Applied Mechanics of Solids (A.F. Bower) Sect 10.3: Simplified theories for Rods and Beams

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 $ρ$ , 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 $u = u_{1} (x_{3}) e_{1}$ .

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

1. The curvature of the string can be approximated as $κ \approx \frac{d^{2} u_{1}}{d x_{3}^{2}} e_{2}$

2. The stretch of the string can be approximated as $\frac{d s}{d x_{3}} = 1 + T_{0} / E A$

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 $κ_{2} T_{3} + p_{1} = ρ A a_{1} \frac{d T_{3}}{d s} = 0$

5. Combining 1-4 shows $T_{3} = T_{0}$ and gives the equation of motion for the stretched string

$T_{0} \frac{d^{2} u_{1}}{d x_{3}^{2}} + p_{1} = ρ A \frac{d^{2} u_{1}}{d t^{2}}$

6. 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 is subjected to transverse forces per unit length $p = p_{1} e_{1} + p_{2} e_{2}$ . The beam has Young’s modulus $E$ and mass density $ρ$ , and its cross-section has area A and area moments of inertia $(I_{11}, I_{12}, I_{22})$ defined as

$\begin{array}{l} I_{11} = \int_{A} {(x_{2} - {\bar{x}}_{2}^{})}^{2} d A \\ I_{22} = \int_{A} {(x_{1} - {\bar{x}}_{1}^{})}^{2} d A \\ I_{12} = \int_{A} (x_{1} - {\bar{x}}_{1}^{}) (x_{2} - {\bar{x}}_{2}^{}) d A \end{array}$

where

$\bar{r} = {\bar{x}}_{1} e_{1} + {\bar{x}}_{2} e_{2} = \frac{1}{A} \int_{A} (x_{1} e_{1} + x_{2} e_{2}) d A$

is the position of the centroid of the cross-section. Its ends may be constrained in various ways, as described in more detail below, but in this section we assume that no axial force or twisting moment is applied to the ends of the bar. We suppose that the beam experiences a small transverse displacement $u = u_{1} e_{1} + u_{2} e_{2}$ , and wish to calculate $u_{1}, u_{2}$ as functions of time, given appropriate initial conditions. Since deflections are small, we can assume that the basis vectors normal and transverse to the deflected beam remain parallel to the initial coordinate directions $m_{i} \approx e_{i}$ .

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

1. The stretch of the bar and the curvature vector may be approximated by

$\frac{d s}{d x_{3}} \approx 1 κ \approx - \frac{d^{2} u_{2}}{d x_{3}^{2}} e_{1} + \frac{d^{2} u_{1}}{d x_{3}^{2}} e_{2}$

2. The internal forces can be characterized by the internal force $T \approx T_{1} e_{1} + T_{2} e_{2}$ and internal moment $M \approx M_{1} e_{1} + M_{2} e_{2}$ . 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

$M = E I κ [\begin{matrix} M_{1} \\ M_{2} \end{matrix}] = E [\begin{matrix} I_{11} & - I_{12} \\ - I_{12} & I_{22} \end{matrix}] [\begin{matrix} κ_{1} \\ κ_{2} \end{matrix}]$

4. Equations of motion can be reduced to

$\frac{d T_{1}}{d x_{3}} + p_{1} \approx ρ A a_{1} \frac{d T_{2}}{d x_{3}} + p_{2} \approx ρ A a_{2}$

$\frac{d M_{1}}{d x_{3}} - T_{2} \approx 0 \frac{d M_{2}}{d x_{3}} + T_{1} \approx 0$

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

$\begin{array}{l} E (I_{22} \frac{d^{4} u_{1}}{d x_{3}^{4}} + I_{12} \frac{d^{4} u_{2}}{d x_{3}^{4}}) + ρ A \frac{d^{2} u_{1}}{d t^{2}} = p_{1} \\ E (I_{12} \frac{d^{4} u_{1}}{d x_{3}^{4}} + I_{11} \frac{d^{4} u_{2}}{d x_{3}^{4}}) + ρ A \frac{d^{2} u_{2}}{d t^{2}} = p_{2} \end{array}$

6. The stresses and strains in the beam are related to its curvature by

$ε_{33} = - κ_{2} (x_{1} - {\bar{x}}_{1}) + κ_{1} (x_{2} - {\bar{x}}_{2})$

$σ_{33} = - E κ_{2} (x_{1} - {\bar{x}}_{1}) + E κ_{1} (x_{2} - {\bar{x}}_{2})$

where $({\bar{x}}_{1}, {\bar{x}}_{2})$ are the coordinates of the centroid of the cross-section.

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

1. The ends of beam may be subjected to prescribed displacements $u (0) = u^{*} u (L) = u^{*}$ or prescribed forces $T = - P x_{3} = 0 T = P x_{3} = L$ (we assume here that $P_{3} = 0$ on both ends of the bar). The two transverse forces are related to the displacements by

$\begin{array}{l} - E (I_{22} \frac{d^{3} u_{1}}{d x_{3}^{3}} + I_{12} \frac{d^{3} u_{2}}{d x_{3}^{3}}) = T_{1} \\ - E (I_{12} \frac{d^{3} u_{1}}{d x_{3}^{4}} + I_{11} \frac{d^{3} u_{2}}{d x_{3}^{4}}) = T_{2} \end{array}$

2. The ends of the beam may be subjected to prescribed rotations $d u_{1} / d x_{3} = θ_{2}^{*} - d u_{2} / d x_{3} = θ_{1}^{*}$ or prescribed moments $M = - Q (x_{3} = 0) M = Q (x_{3} = L)$ . We assume here that $Q_{3} = 0$ on both ends of the bar. The moment is related to the displacement by

$M_{1} = - E (I_{11} \frac{d^{2} u_{2}}{d x_{3}^{2}} + I_{12} \frac{d^{2} u_{1}}{d x_{3}^{2}}) M_{2} = E (I_{12} \frac{d^{2} u_{2}}{d x_{3}^{2}} + I_{22} \frac{d^{2} u_{1}}{d x_{3}^{2}})$

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 an initially straight beam. The theory can also be used to calculate buckling loads for beams, but does not accurately model their deflection 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 $p = p_{1}^{(e)} e_{1} + p_{2}^{(e)} e_{2} + p_{3}^{(e)} e_{3}$ . The superscript on the components of force has been introduced to clarify that the forces do not change their direction with the rotation of the beam. The beam has Young’s modulus $E$ and mass density $ρ$ , and its cross-section has area A and moments of inertia $(I_{11}, I_{12}, I_{22})$ defined in Section 10.3.2. Its ends may be constrained in various ways, as described below. We assume that a large axial internal force $T_{3} m_{3}$ is developed in the beam, either by a horizontal force per unit length $p_{3}^{(e)}$ or horizontal forces $P_{3}^{(e)} e_{3}$ acting at the ends of the beam. Twisting of the beam is neglected. We suppose that the beam experiences a small displacement $u = u_{1} m_{1} + u_{2} m_{2} + u_{3} m_{3}$ , which we wish to calculate as a function of time, given appropriate initial conditions. In the analysis to follow we assume that the axial internal and external forces $p_{3}^{(e)}$ and $T_{3}$ are large, so that their products with the displacement components and their derivatives must be included in the equations of motion. The transverse forces, displacements and accelerations are small, so their products are neglected.

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

1. The stretch of the bar and the curvature vector are

$\frac{d s}{d x_{3}} \approx 1 + \frac{d u_{3}}{d x_{3}} κ \approx - \frac{d^{2} u_{2}}{d x_{3}^{2}} m_{1} + \frac{d^{2} u_{1}}{d x_{3}^{2}} m_{2}$

2. The internal forces can be characterized by the internal force $T = T_{1} m_{1} + T_{2} m_{2} + T_{3} m_{3}$ and internal moment $M = M_{1} m_{1} + M_{2} m_{2}$ . 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

$M = E I κ [\begin{matrix} M_{1} \\ M_{2} \end{matrix}] = E [\begin{matrix} I_{11} & - I_{12} \\ - I_{12} & I_{22} \end{matrix}] [\begin{matrix} κ_{1} \\ κ_{2} \end{matrix}]$

4. Equations of motion may be approximated by

$\begin{array}{l} \frac{d T_{1}}{d x_{3}} + T_{3} \frac{d^{2} u_{1}}{d x_{3}^{2}} + p_{1}^{(e)} \approx p_{3}^{(e)} \frac{d u_{1}}{d x_{3}} + ρ A a_{1} \\ \frac{d T_{2}}{d x_{3}} + T_{3} \frac{d^{2} u_{2}}{d x_{3}^{2}} + p_{1}^{(e)} \approx p_{3}^{(e)} \frac{d u_{2}}{d x_{3}} + ρ A a_{2} \end{array}$

$\frac{d T_{3}}{d x_{3}} + p_{3}^{(e)} = ρ A a_{3}$

$\frac{d M_{1}}{d x_{3}} - T_{2} \approx 0 \frac{d M_{2}}{d x_{3}} + T_{1} \approx 0$

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

$\begin{array}{l} E (I_{22} \frac{d^{4} u_{1}}{d x_{3}^{4}} + I_{12} \frac{d^{4} u_{2}}{d x_{3}^{4}}) + p_{3}^{(e)} \frac{d u_{1}}{d x_{3}} + ρ A \frac{d^{2} u_{1}}{d t^{2}} = T_{3} \frac{d^{2} u_{1}}{d x_{3}^{2}} + p_{1}^{(e)} \\ E (I_{12} \frac{d^{4} u_{1}}{d x_{3}^{4}} + I_{11} \frac{d^{4} u_{2}}{d x_{3}^{4}}) + p_{3}^{(e)} \frac{d u_{2}}{d x_{3}} + ρ A \frac{d^{2} u_{2}}{d t^{2}} = T_{3} \frac{d^{2} u_{2}}{d x_{3}^{2}} + p_{2}^{(e)} \end{array}$

Boundary conditions: The boundary conditions for this case reduce to:

1. Prescribed displacements $u (0) = u^{*} u (L) = u^{*}$ or

Prescribed forces

$\begin{array}{l} T_{1} = - P_{1}^{(e)} + P_{3}^{(e)} (d u_{1} / d x_{1}) T_{2} = - P_{1}^{(e)} + P_{3}^{(e)} (d u_{2} / d x_{1}) T_{3} = - P_{3}^{(e)} (x_{3} = 0) \\ T_{1} = P_{1}^{(e)} - P_{3}^{(e)} (d u_{1} / d x_{1}) T_{2} = P_{1}^{(e)} - P_{3}^{(e)} (d u_{2} / d x_{1}) T_{3} = P_{3}^{(e)} (x_{3} = L) \end{array}$

where the two transverse forces are related to the displacements by

2. Prescribed rotations $d u_{1} / d x_{3} = θ_{2}^{*} - d u_{2} / d x_{3} = θ_{1}^{*}$ or

Prescribed moments $M = - Q (x_{3} = 0) M = Q (x_{3} = L)$ where the moment is related to the displacement by

$M_{1} = - E (I_{11} \frac{d^{2} u_{2}}{d x_{3}^{2}} + I_{12} \frac{d^{2} u_{1}}{d x_{3}^{2}}) M_{2} = E (I_{12} \frac{d^{2} u_{2}}{d x_{3}^{2}} + I_{22} \frac{d^{2} u_{1}}{d x_{3}^{2}})$