The reader will find the cases of the deflection of beams

developed from the above fundamental formula in Mr. Moseley's Mechanical Principles of Engineering.' The following examples are all that our limits permit :

Ex. 498. If a hollow cylinder the radii of whose section are r1 and r be supported horizontally at two points whose distance is a; show that when it sustains a weight w at its middle point, the radius of curvature of the neutral line at a point distant & from the middle is given by the formula

and the deflection at the middle point by the formula

Ex. 499. If in the last Example the cylinder sustains throughout its length a uniform pressure of w lbs. per inch, then

Ex. 500. If an iron girder* has a section of the form shown in the annexed diagram, of the following dimensions, AE=C1, AB=b1, CF=C, CD=b, the lower end GH being of the same dimensions as FIG. 135. the upper, show that when this girder sustains a uniform force throughout the whole of its length the deflection at the B CF

A

G

D

E

H

Ex. 501. If there are two beams containing the same amount of materials, of the same length and the same depth, and sustaining the same weight, the one has a rectangular section, the other a section of the form shown in the last Example; given that b = 4 in., c=1 in., b1=1 in.,

* In practice the lower flange is commonly made much larger than the upper, since cast iron is much stronger in resisting a crushing pressure than a strain, and of course the greatest economy of materials is effected when the pressure that would tear the lower flange would also crush the upper. To discuss this question would lead us beyond our present limits. See Mr. Moseley's Mechanical Principles, p. 556; Mr. Rankine's Applied Mechanics, p. 319; see also Mr. Fairbairn's Useful Information, Append. I.

c1=4 in., show that the deflection of the rectangular beam will be of the deflection of the other beam.

99. Rupture of a Rectangular Beam. Referring to Prop. 21, it has been remarked that the curvature of the beam becomes progressively greater in going from L1 to L, consequently the extension of the substance is greatest at B, and when rupture occurs it must result from the extension of the substance at

FIG. 136.

D

B being greater than it can bear. Let us suppose that a pressure of s lbs. per square inch will produce just that degree of extension at which rupture ensues, and let us examine the state of

a small portion of the beam at BC, the natural length of which is l; construct a figure similar to that in Prop. 21, and use the same notation; suppose BL to be divided into a number of parts each equal to 8z; now, as the lamina at BK is on the point of breaking, it must be stretched by a pressure of s lbs. per square inch, and if its

extension is dl we shall have dl=

sl

Ecdz

If we consider the extension d'l of any other lamina m n, whose distance Ln from L equals z, we shall have

But the contractile force of this lamina (q) is given by the equation

and the expansive force of any lamina between Land c will be given by the same formula. Now, the moment of Pround L must equal the sum of the moments of the contractile forces of the lamine between B and L and

1

those of the expansive forces of the laminæ between L and c; these moments are respectively pa, Tscb2 and Ts cb2 (by Ex. 471), and therefore the force P producing rupture is given by the equation

The coefficient s is termed the modulus of rupture; it is not the same as the tenacity of the substance, but is closely related to it. The following table * gives the value of s for certain substances :

Ex. 502. If a horizontal beam, whose weight is neglected, is supported at its extremities and subjected to the action of a vertical force Pat its middle point, it will break (across its middle section) when

Ex. 503. If a horizontal beam is supported at one end, and every inch of its length sustains a pressure w, show that the beam is on the point of breaking when

Ex. 504. If in the last Example the beam had been supported at both ends, show that

* From Mr. Moseley's Mechanical Principles of Engineering, p. 622. For further information on the subject of the text the reader is referred to that work and to Mr. Rankine's Applied Mechanics.

Ex. 505. What load applied at the centre of a beam of oak 20 ft. long, 3 in. deep, and 4 in. wide will be sufficient to produce rupture, its own weight being neglected? Ans. 1003 lbs.

Ex. 506. A wall of brickwork 9 in. thick rests on a beam of oak 6 in. wide, 1 ft. deep, supported on two points 10 ft. apart: under what height of wall would the beam break? Ans. 114 ft.

Ex. 507. A beam of larch 1 ft. square has its end firmly embedded in masonry from which it projects 7 ft.; to what height could a wall of brickwork 2 ft. thick resting on this beam be carried without producing rupture? Ans. 21.8 ft.

Ex. 508. A beam whose weight is w, when supported at the ends in a horizontal position, will just break under a pressure e applied at its middle point; show that

Ex. 509. If a beam A B whose length is a is supported at its ends in a horizontal position and sustains a force of plbs. at a point c such that AC=a1 and BC=a2, and if x is any section at a distance x from B, show that the moment tending to produce rupture round x equals Pxa1 when x is

a

between B and c, and equals P (a-x) as when x is between a and c;

a

show also that the moment tending to produce rupture round c equals PA1 A2

a

Ex. 510. Show that in the last Example the force which acting at c will produce rupture is given by the formula

and that the smallest force that can produce rupture must act at the middle point of the beam.

Ex. 511. Given a cylindrical log of wood, show that the strongest rectangular beam that can be cut out of it is one whose sides are in the proportion of 1: 2.

Ex. 512. A beam of oak is supported in a horizontal position on points 20 ft. apart, it is 3 in. deep and 4 in. wide; determine the weight that can be suspended at a distance of 6 ft. from one point of support without breaking it. What would be the magnitude of the weight if the depth were 4 in. and breadth 3 in.? Ans. (1) 1128.6 lbs. (2) 1504.8 lbs.

Ex. 513. What must be the depth of a beam of Riga fir 4 in. wide, 30 ft. long, that will just sustain a weight of a ton at its middle, taking into account its own weight? Ans. 5 in.

CHAPTER X.

VIRTUAL VELOCITIES-MACHINES IN A STATE OF UNIFORM MOTION TOOTHED WHEELS.

100. Definition of the virtual Velocity and virtual Moment of a Force. - Let P be a force acting at the point a along the line AP, and let

FIG. 137.

B

P

C

it be represented by AC (Art. 25). Suppose P's point of application to be shifted through an indefinitely A small space to B, draw on at right angles to ac or ca produced, and let an be denoted by p, which is commonly reckoned positive when n falls between A and c, and negative when it falls on ca produced, then p is called the virtual velocity of P, and pp its virtual moment.

101. The principle of virtual Velocities. This principle is as follows: If a system of forces in equilibrium act on any machine which receives any small displacement-consistent with the connection of the parts of the machine-the algebraical sum of the virtual moments of the forces will equal zero.

P3

....

If P1, P2, P3..... are the separate forces, and P1, P2, their virtual velocities, the principle is expressed algebraically by the following equation, which is commonly called the equation of virtual velocities :

P1P1+P2P2+P3P3+....=0

It must be remarked that in the definition of Art. 100 the line AB is considered a small quantity of the first order