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methods of applying through stays. One of the most common, especially for land boilers, is to allow the ends of the rod to project through the plates to be stayed, and holding them in place by a nut and copper washer, both inside and outside the plate. Another and still better plan is to rivet 6-in. channel bars across

the head, inside above the tubes, the number of bars depending upon the height of the segment to be stayed.

The channel bars

are drilled to correFIGURE 106.

spond with the holes

that are drilled in the plate to receive the stay rods, which latter are then secured by inside and outside nuts and copper washers.

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These channel bars act as girders and serve to greatly strengthen the head or flat plate. Fig. 107 will serve to illustrate this method.

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Sometimes a combination of channel bar and diagonal crow foot braces is used, as shown by Fig. 108.

A good form of diagonal crow foot stay is obtained by using double crow feet, made of pieces of boiler plate bent as shown by Fig. 109 and riveted to the plate by four rivets. A hole is drilled through the body of the

00000jos OOO crow foot, and a bolt pass

FIGURE 108. ing through this secures the forked end of the stay.

Another method of securing through stays to the heads is shown by Fig. 110 and is applied where too

many stay rods would be required to connect all the points to be stayed. A tee iron is first riveted to the flat plate to be stayed, and two V-shaped forgings are bolted to it as shown. The through stay

is then bolted to the forgings, and FIGURE 109. thus two points in the Aat head

are supported by one stay. It will readily be seen that this method will reduce the number of through stay rods required.

Calculating the Strength of Stayed Surfaces. In calculations for ascertaining the strength of stayed sur

faces, or for finding the number of stays required for any given flat surface in a boiler, the working pressure being known, it must be remembered that each stay is subjected to the pressure on an area bounded by lines drawn midway between it and its neighbors. Therefore the area in square inches, of the surface to

FIGURE 110.

be supported by each stay, equals the square of the pitch or distance in inches between centers of the points of connection of the stays to the flat plate. Thus, suppose the stays in a certain boiler are spaced 8 in. apart, the area sustained by each stay = 8 x 8 = 64 sq. in., or assume the stay bolts in a locomotive fire box to be pitched 42 in. each way, the area supported by each stay bolt 412 x 472 2014 sq. in. Again taking through stay rods, suppose, for example,

the through stays shown in Fig. 107 to be spaced 15 in. horizontally and 14 in. vertically, the area supported by each stay = 15 x 14 = 210 sq. in.

The minimum factor of safety for stays, stay bolts and braces is 8, and this factor should enter into all computations of the strength of stayed surfaces.

The pitch for stays depends upon the thickness of the plate to be supported, and the maximum pressure to be carried.

In computing the total area of the stayed surface it is safe to assume that the flange of the plate, where it is riveted to the shell, sufficiently strengthens the plate for a distance of 2 in. from the shell, also that the tubes act as stays for a space of 2 in. above the top row. Therefore the area of that portion of the flat head or plate bounded by an imaginary line drawn at a distance of 2 in. from the shell and the same distance from the last row of tubes is the area to be stayed. This surface may be in the form of a segment of a circle, as with a horizontal cylindrical boiler, or it may be rectangular in shape, as in the case of a locomotive or other fire box boiler. Other forms of stayed surfaces are often encountered, but in general the rules applicable to segments or rectangular figures will suffice for ascertaining the areas.

The method of finding the area of the segmental portion of the head above the tubes is given in Chapter I, pages 22 and 23, and will not be enlarged upon here, except to add Table 19, which covers a much greater number of segments than Table 1, page 22, does. The diameter of the circle and the rise or height of the segment being known, the area of the segment may be found by the following rule:

Rule. Divide the height of the segment by the

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