Analytical and Mathematical Analysis

The amount of heat input to the weld at its rate determines the geometry of the weld bead deposited and the width of the heat affected zone. It also affects the microstructure of the weld and heat affected zone, which in tern affects the mechanical properties of the joints obtained. In the following paragraphs we shall be discussing the factors like the determination of heat input to the weld, maximum heat input rate, in fusion welding of plates and resistance weld­ing of thin sheets.

The discussion will also include the heat flow in welding peak temperatures reached adjascent to the weld and in the HAZ, estimation of the width of HAZ and the effect of pre-heat of this width. Determination of cooling rates has also been included in the discussion as it affects the weld microstructure and consequently the mechanical properties of the welds.

The following sections provide practical working equations for consumable electrode welding applications and other weld processes. The following important quantities can be estimated using the heat flow equations :

1. Peak temperatures

2. Width of HAZ

3. Cooling rates

4. Solidification rates.

Before going into the details of the above equations, let us first concentrate on the heat input to the weld.

6.1. HEAT INPUT TO THE WELD

The heat input, Q in watts, in the case of arc welding is given by,

Q = VI J/S...(6.1)

For the melting of the weld at the joint, the exact amount of heat that enters the joint can be calculated (for an electrode moving at a speed of Sw mm/s) using the following relation.

Q

H = J/mm...(6.2)

Sw

But the actual heat utilized by the joint depends upon how effectively this heat is trans­ferred from electrode tip to the joint. Hence heat transfer efficiency factor f enters the calcu­

lations of net heat available at the joint.

fV

Hnet = 1 J/mm...(6.3)

Sw

All of this net heat is not used for melting since part of it is conducted away to the base plate. The heat actually used for melting Hm can be obtained by another efficiency factor f2

f f2 VI

Hm = ^ ...(6.4)

Sw

Heat required to melt the joint

where f2 =

Net heat suplied.

Ex. 1. Calculate the melting efficiency in the case of arc welding of steel with a current of 200 A at 20 V. The travel speed is 5 mm/s, and the cross-sectional area of the joint is 20 mm2. Heat required to melt steel may be taken as 10 J/mm3 and heat transfer efficiency is 0.85. Volume of base metal melted = 20 x 5 = 100 mm3/s Heat required for melting = 100 x 10 = 1000

1000 1000 f2 = f1 VI = 0.85 x 20 x 200 = °'2941 = 29-41%

Оставить комментарий