Stiffness for an indicator mount

[Neil LickfoldParticipant @neillickfold44316]

Cast Iron makes for the best indicator stands as it generally has the highest amount of rigidity . The mentioned indicator I think is for comparator stands for comparing components and not to be used as an indicator in a lathe for example. For a part indicator I have always brought the ones with the very low pressure stylis. These have the least deflecting influence when indicating parts to as high a precision as you can. A lot of indicator stands actually will not allow you to indicate a part to 0.01 mm TIR as the indicator stylis load and stand flex is such that it can not actually be achieved. Once you start to getting to less than .01mm it really does take a lot more attention to detail especially when you are in the 0.001mm range. Less than 0.001mm is a totally different world to be working in.

Neil

[Iain DownsParticipant @iaindowns78295]

Following the advice of posters who clearly know more than me, I've dragged my armchair into the shed and commenced some passive engineering. (well, I stayed in the house, really. Warmer and better internet!)

I've revised bending. I've tried to do some calculations and got some somewhat conflicting results. Interesting, but I might need some help.

The lazy (armchair) approach is to find a calculator online, plug in the numbers and claim personal genius. Sadly you have to understand something of the matter to know what to plug in.

The formulae I've found for my round bar are as follows.

The 'moment of inertia' – which is more or less how much the bar resists bending is PI * r ^ 4 / 4. As Dinosaur indicated.

This then gets plugged into a bending formula which shows the deflection as follows:-

deflection = Force * Length^3 /(3 * Youngs Modulus * Moment of Inertia).

In short, the deflection is proportion to the cube of the length of the bar – doubling the length increases the deflection at the far and by 8.

for my 0.005m radius steel bar I got a moment of inertia of 4.9 * 10^-10 or (careful counting required) 0.00000000049.

Plugging that into the other equation, with Steel Young's Modulus at 200GPa and a100mm bar I get a deflection of 0.016mm or about 2/3rds of a thou. (assuming my indicator + mount is about 500gms – a bit high, but…)

at 200mm it's .13mm or 4 thou.

Sadly, if I use an online calculator I get a slightly different result. There seem to be no metric calculators so Iv'e had to convert to imperial.

That gives me a 2 thou or .05mm deflection at 100mm, which is clearly worse. I'm inclined to trust the anonymous computer calculation, but I can't see what's wrong with mine. Marking in red pen by an expert would be appreciated!

But broadly with a 100mm long 10mm bar and my indicator, over the 5mm range of measurement I would get an error of several thou as the load on the indicator was relieved by the spring in the needle.

Next I want to try and work out how well this would fare in similar circumstances.

The idea is to have the indicator and mount bolted to the vertical bar – in either orientation. when as shown above there is a natural balance. If I turned the indicator round I'd need to put a weight on the base to stop it falling over. (I want to keep the base thin so I can slide it things)

I can see there is a compressive force down on the bar, but also a moment which will tilt it forward and reduce the height a tiny amount. My gut feel is that the effect will be small, but the mathematics may prove me wrong.

However, that effort will require some cooling of the brain, a cup of tea and, most likely getting out of the armchair for a while.

Iain

[Michael GilliganParticipant @michaelgilligan61133]

Iain,

This might help : **LINK**

http://www.clag.org.uk/beam.html

But, once you get beyond fairly simple individual beams, the best approach is finite element analysis.

… [or lots of prototyping]

MichaelG.

[Iain DownsParticipant @iaindowns78295]

Hi, Michael – that's one of the ones I found thanks to Google, but thanks for the link..

For my mount as described above, here are my calculations.

Assuming the indicator mounted 200mm up the vertical.

compressive strain difference with the 200gms of indicator loaded or not, with a 50x6mm steel bar amounts to about a nanometer (6.67*10^-10). Nothing to worry about there!

In terms of the effect of the torsion because the indicator hangs 50mm off the centre of the bar, i *think* the correct way to calculate this is to work out the moment of the indicator (2 N * .05 m) and then apply this moment to the vertical bar to see what the sideways force is. Then calculate the deflection of the vertical bar based on the force.

If this is right, the sideways force amounts to 0.5N and the deflection (horizontal) around 1 micron. the change in effective height due to this is stupid small (10E-12 – a picometer?).

These values are pretty similar if I use Al instead of steel – by similar I mean totally negligable.

As I have some Ally of the right size I will probably give it a go in that rather than wait for some steel.

Iain

[Michael GilliganParticipant @michaelgilligan61133]

Posted by Iain Downs on 12/11/2016 11:32:01:

…

These values are pretty similar if I use Al instead of steel – by similar I mean totally negligable.

.

Good to see we're on the same wavelength now, Iain

[see item 4 in my post of 08-Nov.]

You may also find this brief discussion interesting: **LINK**

http://www.spaceflight.esa.int/impress/text/education/Mechanical%20Properties/Question_Mechanical_Properties_12.html

MichaelG.

[Author]

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