ) ( ) ( )[ ]e e ek q f
e e
e e
k k
k k
9
SYSTEM OF SPRINGS cont.
• Element equation
and assembly
F2
F3
F41 2
3
4
5
6
1
2
3
4 5
(1)1 1 1 1
(1)1 1 2 2
k k u f
k k u f
(1)11 1 1
(1)21 1 2
3
4
5
0 0 0
0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
uk k f
uk k f
u
u
u
(2)2 2 2 2
(2)2 2 4 4
k k u f
k k u f
(1)11 1 1
(1)21 1 2
3
4
(2)2 2 2
(2)2 2 4
5
0 0 0
0 0
0 0 0 0 0 0
0 0 0
0 0 0 0 0 0
k k f
uk k f
uk k f
u
uk
u
k f
4
10
SYSTEM OF SPRINGS cont.
(3)3 3 2 2
(3)3 3 3 3
k k u f
k k u f
(4)14 4 1
(4)34 4 3
uk k f
uk k f
(5)5 5 3 3
(5)5 5 4 4
k k u f
k k u f
(3)(2)2 2 2
(1)11 1 1
(1)3 3 2
(3)3 3
(
2
2)2 2 4
1
3
4
5
3
1 2
0 0 0
0
0 0 0
0 0 0
0 0 0 0 0 0
uk k f
uk k f
u
u
u
k k f
k k
k k
f
k k f
f
(1)11 1 1
(1)21 1 2
3
4
5
(2)2 2 2
(2)2
(4)4 4 1
(4)4 4
(3)3 3 2
(3)3 3 3
2
3
4
0 0
0
0 0
0 0 0
0 0 0 0 0 0
uk k f
uk k f
u
u
k k f
k k
k k f
k k
k k f
k
f
k
u
ff
(2)2 2 2
(2)2 2
(4)4 4 1
(4)4 4
(5)5 5 3
(5)5 5 4
(1)11 1 1
(1)21
(3)3 3 2
(3)3 3
1 2
3
4
3
5
3
4
0 0
0
0
0 0
0 0 0 0 0 0
k k f
k k k k f
k
k k f
k
uk k f
uk
k f
k fk k
u
u
u
f
f
k fk k f
11
SYSTEM OF SPRINGS cont.
(6)6 6 4 4
(6)6 6 5 5
k k u f
k k u f
(5)5 5
(3)3 3 2
(3
(6)
3
(5)5
(1)1 1
5
(2)2 2
6
2
(2
(4)4
)
1 1
(1)1 1 2 2
3
4
)3
6
4 1
(4)3
42 2
34 4 3
4
(6)6 5
4
6 5
0 0
0
0
0
0 0 0
k k u f
k k u f
u
u
u
k k f
k
k k f
k k
k k f
k
k k f
k k f
k k
k k
f
f
k ff
f
k
F2
F3
F41 2
3
4
5
6
1
2
3
4 5
12
SYSTEM OF SPRINGS cont.
• Relation b/w element
forces and external force
• Force equilibrium
• At each node, the summation of
element forces is equal to
the applied, external force
F2
F3
F41 2
3
4
5
6
1
2
3
4 5
F3
3
(5)3f
(3)3f
(4)3f
1
1
0
, 1,...
e
e
ie
i i
e
ie
i i
e
F f
F f i ND
(1) (4)11 1
(1) (2) (3)22 2 2
(3) (4) (5)33 3 3
(2) (5) (6)44 4 4
(6)55
Rf f
Ff f f
Ff f f
Ff f f
Rf
5
13
SYSTEM OF SPRINGS cont.
• Assembled System of Matrix Equation:
• [Ks] is square, symmetric, singular and positive semi-definite.
• When displacement is known, force is unknown
R1 and R5 are unknown reaction forces
1 4 1 4 1 1
1 1 2 3 3 2 2 2
4 3 3 5 4 5 3 3
2 5 2 5 6 6 4 4
6 6 5 5
0 0
0
0
0
0 0 0
k k k k u R
k k k k k k u F
k k k k k k u F
k k k k k k u F
k k u R
[ ]{ } { }s s sK Q F
1 5 0u u
14
SYSTEM OF SPRINGS cont.
• Imposing Boundary Conditions
– Ignore the equations for which the RHS forces are unknown and strike
out the corresponding rows in [Ks].
– Eliminate the columns in [Ks] that multiply into zero values of
displacements of the boundary nodes.
1 4 1 4 1 1
1 1 2 3 3 2 2 2
4 3 3 5 4 5 3 3
2 5 2 5 6 6 4 4
6 6 5 5
0 0
0
0
0
0 0 0
k k k k u R
k k k k k k u F
k k k k k k u F
k k k k k k u F
k k u R
15
SYSTEM OF SPRINGS cont.
• Global Matrix Equation
• Global Stiffness Matrix [K]
– square, symmetric and positive definite and hence non-singular
• Solution
• Once nodal displacements are obtained, spring forces can be
calculated from
1 2 3 3 2 2 2
3 3 4 5 5 3 3
2 5 2 5 6 4 4
k k k k k u F
k k k k k u F
k k k k k u F
[ ]{ } { }K Q F
1{ } [ ] { }Q K F
e e e e
j iP k k u u
6
16
UNIAXIAL BAR
• For general uniaxial bar, we need to divide the bar into a set
of elements and nodes
• Elements are connected by sharing a node
• Forces are applied at the nodes (distributed load must be
converted to the equivalent nodal forces)
• Assemble all elements in the same way with the system of
springs
• Solve the matrix equation
for nodal displacements
• Calculate stress and strain
using nodal displacements
Fp(x)x
Fp(x)
Statically indeterminate
Statically determinate
17
1D BAR ELEMENT
• Two-force member
• Only constant
cross-section
• Element force is
proportional to
relative displ
• First node: i
second code: j
• Force-displacement relation
x
f1 f2
L A
Node i Node j
ui uj
K=EA/L ( )eif
( )ejf
( )
( )
( )
( ) ( )
( )
( )
e
e
i i j
e
e e
j i j i
AEf u u
L
AEf f u u
L
Similar to the spring element
18
1D BAR ELEMENT cont.
– Matrix notation
– Either force or displacement (not both) must be given at each node.
– Example: ui = 0 and fj = 100 N.
– What happens when fi and fj are given?
• Nodal equilibrium
– Equilibrium of forces acting on Node I
– In general
( )( )
( )
1 1
1 1
eeii
ejj
uf AE
uf L
Node i Node j
ui uj
K=EA/L ( )eif
( )ejf
( ) ( ) ( ){ } [ ]{ }e e ef k q
Element e+1Element e
Node i
Fi
fi(e)
fi(e+1)
( ) ( 1) 0e e
i i iF f f ( ) ( 1)e e
i i if f F
( )
1
eie
i i
e
F f
7
19
1D BAR ELEMENT cont.
• Assembly
– Similar process as spring elements
– Replace all internal nodal forces with External Applied Nodal Force