We must keep in mind that, although a rotor is statically balanced, it may have a dynamic imbalance. Dynamic unbalance occurs only when the rotor is rotating.

In the figure we see a rotor where we assume that 2 equal masses are placed at opposite ends. The masses exert a force so that the rotor remains statically balanced.

But, when the rotor is rotating, the inertial forces due to the masses give rise to a couple of forces that causes a dynamic imbalance.

Therefore, we can conclude that static balancing is not enough to balance a rotor and therefore we must perform dynamic balancing, that is, in two planes.

To determine both the magnitude and the distance to which we owe corrections in a dynamically unbalanced rotating axis we will use the vector equations for mechanical systems in equilibrium:.

(equation 1).

(equation 2).

The forces that will appear in the first equation will be inertial forces (centrifugal forces), which will act in a radial direction (from the center of rotation towards each of the masses). These forces are proportional to the mass, to the eccentricity of each of these masses (R) and to the angular velocity squared (ω²):.

Fcen,i = e Ri ω².

The moments that will appear in the second equation are those generated by the forces of inertia.

Since the angular velocity is the same for all inertial forces, we will scale the force and moment polygons with the factor 1/ω² to facilitate calculations (the angular velocity term will not appear).

The steps to follow for the example in the figure are:.

1.- Sum the moments (equation 2) of the centrifugal forces with respect to point A or point B (points of intersection of the balance planes with the axis) to prevent the forces from appearing in bearing A and in bearing B respectively. In this case it is done with respect to point A.

2.- With this we build the moment polygon (figure c). The vectors that constitute this polygon have the direction of the position vectors Ri in figure a. These vectors would have to be multiplied by w² and rotated 90° so that they really were the moments of the forces of inertia. The polygon closure vector (in blue) directly provides us with the value that dynamically balances the system. Thus, once the imbalance is known, we can compensate the system by choosing one of the two magnitudes and obtaining the other from the value of the calculated imbalance.

3.-Given the balance between the existing forces (equation 1), we construct the force polygon (figure d) and solve for the correction that we must apply in the second plane (correction plane D).

El desequilibrio se ha medido, por costumbre, en onza-pulgada, gramo-centímetro o gramo-pulgada. Pero, si usamos el sistema internacional, la unidad más apropiada es el miligramo-metro.

El equilibrado es el hecho de determinar y corregir el desequilibrio. Como hemos visto anteriormente, cuando la masa se encuentra en un solo plano de rotación, ruedas, discos, etc.; basta con el equilibrado estático. Sin embargo, en rotores las fuerzas centrífugas que aparecen por el desequilibrio dan lugar a un par de fuerzas. El objetivo del equilibrado es medir ese par y realizar otro de la misma magnitud pero sentido contrario. Esto se puede hacer añadiendo o eliminando masas en dos planos cualesquiera llamados planos de corrección.

Generalmente, los rotores suelen estar desequilibrados tanto estática como dinámicamente. Por tanto, para equilibrarlos necesitaremos conocer tanto la cantidad de masa como su ubicación en cada uno de los planos de corrección.

Para medir estas dos magnitudes y corregir el desequilibrio pueden utilizarse tres métodos de uso general: bastidor basculante, punto nodal y compensación mecánica.

Tilting frame

In the figure we can see a rotor to be balanced mounted on support rollers that are attached to a tilting frame.

We choose the correction planes that will coincide with the pivots. The two pivots will never be working. First a pivot is released and the rotor is rotated. The magnitude and angle of location of the correction are measured. Then the same thing is done but releasing the other pivot, since the measurements are not affected by the moments in the plane of the fixed pivot. Therefore, the balances measured with the fixed right pivot will be corrected in the left correction plane and vice versa.

nodal point

This method consists of finding the zero vibration point.

To do this, we place the rotor to be balanced on bearings on a support known as a nodal bar. We assume that the axis is balanced in the left correction plane but there is an imbalance in the right plane.

If the rotor is rotated, the assembly vibrates and the nodal bar rotates around some point. To know what that point is, we slide a dial indicator over the nodal bar and see when the movement is zero. That point will be the nodal or null point.

We must remember that we assumed at the beginning that there was no imbalance in the plane of correction on the left. Therefore, if the magnitude of the imbalance existed, it would be marked by the dial gauge located at the nodal point calculated above regardless of the imbalance that existed in the plane on the right.

Mechanical compensation

This method will be used to ensure that a shaft rotates smoothly, without vibrations due to imbalances. Additionally, the rotor will rotate smoothly at all rotation speeds. The rotor can be driven with a belt, a universal joint, or it can be self-driven if it is, for example, a motor.

To find the magnitude and direction of the forces that create the imbalance, we attach two masses (m) to the rotor that rotate in solidarity with it. These masses may be distanced by an angle β/2 each with respect to the common axis.

Through two controls we will obtain the magnitude and angular offset of the imbalance M:.

• Magnitude control:

By varying the angle β we will obtain the magnitude of the imbalance. Note that if β=0º the force created by the compensating masses is maximum, while if β=180° both counteract each other and the only force left in the rotor is that of unbalance.

• Location control:

The angular position of the compensating weights with respect to the imbalance, given by α, will allow us to find the direction in which the decompensation acts, that is, the angular offset of the imbalance.

Even if a rotor is balanced in the workshop where it is manufactured, transportation or use itself generates new imbalances, which makes it necessary to rebalance it. Failure to balance it would lead to undesirable effects such as premature wear of the bearings, breakages and fractures due to fatigue, deformation of the axles and could even pose a danger to the operators in the event of complete failure of the machine.

To do this we will use “in situ” balancing, using a programmable calculator and complex algebra (methods developed by Rathbone and Thearle). This consists of balancing a single plane at a time, doing it two or three times for each plane to avoid cross-effects and interference in the results.

If we use bold letters to represent complex numbers:

R = ( R , θ ).

After locating the imbalances in the two reference planes (MI = (MI, φI) and MD = (MD, φD)) we will measure the amplitudes and phase shifts of the disturbances they generate in bearings A and B with commercial “in situ” balancing equipment that we will designate by the letter X=(X, q). For this, the three-race method (three different tests) is used:

First test: the disturbances XA=(XA, θA) and XB=(XB, θB) are measured in the bearings A and B due to the original imbalances MI = (MI , θA) and MD = (MD, θB).

Second test: a test mass is introduced to the left correction plane () and the imbalances are measured ().

Third test: the previous test mass is removed and the second test is repeated by placing a test mass on the right correction plane (with I will obtain ).

If we define new magnitudes that are also complex (A and B) which we will call complex stiffnesses and which reflect the contribution of the test masses to the forces created in the bearings:

Doing the same for the second test mass (third run), we can solve for the four complex rigidities and substituting in the following equations we will obtain the magnitude and position of the imbalances (remember that they are complex variables):

• - Engine balancing.

• - Vibration.

• - Force of inertia.

• - Material fatigue.

• - Degree of balancing and tolerances Archived March 26, 2008 at the Wayback Machine.

• - Maintenance against imbalances and vibrations (companies).

• - Maintenance against imbalances and vibrations (machines).

Contenido

Para ver si un disco está en equilibrio estático, se pueden hacer unos sencillos experimentos:

Se suponen un disco y un eje, apoyado este último en unos rieles rígidos, de manera que el eje pueda rodar sin ningún tipo de rozamiento. Se establece un sistema de referencia fijo en el disco que gire solidario con él.

Pasos del experimento:.

• - Se empuja el disco con la mano y se deja rodar libremente el sistema disco-eje, hasta que se pare y entonces se marca con un lápiz el punto más bajo de la pieza.

• - Repetir esto 4 o 5 veces.

• - Ahora se analizan las marcas que hemos hecho:

• Si éstas están desperdigadas en distintos puntos por el contorno del disco y no coinciden, el disco estará equilibrado estáticamente.

• Si en cambio están todas en el mismo punto, es decir, si coinciden, podremos decir que el disco está estáticamente desequilibrado. Esto significa que el centro de masas del disco y el eje no coinciden.

La posición de las marcas con respecto a los ejes x e y indica la localización angular del desequilibrio, pero no la magnitud. No es probable que las marcas queden unas a 180° de otras.

El desequilibrio se puede corregir eliminando material en los puntos donde hemos hecho las marcas o si se prefiere añadiendo material a 180° de ellas. Como no se conoce la magnitud del desequilibrio las correcciones deberán hacerse tanteando.

Si queremos precisar la corrección que hay que introducir, podemos añadir una masa de prueba m:.

Al añadir esta masa de prueba m (conocida), el disco girará un ángulo φ y luego se detendrá otra vez. Ese ángulo será fácil de determinar.

Las dos masas (la de prueba y la del centro de masas del disco) provocarán una fuerza cada una (el peso de cada una de ellas) que a la vez harán que haya dos momentos. Para calcular el desequilibrio plantearemos el equilibrio de momentos como se puede ver en la figura.

(ecuación 1).

(ecuación 1.1).

(ecuación 2) donde es el Desequilibrio.

Para equilibrar el sistema habrá que colocar una masa en el punto A', es decir, a 180° de la marca hecha.

The equation of motion

When mounting an unbalanced shaft-disc system on bearings A and B, if these bearings are rotated, a centrifugal force will appear. The centrifugal force will act on the shaft and cause rotating reactions in the bearings A and B, as can be seen in the figure. To describe the observations, the following notation is introduced:

Assuming a reference system XYZ and taking any coordinate of the X axis in any direction normal to the axis of rotation, we apply the balance of forces at point O:.

By solving this differential equation we will obtain the vibration movement of the point or axis:

with phase angle which is the angle between the centrifugal force and the amplitude If in the equation of the amplitude of The value that makes it zero is called natural angular velocity, critical velocity or natural circular frequency:.

For the value of it has been studied that no vibrations occur, except for a damped displacement that tends to zero. This value is the critical damping.

The damping ratio is defined as the ratio between the real and critical damping.

(equation 6).

For almost all mechanical systems, if damping has not been intentionally introduced, its value will be within the following range:.

If we call X the amplitude of the cosine we will have:

(equation 7).

Then we can express the equation of motion of the point or axis in the following way:

(equation 8).

Dividing the numerator and denominator of the amplitude

which gives us the ratio of vibration amplitudes of a rotating disc-shaft system.

Without considering the damping (that is, ) the unbalance masses and the total masses are the same, and if we also substitute , we will get:.

where X is the amplitude of any frequency ratio.

If we plot the amplitude against the frequency ratio, interesting deductions can be obtained, such as amplitude and phase relationships. When the system is started, the vibration amplitude is very small and as the shaft speed increases, the amplitude also increases, becoming infinite at the critical speed. This is what we call resonance. When the shaft passes the critical speed, the amplitude changes to a negative value and will decrease as the shaft speed continues to increase. The amplitude of the movement will reach a limit value of , in which case the disc will rotate around its center of gravity (which will coincide with the line of the axis). We can then conclude after having seen all this that when a rotating system is statically unbalanced, it will produce vibrations and unwanted rotational reactions in the bearings.

Static balancing machines

A static balancing machine is used to see, as its name indicates, whether a piece is statically balanced or not, and if it is not, calculate the magnitude and location of the imbalance, that is, it is used to measure the imbalance.

These machines are only used to calculate imbalances of parts whose axial dimensions are very small, such as: gears, cams, pulleys, wheels, fans, flywheels, impellers... Sometimes, it can be considered that the mass of the parts is concentrated in a single plane, therefore these machines are usually called single-plane balancing machines.

When mounting more than one wheel on an axle, each wheel must first be statically balanced individually, and after mounting them the entire assembly can be balanced.

In practice, the process of static balancing of a disc is a weighing process. There are two balancing methods, depending on the type of force applied to the piece. The force may be gravity or centrifugal.

In the example seen previously of the disc-axle assembly, the force that was used to find the imbalance was that of gravity. Another way to do it would be to spin the disc at a certain speed. Thus, the reactions in the bearings would be measured, thus using their magnitudes to calculate the magnitude of the imbalance. A stroboscope is used to give the location of the correction, since the piece rotates when measurements are made.

When manufacturing large quantities of parts, what is needed is a quick way to measure the imbalance and tell what the correction is. If you avoid rotating the piece, you save time, so the most used method in these cases would be to apply a force of gravity.

A pendulum supported on a pivot is usually used, on which the piece is placed. Damping is used to prevent swinging of the pendulum. This will be inclined at an angle and will logically go down in the radial direction in which the imbalance is found, as we can see in the figure. Then the direction of the inclination will give us the location of the imbalance and the angle the magnitude.

To make unbalance measurements correctly, a universal level like the one in the figure is mounted on the platform of the unbalance machine.

A bubble placed in the center moves with the imbalance and will show the location and magnitude of the imbalance's correction.

We must keep in mind that, although a rotor is statically balanced, it may have a dynamic imbalance. Dynamic unbalance occurs only when the rotor is rotating.

In the figure we see a rotor where we assume that 2 equal masses are placed at opposite ends. The masses exert a force so that the rotor remains statically balanced.

But, when the rotor is rotating, the inertial forces due to the masses give rise to a couple of forces that causes a dynamic imbalance.

Therefore, we can conclude that static balancing is not enough to balance a rotor and therefore we must perform dynamic balancing, that is, in two planes.

To determine both the magnitude and the distance to which we owe corrections in a dynamically unbalanced rotating axis we will use the vector equations for mechanical systems in equilibrium:.

(equation 1).

(equation 2).

The forces that will appear in the first equation will be inertial forces (centrifugal forces), which will act in a radial direction (from the center of rotation towards each of the masses). These forces are proportional to the mass, to the eccentricity of each of these masses (R) and to the angular velocity squared (ω²):.

Fcen,i = e Ri ω².

The moments that will appear in the second equation are those generated by the forces of inertia.

Since the angular velocity is the same for all inertial forces, we will scale the force and moment polygons with the factor 1/ω² to facilitate calculations (the angular velocity term will not appear).

The steps to follow for the example in the figure are:.

1.- Sum the moments (equation 2) of the centrifugal forces with respect to point A or point B (points of intersection of the balance planes with the axis) to prevent the forces from appearing in bearing A and in bearing B respectively. In this case it is done with respect to point A.

2.- With this we build the moment polygon (figure c). The vectors that constitute this polygon have the direction of the position vectors Ri in figure a. These vectors would have to be multiplied by w² and rotated 90° so that they really were the moments of the forces of inertia. The polygon closure vector (in blue) directly provides us with the value that dynamically balances the system. Thus, once the imbalance is known, we can compensate the system by choosing one of the two magnitudes and obtaining the other from the value of the calculated imbalance.

3.-Given the balance between the existing forces (equation 1), we construct the force polygon (figure d) and solve for the correction that we must apply in the second plane (correction plane D).

El desequilibrio se ha medido, por costumbre, en onza-pulgada, gramo-centímetro o gramo-pulgada. Pero, si usamos el sistema internacional, la unidad más apropiada es el miligramo-metro.

El equilibrado es el hecho de determinar y corregir el desequilibrio. Como hemos visto anteriormente, cuando la masa se encuentra en un solo plano de rotación, ruedas, discos, etc.; basta con el equilibrado estático. Sin embargo, en rotores las fuerzas centrífugas que aparecen por el desequilibrio dan lugar a un par de fuerzas. El objetivo del equilibrado es medir ese par y realizar otro de la misma magnitud pero sentido contrario. Esto se puede hacer añadiendo o eliminando masas en dos planos cualesquiera llamados planos de corrección.

Generalmente, los rotores suelen estar desequilibrados tanto estática como dinámicamente. Por tanto, para equilibrarlos necesitaremos conocer tanto la cantidad de masa como su ubicación en cada uno de los planos de corrección.

Para medir estas dos magnitudes y corregir el desequilibrio pueden utilizarse tres métodos de uso general: bastidor basculante, punto nodal y compensación mecánica.

Tilting frame

In the figure we can see a rotor to be balanced mounted on support rollers that are attached to a tilting frame.

We choose the correction planes that will coincide with the pivots. The two pivots will never be working. First a pivot is released and the rotor is rotated. The magnitude and angle of location of the correction are measured. Then the same thing is done but releasing the other pivot, since the measurements are not affected by the moments in the plane of the fixed pivot. Therefore, the balances measured with the fixed right pivot will be corrected in the left correction plane and vice versa.

nodal point

This method consists of finding the zero vibration point.

To do this, we place the rotor to be balanced on bearings on a support known as a nodal bar. We assume that the axis is balanced in the left correction plane but there is an imbalance in the right plane.

If the rotor is rotated, the assembly vibrates and the nodal bar rotates around some point. To know what that point is, we slide a dial indicator over the nodal bar and see when the movement is zero. That point will be the nodal or null point.

We must remember that we assumed at the beginning that there was no imbalance in the plane of correction on the left. Therefore, if the magnitude of the imbalance existed, it would be marked by the dial gauge located at the nodal point calculated above regardless of the imbalance that existed in the plane on the right.

Mechanical compensation

This method will be used to ensure that a shaft rotates smoothly, without vibrations due to imbalances. Additionally, the rotor will rotate smoothly at all rotation speeds. The rotor can be driven with a belt, a universal joint, or it can be self-driven if it is, for example, a motor.

To find the magnitude and direction of the forces that create the imbalance, we attach two masses (m) to the rotor that rotate in solidarity with it. These masses may be distanced by an angle β/2 each with respect to the common axis.

Through two controls we will obtain the magnitude and angular offset of the imbalance M:.

• Magnitude control:

By varying the angle β we will obtain the magnitude of the imbalance. Note that if β=0º the force created by the compensating masses is maximum, while if β=180° both counteract each other and the only force left in the rotor is that of unbalance.

• Location control:

The angular position of the compensating weights with respect to the imbalance, given by α, will allow us to find the direction in which the decompensation acts, that is, the angular offset of the imbalance.

Even if a rotor is balanced in the workshop where it is manufactured, transportation or use itself generates new imbalances, which makes it necessary to rebalance it. Failure to balance it would lead to undesirable effects such as premature wear of the bearings, breakages and fractures due to fatigue, deformation of the axles and could even pose a danger to the operators in the event of complete failure of the machine.

To do this we will use “in situ” balancing, using a programmable calculator and complex algebra (methods developed by Rathbone and Thearle). This consists of balancing a single plane at a time, doing it two or three times for each plane to avoid cross-effects and interference in the results.

If we use bold letters to represent complex numbers:

R = ( R , θ ).

After locating the imbalances in the two reference planes (MI = (MI, φI) and MD = (MD, φD)) we will measure the amplitudes and phase shifts of the disturbances they generate in bearings A and B with commercial “in situ” balancing equipment that we will designate by the letter X=(X, q). For this, the three-race method (three different tests) is used:

First test: the disturbances XA=(XA, θA) and XB=(XB, θB) are measured in the bearings A and B due to the original imbalances MI = (MI , θA) and MD = (MD, θB).

Second test: a test mass is introduced to the left correction plane () and the imbalances are measured ().

Third test: the previous test mass is removed and the second test is repeated by placing a test mass on the right correction plane (with I will obtain ).

If we define new magnitudes that are also complex (A and B) which we will call complex stiffnesses and which reflect the contribution of the test masses to the forces created in the bearings:

Doing the same for the second test mass (third run), we can solve for the four complex rigidities and substituting in the following equations we will obtain the magnitude and position of the imbalances (remember that they are complex variables):

• - Engine balancing.

• - Vibration.

• - Force of inertia.

• - Material fatigue.

• - Degree of balancing and tolerances Archived March 26, 2008 at the Wayback Machine.

• - Maintenance against imbalances and vibrations (companies).

• - Maintenance against imbalances and vibrations (machines).