HOW TO CALCULATE THE CONSUMPTION OF FILLER METAL
The following tables will help you estimate electrode quantity and cost for a variety of joints. The bases for the tabulations are explained below. Should you encounter a variation in conditions or joint preparation that is not shown in the tables, substitute appropriate figures in the W =D formula and calculate it. Electrode requirements have been calculated as follows: 1-L
W =D
1-L W=Weight of electrodes required
D=Weight of steel deposited
L=Total electrode losses
To arrive at the weight of steel deposited, it is necessary to calculate first the volume of deposited metal (area of the groove multiplied by the length). Then this volumetric value is converted to weight by the factor 0.283 pounds per cubic inch for steel. Where weld reinforcement is involved, it is added to the requirements for net, unreinforced welds. These figures are based on the efficiency of the process used. On square and "V" groove joints, the figures are based on stick electrode efficiency. For tubular wire results, divide steel deposit weight by .80 and for solid wire divide by .90.
S
E
E
F
I
G
U
R
E
1
Size of fillet
L
(in inches)
Steel deposited
per linear foot
of weld (lbs.)
Pounds of electrodes required
per linear foot of weld* (approx.)Stick*TubularSolid
*Includes scrap end and spatter loss
S
E
E
F
I
G
U
R
E
2Joint dimensions (in inches)Steel deposited per linear foot of weldPounds of electrodes required per linear foot of weld* (approx.)
MTL. THICK TBEAD WIDTH BROOT OPEN GWithout rein- forcement(lbs.)With rein- forcement(lbs.)Without reinforcementWith reinforcement**
3/16
1/4
5/163/8
7/16
1/2
------ 0.020
0.027
0.039
0.033
0.0500.088
0.109
0.129
0.143
0.153
0.170------ 0.04
0.05
0.07
0.06
0.090.16
0.20
0.23
0.26
0.27
0.301/8
3/16
1/41/4
3/8
7/16
0.020
0.040
0.053
0.0800.119
0.132
0.199
0.218
0.261
0.288------ 0.03
0.04
0.07
0.10
0.140.21
0.24
0.36
0.39
0.47
0.53
S
E
E
F
I
G
U
R
E
3
Joint Dimensions (in inches)
Steel deposited per linear foot of weldPounds of electrodes required per linear foot of weld* (approx.)
MTL. THICK
T
BREAD WIDTH
B
ROOT OPEN
G
Without rein- forcement(lbs.)
With rein- forcements**(lbs.)Without rein- forcementWith rein- forcement**1/4
5/16
3/8
1/2
5/8
3/4
10.207
0.311
0.414
0.558
0.702
0.847
1.1381/16
3/32
1/8
1/8
1/8
1/8
1/80.085
0.173
0.282
0.489
0.753
1.088
1.9300.143
0.258
0.394
0.641
0.942
1.320
2.240
*Includes scrap end and spatter loss. **R=Height of reinforcement.
Figure 1 Figure 2
Use Hastaloy X
Heat expands, cold contracts... When you weld, the 2 metals will expand toward each other. once your done, they will try to pull apart. not enough filler rod, not clamping, rapid cooling, undercut, wrong filler rod, are just a few causes of cracks
SFA stands for Steel framing alliance Cheers T. Mathew mathew@harimachines.in
D*d/162 = 25*25/162 = 3.85kg/m
yes, with a cast iron rod and an oxy-fuel set up or with a nickel rod and an arc machine
YES
2550 degree Fahrenheit
Use Hastaloy X
The flux is there to remove any oxidation from the surfaces of the metals to be welded.
0.125 rod = 0.000390625 mile
one of the method is manually dipping the filler rod in a canister of loose flux as the weld proceeds
It can be either including the wire in MIG (GMAW). In stick (SMAW) the electrode. Sometimes the coating adds small amounts of filler.
Formula of hexagonal ms rod
for steel structure building can i use: -size 2.5mm of 7016 welding rod as a filler for welding IPE240 to plates -size 2.5mm of 7016 welding rod as 2nd pass
calculate amp usage
Heat expands, cold contracts... When you weld, the 2 metals will expand toward each other. once your done, they will try to pull apart. not enough filler rod, not clamping, rapid cooling, undercut, wrong filler rod, are just a few causes of cracks
Cast iron welding rod is designed for the SMAW process which is built by flux and a filler metal that suits the base metal's chemical & mechanical properties.