Realización de casos de uso del negocio

The electrical resistance could be used as a source of heat. It could be

(a) contact resistance of interfaces or (b) Resistance of molten flux and slag

Resistance of each hemispherical constriction R = ρ(r₂ – r₁)/S

where ρ = resistivity of material (r₂ – r₁) = length of current path

S = geometric mean area of the two hemispheres of radii r₁ and r₂ respectively.

= (2πr₂²)(2πr₁²) = 2π r₁r₂

Total constriction resistance R_c of n such spheres/unit area R_c = ¹

This approximation does not cause an error of more than 15%

Thus R_c = 0.85 ρ/nπr₁

Heat generation rate by this contact resistance with an applied voltage of V is Q = V²/R_C per unit area.

However after a very short time (≈ .001 sec) the contact resistance drops to 1

10th of its original value. Due to softening of material due to increase in temperature.

Example. In a resistance welding process applied voltage = 5 V

Bridges formed n = 25/cm²

Bridge radius r₁ = 0.1 mm. = 0.01 cm

r₂

r₁

Fig. 6.4

resistivity of material ρ = 2 × 10^–5 ohm-cm. Rate of heat generated/unit area

Q = V

Example 1. Two different pairs of sheets of the same material have to be spot welded. In one pair, there are 25 bridges/cm² and the average radius of each bridge is 0.1 mm. The other pair of sheets contains 50 bridges/cm² with the same average radius of each bridge. Determine the ratio of the voltages to be applied in these two cases to generate the same rate of heating/unit area.

The rate of heat generated by contact resist-ance with an applied voltage V is V

R_C

Case 1. Rate of heat generated/unit area

= V

Case 2. Rate of heat generated/unit area

= V For equal heat to be generated

V₁² 25 r V₂² r

Example 2. The voltage-arc length characteristic of a dc arc is given by : V = (20 + 4l) volts.

where l is the arc-length in mm. During a welding operation it is expected that the arc length will vary between 4 mm and 6 mm. It is desired that the welding current be limited to the range 450–550 A. Assuming a linear power source characteristic, determine the open circuit voltage and short circuit current of the power source.

ρ = resistivity of the material V = applied voltage

R_c = constriction resistance n = number of bridges/cm²

r = radius of bridge (average)

D.C. Arc voltage V = 20 + 41

Arc length varies between 4 mm and 6 mm

It is desired that welding current should be between 450 to 550 A (difference 100 A) Assume a linear power source characteristics

Find open circuit voltage and short circuit current voltage variation range : V = 20 + 4 × 4 = 36 V to

20 + 4 × 6 = 44 V

80 V

V

8 V 100 A

1000 A I

Fig. 6.5 current range (450 – 550) _~− 100 Amp.

Slope = 8

100 = 0.08 V = C – mI = C – 80

100 I 36 = C – 80

100 × 550

C = 80 Thus V = 80 – 0.08 I V = C –.08 I

V = 80 – 0.08 I When V = 0

I = 80 08

. = 1000 A Short circuit current = 1000 A Open circuit voltage = 80 V

UV W

Example 3. During an experimental investigation the arc-voltage has been found to be related with arc-length as V = (22 + 4l) volts. The power source characteristics is as follows

V volts and I₀ = 1000 Amp. What will be the values of welding currents for arc lengths of 3 mm and 5 mm with corresponding arc voltage of 30 volts and 40 volts.

Solution. Using the data given 30

The values of welding currents are 444.44 Amp and 400.61 Amp corresponding to arc-voltages of 30 and 40 volts respectively.

QUESTIONS

6.1 Briefly discuss how residual stresses and distortions occur in welded structures. How these stress could be minimised and eliminated?

6.2 By means of neat sketches discuss transverse shrinkage in V-butt welds. How can trans-verse shrinkage be calculated (estimated) in butt welds, fillet welds and T-welds.

6.3 How residual stresses occur in welds? Briefly explain stress-relieving treatment of welds.

+ ^0)26-4  %

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Some materials are easily weldable while certain others require special procedures to weld them. These materials are called difficult to weld materials. The welding of the following such materials will be discussed in this chapter.

1. Welding of cast irons

2. Welding of aluminium and its alloys 3. Welding of low carbon HY pipe steels 4. Welding of stainless steels

In addition to the above, the welding of dissimilar metals and the hardfacing and clad-ding will also be discussed.

7.1 WELDING OF CAST IRONS

7.1.1 Composition of Cast Irons

Element Gray C.I. Malleable C.I. Nodular C.I.

Carbon 2.5–3.8 2–3 3.2–4.2

Silicon 1.1–2.8 0.6–1.3 1.1–3.5

Manganese 0.4–1.0 0.2–0.6 0.3–0.8

Sulphur 0.1 0.1 0.02

Phosphorus 0.15 0.15 0.08

7.1.2 Oxy-Acetylene Welding of Gray and Nodular Cast Irons

• Grey cast iron contains much of carbon in flake form. This flake carbon distribution causes it to be brittle and, therefore, the standard set for its welding is not very high.

• Nodular Iron is cast with magnesium, nickel or rare earth addition, the graphite is in the form of spheroids with ferrite or pearlite matrix. This iron has ductility in as cast state upto 4% and on annealing-upto 15–25%. Its weldability is better than that of Grey cast iron as S and P are at low level. Thus the risk of hot tearing in weld metal is reduced. Welding steps are given below.

Welding of Materials

• A 60 – 90 Vee grove is prepared.

• When repairing a crack a hole should be drilled at each end of the crack to arrest it.

• The job before welding is preheated to 300–650 C in a furnace then covered with asbestos cloth, exposing only the cavity to be welded.

• If furnace is not available the casting is covered with asbestos cloth and locally heated by gas flame. Thick sections should be preheated in a furnace.

• Filler material should have the same composition as the base metal with minimum S and P. Special rods containing Ti and high Si content are also sometimes used.

• Welding rods are square or round cast bars.

• Fluxes for grey iron filler rods are composed of borates, soda ash, and small amounts of ammonium sulphate, iron oxide, etc.

• Torch tip is one size larger than that required for steel of the same thickness.

• Adjust the torch to a neutral flame.

• Move the flame along the groove untill the entire joint is preheated to dull red.

• Concentrate the flame at the bottom of the vee with tip of inner cone about 3.0 to 6.0 mm from the metal surface. As the bottom fuses thoroughly move the flame from side to side to let the liquid metal run down to the pool and rotate the torch to mix the molten metal from side walls to mix with the metal in the pool.

• If metal gets too fluid and runs down raise the flame.

• After the weld pool is formed, heat the filler rod end by outer envelop of the flame, dip the rod into the flux.

• Introduce the Flux coated (dipped) filler rod into the molten pool and apply flame to the tip of the filler rod and the welding is carried out.

• As the weld completes, cover it with asbestos and allow it to cool slowly.

• Post welding stress relieving be carried out for complex shapes. For this purpose keep casting in a furnace at 650°C for one hour per 25 mm thickness and cooled to 260°C or below at a rate not faster than 28°C per hour.