The strains which the various stresses produce, as hereinbefore enumerated, can generally be calculated arithmetically and graphically, to check each other; and it is intended in this short chapter to briefly and concisely put before the student a few of the methods in vogue for that purpose.

Assammg that he thoroughly understands everything in the previous chapter, we will first deal with plain cantilevers of rectangular forms. Fig. 948 represents a cantilever 12 inches deep and 3 feet projection, loaded at its end by a given weight of 4 tons. The formula by which the greatest strain is calculated is (wl)/(ld), w representing the weight, l the projection or length, and & the depth. Therefore we have:

(wl)/(ld)= (4 tons x 3 feet)/(1x 12 inches)=12 feet/1 foot =12 tons strain, which can be graphically proved as illustrated in Fig. 949 and the following. "Draw the cantilever to scale as the Fig.ure, join A B diagonally across the member, and set up from C, at right angles to C B, C D, equivalent to 4 tons (1 ton or foot being of the same scale); then from the point D draw D E parallel to A B, cutting C B produced in E, and C E will give the amount of strain which, when reduced according to scale, will equal 12 tons."

As it is well known that the strain is greatest at the point of support, and diminishes to nothing immediately under the weight, the strains at intermediate points will graduate between the two external points of support and application of the load, so that such strains at the different intermediate points can be ascertained according to Fig. 950, the perpendiculars representing the strains at the points from which they are raised.

On Fig. 950 the strain line C E on Fig. 949 is raised up to the perpendicular as C F, and the points F and B are joined, forming a triangle, as F C B, which represents the form of the whole strains on the cantilever, and the intermediate perpendiculars indicate what is required.

Having ascertained these strains, we can calculate the sectional area required to resist them by dividing the amount of strain by the "power of resistance" of the material to be employed; always remembering that the top flange, or part, has to resist the amount of one strain (tension) and the bottom part the same amount of the contrary strain (compression), to one or the other of which every part is subjected, according to circumstances, as hereinbefore explained.

 Wrought iron will resist safely 4 tons compression per superficial inch. " " 5 " tension " Cast iron " 1 1/2 " " " " " 8 " compression " Steel " 6 " " " " " 6 " tension " Oak " (about) 16 cwt. " " " " " 13 " compression " Memel " „ 10 " " " " " " 12 " tension " Fir " " 10 " " " " " " 7 " compression "

Therefore, according to the table, the top flange, if in wrought iron, should be of an area of 2 2/5 superficial inches, to resist 12 tons of tension; and the bottom flange 3 inches in area, to resist the same amount of compression. If fir were used it would require to be 12 inches deep (a fixed depth according to your calculations, which assume it as 12 inches), x 5 inches broad, giving an area of 60 square inches, 24 of which are required (at 10 cwt. per square inch) to resist 12 tons of tensior, while the remaining 36 square inches would be required to resist 12 tons of compression (at 7 cwt. per square inch) on the lower part.

Theoretically the cantilever might be constructed as Fig. 951 or Fig. 952; but this is not done in practice, as a rule.