Materials Science and Manufacturing Engineering Miscellaneous
- In arc welding of a butt joint, the welding speed is to be selected such that highest cooling rate is achieved. Melting efficiency and heat transfer efficiency are 0.5 and 0.7, respectively. The area of the weld cross section is 5 mm2 and the unit energy required to melt the metal is 10 J/mm3. If the welding power is 2 kW, the welding speed in mm/s is closest to
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Welding power applied = Heating power needed
⇒ P × ηm × ηH.T. = EU × (A) × f
⇒ 2 × 103 × 0.5 × 0.7 = 10 × 5 × f
⇒ f = 14 mm/sec.Correct Option: B
Welding power applied = Heating power needed
⇒ P × ηm × ηH.T. = EU × (A) × f
⇒ 2 × 103 × 0.5 × 0.7 = 10 × 5 × f
⇒ f = 14 mm/sec.
- A direct current welding machine with a linear power source characteristic provides open circuit voltage of 80 V and short circuit current of 800 A. During welding with the machine, the measured arc current is 500 A corresponding to an arc length of 5.0 mm and the measured arc current is 460 A corresponding to an arc length of 7.0 mm.
The linear voltage (E) - arc length (L) characteristic of the welding arc can be given as (where E is in Voit and L is in mm)
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The power source characteristic can be written analytically as
E = 80 − 80 I (a) 800
The arc characteristic is given as I = aL + b
where a and b are constant Given,
When I = 500A then L = 5.00 mm
∴ 500 = 5a + b ...(i)
when I = 460A
then L = 7.00 mm
∴ 460 = 7a + b ...(ii)
Solving Eqs. (i) and (ii, we get a = –20
and b = 600
Then arc characteristic equation I = –20L + 600
From equations (a) and (b)E = 80 − 80 (−20L + 600) 800
= 80 – 0.1(–20L + 600)
= 80 + 2L – 60
= 20 + 2LCorrect Option: A
The power source characteristic can be written analytically as
E = 80 − 80 I (a) 800
The arc characteristic is given as I = aL + b
where a and b are constant Given,
When I = 500A then L = 5.00 mm
∴ 500 = 5a + b ...(i)
when I = 460A
then L = 7.00 mm
∴ 460 = 7a + b ...(ii)
Solving Eqs. (i) and (ii, we get a = –20
and b = 600
Then arc characteristic equation I = –20L + 600
From equations (a) and (b)E = 80 − 80 (−20L + 600) 800
= 80 – 0.1(–20L + 600)
= 80 + 2L – 60
= 20 + 2L
- In an arc welding process, the voltage and current are 25 V and 300 A respectively. The arc heat transfer efficiency is 0.85 and welding speed is 8 mm/s. The net heat input (in J/mm) is
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V = 25 Volt, I = 300 A, η = 0.85
Weld speed = 8 mm/s.
Power Generated at arc = V.I
= 25 × 300 = 7500 J/S
Heat transferred = 0.85 × 7500
= 6375 J/S
So net Heat Input = 6375/8 = 797 J/mm.Correct Option: B
V = 25 Volt, I = 300 A, η = 0.85
Weld speed = 8 mm/s.
Power Generated at arc = V.I
= 25 × 300 = 7500 J/S
Heat transferred = 0.85 × 7500
= 6375 J/S
So net Heat Input = 6375/8 = 797 J/mm.
- Consider the following statements related piercing and blanking
1. Shear on the punch reduces the maximum cutting force
2. Shear increases the capacity of the press needed
3. Shear increases the life of the punch
4. Total energy needed to make the cut remains unaltered due to provision of shear.
Which of the above statements are correct?
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Over all Energy requirement Unaltered & Maximum Force decreased.
Correct Option: B
Over all Energy requirement Unaltered & Maximum Force decreased.
- A hydraulic press is used to produce circular blanks of 10 mm diameter from a sheet of 2 mm thickness. If the shear strength of the sheet material is 400 N/mm2, the force required for producing a circular blank is
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Punching force = π Dt τ
= π × 2 × 10 × 400 = 25.13 KNCorrect Option: B
Punching force = π Dt τ
= π × 2 × 10 × 400 = 25.13 KN
