Materials Science and Manufacturing Engineering Miscellaneous
- Assertion (A) : The ratio of uncut chip thickness to actual chip thickness is always less than one and is termed as cutting ratio in orthogonal cutting.
Reason (R) : The frictional force is very high due to the occurrence of sticking friction rather than sliding friction.
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Sliding friction is very higher
Correct Option: C
Sliding friction is very higher
- Match List-I (Cutting tool materials) with ListII (Manufacturing methods) and select the correct answer using the codes given below :
List-I List-II A. HSS 1. Casting B. Satellite 2. Forging C. Cemented carbide 3. Rolling D. UCON 4. Extrusion 5. Powder metallurgy
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HSS – Forging Stellite – Casting Cementite – Powder metallurgy UCON – Rolling
Correct Option: D
HSS – Forging Stellite – Casting Cementite – Powder metallurgy UCON – Rolling
- Match List-I with List-II and select the correct answer using the codes given below :
List-I List-II A. Plane approach angle 1.Tool face B. Rake angle 2.Tool flank C. Clearances angle 3.Tool face and flank D. Wedge angle 4.Cutting edge 5.Tool nose
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Correct Option: C
- Consider the following statements: During the third stage of tool wear, rapid deterioration of tool edge takes place because
1. Flank wear is only marginal
2. Flank wear is large
3. Temperature of the tool increases gradually
4. Temperature of the tool increases drastically
Which of the statements given above are correct?
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Correct Option: B
- In a machining operation chip thickness ratio is 0.3 and the back rake angle of the tool is 10°. What is the value of the shear strain?
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r = 0.3 = t1 , α = 10° t2 tanφ = rcosα 1 - rsinα tanφ = 0.3cos10° 1 - 0.3sin10°
φ = 17.3°
Shear strain = tan (φ – α) + cot φ
Shear strain = tan (17.3 – 10) + cot 17.3
Shear strain = 3.34Correct Option: D
r = 0.3 = t1 , α = 10° t2 tanφ = rcosα 1 - rsinα tanφ = 0.3cos10° 1 - 0.3sin10°
φ = 17.3°
Shear strain = tan (φ – α) + cot φ
Shear strain = tan (17.3 – 10) + cot 17.3
Shear strain = 3.34