4.
Prismatic Rods
4.1
Formulation for Prismatic Rods
Prismatic Rods
The case of prismatic rods virtually covers all structural forms studied in the course of
Strength of Materials. The issue is revisited here by the rigorous and systematic
approach of elasticity. Beside linear elasticity assumption and occasional use of Saint
Venant principle, no assumptions will be made (in contrast to the approach of Strength
of Materials) to get the solution.
The geometry of a prismatic rod is illustrated in Fig. 4.1. The cross-section of the rod
remains unchanged through the whole axis of the rod. Two ends of the rod are planar,
and the side surface of the rod is cylindrical. The length of the rod, L, is much larger
than any cross-section length measure. Hereafter, all the Latin indices have the range
from 1 to 3, and all the Greek indices have the range from 1 to 2.
T
R
end
Z
L
Y
X
side
T
A
side
V
end
A
Figure 4.1 A prismatic rod.
The following problem will be formulated and solved:
1. The three-dimensional quasi-static elasticity equation is observed in V occupied
by the prismatic rod.
1
2. Traction Ti
side
independent of z is prescribed along the side surface A side .
3. Two ends of the rod are subjected to arbitrary but self-equilibrated traction Ti
end
.
Decomposition
The principle of superposition allows a decomposition of the stated problem into three
sub-problems.
1. Plane strain problem
The plane strain problem takes care of the lateral traction components T
side
along
the side surface. In a plane strain problem, all field variables are independent of z such
as
ui  ui x, y  and  ij   ij  x, y  . The displacement directed to the rod axis also
vanishes, namely u z  0 , x, y, z . These requirements naturally guarantee the plane
strain conditions of
 xz   yz   zz  0 .
(4.1)
The consequence of the plane strain condition (4.1) for stress under a generalized
Hooke’s law is
 xz   yz  0 ,  zz  v xx   yy  .
(4.2)
The remaining field variables u ,  a and   are determine by plane elasticity
that will be discussed extensively in the next two chapters, with the following
boundary conditions
  v   T side on A side ,
(4.3)
where v  is the outward unit normal of the side surface.
2. Anti-plane problem
The traction prescribed on the side surface of the rod along z direction
 3 v   T3 side on A side
(4.4)
could be formulated by an anti-plane problem where all field variables are
independent of z and
u  0 , u3  u3 x, y  ,     33  0 .
2
(4.5)
The problem will be formulated and solved in Section 4.6.
3. Prismatic rod with a traction free side surface
The third problem concerns a prismatic rod of a traction free side surface, namely
 i v   0
on A side .
(4.6)
The end traction is given by
 3i  Ti  Ti
end

 312 


 322 
v  1   1
22
 11







on A end ,
(4.7)
where the second term on the right hand side incorporates the influence of the
previous two problems. The combination of three problems listed above solves the
original problem of a prismatic rod proposed earlier, as can be easily verified.
Further Decomposition for End-loaded Prismatic Rods
Attention is now directed to the last problem of a prismatic rod with traction free side
surface. A Cartesian coordinate system such as shown in Fig. 4.2 can be set up in any
cross-section A (including the ends) of the rod. The axes x and y are taken as the
inertia principal axes originated at the centroid of the cross-section. Accordingly, one
has
 xdA   ydA   xydA  0
A
A
(4.8)
A
Y
X
A
Figure 4.2 Cartesian coordinate system in a cross-section.
Under these coordinates, the boundary condition (4.7) can be further decomposed as
~
 3i  Ti  Ti 0  Bi x  Ti on Aend
3
(4.9)
where Ti 0 and Bi are constants that do not vary with the end coordinates x and y.
These constants are determined by
Ti 0 
1
Ti dA
A A
eik I  Bk   x  T dA  M i ,
(4.10)
A
where
I 
signifies the moments of inertia of the cross-section A. The
self-equilibrium condition for the end tractions would restrict the first term, the axial
force Ti 0 , as a simple tension solution, as will be discussed in the next section. The
end moments M i would induce bending and torsion of the prismatic rod. The
remaining term in (4.9) results in neither a net force nor a net moment, namely
~
~
(4.11)
 Ti dA  0 ,  x  TdA  0 .
A
4.2
A
Uniaxial Tension and Pure Bending
Uniaxial Tension
Consider the first term in (4.9) for the traction at the end of a rod, one has
 3i  Ti  Ti
end

 312 


 322 
v  1   1
22
 11



0
end
  Ti on A



2 
2 
The remains of the anti-plane solution  31
and  32
are constant over A end . So is
Ti 0 according to (4.10). One concludes that Ti
end
is also constant over A end . The
reciprocal theorem of shear stresses for the original prismatic rod problem leads to
T
end
  32  , namely T
end
is exactly the end shear traction balanced with the
anti-plane problem. These arguments finally give the following boundary condition
0
 
 3i   0  on Aend .
T 0 
 3
That is exactly the uniaxial tension problem. The stresses are exactly solved by
4
(4.12)
0 0 0 
 ij  0 0 0  .
0 0  
(4.13)
The use of the generalized Hooke’s law delivers a strain tensor as follows
 v
1
1
 ij  1  v  ij  v kk ij  
E
E



 v

.

 
(4.14)
One may integrate various strain components to get displacements. Recall
u1,1  u2, 2  
v

 and u3,3  , it is a straightforward matter to observe a possible
E
E
set of displacements as:
u1 
v
v

 x , u 2    y , u3  z .
E
E
E
(4.15)
For the present case with all-around traction boundary conditions, the displacements
are unique except to a rigid body movement. The uniqueness theorem then gives rise
of the general form of the displacements as
ui   ij x j  ci  eijk j xk .
(4.16)
Three remarks can be made here. First, if the translation and rotation at x＝0 are fixed,
namely ui x 0  0 and ij
x 0

1
ui, j  u j ,i   0 , one may conclude that ci  0
2
x0
and i  0 . Second, the solution given here can be applied to prismatic rods of length
L of any cross-section. In any cross-section, the u1 component is constant provided
the rigid body rotation is excluded. That is exactly the plane assumption in the
Strength of Materials. Third, the total elongation of the rod is L 

E
L
FL
.
EA
Pure Bending
The problem of pure bending corresponds to an end traction of
 33  B3 x and  3  0 .
By simple observation, one finds that
5
(4.17)
0 
0 0

 ij  0 0
0 
0 0 B3 x 
(4.18)
satisfies the equilibrium, the traction free boundary along the side and the end traction
(4.17).
A generic longitudinal fiber is subjected to uniaxial tension. The corresponding
moments at the ends of the rod are M i  ei 3 I B3 . That gives a null torque, M 3  0 ,
and two bending moment components M1  I 2 B3 and M 2   I 1B3 . As stated
earlier, one can always choose the x-y coordinates as the centroid principal axes of
inertia. Then the tensor of moment of inertia is reduced to a diagonal form of
 I xx
I  


M
M
. The B coefficients can be evaluated as B31   y and B32  x .

I yy 
I yy
I xx
The stresses, when written in a form convenient to use, are
 33  
My
I xx
x
Mx
y.
I yy
(4.19)
The nature of uniaxial tension for each individual longitudinal fiber grants a similar
strain expression
 v 33

1
.
 ij  
 v 33

E

 33 
(4.20)
Translating the above information into the displacement gradient, one has
u3,3 
1
v
B3 x , and u2, 2  u1,1   B3 x . Integrated each expression once to get the
E
E
following form of the displacement:
u1  
v 1

2
 B31x1  B32 x1 x2   f1  x2 , x3  ,
E2

u2  
v
1
2
 B31x1 x2  B32 x2   f 2 x1 , x3  ,
E
2

The unknown functions
u3 
1
B3 x x3  f 3  x1 , x2  .
E
fi
arise from the partial differentiations. Their
6
determination relies on the vanishing values of all shear strain components of this
problem, namely
12  0 
f 2 f1 v

 B31x2  B32 x1  ,
x1 x2 E
13  0 
1
f
f
B31x3  3  1  0 ,
E
x1 x3
 23  0 
1
f
f
B32 x3  3  2  0 .
E
x2 x3
(*)
The difference between the partial derivative with respect to x1 of the third
expression and the partial derivative with respect to x 2 of the second expression
leads to
f 2 f1
  f 2 f1 

  0 , or

 g x1 , x2  . Combining this expression

x1 x2
x3  x1 x2 
with the previous ones, one has
g x1 , x2  
v
B32 x1  B31x2  ,
E
v
f 2 v
B32 x12  h2  x3  ,
 B32 x1 ， f 2 
2E
x1 E
v
f1 v
B32 x22  h1  x3  .
 B31x2 ， f1 
2E
x2 E
The substitution of them into the last two expressions of (*) gives
B
f 3
1
  B31x3  h1x3   h1   31 x3 ,
E
x1
E
B
f 3
1
  B32 x3  h2 x3   h2   32 x3 .
x2
E
E
That consequently leads to the complete determination of unknown functions f i as
v
1
B31x22 
B31x32 ,
2E
2E
v
1
f2 
B32 x12 
B32 x32 ,
2E
2E
f1 
f3  0 .
7
The displacement fields under pure bending are given by






B31
B
v x12  x22  x32  32 vx1 x2 ,
2E
E
B
B
u2  32 v x12  x22  x32  31 vx1 x2 ,
2E
E
1
u3  B3 x x3 .
E
u1 
(4.21)
The last expression verifies the plane assumption engaged in the solution of Strength
of Materials.
4.3
Corrective Solution (Saint Venant Decay)
We discuss the corrective solution that corresponds the last term in the end traction
expression (4.9). To avoid unnecessary complications, its decaying behavior is only
explored for the case of two-dimensional bending of a beam, as shown in Fig. 4.3.
L
end
end 2H
Figure 4.3 Saint Venant decay of a two-dimensional beam.
Following the discussion after (4.9), the self-equilibrium traction along each end
results in neither a force nor a moment. One anticipates that the solution would decay
from the end inward with a decaying length (shaded areas in Fig. 4.3) of
  H  L . For this corrective problem, the decaying layer assembles the
boundary layer solution in fluid mechanics, though the former was solved much
earlier in the history. As shown in Fig. 4.3, two decaying layers associated with each
end are non-interacting. Accordingly, the solution for each decaying layer can be
viewed as the one for a semi-infinite beam.
The self-equilibrium traction along the end is portrayed in Fig. 4.4. The traction can
be decomposed into a symmetric part and an anti-symmetric part, each resulting in
neither total force nor total moment. The aim of the following development is to show
the solution under such specification can only be of decaying nature.
8
Y
Y
X
X
anti-symmetry
symmetry
Figure 4.4 Symmetric and anti-symmetric distributions of a self-equilibrium traction.
To solve the problem, the Airy stress function (its full account will be given in the
chapter to follow)
 xx 
 2
 2
 2
,
,
,





yy
xy
x 2
y 2
xy
(4.22)
is adopted that automatically satisfies the two dimensional equilibrium equation. The
traction free condition along the side surfaces requires
 yy x, H    xy x, H   0
x .
One can translate them to the following boundary conditions for the Airy stress
function
2
2



x
,

H

 x, H   0
x 2
xy
x .
We search for the Saint Venant decaying solution of the following form
 x, y   g  y ex .
For the two-dimensional case, one should have  2   0 . That requirement
translates to ,  0 for the Airy stress function. Substituting the Saint Venant
decaying solution into this two dimensional bi-harmonic equation, one has
2
d 4g
2 d g

2

 4 g  0 ,
dy 4
dy 2
9
whose general solution is given by
g  y    A1  A2 y cos y   A3  A4 y sin y .
For the symmetric self-equilibrated traction, one has
A2  A3  0 ; for the
anti-symmetric self-equilibrated traction, one has A1  A4  0 . We first discuss the
former case, then
g S  y   A1 cosy   A4 y sin y  .
The symmetry of g S simplifies the boundary conditions as
g S H   g S H   0 .
The substitution of the solution for g S into the boundary conditions gives rise to the
following linear and homogeneous algebraic system
 cos H
  sin H

H sin H
  A1  0
.

sin H  H cos H   A4  0
A necessary condition for existing a non-trivial solution is the vanishing of the
coefficient determinant, namely cos H sin H  H cos H   H sin 2 H  0 or
2H  sin 2H  0 .
The only non-decaying case refers to the root of   0 , and that leads to g S  y   A1
and consequently a trivial stress field.
For the anti-symmetric problem, one has
g A  y   A2 y cos y  A3 sin y .
Following the same procedure, the condition for a non-trivial solution is
2H  sin 2H  0
唯一的实根为   0  g A  y   A2 y
合力矩为零
H
 g  y ydy  0 排除
A
H
其余的解均为衰减解
10
4.4
Free Torsion of a Prismatic Rod
Navier Solution
We next turn to the case of free torsion of a prismatic rod. One is familiar with the
Navier solution of the problem
u  zy , v  zx, w  0. .
(4.23)
The solution has the image of planar rotation of flat slices, like a string of ancient
Chinese coins under twist. The twisted angle is proportional to the axial coordinate z
with  being the angle twisted in unit axial length. In terms of the polar coordinates
shown on the left of Fig. 4.5, the Navier solution can be cast in a simple form of
u  zri .
(4.23)’
M
Z
X
i
centre of r
figure
Y
f r e se i d e
v
L
ir
iz
Figure 4.5 Free torsion of a prismatic rod.
Using the coordinates portrayed in Fig. 4.5, the strain and stress tensors for the Navier
solution can be written as
ε
r
2
i z  i
 i  i z  , σ  r i z  i  i  i z  .
(4.24)
The stress tensor so obtained is divergence free. The side traction can be evaluated as
11
σ  ν  ri z i  ν  . Accordingly, the vanishing of the side traction σ  ν  0
necessarily requires i  ν  0 , the latter can not be satisfied for any non-circular
cross-sections, as delineated in Fig. 4.6.
i
v
A
C
Figure 4.6 Contradicting boundary condition for the Navier solution.
For a circular cross-section, the traction at the end plane is given by
T  σ  i z  ri . The torque induced by such a traction distribution is
R 2
M p    r 3 d dr 
0 0

2
R 4  I p
where the twisted angle per unit length is   M p I p with I p 
(4.25)

2
R 4 being the
polar moment of inertia.
Saint Venant Solution
Saint Venant proposed a method, the semi-inverse method, to tackle the problem of
free torsion of a prismatic rod of non-circular cross-section. He modified the Navier
solution (4.23)’ to include a warping in the axial direction, namely
u   zri  wx, y i z 
(4.26)
where wx, y  is termed the warping function. Thus, the torsion of a prismatic rod is
similar to the twisting of a string of ancient Chinese coins, with each coin warped in
the same amount and rotated in the same manner. The trial function w is to be found
by satisfying the field equations and the boundary conditions.
Equation (4.26) clearly indicates that   u  0 , or the Saint Venant displacement
field is incompressible. The strain and stress tensors can be expressed via the Saint
Venant displacement field (4.26) as
12
2ε   i z  s  s  i z  and σ  i z  s  s  i z 
(4.27)
s  ri  w .
(4.28)
with
The equilibrium equation requires that   σ  i z   s   0 . Thus equilibrium
translates to the divergence free of the s vector. That is equal to
2w  0 ,
(4.29)
or the warping function has to be harmonic.
We next consider the implication of the boundary condition σ  ν  i z s  ν   0 .
That gives, unlike the condition of i  ν  0 for the Navier solution, a condition of
s  ν  0 . Writing this for the unknown warping function w, one arrives at
ν  w 
w
  ri   ν

on A side .
(4.30)
The harmonic equation (4.29) and the boundary condition (4.30) describe a Neumann
type boundary value problem for w.
We next pursue an expression for the resultant force and torque due to the traction
over the end plane. The traction can be evaluated by the second expression of (4.27)
as
T  σ  i z  s
on A end .
(4.31)
To facilitate our evaluation, let us prove the following lemma.
Lemma: If   s  0 in A and s  ν  0 on C= A , then
 s  f dA  0
f .
A
Proof:
Since s  ν  0 on C, one has
 fs  νds  0
f . The theorem of Gauss leads to
C
 s  f  f  sdA  0 .
A
Moreover, the vector s is divergence free,   s  0 , then
 s  f dA  0
f , QED.
A
Two applications of the lemma are now in the order. First, if the arbitrary function f is
13
taken as x, one concludes that
 s  xdA   sdA  0 , namely the resultant force
A
A
over the end plane is zero. Second, the torque due to the traction over the end plane is
M p  i z  M    i z  r  sdA . From the lemma, one has
 s  wdA  0 .
A
That,
A
combined with (4.28), provides the following expression for the torque


2
M p    I p   w dA .
A


(4.32)
When compared with the torque expression (4.25) for the Navier solution, the
introduction of a warping function in the Saint Venant solution reduces the torsion
rigidity of the rod. That should come as no surprise, since the additional flexibility of
warping removes one constraint imposed by Navier assumption.
Let us sum up the solution procedure under the Saint Venant method:
1. Solving the warping function w from  2 w  0 in A and
w
  ri  ν on C.



2
2. Find the torsion rigidity from Dt   I p   w dA .
A


3. Calculate the twist angle per unit length from   M p Dt .
We exemplify that procedure for the special case of a circular cross-section, then
i  ν  0 on C. The homogeneous equation  2 w  0 and the homogeneous
boundary condition
w
0
r r  R
necessarily lead to the trivial solution of w  0
provided the rigid body motion is eliminated. The next two steps give directly
Dz  I p and  
Mp
I p
.
Prandtl Solution
We next presents an alternate, but equally elegant, solution for the torsion problems
due to Prandtl. As a departure from the displacement approach of Saint Venant,
Prandtl started from the stress consideration. The only non-trivial stress components
under torsion are  zx and  zy . The equilibrium leads to a relation
14
 zx  zy

 0.
x
y
That equation can be satisfied by choosing the so-called Prandtl stress function 
such that
 zx  

,
y
 zy  

.
x
(4.33)
Let the shear stress vector be denoted as τ   zx i x   zy i y . The torsion problem in
terms of the shear stress vector can be phrased as
  τ  0 in A, and τ  ν  0 on C.
(4.34)
The boundary condition implies that the shear stress vector (or the shear flow) should
be parallel to the boundary.
Equation (4.33) indicates that
τ    0 .
(4.35)
Namely τ is perpendicular to  , the gradient direction of the Prandtl stress
function. In another word, the contours of constant  denote the stress trajectories as
shown in Fig. 4.7.
Comparing
two
stress
expressions
(4.27)
and
(4.33),
one
has
 zx  , y   w, x  y  and  zy  , x   w, y  x  . Taking the partial
derivative of the first expression with respect to y and adding to the partial derivative
of the second expression with respect to x, one obtains the following Poisson’s
equation for the Prandtl stress function
 2  2 .
15
(4.36)
Figure 4.7 Trajectories of shear stress for different cross-sections.
The second expression of (4.34), namely the boundary condition, can be phrased in
 
 
  0 . That is equivalent to
terms of the stress function as τ  ν   x
 y
x 
 y

 0 or   const on C.
t
(4.37)
For simply connected region, the constant can be taken as zero, one then arrives at
 0
on
a single contour C.
(4.38)
Comparing with the governing equation (4.29) and the boundary condition (4.30) for
the warping function w under the Saint Venant formulation, one finds the stress
function formulation by Prandtl gives an inhomogeneous governing equation (4.36)
but a much simpler and homogeneous boundary condition (4.38).
Soap Film Analogy (Simply Connected Regions)
The governing equation (4.36) and the boundary condition (4.38) have an analogy that
may help us to visualize the shape of the stress function. Consider a planar rigid frame
identical in shape to the curve C that encloses the cross-section of the rod under
torsion, as shown in Fig. 4.8. Covered on the frame is a flat thin film under constant
stretching tension of  (or a soup film under the surface tension  ). Then a
uniform pressure p is applied to the thin film so that it bulges out under the constraint
of the rigid frame. If the ratio of  / p is chosen as 2, the profile of the thin film (or
16
the soup film) would resemble that of  . One may conclude from this analogy that
 would always be positive within A, and the maximum value of

occur along

the boundary.
rigid frame

soap bubble
uniform pressure P
Figure 4.8 Soap film analogy for simply connected regions.
Cross-section with Holes
Next consider a cross-section of the rod that contains holes, as shown in Fig. 4.9.
  1
  0  0
 2  2
  i
Figure 4.9 Cross-section containing holes.
The governing equation (4.36) should be observed in the region where holes are
excluded. Along the circumferences of holea, the boundary condition (4.37) requires
that the value of the stress function take a constant value, i , for the i-th hole. To
determine the values of i , the strain compatibility condition should be employed.
For each hole in the cross-section, the global compatibility conditions for a
multiply-connected region should be enforced,
17
 w
,x

dx  w, y dy  0 .
(4.39)
If the integration contour in (4.39) is shrunk to the rim of the i-th hole, and the
relation between the warping function w and the stress function  is utilized, one
arrives at the following relation
 
,y

 y dx  , x  x dy  0 .
Ci
i-th hole
v
Ci
Figure 4.10 Integration contour and outward normal of the i-th hole.
Take the outward normal of the bounding curve Ci of the i-th hole (actually directed
inward to the hole center) as ν , as shown in Fig. 4.10, then dy   1ds
and dx   2 ds
 
,y
along
the
contour.

 y  2  , x  x  1 ds  0 . Since
Ci
The
 x
1
contour
integral
then
becomes
 y 2 ds  2 Ai , with Ai being the
Ci
area of the i-th hole, one obtains the following condition to determine the boundary
values i

 ν  ds    ds  2 A .
(4.40)
i
Ci
Ci
Thin Film Analogy (Multiply Connected Regions)
thin film
frame
flat plate

v
P
18
Figure 4.11 Thin film and plate analogy of multiply connected region.
The case of multiply connected cross-section of a rod under torsion also possesses a
thin film analogy, as shown in Fig. 4.11. The condition (4.40) to determine i can be
viewed as an equilibrium condition in the analogy, namely the upward force by
pressure multiplied by the area of the hole should be balanced by stretching force
along the perimeter of the hole. Therefore, a multiply connected cross-section can be
viewed as composed of several flat plates that occupy the hole areas and a connected
thin film that covers the area between the outer frame the plates.
Consider the resultant forces and moments over a multiply connected cross-section.
The x-component of the resultant force can be phrased conveniently as
Fx    , y dA     2 ds . For a simply connected region, one has   0 on
A
C
C
and that gives Fx  0 . For multiply connected region,   k on Ck . Since Ck is a
closed contour so that
 
2
ds   k  2 ds  0 , one also arrives at Fx  0 . Similar
Ck
Ck
argument can be used to show that Fy  0 . Therefore, the total force over a multiply
connected cross-section vanishes. Next consider the total torque over the cross-section.
The torque can be computed via


M p   x zy  y zx dA .
A
By means of the stress function representation, one has
Mp

   x, x  y, y dA  2  dA   ν  xds .
A
A
C
The unit normal above is again defined as the normal directed toward the center of the
hole, as shown on the right of Fig. 4.12. The second term can be evaluated as
simply connected
0
n
n
n
C  xds    ν  xds  i  ν  xds   2 Aii with n holes .
 i 0 C
i 0
i 1
Ci
i

19
v
  const  i
dS
P
Figure 4.12 Torque of multiply connected region.
A simple geometric interpretation for the torque exists. For a simply connected region,
the torque is proportional to the volume covered by the thin film with a profile  . For
a multiply connected region, the torque is also proportional to the volume covered by
the thin film and the plates, it is precisely
n


M p  2  dA  i Ai  .
i 1
A

4.5
(4.41)
Inverse and Semi-inverse Solutions
Inverse and Semi-inverse Methods
The inverse method would give an anticipated function form of the basic field
variable, such as the warping function w or the stress function  in the problems of
torsion. The unknown coefficients will be determined through the solution process. In
a torsion problem, it is sometimes easy to guess a family of harmonic functions for w,
or a family of Poisson functions for  . The coefficients unknown in the guesswork
can be fixed by substituting in the boundary conditions.
The semi-inverse method, on the other hand, involves an undetermined function
rather than the undetermined coefficients.
We will demonstrate these methods by the examples to follow.
Elliptical Cross-section
20
Y

a
b
X
Figure 4.12 Torsion of an elliptical cross-section.
The torsion of a rod with an elliptical cross-section can be solved by inverse method.
Motivated by the thin film analogy, the profile of the stress function  can be
described by a quadratic function
 x2 y 2 
 2  1
2
b
a

  C 
that automatically satisfies the boundary condition (4.38). Substituting it into the
2
 2
governing equation (4.36), one has  2  C  2  2   2 and that determines the
b 
a
value of C as 
a
2
1
. Thus, the stress function for torsion of an elliptical rod is
 b 2

 x2 y 2 
1
1    .
a  2  b 2  a 2 b2 
(4.42)
The torque required to twist the rod to an angle  per length can be found through
(4.41) as
2
M p  2
a  b 2
 x2 y2 
a 3b 3
A 1  a 2  b 2 dA  a 2  b 2 .
For the special case of a  b , one has M p  
a 4
2
(4.43)
, identical with the formula of a
circular rod obtained in Strength of Materials.
Please note the area of an ellipse is given by A  ab , and the polar moment of
21
inertia for an elliptical cross-section is I p 

4


ab a 2  b 2 . Consequently, formula
(4.43) can be interpreted as
A4
Mp 
.
4 2 I p
(4.44)
That is termed the Saint-Venant formula for torsion, and it may serve as an
approximate formula for torsion of any simply connected cross-section.
Thin Wall Tubes
h(S )
S
  h
S 0
  h
X
S
SL
 0
 0
Figure 4.13 Thin wall tubes of closed cross-section.
Consider a thin-wall tube of closed cross-section, as shown in Fig. 4.13. The tube
thickness h is assumed to be much less than the radius of curvature of the tube. The
shear flow within the tubular cross-section, as shown in the right inset of Fig. 4.13, is
more or less uniform. If the variation of  along x, namely across the thickness of
the tube, is much faster than that along s (the circumferential direction of the tube),
one may assume   h
x
to satisfy the boundary conditions marked on the left
hs 
graph of Fig. 4.13. The shape of the stress function  under this assumption is
portrayed on the bottom graph of Fig. 4.13.
22
Nevertheless, the above approximation can not satisfy the field equation  2  2 .
This can be remedied by including a parabolic correction term that vanishes along the
inner and outer boundaries
  h
x
 xh  x  .
hs 
(4.45)
2
The governing equation (4.36) is satisfied by    2   2 . The torque can be
x
2
evaluated via (4.41) as


1


M p  2  h Ah   dA   2h Ah  h  hs ds   h 3 s ds  ,
3


A


(4.46)
where Ah denotes the cross-section of the hole. The coefficients h can be
determined by the single-valued-ness conditions of the displacement. According to the
right graph of Fig. 4.11, this condition also has the physical significance of balancing
the total pressure force on the plate with the membrane force along the circumference,
 ν  ds  2 A
,
h
ν   
 

 x
where

xh
h
h
the
 h . Define
integrand
ds
h

L
,
h
can
 hd s  A
m
be
simplified
as
and
 h sds  Lh
3
3
,
with Am being the cross-section area of the thin wall tube, h its average thickness,
one arrives at h 
h
2 Ah  Am  . Substituting this result into (4.46), we have
L
h

1
M p    2 Ah  Am   Lh 3  .
3
L

(4.47)
If Ah  Am , L2  h 2 and L2  h 2 , the formula in the Strength of Materials
Mp 
4h 2
Ah
L
is recovered.
Thin Wall Tubes of Open Cross-section
23
X
S
 0
 0
Figure 4.14 Thin wall tubes of open cross-section.
We next consider the case of thin wall tubes of open cross-section. The previous
solution (4.45),   h
x
 xh  x  , is still valid, except that the simply connected
hs 
nature of the cross-section geometry requires h  0 . Accordingly, one has
  xhs   x .
(4.48)
The resulted expression for its torque
Mp 

3
Lh 3
is much smaller than that for a thin wall tube of closed cross-section.
Rectangular Cross-section
24
(4.49)
Y
b
a
a
X
b
Figure 4.15 Torsion of a rectangular cross-section.
The torsion problem for a rectangular cross-section can be solved by the semi-inverse
method. Consider the contour plot in Fig. 4.15. When a   , the stress function 
would have a parabolic surface. That suggests a form of
  b2  y 2   x, y  ,
(4.50)
where the first term b 2  y 2 denotes the parabolic surface just mentioned, and the
second correction term satisfies
 2  0 ，  x,b  0   a, y   y 2  b2 .
Consider a variable separation solution of the form of    f n xgn  y  . Its
substitution into the harmonic equation gives
f nx  f n x    g n y  g n x   n ,
where the eigen-values  n is determined by solving the eigen-problem for g n  y 
g n y   n g n  y   0 ,
g n  b   0 .
The solution is
1

g n  y   cos n  y / b ,
2

The other function is governed by
f nx   2n f n x  .
The symmetry of the problem dictates that
25


1
2
 n   n   / b .

 x, y    cn cosn y 
0
cosh n x 
.
cosh n a 
The boundary conditions for  x, y  along x   a can be written as

c
n
cos n y   y 2  b 2 ,
0
where cn are the coefficients of Fourier cosine expansion for y 2  b2 . Multiplying
both sides by cosn y  , and integrating over the interval  b, b, one obtains
c n   
4
n 1
b3n
by the orthogonality of the Fourier series. The torque can be evaluated as
Mp



 2  b 2  y 2   dxdy 
16 3
ab  2  dxdy
3

1  a  

tanh  n     


16
2  b  
 192 b

 ab 3 1  5 

3
2n  15
  a n 0



4.6
.
(4.51)
Formulation of Anti-plane Problems
Problems of Anti-plane Shear
Recall that the problem of a prismatic rod is decomposed into the plane strain, the
anti-plane shear and a prismatic rod with free side surface. The latter is again
decomposed into tension, pure bending, torsion and the Saint-Venant corrective
solution. We conclude this chapter by discussing the anti-plane shear problems, which
resemble the torsion problems discussed in the previous two sections.
26
dislocation
S
crack
X2
A
s
X3
X1
v
Figure 4.16
A cylindrical body under anti-plane loading.
The general case of an anti-plane shear problem, as shown in Fig. 4.16,
involves
only such field variables such as u3 ,  3 and  3 . They are only the functions of
x1 and x2 . In the quasi-static case, the basic field equations are
 3 ,  0
(4.52)
for equilibrium, and
1
2
 3  u3,
(4.53)
 3  2G 3
(4.54)
for kinematics, and
for the constitutive law. Combining (4.53) and (4.54), one has  3 ,  Gu3, .
Substituting this stress expression into the equilibrium equation (4.52), one yields
 2 u3  0 .
(4.55)
Namely the out-of-plane displacement, u3 , similar to the warping function in a
torsion problem, has to be a harmonic function.
Complex Variable Approach
As a harmonic function, the out-of-plane displacement u3 can be expressed by either
27
the real or the imaginary part of an analytical function of a complex variable
z  x1  ix2 , with i   1 . Even better, one can construct an analytical function
 z   u3  i
(4.56)
with u3 and  individually as real functions. The complex function  z  is
termed the complex potential for anti-plane shear problems.
The condition of  z  being an analytical function is very strong: not only u3 and
 are harmonic, but also they are linked by the following Cauchy-Riemann
conditions:
u3 
u


， 3 
.
x1 x2
x2
x1
(4.57)
Accordingly, one has  31   , 2 ,  32    ,1 and 2  0 . The function  in
(4.56) resembles the Prandtl stress function in a torsion problem. In terms of the
complex potential, the complex stress and strain can be expressed as
 31  i 32   z  ，  31  i 32    z  .
1
2
(4.57)
Frequently it is easier to adopt a polar coordinate system. The stress components
under polar coordinate system are  3r  i 3  ei  31  i 32  .
Next consider the boundary conditions. In a displacement boundary Su, one has
Re   u 3
on
Su.
(4.58)
In a traction boundary St, one instead has  3   t3 . The stress displacement
gradient relation allows a further simplification of  3   u3,   
u3

,


s
where the definition of directional derivative and the generalized Cauchy-Riemann
relation are employed consecutively. One finally arrives at
Im  
1

t 3 ds
on
St.
(4.59)
The general boundary value problem of the anti-plane shear is illustrated in Fig. 4.17.
28
The complex potential  z  is analytical within the domain, and its real part (or
imaginary) part is prescribed along the displacement (or traction) boundary.
given Im
zis analytical
Figure 4.17
given Re
Formulation of the boundary value problem under anti-plane shear.
The strain energy density within the body can be computed as
W   3  3 

2
  z  .
2
(4.60)
A Semi-infinite Crack
Consider a semi-infinite crack within an otherwise infinite body under anti-plane
loading, as shown in Fig. 4.18. Choose the crack tip as the origin of the complex plane.
We are looking for a solution of the following complex series form:

  z    cn z  ， n 1  n real.
n
n 1
The function is analytic for the entire plane cut along the crack. Especially, its
complex stress can be written as

 31  i 32   z     c n n z 
n 1
29
n 1
.
Z  re i
r

Figure 4.18
A semi-infinite crack within an infinite body under anti-plane loading.
Consider the traction free boundary condition along the crack faces, phrased as
 32     0 .
Taking the imaginary part of the complex stress, one has
 32 





n 1
i
i 
i 
  z    z    n c n z n 1  c n z
   n r n 1 c n e i n 1  c n e i n 1
2
2 nm
2 nm
where polar coordinates r and  are illustrated in Fig. 4.18. Substitution of the above
expression into the boundary condition leads to
cn e 2in  cn  0 .
(A)
The above homogeneous equation can accommodate a non-trivial solution only if
e 4i n  1 .
This eigen-value problem gives the following eigen-values
n 
n
,
2
n＝1,2,… .
The eigen-values reduce the coefficient relations (A) as cne  ni  cn  0 , and
consequently
cn  i n n ，  n real.
The complex stress function for a crack problem is

n
2
  z     ni z .
n
(4.61)
n 1
Consider the leading term of the above series. To follow the convention, we relabel
 1 as 
2 KⅢ
 
where K Ⅲ denotes the stress intensity factor of a mode Ⅲ
(meaning anti-plane shear) crack. Accordingly, the first term in (4.61) can be written
30

as
 z   i
2 KⅢ
 
z.
Straightforward calculation gives the following expression
u 3  Re  
KⅢ
2r


sin

(4.62)
2
for the displacement in the vicinity of the crack tip, and a complex stress of
 31  i 32  i
KⅢ
. Separating the real and the imaginary parts, the near tip stresses
2z
are given by



sin

 31  K Ⅲ 
2 .
 



2r  cos 
 32 

(4.63)
2 
One may notify that the stresses are square root singular near the crack tip, while the
out-of-plane displacement approaches zero at the crack tip like
r.
Corner Region
Next consider a corner region as shown in Fig. 4.19 under anti-plane shear. The
corner spans an angle  . The governing equation 2  0 and the traction free
boundary condition   0 is exactly the same as the governing equation and the
boundary condition for the potential flow of non-viscous fluid through the corner
region. The flow lines drawn in Fig. 4.19 are easy to visualize. The complex potential
for this problem is given by
  Cz  /  .
(4.64)



1 i  1 

with C being a real constant. That leads to    Cr  e    , and

 3r  i 3
 e  z  
i


1 i 

Cr  e  .

31

Figure 4.19
A corner region under anti-plane loading.
The stress components are given by
 3r  
   C r
 3  

1


  
 cos    

 .




 sin   

  
(4.65)
One may check the circumferential stress satisfies the traction free boundary
conditions  3   0,  0 . An upper plane constitutes a special case of the corner
region, where    , and   Cz according to (4.64). Only a constant shear stress
 31  C results in this case.
Point Force
Consider an out-of-plane force F that acts at the origin. The angular symmetry leads
to axisymmetric contours of displacement u3 , while the contours of constant  ,
being perpendicular to the u3 -contours, are the rays that denote the shear flows away
from the center. The contour plot is shown in Fig. 4.20. The complex potential for this
case is:

F
1
F  1

ln 
ln  i  ,

2 z 2  r

and the corresponding stress is given by  31  i 32 
(4.66)
F
. Transforming it to polar
2z
coordinates, one has
 3r  
F
,
2r
 3  0 .
(4.67)
The first expression indicates that the upward point force F is balanced by the uniform
32
shear stress along every concentric circle, the second expression results from the
axi-symmetry of the problem.
w  const
  const
Figure 4.20 Contours for u3 and  by an out-of plane point force at the origin.
Screw Dislocations
Consider a screw dislocation of an amplitude b. The solution is given by the following
complex potential

ib 1 ib  1

ln 
 ln  i  .
2 z 2  r

(4.68)
The real part of the expression furnishes the desired out-of-plane displacement
u3 
b
.
2
(4.69)
The expression (4.69) exactly characterizes the geometric aspect of a screw
dislocation. Namely the displacement field is multi-valued and exhibits a helix form.
It increases an amount of b after a complete circle. The stress function is given by
 
b 1
ln .
2 r
The combination of an out-of-plane point force and a screw dislocation can be
characterized by the following potential

1
2
F
 1
  ib  ln .

 z
33
From which the stress can be calculated as  3r  i 3   e i  
F  ib
.
2r
Separating different stress components, one obtains
 3r  
b
F
,  3 
.
2r
2r
(4.70)
Both components have inverse singularity. The strain energy density can be computed
as
W

2
 
2
 F2


 b 2  .
2 2 
8 r  

1
A notorious behavior concerns the strain energy stored in the body, say
U   Wrddr . Calculation indicates that U is unbounded. An inner cutoff core and
an outer cutoff radius have to be selected.
Strips Containing Cracks and Holes
Consider a strip of material loaded by out-of-plane shear along two horizontal ends of
the strip, as shown in Fig. 4.21.
X2
X1

Figure 4.21 Strip loaded by out-of plane shears at two ends.
The uniform shear stresses  31   and  32  0 within the strip give rise of the
following complex potential
  z
(4.71)
where shear strain    /  . Following the analogy between anti-plane shear and
incompressible and non-viscous flow, one observes that the shear flow resembles the
uniform flow of fluid through a channel of uniform width.
Let us introduce a crack in the horizontal direction, as shown in the upper graph of
Fig. 4.22. Since the crack, and henceforth the obstacle in the fluid mechanics analogy,
is parallel to the flow direction, its presence hardly makes a difference. The complex
34
potential is still given by (4.71).
X2

C
C
X1
X2
a
X1

Figure 4.22 Strip with a crack or a hole loaded by out-of plane shears at two ends.
Next consider the situation in the lower graph of Fig. 4.22 where the crack is replaced
by a hole. The fluid mechanics analogy of this problem is feature by the far-field
uniform flow perturbed by passing a cylinder. However, the geometry in the lower
graph can be transformed to that of the upper graph by Rokovsky transformation
cz
2a
a
z
    ,
(4.72)
where a denotes the radius of the cylinder, and c relates to the half-length of the crack.
Using the solution (4.71) for the crack problem and adopting the transform (4.72), one
has
   
c z
a
  .
2 a z
The asymptotic condition at the far ends,  z    z , determines c as 2a.
Accordingly, the complex potential for the hole problem is

    z 

a2 
.
z 
(4.73)
The polar components of stresses of the hole problem can be computed as
35
 3r  i 3
 i  a  2 i 
 a2 
 e    e  1  2     e    e  , that gives


z 
r



i
 3r
i
  a 2 
  1     cos ,
 r 


 3
  a 2 
  1     sin  .
 r 


(4.74)
Along the rim of the hole ( r  a ), one has  3r  0 (traction free boundary) and
 3  2 sin  . The amplitude of shear stress maximizes at   
minimizes at   0,  as zero.
36

2
as 2 , and