moment of the second; then in order to find the amount and direction of the greatest principal stress, we require to combine the greatest direct stress due to bending with the greatest shearing stress due to twisting; this is done by the method of the ellipse of stress. At the point H, let p be the intensity of thrust (or tension) due to the bending moment M,, and q the intensity of shearing stress due to the twisting couple M,; then we have Let p, be the greatest intensity of stress (thrust) at the point, then fig. 136 shows the construction required to find its amount; in that figure OL = p, OR' =q, and we have (2) the greatest intensity of shearing stress is represented by MR = √2 + q2: +q2; (3) and the angle made by the greatest stress p, with the axis of the shaft is given by the equation By substituting for p and q the values given in eqn. 1, we have A very important example of this principle is that of a shaft with a crank attached; in this case we have a force applied to the centre of the crank pin, and resisted by the equal and opposite force at the bearing S. If P represent the force, then the moment of the couple is M = P.SP; (7) this couple may be resolved into two couples, one a bending couple The greatest intensity of stress is found by eqn. 5. If instead of p, we put ƒ the resistance to tension or thrust (the smaller), we get which enables us to calculate the diameter required for the shaft. If we put ƒ for the greatest intensity of the shearing stress, we have which also enables us to calculate the diameter required; and the greater of the two, one got from eqn. 11, the other from eqn. 12, is to be adopted. The angle made by the principal stress with the axis of the shaft is given by eqn. 4. 195. A shaft 9 inches diameter and 12 feet long is supported at its two ends, and loaded at the two points which divide its length into three equal parts with 4 tons at each point; a twisting moment of 20 foot-tons is applied to one end of the shaft while the other is held fixed. Find the greatest intensity of the thrust, tension, and shearing stress; and the angle that the line of greatest principal stress makes with the axis of the shaft. At any point between the two loads, the bending moment M, 16 foot-tons 192 inch-tons; the twisting moment M. 240 inch-tons; 2 = = = 7.24 = 1·35+ +2·80 = 3.5 tons per sq. inch 4 Ρ is the greatest value of the intensity of the thrust at the upper point, and of the tension at the lowest point of the skin near the middle of the length of the shaft; the greatest intensity of the shearing stress is 2:15 tons per sq. inch, and it is situated at the points just mentioned. 196. The crank shaft of an engine is 5 in. diameter; the distance from the centre of the bearing to the point opposite the centre of the crank pin, NS in fig. 148, is 12 inches; the half stroke, NP in figure, is 16 inches; and the pressure the applied to the crank pin is 5000 lbs. Find the greatest intensity of thrust, tension, and shearing stress; and angle made by the line of principal stress with the axis of the shaft. = M1 5.1 125 60,000, and M1 = 80,000. P1 100,000(1 +§) = 6530 lbs. per sq. inch, = the greatest intensity of thrust and of tension, at the bearing, the one being at the one side and the other being at the other side of the shaft. The greatest intensity of shearing stress is 51,100,000 = 4080 lbs. per sq. inch. The angle 125 THRUST OR TENSION COMBINED WITH TORSION. Let the shaft shown in fig. 149 be acted upon by a thrust (or tension) P and a pair of twisting couples of moment M; the stress due to P is uniformly distributed, and that due to M is greatest at the skin; the greatest intensity of stress will therefore be at the skin. If under thrust, the length of the shaft is to be so short compared with its diameter, that the bending action need not be taken into account. At the point H we P 29 πε and a shearing stress proceeding as at fig. 136, we have— The greatest intensity of thrust (or tension) is the angle (3) made by p, with the axis of the Fig.143. 197. A shaft 8 inches diameter is subjected to a thrust of 100 tons uniformly distributed over its two ends, and a twisting moment of 30 foot-tons. Find the greatest intensity of thrust and shearing stress, and the angle made by the line of principal stress with the axis of the shaft. the greatest intensity of thrust; the greatest intensity of shearing stress is 3.71 tons per sq. inch; and 198. Find the diameter of a malleable iron shaft capable of bearing a tension of 50 tons, and a twisting couple whose moment is 25 foot-tons; the resistance of the material to tension and shearing being 5 and 4 tons per sq. inch respectively. 600 191 50 15.92 = πη πρ from which we find, 3.53 inches. sity of shearing stress 4 = = we find r, 364 inches; and since this is greater than the former result, it is to be adopted; that is, the diameter required for the shaft is 71⁄2 inches nearly. |