Upload
mustafa-yilmaz
View
229
Download
0
Embed Size (px)
Citation preview
8/21/2019 Finite Element Methods Lectures_Lui.pdf
1/169
Lecture Notes: Introduction to Finite Element Method Chapter 1. Introduction
© 1998 Yijun Liu, University of Cincinnati 1
Chapter 1. I ntroduction
I . Basic Concepts
The finite element method (FEM), or finite element analysis
(FEA), is based on the idea of building a complicated object with
simple blocks, or, dividing a complicated object into small and
manageable pieces. Application of this simple idea can be found
everywhere in everyday life as well as in engineering.
Examples:
• Lego (kids’ play)
• Buildings
• Approximation of the area of a circle:
Area of one triangle: S Ri i=1
2
2sinθ
Area of the circle: S S R N N
R as N N ii
N
= =
→ → ∞=
∑1
2 21
2
2sin
ππ
where N = total number of triangles (elements).
R
θi
“Element” S i
8/21/2019 Finite Element Methods Lectures_Lui.pdf
2/169
Lecture Notes: Introduction to Finite Element Method Chapter 1. Introduction
© 1998 Yijun Liu, University of Cincinnati 2
Why F ini te Element M ethod?
• Design analysis: hand calculations, experiments, and
computer simulations
• FEM/FEA is the most widely applied computer simulationmethod in engineering
• Closely integrated with CAD/CAM applications
• ...
Appli cations of FEM in Engineer ing
• Mechanical/Aerospace/Civil/Automobile Engineering
•Structure analysis (static/dynamic, linear/nonlinear)
• Thermal/fluid flows
• Electromagnetics
• Geomechanics
• Biomechanics
• ...
Examples:
...
8/21/2019 Finite Element Methods Lectures_Lui.pdf
3/169
Lecture Notes: Introduction to Finite Element Method Chapter 1. Introduction
© 1998 Yijun Liu, University of Cincinnati 3
A Brief H istory of the FEM
• 1943 ----- Courant (Variational methods)• 1956 ----- Turner, Clough, Martin and Topp (Stiffness)
• 1960 ----- Clough (“Finite Element”, plane problems)
• 1970s ----- Applications on mainframe computers
• 1980s ----- Microcomputers, pre- and postprocessors
• 1990s ----- Analysis of large structural systems
8/21/2019 Finite Element Methods Lectures_Lui.pdf
4/169
Lecture Notes: Introduction to Finite Element Method Chapter 1. Introduction
© 1998 Yijun Liu, University of Cincinnati 4
FEM in Structural Analysis
Procedures:
• Divide structure into pieces (elements with nodes)
• Describe the behavior of the physical quantities on eachelement
• Connect (assemble) the elements at the nodes to form anapproximate system of equations for the whole structure
• Solve the system of equations involving unknownquantities at the nodes (e.g., displacements)
• Calculate desired quantities (e.g., strains and stresses) atselected elements
Example:
8/21/2019 Finite Element Methods Lectures_Lui.pdf
5/169
Lecture Notes: Introduction to Finite Element Method Chapter 1. Introduction
© 1998 Yijun Liu, University of Cincinnati 5
Computer Implementations
• Preprocessing (build FE model, loads and constraints)
• FEA solver (assemble and solve the system of equations)
• Postprocessing (sort and display the results)
Available Commercial FEM Software Packages
• ANSYS (General purpose, PC and workstations)
• SDRC/I-DEAS (Complete CAD/CAM/CAE package)
• NASTRAN (General purpose FEA on mainframes)
• ABAQUS (Nonlinear and dynamic analyses)
• COSMOS (General purpose FEA)
• ALGOR (PC and workstations)
• PATRAN (Pre/Post Processor)
• HyperMesh (Pre/Post Processor)
• Dyna-3D (Crash/impact analysis)
• ...
8/21/2019 Finite Element Methods Lectures_Lui.pdf
6/169
Lecture Notes: Introduction to Finite Element Method Chapter 1. Introduction
© 1998 Yijun Liu, University of Cincinnati 6
Objectives of This FEM Course
• Understand the fundamental ideas of the FEM
• Know the behavior and usage of each type of elementscovered in this course
• Be able to prepare a suitable FE model for given problems
• Can interpret and evaluate the quality of the results (knowthe physics of the problems)
• Be aware of the limitations of the FEM (don’t misuse theFEM - a numerical tool)
8/21/2019 Finite Element Methods Lectures_Lui.pdf
7/169
Lecture Notes: Introduction to Finite Element Method Chapter 1. Introduction
© 1998 Yijun Liu, University of Cincinnati 7
I I . Review of Matrix Algebra
L inear System of Algebraic Equations a x a x a x b
a x a x a x b
a x a x a x b
n n
n n
n n nn n n
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
+ + + =
+ + + =
+ + + =
...
...
.......
...
(1)
where x1 , x2 , ..., xn are the unknowns.
In matrix form:
Ax b= (2)
where
[ ]
{ } { }
A
x b
= =
= =
= =
a
a a a
a a a
a a a
x
x
x
x
b
b
b
b
ij
n
n
n n nn
i
n
i
n
11 12 1
21 22 2
1 2
1
2
1
2
...
...
... ... ... ...
...
: :
(3)
A is called a n×n (square) matrix, and x and b are (column)
vectors of dimension n.
8/21/2019 Finite Element Methods Lectures_Lui.pdf
8/169
Lecture Notes: Introduction to Finite Element Method Chapter 1. Introduction
© 1998 Yijun Liu, University of Cincinnati 8
Row and Column Vectors
[ ]v w= =
v v v
w
w
w1 2 3
1
2
3
Matrix Addition and Subtraction
For two matrices A and B, both of the same size (m×n), the
addition and subtraction are defined by
C A B
D A B
= + = +
= − = −
with
with
c a b
d a b
ij ij ij
ij ij ij
Scalar M ul tipli cation
[ ]λ λA = a ij
Matrix Multiplication
For two matrices A (of size l×m) and B (of size m×n), the
product of AB is defined by
C AB= = ∑=
with c a bij ik
k
m
kj
1
where i = 1, 2, ..., l ; j = 1, 2, ..., n.
Note that, in general, AB BA≠ , but ( ) ( )AB C A BC=(associative).
8/21/2019 Finite Element Methods Lectures_Lui.pdf
9/169
Lecture Notes: Introduction to Finite Element Method Chapter 1. Introduction
© 1998 Yijun Liu, University of Cincinnati 9
Transpose of a Matri x
If A = [aij], then the transpose of A is
[ ]AT
jia=
Notice that ( )AB B AT T T = .
Symmetr ic Matrix
A square (n×n) matrix A is called symmetric, if
A A=T or a aij ji=
Uni t (I dentity) Matrix
I =
1 0 0
0 1 0
0 0 1
...
...
... ... ... ...
...
Note that AI = A, Ix = x.
Determinant of a Matri x
The determinant of square matrix A is a scalar number denoted by det A or |A|. For 2×2 and 3×3 matrices, their
determinants are given by
deta b
c d ad bc
= −
8/21/2019 Finite Element Methods Lectures_Lui.pdf
10/169
Lecture Notes: Introduction to Finite Element Method Chapter 1. Introduction
© 1998 Yijun Liu, University of Cincinnati 10
and
det
a a a
a a a
a a a
a a a a a a a a a
a a a a a a a a a
11 12 13
21 22 23
31 32 33
11 22 33 12 23 31 21 32 13
13 22 31 12 21 33 23 32 11
= + +
− − −
Singular Matrix
A square matrix A is singular if det A = 0, which indicates
problems in the systems (nonunique solutions, degeneracy, etc.)
Matr ix I nversion
For a square and nonsingular matrix A (detA ≠ 0), itsinverse A-1 is constructed in such a way that
AA A A I
− −
= =
1 1
The cofactor matrix C of matrix A is defined by
C M ij
i j
ij= − +( )1
where M ij is the determinant of the smaller matrix obtained by
eliminating the ith row and jth column of A.
Thus, the inverse of A can be determined by
AAC
− =11
det
T
We can show that ( )AB B A− − −=1 1 1.
8/21/2019 Finite Element Methods Lectures_Lui.pdf
11/169
Lecture Notes: Introduction to Finite Element Method Chapter 1. Introduction
© 1998 Yijun Liu, University of Cincinnati 11
Examples:
(1)a b
c d ad bc
d b
c a
=
−
−
−
−11
( )
Checking,
a b
c d
a b
c d ad bc
d b
c a
a b
c d
= −
−
−
=
−11 1 0
0 1( )
(2)
1 1 0
1 2 1
0 1 2
1
4 2 1
3 2 1
2 2 1
1 1 1
3 2 1
2 2 1
1 1 1
1−
− −
−
=− −
=
−
( )
T
Checking,
1 1 0
1 2 1
0 1 2
3 2 1
2 2 1
1 1 1
1 0 0
0 1 0
0 0 1
−
− −
−
=
If det A = 0 (i.e., A is singular), then A-1 does not exist!
The solution of the linear system of equations (Eq.(1)) can be
expressed as (assuming the coefficient matrix A is nonsingular)
x A b= −1
Thus, the main task in solving a linear system of equations is to
found the inverse of the coefficient matrix.
8/21/2019 Finite Element Methods Lectures_Lui.pdf
12/169
Lecture Notes: Introduction to Finite Element Method Chapter 1. Introduction
© 1998 Yijun Liu, University of Cincinnati 12
Solution Techniques for L inear Systems of Equations
• Gauss elimination methods
• Iterative methods
Positive Defini te Matr ix
A square (n×n) matrix A is said to be positive definite, if for
any nonzero vector x of dimension n,
x AxT >
0 Note that positive definite matrices are nonsingular.
Di ff erentiation and I ntegration of a Matr ix
Let
[ ]A( ) ( )t a t ij=then the differentiation is defined by
d
dt t
da t
dt
ijA( )
( )=
and the integration by
A( ) ( )t dt a t dt ij
=
∫ ∫
8/21/2019 Finite Element Methods Lectures_Lui.pdf
13/169
Lecture Notes: Introduction to Finite Element Method Chapter 1. Introduction
© 1998 Yijun Liu, University of Cincinnati 13
Types of F inite Elements
1-D (L ine) Element
(Spring, truss, beam, pipe, etc.)
2-D (Plane) Element
(Membrane, plate, shell, etc.)
3-D (Sol id) Element
(3-D fields - temperature, displacement, stress, flow velocity)
8/21/2019 Finite Element Methods Lectures_Lui.pdf
14/169
Lecture Notes: Introduction to Finite Element Method Chapter 1. Introduction
© 1998 Yijun Liu, University of Cincinnati 14
I I I . Spr ing Element
“Everything important is simple .”
One Spring Element
Two nodes: i, j
Nodal displacements: ui , u j (in, m, mm)
Nodal forces: f i , f j (lb, Newton)
Spring constant (stiffness): k (lb/in, N/m, N/mm)
Spring force-displacement relationship:
F k = ∆ with ∆ = −u u j i
k F = / ∆ (> 0) is the force needed to produce a unit stretch.
k
i
u juii j
∆
onlinear
Linear
k
8/21/2019 Finite Element Methods Lectures_Lui.pdf
15/169
Lecture Notes: Introduction to Finite Element Method Chapter 1. Introduction
© 1998 Yijun Liu, University of Cincinnati 15
We only consider linear problems in this introductory
course.
Consider the equilibrium of forces for the spring. At node i,
we have
f F k u u ku kui j i i j= − = − − = −( )
and at node j,
f F k u u ku ku j j i i j
= = − = − +( )
In matrix form,
k k
k k
u
u
f
f
i
j
i
j
−
−
=
or,
ku f =
where
k = (element) stiffness matrix
u = (element nodal) displacement vector
f = (element nodal) force vector
Note that k is symmetric. Is k singular or nonsingular? That is,
can we solve the equation? If not, why?
8/21/2019 Finite Element Methods Lectures_Lui.pdf
16/169
Lecture Notes: Introduction to Finite Element Method Chapter 1. Introduction
© 1998 Yijun Liu, University of Cincinnati 16
Spring System
For element 1,
k k
k k
u
u
f
f
1 1
1 1
1
2
1
1
21
−
−
=
element 2,
k k
k k
u
u
f
f
2 2
2 2
2
3
1
2
2
2
−
−
=
where f im
is the (internal) force acting on local node i of elementm (i = 1, 2).
Assemble the stiffness matrix for the whole system:
Consider the equilibrium of forces at node 1,
F f 1 1
1=
at node 2,
F f f 2 2
1
1
2= +
and node 3,
F f 3 2
2=
k 1
u1, 1
k 2
u2, F 2 u3, 3
1 2 3
8/21/2019 Finite Element Methods Lectures_Lui.pdf
17/169
Lecture Notes: Introduction to Finite Element Method Chapter 1. Introduction
© 1998 Yijun Liu, University of Cincinnati 17
That is,
F k u k u
F k u k k u k u
F k u k u
1 1 1 1 2
2 1 1 1 2 2 2 3
3 2 2 2 3
= −
= − + + −
= − +
( )
In matrix form,
k k
k k k k
k k
u
u
u
F
F
F
1 1
1 1 2 2
2 2
1
2
3
1
2
3
0
0
−
− + −
−
=
or
KU F=
K is the stiffness matrix (structure matrix) for the spring system.
An alternative way of assembling the whole stiffness matrix:
“Enlarging” the stiffness matrices for elements 1 and 2, we
have
k k
k k
u
u
u
f
f
1 1
1 1
1
2
3
1
1
2
1
0
0
0 0 0 0
−
−
=
0 0 0
0
0
0
2 2
2 2
1
2
3
1
2
2
2
k k
k k
u
u
u
f
f
−
−
=
8/21/2019 Finite Element Methods Lectures_Lui.pdf
18/169
Lecture Notes: Introduction to Finite Element Method Chapter 1. Introduction
© 1998 Yijun Liu, University of Cincinnati 18
Adding the two matrix equations ( superposition), we have
k k
k k k k k k
u
uu
f
f f f
1 1
1 1 2 2
2 2
1
2
3
1
1
2
1
1
2
2
2
0
0
−
− + −
−
= +
This is the same equation we derived by using the force
equilibrium concept.
Boundary and load conditions:
Assuming, u F F P 1 2 30= = =and
we have
k k
k k k k
k k
u
u
F
P
P
1 1
1 1 2 2
2 2
2
3
10
0
0−
− + −
−
=
which reduces to
k k k
k k
u
u
P
P
1 2 2
2 2
2
3
+ −
−
=
and
F k u1 1 2= −
Unknowns are
U =
u
u
2
3
and the reaction force F 1 (if desired).
8/21/2019 Finite Element Methods Lectures_Lui.pdf
19/169
Lecture Notes: Introduction to Finite Element Method Chapter 1. Introduction
© 1998 Yijun Liu, University of Cincinnati 19
Solving the equations, we obtain the displacements
u
u
P k
P k P k
2
3
1
1 2
2
2
=+
/
/ /
and the reaction force
F P 1 2= −
Checking the Resul ts
• Deformed shape of the structure
• Balance of the external forces
• Order of magnitudes of the numbers
Notes About the Spring Elements
• Suitable for stiffness analysis
• Not suitable for stress analysis of the spring itself
• Can have spring elements with stiffness in the lateraldirection, spring elements for torsion, etc.
8/21/2019 Finite Element Methods Lectures_Lui.pdf
20/169
Lecture Notes: Introduction to Finite Element Method Chapter 1. Introduction
© 1998 Yijun Liu, University of Cincinnati 20
Example 1.1
Given: For the spring system shown above,
k k k
P u
1 2 3
4 0
= = =
= = =
100 N / mm, 200 N / mm, 100 N / mm
500 N, u1
Find : (a) the global stiffness matrix
(b) displacements of nodes 2 and 3
(c) the reaction forces at nodes 1 and 4
(d) the force in the spring 2
Solution :
(a) The element stiffness matrices are
k 1
100 100
100 100=
−
−
(N/mm) (1)
k 2 200 200
200 200= −−
(N/mm) (2)
k 3
100 100
100 100=
−
−
(N/mm) (3)
k 1 k 2
1 2 3
k 3
4
8/21/2019 Finite Element Methods Lectures_Lui.pdf
21/169
Lecture Notes: Introduction to Finite Element Method Chapter 1. Introduction
© 1998 Yijun Liu, University of Cincinnati 21
Applying the superposition concept, we obtain the global stiffness
matrix for the spring system as
u u u u1 2 3 4
100 100 0 0
100 100 200 200 0
0 200 200 100 100
0 0 100 100
K =
−
− + −
− + −
−
or
K =
−
− −
− −
−
100 100 0 0
100 300 200 0
0 200 300 100
0 0 100 100
which is symmetric and banded .
Equilibrium (FE) equation for the whole system is
100 100 0 0
100 300 200 0
0 200 300 100
0 0 100 100
0
1
2
3
4
1
4
−
− −
− −
−
=
u
u
u
u
F
P
F
(4)
(b) Applying the BC (u u1 4
0= = ) in Eq(4), or deleting the 1st and
4th rows and columns, we have
8/21/2019 Finite Element Methods Lectures_Lui.pdf
22/169
Lecture Notes: Introduction to Finite Element Method Chapter 1. Introduction
© 1998 Yijun Liu, University of Cincinnati 22
300 200
200 300
02
3
−
−
=
u
u P (5)
Solving Eq.(5), we obtainu
u
P
P
2
3
250
3 500
2
3
=
=
/
/( )mm (6)
(c) From the 1st and 4th equations in (4), we get the reaction forces
F u1 2100 200= − = − (N)
F u4 3
100 300= − = − ( ) N
(d) The FE equation for spring (element) 2 is
200 200
200 200
−
−
=
u
u
f
f
i
j
i
j
Here i = 2, j = 3 for element 2. Thus we can calculate the spring
force as
[ ]
[ ]
F f f u
u j i
= = − = −
= −
=
200 200
200 2002
3
200
2
3
(N)
Check the results!
8/21/2019 Finite Element Methods Lectures_Lui.pdf
23/169
Lecture Notes: Introduction to Finite Element Method Chapter 1. Introduction
© 1998 Yijun Liu, University of Cincinnati 23
Example 1.2
Problem : For the spring system with arbitrarily numbered nodes
and elements, as shown above, find the global stiffness
matrix.
Solution :
First we construct the following
which specifies the global node numbers corresponding to the
local node numbers for each element.
Then we can write the element stiffness matrices as follows
k 1
k 242
3
k 3
5
2
1k 4
1
1
2 3
4
Element Connectivity Table
Element Node i (1) Node j (2)
1 4 2
2 2 3
3 3 5
4 2 1
8/21/2019 Finite Element Methods Lectures_Lui.pdf
24/169
Lecture Notes: Introduction to Finite Element Method Chapter 1. Introduction
© 1998 Yijun Liu, University of Cincinnati 24
u u
k k
k k
4 2
1
1 1
1 1
k = −
−
u u
k k
k k
2 3
2
2 2
2 2
k = −
−
u u
k k
k k
3 5
3
3 3
3 3
k = −
−
u u
k k
k k
2 1
4
4 4
4 4
k = −
−
Finally, applying the superposition method, we obtain the globalstiffness matrix as follows
u u u u u
k k
k k k k k k
k k k k
k k
k k
1 2 3 4 5
4 4
4 1 2 4 2 1
2 2 3 3
1 1
3 3
0 0 0
0
0 0
0 0 0
0 0 0
K =
−
− + + − −
− + −
−−
The matrix is symmetric, banded , but singular .
8/21/2019 Finite Element Methods Lectures_Lui.pdf
25/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 25
Chapter 2. Bar and Beam Elements.
L inear Static Analysis
I . Linear Static Analysis
Most structural analysis problems can be treated as linear
static problems, based on the following assumptions
1. Small deformations (loading pattern is not changed dueto the deformed shape)
2. Elastic materials (no plasticity or failures)
3. Static loads (the load is applied to the structure in a slow
or steady fashion)
Linear analysis can provide most of the information about
the behavior of a structure, and can be a good approximation for
many analyses. It is also the bases of nonlinear analysis in most
of the cases.
8/21/2019 Finite Element Methods Lectures_Lui.pdf
26/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 26
I I . Bar Element
Consider a uniform prismatic bar:
L length
A cross-sectional area
E elastic modulus
u u x= ( ) displacement
ε ε= ( ) x strain
σ σ= ( ) x stress
Strain-displacement relation:
ε =du
dx
(1)
Stress-strain relation:
σ ε= E (2)
i i j
ui u j
,E
8/21/2019 Finite Element Methods Lectures_Lui.pdf
27/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 27
Stif fness Matr ix --- D irect Method
Assuming that the displacement u is varying linearly along
the axis of the bar, i.e.,
u x x
Lu
x
Lu
i j( ) = −
+1 (3)
we have
ε = −
=u u
L L
j i ∆(∆ = elongation) (4)
σ ε= = E E L
∆(5)
We also have
σ = F A
( F = force in bar) (6)
Thus, (5) and (6) lead to
F EA
Lk = =∆ ∆ (7)
where k EA
L= is the stiffness of the bar.
The bar is acting like a spring in this case and we concludethat element stiffness matrix is
8/21/2019 Finite Element Methods Lectures_Lui.pdf
28/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 28
k = −
−
=
−
−
k k
k k
EA
L
EA
L EA
L
EA
L
or
k = −−
EA
L
1 1
1 1(8)
This can be verified by considering the equilibrium of the forces
at the two nodes. Element equilibrium equation is
EA
L
u
u
f
f
i
j
i
j
1 1
1 1
−−
=
(9)
Degree of Freedom (dof)
Number of components of the displacement vector at a
node.
For 1-D bar element: one dof at each node.
Physical Meaning of the Coefficients in k
The jth column of k (here j = 1 or 2) represents the forces
applied to the bar to maintain a deformed shape with unit
displacement at node j and zero displacement at the other node.
8/21/2019 Finite Element Methods Lectures_Lui.pdf
29/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 29
Stif fness Matrix --- A F ormal Approach
We derive the same stiffness matrix for the bar using a
formal approach which can be applied to many other more
complicated situations.
Define two linear shape functions as follows
N N i j( ) , ( )ξ ξ ξ ξ= − =1 (10)
where
ξ ξ= ≤ ≤ x
L , 0 1 (11)
From (3) we can write the displacement as
u x u N u N ui i j j( ) ( ) ( ) ( )= = +ξ ξ ξ
or
[ ]u N N uui j
i
j
=
= Nu (12)
Strain is given by (1) and (12) as
ε = =
=du
dx
d
dxN u Bu (13)
where B is the element strain-displacement matrix, which is
[ ] [ ]B = = •d
dx N N
d
d N N
d
dxi j i j( ) ( ) ( ) ( )ξ ξ
ξ ξ ξ
ξ
i.e., [ ]B = −1 1/ / L L (14)
8/21/2019 Finite Element Methods Lectures_Lui.pdf
30/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 30
Stress can be written as
σ ε= = E E Bu (15)
Consider the strain energy stored in the bar
( )
( )
U dV E dV
E dV
V V
V
= =
=
∫ ∫
∫
1
2
1
2
1
2
σ εT T T
T T
u B Bu
u B B u
(16)
where (13) and (15) have been used.
The work done by the two nodal forces is
W f u f ui i j j
= + =12
1
2
1
2u f
T (17)
For conservative system, we state that
U W = (18)
which gives
( )1
2
1
2u B B u u f
T T T E dV
V
∫
=
We can conclude that
( )B B u f T E dV V
∫
=
or
8/21/2019 Finite Element Methods Lectures_Lui.pdf
31/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 31
ku f = (19)
where
( )k B BT= ∫ E dV V (20)
is the element stiffness matrix.
Expression (20) is a general result which can be used for
the construction of other types of elements. This expression can
also be derived using other more rigorous approaches, such as
the Principle of Minimum Potential Energy, or the Galerkin’s Method .
Now, we evaluate (20) for the bar element by using (14)
[ ]k = −
− = −−
∫
1
11 1
1 1
1 10
/
// /
L
L E L L Adx
EA
L
L
which is the same as we derived using the direct method.
Note that from (16) and (20), the strain energy in the
element can be written as
U =1
2u ku
T (21)
8/21/2019 Finite Element Methods Lectures_Lui.pdf
32/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 32
Example 2.1
Problem: Find the stresses in the two bar assembly which is
loaded with force P , and constrained at the two ends,
as shown in the figure.
Solution: Use two 1-D bar elements.
Element 1,
u u
EA
L
1 2
1
2 1 1
1 1k =
−−
Element 2,
u u
EA
L
2 3
2
1 1
1 1k =
−−
Imagine a frictionless pin at node 2, which connects the twoelements. We can assemble the global FE equation as follows,
EA
L
u
u
u
F
F
F
2 2 0
2 3 1
0 1 1
1
2
3
1
2
3
−− −
−
=
1
2 A,E
2 3
,E 1 2
8/21/2019 Finite Element Methods Lectures_Lui.pdf
33/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 33
Load and boundary conditions (BC) are,
u u F P 1 3 2
0= = =,
FE equation becomes,
EA
Lu
F
P
F
2 2 0
2 3 1
0 1 1
0
0
2
1
3
−− −
−
=
Deleting the 1st row and column, and the 3rd row and column,
we obtain,
[ ]{ } { } EA
Lu P 3
2 =
Thus,
u PL
EA2
3=
and
u
u
u
PL
EA
1
2
3
3
0
1
0
=
Stress in element 1 is
[ ]σ ε1 1 1 11
2
2 1
1 1
30
3
= = = −
= −
= −
=
E E E L Lu
u
E u u
L
E
L
PL
EA
P
A
B u / /
8/21/2019 Finite Element Methods Lectures_Lui.pdf
34/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 34
Similarly, stress in element 2 is
[ ]σ ε2 2 2 22
3
3 2
1 1
03 3
= = = −
=
−= −
= −
E E E L Lu
u
E u u
L
E
L
PL
EA
P
A
B u / /
which indicates that bar 2 is in compression.
Check the results!
Notes:
• In this case, the calculated stresses in elements 1 and 2are exact within the linear theory for 1-D bar structures.
It will not help if we further divide element 1 or 2 into
smaller finite elements.
• For tapered bars, averaged values of the cross-sectionalareas should be used for the elements.
• We need to find the displacements first in order to findthe stresses, since we are using the displacement based
FEM .
8/21/2019 Finite Element Methods Lectures_Lui.pdf
35/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 35
Example 2.2
Problem: Determine the support reaction forces at the two ends
of the bar shown above, given the following,
P E
A L =
= × = ×
= =
6 0 10 2 0 10
250 150
4 4
2
. , . ,
,
N N / mm
mm mm, 1.2 mm
2
∆
Solution:
We first check to see if or not the contact of the bar with
the wall on the right will occur. To do this, we imagine the wallon the right is removed and calculate the displacement at the
right end,
∆ ∆0
4
4
6 0 10 150
2 0 10 25018 12= = ×
× = > = PL
EA
( . )( )
( . )( ). .mm mm
Thus, contact occurs.
The global FE equation is found to be,
EA
L
u
u
u
F
F
F
1 1 0
1 2 1
0 1 1
1
2
3
1
2
3
−− −
−
=
1
,E
2 3
1 2
∆
8/21/2019 Finite Element Methods Lectures_Lui.pdf
36/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 36
The load and boundary conditions are,
F P
u u
2
4
1 3
6 0 10
0 12
= = ×
= = =
.
, .
N
mm∆FE equation becomes,
EA
Lu
F
P
F
1 1 0
1 2 1
0 1 1
0
2
1
3
−− −
−
=
∆
The 2nd equation gives,
[ ] { } EA
L
u P 2 1
2−
=∆
that is,
[ ]{ } EA
Lu P
EA
L2
2 = +
∆
Solving this, we obtain
u PL
EA2
1
215= +
=∆ . mm
and
u
u
u
1
2
3
0
15
12
=
.
.
( )mm
8/21/2019 Finite Element Methods Lectures_Lui.pdf
37/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 37
To calculate the support reaction forces, we apply the 1st
and 3rd equations in the global FE equation.
The 1st equation gives,
[ ] ( ) F EA
L
u
u
u
EA
Lu
1
1
2
3
2
41 1 0 50 10= −
= − = − ×. N
and the 3rd equation gives,
[ ] ( ) F EA
L
uu
u
EA
Lu u
3
1
2
3
2 3
4
0 1 1
10 10
= −
= − +
= − ×. N
Check the results.!
8/21/2019 Finite Element Methods Lectures_Lui.pdf
38/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 38
Distr ibuted Load
Uniformly distributed axial load q (N/mm, N/m, lb/in) can
be converted to two equivalent nodal forces of magnitude qL/ 2.
We verify this by considering the work done by the load q,
[ ]
[ ]
[ ]
W uqdx u q Ld qL
u d
qL N N
u
ud
qLd
u
u
qL qL u
u
u uqL
qL
q
L
i j
i
j
i
j
i
j
i j
= = =
=
= −
=
=
∫ ∫ ∫ ∫
∫
1
2
1
2 2
2
21
1
2 2 2
1
2
2
2
0 0
1
0
1
0
1
0
1
( ) ( ) ( )
( ) ( )
/
/
ξ ξ ξ ξ
ξ ξ ξ
ξ ξ ξ
i
q
qL/2
i
qL/2
8/21/2019 Finite Element Methods Lectures_Lui.pdf
39/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 39
that is,
W qL
qLq
T
q q= =
1
2
2
2u f f with
/
/(22)
Thus, from the U=W concept for the element, we have
1
2
1
2
1
2u ku u f u f
T T T
q= + (23)
which yields
ku f f = +q (24)
The new nodal force vector is
f f + = +
+
q
i
j
f qL
f qL
/
/
2
2(25)
In an assembly of bars,
1 3
q
qL/2
1 3
qL/2
2
2
qL
8/21/2019 Finite Element Methods Lectures_Lui.pdf
40/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 40
Bar Elements in 2-D and 3-D Space
2-D Case
Local Global
x, y X, Y
u vi i' ', u vi i,
1 dof at node 2 dof’s at node
Note: Lateral displacement vi’ does not contribute to the stretch
of the bar, within the linear theory.
Transformation
[ ]
[ ]
u u v l m
u
v
v u v m l u
v
i i i
i
i
i i i
i
i
'
'
cos sin
sin cos
= + =
= − + = −
θ θ
θ θ
where l m= =cos , sinθ θ .
i
ui’ Y
θ
uivi
8/21/2019 Finite Element Methods Lectures_Lui.pdf
41/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 41
In matrix form,
u
v
l m
m l
u
v
i
i
i
i
'
'
=−
(26)
or,
u Tui i' ~=
where the transformation matrix
~T
= −
l m
m l (27)
is orthogonal , that is,~ ~T T
− =1 T .
For the two nodes of the bar element, we have
u
v
u
v
l m
m l
l m
m l
u
v
u
v
i
i
j
j
i
i
j
j
'
'
'
'
= −
−
0 0
0 0
0 0
0 0
(28)
or,
u Tu' = with T
T 0
0 T=
~
~ (29)
The nodal forces are transformed in the same way,
f Tf ' = (30)
8/21/2019 Finite Element Methods Lectures_Lui.pdf
42/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 42
Sti f fness Matri x in the 2-D Space
In the local coordinate system, we have
EA L
uu
f f
i
j
i
j
1 11 1
−−
=
'
'
'
'
Augmenting this equation, we write
EA
L
u
v
u
v
f
f
i
i
j
j
i
j
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0
0
−
−
=
'
'
'
'
'
'
or,
k u f ' ' '=
Using transformations given in (29) and (30), we obtain
k Tu Tf ' =
Multiplying both sides by TT and noticing that TT T = I, we
obtain
T k Tu f T ' = (31)
Thus, the element stiffness matrix k in the global coordinate
system is
k T k T= T ' (32)
which is a 4×4 symmetric matrix.
8/21/2019 Finite Element Methods Lectures_Lui.pdf
43/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 43
Explicit form,
u v u v
EA
L
l lm l lmlm m lm m
l lm l lm
lm m lm m
i i j j
k =− −− −
− −
− −
2 2
2 2
2 2
2 2
(33)
Calculation of the directional cosines l and m:
l X X L
m Y Y L
j i j i= =
−= =
−cos , sinθ θ (34)
The structure stiffness matrix is assembled by using the element
stiffness matrices in the usual way as in the 1-D case.
Element Stress
σ ε= =
= −
E E u
u E
L L
l m
l m
u
v
u
v
i
j
i
i
j
j
B
'
'
1 1 0 0
0 0
That is,
[ ]σ = − −
E
Ll m l m
u
v
u
v
i
i
j
j
(35)
8/21/2019 Finite Element Methods Lectures_Lui.pdf
44/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 44
Example 2.3
A simple plane truss is made
of two identical bars (with E, A, and
L), and loaded as shown in thefigure. Find
1) displacement of node 2;
2) stress in each bar.
Solution:
This simple structure is usedhere to demonstrate the assembly
and solution process using the bar element in 2-D space.
In local coordinate systems, we have
k k 1 2
1 1
1 1
' '= −
−
= EA
L
These two matrices cannot be assembled together, because they
are in different coordinate systems. We need to convert them to
global coordinate system OXY.
Element 1:
θ
= = =45
2
2
o
l m,
Using formula (32) or (33), we obtain the stiffness matrix in the
global system
Y 1
2
45o
45o
3
2
1
1
2
8/21/2019 Finite Element Methods Lectures_Lui.pdf
45/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 45
u v u v
EA
L
T
1 1 2 2
1 1 1 12
1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1
k T k T= =
− −
− −
− −
− −
'
Element 2:
θ = = − =1352
2
2
2
o l m, ,
We have,
u v u v
EA
L
T
2 2 3 3
2 2 2 22
1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1
k T k T= =
− −
− −
− −
− −
'
Assemble the structure FE equation,
u v u v u v
EA
L
u
v
u
v
u
v
F
F
F
F
F
F
X
Y
X
Y
X
Y
1 1 2 2 3 3
1
1
2
2
3
3
1
1
2
2
3
3
2
1 1 1 1 0 0
1 1 1 1 0 0
1 1 2 0 1 1
1 1 0 2 1 1
0 0 1 1 1 1
0 0 1 1 1 1
− −
− −
− − −− − −
− −
− −
=
8/21/2019 Finite Element Methods Lectures_Lui.pdf
46/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 46
Load and boundary conditions (BC):
u v u v F P F P X Y 1 1 3 3 2 1 2 2
0= = = = = =, ,
Condensed FE equation,
EA
L
u
v
P
P 2
2 0
0 2
2
2
1
2
=
Solving this, we obtain the displacement of node 2,
u
v
L
EA
P
P
2
2
1
2
=
Using formula (35), we calculate the stresses in the two bars,
[ ] ( )σ11
2
1 2
2
21 1 1 1
0
0 2
2= − −
= + E
L
L
EA P
P
A P P
[ ] ( )σ2
1
2
1 2
2
21 1 1 1
0
0
2
2= − −
= − E
L
L
EA
P
P
A P P
Check the results :
Look for the equilibrium conditions, symmetry,
antisymmetry, etc.
8/21/2019 Finite Element Methods Lectures_Lui.pdf
47/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 47
Example 2.4 ( Multipoint Constraint )
For the plane truss shown above,
P L m E GPa
A m
A m
= = =
= ×
= ×
−
−
1000 1 210
6 0 10
6 2 10
4 2
4 2
kN,
for elements 1 and 2,
for element 3.
, ,
.
Determine the displacements and reaction forces.
Solution:
We have an inclined roller at node 3, which needs special
attention in the FE solution. We first assemble the global FE
equation for the truss.
Element 1:
θ = = =90 0 1o l m, ,
Y
45o
3
2
1
3
2
1
’
’
8/21/2019 Finite Element Methods Lectures_Lui.pdf
48/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 48
u v u v1 1 2 2
1
9 4210 10 6 0 10
1
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
k =
× × −
−
−( )( . )( ) N / m
Element 2:
θ = = =0 1 0o l m, ,
u v u v2 2 3 3
2
9 4210 10 6 0 10
1
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
k = × ×
−
−
−( )( . )( ) N / m
Element 3:
θ = = =451
2
1
2
o l m, ,
u v u v1 1 3 3
3
9 4210 10 6 2 10
2
05 0 5 05 05
05 0 5 05 05
0 5 05 0 5 0 5
0 5 05 0 5 0 5
k = × ×
− −
− −
− −− −
−( )( )
. . . .
. . . .
. . . .
. . . .
( ) N / m
8/21/2019 Finite Element Methods Lectures_Lui.pdf
49/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 49
The global FE equation is,
1260 10
0 5 05 0 0 05 05
15 0 1 05 05
1 0 1 0
1 0 0
15 05
05
5
1
1
2
2
3
3
1
1
2
2
3
3
×
− −
− − −
−
=
. . . .
. . .
. .
.Sym.
u
v
u
v
u
v
F
F
F
F
F
F
X
Y
X
Y
X
Y
Load and boundary conditions (BC):
u v v v
F P F X x
1 1 2 3
2 3
0 0
0
= = = =
= =
, ,
, .
'
'
and
From the transformation relation and the BC, we have
vu
vu v
3
3
3
3 3
2
2
2
2
2
20' ( ) ,= −
= − + =
that is,
u v3 3
0− =
This is a multipoint constraint (MPC).
Similarly, we have a relation for the force at node 3,
F F
F F F
x
X
Y
X Y 3
3
3
3 3
2
2
2
2
2
20
'( ) ,=
= + =
that is,
F F X Y 3 3
0+ =
8/21/2019 Finite Element Methods Lectures_Lui.pdf
50/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 50
Applying the load and BC’s in the structure FE equation by
‘deleting’ 1st, 2nd and 4th rows and columns, we have
1260 10
1 1 0
1 15 05
0 05 05
5
2
3
3
3
3
×
−
−
=
. .
. .
u
u
v
P
F
F
X
Y
Further, from the MPC and the force relation at node 3, the
equation becomes,
1260 10
1 1 0
1 15 05
0 05 05
5
2
3
3
3
3
×
−
−
=−
. .
. .
u
u
u
P
F
F
X
X
which is
1260 10
1 1
1 2
0 1
5 2
3
3
3
×
−
−
=
−
u
u
P
F
F
X
X
The 3rd equation yields,
F u X 3
5
31260 10= − ×
Substituting this into the 2nd equation and rearranging, we have
1260 101 1
1 3 05 2
3×
−
−
=
u
u
P
Solving this, we obtain the displacements,
8/21/2019 Finite Element Methods Lectures_Lui.pdf
51/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 51
u
u
P
P
2
3
5
1
2520 10
3 0 01191
0 003968
=×
=
.
.( )m
From the global FE equation, we can calculate the reactionforces,
F
F
F
F
F
u
u
v
X
Y
Y
X
Y
1
1
2
3
3
5
2
3
3
1260 10
0 0 5 05
0 0 5 05
0 0 0
1 15 05
0 05 05
500
500
0 0
500
500
= ×
− −
− −
−
=
−
−
−
. .
. .
. .
. .
. ( )kN
Check the results!
A general multipoint constraint (MPC) can be described as,
A u j j
j
=∑ 0
where A j’s are constants and u j’s are nodal displacement
components. In the FE software, such as MSC/NASTRAN ,
users only need to specify this relation to the software. The
software will take care of the solution.
Penalty Approach for Handli ng BC’ s and MPC’ s
8/21/2019 Finite Element Methods Lectures_Lui.pdf
52/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 52
3-D Case
Local Global
x, y, z X, Y, Z
u v wi i i' ' ', , u v wi i i, ,
1 dof at node 3 dof’s at node
Element stiffness matrices are calculated in the localcoordinate systems and then transformed into the global
coordinate system ( X, Y, Z ) where they are assembled.
FEA software packages will do this transformation
automatically.
Input data for bar elements:
• ( X, Y, Z ) for each node
• E and A for each element
i
Y
Z
8/21/2019 Finite Element Methods Lectures_Lui.pdf
53/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 53
I I I . Beam Element
Simple Plane Beam Element
L length
I moment of inertia of the cross-sectional area
E elastic modulus
v v x= ( ) deflection (lateral displacement) of theneutral axis
θ =dv
dx rotation about the z-axis
F F x= ( ) shear force
M M x= ( ) moment about z-axis
Elementary Beam Theory:
EI d v
dx M x
2
2 = ( ) (36)
σ = − My
I (37)
i
v j , F j
,I θi , M i θ j , M j
vi , F i
8/21/2019 Finite Element Methods Lectures_Lui.pdf
54/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 54
Di rect Method
Using the results from elementary beam theory to compute
each column of the stiffness matrix.
(Fig. 2.3-1. on Page 21 of Cook’s Book)
Element stiffness equation (local node: i, j or 1, 2):
v v
EI
L
L L
L L L L
L L
L L L L
v
v
F
M
F
M
i i j j
i
i
j
j
i
i
j
j
θ θ
θ
θ
3
2 2
2 2
12 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4
−−
− − −
−
=
(38)
8/21/2019 Finite Element Methods Lectures_Lui.pdf
55/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 55
Formal Approach
Apply the formula,
k B B= ∫ T L
EI dx
0
(39)
To derive this, we introduce the shape functions,
N x x L x L
N x x x L x L
N x x L x L
N x x L x L
1
2 2 3 3
2
2 3 2
3
2 2 3 3
4
2 3 2
1 3 2
2
3 2
( ) / /
( ) / /
( ) / /
( ) / /
= − +
= − +
= −
= − +
(40)
Then, we can represent the deflection as,
[ ]
v x
N x N x N x N x
v
v
i
i
j
j
( )
( ) ( ) ( ) ( )
=
=
Nu
1 2 3 4
θ
θ
(41)
which is a cubic function. Notice that,
N N
N N L N x
1 3
2 3 4
1+ =+ + =
which implies that the rigid body motion is represented by the
assumed deformed shape of the beam.
8/21/2019 Finite Element Methods Lectures_Lui.pdf
56/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 56
Curvature of the beam is,
d v
dx
d
dx
2
2
2
2= =Nu Bu (42)
where the strain-displacement matrix B is given by,
[ ]B N= =
= − + − + − − +
d
dx N x N x N x N x
L
x
L L
x
L L
x
L L
x
L
2
2 1 2 3 4
2 3 2 2 3 2
6 12 4 6 6 12 2 6
" " " "( ) ( ) ( ) ( )
(43)
Strain energy stored in the beam element is
( ) ( )
U dV My
I E
My
I dAdx
M
EI
Mdxd v
dx
EI d v
dx
dx
EI dx
EI dx
T
V A
L T
T
L T L
T
L
T T
L
= = −
−
= =
=
=
∫ ∫ ∫
∫ ∫ ∫
∫
1
2
1
2
1
1
2
1 1
2
1
2
1
2
0
0
2
2
2
2
0
0
0
σ ε
Bu Bu
u B B u
We conclude that the stiffness matrix for the simple beam
element is
k B B= ∫ T L
EI dx
0
8/21/2019 Finite Element Methods Lectures_Lui.pdf
57/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 57
Applying the result in (43) and carrying out the integration, we
arrive at the same stiffness matrix as given in (38).
Combining the axial stiffness (bar element), we obtain the
stiffness matrix of a general 2-D beam element ,
u v u v
EA
L
EA
L EI
L
EI
L
EI
L
EI
L EI
L
EI
L
EI
L
EI
L EA
L
EA
L EI
L
EI
L
EI
L
EI
L
EI L
EI L
EI L
EI L
i i i j j jθ θ
k =
−
−
−
−
− − −
−
0 0 0 0
012 6
012 6
06 4
06 2
0 0 0 0
012 6
012 6
0 6 2 0 6 4
3 2 3 2
2 2
3 2 3 2
2 2
3-D Beam Element
The element stiffness matrix is formed in the local (2-D)
coordinate system first and then transformed into the global (3-D) coordinate system to be assembled.
(Fig. 2.3-2. On Page 24)
8/21/2019 Finite Element Methods Lectures_Lui.pdf
58/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 58
Example 2.5
Given: The beam shown above is clamped at the two ends and
acted upon by the force P and moment M in the mid-span.
Find : The deflection and rotation at the center node and the
reaction forces and moments at the two ends.
Solution: Element stiffness matrices are,
v v
EI
L
L L
L L L L
L L
L L L L
1 1 2 2
1 3
2 2
2 2
12 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4
θ θ
k =
−
−
− − −
−
v v
EI
L
L L
L L L L
L L
L L L L
2 2 3 3
2 3
2 2
2 2
12 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4
θ θ
k =
−−
− − −
−
1 2 ,I
Y
3
1 2
8/21/2019 Finite Element Methods Lectures_Lui.pdf
59/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 59
Global FE equation is,
v v v
EI
L
L L L L L L
L L
L L L L L
L L
L L L L
v
v
v
F M
F
M
F
M
Y
Y
Y
1 1 2 2 3 3
3
2 2
2 2 2
2 2
1
1
2
2
3
3
1
1
2
2
3
3
12 6 12 6 0 06 4 6 2 0 0
12 6 24 0 12 6
6 2 0 8 6 2
0 0 12 6 12 6
0 0 6 2 6 4
θ θ θ
θ
θ
θ
−
−
− − −
−
− − −
−
=
Loads and constraints (BC’s) are,
F P M M
v v
Y 2 2
1 3 1 30
= − =
= = = =
, ,
θ θ
Reduced FE equation,
EI
L L
v P
M 3 22
2
24 0
0 8
=
−
θ
Solving this we obtain,
v L
EI
PL
M
2
2
2
24 3θ
= −
From global FE equation, we obtain the reaction forces andmoments,
8/21/2019 Finite Element Methods Lectures_Lui.pdf
60/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 60
F
M
F
M
EI
L
L
L L
L
L L
v
P M L
PL M
P M L
PL M
Y
Y
1
1
3
3
3
2
2
2
2
12 6
6 2
12 6
6 2
1
4
2 3
2 3
=
−
−
− −
=
+
+
−
− +
θ
/
/
Stresses in the beam at the two ends can be calculated using the
formula,
σ σ= = − x My
I
Note that the FE solution is exact according to the simple beamtheory, since no distributed load is present between the nodes.
Recall that,
EI d v
dx M x
2
2 = ( )
and
dM
dxV V
dV
dxq q
=
=
(
(
- shear force in the beam)
- distributed load on the beam)
Thus,
EI d v
dxq x
4
4 = ( )
If q( x)=0, then exact solution for the deflection v is a cubic
function of x, which is what described by our shape functions.
8/21/2019 Finite Element Methods Lectures_Lui.pdf
61/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 61
Equivalent Nodal L oads of Distr ibuted Transverse Load
This can be verified by considering the work done by the
distributed load q.
i
q
qL/2
i
qL/2
qL2 /12qL2 /12
q
qL qL/2
qL2 /12
8/21/2019 Finite Element Methods Lectures_Lui.pdf
62/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 62
Example 2.6
Given: A cantilever beam with distributed lateral load p as
shown above.
Find : The deflection and rotation at the right end, the
reaction force and moment at the left end.
Solution: The work-equivalent nodal loads are shown below,
where
f pL m pL= =/ , /2 122
Applying the FE equation, we have
1 2 ,I
1 2 ,I
m
8/21/2019 Finite Element Methods Lectures_Lui.pdf
63/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 63
EI
L
L L
L L L L
L L
L L L L
v
v
F
M
F
M
Y
Y
3
2 2
2 2
1
1
2
2
1
1
2
2
12 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4
−
−
− − −
−
=
θ
θ
Load and constraints (BC’s) are,
F f M m
v
Y 2 2
1 10
= − =
= =
,
θ
Reduced equation is,
EI
L
L
L L
v f
m3 22
2
12 6
6 4
−
−
= −
θ
Solving this, we obtain,
v L
EI
L f Lm
Lf m
pL EI
pL EI
2
2
2 4
3
6
2 3
3 6
8
6θ
= − +
− +
= −
−
/
/
(A)
These nodal values are the same as the exact solution.
Note that the deflection v( x) (for 0
8/21/2019 Finite Element Methods Lectures_Lui.pdf
64/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 64
The errors in (B) will decrease if more elements are used. The
equivalent moment m is often ignored in the FEM applications.
The FE solutions still converge as more elements are applied.
From the FE equation, we can calculate the reaction forceand moment as,
F
M
L
EI
L
L L
v pL
pL
Y 1
1
3
2
2
2
2
12 6
6 2
2
5 12
= −
−
=
θ
/
/
where the result in (A) is used. This force vector gives the total
effective nodal forces which include the equivalent nodal forcesfor the distributed lateral load p given by,
−
−
pL
pL
/
/
2
122
The correct reaction forces can be obtained as follows,
F
M
pL
pL
pL
pL
pL
pLY 1
1
2 2 2
2
5 12
2
12 2
=
−
−
−
=
/
/
/
/ /
Check the results!
8/21/2019 Finite Element Methods Lectures_Lui.pdf
65/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 65
Example 2.7
Given: P = 50 kN, k = 200 kN/m, L = 3 m,
E = 210 GPa, I = 2×10-4 m4.
Find : Deflections, rotations and reaction forces.
Solution:
The beam has a roller (or hinge) support at node 2 and a
spring support at node 3. We use two beam elements and one
spring element to solve this problem.
The spring stiffness matrix is given by,
v v
k k
k k s
3 4
k = −
−
Adding this stiffness matrix to the global FE equation (see
Example 2.5), we have
12
,I
Y
3
1 2
k
4
8/21/2019 Finite Element Methods Lectures_Lui.pdf
66/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 66
v v v v
EI
L
L L
L L L
L
L L L
k L
L
k
Symmetry k
v
v
v
v
F
M
F
M
F
M
F
Y
Y
Y
Y
1 1 2 2 3 3 4
3
2 2
2 2
2
1
1
2
2
3
3
4
1
1
2
2
3
3
4
12 6 12 6 0 0
4 6 2 0 0
24 0 12 6
8 6 2
12 6
4
0
0
0
0
0
θ θ θ
θ
θ
θ
−
−
−
−
+ − −
=
' '
'
in which
k L
EI k '=
3
is used to simply the notation.
We now apply the boundary conditions,
v v v M M F P
Y
1 1 2 4
2 3 3
00
= = = == = = −θ ,
,
‘Deleting’ the first three and seventh equations (rows and
columns), we have the following reduced equation,
EI
L
L L L
L k L L L L
v P 3
2 2
2 2
2
3
3
8 6 2
6 12 62 6 4
0
0
−
− + −
−
= −
'
θ
θ
Solving this equation, we obtain the deflection and rotations at
node 2 and node 3,
8/21/2019 Finite Element Methods Lectures_Lui.pdf
67/169
Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements
© 1998 Yijun Liu, University of Cincinnati 67
θ
θ
2
3
3
2
12 7
3
7
9
v PL
EI k L
= −+
( ' )
The influence of the spring k is easily seen from this result.
Plugging in the given numbers, we can calculate
θ
θ
2
3
3
0 002492
0 01744
0 007475
v
=
−
−
−
.
.
.
rad
m
rad
From the global FE equation, we obtain the nodal reaction
forces as,
F
M
F
F
Y
Y
Y
1
1
2
4
69 78
69 78
1162
3488
=
−
− ⋅
.
.
.
.
kN
kN m
kN
kN
Checking the results : Draw free body diagram of the beam
1 2
50 kN
3
3.488 kN116.2 kN
69.78 kN
69.78 kN⋅m
8/21/2019 Finite Element Methods Lectures_Lui.pdf
68/169
Lecture Notes: Introduction to Finite Element Method Chapter 3. Two-Dimensional Problems
© 1998 Yijun Liu, University of Cincinnati 75
Chapter 3. Two-Dimensional Problems
I . Review of the Basic Theory
In general, the stresses and strains in a structure consist of
six components:
σ σ σ τ τ τ x y z xy yz zx, , , , , for stresses,
and
ε ε ε γ γ γ x y z xy yz zx, , , , , for strains.
Under contain conditions, the state of stresses and strains
can be simplified. A general 3-D structure analysis can,
therefore, be reduced to a 2-D analysis.
z
x
σ y
σ z
yz
zx
xy
8/21/2019 Finite Element Methods Lectures_Lui.pdf
69/169
Lecture Notes: Introduction to Finite Element Method Chapter 3. Two-Dimensional Problems
© 1998 Yijun Liu, University of Cincinnati 76
Plane (2-D) Problems
• Plane stress:
σ τ τ ε z yz zx z = = = ≠0 0( ) (1)
A thin planar structure with constant thickness and
loading within the plane of the structure ( xy-plane).
• Plane strain:ε γ γ σ z yz zx z = = = ≠0 0( ) (2)
A long structure with a uniform cross section and
transverse loading along its length ( z-direction).
z
z
8/21/2019 Finite Element Methods Lectures_Lui.pdf
70/169
Lecture Notes: Introduction to Finite Element Method Chapter 3. Two-Dimensional Problems
© 1998 Yijun Liu, University of Cincinnati 77
Stress-Strain-Temperature (Consti tutive) Relations
For elastic and isotropic materials, we have,
εε
γ
ν ν
σσ
τ
εε
γ
x
y
xy
x
y
xy
x
y
xy
E E E E
G
=−
−
+
1 01 0
0 0 1
0
0
0
/ // /
/
(3)
or,
ε σ ε= +−E 1 0
where ε0 is the initial strain, E the Young’s modulus, ν the
Poisson’s ratio and G the shear modulus. Note that,
G E
=+2 1( ) ν
(4)
which means that there are only two independent materials
constants for homogeneous and isotropic materials.We can also express stresses in terms of strains by solving
the above equation,
σ
σ
τ ν
ν
ν
ν
ε
ε
γ
ε
ε
γ
x
y
xy
x
y
xy
x
y
xy
E
=
−−
−
1
1 0
1 0
0 0 1 2
2
0
0
0( ) /
(5)
or,
σ ε σ= +E 0
where σ ε0 0= −E is the initial stress.
8/21/2019 Finite Element Methods Lectures_Lui.pdf
71/169
Lecture Notes: Introduction to Finite Element Method Chapter 3. Two-Dimensional Problems
© 1998 Yijun Liu, University of Cincinnati 78
The above relations are valid for plane stress case. For
plane strain case, we need to replace the material constants in
the above equations in the following fashion,
E E
G G
→−
→−
→
1
1
2 ν
ν ν
ν(6)
For example, the stress is related to strain by
σ
σ
τ ν ν
ν ν
ν ν
ν
ε
ε
γ
ε
ε
γ
x
y
xy
x
y
xy
x
y
xy
E
=+ −
−−
−
−
( )( ) ( ) /
1 1 2
1 0
1 0
0 0 1 2 2
0
0
0
in the plane strain case.
Initial strains due to temperature change (thermal loading)
is given by,
ε
ε
γ
α
α
x
y
xy
T
T
0
0
00
=
∆∆ (7)
where α is the coefficient of thermal expansion, ∆T the changeof temperature. Note that if the structure is free to deform under
thermal loading, there will be no (elastic) stresses in the
structure.
8/21/2019 Finite Element Methods Lectures_Lui.pdf
72/169
Lecture Notes: Introduction to Finite Element Method Chapter 3. Two-Dimensional Problems
© 1998 Yijun Liu, University of Cincinnati 79
Strain and Displacement Relations
For small strains and small rotations, we have,
ε ∂∂
ε ∂∂
γ ∂∂
∂∂
x y xyu x
v y
u y
v x
= = = +, ,
In matrix form,
ε
ε
γ
∂ ∂
∂ ∂
∂ ∂ ∂ ∂
x
y
xy
x
y
y x
u
v
=
/
/
/ /
0
0 , or ε = Du (8)
From this relation, we know that the strains (and thus
stresses) are one order lower than the displacements, if the
displacements are represented by polynomials.
Equil ibrium Equations
In elasticity theory, the stresses in the structure must satisfy
the following equilibrium equations,
∂σ
∂
∂τ
∂
∂τ
∂
∂σ
∂
x xy
x
xy y
y
x y f
x y f
+ + =
+ + =
0
0
(9)
where f x and f y are body forces (such as gravity forces) per unit
volume. In FEM, these equilibrium conditions are satisfied in
an approximate sense.
8/21/2019 Finite Element Methods Lectures_Lui.pdf
73/169
Lecture Notes: Introduction to Finite Element Method Chapter 3. Two-Dimensional Problems
© 1998 Yijun Liu, University of Cincinnati 80
Boundary Conditions
The boundary S of the body can be divided into two parts,
S u and S t . The boundary conditions (BC’s) are described as,
u u v v S
t t t t S
u
x x y y t
= == =
, ,
, ,
on
on(10)
in which t x and t y are traction forces (stresses on the boundary)
and the barred quantities are those with known values.
In FEM, all types of loads (distributed surface loads, bodyforces, concentrated forces and moments, etc.) are converted to
point forces acting at the nodes.
Exact Elastici ty Solution
The exact solution (displacements, strains and stresses) of a
given problem must satisfy the equilibrium equations (9), the
given boundary conditions (10) and compatibility conditions
(structures should deform in a continuous manner, no cracks and
overlaps in the obtained displacement fields).
t x
t y
S u
S t
8/21/2019 Finite Element Methods Lectures_Lui.pdf
74/169
Lecture Notes: Introduction to Finite Element Method Chapter 3. Two-Dimensional Problems
© 1998 Yijun Liu, University of Cincinnati 81
Example 3.1
A plate is supported and loaded with distributed force p as
shown in the figure. The material constants are E and ν.
The exact solution for this simple problem can be found
easily as follows,
Displacement:
u p
E x v p
E y= = −, ν
Strain:
ε ε ν γ x y xy
p
E
p
E = = − =, , 0
Stress:σ σ τ x y xy p= = =, ,0 0
Exact (or analytical) solutions for simple problems are
numbered (suppose there is a hole in the plate!). That is why we
need FEM!
8/21/2019 Finite Element Methods Lectures_Lui.pdf
75/169
Lecture Notes: Introduction to Finite Element Method Chapter 3. Two-Dimensional Problems
© 1998 Yijun Liu, University of Cincinnati 82
I I . F ini te Elements for 2-D Problems
A General Formula for the Stiff ness Matrix
Displacements (u, v) in a plane element are interpolated
from nodal displacements (ui, vi) using shape functions N i as
follows,
uv
N N N N
u
v
u
v
=
=1 2
1 2
1
1
2
2
0 00 0
L
L
M
or u Nd (11)
where N is the shape function matrix, u the displacement vector
and d the nodal displacement vector. Here we have assumed
that u depends on the nodal values of u only, and v on nodalvalues of v only.
From strain-displacement relation (Eq.(8)), the strain vector
is,
ε ε= = =Du DNd Bd, or (12)
where B = DN is the strain-displacement matrix.
8/21/2019 Finite Element Methods Lectures_Lui.pdf
76/169
Lecture Notes: Introduction to Finite Element Method Chapter 3. Two-Dimensional Problems
© 1998 Yijun Liu, University of Cincinnati 83
Consider the strain energy stored in an element,
( )
( )
U dV dV
dV dV
dV
T
V
x x y y xy xy
V
T
V
T
V
T T
V
T
= = + +
= =
=
=
∫ ∫
∫ ∫
∫
1
2
1
2
1
2
1
2
1
2
1
2
σ ε σ ε σ ε τ γ
ε ε ε εE E
d B EB d
d kd
From this, we obtain the general formula for the element
stiffness matrix,
k B EB= ∫ T V
dV (13)
Note that unlike the 1-D cases, E here is a matrix which is given by the stress-strain relation (e.g., Eq.(5) for