Lecture 17: Plane Stress & Strain (the CST element).mech420/Lecture17.pdf · 2012-06-26 · Plane...

MECH 420: Finite Element Applications

n  Chapter 6: Development of the Plane Stress and Plane Strain Stiffness Equations. n  Up until this point we have discussed line or filament elements. n  To solve plane stress or strain problems we now have to consider a

displacement field of dimension 2. n  We will have planar elements that connect along edges. n  We must choose displacement fields that ensure compatibility

along the edges. n  A compatible element:

n  Is conforming – the displacement function ensures displacement continuity within and across elements.

n  Is complete – the necessary level of differentiation is possible. We have spent the last two classes discussing how we manipulate DE’s so that our element type is “complete.”

Lecture 17: Plane Stress & Strain (the CST element).


n  Chapter #6 focuses on developing the ee’s for the constant strain triangular (CST) element. n  Simplest derivation of the planar elements. n  Element exhibits constant strain (what does this say about the

displacement function?). n  Logan applies a variational (stationary potential energy

formulations) to produce the CST element equations. n  We will follow this approach for now. n  The CST element equations are re-used in Chapters 13 and 14 for

heat transfer and fluid flow respectively. n  We will apply a weighted residuals formulation in those later

chapters.



n  §6.1. Basic Concepts of Plane Stress and Plane Strain. n  Variety of three-dimensional problems can be analyzed using a

planar depiction of the problem. n  Two cases where this happens:

1.  Plane stress problems – the original 3-D elastica is thin and is loaded in the x-y plane.

n  Geometry is reduced to a slice. 2.  Plane strain – the body and loading are extruded a great

distance out of the plane. n  We can consider a sample “slice” of the original problem.



n  Plane Stress:

n  Plane Strain:

σ z = τ xz = τ yz = 0

∴σ = σ x σ y τ xy{ }

T

↓

ε = εx ε y γ xy{ }

T

εz = γ xz = γ yz = 0

∴ε = εx ε y γ xy{ }

T

↓

σ = σ x σ y τ xy{ }

T



n  In either case, the strains and stresses must be related by an elasticity matrix (or constitutive matrix).

n  Plane stress:

n  Plane strain:

σ = Dε ; D = E

1−ν 2

1 ν 0ν 1 0

0 0 1−ν2

"

#

$$$$$

%

&

'''''

εx =σ x

E-νσ y

E-ν (0)E

ε y = −νσ x

E+σ y

E-ν (0)E

εz = −νσ x

E−ν

σ y

E+

(0)E→εz is not independent.

σ = Dε ; D = E

(1+ν )(1− 2ν )

1−ν ν 0ν 1−ν 0

0 0 1− 2ν2

"

#

$$$$$

%

&

'''''

εx =σ x

E-νσ y

E-νσ z

E

ε y = −νσ x

E+σ y

E-νσ z

E

0 = −νσ x

E−ν

σ y

E+σ z

E→σ z is not independent.



n  Logan reviews some of the principles of applied mechanics that we applied in our variational formulations (minimum potential energy). n  These relations are limited to the planar case. n  For example, relations between the displacement field and strain:

ˆ ˆ ˆ ˆ, ( , )x u x y

ˆ ˆ ˆ ˆ, ( , )x v x y

εx =∂udx

ε y =∂vdx

γ xy =∂udy+∂vdx



n  Cauchy’s motion equation is also presented for the cases of plane stress and strain (eq. 6.1.11). n  “motion” of a hypothetical particle inside the continuum is

contained to the x-y plane. n  Cauchy’s equation for the plane stress case is given on pg.# 334:

2 2 2 2

2 2 2

2 2 2 2

2 2 2

ˆ ˆ ˆ ˆ1ˆ ˆˆ ˆ ˆ2

ˆ ˆ ˆ ˆ1ˆ ˆˆ ˆ ˆ2

u u u vx ydx dy y

v v v ux ydx dy x

ν

ν

⎛ ⎞∂ ∂ + ∂ ∂+ = −⎜ ⎟∂ ∂∂⎝ ⎠

⎛ ⎞∂ ∂ + ∂ ∂+ = −⎜ ⎟∂ ∂∂⎝ ⎠

Given these governing differential equations you could apply a weighted residuals approach. We won’t until Chapter #13.



n  The constant strain triangular element models the behaviour of the hypothetical triangular slice of material that we look at in either the plane stress or strain case.

n  Element equations must relate the displacement of the planar region in reaction to loads applied… n  over the surface/perimeter of the element. n  over the body/interior of the element. n  at the nodes/keypoints that define the element.

n  A triangular planar region is used as it can be assembled to form a variety of convex or concave geometries.



n  Example…



n  §6.2. Derivation of the Constant-Strain Triangular Element Stiffness Matrix and Equations. n  Step 1: Set the element type. n  The idea of compatibility has to be expanded upon from when we

considered line/filament type elements: Lettering scheme

1 2

Common edge experiences same

displacement for both elements 1 and 2.



n  Step 2: Select a displacement function. n  For constant strain we know that a linear displacement field is

compatible. n  The blend of node displacements in the x and y directions must

follow the same general linear form. u(x, y) = a1 + a2x+ a3yv(x, y) = a4 + a5x+ a6 y

u =a1 + a2x+ a3ya4 + a5x+ a6 y

!"#

$#

%&#

'#

ui =uivi

!"#

$#

%&#

'#

a1 + a2xi + a3yia4 + a5xi + a6 yi

!"#

$#

%&#

'# , u j =

u jv j

!"#

$#

%&#

'#

a1 + a2x j + a3y ja4 + a5x j + a6 y j

!"#

$#

%&#

'# , um =

umvm

!"#

$#

%&#

'#

a1 + a2xm + a3yma4 + a5xm + a6 ym

!"#

$#

%&#

'#

6 constraint equations in terms of 6 unknowns a1, .., a6



n  The finite element approximation to the stress and strain in the region has to be formed from an approximate displacement field.

n  We factor the linear approximation to the displacement field to isolate the shape functions and the nodal displacements.

ψ =u = Na

uv

!"#

$#

%&#

'#=

Ni 0

0 Ni

(

)

**

+

,

--

N j 0

0 N j

(

)

**

+

,

--

Nm 0

0 Nm

(

)

**

+

,

--

(

)

***

+

,

---

dixdiy

!

"#

$#

%

&#

'#

d jxd jy

!

"#

$#

%

&#

'#

dmxdmy

!

"#

$#

%

&#

'#

!

"

######

$

######

%

&

######

'

######

Each of the three shape functions acts to blend

the nodal displacements over the region to create

a displacement field.

di =

dixdiy

!

"#

$#

%

&#

'#

ur



n  The shape functions can be constructed from the equilibrium values of the node positions.

Ni =12A

αi +βi x +γ i y( )

N j =12A

α j +β j x +γ j y( )

Nm =12A

αm +βm x +γm y( )

αi = x j ym − y jxm α j = xm yi − ymxi αm = xi y j − yix jβi = y j − ym β j = ym − yi βm = yi − y jγ i = xm − x j γ j = xi − xm γm = x j − xi

A is the area of the triangle

A≡ Area of the triangular element

2A= xi y j − ym( )+ x j ym − yi( )+ xm yi − y j( )



n  Step 3: Define the stress-strain relationships. n  Given an approximation to the displacement field one can evaluate

the approximate strains. n  Stresses are obtained by applying the appropriate elasticity matrix.

ε =

εxε y

γ xy

!

"##

$##

%

&##

'##

=

∂u∂x∂v∂y

∂u∂y

+∂v∂x

!

"

####

$

####

%

&

####

'

####

= B d = 12A

βi 0

0 γ iγ i βi

)

*

++++

,

-

.

.

.

.

12A

β j 0

0 γ j

γ j β j

)

*

++++

,

-

.

.

.

.

12A

βm 0

0 γmγm βm

)

*

++++

,

-

.

.

.

.

)

*

+++++

,

-

.

.

.

.

.

did jdm

!

"##

$##

%

&##

'##

d =

did jdm

!

"##

$##

%

&##

'##

B



n  Step 4: Derive the element stiffness matrix and equations. n  We have seen how element equations (in structural problems) are

actually a system of algebraic equations that are trying to force a state of minimal potential energy.

n  This state is obtained by applying the best set of nodal displacements.

n  The potential energy in the system can be attributed to: n  Strain energy. n  Distributed body forces. n  Surface tractions. n  Point loads at the node points.

Ω

U



n  The total potential energy is given by:

n  When the potential energy in the system is minimized…

n  To calculate the strain energy we apply:

π P =U +ΩB +ΩS +ΩP

∂π P∂ d{ }

= 0

σ = D B d{ }

Two possibilities depending on case considered.



n  Recall some of the assumptions in Step 1: n  Very thin slice – constant thickness t. n  The shape functions were linear in x and y. n  The strains are functions of first order differentials of the shape

functions. n  All 3 strains are constant over the element. n  There is no dependence of B on x and y.

ε = B d = 1

2A

βi 0

0 γ iγ i βi

!

"

####

$

%

&&&&

12A

β j 0

0 γ j

γ j β j

!

"

####

$

%

&&&&

12A

βm 0

0 γmγm βm

!

"

####

$

%

&&&&

!

"

#####

$

%

&&&&&

did jdm

'

())

*))

+

,))

-))

Dependent on the original (undeformed) locations of the nodes.



n  The stiffness matrix becomes:

k = (t ⋅ A) BTDB"#

$%

Area of the triangular element face.2 2( ) ( ) ( )i j m j m i m i jx y y x y y y

Ax y

≡ ×

= − + − + −



n  Looking at the body fixed loads and the surface tractions.

ΩB = −uT

V (e)∫

b ⋅dV = −

ψT

V (e)∫

Xb ⋅dV = −d T N T

V (e)∫

Xb ⋅dV

f B = − dΩB

d d{ }= NT

V (e)∫

Xb ⋅dV

ΩS = −uT

A(e)∫

t ⋅dA= −

ψT

S (e)∫

Ts ⋅dS = −d

T N T

S (e)∫

Ts ⋅dS

f S = −dΩS

d d{ }= NT

S (e)∫

Ts ⋅dS Since the loading is planar, the

surface area of the element is the area of the three edges. The shape functions are simply the bar element shape functions over these edges.



Along edge ij, Ni varies linearly between 1 at node i and 0 at node j. The thickness of edge ij is constant ‘t.’



n  Step 5: Assembly. n  When we assemble the element equations we will likely have

different reference frames used for each set of element equations.

Element #1 Element #2

nnyf

nxf



n  If we decided to adopt the element #1 coordinates system we could transform the element #2 equations.

{ }{ } { }{ }

(2)

(2) (2) (2)

(2) (2)

ˆ ˆ ˆ

ˆ

ˆ T

f k d

T f k T d

f T k T d

=

=

⎡ ⎤= ⎣ ⎦

zSense of θ defined by the RHR

θ brings the global frame into alignment with the particular element frame

(2)

(2) (2)

(2)

0 00 00 0

Tz

Tz

Tz

RT R

R

⎡ ⎤⎢ ⎥

= ⎢ ⎥⎢ ⎥⎣ ⎦

(2)

(2) (2)(2)

(2) (2)

ˆ

ˆnx nx

zny ny

z

d dR

d d

c sR

s cθ θ

θ θ

⎧ ⎫⎧ ⎫ ⎪ ⎪=⎨ ⎬ ⎨ ⎬

⎩ ⎭ ⎪ ⎪⎩ ⎭

⎡ ⎤−= ⎢ ⎥⎣ ⎦


Lecture 17: Plane Stress & Strain (the CST element).mech420/Lecture17.pdf · 2012-06-26 · Plane...

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Transcript of Lecture 17: Plane Stress & Strain (the CST element).mech420/Lecture17.pdf · 2012-06-26 · Plane...