# Beam Deflection

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.

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 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 $$3$$).

The sum of the projections of all forces on the $$z$$-axis is zero:

${ – Q – qdx }+{ Q + dQ }={ 0.}$

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 or tap a problem to see the 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 $$a.$$
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Concept
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Problems 1-2