Machining Analysis MCQ Quiz - Objective Question with Answer for Machining Analysis - Download Free PDF

Explanation:

Machinability Criteria in CNC:

Machinability refers to the ease with which a material can be machined to meet specified requirements. In CNC (Computer Numerical Control) machining, various criteria are considered to determine machinability, including surface finish, chip type, tool life, and power use. These criteria help in optimizing the machining process for better efficiency, quality, and cost-effectiveness.

Let's analyze the criteria and their importance in CNC machining:

• Surface Finish: The surface finish of a machined part is crucial as it affects the part's performance, aesthetics, and functionality. A good surface finish ensures that the part meets the required specifications and standards. In CNC machining, achieving a high-quality surface finish is essential for applications where precision and appearance are important.
• Chip Type: The type of chips produced during machining can indicate the machinability of the material. Continuous chips usually signify good machinability, while discontinuous or segmented chips may indicate poor machinability. Chip type is important for determining the efficiency and smoothness of the machining process.
• Tool Life: Tool life refers to the duration a cutting tool can effectively perform before needing replacement. Longer tool life means reduced downtime and lower costs for tool replacement. In CNC machining, optimizing tool life is important for maintaining productivity and cost-efficiency.
• Power Use: Power use in machining refers to the energy consumed during the machining process. Efficient power use means lower operational costs and environmental impact. While power use is a consideration, it is often less critical compared to surface finish, chip type, and tool life.

Considering the above criteria, the correct decreasing order of importance in CNC machining is:

Option 2: (ii) Chip type, (i) Surface finish, (iv) Power use, (iii) Tool life

The reasoning behind this order is:

• Chip Type: Chip type is the most critical criterion because it directly affects the machining process's efficiency, smoothness, and overall machinability. Understanding the chip type helps in optimizing cutting conditions and tool selection.
• Surface Finish: Surface finish is the next important criterion as it impacts the final product's quality and functionality. Achieving the desired surface finish is crucial for meeting specifications and ensuring the part's performance.
• Power Use: Power use is considered after chip type and surface finish because, while important for operational costs and energy efficiency, it is secondary to the machining process's quality and effectiveness.
• Tool Life: Tool life is the least critical among the criteria listed because it primarily affects cost and downtime. While important, it does not directly impact the machining process's quality and efficiency as much as the other criteria.

Important Information:

To further analyze the other options:

• Option 1: (i) Surface finish, (ii) Chip type, (iv) Power use, (iii) Tool life
  • This option places surface finish as the most important criterion, which is not accurate because chip type has a more direct impact on machinability.
• Option 3: (i) Surface finish, (ii) Chip type, (iii) Tool life, (iv) Power use
  • This option places tool life before power use, which is not ideal since power use impacts overall efficiency and cost more directly than tool life.
• Option 4: (ii) Chip type, (i) Surface finish, (iii) Tool life, (iv) Power use
  • This option places tool life before power use, similar to Option 3. While chip type and surface finish are correctly prioritized, tool life should be considered after power use.

In conclusion, the correct decreasing order of importance in CNC machining criteria is best represented by Option 2: (ii) Chip type, (i) Surface finish, (iv) Power use, (iii) Tool life. This order ensures that the most critical factors affecting machinability are prioritized, leading to optimal machining performance and efficiency.

Concept:

From the merchant first analysis

\(2\phi + \beta - \alpha = \frac{\pi }{2}\)

Where, ϕ = shear angle, β = friction angle, α = cutting rake angle

There are several theories for the relationship between α (rake angle), β (friction angle), and ϕ (shear angle).

Ernst – Merchant theory:

\(\phi = \frac{\pi }{4} - \frac{\beta }{2} + \frac{\alpha }{2}\)

Stabler theory (Modified Ernst – Merchant theory):

\(\phi = \frac{\pi }{4} - \beta + \frac{\alpha }{2}\)

Lee and Shaffer theory:

Concept:

Different Forces can be related using the merchant circle

\(μ =\frac{f}{N}\)

Friction angle (β) = tan-1(μ)

Calculation:

Given:

FC = 1200 N, FT = 500 N

Now, for zero rake angle, 

FC = R cos β 

⇒ FT = R sin β 

\(⇒ \frac{F_T}{F_c}=tanβ\)

\(⇒ tanβ=\frac{5}{12}\)

⇒ β = 22.619° 

Since,  α = 0°

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

Chip thickness ratio,

\(\left( r \right) = \frac{t}{{{t_c}}} = \frac{{0.25}}{{0.75}} = 1/3\)

\(\tan \phi = \frac{{\left( {\frac{1}{3}} \right)\cos 0^\circ }}{{1 - \left( {\frac{1}{3}} \right)\sin 0^\circ }}\)

\(\tan \phi = \frac{1}{3}\)

ϕ = 18.435°

Explanation:

Merchant’s Circle Diagram (MCD) is an easy way to analyze the cutting forces acting on the orthogonal plane as well as the different angles while removing unwanted material from the workpiece.

In the orthogonal plane, the propagation on the cutting tool is perpendicular to the workpiece.

MCD is used to determine forces and angles, which in turn aids in the selection of machinery and the power supply and whether the workpiece can withstand the threshold of the cutting forces, as they can be predicted using this theory.

Merchant’s Theory is also used to consider properties and parameters of cutting tools to decrease wear and optimize efficiency and quality.

The following assumptions were made by Merchant, and contribute to the Principle of Minimum Energy. This principle was used to consolidate inconsistencies in Merchant’s Theory that arose when the geometrical calculations did not agree with actual experimental data.

• The cutting edge is perpendicular to the direction of propagation of the cutting tool.
• The tool is perfectly sharp. (Not realistic)
• The chip doesn’t move to either side.
• The depth and velocity of the chip are constant and uniform.
• The workpiece has a uniform velocity.
• The chip width remains constant.
• The thickness of the uncut chip is constant.
• The workpiece is rigid and perfectly plastic.
• There is a uniform distribution of normal and shear stress across the shear plane.
• The friction angle is uniform and independent of the shear angle.
• The work material undergoes deformation across a thin shear plane.
• The width of tool is greater than the width of the workpiece.

Calculation:

Given: t₁ = 0.2 mm, t2 = 0.4 mm, V = 2 m/s

Since the friction force vector is ⊥ to cutting velocity vector so the tool back rake angle,

⇒ α = 0° 

Also, \(\frac{{Friction\,force}}{{Normal\,force}} = 1\)

⇒ β = 45° 

Now, using,

\(\tan ϕ = \frac{{r \cos α }}{{1 - r\sin α }}\)

\(r = \frac{t_1}{{{t_2}}} = \frac{{0.2}}{{0.4}} = 0.5\)

\(⇒\tan ϕ = \frac{{0.5\cos \left( 0 \right)}}{{1 - 0.5 × \sin \left( 0 \right)}} = 0.5\)

ϕ = 26.56° 

Now, the Merchant circle can be drawn as,

\(\tan 45 = \frac{f}{N} = 1\)

⇒ F = N = 402.5

\(⇒ R = \sqrt {{f^2} + {N^2}} = \sqrt {{{402.5}^2} + {{402.5}^2}} \)

⇒ F = 569.22 N

So, the shear force,

F_s = R cos (β  - α + ϕ) = R cos (45°- 0° + ϕ) = 569.22 cos (45° + 26.56°)

⇒  F_s = 180.05 N

The rate of heat generation at the primary shear plane,

Power = F_s.V_s

The relation between the sear and cutting velocity,

\(⇒{V_s} = \frac{{V\cos \left( α \right)}}{{\cos \left( {ϕ - α } \right)}} = \frac{{V\cos \left( 0° \right)}}{{\cos \left( {ϕ - 0°} \right)}} = \frac{V}{{\cos \left( ϕ \right)}}\)

\({V_s} = \frac{2}{{\cos \left( { 26.56°} \right)}} = 2.24 \ m/s\)

⇒ P =180.05 × 2.24 

⇒ P = 402.5 W

Concept:

Chip velocity ‘VC’:

• The velocity with which the chip moves over the rake face of the cutting tool. Also represented as V_f
• The chip velocity V_c is the velocity of the chip relative to the tool and directed along the tool face.
• Since chip velocity is the relative velocity between tool and chip hence, If we assume the chip to be stationary then this chip velocity can be considered as the velocity of the tool along the tool rake face.

Shear velocity ‘VS’: The velocity with which metal of the work-piece shears along the shear plane. It is also called the velocity of the chip relative to work-piece.

Cutting velocity ‘V’: The velocity with which the tool moves relative to the work-piece.

The relationship b/w these velocities are:

From velocity triangle-

V = VC + VS

and from sine rule

The velocity relationship is given by the equation:

\(\frac{v}{{{\rm{cos}}\left( {\phi - \alpha } \right)}} = \frac{{{{\rm{v}}_{\rm{c}}}}}{{\sin \phi }} = \frac{{{v_s}}}{{\cos \alpha }}\)

Here is v_s the velocity of chip sliding along shear plane.

\({v_s} = \frac{v}{{{\rm{cos}}\left( {\phi - \alpha } \right)}} \times cos\;\alpha =\frac{{v\cos \alpha }}{{\cos \left( {\phi - \alpha } \right)}} \)

/Concept:

The relation between shear angle (ϕ) and orthogonal rake angle (α) is:

\(\tanϕ=\frac{r\cosα}{1\;-\;r\sinα}\)

where r = cutting ratio.

The relation between Cutting force (F_C), Axial force (F_T), and Friction force (F) is given by:

F = F_T  cos α + F_C sin α

where α = Orthogonal rake angle.

The relation between Cutting force (F_C), Axial force (F_T) and Shear force (F_S) is given by:

F_S = F_C cos ϕ - F_T sin ϕ  

Calculation:

Given:

α = 12°, F_C = 1000 N, F = 600 N, t = 0.75 mm, t_c = 1.5 mm, F_S = ?

\(r=\frac{uncut\;chip\;thickness}{chip\;thickness}=\frac{t}{t_c}\)

\(r=\frac{t}{t_c}=\frac{0.75}{1.5}=0.5\)

We know that

\(\tanϕ=\frac{r\cosα}{1\;-\;r\sinα}\)

\(\tanϕ=\frac{0.5\;×\;\cos 12^\circ}{1\;-\;(0.5\;×\;\sin 12^\circ)}=0.5458\)

∴ ϕ = 28.626°

F = FT cos α + FC sin α

600 = (F_T × cos 12°) + (1000 × sin 12°)

∴ FT = 400.848 N.

FS = FC cos ϕ - FT sin ϕ 

F_S = (1000 × cos 28.626°) - (400.848 × sin 28.626°)

F_S = 685.72 N

Concept:

For single point cutting tool, the uncut chip thickness,

⇒ t = f sin λ  

∵ λ + c_s = 90° 

⇒ t = f cosc_s      ...(1) 

Where,

λ = The approach angle. cs = The side cutting edge angle, f =  feed

Calculation:

Given: 

f = 0.2 mm/rev, Cs = 60°, d = 0.5 mm

Using equation (1),

⇒ t = 0.2 × sin (90-60)

⇒ t = 0.1 mm

Explanation:

Concept:

\( \mu = \tan β\)

where β is the friction angle

\( \phi = 45^\circ + \frac{β}{2} - \frac{α}{2} \)

where:

• \( \phi = 25^\circ \) (shear angle)
• \( α = 0^\circ \) (orthogonal rake angle)

Calculation:

Substituting values:

\( 25^\circ = 45^\circ + \frac{β}{2} - \frac{0}{2} \)

⇒ β = -40°

Now, calculating the coefficient of friction:

\( \mu = \tan(-40^\circ) \)

\( \mu = 0.80 \)

Explanation:

Primary Shear Zone (PSZ):

• The primary shear zone lies between the workpiece and metal chip.
• In the PSZ when shearing action is taking place the atomic bond present between the atoms of the material is getting breaking.
• For breaking the atomic bond it needs to supply a certain amount of energy but during the breaking of the atomic bond they release an equal amount of energy in the form of heat energy.
• Out of the heat generated the maximum (60 to 65%) amount of the heat is carried away by the chip.

Secondary Shear Zone (SSZ):

• It lies between the metal chip and cutting tool.
• In SSZ the energy supplied is converted into heat energy because of the presence of friction at the chip tool interface.
• About 30 to 35% of the energy supplied is converted into heat energy in the SSZ.
• Out of the heat generated the maximum amount of the heat is carried away by the chip Only a small amount is transferred to the tool.
• This is because the thermal conductivity of the tool is less than the chip.

Tertiary Shear Zone(TSZ):

• It lies between workpiece and tool.
• In TSZ the energy supplied is converted into heat energy is due to the presence of friction of the tool work interface.
• About 5 to 10% of the energy supplied is converted into heat energy in this zone.

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):

• The primary shear zone lies between the workpiece and metal chip.
• In the PSZ when shearing action is taking place, the atomic bond present between the atoms of the material is getting breaking.
• For breaking the atomic bond it needs to supply a certain amount of energy but during the breaking of the atomic bond, they release an equal amount of energy in the form of heat energy.
• Out of the heat generated the maximum (60 to 65%) amount of the heat is carried away by the chip.

Secondary Shear Zone (SSZ):

• It lies between the metal chip and cutting tool.
• In SSZ the energy supplied is converted into heat energy because of the presence of friction at the chip tool interface.
• About 30 to 35% of the energy supplied is converted into heat energy in the SSZ.
• Out of the heat generated the maximum amount of the heat is carried away by the chip, Only a small amount is transferred to the tool.
• This is because the thermal conductivity of the tool is less than the chip.

Tertiary Shear Zone (TSZ):

• It lies between workpiece and tool.
• In TSZ, the energy supplied is converted into heat energy is due to the presence of friction of the tool work interface.
• About 5 to 10% of the energy supplied is converted into heat energy in this zone.

Overall the heat is dissipated as:

• A major portion of the total heat is carried away by the flowing chips in the primary shear zone.
• 5–15 % of the total heat goes into a cutting tool in the secondary deformation zone.
• 3–5 % of the total heat goes into the work-piece at the flank wear zone.
• The rest heat goes into the atmosphere.

Concept:

Merchant's Criteria:

For a given workpiece material machining with a given tool under a given condition of machining, the work done during the machining is mainly depends on the shear angle.

The relation between various angles is given by:

2ϕ + β - α = 90°

This is called as optimum shear angle relation for minimum energy or minimum work done or minimum force are also called as Merchant's criteria.

Calculation:

Given:

α = 20°, β = 40° , ϕ =?

Now, we have

2ϕ + β - α = 90°

2ϕ + 40° - 20° = 90°

∴ ϕ = 35° 

Lee and Shaffer Criteria:

The Lee and Shaffer shear angle relation for minimum energy criteria of machining is given by

ϕ + β - α = 45° 

Stabler's Criteria:

Stabler's shear angle relation is given by

ϕ + β - α /2 = 45° 

Concept:

From Merchant cycle, 

F = F_C sin α + F_T cos α 

N = F_C cos α - F_T sin α

Thus \(\;\mu = \frac{F}{N} = \frac{{{F_c}\;sin\alpha \; + \;{F_{t\;}}cos\alpha }}{{{F_c}\;cos\alpha \; - \;{F_{t\;}}sin\alpha }}\)

Also, μ = tan β or β = tan^-1 μ

Where, F = friction force in N, N = normal force in N, F_C = cutting force in N

F_T = thrust force in N, μ = coefficient of friction, and α, β = rake angle and friction angle respectively. 

Calculation: 

Given, α = rake angle = 32°, Cutting force F_C = 3000 N, Feed force F_t = 1200 N

F = F_C sin α + F_T cos α = 3000 × sin 32° + 1200 × cos 32° = 2610 N

N = F_C cos α - F_T sin α = 3000 × cos 32° - 1200 × sin 32° = 1914 N

\(\mu = \frac{F}{N} = \frac{{2610}}{{1914}} = 1.36\)

Concept:

Shear strain during turning is given by

\(s=cot\ \phi +tan\ (\phi-α)\)

Calculation:

Given:

α = 0°

Shear strain 

\(∴ s=cot\ \phi +tan\ \phi\)

for minimum value, \({\frac{ds}{d\phi}} = 0\)

\(\Rightarrow - cose{c^2}\ \phi + {\sec ^2}\ \phi = 0 \Rightarrow - \frac{1}{{{{\sin }^2}\phi}} + \frac{1}{{{{\cos }^2}\phi}} = 0 \Rightarrow {\cos ^2}\phi - {\sin ^2}\phi = 0 \Rightarrow {\tan ^2}\phi = 1 \Rightarrow \phi= \frac{{\rm{\pi }}}{4}\)

again \(\frac{{{d^2}s}}{{d\phi^2}} = - 2cosec\ \phi \times \left( { - {\rm{cosec\ \phi \times cot\ \phi}}} \right) + 2\sec\ \phi\times \left( {\sec\ \phi \tan\ \phi } \right)\)

\(At\ \phi= \frac{{\rm{\pi }}}{4},\)

\(∴\frac{{{d^2}s}}{{{d\phi^2}}} = 2cose{c^2}\left( {\frac{\pi }{4}} \right)\cot \left( {\frac{\pi }{4}} \right) + 2\ {\sec ^2}\frac{\pi }{4}\ \tan \frac{\pi }{4}\)

∴ (2 × 2 × 1) + (2 × 2 × 1) = 8 > 0

So, minimum value at \(\phi = \frac{\pi }{4}\)

\(\therefore{s_{min}} = \cot \frac{\pi }{4} + \tan \frac{\pi }{4} = 1 + 1 = 2\)