- An orthogonal cutting operation is performed on aluminum with a cutting speed of 5.6 m/s, feed of 0.25 mm/rev, and depth of cut of 2.0 mm. - The lathe has a mechanical efficiency of 0.85. - The specific energy values for aluminum need to be determined to calculate the power required and power drawn from the lathe. - Given the cutting parameters and material, the material removal rate, cutting power required, and power drawn from the lathe accounting for efficiency can be calculated using the specific energy values and equations provided.

1. Dr. Shailesh Kumar Dewangan
Assistant Professor
Department of Production Engineering,
BIT Mesra Ranchi
Single Point Cutting Tool
Numerical Problem

2. β = Shear Angle
γ = Rake Angle
 = Friction Angle
Fc = Cutting Force, FT = Thrust Force or Feed Force
Fs = Shear Force, Ns = Normal Shear Force
F = Friction Force (Resistance) , N = Normal friction force
Ft
Fc
F
R
γ
γ

-γ
N
V
90-
Fs
Ns β
The Merchant Circle

3. The Merchant Equation
• Of all the possible angles at which shear deformation can occur,
the work material will select a shear plane angle β that
minimizes energy
β = Shear Angle
 = Friction Angle
𝜸 = Rake Angle
• Toincrease shear plane angle
 Increase the rake angle
 Reduce the friction angle (or coefficient of friction)
 Higher shear plane angle means smaller shear plane which means
lower shear force
 Result: lower cutting forces, power, temperature, all of which mean
easier machining
Lee and Shaper Theory
𝜷 + 𝜼 − 𝜸 = 𝟒𝟓°
Stabber Theory
𝜷 + 𝜼 −
𝜸
𝟐
= 𝟒𝟓°
𝟐𝜷 + 𝜼 − 𝜸 = 𝟗𝟎°

4. Formula used in Merchant circle
• Equations to relate the forces
𝐹 = 𝐹𝑐 sin 𝛾 + 𝐹𝑇 cos 𝛾 tan 𝛽 =
𝑟 cos 𝛾
1−𝑟 sin 𝛾
𝑁 = 𝐹𝑐 cos 𝛾 − 𝐹𝑇 sin 𝛾
𝐹𝑠 = 𝐹𝑐 cos 𝛽 − 𝐹𝑇 sin 𝛽 𝑟 =
𝑡0
𝑡1
𝑁𝑠= 𝐹𝑐 sin 𝛽 + 𝐹𝑇 cos 𝛽 𝜖𝑠 = tan(𝛽 − 𝛾) + cot 𝛽
𝐹𝑇
𝐹𝑐
= tan(𝜂 − 𝛾)
𝐹
𝑁
= tan 𝜂 = 𝜇
𝜁𝑠=
𝐹𝑠 sin 𝛽
𝑡0𝑤
Fs = ζs As
ζs= Shear Stress , Fs = Shear Force
As = Shear Area
MRR= v f d

5. Q.1 In an orthogonal cutting operation, the tool has a rake angle = 15°. The chip thickness
before the cut = 0.30 mm and the cut yields a deformed chip thickness = 0.65 mm.
Calculate (a) the shear plane angle and (b) the shear strain for the operation.
Numerical Problem
Solution :
Given Data rake angle (γ) = 15°.
chip thickness before the cut (t0) = 0.30 mm
chip thickness (t1) = 0.65 mm.
Finding (a) shear plane angle (β) = ? , (b) shear strain (𝜖𝑠) =?
r =to/t1
= 0.30/0.65 = 0.4615
𝛽 = 26.85°
𝜖𝑠 = 2.185
tan 𝛽 =
𝑟 cos 𝛾
1−𝑟 sin 𝛾
𝜖𝑠 = tan(𝛽 − 𝛾) + cot 𝛽

6. Q.2 In an orthogonal cutting operation, the tool has a rake angle = 15°. The chip thickness
before the cut = 0.30 mm and the cut yields a deformed chip thickness = 0.65 mm.
Suppose the rake angle were changed to 0°. Assuming that the friction angle remains the
same, determine (a) the shear plane angle, (b) the chip thickness, and (c) the shear strain
for the operation.
Solution:
(a) From Problem Q.1, rake angle (γ) = 15°, and β (Shear Angle) = 26.85°.
Using Merchant Equation 2𝜷 + 𝜼 − 𝜸 =90°
2x 26.85+ 𝜼- 15 = 90
𝜼 = 51.3° (friction Angle)
(b) Now, with γ = 0 and 𝜼 remaining the same at 51.3°, 2x 𝜷 + 51.3 - 0 =90, 𝜷 =19.35°
(c) Chip thickness at If (γ = 0) tan 𝛽 =
𝑟 cos 𝛾
1−𝑟 sin 𝛾
tan 𝛽 = 𝑟 𝑟 =
𝑡0
𝑡1
So 𝑡1=
𝑡0
tan 𝛽
=
0.30
tan 19.35
t1= 0.854 mm
(c) Shear strain 𝜖𝑠 = tan(𝛽 − 𝛾) + cot 𝛽
𝜖𝑠 = tan (19.35 - 0)+ cot 19.35 = 2.848 + 0.351 = 3.199

7. Q.3 In a turning operation, spindle speed is set to provide a cutting speed of 1.8 m/s. The
depth of cut and feed of cut are 2.6 mm and 0.30 mm, respectively. The tool rake
angle is 8°. After the cut, the deformed chip thickness is measured to be 0.49 mm.
Determine (a) shear plane angle, (b) shear strain, and (c) material removal rate. Use
the orthogonal cutting model as an approximation of the turning process.
Solution:
t0 = Feed
(a) r = to/t1 = 0.30/0.49 = 0.612
𝛽 = tan-1(0.612 cos 8/(1 – 0.612 sin 8)) = tan-1(0.6628) = 33.6°
(b) 𝜖𝑠 = tan (33.6 - 8) + cot 33.6 = 1.509 + 0.478 = 1.987
(c) V = 1.8m/s = 1.8X 103 mm/s
MRR = (1.8 m/s x 103 mm/m)(0.3)(2.6) = 1404 mm3/s
tan 𝛽 =
𝑟 cos 𝛾
1−𝑟 sin 𝛾
𝜖𝑠 = tan(𝛽 − 𝛾) + cot 𝛽
MRR= v f d

8. Solutions: Given data
Fc = 1470 N, Ft = 1589 N
γ= 5°, w= 5.0 mm, t0 = 0.6, r = 0.38
Q.4 The cutting force and thrust force in an orthogonal cutting operation are 1470 N and
1589 N, respectively. The rake angle = 5°, the width of the cut = 5.0 mm, the chip
thickness before the cut = 0.6, and the chip thickness ratio = 0.38. Determine (a) the
shear strength of the work material and (b) the coefficient of friction in the
operation.
Fs/As = ζs 𝜁𝑠=
𝐹𝑠 sin 𝛽
𝑡0𝑤
tan 𝛽 =
𝑟 cos 𝛾
1−𝑟 sin 𝛾
𝐹𝑠 = 𝐹𝑐 cos 𝛽 − 𝐹𝑇 sin 𝛽
𝛽 = tan-1(0.38 cos 5/(1 - 0.38 sin 5)) = tan-1(0.3916) = 21.38°
Fs = 1470 cos 21.38 – 1589 sin 21.38 = 789.3 N
𝜻𝒔 = 95.9 Mpa
μ = tan  =
𝑭
𝑵
μ =1.29
𝟐𝜷 + 𝜼 − 𝜸 = 𝟗𝟎°
𝐹 = 𝐹𝑐 sin 𝛾 + 𝐹𝑇 cos 𝛾
𝑁 = 𝐹𝑐 cos 𝛾 − 𝐹𝑇 sin 𝛾

9. Q.5 The cutting force and thrust force have been measured in an orthogonal cutting operation
to be 300 N and 291 N, respectively. The rake angle = 10°, width of cut = 0.200 mm, chip
thickness before the cut = 0.015 mm, and chip thickness ratio = 0.4. Determine
(a) the shear strength of the work material and (b) the coefficient of friction in the operation.
𝛽 = tan-1(0.4 cos 10/(1 - 0.4 sin 10)) = tan-1(0.4233) = 22.94°
Fs = 300 cos 22.94 - 291sin 22.94 = 162.9 N
𝜻𝒔 = 162.9/0.0077 = 21,167 N/mm2
𝜼 = 2(45) + α - 2(φ) = 90 + 10 - 2(22.94) = 54.1°
μ = tan 54.1 = 1.38

10. Q.6 An orthogonal cutting operation is performed using a rake angle of 15°, chip thickness
before the cut = 0.012 mm and width of cut = 0.100 mm. The chip thickness ratio is
measured after the cut to be 0.55. Determine (a) the chip thickness after the cut, (b) shear
angle, (c) friction angle, (d) coefficient of friction, and (e) shear strain.
Solution:
(a) r = to/t1, t1 = to/r = 0.012/0.55 = 0.022 mm
(b) β = tan-1(0.55 cos 15/(1 - 0.55 sin 15)) = tan-1(0.6194) = 31.7°
(c)  = 90 + 15 - 2(31.8) = 41.6°
(d) μ = tan 41.5 = 0.88
(e) 𝜖𝑠 = cot 31.8 + tan(31.8 - 15) = 1.615 + 0.301 = 1.92
tan 𝛽 =
𝑟 cos 𝛾
1−𝑟 sin 𝛾 𝟐𝜷 + 𝜼 − 𝜸 = 𝟗𝟎° μ = tan 
𝜖𝑠 = tan(𝛽 − 𝛾) + cot 𝛽

11. Q 6.1 An orthogonal cutting operation is performed using a rake angle of 15°, chip thickness
before the cut = 0.012 mm and width of cut = 0.100 mm. The chip thickness ratio is
measured after the cut to be 0.55. Work material shear strength is 40,000 N/mm2. Compute
(a) the shear force, (b) cutting force, (c) thrust force, and (d) friction force.
Shear strength ( 𝜻𝒔) = 40,000 N/mm2
Fs =?, Fc= ?, Ft= ? , F= ?
𝜁𝑠=
𝐹𝑠 sin 𝛽
𝑡0𝑤
𝐹𝑠 = 91.2 N
R=
𝑭𝒔
cos(𝜷+𝜼−𝜸)
R= 173.82 N
Ft = R Sin (𝜂 − 𝛾)
Fc= R Cos (𝜂 − 𝛾)
F = R Cos (90- 𝜂)
Ft = R Sin (41.6−15) = 77.83 N
Fc= R Cos (41.6−15) = 155.42 N
F = R Cos (90- 41.6) = 115.40 N
Ft
Fc
F
R
γ
γ

-γ
N
V
90-
Fs
Ns β

12. Q.7. During an orthogonal operation on mild steel workpiece. The result are
obtained, uncut chip thickness = 0.25 mm, chip thickness = 0.75 mm, width = 2.5
mm, orthogonal rake angle =0°, cutting force (Fc) = 950 N and feed/thrust force
(Ft) = 475 N. Determine coefficient of friction between tool and workpiece and
shear stress.
Solution: Given data
to = 0.25 mm, t1 = 0.75 mm, width (w) = 2.5 mm
rake angle (γ) =0°
Fc = 950 N, Ft = 475 N
μ = tan  =
𝑭
𝑵
V
Ft
Fc
F
R γ=0
γ

-γ
N
90-
Fs
Ns β
𝜁𝑠=
𝐹𝑠 sin 𝛽
𝑡0𝑤
tan 𝛽 =
𝑟 cos 𝛾
1−𝑟 sin 𝛾
𝛽= 18.4 °
𝜁𝑠 = 379.4 N/mm2
𝐹𝑠 = 𝐹𝑐 cos 𝛽 − 𝐹𝑇 sin 𝛽
𝐹𝑠 = 751.3 𝑁

13. Power and Energy Relationships
Power is the product of force and velocity.
• A machining operation requires power
• The power to perform machining can be computedfrom:
Pc = Fc v
where Pc = cutting power, N-m/s or W (ft-lb/min); Fc = cutting force, N (lb); and v = cutting
speed, m/s (ft/min)
Horsepower = HPc= Pc/33000
This power is dissipated mainly in the shear zone (due to the energy required to shear the
material) and on the rake face of the tool (due to tool-chip interface friction).
Power dissipated in the shear plane is Power for shearing Ps= FSVS. Similarly, the
power dissipated in friction is Power for friction Pf= FVC

14. Unit Power in Machining
• It is Useful to convert power into power per unit volume rate of metal
cut called unit power, Pu
MRR
Pc
Us (Sp. Energy)=
where RMR = material removal rate
• Unit power is also known as the specific energy U
Where Units for specific energy are typically N-m/mm3 or J/mm3(in-lb/in3)
vct w
MRR
Pc Fc v
=
U s =
s
specific energy for shearing, U =
Fcv
specific energy for friction, Uf=
vt w
vtow
F v
o
total specific energy, u, is thus
U= Us +Uf
𝑃𝑐
𝑃𝑎
= ηeff

15. Q.8. In a turning operation on stainless steel with hardness = 200 HB, the cutting
speed = 200 m/min, feed = 0.25 mm/rev, and depth of cut = 7.5 mm. How much
power will the lathe draw in performing this operation if its mechanical efficiency =
90%. Specific energy value is 2.8 N-m/mm3
Solution:
Us = 2.8 N-m/mm3
Vc = 200 m/min (convert to mm/sec) = (200x103)/60
MRR = Vcf d
MRR = 6250 mm3/s
Pc = (6250 mm3/s)(2.8 J/mm3) = 17,500 J/s = 17,500 W = 17.5 kW
Accounting for mechanical efficiency, Pa = 17.5/0.90 = 19.44 kW
Us (Sp. Energy)=
Pc
MRR

16. Q.9 A turning operation is carried out on aluminum (100 BHN). Cutting speed = 5.6 m/s,
feed = 0.25 mm/rev, and depth of cut = 2.0 mm. The lathe has a mechanical efficiency =
0.85. The specific energy valuesfor aluminum is 0.7 N-m/mm3, determine (a) the cutting
power and (b) gross power in the turning operation, in Watts.
Solution:
(a)
U = 0.7 N-m/mm3
MRR = vfd = 5.6(103)(.25)(2.0) = 2.8(103) mm3/s.
Pc = U MRR = 0.7(2.8)(103) = 1.96(103) N-m/s = 1960 W
(b) Gross power Pg = 1960/0.85 = 2306 W