Obs. The formulæ for steam boilers will be given on more elementary principles in a Note at the end.

Examples.

(1.) A force of 560 lbs. acts on the piston of the forcing pump of a Bramah press, communicating pressure to a cylinder of 6 inches radius; find the thickness of metal (cast-iron) of the cylinder (allowing 3 of an inch for safety) in order that it may sustain the pressure without injury, the diameter of the piston being half an inch, and the coefficient of resistance of the metal 16,500 lbs.

First, to find the pressure p on a unit of surface of the piston, we have

(2.) Find what pressure per square inch would produce rupture in a cylindrical steam boiler one foot in diameter, the thickness of the sheet iron being of an inch, and the coefficient of resistance of the metal 51,000 lbs. Ans. 3,400 lbs.

E

(3.) Determine the thickness of the cylinder of a high pressure engine of 31 inches bore, whose surface is exposed to a pressure of 100 lbs. per square inch, allowing of an inch of thickness for casting, &c., and taking the coefficient of safety of the coefficient of ultimate strength, which is 34,000.

Ans. 99 of an inch nearly.

(4.) In the last example find the total pressure on the section of the cylinder passing through its axis, the height being 2 feet. Ans. 74,400 lbs.

(5.) A force of 500 lbs. acts on a piston of a forcing pump which communicates pressure to a hollow sphere of 10 inches radius; find the thickness of metal of the sphere in order that it may sustain the pressure without injury, the radius of the piston being of an inch, and the coefficient of safety of the metal 16,500 lbs. Ans. 2.7 inches.

(6.) In the last example find the thickness of metal on the supposition that the pressure is communicated by the piston to the interior surface of a cylinder of 10 inches radius.

Ans. 6.1 inches.

(7.) In the last example what additional thickness of metal must be added to the cylinder in order that it may sustain an additional pressure of 100 lbs. on the piston?

Ans. 1.3 of an inch.

CHAPTER IV.

RESISTANCE TO FLEXURE,

Section 1.

CENTRE OF GRAVITY OF CROSS SECTION IN NEUTRAL AXIS.

19. It is found by experiment that a solid body under the action of a straining force not in the direction of its axis, assumes a certain curvature, so that the fibres of the body on the convex part are elongated or extended, whilst those on the concave part are compressed or shortened. There must then exist at the separation of the elongated and compressed fibres a layer of invariable fibres which are neither lengthened nor shortened. The part of the solid which contains the invariable fibres is called the neutral surface of the solid, and the line of intersection of this surface with a given cross section of the body is called the neutral axis of that section.

If we conceive the body to be divided into two symmetrical portions by a plane applied in the direction of the length of the solid, the intersection of this plane with the neutral surface is called the line or curve of invariable fibres, and the point in which the line of invariable fibres meets the neutral axis is called the neutral point.

If the straining force act at right angles to the axis of the solid before flexure, this force will remain sensibly perpendicular to the same after the curvature takes place, provided such force be kept within the limits of safety.

20. PROP. When a prismatic body is acted upon by a straining force which remains sensibly perpendicular to its axis; then, if the resistances to extension and compression be proportional to the forces of extension and compression, the neutral axis

of any given section of the prism will pass through the centre of

gravity of that section.

H

M

f F

G

L

K

E

e

Fig. 3.

For let EF, KH, (fig. 3,) be two cross sections, indefinitely near to each other, of a prismatic body fixed in a wall A B, and bent by a straining force P, which remains sensibly perpendicular to the axis of the prism. Produce FE, HK, to meet in O, and let GL be a portion of the line of neutral fibres intercepted between EF and HK.

Then because EF and HK are supposed to be indefinitely near to each other, OG or OL will denote the radius of curvature of the line of neutral fibres at the point G.*

Equilibrium being established between the straining force P and the interior forces brought into action in the section HK, the components of the forces in a direction perpendicular to that section will involve the interior forces only, as P is supposed to act in a direction parallel to the section. Now the elongations and compressions of the fibres of the body are evidently proportional to their distances above and below the neutral surface, and therefore, within the limits of elasticity, the tensions and pressures of the different points of the cross section HG are

*The circle of curvature is defined in this way :

Let a c, cb, (fig. 2,) be two consecutive elements of a curve, and m O, n O, two normals drawn from the middle points of a c, cb, and meeting each other in O. Then O is evidently the centre of a circle passing through the points a, b, and c. When these elements, moreover, are diminished indefinitely, the circle passing through the points a, b, and c, is then confounded with the curve, and is called the osculating circle, or circle of curvature, of the curve.

The osculating circle of a curve may hence be regarded, in the language of the differential calculus, as passing through three points in the curve indefinitely near to each other.

proportional to their distances from the neutral axis of that section. Wherefore the sum of the forces such as p, which tend to shorten the elongated fibres, will be equal to the sum of the forces such as q, which tend to elongate the compressed fibres. Hence if we regard the forces p as positive, and the forces q as negative, we shall have the condition

Σ (p) = 0 ;

the symbol comprehending all the points of the section. Put GL = s, and draw eLf parallel to HK, meeting the extreme fibre of the body in f, and the direction of p in n. Denote GM the distance of p from GL by x, and the radius of curvature at G by r; then by the similar triangles GLO and Ln N, we have

But before flexure, all the fibres comprehended between EF and HK, had for common length GL = s; whence it follows that the fibre in the direction MN has been elongated by Nn, and therefore the force p, to which this elongation is due, has for its value (Art. 6),

a denoting the area of the section of the fibre, and E the modulus of elasticity.

For a second fibre we should have in like manner,