Hope for the best, pack for the worst. How one welding symbol on a drawing doubled the time and cost of our last tower modification job.

 We recently got a call from a tower modification general contractor we had submitted a bid to informing us they won the bid and we should plan to head out a week from today to perform the elevated certified welding and certified welding inspection (CWI). As usual, I assigned one of our three CWIs to the job and he started prepping. During a second review of the drawings to determine the amount of filler material required for the job, he noticed the 3/16 in. fillet weld symbol (40 linear feet worth) called out may not be the actual configuration based on the base material size and layout. We looked through it together, and drew a scaled version of what it should be. Yep, he was right, instead of a simple single pass 3/16 in. fillet weld, we may be welding a flare bevel with approx. a ½ in. depth. Anybody who has experience with flare bevels, knows getting code acceptable flare bevel welds takes much more care and consideration than a simple fillet weld, from the torch angle, to the method of verifying the weld size, flare bevels are just plain slow when compared to fillet welds.

The tower mod engineer was calling out angle iron to reinforce the cell tower legs instead of the typical half pipe mod. These fillet welds would connect the existing pipe leg and the new reinforcement angle iron at the top and bottom, referred as terminal welds. We decided to hope for the best and assume the field configuration would in fact only require a 3/16 in. fillet weld, but plan for the worst and pack enough filler material and measuring instruments to complete a flare bevel, welded flush. Using the Shielded Metal Arc Welding (SMAW) process, we calculated the estimated weld filler material consumption and weld times as follows:

3/16 Fillet Weld3 16 fillet weld

Cross Section = [(3/16)2 / 2]* 1.25 (convexity multiplier) = 0.022 in.2

Weld Volume = Cross Section * Length = (0.022 in.2)*(40 ft. * 12 in./ft.) = 10.6 in.3

Deposited Weld Metal Required = Weld Volume * Weld Metal Density = 10.6 in.3 * 0.284 lb./in.3 = 3.0 lb.

Deposition Efficiency = 45%, out of position welding, 3-1/2 in. stub loss, flux loss.

Pre-Weld Metal Required = Deposition Efficiency *  Deposited Weld Metal Required = 3.0lb. / 0.45 = 7 lb.

Deposition Rate, assuming 1/8 in. electrodes @120 Amps = 2 lbs. / hr.

Weld Time = (Deposited Weld Metal Required/Deposition Rate)/Arc Effeciency = (3 lb./2lbs./hr.)/(25%) = 6 hrs. weld time not including climbing and positioning.

Flare Bevel Welded Flushflare bevel

Cross Section = [(0.55in.*.37in.)/2]* 1.25 (convexity multiplier) = 0.127 in.2 , assuming a Z loss of 3/8 in.

Weld Volume = Cross Section * Length = (0.127 in.2)*(40 ft. * 12 in./ft.) = 60.96 in.3

Deposited Weld Metal Required = Weld Volume * Weld Metal Density = 60.96 in.3 * 0.284 lb./in.3 = 17.3 lb.

Deposition Efficiency = 45%, out of position welding, 3-1/2 in. stub loss, flux loss.

Pre-Weld Metal Required = Deposition Efficiency *  Deposited Weld Metal Required = 17.3lb. / 0.45 = 38.5 lb.

Deposition Rate, assuming 1/8 in. electrodes @120 Amps = 2 lbs. / hr.

Weld Time = (Deposited Weld Metal required/Deposition Rate)/Arc Efficiency = (17.3 lb./2lbs./hr.)/(25%) = 35 hrs. weld time not including climbing and positioning.

As you can see from the math above, one incorrect weld symbol callout on the drawing can fool even the best welding estimators. The job was originally scheduled for two 8 hr. days, after mob and demob, but ended up taking four 10 hr. days. The Engineering firm ended up paying for this mistake, but that’s not always the case.