Metal Cutting Processes MCQ Quiz in मल्याळम - Objective Question with Answer for Metal Cutting Processes - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

Chip reduction coefficient is given by 

k = 1/r = t₂/t₁

where t₂ is cut chip thickness, t₁ is uncut chip thickness

Calculation:

Given:

k = 2, t₁ = 0.2 mm

Cut chip thickness is given by:

t₂ = k × t₁ = 2 × 0.2 = 0.4 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.

Explanation:

Cutting fluid

Cutting fluid is a type of coolant and lubricant designed specifically for metalworking processes, such as machining and stamping. There are various kinds of cutting fluids, which include oils, oil-water emulsions, pastes, gels, aerosols (mists), and air or other gases.

Various functions of cutting fluids are

• Cutting fluid cools the workpiece and tool by carrying away the heat generated during machining
• It acts as a lubricant at the friction zones, hence tool life increases
• It prevents the corrosion of chips and machine
• As friction get reduced, the forces and electric power consumption decreases
• It causes to break the chips into small pieces
• It washes away the chips from the tool
• Improves dimensional accuracy and control on the workpiece

Explanation:

In metal cutting, Lee and Shaffer's theory is based on slip line field theory.

It is based on the assumption that:

• The material being cut behaves like an ideal plastic with no strain hardening.
• The shear plane represents a direction of maximum shear stress.

Lee and Shaffer theory:

ϕ = \(\frac{\pi }{4}\;\)+ ⍺ - β

Additional Information

Ernst merchant Theory:

2ϕ + β - ⍺ = \(\frac{\pi }{2}\;\)

Stabler Theory:

ϕ = \(\frac{{\rm{\pi }}}{{4{\rm{}}}}{\rm{\;}} + {\rm{\;}}\frac{\alpha }{2}-\beta\)

where, ϕ = shear angle, ⍺ = rake angle, β = friction angle

Concept:

Where V = cutting speed (in / min), f = feed (in / rev), d = depth of cut (in)

Cutting speed V = \(\frac{{\pi DN}}{{60}}\)

Where D = diameter of shaft

Calculation:

Given:

Cutting speed (V) = 300 ft/min, Feed (f) = 0.010 in/rev, and Depth of cut (d) = 0.100 in

∴ Material removal rate, MRR = V × f × d

⇒ MRR = (300 × 12) × 0.01 × 0.1 = 3.6 in³ / min

Explanation:

• During machining, based on the direction of chip flow and orientation of cutting edge, the machining operation is broadly divided into two types.
• They are as follows:

Concept:

Orthogonal cutting is a type of metal cutting in which the cutting edge of the wedge-shaped cutting tool is perpendicular to the direction of tool motion.

Friction force is given as, F = Fc sinα + Ft cosα

Normal force is given as, N = Fc cos α – Ft sin α

Shear Force is given as, Fs = Fc cos ϕ - FT sin ϕ 

Calculation:

GIven:

Rake angle (α) = 0°, F_c = 1000 N, F_t = 500 N

F = F_c sinα + F_t cosα

F = 1000 sin0 + 500 cos0 = 500 N

N = F_c cos α – F_t sin α = 1000 cos 0 = 1000 N

Concept:

Power of cutting is given as

Power = energy of cutting (J/mm³) × Material removal rate (mm³/s)

Calculation:

Given:

specific energy U = 6 J/mm³, axial feed = 1 m/min = 1000/60 mm/sec

tube thickness t = 1mm, diameter = 100 mm

Area of cut = πdt = 3.14 × 100 × 1 = 314 mm²

Material removal rate  = axial feed × area removed

MRR = 314 × 1000/60 mm³/s 

Power = energy × MRR 

Power = 6 × 314 × 1000/60 = 31400 J/s

P = 31.4 kW

HintIn problem like this if we do not remember formulae then we can calculate it by balancing the unit on both side.

unit of Power is Watt(W) = J/s

Power(J/s) = energy(J/mm³) × axial feed ( mm /sec) × area (mm²)

Explanation:

There are two types of cutting done mainly which are described in the table below.