The chip thickness ratio 'r' in Orthogonal metal cutting operation is always:
Explanation:
Chip thickness ratio / Cutting ratio (r):
It is the ratio of chip thickness before cut (t_{1}) to the chip thickness after cut (t_{2}).
\(r=\frac{Chip\;thickness\;before\;cut\;(t_1)}{Chip\;thickness\;after\;cut\;(t_2)}\Rightarrow \frac{uncut\;chip\;thickness}{chip\;thickness}\)
chip thickness after the cut (t_{2}) is always greater than the chip thickness before the cut (t_{1}), ∴ r is always < 1, i.e. the uncut chip thickness value is less than the chip thickness value.
Assuming volume to be constant:
t_{1}b_{1}L_{1} = t_{2}b_{2}L_{2}
\(\frac{t_1}{t_2}=\frac{L_2}{L_1}\;\;\;\;(\because b_1=b_2)\)
as t_{2} > t_{1}, ∴ L_{2} < L_{1} i.e. length after cut is less than the length before the cut.
Assuming discharge to be constant:
t_{1}b_{1}V = t_{2}b_{2}V_{c}
\(\frac{t_1}{t_2}=\frac{V_c}{V}\;\;\;\;(\because b_1=b_2)\)
as t_{2} > t_{1}, ∴ V > V_{c} i.e. cutting velocity is greater than the chip velocity.
Concept:
We know that,
\(r = \frac{t}{{{t_c}}} = \frac{{{V_c}}}{V}\)
r = chip thickness ratio, t = chip thickness before cutting/(uncut chip thickness) (mm), t_{c} = chip thickness after cutting (mm), V = cutting speed (m/s), V_{c} = chip velocity (m/s)
Calculation:
Given:
V = 2 m/s, Depth of cut = 0.5 mm
In orthogonal cutting,
t = d = 0.5 mm
t_{c} = 0.6 mm
\(\frac{t}{{{t_c}}} = \frac{{{V_c}}}{V}\)
\(\frac{{0.5}}{{0.6}} = \frac{{{V_c}}}{2} \Rightarrow {V_c} = 1.66\;m/s\)
Concept:
Chip thickness ratio / Cutting ratio (r):
It is the ratio of chip thickness before cut (t1) to the chip thickness after cut (t2).
\(r=\frac{Chip\;thickness\;before\;cut\;(t_1)}{Chip\;thickness\;after\;cut\;(t_2)}\Rightarrow \frac{uncut\;chip\;thickness}{chip\;thickness}\)
chip thickness after the cut (t2) is always greater than the chip thickness before the cut (t1), ∴ r is always < 1, i.e. the uncut chip thickness value is less than the chip thickness value.
Assuming discharge to be constant:
t1b1V = t2b2Vc
\(\frac{t_1}{t_2}=\frac{V_c}{V}\;\;\;\;(\because b_1=b_2)\)
as t2 > t1, ∴ V > Vc i.e. cutting velocity is greater than the chip velocity.
[NOTE: In an orthogonal metal cutting depth of cut = uncut chip thickness]
Calculation:
Given:
V_{c} = 2 m/s, depth of cut = t_{1} = 0.5 mm, t_{2} = 0.75 mm.
\(\frac{t_1}{t_2}=\frac{V_c}{V}\;\;\;\;(\because b_1=b_2)\)
\(\frac{0.5}{0.75}=\frac{V_c}{2}\)
∴ V_{c} = 1.33 m/s
Concept:
The material removal rate in turning operation is given by:
MRR = Area of cut × Longitudinal velocity
\(MRR=\frac{\pi}{4}(d_1^2\;\;d_2^2)fN\)
\(MRR={\pi}\left(\frac{d_1\;+\;d_2}{2}\right)\left(\frac{d_1\;\;d_2}{2}\right)fN\)
\(MRR={\pi}d_{avg}dfN\)
where, d_{1} = Diameter before cutting, d_{2} = Diameter after cutting
d is depth of cut, f is the feed in mm/rev and N = angular speed of spindle in r.p.m
Calculation:
Given:
d_{1} = 10 mm, d_{2} = 9 mm, V = fN = 175 mm/min, N = 360 r.p.m.
\(Depth\;of\;cut(d)=\frac{d_1\;\;d_2}{2}=\frac{10\;\;9}{2}=0.5\;mm\)
\(Average\;diameter(d_{avg})=\frac{d_1\;+\;d_2}{2}=\frac{10\;+\;9}{2}=9.5\;mm\)
\(MRR={\pi}d_{avg}dfN\)
\(MRR={\pi}\times\;9.5\;\times\;\times 0.5\;\times \;175=2611.4488\;mm^3/min\)\
Concept:
During orthogonal turning process, the cutting ratio is given as
\(\left( r \right) = \frac{t}{{{t_c}}} = \frac{{{l_c}}}{l}\)
Where t  cut chip thickness, t_{c} uncut chip thickness
Given
Feed = 0.2 mm/rev ⇒ t_{c} = 0.2 mm; t = 0.32 mm
Cutting ratio will be
\(\left( r \right) = \frac{0.32}{{0.2}} = 1.6\)
Explanation:
Orthogonal Cutting 
Oblique Cutting 
In orthogonal cutting, the cutting edge is perpendicular to the direction of the tooltravel. 
In oblique cutting, the angle between the cutting edge and direction of tooltravel is less than 90°. 
The chip flow angle is zero. 
The chip flow angle is more than zero. 
The tool life is less. 
Tool life is more. 
Two components of forces.

Three components of forces.

Poor surface finish. 
Good surface finish. 
Used in slotting, parting, grooving, pipe cutting. 
Used in turning, milling, drilling, grinding. 
Which pair of following statement is correct for orthogonal cutting using a singlepoint cutting tool ?
P. Reduction in friction angle increases cutting force
Q. Reduction in friction angle decreases cutting force
R. Reduction in friction angle increases chip thickness
S. Reduction in friction angle decreases chip thickness
Explanation:
In orthogonal cutting,
(i) Reduction in friction angle decreases cutting force
(ii) Reduction in friction angle decreases chip thickness
According to Merchant theory,
\(ϕ = \frac{{\rm{\pi }}}{4} + \frac{{\rm{α }}}{2}  \frac{{\rm{β }}}{2}\) where, ϕ = shear angle, α = rake angle, β = friction angle
As β decreases. ϕ increases because α is constant.
Now \(\frac{t}{{{t_c}}} = \frac{{sinϕ }}{{\cos \left( {ϕ  α } \right)}}\)
\({t_c} = \frac{{\cos \left( {ϕ  α } \right)}}{{sinϕ }} \times t\)
As ϕ increases, cos (ϕ α) will decrease and Sinϕ will increase. This chip thickness will decrease.
The main cutting force acting on a tool during the turning (orthogonal cutting) operation of a metal is 400 N. The turning was performed using 2 mm depth of cut and 0.1 mm/rev feed rate. The specific cutting pressure (in N/mm^{2}) is
Concept:
For orthogonal turning, t = f
Because t = f sinλ and λ = 90
\({\rm{P}} = \frac{{{{\rm{F}}_{\rm{c}}} \times {\rm{V}}}}{{{\rm{fdV}}}}\)
Where, P = cutting pressure, F_{c} = cutting force, V = velocity of tool, f = feed rate, d = depth of cut,
λ = principle cutting edge angle
Given: F_{c} = 400 N, d= 2 mm, f = 0.1 mm/rev
Calculation:
\({\rm{P}} = {\rm{}}\frac{{400}}{{0.1 \times 2}} = 2000{\rm{}}\,\,{\rm{N}}/{\rm{m}}{{\rm{m}}^2}\)
Concept:
Time for machining \(= \frac{L}{{f\times N}}\)
where L is job length (mm), f is feed (mm/rev), N is job speed (rpm)
Calculation:
Given:
f = 3 mm/rev, N = 600 rpm, L = 300 mm
Time for machining \(= \frac{L}{{f.N}} = \frac{ 300}{3×600} \) = 0.1666 minutes = 0.1666 × 60 = 10 sec
The total machining time to turn the cylindrical surface is 10 sec.
Concept:
Cutting speed of the milling tool is given by
\(V = π DN\)
where V = Cutter speed in m/min, D = Diameter of cutter in meter, N = rpm
Calculation:
Given:
V = 16.5 m/min, N = 250 rpm, \(π = \frac{22}{7}\)
\(16.5 = \frac{22\times D\times 250}{7}\)
\(D=\frac{16.5\times 7}{22\times250 }\)
D = 0.021 m
D = 21 mm.
The diameter of the cutter will be 21 mm.
Concept:
The thrust force is given as, FT = Fc tan (β  α)
The cutting force is given as, Fc = R cos (β  α)
The shear force is given as, Fs = R cos (ϕ + β  α) and \({F_s} = {\tau _s} \times \frac{{b{t_c}}}{{\sin ϕ }}\)
where, α = rake angle, β = friction angle, ϕ = shear angle, t_{c} = chip thickness, b = width
Calculation:
Given:
α = 5°, t_{c} = 0.25 mm, b = 4 mm, τ_{s} = 350 N/mm^{2}, μ = 0.5
Friction angle (β):
β = tan^{1}μ
= tan1 (0.5) = 26.57°
Shear angle (ϕ):
\(\phi = \frac{{90 + \alpha  \beta }}{2}\)
\(\phi = \frac{{90 + 5  26.57}}{2} = 34.22^\circ \)
Shear force:
\({F_s} = {\tau _s} \times \frac{{b{t_c}}}{{\sin ϕ }}\)
\({F_s} = \frac{{350 \times 4 \times 0.25}}{{sin~34.22}}\)
F_{s} = 622.37 N
Cutting Force (F_{c}):
\({F_c} = \frac{{{F_s}\cos \left( {\beta  \alpha } \right)}}{{\cos \left( {\phi + \beta  \alpha } \right)}}\)
\( = \frac{{622.37\cos \left( {26.57  5} \right)}}{{\cos \left( {34.22 + 26.57  5} \right)}}\)
F_{c} = 1029.45 N
Thrust force (F_{T}):
F_{T} = F_{c} tan (β  α)
= 1029.45 tan (26.57 – 5)
F_{T} = 406.96 N
Given that axial feed = 0.4 m/min, N = 400 rpm
then uncut chip thickness, \(t= \frac{f}{N}=\frac{0.4\times10^3}{400}= 1\ mm/rev\)
Chip thickness ratio \(r = \frac{t}{{{t_c}}} = \frac{1}{3} = 0.333\)
And \(\tan \phi = \frac{{r\cos \alpha }}{{1  r\sin\alpha }} = \frac{{0.333\cos5^\circ }}{{1  0.333\sin5^\circ }} = 0.342\)
\(\Rightarrow \phi = {\tan ^{  1}}\left( {0.342} \right) = 18.86^\circ\)Concept:
Cutting speed is given by
V = πDN
Where V = cutting speed, D = diameter, N = speed in RPM
Calculation:
Given:
V = 20 m/min , D = 40 mm = 0.04 m
V = πDN
20 = π × 0.04 × N
N = 160 rpm
Concept:
Shear angle:
The angle made by the shear plane with respect to the cutting velocity vector is known as the shear angle
The expression of the shear angle is given by:
\(ϕ = \;{\tan ^{  1}}\left[ {\frac{{{\bf{r}}\cos {\bf{α }}}}{{1  {\bf{r}} \sin {\bf{α }}}}} \right]\)
From merchant theory
2ϕ + β  α_{0} = 90°
and μ = tan β
where, ϕ = shear angle, α = Orthogonal rake angle, β = friction angle, r = chip thickness ratio = \(\frac{t_1}{t_2}\), \(t_1\) = uncut chip thickness, \(t_2 \) = deformed chip thickness.
Calculation:
Given:
α = 10°, t = 0.125 mm, t_{c} = 0.22 mm
Chip thickness ratio \(r = \frac{t}{{{t_c}}} = \frac{{0.125}}{{0.22}} = 0.568\)
\(\because \tan ϕ = \frac{{r\cos \propto }}{{1  r\sin \propto }} = \frac{{0.568\cos 10^\circ }}{{1  0.568\sin 10}}\)
ϕ = 31.83°
From merchant theory,
\(ϕ = \frac{\pi }{4} + \frac{ \propto }{2}  \frac{β }{2}\)
\(\Rightarrow 31.83= 45^\circ + \frac{{10}}{2}  \frac{β }{2} \Rightarrow β = 36.34^\circ \)
∵ μ = tan β = tan (36.34°)
∴ μ = 0.735≈ 0.74Concept:
\(\tan \phi = \frac{{r\cos \alpha }}{{1  r\sin \alpha }}\)
Calculation:
Gven:
α = 9°
t = 0.2 mm = uncut thickness
\({r_1} = \frac{{0.2}}{{0.25}} = 0.8\)
\(\tan {\phi _1} = \frac{{0.8\cos 9}}{{1  0.8\sin 9}} = 0.903\)
\({\phi _1} = {\tan ^{  1}}\left( {0.903} \right) = 42.087^\circ \)
\({r_2} = \frac{{0.2}}{{0.4}} = 0.5\)
\(\tan {\phi _2} = \frac{{0.5\cos 9}}{{1  0.5\sin 9}} = 0.5357\)
\({\phi _2} = {\tan ^{  1}}\left( {0.5357} \right) = 28.18^\circ \)
\(\frac{{{\phi _1}}}{{{\phi _2}}} = \frac{{42.087}}{{28.18}} = 1.49\)
Concept:
When cutting force F_{C} and friction force F becomes perpendicular, orthogonal rake angle α becomes 0°.
Under this condition, the Merchant's circle diagram looks like:
In this situation, forces F, N, F_{C }and F_{T} form a rectangle.
where F = F_{T} and F_{C} = N.
∴ N = F_{C} ⇒ 1500 N.
Alternate method:
The relation between Normal force N and Cutting force F_{C} and Axial force F_{T} is:
N = F_{C}cosα  F_{T}sinα
∴ N = 1500 × cos 0
∴ N = 1500 N.
Concept:
The above figure represents Merchant's circle diagram.
F_{C }= Cutting force, F_{T }= Axial force, F = Friction force, N = Normal force, F_{S} = Shear force and F_{n} = Shear normal force.
α = Orthogonal rake angle, β = Friction angle, ϕ_{s }= Shear plane angle.
Calculation:
Given:
∠ between F_{C }and F = 90°
∴ (β  α) + (90  β) = 90
⇒ α = 0°
∴ as cutting force F_{C} and friction force F becomes perpendicular, orthogonal rake angle α become 0°.
Concept:
From Merchant’s theory, 2ϕ + λ  α = 90^{o}
μ = tanλ and \(\mu=\frac{F}{N}\)
Calculation:
Given:
∴ λ = 90^{o} – 50^{o} = 40^{o}
\(\begin{array}{l} \therefore μ = \frac{F}{N} = \frac{{Frictionforce}}{{Normalforce}}\\ {\rm{μ }} = tanλ = \tan {40^o} = 0.84 \end{array}\)
So, the nearest possible choice is 0.80.
Concept:
Power = Fc × V = MRR × Specific energy
\(\Rightarrow {F_c} = \frac{{MRR \times Specific\;energy}}{V}\)
MRR = fdv
Given:
Specific machining energy = 2.0 J/mm^{2}, V = 120 m/min, f = 0.2 mm/rev, d = 2 mm
MRR = fdv = 120 × 0.2 × 2
\({F_c} = \frac{{\left( {120 \times 0.2 \times 2} \right) \times 2}}{{120}}\)
= 0.8 J/mm
= 800 N
F_{c} = 800 NIn machining processes, the percentage of total heat generated in shear action is carried away by the chips to the extent of
80 %
Explanation:
The development of high cutting temperature is a major concern in machining. Figure given below shows the sources of heat generation in the cutting zone during machining 
Primary Shear Zone (PSZ):
Secondary Shear Zone (SSZ):
Tertiary Shear Zone (TSZ):
Overall the heat is dissipated as: