From: atm{at}misterg.fsnet.co.uk (Andy Gray)
To: atm{at}shore.net
Reply-To: atm{at}misterg.fsnet.co.uk (Andy Gray)


On Thu, 31 Jul 2003 14:44:06 -0400,  Aplanatic{at}aol.com wrote:

>A few years ago I also needed to know the elastic modulus of RTV. [snip]
>so I did an experiment. [snip]
>was able to estimate the bulk modulus of the RTV at 170 PSI.  This is just a
ball-park estimate for clear GE Silicone II. [snip]
>Dave Rowe

This tallies well with the value of 130N/cm^2 I obtained from extrapolation
of the slope for small strains on a graph in an obscure paper on silicone
used as a road sealant found on the web.

130 N/cm^2 = 130 / 10 x 2.2 x 2.54 x 2.54 = 185PSI.

For those trying to get this figure by tensile strength and elongation at
failure - DON'T. The elongation is highly non linear (strain dependent)  -
silicone undergoes several hundred percent strain at a stiffness (modulus)
that is a good order of magnitude less than that for small strains. While
there was considerable variation in the large strain characteristics of the
information I could find, the initial slope seemed pretty consistent. I was
surprised how difficult it is to find this information.

FWIW, my calculations (ignoring shear) are as follows. Shear is possibly
the most important component, but I need to check back at a text book to be
sure how to calculate it.

(from an earlier post - apologies if you've seen this)

Consider a 2kg (20N) mirror (200mm x 30mm) mounted by three 20mm diameter
blobs 3mm thick at 50% radius (for the sake of argument).

At zenith, the weight of the mirror rests on the pads, resulting in a
stress in the pads of:

 20N / (3 pads x 3.1cm^2 each) ~ 2N/cm^2

and a consequent compression of the pads by 2 / 130 = 1.5% or 1.5% x 3mm ~0.05mm.

At the horizon, the mirror will pull on an upper pad, and push on a lower
one with a moment roughly equal to its weight x 1/2 thickness: 20N x 1.5cm
= 30Ncm. This will leave the centre of the mirror in approximately its
neutral position, but with a pointing error:

[consider one mounting pad is horizontally in line with the centre of the
mirror. The vertical separation of the other 2 is then mounting radius
(20cm x 50%)/2 x sqrt(3) = 8.6cm. I bet anyone a beer that the results work
out the same for arbitrary pad angles, but the math is easier this way]

Force in upper & lower pads = moment / spacing = 30Ncm / 8.6cm ~3.5N,
over 3.1cm^2 pad area ~ 1N/cm^2. Giving a length change of ~0.75% (half of
previous answer), or a difference in height of 0.05mm over 86mm - about 2'
pointing error.

So, a change in focal length of 0.05mm, and a change in pointing of 2'
between zenith & horizon. There will also be a displacement due to
shear of the mountings which my gut feel says will be a bigger effect.

Regards,

Andy

Andy Gray, N. Wales, UK.

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