Download - Solid Mechanics review 061904
1 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Review Solid Mechanics
&Finite Elements
Peter AvitabileMechanical Engineering DepartmentUniversity of Massachusetts Lowell
[ K ]n
[ M ]n [ M ]a[ K ]a [ E ]a
[ ω ]2
Structural Dynamic Modeling Techniques & Modal Analysis Methods
2 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Solid Mechanics Background
Basic understanding of Solid Mechanics is reviewed
This material is needed for refresher to support finite element model development
3 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Stress
The state of stress in an elemental volume is given as
If the coordinates are principal axes then
{ } [ ]xzyzxyzyxT τττσσσ=σ
{ } [ ]000321T σσσ=σ
4 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Strain
The state of strain in an elemental volume is given as
If the coordinates are principal axes then
{ } [ ]xzyzxyzyxT γγγεεε=ε
{ } [ ]000321T εεε=ε
5 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Strain-Displacement Equations
The strain displacement relations are
xw
zw
xv
zv
xu
zu
xw
zu
zw
yw
zv
yv
zu
yu
zv
yw
yw
xw
yv
xv
yu
xu
yu
xv
zw
zv
zu
21
zw
yw
yv
yu
21
yv
xw
xv
xu
21
xu
zx
yz
xy
222
z
222
y
222
x
∂∂
∂∂
+∂∂
∂∂
+∂∂
∂∂
+∂∂
+∂∂
=γ
∂∂
∂∂
+∂∂
∂∂
+∂∂
∂∂
+∂∂
+∂∂
=γ
∂∂
∂∂
+∂∂
∂∂
+∂∂
∂∂
+∂∂
+∂∂
=γ
∂∂
+
∂∂
+
∂∂
+∂∂
=ε
∂∂
+
∂∂
+
∂∂
+∂∂
=ε
∂∂
+
∂∂
+
∂∂
+∂∂
=ε
6 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Strain-Displacement Equations
Retaining only the first order (or linear) terms and neglecting the second order terms gives
zwyvxu
z
y
x
∂∂
=ε
∂∂
=ε
∂∂
=ε
yw
zv
xw
zu
xv
yu
yz
xz
xy
∂∂
+∂∂
=γ
∂∂
+∂∂
=γ
∂∂
+∂∂
=γ
7 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Linear Constitutive Equations
The generalized Hooke's Law can be written as
{ } [ ]{ }
γγγεεε
=
τττσσσ
⇒ε=σ
yz
xz
xy
z
y
x
666564636261
565554535251
464544434241
363534333231
262524232221
161514131211
yz
xz
xy
z
y
x
CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
C
8 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Linear Constitutive Equations
The generalized Hooke's Law can be written as
{ } [ ]{ }
τττσσσ
=
γγγεεε
⇒σ=ε
yz
xz
xy
z
y
x
666564636261
565554535251
464544434241
363534333231
262524232221
161514131211
yz
xz
xy
z
y
x
DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD
D
9 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Linear Constitutive Equations
The [C] and [D] matrices are symmetric and therefore only 21 constants are required to define a material in general
γγγεεε
=
τττσσσ
yz
xz
xy
z
y
x
66
5655
464544
36353433
2625242322
161514131211
yz
xz
xy
z
y
x
CCCCCCCCCCCCCCCCCCCCC
10 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Linear Constitutive Equations
Certain materials exibit symmetry with respect to certain planes within the body so that the number of material constants can be reduced from the general number of 21 material constants required for the anisotropic case.
11 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Linear Constitutive Equations
For instance, an orthotropic material can be expressed using only 9 constants
γγγεεε
=
τττσσσ
yz
xz
xy
z
y
x
66
55
44
33
2322
131211
yz
xz
xy
z
y
x
C0C00C000C000CC000CCC
12 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Linear Constitutive Equations
The stress-strain relations for an orthotropicmaterial may be written in terms of Young's Modulus and Poisson's Ratio as
zx
zxzx
yz
yzyz
xy
xyxy
yy
yzx
x
xz
z
zz
zz
zyx
x
xy
y
yy
zz
zxy
y
yx
x
xx
G,
G,
G
EEE
EEE
EEE
τ=γ
τ=γ
τ=γ
συ
−συ
−σ
=ε
συ
−συ
−σ
=ε
συ
−συ
−σ
=ε
13 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Linear Isotropic Elasticity
Simpliest form of the generalized Hooke's Law where the material is linear, elastic and isotropic
( )( )
( )( )
( )
τττσσσ
υ+
υ+
υ+ν−ν−
ν−ν−ν−ν−
=
γγγεεε
γγγεεε
ν−
ν−
ν−
ν−νννν−νννν−
ν−ν+=
τττσσσ
yz
xz
xy
z
y
x
yz
xz
xy
z
y
x
yz
xz
xy
z
y
x
yz
xz
xy
z
y
x
E)1(2
E)1(2
E)1(2
E/1E/E/E/E/1E/E/E/E/1
221
221
221
11
1
211E
14 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Two Dimensional Elasticity - Plane Strain
Typically used for very long bodies where the loading and boundary conditions do not vary in the longitudinal direction and that there are no displacements in the longitudinal direction. For this case,
xv
yu,
yv,
xu
xyyx ∂∂
+∂∂
=γ∂∂
=ε∂∂
=ε
15 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Two Dimensional Elasticity - Plane Strain
The constitutive law reduces to
( )( )
( )( )
γεε
ν−
ν−ννν−
ν−ν+=
τσσ
xy
y
x
xy
y
x
221
11
211E
16 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Two Dimensional Elasticity - Plane Stress
A condition of plane stress exists when the longitudinal direction is very small in comparison to the other two directions with only inplane loading considered. For this case,
0;0);(1 zzxyzyxz =σ=τ=τε+ε
υ−υ
−=ε
17 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Two Dimensional Elasticity - Plane Stress
The constitutive law reduces to
( )
γεε
ν−
νν
ν−=
τσσ
xy
y
x
2
xy
y
x
21
11
1E
18 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Elementary Plate Theory
When the thickness is small to the other two dimensions assumed displacements can be approximated by
)y,x(ww,ywz)y,x(vv,
xwz)y,x(uu ooo =
∂∂
−=∂∂
−=
19 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Elementary Plate Theory
The strain-displacement equations becomes
yxwz2
yxwz2
xv
yu
ywz
ywz
yv
xwz
xwz
xu
2oxy
2oo
xy
2
2
y2
2o
y
2
2
x2
2o
x
∂∂∂
−γ=∂∂
∂−
∂∂
+∂
∂=γ
∂
∂−ε=
∂
∂−
∂∂
=ε
∂
∂−ε=
∂
∂−
∂∂
=ε
20 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Elementary Plate Theory
Convention is to express the stress-strain relations as
∫∫∫
∫∫∫∫∫
−−−
−−−−−
τ=σ−=σ−=
τ=τ=τ=σ=σ=
2/t
2/txyxy
2/t
2/tyy
2/t
2/txx
2/t
2/tyzy
2/t
2/txzx
2/t
2/txyxy
2/t
2/tyy
2/t
2/txx
zdzM,dzzM,dzzM
dzQ,dzQ,dzN,dzN,dzN
21 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Finite Element Types
An assortment of different element types exist
L
A, EF F
u i
i j
u j
L
J, GT T
θ i
i j
θ j
L
E, I
F F
θ i
i j
θ j
ν i ν j
22 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Finite Element Types - TRUSS
Slender element (length>>area) which supports only tension or compression along its length; essentially a 1D spring
The truss strain is defined asThe stiffness and lumped/consistent mass matrices
L
A, EF F
u i
i j
u j
dxdu=ε
[ ] [ ] [ ]
ρ=
ρ=
−
−=
3/16/16/13/1
ALm;2/1
2/1ALm;
1111
LAEk cl
23 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Finite Element Types - TRUSS
Similar to truss but supports torsion
The torsional stiffness isL
J, GT T
θ i
i j
θ j
[ ]
−
−=
1111
LJGkt
24 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Finite Element Types - BEAM
Slender element whose length is much greater that its transverse dimension which supports lateral loads which cause flexural bendingBeam assumptions are
- constant cross section- cross section small compared to length- stress and strain vary linearly across
section depth
25 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Finite Element Types - BEAM
The beam elastic curvature due to lateral loading is satisfied by
The longitudinal strain is proportional to the distance from the neutral axis and second derivative of the elastic curvature given as
qdx/dEI 44 =υ
22 dx/dy υ=ε
L
E, I
F F
θ i
i j
θ j
ν i ν j
26 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Finite Element Types - BEAM
The stiffness and consistent mass matrices are
L
E, I
F F
θ i
i j
θ j
ν i ν j
[ ]
−−−−
−−
=
L4L6L2L6L612L612L2L6L4L6L612L612
LEIk
2
22
3 [ ]
−−
−−−−
ρ=
22
22
L4L22L3L13L22156L1354L3L13L4L22L1354L22156
420ALm
27 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Finite Element Types - BEAM
The full beam stiffness matrix can be assembled using the truss, torsion and two planar beam elements
−
−
−
−
−−−
−
−
−
−
−−
−−−
−
LEI4
LEI6
LEI2
LEI6
LEI4
LEI6
LEI2
LEI6
LJG
LJG
LEI6
LEI12
LEI6
LEI12
LEI6
LEI12
LEI6
LEI12
LAE
LAE
LEI2
LEI6
LEI4
LEI6
LEI2
LEI6
LEI4
LEI6
LJG
LJG
LEI6
LEI12
LEI6
LEI12
LEI6
LEI12
LEI6
LEI12
LAE
LAE
Z2ZZ
2Z
Y2YY
2Y
2Y
3Y
2Y
3Y
2Z
3Z
2Z
3Z
Z2ZZ
2Z
Y2YY
2Y
2Y
3Y
2Y
3Y
2Z
3Z
2Z
3Z
28 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Finite Element Types – PLANE STRAIN
Element whose geometry definition lies in a plane and applied loads also lie in the same plane
Stress-strain relations are
∆ x ∆ x ∆ x
∆ y ∆ y ∆ y
∆ u
∆ u
∆ v
∆ v∆ u∆ y
∆ v∆ x
xv
yu,
yv,
xu
xyyx ∂∂
+∂∂
=γ∂∂
=ε∂∂
=ε uvu
x/y/y/0
0x/
xy
y
x
∂=ε⇒
∂∂∂∂∂∂
∂∂=
γεε
29 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Finite Element Types – PLANE STRAIN
Displacements in the finite element are interpolated from nodal displacements using the element shape function
=
M
L
L
2
2
1
1
21
21
vuvu
N0N00N0N
vu
30 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Finite Element Types – PLANE STRAIN
Strain displacement matrix is
The element matrices are then given by
NBwherevuvu
Bvuvu
Nu
2
2
1
1
2
2
1
1
xy
y
x
∂=
=
∂=
γεε
⇒∂=ε
MM
[ ] [ ] [ ]
[ ] [ ] [ ][ ] VBCBK
VNNM
TV
TV
∂=
∂ρ=
∫∫∫∫∫∫
31 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Finite Element Types – PLANE STRAIN
Constant Strain Triangle - Probably one of the simplest and first of finite elements formulatedThe displacement field is given by
The resulting strain field is given by
yxvyxu
654
321
β+β+β=β+β+β=
53xy6y2x ,, β+β=γβ=εβ=ε
1
2
3
u1
u2
u3
v 1
v 2
v 3
x,u
y,v
32 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Finite Element Types – PLANE STRAIN
Constant Strain Triangle –The strain field can be expressed in terms of shape functions as 1
2
3
u1
u2
u3
v 1
v 2
v 3
x,u
y,v
−−−−−−−−−
−−−=
γεε
3
3
2
2
1
1
211213313223
123123
211332
xy
y
x
vuvuvu
yyxxyyxxyyxxxx0xx0xx0
0yy0yy0yy
A21
33 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Finite Element Types – PLANE STRAIN
Linear Strain Triangle - Adding midside nodes to the constant strain triangle provides for a quadratic displacement field
and a resulting strain field is given by
1
2
3
u1
u2
u3
v 1
v 2
v 3
x,u
y,v
4
56
u4
v4
u5v 5
u6v 6
21211
210987
265
24321
yxyxyxv
yxyxyxu
β+β+β+β+β+β=
β+β+β+β+β+β=
y)2(x)2(
y2xyx2
11610583xy
12119y
542x
β+β+β+β+β+β=γ
β+β+β=εβ+β+β=ε
34 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Finite Element Types – PLANE STRAIN
Bilinear Quadrilateral - Extending from the triangular element to a 4 noded quadrilateral provides for a bilinear displacement field
and a resulting strain field is given by
1u1
v 1
x,u
y,v
u4
u2
u3
v 2
v 3v 4
2
34
xyyxvxyyxu
8765
4321
β+β+β+β=β+β+β+β=
yx
xy
8463xy
87y
42x
β+β+β+β=γ
β+β=εβ+β=ε
35 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Bilinear Quadrilateral - Extending from the 4 noded to 8 noded quadrilateral provides for a bilinear displacement field
and a resulting strain field is given by
1u1
v 1
x,u
y,v
u4
u2
u3
v2
v 3v 4
2
34
5
6
7
8
Finite Element Types – PLANE STRAIN
216
215
21413
21211109
28
27
265
24321
xyyxyxyxyxv
xyyxyxyxyxu
β+β+β+β+β+β+β+β=
β+β+β+β+β+β+β+β=
216158
27136125103xy
162
15141311y2
87542x
yxy)(2xy)2(x)2(
xy2x2y2x;yxy2yx2
β+β+β+β+β+β+β+β+β+β=γ
β+β+β+β+β=εβ+β+β+β+β=ε
36 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Plane Strain Constant Strain Triangle* good in region where little strain gradient exists* otherwise element does not behave very well* poor element for bending applications* not considered a good general element
Plane Strain Linear Strain Triangle* better than the constant strain triangle* but not a particularly good general element
Finite Element Types – PLANE STRAIN
37 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Plane Strain Bilinear Quadrilateral* does not exactly model a cantilever beam* can not model a state of pure bending very well
due to shear effects* very stiff in bending* bending stiffness improved through incompatible
displacement effects
Plane Strain Quadratic Quadrilateral* good for modeling all states of constant strain* good for modeling pure bending using
rectangular elements
Finite Element Types – PLANE STRAIN
38 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
A general 3 dimensional solid that is relativelunrestricted with respect to shape, loading, material properties, boundary conditions, etc.Displacements are interpolated from nodal displacements from shape function
Finite Element Types – 3D SOLID
=
M
L
L
L
2
2
2
1
1
1
21
21
21
wvuwvu
N00N000N00N000N00N
wvu
39 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Strain displacement matrix is
The element matrices are then given by
Finite Element Types – 3D SOLID
NBwherevuvu
Bvuvu
Nu
2
2
1
1
2
2
1
1
xy
x
∂=
=
∂=
γ
ε
⇒∂=ε
MMM
M
[ ] [ ] [ ]
[ ] [ ] [ ][ ] VBCBK
VNNM
TV
TV
∂=
∂ρ=
∫∫∫∫∫∫
40 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Constant Strain Tetrahedron - Extension of constant strain triangle plain strain element to 3D
Also, Linear Tetrahedron, Trilinear Tetrahedron and Quadratic Tetrahedron
Finite Element Types – 3D SOLID
zy11xwzyxvzyxu
12109
8765
4321
β+β+β+β=β+β+β+β=β+β+β+β=
1
2
3
4
1 2
34
5 6
78
41 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Classical Thin Plate – Kirchoff Thin PlateVery similar to the beam in that flexure occurs -but in two directions. Geometry lies in the plane with loads acting normal to the plane. A two dimensional state of stress exists similar to that of plane stress with the exception that there is a variation of tension to compression across the plate thickness.
Finite Element Types – PLATE
w
u=-z
∆ wx∆
∆ wx∆
∆ wx∆
x,u
w
z
42 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Classical Thin Plate – Kirchoff Thin Plate
Governing equations are
The strain is given by
Finite Element Types – PLATE
y/wzv;x/wzu ∂∂−=∂∂−=
yx/wz2x/vy/u
y/wzy/v;x/wzx/u2
xy
22y
22x
∂∂∂−=∂∂+∂∂=γ
∂∂−=∂∂=ε∂∂−=∂∂=ε
43 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Classical Thin Plate – Kirchoff Thin PlateFor thin plate theory, the governing partial differential equation is given by
For an isotropic material,
Finite Element Types – PLATE
( )( )234 112/Et/qw υ−=∇
yxwzG2;
y/wx/w
11
1Ez
2
xy22
22
2y
x
∂∂∂
−=τ
∂∂∂∂
υ
υ
υ−−=
σσ
44 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Classical Thin Plate – Kirchoff Thin PlateThese stresses are very much like those found in simple beam bending. Flexural stresses vary linearly through the thickness while transverse shear stresses vary quadratically. The plate moments are given by
* transverse shear deformation is neglected* transverse shear can be significant in thick plates
(10:1 ratio applies)* cross section is not distorted by deformation
Finite Element Types – PLATE
∫∫∫+
−
+
−
+
−
τ=σ=σ=2/t
2/txyxy
2/t
2/tyy
2/t
2/txx dzzM;dzzM;dzzM
45 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Thick Plate - Mindlin PlateExtension of thin plate theory to account for transverse shear deformations.
Finite Element Types – PLATE
x,u
w
y,v
w
46 Dr. Peter AvitabileModal Analysis & Controls Laboratory22.515 – Solid Mechanics Review
Thick Plate - Mindlin PlateCombining the displacements for thin plate with those shown constitute the thick plate element.
* transverse shear deformation is included* cross section does not remain the same due to
shear deformation
Finite Element Types – PLATE
x
y
zv
zu
θ−=
θ=
yz
xz
xy
yx
∂∂θ
−=ε
∂
∂θ=ε
yyzx
xyz
xyxy
xwyw
xyz
θ+∂∂
=γ
θ−∂∂
=γ
∂
∂θ−
∂
∂θ=γ