Solid Mechanics review 061904 - Faculty Server...

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


Solid Mechanics Background

Basic understanding of Solid Mechanics is reviewed

This material is needed for refresher to support finite element model development


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


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


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


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


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



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



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



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.



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



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


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


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


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


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


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


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 =

∂∂

−=∂∂

−=



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



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


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


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


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


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



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



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



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


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



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



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=

∫∫∫∫∫∫



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



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



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



Bilinear Quadrilateral - Extending from the triangular element to a 4 noded quadrilateral provides for a bilinear displacement field


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


Bilinear Quadrilateral - Extending from the 4 noded to 8 noded quadrilateral provides for a bilinear displacement field


1u1

v 1

x,u

y,v

u4

u2

u3

v2

v 3v 4

2

34

5

6

7

8


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


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



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



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


Strain displacement matrix is

The element matrices are then given by


NBwherevuvu

Bvuvu

Nu

2

2

1

1

2

2

1

1

xy

x

∂=

=

∂=

γ

ε

⇒∂=ε

MMM

M

[ ] [ ] [ ]

[ ] [ ] [ ][ ] VBCBK

VNNM

TV

TV

∂=

∂ρ=

∫∫∫∫∫∫


Constant Strain Tetrahedron - Extension of constant strain triangle plain strain element to 3D

Also, Linear Tetrahedron, Trilinear Tetrahedron and Quadratic Tetrahedron


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


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


Classical Thin Plate – Kirchoff Thin Plate

Governing equations are

The strain is given by


y/wzv;x/wzu ∂∂−=∂∂−=

yx/wz2x/vy/u

y/wzy/v;x/wzx/u2

xy

22y

22x

∂∂∂−=∂∂+∂∂=γ

∂∂−=∂∂=ε∂∂−=∂∂=ε


Classical Thin Plate – Kirchoff Thin PlateFor thin plate theory, the governing partial differential equation is given by

For an isotropic material,


( )( )234 112/Et/qw υ−=∇

yxwzG2;

y/wx/w

11

1Ez

2

xy22

22

2y

x

∂∂∂

−=τ

∂∂∂∂

υ

υ

υ−−=

σσ


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


∫∫∫+

−

+

−

+

−

τ=σ=σ=2/t

2/txyxy

2/t

2/tyy

2/t

2/txx dzzM;dzzM;dzzM


Thick Plate - Mindlin PlateExtension of thin plate theory to account for transverse shear deformations.


x,u

w

y,v

w


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


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

Solid Mechanics review 061904 - Faculty Server...

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