View
478
Download
5
Category
Preview:
Citation preview
SOLVING LINEAR ELASTICITY BY FEM
SOLID MECHANICS
CONTINUUM MECHANICS
3D (SOLID)
2D (TRUSS)
1D (BAR)
ELEMENT FORMULATION FOR
FEM
ASSUMPTIONS
• Deformations are small• Behavior of material is linear• Dynamic effects are neglected• No gaps or overlaps occur during the
deformation of the solid
1 DIMENSIONAL
Ex: Elastic bar- (linear stress analysis/ linear elasticity)
? To find stress distribution(σ(x)) in bar deformation ε(x)
displacement of points (u(x))
STRONG FORM• Equation and boundary for physical system• Partial differential equations (ordinary
differential equation-1D)
WEAK FORM• Integral equation of strong form• Formulate FEM
STRONG FORM WEAK FORM
DISCRETE EQUATIONS
APPROXIMATION OF FUNCTIONS
• Bar condition1. Equilibrium2. Elastic stress-strain law3. Displacement compatible4. Satisfy strain displacement equation
• Equilibrium eq. of int. force(p(x)) and ext force(b(x)) in x axis direction
lbdx
duAE
x
xxExx
u
x
xuxxux
xxAxpxA
xpx
xbx
xp
xxb
x
xpxxp
xxpxx
xbxp
x0 0)(
)()()(
)()()(
)()()(- )(
)()(
0)()(
0)2
()()(
0)()2
()(
Divide with △x
Limit △x=0
stress
strain
Hook’s Law
2nd order ordinary diff. eq
STRONG FORM--SOLUTIONS• CLUE - either LOAD or DISPLACEMENT at boundary (2 ends)• Essential boundary condition/ Displacement boundary
condition x=l or u(x=l)=u
• Natural boundary condition/traction boundary conditionx=0
unknown is u(x)
(data)given are and ,
)(
)0(
0)(
0
but
ulxu
tdx
duEx
bdx
duAE
dx
d
x
STRONG FORM
tA
p
dx
duE
x
)0(
)0()0(
0
Traction (stress) bound. Cond. (natural)
Disp. bound. Cond. (essential)
Governing strong form eq.
WEAK FORM• What?
- Restated a partial diff eq. into integral form (a trial solution)• How to create ??1. For governing eq. Multiply w(x) and integrate with domain(0,l)2. For trace boundary multiply w(x)A at x=0 (no integral bcoz only
holds at point) w(x) is weight /test/arbitrary function where w(l) =0
wtdx
duEwA
wdxbdx
duAE
dx
dw
x
l
0
0
0
0 (1)
Some restriction on weak form
• denotes that w(x) is an arbitrary fn• weight fn- enforcer to be zero by its arbitrariness• we did not enforce boundary cond. It is easy to construct trial or candidate
solutions u(x) that satisfy displacement condition• Now, transform into the form containing first derivatives. That will lead to a
symmetric stiffness matrix and simplify the treatment of traction boundary• Second derivation of u(x) need smooth trial solution(easy to construct by 1
D) • The trial solution that satisfy WEAK FORM is the solution of the STRONG
FORM• A trial solutions that is smooth and satisfies the essential boundary cond.
admissible• A weight function that is smooth and vanishes on essential bound.
admissible•
w
MATH
l
xlx
ll
l
l ll
dxdx
dwfwfwfdx
dx
dwfwfdx
dx
dfw
dxdx
dwfdxwf
dx
ddx
dx
dfw
dx
dwfwf
dx
d
dx
dfw
dx
dwf
dx
dfwwf
dx
d
0
0
00
0
0 00
)()()(
)(
)( -- )(
Rearrange the Weak Form EQ where f=AE(du/dx)
lll
dxdx
duAE
dx
dw
dx
duwAdx
dx
duAE
dx
dw
000
Rewrite and subtitute stress, σ=AEdu/dx
ll
xlx lwwbdxdxdx
duAE
dx
dwwAwA
00
0 0)( with w 0)()(
For w(l)=0, and Second term of eq (1), (wAt)x=0
l
x
l
lwwbdxtwAdxdx
duAE
dx
dw
0
0
0
0)( with w )(
σ(x=0)=-t
WEAK FORM
Arbitrary Boundary conditionBoundary of 1D which consists of two end points is denoted by ΓBoundary for displacement Γu
Boundary for traction Γt
The traction and displacement both cannot be prescribed at the same boundaryΓu U Γt=0 Γu ∩ Γt=0
Unit normal to the body as, n where n=-1 at x=0 and n=+1 at x=l
Notation of Ω indicates any limits of integration Admissible weight Fn.And H is infinite dimensional set
uon
on
0 0)(
uu
tdx
duEnn
lxbdx
duAE
dx
d
t
l
oUwwbdxtwAdxdx
duAE
dx
dwt
0
)(
uwHxwxwU on 0,)()( 10
STRONG FORM
WEAK FORM
DISCRETIZATION
lT
Tl
lwwbdxwdxdx
duAE
dx
dw
0
0xT
0
0)( with A)t(w-
The procedure1. Evaluate the weak form for FE trial solutions and weight fn.2. By invoking the arbitrariness of the weight fn. Deduce a set of linear algebraic eq.3. Use transpose of weight fn as w(x) is scalar and for consistency
Global The FE weight fn. w(x)≈wh(x)=N(x)w Where N(x)= matrix shape fn w(x)=w1N1(x)+w2N2(x)+w3N3(x)
The trial solution u(x) ≈ uh(x)=N(x)d Where u(x)=u1N1+u2N2+u3N3
Replace the weak form over the entire domain by sum of integrals over the element domains.
02
1
2
1
1
e
e
e
e
el
x
x
eeTeTe
ee
Tx
x
en
eAtwbdxwdx
dx
duEA
dx
dw0x )(-
E to indicate element
element ,e, weight fn. and trial fn
eTeT
TeeTeTeT
eee
eee
dx
dww
dx
duxu
Bw Nw
dB dN
,
,)(
Substitute element weight fn. and trial fn
0)( 01
2
1
2
1
xeeT
x
x
eTeeeex
x
eTn
e
eT
e
e
e
e
e
e
e
el tAbdxdxEA f
fK
NNdBBw
e
e
e
dxEAdxEA eeeeTeeex
x
eTe BBBBK2
1 e
et
e
e
e
tAbdxtAbdx eeTeTx
eeTx
x
eTe
f
f
NNNNfe
)()( 0
2
1
ELEMENT STIFNESS MATRIX & ELEMENT EXTERNAL FORCE MATRIX
Element stiffness matrix Element external force matrix
Element & Global matrix relationship
Element and global matrix relation we=Lew de=Led
Substitutes into weak form of element domains;
01 1
el eln
e
n
e
eeTeeeTT fLdLKLw
Global/assembled stiffness matrix Global/assembled external force matrix
eln
e
eeeT
1
LKLK
eln
e
eeT
1
fLf
Re-subtitute into eq
00
0)(0)(
1
1
ww
lwwwT
T
except rw ,f Kdr
except fKdw
Where r is called the residual
2 Dimensional KINEMATICS (STRAIN –DISPLACEMENT RELATION)
Displacement in matrix and vector form,
juiuuu
uyx
y
x
u ,
Strain and shear strain
2
1
,
y
u
x
ux
u
x
u
xyxy
yyy
xxx
strain vs displacement eq.
y
xss
xy
yy
xx
x u
uu
x
y
y
x
isWhere
S
s
/
/
0
/
0
/
operator matrix gradient symmetric a
STRESS AND TRACTION
Stress component matrix for 2D
yy
xy
xy
xx
Unit normal to the surface, njninn yx
The force equilibrium of triangular body shown in Fig. 9.14
0
dxdydt yx
Dividing by dΓ and noting that dy=nxdΓ and dx=nydΓ
0
yyxx nnt
Multiply by unit vectors i and j
nnnt
nnnt
yyyyxxyy
xyxyxxxx
.
.
Stress and Traction in matrix form
τnt
EQUILIBRIUM
Figure 9.5 shows body force and the surface traction acting in the xy-planeBody force and traction vectors are written as
0),(2
,2
,,2
,2
yxyxbxx
yxxx
yxyyx
xyyx
x yyxx
The equilibrium equation is given by
jtittjbibb yxyx
,
Dividing the above eq. by x y, taking limit as △ △ △x 0, △y0,
0,0
0
yyyx
yx
bb
j ibyx
form; vector the in
and vector unit by multiple and
or 0bσ TS
Equilibrium Eq.
CONSTITUTIVE EQUATION (STRESS & STRAIN EQ)
Hooke’s LawDεσ
Matrix D will depends on whether one assumes a plane stress or a plane strain.
Plane strainthe body is thick relative to xy plane strain normal to plane εz, is zero shear strain that involves angles normal to plane, γxz & γyz are assumed to vanish
Plane stressthe body is thin relative to dimensions in xy plane. no loads are applied on z faces, stress normal to xy plane, σzz is vanish
For isotropic 2D
2/)21(00
01
01
)21)(1(
2/)1(00
01
01
1 2
E
E
D
DPlane stress
Plane strain
STRONG & WEAK FORMSEquilibrium Eq., Kinematics Eq., Constitutive Eq. with a boundary condition as
ttuΓ
uu and on uu ,0
u
tyx
T
on d)
on and c)uD b)
on a)
uu
tntn
bb
yx
yyyx
0,0
STRONG FORM
To obtain weak form, define weight fn. and trial fn. Using Green’s theorem and finally
,,
,,1
0
1
0
u
u
wHU
HUwhere
Uwwww
on 0uu
on uuuu
bddtudD)( TT
s
T
Ωs
t
WEAK FORM
DISCRETIZATIONGlobal nodal displacement matrix
npnp ynxnyxyx uuuuuu ...2211d
Element trial solution
Element weight fn.
Element shape fn.
Weak form over element domains Ωe
Express the strain in terms of element shape fn. and nodal displacement
eeee (x,y)yxyxyx dNuu ),(),().(
eeeTeTT (x,y)yxyxyx Nwww ),(),().(
enen
ee
enen
ee
NNN
NNN
0...00
0...00
21
21N
01
nel
e
eTeTes
eeTs
e et
e
ddd bwtwuDw
eeees
es
e
xy
yy
xx
dBdNu
Strain- displacement matrix, Be
Element stiffness matrix
Element external force matrix
Weak form
e
dEA eeeeTe BBK
eΓ
et
e
f
Γ
f
tNNf
dΓbdx eTeTe
e
F
n
e
eeTn
e
eeeTTelel
-LKL w fLdw
fK
011
x
N
y
N
x
N
y
N
x
N
y
N
x
N
y
N
y
Nx
N
x
N
x
N
en
en
eeee
en
ee
en
ee
ese
enen
en
en
...
0...00
0...00
2211
21
21
NB
Where, weight fn areeTeTTeeTe
s BwwBw )()(
FT
FT
w
w
rw ,f Kdr
fKdw
0
0)(
General Equation
3D2D1DEntity DomainΩ Boundary ΓOne Dimension(1D)Two Dimensions(2D)Three Dimension(3D)
Line SegmentTwo dimensional areaVolume
Two end pointsCurveSurface
3D
0bσ
Ts
yz
xz
xy
zz
yy
xx
Ts
yxz
zxy
zyx
000
000
000
z
y
x
z
y
x
zzzyzx
yzyyyx
xzxyxx
n
n
n
t
t
t
τ
n t
nt T
Equilibrium
Stress-traction relation
Strain-displacement relation
Isotropic Hooke’s Law
zyxT
yxxzxyzzyyxx
s
uuu
u
u
D
Dεσ
2
2100000
02
210000
002
21000
0001
0001
0001
)21)(1(
E
Recommended