# Thread: Optimal design of a steel drum?

1. ## Optimal design of a steel drum?

Hey guys. I have a problem due tomorrow for calculus class that is very difficult for me. To be honest I have no idea where to begin. The only thing I know is that the relevant information comes near the end of the problem. Any help would be greatly appreciated.

A 55-gallon Tight Head Steel Drum is constructed by attaching 18 gage (i.e. .0428 inches thick) steel disks to the top and bottom of a cylinder created by rolling up a 20 gage (i.e. .0324 inches thick) steel sheet.

The vertical seam on the cylinder is welded together and the top and bottom are attached by a pressing/sealing machine. The pressing/sealing process requires approximately -1^3 i•n ches from the cylinder and J inches from the disk to be curled together and hence these inches are lost in the final dimensions. In addition, the top and bottom are set down r inches into the cylinder. For clarification, refer to the American National Standard (ANSI) specification diagram below.

Steel can be purchased in coils (rolls) of any specified width. Construction costs can besummarized as follows:

18 gage steel is 45 cents/square foot

20 gage steel is 34 cents/square foot

welding and pressing/sealing cost 10 cents/foot

cutting steel costs 2 cents/foot.

Is the ANSI specified drum the most efficient use of material in order to obtain the required 57.20 gallon minimum volume capacity of a 55 gallon drum? Fully justify your answer.

2. There's a lot of data given in the problem, but conceptually, it's not any different than any other optimization problem. Let r be the diameter of the two disks and h the height of the cylinder.

I'm not sure what is happening in the middle paragraph - you take "-1^3" inches from the disk and "J" inches from the cylinder, and then "r" inches from the height. These should probably be numbers.

Anyway, you calculate the volume as $\pi$ times the radius (minus the "-1^3" lost to pressing/sealing) squared times the height (minus 2 times the "J" and the "r" lost on the top and bottom). Then you would set the volume to 57.20 gallons (which will need to be converted to cubic feet), and solve for h.

And you calculate the construction cost with the unit costs you specified (as a function of r). Then take the derivative with respect to r and set it to zero, and solve for r.

Post again in this thread if you're still having trouble.

3. Hey, sorry for the mistake up above. I guess my scanner didn't do the fractions correctly. "-1^3" should be 13/16 inches and j is simply 3/4 inches.

But if I understand you correctly, should I be making an equation like this:

$v=((pi*r-/frac[13/16])^2)(h-(6/4))$

4. Hey, sorry for the mistake up above. I guess my scanner didn't do the fractions correctly. "-1^3" should be 13/16 inches and j is simply 3/4 inches.

But if I understand you correctly, should I be making an equation like this:

$v=((pi*r-(13/16)^2)(h-(6/4))$

You say to solve for h, but I do not know r either. What should I do?

5. The volume is only the first part of the problem - once you do that, you'll need to calculate the cost.

But let's concentrate on the volume first. The volume of a cylinder is:

$V=\pi(\text{radius})^2(\text{height})$

The radius is r-(1/12)(13/16) = r-(13/192), because the pressing/sealing process takes away 13/16 of an inch off the original disk (note that we have to convert inches to feet). There are two things to subtract from the height: 3/4 of an inch on each end for the pressing/sealing process and an additional amount because the top and bottom are set into the cylinder. Your original post calls this amount "r", but it should be a number - I'll use 5/8 since that's what's in the ANSI diagram. So the height is:

$h-(2)(\frac{3}{4*12})-(2)(\frac{5}{8*12})=h-11/48$

and the volume is:

$V=\pi(r-\frac{13}{192})^2(h-11/48)$, which needs to be equal to 57.20 gallons or 57.20/7.48 = 7.65 cubic feet. You can solve this for h in terms of r.

The construction costs are pretty easy to calculate.

18 gage steel (top and bottom): $(45)*(2)*(\pi{r^2})$, assuming I only have to pay for the steel I actually use when buying a circular sheet of steel
20 gage steel (side): $(34)*(2\pi{rh})$
welding and pressing/sealing: $(10)*(2)*(2\pi{r})$
cutting steel: $(2)((2)*(2\pi{r})+h)$, assuming I pay to cut out the circumference when buying a circular sheet of steel

So add up the costs and substitute the formula for h, and you have cost as a function of r. To minimize cost, set the derivative to zero and solve for r.

Post again in this thread if you're still having trouble.

- Hollywood

6. ## Re: Optimal design of a steel drum?

So I am another student having serious problems with the problem. My partners and I have been working on it for a project for a while and are making little headway. We found this post but are having trouble simplifying the equation to solve for h in terms of r. Any help would be much appreciated

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### 55 gallon drum weight dimensions

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