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N„caeſº
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¿ſyg º ARY
THE DESIGN QE SPECIAL, BLOOM S.HEARS
A THESIS
FOR THE DEGREE OF
MECHANICAL ENGINEER,
UNIVERSITY of MICHIGAN.
1916.
Guy D. Newton.

l.
The invention of a mill for rolling
steel car-wheels and gear-blanks made it necessary
to build a suitable shear for cutting the stock for
the mill. The pieces must be cut so as not to leave
a fin, and with as little distortion as possible.
The ordinary type of bloom shear, with one stationary,
and one moving blade which crushes its way thru the
metal will cut, the blooms very rapidly and cheaply,
but the pieces cut are so badly distorted that they
cannot be handled in a wheel-mill. To overcome this
difficulty several styles of shears were designed,
but only one built, so far as the writer knows.
In the following pages three of these designs Will
be discussed: a rotary shear with motor drive, a
spiral shear with a hydraulic drive, and a hydraulic
plunger shear. We will first take up the rotary shear,
as one working upon this principal was built, and is
now in use.
- We shall confine ourselves to the
principals involved, such as the power required,
choice of motor and fly-wheel, etc., and not attempt
to work out all of the details.

R O T A R Y S H E A R.
The operation of this machine Will
be readily understood from a study of the diagram,
Plate, 3., and of the assembled drawing of the shear,
Plate, 4., The bloom is revolved between the two
rotating knives, each of which makes a spiral shaped
cut as shown. The right hand knife cutting the cross-
hatched portion of the bloom and the left hand knife
cutting the blank portion. For this shear the blooms
must of course be cylindrical, and must be carried
on a table which will allow them to rotate freely.
It is not necessary however, to provide means for
rotating the blooms as the drag of the knives has
been found sufficient for this. The table for carry-
ing the bloom consists simply of a 12" steel pipe
supported by rollers to allow free rotation. The
bloom lies on the pipe ana is prevented from rolling
off by a horizontal bar of steel on either side.
This arrangement is shown dotted in connection with
the spiral shear, Plate, 3. A hydraulic cylinder
acting against the end of the bloom forces it into
the shear and the pieces as they are cut off, are
allowed to fall into a pit, beneath the machine from
which they are removed by any convenient, means.

The first question to be considered
in attacking a problem of this kind should be that
of the power required. The amount of published data
on the subject, is not, extensive. we were unable to
find any data of actual shearing tests of steel at
high temperatures. - -
The Iron Age, of Feb. 15th., 1916.
publishes an abstract of a paper by A. Schwarze,
describing an electric bloom shear in which it, is
stated that "the knife forces itself into the hot
metal at a considerably lower pressure than e.coo.
pounds per square inch."
On page 299, in Machinery's Handbook,
in a numerical example, 6,000 pounds per Sq. in-, is
assumed as the shearing strength of hot steel slabs.
# Mr. Wm. C. Coryell writes that steel
direct from the furnace has a shearing strength of
about 3, COO pounds per sq. in. If it has been out,
of the furnace about one minute 8,000 pounds, and
if it has been out of the furnace about, four min-
utes 16,000 pounds per Sq. inch.
none of the above figures were based
on actual tests so far as the writer knows.
* Mr. Coryell was formerly steam
engineer at the Youngstown branch of the Carnegie
Steel Co.
In order to obtain some definite
information on the subject a small experimental
rotary shear was built, and a series of tests made.
This shear is shown by the blue print, Plate, 1-,
and by the two photographs on Page, 5. The fly-wheel
is six feet in diameter, and its moment of inertia is
217.875. The blades are geared down as shown to one
twentieth of the speed of the fly-wheel, and were
designed to cut bars of any size up to 3" in diameter.
The cutting is done in exactly the same manner as
by the large shear, each blade cutting a spiral strip
3/8" wide as the bar revolves due to the pull of the
blades which move in a tangential direction.
In order that the blades should not
be burned off they were made rather blunt, as shown
by the sketch, Fig. 3.
As it was necessary to place the
blades exactly opposite each other in order to get

A-247, 2

the desired shape of cut, it will be seen that the
action is not strictly a shear. The blades really
crºush their way thru the metal. •
The first, blades were steel. castings,
but, as these soon gave out the edges were turned Gºff
and "bits", forged from tool steel were bolted on as
shown by the photograph, Fig. 5. , page, 7-
In order to center the piece to be
Gut, the plate shown by Fig. 6. , Page, 7. , was bored
to fit over the ends of the blade shafts. The hole
in the center of the plate is for inserting the stock.
The shear was driven by a belt from
a pulley on the line shaft, to a clutch pulley on
the fly-wheel shaft of the shear, as shown by blue
print, Plate, l. The speed of the fly-wheel was rec-
orded by means of a chronograph. A projecting key on
the fly-wheel shaft, raised the end of the lever a,
Figs. l and 2. Page, 5., making a contact at b, thus
operating the pen of the chronograph at each revolution
of the fly-wheel. By throwing off the power and rec-
ording the number of revolutions made by the wheel
in slowing down, we were able to compute the friction
of the machine when running light, and also the power
used in shearing.




It was thought that we would be able
to scale the time of each revolution during the cut-
ting, and as we could determine the position of the
blades at any time, compute the energy expended in
making each portion of the cut. The drum of the chro-
nograph revolved very slowly however, thus bringing
the revolution 5 So close together on the chronograph
sheet that we were unable to get any consistant, fig-
ures in that Wray. Another thing that contributed to oz//*
failure was the fact, that the stock did not always
rotate in the same Way, so we were not able to tell
just what area was sheared during each revolution of
the fly-wheel. Fig. 7. , Page 7. , is a photograph of
cut #40 which shows that the stock made only 1/2 of
a revolution and then stopped and the blades crushed
their way thru. In this case the piece was not cut.
completely off. A small wedge shaped piece about 1/20
of a square inch at e, and d, being uncut.
The steel was heated in a gas furnace
which was located some distance from the shear, so
that at least so seconds were reqûired to draw the
stock from the furnace and place it in the shear.
The question of determining the tem-
perature of the steel at the time of shearing gave
some trouble at first. Our first idea was to wire
a thermo-couple to a piece of steel of the same
size as the one to be cut, and withdraw them from
the furnace at the same time and read the temper-
ature of the one at the time the other was being
cut. It was found however, that the couple cooled
Off much more rapidly than the steel.
- We next, tried inserting the couple
in a small iron pipe, and this pipe in a larger ©flé,
etc . , thus making a mass of iron as nearly as poss-
ible equal to that of the piece to be sheared. This
arrangement gave better results but still was not
satisfactory.
So We finally thought of the plan
of artiling a hole in the steel and inserting the
thermo-couple, as shown in the photograph, Fig. 4.
Page, 7. It is believed that this arrangement, gave
fairly accurate results. Our only trouble with this
was due to the fact, that the thermo-couple wires
break very easily at a temperature around 2000 deg-
rees Fah. Cold they will stand a great deal of
bending.
In making a test, we proceed as follows, -
The two pieces of steel of the same diameter are
heated at once, the thermo-couple being inserted in
**
10.
one of them. When the steel is hot the shear is
started and we are ready to make the cut. The piece
of paper which is usually kept between the two
contact points at b, Fig. 1 . , Page 5. , is removed
for exactly one minute, thus recording the speed
of the fly-wheel on the chronograph sheet. Then
the steel to be sheared is brought, as quickly as
possible and just before it is inserted between
the blades of the shear the paper is again removed
from between the contact points, and the clutch
thrown out, the bar inserted, and the fly-wheel
allowed to come to rest. We now have a record of the
speed of the fly-wheel and the number of revolutions
it makes in coming to rest after making the cut-
Plate 2., is a sample chronograph sheet with this
data. This will allow us to compute the power tised
in shearing the steel if we know the amount of energy
used up in friction in the machine. Frequent friction
tests were made as the friction gradually decreased
as the bearings became worn. Several of the earlier
shearing tests were thrown out due to the uncertainty
thfººthe friction load. It was also found that, the
friction could be considerably reduced by frequent,
oiling.
In order to know the temperature of
ll.
the steel at the time of shearing, the pyrometer
is read at the instant the steel is removed from
the furnace. And the time at, which the steel leaves
the furnace and the time at which it enters the
shear are noted. As soon as the shearing has been
done, the piece of steel with the pyrometer attached
is removed from the furnace and the temperature is
read every 5 or 10 seconds as it, Gools off. After
several. trials we are able to plot, a curve of ten-
peratures, on time as abscissa, by which we can tell
very nearly the temperature of the piece of steel
which is being sheared. It is only necessary to know
the temperature at which it leaves the furnace and
the time that has elapsed until it, enters the shear.
Curves 1,2,3,4, and 5 are the temperature curves for
the various sizes of steel used.
The triction of the shear was deter-
mined each day in the following manner:-The shear
and chronograph were both started, and the piece of
paper removed from between the contact points for
exactly one minute in order to record the speed of
the fly-wheel. After waiting a few seconds to cause
a break in the chronograph record, the paper was
again removed and at the same time the clutch was
thrown out and the fly-wheel allowed to come to rest.
From the number of notches in the chronograph record
we were able to compute the friction of the shear,
thus.- The kinetic energy of the fly-wheel = 1/2 I w:
I = the moment of inertia, – 217.875 (Computed) -
w = angular velocity = 37N radians per second-
&
*
N = number of revolutions per minute.
Example :- N = ll.5 R. P. M.
W = 3.1416 x 115 x 2 = 1.2.04
2 a 60 : º ºr
WT = 1.2. Q4 = 145.03

Then the kinetic energy = £1.7-2.É a lag.cº.
: 15606 foot, pounds.
Then the kinetic energy of the fly-wheel, divided by
the number of revolutions the wheel makes in coming
to rest, when running light, will be the friction of
the shear in foot, pounds per revolution. The average
of several tests was usually taken as the true fric-
tion load for each days run.
The power required to shear the bar
was determined as follows:-
On the chronograph record count the
revolutions made in one minute, and also the number
of revolutions made while coming to rest after the
cut has been made. Then the shearing power is equal
to the kinetic energy of the fly-wheel at the initial
13.
*
speed, minus ( the friction per revolution multi-
plied by the number of revolutions made while coming
to rest).
Example :- Run. 27. (see tabulated results, Page, 14)
At 118 R. P.M. the kinetic energy = 16640. ft. lbs.
The friction at that time was 39, ft. lbs. per rev.
The fly-wheel made 264 revolutions in coming to
rest, after making the cut.
Then the shearing power = 16640-(39x264)=6340 ft. lbs.
:
The area of the bar cut - 3 - 1.4 sq. inches.
Therefore the shearing power per sq. in. = 3342 =301.9ft. lbs.
3. i4
The force required to shear one sq.in. (shearing
strength) would then be 2010 -- 1/12 = 24.00 pounds.
The tabulated results of the tests will
be found on Page 1.4. -
On curve sheet #7 the shearing powers
per sq. in. are plotted on temperatures as abscissa.
Curve l , shows the results of tests
#9, #1C, #ll and #12. The steel sheared in this case was
a one inch bar of mild steel having a tensil strength
when cold of about 50,000. lbs. per sq.inch.
Curve 3 - , shows the results of tests
#13, #14 and #15. In this case the pieces sheared
were from a 1 1/4” bar of high carbon steel having

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a tensil strength when cold of about 85,000 lbs.
per sq. inch-
Curve 2. , shows the results of tests
#16 to #42 inclusive. These bars were cold rolled
steel shafting, 1 7/16", 2", and 2 1/2" in diameter.
The 2 1/2" pieces, (#40, #41 and #42)
were not completely cut off, but the error is slight.
Two cuts were made on 3" shafting but,
the results are not included in the table because the
cuts were very incomplete.
Our range of temperatures is not, as
great as we had hoped. We were unable to get any
successful cut at a temperature above 1950 degrees Fah.
A3 the melting point of steel is about,
2500 degrees Fah., we have indicated by dotted lines
what seems to be the probable path of the curves.
They should all fall to zero at the melting point.
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per sq. iri, is plotted on temperature as abscissa.
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computed on the assumption that if it reqāires 2000.
ft. lbs. of energy to shear one sq. in. of steel, that
is a force of 2000lbs. acting thru one ff. , it will
require a force of 12x 2000. = 24,000 lbs. acting

thru a distance of one inch to shear the same area.

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17.
We are now able to determine what
power is required to shear a 12" bloom if we know
the temperature of the bloom.
Going back to our temperature curves
again we have computed the amount of heat radiated
from each square inch of surface of the metal in one
half minute, at various temperatures. The method and
tabulated results are given on blue print Page, 16.
These results are not strictly correct, as we have
assumed that the radiation remains Constant, during
a half minute. It really decreases as the temperature
of the steel drops. However the error is not great.
On curve sheet #6. , we have plotted
the mean radiation per sq. inch, as given in the
table Page 16. , except that we have multiplied the
figures by two, plotting the radiation per minute
instead of per half minute. ... "
The dotted curve on curve sheet, ſº
shows the radiation per sq. in. of surface, per minute,
computed by Stefan's law. R = crº Where R = radiation,
and 1 = absolute temperatures.
These two curves do not lie a great
distance apart, but they curve in opposite directions.
Now that we know approximately what
heat is radiated per sq. in. of surface we can know

18.
about what the temperature of our lz" bloom will be
when we wish to shear it... Referring again to Plate
3. , it will be seen that the cutting takes place
during about two fifths of a revolution of the knife.
If we allow four seconds for the cutting, our knife
must make a complete revolution in 10 seconds, and
we will have six seconds in which to advance the **:
bloom for the next cut. A 20 ft. bloom will usually ---
have to be cut into about 25 pieces. At six cuts
per minute this will take about four minutes, § {} º
to be safe we will say the steel will be out of
the furnace five minutes before it is completely
out up. It will be at a temperature of at least,
2200 degrees Fah. when it, leaves the furnace. We
will now compute the drop in temperature of a 12"
bloom in five minutes, using a section one inch
long as a unit.
Weight of sec. 1" long, - 31.7 pounds.
Area of sec. l" * = 37.7 sq.inches.
Specific heat, of steel, O. 113
The total heat, per inch length, in the bloom at
2200 degrees Fah. of 2660 deg. absºlute temperatute
will be, 2660 x .118 x 31.7 =9950 B.T. U.
Radiation per sq.in. per min. from curve #6= 3.9 B.T.U.
Total radiation = 3.9 x 37.7 = 147.2 B.T. U-
Total heat, in the bloom after one minute will
19.
be, 9950 - 147.2 = 98.02.8 B.T. U.
9832.8
Q. 113 x 3 l = 7
bloom after one minute. Continuing in this way we
= 2160 = the temperature of the
find that the temperature of the bloom after five
minutes is 2010 degrees Fah. Referring to curve
sheet #8, curve #2. , we find the shearing strength
per sq. in. at 2010 deg. Fah, is 16100 pounds.
Having determined the shearing strength
of the steel we are now in position to discuss the
design of a 12" shear. The total area of a 12" bloom
is ll:3.1 sq. inches, and each knife cuts an area of
56.55 sq. inches. As may be seen by the diagram on
Plate 3. , the cutting is accomplished during a stroke
of 43 inches, and the resistance to the knife is about
the same as would be offered by a flat plate 43" wide
of the same area, if the knife starts at one edge and
cuts across the plate. The average force on each knife
Would be *# 101.22 = 21175 pounds. The maximum
force is of course greater as the blade does not
take a full "bite" until the bloom has made one half
of a revolution. And also as the cut progresses the
distance of the working portion of the blade from
the center of the knife increases, thus increasing
the resisting moment. To be safe then, we should
design our shear for a maximum of 40,000 lbs. resis-

20.
tance on each knife during the cutting strokes
because a delay of two or three minutes during the
cutting would increase the shearing strength of
the steel considerably.
Referring to the diagram, Plate 3. ,
we have assumed point c, as the center of resistance
to the blade and drawn the line ch, to represent, the
tangential force of 40,000 lbs. Line oc, will then
be the direction of resistance of the bloom, And
ca, the direction of the thrust against the knife
shafts. Completing the triangle of forces we find
shak the radial thrust against the knife shaft to
be 49500 lbs. 9ſ course this might be changed con-
siderably if it were shown that the center of res-
istance is not at the point c, but at some point
farther out.
There is also a frictional regist-
ance due to the thrust of approximately 50,000%
against the bearings carrying the knives. With the
knives mounted on 9" shafts, and driven by a 36"
gear, this force figured at the pitch circle of
the driving gear would be, if the coefficient of
friction is o.o.B, * 52.22: 25 it. = 625 lbs.,

18
for each knife. It is believed however that we

21 -
can safely assume that 40,000 lbs. tangential force
will-be our maximum load.
In addition to the force required to
do the cutting, there is also a certain amount of
energy expended in setting the bloom in motion.
Therefore, during the first quarter of the stroke
we should add a force sufficient to accelerate a
bloom 20 ft. long by 12" in diameter, from rest tºp
a velocity of 1/2 revolution per sec. in one sec.
2
This will require an energy, E = 1/2 I 09.
A = ***** 2: 22–2–432 z_2 *-B = 2
2 1.44 x 32.2 x 2
40 = 3.1416 radians per second.

£8.2 × 2:13.6 ± 2-liig
2
1 = * x 9.5
g

Then E = = 291 ft. lbs.
This force acts thru a distance of about 1 - 33 feet,
and one half of the load is carried by each knife.

The tangential force on each knife due to this 10ad
would be, *-*-*4 = 102.1bs.
sº
As this takes place at the beginning of the stroke
before the knife gets a full "bite", it can be
neglected. -
This shear is to be driven by an
electric motor and to reduce the speed from that

of the motor to 6 RPM. , the speed of the knives,
will require 3 pairs of gears. (see Plate 4-)

The first, reduction will be from the motor to the
fly-wheel shaft. For the second reduction a pinion
on the fly-wheel shaft will drive the gear (2), on
the auxiliary shaft. And for the third reduction a
pinion (3), on the auxiliary shaft will drive the
two gears (4), one of which is keyed to the axle
of each knife. The knife runs 6 RPM., and if we
make the auxiliary shaft, run 14 RPM, , a 7 to 1. red-
uction from the fly-wheel shaft will give us a speed
of 98 RPM., for the fly-wheel.
FLY-WHEEL.
In a machine where the load fluct-
uates from friction load to a maximum several times
a minute as in a shear, a suitable fly-wheel is
absolutely essential in order to equalize the load
upon the motor as much as possible. The best condi-
tion so far as the motor is concerned, would be
for the load on the motor to be constant and equal
to the average of all the loads. As this can hardly
be attained, the usual method is to let the fly-
wheel do all the work of cutting, leaving for the
motor only the friction load and the work of speeding
up the fly-wheel between cuts. The work of cutting
a 12" bloon at 2010 des.faa..which is the lowest

temperature at which we ordinarily expect to work,
is equal to the area of the bloom times the shearing
power per sq. in. , as given by curve 2. on gurve
sheet #7. Or the work = 113.1 x 1390 = 157000 ft.lbs.
This must equal the kinetic energy which the fly-
wheel looses during the cutting, or must equal the
difference between the kinetic energy of the fly-
wheel at 98 RPM. and 30 RPM. if we allow about 20%
reduction in speed.
2.
o = 1/2 I wº. 1/2 14,
Then 157, QC
W. K. -
I : TET, where K = the radius of gyrations
w = e−4+2 = 10.26, and w”- 105.

30 - . . .
4. – 22–4 =# =====
4 * 6C
2
If K = 5ft, . K = 25.
2
8, 37, and 4, : 7Q.
2
5
Arid I = W #.a = ºf x Q, 78
E = 1/2 I wº- 1/2 w x 0.78wº o.39W wº
Therefore, 157,000 =(0.39 x 105w)-(0.39×70w)= 13.66w.
157, QQQ
w = *:::::= = 11500 lbs. ww.of fly-wheel.
As we have assumed a rather large
fluctuation, inspeed we will be justified in making
our fly-wheel lift. in diameter outside, with a
rim weight of 12,000 lbs. This will give us a per-
ipheral velocity of 98 x 11 x 3.1416 = 3386 ft - per
minute, which is well within the limits of safety •

MOTOR.
In addition to the work of shearing
the bloon the motor must also carry the friction
load. As this cannot be exactly determined it is
customary to allow 25% of the motor rating for the
friction load. If the work of cutting is 157,000
ft. lbs • , and six cuts are made per minute, the
cutting will require 6 x 1570.00 =942,000 ft. lbs.
942OOO -
- --- 2: 2 - -
33000 3. 6 HP
And the motor rating should be 4/3 x 23.6 = 38. HP.
of energy per minute, or
But, as the motor is running at friction load over
half of the time we should probably use a 60 HP,
motor. And the speed characteristic should be such
that the motor will slow down as the load increases,
add will increase in speed as the load decreases.
If we choose a compound wound motor having a speed
variation from light to full load of 15%, and a speed
variation from 25% friction load to full load of
about 10%, the load condition would be somewhat as
shown by the curve, Plate 5., in which the ordinates
represent, the torque and the abscissa the time of
shearing, in seconds. The Motor is running at friction
at £riºtism speed with the torque represented by the
ordinate of A, until the peak load is reached. This
is represented by the area ABCD, and should equal

the torque required to shear the bloom. The speed
now rapidly falls off and the torque curve of the
motor follows the line AE. At E, the shearing has
been finished and the fly-wheel begins to speed up
again. We designed our fly-wheel to do all of the
work of gutting with a reduction of about 20% in
speed, but the torque represented by area AED., is
delivered by the motor, and only the portion ABCE,
by the fly-wheel. So the speed will probably not
be reduced more than 12%. The tárque represented
by the area DEF is that required to speed up the
fly-wheel again. Area DEF must of course equal area
ABC.E.
GEARS.
In designing the gears we have used
the formula, W = spfy, in which one tooth is
made to take the whole load, which is considered
as distributed over the entire length of the tooth.
The tooth is figured as a beam loaded at one end.
W = load transmitted in pounds.
p = pitch in inches.
f = face in inches.
s = the safe working stress,
y = a factor depending on the shape of tooth-
We will start with the last reduction,


gear (4), and pinion (3) as we have decided that
these gears should be designed for a load of
40,000 lbs. at the pitch line. We will assume 36, 18 tº
inches and 15.5 inches as the pitch diameters of
the gear and pinion respectively. Making the pitch
4.06 inches we will have 28 teeth on the gears and
12 teeth on the pinion. If we assume f as ll" and
apply our formula , its we will get a fiber strain
of 8,930 lbs. for the gear, and 17,200 lbs. for the
pinion. These fiber strains are rather large but
still safe for steel gears at a peripheral speed
of less than 100 ft. per minute.
As gears (4) make 6 revolutions
per minute, pinion (3) must make ** = 14 RPM.
And as pinion (3) drives both gears (4), it must
transmit, a force at the pitch line of 2 x 40,000
of 80,000 lbs.
For the gears (2) which drives the
auxiliary, we will assume a diameter of 84 inches,
a face of 8 inches, and a pitch of 3.1416 inches,
which will give us 84 teeth. The force transmitted
Will be *; x 15.5 = 14,760 lbs. And the perip-
heral speed will be 7 x 3.1416 x 14 = 308 ft-per min.
Applying the formula we get a fiber strain of 5090 lbs.
for gear (2), and for pinion (l) which drives gear
(2) a fiber strain of 6785 lbs.



27.
As these Strairis are higher than would be safe
for cast iron, at a peripheral speed of 308 ft. per
minute, we will make this pair of gears of steel
also. And as gear (2) makes 14 RPM, , pinion (1)
Willmake ++++ = 98 RPM., which is the speed
for which we designed our fly-wheel.
SHAFTII.G.
The greatest strain upon the shafts
carrying the knives is the combined bending and
tortional strees due to the force exerted upon
the gear (4) by the pinion ts). This is 40,000lbs.
at the pitch line of gear (4). If 1 = the distance
from the center of the gear to the center of the
bearing, and r = the radius of the gear, then M,
5
= Pl, - 40,000 x 14 = 560,000 inch lbs. And
7
T = Pr, 40,000 x 13 = 720,000 inch lbs.
And the combined strain is equivalent to that, which
would be produced by a single twisting moment. Te ,
where Te = Mr Vºrº. (See Unwin, Page 215)
On page 218. Unwin gives a table of shaft diameters
figured for a stress of 9,000 lbs. per sq-in, for
various values of T, and k, where k = #.
560.000
In this case k = ± = 0.77 From the table
720,000
we find that a shaft about 9.25 inches in diameter
will be required. We will use a 9 inch shaft.
The auxiliary shaft must sustain a
combined tortional and bending stress due to the
two forces of 40,000 lbs. each, at the pitch line
of pinion (3). -
1.5 inches. -
In this case + =
r = 7.75 "
P = 80,000 lbs.
º And P’ = 47,000 lbs. (See stress diagram
on Plate, 6.)
Then T = Prº = 80,000 x 7.75 = 620,000 in-lbs.
And M = P'1 = 47,000 x 15 = 705,000 " "
M 705, QQQ
*- T 620,000 1-13
From the table we find that the auxiliary shaft,
will also have to be 9" in diameter.
The fly-wheel shaft must carry the
combined stresses of the bending moment M due to
the weight of the fly-wheel, and the tortional stress
T due to the load transmitted, which is the same as
that transmitted by pinion (1) •
H = 16,000 x 16 = 256,000.
T = 14,760 x 6 = 88,560.
- 256,000
And k 68,560
This calls for a 7 inch shaft.
2.89


-------
MOTOR PINION & GEAR.
Taking the motor speed at 500 RPM.
at friction load, it will require a 5 to i reduction
to the fly-wheel shaft. A lot pinion on the motor
will call for 50" gear on the ris-me- shaft.
The peripheral speed will be * #444 #–822– 1309
feet per minute. With a pitch of 2.094 inches, we
will have is teeth on the pinion ana's teeth on the
gear. While it is not expected that these gears
will take the shock of shearing the bloom, it is

thought best to design them heavy enough for the task.

The load transmitted will be ###2+* = 3540 lbs.
25
Assuming a face of 5", and applying our formula we
get a fiber strain for the gear of 29.30 lbs, and
for the pinion of 4510 lbs. While these fiber strains
are rather high for a peripheral speed of 1309 ft-per
minute, they are still safe for steel gears.
BEARINGS,
The bearings are all made large, so
we need figure only one of them. We will take the
worst case, that of the knife shaft bearing on the
gear end, where there is a load equal to the resul-
tant of three forces: the thrust of the knife due
to the resistance of the bloom, the reaction due
to the turning force of 40,000 lbs. upon the gear,
and the reaction due to the tertional resistance
of the bloom to the knife.
The thrust of the bloom against the
knife is carried equally upon each bearing and is
equal to gºod = 30,000 lbs. The reaction due to
the tortional resistance upon the knife is also
divided equally between the bearings, and is equal
to *222 = 20,000 lbs. And by taking moments
about the center of the cutside bearing we have the
reaction due to the turning force upon the gear
equal to *** * * = 56,000 lbs.
By constructing a stress diagram as shown, Plate, 6.,
we find the pressure on the bearings to be 70,000 lbs.
The bearing is 18" long by 9" in diameter, and has
a projected area of 162 sq. inches. The pressure
pervsq.inch , then is, #º = 4:32 lbs. As the
pressure is intermittant and the speed of rotation
is slow, this is a safe pressure and babbitted
bearings should give good service-
The qaestion of the most suitable
blade to use can only be settled by experience.
It must not be so thin as to become red hot, while
making a cut as it would wear away too rapidly-

31.
The wear in any case will be great and also there
will be considerable tendency to warp, as the knives
must be water coolea continually during the operat-
ion of the shear. so the design should be such
that the worn blades may be easily removed and new
ones put in place. The construction shown on Plate,
3., is only a suggestion, and may not be found the
best in practice. It is probable that a so called
"high speedt steel, that can be heated up to around
600 degrees . . , before its structure changes, will
be found the most suitable material for the blades.


32 *
H Y D R A J L I C S P T R A L. S. H. E. A. R.
While electric power is largely used
in steel mills, there are some machines which are
perhaps better operated by hydraulic power. Among
these are heavy shears and punches.
The spiral shear shown in Plate, 7
is the result of an effort to design a hydraulic
shear to work on the same principal as the rotary
shear. As will be seen the cutting action is the
same, the blades are mounted on levers which are
tied together by a link. The ends of this link are
connected to the cylinders as shown. As the plunger
of the large cylinder moves out, the left hand knife
moves up and the right hand knife moves down, the
bloon is caught between the blades and the erreet
is the same as in the rotary shear. When the stroke
has been completed the blades are brought back to -
their original position by the small cylinder. -
The general arrangement of this shear
may not be the best that could be devised, as the
same result might have been obtained in a number
of ways. The cylinders could be connected at any angle.
The machine wauld have been shortened considerably

by placing the cylinders in a vertical position,
one at either side of the knives as could have
been easily done. or the small cylinder could have
been omitted if we had made the large cylinder
aouble acting, but this would call for insiae pack-
ing which is always a source of trouble. -
In the arrangement as We hatre it. the
Small cylinder need only be large enough to overcome
the friction of the machine. But as the pressure
will be on this cylinder all of the time, the water
being forced back into the supply main during the
cutting stroke, the large cylinder must be large
enough to do the work of shearing the bloom and in
addition to overcome the reletional resistance and
the force exerted by the small cylinder .
- The "bite" is the same, and the
length of stroke is the same as that of the rotary
shear, so the energy required to make the cut will
be the same. We might figure then, as in the rotary
shear, a force of 21175 lbs • acting thru a distance
of 3.6 feet, for each knife, as the energy required
to shear the bloom. But in this case we have no fly-
wheel to carry us over the peak load so we must
assign our cylinders for the maximum load, which we
should expect to be as high as 30,000 lbs. at times.


We need not, however design the other parts of the
machine for a load greater than the cylinders can
exert. If we construct a force polygon (see plate 6).
making the assumption that the tangential force at
the blades will be 30,000 lbs., we find that we
have a thrust of 36,000 lbs. against the blade sharts.
We shall not go into the design of the details of
$his shear except to figure the diameters of the
cylinders. In the worst case the tangential force
of 30,000 lbs. will act at a distance of 48" from
the center of rotation, while the force exerted by
the cylinders is applied at a distance or se inches
from the center. The force exerted by the cylinder
36
80,000 lbs. If we figure the customary 25% for
for the shearing alone alone must be
friction, we will have 26,670 lbs. as the force
that the small cylinder must exert during the return
stroke. And with a water pressure of 500 lbs • per sq.
inch, which is about what is commonly maintained
in steel mills, the plunger must have an area of
# = 53,3 sq. inches, or approximately an -
8 inch plunger's
The large cylinder must exert a
pressure of 80,000 lbs. to shear the bloom, plus
26,670 lbs. to overcome the friction, plus 26,670 lbs.

35.
to overcome the resistance of the small cylinder,
or a total of about 134,000 lbs. The large plunger
134,228
5QC
A diameter of 18 l/2 inches will give us an area of
then, must have an area of = 268 sq. inches.
268 - 8 sq. inches.
The stroke of the knife is 43 inches,
and if allow one inch at each end for clearance, the
stroke of the cylinder will be # * = 34 inches,
or 2.833 feet, .
As the pressure is on the small cyl-
inder all of the time it will balance an equal area
of the large cylinder during the cutting stroke,
and the work done during the cutting stroke is the
same as would be done by a plunger having an area
equal to the difference between the areas of the
two plungers. On the return stroke however, the
small plunger does work against the friction of the
machine • so the total work done in cutting one
piece is equal to the area of the large plunger in
sq - inches, times the pressure per sq. inch, times
the length of the stroke in feet. This will be
268.8 x boo x 2.833 = 380,750 foot lbs. And if 6
cuts are made per minute, the power consumed will
be *####–4 = 69.2 horse power.

36.
This is more than twice the power
consumed by the motor driven shear due to the con-
dition that the full stroke must always be made,
and the pressure is constant, throughout the stroke
and is sufficient to overcome the maximum resistance.
r
37,
H Y D R A J L I C P L U N G E R S H E A. R.
The drawing on Plate, 9 illustrates
What might naturally be the first method considered
in designing a shear for this purpose. No attempt
has been made to develop the design further than
is necessary to bring out, the difficulties encoun-
tered, and enable us to make a comparison of this
machine with the two machines previously discussed.
The usual method of figuring the
energy required to shear a bloom is to multiply
the area of the bloom in sq. inches, by the shearing
strength, and by the stroke in feet. Quoting from
Machinery's Handbook, Page 290. ,- "A hot slab shear
cuts slabs 4" by 15" at a shearing stress of 6000 lbs.
per sa. INch. This gives a pressure between the
knives of 4 x 15 x 6000 = 360,000 lbs. The total
energy required is *#2 *-* = 120,000 ft-lbs."
It, Wiil. be seen that if the slab is turned on edge,
we will have the same pressure between the knives
and a stroke of 1.5", so the energy required to shear
the slab on edge will be ******* = 450,000 ft.lbs.
And to shear a plate l" thick by 60" wide, which
will have the same area, the energy required will

38.
be ******* * 30,000 ft. lbs. This latter figure
is evidently correct, or nearly so, but for a long
stroke as in shearing a 4" by 15" slab on edge the
result, obtained by this formula Beems doubtful. . For
if the pressure between the knives is equal to the
area of the slab times the shearing strength, it
Will not be constant throughout the stroke, but
will decrease as the area to be sheared decreases,
being a maximum at the beginning of the 3troke and
zero at the end of the stroke. Also in shearing a
hot slab on edge it is doubtful if the pressure
required is ever equal to the area times the shearing
strength as the metal at the edges will give way
before the remote portions at the center of the
slab are effected at all.
While it, is very probable that our
results will be higher than they should be, we will
figure the power required to shear the 12" bloom
in this shear, by the usual method. The area of
the bloom is 113.1 sq. inches. The shearing strength
we have found to be 16, 100 lbs per sq.inch. The
pressure between the knives will then be 16, 100 x
liº. 1 = 1, 220,910 lbs. And the energy expended
in shearing the bloom, if the stroke of each piston
is 7" to allow one inch clearance between each

39.
blade and the bloom when the blades are withdrawn,
will be ******** = 2,124,500 Ft-Lbs.
i.º -
And the horsepower expended if & cuts are made per º
2, 124,500 x 6 — a
- Tº * = 336 HP. Arld in thi
33000 336 HP. Arld in this
minute, Will be -
case we have not added the usual 25% for friction.
40.
CONCLUSIONS.
Each stroke of a hydraulic cylinder
consumes a fixed amount of energy. This energy is
the product of the stroke in feet, by the area of
the piston in sq. inches, and the water pressure in
pounds per sq- inch. So in the case of our shear
this is constant regardless of the size or temper-
ature of the bloom to be cut.
The hydraulic plunger shear as we
have designed it will consume 2,124,500 ft. lbs.
of energy in making one complete stroke, and this
whether it shears a bloom or whether it meets no
*ēs; stance whatever. This figure is doubtless much
too high as a much smaller cylinder would probably
do the work. This is a subject, that should be inves-
tigated further.
We found, opage 35. , that the spiral
shear would consume 360,750 ft.lbs. of energy at
each stroke, and this when once established is
constant for each stroke of the machine.
With the motor driven shēafº we have
a very different condition. The fly-wheel is to
furnish the energy required to make the Gutt, and
it will give up just the energy required to overcome

the resistance met. On page 23. , we find that,
157,000 ft - lbs. of energy are required to make a
cut, not including the energy consumed by friction,
which is taken care of by the motor. This is the
Worst, case, that is, it is for a 12 inch bloom
and the last cut made when the shearing strength
has been found to be l6, lºo lbs. per sq.inch. The
first, cut, when the bloom has been out of the fur-
nace not, more than one minute would require very
much less energy. Also if a smaller bloom is out
the energy consumed will be less •
It will be seen that, the advantage
lies largely with the motor driven shear. However,
on account of its convenience, the hydraulic
spiral shear might be chosen in some cases, espec-
ially if the blooms do not fodilow each other closely ,
and there is time lost, , or if it is found necessary
to wait for the blades to cool off. In such cases
the hydraulic shear might, be found more economical
as the motor runs continuously and must carry the
friction load while the hydraulic shear uses power
only while actually doing useful work-

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