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# Strength of materials

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bending moment. This, in turn, is then combined with the torsional
moment in the manner indicated in the following.

It should be noted that in the case of shafting, the location and
direction of the tooth loads, belt pulls, etc., which produce bending,
remain fixed while the shaft rotates. The bending stresses are thus
constantly varying in direction, and a greater factor of safety should
be used than for a beam subjected to a load in one direction only.

Combined Bending- and Torsion

If a bar or beam is subjected to both bending and torsion at the
same time, the moment to which it is subjected is known as the com-
bined moment and an empirical formula for this condition may be
written:

Combined moment = \/M b 2 + M t 2 = SZ

in which S = permissible working stress in pounds per square inch;

Z = section modulus for bending (moment of resistance);

Mb = maximum bending moment in inch-pounds;

M t = maximum torsional moment in inch-pounds.

This formula is entirely empirical. It was published in 1900 by J.

J. Guest as the result of experiments made by him. It is apparently

not applicable in cases where the torsional moment is very large

compared with the bending moment, but for most practical conditions,

where it is likely that both moments are of appreciable magnitude,

or especially where the bending moment is larger, the formula gives

satisfactory values in cases where shafts, beams and machine parts are

subjected to combined bending and torsional stresses. The formula

has been proved by experiments to be especially applicable to mild

steel (machine and structural steel), and as the machine designer is

almost exclusively concerned with this material, when dealing with

questions of combined bending and twisting stresses, the formula may

be accepted as safe and correct for all practical purposes. The safe

stress S in this formula may be assumed to be equal to the safe stress

in bending.

Example. Assume that a square bar is subjected to a combined
bending and 'torsional moment. The bending moment is caused by a
load of 3000 pounds supported at the outer free end of the bar, the
length of which is 30. inches. The torsional moment is 90,000 inch-
pounds. Find the size of square bar made from structural steel which
will stand the combined moment with a factor of safety of 5.
In this case,

M b = 3000 X 30 = 90,000 inch-pounds;
M t = 90,000 inch-pounds;

34 No. 136 STRENGTH OF MATERIALS

8.== 60,000 -f- 5 = 12,000 pounds per square inch;

Z for a square section (see MACHINERY'S HANDBOOK, page 308)

= a 3 -f- 6, in which a = the side of the square.
Then:

Combined moment = |/90,000 2 + 90,000 2 = 127,000 approx. and

a 3
127,000 = SZ = 12,000 X = 2000 a 3 ;

6

127,000
a 3 = - = 63.5;

2000
a = 4 inches, very nearly.

Combined Torsion and Compression

Propeller shafts of steamers and vertical shafts carrying considerable
weight, are subjected to combined torsion and compression. Let P t =
maximum resultant compressive stress; P 2 = maximum resultant shear-
ing stress; C = the compressive stress due to the thrust; 8 = the
shearing stress due to the twisting moment. Then,

PI = 2 (C + \/C' 2 + 4S 2 ); p 2 =
It is evident that the safe compressive stress must not exceed P u
and the safe shearing or torsional stress must not exceed P 2 .

Torsional Deflection of Shafting-

Shafting must be proportioned not only so that it has the required
strength for transmitting a given amount of power, but so that it can-
not be twisted through a greater angle than has been found satisfactory
by previous experience. Ordinarily, it is assumed by many authorities
that the allowable twist in degrees should not exceed five minutes or
about 0.08 degree per foot length of the shaft. The following formula
gives the angle of torsional deflection of a cylindrical shaft:

584 Tl
a = -

d*G

in which a = angle of torsional deflection in degrees;
T = twisting moment in inch-pounds;
I = length of shaft being twisted in inches;
d = diameter of shaft in inches;
G = torsional modulus of elasticity, generally assumed as

12,000,000 for steel shafting.

It will be seen from an inspection of the formula above that in the
case of torsional deflection the length of the shaft enters in the calcu-
lation. From the previous formulas relating to the strength of the
shaft, it will be seen that when mere strength is calculated, the length
of the shaft subjected to torsion only does not influence its strength.
Example. Find the torsional deflection for a shaft 4 inches in
diameter and 48 inches long, subjected to a twisting moment of
24,000 inch-pounds.

584 X 24,000 X 48

a = = 0.22 degree, or 13 minutes.

4 4 X 12,000,000

SHAFTING 35

The deflection per foot, then, equals 13 -=- 4 = S 1 ^ minutes, which
is within the limits generally allowed.

If it is required to find the diameter of shaft which will give a cer-
tain angle of torsional deflection, the following formula may be used:

J32 X 12 X
Nl 3.14 X G

X L X 360 X PR
d

X a X 2 X 3.14
in which d = diameter of shaft;

P = force acting on the shaft, producing rotation, in pounds;
R = length of lever arm of force P, in inches;
a = angle of torsional deflection in degrees;
L = length of shaft being twisted in feet;
O = torsional modulus of elasticity (=12,000,000).
For an angle of deflection equal to 0.08 degree per foot length of the
shaft, or a total angle a of 0.08 L degrees,

4 I H ' P -

d = 0.29>y PR = 4.6 v .

>/ n

Example. Find the diameter of a lineshaft to transmit 10 horse-
power at 150 revolutions per minute, with a torsional deflection not
exceeding 0.08 degree per foot of length.

I 10

* = 2.;

N 150

d = 4.6 * I = 2.35 inches.

150

It will be seen, by comparing with the section, "Application to
Shafting," that a larger diameter is required, in this case, to prevent
excessive torsional deflection than is required by mere considerations
of strength. For short shafts, it is unnecessary to calculate for the
angular deflection. It is only in the case of long shafts 'that this is
necessary, and even then only if the torsional deflection would be
objectionable.

Linear Deflection of Shafting-
Shafting is subjected to combined bending and twisting moments,
the -twisting being caused by the forces which give it rotary motion,
while the bending is caused partly by the weight of the shaft itself
between the bearings and partly by the load placed upon it in the
form of pulleys, gears, etc. In the case of shafting, the deflection due
to bending must be considered, as well as the torsional deflection and
torsional strength. It is considered good practice for line-shafting to
limit the deflection to a maximum of 0.010 inch per foot of length. The
maximum distance in feet between bearings for average conditions, in
order to avoid excessive linear deflection, is determined by the
formulas :

L = ^TlOdTfor bare shafts;

L = ]<Kl40d 2 for shafts carrying pulleys, etc.,
in which d = diameter of shaft in inches;

L = maximum distance between bearings in feet.

36 No. 136 STRENGTH OF MATERIALS

It is understood that to avoid excessive deflection due to bending,
pulleys and gears should be placed as close to the bearings as possible.

Hollow Shafts

For the same weight per linear foot, or, which is the same, for the
same area of cross-section or the same amount of material, a hollow
shaft is stronger than a solid shaft, because the section modulus for
an annular ring is greater than for a "solid" circle of the same area.
The hollow shaft, of course, has a greater diameter than the solid shaft
Calculations will easily show the above statement to be true.

The use of hollow shafts not only reduces 'the weight of a shaft for
a given strength, but increases the reliability of the shafting, on ac-
count of the removal of the metal from the core of the shaft This
applies especially to shafts of large diameters, as in large steel ingots
the central core is likely to be less dense than the outer portion and
to show shrinkage cavities near the center. If the ingot is bored out,
the spongy or "piped" portion will be removed, and 'the metal remaining
will be superior in quality to that in a solid shaft. Ingots for shafting
should, however, not be cast hollow, but be bored out after having been
cast solid.

The following is a simple method given by Mr. E. Hammarstrom
in MACHINERY for finding the dimensions of a hollow shaft which can
be substituted for a solid shaft of equal strength to resist bending or
torsion:
Let A = diameter of solid shaft;

D = outside diameter of hollow shaft;
d = inside diameter of hollow shaft;
t = (D d)= thickness of metal of hollow shaft;
k = d -r- D = ratio of diameters of hollow shaft.

As the hollow shaft is to have the same strength to resist bending
as the solid shaft, the moment of resistance of both must be equal.
Hence :

w(D 4 d 4 ) TrD, 3 d 4

, from which D 3 = D, 3 ( 1 )

32 D 32 D

If kD is substituted for d in Equation (1):

D 3 ~

D 3 D 3 k 4 =A 3 , from which = A | (2)

1 k 4
d

In a similar manner, by substituting for D in Equation (1) :

k

-*=* A (- (3)

D, \ 1 k 4

Further, as t = \$ (D d), Formula (4) is found by substitution and
simplification:

t l k

(4)

D, \ 1 k 4

In the accompanying table the values of the factors containing k in

SPRINGS

Equations (2), (3), and (4) are calculated for certain values of fc.
The bottom line of the table gives the weight of the hollow shaft in
per cent of that of the solid.

It is evident that Equation (1) would be the same, if it were derived
under the assumption that the hollow shaft had the same torsional
strength as the solid one, instead of having the same strength against
bending, as assumed. The table will therefore hold true for shafts
subjected to bending or torsion, or both.

Assume, as an example, that a solid shaft 3 inches in diameter is

TABLE OF FACTORS FOB FINDING DIMENSIONS OF HOLLOW SHAFTS
TO REPLACE SOLID SHAFTS

j

1

K D H

Ratio d -*- D = k

Ratio of

0.50

55

0.60

0.65

0.70

0.75

0.80

0.85

0.90

D+- D l =

1.021

1.032

1.047

1.067

1 095

1.135

1.192

1.279

1 427

d -r- D* =

0.510

0.567

0.628

0.694

0.767

851

0.951

1.087

1 284

t-*- D! =

0.257

0.232

0.209

0.186

0.164

0.141

0.119

0.096

0.071

Weight of hol-
low shaft*....

78.3

74.35

70.2

65.8

61.3

56.4

51.6

45.4

38.7

* Weight of hollow shaft is given in per cent of weight of solid shaft.

to be replaced by a hollow, shaft, ratio fc being 0.5. Then, by inserting
the value found from the table in Equation (2):

D

= 1.021 and D = 3 X 1.021 = 3.063 inches,

D,
d = Q.5D = 1.532 inch.

Application to Helical Springs

In helical springs the safe load W that may be placed upon a spring
made from a round wire or rod may be found from the following
formula:

0.4 Sd*

W =

D d

in which S = safe shearing strength of material in pounds per square

inch;

d = diameter of wire or bar from which spring is made;
D = outside diameter of helical spring.

The deflection in one coil of a helical spring may be found for a
spring made from round wire or rod from the following formula:

3.14

Gd 4

Gd

38 No. 136 STRENGTH OF MATERIALS

in which F = deflection in one coil in inches;
W = safe load in pounds;

D = outside diameter of helical spring in inches;
G = torsional modulus of elasticity (12,600,000 for spring

steel) ;
8 = safe shearing strength of material in pounds per square

inch;
cl = diameter of wire or bar from which spring is made, in

inches.

Complete formulas covering all classes of springs will be found in
standard handbooks. (See MACHINERY'S HANDBOOK, page 412.)

Maximum Safe Stresses in Coil Spring's

The following values may be used for the torsional or shearing
stresses in coil springs made from a good grade of steel. Assume
the ratio of the mean diameter of the spring to the diameter of the
bar to equal R; then:

For bars below % inch diameter:

R = 3 & = 112,000 pounds per square inch.
R = 8 8 = 85,000 pounds per square inch.
For bars 7/16 to 3/4 inch in diameter:

R = 3 8 = 110,000 pounds per square inch.
R = 8 8= 80,000 pounds per square inch.
For bars from 13/16 to 1 1/4 inch in diameter:

R = 3 8 = 105,000 pounds per square inch.
R = 8 8 = 75,000 pounds per square inch.

For bars over 114 inch in diameter a stress of more than 100,000
should not be used. Where a spring is subjected to sudden shocks, a
smaller value of 8 is necessary.

These values are applicable to compression springs with open coils.
Experience has shown that in close-coiled springs and extension springs
the safe value of the stress per square inch, 8, is only about two-thirds
of that for open-coiled compression springs of the same dimensions.
The safe torsional or shearing strength for spring brass and phosphor-
bronze may be taken as 25,000 pounds per square inch. The torsional
modulus of elasticity may be taken as 6,000,000 for spring brass and
phosphor-bronze, and 12,600,000 for steel.

The best proportions for coil springs is to use an outside diameter
of the spring equal to from five to eight times the diameter of wire or
bar from which the spring is made; under no circumstances should the
outside diameter be made less than four times the diameter of the
wire. The effective number of coils in a compression spring may be
considered as 2 less than the actual number, owing to the squared ends
of the spring. Springs of small diameter may be safely subjected to a
higher unit stress than those of large diameter.

Materials Used for Springs

Steel containing about one per cent carbon and comparatively free
from phosphorus and sulphur, generally known as spring steel, is
ordinarily used for springs.

SPRINGS 39

For small springs, music wire is used to a great extent and is the
best material obtainable for this purpose. It is especially recommended
for devices where the spring is compressed frequently and suddenly.
Vanadium steel has recently come into use to a considerable extent
for springs. The addition of a small percentage of vanadium to steel
increases the elasticity of the material, but the cost of springs made
from this material is considerably higher.

Brass and phosphor-bronze should be used for springs that must
resist moisture. These springs, however, are much more expensive
than steel springs, bo>th on account of the higher cost of the material,
and because the permissible stress is less, thus making larger sizes
of these springs necessary for the same capacity.

Factor of Safety in Springs Frequently Compressed
When a spring acts only occasionally it can be safely designed to
carry a load which causes a fiber stress nearly equal to the elastic
limit of the spring, but when the compressions or extensions are
frequent, a larger factor of safety must be used. A valve spring in an
automobile motor, for example, which operates, say, 200 times a min-
ute, should have a factor of safety of at least 4. In other words, a
spring made of %-inch wire, which ordinarily could be designed for a
torsional stress of 100,000 pounds per square inch, should be designed
to work at a stress not over 25,000 pounds per square inch when used
in service of the kind mentioned.

High-class springs, such as valve springs, should have the ends
squared and ground at right angles to their axis; the outside diameter
should be at least one-third of the length, and it should be supported
its entire length, unless it is very short, in order to prevent buckling,
which introduces bending and twisting strains. High-class valve
springs when placed on end on a flat plate should not vary more than
% degree from the perpendicular to the plate. These springs should
be protected from rusting by a good coat of japan, baked on, or by
electro-galvanizing.

CHAPTER V

MISCELLANEOUS APPLICATIONS

Strength of Columns or Struts

When a member subjected to compression stresses has considerable
length in proportion to its width, depth or diameter, the ordinary
formulas for compression are not applicable, because bending stresses
are set up on account of the length of the column or strut, and these
stresses tend to bend or bulge the member. A number of empirical
formulas have been devised for calculating the strength of columns.
These formulas are all based upon what is known as the Gordon
formula. In the formulas given in the following,

p = ultimate load in pounds per square inch;
1 = length of column or strut in inches;
r = radius of gyration in inches.

The radius of gyration r is found from the moment of inertia and
the area of section as follows:

moment of inertia

area of section

To find the safe load for a given section from the formulas given
in the following, it is necessary to multiply the value of p, as found
from the formulas, by the area of the section, thus finding the total
ultimate load for the whole section; then divide this load by the
factor of safety to find the safe load that may be placed on the column
or strut. Formulas for seven different cases will be given.

1. Assume that a steel column has both ends fixed or resting on
a flat support, preventing any sidewise motion. The formula for this
case is:

50,000

1 +

36,000 r 2

2. For a steel column with one end fixed or resting on a flat support
and with the other end round or hinged, the formula is:

50,000

1 +

24,000 r 2

3. For a steel column with both ends round or hinged, the formula is:

50,000

1 +

18,000 r 2

COLUMNS 41

4. For a round cast-iron column, solid, having both ends fixed or
resting on a flat support, and where d is the diameter of the column,
the formula is:

80,000

I 2

i _j ____

800 cl 2

5. For. a cast-iron column, circular in cross-section, but cast hollow
and having both ends fixedi or resting on flat supports, the formula is
(d = outside diameter of column) :

80,000

1 +

800 cl-

6. For a cast-iron column, square cross-section, cast with a hollow
square in the center and having both ends fixed or resting on flat sup-
ports, the formula is (s = outside dimension of square):

80,000
p =

1000 s 2

7. For a square wooden column with flat supports or with both ends
fixed, in which the side of the square is s, the formula is:

5000

1 +

250s 2

Example: What would be the load that could safely be carried by
a steel column of bar stock, 2 inches in diameter, 5 feet long, with a
factor of safety of 4, assuming that both ends of the bar are hinged?

50,000 50,000

P = = - = 27,800 pounds per square inch.

(5 X 12) 2 1 + 0.8

18,000 X (0.5) 2
Safe load on column:

27,800 X 3.14

= 21,800 pounds.
4

Pipe Columns

The allowable compressive stress for steel pipe columns may be
determined from the formula:

S = 15,200 58 L -r- R

in which S = allowable compressive stress in pounds per square inch;
L = length of column in inches; R-= radius of gyration in inches. This
formula is applicable to steel pipe columns with flat ends. No columns
should be. used having an unsupported length greater than 120 times its
radius of gyration. The formula is based upon the requirements of the
New York Building Code.

42 No. 136 STRENGTH OF MATERIALS

A similar formula, based upon the Chicago Building Ordinances, is:

S = 16,000 70 L -f- R

in which the letters denote the same quantities as in the previous
formula.

Flat Stayed Surfaces

In many cases, large flat areas are held against pressure by stays
distributed at regular intervals over the surface. In boiler work, these
stays are usually screwed into the plate and the projecting end riveted
over to insure steam tightness. Ttie U. S. Board of Supervising In-
spectors and the American Boiler Makers' Association rules give the
following formula for flat stayed surfaces:

C X t*

S 2
in which P = pressure in pounds per square inch;

C = a constant which equals 112, for plate 7/16 inch and
under; 120, for plates over 7/16 inch thick; 140, for
plates with stays having a nut and bolt on the inside
and outside; and 160, for plates with stays having
washers of at least one-half the thickness of the plate,
and with a diameter at least one-half of the greatest
pitch.

t = thickness of plate in sixteenths of an inch (thickness =
7/16, * = 7);

8 = greatest pitch of stays in inches.

Strength of Flat Plates

The machine designer is often called upon to carry out designs
consisting in part of flat surfaces, such as plates supported or fixed
at the edges, with or without intermediate supports or ribs. Exact
formulas for finding the bending moments of flat plates supported
along their edges and subjected to stresses created by pressures
normal to their surfaces have not been determined. The formulas
given by different authorities are founded on assumptions and should
be considered as approximations only; they should be used with
caution, as the results obtained are not likely to be very accurate.

A square cast-iron plate rigidly held at the edges and loaded with
a uniformly distributed load, or a load concentrated at the center,
would be likely to fail as shown in Fig. 29. It would first fracture
along the diagonal lines from A to B and then fail at or near the
fixed edges along lines BB. The plate might also shear off along the
edges BB, depending upon the method of loading and the thickness
of the plate. If the plate were merely supported along all the four
edges, but not rigidly held, it would be likely to fail by breaking
along the diagonal lines AB only.

In Fig. 30 is illustrated the probable manner of failure of a flat
rectangular plate of cast iron, loaded with a uniformly distributed
load. The plate, if secured along all the four edges, would probably
fail by fracturing along the center line AA of the long axis of the

FLAT PLATES

43

plate and along the diagonal lines AB, and then fail at or near the
edges of the support along the lines BB. If the plate were merely
supported along all four edges, it would fail simply by fracturing
along the center line AA and the diagonal lines AB. A plate firmly
secured at the edges offers greater resistance to the stress created
by the load than does a plate merely supported at the edges.

Approximate formulas for round, square and rectangular plates will
be found in standard handbooks. An unusually complete set of
formulas will be found in MACHINERY'S Data Book No. 17, "Mechanics
and Strength of Materials," and also in MACHINERY'S HANDBOOK, page
363. While the formulas given are approximate only, it is important
that formulas be deduced and used for designs of this character,

i

he -

-. 29

Fig. 30

because they indicate, in a general way, the dimensions required,
and the factor of safety assumed will always be taken large enough
so that, practically, the approximate nature of the formulas does not
detract from their value.

Spherical Shells Subjected to Internal Pressure

In the following formulas for spherical shells subjected to internal
pressure let,

D = internal diameter of shell in inches;
P = internal pressure in pounds per square inch;
8 = safe tensile stress per square inch;
t = the thickness of metal in the shell in inches. Then :

irD 2

P - =
4

PD

, and t = -
4/8

This formula also applies to hemi-spherical shells, such as the heml*
spherical head of a cylindrical container subjected to internal
pressure, etc.

Example. Find the thickness of metal required in the hemi-
spherical end of a cylindrical vessel, 2 feet in diameter, subjected to
an internal pressure of 500 pounds per square inch. The material
Is mild steel and a tensile stress of 10,000 pounds per square inch is
allowable.

44 No. 136 STRENGTH OF MATERIALS

500 X 2 X 12

t = - = 0.3 inch.

4 X 10,000

If the radius of curvature of the dome head of a boiler or container
subjected to internal pressure is made equal to the diameter of the
boiler, the thickness, of the cylindrical shell and of the spherical head
should be made the same. For example, if a boiler is 3 feet in
diameter, the radius of curvature of its head sho'uld be made 3 feet,
if material of the same thickness is to be used and the stresses are
to be equal in both the head and cylindrical portion.

Collapsing- Pressures of Cylinders and Tubes Subjected to
External Pressures

The following formulas may be used for finding the collapsing pres-
sures of modern lap-welded Bessemer steel tubes:

t

p = 86,670 1386 (1)

D

P = 50,210,000 ( ) (2)

in which P = collapsing pressure in pounds per square inch; D =

1 2 3 4 5

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