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STAYING FOR CONSTRUCTION OF A BOILER As large a portion as possible of the shell of a boiler is made cylindrical, for in this form plates can be made sufficiently strong without the aid of stays or braces. But all flat surfaces must be stayed ; not only to prevent rupture, but also to provide against distortion and grooving. The theoretical investigation of the strength of flat surfaces, can be worked out only with higher mathematics. From the formula deduced, the solid end plate would have to be about 2 inches thick for a boiler only 3 feet in diameter with plates inch thick. It is evident that the flat ends if of ordinary thickness must be strengthened by stays or braces. The calculation of stresses in a flat plate, supported by stays, can be calculated only when the supported points are in rows thus dividing the surface into equal squares. Even when the stays are not to be placed in rows forming squares, it is well to make the calculation for a standard.

The equation for finding the area supported by a stay rod is, a^2 =(9*(t^2)*s)/(2p) in which a^2 = the area supported, t = the thickness of the end plate, S = the allowable stress on the area of the rod, and p the working steam pressure. Let us find the area supported in our multitubular boiler. In order to provide for future corrosion we will use a factor of safety of 12 and assume, in the absence of exact knowledge, the ultimate breaking strength of the rod to be 60,000 pounds per square inch. It is usual to make the diameter of the rods one to two inches, so we will make ours inches in diameter with an area of 1.767 square inches. Then the stress per rod is 5,000 X 1.767 = 8,835 pounds.

a^2 = = (9*(1/4)*8.835)/(2*75) = 132.5 square inches.

Then as the rod supports 132.5 square inches and the segment of the steam space is 547.2 square inches, the number of rods will be 5472/1325 = 4.

The same formula will apply in finding the number of short screw stay -bolts of the fire box.

Suppose we wish to use a diagonal or crow foot stay, making an angle of 20° with the shell. If the rod is 1 inch in diameter and the stress is limited to 7,000 pounds, then it will carry a pull . of .7854 X 7,000 = 5497.8 pounds, and since it makes an angle of 20°, the pull perpendicular to the head will be 5497.8 X cos. 20° = 5497.8 X .9397* = 5,166 pounds. If the end is fastened by two rivets or bolts each will carry 2,583 pounds. If each rivet or bolt supports a square with a side equal to a, then 5,166 = 75 a^2 = 5166/75 = 68.9 square inches (nearly).

• NOTE. Taken from a table of cosines.NoteE. Taken from a table of cosines.