BIZOMAT MAKEOS · Fabrication

Shear Spinning Blank Thickness and Force

How thick must the blank be for this shear-spun cone, and what force and power does the roller need?

Shear (cone) spinning by the sine law: starting blank thickness for a required wall thickness and cone half-angle, the ideal tangential roller force from the material’s average flow stress, thickness reduction and feed, and the power at the roller.

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How this tool works

Inputs, units and where to find them

InputUnitTypical source
Wall thickness of the spun partmmRequired wall thickness of the cone.
Half angle of the cone°Half angle of the cone (between the wall and the axis).
Average flow stress of the materialMPaAverage flow stress over the strain range (from a true stress–strain calculation).
Roller feed per revolutionmm/revAxial feed of the roller per revolution.
Diameter at the rollermmWork diameter at the roller contact.
Spindle speedrpmSpindle speed.

What each input means

Open an input to see what it is, its units, the worked example and the common mistakes, with a small diagram.

What this means: Wall thickness of the spun part
Illustration: Wall thickness of the spun partThe cross-section of a plate or wall with a dimension across its thickness, marking Wall thickness of the spun part.Twsection through the materialExample: 2 mmWall thickness of the spun part
What it is
The wall thickness the finished spun cone must have, measured square to the cone wall.
Units and format
Enter a number in mm: millimetres. The form accepts 0.000001 mm to 100 mm. Required: the tool does not calculate until every input is valid.
Example
The form opens with 2 mm as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
Effect on the result
Blank thickness is T_w ÷ sin α, so it scales directly with the wall thickness.
Get it right
Measure normal to the wall, not along the axis.
What this means: Half angle of the cone
Illustration: Half angle of the coneTwo lines meeting at a point with an arc between them, marking the angle Half angle of the cone.αExample: 30°Half angle of the cone
What it is
The half angle of the finished cone: the angle between the cone wall and the spindle axis.
Units and format
Enter a number in °: degrees of angle. The form accepts 5° to 89°. Required: the tool does not calculate until every input is valid.
Example
The form opens with 30° as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
Effect on the result
A smaller half angle needs a thicker blank and more thickness reduction, since T = T_w ÷ sin α.
Get it right
Half angle, not the included angle of the cone.
What this means: Average flow stress of the material
Illustration: Average flow stress of the materialA test bar pulled from both ends: stress Average flow stress of the material is the force carried by each square millimetre of its cross-section.σfstress = force ÷ areaExample: 300 MPaAverage flow stress of the material
What it is
The average true stress the material needs to keep deforming over the strain range of the spinning pass, from a true stress–strain curve or material data.
Units and format
Enter a number in MPa: megapascals (1 MPa = 1 N/mm²). The form accepts 0.000001 MPa to 3000 MPa. Required: the tool does not calculate until every input is valid.
Example
The form opens with 300 MPa as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
Effect on the result
Roller force and power are directly proportional to it.
Get it right
Use the average flow stress over the strain range, not the yield strength at the start of deformation.
What this means: Roller feed per revolution
Illustration: Roller feed per revolutionA cutting tool moving along a turning workpiece; Roller feed per revolution is how far the tool advances for each revolution or tooth.fadvance perrev or toothExample: 0.5 mm/revRoller feed per revolution
What it is
How far the roller advances along the cone axis in one spindle revolution.
Units and format
Enter a number in mm/rev: millimetres per revolution. The form accepts 0.000001 mm/rev to 10 mm/rev. Required: the tool does not calculate until every input is valid.
Example
The form opens with 0.5 mm/rev as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
Effect on the result
Roller force and power rise in direct proportion to the feed.
Get it right
Axial feed per revolution, not per minute.
What this means: Diameter at the roller
Illustration: Diameter at the rollerA circle with a line straight through its centre from edge to edge, marking the diameter Diameter at the roller.Ømeasured acrossthe centreExample: 200 mmDiameter at the roller
What it is
The work diameter where the roller is in contact, which sets the surface speed for the power estimate.
Units and format
Enter a number in mm: millimetres. The form accepts 0.000001 mm to 5000 mm. Required: the tool does not calculate until every input is valid.
Example
The form opens with 200 mm as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
Effect on the result
Power rises in proportion to this diameter; blank thickness and force do not depend on it.
Get it right
The diameter changes along the cone; use the largest one in the pass for the highest power.
What this means: Spindle speed
Illustration: Spindle speedA turning disc with a marker; Spindle speed is how many full turns it makes in each minute.turns of the shaftper minutenExample: 400 rpmSpindle speed
What it is
The rotational speed of the spindle and mandrel during the spinning pass.
Units and format
Enter a number in rpm: revolutions per minute. The form accepts 0.000001 rpm to 100000 rpm. Required: the tool does not calculate until every input is valid.
Example
The form opens with 400 rpm as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
Effect on the result
Power rises in proportion to speed; force and blank thickness are unchanged.

Method and formulas

  1. Sine law: T = T_w ÷ sin α (the flange keeps its blank diameter, the wall thins).
  2. Ideal tangential force F_t = σ_f (T − T_w) f.
  3. Power = F_t × surface speed at the roller.
  4. Example: a 2 mm wall at 30° half angle needs a 4 mm blank.

Limitations

  • Ideal force; friction and redundant work add to it.
  • Very small half angles exceed the material’s spinnability limit; confirm with trials.
  • Example values are illustrations. Replace them with values for your actual task.

Typical uses

  • Choose blank thickness for a spun cone
  • Estimate roller force and machine power
  • Check how steep a cone can be spun in one pass

Who it is for

  • Forming engineers
  • Spinning machine operators
  • Manufacturing engineers

What you get

  • Starting blank thickness
  • Tangential roller force
  • Power at the roller

Interpreting the result

  • Thickness reduction (1 − sin α) is the measure of severity; stay within the material’s tested limit.

References

  • Machinery’s Handbook, 29th edition (Industrial Press, 2012), printed pp. 1362–1364. Topic reference; no source tables or recommended settings are reproduced.

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Open the Shear Spinning Blank Thickness and Force calculator