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Chapter 8 Two-Dimensional Problem Solution. 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

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chapter 8 two dimensional problem solution

Chapter 8 Two-Dimensional Problem Solution

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
Cartesian Coordinate Solutions Using Polynomials

In Cartesian coordinates we choose Airy stress function solution of polynomial form

where Amn 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. This 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  1do 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  3will automatically satisfy the biharmonic equation for any choice of constants Amn. However, for higher order terms, constants Amn will have to be related in order to have the polynomial satisfy the biharmonic equation.

example 8 1 uniaxial tension of a beam
Example 8.1 Uniaxial Tension of a Beam

Displacement Field (Plane Stress)

Stress Field

Boundary 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 A02 = T/2, and all other boundary conditions are identically satisfied. Therefore the stress field solution is given by

. . . Rigid-Body Motion

“Fixity conditions”needed to determine RBM terms

example 8 2 pure bending of a beam
Example 8.2 Pure Bending of a Beam

Stress Field

Displacement Field (Plane Stress)

Boundary Conditions:

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

Moment boundary condition implies that

A03= -M/4c3, and all other boundary conditions are identically satisfied. Thus the stress field is

“Fixity conditions”to determine RBM terms:

slide5
Example 8.2 Pure Bending of a BeamSolution 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

Two solutions are identical, with the exception of the x-displacements

example 8 3 bending of a beam by uniform transverse loading
Example 8.3 Bending of a Beam by Uniform Transverse Loading

Stress Field

Boundary Conditions:

BC’s

slide7
Example 8.3 Beam ProblemStress Solution Comparison of Elasticity with Elementary Mechanics of Materials

Elasticity Solution

Mechanics of Materials Solution

Shear stresses are identical, while normal stresses are not

slide8
Example 8.3 Beam ProblemNormal Stress Comparisons of Elasticity with Elementary Mechanics of Materials

x – Stress at x=0

y - Stress

Maximum difference between the two theories is wand 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.

Maximum 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.

example 8 3 beam problem normal stress distribution on beam ends
Example 8.3 Beam ProblemNormal Stress Distribution on Beam Ends

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

example 8 3 beam problem
Example 8.3 Beam Problem

Displacement Field (Plane Stress)

Choosing Fixity Conditions

Strength of Materials:

Good match for beams where l >> c

cartesian coordinate solutions using fourier methods
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
Example 8.4 Beam with Sinusoidal Loading

Stress Field

Boundary Conditions:

example 8 4 beam problem14
Example 8.4 Beam Problem

Displacement Field (Plane Stress)

For the case l >> c

Strength of Materials

example 8 5 rectangular domain with arbitrary boundary loading
Example 8.5 Rectangular Domain with Arbitrary Boundary Loading

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

Boundary Conditions

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

S

R

y

r

x

Polar Coordinate FormulationAiry Stress Function Approach  = (r,θ)

Airy Representation

Biharmonic Governing Equation

Traction Boundary Conditions

general solutions in polar coordinates michell solution
General Solutions in Polar CoordinatesMichell 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
Axisymmetric Solutions

Stress Function Approach:=(r)

Navier Equation Approach:u=ur(r)er

(Plane Stress or Plane Strain)

Gives Stress Forms

Displacements - Plane Stress Case

Underlined terms represent

rigid-body motion

  • a3term leads to multivalued behavior, and is not found following the displacement formulation approach
  • Could also have an axisymmetric elasticity problem using  =a4 which gives r =  = 0 andr= a4/r  0, see Exercise 8-14
example 8 6 thick walled cylinder under uniform boundary pressure
Example 8.6 Thick-Walled Cylinder Under Uniform Boundary Pressure

General Axisymmetric

Stress Solution

Boundary Conditions

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

example 8 6 cylinder problem results internal pressure only

r1/r2 = 0.5

Dimensionless Stress

θ/p

r /p

r/r2

Dimensionless Distance, r/r2

Example 8.6 Cylinder Problem ResultsInternal Pressure Only

Thin-Walled Tube Case:

Matches with Strength

of Materials Theory

special cases of example 8 6
Special Cases of Example 8-6

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

Pressurized Hole in an Infinite Medium

example 8 7 infinite medium with a stress free hole under uniform far field loading
Example 8.7 Infinite Medium with a Stress Free Hole Under Uniform Far Field Loading

Boundary Conditions

Try Stress Function

superposition of example 8 7 biaxial loading cases
Superposition of Example 8.7Biaxial Loading Cases

T2

T1

T1

T2

Tension/Compression Case

T1 = T , T2 = -T

Equal Biaxial Tension Case

T1 = T2 = T

slide29
2-D ThermoelasticStress 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/oF(11.7m/m/oC),

qa/k= 100oF (37.7oC), the maximum stress becomes 19.5ksi (88.2MPa)

slide30

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

Two Dimensional Case:(r,/2)/S

Three Dimensional Case:z(r,0)/S ,  = 0.3

Three Dimensional Stress Concentration Problem – Chapter 13

Normal Stress on the x,y-plane (z = 0)

wedge domain problems
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
Half-Space ExamplesUniform Normal Stress Over x 0

Boundary Conditions

Try Airy Stress Function

Use BC’s To Determine Stress Solution

half space under concentrated surface force system flamant problem
Half-Space Under Concentrated Surface Force System (Flamant Problem)

Boundary Conditions

Try Airy Stress Function

Use BC’s To Determine Stress Solution

flamant solution stress results normal force case
Flamant Solution Stress ResultsNormal Force Case

or in Cartesian

components

y = a

flamant solution displacement results normal force case
Flamant Solution Displacement ResultsNormal Force Case

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

On Free Surface y = 0

comparison of flamant results with 3 d theory boussinesq s problem
Comparison of Flamant Results with 3-D Theory - Boussinesq’s Problem

Cartesian Solution

Free Surface Displacements

Cylindrical Solution

Corresponding 2-D Results

3-D Solution eliminates the unbounded far-field behavior

notch crack problem
Notch/Crack Problem

Try Stress Function:

Boundary Conditions:

At Crack Tip r 0:

Finite Displacements and Singular Stresses at Crack Tip1<  <2  = 3/2

notch crack problem results
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)

ExperimentalPhotoelasticIsochromaticsCourtesy of URI Dynamic Photomechanics Laboratory

Crack Problem ResultsContours of Maximum Shear Stress
mode iii crack problem exercise 8 32
Mode III Crack Problem – Exercise 8-32

Anti-Plane Strain Case

z - Stress Contours

Stresses Again

curved cantilever beam

Theory of Elasticity Strength of Materials

Dimensionless Stress, a/P

 = /2 b/a = 4

Dimensionless Distance, r/a

Curved Cantilever Beam

P

r

a

b

disk under diametrical compression
Disk Under Diametrical Compression

P

D

=

P

Flamant Solution (1)

+

+

Radial Tension Solution (3)

Flamant Solution (2)

disk problem results
Disk Problem – Results

x-axis (y = 0)

y-axis (x = 0)

applications to granular media modeling contact load transfer between idealized grains

P

P

P

P

(Courtesy of URI Dynamic Photomechanics Lab)

Applications to Granular Media ModelingContact Load Transfer Between Idealized Grains

Four-Contact Grain