Aluminio-silicate scraps

[saefroch]

I found a little baggie of some ceramic scraps from a while ago, and if I recall correctly, they're high-temperature feed-throughs. In any case, I now have some little tubes of 75% alumina, 25% silicate ceramic, which are about 3/32" in diameter with a 1/32" hole in the middle.

Does anyone know what the shear strength is, or how to find it?

Any ideas for applications? The only way I can shape them is with 100% alumina sandpaper, anything else goes to bits quicklike.

[metalmeltr]

Testing shear strength shouldn’t be too difficult. Two metal plates with a hole the same size as the OD of the tube. Place the plates against each other, anchor one plate. On the other plate get a big fish scale and some threaded rod. Attach the scale to the second plate and connect the all thread to the scale, and then you need a piece of angle iron with a hole in it. Put the all thread though this hole; tighten a nut against this plate to apply tension to the setup. Video the test, specifically the scale to determine the failure point.


Edit picture added.

[saefroch]

Why angle iron?

[saefroch]

Testing should occur tomorrow.

[metalmeltr]

That's the basic idea, but im not sure if it will work. What is the purpose of the spring? How are you going to measure shear strenght? Can the plates slide past each other? The picture I posted was a top view.

Edit add picture and spelling.

[saefroch]

I am aware that what I just built needs to be modified to work (Doh!). Unfortunately, no fish scale that I know of can simply be attached to a steel plate and a threaded rod, so that's the spring from one.

EDIT: And that is why I'm waiting until tomorrow to test. I can't think straight at ALL.

[Gippeto]

Shear strength of the material is calculated by multiplying the UTS (Ultimate Tensile Strength) by .6.

ie.. 100 000psi UTS....60 000psi Ts


Then multipy the shear strength of the material by the cross sectional area of the "thing" which is in shear. ie...minor diameter of a bolt (assumes the threaded portion of the bolt is in shear)

If you're going to use a pin in shear, you should also consider calculating the bearing stress, and tear out stress.

[saefroch]

I can't find any data on my particular ceramic though...

bearing stress, and tear out stress.

Ummmmmmmmmm explain?

This apparatus will now actually test shear strength. Yay! I'll probably have a number tomorrow.

[Gippeto]

I'll suggest that you use yield strength in calculations rather than tensile strength, and a reasonable safety factor of 3...just how I do it....aircraft are sf=1.5, but they're working with certified materials as well. Your choice.

Bearing stress;

Simple method;

For a round pin, multiply the diameter of the pin, by the wall thickness of the tube. Multiply the result, by the yield strength of the material.(assuming the pin is holding something in a pressurized tube, which places the pin in shear)

This is the point at which the pin hole will begin to deform...assumes a "perfect" fit.

An alternate method is (pi*d)/3*t *Ty

t=wall thickness

Ty=yield strength


Tear out;

The formula for tear out stress is F = allowable stress *2*e*t

e is the distance from the edge of the hole to the end of the tube

t is thickness

Allowable stress for our purpose is the shear strength of the material calculated using yield strength of the material. This will be the point of failure.

Once you have the bearing stress at reasonable limits...for intents and purposes, if you place the center of the hole for the pin a minimum of 1.5 times the hole diameter from the end of the tube, you shouldn't have to worry about pull out/tear out. I use 2*d on my pcp builds.

It only takes a minute to "run the numbers", so it's easy and a good idea.



Hope that helps.

[saefroch]

I'm about 3d from the end so I think I'll be okay on tear out stress.

Results from first trial: No break under 22.87 pounds (edited because I can't read a ruler ), so making the assumption that pinning entirely through both sides will distribute the load over twice the area of the area and double the yield strength, I have an approximate safety factor of 2.33.

EDIT: Found the original design to be ineffective, so here is what I'm using now:

No idea how I got the first result, later trials with version 1 showed a break at 13.09 pounds, but the square bar stock was tilted significantly under the load, so I think they were invalid.

Applied a load of about 26.18 pounds, no break, spring nearly fully compressed. Looking into a stronger spring. I really don't mean to blow off calculations here, I'm a big fan of numbers for certain. I simply can't find any data on my particular ceramic.