8/13/2019 Group_H_10
1/8
April 12, 2013
To: Paul C. Lynch
From: Lab Group HSubject: Lab 10 (Spot Welding)
Dear Paul:
Resistance spot welding (RSW) is a process where two pieces of
metal are joined by theheat generated from resistance to an
electrical current. Work-pieces are held together throughpressure
applied by electrodes. The heat then melts the two pieces of metal
at the weld spot and a
weld-nuggetis formed. The nugget size typically shows the
strength of the weld. Spot welding
is most commonly known for its use in the automotive
manufacturing industry, welding sheetmetal together to form the
shell of a car. The basis for Lab #10 was to determine which
variables
in the spot welding process affect the strength of a spot
weld.
During this lab, thirty two different RSWs were used to weld
together 2 pieces of steel in
each repetition of the process. The variables that were tested
were air pressure (psi), dial current(%), and time (s). Every piece
was tested at air pressures of 50 and 60 psi, dial currents of 3,
4, 5,
and 6%, and times of 0.4, 0.6, 0.8, 1.0 seconds. Once the welds
were secure, the pieces were all
tested in a tensile machine for the shear forces at failure.
These forces were then recorded. Onceall the data was collected, a
linear regression was performed on the data to see which
variables
were statistically significant predictors of the weld strength.
From this lab,
In this lab, by increasing the dial current and the weld time
had an effect on both the size
of the weld nugget and the size of the heat affected zone. In
both cases, increasing the currentand weld time increased the size
of the nugget and heat affected zone. A weld that lasts longer
with more energy heats the metal more dramatically, causing
larger nuggets and affecting a
greater portion of the metal around the nugget.In most cases, as
the weld time increases, so does the shear force of the welded
metal. Additionally, a higher dial current also resulted in a
higher shear force. Thus, the
mechanical properties of the weld are highest at longer weld
times and larger currents (when the
nugget is large).From the regression analysis, not all variables
are significant to the shear force. Because
the p-value of the air pressure variable is 0.25 (greater than
alpha = 0.05), air pressure is not a
significant predictor of the shear force regression variable.
Likewise, because the p-values ofboth the dial current and weld
times are 0 (less than alpha = 0.05), these two variables are
significant predictors of the shear force regression
variable.
The R-square value gives an idea of how well the shear force is
predicted using thecalculated least-squares regression equation.
The closer this value is to 1.0, the better the shear
force is explained by the three welding variables. The
calculated R-square value of .753 (75.3%)
shows that shear force has a fairly linear relationship with the
air pressure, dial current, and weld
time as explained by the least-squares regression equation.Each
scatter plot (Figures 4-6) gives an idea of how well shear force
follows a linear
trend with each individual variable. Both the dial current and
the weld time show a positive
linear relationship with shear force. Thus, as each current or
weld time increases, so does shear
force value range. This is demonstrated by the positive slope of
the regression line of eachscatterplot. However, air pressure does
not indicate a linear relationship with shear force. As air
pressure increases from 50 to 60, there is not much of an
increase in shear force. This is
exemplified by the fairly flat regression line of the
scatterplot.
8/13/2019 Group_H_10
2/8
The regression equation can only be used to estimate the
strength of the weld for time
values slightly higher than those used in the lab. It is logical
that when the time of each weld
goes up by a significant amount, the shear force would start to
decrease because the weld qualitywould decreases. Too much energy
would be put into the metal, and the metal would be
potentially harmed because of it. For example, if we allowed the
weld to go on for 10 seconds
with air pressure of 50 and dial current of 5, we would get a
shear force value of:
Shear Force = - 2269 + 15.1(50) + 454(5) + 1347(10) = 14,226
This value is unlikely to be the true value of the shear force
because the metal would be degradedif the weld was allowed to
continue for so long.
At first glance of the data, it was noticeable that both dial
current and weld time had a
positive relationship with the shear force. It was also
predicted that the air pressure also had apositive relationship
with the shear force. While the data gathered from regression
analysis did
make the shear forces linear relationship with dial current and
weld time more concrete, it also
illustrated that there was no solid relationship between the
shear force and air pressure. Thus, thestatistical analysis was
useful for interpreting the data as it is concrete evidence of
the
relationships between the predictor and response variables.
Sincerely,
Lab Group H
Tim Kovacs #33, Ben Messer #40, Student #2, Student #3, Student
#4, Student #5 (Names innumerical order)
Sources
1) http://people.richland.edu/james/ictcm/2004/weight.html2)
http://www.pindling.org/Math/Statistics/Textbook/Chapter10_ANOVA/ANOVA.htm
8/13/2019 Group_H_10
6/8
Regression Analysis: Shear Force versus Air Pressure, Dial
Current, ...
The regression equation isShear Force = - 2269 + 15.1 Air
Pressure + 454 Dial Current + 1347 Weld Time
Predictor Coef SE Coef T P
Constant -2268.8 784.8 -2.89 0.007
Air Pressure 15.14 12.90 1.17 0.250Dial Current 454.49 57.70
7.88 0.000
Weld Time 1347.4 288.5 4.67 0.000
S = 364.938 R-Sq = 75.3% R-Sq(adj) = 72.6%
Analysis of Variance
Source DF SS MS F P
Regression 3 11350763 3783588 28.41 0.000
Residual Error 28 3729041 133180Total 31 15079804
Source DF Seq SS
Air Pressure 1 183467
Dial Current 1 8262356Weld Time 1 2904941
Unusual Observations
Air ShearObs Pressure Force Fit SE Fit Residual St Resid
13 50.0 1057.0 1754.3 152.7 -697.3 -2.10R
R denotes an observation with a large standardized residual.
