A shaft is a part of a machine or system of machines, and is used to transmit power by virtue of its torsional strength, or resistance to twisting. Shafts are almost always made of metal and are usually circular in cross-section, being sometimes made hollow.

90. Twisting Moment. Let AF, Fig. 54, represent a shaft with four pulleys on it. Suppose that D is the driving pulley and that B, C and E are pulleys from which power is taken off to drive machines. The portions of the shafts between the pulleys are twisted when it is transmitting power; and by the twisting moment at any cross-section of the shaft is meant the algebraic sum of the moments of all the forces acting on the shaft on either side of the section, the moments being taken with respect to the axis of the shaft. Thus, if the forces acting on the shaft (at the pulleys) are P1, P2, P3, and P4 as shown, and if the arms of the forces or radii of the pulleys are a1 a2 a3, and a4 respectively, then the twisting moment at any section in Fig. 54.

* Note. The structural steel handbooks contain extensive tables by means of which the design of columns of steel or cast iron is much facilitated. The difficulties encountered in the use of formulae are well illustrated in this example.

BC is P1 a1,

CD is P1 a1 + P2 a2,

DE is P1 a1 + P2 a2 - P3 a3.

Like bending moments, twisting moments are usually expressed in inch-pounds.

Example. Let a1 , = a2 = a4 = 15 inches, a3 = 30 inches, P1 = 400 pounds, P., = 500 pounds, P3 = 750 pounds, and P4 = 600 pounds.* What is the value of the greatest twisting moment in the shaft?

At any section between the first and second pulleys, the twisting- moment is

400 X 15 = 6,000 inch-pounds; at any section between the second and third it is

400 X 15 + 500 X 15 = 13,500 inch-pounds; and at any section between the third and fourth it is

400 X 15 + 500 X 15 - 750 X 30 = - 9,000 inch-pounds. Hence the greatest value is 13,500 inch-pounds.

91. Torsional Stress. The stresses in a twisted shaft are called "torsional" stresses. The torsional stress on a cross-section of a shaft is a shearing stress, as in the case illustrated by Fig. 55, which represents a flange coupling in a shaft. Were it not for the bolts, one flange would slip over the other when either part of the shaft is turned; but the bolts prevent the slipping. Obviously there is a tendency to shear the bolts off unless they are screwed up very tight; that is, the material of the bolts is subjected to shearing stress.

Just so, at any section of the solid shaft there is a tendency for one part to slip past the other, and to prevent the slipping or shearing of the shaft, there arise shearing stresses at all parts of the cross-section. The shearing stress on the cross-section of a shaft is not a uniform stress, its value per unit-area being zero at the center of the section, and increasing toward the circumference. In circular sections, solid or hollow, the shearing stress per unit-area (unit-stress) varies directly as the distance from the center of the section, provided the elastic limit is not exceeded. Thus, if the shearing unit-stress at the circumference of a section is

* Note. These numbers were so chosen that the moment of P (driving moment) equals the sum of the moments of the other forces. This is always the case in a shaft rotating at constant speed; that is, the power given the shaft equals the power taken off. Fie. 55.

1,000 pounds per square inch, and the diameter of the shaft is 2 inches, then, at ½ inch from the center, the unit-stress is 500 pounds per square inch; and at ¼ inch from the center it is 250 pounds per square inch. In Fig. 55 the arrows indicate the values and the directions of the shearing stresses on very small portions of the cross-section of a shaft there represented.

92. Resisting Moment. By "resisting moment" at a section of a shaft is meant the sum of the moments of the shearing stresses on the cross-section about the axis of the shaft.

Let Ss denote the value of the shearing stress per unit-area (unit-stress) at the outer points of a section of a shaft; d the diameter of the section (outside diameter if the shaft is hollow); and d1 the inside diameter. Then it can be shown that the resisting; moment is:

For a solid section, 0.1963 Ss d3;

For a hollow section, [0.1963 Ss(d4-d14)]/d.

93. Formula for the Strength of a Shaft. As in the case of beams, the resisting moment equals the twisting moment at any section. If T be used to denote twisting moment, then we have the formulas:

For solid circular shafts, 0.1963 Ss d3 = T;

For hollow circular shafts, [0.1963 Ss (d4-d14)] /d= T.

(5)

In any portion of a shaft of constant diameter, the unit-shearing stress Ss is greatest where the twisting moment is greatest. Hence, to compute the greatest unit-shearing stress in a shaft, we first determine the value of the greatest twisting moment, . substitute its value in the first or second equation above, as the case may be, and solve for Ss. It is customary to express T in inch-pounds and the diameter in inches, Ss then being in pounds per square inch.

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 Fig. 54, assuming values of the forces and pulley radii as given in the example of article 90. Suppose 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 Ss X 23 = 6,000; or, Ss = 6,000/ [0.1963x8] = 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 Ss, d, and d1 in the second of the above formulas 15, and solve for T. Thus,

T = [0.1963 x10,000 (164-84)] / 16 = 7,537,920 inch pounds.

## Examples For Practice

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 9.G 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 shearing 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 II denote the amount of horse-power; Ss the shearing working strength in pounds per square inch; d the diameter (outside diameter if the shaft is hollow) in inches; d1 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, H=Ssd3n/321,000;

For hollow shafts, H=Ss(d4-d14n/321,000d

(16)

Examples. 1. What horse-power 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 formulas 16 above, and reduce. Thus,

H=[10,000/(164-84)50] / 321,000 x 16 = 6,000 horse-power(nearly).

2. What size of solid shaft is needed to transmit 6,000 horsepower 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 first of the two formulas 16, and solve for d. Thus,

6,ooo = 10,000 X d3 x 50/321,000; therefore d3=(6,000 X 321,000) / (10,000 X 50) =3,852; 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 horsepower 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 Ss. Thus,

200 = Ss X 43 X 200 / 321,000 or, S = [200 X 321,000] /[ 43 X 200] = 5,010 pounds per square inch.

## Examples For Practice

1. What horse-power can a solid shaft 9.6 inches in diameter safely transmit at 50 revolutions per minute, if its shearing working strength is 10,000 pounds per square inch?

Ans. 1,378 horse-power.

2. What size of solid shaft is required to transmit 500 horsepower 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 horse-power 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.