fundamentals of elasticity theory professor m. h. sadd reference: elasticity theory applications and...
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Fundamentals of Elasticity Theory
Professor M. H. Sadd
Reference: Elasticity Theory Applications and Numerics, M.H. Sadd, Elsevier/Academic Press, 2009
Theory of ElasticityBased Upon Principles of Continuum Mechanics, Elasticity Theory Formulates Stress Analysis Problem As Mathematical Boundary-Value Problem for Solution of Stress, Strain and Displacement Distribution in an Elastic Body.
Governing Field Equations Model Physics Inside Region (Same For All Problems)
Boundary Conditions Describe Physics on Boundary(Different For Each Problem)
R
Su
St
Value of Elasticity Theory
- Develops “Exact” Analytical Solutions For Problems of Limited Complexity
- Provides Framework for Understanding Limitations of Strength of Materials Models
- Establishes Framework for Developing Linear Finite Element Modeling
- Generates Solutions for Benchmark Comparisons with FEA Solutions
Deformation and Strain
u(x,y)
u(x+dx,y)
v(x,y)
v(x,y+dy)
dx
dy
A B
C D
A'
B'
C'
D'dy
y
u
dxx
v
x
y
xyxy
y
x
x
v
y
ue
y
ve
x
ue
2
1
2
1
Strain Displacement Relations
zzyzx
yzyyx
xzxyx
eee
eee
eee
][eeThree-Dimensional Theory
Two-Dimensional Theory
Deformation and Strain Example
AxCyz
u
x
we
CxCxy
w
z
ve
BxBxx
v
y
ue
z
we
Byy
ve
Azx
ue
CBACxywyxBvAxzu
zx
yz
xy
z
y
x
2
1
2
1
2
10
2
1
2
1
202
1
2
1
0
2
_________________________________________
constants are ,, where, ,)(,
fieldnt displaceme following for thestrain of components theDetermine22
Rigid Body MotionTwo-Dimensional Example
uo
vo
dx
dy
A B
C D
y
uz
x
vz
x
y
xvv
yuu
zo
zo
*
*
Fieldnt Displaceme of Form General
Zero Strains!
Strain Compatibility
yx
e
x
e
y
e xyyx
2
2
2
2
2
2
2
3
1
4
Undeformed Configuration
2
3
1
4
Deformed ConfigurationContinuous Displacements
2
3
1
4
Deformed ConfigurationDiscontinuous Displacements
Discretized Elastic Solidx
yCompatibility Equation
Strain Compatibility Example
BA
BABA
yxBAxAyyx
e
x
e
y
e
yxBxyeAxeAye
xyyx
xyyx
3
2ith equation w satisfiesonly
3
246
)22(2662
__________________________________________
equation ity compatibil ldimensiona- two thesatisfies
)(,,
fieldstrain following theif see Check to
2
2
2
2
2
33
Body and Surface Forces
Sectioned Axially Loaded Beam
Surface Forces: T(x)
SCantilever Beam Under Self-Weight Loading
Body Forces: F(x)
Traction and Stress
F
n
A
(Sectioned Body)P1
P2
P3
p
(Externally Loaded Body)
AA
FnxT n
0lim),(
Traction Vector
Note that ordinary elasticity theory does not include nor allow concentrated moments to exist at a continuum point
Stress Components
3213
3212
3211
eeeenxT
eeeenxT
eeeenxT
zzyzx
yzyyx
xzxyx
),(
),(
),(
n
n
n
zzyzx
yzyyx
xzxyx
][
3
2
1
e
e
eT
)σττ(
)τστ(
)ττσ(n
zzyyzxxz
zzyyyxxy
zzxyyxxx
nnn
nnn
nnn
x
z
y
y
x
yx
z
xy
xz
zy
yz
zx
Stress Transformation
333
222
111
),cos(
nml
nml
nml
xx ji
)()()(
)()()(
)()()(
)(2
)(2
)(2
131313131313131313
323232323232323232
212121212121212121
33333323
23
23
22222222
22
22
11111121
21
21
nllnmnnmlmmlnnmmll
nllnmnnmlmmlnnmmll
nllnmnnmlmmlnnmmll
lnnmmlnml
lnnmmlnml
lnnmmlnml
zxyzxyzyxzx
zxyzxyzyxyz
zxyzxyzyxxy
zxyzxyzyxz
zxyzxyzyxy
zxyzxyzyxx
e1
e3 e2
e3
e2e1
x3
x1
x2
x1
x2
x3
)sin(coscossincossin
cossin2cossin
cossin2sincos
22
22
22
xyyxxy
xyyxy
xyyxx
Two-Dimensional Transformation
Three-Dimensional Transformation
x
y
x'
y'
0 10 20 30 40 50 60 70 80 90-0.5
0
0.5
1
(degrees)
Dim
ensi
on
less
Str
ess
Stress Transformation Example
xx
cossin
cos2
x
x
2cos/ x
cossin/ x
Principal Stresses and Directions
00
)(
)(
)(
Solution Trival-Non Equations, Algebraic of System sHomogeneou
0
)(
)(
)(
0)(
0)(
0)(
322
13
3
2
1
321
321
221
III
n
n
n
nnn
nnn
nnn
zyzxz
yzyxy
xzxyx
zyzxz
yzyxy
xzxyx
zyzxz
yzyxy
xzxyx
Roots of the characteristic equation are the principal stresses Corresponding to each principal stress is a principal direction nnn
that can be used to construct a principal coordinate systemy
(General Coordinate System)
1
3
2
(Principal Coordinate System)
n
n
n
x
z
y x
yx
z
xy
xz
zy
yz
zx
1
3
2
Ii = Fundamental Invariants
Equilibrium Equations
yxxy
yyxy
y
xyxx
x
M
Fyx
F
Fyx
F
0
00
00
x
y
dxx
xx
xy
yx
dyy
yy
dxxxy
xy
dyyyx
yx
xF
yF
Body Forces
Equilibrium Equation Example
0000
02
3
2
30
_______________________________________
)1(4
3,0,
22
3
equations mequilibriu esatisfy th stresses
following that theshow forces,body no Assuming
33
2
2
3
yx
c
Py
c
Py
yx
c
y
c
P
c
N
c
Pxy
yxy
yxx
xyyx
Hooke’s Law
zxyzxyzyxzx
zxyzxyzyxyz
zxyzxyzyxxy
zxyzxyzyxz
zxyzxyzyxy
zxyzxyzyxx
eCeCeCeCeCeC
eCeCeCeCeCeC
eCeCeCeCeCeC
eCeCeCeCeCeC
eCeCeCeCeCeC
eCeCeCeCeCeC
666564636261
565554535251
464544434241
363534333231
262524232221
161514131211
222
222
222
222
222
222
Isotropic Homogeneous Materials
zxzx
yzyz
xyxy
zzyxz
yzyxy
xzyxx
e
e
e
eeee
eeee
eeee
2
2
2
2)(
2)(
2)(
zxzxzx
yzyzyz
xyxyxy
yxzz
xzyy
zyxx
Ee
Ee
Ee
Ee
Ee
Ee
2
11
2
11
2
11
)(1
)(1
)(1
= Lamé’s constant = shear modulus or modulus of rigidityE = modulus of elasticity or Young’s modulusv = Poisson’s ratio
Orthotropic Materials(Three Planes of Material Symmetry)
xy
zx
yz
z
y
x
xy
zx
yz
z
y
x
e
e
e
e
e
e
EEE
EEE
EEE
2
2
2
1
01
001
0001
0001
0001
12
31
23
32
23
1
13
3
32
21
12
3
31
2
21
1
Nine Independent Elastic Constants for 3-DFour Independent Elastic Constants for 2-D
Physical Meaning of Elastic Constants
(Simple Tension)
(Pure Shear)
p
p
p
(Hydrostatic Compression)
000
000
00
ij
E
E
E
eij
00
00
00
000
00
00
ij
000
002/
02/0
ije
xyxye /2/xeE /
ijij p
p
p
p
00
00
00
pE
pE
pE
eij
2100
021
0
0021
pE
ekk
)21(3 kp
ModulusBulk )21(3
pE
k
Relations Among Elastic Constants
Typical Values of Elastic Constants
Basic FormulationFundamental Equations (15) - Strain-Displacement (6) - Compatibility (3) - Equilibrium (3) - Hooke’s Law (6)
Fundamental Unknowns (15) - Displacements (3) - Strains (6) - Stresses (6)
Displacement Conditions Mixed ConditionsTraction Conditions
R
S
R
Su
St
T(n)
R
S
u
Typical Boundary Condtions
Basic Problem Formulations
Problem 1 (Traction Problem) Determine the distribution of displacements, strains and stresses in the interior of an elastic body in equilibrium when body forces are given and the distribution of the tractions are prescribed over the surface of the body.
Problem 2 (Displacement Problem) Determine the distribution of displacements, strains and stresses in the interior of an elastic body in equilibrium when body forces are given and the distribution of the displacements are prescribed over the surface of the body.
Problem 3 (Mixed Problem) Determine the distribution of displacements, strains and stresses in the interior of an elastic body in equilibrium when body forces are given and the distribution of the tractions are prescribed over the surface St and the distribution of the displacements are prescribed
over the surface Su of the body.
Displacement Conditions Mixed ConditionsTraction Conditions
R
S
R
Su
St
T(n)
R
S
u
Basic Boundary Conditions
r
r
r
r
rx
xy=Tx
y=Ty
x
y x=Tx
xy=Ty
y
(Cartesian Coordinate Boundaries) (Polar Coordinate Boundaries)
Coordinate Boundary Examples
Non-Coordinate Boundary Example
x
y
),()( yxFnnT xyxyxxn
x
),()( yxFnnT yyyxxyn
y n = unit normal vector
Boundary Condition Examples
Fixed Conditionu = v = 0
Traction Free Condition
x
y
a
b S
Traction Condition
0)( nyT
x
y
l
0)( nxT
Fixed Conditionu = v = 0
Traction Condition
(Coordinate Surface Boundaries) (Non-Coordinate Surface Boundary)
Traction Free Condition
S
0, )()( xyn
yxn
x TSTSTT y
nyxy
nx )()( ,0
0,0 )()( yn
yxyn
x TT
Symmetry Boundary Conditions
Symmetry Line
0
0)(
n
yT
u
x
y
Rigid-SmoothBoundary Condition
Example Solution – Beam Problem
x - Contours
x
Saint-Venant’s PrincipleThe Stress, Strain and Displacement Fields Due to Two Different Statically Equivalent Force Distributions on Parts of the Body Far Away From the Loading Points Are Approximately the Same.
x
y
P
x
y
P/2 P/2
xy
y
x
xy
y
x
Stresses Approximately Equal
Strain Energy
xxxx
x
eEe
Edxdydz
dUU
dxdydzE
dxdydzE
ddxdydz
x
ud
dydzdudydzdxx
uuddU
xx
xx
2
1
22
Volume
EnergyStrain
2
)(
)(
22
2
00
00
dxu
dz
dxx
uu
dy
x
y
z
Strain Energy = Energy Stored Inside an Elastic Solid Due to the Applied Loadings
One-Dimensional Case
Three-Dimensional Case
0)2
1
2
1
2
1()(
2
12
1)(
2
1
2222222
zxyzxyzyxzyx
ijijzxzxyzyzxyxyzzyyxx
eeeeee
eeeeU
Principle of Virtual Work
0
WUdVuFdSuTdVU TiV iS i
niV t
Change in Potential Energy (UT-W) During a Virtual Displacement from Equilibrium is Zero.
V zxzxyzyzxyxyzzyyxx
V ijijT
dVeee
dVeU
)(
Energy Strain Virtual
The virtual displacement ui = {u, v, w} of a material point is a fictitious
displacement such that the forces acting on the point remain unchanged. The work done by these forces during the virtual displacement is called the virtual work.
dVuFdSuT
W
iV iS in
it
sBody Force andSurface by Done Work Virtual
dVuFdSuTdVe iV iS in
iijV ijt
Virtual Strain Energy = Virtual Work Done by Surface and Body Forces