ES240
Solid Mechanics
Z. Suo
Plane Elasticity Problems
Main Reference: Theory of Elasticity, by S.P. Timoshenko and J.N. Goodier, McGrawHill, New York. Chapters 2,3,4.
A thin sheet subject to loads in its plane. A thin sheet of an isotropic material lies
in the plane (x, y ) , and is subject to load in the plane of the sheet. The top and the bottom
surfaces of the sheet are traction-free. The edge of the sheet is subject to two kinds of the
boundary conditions: at each material particle on the edge either the
displacement or the traction is prescribed. In the latter case, we write
t
σ xx n x + τ xy ny = t x
τ xy n x + σ yy ny = t y
where t x and t y are components of the traction vector prescribed on
the edge of the sheet, and n x and ny are the components of the unit
St
y
x
vector normal to the edge of the sheet. The above equations provide
Su z
two conditions at each point on the edge of the sheet.
Semi-inverse method. We next go into the interior of the
sheet. We have already obtained a full set of governing equations of
linear elasticity. However, no general approach exists to solve partial differential equations
analytically. Numerical methods are now readily available to solve any elasticity problem that
you can pose. In this introductory course, to gain insight into solid mechanics, we will make
reasonable guesses of solutions, and see if they satisfy all the governing equations. This trialand-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: σ xx ,σ yy ,τ xy , and that the components out of the plane vanish:
σ zz = τ xz = τ yz = 0 . Furthermore, we guess that the in-plane components of stress may vary with
x and y, but are independent of z. That is, the stress field in the sheet is described by three
functions of two variables:
σ xx (x , y ), σ yy (x , y ), τ xy (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 components of stress.
Stress-strain relations. Given the guessed stress field, the 6 components of the strain
field are
σ
σ
2(1 +ν )
σ
σ
ε xx = xx −ν yy , ε yy = yy −ν xx , γ xy =
τ xy
E
E
E
E
E
ν
(σ xx + σ yy ), γ xz = γ yz = 0 .
E
Strain-displacement relations. Recall the 6 strain-displacement relations:
∂u
∂v
∂u ∂v
ε xx = , ε yy = , γ xy =
+
∂x
∂y
∂y ∂x
∂w
∂u ∂w
∂v ∂w
+
+
ε zz =
, γ xz =
, γ yz =
.
∂z
∂z ∂x
∂z ∂y
ε zz = −
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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
∂w ∂w
=
=0.
∂x ∂y
Thus, w is independent of x and y, and can only be a function of z. If we insist that ε zz be
independent of z, and from ε zz = ∂w / ∂z , then ε zz 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.
Plane stress problem. 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 =
σ xx
E
−ν
σ yy
E
, ε yy =
∂τ xy
∂x
σ yy
E
−ν
+
∂σ yy
σ xx
E
∂y
=0
, γ xy =
2(1 +ν )
τ xy
E
∂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.
Plane strain problem. Consider an infinitely long cylinder with
the axis in the z-direction, and a cross section in the (x, y ) plane. The
cylinder is constrained from deforming in the z-direction, and the loading
is 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:
∂u ∂v
∂v
∂u
.
ε xx (x , y ) = , ε yy (x , y ) = , γ xy (x , y ) =
+
∂y ∂x
∂y
∂x
The three out-of-plane strains vanish: ε zz = γ xz = γ yz = 0 .
ε xx =
y
x
z
We assume the material is isotropic and linearly elastic. 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 that
σ zz = ν (σ xx + σ yy ) .
Thus,
ε xx =
2
1
(σ xx −νσ yy −νσ zz ) = 1 −ν ⎛⎜σ xx − ν σ yy ⎞⎟
E
E ⎝
1 −ν
⎠
2
1
(σ yy −νσ xx −νσ zz ) = 1 −ν ⎛⎜σ yy − ν σ xx ⎞⎟
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:
ε yy =
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E
ν
, ν =
.
2
1 −ν
1 −ν
The quantity E is called the plane strain modulus.
A theorem in calculus. Given two functions f (x , y ) and g(x , y ) , there exists a
function P (x , y ) , such that
∂P (x , y )
∂P (x , y )
f =
, g=
,
∂x
∂y
if and only if the two functions satisfy the condition
∂f ∂g
=
.
∂x ∂y
Proof. In calculus we know that the order of partial differentiation can be changed, such
as
∂ ⎛ ∂A(x , y ) ⎞ ∂ ⎛ ∂A(x , y ) ⎞
⎜
⎟=
⎜
⎟.
∂x ⎜⎝ ∂y ⎟⎠ ∂y ⎝ ∂x ⎠
Consequently, the conditions
∂P (x , y )
∂P (x , y )
, g=
f =
∂y
∂x
imply
∂f ∂g
=
.
∂x ∂y
E=
We next prove the converse is also true. For a given function f (x , y ) , we can always find
a function A(x , y ) , such that
∂A(x , y )
f =
.
∂y
Indeed, we can obtain A(x , y ) by integration:
A(x , y ) =
∫ f (x , y)dy + h(x ) ,
where h(x ) is an arbitrary function. From the prescribed condition
∂f ∂g
=
,
∂x ∂y
we obtain that
∂g ∂ ⎛ ∂A(x , y ) ⎞
⎜
⎟.
=
∂y ∂x ⎜⎝ ∂y ⎟⎠
Integrating with respect to y, we obtain that
∂A(x , y )
+ k (x ) ,
∂x
where k (x ) is another arbitrary function. Let
g=
P = A(x , y ) + k (x )dx ,
∫
so that
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f =
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∂P (x , y )
∂P (x , y )
, g=
.
∂x
∂y
The Airy stress function. We now apply the above theorem to the equilibrium
∂σ xx ∂τ xy
+
= 0 , we deduce that there exists a function A(x , y ) , such that
equations. From
∂x
∂y
∂A
∂A
σ xx =
, τ xy = −
.
∂y
∂x
∂τ xy ∂σ yy
+
= 0 , we deduce that that there exists a function B(x , y ) , such that
From
∂x
∂y
∂B
∂B
σ yy =
, τ xy = −
.
∂y
∂x
Note that is the above the shear stress τ xy have been expressed in two ways, so that
∂A ∂B
.
=
∂x ∂y
From this equation we deduce that that there exists a function φ (x, y ) , such that
∂φ
∂φ
, B=
.
A=
∂x
∂y
The function φ ( x, y ) is known as the Airy (1862) stress function. The three components
of the stress field can now be represented by the stress function:
∂ 2φ
∂ 2φ
∂ 2φ
,
,
.
=
−
σ
τ
=
yy
xy
∂y∂x
∂x 2
∂y2
σ xx =
Recall that we have started with the two equilibrium equations governing the three stresses.
Using the stress-strain relations, we can also express the three components of strain field
in terms of the Airy stress function:
∂ 2φ ⎞
∂ 2φ ⎞
1 ⎛ ∂ 2φ
1 ⎛ ∂ 2φ
2(1 + ν ) ∂ 2φ
⎜⎜ 2 −ν 2 ⎟⎟ , ε yy = ⎜⎜ 2 −ν 2 ⎟⎟ , γ xy = −
.
∂x ⎠
∂y ⎠
E ⎝ ∂y
E ⎝ ∂x
E ∂x∂y
Compatibility equation. Recall the strain-displacement relations:
ε xx =
ε xx =
∂u
∂v
∂u ∂v
, ε yy =
, γ xy =
+ .
∂x
∂y
∂y ∂x
Eliminate the two displacements in the three strain displacement relations, and we obtain that
2
2
∂ 2ε xx ∂ ε yy ∂ γ xy
.
+
=
∂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 2 ∂y 2 ∂y 4
∂x 4
This equations can also be written as
⎛ ∂2
∂ 2 ⎞⎛ ∂ 2φ ∂ 2φ ⎞
⎜⎜ 2 + 2 ⎟⎟⎜⎜ 2 + 2 ⎟⎟ = 0 .
∂y ⎠
∂y ⎠⎝ ∂x
⎝ ∂x
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It is called the biharmonic equation.
Thus, once φ (x, y ) is solved from the above PDE, we can calculate stresses and strains.
After the strains are obtained, the displacement field can be obtained by integrating the straindisplacement 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 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 a load is applied in a region small compared to the
overall size of a body, and the load has a vanishing resultant force and resultant moment, then
the stress field decays rapidly away from the region where the load is applied.
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, 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
3
1 ⎛a⎞
⎛a⎞
σ rr = − p⎜ ⎟ , σ θθ = p⎜ ⎟ .
2 ⎝r⎠
⎝r⎠
Another example that supports Saint-Venant’s principle is give below.
A half space subject to periodic traction on the surface. An elastic body
occupies a half space, x > 0 . On the surface of the body, x = 0 , the traction is prescribed as
σ xx (0, y, z ) = σ 0 cos ky, τ xy (0, y, z ) = τ xz (0, y, z ) = 0 .
The applied load is sinusoidal, with amplitude σ 0 , and wave number k. The wave number k
relates to the wavelength as k = 2π / λ .
The applied load is localized in a region of length λ . Within a period, the load has a
vanishing resultant force and a vanishing resultant moment. We expect the field of stress in the
body decays over the length scale λ . To ascertain this expectation, we wish to determine the
stress field inside the body.
The body clearly deforms under the plane strain conditions. Recall that
∂ 2φ
∂ 2φ
∂ 2φ
.
σ xx = 2 , σ yy = 2 , τ xy = −
∂y
∂y
∂y∂x
Inspecting the boundary conditions, we guess that the Airy function takes the form
φ (x , y ) = f (x )cos ky .
The biharmonic equation becomes
d4 f
d2 f
− 2k 2
+ k4 f = 0 .
4
2
dx
dx
This is a homogenous ODE with constant coefficients. A solution is of the form
f (x ) = e αx .
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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
σ xx = 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
D
∂φ ⎛
⎞
σ yy = 2 = ⎜ Be −kx − 2 e −kx + Dxe −kx ⎟k 2 cos ky
k
∂x
⎝
⎠
Recall the boundary conditions
σ xx (0, y ) = σ 0 cos ky, τ xy (0, y ) = 0 .
τ xy = −
We find that
B = −σ 0 / k 2 , D = −σ 0 / k .
The stress field inside the material is
σ xx = σ 0 (1 + kx )e − kx cos ky
τ xy = −σ 0 kxe −kx sin ky
σ yy = σ 0 (1 − kx )e −kx cos ky
The stress field decays exponentially. This decay once again provides an example of SaintVenant’s principle.
Change of the components of a state of stress 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 stress are σ xx ,σ yy ,τ xy . If we represent the same
material particle under the same state of stress by a square in the (r ,θ ) coordinates, the
components of stress are σ rr ,σ θθ ,τ rθ . The two sets of the components of the same state of
stress are related as
σ rr =
σ θθ =
σ xx + σ yy
2
σ xx + σ yy
τ rθ = −
+
−
2
σ xx − σ yy
2
σ xx − σ yy
2
σ xx − σ yy
2
cos 2θ + τ xy sin 2θ
cos 2θ − τ xy sin 2θ
sin 2θ + τ xy cos 2θ
These relations are derived in your course on the strength
of materials. They are a consequence of force balance. For
example, consider the force balance of the triangle in the
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σ xx
τ xy
σ rr
τ rθ
θ
b
a
τ xy
c
σ yy
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radial direction. We obtain that
σ rr c = σ xx b cos θ + σ yya sin θ + τ xyb sin θ + τ xya cos θ .
Note that a = c sin θ and b = c cos θ . Recall the identities of trigonometry:
1 − cos 2θ
1 + cos 2θ
cos θ cos θ =
, sin θ sin θ =
, 2 sin θ cos θ = sin 2θ .
2
2
Governing 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:
σ rr =
1 ∂φ
∂ 2φ
,
+
2
2
r ∂θ
r ∂r
σ θθ =
∂ ⎛ ∂φ ⎞
∂ 2φ
, τ rθ = − ⎜
⎟
2
∂r ⎝ r∂θ ⎠
∂r
The biharmonic equation is
⎛ ∂2
∂ 2φ ⎞
∂ 2 ⎞⎛ ∂ 2φ ∂φ
∂
⎜⎜ 2 +
+ 2 2 ⎟⎟ = 0 .
+ 2 2 ⎟⎟⎜⎜ 2 +
r∂r r ∂θ ⎠⎝ ∂r
r∂r r ∂θ ⎠
⎝ ∂r
The stress-strain relations in polar coordinates are similar to those in the rectangular
coordinates:
σ
σ
σ
2(1 + ν )
σ
ε rr = rr − ν θθ , ε θθ = θθ −ν rr , γ rθ =
τ rθ .
E
E
E
E
E
The strain-displacement relations are
∂u
∂u
u
∂u
∂u
u
ε rr = r , ε θθ = r + θ , γ rθ = r + θ − θ .
∂r
r r∂θ
r∂θ ∂r
r
Stress field symmetric about an axis. Consider a very special case: the stress
function be φ (r ) . The biharmonic equation becomes
⎛ d 2 1 d ⎞⎛ d 2φ 1 dφ ⎞
⎟=0 .
⎜⎜ 2 +
⎟⎜
+
r dr ⎟⎠⎜⎝ dr 2 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 = rm .
Inserting into the biharmonic equation, we obtain that
2
m 2 (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
1 ∂φ A
∂ 2φ
= + B (1 + 2 log r ) + 2C ,
σ rr = 2 2 +
r ∂θ
r ∂r r 2
∂ 2φ
A
σ θθ = 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.
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Lame problem. The contributions due to A and C are familiar: they are the same as
the Lame problem:
A
σ rr = C + 2 ,
r
∂ 2φ
A
σ θθ = 2 = C − 2 .
∂r
r
The constants A and C are determined by using the boundary conditions. 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
⎡ ⎛ 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). The stress function is
φ (r ) = Br 2 log r .
The corresponding stresses are
σ rr = B(1 + 2 log r ) , σ θθ = B(3 + 2 log r ) , τ rθ = 0 .
The strains are
1
B
ε rr = (σ rr −νσ θθ ) = [(1 − 3ν ) + 2(1 −ν ) log r ]
E
E
B
1
ε θθ = (σ θθ −νσ rr ) = [(3 −ν ) + 2(1 −ν ) log r ]
E
E
γ rθ = 0
So far everything is straightforward. The next idea requires some imagination, and
surprised me when I first learned about the idea. Even though the field of stress is axisymmetric,
the filed of displacement need not be axisymmetric. If the field of displacement is allowed to be
non-axisymmetric, the field has two components, ur (r ,θ ) and uθ (r ,θ ) . Imagine a ring, with a
wedge of angle α cut off. The ring with the missing wedge was then welded together. In the
welded ring, the stress field is axisymmetric, but the displacement field is not axisymmetric.
To obtain the displacements, recall the strain-displacement relations
∂u
u ∂u
∂u
∂u u
ε rr = r , ε θθ = r + θ , γ rθ = r + θ − θ .
∂r
r r∂θ
r∂θ ∂r
r
Integrating ε rr , we obtain that
ur (r ,θ ) =
B
[2(1 −ν )r log r − (1 +ν )r ] + f (θ ) ,
E
where f (θ ) is a function still undetermined. Integrating ε θθ , we obtain that
θ
4 Brθ
uθ (r ,θ ) =
− ∫ f (q )dq + g(r ) ,
E
0
where g(r ) is another function still undetermined. Inserting the two displacements into the
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expression
γ rθ =
and we obtain that
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∂ur ∂uθ uθ
+
−
= 0,
r∂θ ∂r
r
dg(r ).
df (θ )
+ ∫ f (q )dq = g(r ) − r
dr
dθ
0
θ
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,
df (θ )
+ ∫ f (q )dq = G
dθ
0
θ
g(r ) − r
dg(r ).
=G
dr
where G is a constant. The solutions to the two ODEs take the forms:
f (θ ) = H sin θ + K cos θ
g(r ) = Fr + G
Substituting these forms into the ODEs, we obtain that G = H .
The two displacements are
B
ur = [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
Inspecting the above equations, we observe that F represents a rigid-body rotation, and H and K
represent a rigid-body translation. The rigid-body rotation and translation will not generate any
stress.
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
B=
αE
.
8π
This cut-and-weld operation clearly introduces a stress field in the ring. The field of stress is
axisymmetric, but the field of displacement is non-axisymmetric.
An alternative approach. We can also solve the cut-and weld problem directly in terms
of displacement field. Upon expecting the geometry of the operation, we can readily convince
ourselves that the hoop displacement must be
uθ (r ,θ ) =
αrθ
.
2π
The inspection should also convince us that the radial displacement must be independent of θ ,
namely,
ur (r ,θ ) = f (r ) .
We can then insert the above expressions into the general equations of linear elasticity and
derive the ODE for f (r ) , which can be solved once appropriate boundary conditions are
prescribed.
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Separation of variables. We next consider the stress function as a function of two
variables, φ (r ,θ ) . One can obtain many solutions by using the procedure of separation of
variable, assuming that
φ (r ,θ ) = f (r )g(θ ) .
Formulas for stresses and displacements can be found on p. 205, Deformation of Elastic Solids,
by A.K. Mal and S.J. Singh.
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 , σ xx = σ yy = 0 .
The remote stresses in the polar coordinates are
σ rr = S sin 2θ , σ θθ = − S sin 2θ , τ rθ = S cos 2θ .
Recall that
∂ 2φ
∂ ⎛ ∂φ ⎞
1 ∂φ
∂ 2φ
σ rr = 2 2 +
, σ θθ = 2 , τ rθ = − ⎜
⎟.
∂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
4 ⎞⎛ ∂ 2 f
∂f 4 f ⎞
d
⎟=0.
⎜⎜ 2 +
−
− 2 ⎟⎟⎜⎜ 2 +
rdr r ⎠⎝ ∂r
r∂r r 2 ⎟⎠
⎝ 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
(
)
1 ∂φ
6C 4 D ⎞
∂ 2φ
⎛
+
= −⎜ 2 A + 4 + 2 ⎟ sin 2θ
2
2
r ∂θ
r ∂r
r
r ⎠
⎝
2
6C ⎞
∂φ ⎛
σ θθ = 2 = ⎜ 2 A + 12 Br 2 + 4 ⎟ sin 2θ
∂r
r ⎠
⎝
6C 2 D ⎞
∂ ⎛ ∂φ ⎞ ⎛
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:
Remote from the hole, namely, r → ∞ , σ rr = S sin 2θ , τ rθ = S cos 2θ ,
A = − S / 2, B = 0 .
σ rr =
•
•
giving
On the surface of the hole, namely, r = a, σ rr = 0, τ rθ = 0 , giving D = Sa 2 and
C = − Sa 4 / 2 .
The stress field inside the sheet is
4
2
⎡
⎛a⎞
⎛a⎞ ⎤
σ rr = S ⎢1 + 3⎜ ⎟ − 4⎜ ⎟ ⎥ sin 2θ
⎝r⎠
⎝ r ⎠ ⎦⎥
⎣⎢
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Z. Suo
4
⎛a⎞ ⎤
⎝ r ⎠ ⎦⎥
σ θθ = − S ⎢1 + 3⎜ ⎟ ⎥ sin 2θ
⎣⎢
τ rθ
4
2
⎡
⎛a⎞
⎛a⎞ ⎤
= S ⎢1 − 3⎜ ⎟ + 2⎜ ⎟ ⎥ cos 2θ
⎝r⎠
⎝ r ⎠ ⎦⎥
⎣⎢
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 biaxial 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. An elastic material occupies the
half space x > 0 , and is subject to a line force on its surface in the direction of x. Let P be the
force per unit length. Recall that the field equations have no length scale. The boundary
conditions in this problem also have no length scale. This consideration, along with linearity,
requires that the stress field in the half space take the form
σ ij (r ,θ ) =
P
gij (θ ) ,
r
where gij (θ ) are dimensionless functions of θ . Recall that
∂ 2φ
∂ ⎛ ∂φ ⎞
1 ∂φ
∂ 2φ
+
σ
=
,
, τ rθ = − ⎜
⎟
θθ
2
2
2
∂r ⎝ r∂θ ⎠
r ∂θ
r ∂r
∂r
We guess that the stress function takes the form
φ (r ,θ ) = rPf (θ ) ,
σ rr =
where f (θ ) is a dimensionless function of θ . A quick note: a homework problem will show
that this guess is incomplete for a line force in an infinite block. This guess, however, suffices for
the present problem.
Inserting this guess of the stress function into the biharmonic equation, we obtain an
ODE for f (θ ) :
f +2
d2 f d4 f
+
=0.
dθ 2 dθ 4
The general solution to the ODE is
f (θ ) = A sin θ + B cos θ + Cθ sin θ + Dθ cos θ ,
so that
φ (r ,θ ) = rP ( A sin θ + B cos θ + Cθ sin θ + Dθ cos θ ) .
Observe that r sin θ = y and r cos θ = x 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 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:
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2CP cos θ
, 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
σ rr =
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 σ rr cos θrdθ . The force balance of the half cylinder requires that
π /2
P+
∫ σ rr cos θrdθ = 0 .
−π / 2
Integrating, we obtain that C = −1 / π .
The stress components in the x-y coordinates are
σ xx = −
2P
2P
2P
cos 4 θ , σ yy = −
sin 2 θ cos2 θ , τ xy = −
sin θ cos 3 θ
πx
πx
πx
The displacement field is
(1 −ν )P θ sin θ
2P
cos θ log r −
πE
πE
(1 −ν )P θ cos θ − (1 −ν )P sin θ
2νP
2P
uθ = −
sin θ +
sin θ log r −
πE
πE
πE
πE
ur = −
The solution of a line force on the surface of a half space can used to construct solutions
to other problems by linear superposition. We illustrate this idea using the following example.
An application. 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 treating
laminates, we have so far ignored the edge effect. This approximation has troubled us because
we know that the edge often initiates failure. 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 σ yy 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, as illustrated in Fig. 3. Let σ M 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 σ M on the edge of the thin layer, so that the thin layer is
under the uniform biaxial stress, and the two thick blocks are stress-free. In problem B, we
remove thermal expansion misfit, but apply a tensile traction on the edge of the thin layer. The
original problem is the superposition of Problem A and Problem B. In particular, the stress field
σ yy in the original problem is the same as the stress σ yy in Problem B.
Because the thickness of the thin layer is small compared to its lateral dimensions, the
edge of the layer in Problem B is approximately under the plane strain conditions. With
reference to the inset in Fig. 4, let us calculate the stress distribution σ yy (x ,0) . Recall that when
a half space is subject to a line force P, the stress is given by
2P
σ yy = −
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 .
π
x
−t / 2
∫
We next evaluate this integral.
Note that η = x tan θ , and let tan β = t / 2 x .
x
Consequently, dη =
dθ , and the integral becomes that
cos 2 θ
β
β
2σ
2σ
1 − cos 2θ
σ yy (x ,0) = M sin 2 θdθ = M
dθ
π −β
π −β
2
∫
∫
Integrating, we obtain that
2σ M ⎛
1
⎞
⎜ β − sin 2 β ⎟ .
π ⎝
2
⎠
Consider two limiting cases. At the edge of the layer, x / t → 0 and β = π / 2 , so that
σ yy (x ,0) =
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σ yy (0,0) = σ M .
This stress is tensile, and causes the cracking of the thin layer.
Far from the edge, t / x → 0 , recall the Taylor expansion:
sin 2 β = 2 β −
(2β )
3
6
+ ...
We obtain that
σM
6π
3
⎛t ⎞
⎜ ⎟ .
⎝x⎠
Thus, this component of stress decays as x −3 . We may regard this decay as an example of SaintVenant’s principle.
σ yy (x ,0) →
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