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Strength of Boilers for Construction of Boilers

pressure and rupture

STRENGTH OF BOILERS FOR CONSTRUCTION OF BOILERS According to Pascal's Law, liquids and gases exert pressure equally in all directions. Steam in a boiler exerts the same pressure on all portions of the shell. As the pressure inside a boiler is considerably greater than that outside (the atmospheric pressure), there is a tendency to burst the . shell. This tendency is resisted by the plates of the boiler.

A sphere is the strongest form to resist pressure, for since pressure is equal in all directions, there is a tendency towards enlarging the sphere and not to rupture. But a sphere has the smallest area for a given volume and, as a large heating surface is desirable, and on account of mechanical difficulties, a spherical boiler is never used. The boiler is made cylindrical to obtain greater heating surface and the loss in strength is made up by staying.

In the consideration of the strength of cylinders it is usual to divide the rupturing strains into two classes ; those which tend to rupture the cylinder longitudinally and those which tend to rupture it circumferentially or transversely.

Let us examine them separately. The tendency to cause longitudinal rupture or to rend the cylinder in lines parallel with the axis, may be considered as the pressure exerted on a semicircumference, and tending to rupture the cylinder in a plane through the diameter. Since pressure acts equally in all directions, the whole amount exerted on a semi-circumference is not exerted directly upwards and downwards. But all these forces may be resolved into their vertical and horizontal components. If we take the plane as horizontal, it is evident that the horizontal components have no tensional effect at the points of rupture. By taking the vertical components at an infinite number of points it can be proved that their sum is equal to the full pressure exerted on a rectangular plane equal to the projection of the cylindrical surface. In this case the projection is the plane through the diameter and has an area equal to the product of the length of the cylinder multiplied by the diameter of the cylinder. Then the force tending to rupture would be the pressure per square inch multiplied by the area. Let p = pressure in pounds per square inch, D = diameter of boiler, t = thickness of plate, L = length of boiler, S = tensile strength, E = efficiency of joint, and f = factor of safety. The force tending to rupture longi

tudinally will be, pLD. The strength of the cylinder to resist this rupturing force is represented by the tensile strength of the material multiplied by the areas of sections of metal. Or expressed algebraically is 2tLS. When rupture is about to take place the rupturing force and the strength are equal, or pDL = 2tLS or pD = 2tS from which p = 2tS and t = (pD)/(2S)which are the formulas for pressure and thickness and for longitudinal strength.

The extra pressure due to increased length is balanced by the increase of metal as is shown by the elimination of the factor L ol the equation.

The tendency to rupture circumferentially is evidently represented sented by the area of the end or (pi*D*D)/4 multiplied by the pressure per square inch. The strength to resist this force is the area ol metal to be ruptured multiplied by the tensile strength or pi*DtS: ((pi*D*d)/4)*pi = pi*DtS Dp = 4tS By comparing these two formulas we see that with the same internal pressure, diameter and thickness of shell, a cylindrical boiler is twice as strong transversely as it is longitudinally, hence the greatest tendency to rupture is along the longitudinal seams.

Therefore, in designing the thickness of shell we use the formula for longitudinal rupture, pD = 2tS or t = or, inserting the factors for efficiency of joint and factor of safety, pD = (2tSE)/f For allowable pressure p = (2tSE)/(fD)For thickness of shell t = (fDp)/(2SE) Now let us find the thickness of the boiler that we are designing. Suppose after testing our material we find that its ultimate tensile strength is 54,000 pounds per square inch. In this case 6 will be sufficiently large for a factor of safety. This factor can be reduced if the efficiency of the joint is large. Let us assume that our joint has an efficiency of 70%. This is merely a supposition because we have not yet constructed the joint ; but we assume a factor in order to find a trial thickness.

Then t = (fDp)/2SE = (6 X 54 x 75)/(2 x 54000 x .7) = .32 or about 5/16 inches.