Machining Analysis MCQ Quiz in বাংলা - Objective Question with Answer for Machining Analysis - বিনামূল্যে ডাউনলোড করুন [PDF]

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°

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

Shear force, \({F_s} = \frac{{{\tau _s} \cdot w \cdot {t_1}}}{{\sin \phi }} = \frac{{400 \times 2 \times 0.2}}{{\sin \left( {36.7} \right)}} = 267.72N\) 

∵ F_s­ = R.cos(ϕ + β – α)

\(\therefore R = \frac{{267.72}}{{\cos \left( {36.7 + 26.7 - 10} \right)}} = 448.9N\) 

Where R = resultant force of cutting force and thrust force

∴ Thrust force, F_T = R sin (β - α)

= 448.9 sin (26.57 - 10) = 128.02 N

Concept:

The relation between shear angle (ϕ), chip thickness ratio (r) and rake angle (α) is given by,

\(tan~{ϕ}=\frac{r~cos{α}}{1-rsinα}\) and \(r=\frac{t}{t_c}\)

where t = uncut thickness, tc = chip thickness

Calculation:

Given:

α = 12°, t = 0.81 mm, tc = 1.8 mm 

Therefore, \(r=\frac{t}{t_c}=\frac{0.81}{1.8}=0.45\)

⇒ \(tan~{ϕ}=\frac{0.45\times cos12}{1-0.45\times sin12}\)

tan ϕ = 0.485

Therefore, ϕ = 25.90° ≈ 26°  

Explanation:

The development of high cutting temperature is a major concern in machining. The figure given below shows the sources of heat generation in the cutting zone during machining -

Tertiary Shear Zone (Tool and chip Interface):

• 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.

So, we can say that the maximum temperature is lies between the Tool and Chip interface.

Additional Information

Primary Shear Zone (PSZ):

• The primary shear zone lies between the workpiece and the 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.

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 workpiece at the flank wear zone.
• The rest heat goes into the atmosphere.

Due to the actual plastic deformation of the chip in the shear zone 70-75% of the total heat is generated at the chip and carried away by the chip.

Due to friction between the chip and the tool at the rake surface, 15-20% of the total heat generated on the tool portion

and due to rubbing action between the tool and the workpiece, 5-10% heat is generated on the workpiece.

Concept:

The thrust force is given by: F_t = R sin (β – α)

Cutting force is given by: F_c = R cos (β – α)

Here, β and α are friction and rake angle

And the friction force is: F = R sin β

Calculation:

\(\tan \left( {\beta - \alpha } \right) = \frac{{{F_t}}}{{{F_c}}} \Rightarrow \;\beta = \alpha + {\tan ^{ - 1}}\left( {\frac{{{F_t}}}{{{F_c}}}} \right)\)

\(\beta = 5 + {\tan ^{ - 1}}\left( {\frac{{450}}{{900}}} \right) = 31.565^\circ \)

\(R = \sqrt {F_t^2 + F_c^2} = \sqrt {{{450}^2} + {{900}^2}} = 1006.23\;N\)

Thus, friction force will be:

\(F = R\sin \beta = 1006.23 \times \sin 31.565 = 526.726\;N\)

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°.

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

α = 0, F_c = 500 N, F_T = 250 N

μ = ?

Calculation:

\(\mu = \frac{F}{N} = \frac{{{F_c}\sin \alpha + {F_T}\cos \alpha }}{{{F_c}\cos \alpha - {F_T}\sin \alpha }}\)

sin 0 = 0, cos 0 = 1

\(\therefore \mu = \frac{F}{N} = \frac{{{F_T}}}{{{F_C}}} = \frac{{250}}{{500}} = \frac{1}{2}\)

Merchant circle diagram

For α ≠ 0

ϕ → shear plane angle

For α = 0

Merchant circle diagram is a rectangle i.e. F_t = F and F_c = N

Alternate method

\(\because\mu = \tan \beta \; = \frac{f}{N} = \frac{{{F_T}}}{{{F_c}}}\)     [∵ from merchant circle diagram for α = 0]

\(\mu = \frac{{250}}{{500}} = \frac{1}{2}\)

Concept:

\(Chip\;thickness\;ratio\left( r \right) = \frac{t}{{{t_c}}} = \frac{{{l_c}}}{l} = \frac{{{v_c}}}{v} = \frac{{\sin \phi }}{{\cos \left( {\phi - \alpha } \right)}}\)

Where,

t = uncut chip thickness

t_c = cut chip thickness

l_c = length of cut chip thickness

v_o = velocity of tool relative to work piece

v_s = velocity of chip relative to the workpiece

v_c = velocity of chip relative to tool

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

From velocity triangle,

Applying sine rule,

\(\frac{V}{{\sin \left( {90 - \left( {\phi - \alpha } \right)} \right)}} = \frac{{{V_c}}}{{\sin \phi }} = \frac{{{V_s}}}{{\sin \left( {90 - \alpha } \right)}}\)

\(\therefore \frac{V}{{\cos \left( {\phi - \alpha } \right)}} = \frac{{{V_c}}}{{\sin \left( \phi \right)}} = \frac{{{V_s}}}{{\cos \alpha }}\)

Calculation:

l_c = 50 mm

Diameter (D) = 25 mm

v = 10 m/min

depth of cut (d) = 0.12 mm

\(r = \frac{{{l_c}}}{l} = \frac{{{l_c}}}{{\pi D}} = \frac{{50}}{{\pi \times 25}} = 0.637\)

Now, \(\tan \phi = \frac{{r\cos \alpha }}{{1 - r\sin \alpha }} = \frac{{0.637\cos 10}}{{1 - 0.637\sin 10}} = 0.705\)

∴ ϕ = 35.19°C

\(V_s= \frac{{10\cos 10}}{{\cos \left( {35.19 - 10} \right)}} = 10.88\;m/min\)

Concept: 

From the orthogonal cutting

Chip thickness ratio \(r = \frac{t}{{{t_c}}}\)

Where t is the uncut chip thickness, tc is the cut chip thickness

\(r = \frac{{{V_c}}}{V} = \frac{{{\rm{Chip\;velocity}}}}{{{\rm{Cutting\;speed}}}} \)

∴ Vc = V × r 

Additional Information

Diagram

V = cutting velocity, Vs = shear velocity, Vc = chip velocity

From triangle, we can write as

Using the sine rule:

\(\frac{V}{{\cos \left( {\phi - \alpha } \right)}} = \frac{{{V_c}}}{{\sin \phi }} = \frac{{{V_s}}}{{\cos \alpha }}\)

• The velocity of tool relative to the workpiece is called as Cutting velocity (V).
• The velocity of chip relative to the workpiece (Shear plane) is called as Shear velocity (Vs).
• The velocity of chip relative to the tool is called as Chip velocity (Vc).
• The velocity of chip sliding along the shear plane is given by, \(V_s=\frac{{V\cos \alpha }}{{\cos \left( {\phi - \alpha } \right)}}\)

Concept:

Shear force (F_s) is calculated by;

\({{F}_{s}}=\frac{\tau bt}{\sin \phi }\Rightarrow \tau =\frac{{{F}_{s}}\sin \phi }{bt}\)     …(1)

Here, b = width of cut (mm)

t = uncut chip thickness (mm)

for orthogonal turning; λ = 90°

\(b=\frac{d}{\sin \lambda },t=f\sin \lambda\)

⇒ b = d, t = f

\(\tau =\frac{{{F}_{s}}\sin \phi }{df}\)    ...(2)

F_s = F_H cos ϕ - F_v sin ϕ

Where, F_H = Tangential face, F_V = Axial force/feed force

ϕ = shear angle

\(\tan \phi =\frac{rcos\alpha }{1-r\sin \alpha }\)    ...(3)

Calculation:

Given data;

t_c = 0.3 mm, f = 100 mm/min, N = 1000 rpm.

\(r=\frac{t}{{{t}_{c}}}=\frac{f}{{{t}_{c}}}=\frac{100}{\left( 100 \right)\left( 0.3 \right)}=\frac{1}{3}\) 

r = 0.333, α = 10°

\(\Rightarrow \tan \phi =\frac{\left( 0.3333 \right)\cos \left( 10{}^\circ \right)}{1-\left( 0.333 \right)\sin \left( 10{}^\circ \right)}\Rightarrow \phi =19.21{}^\circ \) 

F_H = 1601 N_, F_V = 1259 N

⇒ F_S = (1601) cos(19.21°) – (1259) sin(19.21°) = 1097.62 N

\(d=5mm,~f=\frac{100}{1000}=0.1~mm/rev\) 

\(\Rightarrow \tau =\frac{\left( 1097.62 \right)\sin \left( 19.21{}^\circ \right)}{\left( 5 \right)\left( 0.1 \right)}=722.302~MPa\) 

Key points:

Always remember relations between orthogonal machining and turning.

\(width~of~cut\left( b \right)=\frac{Depth~of~cut~\left( d \right)}{\sin \lambda }\) 

λ = approach angle = 90° - C_S

C_S side cutting edge angle

For orthogonal turning; λ = 90° ⇒ C_S = 0°

Uncut chip thickness (t) = f sin λ

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 °