of tH, if S be the coefficient of resistance of the sheet-iron, the total resistance P is

Hence, as the values of P and P1 must be equal, we have by (4) and (5),

Let

It is usual in practice to increase this thickness by a small quantity, in order to provide for unforeseen contingencies. t' be the additional thickness; then,

13. Mr. J. C. Trautwine, Civil Engineer, has given the following empirical rule for values of t' of last article in reference to cast-iron pipes :—

14. According to Fairbairn, the simple rivetings reduce by one-half the resistance of plate-iron. The value of S, therefore, must be taken one-half its value in case of riveting.

15. PROP. To find the resistance of the base of a cylindrical steam boiler.

r

Let be the interior radius of the cylinder, and t the thickness of metal; then the exterior radius will be r + t. Wherefore, if p be the unit of pressure, as in Art. 12, which the steam exerts

upon the base of the cylinder, the total pressure on the base will be

Again, as the difference between the exterior and interior circular areas of its base is

16. PROP. To find the resistance of a hollow sphere subjected to an internal pressure.

If p be the pressure in pounds avoirdupois per square inch on the interior surface of the sphere, then the normal pressure on an element a of the interior surface will be Ρ α and if a be the distance of this element from a diametral plane of the sphere, the component of pa, estimated perpendicularly to this diametral plane, will be

r being the radius of the hollow sphere.

Wherefore the sum P of all the pressures corresponding to (1) for all the points of the half surface of the sphere will be

But if x be the

p and r being the same for every point.

distance of the diametral plane in question from the centre

of gravity of the half surface of the sphere, then by the property of the centre of gravity,

2πr2 being the semi-surface of the sphere. Hence, eliminating (ax) from (2) and (3), we get

This is the whole pressure at right angles to a diametral plane of the sphere.

Next, to find the resistance of the hollow sphere, let t be the thickness of metal; then the section of the sphere which resists the pressure P being

*This expression is easily deduced thus: Let y be the radius of a circular slice of the sphere parallel to the diametral plane, and at a distance r from the centre; then the convex surface of this elementary slice is

a = 2πу. ds.

Hence

≥ (ax) = 2 x fx

xy.ds.

But by the equation of the generating circle, viz.,

Integrating this equation betweeen the limits a = r and x =0, we get

if S be the coefficient of resistance, the total resistance will be

This gives for p, the pressure per unit of surface on the interior surface of the hollow sphere, the value

COEFFICIENTS FOR STEAM BOILERS AND WATER PIPES.

17. Fairbairn estimates the ultimate strength of wrought-iron steam boilers at 34,000 lbs. per square inch, and that of castiron water pipes at 16,500 lbs. per square inch. Now, as it is considered prudent in steam boilers to take the coefficient of safety one-eighth, and in cast-iron water pipes one-sixth, of the coefficient of ultimate strength, the following are therefore the respective coefficients for steam boilers and water pipes :—

18. The sides of locomotive fire-boxes and the ends of cylindrical boilers are often made of flat plates connected together by tie bars, called bolts when short, and stays when

long. Each of these bolts or stays sustains a portion of the pressure of the steam against the plates,—which pressure it is important to estimate. The following example will illustrate the method of calculation in such cases :

PROBLEM. How many bolts must be put at the bottom of a steam cylinder in order to resist a given interior pressure?

Let p be the given pressure per square inch, r the radius of a bolt in inches, and n the number of them; then the pressure sustained by each bolt, R being the radius of the cylinder in inches, will be

If S be the pressure per square inch which the metal of a bolt is capable of resisting with safety, the total resistance of a bolt will be

Hence we get the equation

In order to determine n, the number of bolts, we require another independent condition.

If we divide the circumference of the base of the cylinder into n equal parts, it is clear that each of these will be greater than the diameter of each bolt. Hence, m being an arbitrary quantity greater than 2, let us assume

Eliminating between (1) and (2), we get

This formula and the equation (1) will solve the proposed

problem.