Differential Equations

Nth Order Equations

Beam Deflection

Page 1
Theory
Page 2
Problems 1-2

A beam is a constructive element capable of withstanding heavy loads in bending. In the case of small deflections, the beam shape can be described by a fourth-order linear differential equation.

Consider the derivation of this equation. For a bending beam, the angle \(d\theta\) appears between two adjacent sections spaced at a distance \(dx\) (Figure \(1\)).

The deformation \(\varepsilon \) at each point is proportional to the coordinate \(y,\) which is measured from the neutral line. The length of the neutral line is unchanged.

A bending beam

Figure 1.

It follows from the geometry of Figure \(1\) that

\[\varepsilon = \frac{y}{R},\]

where \(R\) is the radius of curvature of the beam.

The magnitude of the normal stress \(\sigma\) in the cross section will also depend on the coordinate \(y.\) It can be estimated by Hooke’s law:

\[\sigma = \varepsilon E = \frac{E}{R}y,\]

where \(E\) is the modulus of elasticity of the beam.

The normal stress sigma in the cross section of a beam

Figure 2.

The bending moment \(M\left( x \right)\) for a section of the beam relative to the \(z\)-axis is given by

\[
{M\left( x \right) = {M_z} }={ \int\limits_A {\sigma ydA} }
= {\frac{E}{R}\int\limits_A {{y^2}dA} }
= {\frac{E}{R}I,}
\]

where \(I\) is the moment of inertia of the cross section with respect to the neutral \(z\)-axis (Figure \(2\)).

Hence, we obtain the following expression for the radius of curvature of the beam:

\[R = \frac{{EI}}{{M\left( x \right)}}.\]

It is known that the radius of curvature is given by

\[R = \frac{{{{\left[ {1 + {{\left( {y’} \right)}^2}} \right]}^{\large\frac{3}{2}\normalsize}}}}{{y^{\prime\prime}}}.\]

Assuming that the deflection of the beam is sufficiently small, we can neglect the first derivative \(y’.\) Then the differential equation of the elastic line can be written as follows:

\[{y^{\prime\prime} = \frac{{M\left( x \right)}}{{EI}}\;\;\text{or}\;\;}\kern-0.3pt{\frac{{{d^2}y}}{{d{x^2}}} = \frac{{M\left( x \right)}}{{EI}}.}\]

The bending moment \({M\left( x \right)}\) can be expressed in terms of the known external load \({q\left( x \right)}\) acting on the beam. Indeed, we choose a small element \(dx\) and consider the conditions of its equilibrium (Figure \(3\text{).}\)

The sum of the projections of all forces on the \(z\)-axis is zero:

\[{ – Q – qdx }+{ Q + dQ }={ 0.}\]
The forces and moments acting on the beam

Figure 3.

The sum of the moments of all forces about, for example, the right edge of the element \(dx\) (the point \(B\) in Figure \(3\)) is also equal to zero:

\[{ – M + M + dM }-{ Qdx – q\frac{{{{\left( {dx} \right)}^2}}}{2} }={ 0.}\]

This implies the relationship

\[{\left\{ \begin{array}{l}
\frac{{dQ}}{{dx}} = q\\
\frac{{dM}}{{dx}} = Q
\end{array} \right.\;\;}\kern-0.3pt{\text{or}\;\;\frac{{{d^2}M}}{{d{x^2}}} = q.}\]

With this expression the differential equation becomes:

\[
{M\left( x \right) = EI\frac{{{d^2}y}}{{d{x^2}}},\;\;}\Rightarrow
{\frac{{{d^2}M}}{{d{x^2}}} }={ \frac{{{d^2}}}{{d{x^2}}}\left( {EI\frac{{{d^2}y}}{{d{x^2}}}} \right) }={ q.}
\]

This equation is called the Euler-Bernoulli differential equation. If the values of \(E\) and \(I\) are constant along the \(x\)-axis, we get a fourth-order equation:

\[EI\frac{{{d^4}y}}{{d{x^4}}} = q.\]

This equation under the appropriate boundary conditions determines the deflection of a loaded beam.

Solved Problems

Click on problem description to see solution.

 Example 1

Determine the deflection of the beam rigidly clamped at both ends and loaded by a uniformly distributed force (Figure \(4\)).

 Example 2

A thin cylindrical shaft of length \(L\) rotates with angular velocity \(\omega.\) At what speed \(\omega\) the shaft is destroyed? The elastic modulus of the material is \(E,\) the mass of the shaft is \(M,\) and the radius of the section is \(a.\)

Page 1
Theory
Page 2
Problems 1-2