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From Cutting Tool Engineering

Boring: Calculating deflection: Turning Performance

Eliminating chatter, especially when boring deep holes, is one of the greatest machining challenges.

July 15, 2019By Edmund Isakov, Ph.D.

To help ensure a successful boring operation, the overhang should be as short as possible to minimize boring bar deflection.

Boring is an internal turning operation performed with a boring bar to enlarge a previously drilled hole to form an internal shape of specified dimensions. Boring operations range from semiroughing to finishing.

A boring bar has three basic elements: an indexable cutting insert, a shank and an anchor. The designation system for indexable inserts is the same as for turning. The anchor is the clamping portion of the shank that is held in the tool block, and the minimum clamping length is approximately three to four diameters of the shank. The distance the boring bar extends beyond the tool block, which is called overhang, determines the cutting depth. The overhang is the unsupported portion of the boring bar. Long overhang causes excessive deflection of the shank and generates vibration, or chatter, which deteriorates the surface finish of the bore.

Eliminating chatter, especially when boring deep holes, is one of the greatest challenges faced by manufacturers and users of boring bars. Deflection of a boring bar depends on the mechanical properties of the shank material, the length of the overhang and the cutting conditions.

The following equation is used to calculate deflection (y) of a boring bar:

y = FL3/3EI (1)

Where

F is the cutting force, lbf or N.

L is the unsupported length of a boring bar (overhang), in. or mm.

E is the modulus of elasticity (in tension) of a boring bar material, psi or N/mm2.

I is the moment of inertia of a boring bar cross section area, in.4 or mm4.

Boring Bar Deflection Calculator





function calculateDeflection() { let F = parseFloat(document.getElementById(‘force’).value); let L = parseFloat(document.getElementById(‘length’).value); let E = parseFloat(document.getElementById(‘elasticity’).value); let I = parseFloat(document.getElementById(‘inertia’).value); let deflection = (F * Math.pow(L, 3)) / (3 * E * I); document.getElementById(‘deflectionResult’).innerHTML = “Deflection (y) = ” + deflection.toFixed(4);}

Formula: y = (F * L³) / (3 * E * I)

Cutting Force

Cutting force (F) expressed in customary U.S. units of measure, calculated by formula (2):

F = 396,000 × d × f × Kp × C (2)

Where

The number 396,000 is expressed through:

A unit of power equal to 550 ft.-lbs./sec.: 550.

A unit of power equal to 550 ft.-lbs./min.: 550 × 60 = 33,000.

A unit of power equal to 550 in.-lbs./min.: 33,000 × 12 = 396,000.

d is DOC, in.

f is a feed rate, ipm.

Kp is a power constant, hp/in.3/min.

C is the feed rate factor for the power constant adjustment.

Example of calculating the cutting force (F)

Given:

The workpiece is AISI 4140 chromium-molybdenum steel, 220 to 240 HB.

DOC, d = 0.08″.

Feed rate, f = 0.008 ipr.

Power constant, Kp = 0.76 hp/in.3/min.

Feed factor, C = 1.08.

Adjusted power constant, Kpa = Kp × C = 0.76 × 1.08 = 0.82.

Calculating:

F = 396,000 × d × f × Kpa = 396,000 × 0.08 × 0.008 × 0.82 = 207.8 lbf

Cutting force (F) expressed in metric units of measure, calculated by formula (3).

F = 60,000 × d × f × Kp × C (3)

Where

The number 60,000 is expressed through:

A unit of power equal to 1kW × m/sec.: 1.

A unit of power equal to 1kW × m/min.: 1 × 60 = 60.

A unit of power equal to 1kW × mm/min.: 1,000 × 60 = 60,000.

d is DOC, mm.

f is a feed rate, mm/min.

Kp is a power constant, kW/cm3/min.

C is the feed rate factor for the power constant adjustment.

Example of calculating the cutting force (F)

Given:

The workpiece is AISI 4140 chromium-molybdenum steel, 220 to 240 HB.

DOC, d = 2.03mm.

Feed rate, f = 0.2mm/rev.

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