_Answer._ They are inserted into the holes drilled to receive them, and the ends are allowed to project about a quarter of an inch beyond the tube-sheets. A tapered plug, fig. 46, is then driven into the tube, to expand it so that it will fit the hole. A tool is used called a _tube-expander_, fig. 47, which is what might be called an expanding plug, consisting of a number of sections, _a_, _b_, _c_, _d_, _e_, _f_, _g_, _h_, held together by a spring clasp _s_, which embraces them, as shown in the engraving. This plug when the sections are drawn together is inserted into the mouth of the tube, and the tapered plug _p p_, is then driven into the opening left in the center of the cluster of sections, which are thus expanded. By this means, the ridge _a b c d_ expands the tube at the inner edge of the tube-sheet, forming a ridge or corrugation, as shown at _c c_, fig. 48. At the same time the shoulder _j k l_ on the tool expands the outer edge of the tube somewhat as is shown at _f f_, fig. 48. By repeating this process, and slightly turning the expander each time, the tubes can be made perfectly water-tight. There are other forms of tube-expanders, but the one described, known as Prosser’s expander, is more generally used than any other. In many cases, after the tubes are expanded with the tool described, the outer edge is turned over still more with what is called a _thumb-tool_, fig. 49, probably from its resemblance in form to a man’s thumb. By placing the curved shoulder _a_ on the end _f_, fig. 48, of the tube it is turned over, somewhat in the form shown in the engraving, by repeated blows of a hammer on the end of the tool. Copper ferrules, represented by the black shading, _a a_, are also much used now on the outside of locomotive tubes, and it is said that with them the joints can be kept tight much easier than without. By turning over the outside edge of the tube as shown in fig. 48, it not only protects the copper ferrule, but, as the tubes must act as braces to sustain the pressure of steam in the flat tube-sheets, it gives the joints the requisite strength for resisting such strains. [Illustration: Fig. 48. Scale ¹⁄₄.] [Illustration: Fig. 49.] QUESTION 98. _How can the strain on the cylindrical part of a boiler be calculated?_ _Answer._ BY MULTIPLYING THE DIAMETER IN INCHES BY THE LENGTH IN INCHES AND THE PRODUCT BY THE STEAM-PRESSURE PER SQUARE INCH. Thus for a boiler 48 inches in diameter and 10 feet long with 100 pounds pressure the calculation would be 48 × 120 × 100 = 576,000 lbs. QUESTION 99. _Why do we multiply the diameter, instead of the circumference, by the length, to get the strain on the cylindrical part?_