EN0175
09 / 07 / 06
1.2 Introduction to FEM Finite element method is a very powerful tool for numerical analysis in solid and structural mechanics as well as in many other engineering disciplines. Here we present an introduction to the basic concepts of F
EN0175
10 / 26 / 06
Plastic material behavior
Yield condition: Plastic loading:
= Y = Y , d > 0
Y Y0
E
1
E
1
0 P
We will now denote the initial yield stress as
E
Y 0 and the current yield stress as
Y ; see above figure.
Decompose strain into elastic &
EN0175
11 / 02 / 06
Remarks on plastic material behavior 1) Yield surfaces (a surface in the stress space representing the condition/criterion whether the solid responds elastically or plastically to the applied load) The von Mises and Tresca yield condit
EN0175
11 / 07 / 06
Chap. 6
Boundary value problems in linear elasticity
& Equilibrium equations: ij , j + f i = ui
Kinematic equations: ij = Hookes law: ij =
(1) (2) or
1 + ij kk ij E E
1 (ui, j + u j ,i ) 2
ij = 2 ij + kk ij
(3)
For the most general pr
EN0175
11 / 09 / 06
Continue on the Airy stress function method in elasticity:
2 2 = 0 ( : Airy stress function)
xx =
2 2 2 , xy = , yy = y 2 xy x 2
Example 4:
P
0
y
thickness : 1
Consider bending of a beam (height: 2c ; thickness: 1) caused by a conce
EN0175
11 / 14 / 06
Linear elasticity solution in polar coordinates Typical problems: Stress around a circular hole in an elastic solid.
0
a
Boundary conditions: Traction free @ r = a :
0
er = rr er + r e = 0 , i.e. rr = 0 , r = 0
v
v
v
b
a
Boundary cond
EN0175
11 / 16 / 06
Continue on the problem of circular hole under uniaxial tension (remote). Stress concentration occurs at r = a , =
2
.
3T
a
T
Governing equation is:
2
2
=0
The stress components in polar coordinates are:
rr =
1 1 2 + r r r 2 2
1 r
EN0175
11 / 21/ 06
Principle of minimum potential energy (continued) The potential energy of a system is
V = wdV f i ui dV ti ui dS
V V S
Principle of minimum potential energy states that for all kinematically admissible u i , the actual displacement fiel
EN0175
11 / 29/ 06
Chap. 9 Finite element method (Read Chap 7 of Prof Bowers notes) Principle of virtual work:
V
ij
ij dV = f i ui dV + ti ui dS
V AT
Principle of minimum potential energy
1 Min V = ij ij dV f i ui dV ti ui dS V 2 V AT
ti
We can represen
EN0175
11 / 30/ 06
Finite element method (continued) Summary of essential concepts: FEM analysis begins with calculations on the element level. Element
#3
#1
Element nodal displacement:
#2
u1#1 #1 u2 u1# 2 u el = # 2 u 2 u1#3 #3 u2
Element interpolation
EN0175
12 / 05/ 06
9. Boundary value problems in plasticity Slip line theory An important theory in the plane problems of plasticity is the slip line theory. This theory simplifies the governing equations for plastic deformation by making several assumpti
EN0175
10 / 24 / 06
Review of deformation tensors:
F , C , B , U , V , R , E , E*
Given F , one can follow the following standard procedure to determine the other strain measures. 1) Most simply, C = F F , B = F F , E =
T T
1 (C I ) , E * = 1 I B 1 2 2
(
EN0175
10 / 19 / 06
Mechanical Behavior of Solids Linear Elastic solids
= E (1D) ij = Cijkl kl or ij = S ijkl kl (3D)
where C is sometimes called the stiffness tensor and S is sometimes called the compliance tensor. Both of them are 4th order elastic mod
EN0175
10 / 17 / 06
Summary of elementary strain concepts and their generalizations to 3D tensors:
1D
3D
(1) =
l0
=
l l0 = 1 l0
(U I )
(2 ) =
l
=
l l0 = 1 1 l
(I V )
1
l 2 l02 1 2 (3 ) = = 1 2l02 2
(
)
)
1 (C I ) 2 1 1 I B 2
(4 ) =
l 2 l02 1 = 1 2 2l 2
EN0175
09 / 12 / 06
Intro to FEM (continued) Examples of using FEM to solve a problem and comparison with exact solution: We consider a problem already discussed in the previous class:
x
g
Solution by exact method:
x (L x ) 2E g (L 2 x ) = 2 u=
g
u
gL
2
EN0175
09 / 14 / 06
Lecture 4 Announcement: ABAQUS tutorial will be given between 6-8 PM on September 18 (Monday). FEM interpolation functions: 1D FEM:
w( x ) = a1 + a 2 x (linear interpolation)
Element types in 2D:
y
3 noded triangle (linear element)
w(
EN0175
09 / 19 / 06
Today we introduce the application of index notations in vector and tensor algebra. Index notation:
1 0 0 1, i = j e.g. Kronecker delta: ij = . In matrix form: 0 1 0 0, i j 0 0 1
Dot product of vectors: a b = (ai ei ) b j e j = ai b j
EN0175
09 / 21 / 06
We continue on the mathematical background. Base tensors: ei e j (dyadic form)
v
v
1 0 0 0 1 0 0 0 1 v v v v v v , e e = 0 0 0 , e e = 0 0 0 e1 e1 = 0 0 0 1 2 3 1 0 0 0 0 0 0 0 0 0
Note that: ei e j e j ei , ei e j e j ei
v
v
v
v
v
v
EN0175
09 / 26 / 06
Review on coordinate transformation (change of basis) for tensors.
v e3
v' e2
v e1
v v v' u = ui ei = u 'p e p v v' v v' u 'p = u e p = ui ei e p v v v ' v ui = u ei = uq eq' ei
v v u e'
1
v e2
v' e3
(
)
(
)
v v' ' v v ' u 'p = uq ei e
EN0175
09 / 28 / 06
Continuing on concepts of stress in a continuum: Traction on an arbitrary plane with normal vector n
v
t j = ij ni v Tv t = n
This equation indicates that the normal stress on a plane with normal vector n is
v
n = t n = n T n
Accordin
EN0175
10 / 03 / 06
Stress in a solid Examples of common stress states: 1. Uniaxial stress
v e2
v e3
v e1
= 0 0 0 0 0 0 0 0
Principal stresses:
I = , II = III = 0
v v
v
Principal directions: e1 , e2 , e3 ( e2 , e3 can be replaced by any direction in 2-
EN0175
10 / 05 / 06
Maximum and minimum shear stresses in a solid An general stress state
11 12 13 = 21 22 23 , 31 32 33
when expressed in the principal directions, becomes diagonalized as
I =0 0
0
II
0
0 0 III
v e3 III
s
v t
v n
n
v e2 II
v e1 I
Th
EN0175
10 / 10 / 06
Strain in a solid
2
v dx2 v dx1
v u
v d y2
v d y1
v x
v y
1
3
Consider an arbitrary fiber within the elastic body, In the undeformed configuration, we can represent the fiber as a small vetor: dx = mdl0 where
v
v
v dl 0 is the length a
EN0175
10 / 12 / 06
Announcements: Mid-term (Tuesday, Oct 31) (1 page double sided notes allowed) Proposal/team for ABAQUS project (Thursday, Nov 9) Review of strain concepts
3
deform
v x
v y
2
1
v v v y = y (x , t ) , yi = yi ( x1 , x2 , x3 , t ) = xi +