Materials Science and Manufacturing Engineering Miscellaneous
- The process of hot extrusion is used to produce
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Aluminium is ductile at higher temperature.
Correct Option: A
Aluminium is ductile at higher temperature.
- For wire drawing operation, the work material should essentially be
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For wire drawing, work material should be ductile in nature.
Correct Option: A
For wire drawing, work material should be ductile in nature.
- During open die forging process using two flat and parallel dies, a solid steel disc of initial radius (RIN) 200 mm and initial height (HIN) 50 mm attains a height (HFN) of 30 mm and radius of RFN,
Along the die-disc interfaces,
(i) The coefficient of friction (μ) is:
μ = 0.35 [1 + e–RIN/RFN]
(ii) In the region Rss ≤ r ≤ RFN sllding friction prevails and P = √3 K.e2μ(RIN–r)/HFN and ι = μp,
sllding friction prevails and
Where p and ι are the normal and the shear stresses respectively; K is the shear yield strength of steel and r is the radial distance of any point.
(i) In the region 0 ≤ r ≤ RIN, sticking condition prevails.
The value of Rss (in mm), where sticking condition changes to sliding friction is
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At RSS,
π d02 × h0 = πR2FN × h1 4
RFN
Shear stress sticking (k) = Shear stress in sliding (μPss)∴ RSS = RFN - h In 1 2μ √3μ = 258.2 - 30 In 1 2 × 0.51 √3 × 0.51
= 254.55 mmCorrect Option: B
At RSS,
π d02 × h0 = πR2FN × h1 4
RFN
Shear stress sticking (k) = Shear stress in sliding (μPss)∴ RSS = RFN - h In 1 2μ √3μ = 258.2 - 30 In 1 2 × 0.51 √3 × 0.51
= 254.55 mm
- Hot die steel, used for large solid dies in drop forging, should necessarily have
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High toughness for hammering action and low thermal conductivity for retaining higher temperature.
Correct Option: C
High toughness for hammering action and low thermal conductivity for retaining higher temperature.
- By application of tensile force, the cross sectional area of bar P is first reduced by 30% and then by an additional 20%. Another bar Q of the same material is reduced in cross sectional area by 50% in a single step by applying tensile force. After deformation, the true strain in bar P and bar Q will, respectively be
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True strain
ε = In Ai Af εT = In Ai 0.8 × 0.7Ai εT = In 1 0.56
εT = 0.58
True strain in single step of 50%εT = In Ai Af = In Ai 0.5Ai
εT = In(2) = 0.693Correct Option: B
True strain
ε = In Ai Af εT = In Ai 0.8 × 0.7Ai εT = In 1 0.56
εT = 0.58
True strain in single step of 50%εT = In Ai Af = In Ai 0.5Ai
εT = In(2) = 0.693