Clutches, brakes, and inertia

Oct. 1, 2006
The optimal size for a clutch or brake is determined by three things: Required torque, thermal horsepower per engagement, and average required thermal

The optimal size for a clutch or brake is determined by three things: Required torque, thermal horsepower per engagement, and average required thermal horsepower. Here, we'll discuss the first, learning to calculate torque capacity under static and dynamic conditions.

Dynamic torque

To accurately determine the torque required during acceleration or deceleration, it is necessary to know total inertia, component efficiency, total load torque, and the amounts reflected back to the clutch/brake output shaft. A major consideration is inefficiencies; individual drive components and their power losses during acceleration or deceleration put significant burden on a system — which the clutch or brake must overcome. The total inertial torque is the sum of all the torques of the individual drive components. The dynamic torque is then found by adding total inertial torque and load torque together. With drive efficiency E, load torque TL, and inertial torque Ti dynamic torque is calculated:

Note that efficiency and load are always factors in a system, but the inertial term only applies when the system is accelerating or decelerating.

Determining inertia

Inertia is the measure of an object's resistance to changes in motion. Rotational inertia is a function of an object's mass and how it's distributed about the rotating axis. The effective radius is where the entire mass of the object is taken to be concentrated. Called the radius of gyration, this is determined by an object's geometry and is designated K. The first step in calculating dynamic torque is to determine the inertia (in the system) that must be accelerated or decelerated. The term used to quantify it is WK2. To obtain the WK2 in a form that can be used, calculate the inertia of each component in the system that will be cycled — and the inertia reflected from each cycled component back to the clutch and brake. An equivalent WK2 can then be attached to the output of the clutch/brake to represent the inertia of all cycled components in the system.

Learning by example

Here we calculate the inertia of several objects involved in a motion system; then we calculate how much of that inertia is seen by the clutch/brake. Assume for the conveyor system on page 26 that:

Acceleration time = 0.4 sec

Deceleration time = 0.13 sec

Cycles per min. = 10

Conveyor efficiency = 0.8

Chain drive efficiency = 0.9

Reducer efficiency = 0.8

Maximum pressure = 60 psig

The procedure to obtain the equivalent or reflected inertia is based on the principle that the total energy in the system is conserved. This means that the reflected inertia of an object is equal to the actual kinetic energy it possesses in the drive system. Since the kinetic energy varies with the square of the speed, the reflected inertia (for the boxes in our example) is the object's actual inertia on the square of the ratio of the operating speed to clutch/brake speed.

Because the weight of each box in our example system is significant, the inertia they reflect to the clutch/brake is also significant. In contrast, the amount of weight the actual conveyor belt sends to the clutch/brake will be small. Nonetheless, the steel pulleys can still affect the overall system.

The sprockets on our system, though attached to the pulleys, will have a different affect on the clutch/brake because of their larger geometry. Using the same general formulas, its calculated weight is 266.72 lbs. and its inertia 0.232 lb-ft2. Its inertia is reflected through the chain drive and reducer; its larger diameter reflects more inertia back to the clutch/brake. In contrast, the smaller sprocket on our system contributes significantly less inertia: From the same formulas used to calculate pully and large sprocket values, the small sprocket weighs 66.65 lb with an inertia reflected of 0.06 lb-ft2.

After these inertia values are calculated, those of other standard system components can often be obtained from manufacturers. Assuming typical inertia values for reducers, couplings, and clutch/brakes involved, the values for WK2 might be 0.17, 0.78, and 0.10 lb-ft2 respectively. From these values, the final, total system-reflected inertial torque for our example system is 0.868 + 0.232 + 0.06 + 0.17 + 0.78 + 0.10 = 2.27 lb-ft2.

Torque capacities

To simplify the complex procedure of determining total torques, one technique is to use a dynamic torque analysis table. See page 25; charting torque values in tabular form allows easy summation of forces to see what maximum values must be met. But how is such a chart filled out?

First, all cycled components should be listed. It's easiest if the system pieces are listed starting from the load, and working to the clutch/brake. Then, enter the reflected inertia for each componen; these are the values we calculated using their weights and K values. Next, enter the drive efficiency associated with each component. (Many of these values can be obtained from vendors; otherwise, good engineering judgement returns acceptable estimates.)

These efficiency values are then used to determine efficiency factors — numbers expressing the relative effect each component has on the system. They're determined by multiplying all the efficiency values at or below the component considered on the table. For the boxes in our example, the value is 1 × 0.8 × 1 × 0.9 × 0.8 × 1 × 1 = 0.58. For our reducer, it's 0.8 × 1 × 1 =0.8.

The first torque to calculate is load torque TL. For our conveyor example, this is force seen at the drive pulley — required to move the system at constant velocity. It can be found by solving for static equilibrium, made easier with free-body diagrams for key points of force interaction: At the box, slider and belt, and head pulley. (Loads in our example are due to weight of the boxes and frictional forces.) After determining the load requirement, we apply the associated efficiency factor and reflect it back to the clutch/brake. So load torque at the clutch/brake equals:

The next step is to calculate the inertial clutch torque for each component:

Once entered into the table, the Tic values are added together to determine the total inertial clutch torque required.

The sum of the inertial clutch and load torque is the dynamic clutch torque required: Tdc = Tic + TL. In our case, this is 552 + 584 = 1,136 in.-lb. This is the value that should be used in selecting a clutch.

A similar process is used to determine braking requirements. The first step: Find dynamic braking requirements by calculating the actual inertial torque of each component using:

Speed change is actually negative here. Once again, enter data into the table and sum Tib values for total inertial brake torque required. The sum of the inertial brake torque and the load torque is the dynamic brake torque required Tdb =Tib + TL — in our case, -1032 + 584 = -448 in.-lb. This is one of the values that should be used in selecting a brake.

Note that inertial brake torque is that required to deceleration the system — not necessarily equal to holding torque. In fact, the former is often much larger. Also, load torque is the same for acceleration as it is for deceleration. Why? Once the boxes are stopped, friction F2 acts in the other direction — because friction always opposes motion. If this total dynamic brake requirement is the same sign as the clutch/torque it indicates that the system will decelerate in less time than required, and no brake is required to stop the system — though one may still need a holding brake.

The final step is calculating holding brake torque requirements. This is torque required to keep the system stopped; all the inertial torques disappear here. It's similar to load torque, but different because frictional forces act in the opposite direction when the system is stopped.

Sign convention

During dynamic torque analysis, it's important to adhere to one sign convention. Each (inertial and load) torque is considered separately. Load torque is found by solving for static equilibrium. To prevent the sign for load torque from changing during the analysis:

  1. The direction of torque required to accelerate the mass of the system is always considered positive.

  2. If load torque acts in the direction of the acceleration or inertial torque, then it is considered to be positive. Static free-body diagrams make this orientation easier to identify.

  3. The sign of the deceleration inertial torque is opposite the sign for the acceleration torque.

Load torque is generally considered to be positive, especially if the load is predominately a friction or inertia load. It is possible, however, for the load torque to be negative. This could happen if the weight of the load, or some other kind of stored energy like a compressed spring, helps the system accelerate.

For more information, call (513) 868-0900 or e-mail [email protected].

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