Alan-Tucson,
The 1000 ft radius would say that the bend could be approximated by individual segments 50 ft length (add nodes between support points to get midpoint deflection / droop information). Bend SIF would be esentially same as pipe. Basic trigonometry 300 ft over 1000 ft gives arc of approximately 16 degrees. Each segment (300/50 = 6) would be rotated by increment of 16/5 = 3.2 deg to make the 16 deg bend. The CAESARII elements could be defined by a straight line of 6 spans and then rotating each following span by 3.2 deg, or you could do the trigonometry to input the x and z coordinates of each node. (approx 2.8ft offset for each 50 ft span). You could 'smooth' out the bend by rotating also at each of the midpoint nodes.
Your decision.
If the pipe is supplied in double random or triple random lengths of 40 / 60 ft, then the weld joints will be typically away from the welds and not be necessary to include the welds in the model. Locate the supports with your 50 ft spacing. I would expect the field to be able to bend the pipe into the 1000 ft radius with proper equipment (Caterpillar D50 dozers?).
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R Yee