Production Engineering Test 6

Concept:

Rake angle is an important parameter for the surface finish of the product. Rake angle decides the cutting forces and power needed for cutting a soft and a hard material.

A positive rake angle is used for cutting soft and ductile material.

A negative rake angle is used for cutting a very hard or a brittle material.

A zero rake angle is used for machining brass material.

At low positive rake angle a higher cutting force is required but with the increase of positive rake angle the lesser amount of cutting forces and power is required which gives better surface finish. But the tool becomes thin and weak.

Concept:

The Taylor’s tool life equation for flank wear on the tool is

VTⁿ = C also V₁T₁ⁿ = V₂T₂n

where, V is cutting speed in m/min, T is tool life in min,

C = machining constant(depends on both tool and workpiece material) and n = tool life exponent (depends only on the tool material) 

Calculation:

Given:

n = 0.25, V₂ = V₁/2 

From, the Taylor’s tool life equation

VTⁿ = C 

\(\frac{{T_2 }}{{T_1 }} = (\frac{{V_1}}{{V_2}})^{\frac{{1}}{{n}}}\)

⇒ \(\frac{{T_2 }}{{T_1 }} = (2)^{\frac{{1}}{{0.25}}}\)= 2⁴ =16

⇒ T₂ = 16 T₁

Explanation:

Given:

t₁ = 0.25, V_c = 1 m/sec, α = 0, t₂ = 0.75, F_C = 900 N, F_T = 450 N

\(r = \frac{{{t_1}}}{{{t_2}}} = \frac{{0.25}}{{0.75}} = \frac{1}{3}\)

\(\phi = {\tan ^{ - 1}}\left( {\frac{{r\cos \alpha }}{{1 - \sin \alpha }}} \right) = 18.43^\circ \)

Now,

Power (P) = F_C × V_c

Power = 900 × 1 = 900 W

∴ Power = 0.9 kW

Now,

\(\tan \left( {\beta - \alpha } \right) = \frac{{{F_T}}}{{{F_C}}}\)

\(\beta = \alpha + {\tan ^{ - 1}}\left( {\frac{{{F_T}}}{{{F_C}}}} \right)\;\)

\(\beta = 0 + {\tan ^{ - 1}}\left( {\frac{{450}}{{900}}} \right)\)

∴ β = 26.56°

∴ Coefficient of friction = μ = tan β = 0.5

Explanation:

Given:

Tool change time (T_c) = 5 min, n = 0.4, c = 250, Tool-regrind time = 10 min,

Machining cost (C_m) = Rs. 0.5/min

Calculation:

Tooling cost (c_t) = 0.5 × 10 = Rs. 5

Since,

\({T_0} = \left( {{T_c} + \frac{{{C_t}}}{{{C_m}}}} \right)\left( {\frac{{1 - n}}{n}} \right)\)

\({T_0} = \left( {5 + \frac{5}{{0.5}}} \right)\left( {\frac{{1 - 0.4}}{{0.4}}} \right)\)

∴ T₀ = 22.5 min

Also,

V₀T₀ⁿ = C

V₀(22.5)^0.4 = 250

V₀ = 71.95 m/min

∴ V₀ = 1.199 m/sec

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

Explanation:

Given:

n = 0.5, C = 550

Calculation:

Let initial speed of tool = V₁

And, initial tool life = T₁

Now, cutting speed is reduced by 35%,

Thus, V₂ = (0.65) V₁

Taylor’s equation for tool wear,

VTⁿ = C

VT^0.5 = 550

V₁T₁^0.5 = 550     …1)

Also, V₂T₂^0.5 = 550

(0.65 V₁) T₂^0.5 = 550

V₁T₂^0.5 = 846.15     …2)

Dividing 2) by 1)

\(\frac{{{V_1}T_2^{0.5}}}{{{V_1}T_1^{0.5}}} = \frac{{846.15}}{{550}}\)

\(\left( {\frac{{{T_2}}}{{{T_1}}}} \right) = 2.366\)

Now,

\(\frac{{{T_2} - {T_1}}}{{{T_1}}} = 1.366\)

Percentage increase in tool life is

\(\left( {\frac{{{T_2} - {T_1}}}{{{T_1}}}} \right) \times 100 = 1.366 \times 100\)

∴ Percentage increase in tool life = 136.6%

Explanation:

Given:

L₂ = 84 mm, L₁ = 200 mm, V = 30 m/min, t₁ = 0.5 mm, α = 20°

Now,

Volume before machining = volume after machining

L₁ × b × t₁ × = L₂ × b × t₂

200 × 0.5 = 84 × t₂

\({t_2} = \frac{{200\; \times \;0.5}}{{84}}\)

∴ t₂ = 1.19

Now,

\(r = \frac{{{t_1}}}{{{t_2}}} = \frac{{0.5}}{{1.19}}\)

∴ r = 0.42

Now,

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

\(\phi = {\tan ^{ - 1}}\frac{{\left( {0.42\cos 20} \right)}}{{1 - 0.42\sin 20}}\)

∴ ϕ = 24.75°

Now,

Condition for minimum strain ⇒ 2ϕ - α = 90°

∴ Optimum shear angle

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

∴ ϕ = 55°

Concept:

Total life for maximum production rate is given by:-

\({T_{max}} = \left( {\frac{1}{n} - 1} \right){T_c}\)

Where n = Taylor’s tool exponent, T_c = tool change time

Calculation:

Given:

n = 0.2, T_c = 8

Now,

\({T_{max}} = \left( {\frac{1}{n} - 1} \right){T_c}\)

\({T_{max}} = \left( {\frac{1}{{0.2}} - 1} \right) \times 8.0\)

Tmax = (5 - 1) × 8

∴ T_max = 32 min

Now,

Apply Taylor’s tool life equation

\(VT_{max}^n = c\)

\(V{\left( {32} \right)^{0.2}} = 120\)

V × 2 = 120

∴ V = 60 m/min