University of Engineering and Technology,
Peshawar
NAME: Wajid Ur Rehman
DEPARTMENT: ENERGY ENGINEERING
REG NO: 24PWCEE0129
SUBMITTED TO: Dr. Khurshid Ahmed
SUBMITTED DATE : 20-05-2025
Question: Welding Process Analysis: SMAW, FCAW, and Robotic GMAW Comparison A
manufacturing work cell performs Shielded Metal Arc Welding (SMAW) using a fitter and a
welder. The fitter takes 5.5 minutes to place unwelded components into the welding fixture at
the beginning of the cycle and 2.5 minutes to unload the completed weldment at the end. The
total weld seam length is 2000 mm, and the welder operates at an average travel speed of 400
mm/min. During welding, the electrode stick must be changed every 750 mm of weld length,
which takes 0.8 minutes. The work cell operates sequentially, meaning the fitter and welder
cannot work simultaneously - while the fitter is working, the welder is idle, and vice versa.
Considering the SMAW process: (a) Determine the average arc time (actual welding time) per
cycle. (b) Calculate the production rate in weldments completed per hour. Now, Consider the
welder would now use a continuously fed electrode, eliminating stick changes. However, the
flux-cored wire spool must be changed every five weldments, taking 5.0 minutes. The travel
speed remains 400 mm/min. (c) Determine the improvement in arc time compared to SMAW.
(d) Calculate the new production rate with FCAW. If a new configuration includes an industrial
robot and two welding fixtures. This allows parallel operations - the robot welds at one fixture
while the fitter simultaneously loads/unloads the other fixture, with roles switching after each
cycle. The electrode wire spool requires changing every five weldments, taking 5.0 minutes
(performed by the fitter). (e) Determine the arc time per cycle in this robotic configuration. (f)
Calculate the production rate for the robotic work cell. Discuss the results.
Given Data:
Fitter loading time = 5.5 minutes
Fitter unloading time = 2.5 minutes
Total weld length = 2000 mm
Travel speed = 400 mm/min
Electrode stick change (SMAW) = every 750 mm, taking 0.8 minutes
Wire spool change (FCAW/Robotic) = every 5 weldments, taking 5 minutes
SMAW (Shielded Metal Arc Welding)
(a) Average arc time per cycle:
Arc welding time:
2000 mm400 mm/min=5.0 min\frac{2000 \, \text{mm}}{400 \, \text{mm/min}} = 5.0 \, \
text{min}400mm/min2000mm=5.0min
Number of electrode changes:
⌊2000750⌋=2 changes\left\lfloor \frac{2000}{750} \right\rfloor = 2 \, \text{changes}⌊7502000
⌋=2changes
Time for stick changes:
2×0.8=1.6 min2 \times 0.8 = 1.6 \, \text{min}2×0.8=1.6min
Total arc time:
5.0+1.6=6.6 min5.0 + 1.6 = 6.6 \, \text{min}5.0+1.6=6.6min
✅ Answer (a): Average arc time per cycle (SMAW) = 6.6 minutes
(b) Production rate (SMAW):
Total cycle time (sequential work):
5.5 (load)+6.6 (weld)+2.5 (unload)=14.6 min5.5 \, (\text{load}) + 6.6 \, (\text{weld}) + 2.5 \, (\
text{unload}) = 14.6 \, \text{min}5.5(load)+6.6(weld)+2.5(unload)=14.6min
Production rate:
6014.6≈4.11 weldments/hour\frac{60}{14.6} \approx 4.11 \, \text{weldments/hour}14.660
≈4.11weldments/hour
✅ Answer (b): Production rate = 4.11 weldments/hour
FCAW (Flux-Cored Arc Welding)
(c) Improvement in arc time:
No stick changes; arc time only:
2000400=5.0 min\frac{2000}{400} = 5.0 \, \text{min}4002000=5.0min
Improvement:
6.6−5.0=1.6 min6.6 - 5.0 = 1.6 \, \text{min}6.6−5.0=1.6min
✅ Answer (c): Arc time improvement = 1.6 minutes
(d) Production rate (FCAW):
Cycle time without electrode change:
5.5+5.0+2.5=13.0 min5.5 + 5.0 + 2.5 = 13.0 \, \text{min}5.5+5.0+2.5=13.0min
Add 5-minute spool change every 5 weldments:
55=1.0 min per cycle\frac{5}{5} = 1.0 \, \text{min per cycle}55=1.0min per cycle
Total cycle time:
13.0+1.0=14.0 min13.0 + 1.0 = 14.0 \, \text{min}13.0+1.0=14.0min
Production rate:
6014.0≈4.29 weldments/hour\frac{60}{14.0} \approx 4.29 \, \text{weldments/hour}14.060
≈4.29weldments/hour
✅ Answer (d): Production rate = 4.29 weldments/hour
Robotic GMAW with Parallel Loading
(e) Arc time per cycle (same as FCAW):
2000400=5.0 min\frac{2000}{400} = 5.0 \, \text{min}4002000=5.0min
✅ Answer (e): Arc time per cycle = 5.0 minutes
(f) Production rate (Robot):
Fitter time = 5.5 + 2.5 = 8.0 min
Parallel operation:
Cycle time=max (5.0,8.0)=8.0 min\text{Cycle time} = \max(5.0, 8.0) = 8.0 \, \
text{min}Cycle time=max(5.0,8.0)=8.0min
Add wire spool change every 5 weldments:
55=1.0 min\frac{5}{5} = 1.0 \, \text{min}55=1.0min
Total cycle time = 8.0 + 1.0 = 9.0 min
Production rate:
609.0≈6.67 weldments/hour\frac{60}{9.0} \approx. 6.67 \, \text{weldments/hour}9.060
≈6.67weldments/hour
✅ Answer (f): Production rate = 6.67 weldments/hour
Discussion:
SMAW has the slowest production rate due to frequent manual electrode stick changes.
FCAW improves arc time by 1.6 minutes, translating into a modest production gain.
Robotic GMAW maximizes efficiency by allowing parallel operations and eliminates
welder idle time, giving a 61% increase in production rate over SMAW.
The robot reduces the cycle time significantly, despite the same arc speed, by overlapping
loading and welding processes.
Summary Table:
Method Arc Time (min) Total Cycle Time (min) Prod. Rate (weldments/hour)
SMAW 6.6 14.6 4.11
FCAW 5.0 14.0 4.29
Robotic GMAW 5.0 9.0 6.67
