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
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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
- Orthogonal turning of a mild steel tube with a tool of rake angle 10° carried out at a feed of 0.14 mm/rev. If the thickness of the chip produced is 0.28 mm, the values of shear angle and shear strain will be respectively
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r = 0.14 = 0.5 0.28 tanΦ = r cos α 1 - rsin α
∴ Φ = 28.3345º or 28º 20'Correct Option: A
r = 0.14 = 0.5 0.28 tanΦ = r cos α 1 - rsin α
∴ Φ = 28.3345º or 28º 20'
- In a machining operation, if the generatrix and directrix both are straight lines, the surface obtained is
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The surface obtained is plane.
Correct Option: C
The surface obtained is plane.
- A single point cutting tool with 0° rake angle is used in an orthogonal machining process. At a cutting speed of 180 m/min, the thrust force is 490 N. If the coefficient of friction between the tool and the chip is 0.7, then the power consumption (in kW) for the machining operation is_____.
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α = 0
so
F = FT
N = FcF = FT N Fc 490 = 0.7 Fc
Fc = 700 NPower = Fc × V = 700 × 180 = 2.1kw 60 Correct Option: A
α = 0
so
F = FT
N = FcF = FT N Fc 490 = 0.7 Fc
Fc = 700 NPower = Fc × V = 700 × 180 = 2.1kw 60
- An orthogonal turning operation is carried out under the following conditions; rake angle = 5°, spindle rotational speed = 400 rpm; axial feed = 0.4 m/min and radial depth of cut = 5 mm. The chip thickness tc, is found to be 3 mm. The shear angle (in degree) in this turning process is______.
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µ= 5° N = 400 rpm f = 0.4m/min
f = 0.4 × 1000 = 1 mm/res = t1 400
t2 = 3 mmtanΦ = r cos ∝ 1 - tsin ∝ r = t1 = 1 t2 3
tanΦ = 0.342
Φ = 18.88Correct Option: C
µ= 5° N = 400 rpm f = 0.4m/min
f = 0.4 × 1000 = 1 mm/res = t1 400
t2 = 3 mmtanΦ = r cos ∝ 1 - tsin ∝ r = t1 = 1 t2 3
tanΦ = 0.342
Φ = 18.88