Materials Science and Manufacturing Engineering Miscellaneous
- The blank diameter used in thread rolling will be
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NA
Correct Option: D
NA
- The thickness of a sheet is reduced by rolling (without any change in width) using 600 mm diameter rolls. Neglect elastic deflection of the rolls and assume that the coefficient of friction at the roll-workpiece interface is 0.05. The sheet enters the rotating rolls unaided. If the initial sheet thickness is 2 mm, the minimum possible final thickness that can be produced by this process in a single pass is ____ mm (round off to two decimal places).
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(∆h)max = μ² . R
⇒ hi – hf = μ² . R
⇒ 2 – hf = (0.05)2 × 300 ⇒ hf = 2 – 0.75
⇒ hf = 1.25 mmCorrect Option: D
(∆h)max = μ² . R
⇒ hi – hf = μ² . R
⇒ 2 – hf = (0.05)2 × 300 ⇒ hf = 2 – 0.75
⇒ hf = 1.25 mm
- The cold forming process in which a hardened tool is pressed against a workpiece (when there is r elative mot ion bet ween t he tool and the workpiece) to produce a roughened surface with a regular pattern is
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Knurling : It is a process of impressing a diamond shaped or straight line pattern on the surface by a workpiece by using specially shaped hardened metal wheels to improve its appearance and to provide a better gripping surface.
Correct Option: B
Knurling : It is a process of impressing a diamond shaped or straight line pattern on the surface by a workpiece by using specially shaped hardened metal wheels to improve its appearance and to provide a better gripping surface.
- A strip of 120 mm width and 8 mm thickness is rolled between two 300 mm diameter rolls to get a strip of 120 mm width and 7.2 mm thickness. The speed of the strip at the exit is 30 m/min. There is no front or back tension. Assuming uniform roll pressure of 200 MPa in the roll bite and 100% mechanical efficiency, the minimum total power (in kW) required to drive the two rolls is _______.
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Velocity of Neutral plane
By applying continuity Equation
Hn = H2 + 0.25∆H
Hn = 7.2 + 0.25 × 0.8 = 7.4 mm
VnHn = V2H2
V2 → Exit velocity
H2 → Final Height
Vn = 29.19 m/minVn = πDN 60 N = 29.19 × 60 = 30.98 rpm π × 300
Total power = 2TωTotal power = 2 × 1439.9 × 2π × 30.98 = 9.34 kW 60 Correct Option: D
Velocity of Neutral plane
By applying continuity Equation
Hn = H2 + 0.25∆H
Hn = 7.2 + 0.25 × 0.8 = 7.4 mm
VnHn = V2H2
V2 → Exit velocity
H2 → Final Height
Vn = 29.19 m/minVn = πDN 60 N = 29.19 × 60 = 30.98 rpm π × 300
Total power = 2TωTotal power = 2 × 1439.9 × 2π × 30.98 = 9.34 kW 60
- In a rolling operation using rolls of diameter 500 mm if a 25 mm thick plate cannot be reduced to less than 20 mm in one pass, the coefficient of friction between the roll and the plate is _____.
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μ = √ (∆h)max R
⇒ (∆h)max = 25 – 20 = 5mm
R = 250 mm∴ μ = √ 5 = 0.1414° 250 Correct Option: C
μ = √ (∆h)max R
⇒ (∆h)max = 25 – 20 = 5mm
R = 250 mm∴ μ = √ 5 = 0.1414° 250