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Re: [PSUBS-MAILIST] ASME - say what??
Hi Sean,
Thanks for the information. I have PVHO-1-2007 which appears to be
setup a bit different since I don't see the references you mentioned on
Page 36, however I know which illustration you are describing and it is
on page 56 of the 2007 document. While the math is easy enough, I'm not
sure what the value represents when I arrive at an answer. Using the
K350 as a known example, I've got a 1/4inch stiffened shell for the
hull. The viewport dimensions are Do = 8, Df = 6, t=1.25.
Additionally, lets assume maximum external pressure of 350psi giving me
a bit of an operational safety factor. I know that the K350 viewport
seat has 1/2inch thick sides and 3/4 inch thick seat.
For shear loading:
Viewport circumference = 8 * pi = 25.132
Viewport area = 50.265
k = ? How do I calculate minimum thickness for k? Is this just the
same thickness used for the hull (.25)?
Shear area = 25.132 * .25 = 6.283
Shear area = 25.132 * .75 = 18.849 (known thickness of viewport seat for
K350)
Total load = 350 * 50.265 = 17592.75
Shear stress for .25 thick seat (k) = 17529.75 / 6.283 = 2747.05
Shear stress for .75 thick seat (k) = 17529.75 / 18.849 = 930.00
So what exactly do the numbers 2747.05 and 930 indicate? Is it the psi
carried by seat (k) material?
Jon
Sean T. Stevenson wrote:
Jon - assuming you have PVHO-1-2002, look at figure 2-2.20 (c), on
page 36. This gives you basic dimensional constraints on the seat
cavity geometry. I am assuming that you have already designed the
window itself (i.e. required diameter and thickness), as this will
dictate the cavity geometry. D_o is the outer diameter of your
acrylic window. D_s is set by section 2-2.12.9, D_f is constrained by
the relation in the figure (1.250 <= D_o/D_f). You will note in that
figure that the dimension "K" in that figure is selected on the basis
of structural analysis - this must be so, as there is no way for the
standard to anticipate the shell loads that must be accommodated by
the viewport seat in the absence of the shell material in the hole
(since the acrylic viewport is specifically prohibited from bearing
loads in this manner).
The necessary analysis is not that difficult. Perhaps the simplest
one to start with is the direct shear, as the shear area will be
simply the circumference of the viewport (pi multiplied by D_o in the
figure) multiplied by the minimum thickness K. Calculate the total
load (pressure x area of the viewport) and then divide by the shear
area above to arrive at the shear stress, and then adjust K to get
your desired safety factor.
Next, examine the bending stress. Simplify the problem by analyzing
it as a beam in 2D, just as it is presented (in section view) in the
figure. To put that another way, imagine cutting the ring radially at
one location, and then straightening it so you end up with a bar with
a ledge on it - i.e. nothing more than a cantilevered rectangular
beam, with width (conservatively assuming the smallest diameter) b =
pi * D_f, height K, and section modulus Z = (b*K^2)/6. Conservatively
assuming that the entire load will occur at the inner diameter of the
cavity (D_f), the maximum bending stress at the corner will be the
load W (pressure * window area), multiplied by the length of the beam
l ((D_s - D_f)/2), divided by the section modulus. Obviously, the
fact that it is not a beam but a circular object will change the
assumed geometry, and the fact that the load is distributed and not a
point load will change the actual stress, but since all of the
assumptions made above are in the direction of increased conservatism,
that gives the worst-case scenario. Again, check this stress against
the material maximum, and if not sufficient, adjust K to obtain your
desired safety factor.
Torsional stress doesn't really need to be considered in this case,
since there is no torsional loading. The only other stresses
experienced by the viewport cavity are those imposed upon it from the
shell. If you have not already included reinforcement in the adjacent
shell, then the viewport cavity insert could be considered as the
necessary reinforcement as long as it meets the requirement for
sufficient material replaced. I would suggest incorporating this seat
into the shell in a manner which avoids abrupt geometry changes (i.e.
taper the outer diameter of the seat into the shell through some
combination of machining and/or weld deposition), just as with any
transition between plates of differing thickness.
-Sean
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