ED'S
SIMPLIFICATION OF
MIG & FLUX CORED WELD COSTS.
Followed
by Metric
Weld Conversion Data.
What's the actual weld cost per part?
Global MIG & Flux Cored
WELD COSTS:
You may not want to walk into a welding shop, examine that very
common 6 mm MIG or flux cored fillet weld and ask the shop manager or supervisor,
what does one meter of that fillet weld cost?
Its
difficult in the global weld industry to find weld personnel that have full control
or completely understand their weld costs.
The majority of manufacturing
mangers or supervisors will know the cost of the weld wire or gas,
few
will be aware of their MIG and Flux Cored weld deposition potentials or the hourly
deposition rates being achieved by the welders or robots they walk by each day.
|
2008: WHEN YOU HAVE A GLOBAL
EMPATHY TOWARDS WELD COST
CONTROLS, YOU KNOW IT'S TIME FOR CHANGE:
In
the world where managers and supervisors believe that welders have to “play
around” with their MIG and flux cored weld controls, it’s not hard to
understand why many managers fail to understand the fundamental weld process correlation
between the welds, the part thickness, the weld size, the weld transfer mode,
the wire feed settings, the weld deposition rates and the weld costs. Ed's low
cost, process control resources simplify
this.
Tremendous
weld cost reduction potential is always
found in ship yards, oil platforms
and auto / truck plants:  
YOU
WOULD THINK THAT IN THE INDUSTRIES THAT DO THE MOST WELDS THERE WOULD BE GREAT
FOCUS ON WELD COST CONTROLS.
[]
WELD COST FACT: NO ONE USES MORE FLUX
CORED WIRE THAN A SHIP YARD.
[]
WELD COST FACT: NO ONE HAS GREATER WELD
REPAIR COSTS THAN A SHIP YARD.
 2007
WELD REALITY: WITH
WELD REWORK COST PER-SHIP FREQUENTLY MEASURED IN THE MILLIONS OF DOLLARS, THE
MAJORITY OF SHIP YARDS HAVE YET TO ESTABLISH
BEST MIG / FLUX CORED WELD PRACTICES AND PROVIDE THEIR EMPLOYEES WITH MIG AND
FLUX CORED WELD PROCESS CONTROL TRAINING:
IT'S
EASY TO GENERATE MULTI-MILLION
DOLLAR COST SAVINGS FOR A SHIP YARD: During
the first four months of 2007, Ed  and
Tom O'Malley presented one of Ed's unique, manual, flux cored process control
training programs to Aker Kvaerner. Aker is an international ship builder located
at the the Philadelphia Naval Ship Yard.
The
300 plus welders in the yard used E71T-1 (1.2 mm)
flux cored wires to weld all position, vee groove, 12 to 25mm, steel joints with
ceramic backing.
Like
any ship yard, the Aker weld focus was on welder "skills".
To work at the yard, all the welders had to pass the all position, flux cored
weld qualification tests and weld in accordance with the ABS and the pre-qualified
weld procedures. However the welder qualification test had little in common with
the weld process control requirements and application variables typically found
in any ship yard. A written process control weld test did reveal that all those
who passed the weld qualification test, lacked flux cored weld process / consumable
expertise and lacked the awareness of the unique requirements to attain consistent
optimum weld quality for vee groove, ceramic backed welds. This is common in all
ship yards where typically you will find weld rework costs per-ship measured between
one and ten million dollars per-ship.
 The
Aker ship yard contracted with Ed to reduce the weld rework costs. All the welders
and supervisors in the yard participated in Ed's unique Flux Cored Weld Process
Control Training Programs.
The
training program focussed on flux cored Weld Process Controls,consumable knowledge
and optimum weld process techniques for all position, vee groove welds with ceramic
backing.
In a time of MIG and flux cored welder shortage, when many companies
find it difficult to interrupt their daily productivity, management take note.
Ed's unique weld process training program required only eight
hours, four hours classroom and four hours hands on. In a few weeks the
training for the 300 plus welders was complete and 
the
ship yard QA department personnel started to analyze the results.
Three
months after Ed's flux cored weld process control the training, the ship yard
NDT results indicated a 50% reduction in the required weld rework per-ship. The
ship yard management reported that the reduced weld rework, labor and NDT costs,
would resulted in a cost savings of approx. 4 million dollars per-ship.
THE
COST BENEFITS FROM HIGH IMPACT, ONE DAY TRAINING:
Examine
the real world ship yard weld cost reduction and benefits for Ed's unique process
control training program. The training program required 300 x 8 man/hr. = 2400
man hours at an approx. $30/hr, base labor cost for the ship yard. $72,000. For
the training. To this add the actual training costs of approx. $100,000 = for
a total training costs for the 300 welders of $172,000.
Savings, four million dollars
per-ship and training costs $172,000.
An
unreported fact from this yard was the changes that Ed also established in the
development of new weld procedures. The increased flux cored wire feed rates,
(weld deposition rates) in the new procedures increased the daily weld productivity
per-man in a range from 20 to 40%.
 Some
of you may wonder what's the difference
between this type of weld training program and the MIG and flux cored weld training
you could expect in any North American, Korean, Chinese, Japanese, European ship
yard or any manufacturing facility?
For
decades conventional training in ship yards or  manufacturing
plants has focussed on the welders skills, especially stick welding skills which
has nothing in common with the requirements for MIG or flux cored. It's not unusual
for weld personnel to have weeks of hands on training at the ship yards and then
find that when it comes to MIG and flux cored welds, the welders will;
[a]
Play with the MIG weld equipment controls and rarely dial in optimum settings
for the hot pass, fill pass and cap passes.
[b] Utilize the weld controls
in a very limited manner. One setting for all welds is a common global practice.
[c] Not utilize optimum weld techniques for the MIG and flux cored process.
Typically stick weld techniques were
common with the MIG and flux cored process.
[d] Lack awareness of the weld
deposition rate potential for a specific weld. This of course limits the daily
weld productivity potential the welders could achieve.
The
process training I provide
which is available in CD format, as with
all my training programs, enables each individual to achieve flux
 cored
weld process optimization for the consumables and for the all position, vee groove,
ceramic backed, or open root applications. For the welder who took the eight 8
hour process control training, the program enabled that individual the ability
to instantly set optimum parameters for the consumable, and weld joint variables,which
provided instant dramatic improvements in their weld ability.
As
you can see, on the left picture we have a weld made by an individual with poor
skills, poor techniques and poor settings. These two vertical up, 15 mm vee groove,
weld samples using E71T-1 flux cored wires and straight CO2, were made by the
same welder on the day of training. On the left before and on the right after
the 8 hours of process control training.
Even
when the welder had good skills, process control training will increase the welders
weld quality and productivity capability. What was also important, each welder
became aware of the unique flux cored weld parameters and technique requirements
to address the variable root gaps over the ceramic. Ceramic is rarely used outside
ship yards and welding oversized root gaps across a none conductive ceramic requires
unique process and technique requirements. The weld training results were dramatic,
with
all weld personnel attaining an instant
reduction in lack of weld fusion, slag entrapment and porosity defects.
Training.
2400 hours at $172.000 versus 12000 hrs and $500,000 costs  The
improvements by all the welders was immediately noticed by the ship yard QA management
who daily measured the dramatic improvements evident with NDT and radiographs
and the allocated man hours required for weld rework .
By
the way, few global ship yards or manufacturing facilities examine the cost effectiveness
of the training programs they develop. Instead of an eight hour training program,
many yards would not think twice about providing a forty hour welder training
program. For 300 welders in a North American facility, that 40 hours training
with labor and associated training costs would be approx. $500,000.
Plus the facility will have lost 12000 production hours.
Many
thanks to the Aker Kvaerner
management for recognizing the Best Practices / Process
Controls training necessary for their organization, and special thanks to Tom
O'Malley the owner of Excell. Tom's company is the prime weld products supplier
to the Philadelphia Naval Ship yard. Tom provided the facilities and equipment
for the training. Tom also assisted Ed with the program in both the classroom
and hands on training. Tom is one of those rare owners of a weld supply company
that actually spends many hours per-week evaluating weld processes equipment and
consumables.
"THERE
IS ONE WAY TO QUICKLY ATTAIN BEST WELD PRACTICES
AND WELD PROCESS CONTROL
EXPERTISE".
Ed
took > 2000 hours to develop this program. The flux cored weld process control
program is now available in CD Power
Point format for approx. $300
 ROBOTS
AND WELD COSTS ....When a manufacturing company
invests in costly robots and fixtures, the robot purchase is in most instances
intended to reduce the costs of manual weld production.
The
sad reality with many companies, is their robots rarely provide their real weld
their production potential and too many robot welds
require extensive manual weld repairs.
The
prime factors that control the robot weld production efficiency potential are;
[1]
optimizing the robot wire feed rates utilized,
[2] controlling the part
fit,
[3] controlling the weld size and length,
[4] optimizing
the robot movement / motion times,
[5] maintaining the robot arc on times,
[6]
eliminating the causes of the robot down time,
[7] eliminating the robot
weld rework.
The
primary influence on robot weld efficiency is the required robot “arc on
time” a time regulated by the wire feed rate attained. The wire feed rate
controls the robot weld speed.
It's unfortunate that some Japanese robots
do not provide the robot wire feed rates. These robots will have the robot programmer
select "weld amps". I have been in numerous Panasonic robot weld cells
and the companies do not have a clue as to the weld wire feed rates produced.
So much for manufacturing and Japanese focus on weld productivity.
To control MIG weld costs, someone in the plant has to understand
the relationship between the desired welds, the weld wire size, the wire feed
rates and weld deposition attained. This process subject is part of the fundamental
weld process expertise that should be common knowledge in all plants that MIG
and flux cored weld.
The
following book and CD training resources developed by Ed, provide the manual and
robot process control data that all engineers, managers, supervisors and technicians
need to control weld costs: ASK
A WELD SHOP MANAGER OR SUPERVISOR THE COST OF A MIG WELD AND THEY WILL TELL YOU
THEY DON'T KNOW, BUT THEY WILL TELL YOU THE COST OF THE WELD WIRE OR GAS.
Weld
process control knowledge is the key component to implementing effective BEST
WELD PRACTICES AND WELD PROCESS CONTROLS. This sounds logical, yet few managers
look for individuals that have this process expertise.
To effectively manage a weld shop and maximize the daily robot weld quality
and productivity, management, supervisors and engineers must provide their employees
with a weld process control training program that provides;
[a]
WELD PRODUCTIVITY POTENTIAL: The
weld process fundamentals that focus on the application's weld deposition rate
potential for the part thickness, weld size, wire size and weld transfer mode
utilized.
[b] WELD QUALITY POTENTIAL: The
weld process fundamentals required to
attain "consistent" optimum weld quality.
[c] CONTROLLING WELD EQUIPMENT / CONSUMABLES
COSTS: The
weld process knowledge necessary to implement effective, uniform, robot / manual
weld best practices.
[d]
WELD PROCESS CONTROLS:
The weld process knowledge necessary for weld personnel to stop playing with weld
controls. A knowledge that enables weld personnel to optimize weld parameters
to meet the day to day variables common to all weld shops.
| WHAT
IS THE PURPOSE OF A WELD MANAGER OR SUPERVISOR?
 This
week in the global weld industry, its unlikely you will find one manager in a
hundred who takes the time to have a discussion with their employees on the subject
of controlling the weld department costs.
The weld reality is most weld
supervisors are more interested in ensuring the weld personnel are not hiding
in the wash room, than in how to minimize their weld rework or maximize the daily
weld deposition rates.
While weld cost focus is too often on weld consumable
costs, the MIG / flux cored wire and gas typically account for only 12 to 16%
of the total cost of a carbon steel weld.
|
HOW
CAN ANY WELD MANGER, ENGINEER, SUPERVISOR OR TECHNICIAN,
WORK IN AN ENVIRONMENT
IN WHICH SO CALLED SKILLED WELD
PERSONNEL "PLAY AROUND" WITH THE
WELD CONTROLS?
He is a weld supervisor, we have to
cut
out his "play around organ"
The
first step in controlling MIG weld costs
is
understand the "wire feed control".
 USING
ED'S UNIQUE, WELD CLOCK METHOD, WELD PROCESS CONTROL
IS SIMPLE. WITH A TRADITIONAL "NONE DIGITAL" WIRE FEED CONTROL SIMPLY
DIVIDE THE CONTROL INTO TEN CLOCK SETTINGS BETWEEN 7 AND 5 O'CLOCK.
EACH WIRE FEED TURN DELIVERS APPROXIMATELY 70 in./min PER WIRE FEED TURN.
IN EUROPE I USE TWO METERS PER TURN
WITH AN 0.035 (1mm) STEEL OR STAINLESS
WIRE, EACH WIRE FEED TURN WILL DELIVER 1.1 POUNDS PER TURN: SET THE WIRE FEED
CONTROL AT THE 3 O'CLOCK POSITION, EIGTH TURN AND THE WIRE FEEDER DELIVERS 8 TO
9 LB/HR...
 With
the 0.045 (1.2 mm) wire, each wire feed turn is approx. 2 lb/hr. Using a standard,
”none digital” wire feeder, to set a low spray setting, set
the wire feed at the 12 o'clock position. 12 o'clock is the fifth turn. 5 x 70
= 350 inch/min.This wire feed position with the 0.045 (1.2 mm) wire would deliver
approximately 10 lbs/hr.
The robot set at 350 inch/min has an arc on time of 30 minutes
per/hr, so the robot uses 5 lbs/ of wire per hour, or 40 lb per-shift. Lets say
this application could be welded with another turn 12 lb/hr, the robot weld time
would be reduced by 20%. Are you getting how easy the weld clock method is?
Check
out Ed's unique, simple weld clock process
control method, it's in his CD training resources and process control books. Ed
spent 30 years on simplifying the MIG/MAG flux cored processes and you know how
important the KIS principle is to a weld shop. This method is easy to teach, because
it’s easy to remember.
ROBOTS
AND MANUAL WELD QUALITY:
The
number one controlling factor of setting optimum quality MIG welds for any weld
weld transfer mode is to understand where you set the MIG wire feed setting.
 ROBOTS
AND AND WELD COSTS:
The
number one method of controlling MIG and flux cored weld costs is understanding
the relationship between wire feed setting and the weld deposition rate attained.
The majority of welds produced daily in the industrial world are based
on three simple fillet weld sizes. 3/16 - 1/4 - 5/16 (4 - 6 - 8 mm). Is it therefore
reasonable to expect that all the weld personnel at your company should be aware
of the optimum wire feed settings and weld deposition rate potential for these
weld sizes?
HOW
CAN COSTS BE IN CONTROL
WHEN FUNDAMENTALS ARE MISSING?
THE
MOST COMMON MIG WELD IN THE SHOP IS A 1/4 (6 mm) FILLET. The
welds are made with 0.045 (1.2 mm) wire. On the conventional wire feeders, the
manual welders have a scratch mark set at the one o'clock position. In an other
area of the shop the the digital feeders are set at 420 inch/min, (10.5 m/min).
 Over in
the robot cell the weld data is also set at 420 inch /min.
You ask three
of the most experienced weld personnel in your organization, what weld deposition
rate per/hr is achieved with the wire feed set at 420 inch/min and would we be
better of with an 0.052 (1.4 mm) wire set at 350 inch/min? Listen to the confusion
in their replies.
Now
ask yourself, when are you going to get serious about being in the welding business?
When are you going to get a grip of your weld costs? When are you going to arrange
weld process control training for your
employees?
MY
PROCESS CONTROL BOOKS SIMPLIFY THE TASK OF ASSOCIATING WIRE
FEED SETTINGS WITH WELD WIRES / WELD SIZES AND WELD DEPOSITION RATES. USE THESE
BOOKS, OR MY ROBOT WELD PROCESS CONTROL TRAINING PROGRAMS TO TAKE YOUR
ORGANIZATION UP A NOTCH. THE PROCESS TRAINING WILL ENSURE ALL YOUR WELD PERSONNEL
ARE AWARE OF THE MANUAL OR ROBOT WELD DEPOSITION RATES THEY SHOULD STRIVE FOR
WHEN WELDING YOUR PARTS.
Weld
Co$ts per-part:
WELD DEPOSITION RATES:
In the following example MIG welding 24 parts an hour,
the parts are made out of 4 - 6 mm carbon steel. We learn from the clock method
that the average weld deposition rate attained by the welder to spray transfer
the 5 mm fillet weldAs, using an 0.045 (1.2mm) wire, is 9
lb/hr (4.5kg/hr). We also lean that deposytion drives costs that
why more focus is required on this important subject
 THE
MIG WIRE COSTS: If the manual
MIG welders arc on time per-hour is 20 minutes, the welder deposits 3 lb/hr. The
carbon steel, MIG wire cost IS $1 per/lb. The hourly MIG wire cost was
$3
an hour.
 THE
MIG WELD GAS COSTS: The
argon CO2 cylinder mix was $40.00 per-cylinder. A typical
full size gas cylinder will deliver on average approximately 300 cuft. The cylinder
gas cost is
13 cents per-cubic foot.
The MIG gas flow rate per/hour
is 30 cuft/hr. The welder average an arc on time of 20 minutes per hour results
in a gas use of 10 cuft/hr.
10 cuft x 13 cents = a MIG gas cost of $1.30
an hour.
THE
MIG WELD LABOR COSTS: The average, 2007
hourly "welder's wage" in the US is a sad $13 an hour, with benefits
it’s approx. $20 an hour.
Note; some companies when trying to evaluate welding costs, like to add the total
white collar overhead including the kitchen sink and the coffee machines on the
backs of the blue collar labor cost. In weld cost calculations this overhead is
an unnecessary distraction to the real world weld cost formula.
With an overhead of $20 / hr, plus $3
/ hr for wire and $1.30 hr for weld gas, the total hourly cost per welder is $24
hour. Producing welds at 3 lb/hr results in a weld cost per pound of weld
metal deposited at $8 per/lb. Producing 24
parts an hour, the weld cost per-part is
a $1.
Weld
Reality.
If you cannot work out this stuff in your head you are not in control of
your weld costs. An engineer, manager or supervisor who had utilized my process
control CDs or books would be aware that
for the 5 mm fillet weld, the welders could increase their wire feed control to
readily attain a weld deposition rate of 12 lb/hr
(5.4 kg/hr).
The 25% an hour increase in weld deposition allowed
25% more parts per-hour. Productivity increases from 24 to 30 parts. With the
12 lb/hr, the 20-minute arc results in a weld deposition rate of 4
lb/hr (1.8 kg/hr). The overhead costs are increased by a dollar for the
extra one pound of wire utilized. Labor, weld wire and gas therefore cost $25
Divide the $25 by 30 parts and you have approximately $0.83 per - part for a saving
of 0.17 cents (17 %) per-part, all from a process
awareness and simple single turn of the wire feed control.
|
PROCESS KNOWLEGE: A
simple turn of the MIG wire feed control, a change in wire diameter or from short
circuit to globular, globular to spray or pulsed to spray and most weld shops,
can typically reduce their weld costs by 15 to 50%.
|
WELD
CO$T FOCUS. A focus by management and
supervision on process capability, wire feed rates and weld deposition rate potential
typically will ensure dramatic weld cost reduction, however this is not likely
to happen with"PLAY AROUND" with the MIG/MAG control employees.
To make weld process changes, requires weld personnel have process confidence.
This is one good reason to think a little less about spending thousands on over
priced pulsed MIG welding equipment and a little more about investing a few hundred
dollars for providing Ed's MIG process-training
CDs.
Quoting
on this manual MIG weld trailer job was simple.
 Five
hundred trailers were required by the Smith Alloy Corporation. Each trailer had
a total of forty feet (12 m) of 1/4 (6mm) fillet weld. The MIG weld wire used
was an E70S-3. The wire size 0.045 (1.2 mm). The cylinder weld gas, argon - 10%
CO2.
The company that got the contract to build the trailers had five
MIG welders who each work an 8-hour day. The welder labor overhead was $25 an
hour. How long will it take to complete the job and what will be the welding cost?
Using my robot and manual MIG weld process control
training resources this is how you would
take 8 simple steps
to approach this task.
[1] How
much MIG weld wire is required? The 1/4
(6mm) fillet requires 0.11 lbs of weld per foot of weld. That's 0.11 x 40 feet
= 4.4 lbs of filler metal required per trailer x trailers = 2.200 pounds of MIG
wire for all the trailers.
[2]
How much will the weld wire cost?
The 0.045 filler metal cost $0.90 cents / lb x 4.4 lbs/part = $3.96 weld wire
costs per trailer, or the wire cost for 500 trailers is $1980.00.
[3]
How many man-hours are required?
Using my simple weld clock parameter method, you would be aware of where the welders
will have to set the 0.045 wire feed to weld the 6 mm fillet and that the weld
deposition rate would be approx. 12 lb/hr. However the manual arc on time per/hr
for the average welders is only 20 minutes. The welders deposited approximately
4 pounds per/hr. The 500 trailers will use 2200 pounds, divide by the 4 lb/hr,
the welding job will require
550 man hours.
[4]
The weld gas costs, and how many cylinders required?
A cylinder of argon CO2 costs $40 cyl. The cyl contains
330 cuft ($0.12 per cu/ft). The gas flow is 30 cuft/hr but the welders arc on
time is 20 min, so the gas used is only 10 cuft/hr x 0.12 cents = $1.20 / hr for
the gas. The job requires 550 hrs x $1.20 = $660 for the gas, (550 hrs x 10 cuft
= 5500 cuft divide by a cyl 330 cuft = 17-18 cylinders required for the project.
[5]
The total weld consumables cost per trailer: The cost of consumables for the job
is weld wire $1980, + gas $660, = $2640 divide by 500 trailers or consumable costs
$5.28 per trailer.
[6]
The labor cost per / trailer. The cost of labor is 550 man hours required x $25/hr
= $13.750 for the 500 trailers or labor cost per trailer $27.50.
[7]
Total weld cost per trailer is $5.28 for consumables + $27.50 for labor
=
$33 per trailer.
[8]
The time required to weld all the parts. The time required to do the job with
5 welders (40 hour / day). The job requires 550 man/hours divide by 40 requires
14 days.
Ed
has a point. The bottom line, this weld cost approach is simple. Last time I saw
Miller, or was
it the Licoln or Hobart rep? I asked for a manual weld cost analysis, these guys
used a computer and delivered pages of confusing weld cost data. Now
I have finally got down to figuring out the real costs of the manual MIG welds,
I will see if we can do anything about weld cost reduction by using robots.
For those individuals making "weld cost decisions" consider
Ed's process control CD training programs
for either MIG and flux cored
For
those interested in how the manual trailer weld costs are
reduced using a
robot for welding 500 trailers, scroll down.
Quoting
this trailer job with a Robot is made simple.
We will now quote the trailer welds
using a robot. Each of the
500 trailers has forty feet (12 m) of 1/4 (6mm) fillet weld. We will weld with
an 0.045 (1.2 mm) MIG wire, and an argon - 10% CO2 cylinder mix.
We
will use one robot with two operators. One robot operator works days, the other
is on the afternoon shift. The weld labor overhead is $25/hr. How long will it
take to complete the job with robot? What's the weld cost per-part? How will the
cost compare to the above manual welding operation?
USING
THE EIGHT STEP METHOD:
[1]
How much weld wire required? A 1/4
6mm fillet requires 0.11 lbs per foot of weld x 40 feet = 4.4 lbs of filler metal
required per part x 500 = 2.200 lbs of MIG wire required.
[2]
How much will the weld wire cost? The 0.045
filler metal cost $0.90 cents/lb x 4.4 lbs/part = $3.96 per part, or the wire
cost for 500 parts is $1980.00.
[3]
How many robot / man hours required?
Using my clock parameter method you would know the robot
is set to deposit the 0.045 wire feed at a weld deposition rate of 12 lb/hr. However
in contrast to the manual welders, the robots weld faster and the arc on time
per/hr is increased to 40 minutes per-hour. With the increased arc on time the
robot deposits 8 pounds per/hr for two shifts. The two shifts (16 hrs a day x
8 lbs / hr = 128 lbs per day). The 500 parts require 2200 pounds of weld metal,
divide by the 8 lb/hr deposited the job will require 275
man hours. (Its actually less
when you look into the weld speed benefits, I have got to save something for my
process control weld books, right).
For
the 500 welded parts, the robot with the increased arc on time will require 275
man hours. The manual welding job as discussed required 550
man hours for the same application . |
[4]
The robot gas costs and how many cylinders required?
A cylinder of argon - CO2 costs $40 cyl. The cylinder contains
330 cuft ($0.12 per cu/foot). The gas flow is 30 cuft/hr, however the robot's
arc on time is 40 min, so the gas used is 20 cuft/hr x 0.12 cents = $2.40/ hr
for the gas. The job requires 275 hrs x $2.40 = $660 for the gas. As the robot
weld speeds will be approx 20% quicker with the robot you can expect a 20% reduction
in the weld gas cost. Gas cost
$528
[5]
The total consumable cost: The cost of consumables for the job, MIG wire
$1980, + gas $528, = $2508 divide by 500 trailers or $5
per part.
[6]
The labor cost per trailer. The
cost of labor is 275 robot / man
x
$25/hr = $6.875.00
or labor per trailer is $13.75.
[7]
Total weld cost per trailer.
$13.75 labor = $5 consumables = $18.75
[8]
The Costs.The
robot costs for welding the trailers was $6.875.00.
The manual labor costs for the same parts was $13.750.00.
|
The
total manual weld cost per trailer was $32.78.
Using a robot reduced
the weld cost to $18.75 per trailer.
|
:
HOW
DO I KNOW HOW MUCH WELDING
WIRE TO ORDER FOR A WELD JOB?
First determine the average fillet or groove weld
size. Use the following guidelines, and when you come to the final MIG weld wire
amount add 10% as a cushion factor. If you are going to use flux cored wires add
20% to the MIG wire recommendations. For the metric users, 1 lb / ft of weld to
1 kg/m, x 1.488.
The
following is the approximate amount of MIG filler metal required per foot of weld.
All the weld cost data you need for any MIG or FCAW application can be found in
my "Management Engineers Guide To MIG"
book.
CARBON
STEEL or STAINLESS WELD WIRE
REQUIRED PER FOOT OF WELD:
A
1/8 ( 3 mm fillet requires 0.03 lbs of weld wire
per foot of weld. A
1/4 (6 mm fillet requires 0.11 lbs of weld wire per foot of weld. A
3/8 (9 mm fillet requires 0.29 lbs of weld wire per foot of weld. A
1/2 (13mm fillet requires 0.42 lbs of weld wire per foot of weld. A
3/4 (19mm fillet requires 1.09 lbs of weld wire per foot of weld. A
1" (25mm fillet requires 1.80 lbs of weld wire per foot of weld.
A
3/8 (9 mm butt 60 degree Single vee 1/8 root 0.55
lb/per foot A
1/2 (13mm butt 60 degree Single vee 1/8 root 0.85 lb/per foot A
1/2 (13mm butt 60 degree Single vee 3/8 root 1.4 lb/per foot A
3/4 (19mm butt 60 degree Single vee 1/8 root 1.64 lb/per foot A
3/4 (19mm butt 60 degree Single vee 3/8 root 2.3 lb/per foot A
1" (25mm butt 60 degree Single vee 1/8 root 2.67 lb/per foot A
1" (25mm butt 60 degree Single vee 1/2 root 4 lb/per foot A
2" (50mm butt 60 degree Single vee 1/8 root 9.6 lb/per foot A
3" (75mm butt 60 degree Single vee 1/8 root 20 lb/per foot A
4" (100mm butt 60 degree Single vee 1/8 root 36 lb/per foot If
45 degree groove is used reduce wire requirements approx 22%
ALUMINUM
HOW MANY POUNDS OF WIRE REQUIRED? Aluminum
1/8 3.2 mm fillet = 0.092 lb/ft 0.03 kg/m Aluminum
3/16 4.8 mm fillet = 0.026 lb/ft 0.04 kg/m Aluminum
1/4 6.4 mm fillet = 0.05 lb/ft 0.07 kg/m Aluminum
3/8 9.5 mm fillet = 0.06 lb/ft 0.09 kg/m
Aluminum
butt weld 13 mm plate 60 degree single V = 0.3 lb/ft
0.43kg/m Aluminum
butt weld 18 mm plate 60 degree single V = 0.4 lb/ft 0.66kg/m Aluminum
butt weld 25 mm plate 60 degree single V = 0.82 LB/FT 1.2kg/m Aluminum
butt weld 38 mm plate 60 degree single V 1.7 lb/ft 2.6 kg/m Aluminum
butt weld 50 mm plate 60 degre single V 2.8 lb/ft 4.2 kg/m
ALUMINUM "ipm" FEED RATE TO WELD
DEPOSITION RATE:
030
0.8 mmwire, ipm x 0.004 = lb/hr 035
0.9 mmwire, ipm x 0.0056 - l/hr 046
1.2mm wire, ipm x 0.0099 = lb/hr 052
1.4mm wire, ipm x 0.012 = lb/hr 062
1.6mm wire, ipm x 0.017 = lb/hr 093
2.4mm wire, ipm x 0.0415 = lb/hr.
ALUMINUM "m/min"FEED RATE TO WELD
DEPOSITION RATE:
0.8mm
wire, m/min x 0.07 = kg/hr 1
mm wire, m/min x 0.09 = kg/hr 1.2
mm wire, m/min x 0.16 = kg/hr 1.4
mm wire, m/min x 0.23 = kg/hr 1.6
mm wire, m/min x 0.306 = kg/hr 2.4
mm wire, m/min x 0.74 = kg/hr.
This
structural steel company wants to MIG weld 5 bridge structures. Each structure
has approximately 5000 feet of 1/4 fillet welds. The welders will use 0.045 MIG
wire. The manager wants a quick quote on the following.
[a]
How much weld wire required?
[b]
How much weld gas required?
[c]
how many man hours to weld?
The above table tells you that a 1/4 6 mm fillet requires 0.11 lb/foot of
weld. 5000 feet x 0.11 requires 550 pounds of weld metal. Add 15% for cushion
factor.
Use my clock method to determine that the spray setting will deposit for this
application 12 lb/hr. Use the average arc on time of 20 minutes per/hr. The manual
MIG welder therefore deposits on average 4 lb/hr.
To deposit the 4 lb/hr
for the 550 lbs/ of weld will require 138 hrs of labor. You now know on average
a manual MIG weld uses 10 cuft.hr of gas x 138 hrs requires 1380 cuft add 15%
as a cushion 1587 cuft. Each gas cylinder requires approx 300 cuft of gas = 6
cylinders per unit.
WELD
COST REDUCTION WITH CLADDING ON
WATER WALL BOILER APPLICATIONS.
 
Ed's
cost contribution to the power and
waste management industry
2007:
Welding Services (WSI) is the largest power industry contractor in North America.
WSI is primarily involved in repairs and refurbishment in the power, waste energy
and nuclear industries. In terms of water wall cladding, WSI has clad approx.
80% of the North American boilers. Each year WSI uses approx one million pounds
of Inconel 625-622 and 300 series stainless MIG wires for cladding water wall
tubes.
While WSI has produced some of the most innovative, automatic MIG
cladding equipment available in North America, WSI did not have a resident MIG
process control expert who had the expertise necessary to make radical improvements
to it's traditional water wall clad MIG welds. Ed was contracted for this work
by the WSI engineering manager. In less than 6 months, Ed not only dramatically
improved the water wall overlay weld quality but also reduced
the amount of overlay typically required by > 28%.
As many in
the power industry are aware, with any cladding application "less weld and
less heat provides the best results. Apart from the high cost of >$20 lb for
the Inconel weld wires, the boilers operate more efficiently when the clad surface
is thinner and the clad weld pass thickness is uniform and with out defects.
With
water wall clad applications, the minimum, single pass
weld clad chemistry required is 20% Chrome. To attain
the minimum Chrome requirements the pulsed MIG weld procedures had to ensure the
weld dilution was less than 8%. The vertical down
welds not only have to attain minimum weld dilution with consistent weld fusion
on the carbon steel boiler tubes. The weld passes I developed were not only thinner
they also supplied an improved transition with the weld pass tie-ins. Of course
from a weld productivity perspective, the two gun weld procedure utilized, enabled
the single operator to deposit > 26 lbs/hr.
2008
Traditional
pulsed MIG clad weld mess.
622 overlay. Miller Equipment
Ed's MIG clad welds have
a smooth finish similar to a laser / powder overlay.
Note:
All photos are untouched and no weld
cleaning except brushing was provided.
With
Clad welds on boilers, less is always better,
unless you sell weld consumables. 
Tremendous
benefits were attained for the boiler
industry with Ed's new, "thinner"
single pass
MIG procedures developed in 2005 - 2006.
 The
vertical down 622 Inconel / stainless clad pulsed MIG welds were derived from
a low cost, six thousand dollar pulsed power source and a MIG gas mix developed
by Ed. (See gas data section). These clad welds required 28% less Inconel or stainless
per square foot of water wall weld.
The welds also required an engineering
manager that believed that there was more to MIG clad welding than asking the
advice of a Lincoln sales rep or a welder pulling a gun trigger. It also helped
that WSI had excellent, well designed automated weld equipment that compensated
for the curves (wire stick out voltage variations) when welding of the boiler
wall tubes. Ed's clad development was complete in 2006. WSI applied for the US
Patent during 2006 and the European patent was approved in 2009. Visit the Clad
weld section for more info.
METRIC
WELD TABLES:
Metric and Weld Deposition Rates. Weld
Deposition Rates: = lb/hr x 0.4536 = kg/hr.
Weld Deposition Rates: = 1 lb/ft
x 1.49 = kg/m
Weld Deposition Rates: = Ib/in x 17.85 = kg/m.
Weld Deposition
Rates: = lb/min x 27.216 = kg/hr.
Weld Deposition Rates: kg/hr x 2.205 = lb/hr.
Metric and
Gas Flow Rates. Flow
cuft x 28.317 = liters
Flow 0.47195 liter/min = 1 cuft/hr.
cuft x 1728
= cubic inch.
cuft x 0.02832 = cubic meter.
cuft/hr x 0.4719 = liters/min
cuft/min x 28.31 = liters/min.
liters/min x 2.119 = cuft/hr.
cu/meters/min
x 0.002119 = cuft/hr.
cuft x 7.4805 = gal US.
liters x 0.03531 = cuft.
gal/hr x 0.06309 = liter/min
gal/hr x 0.13368 = cuft/hr
gal/min x 8.0208
= cu/ft/hr.
gal/min x 3.785 = liter/min
gal x 3.785 = liter
liters/sec
x 127.13 =cuft/hr
Metric and
Thermal Conductivity.
Joules
Per inch = Volts x Amps x 60, Divide by weld speed ipm
Watt per meter kelvin
w/m.k
Heat input joules (j) energy J = watts/second
1 watt = 1 joule/sec
1 Kw/hr = 3,600,000
ft/ib = J x 1.356
1 joule = 0.73756 ft/ib.f
j/inch
x 39.37 = j/m
j/m x 0.0254 = j/inch
kj/inch x 39.37 = kj/m
Btu x 1054.4
= joules
btu/lb x 2.326 = kj/kg
cal/g x 4.1868 = kj/kg
Metric
and Fracture Toughness.
Metric Meganewton meter MN.m-3/2
ksi.in 1/2 x 1.099 = MN.m-3/2
MN.m-3/2 x 0.910 = ksi/in ½
Metric and Electrode Force. Metric
Newton = N
pound force x 4.448ton = newtons
kilogram force x 9.807 =
newtons
newton x 0.2248 = lbf
Metric and Area. in/sq
x 645.2 = mm/sq
mm/sq x 0.001550 = in/sq
in/sq x 6.451 = cm/sq
ft/sq
x 0.09290 = m/sq
Metric and Speed.
ipm
x 0.4233 mm/sec
mm/sec x 2.362 = ipm
in/sec x 0.0254 = m/sec
ft/sec
x 0.3048 = m/sec
ft/hr x 0.00008466 m/sec
ft/min x 0.00508 = m/sec
km/hr x 0.027777 = m/sec
mph x 1.609 = km/h
cm/sec x 1.9685 = ft/min
cm/sec x 0.32808 = ft/sec
m/sec x 196.85 = ft/min
Metric and
Impact Strength. Metric
= Joules = J
1 ft/lb = 1.355818 J
ft/lb x 0.13825728 = kg/m
Metric
and Volume. cu/in
x 16.387 = cu/cm
cu/in x 0.00057870 = cuft
cu/in x 0.000016387 = cu/m
cuft x 0.02831 = cu/m
cu/in x 16390.0 = cu/m
cu/cm x 0.000035315 = cuft
cu/cm x 0.061024 = cu/inch
cu/cm x 0.00026417 = gal US
cu/m x 35,315 =
cuft
cu/m x 0.00061`024 = cu/inch
liquid gallon to cu/m x 0.003785
Weld
Current Density Electrode
Area Square inch Divide by Weld Amp
Weld current density metric = amp per square
millimeter =
A/mm2
WIRE 0.030 area = 0.00071 in/sq
WIRE 0.8 mm area = 0.00458 cm/sq
WIRE 0.030 area = 0.00071 in/sq
WIRE 0.035 area = 0.00096 in/sq
WIRE 1mm
area = 0.006193 cm/sq
WIRE 0.045 area = 0.00160 in/sq
WIRE 1.2mm area
= 0.1032 cm/sq
WIRE 0.062 area = 0.00307 in/sq
WIRE 1.6mm area = 0.0198
cm/sq
Metric and Length. 1
meter = 3.281 ft
1mm = 0.03937 inch
1inch x 25.4mm
mm x 0.03937 =
inch
1 meter x 39.370 inch
ft x 304.8 = mm
mm x 0.003281 = ft
ft x 0.3048 = meter
yard x 0.9144 = meter
Metric and Weight. oz
x 0.02834 = kg
lb x 0.4535 = kg
Short ton 2000 lb x 907 = kg
Long
ton 2240 lb x 1016 = kg
Tensile strength x 1000 = ksi
Metric Megapascal
MPa
I MPa = 145.03 psi
kPa = 0.14503 psi
psi = 0.00689 MPa
mpa
x 0.145 = ksi.
Metric and
Pressure (psi).
Metric
Kilopascal kPa.
psi x 6894.575 = Pa
kPa x 0.1451 = psi
psi x 0.068948
= bar
psi x 6894.8 = newtons/sqm
1 psi = 0.0068 MPa
1 bar x 100000
= Pa
ksi x 6.894 = MPa
Pa x 0.000145 = psi
lb/sqft x 4.788 = Pa
N/sqmm x 1000000 = Pa
Pa x 0.000145 = psi
Pa x 0.02089 = psi.
Pa x
0.02068 = lb/sqft
Pa x 0.000001 N/sqmm
bar = 14.5 psi
ksi x 6.894757
= MPa
psi x 0.006894 = MPa 
Get
those weld costs down and afford that Penthouse
you have always dreamed about. |
