2 Plane Elasticity - Home | Scholars at Harvard · · 2014-07-08ES240 Solid Mechanics Fall 2009...

ES240 Solid Mechanics Fall 2009

9/26/11 Linear Elasticity-1

2. Plane Elasticity Problems

Main Reference: Theory of Elasticity, by S.P. Timoshenko and J.N. Goodier, McGraw-Hill, New York. Chapters 2, 3, 4.

2.1 The plane-stress problem

A thin sheet of an isotropic material is subject to loads in the plane of the sheet. The sheet lies in the (x, y) plane. Both top and the bottom surfaces of the sheet are traction-free. The edge of the sheet may have two kinds of the boundary conditions: displacement prescribed, or traction prescribed. In the latter case, we write

yyyyxxy

xyxyxxx

tnntnn

=+

=+

σττσ

where tx and ty are components of the traction vector prescribed on the edge of the sheet, and nx and ny are the components of the unit vector normal to the edge of the sheet. The above two equations provide two conditions for the components of the stress tensor along the edge.

Semi-inverse method. We next go into the interior of the sheet. We already have obtained a full set of governing equations for linear elasticity problems. No general approach exists to solve these partial differential equations analytically, although numerical methods are readily available to solve most elasticity problem. In this introductory course, in order to gain insight into solid mechanics, we will make reasonable guesses of solutions, and see if they satisfy all the governing equations. This trial-and-error approach has a name: it is called the semi-inverse method.

It seems reasonable to guess that the stress field in the sheet only has nonzero components in its plane: xyyyxx τσσ ,, , and that the components out of plane vanish:

0=== yzxzzz ττσ .

Furthermore, we guess that the in-plane stress components may vary with x and y, but are independent of z. That is, the stress field in the sheet is described by three functions:



( ) ( ) ( )yxyxyx xyyyxx ,,,,, τσσ .

Will these guesses satisfy the governing equations of elasticity? Let us go through the equations one by one.

1. Equilibrium equations. Using the guessed stress field, we reduce the three equilibrium equations to two equations:

0,0 =∂∂

+∂∂

=∂∂

+∂∂

yxyxyyxyxyxx σττσ .

These two equations by themselves are insufficient to determine the three functions.

2. Stress-strain relations. Given the guessed stress field, the 6 components of the strain field are

EEyyxx

xx

σνσε −= ,

EExxyy

yyσν

σε −= , ( )

xyxy Eτνγ += 12

( ) 0, ==+−= yzxzyyxxzz Eγγσσνε .

3. Strain-displacement relations. Recall the 6 strain-displacement relations:

xv

yu

yv

xu

xyyyxx ∂∂+

∂∂=

∂∂=

∂∂= γεε ,,

yw

zv

xw

zu

zw

yzxzzz ∂∂+

∂∂=

∂∂+

∂∂=

∂∂= γγε ,, .

It seems reasonable to assume that the in-plane displacements u and v vary only with x and y, but not with z. From these guesses, together with the conditions that 0== yzxz γγ , we find that

0=∂∂=

∂∂

yw

xw .

Thus, w is independent of x and y, and can only be a function of z. If we insist that zzε be independent of z, and from zwzz ∂∂= /ε , then zzε must be a constant, czz =ε , and bczw += . On the other hand, we also have ε zz = −ν σ xx +σ yy( ) / E , which may not be a constant. This inconsistency shows that our guesses are generally incorrect.



Summary of equations of plane elasticity problems. Instead of abandoning these guesses, we will just call our guesses the plane-stress approximation. If you neglect the inconsistency between czz =ε and ( ) Eyyxxzz /σσνε +−= , at least the following set of equations is self-consistent:

0,0 =∂∂

+∂∂

=∂∂

+∂∂

yxyxyyxyxyxx σττσ

EEyyxx

xx

σνσε −= ,

EExxyy

yyσν

σε −= , γ xy =

2 1+ ν( )E

τ xy

xv

yu

yv

xu

xyyyxx ∂∂+

∂∂=

∂∂=

∂∂= γεε ,, .

These are 8 equations for 8 functions. We will focus on these 8 equations.



2.2 The plane-strain problem

Consider an infinitely long cylinder with axis in the z-direction and a cross section in the ( )yx, plane. We assume that the loading is invariant along the z-direction. Under these conditions, the displacement field takes the form:

u x, y( ) , v x, y( ) , w = 0 .

From the strain-displacement relations, we find that only the three in-plane strains are nonzero: ε xx x, y( ),ε yy x, y( ),γ xy x, y( ) . The three out-of-plane strains vanish: 0=== yzxzzz γγε .

Because γ xz = γ yz = 0 , the stress-strain relations imply that τ xz = τ yz = 0 . From ε zz = 0 and ε zz = σ zz −νσ xx −νσ yy( ) , we obtain further that

σ zz = ν σ xx +σ yy( ) .

Furthermore, we have

ε xx =1E

σ xx −νσ yy −νσ zz( ) = 1−ν2

Eσ xx −

ν1−ν

σ yy⎛⎝⎜

⎞⎠⎟ ,

ε yy =1E

σ yy −νσ xx −νσ zz( ) = 1−ν2

Eσ yy −

ν1−ν

σ xx⎛⎝⎜

⎞⎠⎟ ,

γ xy =2 1+ ν( )

Eτ xy .

These three stress-strain relations look similar to those under the plane-stress conditions, provided we make the following substitutions:

E =E

1−ν 2, ν =

ν1−ν

.

The quantity E is called the plane strain modulus. Finally, we also have the equilibrium equations:



0,0 =∂∂

+∂∂

=∂∂

+∂∂

yxyxyyxyxyxx σττσ .

2.3 Solution of plane problems and the Airy stress function

From the forgoing, it is clear that plane stress and plane strain problems are described by the same equations, as long as one uses the appropriate elastic constants. This also means that the solution technique for both types of problems is the same. We make use of the following calculus theorem:

A theorem in calculus. If two functions f x, y( ) and g x, y( ) satisfy the following relationship

∂f∂x

=∂g∂y

,

then, there exists a function ( )yxA , , such that

f = ∂A∂y, g = ∂A

∂x.

The Airy stress function. We now apply the above theorem to the equilibrium equations. From the equation

∂σ xx

∂x+∂τ xy∂y

= 0 ,

we deduce that there exists a function ( )yxA , , such that

σ xx =∂A∂y, τ xy = −

∂A∂x

.

From the equation

∂τ xy∂x

+∂σ yy

∂y= 0 ,

we further deduce that that there exists a function ( )yxB , , such that

σ yy =∂B∂x, τ xy = −

∂B∂y

.



Finally, from

yB

xA

∂∂=

∂∂ ,

we deduce that that there exists a function ( )yx,φ , such that

A =∂φ∂y, B =

∂φ∂x

.

The function ( )yx,φ is known as the Airy stress function. The three components of the stress field can now be represented by the stress function:

�

σ xx =∂2φ∂y2 ;σ yy =

∂2φ∂x2 ;σ xy =

∂2φ∂x∂y

.

Using the stress-strain relations, we can also express the three components of strain field in terms of the Airy stress function:

ε xx =1E

∂2φ∂y2

−ν ∂2φ∂x2

⎛⎝⎜

⎞⎠⎟

, ε yy =1E

∂2φ∂x2

−ν ∂2φ∂y2

⎛⎝⎜

⎞⎠⎟

, γ xy = −2 1+ ν( )

E∂2φ∂x∂y

.

Compatibility equation. Recall the strain-displacement relations.

ε xx =∂u∂x, ε yy =

∂v∂y, γ xy =

∂u∂y

+∂v∂x

.

We derived the compatibility equation by eliminating the two displacements in the three strain displacement relations to obtain the compatibility equation

∂2ε xx∂y2

+∂2ε yy∂x2

=∂2γ xy

∂x∂y.

Biharmonic equation. Inserting the expressions of the strains in terms of ( )yx,φ into the compatibility equation, we obtain that

∂4φ∂x4

+ 2 ∂4φ∂x2∂y2

+∂4φ∂y4

= 0 .



This equations can also be written as

∂2

∂x2+

∂2

∂y2⎛⎝⎜

⎞⎠⎟

∂2φ∂x2

+∂2φ∂y2

⎛⎝⎜

⎞⎠⎟= 0 .

Because of its obvious similarity to the harmonic equation, it is called the biharmonic equation.

Thus, a procedure to solve a plane stress problem is to solve for ( )yx,φ from the above PDE, and then calculate stresses and strains. After the strains are obtained, the displacement field can be obtained by integrating the strain-displacement relations.

Dependence on elastic constants. For a plane problem with traction-prescribed boundary conditions, both the governing equation and the boundary conditions can be expressed in terms of φ . All these equations are independent of the elastic constants of the material. Consequently, the stress field in such a boundary value problem is independent of the elastic constants. Once we go over specific examples, we will find that the above statement is only correct for boundary value problems in simply connected regions. For multiply connected regions, the above equations in terms of φ do not guarantee that the displacement field is continuous. When we insist that displacement field be continuous, elastic constants may enter the stress field.

2.3.1 Solution of 2D problems in Cartesian coordinates: A half space subject to periodic traction on the surface An elastic material occupies a half space, 0>x . On the surface of the material, 0=x , the traction vector is prescribed

�

σ xx 0,y,z( ) = σ 0 cosky, τ xy 0,y,z( ) = 0, τ xz 0,y,z( ) = 0.

Determine the stress field inside the material.

Solution: The material clearly deforms under the plane strain conditions. It is reasonable to guess that the Airy stress function should take the form

φ x, y( ) = f x( )cosky .

The biharmonic equation then becomes

02 42

22

4

4

=+− fkdxfdk

dxfd .



This is a homogenous ODE with constant coefficients. The solution must be of the form

( ) xexf α= .

Insert this form into the ODE, and we obtain that

α 2 − k2( )2 = 0 .

The algebraic equation has double roots of k−=α , and double roots of k+=α . Consequently, the general solution is of the form

( ) kxkxkxkx DxeCxeBeAexf −− +++= ,

where A, B, C and D are constants of integration.

We expect that the stress field vanish as +∞→x , so that the stress function should be of the form

( ) kxkx DxeBexf −− += .

We next determine the constants B and D by using the traction boundary conditions. The stress fields are

( )

kykDxeekDBe

x

kykDxeekDBe

yx

kykDxeBey

kxkxkxyy

kxkxkxxy

kxkxxx

cos2

sin

cos

22

2

22

22

2

⎟⎠⎞⎜

⎝⎛ +−=

∂∂=

⎟⎠⎞⎜

⎝⎛ −+−=

∂∂∂−=

+−=∂∂=

−−−

−−−

−−

φσ

φτ

φσ

Recall the boundary conditions

( ) ( ) 0,0,cos,0 0 == ykyy xyxx τσσ .

Consequently, we find that

kDkB /,/ 02

0 σσ −=−= .

The stress field inside the material is



( )

( ) kyekx

kykxekyekx

kxyy

kxxy

kxxx

cos1

sin

cos1

0

0

0

−

−

−

−=

−=

+=

σσ

στσσ

The stress field decays exponentially.

We have solved the problem where the traction on the boundary of a half space is given by a simple cosine function. Through application of the superposition principle, which is valid for linear elastic materials, it is now straightforward to extend this analysis to any periodic traction distribution. Indeed, a periodic traction distribution can be written as a Fourier series each term of which is of the form found in the previous problem.

Application: de Saint-Venant’s principle When a load is applied in a small region, and the load has a vanishing resultant force and resultant moment, then the stress field is localized. We used this principle in discussing the laminate problem, where we have neglected the edge effects. While Saint-Venant’s principle cannot be proved in such a loose form, the foregoing is a nice example of the principle: the traction applied to the boundary is self-balancing and hence the stress field associated with the tractions die out. If we had imposed an additional constant traction term, the stress field would quickly decay to a constant stress.



2.3.2 Solution of 2D problems in polar coordinates

1. Transformation of stress components due to change of coordinates. A material particle is in a state of plane stress. If we represent the material particle by a square in the ( )yx, coordinate system, the components of the stress state are xyyyxx τσσ ,, . If we represent the same material particle under the same state of stress by a square in the ( )θ,r coordinate system, the components of the stress state are θθθ τσσ rrr ,, . From the transformation rules, we know that the two sets of the stress components are related as

θτθσσ

τ

θτθσσσσ

σ

θτθσσσσ

σ

θ

θθ

2cos2sin2

2sin2cos22

2sin2cos22

xyyyxx

r

xyyyxxyyxx

xyyyxxyyxx

rr

+−

−=

−−

−+

=

+−

++

=

2. Equations in polar coordinates. The Airy stress function is a function of the polar coordinates, ( )θφ ,r . The stresses are expressed in terms of the Airy stress function:

rrrrr ∂

∂+∂∂= φθφσ 122

2

, 2

2

r∂∂= φσθθ , ⎟

⎠⎞⎜

⎝⎛∂∂

∂∂−=

θφτ θ rrr

The biharmonic equation is

022

2

2

2

22

2

2

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂+

∂∂+

∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂+

∂∂+

∂∂

θφφφ

θ rrrrrrrr.

The stress-strain relations in polar coordinates are similar to those in the rectangular coordinate system:

EE

rrrr

θθσνσε −= , EErrσνσε θθ

θθ −= , ( )θθ τνγ rr E

+= 12

The strain-displacement relations are

ru

ru

ru

ru

ru

ru r

rrr

rrθθ

θθ

θθ θγ

θεε −

∂∂+

∂∂=

∂∂+=

∂∂= ,, .



3. A stress field symmetric about an axis. Let the Airy stress function be ( )rφ . The biharmonic equation becomes

0112

2

2

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛+

drdrdr

ddrd

rdrd φφ .

Each term in this equation has the same dimension in the independent variable r. Such an ODE is known as an equi-dimensional equation. A solution to an equi-dimensional equation is of the form

mr=φ .

Inserting into the biharmonic equation, we obtain that

( )22 2−mm .

The fourth order algebraic equation has a double root of 0 and a double root of 2. Consequently, the general solution to the ODE is

( ) DCrrBrrAr +++= 22 loglogφ .

where A, B, C and D are constants of integration. The components of the stress field are

( ) CrBrA

rrrrr 2log211222

2

+++=∂∂+

∂∂= φθφσ ,

( ) CrBrA

r2log2322

2

+++−=∂∂= φσθθ ,

0=⎟⎠⎞⎜

⎝⎛∂∂

∂∂−=

θφτ θ rrr .

The stress field is linear in A, B and C.

The contributions due to A and C are familiar: they are the same as the cylindrical Lamé problem. For example, for a hole of radius a in an infinite sheet subject to a remote biaxial stress S, the stress field in the sheet is



⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞⎜

⎝⎛+=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞⎜

⎝⎛−=

22

1,1raS

raSrr θθσσ .

The stress concentration factor of this hole is 2. We may compare this problem with that of a spherical cavity in an infinite elastic solid under remote tension:

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞⎜

⎝⎛+=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞⎜

⎝⎛−=

33

211,1raS

raSrr θθσσ .

A cut-and-weld operation. How about the contributions due to B? Let us study the stress field (Timoshenko and Goodier, pp. 77-79)

( )rBrr log21+=σ , ( )rB log23+=θθσ , 0=θτ r .

The strain field is

( ) ( ) ( )[ ]rEB

E rrrr log12311 νννσσε θθ −+−=−=

( ) ( ) ( )[ ]rEB

E rr log1231 νννσσε θθθθ −+−=−=

0=θγ r

To obtain the displacement field, recall the strain-displacement relations

ru

ru

ru

ru

ru

ru r

rrr

rrθθ

θθ

θθ θγ

θεε −

∂∂+

∂∂=

∂∂+=

∂∂= ,, .

Integrating rrε , we obtain that

( ) ( )[ ] ( )θνν frrrEBur ++−−= 1log12 ,

where ( )θf is a function still undetermined. Integrating θθε , we obtain that

( ) ( )rgdfEBru +−= ∫ θθθ

θ4 ,



where ( )rg is another function still undetermined. Inserting the two displacements into the expression

0=−∂∂

+∂∂

=ru

ru

rur

rθθ

θ θγ ,

and we obtain that

( ) ( ) ( ) ( ).'' rrgrgdff −=+ ∫ θθθ

In the equation, the left side is a function of θ , and the right side is a function of r. Consequently, the both sides must equal a constant independent of r and θ , namely,

( ) ( )( ) ( ) Grrgrg

Gdff

=−

=+ ∫'

' θθθ

Solving these equations, we obtain that

( )( ) GFrrg

KHf+=

+= θθθ cossin

Substituting back into the displacement field, we obtain that

( ) ( )[ ]

θθθ

θθνν

θ sincos4

cossin1log12

KHFrEBru

KHrrrEBur

−++=

+++−−=.

Consequently, F represents a rigid-body rotation, and H and K represent a rigid-body translation.

Now we can give an interpretation of B. Imagine a ring, with a wedge of angle α cut off. The ring with the missing wedge was then welded together. This operation requires that after a rotation of a circle, the displacement is

�

uθ 2π( ) − uθ 0( ) = αr

This condition gives

π

α8EB = .



This cut-and-weld operation clearly introduces a stress field in the ring. The stress field is axisymmetric, as given above.

4. A circular hole in an infinite sheet under remote shear. Remote from the hole, the sheet is in a state of pure shear:

0, === yyxxxy S σστ .

The remote stresses in the polar coordinates are

θτθσθσ θθθ 2cos,2sin,2sin SSS rrr =−== .

Recall that

rrrrr ∂

∂+∂∂= φθφσ 122

2

, 2

2

r∂∂= φσθθ , ⎟

⎠⎞⎜

⎝⎛∂∂

∂∂−=

θφτ θ rrr .

We guess that the stress function must be in the form

( ) ( ) θθφ 2sin, rfr = .

The biharmonic equation becomes

04422

2

22

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛−

∂∂+

∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛−+

rf

rrf

rf

rrdrd

drd .

A solution to this equi-dimensional ODE takes the form ( ) mrrf = . Inserting this form into the ODE, we obtain that

m − 2( )2 − 4( ) m2 − 4( ) = 0 .

The algebraic equation has four roots: 2, -2, 0, 4. Consequently, the stress function is

( ) θθφ 2sin, 242 ⎟

⎠⎞⎜

⎝⎛ +++= D

rCBrArr .

The stress components inside the sheet are



θφθφσ 2sin4621

2422

2

⎟⎠⎞⎜

⎝⎛ ++−=

∂∂+

∂∂=

rD

rCA

rrrrr

θφσθθ 2sin6122 42

2

2

⎟⎠⎞⎜

⎝⎛ ++=

∂∂=

rCBrA

r

θθφτ θ 2cos2662 24

2 ⎟⎠⎞⎜

⎝⎛ ++−−=⎟

⎠⎞⎜

⎝⎛∂∂

∂∂−=

rD

rCBrA

rrr .

To determine the constants A, B, C, D, we invoke the boundary conditions:

1. Remote from the hole, namely, ∞→r , θτθσ θ 2cos,2sin SS rrr == , giving 0,2/ =−= BSA .

2. On the surface of the hole, namely, r = a, 0,0 == θτσ rrr , giving 2SaD = and 2/4SaC −= .

The stress field inside the sheet is

θσ 2sin43124

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞⎜

⎝⎛−⎟

⎠⎞⎜

⎝⎛+=

ra

raSrr

θσθθ 2sin314

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞⎜

⎝⎛+−=raS

θτ θ 2cos23124

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞⎜

⎝⎛+⎟

⎠⎞⎜

⎝⎛−=

ra

raSr

5. A hole in an infinite sheet subject to a remote uniaxial stress. Use this as an example to illustrate linear superposition. A state of uniaxial stress is a linear superposition of a state of pure shear and a state of biaxial tension. The latter is the Lame problem. When the sheet is subject to remote tension of magnitude S, the stress field in the sheet is given by

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞⎜

⎝⎛+=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞⎜

⎝⎛−=

22

1,1raS

raSrr θθσσ .



Illustrate the superposition in figures. Show that under uniaxial tensile stress, the stress around the hole has a concentration factor of 3. Under uniaxial compression, the material may split in the loading direction.

6. A line force acting on the surface of a half space. A half space of an elastic material is subject to a line force on its surface. Let P be the force per unit length. The half space lies in

0>x , and the force points in the direction of x. This problem has no length scale. Linearity and dimensional considerations requires that the stress field take the form

( ) ( )θθσ ijij grPr =, ,

where ( )θijg are dimensionless functions of θ . We guess that the stress function takes the form

( ) ( )θφ rPfr = ,

where ( )θf is a dimensionless function of θ . (A homework problem will show that this guess is not completely correct, but it suffices for the present problem.)

Inserting this form into the biharmonic equation, we obtain an ODE for ( )θf :

02 4

4

2

2

=++θθ dfd

dfdf .

The general solution is

( ) ( )θθθθθθθφ cossincossin, DCBArPr +++= .

Observe that yr =θsin and xr =θcos do not contribute to any stress, so we drop these two terms. By the symmetry of the problem, we look for stress field symmetric about 0=θ , so that we will drop the term θθ cos . Consequently, the stress function takes the form

( ) θθθφ sin, rPCr = .

We can calculate the components of the stress field:

0,cos2 === θθθ τσθσ rrr rCP .



This field satisfies the traction boundary conditions, 0== θθθ τσ r at

�

θ = π /2 and

�

θ = −π /2 . To determine C, we require that the resultant force acting on a cylindrical surface of radius r balance the line force P. On each element θrd of the surface, the radial stress provides a vertical component of force θθσ rdrr cos . The force balance of the half cylinder requires that

0cos2/

2/

=+ ∫−

π

π

θθσ rdP rr .

Integrating, we obtain that π/1−=C .

The stress components in the x-y coordinates are

θθπ

τθθπ

σθπ

σ 3224 cossin2,cossin2,cos2xP

xP

xP

xyyyxx −=−=−=

The displacement field is

( )

( ) ( ) θπνθθ

πνθ

πθ

πν

θθπνθ

π

θ sin1cos1logsin2sin2

sin1logcos2

EP

EPr

EP

EPu

EPr

EPur

−−−−+−=

−−−=

7. Separation of variable. One can obtain many solutions by using the procedure of separation of variable, assuming that

( ) ( ) ( )θθφ Θ= rRr, .

Formulas for stresses and displacements can be found on p. 205, Deformation of Elastic Solids, by A.K. Mal and S.J. Singh.

A real-life example.

From: S. Ho, C. Hillman, F.F. Lange and Z. Suo, " Surface cracking in layers under biaxial, residual compressive stress," J. Am. Ceram. Soc. 78, 2353-2359 (1995).

In previous treatment of laminates, we have ignored edge effect. However, we also know that edges are often the site for failure to initiate. Here is a phenomenon discovered in the lab of Fred Lange at UCSB. A thin layer of material 1 was sandwiched in two thick blocks of material 2. Material 1 has a smaller coefficient of thermal expansion than material 2, so that, upon cooling,



material 1 develops a biaxial compression in the plane of the laminate. The two blocks are nearly stress free. Of course, these statements are only valid at a distance larger than the thickness of the thin layer. It was observed in experiment that the thin layer cracked, as shown in Fig. 1.



It is clear from Fig. 2 that a tensile stress yyσ can develop near the edge. We would like to know its magnitude, and how fast it decays as we go into the layer.

We analyze this problem by a linier superposition shown in Fig. 3. Let Mσ be the magnitude of the biaxial stress in the thin layer far from the edge. In Problem A, we apply a compressive traction of magnitude Mσ on the edge of the thin layer, so that the stress field in thin layer in Problem A is the uniform biaxial stress in the thin layer, with no other stress components. In problem B, we remover thermal expansion misfit, but applied a tensile traction on the edge of the thin layer. The original problem is the superposition of Problem A and Problem B. Thus, the residual stress field yyσ in the original problem is the same as the stress yyσ in Problem B.



With reference to Fig. 4, let us calculate the stress distribution ( )0,xyyσ . Recall that when a half space is subject to a line force P, the stress is given by

θθπ

σ 22 cossin2xP

yy −= .

We now consider a line-force acting at η=y . On an element of the edge, ηd , the tensile traction applied the line force ησ dP M−= . Summing up over all elements, we obtain the stress field in the layer:

( ) ∫−

=2/

2/

22 cossin20,t

t

Myy x

dx θθπ

ησσ .

Note that θη tanx= , and let xt 2/tan =β . Consequently,

θθ

η dxd 2cos= ,

and the integral becomes

( ) ∫∫−−

−==β

β

β

β

θθπσθθ

πσσ ddx MM

yy 22cos12sin20, 2 .

Integrating, we obtain that

( ) ⎟⎠⎞⎜

⎝⎛ −= ββ

πσσ 2sin

2120, M

yy x .



At the edge of the layer, 0/ →tx and 2/πβ = , so that ( ) Myy σσ =0,0 . Far from the edge, 0/ →xt ,

( )3

60, ⎟

⎠⎞⎜

⎝⎛→xtx M

yy πσσ .

Thus, this stress decays as 3−x .

2 Plane Elasticity - Home | Scholars at Harvard · · 2014-07-08ES240 Solid Mechanics Fall 2009...

Documents

Transcript of 2 Plane Elasticity - Home | Scholars at Harvard · · 2014-07-08ES240 Solid Mechanics Fall 2009...