"stress = ( pressure * diameter ) / ( 2 * wall thickness )

For a 36" diameter, .312 wall tube, of 40,000 psi yield strength, that would be 693 psi internal pressure to start yielding."

That's good, now I don't have to rethink my basic design:) {which I started to do [& btw, thought of one way it might still work- but it was more complicated]}.
So for the moment not considering the aspect of not having the "infinitely strong" weld....:
16 pipes which were 3' diameter with .312 walls with approx 1' foot wide spacers between (which are .624 plate steel).
Gives a diameter of about 20' (240") at weld point (inside around 18' including bumps).
Assuming a pressure of say 120 psi and using 240":
Stress = 120 * 240 / 1.248
So that's 23,077, which is not having much safety margin nor allowing for any pressure in the 3' pipes if using steel with 40,000 psi yield str.
Hmm, if use 100 psi and include bumps (216"), I get 17,307. Still pretty grim- though interesting. What it means to me at first blush is that a small pipelauncher could probably use steel which had around 40k psi, larger pipelaunchers [over 1000'] will probably require steel with higher psi yield str (which would probably be harder to weld).
Well, got something to mull over, and maybe I'll take that trip to library- for some of that good old fashion learning:)