This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1850 Excerpt: ...its own weight, will be 1112x16-8 = 2.335 tons; 8 there are 535 square inches in the bottom, therefore the strain per square inch is or 436 tons. A ton of load to the foot throughout would be 400 tons, the strain at the centre from this would be Again, tubes are generally constructed with their depth equal to about one-sixteenth of their length, in which case W 16 1= 16, and S=--=2 W; o t. e. in such tubes the total strain at the centre, tending to crush the top or tear asunder the bottom, is merely twice the weight of the tube and its equally distributed load. This is extremely convenient for rapid approximation, but if the depth is not one-sixteenth of the length, as, for example, one-fifteenth, then the strain found as above has only to be diminished in the proportion of 16:15. Example.--Required approximately a section of bottom, section of top, and weight of a tube for 150 feet span, to carry 1 ton per foot. Weight of the model 7 tons, weight of 150 feet span of a similar tube twice as large, = 2s x 7 = 56 tons. Total load = 150 + 56 = 206 tons. Total strain at the centre = 2 x 206 = 412 tons. If, therefore, we intend adopting 5 tons per inch as the extreme strain, 412 The section of the bottom will be = 82 square inches. 5 Section of top one-fourth more =102 1 '50 Depth=f = 9 ft, 4in. lb Width at pleasure. Area of the two sides about one-half the area of the bottom. The weight may then be taken out more accurately, and the dimensions calculated more in detail. Again, if we wanted to apply cast-iron to the top of this tube, assuming the weight of the tube, which is not large as compared with the whole weight, to remain unaltered by the change;--the compression at the top being 412 tons, if we assume 15 tons per square inch as perfectly ...