Metal Cutting Processes MCQ Quiz in తెలుగు - Objective Question with Answer for Metal Cutting Processes - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

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).

chip thickness after the cut (t2) is always greater than the chip thickness before the cut (t1), ∴ r is always , i.e. the uncut chip thickness value is less than the chip thickness value.

Assuming discharge to be constant:

t1b1V = t2b2Vc

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.

∴ V_c = 1.33 m/s

Concept:

Chip reduction coefficient is given as,

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

Calculation:

Given:

Chip thickness,

Feed,

The chip reduction coefficient is,

Perform the division,

The chip reduction coefficient () 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 = + ⍺ - β

Additional Information

Ernst merchant Theory:

2ϕ + β - ⍺ = 

Stabler Theory:

ϕ =

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

Concept:

Material removal rate, MRR = V × f × d

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

Cutting speed V = 

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

Coefficient of friction (μ) = F/N = 500 / 1000 = 0.5

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.