Calculating weld volume and weight...

Author:
Mindaugas Rutė

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*ig.+. rea of an equal leg length fillet weld n as!mmetrical fillet weld is a little more difficult' the area of a triangle is given b! the base times the height + divided b! - so when wh en a fillet weld is deposited with unequal leg lengths the area can be calculated from multipl!ing the throat, a, b! the length of the face and divided b! as illustrated in Fig.2 in Fig.2..

*ig.-. rea of an unequal leg length fillet "urning now to butt welds, the calculations become a little more comple&. "here are three factors that determine the volume of the weld in a single / butt weld. "hese are the angle of the bevel, b, the e&cess weld metal and the root gap, g, as illustrated in *ig 0. "o calculate the area of this weld we need to be able to add together the areas of the four components illustrated in Fig.3.

*ig.0. "he four areas of a single1/ butt weld "he dimension 2c2 is given b! (tan b & t)' the area of a single red triangle is therefore t(tan b & t)3-. "he total area of the two red regions added together can be calculated using the formula -t(tan b & t)3- or t(tan b & t). "he width of the weld cap, w, is given b! W 4 -(tan b & t) 5 g. "he area of the e&cess weld metal is appro&imated b! the formula (W & h)3-. "he area provided b! the root gap b! g & t. "he bevel angles, b, most often used are +6% 4 (tan 6.+78), +$% 4 (tan 6.-89), --.$% 4 (tan 6.#+#) 0-.$% 4 (tan 6.807) and #$% 4 (tan +.66). s will become obvious when the weight is calculated, it is easier to ensure that the decimal point is in the right place if centimetres are used in the calculations rather than millimetres. s a worked e&le, if the weld is in a plate -.$cm thickness, 6.0cm root gap, 8$% included angle (b 4 0-.$6%' tan 0-.$% 4 6.807) and with a cap height of 6.-cm we have:1

+. c 4 tan0-.$ & -.$ 4 6.807 & -.$ 4 +.$;cm -. w 4 -(6.807&-.$) 5 6.0 4 0.#9$cm so the area of the cap 4 (0.#9$&6.-)3- 4 6.0#9 sq. cm. 0. area of the orange area 4 6.0 & -.$ 4 6.7$ sq.cm. #. area of the two red areas 4 - & (+.$; & -.$)3- 4 0.;7sq.cm. "his gives a total area of $.67sq cm. "he volume can then be calculated b! multipl!ing the length of the weld b! the area 1 ensuring that this length is also given in centimetres< Conventionall!, the volume is often e&pressed in cubic centimetres (cu.cm). per metre so in this e&le the volume is $67 cu. cm3metre. "o obtain the weight of weld metal this figure is then multiplied b! the densit! of the allo!. Table 1 gives the densit! of some of the more common allo!s in gm3cu.cm. =ote that with some allo!s the allo!ing elements can change the densit! quite significantl!. Table 1. Densities of some of the more common alloys. Density (gm/cm !

Alloy

iron 6.-$> carbon steel +->Cr steel 06# stainless steel nickel 963-6 =i.Cr 8-$ t!pe allo! copper 76306 brass 7> l brone aluminium l $6$l 767$

7.97 7.98 7.76 7.;9.;6 9.#6 9.## 9.;# 9.$0 7.9; -.76 -.8$ -.9

"he weight of weld metal to fill one metre length of the joint described above would therefore be' in carbon steel ($67 & 7.98) 4 0;9$gms or 0.;9kgs3metre' in a $??? series aluminium allo! ($67 & -.8$) 4 +0#0gms, +.0#kgs3metre. Calculating the weight of weld metal in double sided /1joints uses the same approach b! dividing the weld into its individual 2/2s and a dding the products. @1preparation, however, adds another area into the equa tion' that of the half circle at the root of the weld, see Fig.4. "he formulae given above to calculate 2c2, the area of the two red components and the e&cess weld metal remain unchanged but the width of the cap must be

increased b! -r. "here are also the two areas, 22 and 2B2, to calculate and the two white root radius areas to be added to the total.

*ig.#. Aingle 22 preparation (other notation as in *ig.0) "he relevant formulae are thus: +. the dimension 2c2 is given b! (tan b & (t1r))' the total area of the two red regions is therefore given b! the formula -((t1r)(tan b & (t1r))3- or ((t.1r)(tan b & (t1r)). -. the width of the weld cap, w, is given b! w 4 -(tan b & (t1r)) 5 g 5-r. 0. the area of the e&cess weld metal is given b! the formula (w & h)3-. #. the area 22 is (t1r) & (-r 5g). $. the area 2B2 is g & r. 8. the root radius area is (r -)3# *or a double1 preparation it is necessar! to c alculate the areas of both sides and add these together. Daving calculated the weight of weld metal required to fill a weld preparation it is then possible to calculate the weight of filler metal required (these two figures are not necessaril! the same) and to estimate the time required to deposit this weld metal' both essential in order to arrive at a cost of fabricating the weld. "his will be covered in future @ob Enowledge articles. "his article was written b! "ene #athers.

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