We are building a mandrel to bend a 3" rad(6" dia.) on a 14" long piece of
1/8" 3003. Problem we are encountering is the amount of springback opens the
piece back up to 7 1/4" dia. Even when held in place and heated to relieve
tension the part still has more spring than we would have thought. This
doesn't happen when we spin it but in this particualr case we need to mandrel
bend the part. Any help would be greatly appreciated. We are also planning to
bend other rads so trying to build the right diameter mandrel first time is
essential.

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On Tue, 27 Nov 2012 18:18:02 +0000, randy wrote:

We are building a mandrel to bend a 3" rad(6" dia.) on a 14" long piece of
1/8" 3003. Problem we are encountering is the amount of springback opens the
piece back up to 7 1/4" dia. Even when held in place and heated to relieve
tension the part still has more spring than we would have thought. This
doesn't happen when we spin it but in this particualr case we need to mandrel
bend the part. Any help would be greatly appreciated. We are also planning to
bend other rads so trying to build the right diameter mandrel first time is
essential.

I don't know a general formula and also don't know of a source of
appropriate constants if you had a formula.

Here are some lines from Marvin Klotz's MANDREL.C mandrel size
calculator program, which has wt = 0 for music wire and wt = 1
for phosphorus bronze:

/* MANDREL.C M. W. Klotz 2/05
Computing mandrel size for spring winding. Based on Kozo Hiraoka's
article in "Home Shop Machinist", July/August 1987, pg. 30. */
....
int wt; //wire type index
dbl c0[]={0.980364,0.012436}; //constant coefficient
dbl c1[]={-0.012436,-0.11018}; //first order coefficient
dbl id; //spring internal diameter (in)
dbl ds; //average spring diameter (in)
dbl dw; //wire diameter (in)
dbl fact; //empirical factor (nd)
dbl dm; //mandrel diameter (in)
....
ds=id+dw; //spring average diameter
fact=c0[wt]+c1[wt]*ds/dw; //empirical factor
dm=fact*ds-dw; //mandrel diameter

You could set up a handful of different mandrel sizes,
eg {3, 4, 5, 6, 7} inches, bend a piece around each, and from
the resulting measurements see if the model applies and what
the constants should be. The bend you mentioned, with a 6" diameter
mandrel giving a 7.25" diameter piece, gives a data point with dm=6,
dw=.125, and ds = 7.25 + dw (if the 7.25" was an inside measurement).

I assume you are bending a bar or plate instead of a wire but don't
know if the model needs to be adjusted because of that.

--
jiw

responding to
http://www.polytechforum.com/metalwo...en-560540-.htm
, Randy wrote:
not wrote:

On Tue, 27 Nov 2012 18:18:02 +0000, randy wrote:

We are building a mandrel to bend a 3" rad(6" dia.) on a
14" long piece of
1/8" 3003. Problem we are encountering is the amount of
springback opens the
piece back up to 7 1/4" dia. Even when held in place and
heated to relieve
tension the part still has more spring than we would have thought.
This
doesn't happen when we spin it but in this particualr case we need
to mandrel
bend the part. Any help would be greatly appreciated. We are also
planning to
bend other rads so trying to build the right diameter mandrel
first time is
essential.

I don't know a general formula and also don't know of a source of
appropriate constants if you had a formula.

Here are some lines from Marvin Klotz's MANDREL.C mandrel size
calculator program, which has wt = 0 for music wire and wt = 1
for phosphorus bronze:

/* MANDREL.C M. W. Klotz 2/05
Computing mandrel size for spring winding. Based on Kozo Hiraoka's
article in "Home Shop Machinist", July/August 1987, pg. 30.
*/
...
int wt; //wire type index
dbl c0[]=; //constant coefficient
dbl c1[]=; //first order coefficient
dbl id; //spring internal diameter (in)
dbl ds; //average spring diameter (in)
dbl dw; //wire diameter (in)
dbl fact; //empirical factor (nd)
dbl dm; //mandrel diameter (in)
...
ds=id+dw; //spring average diameter
fact=c0[wt]+c1[wt]*ds/dw; //empirical factor
dm=fact*ds-dw; //mandrel diameter

You could set up a handful of different mandrel sizes,
eg {3, 4, 5, 6, 7} inches, bend a piece around each, and from
the resulting measurements see if the model applies and what
the constants should be. The bend you mentioned, with a 6"
diameter
mandrel giving a 7.25" diameter piece, gives a data point with
dm=6,
dw=.125, and ds = 7.25 + dw (if the 7.25" was an inside
measurement).

I assume you are bending a bar or plate instead of a wire but don't
know if the model needs to be adjusted because of that.

--
jiw

Thanks. We are already making smaller mandrels to find what works. Only
problem with this is it is cost prohibitive when we have to pay a machine shop
to do so. However, if we find some ratio that is common for our part we will
then be able to get other change parts much closer on the first attempt.
Kindest regards, Randy.

On Tuesday, November 27, 2012 9:18:02 PM UTC-5, Randy wrote:

Thanks. We are already making smaller mandrels to find what works. Only

problem with this is it is cost prohibitive when we have to pay a machine shop

to do so. However, if we find some ratio that is common for our part we will

then be able to get other change parts much closer on the first attempt.

Kindest regards, Randy.

Just remember with smaller mandrels , the springback is less. As long as you are using the same rod , you should be able to plot the amount of springback for three mandrels and be able to predict the springback for other sized mandrels. Couldn't you use some stock sized pipe or tubing to see how much springback you get at various diameters to get real close to the right size of mandrel needed. How precision do you need the finished product?

Dan

responding to
http://www.polytechforum.com/metalwo...en-560540-.htm
, randy wrote:
dcaster wrote:

On Tuesday, November 27, 2012 9:18:02 PM UTC-5, Randy wrote:

Thanks. We are already making smaller mandrels to find what works.
Only

problem with this is it is cost prohibitive when we have to pay a
machine
shop

to do so. However, if we find some ratio that is common for our
part we w
ill

then be able to get other change parts much closer on the first
attempt.

Kindest regards, Randy.

Just remember with smaller mandrels , the springback is less. As long
as y
ou are using the same rod , you should be able to plot the amount of
spring
back for three mandrels and be able to predict the springback for other
siz
ed mandrels. Couldn't you use some stock sized pipe or tubing to see
how m
uch springback you get at various diameters to get real close to the
right
size of mandrel needed. How precision do you need the finished
product?

Dan




-- Thanks Dan, but we are bending 1/8" sheet. Piece is 14x18 x 1/8. The
centerline of the bend will be at the halfway point of the 18" side. We will
plot as suggested as there are several different pieces (diameters) we will
eventually need to make. Using some off the shelf pipe is an excellent
suggestion. The shop should have plenty of end cuts laying around.
Randy.

On Wednesday, November 28, 2012 4:28:04 PM UTC-5, randy wrote:
responding to



-- Thanks Dan, but we are bending 1/8" sheet. Piece is 14x18 x 1/8. The

centerline of the bend will be at the halfway point of the 18" side. We will

plot as suggested as there are several different pieces (diameters) we will

eventually need to make. Using some off the shelf pipe is an excellent

suggestion. The shop should have plenty of end cuts laying around.

Randy.

You also should be able to use say 1 inch wide strips of 3003 to get how much springback there is. I see no reason to use a wide sheet.

Dan