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Welding Problems

This document discusses two questions related to welding processes. The first question calculates the power densities within different diameter circles during a welding operation. The second question calculates the portion of heat generated during a resistance spot welding operation that was used to form the weld versus what was dissipated into the surrounding metal. The document also provides information on the functions of pressure in resistance welding and some general advantages and drawbacks of resistance welding.

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aksgupta24
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226 views2 pages

Welding Problems

This document discusses two questions related to welding processes. The first question calculates the power densities within different diameter circles during a welding operation. The second question calculates the portion of heat generated during a resistance spot welding operation that was used to form the weld versus what was dissipated into the surrounding metal. The document also provides information on the functions of pressure in resistance welding and some general advantages and drawbacks of resistance welding.

Uploaded by

aksgupta24
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Problems on welding

Q.1 A heat source is capable of transferring 3000 W to the surface of a metal


part. The heat impinges the surface in a circular area, with intensities
varying inside the circle. The distribution is as follows: 70% of the power id
transferred within a circle of diameter = 5 mm, and 90% is transferred
within a concentric circle of diameter = 12 mm. What are the power densities
in (a) the 5 mm diameter inner circle and (b) the 12 mm diameter ring that
lies around the inner circle?
Solution: (a) The inner circle has an area A = =

(5) 2

4
The power inside this area P = 0.70 x 3000 = 2100 W

= 19.63 mm2

Thus the power density , PD = 2100/19.63 = 107 W/mm


(b) The area of the ring outside the inner circle is A =

The power in this region P = 0.9

(12 2 5 2 )
4

= 93.4 mm2

3000 2100 = 810 W

The power density is therefore , PD = 810/93.4 = 8.7 W/mm2


Observation: The power density seems high enough for melting in the inner
circle, but probably not sufficient in the ring that lies outside this inner circle.
Q.2 A resistance spot-welding operation is performed on two pieces of 1.5
mm thick sheet steel using 12,000 amps for a 0.20 second duration. The
electrodes are 6 mm in diameter at the contacting surfaces. Resistance is
assumed to be 0.0001 ohms, and the resulting weld nugget is 6 mm in
diameter and 2.5 mm thick. The unit melting energy for the metal Um = 12.0
J/mm3. What portion of the heat generated was used to form the weld. And
what portion was dissipated into the surrounding metal?
Solution: The heat generated in the operation is given by eq. below
H = (12,000)2 (0.0001) (0.2) = 2880 J

The volume of the weld nugget (assumed disc-shaped) is V=


((6) 2
2.5 x
= 70.7 mm3
4
The heat required to melt this volume of metal is Hm = 70.7 (12.0) = 848 J
The remaining heat, 2880 848 = 2032J (70.6% of the total), is absorbed onto
the surrounding metal.
Success in resistance welding depends on pressure as well as heat. The principal
functions of pressure in RW are to (1) force contact between the electrodes and
the work parts and between the two work surfaces prior to applying current, and
(2) press the faying surfaces together to accomplish coalescence when the proper
welding temperature has been reached.
There are some general advantages of resistance welding; (1) no filler metal is
required, (2) high production rates are possible, (2) high production rates are
possible,(3) it lends itself to mechanization and automation,(4) operator skill level
is lower than that required for arc welding, and (5) good repeatability and
reliability. Drawbacks are that initial equipment cost is high-usually much higher
than most AW operations, and the types of joints that can be welded are limited to
lap joints for most RW processes.

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