ES240
Solid Mechanics
Z. Suo
Plane Elasticity Problems
Main Reference: Theory of Elasticity, by S.P. Timoshenko and J.N. Goodier, McGraw-Hill, New
York. Chapters 2,3,4. (I have not proof read these notes carefully. If in doubt, please go back to
Timoshenko and Goodier.)
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 plane ( x, y ) . The top and the bottom surfaces of the sheet is 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
t
σxx nx +τxy n y =t x
τxy nx +σyy n y =t y
St
b
where t x and t y are components of the traction vector prescribed on the edge of
y
the sheet, and nx and n y are the components of the unit vector normal to the
x
edge of the sheet. The above two equations provide two conditions for the
Su z
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 partial differential equations analytically. However,
numerical methods are now readily available to solve any elasticity problem that you can pose. In this
introductory course, to gain some 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: σxx ,σyy ,τxy , and the components out of the plane vanish: σzz =τ xz =τ yz = 0 . 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:
σxx ( x, y ), σyy ( x, y ), τxy ( x, y ) .
Will these guesses satisfy the governing equations of elasticity? Let us go through these equations one by
one.
Equilibrium equations. Using the guessed stress field, we reduce the three equilibrium equations
to two equations:
∂τ xy ∂σ yy
∂σ xx ∂τ xy
+
= 0,
+
=0.
∂x
∂y
∂x
∂y
These two equations by themselves are insufficient to determine the three functions.
Stress-strain relations. Given the guessed stress field, the 6 components of the strain field are
σ
σ
2(1 +ν )
σ
σ
τ xy
ε xx = xx −ν yy , ε yy = yy −ν xx , γ xy =
E
E
E
E
E
ν
ε zz = − (σ xx + σ yy ), γ xz = γ yz = 0 .
E
Strain-displacement relations. Recall the 6 strain-displacement relations:
∂u
∂v
∂u ∂v
, ε yy =
, γ xy =
+
∂x
∂y
∂y ∂x
∂w
∂u ∂w
∂v ∂w
=
, γ xz =
+
, γ yz =
+
.
∂z
∂z
∂x
∂z
∂y
εxx =
εzz
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 γ xz = γ yz = 0 , we find that
Thus,
w
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∂w
∂w
=
=0 .
∂x
∂y
is independent of x and y, and can only be a function of z. If we insist that εzz be independent
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of z, and from ε zz = ∂w / ∂z , then εzz must be a constant, ε zz = c , and w = cz + b . 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 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
ε zz = c and ε zz = −ν (σ xx + σ yy ) / E , at least the following set of equations look self-consistent:
∂σ xx ∂τ xy
+
= 0,
∂x
∂y
ε xx =
∂τ xy ∂σ yy
+
=0
∂x
∂y
σ
σ
2(1 +ν )
σ xx
σ
τ xy
−ν yy , ε yy = yy −ν xx , γ xy =
E
E
E
E
E
εxx =
∂u
∂v
∂u ∂v
, ε yy =
, γ xy =
+
.
∂x
∂y
∂y ∂x
These are 8 equations for 8 functions. We will focus on these 8
equations.
y
Plane strain problem. An infinitely long cylinder with the axis
in the
x
z-direction, and a cross section in the ( x, y ) plane. The loading is
z
invariant along the z-direction. Consequently, 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 inplane
strains are nonzero: εxx ( x, y ), εyy ( x, y ), γxy ( x, y ) . The three out-ofplane
strains vanish: ε zz = γ xz = γ yz = 0 .
Because γ xz = γ yz = 0 , the stress-strain relations imply that τxz =τ yz = 0 . From ε zz = 0 and
ε zz = (σ zz −νσ xx −νσ yy ) , we obtain that
σ zz =ν (σ xx + σ yy ) .
Thus,
1
1 −ν 2 
ν

ε xx = (σ xx −νσ yy −νσ zz ) =
σ yy 
σ xx −
E
E 
1 −ν

2
1
1 −ν 
ν

ε yy = (σ yy −νσ xx −νσ zz ) =
σ xx 
σ yy −
E
E 
1 −ν

2(1 +ν )
γ xy =
τ xy
E
These three stress-strain relations look similar to those under the plane stress conditions, provided we
make the following substitutions:
E
ν
, ν =
.
2
1 −ν
1 −ν
The quantity E is called the plane strain modulus.
A theorem in calculus. If two functions f ( x, y ) and g ( x, y ) satisfy
E=
there exists a function A( x, y ) , such that
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∂f
∂g
=
,
∂x
∂y
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f =
Z. Suo
∂A
∂A
, g=
.
∂y
∂x
The Airy stress function. We now apply the above theorem to the equilibrium equations. From
∂σ xx ∂τ xy
+
= 0 , we deduce that there exists a function A( x, y ) , such that
∂x
∂y
σxx =
From
∂A
∂A
, τxy = −
.
∂y
∂x
∂τ xy ∂σ yy
+
= 0 , we deduce that that there exists a function B ( x, y ) , such that
∂x
∂y
σyy =
∂B
∂B
, τ xy = −
.
∂x
∂y
∂A
∂B
=
, we deduce that that there exists a function φ( x, y ) , such that
∂x
∂y
∂φ
∂φ
A=
, B=
.
∂y
∂x
The function φ( x, y ) is known as the Airy (1862) stress function. The three components of the
Finally, from
stress field can now be represented by the stress function:
∂2φ
∂2φ
∂2φ
σxx = 2 , σ yy = 2 , τ xy = −
.
∂y
∂y
∂y∂x
Using the stress-strain relations, we can also express the three components of strain field in terms
of the Airy stress function:
1  ∂2φ
∂2φ 
1  ∂2φ
∂2φ 
2(1 +ν) ∂2φ




−
ν
ε
=
−
ν
γ
=
−
,
,
.
yy
xy
2
2
E
∂x∂y
E
∂x 2 
E
∂y 2 
 ∂y

 ∂x

Compatibility equation. Recall the strain-displacement relations:
ε xx =
εzz =
∂w
∂u ∂w
∂v ∂w
, γ xz =
+
, γ yz =
+
.
∂z
∂z
∂x
∂z
∂y
Eliminate the two displacements in the three strain displacement relations, and we obtain that
2
∂2γ xy
∂2ε xx ∂ ε yy
+
=
.
∂y 2
∂x 2
∂x∂y
This equation is known as the compatibility equation.
Biharmonic equation. Inserting the expressions of the strains in terms of φ( x, y ) into the
compatibility equation, and we obtain that
∂4φ
∂4φ
∂4φ
+
2
+
=0 .
∂x 4
∂x 2 ∂y 2
∂y 4
This equations can also be written as
 ∂2
∂2  ∂2φ ∂2φ 

 ∂x 2 + ∂y 2 

 2 + ∂y 2 
=0.

 ∂x

It is called the biharmonic equation.
Thus, a procedure to solve a plane stress problem is to solve for φ( x, y ) 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-prescribe boundary
conditions, both the governing equation and the boundary conditions can be expressed in terms of φ.
All these equations are independent of elastic constants. 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
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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.
Saint-Venant s principle. When load is applied in a small region, and the load has a vanishing
resultant force and resultant moment, then the stress field is localized.
While Saint-Venant s principle cannot be proved in such a loose form, we can certainly give a few
examples to illustrate the idea.
• We have used this principle in discussing the laminate problem, where we have neglected the edge
effects.
• For a spherical cavity in a block of material under internal pressure, the stress field in the block is
3
σ rr
3
1 a
a
= − p  , σ θθ = p  .
2 r 
r 
A half space subject to periodic traction on the surface. An elastic material occupies a half
space, x > 0 . On the surface of the material, x = 0 , the traction vector is prescribed
σxx ( 0, y , z ) = σ0 cos ky, τ xy ( 0, y , z ) = τ xz ( 0, y, z ) = 0 .
Determine the stress field inside the material.
The material clearly deforms under the plane strain conditions. It is reasonable to guess that the
Airy function should take the form
φ( x, y ) = f ( x ) cos ky .
The biharmonic equation becomes
2
d4 f
2 d f
−
2
k
+ k4 f = 0 .
4
2
dx
dx
This is a homogenous ODE with constant coefficients. A solution is of the form
f ( x ) = eαx .
Insert this form into the ODE, and we obtain that
(α 2 − k 2 )2 = 0 .
The algebraic equation has double roots of α = −k , and double roots of α = +k . Consequently, the
general solution is of the form
f ( x ) = Ae kx + Be −kx + Cxe kx + Dxe −kx ,
where A, B, C and D are constants of integration.
We expect that the stress field vanishes as x → +∞ , so that the stress function should be of the
form
f ( x ) = Be −kx + Dxe −kx .
We next determine the constants B and D by using the boundary conditions.
The stress fields are
∂2φ
σxx = 2 = −( Be −kx + Dxe−kx )k 2 cos ky
∂y
∂2φ
D


=  − Be −kx + e −kx − Dxe −kx k 2 sin ky
∂x∂y 
k

2
∂φ 
D

= 2 =  Be −kx − 2 e −kx + Dxe −kx k 2 cos ky
∂x
k


τ xy = −
σ yy
Recall the boundary conditions
σxx ( 0, y ) = σ0 cos ky, τ xy ( 0, y ) = 0 .
We find that
B = −σ 0 / k 2 , D = −σ 0 / k .
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The stress field inside the material is
σxx =σ0 (1 + kx )e −kx cos ky
τxy = −σ0 kxe −kx sin ky
σyy =σ0 (1 − kx )e −kx cos ky
The stress field decays exponentially.
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 ( x, y ) coordinates, the
components of the stress state are σxx ,σyy ,τxy . If we represent the same material particle under the
same state of stress by a square in the ( r ,θ) coordinates, the components of the stress state are
σrr ,σθθ ,τrθ . The two sets of the components are related as
σ + σ yy σ xx − σ yy
σ rr = xx
+
cos 2θ + τ xy sin 2θ
2
2
σ + σ yy σ xx − σ yy
σθθ = xx
−
cos 2θ − τ xy sin 2θ
2
2
σ − σ yy
τ rθ = − xx
sin 2θ + τ xy cos 2θ
2
Equations in polar coordinates. The Airy stress function is a function of the polar coordinates,
φ( r ,θ) . The stresses are expressed in terms of the Airy stress function:
∂  ∂φ 
∂2φ
1 ∂φ
∂2φ

, σθθ = 2 , τ rθ = − 
σ rr = 2 2 +
∂r  r∂θ 
r ∂θ
r ∂r
∂r
The biharmonic equation is
 ∂2
∂
∂2  ∂2φ ∂φ
∂2φ 



+
+
+
+
 ∂r 2 r∂r r 2∂θ 2  ∂r 2 r∂r r 2∂θ 2 
=0 .



The stress-strain relations in polar coordinates are similar to those in the rectangular coordinates:
2(1 +ν )
σ
σ
σ
σ
τ rθ
ε rr = rr − ν θθ , ε θθ = θθ − ν rr , γ rθ =
E
E
E
E
E
The strain-displacement relations are
∂u
u
∂u
∂u
∂u
u
ε rr = r , εθθ = r + θ , γ rθ = r + θ − θ .
∂r
r r∂θ
r∂θ
∂r
r
Stress field symmetric about an axis. Let the stress function be φ( r ) . The biharmonic
equation becomes
 d2
1 d  d 2φ 1 dφ 



+
 2

 2 +
=0 .
r dr  dr
r dr 
 dr
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
φ =r m .
Inserting into the biharmonic equation, we obtain that
2
m2 ( m − 2) .
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
φ( r ) = A log r + Br 2 log r + Cr 2 + D .
where A, B, C and D are constants of integration.
The components of the stress field are
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∂2φ
1 ∂φ
A
+
= 2 + B(1 + 2 log r ) + 2C ,
2
2
r ∂θ
r ∂r r
2
∂φ
A
σθθ = 2 = − 2 + B( 3 + 2 log r ) + 2C ,
∂r
r
∂  ∂φ 
τ rθ = − 
=0.
∂r  r∂θ 
The stress field is a linear in A, B and C.
The contributions due to A and C are familiar: they are the same as the Lame 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
σ rr =
  a 2 
  a 2 
σ rr = S 1 −   , σ θθ = S 1 +    .
  r  
  r  
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:
  a 3 
 1  a 3 
σ rr = S 1 −   , σ θθ = S 1 +    .
  r  
 2  r  
A cut-and-weld operation. How about the contributions due to B? Let us study the stress field
(Timoshenko and Goodier, pp. 77-79)
σ rr = B(1 + 2 log r ) , σθθ = B( 3 + 2 log r ) , τrθ = 0 .
The strain field is
1
B
ε rr = (σ rr −νσθθ ) = [ (1 − 3ν ) + 2(1 −ν ) log r ]
E
E
1
B
εθθ = (σθθ −νσ rr ) = [ ( 3 −ν ) + 2(1 −ν ) log r ]
E
E
γ rθ = 0
To obtain the displacement field, recall the strain-displacement relations
∂ur
u
∂u
∂u
∂u
u
, εθθ = r + θ , γ rθ = r + θ − θ .
∂r
r r∂θ
r∂θ
∂r
r
ε
Integrating rr , we obtain that
ε rr =
B
[ 2(1 −ν ) r log r − (1 +ν ) r ] + f (θ ) ,
E
where f (θ ) is a function still undetermined. Integrating εθθ , we obtain that
4 Brθ
uθ =
− ∫ f (θ ) dθ + g ( r ) ,
E
where g ( r ) is another function still undetermined. Inserting the two displacements into the expression
ur =
γ rθ =
and we obtain that
∂u r ∂uθ uθ
+
−
=0,
r∂θ
∂r
r
f ' (θ) + ∫ f (θ)dθ = g ( r ) − rg ' ( r ).
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,
f ' (θ ) + ∫ f (θ ) dθ = G
g ( r ) − rg ' ( r ) = G
Solving these equations, we obtain that
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f (θ ) = H sin θ + K cos θ
g ( r ) = Fr + G
Substituting back into the displacement field, we obtain that
B
[ 2(1 −ν ) r log r − (1 +ν ) r ] + H sin θ + K cos θ
E
.
4 Brθ
uθ =
+ Fr + H cos θ − K sin θ
E
ur =
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 weld together. This operation requires that after a rotation of a
circle, the displacement is
v( 2π ) − v( 0 ) = αr
This condition gives
αE
B=
.
8π
This cut-and-weld operation clearly introduces a stress field in the ring. The stress field is axisymmetric,
as given above.
A circular hole in an infinite sheet under remote shear. Remote from the hole, the sheet is in a
state of pure shear:
τ xy = S , σxx = σ yy = 0 .
The remote stresses in the polar coordinates are
σrr = S sin 2θ, σθθ = −S sin 2θ, τ rθ = S cos 2θ .
Recall that
∂  ∂φ 
∂2φ
1 ∂φ
∂2φ
.
, σθθ = 2 , τ rθ = − 
σ rr = 2 2 +
∂r  r∂θ 
r ∂θ
r ∂r
∂r
We guess that the stress function must be in the form
φ( r ,θ ) = f ( r ) sin 2θ .
The biharmonic equation becomes
 d2
d
4  ∂2 f
∂f
4f 
 2 +
− 2  2 +
− 2  = 0 .
rdr r  ∂r
r∂r r 
 dr
A solution to this equi-dimensional ODE takes the form f ( r ) = r m . Inserting this form into the ODE, we
obtain that
( m − 2) 2 − 4 m 2 − 4 = 0 .
The algebraic equation has four roots: 2, -2, 0, 4. Consequently, the stress function is
C


φ ( r ,θ ) =  Ar 2 + Br 4 + 2 + D  sin 2θ .
r


The stress components inside the sheet are
(
)(
)
∂2φ
1 ∂φ
6C 4 D 

+
= − 2 A + 4 + 2  sin 2θ
2
2
r ∂θ
r ∂r
r
r 

2
∂φ 
6C 
σθθ = 2 =  2 A +12 Br 2 + 4  sin 2θ
∂r
r 

∂  ∂φ  
6C 2 D 
2
τ rθ = − 
 =  − 2 A − 6 Br + 4 + 2  cos 2θ .
∂r  r∂θ  
r
r 
To determine the constants A, B, C, D, we invoke the boundary conditions:
σ rr =
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the hole, namely, r → ∞ , σrr = S sin 2θ, τrθ = S cos 2θ , giving
A = −S / 2, B = 0 .
• On the surface of the hole, namely, r = a, σrr = 0, τrθ = 0 , giving D = Sa 2 and C = −Sa 4 / 2 .
The stress field inside the sheet is
•
Remote
from

4
a 
r 
2
a  
r  

σrr = S 1 + 3  − 4   sin 2θ


σθθ
4

a  
= −S 1 + 3   sin 2θ
r  




4
a 
r 
2
a  
r  

τrθ = S 1 − 3  + 2   cos 2θ


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
  a 2 
  a 2 
σ rr = S 1 −   , σ θθ = S 1 +    .
  r  
  r  
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, material may split in the loading direction.
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 x > 0 , 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
P
σ ij ( r , θ ) = g ij (θ ) ,
r
where g ij (θ ) are dimensionless functions of θ . We guess that the stress function takes the form
φ( r ) = rPf (θ ) ,
where f (θ ) 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 f (θ ) :
d2 f d4 f
f +2 2 +
= 0.
dθ
dθ 4
The general solution is
φ( r ,θ ) = rP ( A sin θ + B cos θ + Cθ sin θ + Dθ cos θ ) .
r
sin
θ
= y and r cos θ = x do not contribute to any stress, so we drop these two terms.
Observe that
By the symmetry of the problem, we look for stress field symmetric about θ = 0 , so that we will drop the
term θ cos θ . Consequently, the stress function takes the form
φ( r , θ ) = rPCθ sin θ .
We can calculate the components of the stress field:
2CP cos θ
σ rr =
, σ θθ = τ rθ = 0 .
r
This field satisfies the traction boundary conditions, σ θθ = τ rθ = 0 at θ = 0 and θ = π . 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 rdθ of the surface, the radial stress provides a vertical component of force
σrr cos θrdθ . The force balance of the half cylinder requires that
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P+
π/2
∫σ
rr
Z. Suo
cos θrdθ = 0 .
−π / 2
Integrating, we obtain that C = −1 / π .
The stress components in the x-y coordinates are
2P
2P
2P
σ xx = −
cos 4 θ , σ yy = −
sin 2 θ cos 2 θ , τ xy = −
sin θ cos 3 θ
πx
πx
πx
The displacement field is
2P
(1 −ν ) P θ sin θ
ur = −
cos θ log r −
πE
πE
2νP
2P
(1 −ν ) P θ cos θ − (1 −ν ) P sin θ
uθ = −
sin θ +
sin θ log r −
πE
πE
πE
πE
Separation of variable. One can obtain many solutions by using the procedure of separation of
variable, assuming that
φ( r, θ ) = R ( r )Θ(θ ) .
Formulas for stresses and displacements can be found on p. 205, Deformation of Elastic Solids, by A.K.
Mal and S.J. Singh.
An example. 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.
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It is clear from Fig. 2 that a tensile stress σyy 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 σM 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 σM 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 σyy in the original
problem is the same as the stress σyy in Problem B.
With reference to Fig. 4, let us calculate the stress distribution σyy ( x,0 ) . Recall that when a half
space is subject to a line force P, the stress is given by
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2P
sin 2 θ cos 2 θ .
πx
We now consider a line force acting at y = η . On an element of the edge, dη, the tensile traction
applied the line force P = −σ M dη . Summing up over all elements, we obtain the stress field in the
layer:
t/2
2σ M dη
σ yy ( x,0 ) = ∫
sin 2 θ cos 2 θ .
πx
−t / 2
σ yy = −
Note that η = x tan θ , and let tan β = t / 2 x .
Consequently, dη =
x
dθ , and the integral
cos 2 θ
becomes that
σ yy ( x,0 ) =
2σ M
π
β
2
∫ sin θdθ =
−β
2σ M
π
β
1 − cos 2θ
dθ
2
∫β
−
Integrating, we obtain that
2σ M 
1

 β − sin 2β  .
π 
2

At the edge of the layer, x / t → 0 and β = π / 2 , so that σ yy ( 0,0 ) = σ M . Far from the edge, t / x → 0 ,
σ yy ( x,0 ) =
3
σ t 
σ yy ( x,0 ) → M   .
6π  x 
Thus, this stress decays as x −3 .
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Plane Elasticity Problems-12
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