SME
Manufacturing Processes · Problem type
Resistance spot welding
Use this for “A spot weld runs 10 kA through 0.1 mΩ for 0.2 s; 30% of the heat goes into the nugget (Uₘ = 10 J/mm³). Find the heat and the nugget’s volume.”
Chart
H(10) = 2,000
This is an example. Type your own values to start a new problem.
H (J)
=
(
I (kA)
)
2
×
R (mΩ)
×
t (s)
Vₙ (mm³)
=
%
×
H (J)
Uₘ (J/mm³)
With I in kA and R in mΩ, I²R comes out in kilowatts: the formulas’ 1000 × turns the heat into joules.
hₙ
Nugget thickness
example · mm
dₙ
Nugget diameter
calculated · mm
Formulas
Vₙ
=
π
×
dₙ
2
×
hₙ
4
60
=
π
×
7.1365
2
×
1.5
4
H
=
1,000
×
I
2
×
R
×
t
2,000
=
1,000 ×
10
2
×
0.1
×
0.2
Vₙ
=
(
f
100
)
×
H
Uₘ
60
=
(
30
100
)
×
2,000
10
Assumptions
•
The resistance R stays the same through the whole pulse.
•
The rest of the heat goes into the sheets around the nugget and into the electrodes.
•
The nugget is a flat disc hₙ thick (about twice the sheet), so Vₙ = πdₙ²hₙ ÷ 4.
•
Uₘ comes from the metal’s melting point; for steel it is about 10 J/mm³.
Step-by-step
WE KNOW
I = 10 kA, R = 0.1 mΩ, t = 0.2 s, f = 30%, Uₘ = 10 J/mm³, hₙ = 1.5 mm
FIND
heat (H), nugget volume (Vₙ), nugget diameter (dₙ)
Step 1 · Find heat (H)
Use H = 1,000 × I² × R × t
The
power
I
2
R
for
the
time
t
;
kA²
×
mΩ
gives
kW,
so
×
1000
gives
joules.
H
=
1,000 ×
10
2
×
0.1
×
0.2
H
=
1,000
×
100
×
0.1
×
0.2
H
=
100,000
×
0.1
×
0.2
H
=
10,000
×
0.2
H
=
2,000
J
Step 2 · Find nugget volume (Vₙ)
Use Vₙ = (f ÷ 100) × H ÷ Uₘ
The
share
f
of
the
heat
melts
the
nugget;
each
mm³
of
it
takes
Uₘ.
Vₙ
=
(
30
100
)
×
2,000
10
Vₙ
=
0.3
×
2,000
10
Vₙ
=
600
10
Vₙ = 60 mm³
Step 3 · Find nugget diameter (dₙ)
Use Vₙ = π × dₙ² × hₙ ÷ 4
The
nugget
is
a
flat
disc
hₙ
thick:
its
volume
over
πhₙ
4
is
dₙ
2
,
then
the
root.
dₙ
=
√(4
×
Vₙ
π × hₙ
)
dₙ
=
√(4
×
60
π × 1.5
)
dₙ
=
√(
240
1.5π
)
dₙ
=
√
50.93
dₙ = 7.1365 mm
Check
Put the numbers back in. Both sides should match.
60
=
π
×
7.1365
2
×
1.5
4
✓
2,000
=
1,000 ×
10
2
×
0.1
×
0.2
✓
60
=
(
30
100
)
×
2,000
10
✓
Topic check
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