Orthogonal Cutting MCQ Quiz - Objective Question with Answer for Orthogonal Cutting - Download Free PDF

Explanation:

Parting-Off Operation

• Parting-off is a machining operation performed on a lathe where a thin cutting tool, known as a parting tool, is used to cut off a portion of the workpiece. This operation is commonly used to separate finished components from the raw material or to create grooves or recesses in the workpiece. The parting tool is fed perpendicular to the axis of the workpiece, slicing through it as the lathe spindle rotates.
• In a parting-off operation, the cutting tool is mounted on the tool post, and the tool post is attached to the cross-slide of the lathe. The cross-slide allows the tool to move in a direction perpendicular to the axis of the workpiece. During the operation, the operator manually rotates the cross-slide screw to feed the parting tool into the workpiece. This manual control ensures precision and allows the operator to adjust the feed rate based on the material type, tool geometry, and cutting conditions.

Why Option 2 is Correct:

1. Mechanism of Cross-Slide Feed: The cross-slide screw is a threaded component connected to the cross-slide of the lathe. When the operator rotates the screw, the cross-slide moves along its dovetail ways, carrying the tool closer to or away from the workpiece. This movement is precisely controlled, which is essential for delicate operations like parting-off.

2. Perpendicular Feed: Parting-off requires the tool to move perpendicular to the workpiece's axis to achieve a clean separation. The cross-slide screw facilitates this perpendicular feed, making it the most appropriate choice for the operation.

3. Manual Control: Unlike automated feeds, manual control allows the operator to feel the resistance encountered during cutting. This tactile feedback helps the operator avoid excessive tool pressure, which can lead to tool breakage or deformation of the workpiece.

4. Practical Application: In most lathe operations, especially in small workshops or for custom jobs, manual control of the cross-slide is preferred for parting-off. It provides the flexibility to accommodate variations in material properties and tool wear.

Concept:

The shear plane angle \( \phi \) in orthogonal cutting is calculated using:

\( \tan \phi = \frac{r \cos \alpha}{1 - r \sin \alpha} \)

Given:

Chip thickness ratio \( r = 0.4 \), Rake angle \( \alpha = 20^\circ \)

\( \cos(20^\circ) = 0.94 \), \( \sin(20^\circ) = 0.34 \)

Calculation:

\( \tan \phi = \frac{0.4 \cdot 0.94}{1 - 0.4 \cdot 0.34} = \frac{0.376}{0.864} \approx 0.435 \)

\( \phi = \tan^{-1}(0.435) \)

Explanation:

\({\rm{Shear\;stress\;}}\left( \tau \right) = \frac{{{F_s}}}{{b{t_1}}}\sin \phi \)

[where  Fs = shear force]

\( ⇒ {F_s} = \frac{{{F_c}\cos \left( {\phi + \beta - \alpha } \right)}}{{\cos \left( {\beta - \alpha } \right)}}\)

[From Merchant’s circle diagram]

\(\therefore \tau = \frac{{{F_c}\sin \phi \cos \left( {\phi + \beta - \alpha } \right)}}{{b{t_1}\cos \left( {\beta - \alpha } \right)}}\)

Now,

\(\frac{{{F_c}}}{{b{t_1}}} = {\rm{\;specific\;cutting\;energy\;}}\left( {\rm{U}} \right)\)

\(\therefore \tau = \frac{{U\sin \phi \cos \left( {\phi + \beta - \alpha } \right)}}{{\cos \left( {\beta - \alpha } \right)}}\)

⇒ \(U = \frac{{\tau\cos \left( {\beta - \alpha } \right)}}{{sin \phi \cos \left( {\phi + \beta - \alpha } \right)}}\)

Concept:

Cutting speed is given by

V = πDN

Where V = cutting speed, D = diameter, N = speed in RPM

Calculation:

Given:

N = 400 rpm, D = 12.5 mm = 0.0125 m

V = πDN

V = π × 0.0125 × 400

V = 15.7 m/min 

Explanation:

• The above diagram is called the "Merchant Circle Diagram", which represents the inter-relationship of the different force components in continuous chip formation under orthogonal cutting.
• The term (Fs) represents the "Shear force" which is mainly responsible for the chip separation from the parent body by shearing and is used to determine the yield strength of the work material. This shear force acts along the shear angle (ϕ).
• Feed Force, F = FCsin α + FTcos α 
• Normal Force, N = FCcos α - FTsin α 
• Where, Fc = Cutting Force, α = Rake Angle.

From Merchant circle:

• It can be seen that when the rake angle is 0o, friction force becomes parallel to thrust force and normal force becomes parallel to cutting force.
• Hence friction force, normal force, thrust force, and cutting force forms a rectangle inscribed in the Merchant circle.
• Therefore, at rake angle = 0o, friction force = thrust force and normal force = cutting force.
• Mathematically also, when you put α = 0o, you will get N = Fc and FT = F. 

Calculation:

Given:

Fc = 1500 N and Ft = 1170 N, α = 0°

α = 0o, you will get N = Fc

N = 1500 N

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:
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 = f(feed) 

and b(width) = d(depth of cut)

but in the question, nothing is mentioned about the uncut chip thickness and feed so we have taken the uncut chip thickness equal to the depth of cut. (Official Question of UPPSC AE)

t_c = 0.6 mm

\(\frac{t}{{{t_c}}} = \frac{{{V_c}}}{V}\)

\(\frac{{0.5}}{{0.6}} = \frac{{{V_c}}}{2} \Rightarrow {V_c} = 1.66\;m/s\)

Explanation:

In orthogonal cutting,

(i) Reduction in friction angle decreases the cutting force

(ii) Reduction in friction angle decreases chip thickness

According to Merchant's 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.

• If all other factors remain the same, a higher shear angle results in a smaller shear plane area. Since the shear strength is applied across this area, the shear force required to form the chip will decrease when the shear plane area is decreased.

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.
• Hence from the above options Q and S, both are correct..

Explanation:

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 volume to be constant:

t₁b₁L₁ = t₂b₂L₂

\(\frac{t_1}{t_2}=\frac{L_2}{L_1}\;\;\;\;(\because b_1=b_2)\)

as t₂ > t₁, ∴ L₂ < L₁ i.e. length after cut is less than the length before the cut.

Assuming discharge to be constant:

t₁b₁V = t₂b₂V_c

\(\frac{t_1}{t_2}=\frac{V_c}{V}\;\;\;\;(\because b_1=b_2)\)

as t₂ > t₁, ∴ V > V_c i.e. cutting velocity is greater than the chip velocity.

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:

Cutting speed is given by

V = πDN

Where V = cutting speed, D = diameter, N = speed in RPM

Calculation:

Given:

N = 400 rpm, D = 12.5 mm = 0.0125 m

V = πDN

V = π × 0.0125 × 400

V = 15.7 m/min 

Concept:

In orthogonal turning operation, \({\rm{t}} = {\rm{f\;sinλ }}\)

But λ = 90° where λ  = principle cutting edge angle

∴  t = f where, t = thickness, f = feed

Cutting shear strain \(\left( {\rm{\gamma }} \right) = \cot \left( \phi \right) + \tan \left( {\phi - {\rm{\alpha }}} \right)\)...........(1)

\(\tan \left( \phi \right) = \frac{{{\rm{r}}.{\rm{cos\alpha }}}}{{1 - {\rm{r}}.{\rm{sin\alpha }}}}{\rm{\;}}\)..............(2)

Where, \({\rm{\alpha }} = {\rm{rake\;angle}},{\rm{\;}}\phi = {\rm{shear\;angle}}\), r = chip thickness ratio 

\({\rm{r}} = \frac{{\rm{t}}}{{{{\rm{t}}_{\rm{c}}}{\rm{\;}}}}\)

Given:

f = t = 0.2 mm, tc =0.5 mm, α = 0, \({\rm{r}} = \frac{{0.2}}{{0.5}} = 0.4{\rm{\;}}\)

Calculation: 

Using the values of chip thickness ratio and rake angle in (2) we get shear angle.

\(\phi = {\tan ^{ - 1}}0.4\)

\(\phi = 21.801^\circ \)

Using the above-obtained shear angle in (1) we get shear strain.

\(∴ {\rm{\gamma }} = \cot \left( {21.801} \right) + \tan \left( {21.801 - 0} \right)\)

\({\rm{\;\gamma }} = {\rm{\;}}2.90\)

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:

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 = t₁ = 0.5 mm, t₂ = 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:

Chip reduction coefficient is given as,

\(\xi = \frac{t_c}{f}\)

The chip thickness (\(t_c\)) is 0.32 mm, and the feed (\(f\)) is 0.2 mm/rev.

Calculation:

Given:

Chip thickness, \(t_c = 0.32 \text{ mm}\)

Feed, \(f = 0.2 \text{ mm/rev}\)

The chip reduction coefficient is,

\(\xi = \frac{0.32 \text{ mm}}{0.2 \text{ mm/rev}}\)

Perform the division,

\(\xi = \frac{0.32}{0.2} = 1.6\)

The chip reduction coefficient (\(\xi\)) is 1.6.