Segmental Bend Software (SEGB)- Math first created 24/01/06 - last modified 17/02/06 Page Author: Ty Harness
Unfinished.
Right Segmental bend vs. Oblique segmental Bend. A right segmental bend has two half segments on either end. The benefit is that the input and output diameters
are the same as the cross section of the tube. Thus the tube has a circular cross section. This works well when you
are working with existing tube stock.
Figure XX - Right segmental bend and an oblique segmental bend
An oblique segmental bend suffers from a reduced tube area. By forcing the input and output ends to be circular produces
an elliptical cross section where where D is the major axis diameter and De is the minor axis diameter.
Usually a sheet metal worker uses a minimum of 6 segments and
the reduction in tube area is irrelevent for most jobs.
$D_e = D*cos(M)$
The area of the elliptical area is
$D_e = pi*((D*D_e)/4)$
Finding the surface area of a cylinder with oblique cuts as used for segmental bends.
SEGB software uses an approximate method by summing all the quadrilateral elements around the cylinder.
The finite number of segments always introduces a truncation error.
Figure XX - Quadrilateral Area
The quadrilateral elements are actually trapezium elements in SEGB, but for a more general case we can say the
quadrilateral element is made up of 2 triangles. Because the face vertex information is readily available the use of Heron's
formula is easily applied.
The correct surface of area for a Right Segmental lobster back pattern
Figure XX - A right cylinder with two oblique cuts
From the above figure you can see that a sliced off portion of the cylinder can be welded on top to
form a right cylinder. Therefore it's easy to calculate the outer surface area:
$A_T = 2*pi*R*(a + b)$
That doesn't tell us anything about the curve, but the curve is described by:
$dA = y* dx = b*(1/2 + cos(theta)/2) * R d theta $
$A =b*R int_0^(2 pi) (1/2 + cos(theta)/2) d theta $
$A = [theta/2 - sin(theta)/2]_0^(2*pi) $
$A = pi*b*R $
$A_T = 2*A + 2*pi*R*a = 2*pi*R*(a + b) $
Even with a right segmental the half segments are just half the area of the full segments.
It's useful knowing the internal volume of the segment because with it being a pipe we're concerned with
the fluid inside the pipe. Again using the same technique as above it's very easy to calculate the Volume
of segment.