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Libraiy
ICF
THE ELASTIC STRENGTH OF GUNS
THE
ELASTIC STRENGTH OP GUNS
BY
PHILIP R. ALGER
Professor of Mathematics t {]• S, Navy
THIRD EDITION
1916
BALTIMORE, MO., U. 8. A.
Oopyrigrht. 1904, by
PHILIP R. ALGEB
Profesaor of M«liMin«iloi, U. 8. N.
Copyrifht, 1916, by
MKS. PHILIP R. ALGER
r^
PREFACE
This little book was prepared primarily for use by the midship-
^ men at the U. S. Naval Academy. It essays to present the subject
r\ of the elastic strength of guns as concisely as is consistent with
^ clearness^ and to that end treats only of steel guns of modem con-
struction, built-up or wire-wound.
The hypothesis that permanent set will not occur unless the
resultant strain in some direction exceeds the limit of elastic strain,
regardless of what the stresses may be, is adopted. This hypothesis
appears to the writer to be the only reasonable one, but it is to be
regretted that its truth has never been demonstrated experimentally.
The longitudinal stress is taken to be zero, an assumption made
by Claverino in his first treatise on the "Eesistance of Hollow
Cylinders,^^ published in the " Giomale d^Artiglieria ^' in 1876, and
adopted by Bimie in his exhaustive studies of the resistance and
shrinkages of built-up cannon.
The formulae for wire-wound guns were originally deduced by
the writer some twenty years ago, and were then first published in
the U. 8. Na/vaJ Institute Proceedings,
A number of illustrative examples are solved in the text, and
others, with their answers, follow each chapter.
U. S. Naval Academy,
Depabtment of Mechanics,
November, 1904.
■ *
CONTENTS
CHAPTER I. j.^^
Introductory. — Stress and Strain. — ^Hooks' Law. — The Modulus cf
Elasticity. — The Elastic Limit. — The Principal Strains in a hol-
low cylinder. — ^Examples 1 9
CHAPTER IL
Stress and Strain in Simple Hollow Cylinders. — ^Lamd's Laws. True
Stresses. — The Distribution of True Stresses in Simple Hollow
Cylinders. — Change of Wall Thickness. — ^Examples II 17
CHAPTER III.
The Elastic Strength of Simple Hollow Cylinders under Internal
Pressure. — ^Examples. — Under External Pressure. — Examples. —
Under both Internal and External Pressure. — ^Examples 27
CHAPTER IV.
The Elastic Strength of Compound Cylinders. — Nomenclature. — The
Shrinkage. — ^Formulae for the Case of a Compound Cylinder of
two parts. — Illustrative Example. — Compression of Bore caused
by a given Shrinkage. — ^Examples IV 37
CHAPTER V.
The Elastic Strength of Compound Cylinders, Continued. — Graphic
Illustration of True Stresses at Rest and in Action. — The Princi-
ple of Superposition of Strains and Stresses. — ^The Maximum
Possible Elastic Strength. — The Deduction of the Formulae for
the case of Compound Cylinders of Three Parts. — The method
of Procedure with such Cylinders. — Compression of Bore caused
by given Shrinkages. — Examples V 47
CHAPTER VI.
Applications to Built-up Guns. — ^The Longitudinal Stress. — Propor-
tions of Radii for Maximum Strength. — Conditions affecting the
values assigned for Shrinkages. — ^The Computation of the Elastic
Strength, Shrinkages, and State of Strain at the surfaces of Con-
tact in the case of a 5-inch Gun 58
8 Contents
CHAPTER VII. p^oBs
Wire-Wound Guns. — Comparative Advantages. — Winding with Con-
stant Tension. — Illustrative Example. — Variable Tension such
that the Circumferential Strain is Constant in the State of
Action. — Illustrative Example. — Examples VII 67
CHAPTER VIII.
Gun Design. — General Considerations. — Longitudinal Resistances. —
Gun Projects. — Outline of Gun. — Number of Layers. — Thickness
of Layers 78
CHAPTER IX.
Gun Computations. — Computation Forms. — Adjustment of Shrink-
ages. — Final Computation. — Computation for the Liner. — Ex-
ample of Gun Computations. — Preliminary Computations. — Final
Computations 87
APPENDIX.
The Formulae for Compound Cylinders of four parts 101
THE ELASTIC STRENGTH OF GUNS
CHAPTEB I.
INTEODUCTOEY.
1. Stress and Strain. — We give the name stress to a mutual action
between the parts of a body, or between one body and another, caus-
ing, or tending to cause them to move relative to one another; it is
any pair of equal and opposite actions each of which is what is
called a force.
Thus, if a rope be stretched vertically downwards from A to B,
we speak of the tension T of the rope as the force T acting down-
ward on ^, or as the force T acting upward on B, according as we
are considering A or B; but we speak of the action in the rope,
which tends to break it, as the stress in the rope.
2. We call the change of volume or figure of any solid or liquid
under the action of force a strain.
Thus, if a bar is lengthened or shortened, it is strained ; a com-
pressed liquid is strained; a stone, a piece of metal, or other part
of any structure, is said tg experience a strain if it be bent, or
twisted, or compressed, or dilated, or in any manner distorted.
Furthermore, any change in the configuration of a group of bodies
whose relative positions are subject to fixed conditions is called a
strain. Thus, any structure is said to strain when its different
parte experience relative motion, as, for example, a ship ** strains "
in a seaway.
3. If we imagine any plane area within a strained body as form-
ing a division between the parts of the body on either side of it,
then the force which each of the two parts exerts upon the other is
one of the pair of forces which constitute the stress on the area.
In other words, the stress on any sectional area is the pair of equal
and oppodte actions which hold the area in its state of strain.
4. The intensity of stress is the number of units of force per
unit of area. We shall always express it in tons weight, or pounds
10 The Elastic Strength of Guns
weight, per square inch; and, for brevity, we shall use the word
stress as meaning " intensity of stress,^^ always applying the term
" total stress '^ to the whole force acting on any area. If the inten-
sity of the stress (p) is the same at all points of a given area (A),
the stress on the area is said to be uniformly distributed, and P
p
being the total stress on the area, we have p = t- ^^ ^^^ stress
is not uniformly distributed, its intensity at any point is given
dP
by P = 3-4> where dP is the total stress on the elementary area
dA.
6. Hook's law. — Every stress is accompanied by a strain, and
experiments show that in all solid bodies the strain is proportional
to the stress which causes it, provided the stress does not exceed
certain limits which vary with the material. This is what is known
as Hook's law, — " ut tensio sic vis'* (as the extension so the force).
6. The simplest form of stress is that which exists in a bar of
uniform section to which equal and opposite forces are applied
axially, 4ending to lengthen or shorten it. If the forces act to
lengthen the bar, the stress is called tension, and if they act to
shorten it, the stress is called compression; but mathematically
considered compression is merely negative tension.
The strains accompanying tension are an elongation in the direc-
tion of the pull and a contraction in all directions perpendicular
to it; while the strains accompanying compression are the reverse,
i, e,, a shortening in the direction of the push and an expansion in
all directions perpendicular to it. These strains are elastic, that
is, they disappear with the removal of the forces which caused them,
so long as the tension — or the compression, as the case may be —
does not exceed a value which is called the elastic limit of the mate-
rial. Within that limit the strains follow Hook's law.
7. If P be the total pull (or push) on the bar, and A be the area
of its right section, the total stress on any such section is P, and,
P
since it is uniformly distributed, its intensity is p = -^ . The
elastic limit * is the value of p beyond which the strain ceases to
* Some writers use the term elastic limit to denote the greatest
elastic strain under simple tension or compression, instead of the great-
est stress causing only elastic strains. We shall use the term elastic
limit of strain to distinguish^ the former concept, and shall use elastic
limit to denote the elastic limit of stress.
Introductory 11
be wholly elastic; if this value is exceeded, the bar takes a per-
manent set, t. e., when released it will be found to be longer (or
shorter) than it was originally. With some materials, notably cast
iron, the elastic limit under compression considerably exceeds that
under tension, but in the case of steel the diflference, if it exists, is
not important. The elastic limit of the steel forgings used in
modem gun construction is from 35,000 to 75,000 pounds per
square inch.
8. The Modulus of Elasticity. — ^Within the elastic limit the ratio
of stress to strain is, by Hookas law, a constant, and the value of
this constant for the case of simple tension or compression is called
the modulus of elasticity and is denoted by E. That is to say, if e
p
is the change of length per unit length under the stress p = j-y
tiienE = ^.
e
Since e is the relative, not the total, strain, it is an abstract num-
ber, being, in the case considered, the total change of length of- the
bar (due to its tension or compression) divided by its length when
free. Consequently ^ is a quantity of the same kind as p and its
value depends upon the units in which p is expressed.
When p is given in pounds per square inch, E has the value
29jD00,000 for steel; when p is expressed in tons per square inch,
E has the value 18,000.
Evidently E is the stress which would double the length of a bar
under tension (if it continued to obey Hookas law to that point),
fiinoe when e = l, p=: E.
It must be understood that E is the value of the stress on a right
section of the bar divided by the strain perpendicular to that sec-
tion, or in the direction of the external forces causing the strain;
the strains at right angles to the axis of the bar, though propor-
tional to the principal strain, are less in value, their ratio to it,
determined by experiment, being, in this work, taken to have the
value J.*
9. Example. — ^As an example, suppose a round steel bar, 2 inches
in diameter and 20 inches long, to be under a tension of 60 tons ;
then the stress on a right section of the bar is p = — = 19.1 tons
* This quantity is known as ** Poisson's ratio " from the great French
mathematician. Its value varies for different materials, and for steel
has been taken by different authorities as ^, % and %. The best
modern experiments assign to it a value in the neighborhood of %.
12 The Elastic Strength of Guns
per square inch ; the strain in the direction of the axis of the bar is
•V- = iQQQQ = .00147 ; and the strain at right angles to the axis
00147
is - — g — = .00049. The length of the bar is increased by the
tension 20 X .00147 = .0294 inches, making its strained length.
20.0294 inches; and its diameter is diminished 2 X.00049 = .00098,
making its strained diameter 1.99902 inches.
If the force of 60 tons were applied to compress the same bar, it
would be shortened .0294 inches and its diameter would be increased
.00098 inches.
Under tension the volume of the bar is increased in the ratio
1 to 1.000488; while under compression its volume is diminished
in the same ratio.
10. If more than one pair of equal and opposite forces .act upon
a body, the stress upon any sectional area of the body is the resultant
pf the stresses which would be caused by the pairs of forces acting
separately; and the strain at any point due to the simultaneous
action of all the stresses is obtained by simply superposing the
strains due to the different stresses taken separately.
Thus, taking a rectangular right prism with equal and opposite
forces acting normally upon each pair of its opposite faces, let X, Y
and Z be the forces acting per unit area of the respective faces:
then the stress on each right section perpendicular to the X axis
will be X, the stress on each right section perpendicular to the Y
axis will be Y, and the stress on each right section perpendicular to
the Z axis will be Z. Also, at each point in the prism, the resulting
strains in the directions of the axes will be :
«« = -l(^ 3 3-'
*»~ E\ 3 3 )
1 (y X Y\
(1)
In these expressions e^, «y and e, are the changes of length per
unit length in the directions of the X, Y and Z axes, respectively,
and are plus when they are lengthenings and minus when they are
shortenings, provided the stresses X, Y and Z are given plus signs
when they are tensions and minus signs when they are compres-
sions.
Introductory 13
11. Evidently if either F or Z be of opposite sign to X, the
X
strain in the X direction may be greater than-g, and similarly
the strains in the F or Z directions may either be greater or less
Y Z
than^ ^^^-^ respectively, according as X, Y and Z are unlike or
like forces. If, for example, Z = 15 tons per square inch tension,
and Y and Z are each 15 tons per square inch compression, we have
ex = -^f 15-|--g- + "g- J = 13000 ^^ .001923, and the prism would
lengthen .001923 inches per inch of its free length instead of only
15
-^;= .001154 inches per inch, as would be the case if the stress X
alone acted.
12. In our investigations of the strength of guns we accept the
following principle:
The total strain in any direction due to all the stresses is the
measure of the tendency to yield in that direction, so that the limit
of elastic strength is reached, not when the stress in any direction
equals the elastic limit of the material, but when the strain in any
direction equals the strain which would be caused by the direct
action of a single stress equal to that elastic limit.
If, for example, a steel forging has an elastic limit of 58,000
poimds per square inch, t. e., if 58,000 pounds per square inch is
the greatest simple tensile stress which the steel will withstand
without permanent lengthening, then for the safe use of such a
forging it is necessary, and sufficient, that at no point within it
,11.,.., ,. , 58000 58000 ^^«
shall the stram at any time exceed — -^ — = " oqqqqqoq ^^ -^^^
inches per inch in any direction.
13. At any point in a strained solid there are always three planes,
at right angles to one another, upon each of which the stress is
wholly normal. These three simple stresses (tensions or compres-
sions) are called the principal stresses at the point, and their direc-
tions are called the principal axes of stress.
In the case we are about to investigate — a hollow cylinder under
internal and external fluid pressure — ^the principal axes of stress
are evidently radial, circumferential, and longitudinal (parallel to
the cylinder's axis), and the principal stresses, which we denote by
p, t and q, are illustrated in Figure 1, where one of the elementary
prisms of which we imagine the cylinder to be composed is shown
in equilibrium under their joint action.
14
The Elastic Strength of Guns
The strains in the directions of the principal axes of stress are
called the principal strains; they are simple longitudinal strains
(lengthenings or shortenings), and their relations to the prin-
cipal stresses are those given by equations (1).
14, Since the external pressures with which we are to deal are
compressive forces, it will be convenient to call the radial stress
(p) plus when it acts to compress the material of the cylinder.
Fig. 1.
though continuing to call the circumferential* stress (t) and the
longitudinal stress (g) plus when they produce tension. With this
convention, equations (1) become:
'>p == - -wlf + -s- + -t]
(2)
Tangential stress is synonymous with circumferential stress.
Intkoductory 15
in which et is the strain in the direction of the circumference, ep
the strain in the direction of the radius, and e^ the strain in the
direction of the axis of the cylinder, in each case a plus value indi-
cating extension and a minus value compression.
In the theory of elasticity it is shown that if an ellipsoid be con-
structed with semi-axes representing the principal stresses at a point,
the stress upon any plane at the point is represented In magnitude and
direction by a radius vector of the ellipsoid, which is called the ellip-
soid of stress. Evidently, then, one of the three principal stresses
acting at each point in a strained solid is the greatest stress at the
point In a similar way it is shown that one of the three principal
strains at a point is the greatest strain at the point.
Examples I.
(1) A round steel rod 1 inch in diameter and 6 feet long is
found to stretch .07 inches under a load of 10 tons. What is the
intensity of the stress on its transverse section, and what is the
value of the modulus of elasticity?
12.73 tons per sq. in. ; 13,096 tons.
(2) What length of uniform steel rod, hanging vertically, will
just carry its own weight, if the maximum allowable stress is 8
tons per square inch (steel weighs .283 lb. per cu. in.) ? 6277 ft.
(3) The ends of a steel I beam whose flanges are 8 inches wide
rest on stone supports. If each support takes half the total load of
20 tons, whai should the length of bearing surface be, the safe
compression stress for stone being 300 lbs. per square inch?
9.3 in.
(4) A bar of steel 2 inches in diameter is bent so that its axis
forms the arc of a circle of 372 ft. diameter. What is the greatest
strain at any point of the transverse section, and what is the
greatest stress? (E for steel is 29,000,000 lbs. in.)
.000448 ; 12,992 lbs. per sq. in.
(5) A steel bar, 10 inches long and of square section, 1 inch on
the side when free, is imder 40,000 poimds tension. What are its
dimensions under this stress, which is within the elastic limit?
10.0138 X .99954*,
(6) A copper rod of square cross section, 2 inches on the side,
and 5 feet long, stretches .0375 inches under a load of 40,000
16 The Elastic Strength of Guns
pounds. What is the modulus of elasticity, and what is the cross-
section while the bar is under this stress ?
16,000,000 lbs. in.; 3.9983.
(7) A one-inch square steel bar of 32,000 lbs. elastic limit is
under a tension of 24,000 lbs. ; what pressure per square inch on
all of its sides will cause it to lengthen permanently ? 12,000 lbs.
(8) If a cube be subjected to equal tensions, or compressions, in
each of the three directions normal to its opposite pairs of faces,
what relation must exist between the stress of tension, or compres-
sion, and the elastic limit of the material in order that the cube
may be permanently strained ? p = 3^.
(9) The modulus of elasticity of copper being 16,000,000
(lbs. in.), how much will the length and diameter of a round copper
rod, 20 inches long and 3 inches in diameter when free, change
under a tensile stress of 9000 lbs. per sq. in. ?
.01125 in. ; .00056 in.
(10) In order to bring to the vertical opposite walls which have
fallen away from each other, round steel rods of 1 in. diameter are
stretched from wall to wall and after being heated to 400° C. are
set up taut. What pull will each rod exert when its temperature
has fallen to 200° C, supposing the walls not to have yielded at
all? The coefficient of expansion of steel is .000011 for 1° C.
50,100 lbs,
(11) How much would the steel rod of Example (2), which is
5277 ft. long when free, be increased in length by the stress due to
its own weight? 19.56 in.
CHAPTER II.
STBESS AND STEAIIT DT SDiPLE HOLLOW CYUNDEBS.
15. Consider a horizontal hollow cylinder, open at the ends, which
are faced off in planes normal to the axis ; and let this cylinder be
filled with a fluid which is forced inward by two expanding plungers,
the result being a uniform normal pressure upon the entire internal
surface of the cylinder. Also let the entire outer cylindrical surface
be subjected to a fluid pressure. Then, the ends of the cylinder
being free, and there being no longitudinal stress upon its walls, it
is clear that the cylinder will remain a cylinder under the action of
the pressures, and that each transverse section normal to the axis
will remain a plane normal to the axis. Whatever shortening or
lengthening of the cylinder may result from applying internal and
external fluid pressure to it must be uniform over its whole cross-
section; i. e,, the longitudinal strain must, under the stated con-
ditions, be constant throughout the cylindrical walls.
16. Let be any point (of radius r) within the walls of a
cylinder (Figure 2) whose inner and outer radii are if<, and R^^
and which is subjected to internal and external pressures P© ^^^ Pn
respectively. Also let t, p and q be the circumferential, radial and
longitudinal stresses, and e^, e^ and e^ the circumferential, radial
18 The Elastic Strength of Guns
and longitudinal strains, at the point 0, E being the modulus of elas-
ticity of the material. And let To and T^ be the circumferential
tensions at the inner and outer surfaces, or the values of t when
r = B^ and when r = R^.
In the strained cylinder, the principal stresses at the point are
evidently the radial pressure p, which varies in value from Po at
the inner to P^ at the outer surface ; the circumferential tension t,
which varies from To at the inner to T^ at the outer surface ; and
the longitudinal stress q, which is zero in the particular case con-
sidered but which might be either tension or compression and either
constant or variable. From equations (2), therefore, we obtain as
the values of the principal strains.
K(3)
and since, under the stated conditions, eq is constant,
* t — p = constant = ^• (4)
If the cylinder is cut by a diametral plane, the whole pressure
acting outward upon the section is 2PoRo, and the whole pressure
acting inward upon the section is 2PnRn, so that the total force tend-
ing to burst the cylinder is 2PoRo—2PnRn- This force must be
balanced by the total stress developed in the two sections of the
(R
cylinder walls, each of which is p** tdr,'\ Thus we have
^""tdr^PoRo-PnRn (5)
1R
and, assuming t = f{r), this gives /(r) n" = -^o ^o — -^n ^«>
from which we see that f{r) = — pr ±l constant, so that the value
of < = /' (r) is given by
t = -p-r^l. (6)
♦ It should be noted that this same result, t — p= constant, follows
when q is constant as well as when q is zero.
t We here assume the cylinder to be of unit length.
Stress and Strain in Simple Hollow Cylinders 19
Combining (6) with (4) we obtain
2p+h=-r^ (7)
dp
= -\y'^ ilog(2p+fc)=logi+logA;,
2p'\-Ic
V^+l=^; 2p + k=^ (8)
r^
in which Jc^ is a constant of integration.
Finally, eliminating k from (8) by means of (4),
t + p=-^ (9)
17. Equations (4) and (9) express what are known as Lame's
Laws: *
1. At any point whatever in a cylinder under fluid pressure the
sum of the circumferential tension and the radial pressure varies
inversely as the square^ of the radium,
2. The difference of the circumferential tension and the radial
pressure is the same at all points.
These, then, are the equations which express the relation between
the circumferential tension and the radial pressure at all points
within the cylinder walls :
t—p = i = z—p„ = r,— p„
}(10)
18. Eliminating 2^« between the last parts of equations (10), we
have
and substituting this value of To in the first parts of the same equa-
tions, we have, after combining:
* M—Bl + — I^l—Rl ^ ^ ^
P Bi—R\ + — Ri-K "^ ^ ^
and these equations enable us to determine the values of t and of p
at any point.
* As explained, these laws are only strictly true when the longi-
tudinal stress is constant, or zero.
20 The Elastic Strength of Guns
19. To determine the principal strains at any point, we have only
to substitute in (3) the values of the principal stresses (t and p)
as given in (11) and (12), thus obtaining
'*- "E'La Ri~Rl +"3" lii — R!i r^J ^^^■^
. - 1 [2 P,Rl-P Rl 4 ^MPo-Pn) 1 "1 ....
^" -&L3 Ri — Rl "3" ^^^M "^J ^^
— :rL"3" Ri—Ri J ^^
««
The first two of these equations are the fundamental ones from
which we shall deduce all the formulae used in our -study of the
elastic strength of guns.
The greatest of the three strains given by (13), (14) and (15)
for any point in the cylinder walls must not at any time exceed the
elastic limit of strain of the material of the cylinder. That is,
calling the elastic limit of the material, as determined in a testing
machine, the limiting value for each of the three strains «& Sp and
. e
As Bt and Bp denote the general values of the circumferential and
radial strains (at any radius r), we shall distinguish the values of
the circumferential and radial strains at radius Ro by €t{R^
and ep(Ro), and those at radius i?„ by et(Rn) and «p(-B»),
20. The quantities Eet, Ecp and Eeq, respectively, equal in value
the simple stresses which, acting alone, would cause the strains «<, Bp
and Bq, but these strains are actually caused by the concurrent action
of the two stresses p and t. We shall hereafter designate Eety jEfep
and EBq as the true stresses, circumferential, radial and longitu-
dinal respectively.
21. The distribution of the true stresses throughout the walls of
a simple cylinder under fluid pressure is best shown graphically,
and we will therefore do this for three cases ; first, when the outer
pressure (P^) is zero; second, when the inner pressure (P^) ^s zero;
and third, when both pressures act and P^ is greater than P^. In
each case we assume a cylinder whose outer is three times its inner
radius (P„ = 3 Ro) , so that its walls are a caliber thick.
♦ Note that et is a strain while etXE Is a stress, so that these equa-
tions as they stand represent strain, and may be used to find the stress
by multiplying both sides by E.
--1
Stress and Strain in Simple Hollow Cylinders 21
!• Case L — ^No Exterior Pressure. — Putting Pn= and Rn —
3^0 in (13), (14) and (15), we obtain as the values of the true
stresses :
Eet =
Bep =
Eb^ =
12
12
Po
-¥-)
12
(16)
From these it will be seen that as r increases from Bo to i?„ the
09 its
19 3
circumferential true stress diminishes from-rirP^ to -ttjP.
12
12
11
value midway, where r = 2Bo> ^ciiig-24"-Po^ ^h® radial true stress
Fig. 3.
17 1
diminishes (algebraically it increases) from — y^-Poto — ToP©,
7
its value midway being — "24^0^ while the longitudinal true stress
has the constant value — y^ Po throughout the cylinder wall. Figure 3
illustrates the distribution of the tangential and radial true stresses,
the former on the right and the latter on the left of the section,
the ordinates above the horizontal diameter indicating tensions and
those below it indicating compressions. The figures on the inner,
middle and outer ordinates are the true stresses in tons per square
22
The Elastic Stbength of Guns
inch which would result from an internal pressure of 12 tons per
square inch.
23. Case n. — ^No Interior Pressure. — Putting Po = and Bn »»
3i?o in (13), (14) and (15) we obtain as the values of the true
stresses.
JEet
(17)
From these it will be seen that as r increases from Bo to i?„, the
circumferential true stress diminishes (algebraically it increases)
Fig. 4.
9 11
from — J- Pn to j^ Pny its value midway, where r = 2Bo9 being
9 3 7
— o^Pnl the radial true stress diminishes from + -r- -P* *<> — To-^**'
3
its midway value being g"^»5 while the longitudinal true stress
3
has the constant value +-j-P». Figure 4 illustrates this, the
right-hand curve showing the tangential and the left-hand curve the
radial true stress at each point in the wall thickness, ordinates
Stress and Strain in Simple Hollow Cylinders 23
above the horizontal diameter indicating tensions and those below
it indicating compressions. The figures on the inner^ middle and
outer ordinates are the true stresses in tons per square inch which
would result from an external pressure of 12 tons per square inch.
24. Case m. — ^Exterior Pressure One-half tlie Interior Pressure.
—Putting Pn = Ji'o and iJ„ = 3i?o in (13), (14) and (15), we
obtain as the values of the true stresses :
Bet = +
^(l8#-7)
(18)
From these it will be seen that as r increases from Ro to i?^ the
11 n
circumferential true stress diminishes from gj ^o ^o — "gT, -^o* i^
value midway being — "Im '^ol the radial true stress diminishes
25
24
P«to
9
24
Po» its value
(algebraically it increases) from -
23
midway being — js Pol while the longitudinal true stress has the
constant value + -jj P©. Figure 5 illustrates the distribution of
Fio. 5.
the tangential and radial true stresses, the former on the right and
the latter on the left of the section, the ordinates above the longi-
tudinal diameter indicating tensions and those below it indicating
24 The Elastic Strength of Guns
compressions. The figures on the inner, middle and outer ordinates
are the true stresses in tons per square inch which would result
from an internal pressure of 12 tons per square inch and an exter-
nal pressure of 6 tons per square inch.
25. Comparing Figure 5 with Figures 3 and 4, it will be seen
that the ordinates of the curves in the former are the algebraic sums
of the corresponding ordinates of Figure 3 and half those of Fig-
ure 4; the stresses due to 12 tons internal and 6 tons external pres-
sure acting together are the same as the algebraic sums of the
stresses due to the same pressures acting separately.
26. Since the strains given by equations (13), (14) and (15) are
changes of length per unit length, the change of thickness of the
cylinder wall may be determined in any case by integrating ep dr
between the limits B» and R^. But the change of radius at any
point whose radius is r must be re^, and the difference between the
change of the outer radius {Rn ^(-Bn)) and the change of the
inner radius (i?o h{Rof) must equal the change of thicknesp
Therefore [^ Cp dr = R^ ^(Rn) — Ro et(Ro)y and it will be
found upon trial that this is true of the values of Bp and Bt given
by (13) and (14).
27. The hypothesis made in 15 that there is no longitudinal stress,
is, of course, not true, as a rule, for actual constructions. In the
built-up guns, for example, whose strength we are investigating, one
end of the bore is closed by a breech-block which sustains the in-
ternal pressure and thus causes a total longitudinal stress n R^ Po
which is distributed over the cross-section of one or more of the
cylinders of which the gun is composed. This stress may be taken
account of by assuming that it is uniformly distributed, but, as will
be shown further on, the hypothesis that q is zero accords as well or
better with the facts than any other available one.
Examples II.
(1) Show that in an infinitely thick hollow cylinder (i2,|=o&)
subjected only to internal pressure (P^) the true circumferential
and radial stresses at the inner surface are of equal value but oppo-
site sign. What are their values ? What is the value of the longi-
tudinal stress?
4 4
• s
Stress and Strain in Simple Hollow Cylinders 25
(2) What are the true stresses at the inner surface of an infinitely
thick hollow cylinder subjected only to external pressure (P^) ?
2 2
(3) If the external and internal pressures are equal, what is the
state of stress in the cylinder walls?
2 2
(4) What would be the change of thickness of a hollow cylinder
one diameter thick under internal pressure alone ?
5 PnRr
O ■**o
6 E •
(5) What would be the change of thickness of a hollow cylinder
one diameter thick under external pressure only ?
Pn Rq
(6) A hollow cylinder half a caliber thick is subjected to an
internal pressure of 6 tons per square inch. What is the greatest
true stress resulting and where does it occur? What are the true
stresses at the outer surface ?
I!et(^Ro)= 12 tone per sq. in,
Eet{R^) = 4 ; ^^pCPJ = - H; ^6^ = - If
(7) If the cylinder of Example (6) is only one quarter of a
caliber thick, what are the true stresses at inner and outer surfaces ?
EetCB„) = 17| ; I!e^(Bo) = - Hi 1 „ _
(8) Show that, as the thickness of wall of a cylinder under
internal pressure is made a smaller and smaller fraction of its inner
diameter, the circumferential stress becomes more and more nearly
constant throughout the wall. If the circumferential stress were
constant, what would be the relation between it and the internal
pressure?
^^^'~Ro ^•
(9) A hollow steel tube, radii 3 in. and 6 in., is subjected to an
internal pressure of 13 tons per sq. in. Determine the three prin-
cipal strains at the inner surface. What is the least elastic limit
26 The Elastic Sthength of Guns
of the steel which will permit the application of such a pressure
without permanent set?
e<(i2j =.002; ep{Ro) = —.00166; e^ = —.00022.
26 tons per sq. in.
(10) With the data of Example (9) determine the three princi-
pal strains at the outer surface of the tube.
elR^^ = .00067; e^iR,,) = —.00022; e^ = —.00022.
(11) Show that the change of wall thickness of a cylinder is inde-
pendent of the value of the external pressure in the case where the
outer radius is twice the inner radius.
Change = -?%^.
CHAPTEE III.
THE ELASTIC STSENOTH OF SIMPLE HOLLOW CYIINDEBS.
28. We will denote the elastic limit under tension of the material
of the cylinder by and its elastic limit under compression by p.
In the case of the forged steel used in modem gun construction,
these elastic limits are usually taken to be equal, but with some
materials, notably cast iron, p is considerably greater than $, and
even in the case of steel it is probable that p is always somewhat
greater than $,
In accordance with the principle stated in 12, we consider that
the limit of safety is reached whenever either of the principal strains,
circumferential, radial or longitudinal, attains the value -^ in
extension or the value -^ in compression; in either case we sup-
pose that the strain ceases to be wholly elastic, and though rupture
may not follow, some permanent change of dimensions or distortion
will result.
In order, therefore, to determine the maximum pressure which
a given cylinder will withstand without permanent set, we have
only to equate the greatest strain of extension which results from
the pressure to -w* and the greatest strain of compression to -w-
and the least of the pressures given by solving these two equations
is the greatest pressure which the cylinder can safely be subjected
to. In other words, the limit of the elastic strength of the cylinder
is reached when either the greatest true stress of tension equals the
elastic limit of the material under simple tension, or the greatest
true stress of compression equals the elastic limit of the material
under simple compression.
29. Internal Pressure Only. — ^Putting Pn = in (13) we obtain
This is always plus, showing that the circumferential true stress
is always tension; and its greatest value is when r has its least
28 The Elastic Strength of Guns
value jBo. Hence we find the value of Po which will make the
greatest circumferential true stress equal the elastic limit of the
material by putting Eet = and r = J?© in (19), This gives
Next putting P,:= in (14), we obtain
Be„ = i
^Po Ro (a ^-^n
* - 3(i^ - i?,)
--^) (21)
This is always negative, showing that the radial true stress is
always compression; and its greatest value (numerically) is when
r = Rf,, Hence we find the value of Po which will make the
greatest radial true stress equal the elastic limit of the material
by putting Eep = — p and r = Ro in (21). This gives
-^=3(i2i-4)(^'~^^~)
The determination of the value of the longitudinal true stress is
unnecessary, since it can never exceed, and in all practical cases is
much less than, one or the other of the other two principal true
stresses, the circumferential and the radial.
Now, comparing (20) and (22), since p is always equal to or
greater than 0, and since the denominator of (20) is greater than
the denominator of (22), the value of Po given by (20) will always
be less than the value of Po given by (22). When P^ reaches the
value given by (20), the elastic limit of strain is reached circum-
ferentially, and further increase of P© is inadmissible.
Consequently the maximum internal pressure allowable in the
case of a simple hollow cylinder under no exterior pressure is given
by
in which 6 is the elastic limit of the material under tension.
Evidently equation (20) gives not only the relation between the
maximum allowable internal pressure and the elastic limit of the
The Elastic Strength of Simple Hollow Cylinders 29
material, but equally the relation between any internal pressure
and the greatest resulting true stress (within the elastic limit).
Moreover, by means of (20) the necessary thickness of a cylinder to
safely withstand a given internal pressure is readily determined,
since, solving for i?„, we have Rn = Ro a/ q^ "^ ^ p '
6
*
•«
I«I5 'Qfil 'fl^
2 3 4 S
Fig. 6.
t
Figure 6 shows how the ratio -—- increases with the ratio
e
Rn
Rn
attaining the maximum value f when -w^ = oo, and clearly indi-
cates the small effect upon strength of increasing wall thickness
beyond a caliber.
30. Examples. — (1) What is the limiting value of the internal
pressure which any simple cylinder (regardless of its thickness)
will stand without permanent set, the elastic limit of its material
being Bl f^.
(2) The walls of a 6-inch steel shell are 1.5 in. thick; if the
tensile strength of the steel is 50 tons per sq. in., what powder
pressure will burst the shell? 25 tons per sq. in.
(3) What internal pressure will produce a circumferential elon-
gation of .0015 in the case of a simple steel tube of 3 in. interior
and 6 in. exterior radius? 9.75 tons per sq. in.
(4) What internal pressure will a cast-steel cylinder of 4 in.
internal and 6 in. external radius stand within its elastic limit of
30,000 lbs. per sq. in.? 10,227 lbs. per sq. in.
(5) A nickel-steel cylinder of 7 in. interior radius and 0.5 in.
wall thickness has an elastic limit of 70,000 lbs. per sq. in. What
internal pressure will it withstand? 4713 lbs. per sq. in.
30 The Elastic Strength of Guns
(6) A cylinder of 7 in. interior diameter has walls 3.5 inches
thick. If its elastic limit is 36,000 lbs. per sq. in., what internal
pressure will it stand? How much pressure could it withstand if
its wall thickness were doubled? if trebled?
18,000 ; 22,740 ; 24,550 lbs. per sq. in.
(7) Determine the proper thickness for a cylinder of 6 in.
inner radius which is to stand an internal pressure of 3000 lbs.
per sq. in., the elastic limit of the material being 28,000 lbs. per
sq. in. 0.708 in.
(8) If the radii are 8 in. and 9 in. and the elastic limit is
60,000 lbs. per sq. in., what is the maximum allowable internal
pressure? What would it be if the circumferential stress were
constant throughout the cylinder walls?
6770; 7500 lbs. per sq. in.
(9) What thickness should a cylinder of 4 in. interior radius
have to withstand an internal pressure of 8000 lbs. per sq. in., if
the elastic limit is 40,000 lbs. per sq. in. ? 0.973 in.
(10) What internal pressure will a cylinder of 6 in. interior
radius and 4 in. wall thickness withstand, if the elastic limit is 18
tons per sq. in. ? 7.32 tons per sq. in.
31. External Pressure Only. — Putting P<,= in (13) we obtain
This is always negative, showing that the circumferential true stress
is always compression ; and its greatest value is when r = Ro' Hence
we find the value of Fw which will make the greatest circumfer-
ential true stress equal the elastic limit of the material by putting
JS^i = — p and r = J?<, in (23). This gives
2-r „ Kn
Next, putting Po='0 in (14), we obtain
The Elastic Strength of Simple Hollow Cylinders 31
This is positive when r=:R^ and continues so until r attains the
value Rf,^^, beyond which point it, becomes negative; its greatest
numerical value, however, is when r = B^. Hence, to find the
value of Pn which would make the greatest radial true stress equal
the elastic limit of the material, we would put Eep = $ and r=^Ro
in (25). A comparison of (25) with (23), however, will show that,
for every value of r^ Eet is greater than Eep, so that the elastic
strength of the cylinder depends upon its resistance to circumferen-
tial stress and not upon its resistance to radial stress.*
Consequently the maximum external pressure allowable in the
case of a simple hollow cylinder under no interior pressure is given
by
iZn — -ff(
in which p is the elastic limit of the material under compression.
p
i
/? t 7 **
/ 2 3 4 -5"
-k
Fig. 7.
p
Figure 7 shows the increasia of the ratio — ^ as wall thickness
increases, and clearly indicates how little is gained by going beyond
a thickness of one caliber.
Of course (24) expresses the relation between the external pres-
sure and the greatest resulting true stress within as well as at the
limit of elastic strain.
32. Examples. — (1) What is the limiting value of the external
pressure which any simple hollow cylinder, regardless of its thick-
ness, can withstand without permanent set, the elastic limit of com-
pression being p ? y.
"^Wlth a material like cast Iron, of which the elastic limit of com-
pression greatly exceeds that of tension, the limit of elastic strain rad-
ially (in this ease extension) may in some cases be reached before the
limit of elastic strain circumferentially (in this case compression) is
attained.
32 The Elastic Strength of Guns
(2) What external pressure can a tube of 7.5 in. interior radius
and 1.75 in. thickness of wall withstand, the elastic limit for
compression being 30,000 lbs. per sq. in.? 5139 lbs. per sq. in.
(3) How thick should the tube of Example (2) (22©= 7.5 in.)
be to withstand an external pressure of 10,000 lbs. per sq. in. ?
5.49 in.
(4) The inner and outer radii of a steel tube are 4 in. and 7 in.,
and it is to be subjected to an external pressure of 8.3 tons per
sq. in. What are the circumferential and radial strains at the
inner surface? What is the greatest true stress?
— .001897; +.000632; 24.65 tons per sq. in.
(5) How thick should the walls of a 6-inch shell be to withstand
6 tons per sq. in. external pressure, without passing the elastic limit
of compression of 18 tons per sq. in. ? 1.27 in.
(6) What external pressure will a cylinder of 6 in. interior
radius and 4 in. wall thickness withstand, if the elastic limit is 18
tons per sq. in. ? 5.76 tons per sq. in.
(7) What wall thickness should a cylinder have to withstand
8000 lbs. per sq. in. external pressure, the interior radius being
4 in. and the elastic limit 40,000 lbs. per sq. in.? 1.16 in.
(8) If the interior radius is 8 in., the wall thickness 1 in., and
the elastic limit 60,000 lbs. per sq. in., what is the maximum allow-
able external pressure? What would it be if the circumferential
stress were constant throughout the walls?
6297; 6667 lbs. per sq. in.
33. Both Internal and External Pressure. — ^Which of the true
stresses first reaches the elastic limit depends, in this case, upon
the relation between the two pressures, and we must consider the
three possible cases separately.
Pn>Po
In this case both terms of the value of Esty equation (13), are
negative, showing that the circumferential true stress is always
compression; and its greatest numerical value is when r has its
least value Bo* On the other hand, the two terms of the value of
Eepy equation (14), have opposite signs, showing that the radial true
stress may be either tension or compression, according to which
term preponderates, and also showing that at each point Eet is
The Elastic Strength of Simple Hollow Cylinders 33
greater mimerically than Eep. We therefore obtain an equation
between the values of P© aiid Pn which will make the greatest true
stress resulting from their concurrent action equal the elastic limit
of the material by putting Eet = — p and r = l?© in (13). This
gives
^= 3(7^^=^^!)
Z> 6 P» iS 8(^ Ro)p fc^gr^
Consequently, when Pn exceeds Po, the relation between the inter-
nal pressure and the maximum allowable external pressure is given
by (26), in which p is the elastic limit under compression.
Po>PnhVii PoRl<PnI?n.
In this case the first term of the value of Eei, equation (13),
remains negative, while the second term is positive, so that the cir-
cumferential true stress may be either tension or compression,
according to which term preponderates. But both terms of the
value of Eep, equation (14), are now negative, showing that the
radial true stress is always compression and is numerically greater
than Ee^ at each point; moreover, the maximum numerical value
of Esp is when r has its least value R^. We therefore obtain an
equation between the values of Po and P^ which will make the
greatest true stress resulting from their concurrent action equal
to the elastic limit of the material by putting Eep = — p and r = Ro
in (14). This gives
Pa(4Ri - 2Rl) - 2Pn R]
P =
T> o\^Jin — -n^ ojf^ -f ^-^ /»•**» /OWN
3(Rl - R l)p + 2P^ m
I'M mHq
Consequently, when P^ exceeds P» hut at the same time PoRl
is less than P«i?», the maximum allowable internal pressure is given
by (^'^)> i^ which p is the elastic limit of the material under com-
pression.
Po>PnfindPoRl>PnI^n
In this case both terms of the value of Eet, equation (13), are
positive, showing that the circumferential true stress is always ten-
sion; its greatest value occurs when r =^ RqI and at each point it is
34
The Elastic Strength of Guns
numerically greater than Eep since the two terms which ma«ke up
the latter^fi value are now of different signs. We therefore obtain
an equation between the values of P<, and P» which will make the
greatest true stress resulting from their concurrent action equal the
elastic limit of the material by putting Eet = $ and r = BQ in (13) .
This gives
3(i?i - Mi)
T> _ S(R'n — Jii)ff -\- 6Pn R.
2
n
(28)
Consequently, when Pf, exceeds P« and at the same time Pq Rq
exceeds Pn Rl, the maximum allowable internal pressure is given by
(28), in which 6 is the elastic limit of the material under tension.
Xfi'T'Xp' ' ///)7'
p/
Fia. 8.
Of course equations (26), (27) and (28), each under its appro-
priate conditions, express the relation between the internal pressure,
the external pressure and the greatest resulting true stress within
the elastic limit as well as at that limit.
34. The relation between simultaneous values of Po andP» which
will just bring a given cylinder to the limit of its elastic strength
The Elastic Strength of Simple Hollow Cylinders 35
may be graphically shown by drawing the three straight lines repre-
sented by equations (26), (27) and (28). In Figure 8, values of
Po are represented by the ordinates, and corresponding values of
Pn by the abscissae, and the cases of five different thicknesses of
cylinder wall are illustrated.
Taking (28) first, when P«= 0, the value of P^is ^^^7^} e,
approaching %9 as a limit when the thickness of the cylinder is
indefinitely increased. If Pn is given successively increasing values,
the maximum allowable value of P© also increases, the relation
between them being given by (28) and represented by the full lines
of the diagram. When P» attains the value p^ 6^ P^has the
value %6, regardless of the thickness of the cylinder, and with these
values of the pressures the inner surface of the cylinder is both at
its elastic limit of extension circumferentially and at its elastic
limit of compression radially.
Further increase of P^ is allowable if P» be also increased, but
from the point where P© = %0 the relation between Pq and P« is
given by (27), and represented by the dotted lines of the diagram.
3 8
When Pq reaches the value -^ p, P„ must also equal -q- p, regard-
less of the thickness of the cylinder wall, and no value of P^ will
3
enable P^ to exceed -^ p, nor will any value of Po enable P» to
3
exceed -q- p, without the elastic strength of the cylinder being ex-
ceeded. At this point, where Pq=lP^=i -^ p, the inner surface
of the cylinder is at its elastic limit of compression both radially
and circumferentially.
If now P^be gradually reduced, while P„ is kept as great as
allowable, the relation between the values of Po and Pn will be given
by (26) and is represented by the dash lines of the diagram. When
Po has been reduced to zero, the inner surface of the cylinder being
maintained at the elastic limit of compression circumferentially,
Pn has the value ** ^ ^ p, approaching -^ as a limit when the
thickness of the cylinder wall is indefinitely increased.
It will be observed that the full and dotted lines represent the
36 The Elastic Strength of Guns
relation between P© and Pn when the former is as great as allow-
able ; while the dash lines represent the relation between Po and P»
when the latter is as great as allowable.
36. Examples. — (1) If B« = 25© and Pn = p = 0, what is the
greatest and what the least allowable value of Po ? 17 ^ . 5
5
(2) If Rn = -V- Ro and Pn = p, what are the greatest and
least allowable values of P^? 77 . 41
3
(3) If ^„ = -^ jRo, what value must Pn have in order that Po may
have the value of J^? ^ o ^^ ^ a
-3-<?or-g-^.
5
(4) If -B„ = -J- jRo, what value must P» have in order that Po may
have the value iO? 12^21^
^25~^^^-25-^-
(5) What internal pressure will a cast-steel cylinder of 4 in.
interior and 6 in. exterior radius stand within its elastic limit of
30,000 lbs. per sq. in. if it is under an external pressure of 5000 lbs.
per sq. in. ? 16,363 lbs. per sq. in.
(6) A nickel-steel cylinder of 7 in. interior radius and 1.5 in.
wall thickness has an elastic limit of 70,000 lbs. per sq. in. What
external pressure will it withstand if it is under an internal pressure
of 10,000 lbs. per sq. in. ? 20,190 lbs. per sq. in.
(7) The inner and outer radii of a steel tube are 4 in. and 7 in. ;
what external pressure will enable it to withstand an internal pres-
sure of 20,000 lbs. per sq. in., if the elastic limit of the steel is
36,000 lbs. per sq. in. ? 3390 to 27,630 lbs. per sq. in.
CHAPTEE IV.
THE ELASTIC STBENOTH OF COMPOUND CYIINDEBS.
36. A reference to Figure 3 will show that the outer portions of
a thick simple cylinder play but a small part in resisting internal
pressure. A compound cylinder is one formed by the superposition
of simple cylinders, the object being to utilize to the utmost the
contractile power of the outer parts and thus to increase the resist-
ance to internal pressure beyond what it would be if the entire mass
were in one piece.
If the elementary cylinders are of the same material, or have
equal moduli of elasticity, they must be assembled so that each
exerts an initial pressure upon the one within it. This is accom-
plished by making the interior diameter of each elementary cylinder
(before it is put in place) less than the exterior diameter of the
cylinder upon which it is to be superposed by a certain quantity
which is called the shrinkage, A compound cylinder so assembled
is said to be under initial tension.
If the elementary cylinders are of different materials, and are so
arranged that the modulus of elasticity of each is greater than that
of the one within it, they may be assembled without shrinkage.
Such a cylinder is called a compound cylinder of variable elasticity.
These two principles of variable elasticity and of initial tension
were formerly often employed in combination, the commonest exam-
ples being cast-iron guns with reinforcing hoops of steel, but in
modem gun construction, excepting for certain bronze field pieces,
steel is now used to the exclusion of other metals, and the principle
of initial tension is universally adopted.*
37. In the investigation of the elastic strength of a compound
♦Rodman was to some degree successful in applying the principle
of initial tension to solid guns, the cast iron smooth bore guns known
by his name having been cast hollow and cooled from the interior with
the object of securing compression of the bore and tension of the outer
parts of the finished gun; and the application of essentially the same
process to steel guns, either cast or forged in one piece, has been
shown to be feasible and advantageous.
38 The Elastic Strength of Guns
cylinder, it is necessary to consider its state of strain both wlien the
maximum internal pressure is acting and when the internal pressure
is zero : the first of these two conditions is called the state of action
and the second is called the state of rest.
In the state of action each cylinder except the outer one is sub-
jected to two pressures, one internal and the other external, while
the outer cylinder is subjected to internal pressure only, atmospheric
press\ire being neglected on account of its insignificant value as
compared with the other forces.
In the state of rest the inner cylinder is under external pressure
only, the outer cylinder is under internal pressure only, and each of
the intermediate cylinders is subjected to both an internal and an
external pressure.
38. We adopt the following nomenclature :
Bq and i?i are the inner and outer radii of the innermost or 1st
elementary cylinder, T^^'and R2 of the next, . . . . , E„_i and Rn
of the outermost or nth.
$0 and po, ^1 and p^^, .... On and p» are the elastic limits of the
material of the elementary cylinders in the same order, from the
1st to the nth ; and E is their common modulus of elasticity.
Po> ^1^ .... P«are the radial stresses in the state of action at
the successive surfaces of the elementary cylinders, and Po, P^
. . . . P„ are the radial stresses at the same surfaces in the state
of rest ; they are always plus, excepting that Po^ Pn and Pn, being
only atmospheric pressures, are considered to be zero.*
To, Ti, . . . . T„ are the circumferential stresses in the state of
action, and T^y T^, . . . . 1\ are the circumferential stresses in
the state of rest, at the successive surfaces whose radii are E©,
Ej, . . . . B^'y they are plus when tensions and minus when com-
pressions.
ep{Ro), ep(-Bi), ep{Rn) are the radial strains, and
et{Po), et(Ri), et{Rn) are the circumferential strains
at radii R^,, Ri, .... jB«, in the state of action ; the same symbols
with a dash over each, as ep(Eo), are the corresponding strains in
the state of rest; they are all plus when lengthenings and minus
when shortenings.
* This convention that radial stresses which are compressive shall be
called positive, is explained in 14: it must be remembered, however,
that a radial strain, like all other strains, is called minus when it de-
notes a decrease of length.
The Elastic Strength of Compound Cylinders 39
Since the states of stress and strain on either side of the surface
of contact of two elementary cylinders may be different (must be if
they were assembled with shrinkage), it is necessary to distinguish
between them. A prime mark over any letter or symbol indicates
that it refers to the outer of the two surfaces which are united by
the contact. Thus T^ is the tension at the inner surface of the
second cylinder as distinguished from T^ which is the tension at
the outer surface of the first cylinder; ep^R^') is the radial strain
in the outer of the two surfaces which meet at Bg; ^^t(Ri) and
Eet{Bi) are the circumferential true stresses in the outer and
inner of the two surfaces which meet at R^; and so on. (At Ro
and Rn no prime marks are needed, as there is but one surface at
each.)
Poy Pi, ..../?» are the simultaneous changes in the radial pres-
sures Poy Pi, . . , , Pn resulting from any cause, such, for ex-
ample, as the cessation of the internal pressure Po-
39. Evidently, with any given assemblage of elementary cylinders,
the elastic strength to resist internal pressure will be greatest when
in the state of action each cylinder is strained to its elastic limit.
Moreover, in a compound cylinder so assembled that all the elemen-
tary cylinders reach their elastic limits of strain simultaneously
under the action of the internal pressure P^, that pressure must be
greater than the pressure P^ which acts at the surface of contact of
the two innermost elementary cylinders; and the pressures at the
different surfaces of contact must diminish successively, P^ being
greater than P^, P2 greater than P3, and so on ; for the reason that
each of these pressures is balanced by the contractile force of only
that part of the compound cylinder which is outside of it.
We will first consider a compound cylinder composed of two ele-
mentary cylinders so assembled that each reaches the limit of its
elastic" strength when the internal pressure Po acts.
Then, since the outer cylinder is at its elastic limit of strain
under the sole action of an internal pressure Pi, we have, applying
(20).
_ 3(^ — Sn) . fans
And, since the inner cylinder is at its elastic limit of strain under
the joint action of an internal pressure Po and an external pressure
40 The Elastic Steength of Guns
P^, of which pressures Po is the greater, we have, applying (27)
and (28),
either P,(,) = 3(^-p^' + ^^- M
4^ — 2^
(30)
of which (30) gives the value of Po which will bring the inner
surface to its elastic limit of strain by radial compression, while (31)
gives the value of P© which will bring the inner surface to its elastic
limit of strain by circumferential extension. The least of these two
values of Po is the true value of the maximum allowable internal
pressure, but, since which will be the least depends upon the values
of Pi, Ro and Pj, we have to express both values, and we therefore
distinguish between them as shown.
40. Having ascertained what maximum internal pressure our
assumed compound cylinder will safely withstand, we have next to
determine its condition when the internal pressure is removed, for
no part of it must be overstrained either in the state of action or
in the state of rest.
The state of rest differs from the state of action solely in the ces-
sation of Po ; this must reduce Pj, and consequently the outer cylin-
der, which is subjected to no other pressure than P^, must be under
less strain after the removal of P© than while it acts; the inner
cylinder, however, while under a less external pressure, is no longer
supported by Po and so may be under greater strain in the state of
rest than it was in the state of action. To determine whether this
be so, we must find the value of the external pressure to which the
inner cylinder is subjected after Po has been removed.
Putting r=^Ri in (13), we obtain for the value of the circum-
ferential strain at the outer surface of the inner cylinder (Rn and
Pn becoming R^ and Pj in this case),
.p, 1 r 6P,J?|-fi(4i?l + 2JZ?) "| ,„o^
Also, remembering that the radii of the outer cylinder are fij and
R2, and that it is subjected only to an internal pressure Pj, we
The Elastic Strength of Compound Cylinders 41
obtain for the value of the circumferential strain at the inner surface
of the outer cylinder
^ti^i) - -^ L HBl—iii) J ^^^
These equations, giving the strains caused by the pressures P©
and Pi, will also give the changes of strain resulting from simul-
taneous changes of the pressures (po and p^). But the surfaces
of contact of the elementary cylinders must contract and expand
together, and so the change of circumferential strain at the outer
surface of the inner cylinder must equal that which simultaneously
occurs at the inner surface of the cylinder embracing it. Hence,
substituting po for Po and p^ for P^ in the second numbers of (32)
and (33), and equating them, we obtain the following relation
between simultaneous changes of pressure at r = Ro and r = fij :
6 p, jy, - p, (4ig + 2 Jgf ) p, {2ISi + 4i gi)
s(W^^m - siRi-mi)
Any change of pressure (pn) at the inner surface, where r = Roy
will cause the change of pressure (p^) at the surface of contact,
where r =z JK^, given by (34) ; and, vice versa, any change p^^ will
cause the change /?©, given by (34). Therefore, putting po = — Po
in (34) we have the change in P^ which results from the suppres-
sion of the internal pressure Po, and so P^^zP^ — ^ItdI iSyPo
is the external pressure to which the inner cylinder is subjected in
d2 pa
the state of rest, and this must not exceed — ^ pg ^ po, which has
been shown in 31 to be the greatest external pressure which, acting
alone on the cylinder, is allowable.
41. The Shrinkage. — The excesses of the exterior diameters of the
elementary cylinders, before assemblage, over the interior diameters
of the cylinders which are to embrace them are called the shrinkages,
and are designated by /Si, S^, S^ etc., 8^ being the shrinkage of the
cylinder whose interior radius is Bi, 82 that of the cylinder whose
42
The Elastic Strength of Guns
interior radius is R2 etc.* The differences of diameter per unit of
diameter, oTf"' 2R~^ 2!S" ^^^'^ ^^ called the relative shrinkages,
1x8
and are designated by <^i, <^2> <l>z ^tc.
Eef erring to Figure 9, Oa and Ob represent the inner and outer
radii of the inner of two elementary cylinders, and Oh' and Oc the
inner and outer radii of the outer one, before assembling, so that
2Vh = iSi is the shrinkage; while OA, OB and 00 represent the
inner radius (Ro), the radius of the surface of contact (R^) and
the outer radius {R2) after assemblage. When the internal pres-
sure Po acts, the compound cylinder is expanded, the three radii
becoming 0A\ OR and OC, respectively, and, by hypothesis, in
Fig. 9.
this state the inner surface of the outer cylinder is under the cir-
cumferential true stress 6^; i. e., its circumferential strain is;^»
But the change of the inner radius of the outer cylinder from its
free state to the state of action is OB'— Oh'; therefore Off— 0V=
-—y^ . And the change of the outer radius of the inner cylinder
from its free state to the state of action is Off — 06, and this, by
(32), is B, elR,) = -f- [6^-^Zl|i«_±:M)] . Hence S,
— 2h'h = 2[0B' — Ob' —(OB' — Ob)} is given by
<? 2E,r 6P,Ei-PA4Rl+ 2^) 1 ,«.x
*' - ~u~ l^' 3(i??— i^i) J ^"^^^
42. The formulffi which we have deduced for this case of a com-
• The shrinkages are so small In comparison with the radii that .it is
unnecessary to distinguish iJ, ± S, from B« jB, ± i8» from R, etc., in
the various formulae.
The Elastic Stuength of Compound Cylinders 43
pound cylinder composed of but two elementary cylinders are
grouped together in (36).
(i) Po(<?) 4if^ + 2ni
(b') P (o) - ^m-^o)Po±^rS^
(6 ) ^o(p) 4Rt—2Rl
m—m
(36)
(„^ p(-p igg(i?|-i?!)p\^ig —
.. _ 2B,r. . P,(4Bl-\-2m) — 6P,Rl -[
(d) ^i=-^L<?i+ z(iii—iti) J
To apply these formulae, calculate P^ and the two values of Po
by (a), (b) and (b'), using for $^, So and po the elastic limits of the
material as determined in the testing machine; then, with P^ and
the least of the two values of P^ determine whether the condition
required by (c) is fulfilled ; if it is, calculate 8-^ with the same
values of P^ and Po', if it is not, find new values of P^ and Poy
using the same values of Bo and po but a value of 0-^ sufficiently less
than the first value assigned it to cause the condition of (c) to be
met.
43. As an example, we will determine the strength of a compound
cylinder of steel for which l^o = 3 in., R^zmh in., 2^2 = ^ i^'>
B^ = 24 tons per sq. in., and ^o = /oo = 18 tons per sq. in.
3(64-25)
^' - 256 + 50 X -^^ - ^-l**
Pfo\ 3(25-9) X 18+ 6X25X9.18
/'„(<')= 100 + 18 — ^^-^^
p.- 3 (25 -9 ) X 18 + 2X25X9.18 ,.,„
■PoO')= — loo^TTg 1^-1^
An internal pressure of 16.13 tons per sq. in. will bring the radial
strain of the inner surface to the elastic limit, and so this is the
greatest safe pressure, although the circumferential strain does not
reach the elastic limit unless the internal pressure is raised to 18.99
44 The Elastic Strength of Guns
tons per sq. in. We therefore proceed to see if the condition of
equation (c) is met with the values P^ = 9.18, Po = 16.13.
9.18 — 4.12 < 5. 76
6.06 < 5.76
The external pressure on the inner cylinder in the state of rest is
5.06 tons per sq. in., while it is capable of withstanding 5.76 tons
per sq. in. Therefore the values P^ = 9.18 and Po = 16.13 are
allowable, and we proceed to determine the shrinkage.
a _ 10 r 9A_, 9.18 (36 + 50) — 6 X 16.13 X 9 1
^» - 13000 L '^^ "• 3 (26 — 9) J
The inner diameter of the outer cylinder must be bored to a diam-
eter .01715 inches less than the outer diameter of the inner cylinder,
and, if assembled with this shrinkage, the compound cylinder can
be safely subjected to the internal pressure 16.13 tons per sq. in.
44. If the shrinkage used in assembling the compound cylinder
be known, the resulting strains and elastic strength are determined
as follows :
As shown in 41 and illustrated by Figure 9, the shrinkage is the
sum of the contraction of the inner diameter of the outer cylinder
and the expansion of the outer diameter of the inner cylinder which
would result from disassembling them. In other words, the rela-
tive shrinkage is given by <l>i = et(Rt) — «f(Bi), in which
et(Ri) and et(Ri) are the circumferential strains at the two
surfaces of contact which the pressure between them after assem-
blage (in this case Pi) causes. The values of these two strains
being obtained by applying (13), we have
1 /Pt (2Jg? + 4i^A I 1 / 7i (4g; + 2m _ ■
This equation (37) gives the value of the pressure at the surface
of contact caused by placing a cylinder of radii R^ and B^ over a
The Elastic Strength of Compound Cylinders 45
cylinder of radii Ro and R^ with the relative shrinkage <^i, and the
1 2P IP
resulting circumferential strain at Ro being w- -^2"^^ — 5ji> we have
J^ Hi — Mo
«<{-Ro) = — W 2^2 ^1 (3^)
by which the relative compression of the bore of the inner cylinder
caused by superposing the outer cylinder with the relative shrinkage
<^i may be computed.
Since the only stress at the inner surface in the state of rest is
the circumferential compression, the radial strain is one-third the
circumferential strain given by (38).
Examples IV.
(1) Given Ro = 1.80", R^ = 2.85", R^ = 4.50", $o = po= 18.75
tons, ^1 = pi = 21.50 tons; find Po{0), Po{p) and 8^; also the
compression at Ro in the state of rest.
Po{0)= 17.11; Po{p)= 16.58 tons.
Si = .0074 in.
Fi = 3.61 ; Eet (Ro) = — 12.02 tons.
(2) Given Ro = 2.85", R^ = 4.70", R^ = 7.50", 9o = po = 18.75
tons, ^i = pi = 21.5 tons; find Po{0)y Po(p) and S^; also the
compression at Ro in the state of rest.
Po{0)= 17.88; Po{p)= 15.91 tons.
S^ = .0130 in.
Pi = 4.03 ; Eet (Ro) = — 12.75 tons.
(3) Given Ro = 4.00", R^ = 6.35", R^ = 8.04", Oo = po = 18.5
tons, ^i = pi = 21.0 tons; find Po{0), Po{p) and 8^; also the
compression at Ro in the state of rest.
Po{0)= 12.64; Po(p)= 13.26 tons.
8^ = .0126 in.
Pi = 2.01 ; Eet (Ro) = — 6.67 tons.
(4) Given Ro = 6.00", R^ = 8.70", R^ = 10.46", Oo = po = 18.5
tons, ^i = pi = 21.0 tons; find Po(0), Po(p) and 8^; also the
compression at Ro in the state of rest.
Po(e)= 10.25; Po(p)=^ 11.91 tons.
'iSi = .0148 in.
Pi = 1.37 ; Eet (Ro) = — 5.22 tons.
46 The Elastic Strength of Guns
(5) Given Ro = 4.00", R^ = 5.80", R^ = 7.14", $0 = 90 = 18.5
tons, ^i==pi = 21.0 tons; find Po{B)y Po(p) and 8^; also the
compression at Ro in the state of rest.
Po{e)= 10.60; Po(p)= 12.19 tons.
5i = .0105 in.
Pi = 1.63 ; Eet (Ro) = — 5.83 tons.
(6) Given Ro = 4.00", R^ = 5.80", B^ = 7.14", if the shrinkage
was iSi ^ .0105, what is the pressure at the surface of contact and
what is the compression of the bore (at Ro) in the state of rest?
(Compare result with answers to Example (5).)
Pi = 1.53; Eet{Ro) = — 5.83 tons.
CHAPTER V.
THE ELASTIC STRENGTH OF COMPOUND CYUNDEES.—
CONTINUED.
46. The true stresses, circumferential and radial, at the inner
and outer surfaces of each of the elementary cylinders are readily
calculated by (13) and (14), which, when applied to the case of a
compound cylinder of two parts, become
Circumferential True Stresses,
■»«.(-«.> 8(iJ»-if:)
■®''*-*'' S ( J^ - £■)
(89) (40)
£adial True Btrtttet.
P, (4i2J-22^) - 2P,^
EeiBT) =
3(^-^)
in which, for the state of action, Po and P^ have the values used in
calculating the shrinkage, and, for the state of rest, Po is zero and P^
is the pressure at the surface of contact when Po = 0(Pi).
Applying (39) and (40) to the example worked out in 43, for
which Po = 16.13, Pi = 9.18 and Pj = 5.06, we obtain the results
illustrated in Figure 10, the right-hand side of which represents cir-
cumferential and the left-hand side radial true stresses, full lines
indicating the state of action and dotted lines the state of rest.
It will be seen that in the state of action both cylinders are at the
elastic limit of strain, the inner one radially and the outer one cir-
cumferentially.
46. The fact that the greater the value of Po used in calculating
the shrinkage the less the shrinkage and consequently the less the
stresses in the state of rest, suggests an investigation of the results
of always using Po{0) in (36d) instead of using Po(p) when it is
the smaller of the two values of Po.
In the example of 43 the shrinkage found by using Po(/d)= 16.13
tons was 0.01715"; if we had used Po(6)=^ 18.99 tons, we would
have found the shrinkage to be 0.01468", or nearly 0.0025" less. The
48
The Elastic Strength of Guns
true stresses in the states of action and of rest have been computed
for the greater shrinkage ; we will now determine their values under
the same conditions (P© = 16.13 tons and P© = 0), supposing the
reduced shrinkage to be used.
With the reduced shrinkage the value Pj = 9.18 corresponds to
Po =z 18.99, and so we have first to find the change in P^ which
results from reducing Po from 18.99 to 16.13; this by (34) is
— 0.73, making the value of P^ for our assumed state of action
9.18 — 0.73 = 8.45. Substituting the values Po = 16.13 and P^ =
8.45 in (39) and (40), we obtain the values of the true stresses in
the state of action. For the state of rest we find P^ = 4.33 by
Fig. 10.
Radial True Stresses.
State of action.
" rest.
it
Circumferential True Stresses.
State of action.
** rest.
tt
(36c), getting the same result, of course, whether we use P© == 18.99
and Pj = 9.18 or Po = 16.13 and P^ = 8.45 ; then, putting Po =
and Pi=7\ = 4.33 in (39) and (40), we get the values of the
true stresses in the state of rest.
The following table gives, side by side, the true stresses resulting
from the use of the full and the reduced shrinkages :
Circumferential Stress. BadiaZ Stress.
Full Reduced Full Beduced
Shrink- Shrink- Shrink- Shrink-
age.
age.
age.
age.
rinner cylinder, inner surface
+10.97
+13.26
-18.00
-18.76
State of Action.
outer
+ 1.70
+ 3.01
- 8.73
-8.62
Po = 16.13 tons. "
Outer " inner *'
+24.00
+22.09
-16.16
-14.88
. " '* outer *'
+11.77
+10.88
-8.92
-3.61
["Inner cylinder, inner surface
-16.81
-18.63
+ 6.27
+ 4.51
- « . " " outer
- 9.07
- 7.76
- 1.48
- 1.26
State of Rest. ^ Q^^^ .. ^^^^^ .
+13.23
+11.83
- 8.91
- 7.62
^ " " outer
+ 6.49
+ 6.65
-2.16
-1.86
The Elastic Strength of Compound Cylinders 49
47. It will be seen that the reduced shrinkage, given by adopting
Po{0) instead of Po{p) as the value of Po, results in a slight loss
of elastic strength,* since the internal pressure (16.13 tons) which
with full shrinkage just compressed the inner surface to its elastic
limit of strain radially, with the reduced shrinkage compresses that
surface slightly beyond its elastic limit. As an offset to this, the
smaller shrinkage considerably reduces all the stresses in the state
of rest, and those of the outer cylinder in the state of action.
Moreover, there is. reason to suppose that the elastic strength to
resist radial compression in the case of a cylinder wall confined by
an outer cylinder is greater than would be indicated by the elastic
limit of compression of specimens of its material, so that the value
of Po{p) may probably be exceeded without producing any per-
manent set. At all events, it is not radial compression, but circum-
ferential extension, an excessive value of which will cause enlarge-
ment and ultimately rupture, and we are therefore adopting a
measure of safety when we adjust the shrinkage so as to cause the
elementary cylinders to reach their elastic limits of circumferential
strain simultaneously, even though it be under a pressure greater
than that which will cause the inner one of them to reach its elastic
limit of radial strain.
For these reasons the Ordnance Departments of the United States
Army and Navy have adopted the practice of disregarding the
values of Po(p) and determining the shrinkages for the superposed
cylinders of their built-up steel guns by using the values of Po(0),
We will follow the same method, using Po{0) for computing
shrinkages, but still regarding Po{p), when it is less than Po{0)y
as the upper limit of safe internal pressure.
48. In 40, by equating the simultaneous changes of circumfer-
ential strain of the two surfaces in contact at R^, we found the
relation (34) between simultaneous changes of P© and P^ in the
case of a compound cylinder composed of two elementary cylinders..
The same relation might as readily have been found from the con-
sideration that, within the elastic limit, the stresses and strains
♦ With the reduced shrinkage the internal pressure which will bring,
the inner surface to its elastic limit of radial strain is given by
P^ ^ ^-^ . ^^^"^ \tf^^^ ' t^® ^*^"® ^* ^^^c^ ^or the
example of 43 Is 15.62 tons.
50 The Elastic Strength of Guns
resulting from the application of any force are independent of prior
stresses and strains, so that the effect of an internal pressure is
exactly the same upon a compound cylinder as it would be upon a
simple cylinder of the same dimensions. Thus, putting Pn =:
and substituting R^ for Rn in (12), we obtain for the pressure at
any point in a homogeneous cylinder of radii Ro and Ro under the
sole pressure Po,
^'•)=-|Si(^-i) (41)
and, making r = ^i in this, we find
which is the same as the relation given by (34).
49. The general principle of which the foregoing is an illustra-
tion may be stated as follows :
If any pressure be applied to a compound cylinder, the strain
{or stress) at each point will he the algebraic sum of the strain {or
stress) at the point before the pressure was applied and the strain
(or stress) which the same pressure would cause at the correspond-
ing point in a simple cylinder of the same dimensions as the com-
pound one,
60. An important application of this principle shows that the
maximum strength of any compound cylinder to resist internal
pressure cannot exceed three-fourths the sum of the elastic limits
of tension and compression of its inner elementary cylinder, re-
gardless of the strength of its outer parts. For in the state of rest
the pressure upon the inner cylinder due to the outer ones is limited
to that which will compress the inner surface circumferentially to
its elastic limit of compressive strain -^ ; and in the state of action
the internal pressure is limited to that which will extend the inner
surface circumferentially to its elastic limit of tensile strain -w- ;
therefore the greatest allowable value of Po is that which, acting
upon a simple cylinder of the same dimensions as the compound
one, would cause the circumferential strain -^^~^~^ at its inner
The Elastic Strength of Compound Cylinders 61
surface, and, calling the inner and outer radii Ro and Bn, the value
of this greatest pressure is by (20),
the maximum value of which, when5« = oc, is f (/o^ -f- ^^), or, when
^o = Pay iO.
This maximum possible value of the elastic resistance will here-
after be denoted by [Po], and, since we accept the condition po =
Boy it will be written
This is the maximum possible value of Po{0) ; Po{p) cannot exceed
Po in value.
61. From the formulae for a compound cylinder of two parts,
those for the general compound cylinder (of n parts) may be directly
derived, but as the case of three elementary cylinders is the com-
monest in gun construction, we will deduce the formulae for that
case separately, and explain how they should be used.
We begin by finding the values of the pressures in the state of
action (Pj? Pi and Po)^ supposing the cylinders to have been so
assembled that they reach their elastic limits of circumferential
strain simultaneously.
The outer cylinder being under the sole action of the internal
pressure P2, we have from (20),
The middle cylinder being under the external pressure P^ and
the internal pressure Pj, of which the latter is the greater, we have
from (28),
And the inner cylinder being under the external pressure P^ and
the internal pressure Poy of which the latter is the greater, we have
from (28),
Po(0) - TJSrF2i2^ ^^^^
52 The Elastic Strength of Guns
Before adopting these values of P2, Pi and Po, we must see that
the shrinkages which they require will not over-compress the inner
surface in the state of rest. This is most readily done by com-
puting [Po] by (43) and comparing it with Po{0) ; if the latter be
the greater, the inner surface would be compressed beyond its elastic
limit of circumferential strain when in the state of rest, and so less
values must be assigned to one or both the elastic limits of the
outer cylinders and new values of Pg, P^ and Po computed. When
the assumed values of O29 6^ and Oo are such that [Po] exceeds
Po(0), the inner cylinder will not be too much compressed, and
then the values of P^iO), P^iO) and Po(e) given by (42), (43) and
(44) may be accepted.
52. The formulae for the shrinkages are deduced by the same
method that was explained in 41. The inner surface of the outer
cylinder when in the state of action is, by hypothesis, under the
circumferential strain -w- , so that its diameter is 2R2 -4- greater
than when it was free (before, assembling). If, then, we find the
change of diameter (2R2et(R2)) of the outer surface of the middle
cylinder which would result from the simultaneous removal of the
outer cylinder and suppression of the internal pressure Po, the
shrinkage with which the outer cylinder was assembled will evi-
dently be given by S2 = 2R2 -w + 2526^(^2)-
By substituting R2 for fi«, Pg for P» and iJg ^^^ r in (13) we
obtain the following expression for the circumferential strain at
the outer surface of a cylinder of radii Rq and R2 under internal
pressure P© and external pressure P^ :
1 r 6P,igg~A(4igg + 2igin ,47^
But by the principle laid down in 49 the same expression gives
the change of strain which the application of the same pressures
will cause in a compound cylinder of the same dimensions. There-
fore, putting — Po for Po and — Pj for P2 in (47), we obtain the
change of circumferential strain at R2 due to suppressing Po and
P2, and this multiplied by 21^2 "^U he the change of diameter.
Consequently the shrinkage of the outer cylinder is given by
2i2, r . , P,(4Rg + 2^-6J^oign ,40N
The Elastic Strength of Compound Cylinders 53
Similarly the change of circumferential strain at the outer sur-
face of the inner cylinder due to removing the two outer cylinders
(f. e., suppressing P^) and simultaneously suppressing Po is found
to be
1 [ P, {4Rl + 2m) - 6Po Bn ,49.
M^i) - -^ L 3(i?[— i^) J ^^^^
and so the shrinkage of the middle cylinder is
2B,r P,{4Rl + 2i25)- 6P, RH
^-^- L'^i + sjm^^Bj) — - J (^^)
52. We have, finally, to determine the elastic strength to resist
internal pressure of the system thus assembled. We know that
Po{0) is the pressure which will bring its elementary cylinders
simultaneously to their assumed elastic limits of circumferential
strain, but a less pressure may bring one or more of them to the
elastic limit of radial strain, and, if so, this latter pressure, and not
Po{0), should be taken as the maximum safe pressure.
The outer cylinder being under internal pressure only, P2(0) is
always less than Pzip), as explained in 29. Applying (27) to the
middle and inner cylinders, we obtain the following values for the
respective internal pressures which will bring them to their elastic
limits of radial strain :
P /„^ _ 3(^- i??) Pi + 2P» m « IN
4i^— 2i2?
3 ( P? - Ji^„) p, + 2P, Rj
4K1— 2P»
(62)
If Pi(p) given by (51) is less than the value of Px{6) used in
computing the shrinkages, then the former is to be used for P^ in
(52) instead of the latter, and if Po{p) given by (52) is less than
the value of Po{0) used in computing the shrinkages, it, and not
Po{0)y is the maximum safe pressure. That is, with Po{p) <Po{0),
the former would be a safe pressure if suitable shrinkages were
assigned, but since, for good reasons, we adopt shrinkages based
upon the values of Px(0) and Po(0), the actual maximum safe
64 The Elastic Strength of Guns
pressure is somewhat less than Po{p)- We will call the true maxi-
mum safe pressure P©, thus distinguishing it from Po{p) and
Po{6) ; its value, when it does not equal Po{0), is found as follows :
The pressures in the state of rest are given by (53) and (54),
the negative part of each value being the change of pressure due
to the suppression of Po(0) :
F. = Pm- ^[^Z^) Po{0) (53)
P. = P.W-§f|^|PoW (54)
Then by (14) the radial strain at the inner surface of the inner
1 2P P?
cylinder, in the state of rest, is -p- o (Ty^^__ ^m\ * ^^^ ^^ internal
pressure which will change this radial strain to — -^ ,i- e,, which
will bring the inner surface to its elastic limit of compression radi-
ally, is, by the principle of 49,
p^ 3(J?i— ig|) / 2P,i;! \
^o 2i2J— 4i^\ 3(i2?— iij) ^V
p^ m-mi 3(Jg?-ig|)p, + 2Ai?? ....
^^ Ei — Bl • 4i^— 2i?| ^^
The same method applied to the middle cylinder, which in the
state of rest is acted on by Pg externally and P^ internally, would
determine the internal pressure which would bring its inner surface
to the elastic limit of compression radially,* but this pressure will
practically always be greater than that given by (55), and, accord-
ingly, (55) gives the true elastic strength of the system.
♦The formula is
»2 (739 M\
The Elastic Strength of Compound Cylinders 55
54. The formulae for the case of a compound cylinder composed
of three elementary cylinders are grouped together in (56) :
(a) Pld)= 4j}5^2i^
(«J ^i(i») - - 4i£j — 2i^
(/) Po(/') 4iJ? — 2iJJ
^ \ c? 2i2,r. . P,(g) (4fi| +2ig.) - 6P„ (g ) .Rn
,h\ <f 2i2ir Pi(g)(4P^+ 2 J?f)-6P„(<?).R^J
(A) *!--:£- [<'» + SfPl^^TRI) J
0-) A = p.w-f|fE|-ji'o(^)
w
p _ ^ — ^0 S{m—Bl)p, + 2F,Ri
■^0 - El—Bl • 4P^ - 2P^
(66)
The method of procedure is as follows :
(1) Calculate P^iO), P^{e) and Po{e) by (a), (b) and (c),
using for Bzy ^i ^.nd So the elastic limits of the materials as deter-
mined in a testing machine.
(2) See if the condition (d) is fulfilled. If it is not, find new
values of PgC^)^ ^i(^) and Po{0), using values of 0^ and 0^ (one
or both) suflSciently less than their true values to cause the con-
dition (d) to be met.
(3) Calculate the shrinkages by (g) and (h), using the values of
^2, ^1, Pii^), Pi{0) and Po(0) which satisfied (d).
(4) Calculate Pi(/o) and Po{p) by (e) and (f), using for p©
56 The Elastic Strength of Guns
and pi ihe true elastic limits of the materials, and for Pi in (f)
putting whichever is least, Pt(p) or the value of Pi(^) calculated
with the true values of O2 and 0^^,
(5) If Po(p) is greater than the value* of Po{0) used in comput-
ing the shrinkages, the latter is the true measure of the elastic
strength of the system; if it be less, then Po, calculated by (k), is
the true measure.
55. To find the state of strain at the inner surface (at Ro) caused
by superposing the two outer cylinders with relative shrinkages,
respectively <^i and <^2> ^^ have only to apply (38) to this case, the
strain resulting from the compressive action of both outer cylinders
being merely the sum of the strains caused by their actions consid-
ered separately. Thus we have
et{Ro)^ — jn jj^^i — JM j!^s^i (67)
Moreover, the radial strain at the inner surface in the state of
rest (?p(i?o)) will be one-third the circumferential strain given by
(57).
Examples V.
(1) Given Bo = 7.0", B^ = 9.5", iJ^ = 15.0", Bj, = 21.0"; if
$o=z po=^ 20.0 tons, what is the greatest possible value of the in-
ternal pressure which can be withstood elastically ? If ^o = 20.0,
^1 = 21.4 and 0^ = 22.3 tons, find Po{0), S^ and 8^. What is the
true elastic strength after assemblage with the shrinkages based
on the value ofPo(*)?
[P,] =: 25.26; Po{e) = 24.46 tons.
/Si = .0183; /S2 = .0386.
Pi = 4.28 ; Po = 18.52 tons.
(2) Given Ro = 5.0", R^ = 9.5", R^ = 15.0", R^ = 19.0", Oo =
Pf, = 20.0 tons; what is the limiting value for the internal pres-
sure? If do = 20.0, 0^ = 21.0, and 0^ = 24 tons, find Po{0). If
assembled with the shrinkages corresponding to the value of Po{6)y
what would be the compression at Ro in the state of rest ?
[Po] = 26.99; Po(6) = 28.39 tons.
22.06 tons.
The Elastic Strength of Compound Cylinders 57
(3) Given Ro = 6.0", R^ = 10.3", R^ = 15.0", R^ = 17.7", $o =
17.5 tons, 0^ = 22.0 tons, 0^ = 22.0 tons; find [Po], Po{S), 8^ and
82. [Po] = 21.97; Po(0) = 21.79 tons.
fi'i = .0296 ; 82 = .0400 in.
(4) Given Ro = 4.75", R^ = 7.50", R^ = 11.375", R^ = 14.375",
Oo = 16 tons, ^1 = 17 tons, 0^ = 22.2 tons; find Po(0), 8^ and /Sg-
Po(0)= 20.68 tons; /S^ = .0149; 82 = .0327 in.
(5) Given Ro = 6.0", R^ = 11.0", i^g = 17.0", E3 = 21.0", ^0 =
18.0, 6^ = 19.0, $2 = 21 tons; find [Po], Po{B), 8^, 82 and Po.
[Po] = 23.82; Po(^) = 23.82 tons.
8^ = .0287 ; 8 2 = .0476 in.
R, = 6.32 ; Po = 17.23 tons.
(6) Given Ro = 6.0", B^ = 11.0", R2 = 17.0" and R^ = 21.0",
if the shrinkages were 8^ = .0287 and 82 = .0476, find the circum-
ferential and radial true stresses at the inner surface (at -Bo) in the
state of rest. Then, by the principle of 49, find the internal pres-
sures which will strain the inner surface to the elastic limit (18
tons) first radially and second circumferentially. (Compare results
with answers to example (5).)
Eet(Ro) = 18.09; Eep(Ro) = 6.03 tons.
Po(e) = 23.87; Po(p) = 17.25 tons.
CHAPTEE VI.
APPUCATIONS TO BUILT-UP GUNS.
56. The modem gun is essentially a compound cylinder, but,
being constructed to withstand an internal pressure which dimin-
ishes from the breech end to the muzzle, the number of layers and
the exterior dimensions are correspondingly decreased for economy
of weight, making it necessary to divide the whole length into a
number of sections for each of which a separate computation of the
elastic strength and shrinkages must be made. In United States
guns the inner layer, in which are formed the chamber and bore
proper, is called the tube; the second layer consists of a jacket, in
which the breech block is housed, and chase hoops, which extend
from the front end of the jacket nearly or quite to the muzzle; over
that part of the bore in which the maximum powder pressure acts
a third and sometimes a fourth layer of hoops is placed. With
increase of knowledge and of facilities larger and larger steel forg-
ings of assured good quality have become available, and the number
of separate parts constituting a built-up gun has tended to dimin-
ish, so that at the present time the outer layers, as well as the tube,
are sometimes made in one piece.
In one particular, however, there is an important difference be-
tween a gun and the compound cylinders with free ends which we
have thus far considered; in the latter there is no longitudinal
stress, while in a gun the internal pressure, ax;ting upon the breech
block as well as upon the cylinder walls, gives rise to a longitudinal
stress of very considerable intensity.
67. The .iongitudinal Stress.— If we consider a gun recoiling
freely under the action of the powder pressure on the bottom of its
bore, we see that the total longitudinal stress on any cross-section
of the gun must equal the product of the acceleration by the mass
forward of the section, so that the said stress diminishes rapidly as
we go forward from the front thread of the screw box, where it is a
maximum. When recoil is resisted by a brake of any kind, the
acceleration is reduced and so, to the same extent, is the longi-
tudinal stress on all cross-sections forward of the point of attach-
Applications to Built-up Guns 59
ment of the brake to the gun ; in rear of that point the longitudinal
stress is increased by the action of the recoil brake, the increase
diminishing as the cross-section through the front thread of the
screw box is approached till, at that point, the total longitudinal
stress is practically the same as in free recoil. When, as in most
modem United States naval gun mounts, the pistons of the recoil
cylinders are attached to a yoke around the breech of the gun, the
longitudinal stress is diminished at all sections, its maximum value
then being --™ (TriZj P — F) , in which M is the whole recoiling
mass, M' is that part of it which is forward of the front thread of
the screw box, Ro is the radius of bore and P the maximum powder
pressure, and F is the total resistance * of the recoil brake at the
instant when P acts.
We do not know how the total longitudinal stress is distributed
over the cross-section of the gun. It is not wholly born by the layer
in which the breech block houses (the jacket in United States guns),
for there is an enormous frictional resistance to the longitudinal
motion of any one layer relative to the others ; if it were uniformly
distributed over the jacket alone, its intensity, even at the section
of greatest stress, would seldom exceed 5 or 6 tons per square inch,
and if, as many writers assume, it is uniformly distributed over the
whole cross-section of the gun, its greatest intensity will not exceed
2 or 3 tons per square inch. Probably the latter assumption is
practically true at some distance forward of the breech block and
is not very far from the truth at any point forward of the gas check.
Moreover, this maximum intensity of longitudinal stress only
exists for the infinitesimally small period of time during which the
maximum powder pressure is maintained ; during the greater part
of the time in which the gun is subjected to internal pressure the
longitudinal stress is very small, even at the section where it has its
greatest value.
For these reasons, therefore, we are justified in applying to guns
the formulas which we have deduced for cylinders with free ends.
58. If circumferential strain alone had to be considered in the
case of a compound cylinder, the greatest strength would be ob-
tained by making the successive radii of the elementary cylinders
♦This total resistance of the recoil brake, however, is never more
than a small fraction of the maximum total predsure on the bottom of
the bore of the gun.
60 The Elastic Strength of Guns
increase in geometrical progression, provided their physical char-
acteristics were the same. Thus, for the case of any one cylinder
superimposed upon another, regarding Po(0) in (36b) as a func-
tion of Ri{Ro and Bj constant, and $^ = $o) and putting — ,S ^
= 0, we find, after simplification, R^ = R0R29 which shows that the
maximum value of Po{0) for a given total thickness of a given
material occurs when the radius of the common surface is a mean
proportional between the inner and outer radii. Very nearly the
same proportions will also give the greatest strength as regards
radial strain.
In practice, however, other considerations govern in the propor-
tioning of the layers of which guns are composed. In the first
place the layer in which the breech block is housed, even though
other layers assist it in taking the longitudinal stress, should be of
suificient cross-section to itself safely sustain that stress. Again,
the thickness of the tube over the chamber should be sufficient to
make relining practicable in case erosion wears away the rifling,
and its thickness elsewhere should be sufficient to give ample stiff-
ness. Finally, the necessity for keeping down the weight, which
prescribes a decreasing exterior diameter towards the muzzle, and
the need for avoiding sudden or great changes of the sections of
the different layers, often require dimensions not otherwise desir-
able.
69. In assigning shrinkages for the different parts of a gun,
while as a general rule the maximum attainable strength should be
sought at each section, great changes of shrinkage in passing from
one section to another must be avoided, as they would cause unde-
sirable inequalities of strain. Not only should each of the parts
which make up the outer layers of the gun be assembled so that the
strains at its inner surface, both in the state of rest and in that of
action, do not change abruptly at any point of its length, but the
tube, similarly throughout its length, should be under a compres-
sion in the state of rest, and of extension in the state of action,
which only gradually varies and at no point changes abruptly. Fur-
thermore, as a rule, slack shrinkages should be preferred to exces-
sive ones, to the end that under the action of an excessive pressure
it may be the tube which gives way rather than an outer part.
60. As a simple example of the method of determining the proper
shrinkages, and the elastic strength of a gun, we will consider the
I
i
Applications to Built-up Guns 61
case of the United States naval 5-inch B. L. E. Mark V, which is
shown in Figure 11, with its curves of computed elastic strength
and of strains at rest and in action.
The computations are made separately for each of the sections
indicated on the drawing, but only those for the most important
section, that through the chamber, will be worked out in the text,
the final results of the other computations, which are obtained in
exactly the same way, being merely stated. As it is always neces-
sary to adjust the shrinkages, in accordance with the principle set
forth in 69, it is most convenient to find their values, as well as the
values of the pressures in the state of action, in terms of the elastic
limits of the different layers, afterwards assigning suitable values to
the elastic limits, always, of course, within their true values as
indicated by the testing machine.
6 'inch B. L. R Mark V.
Section L
R^ 8.50 J2;= 12.25
i?j= 5.25 i5=s 27.56
1^3= 8.25 JS«s 68.06
H^ = 10.25 Il\ s 105.06
8 (iZ; - i2J) = 278.48 log 2.44473
2 i?^ + 2?» =222.87 ♦♦ 2.84707
«« 0.09765
^, = p„ = 20.0 tons ^ ^«= 20.0 -j^80108
6>i=/5i = 21.5 " \* IPJ = 25.04 »* 1.89868
e^ = p, =: 22.0 ««
I
That is, 25.04 tons is the greatest possible elastic strength, whatever the
qualities of the jacket and hoop.
8 {S\ - i2|) = 111.00 log 2.04582
4i^ + 2^ = 556.86 ** 2.74586
A^ ♦♦ 9.29996
<^a = 22.0 ** 1.84242
Pa(0)= 4.389 »» .64288
Pa(^) = A^ 0, = [9.29996] 8^
8 (i^ - 2?J) = 121.50 log 2.08458 6i2| = 408.86 log 2.61104
4iZ| + 2 JBJ = 327.36 ** 2.51503 " 2.51508
A^ " 9.569.55 B^ *» .09601
^j = 21.5 " 1.33244 A^ *» 9.29996
7.980 «♦ .90199 B, A^ ;. "~9. 39597
«2= 22.0 " 1.34242
5.475 " .73889
P^id) = 18.455
P^id) = ^j^j + B^A^e^ = [9.56955] 6^ + [9.89597] 0,
•These are the true elastic limits, being the least values given by
any of the specimens taken respectively from the tube, Jacket and hoop.
62 The Elastic Strength of Guns
8<i^-22J)= 45.98 . .log 1.66210 6J2;=:165.86. .log2.31843 ^,ilg. .log9.8959T
4iS+3.^=184.74 .. "2.12950 "2.12950
A^ *« 9.58260 B^ " .08898 *♦ .08898
6^^= 20.0 .. " 1.80108 A^ " 9.56955
6.818.. ** .83363 B^A^ " 9.65848 B^B.A^. . " 9.48490
^, = 21.5 .. " 1.33244
9.793 " .99092 ^- = 22.0 " 1.34242
6.719 " .82732
P/0) = 23.83O
PjiB) = Afi^ + B^A^e^ + B^B^A^^ = [9.53260] e^ + [9.65848] 6^ + [9.48490] (?,
The value of Po{0) for the true values of the elastic limits being
23.33 tons, while [Po]= 25.04 tons, the inner surface is not too
much compressed in the state of rest, and so we proceed to deter-
mine the shrinkages.
6BJ= 73.60 log 1.86629
2Bi= 10.60 loff 1.02119
i; = iaooo.o " 4.11894
-^=.0008077 " 6.90726 "6.90726
4JR«+2B?= 104.12 " 2.01763
*' 8.92478 " 8.77364
3(B?-B|)= 45.93...... " 1.66210 " 1.66210
'* 7.26368. ... log 7.28268 " 7.11144. . .log 7.11144. . . log: 7.11144
uli " 9.68965 ul« " 9.63260
Bi^, " 9.S9697B^j " 9.66848
" 6.83323.... " 6765866 BpBiJL g ** 9.48490
log 6.64404 " 6:76992 " 6.60634
+ . 0008077^1 + .0006796^ + .0004667^8 - .0004406 d„ - .0006887 «i-. 0003948 «,
+ .0006796 - .0003948
+ .0014873 + .0000609
- .0006887
+ .0008986
^1 = .0008986^^1 — .0004406 6^ + .0000609 ^,
6i2;= 73.50 log 1.86629
2i2a= 16.50... log 1.21748
^=13000.00... " 4.11394
2J2
-^ = .0012692..." 7.10354.... " 7.10354
ABl + 225= 185.12. . . " 2.26745
" 9.37099 " 8.96988
3(^_i22)= 167.43... «« 2. 22383 .... ** 2.22383
*t 7.14716 ** 6.74600... log 6.74600... Iog6.74600
Aa " 9.29996 A, " 9.53260
** 6.44712 BA^ ** 9.65848
^X^a • ..." 9.48490
** 6.27860 *» 6.40448 *» 6.28090
+ .0012692^.^ + .0002800(^9 - .00018996^^ - .00025386^1 - .000 1702 (y,
+.0002800
+ .0015492
—.0001702
+.0013790
5j, = .0013790<?, - .0001899^^ - .0002538^,
Applications to Built-up Guns
63
K^
2.70
jSj=: 5.25
R^^ 8.25
jB, = 10.25
Now, substituting the values 20.0, 21.5 and 22.0 for Oo, 0^ and O^y
respectively, we have for the shrinkages which will cause tube,
jacket and hoop to simultaneously reach their elastic limits of
circumferential strain, under the internal pressure Po(^)= 23.33
tons, S^ = .01185 and 8. = .02108.
In exactly the same way as shown for Section I, the values of the
pressures in the state of action and the corresponding shrinkages
are computed for the other sections, the results being as follows :
Section IL
P2(0) = [9.299961/9,
P^iB) = 19.569551^1 + [9.895971(9,
PJ,e) = 19.697691«„ + 19.691701^1 + [9.51812K?,
8^ = .0010876^1 - .0002896 <9^ + .0000983^,
8^ = .0013976^, - .0001508^^ — .0001488^1
Section HI.
P,(I9) = [8.926181 <9j
P^m = [9.569551 ^j + (9.022191 e,
P^W) « [9.71671]^^ + [9.698991 6»j + [9.1516316/,
8^ = .0009470^1 - .0002468^^ + .00005599^,
5, = .0012509 <?, - .00019196/, - .0001842^1
Section IV.
P^iO) = [9.545771^1
P^iS) = [9.7167116/. + [9.6752116/,
5i = . 0009391 0j - .0002468^^
Section F.
P,((?) = [9.466891 (9i
P^id) = [9.6989716/, + [9.591331^1
8^ = .0008698^1 - .00025646/,
Section VL
P^m = [8.9^28101
P,((9) = [9.69897] 6/, + [9.089221^,
8i = .0008007^1 - .00025646/,
2.50
5.25
8.25
9.00
1
2.50
5.00
7.00
i2 =
2.50
5.00
}
2.50 I
i?i=3.8735)
2.50
4.50
[
Section VIL
PJiO) = [9.6989716/,
Section VIIL
P,(6')=: [9.55980161,
Section IX.
P,((?) = [9.65244]^,
64
The Elastic Strength of Guns
61. We adopt as the shrinkages for that part of the gun which is
represented by Section I the values /Si = .0120 and /Sg = -0210,
being (to the nearest thousandth of an inch) those which result
from substituting the true values of 6oy 0^ and O2 in the expressions
for /Si and 82-
If now we compute the shrinkages for Section II with the same
values of Oo, 0^ and do, we find S^ = .0165, So = .0197 and Po{0) =
27.79 tons, and as the increase of strength over the adjoining section
would be valueless, while the great increase of the jacket shrinkage
would cause a very undesirable inequality of strains in the state of
rest, we see that it will be best to assign less values to 0^ and 0^ and
to adopt a correspondingly less shrinkage for this section. If, on
the other hand, we should adopt the same shrinkages for Section II
as for Section I, an internal pressure which would bring the bore
to its elastic limit would only cause a circumferential true stress of
about 16.6 tons at the inner surface of the jacket, thus causing an
undesirable inequality of strains in the state of action, since in the
adjoining section the jacket reaches its elastic limit with the tube.
We therefore compromise between the two extremes, and adopt the
values /Si = .0130 and 80 = .0200 for part of the gun which
Section II represents.
Guided by similar considerations, we assign, to the shrinkages at
the other sections the values stated on the drawing.
62. We might now, by means of the general values of 8^ and 82
which we have computed for each section, find the values of 0^ and
So which, in combination with the value 20.0 for 60, will give the
shrinkages which have been adopted, and then, with those values of
Oo, ^1 and 60, calculate the elastic strength, compression of bore in
the state of rest, etc. A better method, however, is to start afresh
and with the given shrinkages calculate first, bv (57), the circum-
ferential and radial strains at the surface of the bore in the state
of rest and then the internal pressure which will increase each of
those strains to its greatest allowable value. We will do this for
Section II, as an illustration.
i2„= 2.70 iZ;= 7.29
-B, = 6.25 i2J= 27.56
i?a= 8.25 Bl= 68.06
-B, = 10.25 i^ = 105.06
8, = .0130; 01 = -J|- = .0012881
8.
8^ = .0200 ; 0, = -^
1— = .0012121
Applications to Built-up Guns 65
i5-J!^= 40.5 logl.60746 ^-r-j^ = 37.0 ... .log 1.56820
01 = .0012381 " 7.09275 0, = .0012121 ** 7.08354
" 8.70021 ** 8.65174
2a-i2;=» 60.77 " 1.78369 ^—^23 = 97.77 " 1.99021
.0008251 *« 6.91652
.0004587 *» 6.66153
eJ^R^i = - .0012838. ..." 7.10850
^= :3000 .... " 4.11894
EeJ,R^) = - 16.69 " 1.22244
M(Ji)^ f- 5.56
p'
The true circumferential stress at the surface of the bore in the
state of rest is thus found to be — 16.69 tons, while the true radial
stress is + 5.56 tons. Therefore, applying the principle laid down
in 49, the internal pressures which will, respectively, bring the inner
surface to its elastic limits of strain circumferentially and radially,
are found as follows :
SiBI - J2J) = 298.31 log 2.46733 log 2.46733
d^ + 16.69 = 86.69 " 1.56455
p^-\- 5.56= 25.56 " 1.40756
'* 4.03188 " 3.87489
4i2; + 2JP = 434.82 «» 2.63881
4Je — 2Bi = 405.66 " 2.60816
9
P^id) = 24.75 " 1.39357
P^(p) = 18.48 " 1.26673
In the same way at each of the other sections the effect upon the
bore of superposing the outer cylinders with the adopted shrink-
ages is first calculated, and thence the elastic strength of the assem-
bled system is determined, the results being as shown by the curves
in Figure 11.
63. Since the compression of the bore caused by superposing the
XVq "^~ -ftg
hoop with the relative shrinkage </>o is by (38) j^ j p E^^, the
3 O
pressure at the surface of contact in the state of rest must be
p __ ^2 J^o^ ^ 2 rp. /RON
and, since the whole compression of the bore in the state of rest is
p2 p2 p2 r>2
by (57) ^'^^'^«. + ^^^^*.. ^e have similarly
66
The Elastic Strength of Guns
^1 - ~lfRf~ Ul— i2|^*i + Ri — Bl ^1 ^^
Thus we obtain the values of the pressures in the state of rest at
each of the sections of the gun, and from them, together with the
known value of P©, the strains in the state of rest and of action
may be found.
The following table gives the results of the calculations for the
5-inch gun shown in Figure 11 :
SECTTONS.
I'
/
II
III
IIP
IV
V
VI
VII VIII
IX
Po{e)
^28.80
23.40
24.75
23.71
20.84
19.46
17.69
13.63
10.00 7.26
8.98
Po
18.26
18.16
18.48
17.78
17.12
16.64
16.03
13.63
10.00
7.26
8.98
r.
3.76
2.70
2.66
1.24
•
• • • •
• • •
• • • •
• • • •
• • • •
• • • •
• • • •
Fi
6.41
4.83
6.14
6.46
4.64
8.93
3.39
1.28
• • • •
1
• • • •
• • • •
Eei(Ro)
- 16.76
-17.38
- 18.69
-14.09
- 11.74
- 10.16
- 8.76
- 3.41
• • • •
1 • • •
• • • •
Eet{B\)
-f 6.33
+ 3.93
+ 7.59
+ 10.63
+ 11.66
+ 11.18
+ 11.23
+ 13.88
1
1
• • « •
EetiR'id
+ 13.79
+ 13.63
+ 13.34
+ 14.63
• • • ■
• • • •
• • • •
• • • •
• ••• •■■■
* ■ • •
EetiRo)
+ 11.32
+ 11.61
+ 10.71
+ 12.60
+ 14.21
+ 16.63
+ 17.28 + 20.00
+ 20,00 + 20.00
+ 20.00
E6t{R\)
+ 16.07
+ 17.69
+ 16.51
+ 17.34
+ 18.40
+ 17.98
+ 18.90
+ 21.38
•••• •«••
• • • •
Eet(R'i)
+ 18.88
+ 20.09
+ 17.10
+ 18.03
• • • •
• • • •
• • • •
• • • •
•••• •■••
• • • •
64. The method of procedure when there are more than three
layers is exactly the same as has been explained for the cases of two
and three layers respectively, and the formulae already deduced are
easily extended to cover any number of layers whatever. For the
convenience of any one who may wish to use them, the formulae for
the case of four layers are given in full in an appendix.
_ _ /a\46
OF VALUES"OF ^/oj
lASTIC ^THENCr
C
I
CHAPTER VII.
WIEE-WOUND GUNS.
66. A wire-wound gun differs from a built-up gun in that one
or more of the outer layers of the latter are replaced in the former
by steel wire, primarily for the purpose of increasing strength,
steel in the form of wire having a higher elastic limit and tensile
strength than it is practicable to obtain in large forgings. When
first proposed, wire winding was relatively much more advantageous
than it is to-day, when sound steel forgings of great size, and not
greatly inferior in strength to steel wire, are readily procurable.
Moreover, the promise of quicker and cheaper manufacture, often
made for wire-wound systems of gun construction, has not as yet
been fulfilled in practice. There is little doubt but that a wire-
wound gun can be made of greater ultimate strength than a built-up
gun of the same weight, or of equal strength, both elastic and ulti-
mate, on less weight, but since the elastic strength after all depends
upon the quality of the tube, or inner layer, which is the same in
both systems, and since any reduction of weight by increasing the
violence of recoil requires an increased weight for the gun mount
and its supports, it is difficult to see any great advantage to be
gained by substituting wire for solid forgings in gun construction.
However, wire guns are in use, and possibly their use will become
more extensive as experience in their manufacture increases, and
so the principles of wire winding will be briefly discussed.
66. Winding with Constant Tension. — Let 7?„ and E^ be the
radii of the cylinder upon which the wire is to be wound, R^ being
the outer radius of the layers of wire, and let tw be the constant
tension of winding. Then, if Ar be the thickness of the wire, the
application of a layer of wire at radius r will cause the radial pres-
Ar
sure pr = ^to-—- at that radius.
But this pressure, by (24), will cause a circumferential true
stress at the surface of the bore given by
^EelR,) = - ^^^ = - <«, ,^^^ (58)
68 The Elastic Strength of Guns
And the total circumferential compression of the bore due to all
the wire will be the sum of the partial compressions given by (58),
which, since Ar is small, is given with practical exactness by
EelB;) = -\^t„ ^3^^ = _ <„ log. {^E3) (^^)
But since the greatest elastic strength of the system will result
from compressing the inner surface of the tube to its elastic limit,
when in the state of rest, the proper tension of winding is given by
putting — po for Eet{Ro) in (59), whence we have
, _ Po 0.4343 Po
^-- ( Rl—R l^ - f Bl—Rl \ (60)
Then the internal pressure which will bring the inner surface of
the tube to its elastic limit of circumferential strain will be
and the internal pressure which will bring the inner surface of the
tube to its elastic limit of radial strain will bo
^M = P§^Po (62)
of which two values, when 6o = p«, Po{p) will be the smaller and
therefore the one to be accepted, if jBI is greater than f 5J, which
will practically always be the case.*
67. The compression at Ro in the state of rest, due to the wire,
being po, the compression at R^ will be — ^^^^ — ^ po, and so the
6 Hi
true tension of the inner layer of wire in the state of rest will be
tu, ^^^ — - Po, while that of the outer layer will, of course, be tvj-
oHi
In the state of action, Po being the internal pressure, the true
tension of the inner layer will be increased ^jfi'ZTlp) f ^ + ~Wr
op jxi
while the true tension of the outer layer will be increased ^ ^
j^-Ri-
* If the compression of the bore is p (less than p,), (61) and (62) will still
give correct results provided p be put for p, in (61) and -^—^for p^ in (62j.
Wire-Wound Guns 69
If we suppose Po = Po(0)= 4 ST^^ ^^'' + P"^' *® ^^^
tension of the outer layer of wire in action will be ^«, + ^
^S^ + K
(^0 + Po), and this must not exceed the elastic limit of the wire.
68. As an example we will examine the case of a tube of radii
Ro = 5.0" and E^ = 8.0", with elastic limit 6o-=. po'= 18.0 tons,
with four inches of wire, for which $ = 40.0 tons, wound upon it,
the section of the wire being 0.2" wide by 0.1" thick.
0.4343 log 9.63779
Po-= 18.0 " 1.25527
i2i—i?2 =119.0 ... log 2.07555
Ro= 5.0 i?J= 25.0
R,= 8.0 Ri= 64.0
.fi,=12.0 i^= 144.0
i?2_iJ2= 39.0 ..." 1.59106 " .89306
.48449 . . " 9.6352 8
^^= 16.14 tons " 1.20778
« 3.35025
. =36143 lbs " 4.55803
Therefore, the constant tension of winding which will compress the
bore to its elastic limit is 36,143 pounds per square inch, or, the
cross-section of the wire being 0.02 sq. in., 723 pounds on the wire.
e^ = 18.0 log 1.25527 log 1.25527
i^ — ^J = 119.0 « 2.07555 " 2.07555
3.0 '* .47712
2.0 " .30103
" 3.80794 " 3.63185
2i^ + .fi2 ^ 313.0 " 2.49554
2i?5 — i?2 = 263.0 " 2.41996
P^{0) = 20.53 " 1.31240
Po{p) = 16.29 " 1.21189
The least of these two values, Po(p)= 16.29 tons, is the true
elastic strength of the system.
P^ = 18.0 log 1.25527
£2 + 2i2« = 114.0 " 2.05690
« 3.31217
3i?2 = 192.0 " 2.28330
10.69 " 1.02887
70
The Elastic Strength of Guns
The compression at the outer surface of the tube, and the inner
surface of the wire, due to the pressure of the wire in the state of
rest, is 10.69 tons. Therefore the true tension of the inner layer
of wire at rest is 16.14 — 10.69 = 5.45 tons per square inch.
2P^(^) = 32.58 log 1.51295 log 1.51295
R^ = 25.0 « 1.39794 " 1.39794
^ + 2i2| = 352 « 2.54654
dRl = 192.0
ti
a
<t
5.45743
2.28330
i(
Rl — El = 119.0 «
3.17413
2.07555 "
2.91089
2.07555
12.55 " 1.09858
6.84
« .83534
The increases of true tension at the inner and outer layers of wire
caused by the internal pressure Po{p)^ 16.29 tons, are, respectively,
Fig. 12.
12.55 and 6.84 tons, so that the tensions of the inner and outer
wires in the state of action are, respectively, 12.55 + 5.45 := 18.0
tons and 16.14 -j- 6.84 = 22.98 tons per square inch.
Under the internal pressure Po(^)= ^0.53 tons, the tension of
the outer layer of wire would be 24.77 tons, which is still far within
its elastic limit.
The circumferential true stresses for the case just discussed are
graphically represented in Figure 12, in which the plus ordinates
represent tensions and minus ordinates compressions. The dash
Wire-Wound Guns 71
line shows the tension of winding, the dotted lines represent the
state of rest, and the full lines the state of action with Po = 16.29
tons.
69. When, as is usually the case, the number of layers of wire is
such that a relatively small tension of winding compresses the tube
to its elastic limit, and in the state of action the greatest strain is
well within the elastic limit of the wire, a constant tension of wind-
ing serves every purpose, and, being easier of accomplishment than
a varying tension, is naturally used. If, however, economy of ma-
terial and weight is an object, it will be attained by winding the
wire with' a tension which varies from layer to layer in such a man-
ner that in the state of action all the layers simultaneously reach
the elastic limit of strain. The tension of winding which will
bring about this result is determined as follows :
70. Let Ro and E^ be the radii of the tube or cylinder upon which
wire is wound to an outer radius Rn, and let T be the constant value
of the circumferential true tension of each layer of wire in the state
T
of action, so that-n^ is the circumferential strain throughout the
mass of wire when Po acts.
Then at any point r in the wire, the existing extension (— ^)
results from the concurrent action of three forces, namely, the
tension of winding {tyj), the pressure of the outer layers of wire (;?),
and the internal pressure (Po), and if we find the change of exten-
sion at r resulting from the removal of the outer layers of wire and
the suppression of the internal pressure, and apply it to the exten-
T
sion -gr which exists under those forces, the result will be the
extension which was given to the wire in winding, and this multi-
plied by E is the desired tension of winding.
Let p and t be the radial pressure and circumferential tension at
radius r under the internal pressure Pq. Then, by supposition, the
T
circumferential strain being -^ , we have
t=T—^ (63)
72 The Elastic Strength of Guns
fR
* tdr, therefore
rp = \f'{T-i-)dr (64)
from which, by diif erentiating, we get
rdp + P^'^ ^ ~[ ^ — 3 ) ^^
dp df
h>-
4-T - ^
r
»
\,W)
whence,
by
integration,
knowing i
that.
when r =
-Rzy
P =
■■ and
t T,
P =
2 LI r
-f-
-l'
(66)
and this gives the value of the radial pressure at any point r within
the wire in the state of action.
In accordance with the principle laid down in 49, the change of
circumferential strain at radius r, due to simultaneous changes
{po and p) of the internal and external pressures, will, by (13),
be given by
Therefore, by putting in (67) — Po for p„ and — ^\{rf ~ ^
for f, we obtain the value of the change of circumferential strain
at radius r due to the simultaneous removal of the wire beyond r
T
and suppression of Po, and —a- plus this change of strain is the
extension of the wire in winding, so that the tension of winding is
given by
37 '
^A^* _ i] (4ij_^ + 2r«) — 6P,i?«
«» =
3(»^ — Rl)
T [~f (2i^o + r^)- Rli^ T + 2P,) (68)
1^ — R
71. To determine the elastic strength of the cylinder when
Wire-Wound Guns 73
wound with the varying tensions given by (68), since by (66) the
external pressure on the tube in the state of action is
we have
4B[+2^
and
of which two values of P© the smaller should be taken.
72. Since (69) gives the value of the external pressure on the
tube in the state of action, and since by (34) the change of pres-
sure at i?! due to the suppression of Po is — " ^(W ^^^'^ — ^^'
we have as the value of the pressure at R^ in the state of rest :
^'- 2 WrJ —^\-Ri{Rl^Rl)^o (72)
jp 7^2
and this must not exceed — ^^^-^ po, or the bore will be com-
pressed beyond its elastic limit of circumferential strain.
73. Since by (19) the circumferential true stress at R^ caused
by internal pressure Po in a cylinder of outer radius R^ is
3^ (ik iyf ^^ ^^^^ expression gives the value of the
change of stress caused by a change of internal pressure, and there-
fore the true tension of the inner layer of wire, which in the state
of action is T, becomes in the state of rest, when Po is suppressed,
while the tension of the outer layer of wire in the state of rest is,
of course, the tension it was wound with, the value of which, found
by putting r = fig in (68), is
E^B,)=T-^^^^ (74)
Ro- 5.0
Ri= 35.0
R^ = 8.0
B\ = 64.00
i2, = 10.0
R^ = 100.00
74 The Elastic Strength of Guns
74. As an example we will consider the case of the tube discussed
in 67, but with two inches of wire, instead of four inches, woxind
upon it:
e^ = p^ = 18.0 tonal
^, =A>x = 4o.o « ;
If the elastic strength of the wire, with its given number of
layers, does not allow of compressing the tube to its elastic limit in
the state of rest, we take the elastic limit of the wire {0^) for the
value of T, but when, as in this and most other cases, there is
surplus strength in the wire, it is necessary to find a value for T,
less than ^i, such that in the state of rest the tube is at its elastic
limit of compression, as thus the greatest elastic strength is given to
the system.
P2 7p _
Putting 9 />a ^ po for P^m{l%), we obtain the equation
B\-R?o ^ .R um-sp p
which, for this particular case, reduces to P^ = .1875 Po + 5.484.
Either (70) or (71), according to which gives the smaller value
of Po, furnishes a second equation between P^ and Po> and from
the two equations Pj is found, and then, by (69), T. In this case
an examination will show Po{p) to be smaller than Po{0)y and
(71), after substituting in it the values of Ro, Ri and po, reduces
to Pi = 1.6094 Po — 16.453.
We thus find P^ = 8.377 and Po{p)= 15.432 :
P^ = 8.377 log .92309
^ = 1.25 log .09691
C^y = 1.1604 I « .06461
-|-r(-^)* — 1]= -2406 "938130
r= 34.82 " 1.54179
The proper value for T can readily be found by trial instead of
as just shown ; thus, if we try T = ^^ = 40.0, we shall find the
compression of the bore in the state of rest to be 21.6 tons, showing
that T must be reduced about one-sixth (in order to reduce the
Wire-Wound Guns
75
compression to 18.0 tons) ; then, after a second trial, a suitable
value can be assigned to T.
We next find the tension of winding by (68), which in this case
-?^ + 161.61 r^- 3383.0
reduces to tw = ^ 05 * Giving r in this
equation the successive values 8.0, 8.5, 9.0, 9.5 and 10.0, we find, as
the corresponding values of t^, 31.25, 28.79, 26.96, 25.59 and 24.53.
These are the tensions of winding in tons per square inch for the
1st, 5th, 10th, 15th and 20th or outer layer of wire, and, when
reduced to pounds on the wire, become 1405, 1290, 1208, 1147 and
1099 pounds. The tensions for the other layers may either be
calculated as these were or found by interpolation from them.
l-warl
I
I -''' I
\ y
\
FiQ. 13.
To find the tension of the inner wire in the state of rest we
apply (73) :
2Rl= 50.0 log 1.69897
iZJ + 2J21 = 264.0 " 2.42160
P^ = 15.432 « 1.18843
" 5.30900
SRI = 192.0 " 2.28330
" "3.02570
i^-i^= 75.0 " 1.87506
14.146 « 1.15064
76 The Elastic Strength of Guns
The true tension at fij being reduced 14.15 tons when P© is sup-
pressed, the true tension of the inner layer of wire in the state of
rest is 34.82 — 14.15 = 20.67 tons.
The circumferential true stresses for the case just discussed are
represented in Figure 13, in which plus ordinates represent ten-
sions and minus ordinates compressions. The dash line shows the
tension of winding, the dotted lines the state of rest, and the full
lines the state of action with Po = 15.43 tons.
Examples VII.
(1) At what constant tension must 20 layers of steel wire 0.1"
thick be wound on a tube for which Ro = 5.0" and R^ = 8.0" to
compress the bore to its elastic limit, 18 tons? Find Po(0) and
Po{p) and the true tensions of the inner and outer wires both at
rest and when Po(p) acts.
t^ z= 27.53; Po(e) = 18.0; Po(p) = 15.43 tons.
16.84 and 27.53 tons, at rest.
30.98 and 37.82 tons, in action.
(2) A thickness of 2.5" of wire is wound with the constant ten-
sion 32.5 tons per sq. in. on a tube for which Ro = 5.0" and R^ =:
12.5". Find the compression of the bore in the state of rest, and
the true tensions of the inner and outer wires both at rest and
when Po{p) acts, if Oo^= po= 18.0 tons.
Eet(Ro) = 13.69; Po(p) = 15.93 tons.
26.48 and 32.50 tons, at rest.
31.63 and 36.48 tons, in action.
(3) A thickness of 1.25" of wire is wound with the constant ten-
sion 15 tons per sq. in. on a tube for which Ro = 3.0", R^ = 5.25"
and ^o = Po = 18 tons. Find the compression of the bore in the
state of rest, and the true tensions of the inner and outer wires both
at rest and when Po{p) acts.
Eet(Ro) = 8.75; Po{p) = 13.82 tons.
10.18 and 15;00 tons, at rest.
20.32 and 22.48 tons, in action.
(4) If, in the case of Example (2), the wire be so wound as to
be under the constant true tension 34 tons per sq. in., find the
values of Po{0), Po{p)y and the compression of the bore and the
Wise-Wound Guns 77
true tensions of the inner and outer wires in the state of rest. At
what tension must the inner and the outer wires be wound ?
Po{6)= 19.65; Po{p)= 15.91 tons.
£^*(fio)= 13.60; Eet{E^)=.2^Mio-ns,.
tw = 34.85 at i?i ; 30.02 at R^,
(5) Given Ro = 4.0", R^ = 5.5", R^ = 8.0", $o = 30.0, po = 35.0
and T = 40.0 tons (from Ei to iJj wire so wound as to have con-
stant true tension of 40 tons when Po{0) acts) ; find Po{0)y com-
pression of bore at rest, true tension of inner and outer wires at
rest, and tension of winding inner and outer wires.
Po(^)= 28.39; j&ef(i?o)= 31.29 tons.
Eet{Ro) =27.47; Eet{R^) = 7.00 tons.
<v = ^5.83 at 2?i ; 21,07 at R^.
CHAPTER VIII.
ELEMENTARY GUN DESIGN.*
75. General Considerations. — The modern high-powered gun is
espentially a compound cylinder designed to withstand rapidly vary-
ing but not instantaneous internal pressures. The object of the sub-
division of the gun into various elements is twofold : 1st, to increase
the range through which the metal of the gun may be worked and
thus increase the magnitude of the resisting elastic forces by
assembling the elements with shrinkage; and 2d, to insure the
homogeneity of the metal and thus the safety of the guri by its sub-
division into sufficiently small elements. It is a principle of metal-
lurgy, in the present state of the art, that there is a practical limit
to the size of cast steel ingots. If this size, which may be determined
solely by experience for each kind of steel, is exceeded, the ingot
will have unsound areas which no subsequent forging can entirely
cure. This unsound metal, in the forms commonly known as segre-
gations, sand-splits, streaks, and blow holes, must be carefully
avoided during manufacture if the guns are to merit a proper degree
of confidence. Manufacturing processes are undergoing constant
improvement, but at the present time two principles must be in-
variably considered in gun construction : 1st, that in high-powered
guns there should be at least two elements resisting stresses whose
character is definitely known ; and 2d, that a sound forging cannot
be obtained if its wall-thickness, its length, and its diameter are all
very great. Furthermore, the weight of a gun has an important
bearing on its mounting on board ship, and since the weight in-
creases nearly proportionally to the cube of the caliber it is apparent
that this fact and the above two considerations tend to limit the
caliber and power of naval guns.
♦ Written by Lieutenant ( j. g.) R. K. Turner, U. S. Navy, at the Naval
Gun Factory, February, 1916.
Elementary Gun Design 79
If a pressure curve is drawn from tlie formulas of interior
ballistics, it is seen that the whole gun in rear of the base of the
projectile is subjected to the pressure represented by the successive
ordinates passed by the projectile during its travel down the bore.
When the base is opposite the maximum ordinate the whole gun in
rear of this ordinate is subjected to the maximum pressure and
should therefore be cylindrical from the breech to this point. The
forward portion of the gun, however, is subjected to continuously
decreasing pressures and may therefore continuously decrease in
thickness. This decrease in thickness may be theoretically pro-
portional to the decrease in height of the pressure ordinates. For
this reason the gun is made smaller at the muzzle than at the breech
and thus an economy in weighit and cost is effected. The muzzle
itself is flared out in the form of a bell because the metal at that
point is not supported on the forward side and it is thought that the
absence of slightly extra strength might induce splitting. We
know that the resistance formulas do not tell the whole truth, since
they take into consideration neither the supporting nor the shearing
effect due to the continuation of the metal beyond the particular
section considered, but experience has shown that the formulas in
use give the best approximate mathematical measure of the strength
of the gun as a whole, at least relatively to guns of proved worth.
76. Longitudinal Resistance. — In the deduction of the resistance
formulas the gun is considered to be undergoing strains in the
planes normal to the axis only. This assumption does not accord
with the facts, since part of the gun resists for a short time the total
gas pressure on the face of the breech block. Suppose that section
of the gun which takes this pressure, t. e,, those elements to which
the block transmits its stresses, have inner and outer radii of R'o
and R'„, respectively, and the minimum obturator radius is p©, the
bore pressure per square inch being Po. If the gun did not recoil,
the section under consideration would sustain a longitudinal stress T
in addition to the transverse stresses, such that :
and
This stress would exist only to the rear of the plane of attachment
of the gun to the carriage, which is usually a shoulder turned on the
80 The Elastic Strength of Guns
outside near the breech. A yoke to which the piston rods are
secured takes against this shoulder.
As a matter of fact, however, the gun recoils, and in doing so
relieves this stress to a certain extent. Let W be the weight of the
recoiling parts, w^ the weight to the rear of the longitudinal in-
stantaneous center of pressure of the screw-box liner, v the velocity
of recoil, and Re the constant brake resistance; the total effective
thrust, F, on the breech of the gun, neglecting the friction of the
projectile in the bore, will be
IP W dv
9 dt
The total rearward force across any section forward of the breech
diminishes proportionally to the decrease of the mass forward of
that section. Therefore the maximum stress will be in the plane of
the longitudinal instantaneous center of pressure between the screw-
box liner and the gun. The force F' at this point will be :
g dt
and the ratio between the two forces is
F " W ' W
But the total force acting to push the gun to the rear is the differ-
ence between the total gas pressure and the constant brake resist-
ance, or
r = vpo'Po — Rv
and therefore the total stress on the metal of the gun is :
F^= — "^ {irpo'Po-Re)
and the unit longitudinal stress is :
m_ F' _ W-W, irpo^Po-Re .^..
7r{Rn'^-Ro'^) W ^ 7r(Rn'^-Ro'^) ^
This force acts only in the plane of the instantaneous center of
longitudinal pressure of the screw-box liner against the threads of
its housing. From this point forward the stress decreases as far as
the yoke shoulder. At the yoke shoulder it suddenly changes, how-
ever, and the only, force acting becomes that of the inertia of the
mass forward of any section considered. If this mass is taken
Elementary Gun Design 81
equal to—?, where W2 is the weight of the gun forward of the section
considered, the total stress is :
F= !?!? X —
g dt
It is useless to attempt to calculate the exact unit stress in any
layer because the gun is not a homogeneous tube and we cannot
state the relations between the stresses of the various elements. The
work is unnecessary, however, because the total force is small and
may be neglected.
77. Gun Projects. — The preliminary design of a gun is called a
project. It includes tentative sketches and rough computations as
to maximum strength, muzzle velocity, and chamber capacity.
When it has been decided that a gun of a new type is needed the
general requirements of such a type are tentatively fixed and the
project commenced. For instance, suppose that a new gun is
desired, the progress in artillery having reduced the comparative
value of the existing type. Progress being usually along lines of
greater power, reduction of erosion, ease of operation, rapidity of
fire, or increase in striking energy, it is probable that as many im-
provements as possible along each of these lines will be incorporated
in the new gun. The caliber is first settled upon, and then the
approximate length in calibers. In the case of small guns the
muzzle velocity is tentatively fixed, but since erosion is proportion-
ately larger for large guns it usually seems more desirable in the
case of large calibers to fix the limit of pressure and with that
pressure to get as high a velocity as possible. Several sets of com-
putations are made with variations of the chamber capacity and
powder characteristics until a proper combination is secured.
Suppose it is required to design a 12-inch 50-caliber gun. With
the three elements of caliber, length, and powder pressure several
chamber capacities are chosen and calculations made as to the
efl*ects of several powders in them. From previous experience as
to the limits of allowable densities of loading the weight of powder
to be used is approximated and then the various elements varied
until several reasonable combinations of chamber capacity, weight of
charge, muzzle velocity, and maximum pressure have been obtained.
For several years the allowable densities of loading have risen in
value, due to the use of more progressive powders and the tendency
82 The Elastic Strength of Guns
toward a reduction in the size of chambers for a given power. It is
desirable to have a short chamber so as to lose as little of the travel of
the projectile as possible and also to g^t more uniform ignition, and
to have a small chambrage in order that the outside dimensions of
the gun need not be too great. As a general rule, though a rule that
is departed from without hesitation, it may be stated that the length
of the chamber is usually between 6 and 7 calibers, and the cham-
brage is about 1.20. At least the ratio of chamber length to cham-
brage is kept near these approximate proportions.
The general design and method of attachment of the screw-box
liner is selected. Its length has usually been fixed at about one
caliber, but the tendency at present seems to be toward an increase
in this dimension. An attempt is made to eliminate defects that
may have appeared in previous designs.
Several drawings are now made of the project. The length in all
cases is equal to the length in calibers times the caliber plus the
length of the screw box.
So many variables enter into a design that experience, based on a
sound understanding of the principles of gun construction, can be
the only safe guide. The consequences of the bursting of a gun in
service are so grave that all possibility* of such an accident must be
avoided, and yet the gun must not be made excessively heavy nor of
a form that cannot be mounted in turrets that have proved the most
satisfactory. Experience has shown the general form a gun must
take to give the best results with the powders in use at present, and
no radical changes in this form can be made without inviting cer-
tain disaster. With any new design it is attempted to retain the
advantages of previous types and to eliminate any defects tliat have
shown up in service or may seem to be indicated by carefully tested
theories. Therefore, in laying down a gun the previous designs are
closely followed so far as regards the general outline, thickness and
length of elements, mode of attachment of the various parts to each
other, manner of assembly and approved practice in general where
it appears to answer the purpose. The radical change of too many
variables being inadmissible, it follows that progress is necessarily
slow, and that at one stroke all previous defects may not be elimi-
nated and a gun produced that will be perfect for all future time.
With these considerations in mind the outline of the new gun
will follow closely the outline of a previous gun that seems best
adapted to the purpose; changes in the outer dimensions will be
Elementary Gun Design 83
made where it seems necessary and thus the form of the gun will
be arbitrarily fixed. It may be that a gun of the same caliber will
not be chosen as a pattern, but one of a smaller or larger caliber that
seems to have fulfilled certain of the requirements for the new type.
For two reasons the breech cylinder over the powder chamber is
usually larger than the slide, which is also cylindrical. The first
reason is that the chamber diameter under the breech cylinder is
larger than the bore diameter under the slide cylinder, and there
must therefore be an increased outside diameter for strength. The
second reason is that if the gun is heavier at the breech its center of
gravity will be farther from the muzzle and a smaller length need be
put inside the turret. The gun usually has an approximately con-
stant slope from the slide cylinder to the neck cylinder just in rear
of the muzzle ; the muzzle bell is also a f rustrum of a cone similar
to previous types.
The question then arises as to the number of layers of metal to use.
Generally large calibers have either four or five layers: four if the
tube is later to be bored for the insertion of a liner and five if the
liner is to be included in the gun as originally built. This rule is by
no means rigid, however, as witness the 14" Mark IV gun with four
layers, liner included. The practice most in favor at the present
time is to build five-layer guns with a liner tapered from breech to
muzzle for easy removal.
The problem now is to apportion the metal among four layers,
the inner and outer radii being given. For the greatest theoretical
transverse strength the law of thickness requires that if Ro, t\ R^,
R2, and Eg are the respective radii from the bore outward, they must
be connected together by the following relations :
t^zzRqR^ r^ =xt2 R2 ^^R\Rz
These ratios may not be rigidly adhered to for the following reasons :
1. For large caliber guns the breech diameter of the liner must
be great enough to allow for at least three shoulders having a height
of from 0''2 to 0''25 and the proper taper and yet leave sufficient
metal at the muzzle for rigidity and for the prevention of creep due
to the mandrelling effect of the projectile.
- 2. It is desirable to have a heavy tube so as to provide rigidity
for the gun and so prevent droop of the muzzle.
3. The layer carrying the screw-box liner must have enough addi-
tional thickness to provide for taking the longitudinal stresses
84 Tub Elastic Strength of Gons
without impairing the transverse resistance of the gun. The usual
rule is to compute this layer for longitudinal strength and then
make it from 2.6 to 3 times as thick as necessary to carry the longi-
tudinal stress. The extra thickness is taken about' equally from the
contiguous layers on both sides. The calculation for strength is
usually made by equation (76).
4. The thickness of the outside layers must not be so great that it
will be impossible to get good f orgings.
6. Sudden and great changes in the diameter of the gun or its
component parts must be avoided.
It is apparent that in the etise of a large gun with a large number
of elements, as, for instance, the Mark VII 12*' 60-caliber gun,
which is in twelve parts, considerable juggling will be necessary
before the above conditions can be satisfied and yet obtain sufficient
transverse strength.
Having decided upon the various diameters near the breech, at the
forward end of the slide cylinder, and at the neck the related ques-
tions of the manner of assembly and the character of the joints and
shoulders are taken up. The following principles in this connection
must be rigidly observed :
1. Joints must be of such a character as to allow the elements to
be easily assembled.
2. The tube and liner must be locked to prevent crawl, and all
other elements must be locked both ways to prevent movement in
either direction.
The tube and liner are so long that ordinarily the shrinkage
friction will prevent rearward motion, but shoulders must be pro-
vided to keep them from going out at the muzzle.
Locking is accomplished by means of locking rings, locking
hoops, and. shoulders. Locking rings are relatively short and thin
rings either hooked or screwed to the elements of the gun ; they are
not assembled with shrinkage and do not contribute to the trans-
verse strength. Locking hoops ordinarily attach to the other ele-
ments by hook joints and are assembled with shrinkage; they are
longer and heavier than the rings.
Shoulders are turned on an element to prevent relative longi-
tudinal movement between it and the element shrunk over it. The
distance between shoulders varies as experience dictates. Their
height may be from 0''2 to 1''0, the usual height being about 0T5
Element AKY Gvs Design 85
where possible. As a general rule two shoulders are not put in the
same transverse plane, because a plane of rupture is most likely to
form at a shoulder, and it is best to scatter the weakest parts so
that one plane will not include the weak points of several layers.
The same rule is followed in the case of joints.
Butt joints are avoided when it is possible to use a lap joint.
The latter are preferable because they distribute the weakness over
a greater length, they assist locking, and contribute to the stiffness.
Joints at the outside of the gun in particular must be designed so
as to prevent droop, as droop is due partly to stretch of the metal
and partly to working at the joints.
The several drawings are worked up to embody the various ideas
that have been expressed. If there are three drawings, for example,
one may show a heavy gun, one a light gun, and one a gun of medium
weight, and in each the arrangement will be slightly different.
Possibly one drawing will be of a four-layer un lined gun, one of
a four-layer lined gun, and one of a five-layer lined gun. Or, in one
the joints and layers may be arranged according to previous designs
and in one they may be laid down on a new plan. During their
construction the drawings are subjected to continuous criticisms
and change and new ideas are included as they may occur to those
in charge of the project.
Finally, after several weeks^ work, when the various projects seem
to answer the requirements determined upon, the total weight, loca-
tion of the center of gravity, and an approximate strength curve
are computed for each. They are then submitted for decision and
final criticism.
Usually one of the projects is decided upon, though it may be
desirable to make a few minor changes in it. The exact chamber
is definitely selected and, as a rule, the maximum bore pressure and
the muzzle velocity are fixed, together with the desired weight of
charge. Orders are then issued for the definite working up of the
design, and a decision is made as to whether the batteries of one or
more ships are to be built at once or a type gun only. It is the usual
practice to build a type gun when a new caliber is in question or
when the changes have been numerous and radical as compared with
existing guns.
A Mark is then assigned to the design selected.
As a rule, the drawings are worked up in the following order:
1. Shrinkages, strength, velocity, and pressure curves.
86 The Elastic Strength of Guns
2. General arrangement.
3. Details.
4. Chamber and breech.
6. Bough forgings.
6. Shrinkage sheet.
7. Center of gravity for shrinkage pit.
8. Rifling.
Other drawings may sometimes be required, but these drawings
are always made, though not always in the above order.
The breech mechanism drawings and computations constitute an
eritirely separate set.
CHAPTEE TX.
GTIN COMPTITATIONS.*
78. Preliminary Computations. — A pencil drawing of the gun is
laid down and the sections selected for strength computations.
These sections vary in number according to the gun; in some cases
there are as many as 28. In the case of the gun selected for the
purposes of illustration, the Mark VII, Modification 3, 12" 50-
caliber gun, computations were made at 24 sections. The sections
are numbered in Roman numerals, the lowest number being near
the breech.
The principle governing the selection of sections may be generally
stated as follows : '^ Computations must be made for every plane of
the gun having a strength different from that of the planes on either
side of it, and where there is reason to believe a sudden change in
strength occurs, on both sides of the change and close to it.'^ The
plotted results of the computations must give a continuous strength
curve from breech to muzzle.
In order to reduce the immense amount of labor involved in the
case of a gun of large caliber, the computations are made on printed
forms. The Birnie formulas, involving the introduction of sub-
sidiary constants, are used. These are the same as those given in
this book by Professor Alger, but arranged for greater convenience
and known as the *^ Reduced Formulas.'' For a thorough under-
standing of their meaning it would be necessary to deduce each one
from the fundamental equations ; this work is not given here as it is
easy enough, though long.
In considering these forms we find various methods used that are
not those that we have been accustomed to. It is important to know
the formulas on which the forms are based. Logarithms are denoted
by letters only. An expression such as a^ (0^) indicates that flg is to
be multiplied by 0^ ; therefore their logs are to be added. This is
further indicated by a + sign after ^3. This method is used through-
out. Expressions are denoted by letters or numerals and are there-
after always referred to by such letters or numerals. The pressures
in the state of rest are denoted by P' instead of P as in this text-book.
* Written by Lieutenant ( j. g.) R. K. Turner, U. S. Navy, at the Naval
Gun Factory, February, 1916.
88 The Elastic Strength of Guns
Sheets 2, 3, and 4 are used for the preliminary computations and
Sheet 6 for the final computations.
It sometimes happens that the dimensions of the gun as laid down
in the pencil drawing do not given sufficient strength, or that a very
sudden break in the curve is caused by an improperly designed joint.
To correct these faults new dimensions are tentatively assigned or
the faults at the joint in <}uestion corrected. The strength is then
computed with the new dimensions.
79. Computation Forms. — Sheet 1 of the computations is headed
^' Constants depending on fixed radii and constant modulus of
elasticity ^^ and gives the values of the radii and their various com-
binations with each other, together with the logarithms.
Sheets 2 and 3 are headed '* Computations for reduced formulas
and maximum values/^ and " Computations for reduced formulas
and maximum values corrected," respectively, and give the loga-
rithmic forms for computing :
1. The maximum elastic forces Pm{Om) and Pm(pm) (for any
layer m).
2. A function lm{Pm) of the variations of pressures between the
states of action and rest.
3. The pressures, state of rest, Pm^.
4. The initial limiting pressure on the tube, P/.
5. The adopted values of Po(Oo) and Po{po) corresponding to the
minimum of P^'.
6. PnJ corresponding to the minimum P/.
Pm{Om) is the pressure at any surface that will bring the metal
to its limit of elastic tangential strain at that surface, while Pm(pm)
is the pressure that will bring the metal to the elastic limit of radial
strain. If the pressure is greater than Ptn{Om) the metal will actually
be permanently stretched tangentially and may even crack if the
ultimate strength is passed. On the other hand, if the pressure is
greater than Pm(pm) without being greater than Pm(Om) the metal
will crush slightly and so enlarge the bore, but its tangential tenacity,
upon which the actual stability of the gun depends, will in no way
be affected; in other words, Pm{pm) may be exceeded without any
other effect than a slight increase in the diameter of the bore, so long
as the metal at the outside of the layer is not strained in the same
way beyond its elastic limit. This increase in the bore diameter will
be very slight and is therefore never considered in the case of the
inner layer, so that Po{0o) is always used instead of Po{po)f no
Gun Computations 89
matter which is the smaller. It will be otherwise with the other
layers, however, because it is apparent that any increase in the bore
diameter of any layer except the first will reduce the shrinkage of
that layer, and will therefore decrease the possible range of work-
ing of the inner layers, thus reducing the elastic tangential resist-
ance. The only layer this does not apply to is the outer, since there
Pfi_i(pn_i) is always less than P«_i(^«_i). Therefore the following
rule is adopted in computing the successive values of Pm{Om) and
Pm{pm) '- "For computing the successive values of Pm{Om) and
Pm(pm) always use the smaller of the two quantities, Pm+i{Om^i)
andPm+i(pm+i).''
The functions lm{pm) is obtained from the formula:
7 (n \ ^m (/»« /^m-t-i )
and is used for the purpose of computing the variations in pressures,
pm. The latter is computed with the formula :
and the pressure, state of rest, from
•t m ^ •* m Pm
The initial limiting pressure on the tube is determined by that
pressure, Pi', which the tube will sustain in the state of rest without
passing the elastic limit of compression, since it has been shown in
the deduction of the formulas (Alger, equation 24) that the dan-
gerous strain, in a tube subjected to external pressure only, occurs
at the inner surface and is a tangential compressive strain. If the
tube is not to be bored for a liner the first of the two formulas, that
for finding po\ is not used, but the second formula only, po being
taken equal to p„ and Ro equal to the inner radius of the tube. Both
formulas must be used if the tube is to be bored for a liner.
The " Pressures, state of rest, relieving jacket " give first the
computation of the pressures in the state of action at the inner sur-
face of the jacket, using the minimum value of Pj', and then the
pressures in the state of rest in the outer layers of hoops that will
be required to produce the maximum pressures for the state of
action when the variations in pressures have been reduced propor-
tionately to the reduction in Po. In other words :
Pm' = Pm'(max.)-p«
This will subject the two outer layers to the maximum elastic
stress and will reduce the maximum stress in the jacket. Thus all
90 The Elastic Strength of Guns
the layers will not participate proportionately iii the transverse work,
and it may happen, when the working limits on Sheet 4 are figured,
that the metal of the jacket will be found to be strained beyond its
elastic limit in the state of rest. In this case it will be necessary to
reassign values of Pm to the outer hoops to make the proper adjust-
ments. Therefore this is essentially a trial method and may entail
a great deal of additional labor. For this reason the set of approxi-
mate formulas under " Pressures, state of rest, corrected, relieving
hoops ^' were adopted and are • ordinarily used. These formulas
relieve the pressures, Pm, on the outer hoops proportionately to the
reduction in P^ and differ from the theoretically correct pressures
by negligible amounts, a small constant term having been omitted
in the derivation of each of the formulas. When using this method
one may be sure of getting values of Pm that will not over compress
the jacket in the state of rest, though the total maximum resistance
of the gun may be very slightly reduced. The jacket is thus made
to do its proper share of the work, which is desirable, unless there
are special reasons to the contrary, as, for instance, when the screw-
box liner is attached only to a comparatively thin jacket.
Sheet 4 gives the " Computations for reduced formulas, shrink-
ages, and compression of the bore," using the adopted pressures,
state of rest, corrected, Pm* The formulas are self-explanatory.
Sheet 5 is a summary of the reduced formulas and a tabulation
of the values of the subsidiary constants a^ h„, c«, etc. This sheet
is no longer used, however, as it consumes more time than it saves.
In its place has been substituted Sheet 7, with one set of values
omitted, viz., the ^^Belative shrinkages." This space is then used
for writing in the ^'adopted" shrinkages in the adjustment of
shrinkages. As Sheet 7 gives the absolute values of all the quantities
required, instead of their values in terms of the subsidiary con-
stants, it is much easier to visualize all the conditions obtaining at
the various sections and thus gain a clearer viewpoint for the proper
adjustment of shrinkages.
In general the preliminary computations may be considered com-
plete with the completion of Sheets 1, 2, 3, and 4.
80. Adjustment of Shrinkages. — When the preliminary computa-
tions have been finished the values for all sections are tabulated on
Sheet 7 as outlined above. It may be noted that the absolute shrink-
ages come out to six or eight places of decimals, though it is known
that it is necessary to allow a plus or minus tolerance of about .0005
inch, since large machine turning cannot be done more accurately
Gun Computations 91
than that. It is obvious, therefore, that the assigned shrinkage can
only be given to thousandths of an inch and a total tolerance range
allowed of .001 inch.
It will be found that the shrinkages often change their value
abruptly when computed for maximum strength, and since it is not
desirable to cut a large number of shoulders on the various elements
the change must be made gradually in the form of a cone. Also, for
the sake of economy^and accuracy, it is better to have one shrinkage
extend over the greatest possible length of the surface of the ele-
ment. There are many other practical aspects of the subject of
shrinkage, as, for instance, the fact that a heavy shrinkage must not
be put on a thin section either for fear of overstraining the metal
or because it is obvious that it will not hold the shrinkage until the
next envelope is in place. All the various considerations are the
result of experience and therefore the assignment of shrinkages can
follow no definite rules that will be applicable to all cases.
In general, however, shrinkages are assigned the same value over
as long a surface as possible and the value expressed to the nearest
thousandth of an inch below the minimum theoretical shrinkage for
that surface. The various contact surfaces are considered in order
beginning at the inner, and their relation to each other must be
understood in order to assign proper values. For instance, if the
theoretical shrinkages are :
Si = .0016 S2 = .0483 §3 = . 0571
it is at once apparent that Si must be made greater than .0016, be-
cause the tube and jacket will not hold together under so small a
pressure as will result from the use of this shrinkage. Therefore a
larger value of S^ is chosen and Sz and S^ decreased so that the
pressure at the outer surface of the layers in the state of rest will not
be too great. In this case shrinkages might be assigned as follows :
Si = .012 S^ = MO S3 = .047
A reassignment will be made if these values are shown to be unsuit-
able by the computations on Sheet 6.
One other general principle of shrinkage is that it is desirable to
work the tube higher than the outer layers ; in other words, the tube
is to be considered the limiting layer.
81. Final Compiitations. — ^The final computations are made on
Sheet 6 and the results tabulated on Sheet 7, together with the
maximum theoretical pressures found on Sheets 2, 3, and 4. From
92 The Elastic Strength of Guns
these tabulated values are constructed the curves of tangential and
radial resistance and relative compression of the bore.
Sheet 6 shows the assumed values of the shrinkages and gives the
forms for the logarithmic computation of values of the compres-
sions, the pressures Pm and Pm^ and the tangential compression
resulting from the use of the assigned shrinkages.
In addition to the formulas for finding the necessary quantities
are a considerable number of check formulas obtained by the trans-
position of the regular formulas. For instance, there are three sets
of computations for " Working limits/' one for checking the theo-
retical values of PmiJ^m) and Pmipm), one for checking the assumed
values of Pm{Om) and Pm{pm), and one for checking the values of
those quantities after the assumed values of the shrinkages have
been used. The formulas for " Working limits '' give the effective
values of dm and pm, when the various values of Pm{^m) and Pm{pfn)
are used, and in all cases these values must be equal to or lower than
the respective elastic limits of extension and compression for the
layer under consideration, except in the case of the inner layer,
where po may exceed the elastic limit. If, for instance, the true
value of Btn is 60,000 and we get a value of ^m = 50,000 from the
computation of the working limits, the layer could be replaced by a
layer whose elastic limit of extension is 60,000 without reducing the
height of the strength curve, and therefore all the total available
strength of the layer will not be used when the bore pressure becomes
equal to the adopted value of Po- But if the values of dm or pm are
greater than the elastic limits either an error has been made or new
values of Pm must be chosen.
82. Computations for the liner. — When a liner is to be originally
inserted in the gun it is assembled after the rest of the gun has been
built up. In this case computations as to its effect on the other
layers are made, the original computations being essentially what
would be required for a gun with a bore diameter equal to the inner
diameter of the tube.
The formulas are based on the assumption of a two-layer gun, the
tube, jacket, and hoops forming the outer layer and the liner the
inner layer. The formulas may be deduced from the theoretical
formulas for a two layer gun assembled with the shrinkage assigned
for the liner. This shrinkage is small because it is desirable to be
able to remove the liner and insert a new one without having to bore
it out. There is also a possibility of the liner's sticking during
assemblage if the shrinkage is very great^ since the assembled gun
:ial a:
vest-
mpre?
ressio:
^e
le
e
Gun Computations 93
must not be heated to too high a temperature for fear of disassembly
of the elements.
The computations are arranged on two sheets, Nos. 8 and 9. It
must be understood that so far as the constants are concerned, the
liner is treated as a regular layer. Bo being the bore radius and B^
the outside radius of the liner.
Sheets, Assumed Shrinkage = 8^,
Pressures, State of Best, at B^. — A pressure P^' on the outside of
the liner is caused by putting in the liner with a shrinkage S^^.
p /- 5, _ o ^ E(B,'-Bo'){Bn'-B,')
^""/ii ^"^ 2B^^Bn^-Bo^)xD^
ntitift
tracj-
lesei'
uma
eso:
have Change of Pressure in State of Best. — These formulas give the
rj^iye increase of pressure at the various surfaces that result from the
' ) insertion of the liner, this causing a pressure of P^' at the inner
]2an surface of the tube where no pressure existed before.
the P2=Plh P3=P2k pJ = Psh
'^^) Pressures, State of Best. — The addition of the increase of press-
"^ ures in the state of rest to the original pressures before the insertion
of the liner gives the new pressures, state of rest, at the contact
^ surfaces.
P/ = /?2' + P2' (original)
'^ Pa' = P3' + ^3' (original)
P; = p;4-P/(original)
Tangential Compression of Bore, — Tpo is the tangential compres-
sive stress caused at the inner surface of the liner in the state of rest
by an outside pressure of Pj'.
L Strength of Oun Limited by Liner. — The curve drawn through
the plotted values of P© will be the curve of tangential resistance
of the gun when the stress in the liner has a value of Oo.
Variation of Pressures. — When the gun is fired and a pressure P©
brought into existence in the bore it causes the increase of pressure
P» at the other contact surfaces.
Pl = ZaPo P2 = liPi PZ — I2P2 P4 = hPz
Pressures, State of Action. — P^ is the algebraic sum of the
pressures, state of rest, and the increase of pressure /?»,. The latter
94 The Elastic Strength of Guns
quantity is considered positive in the present case, since it is one of
tension with respect to Pm'.
Pr=p,'+p^ p,=p,'+p, p,=p,'+p, p,=p:+p.
Working Limits. — 6^ and pm are the stresses in the various layers
when a pressure Po is caused in the bore.
$.=
PAO.)
a
8
U'2 tvA Co Co
ai «! Ci ^ Ci
^^^ Po(Oo) _p^x K p^^ Po(Po) _p^
X -^
When an old gun is to be relined, computations for the liner are
made in accordance with a somewhat similar set of formulas, the
chief differences being: ]st, that now r represents the outer radius
of the liner and R^ the outer radius of the tube ; and 2d, that several
additional formulas are necessary to show the changes in the press-
ure6, state of rest, at the various contact surfaces that will result
when the tube is bored and the liner inserted.
83. Example of Oun Computations. — The Mark VII, Modifica-
tion 3, 12" 50-caliber gun has been selected for the purposes of illus-
tration, the results being given in the case of the section over the
chamber, number IV. This is an unlined gun, but provision has
been made for the insertion of a conical liner after the inner sur-
face of the tube has been worn out.
It will not be necessary to give the formulas used, as they may be
obtained directly from the computation sheets. The results only
will be given.
In this case :
Bo =15.20 r=: 17.083 R^= 19.75 ^2 = 26.0
7^3 = 34.0 K^ = U.O £7 = 30,000,000
^0=^0 = 55,000 ^i=pi = 60,000 ^o=^8=P8=P3 = 65,000
The elastic limits are the specified values, the actual values not
being used because it would be impossible and undesirable to con-
struct strength curves for each individual gun, one set of curves
being computed that will apply to all, provided they meet the
specifications.
The subscript m will be used to show that a quantity may apply
to any layer.
Gun Computations 95
Preliminary Computations.
SHEET 2. COMPUTATIONS FOR REDUCED FORMULAS AND MAXIMUM
VALUES,
1. Maximum Pressures, — ^Theoretically possible.
P3 (^s) = 15,056
Po ($2) = 33,217 P2 (P2) = 39,293 [Use Ps z= P3 (^3) ]
Pi (tfi) =53,442 Pi(pi)=i)0,095 [P2i^2)<P2{p2) Use P2 = i\'(«2)]
Po {$0) = 70,948 Po (fio) = 60,479 [Pi (pi)< Pi (^1) Use Pi == Pi (pi) ]
2. Working Limits, — Check for accuracy of computations for ( 1 ) .
^3 = 65,000 ^1=60,000 pi = 60,000
^2 = 65,000 p2= 65,000 ^0= 65,000 po = 55,000
3. Variations of Pre^SMre^.—-- Computation of Imipm) .
Zo(po) = (7.73004) Zi(pi) = (7.67236) Z2(/?2) = (7.55872)
SHEET 3. COMPUTATIONS FOR REDUCED FORMULAS AND MAXIMUM
VALUES, CORRECTED.
4. Pressures, State of Rest, — Theoretically possible.
^j = 38,105 {Po{0o) has been used for the reasons given
in §79.)
P2 = 17,920 /?3 = 6,487
Pi'=i'i+(~Pi) = 11,990 P/ = P2+(-P2) =15,297
The minimum values of Pm are used for the reasons given in §79.
The quantity pm carries the minus sign because it is negative as
compared to Pm-
5. Initial Limiting Pressure on Tube, — This is the value of the
maximum P/ that will allow the tube to be bored for the liner with-
out collapsing. The formulas are based on the assumption of a two-
layer gun.
po' = 53,166 Pi' = 10,838
It will be noted that the P/ given by (4) is greater than that found
here, therefore the latter value will be used for correcting the maxi-
mum allowable pressures.
6. Po Corresponding to Pn-i- — Check for " Variations of Press-
ures " and " Pressures, State of Rest.^'
Po = 70,948
7. Po Corresponding to P^-^- ( — f^i). — Computation of subsidiary
constants, check for P/, and computation of radial resistance when
Po{0o) (theoretical) is used.
96 The Elastic Strength of Guns
Po{0o) =70,9^9
PoU) =52,405
It may be noted that the Po(po) found here is less than the theo-
retical Po{po) ; this will always be the case when Po{Oo)>Po{po)
as found on Sheet 2.
8. Pressures Corrected.— This gives the maximum theoretical
Po{0o) and Po{po) using the minimum of the two values of P^, and
the maximum variations in pressures corresponding to the adopted
value of Po. If the initial limiting pressure on the tube is not found,
this computation is unnecessary.
Po(^o)adopted = 67,425 Pp(po) adopted =51,081
p^=: 36,213 Po(^o) (adopted) is used for the reason given in §79.
P2 = 17,030 P8 = 6,165
9. Pressures, State of Rest, Relieving Jacket. — See explanation
of this and the following set of formulas in §79. It may be noted
that only the first of the three following formulas has been used, so
as to obtain the pressure, state of action, at the outer surface of the
tube corresponding to P^ (min.) . The hoops and not the jacket have
been relieved in this gun.
Pi = 47,051
10. Pressures, State of Rest, Relieving Hoops. — ^These now be-
come the preliminarily adopted values of the pressures in the state
of rest on the jacket and C-hoop.
P/= 14,537 P8' = 8,209
SHEET 4. COMPUTATIONS FOR REDUCED FORMULAS, SHRINKAGES,
AND COMPRESSION OF BORE.
11. Shrinkages. — The shrinkages here found are those necessary
to give the adopted pressures in the state of rest.
Si = .0092 153 /S2 = . 039499 ^3 = . 050832
12. Compressions of Bore. — 81, 82, and 83 are the partial relative
bore compressions that result from the successive shrinkage of the
three outer layers, and the final relative compression is their sum.
The same applies to Am, the absolute compressions.
81 = .00029984 Ai = .0045575
82 = .00078837 A2=. 011983
8, = .00068397 A3 = .01 0260
Bo = Si + 8j + 83 = .00177218
Gun Computations
97
13. P/ Corresponding to So. — This is a check for (12).
Pi' = 10,838
14. Tangential Compression,
^=53,166
p should equal the elastic limit of the metal if the theoretical values
of Pj' has been used. If a lower value has been adopted such that
the bore of the tube will not be compressed to the elastic limit in the
state of rest, p will be less than Oq.
16. Working Limits. — The values of 6m and pm are the effective
elastic limits as defined in §81. They are introduced as a check on
the accuracy of the preceding work, and to find the relative partici-
pation of the layers after the outer layers have been relieved so that
the tube will not be over-compressed when bored for the liner. It
may be noted that Bo equals the allowed elastic limit, while po greatly
exceeds this limit; this is in accordance with what has been said in
§79. In the case of the other layers neither 6 nor p may exceed the
proper elastic limits. The ratios of Om found here to the allowed Om
show the relative participation of the various layers.
(90 = 55,000 po=7S,270 (92 = 61,770 ^2=48,649
(9i=41,842 pi = 55,770 ^3 = 61,772
Sheet 5 is no longer used, but Sheet 7 instead. This will be
called Sheet 7a.
Final Computations.
Sheet 7a is for the adjustment of shrinkages. As the shrinkages
are adjusted with relation to the other sections, the shrinkages for
Sections II to IX are tabulated below to show the method used.
THEORETICAL ABSOLUTE SHRINKAGES.
Shrink-
Section.
age.
IL
IIL
IV.
V.
VL
VII.
VIII.
IX.
s.
.0249
J
m m
54
•
.0092313
.039574
.050773
.0092153
.010705
.010428
.039428
<-
.013466
.039075
.049194
.01258"^
.039462
.049550
.012470
.03756?^
s.
.039499
.050S32
.039415
8,
• • •
.050822
.049281
.047608
98
The Elastic Strength of Guns
The small arrows indicate the presence of a shoulder between the
two sections where they occur. Thus the tube has a shoulder between
Sections VI and VII.
From what has been said before it is at once apparent that it
would be impossible to obtain these theoretical shrinkages on account
of their wide variations and it would therefore be useless and bad
practice to assign them. The first thing to do is to examine this
table carefully and then by balancing the various considerations
governing the adjustment of shrinkages finally arrive at a logical
conclusion.
Si for Section II, where there are two layers only, may at once be
given a value of .025.
Prom Sections III to VI S^ varies from .0092313 to .010428. In
no case may these shrinkages be exceeded without over-cotopressing
the tube when bored for the liner, so that a proper value of S^ for
these sections seems to be .009. For the same reasons S^ from Sec-
tions VII to IX is given a value of .012.
Proceeding in this way from one surface to another the shrinkages
are relieved slightly in every case, until finally the shrinkages as
given in the table below are tentatively adopted.
ASSIGNED SHRINKAGES.
Shrink-
Section.
age.
II.
III.
IV.
V.
VI.
VII.
VIII.
IX.
s.
-<-
.025
• • • •
.009
• • • •
->-
• • ■ •
4- •
.012
• • • •
.012
8,
• >
• •
k
• « • •
.039^
• • • •
.037
....
• • • •
• • ■ •
8,
• • •
• • • »
.050
m • • •
<-
• • • •
.047
* • • •
• • • •
The assigned shrinkages must now satisfy the conditions that the
tube will not be over-compressed when bored for the liner and that
no metal in the gun is strained beyond its elastic limit. Whether
it fulfills these conditions is determined in the computations on
Sheet 6.
As a matter of fact the shrinkages finally assigned to Section III
were iSi=:.009, /S2 = .025, and 8^^.025, because the cross strains
due to the heavy longitudinal stresses at that section made it
advisable to relieve the tangential strains that would have occurred
if the values given in the table had been used.
Gun Computations 99
When a large change in the assigned shrinkages occurs at a sec-
tion the change is made gradual by the use of a coned surface. This
method is shown in the small figure in the lower left hand corner
of the drawing of the strength curves, etc.
SHEET 6. COMPUTATIONS FOR ADJUSTED VALUES.
These formulas are similar to those on Sheets 2, 3, and 4, except
that the subsidiary constants do not have to be computed, and the
correction and most of the check formulas may be omitted.
16. Assumed Values.
Si = .009 fi'2 = .039 fif3 = .050
17. Shrinkages and Compressions. — Computations are made
using the assumed value oiS and subsidiary constants from Sheet 4.
<^i = .00045569 8i = .00029283 A^ = .0044510
<^2 = .0015000 82 = .00077841 A^ = .011832
<^3 = .0014706 83 = .00067279 Ag = .010226
80 = .001 74403
18. Pressures, State of Rest, — These are computed from the
relative compressions and certain subsidiary constants obtained
from Sheet 4.
P3' = 8,075 P2' = 14,328 P/= 10,665
19. P/ Corresponding to So. — Check for (17) and (18).
P/ = 10,665
The value of P^ must not be greater than the ^' Initial limiting
pressure on tube ^' found on Sheet 3.
20. Tangential Compression. — This must not be greater than the
elastic limit of compression.
p= 52,320
21. Pressures, State of Action. — The first five of these formulas
are similar to those in (8) ; the remainder are those in (4) reversed.
P^ {60) = 66,898 Po (po) = 50,883
Pi = 35,930 /?2 = 16,897 ^3= 6,117
Pi = 46,595 P2 = 31,225 Pg = 14,192
22. Working Limits. — Eelative participation of layers,
^0 = 55,000 /5o= 77,796
^1 = 41,610 /,! = 55,286
^2 = 61,222 /,2 = 48,164
^8 = 60,989
100
The Elastic Strength of Guns
The values of dm must not be greater than the elastic limit ; 6 will be
equal to the elastic limit of the tube, while in general the values of
6 for the other layers will be less than the elastic limits of those
layers, po may be greater than the elastic limit of the tube, but the
values of p for the other layers must not be greater than the elastic
limit.
SHEET 7 IS THE TABULATION OF THE COMPUTATIONS AND GIVES
THE VALUES OF
]\raximum pressures.
Initial maximum compression (P/ only).
Limiting pressure on tube.
Adjusted pressures.
a. In action.
b. At rest.
Shrinkages.
a. Relative.
b. Absolute.
Compressions of bore.
a. Relative.
b. Absolute.
Working limits.
Tangential compression.
When all the sections have been computed the curves of tangential
resistance, radial resistance, and relative compression of the bore
are drawn with the values of Po{Bo), Po{po), and 8©, 8i, Sg? ^-^^ ^»
computed on Sheet 6. The curves of velocities and pressures in the
bore are then, drawn for purposes of comparison. It will seldom
happen that any of the curves will be changed after all the sections
have been computed, because the strength actually obtained for each
section is compared with the requirements as soon as the computa-r
tions for the section have been finished.
The figure shows the drawing of the "Shrinkages, strength,
velocity, and pressure curves" for the Mark VII, Modification 3.
12" 50-caliber gun.
Formulas
No
. 1.
(C
«
4.
«
«
5.
u
<(
21.
«
66
18.
i<
66
17.
((
66
16.
<(
66
17.
«
66
17.
((
(C
22.
((
(C
20.
oTor
-^t-k%
9^1
Lu»VAJL.CNT pH
« •
.^iH
APPENDIX.
FOBUULiE FOB THE CASE OF COMPOUND CYIINDEBS OF
FOTJB LAYEBS.
(1) P^O)- 4jj2_,.2ii,i
(4) -PoW = -^ AK{ + 2i?l
,M P , . 3(.R| - - Rl) />a + 2P3-BI
W A(^) ~4P?— 2ii|
(A)
(6)
3(i?l - J?j) />! + 2P2i?i
4i^ — 2i?5
m p.. 3(ig?-.R',)/>o+2Pxi?'x
(0 -f^»(i»; 4fi2 _ 2Ei
3(J^-J2^„) (g<, + Po)
If Po{6) is greater than \_Po], the tube will be compressed
beyond its elastic limit of compression (po) by shrinkages deter-
mined with the values of P^ie) P^ie), Pi ((9) and Po{0), and so
the values of one or more of the assumed elastic limits ^3, 6^ and
$1 must be reduced until Po(^) equals, or is less than, [_Po]-
. 2i^ r p^d) (4R| + 2i^) - 6Po(e) Br \
W ^s-^g-L^/sH 3(i^— i^) J
2i2, r P,(<?) (4J?, + 2i^) - 6P„(g) Pg -]
_ 2P, r p,(^) (4BJ + 2P?) - 6P,(<?) pn
(8) ^^--:e-U'+ 3XP? — El) J
In these expressions for the shrinkages, the values of 0^, O2 ^^^ 6^
are not necessarily the real elastic limits, but are the assumed
102 The Elastic Strength of Guns
elastic limits with which the finally accepted values of Ps{0),
P2W, PiW aJid Po(e) were calculated.
(1) Ps = Ps«^)-|^|]p,«?)
(2) A = p^<^)-f|fEi)^oW (C)
These are the pressures at the surfaces of contact in the state of
rest, Ps{0), PziO), Pi(0) and Po(0) being the values of the pres-
sures in the state of action used in calculating the assigned shrink-
ages.
et[^o) Ei — Rl 2R~m—Ri 2Ii^~Rl — £!l 2E^ ^^^
This is the circumferential strain at the surface of the bore
caused by the superposition of the three outer layers with their
respective shrinkages 8^, 82 and 8^, the successive terms being the
three circumferential strains produced by the three successive layers.
2Roet{Ro) is the change of diameter (contraction) of the bore
from its free state to that of complete assemblage of the system,
and — Eet{Ro) is the circumferential compression of the bore in
the state of rest.
The radial strain at the surface of the bore in the state of rest is
ep{Ro)^ — ie'tiRo), so that it is under a true tension radially
one-third as great as its circumferential compression. Therefore
the real elastic strength of the system when assembled with shrink-
ages 81, 82 and 8^ is the least of the two following values of Po -
(2) PS'> = |g^^(/>,-J^e,W)
In these expressions it i-s important to note that et{Ro) is a
negative strain, so that the last factor in each of the two values of
Po is numerically the sum, not the difference, of the elastic limits
(of tension and compression respectively) and the true stresses at
the surface of the bore in the state of rest (circumferential and
radial respectively).
Appendix 103
The pressures in the state of rest may be computed directly from
tbe shrinkages by the following formulae:
^^^ ^^""^ 214 • ni—Ry ^ET
[^) ^2-J^ 2iJi [li^^ — Ri~2R^'^Ri — Ri2lt,) (F)
(<K\ p_j, I^-Iil ( Iil-Rl 8, ,Rl-RI 8,
(i5) r^-JL 2^2 \Rl—Rl 2R, "t" Rf^^ 2R^,
, Rl-R l 8, \
■+"i21 — ^1 2R,)
The terms in the parentheses are the values of the circumferential
strains at Ro caused by the assemblage of the successive layers,
their sum with the negative sign being the total compressive strain
at the surface of the bore as given by equation (D).
Prom the pressures in the state of rest, as given by (F), the
pressures in the state of action may be found by equations (C),
and the true circumferential tension of the inner surface of any
layer can then be found by
J^A^ n-l j Z(Rl—Rl^^) y^^
In this Rn and Bn—i are the outer and inner radii of any layer,
Pn and Pnr-i are the outer and inner pressures (either of action or
of rest) and E^i iRnr-i) is the true circumferential tension at the
inner surface of the layer resulting from the action of Pn and Pn-i.
Similarly, the true radial compression at the inner surface of any
layer, either in the state of rest or of action, is given by
TP^ fT>f \ Pn—\ \^Rn ^Rn—l) ^^nRf i /tt\
r.
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