. A manual of marine engineering: comprising the design, construction, and working of marine machinery. when the twisting moment is 250,000 inch-pounds. The shaft of iron / = 7500 lbs. Here Tj = 40,000 +^40,000^ + 250,000^ = 40,000 + 10,000^42 -i- 252= 293,170 inch-pounds. »,/ f ^ ^ V 7500 182 MANUAL OF MARINE ENGINEERING. Crank Shafts.—These shafts are subject always to twisting, bending, andshearing stresses; the latter are so small compared with the former thatthey are usually neglected directly, but allowed for indirectly by means ofthe factor f, as already stated. The two principal stress
. A manual of marine engineering: comprising the design, construction, and working of marine machinery. when the twisting moment is 250,000 inch-pounds. The shaft of iron / = 7500 lbs. Here Tj = 40,000 +^40,000^ + 250,000^ = 40,000 + 10,000^42 -i- 252= 293,170 inch-pounds. »,/ f ^ ^ V 7500 182 MANUAL OF MARINE ENGINEERING. Crank Shafts.—These shafts are subject always to twisting, bending, andshearing stresses; the latter are so small compared with the former thatthey are usually neglected directly, but allowed for indirectly by means ofthe factor f, as already stated. The two principal stresses vary throughout the revolution, and the maxi-mum equivalent twisting moment can only be obtained accurately by a seriesof calculations of bending and twisting moments taken at fixed intervals,and from them construct a curve of strains. Curve of Twisting Moments.—The twisting moment at any position of thecrank is equal to the pressure on the piston multiplied by the distance inter-cepted by a line through the connecting-rod on a line at right angles to centreline through centre of Fig. 54. Let A B (fig. 54) be the centre line of the engine through the cylinder andshaft centres, A C the position of the crank, B C the connecting-rod, and A Da line at right angles to A B. Produce B C to cut the line A D, and dropfrom A a line A E perpendicular to B C. P is the load on the piston, and Ris the thrust on the connecting-rod. It will easily be proved that the angleD A E is equal to the angle A B D, called for convenience a. Then P = R cos a; and A E = A D cos twisting moment = RxAE = RxAD cos a P X AD. Let the twisting moment be calculated at equal intervals of say 10° ofangular movement of the crank, so that in the whole revolution there will be36 observations, or 18 in the half revolution. Draw a line A B (fig. 55), anddivide it into 18 equal parts, A a^, a^a^, &c.; erect at these points perpen-diculars, and cut off parts a-p^, ajb^, &c., to represent the val
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