Voids in solder joint under high environmental pressure

[Glen Walpert]

Sounds like metal migration which is a completely different kettle than
wetting.

Metal migration is a surface phenomenon, not a bulk phenomenon like
diffusion. The term "wetting" does imply a liquid being involved, when
solid metals are involved it is generally referred to as forming a
metallurgical bond, as is the case in the silver-bonded seal rings.

Yes, quite like corn starch and water. ...and the shape of the "beads"
changes the characteristics of the paint (directionality of the shear).

I'm still not buying the "solid" part.

I don't blame you for not buying the "solids wetting" part, but there are
lots of examples of solid to solid metallurgical bonds made with no
melting, including fabrication of bi-metal sheet by pressure bonding
between rollers, explosive bonding of heavy bi-metal plates, and the
fabrication of aluminum evaporators for refrigerators, where a pattern
for the integral tubing is printed on one clean aluminum sheet, two clean
sheets are pressure bonded, and then the unbonded tube pattern is
hydraulically expanded.

 

[Glen Walpert]

Hmm, seems like a potentially bad idea -- diffusion can leave porosity:
http://en.wikipedia.org/wiki/Kirkendall_effect Maybe not an issue for
hermetic seals.

Interesting link, thanks. I never heard of it before. This does not
seem to be an issue with the silver plated steel seal rings however.
When they are broken apart with a large hammer what you see is a fracture
of a steel-silver alloy, mostly steel (no silver visible, although it
must still be there), with no significant voids, where a leak would be
required for it to be significant. The only leak I ever saw in one of
these was caused by faulty surface preparation.

 

[Tim Williams]

Interesting link, thanks. I never heard of it before. This does not
seem to be an issue with the silver plated steel seal rings however.
When they are broken apart with a large hammer what you see is a
fracture
of a steel-silver alloy, mostly steel (no silver visible, although it
must still be there), with no significant voids, where a leak would be
required for it to be significant. The only leak I ever saw in one of
these was caused by faulty surface preparation.

By eye alone? Seems to be a microscopic phenomenon, probably not visible
on that scale.

Hmm... I find no phase diagram on Ag-Fe in my list, so I have no idea what
intermetallics form (if any). If it's like Cu-Fe, then there would be
only a little solid state solubility, and that's it, which wouldn't
diffuse much. Offhand I'd expect behavior near nickel or palladium (which
do form solid solutions with iron), but not so, wholeheartedly shall we
say.

If the seals tend to fracture on the seam, then there's certainly
something going on, such as voids and brittle intermetallics. Fracture
would seem to suggest a brittle rather than malleable phase (such as Fe-Ni
or Fe-Pd analogs would probably give).

Oh... also on topic, "3000 A.D.":
http://en.wikipedia.org/wiki/Terry_Fugate-Wilcox
He made some other interesting things too.

Also...is that girl in the "Weathering Wood" pic topless... :-?

Tim

 

[Tim Williams]

Ice is a weird material. That doesn't happen with metal, at least not
at ordinary temperatures.

Weird indeed!

Although, albeit at much higher pressures, bismuth and a couple of other
things ought to pull the same trick, no? Expanding on freezing can only
exert so much pressure, thermodynamically speaking.

Tim

 

[George Herold]

On 12/6/2013 7:51 PM, Tom Miller wrote:



Indium isn't quite as low-melting as that.

I use to make low temperature (helium leak tight) seals by taking two flat pieces of metal (mostly copper alloys) and pressing indium onto both sides, and then a light smear of vacuum grease... with a several screws around the perimeter.

George H.

 

[George Herold]

On Thu, 05 Dec 2013 17:12:38 -0500, the renowned Phil Hobbs

Not likely important for the OP, but what we were taught in first
year...

http://en.wikipedia.org/wiki/File:S-N_curves.PNG

Interesting thanks.
Hmm I make little flexure mounts from aluminum (7075) for diffraction gratings. Vibrated at 10 Hz to 1 kHz, should I worry?

George H.

 

[Glen Walpert]

Interesting thanks.
Hmm I make little flexure mounts from aluminum (7075) for diffraction
gratings. Vibrated at 10 Hz to 1 kHz, should I worry?

George H.

That depends on the peak stress and the required life expectancy.

Considering only bending moment (neglecting shear, probably safe)

stress = My/I (lbf/inch^2)

M = bending moment (lbf*inch)
y = maximum distance to the neutral axis, 1/2 thickness for a rectangular
beam. (inch)
I - rectangular moment of inertia (inch^4) = 1/12 width x thickness^3 for
a rectangular beam bent across the thickness.

The peak bending moment is a bit tricky (meaning I don't recall how) for
a resonant beam, which is not simply loaded, but if you can estimate the
minimum radius of curvature r (inches) then:

M = EI/r

E = longitudinal modulus of elasticity (Young's modulus) (lbf/inch^2) =
approx 10E6 for aluminum

Look up your expected fatigue life for the estimated peak stress on the
above wikipedia graph and see if it meets your needs.

 

[Glen Walpert]

That depends on the peak stress and the required life expectancy.

Considering only bending moment (neglecting shear, probably safe)

stress = My/I (lbf/inch^2)

M = bending moment (lbf*inch)
y = maximum distance to the neutral axis, 1/2 thickness for a
rectangular beam. (inch)
I - rectangular moment of inertia (inch^4) = 1/12 width x thickness^3
for a rectangular beam bent across the thickness.

The peak bending moment is a bit tricky (meaning I don't recall how) for
a resonant beam, which is not simply loaded, but if you can estimate the
minimum radius of curvature r (inches) then:

M = EI/r

E = longitudinal modulus of elasticity (Young's modulus) (lbf/inch^2) =
approx 10E6 for aluminum

Look up your expected fatigue life for the estimated peak stress on the
above wikipedia graph and see if it meets your needs.

It occurred to me after posting that if the mass of the grating assembly
is reasonably larger than the the mass of the flexure, then you could
approximate as simply loaded on the end of the flexure and calculate
stress based on peak deflection, which is probably easier to measure than
radius:

M = 3EId/l^2

where d is peak deflection from free state (1/2 total deflection at the
end of the beam) and l is the length of the beam.

 

[George Herold]

On Mon, 09 Dec 2013 06:39:43 -0800, George Herold wrote:





That depends on the peak stress and the required life expectancy.

Considering only bending moment (neglecting shear, probably safe)

stress = My/I (lbf/inch^2)

M = bending moment (lbf*inch)
y = maximum distance to the neutral axis, 1/2 thickness for a rectangular
beam. (inch)
I - rectangular moment of inertia (inch^4) = 1/12 width x thickness^3 for
a rectangular beam bent across the thickness.

The peak bending moment is a bit tricky (meaning I don't recall how) for
a resonant beam, which is not simply loaded, but if you can estimate the
minimum radius of curvature r (inches) then:

M = EI/r

E = longitudinal modulus of elasticity (Young's modulus) (lbf/inch^2) =

approx 10E6 for aluminum

Look up your expected fatigue life for the estimated peak stress on the
above wikipedia graph and see if it meets your needs.

Thanks Glen, My question was only half serious. I'd have to read some more first. maybe this?
http://fp.optics.arizona.edu/optomech/references/OPTI_222/OPTI_222_W9.pdf

George H.

 

[Glen Walpert]

Thanks Glen, My question was only half serious. I'd have to read some
more first. maybe this?
http://fp.optics.arizona.edu/optomech/references/OPTI_222/ OPTI_222_W9.pdf

George H.

That should give you an understanding of pure bending, but isn't really
necessary for estimating the approximate peak stress, and isn't adequate
for an exact solution. The basic stress equations I presented in my
second message, which passed this one in the mail, should be easy to
verify in Marks' Standard Handbook for Mechanical Engineers, any
mechanics of materials textbook, or probably Wikipedia, easy to solve,
and probably adequate, especially with a safety factor of 1.5 or more.

Regards,
Glen