Surface Area of a 5D bend

1. do a google search for "surface area formula bent tubing". Lots of good info.

2. calculate the length of the centerline of the tube.

3. then use that # as the length of a straight tube.

4. circumference is Pi*D

I agree with that calculation by ddk , which depends on the length of centerline L of bend (neither the intrados nor the extrados).

Surface area, A = L _^* ∏ D, where D = Segment Pipe Outside Diameter

Yes a lot of good stuff for typical 1D bends, what I wanted to clear up was a bend of 5D or 7D radius? I also appreciate that a typical 90 degree bend, the outer is cancelled by the inner, so you can safely assume that the areas is as a straight length, this would not be the case on a long Induction heated 5D radius............. well I would think not anyway!

Cheers for your comments.

It is still the case: A = L π D (where L is the center-line arc length of the bend).

Further, L = 5D (degrees of bend) π/180.

Can you clarify for us what you mean by 5D & 7D?

I was assuming that you meant 5 Degrees and 7 Degrees.

To calrify,

5D bend for example is 5 x nominal Diameter bend radius, (7D - 7 x diameter bend radius), e.g 100mm OD with a 5D bend the radius section will be 500mm long. Regardless of the required degree bend, 45 / 65 / 90. When using Induction heating for the bending I was trying to asscertain if there was any dimensional change, I have since worked out that there is minimal outer wall thickness loss/stretch. This is compensated by a small reduction in diameter around the bend. Therefore the bend should in theory have the same surface area as a 1D typical bend. Certainly there is no real % change.

Thanks

I imagine what the op means is "D" is pipe diameter so a 5 D bend is about a mandrel of 2.5 pipe diameters radius. Or 300 = 5D = 60 mm tube/pipe od

It doesn't matter though, what you said, and Rorschach, and Tornado said is correct - in theory.

In fact; the tube/pipe is stretched more than compressed, so becomes of lesser diameter, or ovoid. But the larger the radius the less distortion, so truer to the 'neutral fiber' theory.

CAD of course is "true" only for a perfect bend - which is never achievable, even with an internal mandrel. But if it's just about painting, plating, or anodizing area - all are usually near enough.

Good, point, and that is probably doubly true with an induction heated bend as the material will be more amenable to stretching. But again, as you said, the differnence will be down in the weeds and won't matter for much.

Assuming a 90 degree bend, then you are wanting the surface area of 1/4 of a ring torus. the math works out that the surface area loss on the inside cancels out the surface area gain on the outside, so you can treat it as a cylinder of the circumferential length of the bend through the axis.

http://en.wikipedia.org/wiki/Torus

Most 3D CAD software packages have options for working out surface area. You just model the part in 3D, press a button and the answer pops up on the screen.

Weigh the <...5D bend...>. Measure its centreline length. Weigh a piece of straight pipe the area of which will be 2 x pi x length. Pro-rata the two and "Robert is your mother's brother".