**The ****tragic accident involving a construction crane**** highlights the importance of the stability of technical installations.**

In this article, we aim to introduce you to the fundamentals of stability and explain how to calculate the stability factor using a simple example. The simplified calculation formula presented here is often found in the foundational studies at many universities, where it is applied to almost all types of machinery.

If you want to go beyond the theoretical consideration of stability, for instance, to produce and market a machine, this form of stability verification is not sufficient. In Europe, machines must meet the requirements of the Machinery Directive or a corresponding harmonized standard.

An example of such a standard is DIN EN 280, which provides specific guidelines for the stability verification of a mobile elevating work platform — for instance, for applicable wind loads, payloads and dynamic forces. To develop a machine in compliance with standards, these and other requirements must be considered and verified in the calculations.

## Torque, stability moment and overturning moment: Basic terms for assessing stability

Before we can explain how to assess the stability of a machine, it’s essential to understand the terms torque, stability moment and overturning moment.

### Torque

A torque is the product of force and lever arm. This means: the greater the force or the longer the lever arm, the greater the torque acting on the component.

The relationship between force, lever arm and torque can be easily illustrated with a simple example: loosening a tight screw with a wrench. If the screw is stuck, there are two ways to increase the torque:

**Increase the force:**Press harder on the wrench to loosen the screw.**Extend the lever arm:**Use a longer wrench or attach a pipe as an extension to the wrench.

### Stability moment

The stability moment is the sum of all moments that work to prevent a machine from tipping over. These moments are generated, for example, by counterweights or the machine’s own weight.

Whether the machine’s own weight contributes to the stability moments or the overturning moments depends on the location of the tipping edge and the center of gravity of the machine — but more on that later.

### Overturning moment

The overturning moment is the sum of all moments that tend to tip the machine over. These moments are generated, for example, by wind forces or by loads extending beyond the tipping edge.

## Calculating the stability

A machine is stable whenever the stability moments are greater than the overturning moments. Therefore, to assess the stability of a machine, all stabilizing and destabilizing forces must be taken into account.

To illustrate this, we will explain the determination of stability using a simple cube as an example.

### Cube on a flat surface

In this example, we consider a cube with an edge length of one meter and a mass of ten kilograms. The cube is subjected to a wind pressure of 70 N/m². We assume that the cube has a uniform density, so its center of gravity is exactly at the geometric center.

The following sketch visually illustrates the initial situation:

#### Weight force

To assess stability, we need to determine both the acting forces and their lever arms. Therefore, we first identify where the forces apply and their magnitudes.

Since the cube has uniform (homogeneous) density, its center of gravity is exactly at its geometric center. The weight force always acts at the center of gravity of an object and is directed towards the center of the Earth. This gives us the first point of application. Now we need to calculate the magnitude of the weight force, which can be determined 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:

#### 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.

#### 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 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.

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