Merchant Circle Analysis MCQ Quiz in मल्याळम - Objective Question with Answer for Merchant Circle Analysis - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

Merchant’s condition for minimum power consumption

\(ϕ = \frac{\pi }{4} + \frac{α }{2} - \frac{\beta }{2}\)

ϕ = shear angle, α = rake angle, β = shear angle

Calculation:

Given:

α = 5° 

β = tan-10.5 = 26.56° 

ϕ = 45 + 2.5 -13.28 = 34.22 ° 

Explanation:

The relation between various forces have been worked out by Merchant with a large number of assumptions as follows

1. The chip behaves as a free body in stable equilibrium under the action of two equal opposite and collinear resultant forces.
2. A continuous chip without a built-up edge is produced.
3. The cutting velocity remains constant.
4. The cutting tool has a sharp cutting edge and it does not make any flank contact with the workpiece Merchant suggested a compact and easiest way of representing the various forces inside a circle having the vector F as the diameter.

where, γ = Rake angle of the tool, ϕ = Shear angle, β = Friction angle of tool face

According to modified merchant theory relation between rake angle γ, shear angle ϕ, and friction angle β as shown below

\(ϕ = \frac{π }{4} - \frac{β }{2} + \frac{\alpha }{2}\)

2ϕ - α + β = π/2 

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:

\({\rm{Shear\;angle\;}}(\tan \phi ) = \frac{{r\cos \alpha }}{{1 - \;r\sin \alpha }}\)

Calculation:

\({\rm{r\;}} = {\rm{\;cutting\;ratio\;}} = \frac{{{t_1}}}{{{t_2}}} = \frac{{0.2}}{{0.4}} = 0.5\)

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

\(\tan \phi = \frac{{\left( {0.5} \right)\cos 45^\circ }}{{1 - \left( {0.5} \right)\sin 45^\circ }} = \frac{{\left( {\frac{1}{2}} \right)\left( {\frac{1}{{\sqrt 2 }}} \right)}}{{1 - \left( {\frac{1}{2}} \right)\left( {\frac{1}{{\sqrt 2 }}} \right)}}\)

\(\tan \phi = \frac{1}{{2\sqrt 2 - 1}} = \frac{1}{{3 - 1}} = \frac{1}{2}\)

∴ tan ϕ = 0.5

Concept:

The friction force \(F = F_C sinα +F_T cosα \)

The normal friction force \(N=F_C cosα -F_T sinα \)

\(μ = tanβ =\frac{F}{N} = \frac{{{F_C}sin\;α + {F_T}\;cos\;α }}{{{F_C}cos\;α - {F_T}\sin α }}\)

\(\mu = \frac{{\frac{{{F_T}}}{{{F_C}}} + \tan \alpha }}{{1 - \frac{{{F_T}}}{{{F_C}}}\tan \alpha }}\)

where F_C and F_T are cutting and thrust forces respectively in N.

α is back rake angle

and β is friction angle.

the power consumption P = F_C× V 

where V is cutting speed in m/s.

Calculation:

Given : α  = 0°, V = 180 m/min = 3 m/s F_T = 490 N and μ = 0.7

\(\therefore μ = \frac{{{F_T}}}{{{F_C}}}\)

\(0.7 = \frac{{490}}{{{F_C}}}\)

∴ F_C = 700 N

Power consumption, \(P = {F_C} × {V} = 700 × \frac{{180}}{{60}} × \frac{1}{{1000}}\left( {kW} \right)\)

P = 2.1 kW

Cutting speed, V_c = 200 m/min, α = 10°, w = 2 mm, t = 0.2 mm, μ = 0.5, τ_s = 400 N/mm²

Friction angle, β = tan^-1(μ) = tan^-1(0.5)

β = 26.57°

Using Merchant’s first solution to find out shear plane angle

2ϕ + β - α = 90°

\(\phi = \frac{{90 + 10 - 26.57}}{2} = 36.7^\circ\) 

Concept:

Shear strain (γ) = cot ϕ + tan (ϕ – α)

where ϕ = shear plane angle and α = orthogonal rake angle

For minimum shear strain (γ),

\(\frac{{d\gamma }}{{dϕ }} = 0\)

⇒ -cosec2 ϕ + sec2(ϕ – α) = 0

sec2(ϕ – α) = cosec2 ϕ

sin2ϕ = cos2(ϕ - α)

sinϕ = cos(ϕ – α)

\(\Rightarrow \sin ϕ = \sin \left\{ {\frac{\pi }{2} - \left( {ϕ - α } \right)} \right\}\)

\(\Rightarrow ϕ = \frac{\pi }{2} - ϕ + α\)

\(\Rightarrow2ϕ = \frac{\pi }{2} + α\)

\(\Rightarrow ϕ = \frac{\pi }{4} + \frac{α }{2}\)

When orthogonal rake angle α = 0°.

\(\therefore ϕ = \frac{\pi }{4} =45^{\circ}\)

Explanation:

The relation between various forces have been worked out by Merchant with a large number of assumptions as follows

1. The chip behaves as a free body in stable equilibrium under the action of two equal opposite and collinear resultant forces.
2. A continuous chip without a built-up edge is produced.
3. The cutting velocity remains constant.
4. The cutting tool has a sharp cutting edge and it does not make any flank contact with the workpiece Merchant suggested a compact and easiest way of representing the various forces inside a circle having the vector F as the diameter.

where, γ = Rake angle of the tool, ϕ = Shear angle, β = Friction angle of tool face

According to modified merchant theory relation between rake angle γ, shear angle ϕ, and friction angle β as shown below

\(ϕ = \frac{\pi }{4} - \frac{β }{2} + \frac{\alpha }{2}\)

• shear will take place in a direction in which energy required for shearing is minimum. 
• shear stress is maximum at the shear plane and it remains constant.

we have proved that 

\({F_H} = \frac{{{F_s}{\rm{cos}}\left( {\beta ~- ~\alpha } \right)}}{{{\rm{cos}}\left( {ϕ~ +~ \beta~ -~ \alpha } \right)}}\)

Now Fs = fs × A2

where, Fs = Shear stress, A2 = Area of shear plane = \(\frac{{{b_1}{t_1}}}{{sinϕ }}\)

 \({F_H} = \frac{{{b_1}{t_1}{\rm{cos}}\left( {\beta ~- ~\alpha } \right)}}{{sinϕ {\rm{cos}}\left( {ϕ ~+ ~\beta ~-~ \alpha } \right)}}\)

differentiate w.r.t β 

 \(\frac{{d{F_H}}}{{d\beta }} = \; - {f_s} \times {b_1}{t_1}\cos \left( {\beta ~-~ \alpha } \right) \times \frac{{cosϕ \cos \left( {ϕ ~+ ~\beta ~-~ \alpha } \right) - sinϕ {\rm{sin}}\left( {ϕ ~+ ~\beta ~-~ \alpha } \right)}}{{{{\sin }^2}ϕ {{\cos }^2}\left( {\alpha ~+~ \alpha ~-~ \gamma } \right)}}~=~0\)

In order that ϕ will assume that value which required a minimum force to cut the material.

\(cos\phi \cos \left( {\phi + \beta - \alpha } \right) - sin\phi \sin \left( {\phi + \beta - \alpha } \right) = 0\)

\(\cos \left( {\phi + \phi + \beta - \alpha } \right) = 0\)

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

\(2\phi + \beta - \alpha = c\)

In orthogonal turning

λ = 90°, F ⊥ velocity, So α = 0

F = 402.5 N

\(\frac{F}{N} = 1\) ⇒ Coefficient of friction = 1

So Friction angle (β) = tan^-1(1) = 45°

Uncut chip thickness = 0.2 mm

Cut chip thickness = 0.4 mm.

Cutting velocity v = 2 m/s

\(\begin{array}{l} r = \frac{t}{{{t_c}}} = \frac{{0.2}}{{0.4}} = 0.5\\ \tan \phi = \frac{{r\cos \alpha }}{{1 - r\sin \alpha }} = \frac{{0.5\cos 0}}{{1 - r\sin 0}} = 0.5 \end{array}\)

ϕ = tan^-1(0.5) = 26.56°

from Merchant circle,

F = R sin β

\(\Rightarrow R = \frac{F}{{\sin \beta }} = \frac{{402.5}}{{\sin 45}} = 569.22\;N\)

Shear force F_S = R cos (ϕ + β - α)

= 569.22 cos(26.56 + 45 - 0)

= 180 N

Explanation:

Shear strain (γ) = cot ϕ + tan (ϕ – α)

where ϕ = shear plane angle and α = orthogonal rake angle

For minimum shear strain (γ),

\(\frac{{d\gamma }}{{dϕ }} = 0\)

⇒ -cosec2 ϕ + sec2(ϕ – α) = 0

sec2(ϕ – α) = cosec2 ϕ

sin2ϕ = cos2(ϕ - α)

sinϕ = cos(ϕ – α)

\(\Rightarrow \sin ϕ = \sin \left\{ {\frac{\pi }{2} - \left( {ϕ - α } \right)} \right\}\)

\(\Rightarrow ϕ = \frac{\pi }{2} - ϕ + α\)

\(\Rightarrow2ϕ = \frac{\pi }{2} + α\)

\(\Rightarrow ϕ = \frac{\pi }{4} + \frac{α }{2}\)

When orthogonal rake angle α = 0°.

\(\therefore ϕ = \frac{\pi }{4} =45^{\circ}\)