1
3-year MTech in Mechanical Engineering
Semester 1
(January – April 2024)
Advances in Machining
(MEC 514)
Cutting Forces
Dr. Suman Saha
Assistant Professor
Department of Mechanical Engineering
IIT (ISM) Dhanbad
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Turning process using a single point cutting tool
𝑁
Workpiece
𝐷
𝛼𝑂
𝜋𝑂 𝜋𝑅
𝛾𝑂
• A cylindrical workpiece of original diameter 𝐷 (mm)
• Workpiece is rotated at speed 𝑁 (rpm) 𝛷
𝜋𝐷𝑁
• Peripheral velocity or cutting velocity, 𝑉𝑐 = 1000 (m/min)
𝜋𝐶 𝜋𝑂
• Right-handed single point turning tool (SPTT) Orthogonal machining:
• Tool is fed at a constant rate (feed/rev): 𝑠 (mm/rev) 𝜋𝑅 Chip flows in the
• Radial depth of cut: 𝑡 (mm) orthogonal direction
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Measurable forces during turning
• Cutting forces are measured by dynamometer.
Workpiece
Chip Can we measure forces in any direction?
• Yes. Direction must be known for analysis.
𝑃𝑋𝑌
• It is efficient to measure 𝑃𝑍 , 𝑃𝑋 , and 𝑃𝑦 𝑃𝑍
𝑎2 𝛼𝑂
Do we need to measure all three forces? 𝜋𝑂 𝜋𝑅
Cutting tool 𝛾𝑂
𝑃𝑋𝑌
𝑎1
𝑃𝑍
𝑉𝐶 𝑃𝑍
𝑃𝑋
Workpiece 𝛷 𝑃𝑌 𝛷
𝑃𝑋𝑌
𝜋𝐶
𝜋𝑅
• Conventional machining is one mechanical energy based cutting process.
• This energy is expended to accomplish chip separation by shearing (or brittle
fracture). 𝑃𝑋 = 𝑃𝑋𝑌 sin 𝛷
• Consequently, forces develop and act on both the tool and workpiece in all 𝑃𝑌 = 𝑃𝑋𝑌 cos 𝛷
possible directions. 𝑃𝑋
• The magnitude of cutting forces can vary from one direction to another. = tan 𝛷
𝑃𝑌
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Need for force assessment in different directions
𝐹
Chip 𝑅
𝑎2
𝑃𝑆
Cutting tool
𝑃𝑋𝑌 𝑁
𝑅
𝑎1
𝑃𝑛
𝑅 = 𝐹Ԧ + 𝑁 = 𝑃𝑠 + 𝑃𝑛
𝑃𝑍
𝑉𝐶 Equilibrium of chip segment
under resultant force
Workpiece
Why different force components are desired?
Mechanical energy expended during machining = 𝑃𝑍 𝑉𝐶 + 𝑃𝑋 𝑉𝑓𝑒𝑒𝑑 ≅ 𝑃𝑍 𝑉𝐶 𝑉𝑠
𝑉𝑓
Energy expended for shear deformation = 𝑃𝑆 𝑉𝑆 𝑉𝑐
Energy expended for to overcome friction = 𝐹𝑉𝑓
∴ 𝑃𝑍 𝑉𝐶 = 𝑃𝑆 𝑉𝑆 + 𝐹𝑉𝑓 Velocity triangle
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Merchant Circle Diagram (MCD)
Chip Zero rake tool 𝛾𝑂 = 0°
𝛾𝑂
Cutting tool
𝑃𝑍
𝑃𝑋𝑌 𝐹
𝑃𝑆
𝑃𝑍 𝛽𝑂 𝑅 𝑃𝑍 = 𝑁
𝑁 𝑃𝑋𝑌 = 𝐹
𝑃𝑛
𝑉𝐶
𝑅
𝑃𝑋𝑌 𝐹
𝛾𝑂 < 0°
𝜂 𝛾𝑂 Negative rake
𝑃𝑆
𝑁
Assumptions: 𝑃𝑍
𝐹 = 𝑃𝑍 sin 𝛾𝑂 + 𝑃𝑋𝑌 cos 𝛾𝑂 1. Perfectly sharp tool (no contact at flank)
𝑃𝑋𝑌 𝑃𝑛 𝐹
2. Plain strain (no side flow of material)
𝑁 = 𝑃𝑍 cos 𝛾𝑂 − 𝑃𝑋𝑌 sin 𝛾𝑂 3. Constant cutting velocity
4. Continuous chip without BUE 𝑅
𝑃𝑆 = 𝑃𝑍 cos 𝛽𝑂 − 𝑃𝑋𝑌 sin 𝛽𝑂
5. Free-cutting (no chip deviation) 𝑁
𝑃𝑛 = 𝑃𝑍 sin 𝛽𝑂 + 𝑃𝑋𝑌 cos 𝛽𝑂 6. Resultant force for all 3 sets are colinear.
Merchant solution of forces – Minimum Energy Principle 6
Ernst and Merchant Solution 𝑑𝑃𝑍
𝑃𝑆 𝑏1 =0
𝑃𝑍
(Minimum Energy Requirement) 𝑑𝛽𝑂
𝛽𝑂
2𝛽𝑂 + 𝜂 − 𝛾𝑂 = 90° Merchant’s First Solution
𝑃𝑛 𝑎2
𝑅 𝑏1 𝑃𝑍 = 2𝑡𝑠𝜏𝑠 cot 𝛽𝑂 (Brittle material)
𝑃𝑋𝑌 𝐹
𝑎1 • Dynamic shear strength on shear plane does
𝜂 𝛾𝑂 𝛽𝑂
not remain constant for ductile materials!
𝑁 • Shear angle for brittle material cannot be
easily obtained!
𝑃𝑍 = 𝑅 cos 𝜂 − 𝛾𝑂 𝑎1 𝑠𝑡 Engineering materials are work-hardenable.
Shear plane area, 𝐴𝑠 = 𝑏1 = Dynamic shear strength of the material
sin 𝛽𝑂 sin 𝛽𝑂
𝑃𝑠 = 𝑅 cos 𝛽𝑂 + 𝜂 − 𝛾𝑂
𝑠𝑡 changes linearly under applied stress (this is
Shear force, 𝑃𝑠 = 𝜏𝑠 𝐴𝑠 = 𝜏𝑠 attributed to strength hardening).
sin 𝛽𝑂
𝑃𝑠 cos 𝜂 − 𝛾𝑂 𝑡𝑠𝜏𝑠 cos 𝜂 − 𝛾𝑂
For ductile materials, Lee ∴ 𝑃𝑍 = = 𝜏𝑠 = 𝜏𝑜 + 𝑘𝜎𝑛
cos 𝛽𝑂 + 𝜂 − 𝛾𝑂 sin 𝛽𝑂 cos 𝛽𝑂 + 𝜂 − 𝛾𝑂
Shafer formula based on slip-
line field theory can be applied. 𝑡𝑠𝜏𝑠 cos 𝜂 − 𝛾𝑂
𝑃𝑍 =
𝛽𝑂 + 𝜂 − 𝛾𝑂 = 45° sin 𝛽𝑂 cos 𝛽𝑂 + 𝜂 − 𝛾𝑂 1 − 𝑘 tan 𝛽𝑂 + 𝜂 − 𝛾𝑂
ζ 2𝛽𝑂 + 𝜂 − 𝛾𝑂 = 𝐶 = cot −1 𝑘
𝑃𝑍 = 𝑡𝑠𝜏𝑠 − tan𝛾𝑜 +1 Applicable for semi-
cos 𝛾𝑜 ductile material
𝑃𝑍 = t s τs cotβo + tan(C − βo )
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MCD construction from given machining scenario
Question: During orthogonal turning of mild steel (ductile material) at 4 m/s cutting velocity, the chip flow velocity is
found to be 3 m/s whereas the shear velocity remained 5 m/s. What is the expected ratio between the orthogonal
thrust force (PXY) and main cutting force (PZ)?
Cutting velocity, 𝑉𝑐 = 4 m/s Workpiece material is ductile.
Chip flow velocity, 𝑉𝑓 = 3 m/s So apply Lee Shafer formula
𝛽𝑜 + 𝜂 − 𝛾𝑜 = 45°
Shear velocity, 𝑉𝑠 = 5 m/s
𝑉𝑠
A quick look at the data yields: 𝑉𝑓 Since 𝛾𝑜 = 0°
42 + 32 = 52 𝑃𝑍 𝑉𝑐
∴ 𝛽𝑜 + 𝜂 = 45°
𝑉𝑐2 + 𝑉𝑓2 = 𝑉𝑠2 𝑃𝑋𝑌 𝐹
Velocity triangle is right-angled triangle 𝑅 𝜂 = 45° − 𝛽𝑜
𝑃𝑋𝑌
∴ Orthogonal rake, 𝛾𝑜 = 0° 𝜂 𝑁
=?
𝑃𝑍 tan 𝜂 = tan 45° − 𝛽𝑜
1−tan 𝛽
𝑃𝑋𝑌 𝐹 ∴ tan 𝜂 = 1+tan 𝛽𝑜
𝑜
𝑉𝑓 3 = = tan 𝜂 =?
tan 𝛽𝑜 = = 𝑃𝑍 𝑁
𝑉𝑐 4
𝑃𝑋𝑌 𝐹 1 − 3Τ4
= = tan 𝜂 = = 0.143
𝑃𝑍 𝑁 1 + 3Τ4
