Download Plane Elasticity Problems

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. (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
 xx nx   xy ny  t x
St
 xy nx   yy ny  t y
b
where t x and t y are components of the traction vector prescribed on the
y
x
edge of the sheet, and n and n are the components of the unit vector
x
y
normal to the edge of the sheet. The above two equations provide two
Su z
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 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:  xx , yy , xy , and the components out of the plane vanish:  zz   xz   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:
 xx x, y ,  yy x, y ,  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:
 xy  yy
 xx  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


21   


 xy
 xx  xx  yy ,  yy  yy  xx ,  xy 
E
E
E
E
E

 zz    xx   yy ,  xz   yz  0 .
E
Strain-displacement relations. Recall the 6 strain-displacement relations:
u
v
u v
 xx  ,  yy  ,  xy 

x
y
y x
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w
u w
v w
.
,  xz 

,  yz 

z
z x
z y
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  xz   yz  0 , we find that
 zz 
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  zz be
independent of z, and from  zz  w / z , then  zz must be a constant,  zz  c , and w  cz  b .
On the other hand, we also have  zz    xx   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  zz  c and  zz    xx   yy / E , at least the following set of equations
look self-consistent:
 xx  xy

 0,
x
y
 xx 
 xx
E

 yy
E
,  yy 
 yy
E
 xy
x


 yy
 xx
E
y
0
,  xy 
21   
 xy
E
u
v
u v
,  yy  ,  xy 
 .
x
y
y x
These are 8 equations for 8 functions. We will focus on these 8 equations.
Plane strain problem. An infinitely long cylinder with the axis in
the z-direction, and a cross section in the x, y  plane. 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 in-plane
strains are nonzero:  xx x, y ,  yy x, y ,  xy x, y  . The three out-of-plane
 xx 
y
x
z
strains vanish:  zz   xz   yz  0 .
Because  xz   yz  0 , the stress-strain relations imply that  xz   yz  0 . From  zz  0
and  zz   zz   xx   yy  , we obtain that
 zz    xx   yy  .
Thus,
1
1  2 


 xx   xx  yy  zz  
 yy 
 xx 
E
E 
1 

2
1
1  


 yy   yy  xx  zz  
 xx 
 yy 
E
E 
1 

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21   
 xy
E
These three stress-strain relations look similar to those under the plane stress conditions,
provided we make the following substitutions:
 xy 
E

,  
.
2
1 
1 
The quantity E is called the plane strain modulus.
A theorem in calculus. If two functions f x, y  and g x, y  satisfy
f g
,

x y
there exists a function Ax, y  , such that
A
A
.
f 
, g
y
x
The Airy stress function. We now apply the above theorem to the equilibrium
 xx  xy

 0 , we deduce that there exists a function Ax, y  , such that
equations. From
x
y
A
A
 xx  ,  xy   .
y
x
 xy  yy

 0 , we deduce that that there exists a function Bx, y  , such that
From
x
y
B
B
 yy  ,  xy   .
x
y
A B
Finally, from
, we deduce that that there exists a function  x, y  , such that

x y


.
A
, B
y
x
The function  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:
 2
 2
 2
.
 xx  2 ,  yy  2 ,  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:
E
1   2
 2 
1   2
 2 
21     2
 2  2  ,  yy   2   2  ,  xy  
.
E  y
x 
E  x
y 
E xy
Compatibility equation. Recall the strain-displacement relations:
w
u w
v w
.
 zz 
,  xz 

,  yz 

z
z x
z y
Eliminate the two displacements in the three strain displacement relations, and we obtain that
 xx 
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2
2
 2 xx   yy   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  x, y  into
the compatibility equation, and we obtain that
 4
 4
 4

2

0.
x 4
x 2y 2 y 4
This equations can also be written as
 2
 2   2  2 
 2  2  2  2   0 .
y  x
y 
 x
It is called the biharmonic equation.
Thus, a procedure to solve a plane stress problem is to solve for  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  . 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  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
3
1 a
a
 rr   p  ,    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
 xx 0, y, z    0 cos ky,  xy 0, y, z    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
 x, y   f xcos ky .
The biharmonic equation becomes
2
d4 f
2 d f
 2k
 k4 f  0 .
4
2
dx
dx
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This is a homogenous ODE with constant coefficients. A solution is of the form
f x   ex .
Insert this form into the ODE, and we obtain that


2
 k2  0 .
The algebraic equation has double roots of   k , and double roots of   k . Consequently,
the general solution is of the form
f x   Aekx  Be  kx  Cxekx  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
 2
 xx  2  Be  kx  Dxe kx k 2 cos ky
y
2
 2 
D

   Be  kx  e  kx  Dxe kx k 2 sin ky
xy 
k

2
 
D

 yy  2   Be  kx  2 e  kx  Dxe kx k 2 cos ky
x
k


Recall the boundary conditions
 xx 0, y    0 cos ky,  xy 0, y   0 .
 xy  
We find that
B   0 / k 2 , D   0 / k .
The stress field inside the material is
 xx   0 1  kxe  kx cos ky
 xy   0 kxe kx sin ky
 yy   0 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  xx , yy , xy . If we represent the same material
particle under the same state of stress by a square in the r ,  coordinates, the components of the
stress state are  rr ,  , r . The two sets of the components are related as
 rr 
  
 r
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 xx   yy
2
 xx   yy

 xx   yy
2
 xx   yy
cos 2   xy sin 2

cos 2   xy sin 2
2
2
   yy
  xx
sin 2   xy cos 2
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Equations in polar coordinates. The Airy stress function is a function of the polar
coordinates,  r,  . The stresses are expressed in terms of the Airy stress function:
   
 2
1 
 2
,
 rr  2 2 
   2 ,  r   

r  r 
r 
r r
r
The biharmonic equation is
 2

 2   2 
 2 
 2 
 2 2  2 
 2 2   0 .
rr r   r
rr r  
 r
The stress-strain relations in polar coordinates are similar to those in the rectangular coordinates:




21   
 r
 rr  rr   ,      rr ,  r 
E
E
E
E
E
The strain-displacement relations are
u
u
u
u
u
u
 rr  r ,    r   ,  r  r     .
r
r r
r r
r
Stress field symmetric about an axis. Let the stress function be  r  . The biharmonic
equation becomes
 d 2 1 d  d 2 1 d 
 2 

0 .

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
  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
 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
 2
1  A
 rr  2 2 
  B1  2 log r   2C ,
r 
r r r 2
 2
A
   2   2  B3  2 log r   2C ,
r
r
   
 r   
0.
r  r 
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
  a 2 
  a 2 
 rr  S 1    ,    S 1     .
  r  
  r  
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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 
 rr  S 1    ,    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)
 rr  B1 2 log r ,    B3  2 log r  ,  r  0 .
The strain field is
1
B
 rr   rr     1  3   21   log r 
E
E
1
B
       rr   3     21   log r 
E
E
 r  0
To obtain the displacement field, recall the strain-displacement relations
u
u
u
u
u
u
 rr  r ,    r   ,  r  r     .
r
r r
r r
r
Integrating  rr , we obtain that
B
u r  21   r log r  1   r   f   ,
E
where f   is a function still undetermined. Integrating   , we obtain that
4 Br 
u 
  f  d  g r  ,
E
where g r  is another function still undetermined. Inserting the two displacements into the
expression
u
u
u
 r  r      0 ,
r
r
r
and we obtain that
f '     f  d  g r   rg ' r .
In the equation, the left side is a function of  , and the right side is a function of r.
Consequently, the both sides must equal a constant independent of r and  , namely,
f '     f  d  G
g r   rg ' r   G
Solving these equations, we obtain that
f    H sin   K cos 
g r   Fr  G
Substituting back into the displacement field, we obtain that
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B
21   r log r  1   r   H sin   K cos 
E
.
4 Br 
u 
 Fr  H cos   K sin 
E
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  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2   v0  r
This condition gives
E
B
.
8
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:
 xy  S ,  xx   yy  0 .
The remote stresses in the polar coordinates are
 rr  S sin 2 ,    S sin 2 ,  r  S cos 2 .
Recall that
   
 2
1 
 2
,    2 ,  r   
 rr  2 2 
.
r  r 
r 
r r
r
We guess that the stress function must be in the form
 r,   f r sin 2 .
The biharmonic equation becomes
 d2
d
4   2 f f 4 f 
 2 
  0.
 2  2 

rdr r  r
rr r 2 
 dr
ur 
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 m2  4  0 .
The algebraic equation has four roots: 2, -2, 0, 4. Consequently, the stress function is
C


 r ,    Ar 2  Br 4  2  D  sin 2 .
r


The stress components inside the sheet are


 2
1 
6C 4D 

 rr  2 2 
  2 A  4  2  sin 2
r 
r r
r
r 

2
 
6C 
   2   2 A  12Br 2  4  sin 2
r
r 

    
6C 2 D 
2
 r   
    2 A  6 Br  4  2  cos 2 .
r  r  
r
r 
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To determine the constants A, B, C, D, we invoke the boundary conditions:
Remote from the hole, namely, r   ,  rr  S sin 2 ,  r  S cos 2 , giving
A   S / 2, B  0 .
On the surface of the hole, namely, r = a,  rr  0,  r  0 , giving D  Sa 2 and
C   Sa 4 / 2 .
The stress field inside the sheet is
4
2

a
a 
 rr  S 1  3   4   sin 2
r
 r  

4

a 
    S 1  3   sin 2
 r  

4
2

a
a 
 r  S 1  3   2   cos 2
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 tension of magnitude S, the stress field in the sheet is given by
  a 2 
  a 2 
 rr  S 1    ,    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
 ij r ,    g ij   ,
r
where g ij   are dimensionless functions of  . We guess that the stress function takes the form
 r   rPf   ,
where f   is a dimensionless function of  . (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   :
f 2
The general solution is
d2 f d4 f

 0.
d 2 d 4
 r,   rP A sin   B cos  C sin   D cos .
Observe that r sin   y and r cos  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   0 , so that
we will drop the term  cos . Consequently, the stress function takes the form
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 r,   rPC sin  .
We can calculate the components of the stress field:
2CP cos 
 rr 
,     r  0 .
r
This field satisfies the traction boundary conditions,     r  0 at   0 and    . 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 rd of the surface, the radial stress provides a vertical
component of force  rr cos rd . The force balance of the half cylinder requires that
 /2
P

rr
cos rd  0 .
 / 2
Integrating, we obtain that C  1/  .
The stress components in the x-y coordinates are
2P
2P 2
2P
 xx  
cos 4  ,  yy  
sin  cos 2  ,  xy  
sin  cos 3 
x
x
x
The displacement field is
1   P  sin 
2P
ur  
cos  log r 
E
E
1   P  cos   1   P sin 
2P
2P
u  
sin  
sin  log r 
E
E
E
E
Separation of variable. One can obtain many solutions by using the procedure of
separation of variable, assuming that
 r,   Rr   .
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  yy can develop near the edge. We would like
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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  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  M 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  yy in the original problem is the same as the stress
 yy in Problem B.
With reference to Fig. 4, let us calculate the stress distribution  yy x,0 . Recall that
when a half space is subject to a line force P, the stress is given by
2P
 yy  
sin 2  cos 2  .
x
We now consider a line force acting at y   .
On an element of the edge, d , the tensile
traction applied the line force P   M d . Summing up over all elements, we obtain the stress
field in the layer:
t/2
2 M d
 yy x,0  
sin 2  cos 2  .
x
t / 2
Note that   x tan  , and let tan   t / 2 x . Consequently, d 
x
d , and the integral
cos 2 
becomes that
 yy x,0 
2 M


sin

2
d 
2 M




1  cos 2
d
2
Integrating, we obtain that
2 M 
1

   sin 2  .
 
2

At the edge of the layer, x / t  0 and    / 2 , so that  yy 0,0    M . Far from the edge,
 yy x,0 
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t / x  0,
 t
 yy x,0  M   .
6  x 
3
Thus, this stress decays as x 3 .
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