Chapter 10
Approximate
theories for solids with special shapes:
 rods, beams, membranes, plates and shells
10.2 Motion and Deformation of slender rods
The
figure illustrates the problem to be solved. Â
We suppose that a long, initially straight rod is subjected to forces
and moments that cause it to stretch, bend and twist into a complex three
dimensional shape, which we wish to determine. The initial shape need not necessarily be
stress free. Consequently, we can
solve problems involving a rod that is bent and twisted in its unloaded
configuration (such as a helical spring) by first mapping it onto an
intermediate, straight reference configuration, and then analyzing the
deformation of this shape.
10.2.1 Variables
characterizing the geometry of the rod’s cross-section
The
figure illustrates a generic cross-section of the (undeformed) rod.
Â
We
will characterize the shape of the cross-section as follows:
- We
introduce three mutually perpendicular, unit basis vectors
 ,
with  Â pointing parallel to the axis of the
cylinder, and  Â parallel to the principal moments of
area of the (undeformed) cross section.Â
- We introduce a
coordinate system
 Â within the cross section, with origin
at the centroid of the cross-section.
- The cross-sectional
area of the rod is denoted by

- The principal
moments of area of the cross-section are defined as

- We define a moment
of area tensor H for the
cross-section, with components
 ,
 ,
 Â and all other components zero.
- In calculations to
follow, it will be helpful to note that, because of the choice of origin
and coordinate system,

Principal
moments of area and their directions are listed for a few simple geometries
below. Recall also that area moments
of inertia for hollow sections can be calculated by subtraction.
Areas and area moments of inertia for simple
cross-sections
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10.2.2 Coordinate systems and
variables characterizing the deformation of a rod
 The
orientation of the straight rod is characterized using the  Â basis described in the preceding section.
 The position vector of a material particle
in the reference configuration is  ,
where  Â corresponds to the centroid of the cross
section, and  Â is the height above the base of the
cylinder.
 After
deformation, the axis of the cylinder lies on a smooth curve. The point
that lies at  Â in the undeformed solid moves to a new
position  Â after deformation.
 The
orientation of the cross-section after deformation will be described by
introducing a basis of mutually perpendicular unit vectors  ,
chosen so that  Â is parallel to the axis of the deformed rod,
and is  Â parallel to the line of material points that
lay along  Â in the reference configuration (or, more
precisely, parallel to the projection of this line perpendicular to  Â ).Â
Note that the three basis vectors  Â are all functions of  ,
and if the rod is moving, they are also functions of time.

 The orientation of   can be specified by three Euler angles  ,
which characterize the rigid rotation that maps   onto  .  To visualize the significance of the three
angles, note that the rotation can be accomplished in three stages (i) rotate
the basis vectors through an angle   about the   axis. Â
This results in a new set of vectors  ;
(ii) Rotate these new vectors through an angle  Â about the  Â axis.Â
This rotates the vectors onto a second configuration  ;
(iii) finally, rotate these vectors through angle   about the   direction, to create the   vectors.  Â
 Relationships between the Euler angles and
the curve characterizing the axis of the rod will be given shortly: these
results will show that the angles  Â and  Â can be determined from the shape of the
axis.  The angle   is an independent degree of freedom, and
quantifies the rotation of the rod’s cross-section about its axis. Â
 We let   denote the arc length measured along the
axis of the rod in the deformed configuration.
 The
velocity of the rod is characterized by the velocity vector of its axis, 
 The
rate of rotation of the rod is characterized by the angular velocity   of the basis vectors  . It will be shown in Sect 10.2.3 that the
angular velocity can be related to the velocity v of the bar’s axis and its twist   by

 The
acceleration of the rod is characterized by the acceleration vector of its
axis, 
 The
angular acceleration of the rod is characterized by the angular acceleration   of the basis vectors. It will be shown below that the angular
acceleration can be related to the acceleration a and velocity v of
the bar’s axis and its twist   by

10.2.3 Additional deformation measures
and useful kinematic relations
In
this section we introduce some additional measures of the deformation of the
rod, as well as several useful relations between the various deformation
measures.
 The curve
corresponding to the axis of the deformed rod is often characterized by its tangent, normal and binormal vectors,
together with its curvature, and
its torsion.  These are defined as follows.
- The tangent vector

- The normal vector and curvature
are defined so that
 ,
where n is a unit vector
- The binormal vector is defined as

- The triad of unit vectors
 Â defines the Frenet Basis for the curve
- The torsion
of the curve is defined as
 . Note that the torsion is simply a
geometric property of the curve  it is not necessarily related to the
rod’s twist.
These
variables are not sufficient to completely describe the deformation, however,
since the twist of the rod can vary independently of the shape of its axis.
 The two bases  ,
 Â can be related in terms of the Euler angles
as follows.

These
results can be derived by calculating the effects of the sequence of three
rotations. Note also that since both
sets of basis vectors are triads of mutually perpendicular unit vectors, they
must be related by

where
 Â is a proper orthogonal tensor that can be
visualized as a rigid rotation.  The
rotation tensor can be expressed in several different forms:
- It can be expressed as the sum of three dyadic
products

- It can be expressed
as components in either
 Â or  ,
which can be written in dyadic notation as   or  . Surprisingly, the components both
bases are equal, and are given by  .
The components can be expressed in terms of the Euler angles as a matrix

 In
further calculations the variation of basis vectors   with distance s along the deformed rod will play a central role. To visualize this quantity, imagine that
the basis  Â travels up the deformed rod. The basis
vectors will then rotate with an angular velocity that depends on the
curvature and twist of the deformed rod, suggesting that we can characterize
the rate of change of orientation with arc-length by a vector  ,
analogous to an angular velocity vector.Â
The curvature vector can be expressed as components in the basis   as  . This vector has the following properties
1. The curvature vector is (by definition) related to
the rate of change of   with s
by  ,
which can be expanded out to show that

2. The components  Â quantify the bending of the rod, and are
related the curvature  Â and the binormal vector  Â of the curve traced by the axis of the deformed
rod by  . You can show this result by comparing the
formula for  Â with the formula for b.
3. The curvature vector can also be expressed in terms of the position
vector of the rod’s centroid as

The component of curvature   cannot in general be expressed in terms of r, because the rotation of the rod’s
cross-section about its centroid axis may provide an additional, independent
contribution to  . For the special case where   and   are everywhere parallel to the normal vector
n and binormal b,
respectively, it follows that  . In this case,   is equal to the torsion of the curve.Â
 The
rate of change of   with distance s can also be expressed in terms of the Euler angles. For example, the derivative of   can be calculated as follows

Similar results for  Â and  Â are left as exercises.
 The
bending curvatures  Â and the twist rate  Â are related to the Euler angles by

These
results can be derived from the two different formulas for  ,
together with the equations relating   and   in terms of the Euler angles.Â
Â
 The arc
length s along the rod’s centerline
is related to the position vector of the rod’s axis by

 Some relationships between the time
derivatives of these various kinematic quantities are also useful in
subsequent calculations. The rate of
change in shape of the rod can be characterized by the velocity of the axis  Â and the time rate of change of the
cross-sectional rotation  .
 The time derivative of the tangent vector is
a convenient way to characterize the rate of change of bending of the
rod.  This is related to the velocity
of the rod’s centerline by

If
we express the velocity in components  Â and recall  Â we can write this as

It is important to note that the components  Â are not equal to the time derivatives of the
components of the tangent vector t,
because the basis  Â varies with time.
 The time derivatives of the basis vectors
can also be quantified by an angular velocity vector  ,
which satisfies  .  The components of   are readily shown to be

 The time derivatives of the remaining basis
vectors follow as

 The time derivative of the arc length of the
centerline is related to its velocity as follows

 We shall also require the gradient of the
angular velocity  ,
which quantifies the rate of change of bending. We shall give this vector the
symbol  Â to denote its physical significance: it can
be interpreted (see Appendix E) as the co-rotational time derivative of the
curvature vector, as follows

Evaluating the derivatives of  Â shows that

The co-rotational time derivative of curvature must be used
to quantify bending rate (instead of the time derivative  Â ) to correct for the fact that rigid
rotations and pure stretching do not change bending.
 Finally, to solve dynamic problems, we will
need to be able to describe the linear and angular acceleration of the
bar. The linear acceleration is most
conveniently characterized by the acceleration of the centerline of the bar 
 The angular acceleration of the bar’s
cross-section can be characterized by the angular acceleration   of the basis vectors  . A straightforward calculation shows that

 The second time derivative of the basis
vectors can then be calculated as

10.2.4 Approximating the displacement,
velocity and acceleration in the rod
The position vector after
deformation of the material point that has coordinates  Â in the undeformed rod can be expressed as

This
is a completely general expression. Â
We now introduce a series of approximations that are based on the
assumptions that
- The rod is thin compared with its length;
- The radius of
curvature of the rod (due to bending) is much larger than the
characteristic dimension of its cross section;
- The rate of change
of twist of the rod
 Â has the same order of magnitude as the
bending curvature of the rod.
- The material in the rod
experiences small distorsions
 i.e. the change in length of any
infinitesimal material fiber in the rod is much less than its undeformed
length. Â
With this
in mind, we assume that   can be approximated by a function of the
form

where
the Greek indices  Â can have values 1 and 2, and  Â can be regarded as the first term in a
Taylor expansion of  . The definition of   requires that  . We assume in addition that

for
all possible choices of  . The constants   can be thought of as the components of a
homogeneous in-plane deformation applied to the cross section, while the
function  Â describes the warping of the cross-section.Â
To decouple the warping from the axial displacement of the rod, we
require that

In
addition, for small distorsions, the deformation must satisfy  Â and  ,
the rod curvatures must satisfy  Â for all  ,
and the variation of arc-length s along
the axis of the deformed rod with  Â must satisfy  .
The velocity field in the
bar can be approximated as

where it has been assumed
that  Â and  Â for all  .
Finally, the acceleration
field within the bar will be approximated as

Here,
all time derivatives of   and   have been neglected. This is not so much because they are small,
but because they represent a crude approximation to the distortion of the
cross-section. The time derivatives of
these quantities are associated with short wavelength oscillations in the
bar, which cannot be modeled accurately using the assumed displacement
field.Â
10.2.5 Approximating the deformation
gradient
Based on the assumptions
listed in Section 10.2.3, the deformation gradient in the rod can be
approximated by

The first line of this
expression quantifies the effects of axial stretching, bending and twisting
of the rod. The second line
approximates the distorsion of its cross-section.
The deformation gradient
can also be decomposed as

where
R is the rigid rotation satisfying
 ,
and G and H are deformation gradient like tensors that describe the change
in shape of the rod. These tensors are
most conveniently expressed as components inÂ
 Â and  ,
respectively  we can represent this in diadic notation as   or  . The components can be expressed in matrix
form as

Â
Derivation: The
deformation gradient is, by definition, the derivative of the position vector
of material particles with respect to their position in the reference
configuration, i.e.

To reduce this to the
expression given,
1. Note that 
2. Recall that 
3. Substitute  Â and neglect the derivatives of f and  Â with respect to 
The decomposition   follows trivially by substituting   into the dyadic representation of F and rearranging the result. A similar approach gives  .
10.2.6 Other strain measures
It
is straightforward to compute additional strain measures from the deformation
gradient. Only a partial list will be
given here.
- The
determinant of the deformation gradient follows as

- The components of
the left and right Cauchy-Green tensors can
be computed from
 Â and  ,
where G and H were defined in 10.2.4.  C
and B are most conveniently
expressed as components in  Â and  ,
respectively  we can represent this in diadic
notation as  Â or  .
For small distorsions, the result can be approximated by

- The
Lagrange strain is defined as
 . Its components follow trivially from
the preceding formula. Note that
the matrix of components for E
resembles the formula for the
infinitesimal strain components in a straight bar subjected to axial
stretching, bending and twist deformation.  However, if the bent rod does not lie
in one plane, the twisting measure  Â includes contributions from both the
rotation of the rod’s cross section about its axis, and also from the
bending of the rod.
- The
rate of deformation tensor
 Â will also be required. It is
simplest to calculate the velocity gradient  Â by differentiating the expression given
for the velocity vector in the preceding section.

Substitute  ,
and note that

A tedious set of matrix multiplications shows that
the components of D in  Â are

to
within second order terms in curvature,  Â and  .
10.2.7 Kinematics of rods that are
bent and twisted in the unstressed state
It
is straightforward to generalize the results in sections 10.2.3-10.2.5 to
calculate strain measures for rods that are not straight in their initial
configuration. In this case we must
start by describing the geometry of the undeformed rod. To this end
- We denote the distance measured along the axis
of the initial, unstressed, twisted rod by

- At each point
 Â on the initial rod, we introduce a set
of three mutually perpendicular unit vectors  ,
where  Â is chosen to be tangent to the axis of
the undeformed rod; while  Â are parallel to the principal moments
of inertia of the cross-section.
- Â We also introduce an arbitrary
Cartesian basis
 Â where the unit vectors denote three
fixed directions in space.
- The basis vectors
 Â and  Â together define a set of three Euler
angles  ,
which completely describe the shape of the undeformed rod.
- We define a rotation
tensor
  satisfying   that characterizes the orientation of   with respect to  . The components of   can be found using the formulas in
Section 10.2.3.
- We define three
curvature components
 Â that characterize the bending and
twisting of the initial rod, as follows

The
deformed shape of the rod is characterized exactly as described in Section
10.2.1, except that the axial distance  Â is replaced by the arc-length  Â of the undeformed rod.
Assuming
small distorsions, the deformation gradient can be expressed in dyadic
notation as  ,
where the coefficients   are given below. The deformation gradient can also be
decomposed into two successive rotations and a small distorsion

where
the rotation tensors  Â and  Â satisfy  ,
and the tensors   can be expressed in component form as   . Their components are given by

The
deformation gradient can be written down immediately, by mapping the initial
rod onto a fictitious intermediate configuration in which the rod is
straight, chosen as follows:
1.
The straight
rod has axis parallel to the  Â direction
2. The point at arc-length  Â in the unstressed rod has coordinates  Â in the intermediate configuration.
3.
The principal
axes of the cross section are parallel to  Â in the intermediate configuration
4. The cross-section of the rod has the same shape in
the intermediate configuration as in the undeformed configuration.
The
deformed state can be reached in two steps (i) Deform the rod from the
unstressed configuration to the intermediate configuration, with a
deformation gradient  . The components of   can be calculated as the inverse of the
deformation gradient that maps the intermediate straight rod onto the
undeformed shape. (ii) Deform the rod
from the straight configuration to the deformed configuration, with a
deformation gradient  . The total deformation gradient follows as  .Â
10.2.8 Representation of forces and
moments in slender rods
The
figure shows a generic cross-section of the rod, in the deformed
configuration. To define measures of internal and external force acting on
the rod, we define the following variables
A basis  Â with unit vectors chosen following the
scheme described in 10.2.2. We define
the following vector components in this basis:
The body force acting on the rod  . For simplicity, we shall assume that the
body force is uniform within the cross section (but  Â may vary along the length of the rod).
The tractions acting on the exterior surface of the
rod 
The Cauchy stress within the rod  .
External forces and
moments acting on
the rod are characterized by
- The force per unit length acting on the rod,
 . The force components can be calculated
from the tractions and body force acting on the rod as 
- The moment per unit length acting on the rod,
 . The moment components can be calculated from the tractions
acting on the exterior surface of the rod as as

- The resultant force acting on each end of the
rod. Each force can be expressed
as components as
 . The components are related to the
tractions acting on the end of the rod by  ,
where the area integral is taken over the cross section at the
appropriate end of the rod.
- The resultant moment
acting on each end of the rod.Â
Each moment can be expressed as components as
 . The components are related to the
tractions acting on the end of the rod by

Internal forces and moments in the rod are characterized by the following
quantities:
- The variation of internal shear stress in the
cross section

- The average in-plane stress components

- Three components of a vector bending moment,
defined as

- The axial force on the cross-section

- Two additional
generalized forces
 ,
which represent the transverse shear forces acting on the rod’s cross
section. Unlike the axial force,
however, these forces cannot be directly related to the deformation of
the rod. Instead, they are
calculated from the bending moments, using the equilibrium equations
listed in the next section. Â
The forces  Â and moments  Â define components of a vector force and
moment
 Â is the resultant force acting on an
internal cross-section of the rod;
 Â is the resultant moment (about the
centroid of the cross section) acting on the cross-section.
10.2.9 Equations of motion and
boundary conditions
The internal forces and
moments must satisfy the equations of motion
         
Here,
 ,
T and M are the internal forces and moments in the rod;  Â are the external force and couple per unit
length;  Â is the mass density of the rod; A is its cross-sectional area, H is the area moment of inertia
tensor defined in Sect 10.2.1, while  Â are the acceleration, angular velocity and
angular acceleration of the rod’s centerline. The two equations of motion for
T and M clearly represent linear and angular moment balance for an
infinitesimal segment of the rod.
The
equations of motion for T and M are
often expressed as components in the  Â basis, as


Note
that:
- If the system is in
static equilibrium, the right hand sides of all the equations of motion
are zero.
- In addition, in many
dynamic problems, the right hand sides of the angular momentum balance equations
may be taken to be approximately zero, since the area moments of inertia
are small.  For example, the
rotational inertia may be ignored when modeling the vibration of a beam. The rotational inertia terms can be
important if the rod is rotating rapidly: examples include a spinning
shaft, or a rotating propeller.
Boundary
Conditions: The internal
stresses, forces and moments must satisfy the following boundary conditions

 Â on C
- The ends of the rod
may be subjected to a prescribed displacement.  Alternatively, the transverse or
axial tractions may be prescribed on the ends of the bar: in this case
the internal forces must satisfy
 Â for  Â and  Â for s=0.
- The ends of the rod
may be subjected to a prescribed rotation. Alternatively, if the ends are free to
rotate, the internal moments must satisfy
 Â for  Â and  Â for s=0.
Derivation: Measures of internal force and the equilibrium
equations emerge naturally from the principle of virtual work, which states
that the Cauchy stress distribution must satisfy

for
all virtual velocity fields   and compatible stretch rates  . The virtual velocity field and virtual
stretch rates in the bar must have the same general form as the actual
velocity and stretch rates, as outlined in Section 10.2.4 and 10.2.5. The virtual velocity and stretch rate can
therefore be characterized by   and compatible sets of  . This has two consequences:
 The
virtual work principle can be expressed in terms of the generalized
deformation measures and forces defined in the preceding sections as

 If the
virtual work equation is satisfied for all  Â and compatible sets of  ,
then the internal forces and moments must satisfy the equilibrium equations
and boundary conditions listed above.
It
is straightforward to derive the first result. The Jacobian is approximated as  ;
the components of   follow from the formulas given in Section
10.2.6, and the velocity field is approximated using the formula in
10.2.5. Substituting the definitions
given in Section 10.2.7 for generalized internal and external forces
immediately gives the required result.Â
The algebra involved is lengthy and tedious and is left as an
exercise.
The
equilibrium equations and boundary conditions are obtained by substituting
various choices of  Â and compatible sets of  Â into the virtual work equation.
- Choosing
 Â reduces the virtual work equation to

The
condition  Â follows immediately.Â
- Choosing
 Â reduces the virtual work equation to

Recall
that (by definition)  Â must be chosen to satisfy

Since
the body force is uniform, the term involving  Â is zero.Â
The first integral can then be integrated by parts as follows

Choosing  Â on the boundary yields the equilibrium
equation  ;
choosing any other  Â gives the boundary condition  .
- Choosing
 ,
using  Â as well as  Â yields

where we have integrated by parts to obtain the second
line.  Choosing   to vanish on the ends of the rod yields the
equation of motion  .
Any other choice of   yields the boundary conditions   on the ends of the rod.
- Choosing
 Â and substituting  ,
 ,
where  Â are the components of a virtual rate of
change of the tangent vector  Â reduces the virtual work equation to

To proceed, it is necessary to express  Â and  Â in terms of the virtual velocity components  .
The algebra and the resulting equilibrium equations are greatly simplified if
the tangent vector  Â is regarded as an independent kinematic
variable.  The relationship between t and   must be enforced by a vector valued Lagrange
multiplier  ,
which must satisfy

for
all variations  .
The second integral can be expressed in component form as

This equation can simply be added to the virtual
work equation to ensure that   and   are consistent.  Finally, recall that the curvature rates
and stretch rate are related to  Â  Â by

Substituting these results into the augmented virtual work
equation gives

This equation must be satisfied for all admissible  .
Considering each component in turn, and integrating by parts appropriately
and using  Â gives the last five equations of motion, as
well as the boundary conditions  Â on s=0
and s=L.Â
10.2.10 Constitutive equations
relating forces to deformation measures in elastic rods
The
generalized deformation measures and their conjugate generalized forces are
listed in the table below. Â
Constitutive equations must relate the deformation measures to the
forces. In this section we list the
relationships between these quantities for an isotropic, elastic rod
subjected to small distorsions. For
simplicity, the sides of the rod are assumed to be traction free.
The results depend on the
geometry of the rod’s cross-section, which is characterized as follows.Â
- Introduce a
Cartesian coordinate system
 Â as follows:Â the origin for this coordinate system
is at the centroid of the cross-section, the basis vectors  Â are parallel to the principal axes of
inertia for the cross-section, and   is parallel to the rod’s axis.
- We denote the cross-sectional area of the rod
by A, and the curve bounding
the cross-section by C, and
let
 Â denote the three principal moments of
area of the cross-section (see Sect 10.2.1)
- We introduce a
warping function
 Â to describe the out-of-plane
displacement component   in the cross-section of the rod. The warping function is related to the
out-of-plane displacement  Â by

The warping function depends only on the geometry of
the cross-section, and satisfies the following governing equations and
boundary conditions

You can easily show that this choice of  Â will automatically satisfy the shear stress
equilibrium equation  Â as well as the boundary condition  Â on C.
- Finally we define a modified polar moment of
inertia for the cross section as

Calculating
the warping function is a nuisance, because it requires the solution to a
PDE. In desperation, you can take w=0  this will overestimate the torsional
stiffness of the rod, but in many practical applications the error is not
significant.  For a better
approximation, warping functions can be estimated by neglecting the terms
involving   in the governing equation. A few such approximate warping functions
and modified polar moments of area are listed in the table below.Â
Warping functions and modified polar moments of
area for simple cross-sections
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The force-deformation
relations for the rod are

The
two shear force components   cannot be related to the deformation  they are Lagrange multipliers that enforce
the condition that the rod does not experience transverse shear, as discussed
in the preceding section.
Derivation: These
results can be derived as follows:
- The elastic
constitutive equations for materials subjected to small distorsions, but
arbitrary rotations, are listed in Section 3.3.  They have the form

where  Â are the components of the material stress
tensor, and  Â are the components of the Lagrange strain
tensor. The components of  Â in the basis  Â can be found using the formulas for  Â given in Section 10.2.7, and when
substituted into the constitutive laws give expressions for the components of
material stress  Â in terms of the deformation measures  ,
 ,
 Â and  .
- The Cauchy stress is
related to the material stress by
 . For small distorsions, but arbitrary
rotations, we may approximate this by  ,
so the components of the material stress tensor in  Â can be used as an approximation to the
components of the Cauchy stress tensor in  . Â
- Since we have
assumed that the tractions on the sides of the rod vanish, the in-plane
stress components must satisfy
 . Substituting the formulas for stresses
from (2) and noting that  Â (since that the origin for the
coordinate system coincides with the centroid of the cross section)
shows that  ,
 ,
and

- Substituting the
formula for
 Â into the definitions of  ,
 Â given in Section 10.2.8 and noting that
 Â (because the basis vectors coincide
with the principal axes of inertia) yields

- Recall that the
shear stress components
 Â must satisfy the equilibrium equation
and boundary condition

Substituting the shear stress components from step
(2) into this equilibrium equation and setting  Â gives the governing equation for w

- The shear stresses now follow as

Substituting
these results into the equation defining  Â in Section 10.2.6 gives the last equation

10.2.11 Strain energy of an elastic
rod
The total strain energy of
an elastic rod can be computed from its curvatures as

Derivation:Â The
derivation is similar to the procedure used to compute elastic
moment-curvature relations.
- The strain energy
density in the rod can be computed from the Lagrange strain
  and the Material Stress   as  . The material stress can be related to
the Lagrange strain using the formulas in Section 10.2.10, while the
Lagrange strain can be expressed in terms of of the deformation measures
 ,
 ,
 Â and  Â using the formulas for the deformation
gradient listed in Sections 10.2.7.Â
- The results can be
simplified by recalling that
 ,
 ,
which shows that the strain energy density can be approximated as

where w is
the warping function defined in Section 10.2.9. The two terms in this
expression represent the strain energy density due to stretching and bending
the rod, and twisting the rod, respectively.
- The total strain energy follows by integrating U over the volume of the
rod. Using the measures of
cross-sectional geometry listed in Section 10.2.1, it is straightforward
to show that

- Some additional algebra is required to
calculate the energy associated with twisting the rod. Begin by noting
that

We
need to show that the integral on the right hand side of this expression is
zero.
- To this end, note that

where
we have recalled that the warping function w satisfies  Â in A
as well as  Â on C,
and have used the divergence theorem.
- Secondly, note that

The
sum of (5) and (6) is zero. Using this
result and (4) gives the expression for the strain energy of the rod.
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