Again, when a beam is fixed at one end and loaded at the other, it will only bear of the weight as when supported at both ends and loaded in the middle.

What is the weight requisite to break a deal beam 6 inches broad, 9 inches deep, and projecting 12 feet from the wall?

566 × 6 × 92
12

= 22923 ÷ 4

=

5730.7 lbs.

The same rules apply as well to beams of a cylindrical form, with this exception, that the strength of a round bar (as in the table) is multiplied by the cube of the diameter, in place of the breadth, and square of the depth.

Required the ultimate transverse strength of a solid cylinder of cast iron 12 feet long and 5 inches diameter.

What is the ultimate transverse strength of a hollow shaft of cast iron 12 feet long, 8 inches diameter outside, and containing the same cross sectional area as a solid cylinder 5 inches diameter?

When a beam is fixed at both ends, and loaded in the middle, it will bear one-half more than it will when loose at both ends.

And if a beam is loose at both ends, and the weight laid uniformly along its length, it will bear double; but if fixed at both ends, and the weight laid uniformly along its length, it will bear triple the weight.

PROBLEM II.

RULE.-To find the breadth or depth of beams intended to support a permanent weight.-Multiply the length between the supports, in feet, by the weight to be supported in lbs., and divide the product by one-third of the ultimate strength of an inch bar, (as in the table,) multiplied by the square of the depth; the quotient will be the breadth, or, multiplied by the breadth, the quotient will be the square of the depth, both in inches.

Required the breadth of a cast iron beam 16 feet long, 7 inches deep, and to support a weight of 4 tons in the middle.

What must be the depth of a cast iron beam 3-4 inches broad, 16 feet long, and to bear a permanent weight of four tons in the middle?

When a beam is fixed at both ends, the divisor must be multiplied by 1.5, on account of it being capable of bearing one-half

more.

When a beam is loaded uniformly throughout, and loose at both ends, the divisor must be multiplied by 2, because it will bear double the weight.

If a beam is fast at both ends, and loaded uniformly throughout, the divisor must be multipled by 3, on account that it will bear triple the weight.

Required the breadth of an oak beam 20 feet long, 12 inches deep, made fast at both ends, and to be capable of supporting a weight of 12 tons in the middle.

12 tons 26880 lbs., and

=

26880 × 20 266 × 122 × 1.5

= 9.7 inches.

Again, when a beam is fixed at one end, and loaded at the other, the divisor must be multiplied by 25; because it will only bear one-fourth of the weight.

Required the depth of a beam of ash 6 inches broad, 9 feet projecting from the wall, and to carry a weight of 47 çwt.

5264 × 9

47 cwt. = 5264 lbs., and √379 × 6 × ·25

And when the weight is not placed in the middle of a beam, the effective length must be found as in Problem I.

Required the depth of a deal beam 20 feet long, and to support a weight of 63 cwt. 6 feet from one end.

Beams or shafts exposed to lateral pressure are subject to all the foregoing rules, but in the case of water-wheel shafts, &c., some allowances must be made for wear; then the divisor may be changed from 675 to 600 for cast iron.

Required the diameter of bearings for a water-wheel shaft 12 feet long, to carry a weight of 10 tons in the middle.

10 tons 22400 lbs.,

22400

=

and

4487.65 inches diameter.

And when the weight is equally distributed along its length, the cube root of half the quotient will be the diameter, thus:

Required the diameter of a solid cylinder of cast iron, for the shaft of a crane, to be capable of sustaining a weight of 10 tons;

one end of the shaft to be made fast in the ground, the other to project 6 feet; and the effective leverage of the jib as 12 to 1. 22400 lbs., and

10 tons

=

And 1509 11.47 inches diameter.

The strength of cast iron to wrought iron, in this direction, is as 9 is to 14 nearly; hence, if wrought iron is taken in place of cast iron in the last example, what must be its diameter ?

The strength of bodies to resist torsion, or wrenching asunder, is directly as the cubes of their diameters; or, if square, as the cube of one side; and inversely as the force applied multiplied into the length of the lever.

Hence the rule.-1. Multiply the strength of an inch bar, by experiment, (as in the following table,) by the cube of the diameter, or of one side in inches; and divide by the radius of the wheel, or length of the lever also in inches; and the quotient will be the ultimate strength of the shaft or bar, in lbs. avoirdupois.

2.-Multiply the force applied in pounds by the length of the lever in inches, and divide the product by one-third of the ultimate strength of an inch bar, (as in the table,) and the cube root of the quotient will be the diameter, or side of a square bar in inches; that is, capable of resisting that force permanently.

The following TABLE contains the result of experiments on inch bars, of various metals, in lbs. avoirdupois.

What weight, applied on the end of a 5 feet lever, will wrench asunder a 3 inch round bar of cast iron?

11943 × 33 =5374 lbs. avoirdupois.

60

Required the side of a square bar of wrought iron, capable of resisting the twist of 600 lbs. on the end of a lever 8 feet long.

In the case of revolving shafts for machinery, &c., the strength is directly as the cubes of their diameters, and revolutions, and inversely as the resistance they have to overcome; hence,

From practice, we find that a 40 horse power steam engine, making 25 revolutions per minute, requires a shaft (if made of wrought-iron) to be 8 inches diameter: now, the cube of 8, multiplied by 25, and divided by 40 = 320; which serves as a constant multiplier for all others in the same proportion.

What must be the diameter of a wrought iron shaft for an engine of 65 horse power, making 23 revolutions per minute?

James Glenie, the mathematician, gives 400 as a constant multiplier for cast iron shafts that are intended for first movers in machinery;

200 for second movers; and

100 for shafts connecting smaller machinery, &c.

The velocity of a 30 horse power steam engine is intended to be 19 revolutions per minute. Required the diameter of bearings for the fly-wheel shaft.

Required the diameter of the bearings of shafts, as second movers from a 30 horse engine; their velocity being 36 revolutions per minute.

When shafting is intended to be of wrought iron, use 160 as the multiplier for second movers; and 80 for shafts connecting smaller machinery.

TABLE of the Proportionate Length of Bearings, or Journals for Shafts of various diameters.

Tenacities, Resistances to Compression, and other Properties of the
common Materials of Construction.

Resistance to

Names of Bodies.

Tenacity in lbs. compression Its strength Its extensi- Its stiffness is

per sq. inch.

in lbs. per sq.

is

bility is

in.

It must be understood and also borne in mind, that in estimating
the amount of tensile strain to which a body is subjected, the weight
of the body itself must also be taken into account; for according
to its position so may it approximate to its whole weight, in tend-