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Plane Beam Approximations: Beams Under Axial Loading

Learning Outcomes

  • Reproduce the derivation of the equilibrium equation of a beam under axial loading.
  • Describe the three basic assumptions for the equilibrium equation of beam under axial load.
  • Compute the normal force, the stress distribution, and the strain distribution in a beam by solving the differential equation of equilibrium.

The Euler Bernoulli and the Timoshenko beam formulations described account only for the deformations due to bending, without considering any axial deformation in the neutral axis of the beam. It is possible to augment the Euler Bernoulli and the Timoshenko beams so that they also account for axial deformations of the neutral axis, but because of the assumption of small deformations, the resulting equations of axial-loading deformations and bending deformations are uncoupled. i.e., the axial loading is only related to the axial deformation, and the lateral loading is only related to the lateral deformation. This uncoupling is only valid for small deformations. However, for large deformations, and normal force in an axially loaded beam can affect the lateral deformation as well. For example, phenomena such as buckling can only be modelled when the interaction between the axial loads and the lateral deformations are coupled.

Deformation Assumption

Under an axial load, a beam is assumed to deform such that the horizontal displacement is a smooth function u_1=u_1(X_1) and that cross sections remain plane during the deformation (Figure 7). The effect of Poisson’s ratio is usually neglected. Thus, the position vector function is:

    \[x=\left(\begin{array}{c}X_1+u_1\\X_2\\X_3\end{array}\right)\]

In this case, the displacement function is:

    \[u=x-X=\left(\begin{array}{c}u_1\\0\\0\end{array}\right)\]

Since u_1 is a function of only X_1, the only nonzero component of the strain is \varepsilon_{11}:

    \[\varepsilon=\left(\begin{matrix}\frac{\mathrm{d}u_1}{\mathrm{d}X_1}&0&0\\0&0&0\\0&0&0\end{matrix}\right)\]

On any cross section, X_1 is constant and therefore the values of u_1 and its derivatives are constant. Therefore, \varepsilon_{11} is constant on any cross section perpendicular to X_1.

BeamDeformationNormal
Figure 7. Deformation assumption of beams under axial loading. Cross section perpendicular to the neutral axis move horizontally in the direction of X_1 a distance u_1(X_1).

Equilibrium Equation

The beam is assumed to be under a distributed normal load of value p of units of force/unit length. Figure 8 shows the applied distributed load along with the equilibrium of a beam slide of length \Delta X_1. The equilibrium equation can be written in terms of the normal force N which is the integral of the stress component \sigma_{11} over the cross sectional area:

    \[N=\int\int \! \sigma_{11}\,\mathrm{d}X_2\mathrm{d}X_3\]

Since \sigma_{11} is constant we have:

    \[N=\sigma_{11}A\]

where A is the cross sectional area. Equilibrium can be written by considering the three forces acting on the slice in Figure 8. The first force in N+\Delta N acting on the right cross section and directed to the right. The second force is N acting on the left cross section and directed to the left. The third is due to the distributed load p. Therefore:

    \[\begin{split}F_1&=N+\Delta N = \sigma_{11}A+\frac{\partial N}{\partial X_1}\Delta X_1=\sigma_{11}A+\frac{\partial (\sigma_{11}A)}{\partial X_1}\Delta X_1\\F_2& =N=\sigma_{11}A\\F_3&=p\Delta X_1\end{split}\]

Equilibrium dictates:

    \[F_1-F_2+F_3=0\]

Therefore:

(11)   \begin{equation*}\frac{\partial (\sigma_{11}A)}{\partial X_1}+p=0\end{equation*}

Assuming a linear elastic beam, the differential equation can be written in terms of the displacement by converting the stresses into strains:

    \[\sigma_{11}=E\varepsilon_{11}=E\frac{\mathrm{d}u_1}{\mathrm{d}X_1}\]

Therefore, Equation 11 can be written as follows:

(12)   \begin{equation*}\frac{\mathrm{d} \left(EA \frac{\mathrm{d}u_1}{\mathrm{d}X_1}\right)}{\mathrm{d} X_1}+p=0\end{equation*}

Or equivalently, if E is constant:

    \[E\frac{\mathrm{d} A}{\mathrm{d} X_1}\frac{\mathrm{d}u_1}{\mathrm{d}X_1}+EA\frac{\mathrm{d}^2u_1}{\mathrm{d}X_1^2} +p=0\]

In the case when the cross sectional area A is constant, the equilibrium equation take the simple form:

    \[EA\frac{\mathrm{d}^2u_1}{\mathrm{d}X_1^2} +p=0\]

BeamEquilibriumNormal
Figure 8. The beam is assumed to be under a distributed normal load p of units of load/distance. The equilibrium of a slice of the beam is used to write the equation of equilibrium.

Video

Quiz 26 – Beams Under Axial Loading

Solution Guide

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