shocks. This is due to the fact that the deflection for the same fiber stress for a square bar spring is smaller than that for a round bar spring, the ratio being as 4 to 5. The round bar spring is therefore capable of storing more energy than a square' bar spring for the same stress.

Shocks from Bodies in Motion. The formulas given can be applied, in general, to shocks from bodies in motion. A body of the weight W moving horizontally with the velocity of v feet per second, has a stored-up energy:

This expression may be substituted for Qh in the tables in the equations for unit stresses containing this quantity, and the stresses produced by the energy of the moving body thereby determined.

The formulas in the tables give the maximum value of the stresses, providing the designer with something definite to guide him even in cases where he may be justified in assuming that only a part of the energy of the shock is taken up by the member under stress.

TORSIONAL STRENGTH

- SHAFTING

General Formula for Torsional Strength. — In the following formulas:

P = load applied at end of lever arm R, in pounds;

R = length of lever arm in inches;

Mt torsional or twisting moment in inch-pounds;

=

S permissible working stress in pounds per square inch;
Ip polar moment of inertia of cross-section;

y distance from center of gravity to most remote fiber;

Zp section modulus for torsion, or polar section modulus.

=

(Compare also page 301, "General Formulas for Strength of Materials.")

Polar Moment of Inertia and Polar Section Modulus. - The polar moment of inertia of a surface is the moment of inertia with respect to an axis through the center of gravity, at right angles to the plane of the surface. The polar moment

of inertia equals the sum of two moments of inertia taken with respect to two gravityaxes in the plane of the surface at right angles to each other. Thus, for example, the polar moment of inertia of a circle or a square is equal to two times the moment of inertia with respect to an axis in the plane of the surface through the center of gravity.

The polar section modulus or section modulus of torsion for circular sections equals the polar moment of inertia divided by the distance from the center of gravity to the most remote fiber. This method of obtaining the polar section modulus may also be applied with fair accuracy to sections that are nearly circular. For other cross-sections, the section modulus of torsion does not equal the polar moment of inertia divided by the distance from the center of gravity to the most remote fiber. The accompanying table gives formulas for the polar section modulus for a number of sections, some of which are not circular. In the latter case, the formulas are approximate.

Angle of Torsional Deflection of a Cylindrical Shaft. Let,

L= length of shaft being twisted, in inches;

D= diameter of shaft, in inches;

T = twisting moment, in inch-pounds;

G = torsional modulus of elasticity, generally assumed as 12,000,000 for steel shafting;

Example: - Find the torsional deflection for a shaft 4 inches in diameter and 48 inches long, subjected to a twisting moment of 24,000 inch-pounds.

583.6 X 24,000 X 48

απ

= 0.22 degree, or 13 minutes.

4* X 12,000,000

Strength of Shafting. The twisting strength of a shaft is determined from the formula:

in which T = twisting moment in inch-pounds; P= force acting on the shaft, producing rotation, in pounds; R = length of lever arm of force P, in inches; d= diameter of shaft in inches; S = allowable torsional shearing stress in pounds per square inch; n = number of revolutions per minute; and H.P. horsepower to be transmitted.

=

The allowable stress for ordinary shafting may be assumed as 4000 pounds per square inch for main power-transmitting shafts; 6000 pounds per square inch for lineshafts carrying pulleys; and 8500 pounds per square inch for small, short shafts, countershafts, etc. The horsepower transmitted using these allowable stresses is as follows:

Shafting which is subjected to shocks, sudden starting and stopping, etc., should be given a greater factor of safety than is indicated by the allowable stresses just mentioned.

What would be the diameter of a lineshaft to transmit 10 horsepower?

= 1.53, or, say, I inch.

Example:

What horsepower would a short shaft, 2 inches in diameter, carrying but two pulleys close to the bearings transmit? The shaft makes 300 revolutions per minute.

23 X 300
H.P. =
38

63.

Torsional Deflection of Shafting. The shafting must be proportioned not only so that it has the required strength for transmitting a given power, but so that it cannot be twisted through a greater angle than has been found satisfactory by experience. The allowable twist in degrees should, according to some authorities, not be over 5 minutes or about 0.08 degree per foot length of the shaft.

If G = the torsional modulus of elasticity of the material (= 12,000,000); L = the length of the shaft in feet; a = the angle of torsional deflection; and the other letters denote the same quantities as in the formulas just given for the strength of shafting, then the diameter, as determined from the torsional deflection, will be:

For an angle of deflection equal to 0.08 degree per foot length of the shaft, or a total angle a of 0.08 L,

d = =

4.6 V/H.P.

· 0.29√PR = 4.6

Example:- Find the diameter of a lineshaft to transmit 10 horsepower at 150 revolutions per minute with a torsional deflection not exceeding 0.08 degree per foot of length.

It will be seen, by comparing with the section, "Strength of Shafting," that a larger diameter is required, in this case, to prevent excessive torsional deflection than is required by mere considerations of strength. For short shafts, it is unnecessary to calculate for the angular deflection. It is only in the case of long shafts that this is necessary, and even then only if the torsional deflection would be objectionable.

Linear Deflection of Shafting.

For lineshafting, it is considered good practice to limit the deflection to a maximum of 0.010 inch per foot of length. The maximum distance in feet between bearings, for average conditions, in order to avoid excessive linear deflection, is determined by the formulas:

L=V720 d2 for bare shafts;

L =√140 d2 for shafts carrying pulleys, etc.,

in which d= diameter of shaft in inches; L = maximum distance between bearings in feet. Pulleys should be placed as close to the bearings as possible.

Tables of Horsepower Transmitted by Shafting. -The accompanying table, "Horsepower Transmitted by Shafting made from Medium Steel" gives the relation between the diameter of shaft, revolutions per minute, horsepower transmitted, and maximum distance in feet between bearings. Assume, for example, that it is required to find the diameter of a shaft for transmitting 40 horsepower at a speed of 250 revolutions per minute. The shaft is not subjected to any bending action

except its own weight. From the table, it is found, by locating "40" in the column under 250 revolutions per minute, that the diameter of the shaft required is 2 inches. The maximum permissible distance between the shaft bearings is slightly more than 14 feet. When the exact horsepower cannot be found in the table, it is advisable to take the nearest larger value listed in the table and find the diameter of shafting required to transmit this horsepower. Tables are also given for the horsepower which can be safely transmitted by cold-rolled and turned steel lineshafting. The table for cold-rolled steel shafting is carried up to 5 inches only, because this diameter is the largest which is cold-rolled at the present time. These tables are used by the transmission department of the Jones & Laughlin Steel Co., and are based on the assumption that bearings are placed at intervals of from 8 to 10 feet and all

Between

H.P. H.P. H.P. H.P. H.P. Bearings H.P. H.P. H.P. H.P. H.P. Bearings

pulleys are located as near to the bearings as possible. In these tables, the body part in each gives the number of horsepower to be transmitted. For example, assume that a 3-inch cold-rolled steel lineshaft revolves at a speed of 400 revolutions per minute. Find the power that this shaft can safely transmit. By locating 3 inches in the left-hand column and 400 at the top of the vertical columns, and following the vertical column downward until opposite 3 inches, it is found that under the given conditions 154 horsepower may be safely transmitted.

In general, shafting up to three inches in diameter is almost always made from cold-rolled steel. This shafting is true and straight and needs no turning, but if keyways are cut in the shaft, it must, as a rule, be straightened afterwards, as the cutting of the keyways relieves the tension on the surface of the shaft due to the cold-rolling process. Sizes of shafting from three to five inches in diameter may be either cold-rolled or turned, more frequently the latter, while all larger sizes of shafting must be turned, because cold-rolled shafting is not available in diameters larger than five inches.