Calculating the stability of a machine

• 1. Torque, stability moment and overturning moment: Basic terms for assessing stability
  • 1.1 Torque
  • 1.2 Stability moment
  • 1.3 Overturning moment
• 2. Calculating the stability
  • 2.1 Cube on a flat surface
  • 2.2 Cube on a ramp
• 3. The stability factor

The the tragic accident involving a construction crane highlights how important the stability of technical equipment is.

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.

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.

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:

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

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.

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.

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.

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:

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:

After calculating the weight force, we now need to determine where the wind force acts and how large it is.
From the sketch shown above, it can be observed that the wind exerts an even pressure on the left side of the cube. Instead of considering the surface load caused by the wind, we can assume a single large force that acts precisely at the center of the left surface. This allows us to identify the point of application of the wind force.

The magnitude of the wind force can be calculated based on the dimensions of the cube and the given wind pressure. The wind impact area of the cube is:

Thus, the wind force is:

The updated sketch now shows the weight force and the wind force:

Now that we know both the weight force and the wind force, the question arises as to which lever arms are relevant. To clarify this, we need to determine where the tipping edge of the machine in question — in this case, the cube — is located.

This question can already be answered by examining the sketch. Which force, in our example, attempts to tip the cube over? Correct, the wind force. If the wind force pushes from the left on the cube, the cube will, in the worst-case scenario, tip over the right edge.

Forces can be shifted along their line of action. The lever arm is the shortest distance between the line of action of a force and the tipping edge. In our example, the lever arm of each force, both the gravitational force and the wind force, is exactly half the edge length of the cube.

The torques can therefore be calculated as follows:

As previously mentioned, the wind force attempts to tip the cube over, while the gravitational force presses the cube down onto the ground, thus providing stability. The gravitational force increases the stabilizing moment, while the wind force contributes to the tipping moment.

Since there are no other relevant forces or moments in this example, only the wind force and gravitational force determine the stability of the cube. The stabilizing moment of the gravitational force is greater than the tipping moment of the wind force. Therefore, there is no risk of the cube tipping over.

In the second example, we want to demonstrate how operating conditions affect the stability of a machine. The cube from the first example is no longer on a flat surface but on a ramp with a 30° incline. This situation is relatively common in everyday life, for instance, when a vehicle is parked on a slope.

We assume that the forces and their points of application remain unchanged and are subject to the same conditions as in the first example. This results in the following initial situation:

Since the points of application of the forces are known, the tilt edge and the relevant lever arms must be determined. Once the lines of action of the forces are drawn in the diagram, it quickly becomes clear why a ramp or incline increases the risk of tipping.

While the lever arm of the wind force has not changed, the lever arm of the weight force has become significantly smaller. Up to now, the weight force has prevented the cube from tipping over. Since the torque is proportional to the lever arm, it can already be determined that the cube is at a significantly higher risk of tipping on the incline.

In this case, the calculations yield the following torques:

Since the tipping moment is now greater than the stabilizing moment, the cube would tip over the right edge in this case.

If the ramp becomes steeper, this effect is further intensified. A critical point is reached when the line of action of the weight force no longer lies to the left but to the right of the tipping edge. Up to this point, the weight of the cube has always helped to prevent tipping; however, it would now even promote tipping. Both the wind force and the weight force, with their respective torques, contribute to tipping the cube over the critical edge.

The statement 'the stabilizing moment is greater than the tipping moment' is often not sufficient. One wants to know not only if the risk of tipping is eliminated but also what reserves are available. To enable a more precise assessment of stability, the stability factor has been defined as a basis for evaluation:

In the first example, the safety factor is approximately 1.4:

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

The stability factor as an indicator has the following significance:

Case 1: Stability factor less than 1

With a factor below 1, there is a risk of tipping. The machine will tip over under the current moments.

Case 2: Stability factor equal to 1

Here, the tipping moment and stabilizing moment are equal, meaning the machine stands 'unsteadily.' This situation is unsuitable for practical use, as the risk of tipping cannot be ruled out.

Case 3: Stability factor greater than 1

A stability factor greater than 1 indicates that there is no risk of tipping. The larger the factor, the less likely it is that the machine will tip over.

When calculating the stability of machines, in addition to the static forces considered here, dynamic forces, manual forces and much more must also be taken into account. The precise determination of stability varies according to the type of machine and is described in applicable standards. For example, detailed requirements for mobile elevating work platforms can be found in DIN EN 280.