Why are there 2 angular cases for calculating the setback for bend deduction?

[alewis]

Why are there 2 angular cases for calculating the setback (L1a and L1b) when calculating bend deduction? In other words, why are angles between 90° and 180° calculated as just a 90° removing the tan (β/2) factor?

Thank you.

0° ≤ A ≤ 90° --> L_BA=2 ( ρ+µ ) tan (β/2) - δ

90° < A ≤ 180° --> L_BA=2 ( ρ+µ )  - δ

http://help.autodesk.com/view/INVNTOR/2019/ENU/?guid=GUID-715EE575-76F0-456A-BEC1-48D3401C9255

[johnsonshiue]

Hi! Actually, the Help page explains quite clearly. If you look at the two pictures, the way L1b is calculated between 90 deg and 180 deg is quite different than the one in less than 90 deg. Two different equations are needed.

Many thanks!

[alewis]

Johnson Shiue

Hello. Thank you for your response. I see in the picture that L1b has a different equation. The angular case 0° ≤ β ≤ 90° is more understandable with 2 equal right triangles highlighted in two colors with sides L1b and L2b clearly dimensioned. The angular case 90° < β ≤ 180° is more difficult to see geometrically. The picture does show a right triangle is formed by the (ρ+µ) side with L1b dimensioned. By drawing my own bend, I can see L2b which is also a right-triangle side formed by the corresponding (ρ+µ) side. Does this sound correct so far?

I think my question may be more mechanical. Both cases come to the equation L_BA = L1b + L2b + v, which is simplified from the Bend Allowance equation L_Total = L1 + L2 + BA. Yet, the case 0° ≤ β ≤ 90° using a setback length measured to the virtual sharp and the case 90° < β does not.

What is the reason for this geometric setback difference when all angles up to 180° could be calculated?

Thank you for your time.

Adam

[johnsonshiue]

Hi! When the Bend Angle is less than 90 deg, mechanically the Bend tends to stretch the sheet proportionally less than Bend Angle greater than 90 deg. Two equations are needed to describe the behaviors.

If the Bend Allowance is proportional to Bend Angle, you can simply use K-factor to calculate it. There is no need to have two equations.

Many thanks!

[alewis]

Johnson Shiue

Hello. I don’t understand the purpose of the Bend Deduction type fully. Is the Bend Deduction in the equation L_Total = L1 + L2 – δ intended to be deduced from test bent plate measurements?

Thank You

Adam

[johnsonshiue]

Hi! Yes, the Bend Deduction is the delta in the equation. It is basically the amount to be deducted from the leg length.

Many thanks!

[alewis · Accepted solution]

Johnson Shiue

Hello. I have analyzed the Bend Deduction equation and see there is some error if all angles use

L_BA=2 ( ρ+µ ) tan (β/2) – δ. As the bend angle increases past 90°, the setback also increases. This creates a very long setback. If the legs are shorter than the setback then the total developed length will be negative. So, this leads to another question. Why do angles less than 90° use the equation? The angular case 90° < β ≤ 180° measures the setback to a point on the outside of the bend, where a perpendicular line to the leg is tangent to the bend. This setback point does not work on the angular case 0° ≤ β ≤ 90° because this tangent point would be off the bend. Therefore, this would not equal ρ+µ.

So, my question was a geometry one since the mechanical factors are independent inside of the K-factor. Have I come to a correct conclusion?

Thank you for your help.

Adam

[Hochenauer · Accepted solution]

Because L1 and L2 are defined differently for the two cases. For acute angles they are going to the virtual intersection, which is not directly measurable on the part. For obtuse angles one could use a caliper to measure the leg directly, which is more convenient. 

Because the leg lengths are defined as going to the tangent point on the arc, the tan(β/2) term is no longer relevant for those angles. It would be needed if we still were referring to the virtual intersection.

Kind regards,

Gerald