TJ
25 CENTS
B 3 Dlfl flis
THEORY OF SHRINK- II
AGE AND FORCED FITS
BY WILLIAM LEDYARD CATHCART
WITH TABULATED DATA AND EXAMPLES
FROM PRACTICE
MACHINERY'S REFERENCE BOOK NO. 89
PUBLISHED BY MACHINERY, NEW YORK
MACHINERY'S REFERENCE SERIES
EACH NUMBER IS ONE UNIT IN A COMPLETE LIBRARY OF
MACHINE DESIGN AND SHOP PRACTICE REVISED AND
REPUBLISHED FROM MACHINERY
NUMBER 89
THE THEORY OF SHRINKAGE
AND FORCED FITS
With Tabulated Data and Examples from Practice
By WILLIAM LEDYARD CATHCART
SECOND EDITION
CONTENTS
Introduction - - 3
Preliminary Considerations - 4
Derivation and Application of Lame's Formulas 9
Formulas for Stresses in the Hub - - 16
Formulas for Stresses in the Shaft - - - 19
Shrinkage Allowances - - - 25
Calculating Shrinkage Fits - - 30
Practical Considerations - - :>- 33
Copyright, 1912, The Industrial Press, Publishers of MACHINERY,
49-55 Lafayette Street, New York City
INTRODUCTION
FORCED AND SHRINKAGE PITS
A shrinkage fit is a cylindrical or slightly conical joint between two
machine members, as a crank-web and a shaft, in which the bore of
the outer member or crank is smaller than the diameter of the inner
member or shaft, so that the outer member must be expanded by heat
before it can be set in place, while, in the subsequent cooling, it con-
tracts and grips the inner member with a force which depends on the
character of the metals, on the thickness of the outer member, and
on the difference between the original diameter of the bore and that
of the inner member. This difference is called the allowance for
shrinkage. A forced fit is based on the same principle and is virtually
of the same character, except that the parts are forced together when
cold by hydraulic or other pressure.
These fits have a wide range of application, extending from small
machine parts to built-up crank-shafts for heavy engines and the mass-
ive forgings for high-powered guns. As a rule, the forced fit is re-
stricted to parts of small or moderate size, while shrinkage joints have
no such limitations, being applicable especially where a maximum
"grip" is desired, or, as in ordnance, where accurate results as to
the intensity of the stresses produced in the parts thus united, are
required. With both types, skillful machining and care in assembling
are essential; but the shrinkage joint is compact, has the fewest pos-
sible parts, is secure against slip to the extent for which it was de-
signed, and is tight against fluid pressure.
The fundamental principle governing the construction of the joint
is the same with both types: the bore of the outer hub or other mem-
ber is smaller, and the diameter of the pin or shaft larger, than the
diameter of the finished fit. Hence, the inner member is compressed,
the outer expanded, and the elasticity of the metals produces a radial
pressure at the contact-surfaces of the fit, which pressure gives the fit
its resistance to slip. The same principle is applied in the rolled joints
used in expanding the ends of boiler tubes in place, although, in this
case, the process is reversed, the hollow inner member or tube being
stretched by rolling so that, if free, it would be greater in diameter
than the hole in the tube-sheet or header.
As the integrity of the fit thus depends on the elasticity of the metals
of the members, and as the formulas which follow are based on this
elasticity and on the actions which occur during expansion and com-
pression, it may be well to review these actions briefly and to give the
sense in which the various terms relating to them are used in this
treatise.
CHAPTER I
PRELIMINARY CONSIDERATIONS
Stress Deformation Lateral Contraction
An external force applied to a body acts, partially or wholly, to
change the shape of the latter. A stress is the force acting within the
body to oppose this change of shape. The unit stress is the stress
on a unit of area of the cross-section. Thus, if the upper end of a
steel rod, one inch square, be fixed, and a weight of 10,000 pounds be
suspended from the lower end, the unit-stress on the metal will be
10,000 pounds; if the sectional area of the rod be two square inches and
the weight remain the same, the unit-stress will be 10,000 -=- 2 = 5,000
pounds. Stresses may be either tensile (those that tend to elongate
the body), compressive (those that will shorten it, as in a column), or
shearing (which act to cut across the body, as in punching a rivet
hole). Both tensile and compressive stresses may act at the same time,
in the same line, on the same body, in which case the resultant stress
will be the difference between the two, and in kind like the greater.
Tensile stresses are usually considered as positive, and compressive
stresses as negative, the resultant stress being their algebraic sum.
An external force not only puts the material under stress, but also
causes some, usually slight, change in its shape. This change is called
a deformation, and this deformation may be, under tension, an elonga-
tion; under compression, a shortening; or, under shearing, a detrusion
or thrusting aside of the metal. The unit-deformation is the change
in shape of a unit of the original length of the body. Thus, if a rod,
50 feet (600 inches) long, be stretched one inch by an applied load, the
unit-deformation will be 1/600 of an inch.
A stress, tensile or compressive, has not only full effect in its line
of action, but also produces compression in a direction at right angles
to that line. This action is called lateral contraction. Thus, referring
to Fig. 1, if the short length between the planes ab and cd of a rectangu-
lar bar be subjected to the unit tensile stress T at right angles to the
ends ab and cd, the stress in planes parallel to the line of action of
T will be equal to T; but the stretching of the metal in the direction
of this line causes a contraction in the directions which are perpendicu-
lar to it. This contraction is equivalent to that which would be caused
by a unit compressive stress P^ acting on the sides be and ad, and by
a similar stress P 2 acting on the sides ac and bd. The magnitude of
these induced compressive stresses depends on the metal. For wrought
iron and steel, P! and P 2 are each taken usually as equal to 1/3 T; for
cast iron, the ordinary values are about 1/4 T. This fraction, 1/3 or
1/4, is called the factor of lateral contraction, which factor will be des-
ignated by in the following. "Poisson's ratio," which is a constant
PRELIMINARY CONSIDERATIONS
used to determine the lateral effect of direct stress, refers to the same
action.
If the unit-stress T, Fig. 1, had been compressive instead of tensile,
there would still have been compression on planes parallel to its line
of action, but that compression would then act outward from, instead of
inward toward, the. axis of the body. The lateral effect would be to
elongate, not to contract. So far as is known, the factor of lateral
contraction has the same value in compression as in tension. Thus,
in Fig. 1, assume that P x and P 2 are direct compressive stresses and
that there is no direct tensile stress like T. Then P x and P 2 will each
Machinery.N.Y.
Pig. 1. Lateral Contraction Induced by Direct Tensional Stresses
develop lateral and equivalent tensile stresses, so that the actual unit
stresses will be:
In the direction of T, <j> (P i + P 2 ).
In the direction of P lf P 2 P,.
In the direction of P 2 , P x P 2 .
A stress thus developed by lateral action is identical in effect with
a direct stress of its direction and magnitude. The direct stress, which
does not consider lateral contraction, if the latter exist, is known as
the apparent stress, while the true stress is the algebraic sum of the
apparent stress and the stresses in its direction due to lateral action.
It should be borne in mind that the true stress is the actual stress to
which the body is subjected and by which the deformation is caused.
Merriman says in "Mechanics of Materials," edition of 1899, page 291:
"The true resistance of a body depends upon the actual deformations
produced, and these are measured by the true internal stresses."
6
No. 89 FORCED AXD SHRINKAGE FITS
When there are several direct stresses acting on a body, the use of
a general equation in which all stresses are assumed to be tensile, will
prevent error in ascertaining the true stress in any given direction.
Thus, let there be three direct or apparent tensile stresses, t lt t* and
f s , applied to the three sets of parallel sides of the body in Fig. 2, and
let T lt T a , and T, be the corresponding true stresses. Then:
7 7 1 t t <(> t, <t> f ,
which is the general equation for this stress. If t, had been a com-
pressive stress, the equation would be:
T, = *, ( *,) f, = t, + ( t, *,)
s . * 3 = APPARENT UNIT-STRESSES
T,, Tj,.T 3 = CORRESPONDING
TRUE STRESSES
Machinery.N.Y.
Fig 1 . 2. True and Apparent Stresses
In this way, by writing the general equation for each stress on the
assumption that all are tensile, and then changing the signs of those
which are compressive, the true stresses are readily found.
Elastic Limit-Modulus of Elasticity
The elastic limit is that unit-stress at which the elasticity of the
metal begins to disappear, that is, the stress at which it will not
wholly regain its original form after the removal of the stress, and,
hence, at which some "permanent set" makes its appearance. Theo-
retically, this limit occurs at a definite point, but experimentally it
cannot be sharply marked, and is taken as the stress at which the "set"
becomes fully distinguishable. Within the elastic limit, the deforma-
tion is approximately proportional to the stress producing it; beyond
PRELIMINARY CONSIDERATIONS 7
that limit, this ratio is no longer constant. General values of the elas-
tic limit are: Cast iron, in tension, GOOO, and in compression, 20,000
pounds per square inch; wrought-iron and steel, in either tension or
compression, 25,000 and 50,000 pounds per square inch, respectively.
These values, however, differ considerably for different kinds of steel,
and also depend upon its treatment.
The modulus or coefficient of elasticity, E, is the ratio of a unit-
stress to the unit-deformation which that stress produces. Thus, if
IS is the stress and s the deformation, E = S -=- s. E is a, constant for
each similarly treated metal until the stress reaches the elastic limit.
General values of E, for either tension or compression, are: Cast iron,
15,000,000; wrought-iron, 25,000,000; steel, 30,000,000.
Shrinkage Stresses Approximate Method (Tires)
When the thickness cf the outer member of a shrinkage fit is rela-
tively small as compared with the diameter of the inner member, as is
the case with a locomotive wheel-center and tire, the compression of
the inner member is negligible in practice and the radial pressure on
the fitted surfaces may be considered as expended wholly in producing
stresses in the outer member. In a tire thus shrunk on, there are
two stresses, one radial and compressive, and the other the circumfer-
ential or "hoop" stress which acts tangentially on a diametral plane
to burst the tire. This tangential or hoop stress is the only one re-
quiring consideration.
Let #0 = original internal radius of tire,
R = radius of wheel-center,
t = mean unit tensile hoop stress in tire when expanded,
f t unit-deformation (elongation) due to t,
t
E = = modulus of elasticity,
et
p = unit radial pressure on fitted surfaces of wheel-center and
tire,
fc = width of tire, axially,
T = thickness of expanded tire, radially,
/ =3 coefficient of friction at fitted surfaces.
The deformation or elongation per unit of length of the tire may be
taken as equal to the increase in length of the latter by expansion, di-
vided by the original internal length. Since the length of the circum-
ference is directly proportional to that of its radius, we have:
R R <}
t = Eet = E X -
Ro
The expanded tire is virtually in the condition of a cylinder subjected
at all points internally to the outward pressure p. The force tending
8 No. 89 FORCED AND SHRINKAGE FITS
to rupture such a cylinder on a diametral plane is equal to the pro-
jected area of the cylinder, multiplied by the internal pressure, or:
2R X & X P
and the resistance opposed by the tire to rupture is equal to the
product of its sectional area by the average hoop stress, or:
2& X T X t
Equating the force and resistance, and substituting the value of t,
we have:
Tt ET (RR )
R RR
Multiplying the area of the fitted surface by the radial pressure and
the coefficient of friction, the total resistance to slip is:
As an example, assume that a steel tire, S 1 /^ inches wide and Zy 2
inches thick, is shrunk on a wheel-center 66 inches in diameter. Let
the allowance for shrinkage be about 0.001 inch per inch of diameter,
0.070
or 0.070 inch, total. Then R = 33 inches, R = 33 , = 32.965
2
inches, and, taking E as 30,000,000, the average tensile stress in the
tire is 31,900 pounds per square inch, which is well within an elastic
limit of 50,000 pounds. This value of t gives p = 3380 pounds per
square inch, and, taking / = 0.2, the total resistance to slip is approxi-
mately 385 tons of 2000 pounds each.
This method is approximate for several reasons:
1. As we have assumed no compression in the wheel-center, the
value e t . as given in the first equation, is really the unit-deformation
at the inner surface of the tire, where that deformation is a maximum,
so that the value found for t is, as an average stress, too high, as is
that of p also; thus, the compression of the wheel-center, if considered,
would slightly reduce the average tensile stress.
2. The lateral contraction, due to the radial stress in the tire, is
neglected, and this action would increase the tensile stress, as found
above.
3. The tensile stress is assumed to be uniform over the cross-section
of the tire, while it is really a maximum (see Fig. 5) at the fitted sur-
face. As the thickness of the tire is relatively small as compared with
its diameter, the aggregate error will not be material, if the shrink-
age-allowance is moderate as in this case.
CHAPTER II
DERIVATION AND APPLICATION OP
LAMB'S FORMULAS
When the outer member of a shrinkage fit is relatively thick, as a
wheel-hub or a crank-web, the approximate method given in the previ-
ous section will not serve, and recourse must be had to the formulas
deduced for the investigation of the stresses in thick cylinders sub-
jected to radial pressure this pressure being internal for the outer
member of the fit and external for the inner member. As in the tire,
there are two "apparent" stresses in such a cylinder, the tangential or
"hoop" stress, and the radial stress. The latter is always compressive;
the former, in a shrinkage fit, is tensile in the outer member and com-
pressive in the inner, while, in a gun, built up of superposed cylinders,
it may be either tensile or compressive, as the location of the cylinder
and the magnitude of the powder pressure determine. In any event,
the tangential and radial stresses are interdependent; they affect each
other by lateral contraction; and, through the latter action, they pro-
duce in the outer member a longitudinal compressive stress, parallel to
the axis of the fit.
Various formulas have been proposed for the determination of the
stresses in thick cylinders. Those founded on the principles estab-
lished by Lame have found general acceptance, since they avoid the
assumptions on which others are based. Their close approach to ac-
curacy is shown by their use in the design of high-powered guns, in
which the stresses at the instant of explosion are very near the elas-
tic limit of the metal. Lame's fundamental formula may be deduced
in several ways; the method* given below is due to Professor P. R.
Alger, U. S. Navy, of the Bureau of Ordnance.
Fig. 3 represents a thick, hollow cylinder subjected to internal and
external fluid pressure; the cylinder is assumed to be free at the ends,
in order to prevent direct longitudinal stress.
Let P = internal unit pressure,
P 1 = external unit pressure,
-R = internal radius of cylinder,
.R! = external radius of cylinder,
r = radius of any point within cylinder walls,
t = "apparent" tensile tangential or "hoop" unit stress at ra-
dius r,
p = "apparent" radial .compressive unit-stress at radius r,
Z = "true" longitudinal unit-stress at radius r, due to lateral
contraction.
*Cathcart, "Machine Design: Fastenings," New York,
10
No. S 9 FORCED AND SHRINKAGE FITS
T = "true" tangential stress at inner surface of radius R ,
TJ = "true" tangential stress at outer surface of radius R^ ,
e t unit-deformation due to "true" tangential stress at radius r,
<e p = unit-deformation due to "true" radial stress at radius r,
>e\ =3 unit-deformation due to "true" longitudinal stress at ra-
dius r,
<t> = factor of lateral contraction = 1/3 for steel,
E = modulus of elasticity = 30,000,000 for steel.
In this deduction, it is assumed:
a. That there is no direct longitudinal stress in any layer of the
cylinder walls.
b. That a transverse section of the cylinder when not under pressure,
remains a plane normal to the axis of the cylinder when the latter is
Machinery, &. Y.
Fig. 3. Thick Hollow Cylinder Subjected to Internal and
External Fluid Pressure
under stress, i.e., that the longitudinal stress due to lateral contraction
is uniform over the whole cross-section.
c. That the total or "true" stress in any direction is the measure of
the tendency to yield in that direction.
d. That the factor of lateral contraction is equal to 1/3.
The true stresses in the indefinitely thin cylinder of radius r are:
tangential unit-stress t ( </> p) =t -\
3
radi'al unit-stress = p <p t =
longitudinal unit-stress = <f> t -f /> p
r p
\T~T
By the definition of the modulus of elasticity, the corresponding unit-
deformations are:
LAME'S FORMULAS v 11
e f = (p + #/3) +E (1)
ei = U/3 p/3) -i-^J
Since, by hypothesis, e^ is constant:
f p = constant = k
But,
C Rl td r = Po Po - Pi Ri
and, assuming t = f (r), this gives:
r dp
whence f (r) = pr; and so t = f (r) = p -- ..
dr
Thus, we have
rdp
t p = k, and t + p = --
dr
rdp
whence 2p + k = -- the integration of which gives:
dr
r 2
where k^ is a constant of integration. Combining with t p k, we
k*
have t + p = -- .
r 2
The equations which express the relation between "hoop" or tangen-
tial tension and radial stress at all points within the cylinder walls
are then:
t p = k = T P s= T, P,
(t + p) r 2 =k*=(To + P ) R 2 =(T 1 + P 1 ) R, 2
Eliminating T l between the last parts of these equations, we have:
P (R, 3 + R*) 2R*P,
RS R > RSRS
and substituting this in the first parts of the same equations, we have,
after combining:
PJUPJl? fio'fii'CP. PI) 1
t = - - + - - X (2)
R* R* RSRO* r*
PJVP&* P 2 R? ( Po P* ) 1
p = -- - - + - - X- (3)
RS RJ RSRo' r 3
which are Lame's fundamental formulas for the "apparent" stresses in a
thick cylinder subjected, internally and externally, to fluid pressure.
In deriving these formulas, p has been taken as a compressive stress. If
it had been assumed to be tensile, the signs in Equation (3) would
12
No. 89 FORCED AND SHRINKAGE FITS
liave been reversed. With this change, however, it will be found that,
in the shrinkage fit, this equation will give negative values, showing
that p is a compressive stress. To obtain the "true" or actual stresses,
the value* of t and p from (2) and (3) are modified in the succeeding
equations for the effect of lateral contraction, according to the methods
of Clavarino. ,
Application of Lame's Formulas to Compound Cylinders
The shrinkage fit is applied to a compound cylinder, i. e., to two cyl-
inders, one superposed on the other. The inner cylinder may be solid,
as in the ordinary shaft or hollow, as shafts and large crank-pins of
steel are often made. Fig. 4 represents such a compound cylinder, the
conditions being the same as in Fig. 3, except that the radial pressure
P! is, in Fig. 4, produced by the shrinkage of the outer cylinder of ex-
TANGENTIAL STRESSES
TRUE APPARENT
Machinery, N.Y.
Fig. 4. Compound Cylinder consisting 1 of an Outer
Cylinder shrunk onto an Inner
ternal radius R 2 . There is no external pressure on this cylinder, ex-
cept that of the atmosphere, which is negligible. In the shrinkage fit,
the metals of the inner and outer members may not be the same, and
the tangential stresses in the two cylinders at the contact surface also
differ.
Let E = modulus of elasticity, outer cylinder,
EI = modulus of elasticity, inner cylinder,
= factor of lateral contraction, outer cylinder,
#! =5 factor of lateral contraction, inner cylinder,
t = apparent tangential unit-stress, inner surface of inner cyl-
inder,
*! = apparent tangential unit-stress, outer surface of inner cyl-
inder.
T and T! = corresponding true tangential stresses,
Po and P! = corresponding apparent radial stresses,
* 2 = apparent tangential unit-stress, inner surface of outer cyl-
inder,
T 2 =: corresponding true tangential stress,
2> 2 = corresponding apparent radial stress.
LAME'S FORMULAS
13
It should be observed that, in deriving Equation (2), t was assumed
to be a tensile stress. Therefore, in the deductions by substitution
which follow, if the formula gives a negative value, the stress t or
f,, which represents t for these conditions, is compressive. Similarly
in Equation (3) p is by hypothesis always a compressive stress, and
the formula gives, in the substitutions, simply its numerical value, as
Pi. P-2, etc., for various conditions, and these values, when used in the
equations for the true stresses, should have the minus sign.
Outer Cylinder
In a shrinkage fit, the only important stress in this cylinder is the
true tangential stress at the inner surface, where that stress is a maxi-
Machinery, N. Y.
Fig. 5. Graphical Representation of Stresses produced by Shrinkage Fits
mum. (See Fig. 5). Since, for equilibrium, the pressure P^ from the
outer cylinder must be opposed by an equal and opposite pressure from
the inner cylinder, the former cylinder is virtually under the same
conditions as the latter, except that it is not subjected to external pres-
sure. Hence, Equations (2) and (3) may be applied to the outer cyl-
inder, by changing R to R t , R^ to R 2 , P to P lf and P! to zero. Making.
these substitutions and with r = R we then have the apparent unit-
stresses in the outer cylinder at the inner surface:
R* R*
(5)
Considering lateral contraction, the corresponding true tangential
tensile unit-stress is:
T, = t 2 ( $ p a ) = t, + p 2
(6)
14
o, S 9 FORCED AND SHRINKAGE FITS
Inner Cylinder, Hollow
This cylinder corresponds to a hollow shaft forming the inner mem-
ber of a shrinkage fit. The stresses to be found are the true tangen-
tial stress at the outer surface, which is required to determine the al-
lowances, and the similar stress at the inner surface, since the tangen-
tial stress in such a cylinder is compressive and reaches its maximum
at the bore (See Fig. 6). Equations (2) and (3) are applicable, if P
be made equal to zero, since there is only the atmospheric pressure on
the bore of the shaft.
Machinery. ff.
Fig. 6. Graphical Representation of Stresses produced by Shrinkage Fits
Making r = R 1 , and P = zero, we have the apparent unit-stresses in
the inner cylinder at the outer surface:
P, (R* + fl )
.= -- , -- (7)
The corresponding true tangential compressive stress iS:
5P t = t, ( p,) = t l -f 0, p,
(8)
(9)
> 2 p J
KI -"-o
For the Inner surface, r = -R , and P = zero in Equations (2) and
(3). The apparent stresses, therefore, are:
LAME'S FORMULAS 15
*, = PtX --- (10)
#>' JV
Po=0 (11)
Since p = 0, the true tangential compressive stress is:
2 P, flj 2
T=:*, = - (12)
t -i* Q
which is evidently greater, numerically, than TV
Inner Cylinder, Solid
If the inner cylinder be solid, the conditions will correspond with,
those of a solid shaft forming the inner member of the fit. Equations
(2) and (3) will apply, if R and P be made equal to zero. The only
stress of importance is the tangential stress at the outer surface, which
is required in determining the allowances.
Making these substitutions, the apparent stresses at the outer surface
are:
t 1 = P 1 (13)
Pi = Px (14)
The true tangential compressive stress is, therefore:
T 1 = t, ( 0! Pl ) = ft + X p,
T 1 = ~ Pad 0,) (15)
The values given in Equations (13), (14) and (15) are valid for any
point between the outer surface and the center of a solid shaft, since, if
in Equations (2) and (3), R and P be made equal to zero, tne second
term of the right-hand member of each equation vanishes, no matter
what value may be given to r, the radius of the point considered. In
general, therefore, in a solid shaft subjected to a uniform external
radial pressure, the true radial and tangential compressive stresses are
equal at all points, and the intensity of each is uniform throughout.
CHAPTER III
FORMULAS FOR STRESSES IN THE HUB
As shown in Fig. 5, the tangential tensile stress in the hub reaches
its maximum at the inner surface and decreases rapidly from that sur-
face outward. The true stress at the hore is therefore of primary im-
portance, since the metal is under its greatest stress there. This stress
must not exceed the elastic limit, and is one of the factors which deter-
mine the "grip" of the fit. In Equation (6), the radii are those of the
expanded hub, and the use of these dimensions would make computa-
tion complex. No material error will be caused by the substitution for
them of the corresponding nominal radii, I. e., those of the hub before
expansion, and thus disregarding the allowances which are but a few
thousandths of an inch.
Let A = nominal internal diameter of hub,
D 3 = nominal external diameter of hub,
$ = 1/3 for steel and 1/4 for cast-iron.
Substituting in Equation (6) :
r f = P 1 (o + 0) (16)
4Z> 2 ' + 21V
T a = Pj. X - - for steel, ( 17 )
3 (ZX 2 ZV)
5Z> 2 2 + 3ZV
T., = P 1 X for cast-iron. (18)
4 (D 2 2 IV)
Resistance of Hub to Bursting Load
The relation between the bursting load on the hub, due to the radial
pressure on the fit, and the true tangential stress which resists it, is
shown graphically in Fig. 5. If a cylinder be subjected to the unit in-
ternal radial pressure P,, the force tending to burst it on a diametral
plane is equal, for a section of unit length, to the product of this pres-
sure by the diameter, or P X 2Ri, which is the area of the load-dia-
gram dee'd'. This bursting load is resisted by, and equal to, the sum
of the true tangential stresses in the cylinder-walls, which sum is rep-
resented by the two equal stress-diagrams, abed and a'&'c'd'. Hence:
Load-area dee'd'= 2 X stress-area abed.
The stress-area is laid out by plotting as ordinates on the diameter
the values of the true tangential stress, t + P, as found by the meth-
ods on page 13, and giving r various values from R^ to R 3 . The aver-
STRESSES IN THE HUB 17
age tensile unit-stress in the cylinder-wall, or in the hub in this case,
is equal to the area of the load-diagram, divided by the thickness of
the hub i. e., - .
Fig. 5 shows that it is impossible for the shrinkage-load on the hub
to burst that member, so long as the true hoop stress T 2 at the bore
does not exceed the ultimate tensile stress of the metal. Again, divid-
ing Equation (5) by (4), we have from the apparent stresses:
t 2 tn RJ + -Rl 2
which equation proves that the radial pressure P t at the fit can never
be equal to the apparent hoop stress t 2 in the hub at the bore, even if
t 2 be the ultimate tensile strength and R 2 be increased indefinitely.
This is again shown by the fact that the equation may be transformed
into
Ir+'Pi
from which it appears that if P 1 = t,, R 2 becomes infinite, i. e., no thick-
ness whatever will prevent rupture. This condition fixes the useful
limit of thickness of a cylinder, not reinforced by one or more enclos-
ing cylinders so shrunk on as to put the innermost cylinder under ex-
terior compression. No unsupported cylinder can be made thick
enough to withstand an internal pressure per square inch which is as
great as, or greater than, the ultimate 'tensile strength of the metal.
Rankine gives in "Applied Mechanics," London, 1869, page 293:
P, + 2 P 2
in which T is the ultimate tensile strength of the metal of the cylin-
der. From this equation it follows that if the internal pressure P x is
equal to or greater than the sum T -f 2 P,, of the ultimate strength and
twice the external pressure, no thickness, however great, will enable
the cylinder to resist the pressure.
With regard to the possible intensity of shrinkage-stresses, it should
be borne in mind that shrinkage fits are usually made on the working
parts of machines, and hence that the stresses due to shrinkage may be
increased by others developed by the external forces applied to the
member when the machine is in operation. In such cases, the total
stress which will exist at any time should be considered in determining
the shrinkage-allowances.
Effect of Thickness of Hub on Resistance to Slip
The principle governing the effect of the thickness of hub on the re-
sistance to slip may be seen most readily from the formulas for the
apparent stresses. Thus, Equation (19) shows that if the radius of the
18
No. 89 FORCED AND SHRINKAGE FITS
fit and the tangential stress at the bore of the hub are constant, the
effect of variation in the external radius is simply to change the inten-
sity of the radial pressure P^ at the fit a greater hub-thickness in-
creasing the "grip," and a smaller decreasing it. Thus, if R 2 = 2R lt
P 1 = 0.6 2 ; if R 2 =3R lt P 1 = 0.8 2 , etc.
From Equations (17) and (18), we have:
3 (IV IV)
P! = T 2 x for steel, ( 20 )
4 Z) 2 2 + 2 ZV
TABLE X
Values of Ratio A, as computed from Equation (22).
Pt
p
Ratio of Nomi-
nal Diameters
Ratio A =
T 8
Ratio of Nomi-
nal Diameters
Ratio A =
T a
Da
D 2
of Hub,
of Hub,
Di
Steel
Cast Iron
DI
Steel
Cast Iron
(<*> = *)
(*-*)
Cf-tt
-
1.5
0.341
0.351
2.8
0.615
0.648
1.6
0.382
0.395
8.0
0.632
0.666
1.8
0.449
0.466
3.2
0.645
0.682
2.0
0.500
0.522
3.4
0.657
0.695
2.2
0.539
0.565
3.6
0.666
706
2.4
0.570
0.599
3.8
0.675
0.715
2.6
0.595
0.626
4.0
0.682
0.723
P, = T 2 x
4 (Z) 2 2
for cast-iron,
(21)
5 D 2 2 + 3 D, 2
which give the values of the radial pressure at the fit in terms of the
true tangential stress at the bore of the hub.
From Equation (16) :
P, 1 D.'D>
= - = - - . = 4 (22)
T 2 a + D 2 2 (! + </>) +#i 2 (1 0)
a ratio which is of service in computing the allowances. Table I gives
values of A for various diametral ratios. If the true tangential stress
T 2 is known or assumed for any of the diametral ratios tabulated, the
intensity of P lt and hence the resistance of the fit to slip may be found
by multiplying T 2 by the corresponding value of A.
CHAPTER IV
FOEMULAS FOB STEESSES IN THE SHAFT
The radial and tangential stresses in the inner member are, as shown
previously, both compressive. To both, the same principle applies:
each is a measure of the deformation in its direction only at the point
where the given intensity of stress exists. If, for example, the radial
stress varies from the circumference to the center, its intensity at any
given point will not measure the deformation of the entire radius of
the member, but only the amount of deformation at the point consid-
ered. The only stress which will cover both cases solid and hollow
shafts and give the reduction in the external diameter of the mem-
ber, is, therefore, the true tangential stress at the outer surface, since
the circumference of that surface and its diameter must decrease to-
gether. As with the hub, the nominal diameters may be substituted
for the corresponding dimensions of the compressed shaft.
Let D = nominal internal diameter of hollow shaft,
D! = nominal external diameter of hollow or solid shaft,
R> + R 2 IV + D 2
~ ~
= B
P.
0i = l/3 for steel and 1/4 for cast iron.
Solid Inner Members
Equation (15) gives the true tangential stress at the outer surface.
From that equation:
2\ = 2/3 P! for steel ( 23 )
7\ = 3/4 P x for cast iron. (24)
Since 2\ is a compressive stress:
T!
= 1 a = B for solid inner members (25)
Pi
This ratio is of service in computing the allowances. In a solid
shaft, both the radial and tangential stresses are, as mentioned before,
uniform in intensity from the outer surface to the center, and are equal
at all points.
Hollow Inner Members
Equation (9) gives the true tangential stress at the outer surface.
From that equation:
2\ = P> (ft a ) (26)
= P! X -* - for steel, (27)
3 (IV JD )
20
No. 89 FORCED AND SHRINKAGE FITS
and, since 7\ is compress! ve:
3?,
=h <t>! = B for hollow inner members. (28)
P,
Equation (12) gives the true tangential stress T at the inner surface.
From (12) and (27):
To 3 DS
for steel. (29)
T, D* + 2 ZV
This expression shows the marked increase in the tangential stress
from the outer surface to the bore.
The values of B for hollow steel shafts of various diametral ratios
are given in Table II.
Work Done in Compressing 1 Solid and Hollow Shafts
The compressibilities of solid and hollow shafts differ, the solid shaft
being the stiffer. In a solid shaft under radial pressure, the radial and
TABLE II
Values of Ratio B for hollow steel shafts of external and internal
diameters, D and D , respectively.
D
Do
B-5
Pi
D!
Do
B-,*'
Pi
2.0
2.5
1.333
1.048
3.0
3.5
0.917
0.844
For solid inner members Equation (25), B = 2/3 for steel and 3/4
for cast iron-
tangential stresses are equal at all points, as mentioned, and their in-
tensity is uniform throughout. This can be proved from Equations (2)
and (3) by making R and P equal to zero. The second term of the
right-hand members of both equations will then disappear, and for any
value of r from zero to R lf t = P x and p = P 1 , p being a compressive
stress by hypothesis. These relations are shown graphically in Fig. 6,
where Oa = c& = P l = t = p. The diagram Oa&c, therefore, repre-
sents the total apparent tangential stress in one-half of a solid shaft.
Since this total stress is produced by the total stress in the left side of
the hub, whose tangential value is represented by the diagram cdef,
the two stress-areas are equal, or Oa&c = cdef = P x X RI.
Now, consider the hollow shaft on the right-hand side (Fig. 6), whose
original diameter was sufficiently greater than that of the solid shaft
to make the radius R! of the fit and the radial pressure P x on the latter
the same as before, with the same hub and hub stresses, so that ghkl =
cdef. From Equation (7) it will be seen that the apparent tangential
stress at the outer surface is P t h, and is hence greater than that of a
solid shaft [Equation (13)], since h is always more than unity. Equa-
STRESSES IN THE SHAFT 21
tions (2) and (3), with suitable substitutions, show that the tangential
stress increases rapidly toward the bore, where its magnitude is given
by Equation (10). The area representing the total tangential stress is
Imnq, Fig. 6, and, as before, Imnq = ghkl = cdef = P, X R The
radial stress is no longer uniform as in a solid shaft, but is equal to
P l at the outer surface, and decreases to zero at the bore [see Equa-
tions (8) and (11)].
It will be seen, then, that if two shafts one solid, the other hollow
when subjected to the same external radial pressure P,, are compressed
to the same radius R lt the tangential stresses in the hollow shaft will
be considerably greater than those in the solid shaft. The reason for
this increased effect of P^ on the tangential stress is that the hollow
shaft lacks the support of the solid and compressed cylinder of radius
R which has been removed at the bore. In the solid shaft, at the
layer of radius R , there is an outward radial pressure equal to P lt
while, in the hollow shaft, at this radius, the radial pressure is zero.
These relations can be shown by making P =:P 1 in Equation (2),
when the second term of the right-hand member will disappear, and,
at all radii between R and R lt the tangential stress will be equal to
PL as in a solid shaft. In this assumed case, the outward radial pres-
sure P! at the bore produces the total apparent tangential tensile stress
in the hollow shaft shown by the area qsvl, and, if this be deducted
from the area Imnq, the remainder will be the area Iwxq, corresponding
with that for a solid shaft between the radii R^ and R . The deduc-
tions, as above, apply also to the true tangential stresses, which are the
same in kind as the apparent stresses, although differing in intensity.
Effect of Lateral Contraction
It has been shown that in the outer member of a shrinkage fit, lateral
contraction increases the apparent radial and tangential stresses, each
by an amount equal to one-third for steel, so that the true stresses are
that much greater, and that in the inner member there is the same pro-
portionate, but reverse, effect, which acts to reduce the intensity of the
direct stresses. This action also develops secondary longitudinal stress-
es in both members, which, however, are negligible in a shrinkage fit.
Thus, in the outer member, the tangential tensile stress t produces a
longitudinal compressive stress whose intensity is t, and the radial
compressive stress p causes a longitudinal tensile stress equal to <f>p.
The resultant longitudinal compressive stress at any point of radius r
is then (see Fig. 3) :
l = <j>t + <f>p = $ (t p)
As an extreme example, take a steel hub shrunk on a solid steel shaft,
the external diameter of the hub being 1.5 times that of the shaft. Let
the shrinkage allowances be such as to produce a true tangential tensile
stress of 30,000 pounds per square inch at the bore of the hub. From
Table I we find that the unit radial pressure on the fit is 10,230 pounds.
Applying the formulas previously given:
22 No. 89 FORCED AND SHRINKAGE FITS
Hub at Bore : Apparent Stress True Stress
. Tangential tensile stress 26,598 30,000
rladial compressive stress 10,230 19,096
Shaft at Outer Surface:
Tangential compressive stress 10,230 6,820
Radial compressive stress 10,230 6,820
The stresses given in the table above were calculated as follows:
The true tangential unit stress T, at the bore of the hub is 30,000
R.,
pounds, the ratio of the hub diameter is - = 1.5; from this ratio, jR 2 2 =:
RI
2.25 R?. From Table I, when R 2 -f- R 1 = l. 5, with both members of steel,
ratio A = 0.341. Hence
= 0.341
T z 30,000
P, = 30,000 X 0.341 = 10,230 pounds = unit radial pressure.
Hub at bore. The apparent tangential tensile stress is:
P, (R, 2 + R*)
t, = -- (4)
R.? R*
Substituting the values of P x and R^.
3.25
t t = 10,230 X - = 26,598 pounds.
1.25
The apparent radial compressive stress is:
p 2 =:P 1 = 10,230 pounds. ( 5 )
1
The factor of lateral contraction 0, for steel, is = 0.333. The true
3
tangential stress is:
(6)
The true radial stress is:
/3.25 \
= P! I - + 0.333 I = 30,000 pounds.
\1.25 /
al stress is:
[* (JV + JR,*)"1
l +
R** - #i 2 J
(3.25V
1 + 0.333 X - 1 =19,096 pounds.
1.25/
Shaft at outer surface. The shaft is solid. The apparent tan-
gential (compressive) stress at the outer surface is:
t 1 = P 1 = 10,230 pounds. (13)
The apparent radial (compressive) stress is:
p l = P 1 = 10,230 pounds. (14)
STRESSES IN THE SHAFT 23
The true tangential stress is:
T, = P, (1 0) =10,230 (1 0.333) =6,820 pounds. (15)
The true radial stress is:
P\ = P, (1 0) =6,820 pounds.
It will be seen that the use of the apparent, in place of the true,
stresses introduces errors which, with regard to the hub, may be
serious even in less extreme cases than the above.
Resistance to Slip
The resistance of the fit to slip is theoretically equal to the product
of the area of the contact-surface times the unit radial pressure on
that surface times the coefficient of friction.
Let D! = nominal diameter of fit,
L = length of fit,
P x = unit radial pressure on fitted surfaces,
/ = coefficient of friction,
Q = total resistance to slip.
Then Q = irD 1 XLxP 1 Xf (30)
Since slip begins with the parts at rest, the coefficient of friction
for rest applies in computing the initial resistance. There is con-
siderable variation in the values given for this coefficient. Reuleaux
and Weisbach use 0.2. Rennie, in experiments on metals usually
unlubricated, found the following values for /:
Wrought-iron on cast iron 0.28 to 0.37
Steel on cast iron 0.3 to 0.36
In Professor Wilmore's experiments, the average value of this co-
efficient was 0.102. These tests were made with a series of cast-iron
disks, 4 inches in diameter and 1 inch thick, which were either forced
or shrunk on steel spindles about 1 inch in diameter, the fit being
about 1 inch long. Five sets of these spindles were used, the diam-
eter of the first set being 1.001 inch and the allowances for each
subsequent set increasing by 0.0005 inch. The spindles were pulled
from the disks in the ''tension" tests of the fit and twisted in the
holes in measuring the resistance to slip in torsion. The shrinkage
fits were found to be 1.5 times, and the forced fits 1.3 times, stronger
in torsion than in tension. This result was to be expected, if the
resistance measured was not that to initial slip only, since, in torsion,
the grip is undiminished during progressive slipping, while, in ten-
sion, the area under pressure decreased steadily as the spindles left
the disks.
Let P = force acting to twist a solid shaft,
p = lever arm of P,
/ = polar moment of inertia of shaft,
c = distance of most remote fiber of shaft from axis of latter,
S B = shearing stress at distance c = maximum unit shearing
stress,
Z>i = diameter of shaft.
24 No. 89 FORCED AND SHRINKAGE FITS
Then:
J irDi 3
P x p = S s X = S t
c 16
and from equation (30) :
QD, D l
=iirD l LP l J X
2 2
Taking P t and S a as constant, and equating, we have L = KD, in
which K is a constant. Therefore with a constant radial pressure,
the length of the hub should vary as the diameter of the shaft, in
order to make the grip of the fit proportional to the torsional strength
of a solid shaft. For both practical and theoretical reasons, it is
impossible to make the grip equal to this strength. Hence, with
diameters of 2 inches and upwards, keys should be fitted in addition.
CHAPTER V
SHRINKAGE ALLOWANCES
The total allowance for shrinkage is the difference between the
external diameter of the inner member (shaft) and the internal
diameter of the outer (hub), before shrinkage. The unit shrinkage-
allowance is the allowance per inch of nominal diameter, in either
case, as above; and also, in either case, the wm-deformation of a
given circumference or diameter is the difference between its lengths
before and after shrinkage, divided by its original length. The prin-
ciple which is applied in the derivation of formulas for shrinkage-
allowances, is that the unit-deformation at any point is the quotient
of the unit-stress at that point, divided by the modulus of elasticity.
In a shrinkage fit, the unit-deformations considered are those at the
fit, and the unit-stresses to which these deformations correspond are
manifestly the "true" or actual stresses, and not those which have
been termed "apparent" in this discussion, since, as has been shown,
the effect of lateral contraction is important.
The length of a given circumference varies directly as that of its
diameter. Hence the unit-deformation will be the same for both,
and this deformation when due to the true tangential stress in the
hub at the bore, will be the unit-deformation of the internal diam-
eter of the hub. Similarly, for both solid and hollow shafts, the unit-
deformations of the external diameters are those of the circumfer-
ences of their outer surfaces, produced by the true tangential stresses
there, since that circumference and the external diameter decrease
together. For the unit-deformation of the external diameter of the
inner member, that due to the true radial stress at the outer surface
will serve only for a solid shaft, since in it, as shown in Fig. 6, the
tangential and radial stresses are equal to each other at all points
from the circumference to the center, while, in the hollow shaft, the
intensity of the radial stress varies from Pj at the outer surface to
zero at the bore, and hence the deformation due to this stress at any
given point is that corresponding only with the infinitely small element
of radius in which that stress exists, and not with the average unit-
deformation of the whole radius.
The algebraic methods employed below are those of Reuleaux*,
the true stresses being substituted, since his formulas do not con-
sider lateral contraction, and apply only to solid shafts, as the radial
stress in the inner member is used in their deduction.. As before, let
P! = radial pressure on fitted surfaces,
5T 1 = true tangential compressive stress at outer surface, inner
member,
"The Constructor," Suplee's translation, Philadelphia, Pa., 1895, page 17.
2C
No. 89 FORCED AND SHRINKAGE FITS
T. 2 = true tangential tensile stress at inner surface, outer member,
R, = radius of fit,
R = actual internal radius of outer member before expansion,
R' = actual external radius of inner member before compression,
R' R
S = unit shrinkage-allowance = ,
R
E and = modulus of elasticity and factor of contraction, outer
member,
E! and 0i = modulus of elasticity and factor of contraction, inner
member.
P, T,
A= ; B =; C A X B = T, -*- T 3 .
T, P,
TABLE III
Values of Ratio C for solid steel shafts of nominal diameter Z? t , and
hubs of steel or cast-iron of nominal external and internal diameters D t
and DI, respectively.
T t
Ti
Ratio of Diam-
D 2
C - A X B =
T 2
Ratio of Diam-
D,
C= AXB=
T,
eters
eters
D!
Steel
Cast-iron
Steel
Cast-iron
Hub
Hub
Hub
Hub
1.5-
0.227
0.234
2.8
0.410
0.432
1.6
0.255
0.2K3
3.0
0.421
0.444
1.8
0.299
311
3.2
0.430
0.455
2.0
0.333
0.348
3.4
0.438
0.463
2.2
359
0.377
3.6
0.444
0.471
2.4
0.30
0.399
3.8
0.450
0.477
2.6
0.397
0.417
4.0
0.455
0.482
By the definition of the modulus of elasticity, we have, at the ra-
dius R 1 of the fit, for:
R. R T,
outer member, =
R E
inner member,
R'
Adding, we have:
Dividing by R:
From (31):
RT 2
E
1
R' R T, R' T l
R ERE,.
(31)
(32)
SHRINKAGE ALLOWANCES
27
Substituting this value in (32):
T,
8 =
1
TABLE IV
Values of Ratio C for hollow steel shafts of external and internal
diameters D v and D , respectively, and steel hubs of nominal external
diameter Z) 3 .
D 8
D,
D'
D,
G
.
C
D,
Do
D,
Do
2.0
0.455
2.0
0.820
1 ^
2.5
0.357
O Q
2.5
0.645
8;0
0.313
/* . O
3.0
0.564
3.5
0.288
3.5
0.519
2
0.509
2.0
0.842
1 6
2.5
0.400
q A
2.5
0.662
3.0
0.350
o . U
3.0
0.580
3.5
0.322
3.5
0.533
2.0
0.599
2.0
0.860
1 H
2.5
0.471
39
2.5
0.676
J. . O
3.0
0.412
. &
3.0
0.591
3.5
0.379
3.5
0.544
2.0
0.667
2.0
0.870
2.0
2.5
3.0
0.524
0.459
3.4
2.5
3.0
0.689
0.602
3.5
0,422
3.5
0.555
2.0
0.718
2.0
0.888
2.2
2.5
3.0
0.565
0.494
3.6
2.5
3.0
0.698
0.611
3.5
0.455
3.5
0.562
2.0
0.760
2.0
0.900
2 4
2 5
597
30
2.5
0.707
/v . ^r
C.O
0.523
. o
3.0
0.619
- ':
3.5
0.481
3.5
0.570
2.0
0.793
2.0
0.909
2.6
2 5
3.0
0.624
0.546
4.0
2.5
3.0
0.715
0.625
3.5
0.502
3.5
0.576
The second term of the denominator is so small as to be negligible.
Hence :
T 2 T,
8= + (33)
E E,
This equation is not in a practical form, since for a given value of
8, there are two unknown quantities.
28 No. 89 FORCED AND SHRINKAGE FITS
P!
From Equation (22), A = ; Equations (25) and (28) give
the value of B = . Let A X B = C = . Then T z =
PI 5T 2
and T 1 = CT 2 . Substituting in (33):
Values of Ratio C for hollow steel shafts and cast-iron hubs.
Notation as in Table IV.
D a
D 1
D 3
D,
.
C
C
D 1
Do
D t
Do
2.0
0.468
2.0
0.864
1.5
2.5
3.0
0.368
0.322
2.8
2.5
3.0
0.679
0.594
3.5
0.296
3.5
0.547
2.0
0.527
2.0
0.888
1.6
2.5
3.0
0.414
0.362
3.0
2.5
3.0
0.698
0.611
3.5
0.333
3.5
0.562
2.0
0.621
2.0
0.909
1.8
2.5
3.0
0.488
0.427
3.2
2.5
3.0
0.715
0.625
3.5
0.393
3.5
0.576
2.0
0.696
2.0
0.926
o n
2.5
0.547
3 A
2.5
0.728
9* V
3.0
0.479
. 1
3.0
0.637
3.5
0.441
3.5
0.587
2
0.753
2.0
0.941
2.2
2.5
3.0
0.592
0.518
3.6
2.5
3.0
0.740
0.647
3.5
0.477
3.5
0.596
2.0
0.798
2.0
0.953
2.4
2.5
3.0
0.628
0.549
3.8
2.5
3.0
0.749
0.656
3.5
0.506
3.5
0.603
2.0
0.834
2.0
0.964
2.6
2.5
3.0
0.656
0.574
4.0
2.5
3.0
0.758
0.663
3.5
0.528
3.5
0.610
8 = -- 1
E
8=
1
CT 2
CE
Multiplying (22) by (25), and also by (28), we have,
1 01
for a solid inner member, C = -
(34)
(35)
(36)
SHRINKAGE ALLOWANCES 29
' 0i
for a hollow inner member, (7 = - (37)
a + <
The values of C for various diametral ratios are given, for solid steel
shafts with steel or cast-iron hubs in Table III; and, similarly, for
hollow steel shafts, in Tables IV and V.
Taking the modulus of elasticity for steel as 30,000,000, and for cast
iron as 15,000,000, equations (34) and (35) become, for a cast-iron hub
and a Rteel shaft:
T 2 (2 + C)
8 -- (38)
30,000,000
C X 30,000,000
and, for both hub and shaft of steel:
T.U + C)
(39)
(40)
30,000,000
Ttd + C)
(41)
C X 30,000,000
CHAPTER VI
CALCULATING- SHRINKAGE FITS
In designing shrinkage fits, there are but two main principles to
remember. First, the stress in the hub at the bore, which is the most
important consideration, depends chiefly on the shrinkage-allowances.
If the latter be. too large, the elastic limit will be exceeded and per-
manent set will occur; or, in extreme cases, the ultimate strength of
the metal will be passed and the hub will burst. Second, the inten-
sity of the grip of the fit, and hence the resistance of the latter to
slip, depends mainly on the thickness of the hub. The greater this
thickness, the stronger the grip; and vice versa. Formulas (34) and
(35) and Tables I and III serve all general purposes in practice. In-
formation in detail can be obtained as follows:
a. For a given allowance per inch of diameter, the true tensile
stress T 2 in the hub at the bore can be found from Equations (34),
(38), or (40). These equations hold only up to the elastic limit. It
will be seen that by increasing or decreasing the allowances, any
stress up to this limit can be produced at the bore, and this stress will
be the maximum tensile stress in the hub.
b. When T 2 is assumed at any desired value below the elastic
limit, the corresponding unit-allowances can be found by substituting
in Equation (34).
c. Equations (6) and (22) and Table I show the relation between
the true tensile stress in the hub at the bore and the radial pressure
on the fit. There are several factors which govern the intensity of
this radial pressure: the magnitude of the allowances, the compres-
sibility of the inner member, and the expansibility of the outer. The
two latter depend on the metals; the last is affected by the thickness
of the hub.
d. When T 3 is known, the value of P x can be obtained from Table
I or equation (22).
e. The true tangential compressive stress T^ at the outer surface
of the inner member is usually of minor importance in design; its in-
tensity can be found from (35). The true radial compressive stress
at the surface is equal to the radial pressure P it minus the product of
0! by the value of t u as given by (7) and (13).
/. At the bore of a hollow shaft, the radial pressure is zero. Equa-
tion (12) gives the true tangential compressive stress.
g. The intensity of the apparent stresses is, in general, of academic
interest only. To ascertain their magnitude, the true stresses are first
found from (34) and (35); Equations (25) or (28) will then give the
value of the radial pressure P,, and, by substituting this in the equa-
tions on pages 13 to 15, the apparent stresses can be determined.
CALCULATING SHRINKAGE FITS 1
Examples
Example 1. A steel crank-web, 15 inches least outside diameter, is
to be shrunk on a 10-inch solid steel shaft. Required the allowance
per inch of shaft-diameter to produce a maximum tensile stress in
the crank of 25,000 pounds per square inch, assuming the stresses in
the crank to be equivalent to those in a ring of the diameter given.
D 15
= =1.5; r 2 = 25,000. From Table III, C = 0.227. Substi-
D, 10
tuting in Equation (40), we find /S = 0.001 inch.
Example 2. Let the shaft in Example 1 have a 5-inch axial hole
bored through it, other conditions being the same. Find the unit-
allowance.
D 2 D, 10
=1.5, as before; = = 2; T 2 = 25,000. From Table IV we
DI D 5
find C = 0.455.
Substituting in Equation (40), we find S = 0.0012 inch, the increase,
in the allowance being due to the fact that the hollow shaft is the more
compressible of the two.
Example 3. Let the crank-web in Example 1 be of cast-iron and
the maximum tensile stress in the hub be 4000 pounds per square inch.
Find the unit-allowance.
D 2
=1.5; T., = 4000. From Table III, we find C = 0.234. Sub-
A
stituting in (38) 8 = 0.0003 inch, which, owing to the lower tensile
strength of cast iron, is about one-third of the shrinkage-allowance
in Example 1, although the stress is two-thirds of the elastic limit.
For a forced fit, good practice gives (see Table VI) a unit-allowance
of 0.0013 inch, or one-third greater than that of Example 1. The
stresses which such an allowance would produce are, however, uncer-
tain, as will be further discussed in the following chapter.
Example 4. What is the radial pressure P x in the above examples?
Pi
For Examples 1 and 2, we find from Table I that - = 0.341. Hence,
T 2
P l = 25,000 X 0.341 = 8525 pounds per square inch.
Pi
For Example 3, we find from Table I that = 0.351. In this case
T 2
r a = 4000, hence,
P x = 4000 X 0.351 = 1404 pounds per square inch.
Example 5. What is the resistance to slip per inch of length of hub
in Example 3?
In Equation (30), A = 10, L = l, and from Example 4 we have P t
= 1404; / may be taken as 0.2. Then Q = 8817 pounds, which is the
total resistance of a ring of the hub, one inch in length.
Example 6. Let the crank in Example 3 be 20 inches least diameter,
No. 89 FORCED AND SHRINKAGE FITS
the other dimensions and the tensile stress remaining the same. Find
the increase in the radial pressure P 1? and hence that in the resistance
to slip.
D 2
In this case = 2, Table I gives the ratio A, for this condition,
Di
equal to 0.522, which is 49 per cent greater than the ratio A = 0.351
D 2
for =1.5. This percentage is the increase in radial pressure, and,
D!
hence, that in the resistance to slip.
Example 1. What is the true tangential stress (compressive) at the
bore of the shaft in Example 2?
The radial pressure P t is, from Example 4, 8525 pounds. Substi-
tuting this value, and also R i = 5, and .R = 2.5 in Equation (12), the
true stress T ==22,733 pounds per square inch.
Example 8. What is the intensity of the apparent tangential stresses
in the crank and shaft, Example 1?
The radial pressure P x is, from Example 4, 8525 pounds. Substi-
tuting this value, and also R. 2 = 1.5, and ^ 5 in Equation (4), the
apparent tensile stress t 2 at the bore of the hub is 22,165 pounds per
square inch. The similar compressive stress ^ at the cuter surface
of the shaft is, from Equation (13), equal to PI.
Shrinkag-e Temperatures
The temperature to which the outer member in a shrinkage fit should
be heated for clearance in assembling the parts, depends on the total
expansion required and on the coefficient a of linear expansion of the
metal, i. e., the increase in length of any section of the metal in any
direction for an increase in temperature of 1 degree F. The total ex-
pansion in diameter which is required, consists of the total allowance
for shrinkage and an added amount for clearance.
The value of the coefficient a is, for nickel-steel, 0.000007; for steel
in general, 0.0000065; for cast iron, 0.0000062. As an example, take
an outer member of steel to be expanded 0.005 inch per inch of in-
ternal diameter, 0.001 being the shrinkage allowance and the re-
mainder for clearance. Then:
a X t = 0.005
0.005
t = : = 769 degrees F.
0.0000065
The value t is the number of degrees F. which the temperature of
tlie member must be raised.
CHAPTER VII
PRACTICAL CONSIDERATIONS
Cylindrical and Tapered Fits
The form of the shrinkage fit is usually truly cylindrical and of one
diameter throughout; but both forced and shrinkage fits are, for some
classes of work, either tapered or double-cylindrical, i.e., with part of
the fit of one diameter and part of another. ' The advantages of the
tapered form in forced fits are: The possibility of abrasion of the
fitted surfaces is reduced; less work is required to drive the inner
member home; the drawings may be marked "Pit pin inches from
end of hole," which is the most trustworthy way of measuring the al-
lowances; and the parts are more readily separated, if a renewal of the
fit is desired. On the other hand, the difficulty of securing with ac-
curacy the same form for both fitted surfaces, is somewhat greater;
and the tapered fit is less reliable, since, if slip begins, the entire fit
is virtually free with but little movement. These advantages and dis-
advantages apply also, but in less degree, to the double-cylinder form.
The practice of a prominent shipbuilding company, for both forced
and shrinkage fits in either iron or steel, is: With large fits, both
the inner and outer members have a taper of 1/16 inch to the foot;
the allowances are 0.001 inch per inch of diameter with 0.001 inch
added to the total. If the conditions are such that it is more con-
venient to ream the hole with standard parallel reamers, the inner
member is tapered one half-thousandth inch (0.0005) per inch of
length, unless the fit is so long that this taper would reduce the al-
lowance at the small end to less than one-half that at the other ex-
tremity of the fit.
Differences between Forced and Shrinkage Fits
Lame's formulas, as given in Equations (2) and (3) and as changed
in the subsequent equations for lateral contraction according to the
principles established by Clavarino, are the basis of the ordnance for-
mulas employed by the United States Army and Navy. For economy
in weight, the stresses in the metal of a gun, at the instant of ex-
plosion, approach closely to the elastic limit. It is evident, then, that
the use of these formulas for such work makes their accuracy, for
shrinkage fits in gun-steel, unquestionable. So far as is known, their
fundamental principles are general, and they can be employed with
equal accuracy for similar fits in cast iron. It has been customary to
assume that they could be applied also for the determination of the
stresses in the metals of forced fits. This assumption is, in the au-
thor's opinion, unwarranted, so far, at least, as cast iron outer mem-
bers with large forcing allowances are concerned. There seems to be
considerable evidence in support of this contention.
34 No. 89 FORCED AND SHRINKAGE FITS
The basic principle cf shrinkage and forced fits is the same the
radial pressure on the contact-surfaces produced by the expansion of
the outer member and the compression of the inner; but there is a
radical difference between the methods by which this principle is ap-
plied in the two cases. In the shrinkage fit, the outer member, owing
to its expansion, slips freely into place, giving, in cooling, clean,
smooth, and accurately fitted surfaces. In forced fits, on the contrary,
there may be, in forcing, more or less abrasion, and, further, if the
allowances be large, there may be an axial flow of the metal of the
hub in advance of the entering shaft. It should be noted that, in
forcing allowances, we are dealing with a layer of metal whose thick-
ness is, in general, but 0.001 inch per inch of diameter, so that the
total volume of the metal thus displaced would be very small, while
its removal, with that lost by abrasion, would reduce materially the
amount of the effective allowances, and, in consequence, the stresses
and "grip" of the fit. Taking the elastic limit in tension of cast iron
as 6000 to 7000 pounds and that of steel as 50,000 pounds, and con-
sidering the corresponding values of E, the former will endure, with-
out permanent set, less than one-fourth the deformation of the latter,
yet the forcing allowances of the two metals are often made the same,
and, further, with the same metals and dimensions, some builders
make the allowances for forcing considerably greater than those for
shrinkage fits. In such cases, there must be either permanent set in
the cast-iron hub, or the effective allowances must be materially les-
sened by abrasion, displacement, or both.
In Professor Wilmore's tests, the average resistance of the shrinkage
fit to slip was, for an axial pull, 3.66 times greater than that of the
forced fit, and, in rotation or torsion, 3.2 times greater. In each com-
parative test, the dimensions and allowances were the same for both.
These results imply either permanent set or considerable abrasion or
displacement of the metal of the forced fit. While these experiments
were made on a small scale, they agree with the general estimate of
the comparative strength of forced fits.
Table VI represents the practice of one of the largest builders of
engines and other machinery in the United States, in forcing cast-
iron cranks and wheel-hubs on steel shafts. The allowance for a
crank is greater than that for a wheel-hub, and, with both, the allow-
ance per inch of diameter decreases with increasing diameter. Take
the unit-allowance for a 12-inch wheel-hub which is 0.001 inch. As-
sume the ratio of the external diameter of the hub to that of the shaft
(solid) as 1.8, which gives a hub-thickness of 4.8 inches. If in Equa-
tion (38), -8 = 0.001, and, from Table III, (7 = 0.311, then the true
tensile stress T 2 at the bore of the hub is about 13,000 pounds, or twice
the elastic limit of cast iron. Again, we have here indications of per-
manent set, excessive abrasion, or very considerable displacement of
the metal, so that the effective allowances cannot be those initially
given.
Finally, the following formulas given by Mr. Stanley H. Moore may
PRACTICAL CONSIDERATIONS
35
be cited. In these formulas, d denotes the total allowance, and D is
the diameter of the shaft, in inches.
17
D + 0.5
16
Shrinkage fit d =
Forced fit
1000
2 D + 0.5
1000
These formulas show again a much greater allowance for forcing
than for shrinkage.
Forced fits may be made by levers, screw-jacks, or hydraulic pres-
sure, the latter being the most common. In the drive-fit, the pin is
TABLE VI. ALLOWANCES FOR FORCED FITS
Steel Shaft and Pin to Cast-iron
Cranks. Average pressure re-
quired = 12.5 tons (of 2000 pounds)
per inch of diameter.
Steel Shaft to Cast-iron Wheel-hubs.
Average pressure required = 10
tons (of 2000 pounds) per inch of
diameter.
Diameter of
Allowance per Inch
Diameter of
Allowance per Inch
Shaft, Inches
of Diameter
Shaft, Inches
of Diameter
4
0.0030
12
0.0010
5
0.0024
13
0.0009
6
0.0020
15
0.0008
7
0.0017
17
0.0007
8
0.0015
18
0.0006
9
0.00135
19
0.00055
10
0.0013
22
0.0004
11
0.0012
23
0.00035
12
0.0010
24
0.0003
13
0.0010
26
0.00025
14
0.0010
27
0.0002
15
0.0010
16
0.0009
18
. 0008
20
0.00075
sent home by sledges; the allowances are usually about half that of a
forced fit. With these various methods and the many purposes for
which forced fits can be used, it is natural that the custom as to the
amount of the allowances should differ, as it does, very widely, so that
the practice cited here is not universal. The purpose of this dis-
cussion has been simply to point out that shrinkage formulas will not
give with accuracy the stresses in a cast-iron hub, when the allow-
ances are very large, or in any forced fit. with undue allowances. Such
a fit differs essentially from the shrinkage joint for which the formulas
were constructed.
Cotterill says in his "Applied Mechanics," London, 1895, page 412:
"When the limit of elasticity is overpassed, the formula (Lamp's)
fails, and the distribution of stress becomes different. If the pressure
be imagined gradually to increase until the innermost layer of the
cylinder begins to stretch beyond the limit, more of the pressure is
36 No. 89 FORCED AND SHRINKAGE FITS
transmitted into the interior of the cylinder, so that the stress be-
comes partially equalized. If the pressure increases still further, the
tension of the innermost layer is little altered, and, in soft materials,
longitudinal flow of the metal commences under the direct action of
the fluid pressure. The internal diameter of the cylinder then in-
creases' perceptibly and permanently. This is well known to happen
in the cylinders employed in the manufacture of lead piping, which
are exposed to the severe pressure necessary to produce flow in the
lead. The cylinder is not weakened but strengthened, having adapted
itself to sustain the pressure. Cast-iron hydraulic press cylinders are
often worked at the great pressure of 3 tons per square inch, a fact
which may perhaps be explained by a similar equilization."
Forcing Pressure
When the fit is cylindrical, the forcing pressure varies as the rate
of advance of the inner member, reaching a maximum in continuous
forcing when the pin or shaft is at the inner end of the hole. At this
point, the pressure is theoretically equal to Q, the resistance to slip,
as given in Equation (30), the coefficient of friction / being probably
between 0.12 and 0.2, although it may vary widely. Tables VI and
VIII give values of the forcing pressure, as found in practice. The
assumption above, that the maximum forcing pressure is equal to the
resistance to slip, is true only if that pressure is expended wholly in
overcoming the obstruction to motion produced by the resistance of
the outer member to expansion and of the inner to compression. If
there is abrasion of the surfaces, or axial displacement of the metal
in advance of the entering member, the assumption is not fully jus-
tified.
Applications in Practice
Railway Work. In railway work, steel tires are shrunk on the cast-
iron wheel-centers of driving wheels. The fit is cylindrical; a com-
mon, although not universal, shrinkage-allowance is 0.001 inch per
inch of diameter of the finished wheel. Forced fits are used for se-
curing wheels to axles and crank-pins to driving wheels. In wheel-
fits, the joint is cylindrical; the pressure is usually 9 to 10 tons per
inch of diameter of fit. In removing a wheel after long service, the
total pressure may reach 150 tons.
Stationary Engines. Shrinkage and forced fits the latter more fre-
quently are used for crank-pins, cranks, wheel-hubs, and minor parts.
With different builders, the amount of the unit-allowance has a wide
range, owing to differences in the thickness of hubs, the forcing pres-
sure employed, etc. General practice seems to favor a smaller allow-
ance for shrinkage than for forcing, and, with increasing diameter, a
decreasing unit-allowance. The latter is usually greater for cast iron
than for steel. Table VII, which gives the data for typical fits from
different builders, shows the variation in practice. In Table VIII*
will be found complete data for forced fits from 2 to 9 inches in diam-
eter.
* MACHINERY, May, 1897.
PRACTICAL CONSIDERATIONS
37
Marine Engines. In marine work, built-up crank-shafts are as-
sembled and the casings of propeller shafts are secured by shrinkage
fits. Forced fits have been employed for crank-shafts and are fre-
quently used for smaller parts. In building up a steel shaft, the al-
lowance is usually 0.001 inch per inch of diameter; the cranks and
crank-pins are keyed, in addition to the shrinking. The crank-webs
are heated by gas in a sheet-iron furnace until the expansion is suf-
ficient for a free fit; they are then removed, the pin is pushed home
and keyed, and the webs and pin are cooled with water. The webs
are then set with the bores for the shaft vertical, and one is heated
as before until sufficiently expanded, when the section of the shaft
TABLE VII. EXAMPLES OP TYPICAL FITS, FROM PRACTICE
Diameter of Pin
or Shaft,
Inches
Total Allowance,
Inches
Metals
Shrinkage
Forcing
1.8798
4.2505
8.9
4 to 5
7. 5 to 9
16 to 18
4
8
16
1 to 2
4 to 6
5 to 7
9 to 12
10 to 12
5
5
11
13
0.0031
0.0103
0.0152
0.0090
0.0055
0.0030
0.0120
0.0120
0.0144
0.0010
oioosj'
'oioioo'
0.0050
0.0100
1 !
Shaft, steel. Hub, cast iron
Shaft, steel. Hub, cast iron
Shaft, steel. Hub, cast iron
Cast iron crank
Cast iron crank
Cast iron crank
Crank, cast iron. Shaft, steel
Crank, cast iron. Shaft, steel
Crank, cast iron. Shaft, steel
Shaft, steel. Crank, cast steel
Shaft, steel. Crank, cast iron
Cast iron counter-balance
plates on steel crank-disks
0.0045
0.0027
0.0015
0.0090
0.0156
0.0313
6 '6676*
0.0060
is lowered into place and keyed; the same method is followed with the
other section of the shaft.
Shaft casings are of bronze, usually from % inch to 1 inch thick at
various sections of the shaft. In one case there were two such sections
of casing, each 8 feet long and 20% inches internal diameter. The
shrinkage-allowance, total, was 0.013 inch, or 0.000634 inch per inch
of diameter. Each section was set vertical and heated internally by
gas. When expanded, it was slipped in place on the shaft, and the
inner end was held firmly and cooled with water until it gripped the
shaft.
Gun Construction. When a charge is exploded in the powder-cham-
ber, the principal stress to which a gun is subjected is that due to the
radial pressure of the gases which tends to burst it on an axial plane.
This stress produces tangential (circumferential) tension in the tube,
jacket, and hoops, and, in addition, there is a direct longitudinal stress
38
. 89 FORCED AND SHRINKAGE FITS
in the layer of the tube in which the breech-plug houses. There also
exists at all times, except during explosion, a radial compressive stress
on the inner cylinders of the system, due to the shrinkage pressures
of those outside of them. At the breech, there may be three or four
of these superposed cylinders the tube, the jacket, and one or two
sets of concentric hoops. The radial pressure of the gases would pro-
duce in the tube, if the latter were unsupported, a circumferential
tensile stress which would exceed the elastic limit of the metal. To
TABLE VIII. DATA FOR FORCED FITS, FROM PRACTICE
c
"g
j
1
c
E
5
1
5
,
t
s
4*
u
CK
11
E .8
s
li
3
||
II
H ED
1!
1
ll
i
HI
c
c
"S .
3+3
ll
5
3
5
c
1
I
o 1
l"!
I
P
I
|
s
3
<
>
^
1.8798
6.125
1.8767
0.0031
00.0170
36.0
16.7
2
10
20
1.8819
6.125
1.8770
0.0042
0.00220
36.0
16.7
2
15
23
1.8774
4.375
1.8764
0.0010
0.00052
24.4
13.7
0.5
1
1
2.7455
4.500
2.7387
0.0068
0.00247
38.7
26.5
3
12
25
2.7465
4.500
2.7437
0.0028
0.00100
38.7
26.5
5
12
23
3.2610
5.000
3.2542
0.0068
0.00210
51.0
41.5
5
20
45
3.2625
5.000
3.2555
0.0070
0.00200
51.0
41.5
5
15
30
3.2670
5.000
3.2610
0.0060
0.00180
51.0
41.5
5
15
20
4.2505
6.000
4.2402
0.0103
0.00240
79.8
85.1
5
22
44
4.2388
6.625
4.2478
0.0091
0.00210
78.1
93.4
12
30
60
4.2303
6.500
4.2224
0.0079
0.00190
95.8
91.0
10
60
125
5.9343
4.062
5.9216
0.0127
0.00220
75.7
112.2
6
16
25
5.9381
4.000
5.9252
0.0129
0.00220
74.4
110.4
3
18
35
5.9294
4.125
5.9194
0.0100
0.00170
76.7
113.8
5
15
25
6.8829
5.125
6.8697
0.0132
0.00200
110.7
190.1
8
20
42
6.8890
5.000
6.8785
0.0105
0.00150
108.0
185.9
5
22
45
6.8692
4.875
6.8550
0.0142
0.00210
104.8
180.4
5
35
65
7.8884
5.500
7.8730
0.0154
0.00200
135.9
267.3
5
32
64
7.8715
6.500
7.8575
0.0140
0.00180
160.5
315.9
5
25
50
7.8620
5.625
7.8460
0.0160
0.00200
138.2
272.8
8
40
80
8.9240
6.125
8.9050
0.0190
0.00210
170.8 378.9
20
45
68
8.9000
6.750
8.8848
0.0152
0.00170
188.4 419.9
5
47
96
8.8780
6.500
8.8669
0.0112
0.00130
180.7
401.0
10
45
92
counteract this, the jacket and hoops are shrunk on, each of these
cylinders putting the one which it encases under compression, and
the aggregate of these radial pressures being transmitted to the tube.
The actual tensile stress in the latter, during the burning of the pow-
der, is then the difference between the tensile stress developed by the
gases and the compressive stress due to the jacket and hoops a re-
mainder which is less than, but usually fairly close to, the elastic
limit of the metal.
For maximum economy of material, the relations of the thicknesses
and shrinkage-allowances should be such that the stresses at all points
PRACTICAL CONSIDERATIONS 39
In the walls of the built-up gun will be, during explosion, not only
approximately equal but also the greatest permissible, with due re-
gard to the elastic limit and the factor of safety. The outer layers
of the metal are, therefore, in a state of initial tension, the inner un-
der initial compression, and during explosion all are in tension. The
various thicknesses and allowances for the cylinders of any given gun
can be computed by an extension of the methods shown by Formulas
(2) and (3), and those in (1) for the corresponding unit-deformations
due to the true stresses. The principles involved are, therefore, those
which have been treated herein for shrinkage fits, with the added re-
quirement that the superposed cylinders, during explosion and the sub-
sequent release from pressure, must expand and contract together, so
that each cylinder must have a definite shrinkage-allowance with re-
gard to all the others of the system.
The 16-inch Army rifle, now at Sandy Hook, was designed for a pow-
der-pressure of 38,000 pounds per square inch, a muzzle-velocity of
2500 feet per second, a muzzle-energy of 88,000 foot-tons, a penetration
at the muzzle of 42.3 inches in steel, and a range of 21 miles. The
weight of the gun is 126 tons and its total length is 49 feet 2.9 inches.
At the breech, the gun is built up of a tube, a jacket, and two sets of
hoops, the thicknesses being 5.3, 7.2, 3.7, and 4.3 inches, respectively.
The tube and jacket are of nickel-steel, not fluid-compressed; the hoops
are of fluid-compressed steel containing no nickel. The elastic limits
in tension of the two metals were about 52,000 and 57,000 pounds, re-
spectively, the hoop-metal being thus the harder and stronger. The
forgings, after being rough-turned and bored, were tempered in oil
and annealed. In expanding the jacket or a hoop, it was set vertically
in a cylindrical furnace of fire-brick, and was then encased in a muf-
fle of %-inch boiler steel. The combustion-chamber between the muf-
fle and the furnace-wall was 11 inches wide. The fuel was oil sprayed
with steam through 20 burner openings, the flame striking the muf-
fle at a tangent, so as to give a spiral movement to the gases. The
circulation of the air between the muffle and the hoop kept the tem-
perature of the latter uniform at all points. The heating of the jacket
required 30 hours, and its bore was calipered three times during that
period to determine the expansion.
In shrinking on the jacket, the tube was first set vertical, muzzle-
end down, in a shrinkage-pit adjacent to the furnace; the lower end
was secured in a cast-iron chuck anchored in the concrete foundations
of the pit. Water-connections were made for cooling the interior of
the tube and the exterior of the jacket when seated. The latter, when
removed from the furnace, was measured, centered, and lowered into
place. Water was then applied at the muzzle-end; the cooling con-
tinued for nine hours, the number of encircling "water-rings" or pipes
varying from four, as a maximum, to two at the close of the operation.
The shrinkage of the hoops near the muzzle was effected similarly;
the remainder were assembled with the gun in a horizontal position
in the lathe, each hoop during shrinkage being under the axial pres-
sure of two 30-ton hydraulic jacks.
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valuable results of carefully made experiments are recorded, and
the world's progress in every field of mechanical endeavor is
closely watched.
One of the most valuable features is the four-page monthly
Data Sheet Supplement printed on strong manila paper. These
Data Sheets contain high-grade, condensed mechanical data,
covering machine design, machine operation and kindred subjects.
They are the cream of mechanical information.
THIS BOOK IS DUE ON THE LAST DATE
STAMPED BELOW
AN INITIAL FINE OF 25 CENTS
WILL BE ASSESSED FOR FAILURE TO RETURN
THIS BOOK ON THE DATE DUE. THE PENALTY
WILL INCREASE TO SO CENTS ON THE FOURTH
DAY AND TO $1.OO ON THE SEVENTH DAY
OVERDUE.
DEC 19 1940 M
LD 21-100m-7, '40 (6936s)
YC 53944
w
UNIVERSITY OF CALIFORNIA LIBRARY
CONTENTS OP DATA SHEET BOOKS
Ho. 1. Screw Threads. United States,
Whitworth, Sharp V- arid British Associa-
tion Standard Threads; Briggs Pipe
Thread; Oil Well Casing Gages; Fire Hose
Connections; Acme Thread; Worm
Threads; Metric Threads; Machine, Wood,
and Lag Screw Threads; Carriage Bolt
Threads, etc.
No. 2. Screws, Bolts and Nuts. Fil-
lister-head, Square-head, Headless, Col-
lar-head and Hexagon-h ,'<! Screws; Stand-
ard and Special Nuts; T iuts, T-bolts and
Washers; Thumb Screws and Nuts; A. L.
A. M. Standard Screws and Xu*s; Machine
Screw Heads; Wood Seres, s. Tap Drills;
Lock Nuts; Eye-bolts, etc.
No. 3. Taps and Dies.- Hand, Mn chine,
Tapper and Machine Screv Taps; Taper
Die Taps; Sellers Hobs; Screw Machine
Taps; Straight and Taper Boiler Taps;
Stay-bolt, Washout, and Patch-bolt Taps;
Pipe Taps and H -bs; Solid Square, Round
Adjustable and Spring Screw Threading
Dies.
No. 4. Reamers, Sockets, Drills and
Milling Cutters. Hand Reamers; Shell
Reamers and Arbors; Pip^ Reamers; Taper
Pins and Reamers; Brown & Sharpe,
Morse and Jarrio Taper Sockets and Ream-
ers; Drills; Wire Gages; Milling Cutters;
Setting Angles for Milling Teeth in End
Mills and Angular Cutters, etc.
No. 5. Spur Gearing 1 . Diametral and
Circular Pitch; Dimensions of Spur Gears;
Tables of Pitch Diameter*; Odontograph
Tables; Rollirg Mill Gearing; Strength of
Spur Gears; Horsepower Transmitted by
Cast-iron and Rawhide Pinions; Design of
Spur Gears; Weight of Cnst-iron Gears;
Epicyclic Gearing.
No. 6. Bevel, Spiral and Worm Gear-
ing. Rules and Formulas for Bevel
Gears; Strength rf Bevel Gears; Design
of Bevel Gears; Rules and Formulas for
Spiral Gearing; Tables Facilitating Calcu-
lations; Diagram for Cutters for Spiral
Gears; Rules and Formulas for Worm
Gearing, etc.
No. 7. Shafting, Keys and Keyways.
Horsepower of Shafting; Diagrams and
Tables for the Strength of Shafting;
Forcing-, Driving, Shrinking and Running
Fits; Woodruff Keys; United States Navy
Standard Keys; Gib Keys; Milling Key-
ways; Duplex Keys.
No. 8. Bearings, Couplings, Clutches,
Crane Chain and Hooks. Pillow Blocks;
Babbitted Bearings; Ball and Roller Bear-
ings; Clamp Couplings; Plate Couplings;
Flange Couplings; Tooth Clutches; Crab
Couplings; Cone Clutches; Universal
Joints; Crane Chain; Chain Friction;
Crane Hooks; Drum Scores.
No. 9. Springs, Slides and Machine
Details. Formulas and Tables for Spring
Calculations; Machine Slides; Machine
Handles and Levers; Collars; Hand
Wheels; Pins and Cotters; Turn-buckles,
etc.
No. 10. Motor Drive, Speeds and Peeds,
Change Gearing, and Boring Bars. Power
required for Machine Tools; Cutting
Speeds and Feeds for Carbon and High-
speed Stoel; Screw Machine Speeds and
Feeds; Heat Treatment of High-speed
Steel Tools; Taper Turning; Change Gear-
ing for the Lathe; Boring Bars and Tools,
etc.
No. 11. Milling Machine Indexing,
Clamping Devices and Planer Jacks.
Tables for Milling Machine Indexing;
Change Gears for Milling Spirals; Angles
for setting Indexing Hoad when Milling
Clutches; Jig Clamping Devices; Straps
and Clamps; Planer Jacks.
No. 12. Pipe and Pipe Fitting*. Pipe
Threads ami Gages; "ust-iron Fittit
Bronze Fittings; Pipe Flanges; Pipe
Bends; Pipe '!in,i>.s and Hangers; Dimen-
sions of Pipe f> .r Various Services, etc.
No. 13. Boilers and Chimneys. Flue
Spacing and Bracing f- r Boilers; Strength
of Boiler Joints, Riveting; Boiler Setting;
Chimneys.
No. 14. Locomotive and Bail way Data.
Locomotive Boilers; Bearing Pressures
for Locomotive Journals; Locomotive
Classifications; Rail Sections; Frogs,
Switches and Cross-overs; Tires; Trac'ive
Force; Inertia of Trains; Brake Lev. rs;
Brake Rods, etc.
No. 15. Steam and Gas Engines. Sat-
urated Steam; Steam Pipe Sizes; St<-am
Engine Design; Volume of Cylinders;
Stuffling Boxes; Petting Corliss Engine
Valve Gears: Coii.lc ;ser and Air Pump
Data; Horsepowc of Gasoline Engines:
Automobile Engine Crankshafts, etc.
No. 16. Mathematical Tables. Squares
of Mixed Numbers; Functions of Frac-
tions; Circumference and Diameters of
Circles; Tables for Spa -ing off Circles.
Solution of Triangles; Formulas for Solv-
ing Regular Polygons; Get metrical Pro-
gression, etc.
No. 17. Mechanics and Strength of Ma-
terials. Work; Energy: Centrifi ^al
Force; Center of Gravity: ^ lotion: Fric-
tion; Pendulum; Fail MIS Lo, -^s; Strength
of Materials; Str-i.rth : Wai Plates;
Ratio of Outside a-d Inside Radii of
Thick Cylinders, etc.
No. 18. Beam Formulas and Structural
Design. Beam Formulas; Sectional Mod-
uli of Structural Shapes; Beam Charts;
Net Areas of Structural Angles; Rivet
Spacing; Splices for channels and I-
beams; Stresses in Roof Trusses, etc.
No. 19. Belt, Rope and Chain Drives.
Dimensions of Pulleys; Weights of Pul-
leys; Horsepower of Belting; Belt Veloc-
ity; Angular Belt Drives; Horsepower
transmitted by Ropes; Sheaves for Rope
Drive; Bending Stresses in Wire Ropes;
Sprockets for Link Chains; Formulas and
Tables for Various Classes of Driving
Chain.
No. 20. Wiring Diagrams, Heating and
Ventilation, and Miscellaneous Tables.
Typical Motor Wiring Diagrams; Resist-
ance of Round Copper Wire; Rubber Cov-
ered Cables; Current Densities for Vari-
ous Contacts and Materials; Centrifugal
Fan and Blower Capacities; Hot Water
Main Capacities; Miscellaneous Tables:
Decimal Equivalents, Metric Conversion
Tables, Weights and Specific Gravity of
Metals, Weights of Fillets, Drafting-room
Conventions, etc.
MACHINERY, the monthly mechanical journal, originator of the Reference and
Data Sheet Series, is published in three editions the Shop Edition, $1.00 a year;
the Engineering Edition, $2.00 a year, and the Foreign Edition, $3.00 a year.
The Industrial Press, Publishers of MACHINERY,
49-55 Lafayette Street, New York City, U. S. A.
