MCE 571 Theory of Elasticity

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30 Οκτ 2013 (πριν από 3 χρόνια και 11 μήνες)

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Using the Airy Stress Function approach, it was shown that the plane
elasticity formulation with zero body forces reduces to a single governing
biharmonic equation. In Cartesian coordinates it is given by




and the stresses are related to the stress function by




We now explore solutions
to several specific
problems in both

Cartesian and Polar coordinate systems

Cartesian Coordinate Solutions

Using Polynomials

I
n
Cartesian
coordinates we choose Airy
stress function

solution of polynomial form




where
A
mn

are constant coefficients to be determined. This
method
produces
polynomial stress distributions,
and thus would
not
satisfy
general boundary
conditions. However,
we can modify such boundary
conditions
using Saint
-
Venant’s

principle and replace
a
non
-
polynomial condition
with a statically equivalent
loading.
T
his
formulation
is
most useful for problems with rectangular
domains, and is
commonly based
on the inverse solution concept where we
assume
a
polynomial
solution form and
then try
to find what
problem
it will solve.

Noted
that the three lowest order terms with
m + n


1


do not contribute to the
stresses and will therefore be dropped. It should be noted that second order terms
will produce a constant stress field, third
-
order terms will give a linear distribution of
stress, and so on for higher
-
order polynomials.

Terms
with
m + n


3

will automatically satisfy the biharmonic equation
for
any
choice of constants
A
mn
. However, for higher order
terms, constants
A
mn

will have to
be related in order to have
the polynomial
satisfy
the biharmonic equation.

Example 8.1 Uniaxial Tension of a Beam


B
oundary Conditions:

Since the boundary conditions specify constant
stresses on
all boundaries
,
try
a second
-
order
stress function of the form

The first boundary condition
implies
that
A
02

=
T
/2,
and all other boundary conditions are identically
satisfied. Therefore the stress field solution
is
given by

Displacement Field (Plane Stress)

Stress Field

. . . Rigid
-
Body Motion

“Fixity conditions”

needed
to
determine RBM
terms

Example 8.2 Pure Bending of a Beam

B
oundary Conditions:

Expecting a linear bending stress distribution,
try second
-
order
stress function of the form

M
oment boundary condition implies
that

A
03

=
-
M
/4
c
3
,
and all other boundary conditions
are identically satisfied.
Thus the
stress field
is

Stress Field

“Fixity conditions”

to determine RBM terms:

Displacement Field (Plane Stress)

Example 8.2 Pure Bending of a
Beam

Solution Comparison of Elasticity

with Elementary Mechanics of Materials

Elasticity Solution

Mechanics of Materials Solution

Uses
Euler
-
Bernoulli beam
theory to
find
bending stress and deflection of
beam centerline

T
wo
solutions
are
identical, with the exception of the
x
-
displacements

Example 8.3 Bending of a Beam by
Uniform Transverse
Loading

B
oundary Conditions:

Stress Field

BC’s

Example
8.3 Beam Problem

Stress Solution Comparison of Elasticity

with Elementary Mechanics of Materials

Elasticity Solution

Mechanics of Materials Solution

Shear
stresses are identical, while
normal
stresses are not

Example
8.3 Beam Problem

Normal Stress Comparisons of Elasticity

with Elementary Mechanics of Materials

M
aximum differences
between the two theories
exist
at
top
and
bottom of beam
, and
actual
difference in
stress
values is
w
/5. For
most
beam
problems
where

l

>>
c
, the bending stresses will be much greater than
w
, and thus the differences between elasticity
and
strength of materials will be relatively small.

Maximum
difference between the two theories is
w

and this occurs at the top of the beam. Again this
difference will be negligibly small for most beam
problems where
l

>>
c
. These results are generally true
for beam problems with other transverse
loadings.


x



Stress at x=0


y

-

Stress

Example 8.3
Beam Problem

Normal Stress Distribution on Beam Ends

End stress distribution
does not
vanish and is nonlinear but gives
zero resultant force
.

Example 8.3
Beam Problem

Choosing
Fixity Conditions

Strength of Materials:

Good match for
beams where
l

>>
c

Displacement Field (Plane Stress)

Cartesian Coordinate Solutions

Using
Fourier Methods

A more general solution scheme for the biharmonic equation may be
found
using
Fourier methods
. Such techniques generally use
separation of
variables

along with
Fourier series

or
Fourier integrals
.

Choosing

Example 8.4 Beam
with
Sinusoidal Loading


B
oundary Conditions:

Stress Field

Example 8.4 Beam
Problem


Bending Stress

Example 8.4 Beam
Problem


For the case
l

>>
c


Strength
of
Materials

Displacement Field (Plane Stress)

Example
8.5 Rectangular Domain with
Arbitrary Boundary Loading


B
oundary Conditions

Must use series representation for Airy stress
function to handle general
boundary
loading.

Use Fourier series theory to handle general
boundary
conditions, and this generates a
doubly infinite set of equations to solve for
unknown constants in stress function form.
See text for details

Polar Coordinate
Formulation

Airy Stress Function Approach




(
r,
θ
)

R

S

x


y



r



Airy Representation

Biharmonic Governing Equation

Traction Boundary Conditions

Polar Coordinate
Formulation

Plane Elasticity Problem



Strain
-
Displacement

Hooke’s Law

General Solutions in Polar
Coordinates

Michell

Solution

Choosing the case where
b

=
in
,
n

= integer gives the general
Michell

solution

We will use various
terms from this general
solution to solve
several plane problems
in polar coordinates

Axisymmetric Solutions

Stress Function Approach:


=

(
r
)


Navier

Equation Approach:

u=
u
r
(
r
)
e
r

(Plane Stress or Plane Strain)


Displacements
-

Plane Stress Case

Gives Stress Forms


a
3

term leads to multivalued behavior, and is not found following the
displacement formulation approach


Could
also have an axisymmetric elasticity problem using



=

a
4


which gives

r

=



=
0 and


r


=
a
4
/
r



0, see Exercise
8
-
14

Underlined terms represent

rigid
-
body motion

Example 8.6 Thick
-
Walled Cylinder
Under Uniform Boundary
Pressure

B
oundary Conditions

General Axisymmetric

Stress Solution

Using
Strain Displacement
Relations and
Hooke’s
Law
for plane strain gives the
radial
displacement

Example 8.6
Cylinder Problem Results

Internal Pressure Only

r
1
/r
2

= 0.5

r/r
2



r
/p


θ

/p

Dimensionless Stress

Dimensionless Distance,
r
/
r
2

Thin
-
Walled Tube Case:

Matches with Strength

of Materials Theory

Special Cases of Example 8
-
6

Pressurized Hole in an Infinite Medium

Stress Free Hole in an Infinite Medium
Under Equal Biaxial Loading at Infinity

Example 8.7 Infinite Medium with a Stress
Free Hole Under Uniform Far Field Loading

B
oundary Conditions

Try Stress Function

Example 8.7
Stress Results

Superposition of Example 8.7

Biaxial Loading Cases

T
1

T
2

T
1

T
2

Equal Biaxial Tension Case

T
1

= T
2

= T

Tension/Compression Case

T
1

= T , T
2

=
-
T

Review Stress Concentration Factors

Around Stress Free Holes

K = 2

K = 3

K = 4

=

Stress Concentration Around

Stress Free Elliptical Hole


Chapter 10

Maximum Stress Field

Stress Concentration Around
Stress
Free
Hole in Orthotropic
Material


Chapter
11

2
-
D
Thermoelastic

Stress
Concentration

Problem Uniform Heat Flow Around

Stress
Free Insulation
Hole


Chapter
12

Stress
Field

Maximum
compressive stress on
hot
side of
hole

Maximum
tensile stress on
cold
side

Steel
Plate
:

E

= 30Mpsi (200GPa) and

= 6.5

in/in/
o
F

(
11.7

m/m/
o
C
),

qa
/k

= 100
o
F (37.7
o
C), the maximum stress becomes
19.5ksi (88.2MPa)

Nonhomogeneous Stress Concentration
Around
Stress
Free Hole in a Plane Under Uniform Biaxial Loading
with Radial Gradation of Young’s
Modulus


Chapter 14

Three Dimensional Stress Concentration

Problem


Chapter 13

Normal Stress
on the
x,y
-
plane (
z

= 0)

0
0.5
1
1.5
2
2.5
3
3.5
1
2
3
4
5
Normalized Stress in Loading Direction

Dimensionless Distance, r/a

Two Dimensional Case:



(
r
,

/2)/
S

Three Dimensional Case:


z
(
r
,0)/
S

,


= 0.3

Wedge
Domain
Problems

Use general
stress function
solution to
include
terms that are bounded at
origin
and give
uniform stresses on the
boundaries

Quarter Plane
Example
(


= 0 and


=

/2)

Half
-
Space
Examples

Uniform
Normal Stress Over
x



0

Try Airy Stress Function

Boundary Conditions

Use BC’s To Determine Stress Solution

Half
-
Space Under Concentrated Surface
Force System (
Flamant

Problem)

Try Airy Stress Function

Boundary Conditions

Use BC’s To Determine Stress Solution

Flamant

Solution Stress Results

Normal Force Case

o
r in Cartesian


components

y

=
a

Flamant

Solution
Displacement Results

Normal Force Case

On Free Surface
y

= 0

Note unpleasant feature of 2
-
D model that
displacements become unbounded as
r





Comparison of
Flamant

Results with

3
-
D Theory
-

Boussinesq’s

Problem

Cartesian Solution

Cylindrical Solution

Free
Surface Displacements

Corresponding 2
-
D Results

3
-
D Solution eliminates the
unbounded far
-
field behavior

Half
-
Space Under Uniform Normal
Loading Over

a



x



a


dY

=
pdx

=
prd


/sin



Half
-
Space Under Uniform Normal
Loading
-

Results


max

-

Contours

Generalized Superposition Method

Half
-
Space Loading
Problems

Photoelastic

Contact Stress Fields

Notch/Crack Problem

Boundary
Conditions:

At Crack
Tip
r



0:

Try Stress Function:

Finite Displacements and Singular
Stresses at
Crack Tip



1<


<2




=
3/2

Notch/Crack
Problem Results

Transform to


Variable


Note special singular behavior of stress field O(1/

r
)


A

and
B

coefficients are related to
stress intensity factors
and are useful in fracture
mechanics
theory


A

terms give symmetric stress fields


Opening or Mode I behavior


B

terms give
antisymmetric

stress fields


Shearing or Mode II
behavior

Crack
Problem
Results

Contours of Maximum Shear Stress

Mode I (Maximum shear stress contours)


Mode II (Maximum shear stress contours)

Experimental

Photoelastic

Isochromatics


Courtesy of URI Dynamic
Photomechanics

Laboratory


Mode III Crack Problem


Exercise 8
-
32

Anti
-
Plane Strain Case

Stresses Again


z



-

Stress Contours

Curved Beam Under End Moments

Curved
Cantilever Beam

P

a

b

r



Dimensionless Distance,
r
/
a

Dimensionless Stress
,


a
/
P

Theory of Elasticity
Strength of Materials









=

/2

b/a
= 4

Disk Under Diametrical Compression

+

P

P

D

=

+

Flamant Solution (1)

Flamant Solution (2)

Radial Tension Solution (3)

Disk Problem


Superposition of Stresses

Disk Problem


Results

x
-
axis (
y

= 0)

y
-
axis
(
x

= 0)


Applications to Granular Media
Modeling

Contact Load Transfer Between Idealized Grains

(Courtesy of URI Dynamic Photomechanics Lab)

P

P

P

P




Four
-
Contact Grain