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## Chapter 8 Two-Dimensional Problem Solution

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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

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

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

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

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:

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 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

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 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

Displacement Field (Plane Stress)

Choosing Fixity Conditions

Strength of Materials:

Good match for beams where l >> c

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 Problem

Bending Stress

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

R

y

r

x

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

Biharmonic Governing Equation

Traction Boundary Conditions

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

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 andr= a4/r 0, see Exercise 8-14

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

Dimensionless Stress

θ/p

r /p

r/r2

Dimensionless Distance, r/r2

Example 8.6 Cylinder Problem ResultsInternal Pressure OnlyThin-Walled Tube Case:

Matches with Strength

of Materials Theory

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

Boundary Conditions

Try Stress Function

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

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 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 = 6.5in/in/oF(11.7m/m/oC),

qa/k= 100oF (37.7oC), 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

Two Dimensional Case:(r,/2)/S

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

Three Dimensional Stress Concentration Problem – Chapter 13Normal Stress on the x,y-plane (z = 0)

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 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)

Boundary Conditions

Try Airy Stress Function

Use BC’s To Determine Stress Solution

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

Cartesian Solution

Free Surface Displacements

Cylindrical Solution

Corresponding 2-D Results

3-D Solution eliminates the unbounded far-field behavior

Half-Space Under Uniform Normal Loading Over –axa

dY= pdx = prd/sin

Notch/Crack Problem

Try Stress Function:

Boundary Conditions:

At Crack Tip r 0:

Finite Displacements and Singular Stresses at Crack Tip1< <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

Mode I (Maximum shear stress contours)

Mode II (Maximum shear stress contours)

ExperimentalPhotoelasticIsochromaticsCourtesy of URI Dynamic Photomechanics Laboratory

Crack Problem ResultsContours of Maximum Shear StressTheory of Elasticity Strength of Materials

Dimensionless Stress, a/P

= /2 b/a = 4

Dimensionless Distance, r/a

Curved Cantilever BeamP

r

a

b

Disk Under Diametrical Compression

P

D

=

P

Flamant Solution (1)

+

+

Radial Tension Solution (3)

Flamant Solution (2)

P

P

P

(Courtesy of URI Dynamic Photomechanics Lab)

Applications to Granular Media ModelingContact Load Transfer Between Idealized GrainsFour-Contact Grain

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