Calling All Engineers!

November 24, 2008 12:19 PM

I am stuck on a "design of machine components" project. Anyone care to take a crack? Its actually a pretty simple spring problem.

Overview:

I am messing around with the design for a helical torsion spring but am having trouble finding part of the force.

Specifically the spring that is used on the end of barbells to prevent plates from sliding. I do not know how to determine how much force the spring is putting on the bar.

What i figure is it has to do with the coefficient of friction, number of coils, diameter of the inside of the coil**, etc.

My professor who is a goof ball told me to solve for the angular spring rate, does that sound right?


**as it is a helical spring, when the handles are forced towards each other the numbers of coils increases (i.e. 4.375 turns --> 4.5 turns). As this happens how does the inner diameter change? How does this effect the force when the original diameter is a little smaller then the diameter of the barbell (lets say the barbell is 2 inches and the UNFORCED spring is 1.9, forced its 2.1...)


As you can see I am a little lost....I know the hive mind is usually not so good at engineering questions but I thought I would ask anyway.

posted by Black_Umbrella to education (5 comments total)

I said the helical spring increases in number of rotations as the arms are forced together which is a mistake. As the arms are forced together the number of coils decreases.
posted by Black_Umbrella at 12:21 PM on November 24, 2008

It's been a lot of years since machine design class, but here a thought:

Would the force on the bar be proportional to the amount of torque required to deflect the spring by that distance? For example, if it takes 100 ft-lbs to open your example spring by 0.1 inches, does that mean the spring applies 100 ft-lbs to the bar? Further, could you divide that torque by the radius to get a force, and divide the force by the surface area in contact to provide a pressure?

Most of the results from a google search talk about the torque that the spring applies at its ends, and not so much about the force on something in the middle of the spring.

I doubt the coefficient of friction would do much for you other than help you calculate the force required to spin that spring on its bar.
posted by zompus at 12:55 PM on November 24, 2008

I am a mechanical engineering grad student. I'm into thermal sciences/fluid mechanics/combustion but I took a few mechanical design courses and FEM. IANYMechE.

The helical coil exerts a radial normal force on the bar. The normal force generates friction. Many alternative style clamps have rubber, which has a higher coefficient of friction with steel than the chrome-plated helical jobs you linked to.

For those helical-type clamps, if you try to put the clamp on the bar without pressing the handles together it won't go on because it's too small. When you do press the handles together it opens up the helical coil and when you allow them to spread apart, they spread apart until the ID comes in contact with the bar. Any remaining torsional-"tension" becomes a radial normal (compressive) force on the bar.

It might help for you to separate the problem into two parts. One is a torsional spring that generates a normal force. The other is to "unwrap" the radial normal force into a planar normal force (like you may be more familiar with for block-plane problems) and look at the friction force from there.

My mechanical design textbook has loads of worked examples for questions such as this. It is safely tucked away in my parents basement approximately 1000km from here.

Try Google Books or Amazon Preview for Mechanical Design books by Norton or Shigley. They should be able to give you the set of equations you're looking for.
posted by KevCed at 2:09 PM on November 24, 2008

Upon further examining the picture you posted, and thinking about compressing the handles together, the number of turns decreases when you compress the springs and the diameter increases.

I encourage you to look at the picture and imagine what you would have to do with the handles to "unwind" the helical coil.

Or, to put it another way. The length of metal wire in the coil is the same. If there are fewer turns, those turns must describe a larger diameter. Think of it in reverse. If you took the same coil, magically unbent it, then wrapped it around a pencil (or a die of similar diameter) you'd end up with a lot more coils. Hence, in this case, if you force a larger diameter (by pushing the handles together) you'll end up with fewer coils.

(The reason I'm encouraging so much of that ghastly "thinking" and not just spelling out the answers is because I just finished my undergrad. In my experience, if a homework problem like this comes up and you just print out an answer from MeFi (or similar), it pays to slog through. Keep in mind to imagine what Professor Goof Bal will dream up for the MeFi-free final exam.)

(That and professional ethics.)
posted by KevCed at 2:18 PM on November 24, 2008

I'm with zompus here, I think his/her approach would be my first tack on this problem. I think it should be pretty accurate as a first level approximation.
posted by milqman at 2:28 PM on November 24, 2008