2ZahitMtecitogloufeformulation 2015 03-02-12!36!49
-
Upload
parth-daxini -
Category
Documents
-
view
213 -
download
0
description
Transcript of 2ZahitMtecitogloufeformulation 2015 03-02-12!36!49
13.02.2006 FINITE ELEMENT FORMULATION
Chapter 2FINITE ELEMENT FORMULATION
Zahit Mecitoğlu
FINITE ELEMENT ANALYSIS IN STRUCTURES
© 2006 İstanbul Technical University
FINITE ELEMENT FORMULATION13.02.2006
AIMThe basic aim of solid mechanics problem is to determinethe distribution of displacements and stresses under the loading and boundary conditions.
A mathematical model of the structural problem is necessary to find the desired distributions by using the FEM.
Hence, the basic equations of solid mechanics are summarized in the following sections for ready reference in the formulation of FE equations.
FINITE ELEMENT FORMULATION13.02.2006
BASIC EQUATIONS
For a reasonable solution of a problem, following basicequations must be satisfied for an elastic continua.
Strain-displacement relationsConstitutive relationsEquilibrium equationsCompatibility equationsBoundary conditions
FINITE ELEMENT FORMULATION13.02.2006
FIELD VARIABLESThe loadings and boundary conditions on an elastic body cause the displacements, strains and stresses inside the body.
=PA
σ
P Pσ
L u=
uL
ε
A
FINITE ELEMENT FORMULATION13.02.2006
Strain-Displacement RelationsThe deformed shape of an elastic body under loads can be completely described by the three components of displacement u, v, and w.
The strains induced in the body can be expressed in terms of the displacement components.
Geometric linearity : deformations are small enough so that the strain-displacement relations remain linear.
FINITE ELEMENT FORMULATION13.02.2006
...Strain-Displacement Relations
0 0
0 0
0 0
0
0
0
∂⎡ ⎤⎢ ⎥∂⎢ ⎥
∂⎢ ⎥⎧ ⎫ ⎢ ⎥∂⎪ ⎪ ⎢ ⎥∂⎪ ⎪ ⎢ ⎥ ⎧ ⎫⎪ ⎪ ⎢ ⎥∂⎪ ⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬⎢ ⎥∂ ∂⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎩ ⎭∂ ∂⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎢ ⎥∂ ∂⎪ ⎪ ⎢ ⎥⎩ ⎭
∂ ∂⎢ ⎥⎢ ⎥∂ ∂⎢ ⎥
∂ ∂⎣ ⎦
x
y
z
xy
yz
zx
x
y
uz v
wy x
z y
z x
εε
εγ
γ
γ
{ } [ ]{ }= dε δ
For the 3-D case, the strain-displacement relations can be written as follows,
FINITE ELEMENT FORMULATION13.02.2006
In the case of linearly elastic isotropic three-dimensional solid, the stress-strain relations are given by Hooke’s law.
( ) ( ) ( )( )
( )
12
12
12
1 0 0 01 0 0 0
1 0 0 00 0 0 1 2 0 01 1 20 0 0 0 1 2 0
0 0 0 0 0 1 2
−⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪ ⎪ ⎪⎢ ⎥−⎪ ⎪ ⎪ ⎪⎢ ⎥
−⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬−+ − ⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪−⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥−⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭⎣ ⎦
x x
y y
z z
xy xy
yz yz
zx zx
E
σ εν ν νσ εν ν ν
ν ν νσ εντ γν ν
ντ γντ γ
{ } [ ]{ }= Eσ ε
Constitutive relations
FINITE ELEMENT FORMULATION13.02.2006
Equilibrium EquationsIf we consider an element of material inside the body, it must be in equilibrium due to the internal stresses developed by the loads.
y
x
z
σy
σx
σz
τyx
τxyτyz
τzy
τzxτxz
x
Φ
P1
P2
y
z
(a)
Sσ
S
FINITE ELEMENT FORMULATION13.02.2006
…Equilibrium EquationsTheoretically the state of stress at any point of a body is completely defined in terms of the nine components of stress.
The normal stresses are σx, σy, and σz.
Shear stresses are τxy, τyx, τyz, τzy, τxz, and τzx.
FINITE ELEMENT FORMULATION13.02.2006
…Equilibrium Equations
0
0
0
∂∂ ∂+ + + =
∂ ∂ ∂∂ ∂ ∂
+ + + =∂ ∂ ∂
∂∂ ∂+ + + =
∂ ∂ ∂
xyx xzx
yx y yzy
zyzx zz
bx y z
bx y z
bx y z
τσ τ
τ σ τ
ττ σ
The moment equilibriums about the x, y and z axis give the relations
= = =xy yx xz zx yz zyτ τ τ τ τ τ
The force equilibrium in x, y and z directions give the following equilibrium equations
bx, by , bz : body forces per unit volume.
FINITE ELEMENT FORMULATION13.02.2006
…Equilibrium Equations
0 0 00
0 0 0 00
0 0 0
⎧ ⎫⎡ ⎤∂ ∂ ∂ ⎪ ⎪⎢ ⎥∂ ∂ ∂ ⎪ ⎪ ⎧ ⎫⎢ ⎥ ⎧ ⎫⎪ ⎪∂ ∂ ∂ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪− =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥∂ ∂ ∂ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎩ ⎭⎩ ⎭⎪ ⎪∂ ∂ ∂⎢ ⎥⎪ ⎪⎢ ⎥∂ ∂ ∂⎣ ⎦ ⎪ ⎪⎩ ⎭
x
yx
zy
xyz
yz
xz
x y z bb
y x zb
z y x
σσ
στ
τ
τ
In a compact format
[ ] { } { } { }0− =Td bσ
In matrix format we may write
FINITE ELEMENT FORMULATION13.02.2006
Many problems in theory of elasticity are two dimensional in nature and they can be modeled as plane stress and plane strain.
Plane Stress : A thin plate subjected to in-plane loading acting in its own plane, the state of stress and deformation within the plate is called plane stress. In this case only two dimensions (in the plane of plate) are required for the analysis.
SPECIAL CASES: 2-D CASE
FINITE ELEMENT FORMULATION13.02.2006
PLANE STRESS
x
y
z
x
y
A thin plate under inplane loading.
...SPECIAL CASES: 2-D CASE
FINITE ELEMENT FORMULATION13.02.2006
Plane Strain: If a long body is subjected to transverse loading and its cross section and loading do not vary significantly in the longitudinal direction, a small thickness in the loaded area can be treated as subjected to plane strain.
...SPECIAL CASES: 2-D CASE
FINITE ELEMENT FORMULATION13.02.2006
PLANE STRAIN
A long cylinder under internal pressure.
z
x
y
z
y
p
...SPECIAL CASES: 2-D CASE
FINITE ELEMENT FORMULATION13.02.2006
…SPECIAL CASES: 2-D CASEFor a two-dimensional problem, there will be only three independent stress components (σx, σy, τxy) and three independent strain components (εx, εy, γxy)
0
0
∂∂+ + =
∂ ∂∂ ∂
+ + =∂ ∂
xyxx
xy yy
bx y
bx y
τσ
τ σ
strain-displacement relations0
0
⎡ ⎤∂⎢ ⎥∂⎧ ⎫ ⎢ ⎥
⎪ ⎪ ⎧ ⎫∂⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥∂ ⎩ ⎭⎪ ⎪ ⎢ ⎥⎩ ⎭ ∂ ∂⎢ ⎥
⎢ ⎥∂ ∂⎣ ⎦
x
y
xy
xuvy
y x
εε
γ
equilibrium equations
FINITE ELEMENT FORMULATION13.02.2006
The constitutive equations for plane stress case
0= = =z zx yzσ τ τ ( ) 0= − + = =z x y yz zxEνε σ σ ε ε
( )2
1 01 0
1 10 02
⎡ ⎤⎧ ⎫ ⎧ ⎫⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥−⎪ ⎪ ⎪ ⎪⎢ ⎥−⎩ ⎭ ⎩ ⎭⎢ ⎥⎣ ⎦
x x
y y
xy xy
Eσ ν εσ ν ε
ν ντ γ
...SPECIAL CASES: 2-D CASE
FINITE ELEMENT FORMULATION13.02.2006
w = 0 and at every cross section.0∂=
∂wz
0= = =z zx yzε γ γ ( ) 0= + = =z x y yz zxσ ν σ σ τ τ
( ) ( )
1 01 0
1 1 21 20 0
2
⎡ ⎤⎧ ⎫ ⎧ ⎫⎢ ⎥−⎪ ⎪ ⎪ ⎪⎢ ⎥= −⎨ ⎬ ⎨ ⎬⎢ ⎥+ −⎪ ⎪ ⎪ ⎪⎢ ⎥−⎩ ⎭ ⎩ ⎭⎢ ⎥⎣ ⎦
x x
y y
xy xy
Eσ ν ν εσ ν ν ε
ν νντ γ
The constitutive equations for plane strain case
...SPECIAL CASES: 2-D CASE
FINITE ELEMENT FORMULATION13.02.2006
…SPECIAL CASES: 1-D CASE
Equilibrium equation 0+ =xx
d bdxσ
Constitutive Relation1or= =x x x xEE
σ ε ε σ
Strain-Displacement Relation =xdudx
ε
In one-dimensional problems, only one component of stress, namely, σx will be there.
FINITE ELEMENT FORMULATION13.02.2006
The Number of Basic EquationsType of equations Number of equations
In 3-D problems In 2-D problems In 1-D problems
Strain-Displacement Rel.
6 3 1
Constitutive Relations 6 3 1
Equilibrium Equations 3 2 1
Total Number of Eqs. 15 8 3
FINITE ELEMENT FORMULATION13.02.2006
The Number of UnknownsUnknowns In 3-D problems In 2-D problems In 1-D problems
Displacements u, v, w u, v u
Strains εx, εy, εz, γxy, γyz, γzx εx, εy, γxy εx
Stresses σx, σy, σz, τxy, τyz, τzx σx, σy, τxy σx
Total number of unknowns
15 8 3
FINITE ELEMENT FORMULATION13.02.2006
Summary
[ ]{ } { } { }0Δ − =bσ
{ } [ ]{ }= Eσ ε
{ } [ ]{ }= dε δ
Equilibrium equations
Constitutive relations
Strain-displacement relations
FINITE ELEMENT FORMULATION13.02.2006
POTENTIAL ENERGY
The potential energy of an elastic body Π is defined asΠ = U - W
In this equation U is the strain energy and W is the work done on the body by the external forces.
The strain energy of a linear elastic body is defined as
{ } { } { } [ ]{ }1 12 2
= =∫ ∫T T
V V
U dV E dVε σ ε ε
FINITE ELEMENT FORMULATION13.02.2006
…POTENTIAL ENERGYThe work done by the external forces can be stated as
{ } { } { } { } { } { }1
11=
= − Φ + ∑∫ ∫I
T T Ti i
iV S
W b dV dS Pδ δ δ
{b} is the vector of body forces.{Φ} is the vector of distributed surface force.{P} is the vector of concentrated loads.
FINITE ELEMENT FORMULATION13.02.2006
FORMULATIONS OF SOLID AND STRUCTURAL MECHANICS
Most continuum problems can be formulated according to one of the two methods.
differential equation methodvariational method
FINITE ELEMENT FORMULATION13.02.2006
...FORMULATION METHODSFormulation Methods
Differential equation formulation methods
Variational formulation methods
Displacement method
Force method
Displacement-force method
Principle of minimum potential energy
Principle of minimum complementary energy
Principle of stationary Reissner energy
FINITE ELEMENT FORMULATION13.02.2006
MINIMUM POTENTIAL ENERGY
Most of the equations in this course will be derived from the potential energy theory.
We assume that strains and displacements are small and no energy dissipated in the static loading process.
That is, the external work of gradually applied loads is equal to the energy stored in the structure, and the system is said to be conservative.
FINITE ELEMENT FORMULATION13.02.2006
MINIMUM POTENTIAL ENERGY
The principle says that among all the displacement states of a conservative system that satisfy compatibility and boundary restraints, those that also satisfy equilibrium make the potential energy stationary.
δΠ = δU - δW = 0
Variation is taken with respect to the displacements.
FINITE ELEMENT FORMULATION13.02.2006
F.E. FORMULATION
As a first step, the displacement field of an element is approximated by using interpolation functions and nodal displacements.
{ } [ ]{ }( , , )( , , )( , , )
⎧ ⎫⎪ ⎪= =⎨ ⎬⎪ ⎪⎩ ⎭
u x y zv x y z N qw x y z
δ e
x, u
y, v
q1
q2
q3 q5
q6q4
1
2
3
z, w
4
q7
q8
q9
q10
q11
q12
FINITE ELEMENT FORMULATION13.02.2006
… F.E. FORMULATION
Strain vector can be expressed in terms of nodal displacements {q}, by using strain-displacement relations
{ } [ ]{ } [ ][ ]{ } [ ]{ }== =d d BN q qε δ
{ } [ ]{ } [ ][ ]{ }= = BE qEσ ε
Stress vector can be expresses in terms of nodal displacements, {q}, by using the constitutive relations.
FINITE ELEMENT FORMULATION13.02.2006
… F.E. FORMULATIONThe strain energy of an element can be written in terms of nodal displacements as
{ } [ ]{ }
{ } [ ] [ ] [ ]{ }
{ } [ ] [ ] [ ] { }
{ } [ ]{ }
12
12
12
12
=
=
=
=
∫
∫
∫
e
e
e
T
T T
e
V
VT
T
T
V
U E dV
E dV
q q
q B B q
B E B dV
q qk
ε ε
FINITE ELEMENT FORMULATION13.02.2006
… F.E. FORMULATION
{ } { } { } { }
{ } [ ] { } { } [ ] { }
{ } [ ] { } { } [ ] { }
{ } { } { } { }
1
1
1
1
1
1
= + Φ
= + Φ
= Φ+
= +
∫
∫
∫ ∫
∫
∫
e e
e e
e e
e
V S
V
T T
T TT T
T T
V S
b s
S
T T
T T
W b dV dS
b dV dSq N q N
N b dV N dS
q f
q q
q f
δ δ
The work done by the external forces The work done by the external forces for for an element can be written in terms of nodal displacements as
FINITE ELEMENT FORMULATION13.02.2006
… F.E. FORMULATION
{ } [ ] { } Element body force vector= ∫e
Tb
V
f N b dV
If an element has not a boundary at the surface under distributed load, the contribution of surface forces will be zero for this element.
{ } [ ] { }1
Element surface load vector= Φ∫e
Ts
S
f N dS
[ ] [ ] [ ] [ ] Element stiffness matrix= ∫e
T
V
k B E B dV
The formulae for the element stiffeness and load vector remain the same irrespective of the type of the element. However, the order of the stiffness matrix and the load vector will change for different types of elements.
FINITE ELEMENT FORMULATION13.02.2006
… F.E. FORMULATIONThe potential energy of an element can be written as
{ } { }1=
Π = Π −∑E
T
ec
e FQ
{ }
1
2
1=
⎧ ⎫⎪ ⎪⎪ ⎪= =⎨ ⎬⎪ ⎪⎪ ⎪⎩ ⎭
∑E
e
N
Q q
Q
{Q} is the vector of nodal displacements of the entire structure
Microsoft Word Document
The potential energy of the entire structure can be obtained from the summation of the element energies.
{ } [ ]{ } { } { } { } { }12Π = + = − −T T Te e e
b skW fU q q q fq
FINITE ELEMENT FORMULATION13.02.2006
… F.E. FORMULATIONASSEMBLY: The summation of the equation implies the expansion of element matrices to “structure size” followed by summation of overlapping elements.
{ } [ ]{ } { } { } { } { } { } { }
{ } [ ]{ } { } { } { } { }( )
12
1 1 1
12
= = =Π = −
+
− −
= − +
∑ ∑ ∑E E E
T T T T
e e eb s c
b sT
cT
q q q q Q
Q Q
k f f F
K F F FQ
{ } [ ]{ } { } { }12Π = −T TKQ Q Q F
FINITE ELEMENT FORMULATION13.02.2006
… F.E. FORMULATION
[ ] [ ]1=
= ∑E
eK k
The element stiffness matrix and the global stiffness The element stiffness matrix and the global stiffness matrix are always symmetric.matrix are always symmetric.
Some of the contributions to the load vector {F} may be zero in a particular problem.
Some of the components of the load vectors may be moments if the corresponding nodal displacements represent rotations.
Assembly procedure gives us the global stiffeness matrix and load vector.
{ } { } { }( ) { }1=
= + +∑E
b s ce
F f f F
FINITE ELEMENT FORMULATION13.02.2006
… F.E. FORMULATION
0 1,2,3,Π= = …
ii N
Q∂∂
[ ]{ } { }=K Q FThe required solution for the nodal displacements and element stresses can be obtained after solving above eqs.
The linear algebraic equations of the overall structure can now be expressed as
The static equilibrium equations of the structure can be obtained from the principle of minimum potential energy.
FINITE ELEMENT FORMULATION13.02.2006
The equilibrium equation, [K]{Q}={F}, cannot be solved since the stiffness matrices [k] and [K] are singular, and hence their inverses do not exist.
The physical significance of this is that a loaded structure is free to undergo unlimited rigid body motion (translation and/or rotation) unless some support or boundary constraints are imposed on the structure to suppress the rigid body motion.
These constraints are called boundary conditions.
We solve the equations after incorporating the prescribed boundary conditions to obtain the displacements.
… F.E. FORMULATION
FINITE ELEMENT FORMULATION13.02.2006
INITIAL STRAINS
{ } { }( )0= −eε ε ε
If some initial strains, such as temperature changes, exist in the structure, the elastic portion of the total strain vector can bestated as
{ } [ ] { } { }( )0= −Eσ ε εand stress vector is writen as
P
ΔTΔT
FINITE ELEMENT FORMULATION13.02.2006
...INITIAL STRAINS
{ } { }( ) [ ] { } { }( )
{ } [ ]{ } { } [ ]{ } { } [ ]{ }
0 0
0 0 0
12
1 12 2
= − −
= − −
∫
∫ ∫ ∫e e e
Te
V
T T T
V V V
U E dV
E dV E dV E dV
ε ε ε ε
ε ε ε ε ε ε
The strain energy of an element can be found
The first term yield the element stiffness matrix. Since the last term is constant, its contribution is zero during the minimization of the total potential energy.
FINITE ELEMENT FORMULATION13.02.2006
...INITIAL STRAINS
{ } [ ] [ ]{ }0= ∫e
Tt
V
f N E dVε
{ } [ ]{ }
{ } [ ] [ ]{ }
{ } [ ] [ ]{ }
{ } { }
0
0
0
= −
= −
= −
= −
∫
∫
∫e
e
e
Tet
VT T
VT
VT
T
t
W E dV
q N E dV
q
N E Vq d
f
ε
ε ε
ε
Element nodal force vector due to initial strains
The second term yields the element nodal force vector due to initial strains.