# Beam Deflection

## Solved Problems

### Example 1.

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

Solution.

We assume that the uniformly distributed force \(q\) acts on the beam of length \(L.\) Modulus of elasticity \(E\) and moment of inertia \(I\) of the beam are known.

The beam deflection equation can be written in the form

The "minus" sign in front of \(q\) shows that the force is directed opposite to the positive direction of the \(y\)-axis, i.e. vertically downwards.

When the beam ends are fixed rigidly, the following boundary conditions are valid:

Repeatedly integrating the differential equation, we find the function \(y\left( x \right):\)

It follows from the conditions \(y\left( {x = 0} \right) = 0\) and \({\frac{{dy}}{{dx}}}\left( {x = 0} \right) = 0\) that \({C_3} = {C_4} = 0.\) Along with two other boundary conditions we obtain the following system of equations with unknowns \({C_1}\) and \({C_2}:\)

Solving it, we find that the coefficients \({C_1}\) and \({C_2}:\)

Hence, the deflection of the beam under a uniformly distributed load \(q\) is described by the function

To determine the sag \(\lambda,\) we investigate extreme points of the function \(f\left( x \right) = {x^2}{\left( {x - L} \right)^2}.\) Find the derivative of this function and equate it to zero:

It follows that this function has an extremum at \(x = \frac{L}{2}.\) At this point the function \(f\left( x \right)\) takes the maximum value

Then the sag \(\lambda\) will be determined by the formula

It turns out that the value of sag \(\lambda\) is proportional to the beam length to the fourth power! Such a strong dependence imposes significant limitations on the design of buildings and constructions.

### 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 \(a.\)

Solution.

When the shaft rotates, the centrifugal force appears which is proportional to the deviation of \(y\) from the axis of rotation. With increasing deformation \(y\) at a point, the centrifugal force at this point will also increase, resulting in a further distortion of the shaft. Such an instability occurs at certain frequencies, and can lead to the destruction of the shaft.

We investigate this problem by using a differential equation. The basic equation describing the deformation of the shaft can be written as

where \(f\) denotes the density of the centrifugal force.

The centrifugal force acting on the element \(dx\) is given by

Here the value \({\frac{M}{L}} dx\) corresponds to the mass of the element \(dx,\) \(y\) is the deflection of the shaft equal to the radius of rotation of the element \(dx.\) As a result, our differential equation becomes:

where we introduced the notation \({\alpha ^4} = {\frac{{{\omega ^2}M}}{{EIL}}}.\)

So, we got a nice fourth-order linear differential equation. Find the roots of the auxiliary equation.

The roots have the following values:

Then the general solution is given by

The coefficients \({C_i}\) are determined from the boundary conditions. Assuming that the shaft rotates in two bearings, the boundary conditions are as follows:

- The shaft deflection at \(x = 0\) and \(x = L\) is equal to zero;
- The curvature of the shaft at \(x = 0\) and \(x = L\) is zero.

Hence,

Find the second derivative of the function \(y\left( x \right):\)

Substituting \(y\) and \(\frac{{{d^2}y}}{{d{x^2}}}\) in the boundary conditions, we obtain

The solution of this system has the form:

The case \({C_4} = 0\) corresponds to the trivial solution \(y = 0.\) In this case, the shaft remains undeformed. A meaningful solution exists for \({C_4} \ne 0.\) Then, the following condition is valid:

Note that \(n\) is greater than zero, as for \(n = 0\) we again obtain the trivial solution \(y = 0.\)

Thus, the rotating shaft starts to bend at \(\alpha L = \pi n\) taking the form of the sine wave:

The minimum critical frequency \({\omega _\text{c}},\) which causes the instability is given by the formula

If the shaft is a solid rod with the cross-sectional radius \(a,\) then its moment of inertia about the central axis is

Substituting \(I\) in the previous formula, we find the minimum critical speed of rotation of the shaft: