Forming Force Calculations for PBR Roll Forming

Practical Methods to Estimate Roll Load, Drive Torque, Motor Size, and “Why My Machine Is Struggling”

Forming force in PBR roll forming is hard to “calculate exactly” because it depends on tooling radii, pass design, friction, strain path, and how the profile is distributed across stands. But you can get solid, usable estimates for:

• Forming load per stand (kN)

• Drive torque (N·m)

• Required motor power (kW / HP)

• Why load spikes happen (over-compression, high tensile, poor pass design)

Below are the most practical calculation approaches used in real factories.

1) Key Inputs You Need

Material

• Thickness, t (mm)

• Yield strength, σy (MPa = N/mm²)

• Tensile strength (optional for check)

• Coating/friction behavior (GI/GL/painted)

Geometry

• Coil width (mm)

• Effective “formed length” of bends per meter (or per panel)

• Minimum inside bend radius in the tooling, r (mm)

• Number of stands, pass progression

Process

• Line speed v (m/min)

• Efficiency factor (drivetrain + slip), η (typically 0.75–0.9)

• Friction factor (depends on lubrication/finish)

2) The Most Useful Concept: Forming Force Is Driven by Plastic Work

A very practical way is to estimate power from plastic work and then convert to force/torque.

Step A — Estimate energy per unit length (plastic work per meter)

A simplified estimate:

W′≈K⋅σy⋅t2⋅(∑1ri)W' \approx K \cdot \sigma_y \cdot t^2 \cdot \Big(\sum \frac{1}{r_i}\Big)W′≈K⋅σy⋅t2⋅(∑ri1)

Where:

• W′W'W′ = work per unit length (N) ≈ (J/m)

• KKK = empirical constant (typically 1.0–3.0 depending on profile complexity & friction)

• σy\sigma_yσy in N/mm²

• ttt in mm

• rir_iri are the effective bend radii formed across the profile (mm)

Interpretation: thinner steel drops load fast (t²), high-yield increases load linearly, tight radii increase load.

This isn’t perfect, but it’s extremely useful for comparing scenarios (e.g., 29ga vs 26ga, 345 MPa vs 550 MPa).

3) Quick “Rule-of-Thumb” Load Scaling (Very Practical)

If you already have one known stable setup, load changes can be estimated by ratios:

Thickness change effect (biggest driver)

F2F1≈(t2t1)2\frac{F_2}{F_1} \approx \Big(\frac{t_2}{t_1}\Big)^2F1F2≈(t1t2)2

Yield strength change effect

F2F1≈σy2σy1\frac{F_2}{F_1} \approx \frac{\sigma_{y2}}{\sigma_{y1}}F1F2≈σy1σy2

Combined

F2F1≈(t2t1)2⋅σy2σy1\frac{F_2}{F_1} \approx \Big(\frac{t_2}{t_1}\Big)^2 \cdot \frac{\sigma_{y2}}{\sigma_{y1}}F1F2≈(t1t2)2⋅σy1σy2

• Example (very real-world):
• Moving from 0.46mm to 0.55mm thickness:
• (0.55/0.46)² ≈ (1.1957)² ≈ 1.43 → ~43% more load before you even consider strength changes.

If yield increases from 350 MPa to 550 MPa:
550/350 = 1.57 → ~57% more load

Combined: 1.43 × 1.57 ≈ 2.24 → ~124% more load (more than double)

That’s why machines suddenly trip overload when “the gauge looks close.”

4) Estimating Drive Power from Forming “Line Force”

If you can estimate a net forming resistance force along the line, FFF (N), then:

P=F⋅vηP = \frac{F \cdot v}{\eta}P=ηF⋅v

Where:

• PPP is power (W)

• vvv is line speed (m/s)

• η is overall efficiency (0.75–0.9 typical)

Convert speed:
m/min → m/s by dividing by 60.

Example

Assume net forming resistance F=6,000 NF = 6{,}000 \,NF=6,000N (6 kN), speed 30 m/min = 0.5 m/s, η = 0.8:

P=6000⋅0.50.8=3750W≈3.75kWP = \frac{6000 \cdot 0.5}{0.8}=3750W \approx 3.75kWP=0.86000⋅0.5=3750W≈3.75kW

That’s just forming resistance. Now add:

• drivetrain losses

• acceleration / inertia

• shear hydraulics (if powered from same motor system)

• safety margin

So you may select 7.5–11 kW depending on design.

5) Estimating Torque at the Main Drive

If you know motor power and RPM at the gearbox output:

T=9550⋅PkWnrpmT = \frac{9550 \cdot P_{kW}}{n_{rpm}}T=nrpm9550⋅PkW

Where:

• TTT in N·m

• PkWP_{kW}PkW in kW

• nnn in rpm

Example: 11 kW at 60 rpm:
T = 9550×11/60 ≈ 1751 N·m

If you have multiple drive points (group drives), torque divides, but unevenly if alignment/friction differs.

6) Per-Stand Load Distribution (How Engineers Think About It)

A PBR machine doesn’t take full load at one stand—if pass design is correct.

Typical distribution (conceptual):

• Stands 1–4: entry forming + light shaping (lower load)

• Mid stands: primary rib formation (highest load)

• Final stands: sizing + slight correction (should be moderate, not crushing)

Red flag: if your final stands are hottest / most loaded, you’re usually over-compressing to “fix” shape. That drives:

• bearing heat

• rib distortion

• edge buckling

• tool wear

7) What Makes Loads Spike Beyond Calculations

These are the real-world multipliers:

• Over-tight roll gap (crushing instead of forming)

• Poor strip tracking / side load

• High friction from zinc pickup or rough roll finish

• Coil camber / residual stress

• Too aggressive early passes (thin gauge buckles then “fights” the tooling)

• Misalignment / shaft runout / bearing play (dynamic impact forces)

If your motor current jumps but thickness and strength didn’t change, check these first.

8) Simple “Factory-Usable” Force Check Using Motor Current

If you log motor current (amps) at stable speed:

• Current is a great proxy for forming load.

• Build a baseline chart for each gauge/strength.

Best practice: create a “Normal Operating Current Range” for:

• 29ga low tensile

• 29ga high tensile

• 26ga low tensile

• 26ga high tensile

Then you can detect misalignment or over-compression early.

9) Practical Output: What to Put in a Spec Sheet

If you want your pages to look like real engineering, include:

• Rated thickness range and max yield strength

• Stands count and main shaft diameter

• Main drive motor kW/HP and gearbox ratio

• Typical line speed range by thickness

• Expected motor current range per gauge (if you want to be very pro)

FAQ

Why does load scale with thickness squared (t²)?
Because bending stiffness and plastic section behavior increase strongly with thickness; in practice, thickness changes dominate load.

Why do high-tensile coils cause overload even at same thickness?
Because forming requires plastic deformation; higher yield means higher stress to bend, and springback makes operators tighten rolls, compounding load.

Can I calculate exact kN per stand for PBR?
Not precisely without tooling geometry + pass design data + friction and contact modeling. But you can estimate and compare scenarios accurately enough for motor sizing and troubleshooting.