There are many situations in motion applications where a linear guide or actuator isn’t fully supported along its entire length. In these cases, deflection (due to the component’s own weight and due to applied loads and forces) can affect the running properties of the bearings and cause poor operation, in the form of premature wear and binding.

Products that may be mounted with only end supports, such as linear shafts or actuator assemblies, or in a cantilevered orientation, such as telescoping bearings, will typically have a specification for maximum allowable deflection, and it’s important to check the application and ensure that this maximum deflection is not exceeded. Fortunately, most linear guides and actuators can be modeled as beams, and their deflection can be calculated using common beam deflection equations.

#### Material and design considerations

When calculating deflection, you need to know the properties of the guide or actuator and the conditions of the applied load. In terms of the guide or actuator, the important criteria are the modulus of elasticity and the planar moment of inertia of the component. The modulus of elasticity is a measure of the material’s stiffness, and can typically be found in the product catalog. Moment of inertia describes an object’s resistance to bending and is sometimes provided by the manufacturer of the component. If moment of inertia isn’t specified, it can be reasonably approximated by using the moment of inertia equation for a solid or hollow cylinder (for linear round shaft) or a rectangle (telescoping bearing or linear actuator).

*Modulus of elasticity, also known as Young’s modulus or tensile modulus, can be defined as the ratio of stress (force per unit area) on an axis, to strain (ratio of deformation over a length) along that axis.*

*Planar moment of inertia (also referred to as second moment of area, or area moment of inertia) defines how an area’s points are distributed with regard to an arbitrary plane and, therefore, its resistance to bending. *

From an application and construction standpoint, the criteria that influence beam deflection are the type of support at the ends of the guideway or actuator, the applied load, and the unsupported length. When a component is cantilevered, it can be modeled as a fixed beam, and when it is supported on both ends, it can typically be modeled as a simply supported beam. For cantilevered beams, the maximum deflection will occur when the load is located at the free end of the beam, while for simply supported beams, maximum deflection will occur when the load is located in the center of the beam.

When determining the total deflection, keep in mind that there will be *two loads* that cause deflection: the weight of the guide or actuator itself, and the applied load. The component’s own weight can almost always be modeled as an evenly distributed load, while evaluating the applied load as a point load at the location of maximum deflection (at the free end of a cantilevered beam, or at the center of a simply supported beam) will generally provide the worst-case scenario for total deflection.

#### Deflection of cantilevered beams

Telescoping bearings are often cantilevered, and some Cartesian robot configurations result in a cantilevered actuator on the Y or Z axis. In this case, the beam’s weight, which is reasonably uniform along its length, causes maximum deflection at the end of the beam.

This deflection is calculated as:

*Where:*

*q = force per unit length (N/m, lbf/in)*

*L = unsupported length (m, in)*

*E = modulus of elasticity (N/m ^{2}, lbf/in^{2})*

*I = planar moment of inertia (m ^{4}, in^{4})*

To generate the worst-case deflection scenario, we consider the applied load as a point load (F) at the end of the beam, and the resulting deflection can be calculated as:

Adding the deflection due to the uniform load and the deflection due to the applied (point) load gives the total deflection at the end of the beam:

#### Deflection of simply supported beams

Linear shafts and actuators are often secured at their ends, leaving their length unsupported, much like a simply supported beam. The uniform load on the beam, (the shaft or actuator’s own weight), will induce maximum deflection at the center of the beam, which can be calculated as:

Since this is a simply supported beam, the applied load can be modeled as a point load at the center of the beam for the worst-case scenario.

Deflection due to the in this condition is calculated as:

Total deflection at the center of the beam is:

#### Deflection of shafts with two bearings

When two bearings are used on a simply supported beam, as is typically the case with round shaft guides, the applied load is split between the two bearings, and maximum deflection occurs in two places: at the location of *each* bearing when the bearing assembly (sometimes referred to as a carriage or table) is at the middle of the shaft.

The beam deflection calculation for this condition is:

Again, we must add the deflection due to the beam’s own weight plus the deflection due to the applied load, to get a total deflection of:

There are additional mounting and loading scenarios that may be encountered in some applications, such as an actuator with fixed support on both ends. But like the examples above, even less-common cases can be evaluated using standard beam deflection equations. For a comprehensive list of beam support scenarios and deflection equations, check out this page from Cornell University.

*Feature image credit: wikipedia.org*

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