MANE 4240 & CIVL 4240Introduction to Finite Elements
Introduction to 3D Elasticity
Prof. Suvranu De
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 cases2D (plane stress, plane strain)Axisymmetric body with axisymmetric loading
• Principle of minimum potential energy
1D Elasticity (axially loaded bar)
x
y
x=0 x=L
A(x) = cross section at xb(x) = body force distribution (force per unit length)E(x) = Young’s modulusu(x) = displacement of the bar at x
x
1. Strong formulation: Equilibrium equation + boundary conditions
Lxbdx
d 0;0
LxatFdx
duEA
xatu
00Boundary conditions
Equilibrium equation
F
3. Stress-strain (constitutive) relation : ε(x) E(x) E: Elastic (Young’s) modulus of bar
2. Strain-displacement relationship:dx
duε(x)
Problem definition3D Elasticity
xy
z
Surface (S)
Volume (V)
uv
w
x
V: Volume of bodyS: Total surface of the bodyThe deformation at point x =[x,y,z]T
is given by the 3 components of its displacement
w
v
u
u
NOTE: u= u(x,y,z), i.e., each displacement component is a function of position
3D Elasticity: EXTERNAL FORCES ACTING ON THE BODY
Two basic types of external forces act on a body1. Body force (force per unit volume) e.g., weight, inertia, etc2. Surface traction (force per unit surface area) e.g., friction
BODY FORCE
xy
z Surface (S)
Volume (V)
uv
w
x
Xa dVXb dV
Xc dVVolume element dV Body force: distributed
force per unit volume (e.g., weight, inertia, etc)
c
b
a
X
X
X
X
NOTE: If the body is accelerating, then the inertia force
may be considered as part of X
w
v
u
u
u~ XX
xy
z
ST
Volume (V)
uv
w
x
Xa dVXb dV
Xc dVVolume element dV
py
pz
px
Traction: Distributed force per unit surface area
z
y
x
p
p
p
T S
SURFACE TRACTION
3D Elasticity: INTERNAL FORCES
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.
xy
z
Volume (V)
uv
w
x
Volume element dV
x
y
z
yz
yx
xyxz
zyzx
3D Elasticity
zx
yz
xy
z
y
x
xy
z
x
y
z
yz
yx
xy
xz
zyzx
x, y and z are normal stresses.The rest 6 are the shear stressesConventionxy is the stress on the face perpendicular to the x-axis and points in the +ve y directionTotal of 9 stress components of which only 6 are independent since
xzzx
zyyz
yxxy
The stress vector is therefore
Strains: 6 independent strain components
zx
yz
xy
z
y
x
Consider the equilibrium of a differential volume element to obtain the 3 equilibrium equations of elasticity
0
0
0
czyzxz
byzyxy
axzxyx
Xzyx
Xzyx
Xzyx
Compactly;
0 XT
xz
yz
xy
z
y
x
0
0
0
00
00
00where
EQUILIBRIUM EQUATIONS
(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
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.”
xy
z
Su
ST
Volume (V)
uv
w
x
Xa dVXb dV
Xc dVVolume element dV
py
pz
px
VinXT 0 Equilibrium equations (1)
Boundary conditions
1. Displacement boundary conditions: Displacements are specified on portion Su of the boundary
uspecified Sonuu
2. Traction (force) boundary conditions: Tractions are specified on portion ST of the boundaryNow, how do I express this mathematically?
xy
z
Su
ST
Volume (V)
uv
w
x
Xa dVXb dV
Xc dVVolume element dV
py
pz
px
Traction: Distributed force per unit area
z
y
x
p
p
p
T S
Traction: Distributed force per unit area
z
y
x
p
p
p
T S
n
nx
ny
nz
ST
TS
py
px
pz
If the unit outward normal to ST :
z
y
x
n
n
n
n
Then
zzyzyxxz
zyzyyxxy
zxzyxyxx
nnn
nnn
nnn
z
y
x
p
p
p
nx
ny
ST
In 2D
dy
dx
ds
x
y
n
x
y
nds
dy
nds
dx
cos
sin
TSpy
px
dy
dx
dsx
y
xy
xy
Consider the equilibrium of the wedge in x-direction
yxyxxx
xyxx
xyxx
nnpds
dx
ds
dyp
dxdydsp
Similarlyyyxxyy nnp
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:
x
w
z
uy
w
z
vx
v
y
uz
wy
vx
u
zx
yz
xy
z
y
x
Compactly; u
xz
yz
xy
z
y
x
0
0
0
00
00
00
w
v
u
u
zx
yz
xy
z
y
x
(2)
x
y
A B
CA’
B’
C’
v
udy
dx
dxx
v
xdx
uu
dyy
u
dyy
vv
x
u
x
v
tanβtanβββ)B'A'(C'angle2
π
y
v
dy
dyvdyy
vvdy
AC
ACC'A'
x
u
dx
dxudxx
uudx
AB
ABB'A'
2121
xy
y
x
1
2
In 2D
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
2
2100000
02
210000
002
21000
0001
0001
0001
)21)(1(
E
D
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) stress-strain laws2. 2D elastic problems: 2 situations
PLANE STRESSPLANE 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
x
y
Area element dA
Nonzero stress components xyyx ,,
y
xxy
xy
h
DAssumptions:1. h<<D2. Top and bottom surfaces are free from traction3. Xc=0 and pz=0
PLANE STRESS Examples:1. Thin plate with a hole
2. Thin cantilever plate
y
xxy
xy
Nonzero strains: xyzyx ,,,
Isotropic linear elastic stress-strain law
xy
y
x
xy
y
xE
2
100
01
01
1 2
Hence, the D matrix for the plane stress case is
2
100
01
01
1 2
E
D
D
yxz
1
Nonzero stresses: xyyx ,,PLANE STRESS
PLANE STRAIN: Only the in-plane strain components are nonzero
Area element dA
Nonzero strain components
y
x
xyxy
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
xyyx ,,
x
y
z
PLANE STRAIN Examples:1. Dam
2. Long cylindrical pressure vessel subjected to internal/external pressure and constrained at the ends
1
Slice of unit thickness
x
y
z
y
xxy
xyz
Nonzero stress:
Isotropic linear elastic stress-strain law
xy
y
x
xy
y
xE
2
2100
01
01
211
Hence, the D matrix for the plane strain case is
2
2100
01
01
211
E
D
D
z x y
xyzyx ,,,
Nonzero strain components: xyyx ,,
PLANE STRAIN
Example problem
x
y
3
2 1
4
2
2The square block is in plane strain and is subjected to the following strains
2
2 3
2
3
x
y
xy
xy
xy
x y
Compute the displacement field (i.e., displacement components u(x,y) and v(x,y)) within the block
Solution
Recall from definition
)3(
)2(3
)1(2
32
2
yxx
v
y
u
xyy
v
xyx
u
xy
y
x
Integrating (1) and (2)
)5()(),(
)4()(),(
23
12
xCxyyxv
yCyxyxu
Arbitrary function of ‘x’
Arbitrary function of ‘y’
Plug expressions in (4) and (5) into equation (3)
0)()(
)()(
)()(
)3(
21
322312
3223
12
32
x
xC
y
yC
yxx
xCy
y
yCx
yxx
xCxy
y
yCyx
yxx
v
y
u
Function of ‘y’ Function of ‘x’
)constanta()()( 21 C
x
xC
y
yC
Hence
Integrate to obtain
22
11
)(
)(
DCxxC
DCyyC
D1 and D2 are two constants of
integration
Plug these back into equations (4) and (5)
23
12
),()5(
),()4(
DCxxyyxv
DCyyxyxu
How to find C, D1 and D2?
Use the 3 boundary conditions
0)0,2(
0)0,0(
0)0,0(
v
v
u
To obtain
0
0
0
2
1
D
D
C
Hence the solution is
3
2
),(
),(
xyyxv
yxyxu
x
y
3
2 1
4
2
2
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
U-W
xy
z
Su
ST
Volume (V)
uv
w
x
Xa dVXb dV
Xc dVVolume element dV
py
pz
px
TSS
T
V
T
V
T
dSTudVXu
dV
W
2
1U
Strain energy of the elastic body
V
T
V
T dVDdV 2
1
2
1U
DUsing the stress-strain law
In 1D
L
xVVAdxEdVEdV
0
22
2
1
2
1
2
1U
In 2D plane stress and plane strain
V xyxyyyxx dV
2
1U
Why?
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 exist2. satisfies the boundary conditions on Su
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