Static Thrust Load Calculator

Even when heavy equipment is parked and left alone, components are required to bear their weight. When this weight acts vertically downwards on these components, along the length of pins, bearings or fasteners, they must remain stable and not deform.

This maximum stable axial force is called the static thrust load. This tool calculates it from two values: the strength of the material and the amount of material in the path of the load. If one of the three fields are left blank, then that value will be calculated by the tool.

What is a static thrust load?

Static thrust load is the maximum axial force that a component can withstand when it is stationary or moving slowly enough to avoid fatigue. Axial means along the length of the axis, not across it.

The upper limit is determined by the material's yield strength. This is the point at which a material loses its elasticity and begins to deform permanently. If this limit is exceeded, small depressions called Brinell impressions will form on the raceway of a bearing, a retaining ring may fall out or a pin may shear.

For identical actual components, the static rated load is usually significantly higher than the dynamic rated load. It's easier to keep a load stationary than it is to continually rotate it.

Formula for calculating static thrust.

The static thrust is the product of the material's yield strength and the area exposed to the load.

In this case F_t is the bearing capacity, S_y is the yield strength of the material and A_s is the cross-sectional area of the load path. The units work out automatically. Since a pascal (Pa) is equal to one newton per square meter (N/m²), multiplying a stress in pascals by an area in square meters results in a force in newtons.

Since this equation has three quantities in it, it can be rearranged to solve for any of the missing quantities.

Example calculation:

Let's take a steel rod made of A36 steel as an example. This rod is subjected to an axial load and has a cross-sectional area of 1 square centimeter. This area is equal to 0.0001 square meters. The yield strength of A36 steel is approximately 250 megapascals, which is equivalent to 250,000,000 pascals.

This is approximately the load of a single pin, about the size of a fingernail and weighing as much as two-and-a-half small cars. For this reason engineers almost never specify design values in terms of this quantity. With a safety factor of 2, the allowable load would be around 12.5 kilonewtons.

Formula transformation:

This table shows the same relationship in three different forms, each form handling a possible missing value.

Flow limit, not tensile strength:

Here we use the yield strength and not the tensile strength. With static loads failure is assumed to occur once the yield strength has been exceeded. If a component becomes permanently deformed its shape will no longer conform to design specifications even if it does not break.

The tensile strength is the stress at which a material finally breaks and it is well above the point where an accurate fit would be compromised. By designing dimensions based on the yield strength, we ensure that the component stays within the elastic region and regains its original shape once the load is removed.

Typical material yield strength:

If data sheets are not available these representative yield strengths can be used as a starting point. The actual values will vary with alloy, condition and heat treatment so compare the values to the material actually being used before final design.

Safety factor:

The original bearing capacity is not the design load. By dividing by a safety factor, you get the actual applied load. The size of the safety factor should be chosen according to the degree of knowledge of operating conditions.

For clear and well-predictable stable loads a factor of 2 is usually used. For impact loading, vibration, repeated shocks or significant temperature changes a factor of at least 4 is normally required. In lifting, construction and aviation work the factors laid down in standards must be observed, these values may be even higher. By activating the test of applied load it can be checked where the actual load lies relative to the yield strength and safety factor.

Where do static thrust forces occur?

Static loading occurs wherever a component is subjected to stable axial load. A jet engine on a maintenance stand transfers its weight directly downward through support bolts. The size of the bolts is designed so that there is enough play between the motor's weight and the yield strength.

The gas pressure cylinder of an office chair is supported by internal seals and retaining rings that bear the constant static load caused by body weight. Therefore, in older chairs, sudden failures can occur when sitting down quickly.

Crane hooks, vertical bearings inside elevators and anchor bolts under heavy stationary engines are all designed in the same way.

Tips for reliable results:

Use actual material strength, grade and temperature instead of general textbook values. Apply a safety factor by dividing the capacity by this factor. Use a higher factor until the load is fully stable.

In the case of roller bearings, the calculation of the contact area is omitted and instead the static nominal load value (Coa) specified by the manufacturer is used. This catalog value already takes into account the contact shape between the rolling elements and the raceway, which cannot be captured in a simple surface calculation.