[COPYRIGHTED.]

U. S. NAVAL INSTITUTE, ANNAPOLIS, MD.

ELASTIC STRENGTH OF GUNS.

By LIEUTENANT J. H. GLENNON, U. S. Navy.

Much has been written on the subject of elastic strength of guns in the endeavor to make the subject perfectly clear, with the result that many books on the subject are filled with formulas. Lately the subject has been illustrated with geometrical diagrams quite as complicated as the formulas themselves, so that experts who examine them will probably find themselves wondering if, after all, they know anything about the question. It is thought that the subject may be presented in an elementary way without the use of many formulas, illustrating by numerical examples.

The basisof the modern theory is that within the elastic limit, that is, the limit at which if the load is removed a metal will assume its original dimensions, the strain (stretch or shortening per unit length) is proportional to the stress (or load per unit area). This is probably only approximately true for some metals. For example, some mild steel stretches less just within the elastic limit for a positive increment of stress than it does lower down. It seems to save up, so to speak, for a greatly disproportionate stretch just above the elastic limit; after this excessive stretch, the amount of strain for an increment of stress becomes nearly as small as below the elastic limit and then increases gradually till the ultimate load is obtained, the specimen breaking at a load somewhat lower than the highest point reached.

The highest stress obtained is the tensile strength and the stress that strains a metal to its elastic limit is the elastic strength of the metal (for extension or compression, as the case may be). A distinction should be carefully observed between the elastic strength of a metal and the elastic strength of a gun built of the

metal, the last being the pressure per square inch which, when applied to the gun internally, will permanently deform it by straining some portion of the metal beyond its elastic limit.

Following the ordinary custom, we will consider the strair. within the elastic limit as proportional to the stress in the same direction, when this is the only stress applied. We will, moreover, only consider metals while acting inside their elastic limits, the various fundamental rules given applying only with this condition. Within the elastic limit, then, if a thousand pound pull is given to a specimen of metal, and then a thousand more is added, the stretch for the second will be the same as for the first. This is equivalent to saying that the strain due to any stress is independent of prior stress. A second rule is that when a stress of tension (or compression) is exerted upon a metal, the strains in directions at right angles are of that in the line of the stress and are contractions (or extensions). The coefficient is sometimes used in place of, but this latter is used almost exclusively by the Army and Navy ordnance bureau officers. If two or more stresses at right angles act upon a specimen, the total strains in each direction will be the algebraic sum of the strains in that direction, each component strain being independent of prior strain and each being readily calculated by the rule for the strains produced by a single

stress.

The ratio of the stress to the strain caused by it in the same direction is called the modulus of elasticity of the metal and will be denoted by E.

It is necessary to use calculus only once in the subject of elastic strength of guns, namely, to get the fundamental formulas for tension and radial pressure.

The stresses acting upon a point in the thickness of a gun may be resolved in the three directions of the length, the thickness, and perpendicular to a radius of the gun and the length. Any point at rest is held by equal and opposite forces or otherwise it would move in the direction of the greater force. So that we may represent the radial pressure on any particle by two opposite and equal arrows pointed towards each other along a radius, and the circumferential tension at any point by two equal and opposite arrows pointed tangentially and away from each other. The longitudinal stress would likewise be represented by equal and opposite arrows in the direction of the length.

We will denote the radial stress by p, the circumferential or tangential stress by and the longitudinal stress by q. Similarly, strains in the same directions will be denoted by [p], [t] and [9.]

Ο

If we denote the pressure inside of a closed cylinder by P, outside by P1, the internal radius by R, and the outside by R,, and suppose a plane surface passed through its axis, removing one half of the cylinder, it will be evident that the total tension of the cylinder at the points of junction with the plane must be 2(P ̧R—P ̧R ̧) multiplied by the length of the cylinder, since this will be the total pressure on the flat surface tending to separate it from the half cylinder. But the presence of the flat surface will in nowise alter the forces causing stress in the cylindrical part, and this is evidently therefore the total tension throughout the two thicknesses of the original cylinder.

By dividing this quantity by the area of the section of the metal, we find the mean tension per unit area. A similar method of finding tension will apply to any cylindrical element, so that we readily find a differential expression and then, by integration, the law of radial pressures and tensions throughout the thickness. This is the occasion in the study of elastic strength of guns where we are compelled to use calculus. Such calculus as is necessary is however very simple, but we will not go into it here, and will consider the results only. They are for the varying radial pressure p at any radius r,

and for the circumferential tension at the same point,

(1)

The three conditions necessary to the deduction of these equations (which are absolutely independent otherwise of the metal) are uniform elasticity (or constant modulus), uniform longitudinal stress, and uniform longitudinal strain throughout the thickness. In them, c, and c1 are constants. The longitudinal tension q is, following Clavarino, equal to c1, and according to Birnie, zero. Birnie's assumption will give lower elastic strength generally and therefore being safer, is taken here. It seems moreover to agree more nearly with the facts.

Summing up the strains in the three directions as before indicated, that is, algebraically, we find for the strain [p] in the direction of the radius (due to and f),

(3)

for the strain [] tangentially or circumferentially (due to p and 7),

3E

C1

(4)

None of the resultant strains of the metal must ever exceed the elastic limit determined by dividing the elastic strength by the modulus of elasticity, or else the gun will be permanently deformed. These five equations are all that are necessary in the subject, if judiciously handled.

Example: :-The internal and external radii of a tube are 3" and 4", the elastic strength of the metal for tension or compression (the two are usually taken equal for gun steel) is 60,000 lbs., and the modulus of elasticity (that for steel) 30,000,000. What is the elastic strength?

The Elastic limit is

60,000 30,000,000

The strain used will always be the greatest numerically since the elastic limits for extension and compression are assumed equal numerically.

Unless c, is different in sign from c2, [] will always (see (3) and (4)), be numerically greater than [p]; [] or [p], one of them, will be greater than [9] numerically. The strain used will always be greatest numerically when r is smallest; that is, at the inner surface (see (3) and (4)).

We will first put the elastic limit equal to [], the strain circumferentially at the inner surface of the tube.

The external pressure is the atmospheric, which we will call o, and the greatest internal pressure that the tube can stand will be its elastic strength.

We have, remembering that the external pressure is o, p=0 when r=4. Hence, by (1),

Substituting in (6) we find, c, = 316,096, and c, = 19,756; and finally, from (1), placing r=3 the internal radius, we have 15,366 lbs., which is the elastic strength of the tube (as [1] is the proper strain to use, being greater numerically than [p] because c, and c, are positive).

To summarize the operation performed, we have assumed the internal strain and external pressure and found the corresponding internal radial stress or pressure.

Now suppose this tube is shrunk on another of radii 2" and 3" (about), the condition being that each tube shall stretch internally to its elastic limit when the maximum pressure acts. The internal pressure of the outside tube will be the external pressure of the inside tube.

Suppose the elastic strength of the metal for tension or com60,000

pression of the inside tube to be 60,000. Then

30,000,000

=.002

is the limit of allowable strain. We place then, by (4),

=

(8)

and, remembering that the pressure or p 15, 366 lbs. when r=3, we have, by (1),

=

From these two equations we determine C2 172,417 and c1 = 3791 for the inside tube, and then the internal pressure or elastic

=

4

strength by (1), whence pc, = 39,313 lbs. In this tube c, and c, are both positive and the [] strain (as we find by trial, solving for c, and c1) is the one to use. If the elastic strengths