Materials Science and Manufacturing Engineering Miscellaneous Easy Questions and Answers | Page - 26

Materials Science and Manufacturing Engineering Miscellaneous

A shaft of length 90 mm has a tapered portion of length 55 mm. The diameter of the taper is 80 mm at one end and 65 mm at the other. If the taper is made by tail stock set over method, the taper angle and the set over respectively are

1. 15°32' and 12.16 mm
1. 15°32' and 15.66 mm
1. 11°22' and 10.26 mm
1. 10°32' and 14.46 mm

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Rate of taper,
T = 80 - 65 = 0.27
55

Set over = T × L = 0.27 × 90 = 12.15
2 2

Correct Option: A

Rate of taper,
T = 80 - 65 = 0.27
55

Set over = T × L = 0.27 × 90 = 12.15
2 2

In an orthogonal cutting process the tool used has rake angle of zero degree. The measured cutting force and thrust force are 500 N and 250 N, respectively. The coefficient of friction between the tool and the chip is

1. 0.7
1. 0.6
1. 0.5
1. 0.9

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Because the rake angle is zero,
F = Friction force = F_c = Cutting force = 500 N,
N = normal to friction force = F_t = thrust force = 250 N,
Coefficient of friction = F/N = 250/500 = 0.5

Correct Option: C

Because the rake angle is zero,
F = Friction force = F_c = Cutting force = 500 N,
N = normal to friction force = F_t = thrust force = 250 N,
Coefficient of friction = F/N = 250/500 = 0.5

The tool life equation for HSS tool is VT^0.14f^0.7d^0.4 = Constant. The tool life (T) of 30 min is obtained using the following cutting conditions: V= 45 m/min, f = 0.35 mm, d =2.0 mm. If speed (V), feed (f) and depth of cut (d) are increased individually by 25%, the tool life (in min) is

1. 0.15
1. 1.06
1. 22.50
1. 30.0

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VT^0.14 f ^0.7 d^0.4 = C
⇒ T₁ =30 min,
V₁ =45 m/min,
f₁ =0.35 min,
d₁ =2.0 mm,
⇒ C =V₁ (T₁) ^0.14 (f₁) ^0.7 (d₁) ^0.4
C =45(30)^0.14 (0.35)^0.7 (2)^0.4
C =45.8425
V₂ (T₂) ^0.14 (f ₂) ^0.7 (d₂) ^0.4 = 45.8425
(1.25 × 45)(T₂) ^0.14 (1.25 × 0.35)^0.7 (1.25 × 2)^0.4
= 45.8425
⇒ T₂ =1.06 min

Correct Option: B

VT^0.14 f ^0.7 d^0.4 = C
⇒ T₁ =30 min,
V₁ =45 m/min,
f₁ =0.35 min,
d₁ =2.0 mm,
⇒ C =V₁ (T₁) ^0.14 (f₁) ^0.7 (d₁) ^0.4
C =45(30)^0.14 (0.35)^0.7 (2)^0.4
C =45.8425
V₂ (T₂) ^0.14 (f ₂) ^0.7 (d₂) ^0.4 = 45.8425
(1.25 × 45)(T₂) ^0.14 (1.25 × 0.35)^0.7 (1.25 × 2)^0.4
= 45.8425
⇒ T₂ =1.06 min

The following data is applicable for a turning operation. The length of job is 900 mm, diameter of job is 200 mm, feed rate is 0.25 mm/rev and optimum cutting speed is 300 m/min. The machining time (in min) is_______.

1. 7.539
1. 9.539
1. 6.539
1. 8.539

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L = 900 mm
d = 200 mm
f = 0.25 mm/rev
v = 300 m/min
t =?
v = π DN m/min
N = v = 300 = 478 rpm
πD π × 0.2

t = L = 900 ≈ 7.539 min
f.N 0.25 × 478

Correct Option: A

L = 900 mm
d = 200 mm
f = 0.25 mm/rev
v = 300 m/min
t =?
v = π DN m/min
N = v = 300 = 478 rpm
πD π × 0.2

t = L = 900 ≈ 7.539 min
f.N 0.25 × 478

For a certain job, the cost of metal cutting is Rs. 180 C/V and the cost of tooling is Rs. 270 C/ (TV), where C is a constant, V is the cutting speed in m/min and T is the tool life in minutes. The Taylor's tool life equation is VT^0.25 = 150. The cutting speed (in m/min) for the minimum total cost is________.

1. 93.2
1. 93.9
1. 93.5
1. 93.3

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L_m = 18C
V

C_g = 270C
TV

VT^0.25 = 150
V_opt = V = C n L_m ⁿ
1 - n C_g

= 150 0.25 × 18 T ^0.25
1 - 0.25 270

= 150 1 × 18 × 150⁴ ^0.25
3 270 V⁴

= 150(57.914) × T₁ ^0.25
V⁴

V = 150 × 57.914 × 1
V

V² =150 × 57.914
V =93.2 m/min

Correct Option: A

L_m = 18C
V

C_g = 270C
TV

VT^0.25 = 150
V_opt = V = C n L_m ⁿ
1 - n C_g

= 150 0.25 × 18 T ^0.25
1 - 0.25 270

= 150 1 × 18 × 150⁴ ^0.25
3 270 V⁴

= 150(57.914) × T₁ ^0.25
V⁴

V = 150 × 57.914 × 1
V

V² =150 × 57.914
V = 93.2 m/min