What should be considered when determining welding costs?

Transcription

1 Reasons for Determining Welding Costs Knowing which factors affect welding costs can enable a company to focus its energies on changes that will reduce costs, enabling the business improve its competitiveness and profitability. An accurate cost model can permit comparisons of manufacturing options (for example, comparing the effect of a change of welding processes on overall costs). A correct cost model will permit the estimation of savings that will accrue with automation, so that the projected savings can be used to justify the automation capital investment. Factors Affecting Welding Costs What should be considered when determining welding costs? In reality, every operation resulting from the decision to weld can be legitimately charged to weld fabrication. The greater the number of factors considered when calculating welding costs, the more accurate the results will be. Also, considering all the relevant factors increases the opportunities for cost reduction.when determining whether a specific manufacturing cost should be charged to welding, it is helpful to ask: Would this cost be incurred if the product wasn't welded? When this question is objectively answered, then all of the following factors may be considered to be part of the cost of welding: Time for joint preparation. Time to prepare the material for welding (blasting, removal of oils, etc.). Time for assembly. Time to preheat the joint (when required). Time for tack-up. Time for positioning. Time for welding. Time to remove slag (when applicable).

2 Time to remove spatter. Time for inspection. Time for changing electrodes. Time to move the welder from one location to another. Time to change welding machine settings. Time spent by personnel for personal purposes. Time to repair or re-work defective welds. Costs associated with any required stress relief. Cost of electrodes. Cost of shielding materials. Cost of electric power. Cost of fuel gas for pre-heat (when required). Time the Biggest Cost Of the 20 items identified in the preceding list, 15 begin with the word "time." Unless the application requires unusually expensive alloys, or is a highly automated operation, the time associated with welding operations and the wages that must be paid to skilled personnel will typically dominate welding costs. Simplified Cost Models Two different approaches may be used to determine welding costs: complex and simple. There are the complex, computer-based models that attempt to capture every contributing factor; and there are simplified models. Both have inadequacies, but are useful nevertheless. Only the simplified models will be discussed in this article. In the simplified models, welding costs are estimated based upon: Labor and overhead. (L&O) Cost of welding consumables and shielding materials.

3 In most cases, the cost of power is negligible and accordingly ignored. A variety of costs are often attributed to "overhead," including plant and equipment, supervision, indirect labor, etc. These are significant costs, often exceeding direct labor costs by a factor of 2-4 times. A simplifying assumption is to tie an overhead factor to the direct labor cost. Thus, a single cost per hour is used for "labor and overhead" (L&O) in simplified models. The basic cost-estimating formulas, therefore, will take on the form of: Welding Costs = (L&O) + (Consumables Costs) The Operating Factor A review of the cost factors listed previously reveals there are various "times" listed other than the time required for welding. Any time the welder's arc is not struck represents time that the joining process is not progressing. Since the total hours worked are always more than the total hours spent welding, the ratio of hours spent welding to total hours worked is called the operating factor. As the basis of any cost formula, it must be determined accurately. Since arc time is always divided by a larger number, the ratio is always less than 1.0, and therefore a decimal. For convenience in referring to operating factors, the ratio is multiplied by 100 and expressed as a percentage. Thus, one hears references to operating factors of 30, 40 or 50 percent. When using an operating factor in a cost formula, however, it must be given in the decimal form, so that a 40% operating factor would be expressed as 0.40 in a cost formula. Three Basic Approaches Welding costs can be estimated using one of three basic approaches: Cost per unit. Cost per length. Cost per weight. The application will determine which approach is most appropriate.

4 One caveat: with any of the cost calculation methods, it is critical that the variables used result in an equation that is dimensionally correct. For example, if the wire feed speed is measured in inches per minute, and it is multiplied by the weight of the electrode per length, the weight per length must be in units of pounds per inch. If this is done, the resultant product will be pounds per minute. However, if the weight is measured in pounds per foot, the resultant product would be inch-pounds per foot-minutes. Obviously, this is a meaningless dimension. It can be corrected, however, by multiplying the product by "1 foot/12 inches," returning the dimensions to pounds per minute, and correcting the numerical value by a factor of 12. Cost Per Unit The cost per unit method is most effective when the application involves pieces that move through a workstation. The types and sizes of the welds are immaterial with this method fillet welds, groove welds, plug welds, etc., can all be combined when the cost per unit is determined, since time (the most costly aspect of welding) is measured directly. This makes the per unit method the most accurate of the three approaches. It measures the key cost variable directly, and does not require the use of the operating factor variable. If the process involves wire fed electrode, it is easy to find the cost of the welding materials. It is somewhat harder to calculate consumable costs when SMAW (stick) is used. The cost per unit of production, in dollars per unit, can be estimated using the following formulas: Cost/unit = (L&O/unit) + (filler metal and shielding material cost/unit) L&O/unit = (welding-related time/unit) x (L&O rate) For wire fed processes: Filler metal cost/unit = (wire feed speed) x (welding time) x (weight of electrode/inch) x (electrode cost/pound)

5 Filler metal cost (SMAW) = [{(electrode meltoff rate) x (welding time) x (weight of electrode/inch)}/(% of electrode used)] Shielding gas cost/piece = (flow rate) x (welding time) x (gas cost/ft. 3 ) SAW flux cost/piece = (wt. of flux used) x (cost of flux/lb.) Cost Per Length This method, appropriate for estimating the cost of long welds, is best applied to single pass welds of a prescribed size. The values determined by this method will differ for welds of different sizes. The important variable of time is captured through measurement of travel speed (ft./unit of time). Though good for single pass welds, the method is harder to use for multipass welds. These formulas can be used to estimate the cost per length: Cost/length = (L&O cost/length) + (filler metal and shielding cost/length) L&O cost/length = (L&O rate)/(travel speed)(operating factor) Filler metal cost/length (wire fed processes) = {(wire feed speed) x (wt. of electrode/in.) x (cost of electrode/lb.)}/(travel speed) Filler metal cost/length (SMAW) = (melt off rate) x (wt. of electrode/length) x (cost of electrode/lb.)/(travel speed)(% of electrode used) Shielding gas cost/length = (gas flow rate) x (gas cost/ft. 3 )/(travel speed) Shielding cost/length (flux) = (wt. of weld metal/length) x (ratio of flux to weld metal) x (cost of flux/lb.) Cost per weight Calculating the cost per weight is the easiest cost estimating method, regardless of the welding process. Probably for that reason, it is overused and misapplied. It is best used in applications in which significant volumes of weld metal must be deposited, such as multipass applications. Hardfacing and overlay welding are ideal applications. The variable of time is captured by measuring deposition rate

6 (pounds of deposit per hour). This method is best for estimating the cost of large, multipass welds. Cost per weight is good for evaluating changes in groove joint details. It is not accurate when applied to single pass, small, short welds, and it does not account for overwelding. Cost per weight can be estimated using the following formulas: Cost/lb. = (L&O cost/lb.) + (filler metal and shielding cost/lb.) L&O Cost/lb. = (L&O rate)/{(deposition rate) x (operating factor)} Filler metal cost/lb. (any process) = (cost of filler metal/lb.)/(electrode efficiency) Shielding cost/lb. (gas) = (shielding gas flow rate) x (cost of shielding gas/ft. 3 ) /(deposition rate) Shielding cost/lb. (flux) = (cost of flux/lb.) x (ratio of flux to filler metal) Sample Calculation Cost Per Unit Method A welded subassembly is made in a discrete welding cell. The total cycle time for the part is 2 min.-45 sec. Five welds are made on the part: two 1-in. long fillet welds, two plug welds, and one 3-in. long square edge groove weld. GMAW is used for all the welds, using the same welding procedure, as follows:.035-in. E70S-3 electrode; 300 in./min. wire feed speed; 75% Ar/25% CO2 shielding gas; 35 ft. 3 /hour flow rate gas flow rate. The welding time is 20 sec. each for the two fillet welds, 8 sec. each for the two plug welds, and 18 sec. for the groove weld. Total "arc on" time is 74 sec. The remainder of the welding cycle time involves removal of parts from the bin, cleaning oil off the parts, fixturing the pieces, manipulating the fixture, removing the part, cleaning off spatter, visually inspecting the welds, and stacking the welded components onto a rack. L&O/unit = (welding-related time/unit) x (L&O rate) = (2.75 min.) x (1 hr./60 min.) x ($45/hr.) = $2.0625/piece Filler metal cost/unit (wire fed processes) = (wire feed speed) x (welding time) x

7 (wt. of electrode/in.) x (electrode cost/lb.) = (300 in./min.) x (74 sec.) x (1 min./60 sec.) x ( lb./in.) x (($2.00/lb.) = $0.2035/piece Shielding gas cost/piece = (flow rate) x (welding time) x (gas cost/ft. 3 ) = (35 ft. 3 /hr.) x (74 sec.) x (1 hr./3600 sec.) x ($0.15/ft. 3 ) = $0.108/piece Cost/unit = (L&O/unit) + (filler metal + shielding material cost/unit) = $ $ $0.108 = $2.374/unit Sample Calculation Cost Per Length Method A bridge girder is being fabricated using 5 /16-in. fillet welds for the web-to-flange connection, as well as the stiffenerto- web connections. The girder is 130 ft. long, and 18 ft. deep. Stiffeners are placed every 10 ft. An operating factor of 40% is assumed. All welds will be made with SAW, using the following parameters: 5 /64" EM13K electrode; F7A2 Flux, with a 1.5:1 ratio of flux to electrode; 200 in./min. wire feed speed; and 10 in./min. travel speed L&O cost/length = (L&O rate)/(travel speed)(operating factor) = ($45/hr.) x (1 hr./ 60 min.) / {(10 in./min.) x (1 ft./12 in.) x (0.40)} =$2.250/ft. Filler metal cost/length (wire fed processes) = {(wire feed speed) x (wt. of electrode/in.) x (cost of electrode/lb.)}/(travel speed) = {(200 in./min.) x ( lb./in.) x ($1.75/lb.)}/ (10 in./min.) x (1 ft./12 in.) =$0.5586/ft. Shielding cost/length (flux) = (wt. of weld metal/length) x (ratio of flux to weld metal) x (cost of flux/lb.) = (0.242 lb./ft.) x (1.5) x ($1.20/lb.) =$0.4356/ft. Cost/length = (L&O cost/length) + (filler metal and shielding cost/length) =$2.250/ft. + $0.5586/ft. + $0.4356/ft. =$3.2436/ft. This girder has four web-to-flange welds that are 130 ft. long, and 24 stiffeners (12 on each side). With two 18 ft. stiffener-to-web welds, there are a total of (4 x 130) + (24 x 2 x 18) or 1,384 ft. of weld on each girder. The cost of making the 5 /16-in. fillet welds is estimated therefore at 1,324 ft. x $ or $4490. Sample Calculation Cost Per Weight Method

8 In a weld overlay application, a 1-in. layer is to be applied to a 12-in. dia. roll that is 48-in. long. Two 5 /64-in. dia. electrodes are to be used in a parallel electrode configuration, with the following welding parameters: 200 ipm (per electrode); 32 lb./hr. deposit. The build-up requires a volume of metal that can be estimated a follows: (final volume) - (initial volume) = {(142 /2 x 3.14) x 48} - {(122 /2 x 3.14) x 48} = 1960 in. 2 For steel, this would equate to 566 lb. of weld deposit. L&O Cost/lb. = (L&O rate)/{(deposition rate) x (operating factor)} = {($45/hr.)/(32 lb./hr.) x 40%} = $3.516/lb. Filler metal cost/lb. = (cost of filler metal/lb.)/(electrode efficiency) = $0.80/lb. x 100% = $0.80/lb. Shielding cost/lb. (flux) = (cost of flux/lb.) x (ratio of flux to filler metal) = $0.60 x 1.5 = $0.90/lb. Cost/lb. = (L&O cost/lb.) + (filler metal and shielding cost/lb.) = /lb. + $0.80/lb. + $0.90/lb. $5.215/lb. For 566 lb. of build-up, the cost would be: (566 lb.) x ($5.215/lb.) = $2,952 per roll. NOTE: The welding procedures in the examples, as well as specific numerical values used for labor and overhead cost, and for the welding materials, are illustrative only. They are not presented as accurate for any specific application, and are intended only to demonstrate cost computations. Conclusion Determining the cost of welding is critical as manufacturers struggle to remain competitive in a global economy. Simplified calculations make this task easier, although the simplifi-cation is not without risk. Next month, we'll examine some of the pitfalls that can be encountered when the wrong equation is used, and when assumptions about operating factors and overhead variables are incorrect.

9 Duane Miller is mamager of engineering services for The Lincoln Electric Company, Cleveland, OH