Orthogonal Cutting MCQ Quiz in தமிழ் - Objective Question with Answer for Orthogonal Cutting - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

Chip thickness ratio:

\(r = \frac{t}{{{t_c}}} = \frac{{{V_c}}}{V} = \frac{{{L_c}}}{L}\)

where t, b and V respectively denote the depth of cut, width of cut and cutting velocity. Similarly, t_c, b_c and V_c denote chip thickness, width of chip and chip velocity respectively.

The chip velocity is related to the cutting velocity by: V_c = rV

Calculation:

An orthogonal cutting process produces a chip of thickness 0.75 mm. The depth of cut is 0.5 mm and the cutting velocity used is 2 m/s. The velocity at which the chip will move is:

Given: t_c = 0.75 mm, t = 0.5 mm, V = 2 m/s

\(r = \frac{{0.5}}{{0.75}} = \frac{2}{3}\)

\({V_c} = \frac{2}{3} \times 2 = \frac{4}{3}m/s = 1.33\;m/s\)

Concept:

Rake angle

• The rake angle ground on a tool controls the geometry of chip formation.
• Thereby, it controls the cutting action of the tool.
• The top or back rake angle of the tool is ground on the top of the tool, and it is a slope formed between the front of the cutting edge and the top of the face.
• If the slope is from the front towards the back of the tool, it is known as a positive top rake angle.
• If the slope is from the back of the tool towards the front of the cutting edge, it is known as a negative back rake angle.
• The top rake angle may be ground positive, negative or zero according to the material to be machined.
• When turning soft, ductile materials, which form curly chips, the positive top rake angle ground will be comparatively more than for turning hard brittle metals.
• When turning hard metals with carbide tools, it is the usual practice to give a negative top rake. Negative top rake tools have more strength than tools with positive top rake angles.
• Negative rake angle is recommended to machine Brittle material at high speed.

Explanation:

• Dynamometers are devices used to measure cutting forces in machining operations.
• The cutting force cannot be detected or quantified directly but its effect can be sensed using Transducer.
• For example, a force that can neither be seen nor be gripped but can be detected and quantified respectively by its effect and the amount of those effects (on some material) like elastic deflection, deformation, pressure, strain, etc.
• These effects, called signals, often need proper conditioning for easy, accurate and reliable detection and measurement.

Concept:

Length of uncut chip \(\left( \ell \right) = \frac{\pi }{2}\left[ {{D_i} + {D_f}} \right]\)

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

\(r = \frac{{Le}}{L} = \frac{t}{{{t_c}}}\) (c denotes cut chip quantities)

Calculation:

\(L = \frac{\pi }{2}\left[ {{D_i} + {D_f}} \right] = \frac{\pi }{2}\left[ {80 + 72} \right] = 238.761\;mm\)

ϕ = 0.3314 rad = 18.987°

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

\(r = \frac{{\tan \phi }}{{\left[ {\tan \phi \sin \alpha + \cos \alpha } \right]}} = 0.3267\)

∴ L_c = Lr = 0.3267 × 238.761 = 78.218 mm

Concept:

The thrust force is given as, FT = Fc tan (β - α)

The cutting force is given as, Fc = R cos (β - α)

The shear force is given as, Fs = R cos (ϕ + β - α) and \({F_s} = {\tau _s} \times \frac{{b{t_c}}}{{\sin ϕ }}\)

where, α = rake angle, β = friction angle, ϕ = shear angle, t_c = chip thickness, b = width

Calculation:

Given:

α = 5°, t_c = 0.25 mm, b = 4 mm, τ_s = 350 N/mm², μ  = 0.5 

Friction angle (β):

β = tan^-1μ

= tan-1 (0.5) = 26.57° 

Shear angle (ϕ):

\(\phi = \frac{{90 + \alpha - \beta }}{2}\)

\(\phi = \frac{{90 + 5 - 26.57}}{2} = 34.22^\circ \)

Shear force:

\({F_s} = {\tau _s} \times \frac{{b{t_c}}}{{\sin ϕ }}\)

\({F_s} = \frac{{350 \times 4 \times 0.25}}{{sin~34.22}}\)

F_s = 622.37 N

Cutting Force (F_c):

\({F_c} = \frac{{{F_s}\cos \left( {\beta - \alpha } \right)}}{{\cos \left( {\phi + \beta - \alpha } \right)}}\)

\( = \frac{{622.37\cos \left( {26.57 - 5} \right)}}{{\cos \left( {34.22 + 26.57 - 5} \right)}}\)

F_c = 1029.45 N

Thrust force (F_T):

F_T = F_c tan (β - α)

= 1029.45 tan (26.57 – 5)

F_T = 406.96 N

Concept:

In an orthogonal cutting, the velocity triangle is given as

Applying the sine law,

\( \frac{{{V_c}}}{{sin\;ϕ }}\; = \frac{V}{sin\;[90\;-\;(\phi\;-\;\alpha)]}\)

Where,

v_c = Chip Velocity, V = Cutting or Table Velocity. ϕ = Shear angle, α = Rake angle 

Given:

Calculation:

α = 15°, ϕ = 45°, V = 35 m/min

We know that,

\( \frac{{{V_c}}}{{sin\;ϕ }}\; = \frac{V}{sin\;[90\;-\;(\phi\;-\;\alpha)]}\)

Then,

\({V_c}\; = \;\frac{Vsin\;\phi}{cos\;(\phi\;-\;\alpha)}\)

= \(\frac{35sin\;45^\circ}{cos\;(45^\circ\;-\;15^\circ)}\)

= \(\frac{{35 \;\times\; 0.707}}{{0.866}}\)

= 28.5 m/min

Concept:

Chip thickness ratio / Cutting ratio (r):

It is the ratio of chip thickness before cut (t1) to the chip thickness after cut (t2).

\(r=\frac{cut ~chip~ length~(l_c)}{uncut ~chip ~length ~(l)}=\frac{Chip\;thickness\;before\;cut\;(t_1)}{Chip\;thickness\;after\;cut\;(t_2)}\Rightarrow \frac{uncut\;chip\;thickness}{chip\;thickness}\)

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.

Uncut chip length, l = πD

Mean Uncut chip length, \(l=\frac{\pi (D_1+D_2)}{2}\)

Calculation:

Given:

D₁ = 74 mm, D₂ = 70 mm, l_c = 73 mm

Mean Uncut chip length, \(l=\frac{\pi (D_1+D_2)}{2}=\frac{\pi (74 + 70)}{2}=72 \pi~mm\)

Cutting ratio (r), \(r=\frac{73}{72 \pi}=0.322\)

Given that axial feed, fN = 0.4 m/min, N = 400 rpm, λ = 90, t_c = 3 mm

Therefore, feed, \(f=\frac{0.4 × 1000}{400} = 1~mm/rev\)

then uncut chip thickness, t = f sin λ = 1 × sin (90) = 1 mm

Chip thickness ratio  \(r = \frac{t}{{{t_c}}} = \frac{1}{3} = 0.333\)

And \(\tan \phi = \frac{{r\cos \alpha }}{{1 - r\sin\alpha }} = \frac{{0.333\cos5^\circ }}{{1 - 0.333\sin5^\circ }} = 0.342\)

Explanation:

\({\rm{Shear\;stress\;}}\left( \tau \right) = \frac{{{F_s}}}{{b{t_1}}}\sin \phi \)

[where  Fs = shear force]

\( ⇒ {F_s} = \frac{{{F_c}\cos \left( {\phi + \beta - \alpha } \right)}}{{\cos \left( {\beta - \alpha } \right)}}\)

[From Merchant’s circle diagram]

\(\therefore \tau = \frac{{{F_c}\sin \phi \cos \left( {\phi + \beta - \alpha } \right)}}{{b{t_1}\cos \left( {\beta - \alpha } \right)}}\)

Now,

\(\frac{{{F_c}}}{{b{t_1}}} = {\rm{\;specific\;cutting\;energy\;}}\left( {\rm{U}} \right)\)

\(\therefore \tau = \frac{{U\sin \phi \cos \left( {\phi + \beta - \alpha } \right)}}{{\cos \left( {\beta - \alpha } \right)}}\)

⇒ \(U = \frac{{\tau\cos \left( {\beta - \alpha } \right)}}{{sin \phi \cos \left( {\phi + \beta - \alpha } \right)}}\)

Explanation:

The above diagram is called "Merchant Circle Diagram", which represents the inter-relationship of the different force components in continuous chip formation under orthogonal cutting.

The term (Fs) represents the "Shear force" which is mainly responsible for the chip-separation from the parent body by shearing and is used to determine the yield strength of the work material. This shear force acts along the shear angle (ϕ).

Cutting strain (γ):

The magnitude of  the average strain that develops along the shear plane due to machining action is called "cutting strain".

It is given by-

\(\gamma =\cot ϕ \;+\;\tan \left( {ϕ - α } \right) \)

where ϕ = shear angle and α = rake angle.

For, theoretical minimum possible shear plane angle, 

\(\frac{d \gamma}{d ϕ}=0\)

\(\frac{d \gamma}{d ϕ}=-cosec^2 ϕ+sec^2 (ϕ -α)\)

\(-cosec^2 ϕ+sec^2 (ϕ -12)=0\).................(α = 12°)

cos (ϕ - 12) = sin ϕ

cos (ϕ - 12) = cos (90 - ϕ) 

(ϕ - 12) = (90 - ϕ)

ϕ = 51°