29. RESISTANCE TO SLIDING OR SHEARING.

The resistance of a body to the action of forces parallel to its cross sections, as the component P, in the preceding Article, is called the resistance to shearing.

Experiment shows that this resistance is proportional to the area of the cross section of the body. The subject is treated mathematically in the following manner :

Denoting the original distance between two cross sections of a prismatic body very near to each other by 7, let us suppose that one of these sections slides in its plane, relatively to the other, supposed to be fixed, by a quantity represented by d, which quantity is the same for all the points of the sliding section. An element w of the sliding section passing into w', will tend, by virtue of the elasticity of the body, to resume its original position; the force exerted upon it on this account is found by experiment (see Bresse's Mécanique Appliquée, p. 25) to be of a part proportional to the elementary area w, and of another part

which depends upon the relative sliding Τ Let f be this force.

Then E' being a coefficient constant for the same elementary fibre, but variable from one fibre to another, we have

The quantity E' is called the coefficient of sliding, or coefficient of shearing.

The force f being opposed to the sliding of an element w of the surface of the sliding section, the resultant R of all the parallel forces applied to the different elements of the whole surface, will be found from the equation

If we suppose E' to be the same for every element of the section,

In practice the shearing force of flexion is usually neglected.

30. Table of the resistance of materials to breaking across, in pounds avoirdupois per square inch (values of S), for different materials.

CHAPTER V.

THE EXTERIOR FORCES.

31. In order to apply the results of the preceding chapter, it is necessary to find an expression for the moment of flexure in terms of the exterior forces. In deducing the moment of the straining force of a beam or other body, we must take into account not only the manner of fixing or supporting the beam, but also the manner in which the straining force acts. For instance, the beam may be,

(1). Fixed at one extremity;

(2). Placed on two supports;
(3). Fixed at both extremities;

(4). Fixed at one extremity; the other resting on a support.

In each of these cases the straining force or load may be wholly suspended at a point, or else distributed regularly or irregularly along the length of the beam, whilst the beam itself may be horizontal or inclined to the horizon.

In determining the coefficient of breakage, it would be necessary to take into consideration the small deflections of the beam in finding the moment of the exterior force; but in calculating the dimensions of beams for practical constructions, the small deflections may always be neglected in finding the moment of the exterior force.

32. PROP. A beam being fixed at one extremity in a horizontal position, and loaded at the other, to find the moment of the exterior force on the section of fixture.

Cor. 1. If the load were uniformly distributed along the beam; then w being the straining load for each unit of length, the moment of the strain in this case is

Cor. 2. If instead of a load wl, uniformly distributed along the beam as in Cor. 1, a load wl acted at the distance 7 from the fixed section, the moment of this load would be

the same as that of the load lw uniformly distributed. Hence, a horizontal beam fixed at one extremity and loaded uniformly along its whole length, can bear the double of the load suspended from the free extremity.

Cor. 3. When a beam is fixed at one extremity in a horizontal position, and loaded with a weight W at the other extremity, and also with a weight uniformly distributed along the beam, the moment of the strain on a point at a distance x from the section of fixture is

w being the weight of a unit of the uniformly distributed load, and 7 the length of the beam.

33. PROP. A beam is supported at the ends A and B (fig. 7), in a horizontal position, and loaded by a weight W at a point C between A and B; to find the moment of the straining force upon the point C.

P

с

Let P and Q be the reactions at A and B produced by W. Then if two weights equal to P and Q were placed at A and B and a support at C, the beam would be subjected to the same straining forces as before, because the forces acting upon it would be the same but applied in an opposite direction. And since P at A balances Q at B about C, the beam would still be subjected to the same straining forces if the extremity A were fixed in a wall. moment of the strain on C in this case would be

Fig. 7.

But the

Now in order to find the value of Q, let us take moments

Cor. As the rectangle A C. CB is greatest when the point C is in the middle of the beam, it follows that the strain is greatest when the weight acts at the centre of the beam. The moment of the strain in this case is

Obs. The experiments of Barlow and others show that a beam when fixed in a wall is stronger than when it is merely supported, in the ratio of 3: 2.