Elementary Lessons With Numerical Examples In Practical Mechanics And Machine Design With An Introduction By John Perry

Elementary lessons with numerical examples in practical mechanics

Elementary Lessons with Numerical Examples in Practical Mechanics and Machine Design

Elementary Lessons with Numerical Examples in Practical Mechanics and MacHine Design; with an Introduction by John Perry

This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1896 edition. Excerpt: ...cent. 8. In the case of certain kinds of shafting it is possible to express M as a multiple of T. Thus M = k T. Rankine has found the following values of k: --For such cases as propeller-shafts k varies from-25 to-5. For ordinary light shafting in mills... k -75 to-1. And for crank-shafts and other heavy For each of the cases above mentioned deduce the constant a in the simple formula for the diameter Note.--The values here found agree with values of the multiplier c given at page 99. 9. Find the two equal stresses (tensile and shear) which combined will produce a resultant tensile stress of 66,000 lb. per sq. inch. 10. A cylindric shaft transmits 30 HP at 150 revolutions per minute, and is subjected to a bending moment equal to the twisting moment; find the diameter of the shaft, if the safe shear stress of the material is 9,000 lb. per sq. inch. ANSWEES. 1. Yes. The resultant stress is 5 6 tons per square inch. 2. 6,057-8 lb. per square inch. 3. 6,428 lb. per square inch LESSON XXIV. OVERHUNG CBANK. CRANK-PINS. LENGTHS OP BEARINGS, ETC. A Very good illustration of combined twisting and bending actions is the case of an overhung crank (Fig. 58). Let P be the total push, or pull, of the connectingrod at right angles to the direction of the crank. Fig. 68. Then T = P x A C, and M = P x B C, the resultants of all the forces acting on the journal and crank-pin being assumed to pass through their centres respectively. Then using our rule for combining twisting and bending moments, we have--T( = PxBC + x B C)2 + (P x AC)"2 = PBC + V/b C2 + A C2.-. Te = P (B C + A B). Hence we calculate the twisting moment for an overhung crank as if the crank were of a length equal to the horizontal distance B C, plus the sloping distance A B....