Transcript No Slide Title

Plane problems in FEM analysis
from the real system,
to the mechanical
model, to the
mathematical model
Advanced Design for Mechanical System - Lec 2008/10/09
2
from the real system,
to the mechanical
model, to the
mathematical model
F=KD
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Introduction

Necessary preliminaries from solid mechanics theory are
reviewed.

Plane elements of several types are discussed

Particular attention it is posed to element displacement fields
and what they portend for element behaviour.

Treatment of loads and calculation of stress are discussed

The application of plane element to the frame structural
connection is presented
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Two-dimensional Elements

By definition, a plane body is flat and of constant thickness.
1.
2.
3.
Thin or thick plate elements .
Two coordinates to define position.
Elements connected at common nodes and/or along common
edges.
Nodal compatibility enforced to obtain equilibrium equations
Two types
- Plane stress
- Plane strain
4.
5.
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Plane stress
y
x,y ,xy  0
z ,xz ,yz = 0
T
x
Plate with a Hole
y
T
x
T
z0
Plate with a Fillet
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1.
A state of stress in which normal stress and shear stresses
directed perpendicular to the plane of the body are assumed to
be zero.
2.
If x-y plane is plane of body then only nonzero stresses are:
x,y ,xy
3.
Zero stresses: z ,xz ,yz
4.
Stresses act in the plane of the plate and can be called
membrane stresses. They are constant through the z-direction
5.
The deformation field is tridimensional (z0 ), the thickness is
free to increase or decrease in response to stress in xy plane.
6.
The component of the deformation perpendicular to the plane of
the body is due to the transversal contraction (Poisson effect).
As the thickness of a plane body increase, from much less to greater
than in-plane conditions of the body, there is a transition of behaviour
from plane stress toward plane strain
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Plane strain
y
x ,  y , xy  0
 z , xz , yz = 0
x
y
z
x
z
Pipe subjected to
external pressure
on its surface
z0
Dam Subjected to Horizontal Load
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1.
A state of strain in which normal strain and shear strains
directed perpendicular to the plane of the body are assumed to
be zero.
2.
If x-y plane is plane of body then only nonzero strains are: x ,
 y , xy
3.
Zero strains:  z , xz , yz
4.
Two principal stresses act in plane of the body
5.
The third principal stress, perpendicular to the plane of the
body, depends on the first two principal stresses and on the
Poisson coefficient
6.
The value of the third principal stress guarantee the
deformation perpendicular to plane of the body is zero and
prevent thickness change.
7.
Stresses are constant through the z-direction
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Two-dimensional State of Stress
xy
x
y
xy
dy
dx
xy
y
x
xy
y
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x
 x 


    y 
 
 xy 
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Principal stresses
σ1 
P
σx  σy
2
σx  σy
 σx  σy 
2
  τ xy
 
 2 
2
 σx  σy 
2


σ



τ
xy


x 2
2
 2 
2 τ xy
tan 2 θ p 
σx  σy
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Two-dimensional State of Strain
v
v
dy
y
changein lenght
divided t o
u
dy
y
y, v
t heoriginallenght
D
u
y
dy
B
A
v
dx
x
v
x
v
v
εy 
y
x, u changein t heright
u
dx
u
εx 
x
u
u  dx
x
Displacements and rotations of lines of an
element in the x-y plane
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angle
γ xy
u v


y x
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Constitutive relation
Let the material linearly elastic and isotropic

x 
1  
  1  
 

 y
 
z 
 1 
E
 

0
0
xy  (1 )(1 2)  0
 0
yz 
0
0

 
0
0
 0
zx 
0
0
0
(12)
2
0
0
0
0
0
0
(12)
2
0
 x 
  
 y 
 z 
 
 xy 
 yz 

(12) 
2  zx 
0
0
0
0
0
E = elastic modulus
 = Poisson coefficient
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Plane stress
x,y ,xy  0
z ,xz ,yz = 0
 yz  0
 zx  0
z

 x   y 

1 
1 
E 
 D 
 1
2 
1 
0 0

0 

0 
1  
2 
σ  Dε
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Plane strain
x ,  y , xy  0
 z , xz , yz = 0
1   
 x 
  1  
 

 y
   1 
 z 
E

 
 xy  (1  )(1  2 )  0 0 0
0 0 0
 yz 

 
 xz 
 0 0 0
Advanced Design for Mechanical System - Lec 2008/10/09
0
0
0
0
0
0
(1 2 )
2
0
0
(1 2 )
2
0
0
0   x 



0  y
 
0   0 
 
0   xy 
0  0 
(1 2 )   
 
2   0 
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1  
 x 
E

 
 1 
 y  

(
1


)(
1

2

)
 

0
0
xy
 

0   x 
 
0   y 
(1 2 )   
2   xy 
1 

E

 D 

1 

(1  )(1  2 )
 0
0

0 

0 
(1 2 ) 
2 
σ  Dε
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Strain-displacement relations
FE theory makes extensive use of strain-displacement relations
to obtain the strain field from a displacement field
The strain definitions suitable if the material has small strains
and small rotations are
 u 

 
 x    x    x
   v  
     y   
 0
    y  
 xy    u  v   
y


 y  x 


ε  u
Advanced Design for Mechanical System - Lec 2008/10/09
0 

 u 
 y v 

 
 x
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Displacement interpolation
Displacements in a plane FE are interpolated from nodal
displacement u(x,y), and v(x,y) as follow:
u ( x, y )  N1
u  

 v ( x, y )   0
0
N1
N2
0
0
N2
N3
0
 u1 
v 
 1
0  u2 
 

N 3  v2 
u3 
 
 v3 
u  Nd
where N is the shape function matrix
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According to the previous equation
- u depends only on the ui
- v depends only on the vi
- u and v use the same interpolation polynomials
This is a common arrangement but it is not mandatory
From the strain definition, we obtain
ε  Nd  Bd
where B
B
 N
is the strain-displacement matrix
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The general formula for k
The strain energy per unit volume of an elastic material in terms of
strain and in matrix format is:
1 T
U   ε Dε dV
2
Upon integrating over element volume V and substituting fromε  Nd  Bd
we obtain:
1 T T
1 T
U  d  B DB dV d  d k d
2
2
where the element stiffness matrix is
k   BT DB dV
For a given E, the nature of k depends entirely on B
B  N , the
behavior of an element is governed by its shape functions.
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•Let any element d.o.f. , say the i-th d.o.f., be increased from zero to
the value di.
•This is accomplished by applying to the d.o.f. a force that increases
from zero to Fi..
•The work is Fidi/2, just as it would be stretching a linear spring as
amount di.
•This work is stored as strain energy U.
•The previous equation says that work Fidi/2 is equal to strain energy
in the element when the displacement field is that produced by di and
the element shape functions.
•Example: if di = u1, we can see that the element displacement field is
u(x,y)=N1u1 and v(x,y) =0.
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Nature of the FE approximation
•
Stresses are in general functions of the coordinates, so that each
stress has a rate of change with respect to x and y.
•
In a plane problem the rate of change satisfy the equilibrium
equations
 xy  y
 x  xy

 Fx  0 and

 Fy  0
x
y
x
y
where Fx and Fy are body forces per unit volume.
•
As for deformation, they are called compatible if displacement
boundary conditions are met and material does not crack apart or
overlap itself.
•
If displacement and stress fields satisfy equilibrium, compatibility,
and boundary conditions on stress, then the solution is exact.
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How is the exact solution approached by FEA?
•
Let be elements based on polynomial displacement fields, as for
most elements in common use:
- the compatibility requirement is satisfied exactly within elements
- equilibrium equations and boundary conditions on stress are
satisfied in average sense
- at most point within the FEA model the equilibrium equations are
not satisfied
- as a mesh is repeatedly refined, pointwise satisfaction is
approached more and more closely
NOTE This discussion also applies to three-dimensional elastic
problems.
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Triangular Elements
1.
2.
3.
4.
Two degrees-of-freedom per node.
These are x and y displacements.
ui - x displacement at ith node.
vi - y displacement at ith node.
y
y
T
T
m
i
Thin Plate in Tension
x
j
Discretized Plate Using Triangular Elements
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x
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Constant strain triangle (CST)
The CST element is the earliest and simplest finite element
3 nodes numbered counterclockwise!
6 d.o.f.
3 (x3, y3)
y 1 (x y )
1, 1
2 (x2, y2)
x
Advanced Design for Mechanical System - Lec 2008/10/09
ui 
d i    
vi 
u1 
v 
1

d
 1
  u2 
d   d 2    
d  v2 
 3  u 
3
 
 v3 
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Funzioni di spostamento lineare
In terms of generalized coordinates i, (equal to the d.o.f. of the
element) its displacement field is
ux, y   1  2 x  3 y
vx, y   4  5 x  6 y
1.
2.
3.
4.
Ensures compatibility between elements.
Displacements vary linearly along any line.
Displacements vary linearly between nodes.
Edge displacements are the same for adjacent elements if nodal
displacements are equal.
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Strain field
The resulting strain field ε  u is:
u
x 
x
u
y 
y
 xy
u v


y x
 x  1  y  6  xy  3  5
1.
2.
3.
We see that strain do not vary within the element; hence the
name “constant strain triangle”.
The element also be called “linear triangle”, because its
displacement field is linear in x and y.
Element sides remain straight as the element deform.
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The strain field obtained from the shape function, in the form ε  Bd is:
u1 
v 
1
x 
 y23 0 y31 0 y12 0   
u2 
  1 

0 x32 0 x13 0 x21   
 y  

  2 A  x
v2 

y
x
y
x
y
32
13
31
21

 xy  
23


12
 u3 
 
B
 v3 
Were xi and yi are nodal coordinates (i=1,2,3)
xij = xi-xj
yij = yi-yj
2A= x21y31-x31y21 is twice the area of the triangle
The sequence 123 must go counterclockwise around the element if
the area of the element (A) is to be positive.
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Stiffness matrix
The stiffness matrix for the element is:
k  BT EB tA
Were t is the element thickness
1 
E 
E
 1
2 
1 
0 0

0 

0 
1  
2 
(plain stress conditions)
The integration to obtain K is trivial because B and E contain only
constants
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Observations
The CST gives good results in a region of the FE model where there
is little strain gradient. Otherwise it does not work well.
Stress along the x axis in a beam modeled by CSTs and loaded in pure bending
This is evident if we ask the CSTR element to model pure bending:
-The x-axis should be stress-free because it is the neutral axis;
-The FE model predicts x as a square wave pattern.
The element results enable to represent an x that varies linearly with y
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CST also develop a spurious shear stress when bent:
- v2 creates a shear stress that should not be present
Despite defects of CST, correct results are approached as a mesh
of CST elements is repeatedly refined
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Linear stress triangle
v3
v5
5
v1
1
6 nodes 12 d.of.
u
3
3
v4
u4
v2
u5 4
v6
u1
6
u6
2
u2
u1 
v 
d
 1  1
d  u 
 2  2
d 3  v2 
d       
d 4  u3 
d 5    
   
d 6    
v 
 4
The LST element has midside nodes in addition to vertex nodes.
The d.o.f. are ui and vi at each node i, i=1,2,…6, for a total of 12 d.o.f.
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In terms of generalized coordinates i its displacement field is
u x, y   1   2 x  3 y   4 x 2  5 xy  6 y 2
vx, y   7  8 x  9 y  10 x 2  11 xy  12 y 2
and the resulting strain field is
 x   2  2 4 x  5 y
 y  9  11 x  212 y
 xy  3  8 x  5  210 x  2 6  11 y
The strain field can vary linearly with x and y within the element,
hence the name “linear strain triangle” (LST).
The element may also be called a “quadratic triangle” because its
displacement field is quadratic in x and y
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Shape function
Displacement modes associated
with nodal d.o.f. v2=1 and v5=1
•LST has all the capability of CST, few they are, and more.
•The strain x can vary linearly with y
•If problem of pure bending is solved, exact results for deflection
and stresses are obtained
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Comparison of CST and LST Elements
1 Linear Strain Triangle
6 Nodes
12 D-O-F
4 Constant Strain Triangles
6 Nodes
12 D-O-F
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4 x 16 mesh
480 mm
120 mm
Parabolic Load
40 kN (Total)
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Test
# of Nodes
# of g.d.l.
# of Elements
4 x 16 Mesh
85
160
128 CST
8 x 32 Mesh
297
576
512 CST
2x8
85
160
32 LST
4 x 16
297
576
128 LST
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Run
g.d.l
Tip
deflection
(mm)
1
160
0.45834
51.225
0
120
2
576
0.51282
57.342
0
120
3
160
0.53259
59.145
0
120
4
576
0.53353
60.024
0
120
Exact Solution
Stress
(MPa)
X-location
Ylocation
0.53374
Advanced Design for Mechanical System - Lec 2008/10/09
60.000
0
120
38
Bilinear Quadrilater (Q4)
The Q4 element has 4 nodes and 8 d.o.f.
8 d.o.f.
In terms of generalized coordinates i its displacement field is
ux, y   1  2 x  3 y  4 xy
vx, y   5  6 x  7 y  8 xy
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Strain field
•The name bilinear arise because the form of the expression for u and
v is the product of two linear polynomials
c1  c2 x c3  c4 y
where ci are constants
The element strain field is
u
  2   4 y (linearin y independent of x )
x
v
y 
  7  8 x (linearin x independent of y )
y
u v
 xy 

  3   6    4 x   8 y
x y
x 
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Cannot exactly model a state of pure bending, despite its ability to
represent an x that varies linearly with y.
Correct deformation mode of a
rectangular block in pure bending:
Deformation of a bilinear quadrilater
under bending load:
•Plain section remain plane
•Top and bottom edges remain straight
•Top and bottom edges became arcs
of practically the same radius
•Right angles are not preserved under
pure moment load
•Shear strain xy is absent
•Shear strain appear everywhere (y0)
Q4 element that bent also develop shear strain
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• Cannot model a cantilever beam under transverse shear force where
the moment and the axial strain x vary linearly with x.
Qualitative variation of axial
stress and average
transverse shear stress
Q4 element results too stiff in bending because an applied bending moment
is resisted by spurious shear stress as well as by the expected flexural
stresses (locking)
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Shape functions
If the generalized coordinates i are expressed in terms of nodal d.o.f.,we
obtain the displacement field in the form u  Nd where
Shape function N2
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Strain field
The element strain field is
ε  Nd
ε  Bd
• Equilibrium is not satisfied at every point unless 4 = 8 = 0 (constant
strain)
• The element converges properly with mesh refinement and in most
problem it works better than the CST element which always satisfied
the equilibrium equations
• Non rectangular shape are permitted.
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Comparative examples
a
a
•The test problem chosen here is that of a cantilever beam of unit thickness
loaded by a transverse tip force.
•Plane stress conditions prevail.
•Support conditions are consistent with a fixed end but without restraint of
y-direction deformation associated with the Poisson effect.
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Results
• Simple beam element solves the problem exactly when transverse
shear deformation is included in the formulation.
• As expected, CST elements perform poorly.
• Q4 element are better but not good.
• LST elements give an accurate deflection but a disappointing stress.
Distortion and elongation of elements are seen to reduce accuracy
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Elementi quadrilateri a 8 nodi (Q48)
The Q8 element has midside nodes in addition to vertex nodes.
The d.o.f. are ui and vi at each node i, i=1,2,…8, for a total of 16 d.o.f.
x
a
y

b

16 g.d.l.
In terms of generalized coordinates i its displacement field is
u x, y   a1  a2 x  a3 y  a4 x 2  a5 xy  a6 y 2  a7 x 2 y  a8 xy 2
vx, y   a9  a10 x  a11 y  a12 x 2  a13 xy  a14 y 2  a15 x 2 y  a16 xy 2
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Shape functions (Q8)
The displacement field in terms of shape function is u   Niui v   Ni vi
As examples, two of the eight shape functions are
1
1
1
N 2  1   1     1   2 1     1   1   2 
4
4
4
1
N6  1   1   2 
2
x
a
y

b

The displacements are quadratic in y, which means that the edge
deform into a parabola when a single d.o.f on that edge is nonzero.
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Strain field
The element strain field is :
 u 


  x   x 
    y    v 
    y 
 xy   u v 
 y  x 


  Bd
 x   2  2 4 y  5 y  27 xy  8 y 2
(no term in x2)
 y  11  13 x  214 y  15 x 2  216 xy
….
 xy  3  10   5  212 x  26  13  y 
  7 x 2  28  15 xy  16 y 2
•Q8 element can represent exactly all states of constant strain, and
state of pure bending, if it is rectangular.
•Non rectangular shape are permitted.
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Final comparison
a
a
Advanced Design for Mechanical System - Lec 2008/10/09
Q8 elements are
the best
performers
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