# Calculating the stability of a machine

The tragic accident with a construction crane in the middle of December shows how important the stability of technical plants is. In this article we would like to show you a simple example of how to calculate the stability of a machine.

This article deals with the basics of stability and is intended to explain the calculation of the stability factor using a simple as possible example. The simplified calculation method is often found in the core curriculum of many universities, where it is used for almost all types of machines.

If you would like to go beyond the theoretical examination of stability and, for example, produce a machine and place it on the market, this form of proof of stability is not sufficient, as in Europe you must meet the provisions of the Machinery Directive or a corresponding harmonised standard.
DIN EN 280 provides, for example, precise information for the proof of stability of a mobile aerial work platform. Among other things, specifications are given for wind loads to be applied, payloads and dynamic forces to be taken into account.

## Stability and overturning moment

Before we can explain how the stability of a machine is assessed, the terms stability and overturning moment must be defined.

### Torque

Torque is the product of force and lever arm. The greater the force or the longer the lever arm, the greater the torque acting on the component.
This correlation can easily be seen when you want to loosen a bolt with a wrench. If the bolt sits tight, you have two possibilities:

1. You have to press harder, i.e. you increase the force, or
2. You extend the lever arm by using a longer wrench or by placing a pipe over the wrench.

### Stability moment

We define the stability moment as the sum of all moments that are intended to prevent the machine from tipping over. These moments can be generated by the counterweight or the dead weight of the machine, for example. Whether the dead weight contributes to the overturning or stability moments depends on the position of the tilting edge and the centre of gravity – but more about this later.

### Overturning moment

We define the overturning moment as the sum of all moments that cause the machine to tip over. These moments are generated, for example, by the wind or by loads that project beyond the tilting edge.

## Calculating the stability

A machine always stands securely if the stability moments are greater than the overturning moments. If you want to assess the stability of a machine, you must take all forces that stabilise the machine and those that destabilise it into account.
To keep it as simple as possible, we’ll use a simple cube to explain how stability is determined.

### Cube on a straight surface

The cube has an edge of one metre in length, a mass of ten kilograms and is exposed to wind pressure of 70 N/m². We assume that the density of the cube is homogeneous and that the centre of gravity is therefore exactly in the centre of the cube.

The sketch below represents the initial situation:

#### Weight force

As explained at the beginning, however, we need forces and lever arms to assess the stability and not weight specifications. Therefore we must first calculate how great the forces are and where they act.

Because the density of the cube is the same (homogeneous) everywhere, the centre of gravity is exactly in the centre. The weight force always acts on the centre of gravity of a body and acts in the direction of the centre of the earth. So the first point of action is known, but the amount of the weight force is still missing. This can be calculated using the following formula:

#### Wind force

Once the weight force is known, the next question is where the wind force is applied and how strong it is. In the sketch above, one can clearly see that the wind presses evenly against the left side of the cube. Instead of the surface load caused by the wind, we can therefore also assume a large force that presses exactly against the centre of the left surface. The point of action of the force is therefore known.

The amount of force can easily be calculated using the dimensions of the cube and the given wind pressure. The windage surface of the cube is:

So the wind force is:

The sketch now looks like this:

#### From force to torque

Now that we know the weight force and the wind force, the question is which lever arms are relevant. For this, we need to know where the tilting edge of the machine being examined, or in this case of the cube, is.
This question can already be answered by simply looking at it. In our example, what force will try to tip the cube over? Exactly, the wind force. If the wind force now presses against the cube from the left, over which edge will the cube tip over in the worst case? Exactly, over the right-hand one.

Forces may be displaced along their line of action. The lever arm is the shortest distance between the line of action of a force and the tilting edge. As chance would have it, the lever arm of each force is exactly half an edge length.

The torques can therefore be calculated as follows:

#### Is there a risk of tipping over?

As mentioned before, the wind tries to tip the cube over. The weight force, however, presses the cube against the ground and so counteracts the wind force.
The weight force consequently increases the stability moment, while the wind force contributes to the overturning moment. As there are no other forces and moments present, only the wind force and the weight force determine the fate of the cube in this example. The moment of the weight force and thus the moment of stability is greater than the overturning moment from the wind force. So there is no risk of the cube tipping over.

### Cube on a ramp

I am using the second example to illustrate the influence that the operating conditions of a machine has on its stability. The cube from the first example is no longer on a level surface, but on a 30-degree ramp. In everyday life this situation is found relatively often, e.g. when a vehicle is parked on an slope.

We assume that the forces and points of action have not changed and are the same as in the first example. This presents us with the following situation:

Once the points of action of the forces are known, the tilting edge and the existing lever arms must be determined again. By drawing the lines of action of the forces in the sketch, it quickly becomes clear why a ramp or incline increases the risk of tipping over.

While the lever arm of the wind force has not changed, the lever arm of the weight force has become significantly smaller. We know that the weight force has so far prevented the cube from tipping over. As the torque is proportional to the lever arm, it can already be said that the cube is exposed to a much higher risk of tipping over on the slope.

The calculations show that the following torques exist in this case:

As the overturning moment is greater than the moment of stability, the cube would tip over the right edge in this case. In the case of a cube, this is certainly not tragic, but if it is a vehicle, the chances are good that nobody would be happy.

As the ramp becomes steeper, this effect is further intensified. A special case is when the line of action of the weight force is not on the left of the tilting edge, but on its right. Whereas the weight of the cube has always helped to prevent tipping over up to this point, it would even favour the overturning of the cube from this point on. The wind force and the weight force now generate torques that both try to tip the cube over the critical edge.

## The stability factor

The statement that "the stability moment is greater than the overturning moment" is unsatisfactory in engineering. You do not only want to know whether the risk of tipping over is excluded, but also what reserves the machine has. For this reason, the stability factor has been defined as the basis for assessment:

In the first example the stability factor is about 1.4:

In the second example, the stability factor of 0.51 is less than one.

The following conclusions can be drawn from the stability factor as a parameter:

#### Case 1: Stability factor less than 1

There is a risk of the machine tipping over. At these torques, the machine will tip over.

#### Case 2: Stability factor of 1

In this case, the overturning moment is exactly the same as the stability moment: The machine is "shaky" and it cannot be excluded that it will tip over. This situation is unsatisfactory in practical operating conditions.

#### Case 3: Stability factor greater than 1

There is no risk of this machine tipping over. The greater the stability factor, the less likely the machine is to tip over.

When calculating the stability of machinery, dynamic forces, manual forces and many others must be taken into account in addition to the static forces considered here. The exact determination of the stability differs depending on the type of machine and is described in the applicable standards. For example, detailed specifications are given in DIN EN 280 for mobile aerial work platforms.