Chapter 10

Approximate theories for solids with special shapes:

rods, beams, membranes, plates and shells

10.3 Simplified versions of the general theory of deformable rods

In many practical cases of interest the general equations governing deformation and deformation of generally curved rods can be vastly simplified.  In this section, we summarize the governing equations for a number of special solids, including flexible strings, and various forms of beam theory.

10.3.1 Stretched flexible string with small transverse deflections.

This is the simplest possible version of the general theory outlined in 10.2.   The problem to be solved is illustrated in the figure.  A `string’ with Young’s modulus $E$, mass density $\rho$, cross-sectional area $A$ and negligible area moments of inertia ${I}_{1}={I}_{2}={I}_{3}=0$ is initially straight and parallel to the ${e}_{3}$ direction. The ends of the string are subjected to an axial load ${T}_{0}$ and are prevented from moving transverse to the string.  A force per unit length $p={p}_{1}{e}_{1}$ acts on the string, inducing a small, time dependent, transverse deflection $u={u}_{1}\left({x}_{3}\right){e}_{1}$.

The general governing equations can be simplified to the following form:

1. The curvature of the string can be approximated as $\kappa \approx \frac{{d}^{2}{u}_{1}}{d{x}_{3}^{2}}{e}_{2}$
2. The stretch of the string can be approximated as $\frac{ds}{d{x}_{3}}=1+{T}_{0}/EA$
3. The only nonzero internal force is the axial force ${T}_{3}$ (the moment-curvature relations show that the internal moments vanish; the angular momentum balance equations show ${T}_{1}={T}_{2}=0$ )
4. The equations of motion reduce to ${\kappa }_{2}{T}_{3}+{p}_{1}=\rho A{a}_{1}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{d{T}_{3}}{ds}=0$
5. Combining 1-4 shows ${T}_{3}={T}_{0}$ and gives the equation of motion for the stretched string

${T}_{0}\frac{{d}^{2}{u}_{1}}{d{x}_{3}^{2}}+{p}_{1}=\rho A\frac{{d}^{2}{u}_{1}}{d{t}^{2}}$

1. The boundary conditions are ${u}_{1}=0$ at ${x}_{3}=0,{x}_{3}=L$

Large deflection equations for a flexible string can also be found by substituting ${M}_{1}={M}_{2}={M}_{3}={T}_{1}={T}_{2}=0$ into the general equations of motion for a rod.  The details are left as an exercise.

10.3.2 Straight elastic beam with small deflections and no axial force (Euler-Bernoulli beam theory)

The figure illustrates the problem to be solved: an initially straight beam, with axis parallel to the ${e}_{3}$ direction and principal axes of inertia parallel to ${e}_{1},{e}_{2}$ is subjected to a transverse force per unit length $p={p}_{1}{e}_{1}$.  The beam has Young’s modulus $E$ and mass density $\rho$, and its cross-section has area A and principal moments of inertia ${I}_{1},{I}_{2},{I}_{3}$.  Its ends may be constrained in various ways, as described in more detail below.   We suppose that the beam experiences a small transverse displacement $u={u}_{1}{e}_{1}$, and wish to calculate ${u}_{1}$ as a function of time, given appropriate initial conditions.

The general equations of motion for a rod can be used to deduce that:

1. The stretch of the bar and the curvature vector may be approximated by  $\frac{ds}{d{x}_{3}}\approx 1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\kappa \approx \frac{{d}^{2}{u}_{1}}{d{x}_{3}^{2}}{e}_{2}$
2. The internal forces can be characterized by the internal force $T\approx {T}_{1}{e}_{1}$ and internal moment $M={M}_{2}{e}_{2}$.   These can be interpreted as the force and moment acting on an internal cross-section of the beam which has normal in the ${e}_{3}$ direction.
3. Moment-curvature relations reduce to ${M}_{2}=E{I}_{2}\frac{{d}^{2}{u}_{1}}{d{x}_{3}^{2}}$
4. Equations of motion can be reduced to $\frac{d{T}_{1}}{d{x}_{3}}+{p}_{1}=\rho A{a}_{1}$   $\frac{d{M}_{2}}{d{x}_{3}}+{T}_{1}=\rho {I}_{2}{\alpha }_{2}\approx 0$

5.      Alternatively, the equations in (3) and (4) can be combined to express the equations of motion in terms of displacement

$E{I}_{2}\frac{{d}^{4}{u}_{1}}{d{x}_{3}^{4}}+\rho A\frac{{d}^{2}{u}_{1}}{d{t}^{2}}={p}_{1}$

Boundary conditions: Elementary beam theory offers the following boundary conditions:

1. The end of the beam may be clamped, i.e. rotations and displacement of the end are completely prevented. The transverse displacement must then satisfy ${u}_{1}=\partial {u}_{1}/d{x}_{3}=0$.
2. The end of the beam may be simply supported, i.e. the end cannot move, but may rotate freely.  In this case the transverse displacement must satisfy ${u}_{1}=0$ and the internal moment must satisfy ${M}_{2}=E{I}_{2}{d}^{2}{u}_{1}/d{x}_{3}^{2}=0$
3. The end of the beam may be free, i.e. the end can move and rotate freely.  In this case the internal moment and internal force must satisfy ${M}_{2}=E{I}_{2}{d}^{2}{u}_{1}/d{x}_{3}^{2}=0$, ${T}_{1}=-E{I}_{2}{d}^{3}{u}_{1}/d{x}_{3}^{3}=0$

10.3.3 Straight elastic beam with small transverse deflections and significant axial force

This version of beam theory is used to model beams that are subjected to substantial axial loads (usually due to forces applied at their ends).  The equations can be used to estimate the effects of axial load on the transverse deflection or vibration of a beam. The theory can also be used to calculate buckling loads for beams, but does not accurately model their deformation if the buckling loads are exceeded.

The problem to be solved is illustrated in the figure.   An initially straight beam, with axis parallel to the ${e}_{3}$ direction and principal axes of inertia parallel to ${e}_{1},{e}_{2}$ is subjected to a force per unit length $p={p}_{1}{e}_{1}+{p}_{3}{e}_{3}$.  The beam has Young’s modulus $E$ and mass density $\rho$, and its cross-section has area A and principal moments of inertia ${I}_{1},{I}_{2},{I}_{3}$.  Its ends may be constrained in various ways, as described in the preceding section. We assume that a large axial internal force $N{m}_{3}$ is developed in the beam, either by a horizontal force per unit length ${p}_{3}$ or horizontal forces ${P}_{3}^{\left(0\right)}{e}_{3},{P}_{3}^{\left(L\right)}{e}_{3}$ acting at the ends of the beam.  We suppose that the beam experiences a small transverse displacement $u={u}_{1}{e}_{1}$, and wish to calculate ${u}_{1}$ as a function of time, given appropriate initial conditions.

The general equations of motion for a deformable rod can be approximated as follows:

1. The stretch of the bar and the curvature vector may be approximated by  $\frac{ds}{d{x}_{3}}\approx 1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\kappa \approx \frac{{d}^{2}{u}_{1}}{d{x}_{3}^{2}}{e}_{2}$
2. The internal forces can be characterized by the internal force $T=V{m}_{1}+N{m}_{3}$ and internal moment $M={M}_{2}{e}_{2}$.   These can be interpreted as the force and moment acting on an internal cross-section of the beam which has normal in the ${m}_{3}$ direction.
3. Moment-curvature relations reduce to ${M}_{2}=E{I}_{2}\frac{{d}^{2}{u}_{1}}{d{x}_{3}^{2}}$
4. Equations of motion may be approximated by

$\frac{dV}{d{x}_{3}}+N\frac{{d}^{2}{u}_{1}}{d{x}_{3}^{2}}-{p}_{3}\frac{d{u}_{1}}{d{x}_{3}}+{p}_{1}=\rho A{a}_{1}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{dN}{d{x}_{3}}+{p}_{3}=0$ $\frac{d{M}_{2}}{d{x}_{3}}+V=0$

To interpret these equations, note that (i) the axial force N has been assumed to be much larger than the transverse force V, so that nonlinear terms associated with the axial force have been retained when approximating the equations of motion; (ii) Although the equations of motion are expressed in terms of components of displacement and external force in the $\left\{{e}_{1},{e}_{2},{e}_{3}\right\}$ basis, the equations of motion themselves represent an approximation to the components of the full, nonlinear equations of motion in the $\left\{{m}_{1},{m}_{2},{m}_{3}\right\}$ basis.

5.      The results of 1-4 can be combined to obtain an equation for the transverse deflection of the beam

$E{I}_{2}\frac{{d}^{4}{u}_{1}}{d{x}_{3}^{4}}+\rho A\frac{{d}^{2}{u}_{1}}{d{t}^{2}}+{p}_{3}\frac{d{u}_{1}}{d{x}_{3}}=N\frac{{d}^{2}{u}_{1}}{d{x}_{3}^{2}}+{p}_{1}$

Boundary conditions: Elementary beam theory offers the following boundary conditions:

1. The end of the beam may be clamped, i.e. rotations and displacement of the end are completely prevented. The transverse displacement must then satisfy ${u}_{1}=\partial {u}_{1}/d{x}_{3}=0$.
2. The end of the beam may be simply supported, i.e. the end cannot move, but may rotate freely.  In this case the transverse displacement must satisfy ${u}_{1}=0$ and the internal moment must satisfy ${M}_{2}=E{I}_{2}{d}^{2}{u}_{1}/d{x}_{3}^{2}=0$
3. The end of the beam may be free, i.e. the end can move and rotate freely.  In this case the internal moment and internal force must satisfy ${M}_{2}=E{I}_{2}{d}^{2}{u}_{1}/d{x}_{3}^{2}=0$, while the transverse force must satisfy

$V=\left\{\begin{array}{c}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{P}_{3}^{\left(0\right)}\left(d{u}_{1}/d{x}_{3}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}_{3}=0\\ -{P}_{3}^{\left(L\right)}\left(d{u}_{1}/d{x}_{3}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}_{3}=L\end{array}$

4.      In addition, the axial force N must satisfy

$N=-{P}_{3}^{\left(0\right)}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left({x}_{3}=0\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}N={P}_{3}^{\left(L\right)}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left({x}_{3}=L\right)$