Materials Science and Manufacturing Engineering Miscellaneous
- In a machining operation, doubling the cutting speed reduces the tool life to 1/8th of the original value. The exponent n in Taylor's tool life equation VTn = C. is
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Taylor’s tool life equation,
VTn = C ...(i)
Where, V = cutting speed and T = tool life
When cutting speed is doubled and tool life reduces to 1/8, then2V 1 T n = C ...(ii) 8
Dividing equation (i) by equation (ii), we getVTn = 1 2V T n 8 or VTn = 2V T n = 2Vn2−3n 8
or 1 = 2⋅2−3n
or 1 = 2−3n + 1
or 2−3n + 1 = 20
or –3n + 1 = 0
or 3n = 1or n = 1 . 3 Correct Option: C
Taylor’s tool life equation,
VTn = C ...(i)
Where, V = cutting speed and T = tool life
When cutting speed is doubled and tool life reduces to 1/8, then2V 1 T n = C ...(ii) 8
Dividing equation (i) by equation (ii), we getVTn = 1 2V T n 8 or VTn = 2V T n = 2Vn2−3n 8
or 1 = 2⋅2−3n
or 1 = 2−3n + 1
or 2−3n + 1 = 20
or –3n + 1 = 0
or 3n = 1or n = 1 . 3
- In an orthogonal cutting test on mild Steel, the following data were obtained
Cutting speed : 40 m/min
Depth of cut : 0.3 mm
Tool rake angle : +5º
Chip thickness : 1.5 mm
Cutting force : 900 N
Thrust force : 450 N
Using Merchants analysis, the friction angle during the machining will be
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μ = F = FCsinα + FTcosα N FCcosα − FTsinα = 900sin5º + 450cos5º = 526.72 = 0.614 900cos5º −450sin5º 857.35
∴ λ = tan−1 = 31.5º
Alternative method
From Merchants' Circle
Given: FC = 900 N
FT = 450 N
Nowtan(λ − α) = FT = 450 FC 900 ∴ (λ − α) = tan−1 1 2
[where, α = Rake angle]
Friction angle, λ = 26.5 + 5 = 31.50Correct Option: B
μ = F = FCsinα + FTcosα N FCcosα − FTsinα = 900sin5º + 450cos5º = 526.72 = 0.614 900cos5º −450sin5º 857.35
∴ λ = tan−1 = 31.5º
Alternative method
From Merchants' Circle
Given: FC = 900 N
FT = 450 N
Nowtan(λ − α) = FT = 450 FC 900 ∴ (λ − α) = tan−1 1 2
[where, α = Rake angle]
Friction angle, λ = 26.5 + 5 = 31.50
- Through holes of 10 mm diameter are to be drilled in a steel plate of 20 mm thickness. Drill spindle speed is 300 rpm, feed 0.2 mm/rev and drill point angle is 120°. Assuming drill over travel of 2 mm, the time for producing a hole will be
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Given: Diameter of hole,
d =10 mm
Thickness of steel plate,
t = 20 mmα = 120 = 60º 2
Break through distance,A = d = 10 = 2.8867 mm 2tanα 2tan60º
Total length of tool travel,
l = t + A + 2
= 20 + 2.8867 + 2 = 24.88
Time for drilling= l = 24.88 = 0.4147 min fN 0.2 × 300
= 24.88 sec ≈ 25 sec.Correct Option: B
Given: Diameter of hole,
d =10 mm
Thickness of steel plate,
t = 20 mmα = 120 = 60º 2
Break through distance,A = d = 10 = 2.8867 mm 2tanα 2tan60º
Total length of tool travel,
l = t + A + 2
= 20 + 2.8867 + 2 = 24.88
Time for drilling= l = 24.88 = 0.4147 min fN 0.2 × 300
= 24.88 sec ≈ 25 sec.
Direction: A cylinder is turned on a lathe with orthogonal machining principle. Spindle rotates at 200 rpm. The axial feed rate is 0.25 mm per revolution. Depth of cut is 0.4 mm. The rake angle is 10°. In the analysis it is found that the shear angle is 27.75°.
- In the above problem, the coefficient of friction at the chip tool Interface obtained using Earnest and Merchant theory is
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From Earnest and Merchants' theory
2φ + λ – α = 90
where μ = co-efficient of friction = tan λ
∴ λ = 90 + 10 – 2 × 27.75
∴ = 44.5º
μ= tan (44.5)= 0.98Correct Option: D
From Earnest and Merchants' theory
2φ + λ – α = 90
where μ = co-efficient of friction = tan λ
∴ λ = 90 + 10 – 2 × 27.75
∴ = 44.5º
μ= tan (44.5)= 0.98
- The thickness of the produced chip is
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Given, rake angle, α = 10º
shear angle, φ = 27.75
depth of cut, t1 = 0.4 mm
Let t2 = thickness of chip producedNow, t1 = sinφ t2 cos(φ − α) ⇒ t2 = t1cos(φ − α) sinφ = 0.4cos(27.75 − 10) sin(27.75) = 0.4 cos 17.75 = 0.818 mm sin 27.75 Correct Option: A
Given, rake angle, α = 10º
shear angle, φ = 27.75
depth of cut, t1 = 0.4 mm
Let t2 = thickness of chip producedNow, t1 = sinφ t2 cos(φ − α) ⇒ t2 = t1cos(φ − α) sinφ = 0.4cos(27.75 − 10) sin(27.75) = 0.4 cos 17.75 = 0.818 mm sin 27.75