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 x=0 x
F x=L x
A(x) = cross section at x b(x) = body force distribution
(force per unit length)
E(x) = Young’s modulus u(x) = displacement of the bar at x
1. Strong formulation: Equilibrium equation + boundary conditions d

Equilibrium equation  b

0 ; 0
 x

L dx
Boundary conditions u

0 at x

0
EA du dx

F at x

L

2.
Strain-displacement relationship:

3. Stress-strain (constitutive) relation :

(x)

E ε(x)
E: Elastic (Young’s) modulus of bar

3D Elasticity
Problem definition x z x
Surface (S) u y w v
V: Volume of body
Volume (V)
S: Total surface of the body
The deformation at point x =[x,y,z] T is given by the 3 components of its displacement u
NOTE: u= u(x,y,z), i.e., each displacement component is a function


 u v w

 of position

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 x
Volume element dV z x u y w
X c dV
Body force: distributed
X b dV force per unit volume (e.g.,
X a dV v weight, inertia, etc)
Volume (V)
Surface (S)
X




X
X
X a b c



NOTE: If the body is accelerating, then the inertia force
  u  





 u 
 v 

 may be considered as part of X
X

~
X
   u 

x
SURFACE TRACTION
Volume element dV u
X c dV w
X b dV
X a dV
Volume (V) v p
S
T x z x p z p y
Traction: Distributed force per unit surface area
T
S


 p p p x y z

 y

Volume element dV z x u w v
3D Elasticity:
INTERNAL FORCES
Volume (V)
 z t zx t xz t xy t zy t yz t yx
 x
 y x y
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.

3D Elasticity
 z x z t
 t xz x zx y t xy t zy t yx t yz
The stress vector is therefore

 y
 x
,
 y and
 z are normal stresses
The rest 6 are the shear stresses
.
Convention t xy is the stress on the face perpendicular to the x-axis and points in the +ve y direction

Total of 9 stress components of which only 6 are independent since t xy
 t yx








 t

 t t
 xy yz zx x y z 







 t t yz zx
 t zy
 t xz

Strains: 6 independent strain components
  













 x 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

X a

0

 t x xy 

 y
 y 

 t z yz 
X b

0
 x
 t xz
 x


 t y yz
 y

 z
 
 z z 
X c

0

Compactly;
EQUILIBRIUM
EQUATIONS where
 T  
X
 















 x
 y


0
0

0
 z
0


 x
 z
0

 y
0

0
0

0
 z

 y
 x
















0
(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 u
X c dV w
X b dV
X a dV
Volume (V) v p
S
T x z x
S u p z p y y x
1. Strong formulation of the 3D elasticity problem: “ Given the externally applied loads (on S
T and in V) and the specified displacements (on S u
) we want to solve for the resultant displacements, strains and stresses required to maintain equilibrium of the body .”

Equilibrium equations
 T  
X

0 in V
(1)
Boundary conditions
1. Displacement boundary conditions: Displacements are specified on portion S u of the boundary u
 u specified on S u
2. Traction (force) boundary conditions: Tractions are specified on portion S
T of the boundary
Now, how do I express this mathematically?

x
Volume element dV u
X c dV w
X b dV
X a dV
Volume (V) v p
S
T x z x
S u p z p y
Traction: Distributed force per unit area
T
S


 p p p x y z

 y

T
S n z p z
Traction: Distributed force per unit area
Then n x S
T n y n
If the unit outward normal to S
T
: p x p y p z
  x n x
 t xy n x
 t xz n x
 t
  xy n y y n y
 t zy n y
 t xz n z

 t
 yz n z z n z p y p x n



 n n n x y z



T
S


 p p p x y z



y
In 2D n dy q ds q n x n y dx
S
T x sin q cos q
 dx ds
 n y
 dy ds
 n x
 x t xy dy t xy q ds dx p y
T
S p x
 y
Consider the equilibrium of the wedge in x-direction p x ds
  x dy
 t xy dx
 p x
 p x
 
  x x n dy ds x

 t t xy xy n y dx ds
Similarly p y
 t 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
Stress-strain relationship

2. Strain-displacement relationships :


 xy yz zx

 x
 z


 y

 z
 u

 x v

 y
 w

 u
 z

 y
 v

 z
 u


 x
 w


 v y w x

Compactly;
  













 x y z xy yz zx









   u
 















 x
 y


0
0

0
 z
0


 x
 z
0

 y
0

0
0

0
 z

 y
 x















(2) u



 u v w




 u
 y dy
C’
In 2D v

 v
 y dy y
C dy u

2
A’

1 v
A dx
 x

A' B'

AB
AB
 y

A' C'

AC
AC
 xy


2
 v

π x

 angle

 u x


 dx

 u

 u
 x dx dy
 u


 dx

 dy


 v

 v
 y dx dy


 v


 dy
(C' A' B' )
 β
1
 β
2

 u
 x

 v
 y
 tanβ
1
 tanβ
2
B x u

 u
 x d x
B’
 v
 x dx

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
D

( 1
 
E
)( 1

2

)

1













0
0
0
 
1
 

0
0
0


1
 
0
0
0
0
0
1

0
2

2
0
0
0
0
0
0
1

2

2
0
0
0
0
0
0
1

2

2












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 h
Nonzero stress components t xy
 y t xy  x
 x
,
 y
, t xy
D y x
Assumptions:
1. h<<D
2. Top and bottom surfaces are free from traction
3. X c
=0 and p z
=0

PLANE STRESS Examples:
1. Thin plate with a hole t xy
 y t xy  x
2. Thin cantilever plate

Nonzero stresses :
Nonzero strains :
 x
PLANE STRESS
,

 x y
,
,

 y z
,
, t
 xy xy
Isotropic linear elastic stress-strain law
 
D




 t
 xy x y




E
1
 
2






1
0

1
0
0
1
0
 
2










 x y xy



 z
 
1

 

 x
  y

Hence, the D matrix for the plane stress case is
D

E
1
 
2






1
0

1
0
0
1
0
 
2






PLANE STRAIN: Only the in-plane strain components are nonzero
Area element dA
Nonzero strain components
 x
,
 y
,
 xy
 xy
 y
 xy  x y x
Assumptions:
1. Displacement components u,v functions of (x,y) only and w=0
2. Top and bottom surfaces are fixed
3. X c
=0
4. p x and p y do not vary with z z

PLANE STRAIN Examples:
1. Dam
1
Slice of unit thickness y
 z t xy
 y t xy  x x z
2. Long cylindrical pressure vessel subjected to internal/external pressure and constrained at the ends

PLANE STRAIN
Nonzero
Nonzero stress strain
:
 x
,
 y
,
 z
, t components: xy
 x
,
 y
,
 xy
Isotropic linear elastic stress-strain law
 
D




 t
 xy x y





1
  
E
1

2
 



1




0
 
1
 
0
0
1

0
2

2










 x y xy



Hence, the D matrix for the plane strain case is
D


1
  
E
1

2
 



1




0

1

 
0
0
1

0
2

2





 z

 x
  y


Example problem y
2
3
2
1
4
2 x
The square block is in plane strain and is subjected to the following strains
 x
 y

2 xy

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

Solution
Recall from definition

 x y
 xy


 u
 x
 v
 y

 u
 y

2 xy

3 xy
2

 v
 x
 x
2
( 1 )
( 2

) y
3
( 3 )
Integrating (1) and (2) u ( x , y )
 x
2 y

C
1
( y ) v ( x , y )
 xy
3 
C
2
( x )
( 4 )
( 5 )
Arbitrary function of ‘x’
Arbitrary function of ‘y’

Plug expressions in (4) and (5) into equation (3)
 u
 y


 v

 x
 x
2 y

 x
C
1
2 
( y y
)
3
 y


( 3 ) xy
3 
 x
C
2
( x )

 x
2 

C
1
 y
( y )

C
1
 y
( y )



C
2
 x
( y
3 x )


0

C
2
 x
( x )
 x
2  x
2  y
3 y
3
Function of ‘y’
Function of ‘x’

Hence

C
1
(
 y y )
 

C
2
 x
( x )

C ( a constant )
Integrate to obtain
C
1
( y )

Cy
C
2
( x )

Cx


D
D
2
1
D
1 and D
2 integration are two constants of
Plug these back into equations (4) and (5)
( 4
( 5 )
) v u ( x , y )

( x , y )
 x
2 y

Cy

D
1 xy
3 
Cx

D
2
How to find C, D
1 and D
2
?

Use the 3 boundary conditions u ( 0 , 0 )

0 v ( 0 , 0 )

0 v ( 2 , 0 )

0
To obtain
C

0
D
D
2
1

0

0
Hence the solution is y
3
2 u ( x , y )
 x
2 y v ( x , y )
 xy
3
2
1
4
2 x

Principle of Minimum Potential Energy
Definition: For a linear elastic body subjected to body forces
X=[X a
,X b
,X c
] T and surface tractions T
S
=[p x
,p y
,p z
] 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 T
S
P
U-W

x
Volume element dV u
X c dV w
X b dV
X a dV
Volume (V) v p
S
T x z x
S u y p z p y
U

1
2 V
  T 
W
 
V u
T
X dV dV
 
S
T u
T
T
S dS

Strain energy of the elastic body
Using the stress-strain law
 
D

U

1
2

V
 T  dV

1
2

V
 T
D
 dV
In 1D
U

1
2

V
 dV

1
2

V
E

2 dV

1
2
 x
L

0
E

2
Adx
In 2D plane stress and plane strain
U

1
2

V

 x
 x
  y
 y
 t xy
 xy
 dV
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 S u