Materials Science and Manufacturing Engineering Miscellaneous
- In resistance seam welding, the electrode is in the form of a
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circular disc
Correct Option: D
circular disc
Direction: Resistance spot welding of two steel sheets is carried out in lap joint configuration by using a welding current of 3 kA and a weld time of 0.2 s. A molten weld nugget of volume 20 mm3 is obtained. The effective contact resistance is 200 micro-ohms. The material properties of steel are given as (i) latent heat of melting 1400 kJ/kg, (ii) density 8000 kg/m3, (iii) melting temp 1520°C (iv) specific heat 0.5 kJ/kg°C. The ambient temp is 20°C.
- Heat (Joules) dissipated to the base metal will be (neglecting all other heat losses)
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Heat dissipated = 360 – 344 = 16 J
Correct Option: B
Heat dissipated = 360 – 344 = 16 J
- Heat (in Joules) used for production weld nugget will be (assuming 100% heat transfer efficiency)
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I = 3000 A
t = 0.2 sec
R = 200 μΩ
V = 200 mm3
LH = 1400 kJ/kg
ρ = 8000 kg/m3
C = 0.5 kJ/kg °C
Heat Generation = I2 RT
= 3000 × 200 ×10-6 × 0.2 = 360 Joules.
Heat utilised for production of weld Nugget
= m [LH + C∆T]
800 × 20 × 10-9[1400 + (1500 × 0.5)] × 1000
= 344 JoulesCorrect Option: C
I = 3000 A
t = 0.2 sec
R = 200 μΩ
V = 200 mm3
LH = 1400 kJ/kg
ρ = 8000 kg/m3
C = 0.5 kJ/kg °C
Heat Generation = I2 RT
= 3000 × 200 ×10-6 × 0.2 = 360 Joules.
Heat utilised for production of weld Nugget
= m [LH + C∆T]
800 × 20 × 10-9[1400 + (1500 × 0.5)] × 1000
= 344 Joules
- Aluminium strips of 2 mm thickness are joined together by resistance spot welding process by applying an electric current 6000 A and 0.15 second. The heat required for melting aluminium is 2.9 J/mm3. The diameter and thickness of the weld nugget are found to be 5 mm and 2.5 mm respectively. Assuming the electrical resistance to be 75 micro ohms, the percentage of total energy utilized in forming the weld nugget is
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hn = 2.5 mm, I = 6000 A
R = 75 µΩ, t = 0.15sec
dn = 5 mm Energy/Volume = 2.6 J/mm3
Energy generated = I2 R t
= 60002 × 75 × 10-6 × 0.15 = 405 J
Energy Required = 2.9 × volume of Nugget= 2.9 × π d2nhn 4 = 2.9 × π 522.5 = 142.35 joules 4 Efficiency = 142.35 × 100 ≃ 35% 405 Correct Option: B
hn = 2.5 mm, I = 6000 A
R = 75 µΩ, t = 0.15sec
dn = 5 mm Energy/Volume = 2.6 J/mm3
Energy generated = I2 R t
= 60002 × 75 × 10-6 × 0.15 = 405 J
Energy Required = 2.9 × volume of Nugget= 2.9 × π d2nhn 4 = 2.9 × π 522.5 = 142.35 joules 4 Efficiency = 142.35 × 100 ≃ 35% 405
- Consider the following statements: The magnitude of residual stresses in welding depends upon
1. Metal melted/deposited
2. Design of weldment
3. Support and clamping of components
4. Welding process used
Which of the statements given above are correct?
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Residual stress depends on welding parameters such as current, voltage, speed, current.
Correct Option: B
Residual stress depends on welding parameters such as current, voltage, speed, current.
