Materials Science and Manufacturing Engineering Miscellaneous
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
-
View Hint View Answer Discuss in Forum
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
-
View Hint View Answer Discuss in Forum
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
- A batch of 10 cutting tools could produce 500 components while working at 50 rpm with a tool feed of 0.25 mm/rev and depth of cut of 1 mm. A similar batch of 10 tools of the same specification could produce 122 components while working at 80 rpm with a feed of 0.25 mm/rev and 1 mm depth of cut. How many components can be produced with one cutting tool at 60 rpm?
-
View Hint View Answer Discuss in Forum
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
Now, v1T1n = v2T2n
⇒ 12.5(500 k)n = 20 × (22 k)n⇒ 
500k 
n = 20 122k 12.5 ⇒ 500 n = 1.6 122 ⇒ n ln 500 = ln 122
⇒ n = 0.3332
Hence number of components produced by one cutting tool at 60 rpm is:⇒ 12.5 × 500k 0.3332 = 60 × 0.25 × (nk)0.332 10 ⇒ 12.5 = nk 0.332 60 × 0.25 50k ⇒ n = 0.5786 50
∴ n = 0.5786 × 50 = 28.92 ≈ 29.Correct Option: A
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
Now, v1T1n = v2T2n
⇒ 12.5(500 k)n = 20 × (22 k)n⇒ 500k n = 20 122k 12.5 ⇒ 500 n = 1.6 122 ⇒ n ln 500 = ln 122
⇒ n = 0.3332
Hence number of components produced by one cutting tool at 60 rpm is:⇒ 12.5 × 500k 0.3332 = 60 × 0.25 × (nk)0.332 10 ⇒ 12.5 = nk 0.332 60 × 0.25 50k ⇒ n = 0.5786 50
∴ n = 0.5786 × 50 = 28.92 ≈ 29.
- Tool life testing on a lathe under dry cutting conditions gauge n and C of Taylor tool life equation as 0.12 and 130 m/min. respectively. When a coolant was used, C increased by 10%. The increased tool life with the use of coolant at a cutting speed of 90 m/min is
-
View Hint View Answer Discuss in Forum
n = 0.12, c = 130 m /min
C1 = 130 × 1.1 = 143 m/min
V1 = 90 m /min
V1T1n = C1
90(T1)n = 143T1 = 143 1/0.12 90
T1 = 47.4minCorrect Option: A
n = 0.12, c = 130 m /min
C1 = 130 × 1.1 = 143 m/min
V1 = 90 m /min
V1T1n = C1
90(T1)n = 143T1 = 143 1/0.12 90
T1 = 47.4min
- For turning NiCr alloy steel at cutting speeds of 64 m/min and 100 m/min, the respective tool lives are 15 min-and 12 min. The tool life for a cutting speed of 144 m/min is
-
View Hint View Answer Discuss in Forum
NA
Correct Option: C
NA
