ME698S Machining Dynamics - IITKhome.iitk.ac.in/~mlaw/ME698S_Oblique_cutting_updated.pdf · IIT Kanpur ME698S – Machining Dynamics Dr. Mohit Law [email protected] Oblique Cutting

IIT Kanpur

ME698S – Machining Dynamics

Dr. Mohit Law

[email protected]

Oblique Cutting

mailto:[email protected]

IIT Kanpur

Orthogonal and oblique cutting geometry

2

Cutting velocity is inclined at an acute angle 𝑖 to the cutting edge

Cutting velocity is perpendicular to cutting edge

Altintas, Mfg. Automation

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Oblique cutting geometry

3

Considerations: 1. Cutting edge is perfectly sharp

and no edge rubbing 2. Continuous chip with no built up-

edge 3. Tool subjected to 3D system of

cutting forces because of angle of obliquity, 𝑖

4. Non-plane strain deformation (treated however as a modified plane strain problem)

5. Uniform stress distribution on shear plane

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Recalling orthogonal cutting

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• Plane normal to the cutting edge, and aligned

with the cutting velocity 𝑉 is the normal

plane 𝑃𝑛

• Because of plane strain deformation (and no

side spread), shearing and chip motion are

identical on all normal planes ∥ to 𝑉and ⊥ to

the cutting edge

• Hence, all velocities 𝑉, 𝑉𝑠, and 𝑉𝑐 are all ⊥ to

the cutting edge and lie in the velocity plane

𝑃𝑣 ∥ to or coincident with 𝑃𝑛

• Resultant force 𝐹𝑐, along with other forces

acting on the shear and chip-rake face contact

zone, also lie in the normal plane 𝑃𝑛

• No cutting force ⊥ to 𝑃𝑛, and edge forces = 0

Normal plane 𝑃𝑛

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• Cutting velocity is inclined at an acute

angle 𝑖 to the cutting edge, hence

direction of shear, friction, chip flow, and

resultant force vectors have components

in all there Cartesian coordinates 𝑥, 𝑦, 𝑧

• Plane normal to the cutting edge, and

inclined at an acute angle 𝑖 with the

cutting velocity 𝑉 is the normal plane 𝑃𝑛

• 𝑥 is ⊥ to the cutting edge, but lies on cut

surface 𝑦 is aligned with cutting edge 𝑧 is

⊥ to 𝑥𝑦 plane

• Important planes are shear plane, rake

face, cut surface 𝑥𝑦 normal plane 𝑥𝑧 (or

𝑃𝑛), and the velocity plane 𝑃𝑣

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• Assume that mechanics of oblique

cutting in the normal plane 𝑃𝑛 are

equivalent to orthogonal cutting. Hence

project velocities and forces into normal

plane

• Angle between shear and 𝑥𝑦 plane is the

normal shear angle - 𝜙𝑛

• Shear velocity lies on shear plane but

makes an oblique shear angle - 𝜙𝑖 with

the vector normal to the cutting edge on

the normal plane

• Sheared chip moves over rake face with a

chip flow angle of - 𝜂 measured from a

vector on the rake face but normal to

cutting edge

• Angle between 𝑧 axis and normal vector

on rake face is the normal rake angle - 𝛼𝑛

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Force diagram – oblique cutting

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• Friction force on the rake face 𝐹 𝑢 and

normal force to the rake 𝐹 𝑣 form a resultant

cutting force 𝐹 𝑐 with a friction angle 𝛽𝑎

• This resultant 𝐹 𝑐 has an acute projection angle of 𝜃𝑖 with the normal plane 𝑃𝑛, which in turn has an in-plane angle of 𝜃𝑛 + 𝛼𝑛

with the normal force 𝐹 𝑣 • Where 𝜃𝑛 is the angle between the 𝑥 axis

and the projection of 𝐹 𝑐 on 𝑃𝑛

𝐹𝑢 = 𝐹𝑐 sin 𝛽𝑎 = 𝐹sin 𝜃𝑖

sin 𝜂→ sin 𝜃𝑖 = sin 𝛽𝑎 sin 𝜂

𝐹𝑢 = 𝐹𝑣 tan 𝛽𝑎 = 𝐹𝑣

tan 𝜃𝑛 + 𝛼𝑛

cos 𝜂→ tan 𝜃𝑛 + 𝛼𝑛 = tan 𝛽𝑎 cos 𝜂

Can derive the following geometric relations

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Velocity diagram – oblique cutting

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𝑉 = 𝑉 cos 𝑖 , 𝑉 sin 𝑖 , 0 ,

𝑉𝑐 = 𝑉𝑐 cos 𝜂 sin 𝛼𝑛 , 𝑉𝑐 sin 𝜂 , 𝑉𝑐 cos 𝜂 cos𝛼𝑛 ,

𝑉𝑠 = −𝑉𝑠 cos𝜙𝑖 cos𝜙𝑛 , −𝑉𝑠 sin𝜙𝑖 , 𝑉𝑠 cos𝜙𝑖 sin𝜙𝑛

Defining each velocity vector by its Cartesian components:

Eliminate 𝑉, 𝑉𝑐, and 𝑉𝑠

𝑉𝑠 = 𝑉𝑐 − 𝑉

rearranging, following geometric relation between shear and chip flow directions

tan 𝜂 =tan 𝑖 cos 𝜙𝑛 − 𝛼𝑛 − cos𝛼𝑛 tan𝜙𝑖

sin𝜙𝑛

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Solving for oblique cutting parameters

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sin 𝜃𝑖 = sin 𝛽𝑎 sin 𝜂

tan 𝜃𝑛 + 𝛼𝑛 = tan 𝛽𝑎 cos 𝜂


sin𝜙𝑛

Five unknown oblique cutting parameters that define the directions of resultant force 𝜃𝑛, 𝜃𝑖 , shear velocity 𝜙𝑛, 𝜙𝑖 , and chip flow 𝜂 .

Three equations – what to do? Empirical, model based?

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Minimum energy principle

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Recall – Merchant’s approach to predict shear angle by applying the principle of minimum energy principle to orthogonal cutting

𝑃𝑡𝑐 = 𝑉𝐹𝑡𝑐 Power consumed during cutting: 𝑑𝑃𝑡𝑐

𝑑𝜙𝑐= 0 → 𝜙𝑐 =

𝜋

4−

𝛽𝑎 − 𝛼𝑟

2

𝐹𝑠 = 𝐹𝑐 cos 𝜃𝑛 + 𝜙𝑛 cos 𝜃𝑖 cos𝜙𝑖

+ sin 𝜃𝑖 sin𝜙𝑖

Extend the same principle here for oblique cutting

First need shear force (primary power consumption is during shearing)

Represent shear force as a

projection of resultant 𝐹 𝑐 in the direction of shear:

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𝐹𝑠 = 𝜏𝑠𝐴𝑠 = 𝜏𝑠 𝑏

cos 𝑖

ℎ

sin 𝜙𝑛

Representing shear force as a product of shear stress and shear

plane area 𝐹𝑠 = 𝐹𝑐 cos 𝜃𝑛 + 𝜙𝑛 cos 𝜃𝑖 cos𝜙𝑖 + sin 𝜃𝑖 sin𝜙𝑖

Represent shear force as a projection of resultant

𝐹 𝑐 in the direction of shear: or

𝐹𝑐 =𝜏𝑠𝑏ℎ

cos 𝜃𝑛 + 𝜙𝑛 cos 𝜃𝑖 cos𝜙𝑖 + sin 𝜃𝑖 sin𝜙𝑖 cos 𝑖 sin 𝜙𝑛

𝑃𝑡𝑐 = 𝑉𝐹𝑡𝑐 Power consumed

during cutting: 𝑃𝑡𝑐 = 𝐹𝑐 cos 𝜃𝑖 cos 𝜃𝑛 cos 𝑖 + sin 𝜃𝑖 sin 𝑖 𝑉

In terms of 𝐹𝑐

(a)

(b)

Substitute (a) into (b)

𝑃𝑡′ =

𝑃𝑡𝑐

𝑉𝜏𝑠𝑏ℎ=

cos𝜃𝑛 + tan 𝜃𝑖 tan 𝑖

cos 𝜃𝑛 + 𝜙𝑛 cos𝜙𝑖 + tan 𝜃𝑖 sin 𝜙𝑖 sin𝜙𝑛

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Power 𝑃𝑡′: 𝑃𝑡

′ =𝑃𝑡𝑐

𝑉𝜏𝑠𝑏ℎ=

cos𝜃𝑛 + tan 𝜃𝑖 tan 𝑖

cos 𝜃𝑛 + 𝜙𝑛 cos𝜙𝑖 + tan 𝜃𝑖 sin𝜙𝑖 sin 𝜙𝑛

Minimum energy principle dictates that the cutting power must be minimum for a unique shear angle solution. Since 𝜏𝑠, 𝑏, ℎ, and 𝑉 is constant, and since the direction of shear is characterized

by 𝜙𝑛 and 𝜙𝑖, we have:

𝜕𝑃𝑡′

𝜕𝜙𝑛= 0;

𝜕𝑃𝑡′

𝜕𝜙𝑖= 0




sin𝜙𝑛

Which gives us two additional equations, in addition to the earlier three to solve for five unknowns

Recalling the earlier three

(1)

(2)

(3)

(4-5)

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Solving for oblique cutting parameters

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Solving five equations analytically for five unknowns is not trivial, hence numerical iteration techniques are instead used

Start with 𝜂 = 𝑖 – Stabler’s rule

Altintas, Mfg. Automation

IIT Kanpur

Empirical solutions to oblique cutting parameters

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Armarego & Whitfield

Three equations five unknowns (𝜙𝑛, 𝜙𝑖, 𝜂, 𝜃𝑖, 𝜃𝑛)




sin𝜙𝑛

(1)

(2)

(3)

Assume: 1. Shear velocity is collinear with shear force – from Stabler’s work 2. Chip length ratio in oblique is same as in orthogonal – from experiments

tan 𝜙𝑛 + 𝛽𝑛 =cos𝛼𝑛 tan 𝑖

tan 𝜂 − sin 𝛼𝑛 tan 𝑖 𝛽𝑛 = 𝜃𝑛 + 𝛼𝑛

tan 𝛽𝑛 = tan𝛽𝑎 cos 𝜂 𝜙𝑛 = tan−1𝑟𝑐 cos 𝜂 / cos 𝑖 cos𝛼𝑛

1 − 𝑟𝑐 cos 𝜂 / cos 𝑖 sin 𝛼𝑛

Combine Eq. (1) to (3): (1a)

(2a) (3a)

Solve Eq. (1a) – (3a) numerically to get 𝜂, 𝜙𝑛 and 𝛽𝑛.

Or assume 𝜂 = 𝑖 (Stabler’s rule) for analytical solution

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Prediction of cutting forces

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𝐹𝑟𝑐 = 𝐹𝑐 sin 𝜃𝑖 cos 𝜃𝑖 − cos 𝜃𝑖 cos 𝜃𝑛 sin 𝑖 =𝜏𝑠𝑏ℎ tan𝜃𝑖 − cos 𝜃𝑛 tan 𝑖

cos 𝜃𝑛 + 𝜙𝑛 cos𝜙𝑖 + tan𝜃𝑖 sin𝜙𝑖 sin𝜙𝑛

𝐹𝑡𝑐 = 𝐹𝑐 cos𝜃𝑖 cos𝜃𝑛 cos𝜃𝑖 + sin 𝜃𝑖 sin 𝑖 =𝜏𝑠𝑏ℎ cos𝜃𝑛 + tan 𝜃𝑖 tan 𝑖

cos 𝜃𝑛 + 𝜙𝑛 cos𝜙𝑖 + tan 𝜃𝑖 sin𝜙𝑖 sin𝜙𝑛

𝐹𝑓𝑐 = 𝐹𝑐 cos 𝜃𝑖 sin 𝜃𝑛 =𝜏𝑠𝑏ℎ sin 𝜃𝑛

cos 𝜃𝑛 + 𝜙𝑛 cos𝜙𝑖 + tan 𝜃𝑖 sin𝜙𝑖 sin𝜙𝑛

Cutting force components are derived as projections of resultant cutting force 𝐹𝑐 after subtracting the edge components 𝐹𝑒 from measured resultant 𝐹

Force in direction of cutting speed

Expressing force components as a 𝑓 𝜏𝑠, 𝜙𝑛, 𝜙𝑖 , 𝜃𝑖, 𝜃𝑛 :

Force in direction of thrust

Force in direction of normal

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Prediction of cutting forces

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𝐹𝑡 = 𝐾𝑡𝑐𝑏ℎ + 𝐾𝑡𝑒𝑏; 𝐹𝑓 = 𝐾𝑓𝑐𝑏ℎ + 𝐾𝑓𝑒𝑏;

𝐹𝑟 = 𝐾𝑟𝑐𝑏ℎ + 𝐾𝑟𝑒𝑏;

𝐾𝑡𝑐 =𝜏𝑠 cos 𝜃𝑛 + tan 𝜃𝑖 tan 𝑖

cos 𝜃𝑛 + 𝜙𝑛 cos 𝜙𝑖 + tan 𝜃𝑖 sin𝜙𝑖 sin𝜙𝑛

𝐾𝑓𝑐 =𝜏𝑠 sin 𝜃𝑛


𝐾𝑟𝑐 =𝜏𝑠𝑏ℎ tan 𝜃𝑖 − cos 𝜃𝑛 tan 𝑖


𝐹𝑟𝑐 =𝜏𝑠𝑏ℎ tan 𝜃𝑖 − cos 𝜃𝑛 tan 𝑖


𝐹𝑡𝑐 =𝜏𝑠𝑏ℎ cos 𝜃𝑛 + tan 𝜃𝑖 tan 𝑖

cos 𝜃𝑛 + 𝜙𝑛 cos 𝜙𝑖 + tan 𝜃𝑖 sin𝜙𝑖 sin𝜙𝑛 𝐹𝑓𝑐 =

𝜏𝑠𝑏ℎ sin 𝜃𝑛






Rewriting forces in the convenient form of:

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Prediction of cutting forces – Armarego’s model

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𝐹𝑟𝑐 =𝜏𝑠𝑏ℎ tan 𝜃𝑖 − cos 𝜃𝑛 tan 𝑖


𝐹𝑡𝑐 =𝜏𝑠𝑏ℎ cos 𝜃𝑛 + tan 𝜃𝑖 tan 𝑖

cos 𝜃𝑛 + 𝜙𝑛 cos 𝜙𝑖 + tan 𝜃𝑖 sin𝜙𝑖 sin𝜙𝑛 𝐹𝑓𝑐 =

𝜏𝑠𝑏ℎ sin 𝜃𝑛






Using Armarego’s classical oblique cutting model,

forces are transformed in

terms of 𝑓 𝜏𝑠, 𝛽𝑛,𝜙𝑛𝛼𝑛,𝜂

𝐹𝑡 = 𝑏ℎ𝜏𝑠

sin𝜙𝑛

cos 𝛽𝑛 − 𝛼𝑛 + tan 𝑖 tan 𝜂 sin 𝛽𝑛

𝑐𝑜𝑠2 𝜙𝑛 + 𝛽𝑛 − 𝛼𝑛 + 𝑡𝑎𝑛2𝜂𝑠𝑖𝑛2𝛽𝑛

𝐹𝑓 = 𝑏ℎ𝜏𝑠

sin𝜙𝑛 cos 𝑖

sin 𝛽𝑛 − 𝛼𝑛


𝐹𝑟 = 𝑏ℎ𝜏𝑠

sin𝜙𝑛

cos 𝛽𝑛 − 𝛼𝑛 cos 𝛽𝑛 − 𝛼𝑛 tan 𝑖 −tan 𝜂 sin 𝛽𝑛


Armarego’s force model

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Oblique cutting – force coefficients

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𝐹𝑡 = 𝑏ℎ𝜏𝑠

sin𝜙𝑛



𝐹𝑓 = 𝑏ℎ𝜏𝑠




𝐹𝑟 = 𝑏ℎ𝜏𝑠

sin𝜙𝑛



𝐾𝑡𝑐 =𝜏𝑠

sin𝜙𝑛



𝐾𝑓𝑐 =𝜏𝑠




𝐾𝑟𝑐 =𝜏𝑠

sin𝜙𝑛





Using Armarego’s classical oblique cutting model,

forces are transformed as

Armarego’s force model

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Practical approach – for force prediction

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𝐾𝑡𝑐 =𝜏𝑠

sin 𝜙𝑛



𝐾𝑓𝑐 =𝜏𝑠

sin 𝜙𝑛 cos 𝑖



𝐾𝑟𝑐 =𝜏𝑠

sin 𝜙𝑛



• Evaluate shear angle 𝜙𝑐, friction angle 𝛽𝑎, and shear stress 𝜏𝑠 from orthogonal cutting tests

• Assume orthogonal shear angle equals normal shear angle, i.e. 𝜙𝑐 ≡ 𝜙𝑛

• Assume normal rake angle equals rake angle in orthogonal cutting, i.e. 𝛼𝑟 ≡ 𝛼𝑛

• Assume chip flow angle equals oblique angle (Stabler’s rule), i.e. 𝜂 ≡ 𝑖

• Assume friction angle in orthognal cutting is same as in oblique, i.e. 𝛽𝑎 ≡ 𝛽𝑛

• Assume shear stress remains the same in orthogonal and oblique cutting



Evaluate cutting force coefficients

Predict cutting forces

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Yet another practical approach

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Feed Rate Measured Measured

Test No: h [mm/rev] Ftc [N] Ffc [N] 1 0.050 101 32

2 0.075 140 35

3 0.100 175 39

4 0.125 213 41

Conduct a dedicated series of tests at different feed rates and identify coefficients directly for the tool-workpiece-cutting parameter combination of interest

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Yet another practical approach

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𝐹𝑡 = 𝐹𝑡𝑐 + 𝐹𝑡𝑒; 𝐹𝑓 = 𝐹𝑓𝑐 + 𝐹𝑓𝑒;

Forces are composed of shearing and edge rubbing force components

or

𝐹𝑡𝑐 = 𝐾𝑡𝑐𝑏ℎ; 𝐹𝑓𝑐 = 𝐾𝑓𝑐𝑏ℎ;

Consider only shearing components (since edge forces do not contribute to cutting):

𝐾𝑡𝑐 =𝐹𝑡𝑐

𝑏ℎ; 𝐾𝑓𝑐 =

𝐹𝑓𝑐

𝑏ℎ; 𝐾𝑡𝑒 =

𝐹𝑡𝑒

𝑏; 𝐾𝑓𝑒 =

𝐹𝑓𝑒

𝑏;

Hence the cutting coefficients are:

Pre

dic

tio

n

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Force models

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WZL Aachen

Documents

ME698S Machining Dynamics - IITKhome.iitk.ac.in/~mlaw/ME698S_Oblique_cutting_updated.pdf · IIT Kanpur ME698S – Machining Dynamics Dr. Mohit Law [email protected] Oblique Cutting