385. The work expended upon the deflexion of a beam subjected to the action of pressures applied to its extremities, and to a single intervening point, and also to the action of a system of parallel pressures uniformly distributed over its length.

Let μ represent the aggregate amount of the parallel

pressures distributed over each unit of the length of the beam, and a their common inclination to the perpendicular to the surface; then will uz represent the aggregate of those distributed uniformly over the surface DT, and these will manifestly produce the same effect as though they were collected in the centre of DT. Their moment about the point R is therefore represented by prix cos. a, or by x2 cos. a; and the sum of the moments of the pressures applied to AT is represented by (P,x cos. -μx2 cos. a). Substituting this value of the sum of the moments for P1p, in equation (505), we obtain

386. If the pressures be all perpendicular to the surface of the beam, 1=0, a=0, and I is constant (equation 499); whence we obtain, by integration and reduction,

3

If the pressure P, be applied in the centre of the beam, P1=P2+μa, and a, a, also the whole work U2 of de

3

flecting the beam is equal to 2u,; whence, substituting and reducing,

387. A RECTANGULAR BEAM IS SUPPORTED AT ITS EXTREMITIES BY TWO FIXED SURFACES, AND LOADED IN THE MIDDLE: IT IS REQUIRED TO DETERMINE THE DEFLEXION, THE FRICTION OF THE SURFACES ON WHICH THE EXTREMITIES REST BEING TAKEN INTO ACCOUNT.

It is evident that the work which produces the deflexion of the beam is done upon it partly by the deflecting

1

pressure P, and partly by the friction of the surface of the beam upon the fixed points A and B, over which it moves whilst in the act of deflecting. Representing by the limiting angle of resistance between the surface of the beam and either of the surfaces upon which its extremity rests, the friction Q, or Q, upon either extremity will be represented by P tan. ; and representing by s the length of the curve ca or cb, and by 2a the horizontal distance between the points of support; the space through which the surface of the beam would have moved over each of its points of support, if the point of support had been in the neutral line, is represented by s-a, and therefore the whole work done upon the beam by the friction of each point of support bytan. /Pds.

fpds.

Moreover, D representing the deflexion of the beam under any pressure P, the whole work done by P is represented by

PaD. Substituting, therefore, for the work expended upon

the elastic forces opposed to the deflexion of the beam its value from equation (588), and observing that the directions of the resistances at A and B are inclined to the normals at those points at angles equal to the limiting angle of resistance, we have

SPdD+tan. of Pds=

P2 {a3+(a2 — c2 tan. 24)}}

16Ebc3

ds

But PdD=PapdP; and Pds = PdP=

αξ

2dP

30E-I-SP2¿P by equation (521).

Substituting these values in the above equation, and differentiating in respect to P, we have

̧dD P{a3+(a2— c2 tan. 2)}}

Pdp=

8 Ebc3

Dividing by P, and integrating in respect to P,

388. THE SOLID OF THE STRONGEST FORM WITH A GIVEN

QUANTITY OF MATERIAL.

The strongest form which can be given to a solid body in the formation of which a given quantity of material is to be used, and to which the strain is to be applied under given circumstances, is that form which renders it equally liable to rupture at every point. So that when, by increasing the strain to its utmost limit, the solid is brought into the state bordering upon rupture at one point, it may be in the state bordering upon rupture at every other point. For let it be supposed to be constructed of any other form, so that its rupture may be about to take place at one point when it is not about to take place at another point, then may a portion of the material evidently be removed from the first point without placing the solid there in the state bordering upon rup

ture, and added at the second point, so as to take it out of the state bordering upon rupture at that point; and thus the solid being no longer in the state bordering upon rupture at any point, may be made to bear a strain greater than that which was before upon the point of breaking it, and will have been rendered stronger than it was before. The first form was not therefore the strongest form of which it could have been constructed with the given quantity of material; nor is any form the strongest which does not satisfy the condition of an equal liability to rupture at every point.

The solid, constructed of the strongest form, with a given quantity of a given material, so as to be of a given strength under a given strain, is evidently that which can be constructed, of the same strength, with the least material; so that the strongest form is also the form of the greatest economy of material.

RUPTURE.

389. The rupture of a bar of wood or metal may take place either by a strain or tension in the direction of its length, to which is opposed its TENACITY; or by a thrust or compressing force in the direction of its length, to which is opposed its resistance to COMPRESSION; or each of these forces of resistance may oppose themselves to its rupture transversely, the one being called into operation on one side of it, and the other on the other side, as in the case of a TRANSVERSE STRAIN.

TENACITY.

390. The tenacities of different materials as they have been determined by the best authorities, and by the mean results of numerous experiments, will be found stated in a table at the end of this volume. The unit of tenacity is that opposed to the tearing asunder of a bar one square inch in section, and is estimated in pounds. It is evident that the

tenacity of a fascile of n such bars placed side by side, or of a single bar n square inches in section, would be equal to n such units, or to n times the tenacity of one bar.

To find, therefore, the tenacity of a bar of any material in pounds, multiply the number of square inches in its section by its tenacity per square inch, as shown by the table.

391. A BAR, CORD, OR CHAIN IS SUSPENDED VERTICALLY, CAR

RYING A WEIGHT AT ITS EXTREMITY: TO DETERMINE THE CONDITIONS OF ITS RUPTURE.

First. Let the bar be conceived to have a uniform section represented in square inches by K; let its length in inches be L, the weight of each cubic inch μ, the weight suspended from its extremity W, the tenacity of its material per square inch; and let it be supposed capable of bearing m times the strain to which it is subjected. The weight of the bar will then be represented by μLK, and the strain upon its highest section by LK + W. Now the strain on this section is evidently greater than that on any other; it is therefore at this section that the rupture will take place. But the resistance opposed to its rupture is represented by Kr; whence it follows (since this resistance is m times the strain) that

By which equation is determined the uniform section K of a bar, cord, or chain, so that being of a given length it may be capable of bearing a strain m times greater than that to which it is actually subjected when suspended vertically. The weight W1 of the bar is represented by the formula KLμ,