In the first place, it is obvious that whatever exten sion the cylinders or ring may undergo, there will be still the same quantity of surface, independently of the small change due to compression, in the section of the ring, which area is always proportional to the difference of the squares of the two diameters.
Let I) be the interior diameter before pressure, and D d its diameter when extended by the pressure.
Let also D' be the exterior diameter before pressure, and D' d', its diameter when extended by pressure.
Then from what is stated above we have Or 2 = 2 1) Whence 2 D' d' : 2 1)-1-d : : d : d' Or considering d' and d as very small in comparison with 2 D' and 2 1), this becomes D' : D : : d : d' That is, the extension of the exterior surface is to that of the interior, as the interior diameter to the exterior, But the resistance is as the extension divided by the length, therefore, the resistance of the exterior surface D' D' 2 is to that of the interior, as D . Or : ' • That is, the resistance offered by each successive lamina is inversely as the square of its diameter, or inversely as the square of its distance from the centre; by means of which law the actual resistance due to any thickness is readily ascertained.
Let r be the interior radius of any cylinder, p the pressure per square inch on the fluid, t the whole thickness of the metal, and x any variable distance from the interior surface. Let also s represent the strain exerted, or the resistance sustained, by the in terio• lamina, then by the law last deduced + x)2 sSthe strain at the distance, a from the interior surface, consequently, s d x cor. = sum of all the strains.
(7. + a.) This, when x = t becomes R = s — s r t That is, the sum of all the variable strains or re sistances on the whole thickness t, is equal to the resist ancc that would be due to the thickness— 7.-Ft
acting uniformly with a resistance s.
Let us now suppose (the above law being establish ed) the radius r, and the pressure p per square inch on the fluid, to he given, to find the thickness necessary to resist it, or such that the strain and resistance may be in equilibria the cohesive power of the metal being also given. Let x represent the thickness required, and c the cohesive power of the metal per square inch; then, the greatest strain the area r x can sustain, is r x c r ±x r+x and that which it has to sustain is p r; hence, when these are equal we shall have rp= r c, or p r r-Fx r x= • c—p Hence the following rule in words at length.
To find the thickness of the pres sure per square inch by the radius of the cylinder, and divide the product by the difference between the cohesive power of the metal per square inch, and the pressure per square inch, and the quotient will be the thickness sought.
As an example, let it be required to determine the thickness of metal in two presses, each 12 inches in diameter, in one of which the pressure is lA tons, and in the other, 3 tons per circular inch; the cohesive strength of cast iron being 18,0001bs. per square inch. Here 11 tons per circular incli--=--4278 lbs. per sq. inch, 3 tons =8556 Hence by the rule First 4.278 X 6 1.87 inches thickness.
1800-4278 And 8556 x 6 543 inches thickness, 1800--8556 It appears, therefore, that in this second case, al though the pressure is only double the former, the metal requires to be nearly three times the thickness.