Approximate theories for solids with
rods, beams, membranes, plates and shells
10.4 Exact solutions to simple problems involving
section lists solutions to various boundary and initial value problems
involving deformable rods, to illustrate representative applications of the
equations derived in Sections 10.2.2 and 10.2.3. Specifically, we derive solutions for:
- The natural frequencies and mode shapes for an
initially straight vibrating beam;
- The buckling load for a vertical rod subjected to
- The full post-buckled shape for a straight rod
compressed by axial loads on its ends;
- Internal forces and moments in an initially
straight rod that is bent and twisted into a helix;
- Internal forces, moments, and the deflected shape
of a helical spring.
10.4.1 Free vibration of a straight beam
without axial force
figure illustrates the problem to be solved: an initially straight beam, with
axis parallel to the direction and principal axes of inertia
parallel to is free of external force. The beam has Young’s
modulus and mass density ,
and its cross-section has area A and
principal moments of area . Its ends may be constrained in various ways,
as described in more detail below. We
wish to calculate the natural frequencies and mode shapes of vibration for the
beam, and to use these results to write down the displacement for a beam that is caused to vibrate with
initial conditions ,
at time t=0.
shapes and natural frequencies: The physical significance of the mode shapes and natural frequencies of
a vibrating beam can be visualized as follows:
- Suppose that the beam
is made to vibrate by bending it into some (fixed) deformed shape ;
and then suddenly releasing it. In
general, the resulting motion of the beam will be very complicated, and
may not even appear to be periodic.
- However, there exists
a set of special initial deflections ,
which cause every point on the beam to experience simple harmonic motion
at some (angular) frequency ,
so that the deflected shape has the form .
- The special
frequencies are called the natural frequencies
of the beam, and the special initial deflections are called the mode shapes. Each mode shape has a wave
which characterizes the wavelength of the harmonic vibrations, and is
related to the natural frequency by
- The mode shapes have a very useful property (which is
proved in Section 5.9.1):
mode shapes, wave numbers and corresponding natural frequencies depend on the
way the beam is supported at its ends. A
few representative results are listed below
with free ends:
The wave numbers for each mode are given by the roots
of the equation
The mode shapes are
where are arbitrary constants.
with pinned ends:
The wave numbers for each mode are
The mode shapes are ,
where are arbitrary constants
beam (clamped at ,
free at :
The wave numbers for each mode are given by the roots
of the equation
The mode shapes are
where are arbitrary constants.
of a beam with given initial displacement and velocity
solution for free vibration of a beam with given initial displacement and
velocity can be found by superposing contributions from each mode as follows
Derivation: We will
derive the equations for the natural frequencies and mode shapes of a beam with
free ends as a representative example.
This is a small deflection problem and can be modeled using
Euler-Bernoulli beam theory summarized in Section 10.3.2.
1. The deflection of the beam must satisfy the equation
of motion given in Sect 10.3.2
2. The general solution to this equation (found, e.g. by
separation of variables, or just by direct substitution) is
where the frequency and wave number must be related by
to satisfy the equation of motion.
3. The coefficients and the wave number must be chosen to satisfy the boundary
conditions at the ends of the bar. For
a beam with free ends, the boundary conditions reduce to ,
at . Substituting the formula from (2) into the
four boundary conditions, and writing the resulting equations in matrix form
4. For a nonzero solution, the matrix in this equation
must be singular. This implies that the
determinant of the matrix is zero, which gives the governing equation for
5. Since the equation system in (3) is now singular, we
may discard any one of the four equations and use the other three to determine
an equation relating to . Choosing to discard the last row of the
matrix, and taking the first column to the right hand side shows that
Solving this equation system shows that . Substituting these values back into the
solution in step (2) gives the mode shape.
6. To understand the formula for the vibration of a beam
with given initial conditions, note that the most general solution consists of
a linear combination of all possible
mode shapes, i.e.
for found by substituting ,
multiplying both sides of the equation by and integrating over the length of the
beam. We know that
the result reduces to
formula for is found by differentiating the general
solution with respect to time to find the velocity, substituting ,
and then proceeding as before to extract each coefficient .
10.4.2 Buckling of a column subjected to
problem to be solved is illustrated in the figure. A straight, vertical elastic cantilever beam
with mass density and elastic modulus is clamped at its base and subjected to
gravitational loading. The beam has length L,
cross-sectional area A and principal
moments of area . The straight, vertical rod is always an
equilibrium configuration, but this configuration is stable only if .
objective is to show that the critical buckling length is
number of different techniques can be used to find buckling loads. One of the simplest procedures (which will be
adopted here) is to identify the critical conditions where both the straight
configuration (with ), and also the deflected configuration (with
a small transverse deflection ) are possible equilibrium shapes for the rod.
problem can be solved using the governing equations for a beam subjected to
large axial forces, listed in Section 10.3.3.
For the present case, we note that
forces acting on the rod are ,
where g is the gravitational
is zero (because the rod is in static equilibrium)
equations therefore reduce to
must be solved subject to the boundary conditions
second equation of (3) and using the boundary condition at reduces the first equation of (3) to
6. Integrating this equation with respect to and imposing the boundary condition at shows that
This equation can
be solved for using a symbolic manipulation program, which
are special functions called `Airy Wave
functions of order zero’
boundary conditions are at ,
and at . Substituting (7) into the boundary conditions
and writing the results in matrix form gives
and are Airy wave functions of order 1.
9. For this system of equations to have a nonzero
solution, the determinant of the matrix must vanish, which shows that must satisfy . This equation can easily be solved
(numerically) for . The smallest value of that satisfies the equation is .
10. The buckling length follows as
10.4.3 Post-buckled shape of an initially straight rod subjected to end thrust
figure illustrates the problem to be solved.
An initially straight, inextensible elastic rod, with Young’s modulus E, length L and principal in-plane moments of area (with ) is subjected to end thrust. The ends of the rod are constrained to travel
along a line that is parallel to the undeformed rod, but the ends are free to
rotate. We wish to calculate the
deformed shape of the rod. You are
probably familiar with the simple Euler buckling analysis that predicts the
critical buckling loads. Here, we derive
the full post-buckling solution.
rod is assumed to bow away from its straight configuration as shown: the
deflected rod lies in the plane perpendicular to . The basis and the Euler angle that characterize the rotation of the rod’s
cross sections are shown in the picture; the remaining Euler angles are .
Solution: Several possible equilibrium solutions may exist,
depending on the applied load P.
1. The straight rod, with is always an equilibrium solution. It is stable for applied loads .
2. For applied loads ,
with n an integer, there are n+1 possible equilibrium solutions. One of these is the straight rod; the rest
are possible buckling modes. The shape
of each buckling mode depends on a parameter which satisfies
denotes a complete elliptic integral of the first kind . Note that K
has a minimum value at k=0,
and increases monotonically to infinity as . The equation for has no solutions for ,
and n solutions for ,
as expected. If multiple solutions
exist, only the solution with n=1 is
The shape of the
deformed rod can be characterized by the Euler angle ,
denotes a Jacobi-elliptic function called the `sine-amplitude:’ its second
argument k is called the `modulus’ of
4. The coordinates of the buckled rod can also be
calculated. They are given by
and cn(x,k) denote Jacobi elliptic
functions called the `amplitude’ and `cosine amplitude’, and E(x,k) denotes an incomplete elliptic
integral of the second kind . The shape of the deflected rod for the stable
buckling mode (n=1) is shown in the
is a large deflection problem and must be treated using the general equations
listed in Sections 10.7-10.9.
- The equilibrium
equation immediately shows that T=constant along the rod’s
length. The boundary conditions at
the end of the rod give ,
so that the components of T in follow as .
- Substituting the
expressions for into the moment balance equations shows
- Finally, note that the
curvatures are ,
and recall that ,
so that the angle satisfies
- This is the equation
that governs oscillations of a pendulum, and its solution is well known.
The equation is satisfied trivially by (this is the straight configuration), and
also by two one-parameter families of functions of the form
Here, and are parameters whose values must be determined
from the boundary conditions. The first
of these two functions is called an `inflexional’ solution, because the curve
has points where . The second is called `non-inflexional’
because it has no such points. For the
pendulum, inflexional solutions correspond to periodic swinging motion; the
non-inflexional solution corresponds to the pendulum whirling around the
- The bending moment must
satisfy at both ends of the rod, which requires
that at and . Only the inflexional solution can
satisfy these boundary conditions.
For this case, we have
The cosine amplitude cn is a periodic function (it is
a generalized cosine) and satisfies at . We may therefore satisfy the boundary
conditions by choosing and . This leads to the defining equations for .
- Finally, the formula
for the coordinates follows by integrating and subject to boundary conditions at and at s=0.
- Finally, the (global)
stability of the various solutions can be checked by comparing their
10.4.4 Rod bent and twisted into a helix
consider an initially straight rod with Young’s modulus E and shear modulus . The cross-section of the rod has area ,
principal in-plane moments of inertia and an effective torsional inertia . The
rod is initially straight and unstressed, and is then subjected to
forces and moments on its ends to bend and twist it into a helical shape. The geometry of the deformed rod can be
The radius r of the cylinder that generates the
The number of
turns per unit axial length in the helix
The helix angle ,
which is related to n by
4. The twist curvature ,
which quantify the distorsion induced by twisting the rod about its deformed
axis. For the rod to be in equilibrium, must be constant.
The stretch ratio
. For the rod to be in equilibrium, must be constant.
geometry and forces in the deformed rod are most conveniently described using a
cylindrical-polar coordinate system and basis shown in the figure. In terms of these basis vectors, we may
- The tangent vector to the rod
- The binormal vector is
In terms of these variables:
- The internal moment in the rod is
- The internal force in
the rod is
For the limiting case of an inextensible rod, the
quantity should be replaced by an indeterminate axial
forces acting on the ends of the rod must satisfy and at and at s=0.
variety of force and moment systems may deform the rod into a helical shape,
depending on the twist and stretch. An
example of particular practical significance consists of a force and moment acting at s=L
(with equal and opposite forces at s=0),
This force system is
statically equivalent to a wrench with force and moment acting at r=0.
note that this analysis merely gives conditions for a helical rod to be in
static equilibrium. The configuration
may not be stable.
- We take at s=0,
so that the cylindrical polar coordinates are related to arc-length by .
Note also that the basis vectors satisfy ,
- The position vector of a point on the axis of the
rod can be expressed as ;
- The tangent vector follows as ;
- By definition, the
curvature vector is ,
which can be expressed in terms of the binormal vector as ;
- The moment-curvature relations then give the
internal moment ;
- The equilibrium
equation shows that T=constant. We may
express this constant internal force vector in terms of its components as
- The internal forces
and moments must satisfy the moment equilibrium equation, which shows that
Taking the dot product of both sides of this equation
with shows that .
It then follows that and
- Finally, the
force-stretch relation requires that . This equation can be solved together
with the final result of (7) for the components of internal force in the
10.4.5 Helical spring
behavior of a helical spring can be deduced by means of a simple extension of
the results in the preceding section. We
assume that the spring is made from a material with Young’s modulus E and shear modulus . The cross-section of the rod has principal
in-plane moments of inertia and an effective torsional inertia . The rod is assumed to be inextensible, for
simplicity. The geometry of the
undeformed spring can be characterized as follows
- The length of the rod L, radius of the cylinder that generates the helix;
the height of the spring, the number of turns in the
and the helix angle
- The variables characterizing the undeformed
spring are related as follows
- It is helpful to select a basis to characterize the orientation of the
initial spring. Since we may select and arbitrarily. It is convenient to choose and to be parallel to the normal vector n and binormal vector b of the undeformed spring,
respectively, which gives
- The initial curvature components can be
calculated from the condition that and follow as .
end of the spring at s=0 is held fixed (so it cannot move or rotate). The end of the spring at is subjected to a combination of a force and moment which act at the axis of the helical
solution can be calculated in exactly the same way as the derivation in Section
10.3.5. It can be shown that
- The spring remains
helical: its deformed shape can be characterized by new values of r, and after deformation.
- The spring may tend to
twist about the axis of the helix when it is subjected to load. The twisting can be quantified by the
change in cylindrical-polar coordinates of the point at s=L on the spring. In the undeformed state, these are ;
after deformation, they are . The twisting can be characterized by the
rotation The point where the load is applied
therefore displaces through a distance ,
and rotates through the angle about the axis of the cylinder.
- The displacement and
rotation are related to the rod’s length L, the coil radius r
and helix angle by
- The vectors and after deformation are given by .
- The curvatures after
deformation follow as
- The internal moment
and force in the spring are related to the curvatures and external force
and moment by
- The external force and
moment applied to the axis of the spring are related to the helix angle
and coil radius before and after deformation by
If the spring is
subjected a prescribed force and moment, these equations can be solved for and ,
and the results can be substituted into (3) to calculate the extension and rotation of the spring.
The results cannot be expressed in closed form for large shape changes. If and are small, however, the relations can be
linearized to yield
the spring stiffnesses are