Transcript MANE 4240 & CIVL 4240 Introduction to Finite Elements Prof. Suvranu De Introduction to 3D Elasticity.

MANE 4240 & CIVL 4240
Introduction to Finite Elements
Prof. Suvranu De
Introduction to 3D
Elasticity
Reading assignment:
Appendix C+ 6.1+ 9.1 + Lecture notes
Summary:
• 3D elasticity problem
•Governing differential equation + boundary conditions
•Strain-displacement relationship
•Stress-strain relationship
•Special cases
2D (plane stress, plane strain)
Axisymmetric body with axisymmetric loading
• Principle of minimum potential energy
1D Elasticity (axially loaded bar)
y
F
x
x=0
x=L
A(x) = cross section at x
b(x) = body force distribution
(force per unit length)
x E(x) = Young’s modulus
u(x) = displacement of the bar
at x
1. Strong formulation: Equilibrium equation + boundary
conditions
Equilibrium equation
Boundary conditions
d
 b  0;
dx
0 xL
u  0 at x  0
du
EA
 F at x  L
dx
2. Strain-displacement relationship: ε(x) 
du
dx
3. Stress-strain (constitutive) relation :  (x)  E ε(x)
E: Elastic (Young’s) modulus of bar
Problem definition
Surface (S)
3D Elasticity
V: Volume of body
S: Total surface of the body
Volume (V)
The deformation at point
w
x =[x,y,z]T
is given by the 3
v
u
u 
components of its
 
z
u

v
displacement
x
w 
 
NOTE: u= u(x,y,z), i.e., each
displacement component is a function
y
of position
x
3D Elasticity:
EXTERNAL FORCES ACTING ON THE BODY
Two basic types of external forces act on a body
1. Body force (force per unit volume) e.g., weight, inertia, etc
2. Surface traction (force per unit surface area) e.g., friction
BODY FORCE
Volume
element dV
Xc dV
Xb dV
w
u
z
x
x
y
Xa dV
Volume (V)
v
Surface (S)
Body force: distributed
force per unit volume (e.g.,
weight, inertia, etc)
X a 
 
X  X b 
X 
 c
NOTE: If the body is accelerating,  u 
 



u

then the inertia force
 v 
 w



 
may be considered as part of X
~
X  X   u
SURFACE TRACTION
Volume
element dV
pz
Xc dV
Xb dV p
x
Xa dV
w
Volume (V)
v
ST
u
z
x
x
y
Traction: Distributed
py force per unit surface
area
p x 
 
T S  p y 
p 
 z
3D Elasticity:
INTERNAL FORCES
Volume
element dV
w
u
z
x
Volume (V)
v
z
tzx
txz t
xy
x
tzy
tyz
y
tyx
y
x If I take out a chunk of material from the body, I will see that,
due to the external forces applied to it, there are reaction
forces (e.g., due to the loads applied to a truss structure, internal
forces develop in each truss member). For the cube in the figure,
the internal reaction forces per unit area(red arrows) , on each
surface, may be decomposed into three orthogonal components.
z
3D Elasticity
x, y and z are normal stresses.
tzy
The rest 6 are the shear stresses
tzx
tyz
txz
y Convention
txy
txy is the stress on the face
z
perpendicular to the x-axis and points
tyx
x
in the +ve y direction
Total of 9 stress components of which
y
only 6 are independent since t xy  t yx
x
t yz  t zy
 x 
 
t zx  t xz
y

The stress vector is therefore
 z 
  
t xy 
t yz 
 
t zx 
 x 
 
Strains: 6 independent strain components
 y
  z 
  
 xy 
 yz 
 
 zx 
Consider the equilibrium of a differential volume element to
obtain the 3 equilibrium equations of elasticity
 x t xy t xz


 Xa  0
x
y
z
t xy  y t yz


 Xb  0
x
y
z
t xz t yz  z


 Xc  0
x
y
z
Compactly;
EQUILIBRIUM
EQUATIONS
where

 x

0


0


 y

0


 z
  X 0
T
0

y
0

x

z
0

0

0


z 
0



y 

x 
(1)
3D elasticity problem is completely defined once we
understand the following three concepts
Strong formulation (governing differential equation +
boundary conditions)
Strain-displacement relationship
Stress-strain relationship
Volume
element dV
pz
Xc dV
Xb dV p
x
w
u
z
py
Xa dV
Volume (V)
v
ST
x
Su
y
x
1. Strong formulation of the 3D elasticity problem: “Given the
externally applied loads (on ST and in V) and the specified
displacements (on Su) we want to solve for the resultant
displacements, strains and stresses required to maintain
equilibrium of the body.”
Equilibrium equations
   X  0 in V
T
(1)
Boundary conditions
1. Displacement boundary conditions: Displacements are specified on
portion Su of the boundary
u u
specified
on Su
2. Traction (force) boundary conditions: Tractions are specified on
portion ST of the boundary
Now, how do I express this mathematically?
Volume
element dV
pz
Xc dV
Xb dV p
x
Xa dV
w
Volume (V)
v
ST
u
z
x
Su
x
y
Traction: Distributed
py force per unit area
p x 
 
T S  p y 
p 
 z
TS
nz
pz
n
Traction: Distributed
force per unit area
py
ny
nx
ST
px
n x 
 
If the unit outward normal to ST : n  n y 
n 
 z
Then
p x   x nx  t xy n y  t xz nz
p y  t xy nx   y n y  t yz nz
p z  t xz nx  t zy n y   z nz
p x 
 
T S  p y 
p 
 z
In 2D
n
q
dy ds
y
txy
ny
q
nx
x
dx
ST
x
dx
sin q 
 ny
ds
dy
cos q 
 nx
ds
py
q
dy
txy
ds
dx
TS
px
y
Consider the equilibrium of the wedge in
x-direction
p x ds   x dy  t xy dx
dy
dx
 t xy
ds
ds
 p x   x n x  t xy n y
 px   x
Similarly
p y  t xy nx   y n y
3D elasticity problem is completely defined once we
understand the following three concepts
Strong formulation (governing differential equation +
boundary conditions)
Strain-displacement relationship
Stress-strain relationship
2. Strain-displacement relationships:
u
x
v
y 
y
w
z 
z
u v
 xy 

y x
v w
 yz  
z y
u w
 zx 

z x
x 
Compactly;
 x 
 
 y
  z 
  
 xy 
 yz 
 
 zx 
 u

 x

0


0


 y

0


 z
(2)
0

y
0

x

z
0

0

0


z 
0



y 

x 
u 
 
u  v
w 
 
u
dy
y
In 2D
v
v
dy
y
C’
y
C
2
A’ 
u
1
dy
v
A dx B
B’
x
u


u


u
dx

dx   u 
dx   u 
 dx
x


x
A' B' AB
u



x 
 

AB
dx
x




v
 dy  
  dy

v

dy

v




y
A' C' AC
v




y 


AC
dy
y
π
 xy   angle (C' A' B' )  β1  β 2  t anβ1  t anβ 2
2
v
u


x
x
v
dx
x
3D elasticity problem is completely defined once we
understand the following three concepts
Strong formulation (governing differential equation +
boundary conditions)
Strain-displacement relationship
Stress-strain relationship
3. Stress-Strain relationship:
Linear elastic material (Hooke’s Law)
  D
(3)
Linear elastic isotropic material


1  
 
1 


 

1 

E
0
0
 0
D
(1   )(1  2 ) 
 0
0
0


0
0
 0
0
0
0
0
0
1  2
2
0
0
0
1  2
2
0
0

0 
0 

0 

0 

1  2 
2 
0
Special cases:
1. 1D elastic bar (only 1 component of the stress (stress) is
nonzero. All other stress (strain) components are zero)
Recall the (1) equilibrium, (2) strain-displacement and (3) stressstrain laws
2. 2D elastic problems: 2 situations
PLANE STRESS
PLANE STRAIN
3. 3D elastic problem: special case-axisymmetric body with
axisymmetric loading (we will skip this)
PLANE STRESS: Only the in-plane stress components are nonzero
Area
element dA
Nonzero stress components  x ,  y ,t xy
h
t xy  y
t xy
x
D
y
x
Assumptions:
1. h<<D
2. Top and bottom surfaces are free from
traction
3. Xc=0 and pz=0
PLANE STRESS Examples:
1. Thin plate with a hole
t xy  y
2. Thin cantilever plate
t xy
x
PLANE STRESS
Nonzero stresses:  x , y ,t xy
Nonzero strains:  x ,  y ,  z ,  xy
Isotropic linear elastic stress-strain law   D 
 x 
E
 
 y  
2
1


t 
 xy 

 
1

0

  x 
 1
0   y 

1    
0 0
  xy 
2 

z  
Hence, the D matrix for the plane stress case is


1

0

E 
 1
D
0 
2
1  
1  
0 0

2 


1 

x
y
PLANE STRAIN: Only the in-plane strain components are nonzero
Nonzero strain components  x ,  y ,  xy
Area
element dA
y
x
z
 xy
y
 xy
x
Assumptions:
1. Displacement components u,v functions
of (x,y) only and w=0
2. Top and bottom surfaces are fixed
3. Xc=0
4. px and py do not vary with z
PLANE STRAIN Examples:
1. Dam
Slice of unit
thickness
1
t xy  y
y
z
t xy
x
x
z
2. Long cylindrical pressure vessel subjected to internal/external
pressure and constrained at the ends
PLANE STRAIN
Nonzero stress:  x , y ,  z ,t xy
Nonzero strain components:  x ,  y ,  xy
Isotropic linear elastic stress-strain law   D 

 x 
1 


E
 
 
1 
 y  
t  1   1  2   0
0

 xy 


0   x 
 
0   y 
1  2   
  xy 
2 
Hence, the D matrix for the plane strain case is

1 


E
 
D
1 
1   1  2  
0
 0


0 
0 
1  2 

2 
 z    x   y 
Example problem
y
The square block is in plane strain
and is subjected to the following
strains
2
2
1
2
3
 x  2 xy
4
x
 y  3xy 2
 xy  x2  y3
Compute the displacement field (i.e., displacement components
u(x,y) and v(x,y)) within the block
Solution
Recall from definition
u
 2 xy
(1)
x
v
 y   3xy 2 (2)
y
u v
 xy    x 2  y 3 (3)
y x
x 
Arbitrary function of ‘x’
Integrating (1) and (2)
u ( x, y )  x 2 y  C1 ( y)
(4)
v( x, y)  xy3  C2 ( x)
(5)
Arbitrary function of ‘y’
Plug expressions in (4) and (5) into equation (3)
u v

 x 2  y 3 (3)
y x

 

 x 2 y  C1 ( y )  xy3  C2 ( x)


 x2  y3
y
x
C ( y )
C ( x)
 x2  1
 y3  2
 x2  y3
y
x
C ( y ) C2 ( x)
 1

0
y
x
Function of ‘y’
Function of ‘x’
Hence
C1 ( y)
C2 ( x)

 C (a constant)
y
x
Integrate to obtain
C1 ( y)  Cy  D1
C2 ( x)  Cx  D2
D1 and D2 are two constants of
integration
Plug these back into equations (4) and (5)
(4) u ( x, y )  x 2 y  Cy  D1
(5) v( x, y )  xy3  Cx  D2
How to find C, D1 and D2?
Use the 3 boundary conditions
u (0,0)  0
v(0,0)  0
v(2,0)  0
To obtain
C 0
D1  0
D2  0
Hence the solution is
u ( x, y )  x 2 y
v ( x, y )  xy 3
y
2
2
1
2
3
4
x
Principle of Minimum Potential Energy
Definition: For a linear elastic body subjected to body forces
X=[Xa,Xb,Xc]T and surface tractions TS=[px,py,pz]T, causing
displacements u=[u,v,w]T and strains  and stresses , the potential
energy P is defined as the strain energy minus the potential energy
of the loads involving X and TS
PU-W
Volume
element dV
pz
Xc dV
py
Xb dV p
x
w
u
z
Xa dV
Volume (V)
v
ST
x
Su
x
y
1
T
U     dV
2 V
W   u X dV   u T S dS
T
V
T
ST
Strain energy of the elastic body
Using the stress-strain law   D 
1
1
T
T
U     dV    D  dV
2 V
2 V
In 1D
1
1
1 L
2
U    dV   E dV   E 2 Adx
2 V
2 V
2 x 0
In 2D plane stress and plane strain
1
U    x  x   y  y  t xy  xy  dV
2 V
Why?
Principle of minimum potential energy: Among all admissible
displacement fields the one that satisfies the equilibrium equations
also render the potential energy P a minimum.
“admissible displacement field”:
1. first derivative of the displacement components exist
2. satisfies the boundary conditions on Su