The seat doesn't care what shape the window is, other than the
geometry of the contact faces. The only effect that window shape
has is how the window deflects under pressure, which places the
stress on the inside diameter in the case of flat viewports, and
increases the radial forces in the case of spherical ones. Doing a
very simple analysis off the cuff, In the Stachiw page 526 example,
I'm looking at a projected diameter of 6.932 inches, giving a
surface area of 37.740 in^2. At 4000 psi, that is 150 962 pounds
force. The seat diameter at 6.932 inches and with thickness 0.125
inches, gives a total shear area of 2.722 in^2, so your total shear
stress is 150 962 / 2.722 = 55 460 psi. Within the 80 000 psi
yield, but not a huge safety factor. In any case, this is just an
estimate, and I'm sure that Stachiw knows what he is doing. I
didn't look at your calcs - I don't have sufficient time right now,
but I'll have a look this evening to see what you did.
-Sean
On 14/06/2011 10:58 PM, Jon Wallace wrote:
Sean,
That sounds like the same formula you sent me 4/9/2010 (see below)
when I was looking at flat disk viewports, but it doesn't appear
to work correctly for sphere segments. The results I get using
that formula indicate dimensions that are much larger than what
I've seen in the field.
sigma=(W*l)/Z
=(W)[(D_o-D_f)/2]/[(D_f*pi)(K2)/6]
I'm using values from Stachiw's book, page 526, in which he gives
dimensions for a 120 degree sphere segment used in his testing.
It includes the viewport and the mounting which he tested to
4000psi (see page 527, fig 11.80). The actual seat diameter was
only .125 inches. Using the following dimensions from the book
and the formula above:
Do = 6.932
Df = 5.97
K = .125
h = R * (1 - cos(theta/2)) = 2 (assuming 8 inch sphere)
Surface Area (SA) = Do * (pi) * h = 43.55
P = 4000 psi
Numerator:
W =
4000 * 43.555 = 174220
Do - Df / 2 = (6.932 - 5.87) / 2 = .481
giving
174220 * .481 = 83799.82
Denominator:
Df * (pi) = 5.97 * (pi) = 18.755
K^2 = .125^2 = .0156
giving
18.755 * .0156 / 6 = .0488
Result:
83799.82 / .0488 = 1,717,192
In this case, Stachiw used 80ksi yield steel for the mount and the
numerator alone is above that yield point. I know Stachiw talks
somewhere in the book about the base of sphere segments
contracting inwards under pressure some small amount, but enough
to make a chattering noise if the seat does not have a gasket to
act as a buffer as the acrylic compresses. He describes it as
more than one pilot being startled thinking the acrylic had
cracked. In any event, that seemed to indicate that some amount
of pressure on the segment is radial instead of axial (against the
seat) and using the entire surface area might be very
conservative, and in some cases, too conservative.
Jon
On 6/14/2011 3:28 PM, Sean T. Stevenson wrote:
Jon
- this is exactly how it is done - calculate the projected area
of the dome, find the force, divide by the seat bearing area to
find the stress on the seat. The only caveat is that the stress
will not be evenly distributed, so to find your seat thickness
you need to assume that the entire force acts on the inner
diameter to get the worst-case scenario cantilever.
-Sean
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