GATE Past Year Questions: Cutting Tool Geometry & Tool Life | Manufacturing Engineering - Mechanical Engineering PDF Download

Number of Double stroke = 30
Quick Return Ratio = 0.6 = M
Length of Strike = 250 mm = L
Average cutting velocity = L.N (1 + 19)
= 30 × 250 (1 + 0.6) = 12 m /min

ASA Signature
ψ = 45°, ψ =10°

 ORS signature = 0.0004 mm

Hmax = 0.012 mm

α = 15°
ϕ = 45°
V = 35 m/min

V_c = 25.58 m/min

D = 25 mm,
h = 100 mm
VT^0.25= 55
V = 22 m/min, f = 0.046 mm/ rev

 

D = 31.8mm, t = 100 mm, V = 20m /min
f = 0.3 mm/rev,
Approach length = 4 mm
Over Travel = 9 mm

R_A = 0.6 mm. R_B = 0.33 mm
F_A = 0.2 mm/rev F_B = 0.1 mm/rev

Tool B have higher surface finish them A.

Topt = 36 min

l = 2000 mm
w = 300 mm
f = 0.3 mm/ stroke
No. of stroke = 2 × 10
= 20 stroke / min     {double stroke}
2000 „„„
Total Machining time= 2000/20 =100 min

n = 0.25 V₁ = 20m /min, T₁ = 10 min
V₂ = 40m/min , T₂ = ?

T₂ = 0.625 min

Table feed or Feed rate

L = t + Ap₁
Ap₁ = 0.5 D (holes diameter)
= 10 mm
t = 40 mm

N = 500 rpm; f = 0.2 mm/rev
T_c = 0.585 min or 35.1 seconds

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

Taylor's tool life equation is V^Tn = Constant

Given conditions for both tools:
Tool A: const ant , k₁ = 90 ;
exponential constant, n₁ = 0.45
Tool B: Constant ,  k₂ = 60;
exponential constant, x₂ = 0.3

At point of intersection
v₁ = v₂, and T₁ = T₂ = T

∴  90 = v₁ (14.92)^0.45
⇒ v₁= 26.7 m /min
Above v₁ = 26.7 m/min, tool A will have a higher tool life them too B.

By increasing the cutting speed. Heat dissipation is increased hence there is lower temperature & lower friction coefficient.

Side rake angle is equal to orthogonal rake angle when principle cutting edge angle become 90° and corresponding approach angle SCEA = 0°

Here, ϕ = shear angle = 25º
λ = Friction angle,
α = rake angle = 0º
From Merchant''s theory, 2ϕ + λ - α = 90º
∴ λ = 90º - 50º = 40º

F_C = Cutting force;
V = Cutting velocity
Specific machining energy = 2.0 J/mm³
∴ FC × V =2 × MRR
⇒ F_C = 2 × .2 × 10^–3 × 2 × 10³ = 800 N

Length travelled in forwarded stroke = 640 mm Number of strokes = 100
Time for cutting = 8 min
Return time = 4 min
Total time = 12 min.

Taylor’s tool life equation,
VTⁿ = C ...(i)
Where, V = cutting speed and T = tool life
When cutting speed is doubled and tool life

Given: Diameter of hole,
d =10 mm
Thickness of steel plate, t = 20 mm

Velocity of first cutting tool,
v₁ = 50 × 0.25 = 12.5 mm/min
Velocity of second cutting tool,
v₂ = 0.25 × 80= 20 mm/min
Since tool life = number of component produce × tool constant
∴ Tool life for first tool = T₁ = 500 × k
Took life for second tool = T₂₁ = 122 × k

⇒ n = 0.3332 Hence number of components produced by one cutting tool at 60 rpm is:

n = 0.12, c = 130 m /min
C₁ = 130 × 1.1 = 143 m/min
V₁ = 90 m /min
V₁ T₁ⁿ = C1
90 (T₁)ⁿ = 143

T₁ = 47.4 min

V₁ = 63m/min T₁ = 60 hours = 600 min
C = 257.35

T₂ = 25.6 min

As rake angle a increases, shear plane angle ϕ decreases. So less force is required for cutting.