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I'm trying to avoid doing an extensive if then statement for all possible angle weights.
Does anyone out there have this made already or maybe a formula that, given the leg lengths and thickness, would yield the appropriate lb/ft
for any given angle?
if someone == "yes"
then joe = "grateful"
end if
Solved! Go to Solution.
Yes, the relationship is simply the cross-sectional area * density. The problem is finding the true cross sectional area. The problem is further complicated by the fact that A36 (assuming you are using an American standard since you listed lb/ft.) only specifies minimum radii for the leg radii and inside radius. It's up to the manufacturer to stay within the specified limits but otherwise free to roam within. Finally, from one run to the next, the tool wears so actual dimensions will vary. The published weights are calculated from theoretical dimensions and then rounded. You might be able to call your distributer and find out what they are using for radii, although I've found that it is very difficult to get that information. From there you can write a formula to compute the area and then multiply that by the density to get the lb/ft.
Another approach depends on how accurate you need to be. You can mulitply your values by a fudge factor that approximates the radius. For instance:
L1 = leg 1 in inches
L2 = leg 2 in inches
t = thickness in inches
D= density .28 lb/in^3
X= 12in (for conversion to lb/ft)
F= 1.02
L1*t+(L2-t)*t * D * X * F
Without the fudge factor you will probably get to within 4%. With the fudge factor, you can probably get to within 1-2% and with rounding you may not even tell the difference on most sizes. Otherwise, yes, you will need to write a very lengthy if-then statement.
I calculated the 1.02 fudge factor based on a stock list from Ryerson Tull. I computed the sectional area using thickness and leg lengths and then multiplied by the density. From there I divded by the published values and subtracted that from 1. I took the average over several sizes and got 0.02, so the factor becomes 1.02. Your mileage may vary.
If you do find out manufacturer radii or what they use to compute their areas, I'd love to hear what you or anyone else has found.
Can you provide a sketch or diagram to show what you're looking to calculate?
Yes, the relationship is simply the cross-sectional area * density. The problem is finding the true cross sectional area. The problem is further complicated by the fact that A36 (assuming you are using an American standard since you listed lb/ft.) only specifies minimum radii for the leg radii and inside radius. It's up to the manufacturer to stay within the specified limits but otherwise free to roam within. Finally, from one run to the next, the tool wears so actual dimensions will vary. The published weights are calculated from theoretical dimensions and then rounded. You might be able to call your distributer and find out what they are using for radii, although I've found that it is very difficult to get that information. From there you can write a formula to compute the area and then multiply that by the density to get the lb/ft.
Another approach depends on how accurate you need to be. You can mulitply your values by a fudge factor that approximates the radius. For instance:
L1 = leg 1 in inches
L2 = leg 2 in inches
t = thickness in inches
D= density .28 lb/in^3
X= 12in (for conversion to lb/ft)
F= 1.02
L1*t+(L2-t)*t * D * X * F
Without the fudge factor you will probably get to within 4%. With the fudge factor, you can probably get to within 1-2% and with rounding you may not even tell the difference on most sizes. Otherwise, yes, you will need to write a very lengthy if-then statement.
I calculated the 1.02 fudge factor based on a stock list from Ryerson Tull. I computed the sectional area using thickness and leg lengths and then multiplied by the density. From there I divded by the published values and subtracted that from 1. I took the average over several sizes and got 0.02, so the factor becomes 1.02. Your mileage may vary.
If you do find out manufacturer radii or what they use to compute their areas, I'd love to hear what you or anyone else has found.
Forgot to mention that the above applies to A36 steel. Other materials and standards may vary widely.