Materials Science and Manufacturing Engineering Miscellaneous
- Two identical cylindrical jobs are turned using
(a) a round nosed tool of nose radius 2 mm and
(b) a sharp corner tool having principal cutting
edge angle = 45° and auxiliary cutting edge angle = 10°. If the operation is carried out of the feed of 0.08 mm/rev, the height at micro irregularities on the machined surfaces (in mm) in the two cases will be
-
View Hint View Answer Discuss in Forum
ASA Signature ψ = 45°, ψ =10°
Hmax = f² = 0.08² 8R 8 × 8
ORS signature = 0.0004 mmHmax = f = 0.08 tan ψ + cot ψ tan4548° + cot 10°
Hmax = 0.012 mmCorrect Option: D
ASA Signature ψ = 45°, ψ =10°
Hmax = f² = 0.08² 8R 8 × 8
ORS signature = 0.0004 mmHmax = f = 0.08 tan ψ + cot ψ tan4548° + cot 10°
Hmax = 0.012 mm
- Tool life equations for two tools under consideration are as follows
HSS: VT0.2 = 150
Carbide: VT0.3 = 250
Where V is the cutting speed in m/min and T is the tool life in mm. The breakeven cutting speed above which the carbide tool will be beneficial is
-
View Hint View Answer Discuss in Forum
HSS:
VT0.2 = 150
VT0.3 = 250
150 
1/0.2 = 250 1/0.3 V V
V = 54m/ minCorrect Option: A
HSS:
VT0.2 = 150
VT0.3 = 250150 1/0.2 = 250 1/0.3 V V
V = 54m/ min
- In resistance welding, heat is generated due to the resistance between
-
View Hint View Answer Discuss in Forum
Because of Air Resistance is very high.
Correct Option: B
Because of Air Resistance is very high.
- Spot welding of two 1 mm thick sheets of steel (density = 8000 kg/m³) is carried out successfully by passing a certain amount of current for 0.1 second through the electrodes. The resultant weld nugget formed is 5 mm in diameter and 1.5 mm thick. If the latent heat of fusion of steel is 1400 kJ/kg and the effective resistance in the welding operation is 200 μ-ohms, the current passing through the electrodes is approximately
-
View Hint View Answer Discuss in Forum
ρ = 8000 kJ/m³
t = 0.1 sec dn = 5 mm
hn = 1.5 mm
LH = 1400 kJ/kg R = 200 µΩ
I =?
mass = 8000 × 10-9 × (π/4) 5² × 1.5
= 2.356 × 10-4kg.
Heat = 1400 × 8000 × 2.356 × 10-4
= 329.8 Joules
Heat = I² × 200 × 10-6 × 0.1
I = 4060 Amp.Correct Option: C
ρ = 8000 kJ/m³
t = 0.1 sec dn = 5 mm
hn = 1.5 mm
LH = 1400 kJ/kg R = 200 µΩ
I =?
mass = 8000 × 10-9 × (π/4) 5² × 1.5
= 2.356 × 10-4kg.
Heat = 1400 × 8000 × 2.356 × 10-4
= 329.8 Joules
Heat = I² × 200 × 10-6 × 0.1
I = 4060 Amp.
- Filler material is.... A... in resistance welding and the heat generated in the process is directly proportional to.... B....
-
View Hint View Answer Discuss in Forum
Not used & square of the current H = I² Rt.
Correct Option: E
Not used & square of the current H = I² Rt.
