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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.

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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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diameter of the circle. Then find the decimal opposite this ratio in the column headed "Area." Multiply this area by the square of the diameter. The result is the required area.

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Example. Diameter of circle = 72 in. Height of segment 25 in. 25+ 72.347, which will be found it the column headed "Ratio," and the area opposite this

is .24212. Then .24212 × 72 × 72 = 1,255 sq. in., area of segment.

A boiler is 66 in. in diameter, the working pressure is 100 lbs. per sq. in. The distance from the top row of tubes to the shell is 25 in. Required, the number of diagonal crow foot braces that will be needed to support the heads above the tubes, also the sectional area of each brace. The thickness of the heads is 5% in. and the T.S. 55,000 lbs. per sq. in.

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Assume the head to be sufficiently strengthened by the flange for a distance of 2 in. from the shell, the diameter of the circle of which the segment above the tubes requires to be stayed is reduced by 2 + 2 = 4 in. and will therefore be 66 - 4 62 in. The rise or height of the segment above the tubes is 25-4=21 in. Required, the area.* 2162 = .338. Looking down the column headed "Ratio" in Table 19, area opposite .338 is .23358. Area of segment = .23358 × 62 × 62 = 897.88 sq. in. The total pressure on this area will be 897.88 x 100 = 89,788 lbs.

Assume the braces to be made of 1% in. round steel having a T.S. of 50,000 lbs. per sq. in. and to be designed in such a manner as to allow for loss of material in drilling the rivet holes in the crow feet. Each brace will have a sectional area of .994 sq. in., and using 8 as a factor of safety, the strength or safe holding power of each stay may be found as follows: .994 × 50,000+ 8 = 6,212 lbs., and the number of stays. required = 89,788 lbs. (total pressure) divided by 6,212 lbs. (strength of each stay) = 14.5, or in round numbers 15. If the stays are made of flat bars of steel the sectional area should equal that of the round stays, and the dimensions of the crow feet of all stays should

*See rule for Table 19.

be such as to retain the full sectional area of the body after the rivet holes are drilled.

Each stay is connected to the plate by two %-in. rivets, having a T.S. of 55,000 lbs. per sq. in. and a shearing strength of 45,000 lbs. per sq. in. These rivets are capable of resisting a direct pull of 10,818 lbs., using 5 as a factor of safety; ascertained as follows: 2A x 45,000+ 5 = 10,818= strength of two rivets. They are also subjected to a crushing strain, and the resistance to this is Dx C+5, which expressed in figures is .875 x 90,000+ 5 = 15,750 lbs. ×

The proper spacing comes next, and is arrived at in the following manner:

Area to be stayed 897.88 sq. in.

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Number of stays = 15.

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Area supported by each stay 897.88 15 = 59.8 sq. in.

The square root of 59.8 7.75 nearly, which is the distance in inches each way that the stays should be spaced, center to center.

If through stay rods are used in place of diagonal braces for staying the boiler under consideration, the number and diameter of the rods may be ascertained by the following method:

Assuming the heads to be supported by channel bars, as previously described, and that the stays are pitched 14 in. apart horizontally and 13 in. vertically, each stay would be required to support an area of 14› 13 = 182 sq. in., and the number of stays would be 897.88 182 = 4.9, in round numbers 5. See Fig. 107. The pressure being 100 lbs. per sq. in., the total stress on each stay = 182 × 100 18,200 lbs. Assume the stay rods to be of soft steel having a T.S. of 50,000 lbs. per sq. in., and using a factor of safety of 8, the

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