

Chapter 10
Approximate theories for solids with special shapes: Â rods, beams, membranes, plates and shells
10.4 Exact solutions to simple problems involving elastic rods
This 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.Â Â Specifically, we derive solutions for:
10.4.1 Free vibration of a straight beam without axial force
The figure illustrates the problem to be solved: an initially straight beam, with axis parallel to the Â direction and principal axes of inertia parallel to Â is free of external force.Â The beam has Youngâ€™s modulus Â and mass density , and its crosssection has area A and principal moments of area .Â Its ends may be constrained in various ways, as described in more detail below.Â 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 Â for a beam that is caused to vibrate with initial conditions , Â at time t=0.
Mode shapes and natural frequencies: The physical significance of the mode shapes and natural frequencies of a vibrating beam can be visualized as follows:
The mode shapes, wave numbers and corresponding natural frequencies depend on the way the beam is supported at its ends.Â A few representative results are listed below
Beam withÂ free ends: The wave numbers for each mode are given by the roots of the equation The mode shapes are
where Â are arbitrary constants.
Beam with pinned ends: The wave numbers for each mode are The mode shapes are , where Â are arbitrary constants
Cantilever beam (clamped at , free atÂ :
The wave numbers for each mode are given by the roots of the equation The mode shapes are
where Â are arbitrary constants.
Vibration of a beam with given initial displacement and velocity
The solution for free vibration of a beam with given initial displacement and velocity can be found by superposing contributions from each mode as follows
where
Derivation:Â We will derive the equations for the natural frequencies and mode shapes of a beam with free ends as a representative example.Â This is a small deflection problem and can be modeled using EulerBernoulli 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 Â to satisfy the equation of motion. 3. The coefficients Â and the wave number Â must be chosen to satisfy the boundary conditions at the ends of the bar.Â Â For a beam with free ends, the boundary conditions reduce to , Â at .Â Substituting the formula from (2) into the four boundary conditions, and writing the resulting equations in matrix form yields
4. For a nonzero solution, the matrix in this equation must be singular.Â This implies that the determinant of the matrixÂ isÂ zero, which gives the governing equation for the wavenumber
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 Â to .Â 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 .Â 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.
Formulas for Â found by substituting , multiplying both sides of the equation by Â and integrating over the length of the beam.Â Â We know that
so the result reduces to
The formula for Â 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 gravitational loading
The problem to be solved is illustrated in the figure.Â A straight, vertical elastic cantilever beam with mass density Â and elastic modulus Â is clamped at its base and subjected to gravitational loading. The beam has length L, crosssectional area A and principal moments of area .Â The straight, vertical rod is always an equilibrium configuration, but this configuration is stable only if .
Our objective is to show that the critical buckling length is
A number of different techniques can be used to find buckling loads.Â One of the simplest procedures (which will be adopted here) is to identify the critical conditions where both the straight configuration (with Â ), and also the deflected configuration (with a small transverse deflection Â ) are possible equilibrium shapes for the rod.
This problem can be solved using the governing equations for a beam subjected to large axial forces, listed in Section 10.3.3.Â Â Â For the present case, we note that 1. The external forces acting on the rod are , , where g is the gravitational acceleration; 2. The acceleration is zero (because the rod is in static equilibrium) 3. The equilibrium equations therefore reduce to Â Â Â Â Â Â Â Â Â 4. These equations must be solved subject to the boundary conditions
5. Integrating the second equation of (3) and using the boundary condition Â at Â reduces the first equation of (3) to
6. Integrating this equation with respect to Â and imposing the boundary condition Â at Â shows that
7. This equation can be solved for Â using a symbolic manipulation program, which yields
where Â are special functions called `Airy Wave functions of order zeroâ€™ 8. The remaining boundary conditions are Â at , and Â at .Â Substituting (7) into the boundary conditions and writing the results in matrix form gives
where Â and Â 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 Â must satisfy .Â This equation can easily be solved (numerically) for .Â The smallest value of Â that satisfies the equation is . 10. The buckling length follows as
10.4.3Â Postbuckled shape of an initially straight rod subjected to end thrust
The figure illustrates the problem to be solved.Â An initially straight, inextensible elastic rod, with Youngâ€™s modulus E, length L and principal inplane moments of area Â Â (with Â ) is subjected to end thrust.Â 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.Â Â We wish to calculate the deformed shape of the rod.Â Â You are probably familiar with the simple Euler buckling analysis that predicts the critical buckling loads.Â Here, we derive the full postbuckling solution.
The rod is assumed to bow away from its straight configuration as shown: the deflected rod lies in the plane perpendicular to .Â The basis Â and the Euler angle Â that characterize the rotation of the rodâ€™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 Â is always an equilibrium solution.Â It is stable for applied loads . 2. For applied loads , with n an integer, there are n+1 possible equilibrium solutions.Â One of these is the straight rod; the rest are possible buckling modes.Â The shape of each buckling mode depends on a parameter Â which satisfies
where K denotes a complete elliptic integral of the first kind .Â Note that K has a minimum value Â at k=0, and increases monotonically to infinity as .Â The equation for Â has no solutions for , and n solutions for , as expected.Â If multiple solutions exist, only the solution with n=1 is stable. 3. The shape of the deformed rod can be characterized by the Euler angle , which satisfies
where sn(x,k) denotes a Jacobielliptic function called the `sineamplitude:â€™ its second argument k is called the `modulusâ€™ of the function. 4. The coordinates of the buckled rod can also be calculated.Â They are given by
Here am(x,k) and cn(x,k) denote Jacobi elliptic functions called the `amplitudeâ€™ and `cosine amplitudeâ€™, and E(x,k) denotes an incomplete elliptic integral of the second kind .Â The shape of the deflected rod for the stable buckling mode (n=1) is shown in the figure above.Â
Derivation:Â This is a large deflection problem and must be treated using the general equations listed in Sections 10.710.9.
Here, Â and Â are parameters whose values must be determined from the boundary conditions.Â The first of these two functions is called an `inflexionalâ€™ solution, because the curve has points where . Â The second is called `noninflexionalâ€™ because it has no such points.Â For the pendulum, inflexional solutions correspond to periodic swinging motion; the noninflexional solution corresponds to the pendulum whirling around the pivot.Â
The cosine amplitude cn is a periodic function (it is a generalized cosine) and satisfies Â at .Â We may therefore satisfy the boundary conditions by choosing Â andÂ .Â This leads to the defining equations for .
Â
10.4.4 Rod bent and twisted into a helix
We consider an initially straight rod with Youngâ€™s modulus E and shear modulus .Â The crosssection of the rod has area , principal inplane moments of inertia Â and an effective torsional inertia .Â TheÂ 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.Â The geometry of the deformed rod can be characterized by: 1. The radius r of the cylinder that generates the helix 2. The number of turns per unit axial length Â in the helix 3. 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.Â For the rod to be in equilibrium, Â must be constant. 5. The stretch ratio .Â For the rod to be in equilibrium, Â must be constant.
The geometry and forces in the deformed rod are most conveniently described using a cylindricalpolar coordinate system Â and basis Â shown in the figure.Â In terms of these basis vectors, we may define
In terms of these variables:
For the limiting case of an inextensible rod, the quantity Â should be replaced by an indeterminate axial force .
The forces acting on the ends of the rod must satisfy Â and Â at Â and Â Â at s=0.
A variety of force and moment systems may deform the rod into a helical shape, depending on the twist and stretch.Â An example of particular practical significance consists of a force Â and moment Â acting at s=L (with equal and opposite forces at s=0), where
This force system is statically equivalent to a wrench with force Â and moment Â acting at r=0.
Finally note that this analysis merely gives conditions for a helical rod to be in static equilibrium.Â The configuration may not be stable.
Derivation
Â Â Â Â Â Â
Taking the dot product of both sides of this equation with Â shows that . It then follows that Â and
The behavior of a helical spring can be deduced by means of a simple extension of the results in the preceding section.Â We assume that the spring is made from a material with Youngâ€™s modulus E and shear modulus .Â The crosssection of the rod has principal inplane moments of inertia Â and an effective torsional inertia .Â The rod is assumed to be inextensible, for simplicity.Â The geometry of the undeformed spring can be characterized as follows
The end of the spring at s=0 is held fixed (so it cannot move or rotate).Â The end of the spring at Â is subjected to a combination of a force Â and moment Â which act at the axis of the helical coil.Â
The solution can be calculated in exactly the same way as the derivation in Section 10.3.5.Â It can be shownÂ that
8. If the spring is subjected a prescribed force and moment, these equations can be solved for Â and , and the results can be substituted into (3) to calculate the extension Â and rotation Â of the spring.Â The results cannot be expressed in closed form for large shape changes.Â If Â and Â are small, however, the relations can be linearized to yield
where the spring stiffnesses are


(c) A.F. Bower, 2008 