Home / Science, Engineering & Maths / Engineering / Through Engineers' Eyes - Expanding the Vision: Engineering Mechanics by Experiment, Analysis and Design / Designing a flat belt drive
Designing a flat belt drive
We'll give you all the guidance you'll need for designing this engineering component.
We’ve given ourselves a challenge, see how far a miniature wheeled vehicle can go on the energy from a single AAA battery. We’ll build our vehicle from commonly available items. We have an electric motor that we salvaged from an old toy, it will need a transmission system that will produce the right to rotational speed of the wheels. The motor came with a pulley so a belt drive might be the way to go. Here’s a diagram of the belt drive, it has a small pulley and a large pulley, and a tensioned belt connecting the two. Tension is needed to stop the belt slipping on the pulleys. We’ll use a rubber band in our belt drive.
We want to know how much tension the belt will need. The tension will load the bearings and increase friction, and, if we increase friction, we’ll increase the losses so the vehicle won’t go so far. So that’s our task. Find the tension in the belt that will enable us to transmit the required torque but not generate too much friction in the bearings. Here’s the geometry. To get a good range of a vehicle, we need it to be light so we kept it small. We started with a wheel diameter of 30 millimetres, we found the effective diameter of the pulley on the motor was four millimetres. For our prototype, we decided to try the biggest pulley we could fit on the axle.
This would give us the maximum reduction ratio. The diameter of 25 millimetres seemed a good start. Later, we’ll need to find the reduction ratio but for now we’ll just find the tension we need in the belt. What torque do we need from the belt drive to make the car move? Pause the video and draw a free-body diagram that will show the following. 1, the traction force at the drive wheels, what pushes the car along. Assume it travels from left to right. The gravity load from my car onto the axle.
3, the horizontal force on the axle. 4, the normal force at the road. 5, the torque that the belt drive will apply to axle.
Here is the answer. How did you go? Now we’ll find the required torque. Pause the video and write an equilibrium equation that you can use to find the value of the torque. We know that the wheel diameter is 30 millimetres. By a separate calculation, we estimated the force we will need at the driver wheel is to be 25 millinewtons. A millinewton is a thousandth of a newton. So with this information we can find the numerical value of the required torque.
Here’s the answer. We calculate the required torque, C, as 0.375 millinewton metres. A millinewton metre is a thousandth of a newton metre. That’s the beginning. Now we’ll pause and work out our strategy. We can use the rope around a bollard analysis to find the maximum ratio of the tensions around each pulley. We can use moment equilibrium to find the difference in tensions on either side of each pulley. Solving will give us values for the two tensions from which we can find the additional forces on the bearing. Here’s our belt drive. From the equation for tension ratio of a rope around a bollard, can you tell which pulley will slip first? Pause the video and think about it.
Here is the answer. The small pulley will slip first. Assuming that the coefficient of friction is the same for each pulley. The reason is that the ratio of tensions is the same but the angle of wrap is smaller for this smaller pulley. We need the angle of wrap for the small pulley, it’s just geometry. It is given by this equation. The distance C is the distance between the centres of the two pulleys. For our prototype, we’ll start with C equals 20 millimetres and this gives the angle of wrap for the small pulley as 117 degrees. Now we can find the maximum available tension ratio. It’s given by T2 over T1 equals e to the mu theta.
Our beautiful equation for friction around a bollard.
We have just found theta, so we’ll need to estimate mu. We’ll use a value of 0.3. Putting this into the equation gives us the tension ratio T2 over T1 equals 1.8. Now we need to find the required tension difference. To get this, we’ll take moments about the drive axle. Pause the video and draw an FBD of the large pulley that will show 1, the two tensions. 2, the reaction from the shaft on the pulley that balances the two tensions. 3, the torque from the shaft on the large pulley. We’ll specify the T2 is greater than T1.
Here is the FBD you need. Did you get it? Now we’ll apply equilibrium and find the difference in the tensions. Pause and have a go at it yourself if you want to.
To get the tension difference, we’ll take moments about the drive axle. We’ve already calculated the required torque, that was 0.375 newton metres, but let’s make an allowance, say double it. To find the tension difference, we will use a free-body diagram. It will help us avoid errors. If we sum moments about the centre of the pulley equals naught, we get these equations. And, by manipulating the equations, we will get the tension difference T2 minus T1 equals 0.06 newtons. We have the tension difference and we have the maximum tension ratio. We’re almost there. We can eliminate T2 from these equations, and we’ll find that T1 equals 0.071 newtons, and then we can find a T2 equals 0.131 newtons.
Now to find the extra load on the bearing. We’ll find the rectangular components of each of the belt forces and then combine them into a resultant. First the x-direction. The sum of the belt force components in the x-direction is given by this equation. If we substitute the numbers that we have for our model car, we will find that FBx equals 0.172 newtons. We’ll do the same in the y-direction.
We’ll get this equation, and if we substitute the numbers, we’ll find FBy equals 0.0536 newtons. Combining these two quantities, we find the total force on the bearings is 0.180 newtons, or we could call that 118 millinewtons. Is the bearing friction going to have too much effect? We can get some idea by comparing the extra load from the belt drive to the weight of the vehicle.
The estimate for the vehicle mass is 38 grams, so the bearing load from the vehicle mass is 9.8 times 38 millinewtons, we get 372 millinewtons. The extra load from the belt drive is 180 millinewtons. It’s about 48% of the total load from the vehicle weight.
It will be a noticeable extra load and there will be load on the motor bearing too. Perhaps we should look at alternatives to our belt drive. What do you think? And this is just the start.
We’ll give you all the guidance you’ll need for designing this engineering component.
The video leads you through the process with opportunities to do calculations yourself involving belt tensions, rope around a bollard, traction forces and efficiency. In the end you will find the required tension and determine the implications for efficiency of transmission.
It might help if you download the design specification in the Downloads section below in case you want to refer to it as you go.
If you just watch the video it will take about 11 minutes. If you take the opportunity to do calculations it will take longer; it’s hard to say how much longer because it depends on so many factors, but allow a total of 40 minutes.
If you are stuck (or even if you aren’t) you might like to look at the worked solution that is available from the Downloads section.
Talking points
- How did this work for you?
- Did you like using the equation for ratio of tensions around a pulley?
This article is from the online course:
Through Engineers' Eyes - Expanding the Vision: Engineering Mechanics by Experiment, Analysis and Design
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This article is from the free online
Through Engineers' Eyes - Expanding the Vision: Engineering Mechanics by Experiment, Analysis and Design
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