Spring Constant in Simulation Elements

"Seth Renigar" wrote in message news:38CVe.3243$ snipped-for-privacy@twister.southeast.rr.com...

"lbf" is pound force, as clearly distinguished from pound mass. One real advantage of imperial units over metric is the unity conversion from mass at mean sea level to force due to gravitational attraction. IOW: 1 pound mass at sea level weighs one pound.

If what they are saying is that the force in the spring goes from 3 lbf to 7 lbf in .035 inch then the spring constant K = (7-3)/.035.

The pound is a unit of force, not mass. However, it is commonly used as a unit of mass when weighing things using a balance scale. In this capacity it is called a lbm and is converted to the real imperial mass unit, the slug, by dividing by 32.18 ft/s^2. Where this convention gets confusing is when weighing something with a spring scale on top of a tall mountain ( like Everest) in which case the lb weighed with a spring scale will weigh less than the lb weighed with a balance due to the dimished gravitational acceleration at altitude.

The metric system on the other hand uses the kg as a unit of force (or did for quite a long while.) Such terms as kg-m were used for torque and can still be found in contemporary writing. When I lived in Europe it was rare to find someone who knew what a Newton was.

TOP,

So, based on your simple formula, I come up with 114.3 lbf/in. With your familiarity with this, does this sound like a reasonable number for a 1/4" spring plunger with a 1/8" ball? It seems a bit high to me. But like I said, I am no spring expert.

For a coil spring it sounds a bit high. You can always measure it. Chuck it in the drill press and press down on a scale with a dial indicator to tell you how far.

The 114.3lbs is for 1 inch. Say you only use .05" of travel, multiply .05" X 114.3 to get 5.7lbs. This sound about right. You could verify by measuring as mentioned by someone else.

The problem is how do you make a coil spring with an OD less than .25 inch with that rate?

Say you have a spring 0.120 inches in diameter x 0.024 wire diameter x 0.25 inches in length. From Stock Catalog this spring has a rate of 70.6 lbs/in

Now cut that spring in half and you now have a spring with a rate of 141.2 lbs/in

Tops initial calculations 114,3 lbs/in are correct.

Kman

"Seth Renigar" wrote in message news:YUDVe.3255$ snipped-for-privacy@twister.southeast.rr.com...

Can't measure it! I don't have one. That is the purpose of doing the analysis based on catalog specifications. I am trying to determine the correct ball plunger I need to perform the function that I want.

I just am just trying to figure out what value I need to plug into the Linear Spring Simulation Element feature, based on the catalog information given, in order to simulate it correctly.

As I stated before, there is only .035" travel in the ball plunger. According to the spec's, it takes 3 lb. initial force to move the ball. This tells me that there is pre-load in the spring. Then there is final force 7 lb. at full stroke (the way I interpret it anyway).

So if we take the 114.3 lbf/in. and multiply it by the overall stroke of .035", we get the same difference as subtracting the initial force from the final force, 4 lb. Would this be the correct value to plug into the spring element to simulate it properly? Or is there more to it? I am a bit lost here...

Go to

Spring element in what program again?

SW of course! (may be only in SW Office, not sure)

In an assembly, if you turn on your simulation toolbar, you have the buttons for adding the different physical elements: Linear Motor, Rotary Motor, Linear Spring, & Gravity. In the spring properties (parameters), you can set the spring end points, the free length, and the spring constant. The constant is the property I am having problems figuring out based on the information I have available for the spring I am trying to simulate.