MANE 4240 & CIVL 4240
Introduction to Finite Elements
Reading assignment:
Appendix C+ 6.1+ 9.1 + Lecture notes
Summary:
Prof. Suvranu De
Introduction to 3D
Elasticity
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
• 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
2. Strain-displacement relationship: ε(x) 
du
dx
3. Stress-strain (constitutive) relation :  (x)  E ε(x)
E: Elastic (Young’s) modulus of bar
1. Strong formulation: Equilibrium equation + boundary
conditions
Equilibrium equation
d
 b  0;
dx
Boundary conditions
u0
0 xL
at x  0
du
EA
 F at x  L
dx
1
Problem definition
3D Elasticity:
EXTERNAL FORCES ACTING ON THE BODY
3D Elasticity
Surface (S)
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 
components of its
 
u  v
displacement
w 
 
NOTE: u= u(x,y,z), i.e., each
u
z
x
displacement component is a function
of position
y
x
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
SURFACE TRACTION
BODY FORCE
Volume
element dV
Xc dV
w
u
z
x
x
y
Xb dV
Xa dV
Volume (V)
v
Surface (S)
Body force: distributed
force per unit volume (e.g.,
weight, inertia, etc)
Volume
element dV
Xb dV p
x
w
X a 
 
X  X b 
X 
 c
NOTE: If the body is accelerating,  u 
 
 u   v 
then the inertia force
 w

  
may be considered as part of X
~
X  X   u
pz
Xc dV
u
z
x
x
Xa dV
Volume (V)
v
ST
Traction: Distributed
py force per unit surface
area
p x 
 
T S  p y 
p 
 z
y
2
3D Elasticity:
INTERNAL FORCES
Volume
element dV
w
u
z
Volume (V)
v
x
zx
xz 
xy
x
zy
yz
zy
zx
y
z
yx
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.
 x 
 
 y
  
  z 
 xy 
 yz 
 
 zx 
Consider the equilibrium of a differential volume element to
obtain the 3 equilibrium equations of elasticity
 x  xy  xz


 Xa  0
x
y
z
 xy  y  yz


 Xb  0
x
y
z
 xz  yz  z


 Xc  0
x
y
z
xz
xy
x
x
Strains: 6 independent strain components
3D Elasticity
z
z
yz
yx
y
x, y and z are normal stresses.
The rest 6 are the shear stresses
y Convention
xy is the stress on the face
perpendicular to the x-axis and points
in the +ve y direction
Total
T t l off 9 stress
t
components
t off which
hi h
only 6 are independent since  xy   yx
 x 
 
The stress vector is therefore  y 
 
  z
 xy 
 yz 
 
 zx 
 yz   zy
 zx   xz
Compactly;
EQUILIBRIUM
EQUATIONS
where

 x

0


0


 y

0


 z
  X 0
T
0

y
0

x

z
0
(1)

0

0


z 
0



y 

x 
3
Volume
element dV
3D elasticity problem is completely defined once we
understand the following three concepts
Strong formulation (governing differential equation +
boundary conditions)
u
Strain-displacement relationship
z
p
Stress-strain relationship
pz
Xc dV
Xb dV p
x
Xa dV
w
Volume (V)
v
ST
py
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)
Volume
element dV
pz
Xc dV
Xb dV p
x
w
Boundary conditions
u
z
1. Displacement boundary conditions: Displacements are specified on
1
portion Su of the boundary
u u
specified
Xa dV
Volume (V)
v
ST
Traction: Distributed
py force per unit area
p x 
 
T S  p y 
p 
 z
x
Su
on S u
x
y
2. Traction (force) boundary conditions: Tractions are specified on
portion ST of the boundary
Now, how do I express this mathematically?
4
TS
nz
In 2D
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   xy n y   xz nz
p y   xy nx   y n y   yz nz
p x 
 
T S  p y 
p 
 z
p z   xz nx   zy n y   z nz
n


dy ds
nx
dx
y
ny
ST
x
dx
 ny
ds
dy
cos  
 nx
ds
sin  
py
xy
x
dy
xy

ds
dx
TS
px
y
Consider
C
id the
h equilibrium
ilib i
off the
h wedge
d in
i
x-direction
p x ds   x dy   xy dx
dy
dx
  xy
ds
ds
 p x   x n x   xy n y
 px   x
Similarly
p y   xy n x   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
p
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 
5
u
dy
y
Compactly;
 x 
 
 y
  
  z 
 xy 
 yz 
 
 zx 
 u

 x

0


0


 y

0


 z
In 2D
(2)
0

y
0

x

z
0

0

0


z 
0



y 

x 
u 
 
u  v
w 
 
Strong formulation (governing differential equation +
boundary conditions)
p
Stress-strain relationship
v
dy
y
y
C
C’
2
A’ 
1
dy u
v
A dx B
3D elasticity problem is completely defined once we
understand the following three concepts
Strain-displacement relationship
v

x

y

xy

u

 dx   u   x dx
A' B'  AB


 
AB
dx


v
 dy   v 
dy

y
A' C'  AC


 
AC
dy
π

 angle (C' A' B' )  β 1  β 2
2
v
u


x
x
B’
v
dx
x
x
u


dx
u
  u   dx
x
u



x


  v   dy
v



y
 tan β 1  tan β
2
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 

1  2 
2 
0
0
6
PLANE STRESS: Only the in-plane stress components are nonzero
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)
Area
element dA
h
D
 xy
y
 xy
x
Assumptions:
A
i
1. h<<D
2. Top and bottom surfaces are free from
traction
3. Xc=0 and pz=0
y
x
PLANE STRESS
Nonzero stresses:  x ,  y , xy
Nonzero strains:  x ,  y ,  z ,  xy
PLANE STRESS Examples:
1. Thin plate with a hole
 xy
2. Thin cantilever plate
Nonzero stress components  x ,  y , xy
y
 xy
x
Isotropic linear elastic stress-strain law   D 
 x 
E
 
 y  
2
  1  
 xy 


0   x 
1 
 
 1
0  y 

1    
  xy
0 0

2  
z  

1 

x
y
Hence, the D matrix for the plane stress case is
D


1 
0 
E 


1
0

1  2 
1  
0
0



2 
7
PLANE STRAIN: Only the in-plane strain components are nonzero
PLANE STRAIN Examples:
1. Dam
1
Nonzero strain components  x ,  y ,  xy
Area
element dA
y
x
z
 xy
Slice of unit
thickness
y
 xy
y
x
Assumptions:
A
i
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
The square block is in plane strain
and is subjected to the following
strains
2

0   x 
 
0   y 
1  2   
  xy
2  
 z    x   y 
x
Example problem
2
1
3
4
2

 x 

1  
E
 



1


 y
  1   1  2   0
0

 xy 

 xy
z
2. Long cylindrical pressure vessel subjected to internal/external
pressure and constrained at the ends
y
Isotropic linear elastic stress-strain law   D 
z
x
PLANE STRAIN
Nonzero stress:  x ,  y ,  z , xy
Nonzero strain components:  x ,  y ,  xy
y
 xy
 x  2 xy
x
 y  3 xy 2
 xy  x 2  y 3
Compute the displacement field (i.e., displacement components
u(x,y) and v(x,y)) within the block
Hence, the D matrix for the plane strain case is
D


1  
E
 
1 
1   1  2  
0
 0


0 
0 
1  2 

2 
8
Plug expressions in (4) and (5) into equation (3)
Solution
Recall from definition
u
 2 xy
x 
(1)
x
v
 y   3xy 2 (2)
y
u v
 xy    x 2  y 3 (3)
y x
u v
 x 2  y 3 (3)

y x
Arbitrary function of ‘x’
Integrating (1) and (2)
u ( x, y )  x 2 y  C1 ( y )
(4)
v( x, y )  xy 3  C2 ( x )
(5)

 

 x 2 y  C1 ( y )  xy 3  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’
Arbitrary function of ‘y’
Use the 3 boundary conditions
Hence
C1 ( y )
C ( x)
 2
 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 )  xy 3  Cx  D2
How to find C, D1 and D2?
u (0,0)  0
v(0,0)  0
v(2,0)  0
To obtain
C 0
D1  0
D2  0
y
2
2
1
3
4
2
x
Hence the solution is
u ( x, y )  x 2 y
v( x, y )  xy 3
9
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  is defined as the strain energy minus the potential energy
of the loads involving X and TS
Volume
element dV
x
Su
x
py
Xb dV p
x
Xa dV
w
Volume (V)
v
ST
u
z
U-W
pz
Xc dV
y
U
1
 T  dV
2 V
W   u X dV   u T S dS
T
V
Strain energy of the elastic body
Using the stress-strain law   D 
U
1
1
 T  dV    T D  dV
2 V
2 V
In 1D
U
1
1
1 L
 dV   E 2 dV   E 2 Adx
2 V
2 V
2 x0
T
ST
Principle of minimum potential energy: Among all admissible
displacement fields the one that satisfies the equilibrium equations
also render the potential energy  a minimum.
“admissible displacement field”:
1. first derivative of the displacement components exist
2. satisfies the boundary conditions on Su
In 2D plane stress and plane strain
1
 x x   y  y   xy  xy  dV
2 V
Why?
U
10