2026-04-04
A 6-inch vise delivers 20–40 kN of clamping force. Distributed across two 50.8 mm (2.000″) jaw faces, that is 200–400 N/mm of contact pressure. Whether that deforms the part depends on the wall thickness, the material, and how the force distributes.
Clamping force must exceed cutting force divided by the friction coefficient between the jaw face and the part surface:
F_clamp >= F_cut / muIf the clamping force is too low, the part slips during a cut. The result is a crash — the endmill catches the moving part, the tool breaks, and the part is scrap.
Typical static friction coefficients for jaw-to-part contact:
Lower friction means you need more clamping force to hold the same cutting load. Steel-on-steel is the worst case for smooth jaws — the part wants to slide.
The tangential cutting force on a single-flute engagement:
Ft = Kc * ae * ap * fz
Where Kc is the specific cutting force (N/mm²), ae is radial
depth of cut (mm), ap is axial depth of cut (mm), and fz is feed
per tooth (mm).
Typical Kc values:
Worked example. Roughing 6061-T6 with a 6.35 mm (1/4″) endmill. 3.175 mm radial DOC, 12.7 mm axial DOC, 0.05 mm/tooth chipload:
Ft = 800 * 3.175 * 12.7 * 0.05
Ft = 800 * 2.016
Ft = 1,613 N
This is the peak tangential force on a single flute. With a 3-flute endmill, only one flute
is cutting at a time in a slotting cut (engagement angle < 120°), so
Ft is the instantaneous peak. For conservative clamping calculations, use this
peak value — it is the force that could push the part out of the jaws.
Minimum clamping force from the static equilibrium:
F_clamp >= F_cut / muIndustry standard safety factor: 2x–3x. This accounts for interrupted cuts, engagement spikes when entering material, and uncertainty in the friction coefficient (surface finish, coolant, contamination).
Worked example. The 1,613 N cutting force from above, steel jaw on aluminum (mu = 0.4), safety factor 2.5:
F_clamp >= 1,613 / 0.4 * 2.5
F_clamp >= 4,032.5 * 2.5
F_clamp >= 10,081 NA 6-inch vise at moderate handle torque delivers 20,000+ N. The requirement is met with margin. This is typical — for aluminum in a properly sized vise, slippage is not the failure mode. The failure mode is deformation.
Contact pressure from the clamping force:
P = F_clamp / (jaw_height * contact_length * 2 jaws)Worked example. 20 kN clamping force, 50.8 mm jaw height, 25 mm part contact length per side:
P = 20,000 / (50.8 * 25 * 2)
P = 20,000 / 2,540
P = 7.87 MPaCompare to material yield strengths:
For a solid part with full jaw contact, the surface pressure from a vise is far below yield. The jaw will not mark the part through direct compression.
But thin walls change everything. A thin wall fails in bending, not compression. The clamping force pushes the wall inward. The wall acts as a cantilever — fixed at the base, loaded along the jaw contact height. The bending stress at the base of the wall is:
sigma = (F_wall * h) / (L * t^2 / 6)
Where F_wall is the force on one wall (half the total clamping force),
h is the height where the force acts (half the jaw height for distributed
load), L is the wall length, and t is the wall thickness. The
t² term dominates — halving the wall thickness quadruples the
bending stress.
A 2 mm wall on a 6061-T6 part can deflect visibly at 20 kN of clamping force. The jaw does not mark the surface through compression — it bows the wall inward. The part comes out of the vise with a permanent set. This is why minimum wall thickness (typically 2–3 mm for aluminum) is a critical parameter in automated fixture generation.
The clamping-force-versus-cutting-force model assumes friction-based workholding — a vise or clamp pressing on the part. Other fixturing methods have different force models:
For these methods, the limiting factor is total holding force, not contact pressure. Deformation is rarely a concern because the forces are lower. The tradeoff is that they cannot resist the cutting forces that a vise handles easily.
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