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 elasticity/" data-glossary-id="141729" data-glossary-teaser="Measure of rigidity or stiffness of a metal, defined as a ratio of stress, below the proportional limit, to the corresponding strain. Also known as Young’s modulus." title="Measure of rigidity or stiffness of a metal, defined as a ratio of stress, below the proportional limit, to the corresponding strain. Also known as Young’s modulus." aria-label="Glossary: modulus of elasticity">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

Cutting Force (F): lbf or N
Overhang Length (L): in or mm
elasticity/" data-glossary-id="141729" data-glossary-teaser="Measure of rigidity or stiffness of a metal, defined as a ratio of stress, below the proportional limit, to the corresponding strain. Also known as Young’s modulus." title="Measure of rigidity or stiffness of a metal, defined as a ratio of stress, below the proportional limit, to the corresponding strain. Also known as Young’s modulus." aria-label="Glossary: modulus of elasticity">Modulus of Elasticity (E): psi or N/mm²
Moment of Inertia (I): in⁴ or mm⁴
Calculate Deflection

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 × K_p × 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.

K_p 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, K_p = 0.76 hp/in.³/min.

Feed factor, C = 1.08.

Adjusted power constant, K_pa = K_p × C = 0.76 × 1.08 = 0.82.

Calculating:

F = 396,000 × d × f × K_pa = 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 × K_p × 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.

K_p 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.

Power constant, K_p = 0.0346kW/cm³/min.

Feed factor, C = 1.08.

Adjusted power constant, K_pa = K_p × C = 0.0346 × 1.08 = 0.0374.

Calculating:

F = 60,000 × d × f × K_pa = 60,000 × 2.03 × 0.2 × 0.0374 = 911.1 N

Comparing the data of the cutting forces, we can say that formulas (2) and (3) provide sufficient
accuracy.

By converting the cutting force of 207.8 lbf (customary U.S. units of measure) into metric units of measure, we get:

F = 207.8 × 4.448 = 924.3 N (compare with 911.1 N)

The difference is:

924.3 N - 911.1 N = 13.2 N, or 1.4%

By converting the cutting force of 911.1 N (metric units of measure) into customary U.S. units of measure, we get:

F = 911.1 N × 0.2248 = 204.8 lbf (compare with 207.8 lbf)

The difference is:

207.8 lbf - 204.8 lbf = 3 lbf, or 1.4%

Cutting Force Calculator (U.S. Units)

Depth of Cut (d): in
Feed Rate (f): ipr
Power Constant (Kp): hp/in³/min
Feed Factor (C):
Calculate Cutting Force

Formula: F = 396,000 * d * f * Kp * C

Moduli of Elasticity

Boring bar shanks are made of steel, tungsten-base alloys or cemented carbide. The most frequently used boring bar material is alloy steel. Some boring bar manufacturers use AISI 1144 free-machining medium-carbon steel. Regardless of their grades, all carbon and alloy steels have the same modulus of elasticity:

Customary U.S. units of measure: E = 30 × 10⁶ psi (4)

Metric units of measure: E = 20.6 × 10⁴ N/mm² (5)

A common mistake is to assume that a steel shank with a higher hardness or a steel shank made from higher-quality steel will deflect less. As can be seen from equation (1), the material property that determines deflection is the modulus of elasticity. Hardness does not appear in this equation.

Tungsten heavy alloys for boring bars: E = (45 to 48) × 10⁶ psi (customary U.S. units of measure) and E = (31 to 33) × 10⁴ N/mm² (metric units of measure).

Boring bars made of tungsten heavy alloys will deflect less than steel boring bars of the same diameter and overhang by 50% to 60% when cutting at the same DOC and feed rate.

Cemented carbides for boring bars: E = (84 to 89) × 10⁶ psi (customary U.S. units of measure) and E = (52 to 61) × 10⁴ N/mm² (metric units of measure).

Boring bars made of cemented carbide provide minimum deflection because their moduli of elasticity are higher than those of steel and tungsten heavy alloys.

Modulus of Elasticity Converter

Modulus of Elasticity:
From Unit: psiN/mm²
To Unit: psiN/mm²
Convert

Moment of Inertia

The moment of inertia is a property of areas. Because boring bars are available in various diameters, it is important to calculate the area of a bar cross section using appropriate formulas. A boring bar is usually round with a solid or tubular cross section. The moment of inertia of a solid cross section area is calculated by:

I = π × D₀⁴ ÷ 64 (6)

Where D₀ is a bar OD in in. or mm. Moment of inertia of a tubular cross section area is calculated by:

I = π × (D₀⁴ - D_i⁴) ÷ 64 (7)

Where D_i is a bar ID in in. or mm.

Example of calculating the moment of inertia I (customary U.S. units of measure)

Given: Outside diameter is 1". The moment of inertia is calculated using formula (6).

I = π × 1⁴ ÷ 64 ≈ 0.0491 in.⁴

Example of calculating the moment of inertia I (metric units of measure)

Given: OD is 25.4mm. The moment of inertia is calculated using formula (6).

I = π × 25.4⁴ ÷ 64 = π × 416,231.4 ÷ 64 = π × 6,503.6 = 20,432mm⁴

Moment of Inertia Calculator (Solid Bar)

Diameter (D): in or mm
Calculate Moment of Inertia

Formula for solid bar: I = π * D⁴ / 64

Deflection

Example 1: Unsupported length of a boring bar (overhang), L = 4".

Calculating deflection of the bar (customary U.S. units of measure)

y = FL³ ÷ 3EI = 207.8 × 4³ ÷ 3 × 30 × 10⁶ × 0.0491 = 12,652.8 ÷ 4,419,000 ≈ 0.003"

Example 2: Unsupported length of a boring bar (overhang), L = 101.6mm

Calculating deflection of the boring bar (metric units of measure)

y = FL³ ÷ 3EI = 911.1 × 101.6³ ÷ 3 × 20.6 × 10⁴ × 20,432 = 955,536,257 ÷ 3 × 4,208,992,000 = 0.0757mm

Let's compare: Deflection of the boring bar in example 1 is 0.003", or 0.0762mm, and deflection of the boring bar in example 2 is 0.0757mm, or 0.00298".

To help ensure a successful boring operation, the overhang should be as short as possible to minimize boring bar deflection. In addition, the boring bar should be made of either tungsten heavy alloy or cemented carbide, and the bar diameter should be as large as possible to achieve the maximum moment of inertia.