On Fri, 25 Jun 2010 22:48:29 -0700, Winston
wrote:

Here's what I was on about:

Envision a barrel 22" diameter by 35" high. Let's agree I pressurized
that barrel to 10 PSIG.

Let's arbitrarily say that all seams measure 0.05" in width.
We have a total seam length of ~173" (Two circumferences plus the side
seam). So 173" of seam length x 0.05" of seam width is 8.66 square inches
of total seam area. If we multiply that by our 10 PSIG we get
a force of 86.6 lbs acting against all the seams collectively.
If we divide that force by the seam length, we decide that the force
per inch of seam is about 0.5 lb, yes?

Envision that same barrel only this time, we now have two side
seams and four circumferential seams (instead of just two).
Our new total seam length is now 346.5" for a total seam area
of ~17.3 sq. inches. If we multiply that by our 10 PSIG we get
a force of ~173.2 lbs acting against all the seams collectively.
If we divide that force by the seam length, we decide that the force
per inch of seam is about 0.5 lb, yes?



Where did I go wrong?

The force per unit length resisted by a head seam is pressure times
the area of the *head* divided by the length of the seam. In the
example above:
10 lb/in^2 x 380 in^2 / 69 in = 55 lb/in

The stress in the cylindrical part of the barrel is equal to pressure
x radius / wall thickness -- the classical formula for stress in a
cylindrical shell. Remove the wall thickness and you get the force per
unit length in that seam:
10 lb/in^2 x 11 in = 110 lb/in


--
Ned Simmons