in a wall at one end, and five feet long, it being 4 inches square: what is its deflection?

603 X 32 X 300 6580750 X 4 X 43

1.23 inches the deflection.

If the beam be supported at both ends and loaded in the middle:

length (in inches) x weight

tab. numb. (E, table A) × breadth x depth3

= deflection.

NOTE. When the beam is firmly fixed at both ends, the deflection will be of that given by the rule.

Ex. If a beam of pitch pine, 8 inches broad, 3 inches thick, and thirty feet long, is supported at both ends and loaded in the centre with a weight of 100 lbs.; what is its deflection?

3603 × 100

4900466 × 8 × 33

= 4.407 inches, deflection.

If the beam had been firmly fixed at both ends, the deflection would have been

4.408 X 2 = 2.938 inches.

If the beam had been supported at both ends, and loaded uniformly throughout, the deflection would have been

4.408 × 5

= : 2.754.

To find the ultimate deflection of a beam f timber before it breaks :

length (in inches)"

tab. numb. U (table A) x depth

= ultimate deflection.

What is the ultimate deflection of a beam of ash, 1 foot broad, 8 inches deep, and 40 feet long?

4802

396 × 8

=72.72 inches, the ultimate deflection.

To find the weight under which a column placed vertically will begin to bend, when it supports that weight: tab. numb. E (table A) × least thickness × greatest ⚫2056 length (in inches)2

weight in pounds.-It will be found by the application of this rule, that it will require 40289.22 lbs. to bend a beam of English oak 20 ft. long, 6 in. thick, and 9 in. broad.

BEAMS.

We take the liberty here of introducing a short extract from Messrs. Hann and Dodds' Mechanics, on the subject

of beams.

"In the construction of beams, it is necessary that their form should be such that they will be equally strong throughout. If a beam be fixed at one end, and loaded at the other, and the breadth uniform throughout its length, then, that the beam may be equally strong throughout, its form must be that of a parabola. This form is generally used in the beams of steam engines."

Dr. Young and Mr. Tredgold have considered that it will answer better, in practice, to have some straight-lined figure to include the parabolic form; and the form which they propose is to draw a tangent to the point A of the parabola ACB.

E

D

H

K

M

L

[

G

F

B

Now, if we take CB

To draw a parabola.Let CB represent the length of the beam, and AB the semi-ordinate, or half the base; then, by the property of the parabola, the squares of all ordinates to the same diameter are to one another as their respective abscisses. = 4 feet, and AB 1 foot, we may proceed to apply this property to determine the length of the semi-ordinates corresponding to every foot in the length of the beam, as follow:CB: AB2 :: CF : EF2; that is, 48 12o :: 35: 108 the square root of which is 10.4 nearly And CB AB :: CG: GH2; 48: 122: 24: 72 = GH"; the square root of which is 8.5 nearly CB: AB :: CI : IK2;

48: 122: 12:36

=

EF2;

= EF.

IK2;

the square root of which is 6 inches = IK.

the square root of which is 4.24, which is very near 4 inches LM. Now, if any flexible rod be bent so as just to touch the tops A, E, H, K, M, of the ordinates, and the vertex C, then the form of this rod is a parabola.

To draw a tangent to any point A of a parabola:— From the vertex C of the parabola draw CD perpendicu

ar to CB, and make it equal to AB; then join AD, and the right line AD will be a tangent to the parabola at the point A; that is, it touches the parabola at that point. In the same manner, we may draw a tangent to the parabola at any other point, by erecting a perpendicular at the vertex equal to half the semi-ordinate at that point.

When a beam is regularly diminished towards the points that are least strained, so that all the sections are similar figures, whether it be supported at each end and loaded in the middle, or supported in the middle and, loaded at each end, the outline should be a cubic parabola.

When a beam is supported at both ends, and is of the same breadth throughout, then, if the load be uniformly distributed throughout the length of the beam, the line bounding the compressed side should be a semi-ellipse.

The same form should be made use of for the rails of a wagon-way, where they have to resist the pressure of a load rolling over them.

MODELS. The relation of models to machines, as to strength, deserves the particular attention of the mechanic. A model may be perfectly proportioned in all its parts as a model, yet the machine, if constructed in the same proportion, will not be sufficiently strong in every part; hence, particular attention should be paid to the kind of strain the different parts are exposed to; and from the statements which follow, the proper dimensions of the structure may be determined.

If the strain to draw asunder in the model be 1, and if the structure is 8 times larger than the model; then the stress in the structure will be 83 512. If the structure is 6 times as large as the model, then the stress on the structure will be 63 = 216, and so on; therefore, the structure will be much less firm than the model; and this the more, as the structure is cube times greater than the model. If we wish to determine the greatest size we can make a machine of which we have a model, we have,

The greatest weight which the beam of the model can bear, divided by the weight which it actually sustains = a quotient which, when multiplied by the size of the beam in the model, will give the greatest possible size of the same beam in the structure.

Ex.-If a beam in the model be 7 inches long, and bear a weight of 4 lbs., but is capable of bearing a weight of 26

bs.; what is the greatest length which we can make the corresponding beam in the structure? Here

therefore, 6.5 × 7 = 45.5 inches.

The strength to resist crushing, increases from a model to a structure in proportion to their size, but, as above, the strain increases as the cubes; wherefore, in this case also,. the model will be stronger than the machine, and the greatest size of the structure will be found by employing the square root of the quotient in the last rule, instead of the quotient itself; thus,

If the greatest weight which the column in a model can bear is 3 cwt., and if it actually bears 28 lbs., then, if the column be 18 inches high, we have

wherefore, 3.464 × 18 = 62·352 inches, the length of the column in the structure.

SHAFTS.

THE strength of shafts deserves particular attention; wherefore, instead of incorporating it with the general subject, strength of materials, we have allotted to it a separate chapter under that head.

3

When the weight is in the middle of the shaft, the rule is weight in lbs. × length in feet) diameter in inches. |(weight

500

This is to be understood as the journal of the shaft, the body being usually square.

What is the diameter of a shaft 12 feet long, bearing a weight of 6 cwts., the weight acting at the middle?

If the weight be equally diffused, we have, the weight in lbs.length; extract the cube root and divide by 10; the quotient is the diameter.

Thus, take the last example, then 672 × 12 = 8064; the cube root of which is 20.05, which divided by 10 gives 2.005, the diameter of the shaft.

If a cylindrical shaft have no other weight to sustain be sides its own, the rule is, ✔(007 × length3) = diameter: thus, if a shaft having only the stress of its own weight be 10 feet long;

✔(·007×103) = 2·645 the diameter of the shaft in inches. For a hollow shaft supporting so many times its own weight, we have

012 × length x No. times its own weighty
weight).

1+ inner diameter

outer diameter in inches.

For wrought iron shafts find the diameter by the forego ing rules, which apply to cast iron, then multiply by 935, and for oak shafts the multiplier is 1.83, and for fir 1·716.

Ex. What is the diameter of a cast iron shaft 12 feet long, and the stress it bears being twice its own weight? Here we have,

✓ (012 × 123 × 2) = 6.44 inches. For wrought iron, using the multiplier, 6.44 935 = 6·0215,

and for oak, using the multiplier,

6.44 x 1.83 11.3852,

and for fir, we have

=

6.44 X 1.716 = = 11.05104.

A rule often used in practice, though by no means a correct one, for determining the diameter of shafts is this. The cube root of the weight which the shaft bears taken in cwts. is nearly the diameter of the shaft in inches. It will be found safe in practice, to add one-third more to this result.

If a cast metal shaft has to bear a weight of 11⁄2 ton, that is, 30 cwts., then we have,

330

3.107 inches by this rule; and supposing it 12 feet long, we will apply the other rule, we have,

We have now considered the strength of shafts, so far as regards their power to resist lateral pressure by weight acting on them; we have now to consider their power to resist torsion or twisting.