Examples. 1. Compute the value of the greatest shearing unit-stress in the portion of the shaft between the first and second pulleys represented in Eig. 54, assuming values of the forces and pulley radii as given in the example of article 90. Su'ppose also that the shaft is solid, its diameter being 2 inches.
The twisting moment T at any section of the portion between the first and second pulleys is 6,000 inch-pounds, as shown in the example referred to. Hence, substituting in the first of the two formulas 15 above, we have 0.1963 X = 6,000; 6,000 or Ss = 0.1963 x 8 = 3,820 pounds per square inch.
This is the value of the unit-stress at the outside portions of all sections between the first and second pulleys.
2. A hollow shaft is circular in cross-section, and its outer and inner diameters are 16 and 8 inches respectively. If the working strength of the material in shear is 10,000 pounds per square inch, what twisting moment can the shaft safely sustain ? The problem requires that we merely substitute the values of d, and in the second of the above formulas 15, and solve for T. Thus, 0.1963 X 10,0 — 00 T = 6 inch-pounds. nch-pounds.
1 1. Compute the greatest value of the shearing unit-stress in the shaft represented in Fig. 54, using the values of the forces and pulley radii given in the example of article 90, the diameter of the shaft being 2 inches.
Ans. 8,595 pounds per square inch 2. A solid shaft is circular in cross-section and is 9M inches in diameter. If the working strength of the material in shear is 10,000 pounds per square inch, how large a twisting moment can the shaft safely sustain? (The area of the cross-section is practically the same as that of the hollow shaft of example 2 preceding.) Ans. 1,736,736 inch-pounds.
94• Formula for the Power Which a Shaft Can Transmit.
The power that a shaft can safely transmit depends on the shear ing working strength of the material of the shaft, on the size of the cross-section, and on the speed at which the shaft rotates.
Let H denote the amount of horse-power; the shearing working strength in pounds per square inch; d the diameter (outside diameter if the shaft is hollow) in inches; d, the inside diameter in inches if the shaft is hollow; and n the number of revolutions of the shaft per minute. Then the relation between
power transmitted, unit-stress, etc., is: For solid shafts, n= Ss ; (to) • For hollow shafts, H = 4) I 321,00?) d j Examples. 1. What horsepower can a hollow shaft 16 inches and 8 inches in diameter safely transmit at 50 revolutions per minute, if the shearing working strength of the material is 10,000 pounds per square inch? We have merely to substitute in the second of the two for mulas 10 above, and reduce. Thus, = 10,000 50 H 000 X 16 = 6,000 horsepower (nearly). • 2. What size of solid shaft is needed to transmit 6,000 horse power at 50 revolutions per minute if the shearing working strength of the material is 10,000 pounds per square inch? • 6,000 X 321,000 therefore = = 10,000 x 50 Or, d =15.68 inches.
(A solid shaft of this diameter contains over 25% more material than the hollow shaft of example 1 preceding. There is therefore considerable saving of material in the hollow shaft.) 3. A solid shaft 4 inches in diameter transmits 200 horse power while rotating at 200 revolutions per minute. What is the greatest shearing unit-stress in the shaft? We have merely to substitute in the first of the equations 16, and solve for Thus, 1. What horsepower can a solid shaft 0.6 inches in diameter safely transmit at 50 revolutions per minute, if its shearing work ing strength is 10,000 pounds per square inch ? Ans. 1,378 horsepower.
2. What size of solid shaft is required to transmit 500 horse power at 150 revolutions per minute, the shearing working strength of the material being 8,000 pounds per square inch.
Ans. 5.1 inches.
3. A hollow shaft whose outer diameter is 14 and inner 6.7 inches transmits 5,000 horsepower at 60 revolutions per minute. What is the value of the greatest shearing unit-stress in the shaft? Ans. 10,273 pounds per square inch.