REESE LIBRARY
or Tin:
UNIVERSITY OF CALIFORNIA.
'cessions No. o9 v 6 / . Class A
"^'Lv.
THE
DESIGNING OF DRAW-SPANS,
BY
CHARLES H. WRIGHT,
M. AM. Soc. C. E.,
Chief of Detail and Drafting Department* Edge Moor Bridge Works.
Author, with Prof. Wing, of ^ A Manual of Bridge Drafting."
FIRST EDITION.
FIRST THOUSAND.
NEW YORK :
JOHN WILEY & SONS
LONDON: CHAPMAN & HALL, LIMITED.
1897.
yright, 1897,
BY
CHARLES H. WRIGHT.
ROBERT DRUMMOND, ELHCTROTYPKR AND PRINTER, NEW YORK.
CONTENTS.
PAGES
MOMENTS AND REACTIONS .. 2-19
WEB-STRESSES 19-23
DEFLECTION 23-28
MACHINERY 28- 5
TABLES AND GENERAL DATA -. . 46-69
CUTS OF DRAW-SPANS SHOWING MACHINERY, ETC 70-84
DESIGNING OF DRAW-SPANS.
PART FIRST.
PLATE-GIRDER DRAW-SPANS.
THE following pages aim to give a clear and simple expla-
nation of the methods used in the determination of the
stresses, sections required, and of the deflections produced by
the various conditions of loading assumed. The machinery
necessary for operating the draw is also considered, and the
designing of wedging machinery for raising the ends, latching
devices for preserving perfect alignment when the draw is
closed, methods of raising the rails for clearance when the
draw is opened, and the designing of gears, shafting, andi
bearings are considered in detail. Each point as taken up is
illustrated by examples, as fully as necessary to make the
applications clear. The aim has been to use the simplest
methods, rules, and tables that will give the desired results.
Where formulae derived from the higher mathematics have
been used, full and complete explanations of how they are
used and applied are given.* It is believed the work may be
* The reader is referred to the works of Professor Releaux and Unwin,
from which notes have been taken. The author is also indebted for valu-
able information to Professor Malverd A. Howe of Rose Polytechnic Insti-
tute; the Edge Moor Bridge Works, and to the Pencoyd Bridge Works.
2 DESIGNING OF DRAW-SPANS,,
readily followed and understood by those not having a full
knowledge of the higher mathematics, and that it will prove
of value to any one wishing a practical knowledge ef draw-
spans and their machinery.
PLATE- GIRDER DRAWS.
For spans up to about one hundred and fifty feet the deck
plate girder makes the most satisfactory bridge, and is the
type in most general use. The conditions under whioh the
draw-span works are much more severe than with fixed spans,
and the bridge should be correspondingly heavy and rigid.
Through plate-girder or lattice spans are unsatisfactory for
draw-spans, owing to the small depth usually available below
the floor for the introduction of diagonal bracing necessary
to resist the twisting force produced in turning the draw, and
especially in suddenly stopping or starting. This force is well
illustrated by taking a piece of artist's rubber in the fingers
and twisting. The rubber may be turned through a consid-
erable angle and still a cross-section at any point will be a
perfect rectangle as at first. This shows that any bracing
introduced to resist this twisting action must run diagonally
as in Fig. I and I A . Brace-frames at right angles to the
girders do little good to resist such a force, and the same is
true of bracing in the planes of the chords.
An eighty-five-foot deck plate girder (Fig. i) will be used
as an example to illustrate the methods pursued with girder
draws in general. There are four conditions to be considered,
1st. The span swinging or in position to open, the end wedges
being drawn and all the dead loads being carried by the
centre, no live load acting. 2d. The draw closed and each
arm considered as an independent span for live load; the dead
load not being considered for the present. 3d. The bridge
PLATE-GIRDER DR
considered as two continuous spans for the live load; and 4th,
considered as two continuous spans for the dead load onlv
Cases I and 2 might occur at the same time; also I and 3
or 3 and 4. The one of these combinations giving the
4 DESIGNING OF DRAW-SPANS.
greatest strains is to be used in determining the sections,
required.
If the end wedges are just driven to a bearing but not hard
enough to raise the ends, the dead load would still be carried
by the centre, and the span is still swinging so far as the dead
load is concerned. If both arms were now loaded equally, the
bridge is then a continuous girder of two spans so far as the
live load is concerned. This is not true, however, if a live
load comes on one arm only, unless the other arm be held
down so that it does not raise up off the end support as the
dectd load moves over the first arm. Instead of holding the
unloaded arm down, it may be raised so high by the end
wedges that the deflection produced in the loaded arm will
not be sufficient to raise the unloaded end off the support.
Unless one or the other of these plans is followed there will
be what is called ' hammer ' in the draw. That is, as the
load comes on one end and moves over the bridge, first one
end and then the other will rise off the supports and drop
back again to a bearing. This movement is very noticeable
in some draws, and especially so where the rails are cut just
at the clearance line and a small space left between the ends.
To make sure the rails will clear as the draw turns, this space
may needs be three-eighths or one-half inch. This method,
or lac-k of method, of providing for the continuity of the
rails is now almost entirely superseded by devices which do
not require this clearance. Some of the methods used wi.l
be described later. The amount it is necessary to raise the
ends by means of the wedges or some similar device will be
explained under the deflection of draws.
To determine the strains produced by the dead load swing-
ing, we will assume the weight of the floor (including ties,
guards, rails, bolts, etc.) to be 400 Ibs. per linear foot, and
the weight of the span itself to be 650 Ibs. per linear foot.
400 -f- 650 = 1050 Ibs. = 525 Ibs. for each girder. Only one
arm need be considered if the two arms are equal. If the
PLA TE-GIRDER DRA WS.
two arms are not equal, the shorter one is counterweighted
until they balance, but the strains would have to be consid-
ered separately. The moments may be determined by
assuming the dead load as concentrated at several points;
thus for the moment over the pier we may assume the load
on one arm as concentrated at its centre of gravity, which is
at the centre of the arm (see Fig. 2).
* 4275 H
Arms for Dead-load Moments.
Taking moments at A, we have the dead load of one arm,
525 X 42.5 = 22,310. Assuming this as concentrated at a, the
moment is 22,310 X 21.25 = 474 IO ft.-lbs. This moment
is balanced by forces represented by the arrows and acting in
the flanges of the girder. One force is tension and the other
compression. The depth of the girder at the centre is 7 feet
and the moment 474,100 -r- 7 = 67,750, which is the tension
in the upper flange and the compression in the lower.
The depth assumed (7') should be the depth between the
centres of gravity of the flanges. For the moment at B we
have the load 525 X 31.86 (the distance from the end to E] =
16,730. This multiplied by the distance of the centre of this
load from B, 15.93 feet, = 266,500. Dividing by the depth
at this point, 6.25 feet, we have 266,500-7- 6.25 = 42,800.
At the moment == 525 X 21.25 X 10.62 = 118,470. At D
the moment = 525 X 10.62 X 5-3 1 = 29,600. It is not
6 DESIGNING OF DRA W-SPANS.
necessary to find the chord stresses at each point now. The
moments may be combined with others for live load, and the
areas required for both found at one operation. The moment
at any point for dead load may also be found by means of a
parabola drawn as follows (Fig. 3). Lay off the horizontal
42 ;5 J
NOTE: DOTTED LINE'IS FOR
UNIFORM LOAD SWINGING
METHOD OF DRAWING PARABOLA
line BB' equal to twice the length of one arm of the draw.
From the centre of this line draw a vertical line equal to>
twice the moment at the pier* (in this case 474,000). Any
convenient scale may be used, and the same scale need not
necessarily be used for the horizontal and the vertical lines.
Draw the inclined lines B 12 and B' 12, and divide each of
them into any number of equal parts (12 in the figure). Con-
nect the points i-i, 2-2, 3-3, etc., and the lines so drawn
will be tangents to the required curve, which is now readily
drawn. Only one half the curve is used, as shown by the
figure. The curve being drawn, the moment at any point is
* To find the centre of gravity of any number of loads from any point
(as one of the end loads), multiply each load by its distance from the point,
add the results, and divide by the sum of all the loads. The result will be
the distance of the centre of gravity from the point assumed. Note that
if there is a load at the point from which we start, this load must be in-
cluded in getting the sum of all the loads.
PLATE-GIRDER DRAWS.
found simply by scaling the ordinate between the line B' B
and the dotted curve. Having thus shown two methods for
determining the dead-load moments with the draw swinging,
Fig. 4
Curve of Moments, Dead Load Swinging. 525 Ibs. per lin. ft.
we will now consider the case of the draw closed and each
arm acting as a single span for live load.*
For the live-load moments, each arm acting as a single
span, we should so arrange the loads as to get as many loads
as possible on the span, and the heavier ones as near the
centre as may be. Placing the loads as in Fig. 5, we find the
centre of gravity to be 18.7 feet from wheel No. I, and the
wheels are shifted if need be until the ceritre of the span is
half-way between the centre of gravity and load No. 4. We
now lay off the load line AB, Fig. 5 A , assume a distance
HO = 100,000 on a horizontal line drawn from any point in
AB, and draw the lines AO, BO, etc., connecting the points
found by laying off the loads on AB with the point O. This
figure (5 A ) is called the force polygon. Next, starting from
A' (any point in a vertical line through A) draw the line A' a'
parallel to AO in the force polygon, and from a' draw the line
a'b' parallel to %-O, from b' the line b'c' parallel to 4-O, and
so on until the last \m& f f B f is drawn parallel to BO.
* In drawing the parabola it will be noticed that the moment over the
pier must first be figured. This moment for the load uniformly distributed
is \ivL, L being the length of the arm, and w the dead load of one arm.
(See first method of finding the dead-load moments.)
8
DESIGNING OF DRA W-SPANS.
The line A'a'b'c'-f'B' meeting the vertical lines through
A and B at A' and B' is called the equilibrium polygon. If
the line OR be drawn in the force polygon parallel to A'B'
of the equilibrium polygon, it will divide the load line AB into
Diagram for One Arm as Single Span.
Moment at any point as CC = CC X HO CC X 100.
the two parts AR and RB which represent the reactions at
A and B. Having the equilibrium polygon drawn, the
moment at any point is found by multiplying the ordinate
between the closing line A' B' and the line A'a'b'c' , etc., by
the distance OH m the force polygon. HO being 100,000,
the moment at b, for example, will be bb' multiplied by
100,000. The distance HO is made 100,000 for conveni-
ence. It should be made of such length as will give a good
PLATE-GIRDER DRAWS. 9
depth to the equilibrium curve, so that the ordinates may
be accurately scaled. The distance HO must be measured
to the same scale as the load line AB was laid off, and the
ordinates in the equilibrium polygon must be measured to
the scale used in laying off the half-length of the span (see
Fig. 5). It is not necessary that the two figures be drawn
to the same scale. The moments at as many points as nec-
essary can now be determined. These moments are given
in column 4 of the table of strains. In Fig. 5 the curved
line A 'D and the line A ' B' give the dead-load moments
with the span swinging, A 1 D being a parabola and the
ordinate B' D being the moment at the centre support di-
vided by the distance HO (= 100,000). The signs of the
moments are determined as follows: The loads acting to the
left of the centre support tend to revolve the span downward
in a direction opposite to the movement of the hands of a
clock. These moments are called minus ( ). Considering
the same arm as a single span, the reaction at the left support
tends to revolve the span upward or in the direction of clock
motion. These moments are called plus (+). It is immaterial
which are called plus, provided all moments tending to pro-
duce rotation in the same direction are given the same sign.
The total moment then at any point, as e 1 ', under the two
conditions, dead load swinging and live load discontinuous,
on one arm, would be the ordinate ee' ee* = e*e' multiplied
by the pole distance HO. It might be found that slightly
greater moments would be obtained by placing the loads so
that the centre of the span would be between the centre of
gravity and load number 3, instead of between the centre of
gravity and load number 4 (see Fig. 6). Both positions
should be tried. Having shown how to determine the
moments for the span swinging, and for the condition of one
arm acting as a single span supported at the ends, with live
load only acting, we will now consider the span as a contin-
uous girder under the action of both dead and live load. It
UNIVERSITY
CALIFORH\*
10 DESIGNING OF DRA W- SPANS.
will be noted that in the case of dead load swinging only one
arm was considered. This is sometimes confusing and the
question is asked, * Why can one arm be neglected ? They
must surely both produce strains over the centre.' It is the
old problem of two men pulling at the ends of a rope; each
man pulls one hundred pounds, but the strain on the rope is
not two hundred pounds. One man cannot pull one hundred
pounds unless there is a resistance of this amount opposing
his pull. It makes no difference whether the resistance is
given by a man or by a post at the other end of the line. In
the same way an arm of the draw when open is balanced by
the other arm. And the moment at the centre is the
moment produced by one arm. When the span rests on
three or more supports or the loads are not balanced we can
no longer consider one arm only.
If a load is placed at any point on the span, a greater pro-
portion of this load will be carried to the centre support than
would be the case if the arm on which the load is placed
were considered as a single span resting on two supports.
Just how much more of the load is carried to the centre is
given by the diagram Fig. 9. The figures at the bottom
under the line ' values of k ' are the distances from the
left-hand support to the loaded point, in terms of the length,
and the figures in the line marked ' values of D* give the
per cent of the load going to the left-hand support. Suppose
there is a load at three tenths of the length of the arm from
the left support. From the figure 0.3 in line k l we move up
until this line inteisects the curve marked ' 5, loads in first
arm ' ; from the point where the line through 0.3 intersects this
curve we go over to the left until we reach the line D v , which
is at 0.63. 63 per cent of the load then goes to the left sup-
port. If we wish for the bending moment at this point, we
move up the line through 0.3 in k^ until we meet the curve
marked ' M y loads in first or second arm.' We intersect
PLATE-GIRDER DRAWS.
II
this curve on the horizontal line 0.685,* an d so for any other
point in the span. We will now place the engine-loads on
the span in two or three positions and see which position will
-. -12.8 --*]- S.O^-O- - -7.fr - - 8.1- - '
18. 18. 20. 18 10J2510|25 k]j.2511>25 8". 18. 18. 2b. 18.
O (flafe 6) (5)
FIRST ARM.
k.
c.
CPL.
7.1 -5-42.5 =
12. =
I6. 3
2O.7
33-5 =
38.5
Moment CPL
.167
.282
.384
.487
.788
.906
.0405
.0645
.0819
.0927
.0745
.0400
30,980
49.340
69,610
70,900
32,450
17,420
270,700
279.385
550,085
SECOND ARM.
SECOND ARM
k.
c.
CPL.
6.7
-r-42.5 = .0156
.0380
29,608
n. i
= .261
.0605
51,450
15-4
= .362
.0785
60,030
20.3
= -477
.0920
70,380
28.4
= .668
.0925
31,350
35-9
= -845
.0600
28,680
40-9
= .962
.0165
7.887
279.385
Diagram for Two Spans Continuous. Scales, 20 and 50.
0.685 is the value of C in formula M CPL = moment at any point*
12
DESIGNING OF DRA W-SPANS.
give us the greatest moment over the pier. Arranging the
loads as in Fig. 6, we first find the values of k\ thus for loads
i i
18.
10.26 ID 10.25 11.25 llj.25
(troT (?)
6.6 I 8.1 | 4.9 | 4.3 | 4.4 12.8 j 3.6 i 5.5 5.0 7.5 8.1 4.9 4.3 1 3.6
20.
FIRST ARM.
>&.
c.
CPL.
6.6
14.7
19.6
23-9
28.3
41.1
Moment
-5- 42.5 = .155
= -346
= .461
= .562
= .666
= .967
= CPL
.0380
.0756
.0910
.0963
.0927
.0150
12,929
57,815
69,605
81,850
70,905
6,525
299,629
233,331
532,960
SECOND ARM.
k.
c.
CPL.
3.6-^-42
5 = -084
.0215
18,274
7-9
= .186
.0450
34,422
12.8
= .301
.0685
52,395
20.9
= -492
0933
31,720
28.4
= .668
.0927
44,760
33-4
= .786
.0750
35,862
38.9
= -9J5
.0365
15,898
233.331
Diagram for Two Spans Continuous. Scales, 20 and 50,
PLATE-GIRDER DRAWS. 1
I, 2, 3, 4, 5, and 6 we divide the distances from the left by
the half-span 42.5', and for loads 7, 8, 9, 10, 11, 12, and 1 3 we
divide the distances of the^ loads from the right-hand abut-
Fig. 8 A
Diagram for Uniform Load Continuous. 1000 Ibs. at each eighth point
assumed load in diagram.
ment by the half-span 42.5. The values are given in the
table: .167, .282, etc., for first arm and .0156, .261, etc., for
the second arm. From diagram Fig. 9 we now find the
values of C corresponding. The vertical through k 167
meets the curve of moments on the horizontal line .0405, and
for k = .788 on the line .0745. The values of k for the:
second arm are given from the right abutment, so we find C
exactly as in the first arm. If the distances had been given
from the centre pier, we could have found C in the same
manner, only using the line marked k* in the diagram instead
of line k' \ for example, if a load is .8 the length of the half-
span from the right abutment, it is .2 the half-arm from the
centre pier. o. ik' is over k* = 0.9. It is perhaps a little
simpler to use the line k' all the time, and give the distances
of the loads from the abutments in each case. All values of
C have the same sign. Multiplying each value of C by the
load at that point, and by the length of the half-span, gives
us the moment over the centre pier for that load. CPL =
moment over pier for load P at any point. The values of
14 DESIGNING OF DRAW-SPANS.
these moments for each of the wheel-loads with the engine
placed in the two positions given in Figs. 6 and 7 are given
Fig. 9
in the tables under the figures. Two or three trials will
show how the engine should be placed to give the greatest
PLATE-GIRDER DRAWS. 1 5
moments. By referring to the diagram Fig. 9 it will be
seen that C is greatest for loads near the centre of each arm,
and a little nearer the centre pier than the abutments. The
heavier wheels should then be placed as near these positions
as possible to give the maximum moments. Adding to-
gether the moments produced by all the loads, we have the
total moment. In the two cases given these total moments
are 550,085 and 522,960. It is possible that the uniform
train load might give a greater moment at the pier than the
engines, and this moment should be found.
Before considering the uniform load we will take one more
example of moment from concentrated load to make the
method just described perfectly plain.
Suppose we take wheel No. 1 1 in Fig. 7. The distance of
this wheel from the right abutment is 12. 8. k = 12. 8-^-42. 5
.301. C for k = .301 is .0685, and CPL, the moment, =
52,395. P = iSand L = 425.
Considering now the case of uniform load, span continu-
ous, the Reading loading diagram gives 4000 Ibs. per linear
foot, or 2000 Ibs. on one girder. 2000 X 42.5 = 85,000 =
the load on one arm. The formula for the moment at centre
support with uniform load is \wl* w = tne load per foot,
and / = the length of one arm of the span In this case w =
2000, /= 42.5, wl = 85,000, \wr = 451-562. This is con-
siderably less than the moment from the wheel-loads, which
was 550,085 for one pusition of the loads. It will be noticed
that the moment over the pier, \wr, is just the same as the
moment at the centre of a single span of length equal to one
arm of the draw and covered with the same uniform load;
and is also just one fourth as much as it would be over the
centre support were the draw swinging and covered with the
same load. Note that in moment \wr, wl is load on one
arm. A convenient method of finding tnese moments for
uniform load is to assume a load one pound or one thousand
pounds per foot, find the moments for this loading, and then
1 6 DESIGNING OF DRA W-SPANS.
multiply the results by the ratio of the actual loads to the
one assumed. To make us a little more familiar with the
force and equilibrium polygons, we will divide each arm into,
eight parts and assume a load of 1000 Ibs. at each of these
points and one half load at the ends. The loads at the ends,
coming directly over the supports, may be neglected in the
computation. We lay off then on the vertical line AB, Fig.
8 A , seven spaces representing 1000 Ibs. each. Any scale may
be used, say one-half inch equals 1000 Ibs. Next assume the
point O distant from AB 5000 to the same scale. Note that
the point O may be anywhere in a vertical line which is dis-
tant 5000 from the vertical line AB, and also remember that
we assumed the distance 5000; any convenient distance may
be used. We next connect the point O with each of the
points laid off on AB. Now going to Fig. 8, at any point on
a vertical through A we draw the line A' a' parallel to AO in
Fig. 8 A , and from a' the line a'b' parallel to the next line in
the force polygon, and so on until finally g'B is drawn parallel
to BO in the force polygon. Now connect A' and B' with a
straight line. From B' in Fig. 8 scale off the distance B'-B*
equal to the moment at the centre support divided by the dis-
tance HO 5000 in Fig. 8 A . The distance B ' B' must be laid
off to the same scale as Fig. 8 is drawn to. The moment at
the centre is of course found for the same loading (1000 Ibs.
at each eighth point = Jw/ 2 ). By the diagram Fig. 9 the
values of C for k i, f, f, f, f, f, and -J are:
A' = i = .125 C=.03oS; P = 1000, L 42.5; CPL 1309.00-
A'- = .250 C=.0586; " " =249050
A" =g = .375 C=.o8o6; " " =342550
A' ==.500 "=.0938; " " " =3986.50-
A' = = .625 =.0952; " =4046.00
A' = = .750 C =.0820; " " =3485.00
A'=f = .875 =.0513; " " " =2180.25
.49 2 3 20922 25
The moment \wi, for the same load uniformly distributed
(8000 Ibs. on each arm) is 42,500. The difference by the two
PLATE-GIRDER DRAWS* I/
methods is 655.75 or J f P er cent which shows that the
method is practically correct, and it is merely a question of
reading the diagram correctly to obtain accurate results.
Making a table of the moments (see p. 18), we have first the
column of moments for dead load swinging, the moments being
found by methods shown in Fig. 2 or 4. These moments
are 474,000, 350,000, etc. Next we make the column of
moments for dead load continuous, as shown by Fig. 8,
remembering that the moment at any point is equal to the
moment for the same load, considering the arm as a single
span supported at the ends, less the negative moment at this
point, and that this negative moment is represented by the
ordinate between the lines A'B' and A'B* multiplied by the
pole distance HO ; the ordinate B l E i being the moment
over pier divided by the pole distance HO. Thus the
moment at D equals ordinate dd' minus ordinate dd* (Fig. 8)
multiplied by HO (HO = 5000).
Having the moments tabulated, we now see which com-
binations will give the largest totals. The dead load swing-
ing and live load continuous, case A, give the largest moment
over the centre support, 1,024,000. The same combination
also gives the greatest moment at the -J point. At the quarter
point the dead load swinging and case A live continuous give
a minus moment of 318,000, and live load discontinuous with.
dead load swinging give a plus moment of 187,000, and so at
each of the points f, f, etc., we obtain the results given in
column 8. Dividing these results by the depth of girder
(centre to centre of gravity of flanges), we obtain the results
given in column 10. Dividing these results by the unit
stresses as allowed by the specifications (in this case 8000 Ibs.),
we have the areas required (column 12).* In Fig. I E the
areas required at the several points are laid off to scale, and
the lengths of the cover-plates required readily determined.
* Where the flange-areas are determined for tension, the areas after
deducting rivet-holes must be used.
iS
DESIGNING OF DRA W-SPANS.
d
a
oo M' in co 00 N co M r"
1! II II II II II II II II
1-
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xn c^ M co Tf co c<
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ment.
rt C
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xn xn cT xn cT in xn
xn xn o xn O xn xn
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Dead Load
Continuous.
CO
88 S 2 8 8 8
xn<^ o xnr-^xnco
>-< xn co O O xn co
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Dead Load
Swinging.
CN
r^ \r> o oo c*! o w M
-f co a -. M
1 1 1 1 1 1 1 1
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-
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CU
WEB-STRAINS. 1$
The plates should extend about two feet beyond the points
so determined.
The web is not considered as taking any flange-stress, and
the area in top flange is made up by two 5" X 3%" X -fa"
angles and two 12 X i plates. One of the plates will be too
long to get in one length, and a splice-plate is added to make
up the section at the splice. In the bottom flange two
-f " plates are used.
WEB-STRAINS.
We will next consider the shearing stresses in the web.
The greatest shear at the abutments will be obtained by con-
sidering one arm as a single span for live load and dead load
swinging, no dead reaction at abutment, as the condition of
dead load continuous and live load discontinuous cannot
occur. See combination of strains made. From a table of
'* shears and bending moments ' for this engine we have the
end shear for a span 42. 5 72,650 Ibs. That is, 72, 650 Ibs.
is the upward force exerted by the support at the abutment.
Say the specifications allow 6000 Ibs. per square inch shearing
on webs; then 72,650 -=- 6000 = 12.1 sq. in. required; 48 Xf-
inch web plate gives 18 sq. in. At the quarter point the upward
shear is 46,500 Ibs. From this is to be taken the dead load
between the abutment and this point. This load equals 525
X 10.62 = 5600 Ibs. 42,500 5600 = 36,900 Ibs. Note
that in finding the greatest live-load shears the heavy wheel
at the front of the engine is placed at the point where the
shear is required, and that there is no live load on the span
between the abutment and the point whose shear is being
determined. At the centre of the arm the live shear is
22,700 upward, and the dead-load shear downward is 11,200.
22,700 11,200= 10,500. The greatest shear at the pier
will be with dead load swinging (all dead load carried to the
20
DESIGNING OF DRA W-SPANS.
pier) and with live either continuous or discontinuous. For
discontinuous live load we have the same maximum shear at
the pier as at the abutment, the engine simply headed the
!
I-H
i
i
I
I
1
I
i
i
i
i
2.7
5.6
5.0
7.5
LOADl
8.1
14.750
4.9
4.3
4.4
12.8
CE
nj
E
5.0
5.6
5.0
9.0 i 5.1
Tig. 10
Span.
1
DI
Load.
-s\
2.7
425
0.06
0-93
10,250
9.740
8.3
0.19
0.76
11,250
7.780
13.3
0.31
0.62
11,250
6,975
20.8
0.49
0.41
8,000
3,280
28.9
0.68
0.23
18,000
4,140
33-8
0.79
0.14
18,000
2,520
38.1
0.89
0.06
20,000
0,120
42.5
1.00
o.oo
18,000
0,000
34,355
D,
12.8
17.8
425
0.30
0.42
O.oSg
0.096
10,250
10,250
912
984
23-4
28.4
0.54
0.68
0.091
0.071
11,250
11,250
1,024
782
37-4
0.88
0.024
20,000
481
42.5
1. 00
0.0
000
4,183
114,750 - (34,355 - 4183) = 84,580 .
Shear at Centre, Girder Continuous,
other way. We have then the upward shear live = 72,650 4-
the dead weight of one arm 22,300. 72,650 -f- 22,300 =
94,950.
Considering now the case of live load continuous: it is
clear that a load in any position (as the centre) on one arm
WEB-STRAINS.
21
tends, by causing this arm to deflect, to raise the other arm
off its abutment or end support. This support then has less
to do or the shear is reduced at this point by the load on the
other arm ; it follows therefore that, as all the load on the span
must be carried by the abutments and the pier, if some load
is taken from the abutment it must be added to the load on
the pier, A greater proportion of the load is carried by the
centre pier considering the two arms as continuous than by
considering them as independent spans. And in determining
the shear at any point the loads on both arms must be con-
sidered. By means of diagram Fig. 9 the reactions caused
by loads at any point in either arm are readily determined.
Arranging the loads as in Fig. 10, and finding the values of
k k and Z\, DV we get for the shear just to the left of the
pier 84,580; this added to the dead-load shear gives a total
of 84,580 -f- 22,300 = 106,880. The area of 84 X I web =
31.5 sq. in., against 106,880-^6000= 17.8 required. Stif-
feners should be at intervals of about the depth of the web
apart, with 6 ft. as a maximum.
22 DESIGNING OF DRA W-SPANS.
LATERAL BRACING.
The laterals should be figured for a wind-load of, say, 6oc
Ibs. per lineal foot, the point being to get sections heavy
enough to render the span stiff laterally. Cross-frames should
be used at intervals of ten to fifteen feet. Note that lateral
bracing should be figured to carry strains to the centre, and
that this force, equalling at least 300 Ibs. per foot = 25,500
Ibs. for both arms, should be considered in designing centre
pivot and anchorage.
CROSS-GIRDERS.
When the draw is closed and ready for the passage of
trains the girders are supported at the centre by wedges, so
the cross-girders carry only the dead weight of the span.
This amounts to 44,600 at each side; as there are two cross-
girders, the moment on each is 22,300 X 42 in. = 936,600
in. -Ibs. Using 2O-in. 64-lb. beams, with a moment of
resistance of 114, gives a fibre-stress of 8200, allowing an
ample margin.
CENTRE-POST.
The load on the centre-post is about 90,000 Ibs. The.
base of the post should be large enough to distribute this well
over the masonry and to give the post stability. There
should be anchor- bolts built into the masonry, and their area
should be sufficient to resist the shear from wind-forces, as-
suming for this purpose a wind-pressure of 300 Ibs. per lineal
foot of bridge, and neglecting the friction of the base-plate
on the masonry. This gives a force of 300 X 85 25,500
Ibs. Four ij-in. bolts at 7300 Ibs. each would be ample. A
wrought-steel post is preferable to one of cast iron, as it is
much less liable to break if the bearing on masonry becomes
unequally distributed. The post should be made high enough
DEFLECTION.
to throw the point of suspension into the upper half of the
web ; the girders will then hang better and turn more easily,
as there will be less weight thrown on the trailing-wheels.
DEFLECTION.
Deflection Formulae.
NOTE. These formulae are applicable to spans of any length if the propor-
tions are approximately as given below
r
4.704 WD
D for uniform loao = - - ........ (0
D for load at end = -, ........ (2)
'=58*. *"T?* + K?
i.i66WL*
D for uniform load = - ~i ........ (3)
D for load at end = ". ........ (4)
. ,
I.3I
for uniform load - - ~^i ....... (5)
4.248PZ 9
for load at end = - -r~ - ..... (6)
2 4 DESIGNING OF DRA W-SPANS^
D = deflection ;
h v = height at centre ; h = height for any distance x ;
L = length in inches ;
x = distance from left end in inches ;
P = load at end ; PF= totol load uniformly distributed.
A 'B / 1C ;D
, ! 70 ' 1 >
1-
| 700 ; 1^
1
i
\ r
/ 2-6"x 6*L'
z
III
o
f
Fig. 14.
Fig. 15
The amount the girders will deflect under the various
loads depends upon the length, the depth, and the arrange-
ment of the material in the girders. If the flanges are parallel
and their area of cross-section remain the same or nearly so
DEFLECTION. 2$
throughout their length, the formula for deflection for con-
stant section may be used ; thus for uniform load
wr
D = deflection, W ' = the load on the girder, /= the length
in inches, E = 29,000,000, and 7= the moment of inertia.
For a load at the end
pr
If the flanges are parallel, but the cover-plates are of several
lengths and the girder have about the proportions shown in
Fig. 13, the deflection for uniform load will be
^_ 4.704^ (I)
Eh*
W = total load on arm, and h = depth of girder back to back
of flange-angles. For load at end
Eh:
Number I is equal to
wr
195,500,0007*
and number 2 may be written
- 68,900,0007- ' '
W = total load on one arm in each case, and 7 = the moment
of inertia at the centre support. For girders having approxi-
mately the proportions shown in Fig. 14, which is the span
26 DESIGNING OF DRA W-SPANS.
taken as the example in considering strains, etc., we have for
uniform load
1. 16607* , ,
*>=- ...... (3)
Wl*
.... (3*)
i6i,5oo,ooo7
and for load at end
3-377^/ 3
pr
.... (44
55,700,0007*
Where the girders have the proportions as given in Fig.
15, for uniform load
(5)
132,000,0007
and for a load at end
wr
, . . . . (5*)
4O,8oo,ooo7*
W and 7 as given above. Some one of the formulae would
be applicable to any case likely to occur.
Considering first the case of uniform load : the girder we
have been considering is composed of 84" X i" web, four
5"X Si" X TV" angles, and (neglecting the short splice-plate),
two 12" X i" plates in top flange and two 12" X f" in bottom
flange. To simplify the calculations, we will for the present
consider all cover-plates as \ in. ; if this is not done, we should
first find the centre of gravity of the section, and then the
moment of inertia about this axis. Usually the flange-plates
DEFLECTION. 2f
are the same, and we will obtain nearly correct results by so-
considering them. The moment of inertia of the web about
its centre is equal to -^bk\ (b f , and h = 84.) -fab/i* = T V X
f X 84" = 18,522. The moment of inertia of the cover-plates
and angles about the centre of the web is found by multiply-
ing the area of each by the square of the distance between its
centre of gravity and the centre of the web. Thus the area
of the four -in. plates = 24 sq. in., and the square of the
distance from the centre of web to their centre is (42 -|- J) 2 .
24 X (42.5)' = 43,350. By referring to Carnegie's Pocket-
book we see that the centre of gravity of the 5 X 3i angles is
about i in. from the back of the angle, and that the area of
the four angles is 17.88 sq. in. The half-depth 42 in. I in.
gives 41 in. as the distance from the centre of the web to-
the centre of gravity of the angles. 17.88 X (41)' = 3^^6.
To the moments of inertia thus obtained we add the moments
of inertia of the cover-plates, and the angles about their own
centres of gravity; for the cover-plates ^bh* = T V X 12 X i in.
= i for each flange, and for the angles we have from the
Pocket-book 4.2 for each angle (see page 103, edition of 1893).
4.2 X 4 16.8 + 2 18.8, amount to add for plates and
angles. The total moment is then 18,522+43,350 +
30,056 + 18.8 = 91,946.8. It will be noticed that the
moments of inertia of the plates and angles about their own
axis is very small, and might be neglected without seriously
affecting the result.
Using our formula No. $a, we have
_ wr
161,500,0007*
W = 22,300, as previously found, L = 42.5 ft., and I
91,946.8.
22,300 X 132,651,000
D = - * -2-5- = o. IQ inch = A inch.
161,500,000 X 91,946.8
48 DESIGNING OF DRA W-SPANS.
If each arm is given a camber, this must be considered in
determining the end deflection. Suppose the top chord be
lengthened by adding i in. at a web-splice near the centre of
the arm. If the girder be 5 ft. 6 in. deep at this point, and
the distance to the end be 21 ft., the end will drop i -r- 5.6
X 21 = .94, say f| in. Adding o. 19 + -94 = J -i3 in -
i Jin., end deflection.
MACHINERY.
For Turning. The forces to be overcome in turning the
draw are, first, the inertia of the span itself. That is, there
is a certain mass which has to be revolved through a quarter
of a circle or 90 of an arc in a certain time. Second, there
is the friction on the centre pivot or rollers. Third, the
friction of the trailing-wheels due to the overturning force of
the wind, and the friction on the vertical surface of the pivot
due to the wind-pressure. Fourth, there is the friction of
the trailing-wheels due to any unbalanced load there may be.
Fifth, the friction of the shaft-bearings, etc. Item four
might be considerably increased by the rails on which the
wheels bear being out of level, rough, and with wide openings
at the joints. It is sometimes assumed that the draw shall
turn against a wind-force acting on one arm only of the span.
While this might possibly happen in the case of a long span,
it could hardly occur in the short 85-foot span we are con-
sidering, and this condition will not therefore be treated at
present.
Force required to Overcome Inertia. For convenience
we replace the mass of the bridge by an equivalent mass act-
ing at the rack-circle. This mass is found as follows: Mul-
tiply the weight of the span by the square of half the length
plus the square of half the width, and divide by 96.6 times
the square of the radius of the rack-circle. Putting this in
the form of an equation,
MA CHINEX Y. 29
W(* + ff)
M = 9 6.6R> '
where W = weight of sgan ;
a = half-length of span ;
b = half-width of span ;
R = radius of rack-circle;
M = equivalent mass at rack-circle.
The weight of our span is 89,200 Ibs. = W. a, the half-
length, = 42.5 ft.; b y the half-width, = 3.5 ft.; and R, the
radius of the rack, = 7.85 ft. We have therefore
89,200 X (42. 5' +3-5')
M= ~ 9 6.6 X 7-85^ - 2
If we assume that the draw shall open in two minutes, the
average velocity will be one fourth the circumference of rack
divided by 120 sec. = ^' 3 = 0.103 ft. per second. But
4 X 120
the velocity is not uniform; it increases during the first half
of the turning, and then reduces to o again at the end.
The maximum velocity at the end of 60 seconds is then
twice the average, or 0.206 ft. per second. The rate of
increase is 0.206 -r- 60 = .0034.
The force necessary to give a mass of 27,224 Ibs. a con-
stantly increasing velocity of .0034 ft. per second = 27,224
X 0.0034 92.5 Ibs. We will call this Fm.
Force to Overcome Friction on Centre Bearing. A
Sellers centre is used so the friction from load will be rolling
friction; a coefficient of .003 may be used, and this multi-
plied by the load gives 89,200 X .003 = 267.6 Ibs. This
acts at the centre of the length of the roller, or with a lever-
age of 8 in. or .62 ft. 267.6 X .62 -r- 7.85 =21.1 Ibs. the
force required at rack to overcome it. This force we desig-
nate F p .
Friction on Side of Pivot or End of Rollers for Wind-
pressure. Assuming a wind-load of 300 Ibs. per lineal foot.
30 DESIGNING OF DRAW-SPANS.
there results a total horizontal force of 300 X 85 = 25,500
Ibs. This, whether acting against the ends of the rollers or
on the side of a pivot, will produce sliding friction. Using a
coefficient of o. I, this gives 25,500 X o. i =2550 Ibs. acting
at the end of roller or at circumference of pivot (acting on
vertical surface). Let the radius of end of roller be 9^ in. or
.8 ft., then 2550 X .8 -f- 7.85 = 259.8 Ibs. at rack. We will
denote this by Fw.
Force required to Overcome an Unbalanced Condition
of the Draw. Suppose that from snow or some other cause
there is an unbalanced load on one arm, acting at a point 15
ft. from the centre pivot. The force at the wheel-circle
required to balance this is 15-7-7 (the radius of the wheel-
circle) 2.143 times the load. Assume the load to be 2000
Ibs.; this multiplied by 2.143 gives 4286 as the pressure on
the balance-wheel. The friction caused by this pressure will
be rolling friction and equal to 4286 X .003 = 13 Ibs. Thir-
teen pounds at the wheel-circle will require 13X7-7- 7.85
= n.6 at the rack to overcome it (7 and 7.85 being the
radii of the two circles). This force we will call Fu.
The centre of the surface exposed to the wind, including
ties and guard-rails, is almost exactly in line with the bottom
of the cross-girders, so that the moment of the wind-force
tending to revolve the girders about the centre casting as a
fulcrum is in this case slight and may be neglected. Suppose
the centre of wind-pressure had been one foot above the point
of support for cross-girders, the overturning moment would
then have been 25,500 X i = 25,500 Ibs. ; this divided by the
horizontal distance from the centre support to the centre of
the trail ing- wheel, 7 ft., gives the vertical force acting at wheel
to resist overturning. 25,500 -~- 7 = 3643 Ibs. Using coeffi-
cient of friction .003 gives 10.9 Ibs. 10.9 X 7 -5-7.85 =, say,
9.7, force at rack necessary to overcome it. This will show
how to proceed in cases where this overturning force of the
wind is too great to be neglected.
MACHINERY. 31
Force required to Overcome the Friction of the Shaft.
There will be only one shaft required in the turning
arrangement.
Assuming one man is able to turn the draw, and that he
exerts a pressure of 75 Ibs. horizontally against the top of the
shaft ; assuming for the present also that he works at the end
of a five-foot lever, and that a pinion 8 in. in diameter can be
used in rack, we have a horizontal pressure at foot of shaft
of 75 X 60-^4 = 1125 Ibs. 1125 + 75 = 1200 Ibs., total
pressure on shaft-bearings. The friction caused by this will
be sliding friction, for which the coefficient is 0.05 to o.i.
Multiplying 1200 X o. I = 120 Ibs. as the frictional force
acting at the circumference of the shaft. This we will call Fs.
We have then forces to be overcome as follows: Fm =
92.5, Fp = 21. ij Fw = 259.8, Fu = 1 1.6, and Fs = 120 Ibs.
First we will see how much power is consumed in over-
coming Fs. The radius of the shaft will be assumed as i in.
for the present, then 120 X i H- 60 = 2.5, the power
required at end of turning-lever to balance it. This leaves
us 75 2.5 72.5 Ibs. as available against the other forces
which all act at rack-circle. These equal 92.5 + 21.1 -f- 259.8
-4-11.6 = 384.9 Ibs. Dividing 384.9 by 72.5 gives 5.3,
which is the number of times the power must be multiplied
between the turning-lever and the pinion, or by the two.
We see at once that our power will be greatly in excess of
the amount required. It will multiply as many times as the
radius of the pinion is contained in the length of the turning-
lever, 60 -r- 4 = 15 (using an 8-in. pinion). We might use a
six-inch pinion and four-foot turning-lever. It is well, how-
ever, to have a good excess of power, as machinery may get
out of adjustment, the track become rough, and with gaps at
the joints, the span may become badly unbalanced, etc.
Time for Turning. The man turning the draw will walk
at an average velocity of, say, 3 ft. per second. If he be
moving at the end of a five-foot lever, he will move in a
32 DESIGNING OF DRA W-SPANS.
circle of 31.6 ft. circumference. It will require 31.6 -j- 3 =
10.5 seconds for him to make one complete revolution. The
pinion of course makes one revolution in the same time.
Using a pinion of 25 in. circumference on the pitch-line, and
a rack of 49.3 ft. circumference, the pinion must make
501.6 in. . . c ,, ,
- = 5.0 revolutions in moving over one fourth of
4 X 25
the circumference of the rack, which would be necessary to
open the draw. If one revolution is made in 10.5 seconds,
10.5 X 5.9 = 62 seconds as the time required to open or close
the draw.
Size of Turning-shaft. The man moving at the end of
the turning-lever produces a twisting moment on the shaft of
75 X 60 = 4500 in.-lbs. In addition to this twisting there
is the ben-ding produced by the force acting on the pinion.
Fig. 1 6
Assuming an 8-in. pinion, this force equals 1125 Ibs. ; and
assuming that the lower corner of the tooth is acting, and
that the distance from this corner up to, say, I J in. inside the
journal-bearing equals 6 in., then the bending moment will
be 1125 X 6J = 7312 in.-lbs. By referring to the notes on
shafting we find that the strength of a shaft to resist both
bending and twisting is given by the formula
M= bending moment, and T= twisting moment.
7" = 7312 + 1/53465344+20250000 = 7312 + 8585 = 15897.
MACHINERY. 33
Adding 50 per cent to this to allow for contingencies, we
have 23,846 in.-lbs., requiring a 2-f-in. diam. shaft. Note
that the shaft is weakened by the keyways and the shoulders
for turning-lever.
Proportions of T railing-wheels. The face of the wheel
should be about 4 in., to make sure it always has bearing on
the rail and to keep the bearing back from the edge. Letting
w = width of face, the other proportions would be about as
follows: Thickness of rim = .^W\ thickness of solid web =
.2$w; stiffening-ribs, six in number, thickness = ,2w\ length
of hub, not less than i.$w\ diameter of hub, 1.85 times the
size of axle required.
The side bearings should not be less than the diameter of
the axle, giving total bearing of 2D or more.
In figuring the size of axle required, if a length from the
centre of the wheel to the centre of bearing be used, the unit
stress in bending might be assumed at 30,000 Ibs. per square
inch. The reason for this is that the bearings and hub prac-
tically fix the axle so that it cannot bend until it leaves the
hub or the bearings.
Strength of Teeth in Rack and Pinion. Referring to
the tables and notes on the strength of teeth, we find the
formula for the safe load on cast-iron teeth P = 375/ a . This
formula is for the strength of tooth considering the load as
applied at one corner. We found the pressure on the tooth
to be 1125 Ibs.; then P = 1 125 = 375*''. f- 3, and / =
1.73. (See table of cast-iron teeth.) We find also from the
table that the width of face must be 2\ in. to give the same
strength, assuming the load as uniformly spread over the
length of face. As the speed is slow, we use the value of P l
for 100 ft. per minute or under. It is common practice to
make the breadth of the tooth not less than two to three
times the pitch.
Steel Rollers in Centre Bearing. Making the rollers
hard steel on hard-steel bearing-plates, we can allow a pres-
34 DESIGNING OF DRA W-SPANS.
sure per lineal inch of roller of 1750 Vd\ d being the average
diameter of roller. Calling this average diameter 2.J>", we
have 2765 Ibs. allowed pressure per lineal inch. The weight
of the span is 89,200 Ibs., and this divided by 2765 gives
32.2 lineal inches required. There are 15 rollers, 3 in. long,
giving 45 in. actual.
If a centre-pin is not used, care should be taken to give
the ends of the rollers an even bearing to resist the lateral
pressure as explained above. The plates or rings between
which the rollers move should be thick enough to distribute
the pressure evenly and so that there will be no give or spring
as the span revolves. For three-inch rollers the plates should
not be less than 2f to 3 in. thick.
If a pivot with flat disks had been used (see details of this
form of centre), the coefficient of friction would have been
about o. I (see table of allowed bearing on disks of steel and
bronze). The centre of pressure on pivots is at two thirds
the radius from the centre.
Wedging Arrangement at Centre and Ends. The cen-
tre roller-bearing is supposed to carry dead load only. To
support the span under live load, wedges or some equivalent
device are used under the girders at the centre and at the
ends. The supports at the centre should be driven just hard
enough to bring them to a full solid bearing, but not hard
enough to take the dead load off the centre pivot or rollers.
The amount the end wedges should drive is determined by
the amount of deflection it is found necessary to take out of
the girder so that there shall be no raising of the ends off the
supports as the load passes over one arm. The gears or
levers moving the wedges are easily arranged to give any
desired amount of motion to either set. The amount it is
necessary to raise the ends of the girder will now be consid-
ered. Placing the engine on one arm with the heavier
wheels at the centre, we find the reaction at the end of
unloaded arm to be 7070 Ibs. (see Fig. 9.) This means that
MACHINERY.
a force of 7070 Ibs. must be applied at the end of unloaded
arm to prevent its raising off the support. This force may
be obtained by driving the wedges under the ends of the
girder, and giving it an upward deflection until it is strained
sufficiently to give the reaction required.
Our formula for the deflection from an end load and girder
of varying section is, from page 26, No. 4*2,
.
55,700,0007
We have ^=7070 Ibs., 7=42.5 ft. = 510 in., / =
91,946.8.
7070X132,615,100 ' . nch>
55,700,000 X 91,946.8
Our wedge must then have a vertical movement of some-
thing over T 3 ^ in. If we make the slope of the wedge I in 5, a
horizontal throw of 12 in. will give us ample clearance for
turning.
The horizontal force necessary to drive the wedge will be
7 y (fa being the slope of the wedge) plus the friction of the
top and bottom surfaces of the wedge on their bearings. This
friction we will assume as 236 Ibs. Then 70 6 70 -f- 236= 1414,
which is the horizontal force to be applied. The coefficient of
friction might be as high as o. 10. At this value we have
JLQ B 15 -+ 77 + 77 = 2 59 2 Iks. as against the 1414 Ibs. we are
now using. It will be noticed that the friction is an important
element in determining the actual power to be derived from
the wedge.
The centre wedges should not be driven hard enough to
lift the span off the centre support, but just to a solid bearing.
We will assume, however, for the present that all six wedges
are driven with a force of 1414 Ibs. each. This will give us
an excess of power of about 50 per cent. One man, it was
assumed, could exert a force of 75 Ibs. The power must
then be multiplied between the man and the wedges. 1414
X 6 = 8484 -T- 75 = 113.2 times. Using a 6o-inch lever and
36 DESIGNING OF DRA W-SPANS.
the worm-screw arrangement as shown in Fig. 46, in one
revolution of the shaft the man moves 120 X 3.14 = 376.8 ft.
The pitch of the screw is, say, 2\ in., or there is a vertical
motion of 2j in. Dividing 376.8 by 2| gives 150.7 as the
multiplication of power, against 113.2 required. We do not
then need to increase the power further, and all arms on the
shafts may be of the same length. If the rods connect-
ing centre and end shafts are on one side of the bridge
only, that is, if one set only are used (sometimes one and
sometimes two are employed ; if the bridge is wide, there
should be a set on each side), these rods will carry a strain
of 1414 X 2 2828 Ibs. each. Rods f or f in. round will
be ample. The worm-shaft has a twisting moment of 75 X
60 = 4500 in. -Ibs. ; by the table on shafting we see that this
requires a shaft of, say, I T \ in. diameter. In order to make
a suitable thread for the worm, the shaft ought not to be less
than 5^ or 5} in. diameter. So in this case the worm would
determine the size of shaft to use.
The angle of repose for steel on cast iron is, say, 1 1. The
thread of the worm should then have a slope not exceeding
10 or 12. If the pitch is 2-J in., the thread rises \\ in. in
one half-revolution, and the angle is found by dividing this
rise (ij- in.) by the diameter of screw on the pitch-line.
Assuming this to be 5.8, we have 1.25 -=- 5.8 = .21 = tangent
of 12. Rather than use a shaft of this diameter, it would
be better to make the worm in the form of a sleeve, and
key it to a 2j or 2^ in. shaft. Or a shaft 3^ or 3! in. in
diameter might be used with a worm of ij or ij in. pitch.
The objection to this arrangement for such a light span is
that the time required to operate the machinery is made
unnecessarily great. We will assume that the worm is made
in the form of a sleeve and has a diameter at the centre of
the thread (or pitch-line) of 5.8 in.
Horizontal Shafts. We found that we multiplied our
power between the end of the turning-lever and the sliding-
MA CHINE K Y. 37
or worm-nut on the vertical shaft 150 times. The force
exerted by one man at the end of the turning-lever was
assumed as 75 Ibs. Then 75 X 150 = 11,250 would be the
force exerted upon the sliding-nut, were not a portion of this
used in overcoming the friction of the various parts. We will
first determine what these frictional forces are, up to the
point where the nut-lever acts on the horizontal shaft. These
forces being found and subtracted from 11,250 will give us
the force that the horizontal shaft must carry on to the
wedges.
We have, first, the friction of the bearings of the vertical
shaft; second, the friction on the collars from the thrust of
the vertical shaft; third, the friction of the sliding-nut in its
guides; and fourth, the friction of the sliding-nut on the
thread of the worm-shaft.
These are all sliding frictions for which the coefficient
would be between 0.05 and o. I, depending upon the smooth-
ness of the surfaces and the amount and character of the
lubrication. We will use 0.06.
The horizontal pressure on the journals is the 75 Ibs.
exerted by the man at the lever increased by the leverage due
to the bearing being some distance below the lever. Suppose
the lever to be 42 in. above the box, and that the play in
the box is sufficient so that the lower box might be assumed
as resisting this bending; then we have 75 X 42 -r- 70 45,
as bearing on lower box. There is also the horizontal pres-
sure from the worm-nut in its guides. This is equal to 75 X
60 -r- 8 562.5 Ibs. (Eight inches being the distance from
the centre of the shaft to the centre of bearing of the nut on
38 DESIGNING OF DRA W-SPANS.
its guides.) Forces causing friction on the bearings are then
120 + 45 + 562.5 = 727.5 Ibs. If we use a 2^-in. shaft,
727.5 X .06 X 1.25 -5-60= .91, . . . (i)
the force at end of lever to overcome this friction (1.25
being the radius of the shaft, and 60 the length of the hand-
lever).
Friction of the Collars. The vertical thrust on the shaft
we found to be 11,250 Ibs. This acts on the collars with a
leverage (the distance to the centre of gravity of the ring) of,
say, 1} in. ; then 11,250 X .06 X If -5- 60 = 19.7, the force
at end of hand-lever. This is excessive, and the friction
should be reduced by using a ball bearing in the collars (see
detail of this arrangement in cuts). This reduces the friction
to rolling instead of sliding friction, and the coefficient to
.003 ; we have then
11,250 X .003 X 2 ^- 60 1. 12. ... (2)
Friction of Worm-nut Sliding in its Guides. The hori-
zontal pressure of the nut we found to be 75 X 60 -~- 8 ==
562.5. Then
562.5 X .06 -:- 150 = 0.22 (3)
(The number 150 is the number of times the power is multi-
plied between the hand-lever and the nut.)
Friction on the Worm -thread. The vertical pressure is
11,250; and if the slope of the thread is 12, this gives a force
in the direction perpendicular to the screw-thread of 11,250
-r- 1.022 = I 1, 008. I 1, 008 X .06 = 660.48.
We will assume that the force at end of hand-lever neces-
sary to overcome this friction is 4.4 Ibs. This force is equal
to the friction multiplied by the radius of the worm-thread,
divided by the length of the hand-lever. In some cases the
friction may reduce the efficiency of the worm 40 to 50 per
cent. (See page 86.)
MA CHINER Y. 39
The sum of these frictions is .91 -\- 1.12 -f- 0.22 + 4.4 =
6.65 Ibs. Subtracting this from 75 gives 75 6.65 = 68.35,
the available power at hand-lever. 68.35 X 150 (the number
of times power multiplies) = 10,253, tne power transferred
by worm-nut to the arms on the horizontal shaft.
The horizontal shafts have, in addition to the twisting
moment, the bending due to the distances between the bear-
ings and the various levers which are keyed to the shafts.
On the centre shaft we have the levers or arms working the
struts which draw the centre wedges, the arms driving the
rods to the end wedges, and the arms working into the worm-
nut.
On the end shaft we have the arms working the end
wedges, arms worked by long rods running to centre, and the
cranks which work the rail-lifts. The twisting moment ex-
tends nearly uniformly through the centre shaft if the centre
wedges are only driven to a bearing, and there are rods run-
ning to the end shafts on each side of the bridge. If the rods
are on one side only, the moments of the twisting force will
be greatest between the worm-nut lever and the end of shaft
carrying the rod-arms.
In the end shaft, with one set of driving-rods, the moment
is greatest between the arms driven by the long rods and the
strut driving the end wedge. Then it is reduced by the
amount of the moment on the wedge strut-arm. It is again
reduced by the amount of rail-lift moment when this point
has been passed, and so on to the other end.
With two sets of the driving-rods the moment at the centre
would be o, and increase each way to the ends. For the
bending moments the portion of shaft between bearings will
be considered as a single span, and the bending moments in
each portion combined with the twisting moment (see table
and formulae for shafts).
The distance from one arm or prong of the lever working
in the worm-nut to the nearest bearing is, say, 8 in., and as
40 DESIGNING OF DRA W-SPANS.
each prong carries half the load, the bending moment will be
10,253 ~- 2 X 8 in. = 41,012 in.-lbs. The twisting moment
is, if there are driving-rods on each side, 5126.5 X n =
56,391 in.-lbs. (n being the length of the arm or prong from
the centre of the shaft), If the driving-rods are on one side
only of the bridge and run from the centre to the end on
opposite sides, for opposite ends as in Fig. 19, the moments
ROD
<r
x
CENTER OF <
I
CO
ROD
BRIDGE
ROD
QE
CENTER OF h
BRIDGE
" z
Ui
I
CO
Fig,i8
Fig. 19
are the same; but if the rods are on the same side, as in Fig.
18, the moment will be 10,253 X n = 112,783.
The first arrangement should of course be used, and we
have bending moment = 41,012 and twisting moment
56,391. Our formula (see notes on shafting) is
T\
ST'=4I,OI2+ 1/4,841,928,825 =41,012+69,584= 110,596.
\
By the table a shaft of 4 in. diameter is required for this
moment.
The bearings should be placed as near the points of load-
ing as possible.
The bearings of the horizontal shaft at the centre of the
bridge carry a pressure of twice the vertical force at nut-lever,
or 20,506 Ibs. Using a coefficient of 0.06, and remembering
that the lengths of all levers on this shaft are n in., we have
power lost in friction on this shaft 20,506 X .06 X 2 -=- 1 1 =
223.7 Ibs., and the shafts at the ends of bridge, including
rail-lifts, have about the same amount (it would be figured in
precisely the same manner), 10,253 (224+ 224) = 9805
MACHINERY. 41
Ibs., available for driving wedges, or 1634 Ibs. to a wedge
against 1414 required. As the machinery is liable to get out
of adjustment and the bearings to become dry, t there should
be at least 100 per cent excess of power. The wedges will
a
Fig. 20
stick and more power will be required to start them than will
be necessary to move them when once started.
Special care should be taken to provide ample means for
lubricating the wedges. The surest method is perhaps to
make several deep grooves diagonally across the bearing-sur-
faces. These grooves will retain a large amount of oil and,
as the wedges move, spread it over the surfaces. All oil-holes
should be easy of access and provided with means for exclud-
ing dirt.
The Levers. The lever-arms and the wings on the slid-
ing-nut should be figured as beams fixed at one end and
loaded at the other. For cast iron the fibre-stress should be
about 4000 Ibs., and for cast steel 15,000 Ibs. The bending
moment divided by the fibre-stress gives the momennt of
M
resistance required; thus R = ~-?. M= bending moment,
/ = fibre-stress, and R = moment of resistance. Say we
have a pull of 8000 Ibs. at the end of an arm, and the distance
to the points where the arm joins the hub is 10 in. ; the
42 DESIGNING OF DRA W-SPANS.
moment (M) is then 80,000 in.-lbs. if the arm is cast iron.
M -r- / = 80,000 -T- 4000 = 20. Using a rectangular section,
R ^bJ? (see any table on moments of resistance and inertia).
Assuming b 1.25, then 20 = X 1.25 X /**. Ji 2.1. If
a rectangular section is used, it should be stiffened by ribs on
the sides if the length exceeds six or eight inches. It
must be remembered that these levers are subject to sudden
jars, and should be made amply strong. The hubs are weak-
ened by the keyways, and should not be less than \\ to i-J-
in. thick. The keyways should be cut in the side of hub
next the arm where there is the greatest excess of metal. A
table giving the common sizes of keys used in shafts of differ-
ent diameters is given below,
Elbow-joint. We have found that our power has been
amply multiplied by the turning-lever and the worm, It
may be, and in fact the arrangement of crank on horizontal
shaft and the strut driving-wedge should be, such that they
increase the power two or more times. When the wedges are
driven the crank and strut should be in the same straight line,
or nearly so. As the force on the crank acts tangentially to
the circle described by its end pin, when the crank and strut
are nearly in line this tangential force is capable of exerting
an enormous pressure in the direction of the strut. As the
angle between the strut and crank increases this force de-
creases. It will be noticed that when the crank and strut are
nearly in line there may be a movement of the crank through
a considerable arc and very little movement in the direction
of the strut, so that to get the necessary amount of action in
the wedge the crank must move to a position where it is not
acting to the best advantage. Assuming that the power
necessary to drive the wedge increases regularly from o at the
point where the wedge just takes a bearing to a maximum
when the wedge is fully driven, then Figs. 11 to 14 and the
explanation below them show how a few trials with the wedge
driven to different positions will determine in which one of
MA CHINER Y.
43
them the greatest tangential force is required. And this
greatest force is the one to be used in determining the
moments on the shaft and the power required to turn.
If, when the wedges are driven, the crank and strut stand
at a considerable angle, there may be danger of the wedge
working loose under the action of live load, especially if the
angle of the wedge is steep.
Example of Elbow- or Toggle-joint. Assume that the
horizontal force necessary to raise the end of the girder the
required amount be 2000 Ibs. moved through a distance AB.
The horizontal force is zero when the wedge is drawn out, so
that the point A coincides with the point B and increases as
the distance between A and B increases. Assume that when
the wedge is driven the line of the crank and strut will be
ghk. With wedge moved \ of AB line of lever and strut,
assume line adg. Moved \ of AB, the line becomes beg.
Fig. 23.
Moved | of AB, the line becomes cfg. To find tangentia.
force at end of crank, with end of crank at d, lay off from d
44 DESIGNING OF DRA W-SPANS.
(Fig. 23) a line parallel with ad, also a horizontal line on
which lay off the force on wedge at this point = f of 2000 =
1500. From a' draw a perpendicular to da', intersecting da
at a\ through ^/draw dg, parallel with dg'm Fig. 22. Through
Fig. 24.
a draw ag at right angles to dg. Line ag equals the tangen-
tial force at d. Figs. 24 and 25 are drawn in the same way
and give the tangential force at e and f.
The Latch. Several styles of latch are given in the cuts.
The object of the latch being to hold the bridge in exact line,
it should fit close when driven to place, and it must be strong
Fig- 25-
enough to hold the bridge against the wind ; and if it acts
automatically, it must resist the shock of stopping the bridge
suddenly as it swings in position. Sometimes the latch drives
at the same time as the wedges are driven, but a latch work-
ing independently is more satisfactory.
Rail-splices. The latch should not be relied upon
entirely to keep the rails in line, but a sleeve of some sort,
slipping over the ends of the rails both on the draw and the
abutment, should be used.
Signals. The levers working the latch or the wedges may
also throw danger-signals placed on the abutments, or the
MACHINERY. 45
span as it revolves may be made to throw them. If there are
many attachments to the same set of machinery, some of
them are pretty sure to be out of adjustment most of the time.
And in general the simpler the machinery of a draw-span is
the better. A few heavy, amply strong parts are infinitely
better than a great mass of light, complicated pieces; the one
will be satisfactory in service, the other never will be.
Set-screws. While set-screws may be used in places
where there is little stress, they are not satisfactory in most
places on draw-span machinery. When most needed they
can only be relied upon to fail. Where used they should be
not less than f or f in. diameter. If two are used at one
connection, they should not be placed opposite to each other,
but at right angles.
Care of Draw-spans. To give satisfaction, the best
designed draw must have constant care and attention. Many
complaints of spans not working satisfactorily are due to gross
neglect in their treatment. The writer once went to a draw
that was giving trouble, and found that a coil of old rope left
on the pier some months previously by bridge-carpenters had
become wedged in between the rack and the pinion and
wrapped around the shaft, rendering it almost impossible to
turn the draw. How often had this part of the machinery
been examined in that time ? Not once. In fact, some parts
out of sight and not easy of access had not been oiled in a
year or more. The surest way to insure care in this respect
is to have as few parts as possible, and these easy to be seen
and reached. Other things being equal, the best design is
the one with the fewest parts to keep in repair.
4 6
DESIGNING Of DRAW-SPANS.
TABLES AND GENERAL DATA.
Notes on Spur- and Bevel-gears.
PROPORTIONS OF TEETH.
THICKNESS OF TOOTH ON PITCH LINE PITCH X .45
SPACE BETWEEN "
FROM PITCH LINE TO TOP OF TOOTH
TOTAL DEPTH "
GRANT'S ODONTOGRAPH TABLE FOR EPICYCLOIDAL TEETH.
Number of Teeth.
For One Diametric Pitch.
For i" Circular Pitch.
For any other pitch diameter
divide by that pitch.
For any other pitch multiply
by that pitch.
Exact.
Interval.
Face.
Flank.
Face.
Flank.
12
12
2.OT
.06
00
00
.64
.02
OO
00
1 34
13 to 14
2.04
.07
I5.IO
9-43
.65
.02
4.80
3-00
15^
15 16
2.IO
.09
7.86
3.46
.67
03
2.50
i.io
1?I
17 18
2.14
.11
6.13
2.20
.68
.04
i-95
,70
20
19 21
2.20
.13
5-12
i-57
.70
.04
1.63
50
23
22 24
2.26
15
4-50
1-13
.72
05
1-43
36
27
25 2 9
2-33
.16
4.10
.96
74
05
1.30
.29
33
30 36
2.40
.19
3.80
.72
76
.06
1.20
23
42
37 48
2.48
.22
3-52
63
79
,07
1. 12
.20
58
49 72
2.60
25
3.33
54
83
.08
1. 06
17
97
73 144
2.8 3
.28
3-14
44
.90
.09
1. 00
.14
290
145 rack
2.92
31
3-00
38
93
.10
95
.12
Rads.
Dist.
Rads.
Dist.
Rads.
Dist.
Rads.
Dist.
TABLES AND GENERAL DATA.
47
SPUR GEAR.
P'P' = pitch-circle;
A =face;
B = flank;
C = point;
Fig. 30
D- root;
E height;
F breadth;
RACK.
G = thickness;
H = space;
P = circular pitch.
J-L
Fie:- 31
Double-curve Teeth for Racks and Wheels.
Circle / for face-radius L = P' - of G. Circle N for flank-radius M = P'.
DESIGNING OF DRAW-SPANS.
BEVEL AND MITER GEARS.
Fig. 32.
AOB centre of wheel;
COD = " " pinion;
Qb = largest pitch diameter of pinion;
gh- " " " " wheel;
OiC and OkC= angles of cone pitch-line of pinion;
OjBzn&OkB= " " " " " "wheel;
mr = whole diameter of pinion;
qO " " " wheel;
wt and iO working depths of tooth.
-j^ of wy -f- ab = mr\
Y\j of wx -\-gh- qO.
angle 9 = angle of face of pinion;
angle 6 = angle of face of wheel.
TABLES AND GENERAL DATA.
49
Fig. 27
Fig.
5<D DESIGNING OF DRA W-SPANS.
Gear Teeth. Cast teeth should be made sufficiently
strong to resist the whole force transmitted by a pair of
wheels acting on corner of one tooth, and pitch is determined
as below (see Fig. 28):
Let e = thickness of tooth = -/; F = g= .99*; PG
40
&= .495/1 P= force at point/; moment of flexure = /te;
and greatest stress produced by moment of flexure on section
JSGFis
moment of flexure _ 6Pk
~ moment of resistance ~~ ge* '
which is a maximum when angle PEF 45 and g = 2 &-
<,p
Having then the value S = ~, consequently the proper
thickness for tooth is given by the equation
e
in which 5 may be taken at the values given in the table.
e may be assumed to be thickness on pitch-line = t\ then
40 l~~p 19
/ = A / 3-^, when h = t.
I 9 V 4
The above method of figuring the tooth is independent
of the face of the tooth, and should generally be used when
there is a liability of inaccuracies in the teeth.
If the face of the tooth is to be considered, as in machine-
cut teeth, the pitch can be assumed and the face (b) obtained
from the following.
-p
Fig. 29.
TABLES AND GENERAL DATA.
1 1
j
^>8
SJ
K
.
O D
1-g
1 s. .
1!
1!
.5 cfl t/)
j
oo o
S . u
H 1 ^
N 0\
3 Z
oo *n ** H
<
* 1
8 >
**
aa
rt be
8 u
*tr\ o"
"*- c -
2fc
"
g > II
"3 . -
J
1
||
Is!
TEETH.
|i|
111
C u "
>S 'S
l|
i
L MINUTE.
W
O
1 & n
I" 8 *
4
If
K
OH
h
v 5 IT
w
o
"S o g
8 g
mm
fe
n ts a
N fe
M
g
K^,
STRENG1
lies of /*! and i
ted over the fa
ocity in feet p
ioo Feet
or under.
II
VELOCITY
>! *
k
,v
1
is :
Js Hm
a II II II II II
^ 3
S
T
rON O^O^O ^t-^M ^00 IO fO W
m * 10 10VO t-^00 O ^ O M N ro
-*- iovo t^oo o> o - w -3- <ovo r-.
h
1
C
O ro m t-x o <M >ooo o ff> m t^ o
\o ONNVO O\N lAiow moo w
H
i
c
NVOOO c>^)oO roi'iO^O mt**
O\ - ff, u-ioo O M 10 t^ (N TJ-VO
-<J-vO t^OO ON M r<1 * 10 f^OO ON
c
t^ to N 00 10 (N ON^O ro O VO rr; M
Tl-00 N 10 O< rovo O ^00 t- 10 O-
M w N N M mcOT--<l-Tt-lOlOlO
w
r"
c
O\ ro 10 -<too O -*^o oo M M * t-x
VO *-" 10 O COOO O) *O O IO ON fO t^
10 C^OO ON - N -S- 10 t-xOO 0> M N
HMMMmMMAfil
|
<c
M ^- r-~ ON N TJ-OO o m^c oc M *
t^ - IO ON -^-OO CN t^ i- IO ON -^-QO
<L)
b
c
CNlOO-fONlO^-VO NOO ^nONlOO
VO t~ ON O -4-10 t--00 O - f<1 10
I
<
OO vO ^ ONVO mwOO IONOO OfO
1-1 CNI <N cnco-^-'l-to IONO vo 1 t^ t-
t;
1
00 >O CN) ONNO -^- ^ OO 10 CNI ONVC CO
vo oo O M ro 10 t^oo O P) ro 10 t^
MMMI-M^-C^CJCNIMCN
"C
lOt^ONONM CNI TflOt^ONON*- CNI
QlOO^iHNO-NOMMD-I^N
^ w
roo ot^ioO t^rroo tot^roO
M- CO w ONOO ti. 10 <!- CN] M ONQO t>-
t- ON - CNI -^-vo 00 O CNI * 10 t^ ON
1
c
N t^roON^t-O IOMVO CNIOO roON
IN CM m ro ^- io iovo NO t^ t>-oo oo
S
t(
c
JCSMOOOOCN!wCN)ONOw
OOONOOOOOOONQO
Cj rovo oo N Tf\c 1^ P)
1
c
OMMOOO^WMMOO^
^QVO CNIOO M-QNO CNIOO **-QNC
CNI CO CO TC TT IOVO NO -- t^OO 0\ ON
J
S
c
OO 1000 ONIOONO O fOO OVO N
O t^OO ON M CO TTVO hs O* O ^ CO
222-^??; ^Sr&JoS
c
00 rotxO VOONM-ON rooo M vo O
lONOO IOM t^-rfO t^roOvO ro
CNI rO CO Tt- 10 IONO tv t^OO ON ON O
eet or
ier.
^ *
\O CNI IOIOM IOM ^t^M r^oo <*)
ON CN! ^t VO ON *" ^*-NO OO "- m IOOO
oo - rn jo K o w ^-NO ON - ri\o
fc a
8 3
c
ONtx-^-OOO OfOO t^lOt^OOVC
NO n o P*cr> 5 K * 5 v * o t^.
M en < -<t- iovo NO t^oo oo ON o O
21
c
1151! IMI'H li
M M 0) O1 ^-VO t"* O^ M CO IO t" O
rtH
jjt
^s c
"a
<
R!?? < 5.8 < S.8 > < 8,R,8SS.8
co ^oo H joco rn jooj^^j
.
3
*
DESIGNING OF DRA W-SPANS.
For a pitch t, face b, length of teeth /, and base thickness
of tooth /z, we have for a tooth-oressure / and fibre-stress S
the general formula
and for proportions of teeth given, h being assumed at
(See Table, page 5 I
*-,*4 P=*.
j ID. o
In any case the breadth of face should not be made less,
than i/, and is generally made from 2t to 3/.
It is found that the breadth of face of the tooth should
increase with the increase of /. As the wear on the tooth
depends on the breadth, the tooth should be proportioned so
that -r should not exceed a given amount. For iron r =.
not more than 28,000. n = number of revolutions per minute.
For small forces this constant may be made as low as,
12000 or 6000 without obtaining inconvenient dimensions.
For Hoisting Gears, linear velocity at pitch-circle not
exceeding 100 ft. per minute, 5 may be taken at 42,000.
For Transmission Gears, velocity exceeding 100 ft. per
9600000
minute, take o from table on page 5 I, m which o = ^-
for cast iron. For steel .S may be taken ^S for cast iron.
v = lineal velocity in feet per minute.
Arms of Gears. A good proportion for the arms is ob-
tained when their number A is made as follows: *
Fig. 27*1
Fig. 271
* From Releaux.
TABLES AND GENERAL DATA.
53
= o. 53 \/~Z V 7;
A = 0.73
Z = number of teeth ;
t = pitch.
A = 3 4 $ 6 7 8 10 12
-$o 53 83 119 162 211 330 475
23 36 52
93
146 209
Width of arm // = 2 to 2.5*.
For thickness = '7]-{j}
TABLE OF GEAR-WHEEL ARMS.
h
Value of when
o
t
z
~A = 7
9
12
16
20
25
3
35
40
1.50
0.20
0.28
0.37
0.50
0.62
0.78
0-93
1. 08
1.24
1-75
0.16
0.21
0.27
0-37
0.46
0.57
0.69
0.80
0.91
2.OO
0. 12
o. 16
O.2I
0.28
0-35
0.44
0.53
0.61
0.70
2.25
o. ro
0.12
0.17
0.22
0.28
0.35
0.41
0.48
0-55
2.50
0.08
O. IO
0.13
0.18
0.22
0.28
0.34
0-39
o.45
2-75
0.06
0.08
O. II
0.15
0.18
0.23
0.28
0.32
0-37
3.00
0.05
0.07
0.09
0.12
0.16
0.19
0.23
0.27
0.31
WEIGHT OF GEARS.
The approximate weight G of gear-wheels proportioned
according to the preceding rules may be obtained from the
following:
G = 0.0357^(6.25^ + Q.04Z 2 ).
The following table will facilitate the application of the
G
formula, as it gives the value of -j-^ for the number of teeth
which may be given, and the weight can at once be found by
multiplying the value in the table by bf.
54
DESIGNING OF DRA W-SPANS.
z
c
4
6
8
" 20
5.04
5.60
6.18
6.77
7.38
30
7.99
8.61
9.24
9.89
10.52
40
11.09
1.90
12.59
13-30
14.02
50
14.74
15.43
16.23
17.00
17-77
&
qj
60
18.55
19-35
20.15
20.97
21. 80
6
70
22.65
23-50
24.36
25.24
26.12
H
80
27.02
27-93
28.85
29.79
30.73
0<{ go
31.69
32.66
33.63
34-62
35.63
-i
<u
IOO
36.63
37.67
38.70
39.75
40.81
.Q
120
47.40
48.54
49-69
50.85
52.03
3
MO
59-3
60.56
61.82
63.10
64.27
%
1 60
72.35
73-73
75-10
76.39
77.90
180
86.54
88.03
89-52
91.02
92.54
200
101.88
103.48
104.98
106.70
108.34
I 320
118.36
120.08
122.15
123.52
125.27
For weight of gear-wheels with number of teeth between figures given in
left-hand column use weight given on horizontal line through nearest ten below
the given number of teeth and under the figure in top line nearest last figure in
number of teeth given; thus, 46 teeth = 13.30.
SHAFTING.
Shafting. The formulae and tables given below will be
sufficient to enable the size of shaft required for any case
likely to occur in the consideration of draw-spans to be
readily determined. When the shaft is long and works
through a limited number of revolutions the diameter should
be large, in order that the angular deflection may not be
excessive. The use of too small shafting has been one of the
most common faults in draw-span design, and in many cases
has led to the renewal of machinery that in other respects
would have given satisfactory service.
A deflection of one degree in a length of twenty diameters
is considered good practice in miliwork, but for drawbridge
machinery, if the shaft be long and there are many attach-
ments to it, an angular deflection as great as this may cause
the whole arrangement to work badly. The angular deflection
for any twisting moment may be determined by the following:
OF THB
UNIVERSITY
TABLES AND GENERAL DATA. 5$
formulas : A = the angular deflection in parts of one revolu-
tion, M = the twisting moment in foot-pounds, L = length
of shaft in feet, d= diameter of shaft in inches; then
ML ML
for wrought iron A = 7;, and for steel A = ?
30000^ 36000^*
If the twisting moment M does not exceed M = 50^' for
wrought iron and M = 6od* for steel, the angle of deflection
will not exceed one degree for a length of shaft equal to 20
diameters. Thus if a 3-inch steel shaft have a twisting
moment of M = 6od* = 1620 ft.-lbs., then
A
36000 X 8 1 '
and if the length of shaft be 60 ft., then A = 0.033.
360 X 0.033 I2 - A deflection of one degree in 20
diameters = 12.
Friction of Shaft-bearings. For the slow motion of a
hand-turning draw the friction of the shafts, if well oiled,
would probably be about .025 of the pressure; but as the
conditions of lubrication as well as the state of adjustment
are uncertain, a coefficient of .06 has been used in the exam-
ple considered. As the speed increases the coefficient will
increase, and for higher speeds we may use
3-3
F coefficient of friction, d = diameter of shaft, / = length of
bearing, v = velocity in feet per second. It has been found
that for loads up to 600 or 700 Ibs. per square inch the fric-
tion depends upon the diameter, length of bearing, and
velocity, and is independent of the pressure. With heavy
loads and high speeds a coefficient of o. I 1 should be used.
Collar Friction. For the coefficient of friction on the
collars, 0.06 to o. I (depending upon the method of oiling,
etc.) should be used. This friction should be considered as
56 DESIGNING OF DRA W-SPANS.
acting at the centre of gravity of the ring. For method of
reducing friction of collar, where the thrust is heavy, see cut
of ball-bearings.
General For mules*
T .i$6d 3 s for round shafts; . . . . (a)
T = .2%d s s for square shafts (b)
d = diameter of the shaft in inches;
s = shearing strength in pounds per square inch;
7"=the torsional moment in inch-pounds; that is, the
force in pounds multiplied by the length in inches of the lever
through which the force acts, taking s at 40,000 and 50,000
Ibs. ; working value 9000 and 11,200 Ibs.
T i?6od* for round iron shafts; . . . (c)
T= 22Ood* for round steel shafts; . . . (d)
T= 2$2Od 3 for square iron shafts; (e)
T = $i$od 3 for square steel shafts; .
3 / T
d = A / for round iron shafts ; . (g)
V 1760
'6
d = A / for round steel shafts ; . . . (k)
X/ 2200
=A/
-V.
T f
- lor square iron
- - for square steel shafts.
3150
* Following tables on Strength of Shafting are from Pencoyd Pocket-book.
TABLES AND GENERAL DATA.
57
WORKING PROPORTIONS FOR CONTINUOUS SHAFTING,
IRON OR STEEL.
No Bending Action except its Own Weight.
Maximum
Revolutions per Minute.
Minimum
Diameter
of Shaft in
Inches.
Safe
Torsional
Moment
100
*5
200
250
300
Distance
in Feet
between
in Inch-
pounds.
H. P.
H. P.
H.P.
H.P.
H.P.
Bearings.
i*
5,940
7
IO
14
17
2O
II.7
if
7,552
9
13
17
21
26
12.4
if
9,432
ii
16
21
26
32
13-0
if
II, 602
13
20
26
33
40
13-6
2
14,080
16
24
32
40
48
14-2
2 i
16,892
19
29
38
48
58
I 4 .8
2*
20,048
23
34
46
57
68
15.4
2|
23,580
27
40
54
67
80
16.0
4
27,500
31
47
63
78
94
16.5
2f
36,603
42
62
83
102
124
17-6
3
47,520
54
81
1 08
134
162
18.6
3*
60417
69
103
137
172
206
19.7
3*
75,460
86
129
172
215
258
20.7
3t
92,812
105
158
211
264
316
21.6
4
112,640
128
192
2 5 6
32O
384
22.6
WORKING PROPORTIONS FOR CONTINUOUS SHAFTING,
IRON OR STEEL.
Transmitting Power and subject to Bending Action of Pulleys, Belting, etc.
Maximum
Revolutions per Minute.
Maximum
Diameter
of Shaft in
Inches.
Safe
Torsional
Moment
in Inch-
pounds.
100
H. P.
15
H. P.
200
H. P.
250
H.P.
300
H.P.
Distance
in Feet
between
Bearings.
l|
5,940
5
7
IO
12
14
6.8
l|
7,552
6
9
12
15
18
7-2
If
9>432
8
ii
15
18
22
7-5
'i
11,602
9
14
19
23
28
7-9
2
14,080
ii
17
23
28
34
8.2
*f
16,892
14
21
27
34
42
86
2 T
20,048
16
24
33
41
48
8.9
2f
23,580
19
29
38
48
58
9.2
2l
27,500
22
33
45
55
66
9.6
2|
36,603
24
36
48
60
72
10.2
3
47,520
39
58
77
96
116
10.8
3x
60,417
49
74
98
123
148
11.4
3|
75,46o
61
92
123
153
184
12.0
3i
92,812
75
H3
151
188
226
12.5
4
112,640
9i
137
183
228
274
I3.I
DESIGNING OF DRA W-SPANS.
Shafts having Both Bending and Twisting.
M= bending moment in inch-pounds;
T= twisting moments in inch-pounds;
T' = a new twisting moment which, substituted for T in
equations g to k, will give the desired proportions for the
shaft.
Factor of
Divisor in
Formulae.
Safety.
(j) for Iron.
(A) for Steel.
M . 3 T or less
4 1
1760
22OO
M 6T " "
c
je7O
Io6o
M T " "
5!
1430
I7QO
6
I^IO
1640
Formula for Horse-power.
V " = revolutions per minute;
= 396,000 inch-pounds per minute.
1 =
63,057 HP
36 HP
V
396,000 V
Deflection of Shafting.
I :=V873^ 2 for bare shafts; (/)
/= i/1^5^ 2 for shafts carrying pulleys, etc. ; . (r)
which would be the maximum distance in feet between bear-
ings for continuous shafting subjected to bending stress alone.
If the length is fixed and we desire the diameter of the
shaft, we have
= Y873
for bare shafting ; (s)
TABLES AND GENERAL DATA,
59
d
/ I*
= A / --
for shafting carrying pulleys, etc.
Working Formula.
8 Ao HP
d = \l for bare shafts ;
d= * /70H P
for shafts carrying pulleys, etc. ; (v)
for bare shafts;
1= tyi^pd* for shafts carrying pulleys, etc. .
Shafting- keys.
k = o. 1 6 -f- \d\ k' = o. 16 + ^^
Taper of key, .04 in. to .08 in. in 4 in.
w
- (*)
Shaft 1/2" 5/8" 3/4" i" if 2" a f 3" 3i" 4" 4i" 5
D Keys/32" 1/8" 5/32'' 7/32" 5/16" 7/16" 1/2" 9/16" 9/'6" 5/8" 3/4" 7/8"
From Releaux.
f Fig- 35 j
If we call the diameter of the shaft D, the breadth of the.
key 5, and the middle depth of the key S', we have:
<)O DESIGNING OF DRA W-SPANS.
For draft keys, 5 = 0.24" + ^-; S' = 0.16" + .
/? D
For torsion keys, 5 = o. 16" + : - ; 5 ' = o. 16" + .
5 I0
The taper of such keys is made about T ^.
For the more commonly occurring diameters we have the
following proportions:
D = i 2 3 4 56 789 10
FOR DRAFT KEYS.
S = 3/8" 1/2" 5/8" 13/16" i" if ij" if" if" if"
^' = 1/4" 5/16" 7/16" 1/2" 9/16" 5/8" 3/4" 13/16" 7/8" i"
FOR TORSION KEYS.
S = 3/8" 9/i6" 3/4" i" I T Y' if" ift" if" 2" 2 f"
^'= 1/4" 3/8" 1/2" 9/16" 11/16" 3/4" 7/8" i" I T V IT'S"
For shafts of less diameter than I in. we may make
c_^ c,_^
= r s
If several keys are used, they may be made the same
dimensions as single keys. For hubs which have been forced
on, and hence would be secure without any key, the dimen-
sions for draft-keys may be used.
BEARINGS AND PIVOTS, SPRINGS, CAMS, ETC.
Bearings. The bearings for shafts should be placed as
near the points of loading as possible, and for low speeds and
small loads the length of bearing should be once and one half
to twice the diameter of the shaft. Where the load is heavy
or speed great, the bearings are given a length of twice to
four times the diameter. Where the bearing simply carries
the weight of shaft, a length of once to once and one quarter
the diameter is sufficient. Bearings of brass or a composition
TABLES AND GENERAL DATA. 6}
of metals are used at important points. A bushing of Babbitt
metal is found to give excellent results. The friction is low
and the wearing properties of this metal are good. Two
bearings made in this manner are shown in the cuts. As the
speed increases, the length of the bearing should be increased
about in the ratio given in table below.
N = 100 150 200 250 400 750 1000
l+d = 1.25 1.5 1.75 2.0 2.5 3-5 4-o
N = number of revolutions per minute; / = the length of bearing in inches;
d = diameter of shaft in inches.
Ample provision should be made for keeping the bearing
well oiled, and all oil-holes should be easy of access. To aid
in spreading the oil over the whole bearing-surface small
grooves are often cut spirally around the bearing.
For thickness of metal and proportion of the various parts
see cuts 39 and 40.
Load on Rollers. Setter's Centre. The rotating load per
lineal inch on steel roller should not exceed that given by
the following formula for steel rollers on steel plates:
P pressure per lineal inch of roller;
d = mean diameter of roller in inches.
Load on Wheels. The load per lineal inch of face of
wheel, while span is turning, should not exceed that given by
the following formulae, viz. :
p 705 Vd for a cast-iron wheel on a cast-iron track;
P= 900 Vd " " P" " " wrought-iron track.
For steel wheels use the following formulae as to limit of
pressure per lineal inch of wheel-face while the span is turn-
ing, viz. :
P= 1905 Vd for a steel wheel on a cast-iron track;
P i$i$Vd " " " " wrought-iron track;
P= i 7 $oVd " " " " steel track.
* It is often specified that the load shall not exceed P = 1750 Vd.
62
DESIGNING OF DRA W-SPANS.
In which formulae
P = allowed pressure per lineal inch of face of wheel ;
d = diameter of wheel in inches.
Pivots.
TABLE OF SAFE LOAD FOR
STEEL ON BRONZE.
FORMULAE FOR PIVOTS.
Wrought Iron or Steel on Bronze.
( j* TA22
d.
0.035 ^/P
Slow.
0.05 \ r P.
Under
iso.fi.
Over
iso/?.
Slow-moving pivots -j ^ ~ * * , ^
Load.
Load.
Load.
t jj _ 7OO
I
816
398
204
n or < 150 } ,~ ' A/~B
1.25
i,275
622
319
i a 0.05 y f .
1.50
1,836
895
459
( a = 75.
1.75
2,500
1,219
625
| ^ __ Q^ QQ ^ |// > n
2.00
3,265
1 '59 2
816
2.25
4,132
2,016
1,033
Ca.tf /r0 Bronze.
2.5O
5,102
2,488
1,275
2-75
6,173
3,on
1,543
Slow-moving pivots ^ ~~ ' ,
3.OO
3-25
7,347
8,622
3,494
4,205
1,836
2,155
f * -3 CQ
3.50
10,000
4,877
2,500
w = r < 150 Y d =0 , 07 yp.
3-75
n,479
5,599
2,869
4.00
13,061
6,370
3,265
( a = 75
4-25
M,745
7,192
3,686
C d = 0.006 yPn.
4-50
16 53P
8,063
4,132
Iron or Steel on Lignum Vita.
4-75
5.00
18,418
20,498
8,983
9-954
4,604
5,102
Slow-moving pivots \* '~ 2 44 .
( d = 0.017 y T 1 .
5-25
5-50
5-75
22,140
24,694
26,990
10,974
12,044
13,164
5,535
6,673
6,747
= or < 150 \* = I422 ' -
{</ = 0.035 I// 1 .
6.00
6.25
6.50
29,388
31,890
34,490
14,334
15,630
16,900
7,344
7,972
8,623
>z "> ico -1 ^ :i::: I '^ 22 '
6-75
37,190
18,220
9,298
( rf = 0.035 v^.
7.00
41,690
19,600
10,000
The above table is made from the formula P= 8i6</ 3 for slow speeds, and
P = 8i6</ 2 for high speeds.
n
For cast iron on bronze use one half the above
values and for steel or iron on lignum vitae use double the values given in the
table, n = the number of revolutions per minute,/ = the pressure per square
inch, P = total pressure, d diameter of pivot, and the constant a = 75.
TABLES AND GENERAL DATA. 63
Formulas for Springs.
By GEORGE R. HENDERSON, Mechanical Engineer, N. & W. R. R
For Elliptic Springs. P = maximum static load in
pounds; = corresponding fibre-strain in leaves taken at
80,000 Ibs. ; N ' = number of leaves (in full elliptic), half the
total leaves; B = width of leaves in inches H = thickness of
leaves in inches; L = span (or length) of spring in inches
when loaded ; F = deflection of spring under load P in inches ;
E =. modulus of elasticity taken at 30,000,000. Then
2SNBH*
P=- j. - , and reducing P=
3
For half elliptic F = ^ anc * TG ducing F = .000611-7,.
1 2 PL 3 L?
For full elliptic F= ^ ENBH ^ and reducing F= .00133.
For Helical Springs. P load when spring is down
solid, in pounds; 5 = maximum shearing fibre-strain in bar
taken at 80,000; D = diameter of steel in inches; R =. radius
of centre of coil in inches; L = length of bar before coiling
in inches; G = modulus of shearing elasticity taken at
12,600,000; F= deflection of spring under load, in inches;
H height of spring free in inches; h = height of spring
solid in inches; ^- = 3.1416. Then
and substituting proper constant,
EM Zy D2 T-\*
^ = .o8^-; /T=^(i + .o8^ s ; />= 15.714-^-.
The most generally preferred ratio for size is D = $d,
where D = outside diameter of coil. It is customary to make
the static load about one half the solid load.
64 DESIGNING OF DRAW-SPANS.
Helical Springs.
By D. K. CLARK.
A - - for round steel; . . . (2)
3 fait v d
D=z*/'- - for square steel. ... (3)
E = compression or extension of one coil, in inches;
d = diameter from centre to centre of steel bar composing
the spring, in inches; w =. the weight applied, in pounds;
D = the diameter, or the side of square, of the steel bar of
which the spring is made, in sixteenths of an inch; C a
constant which, from experiments made, may be taken as 22
for round steel and 30 for square steel.
ECCENTRICS.
Eccentrics. An eccentric is nothing more than a crank
in which (if the crank-arm is R and the shaft diameter D) the
crank-pin diameter d' is made so great that it exceeds D -\- 2R y
or is greater than the shaft and twice the throw. The sim-
pler forms of eccentric construction are shown in the illustra-
tions. The most practical of these is that shown in Fig. 37^,
the flanges on the strap, as shown in the section, serving to
retain the oil and insure good lubrication.
The breadth of the eccentric is \\d to 3<^, the same as
that of the equivalent overhung journal subjected to the same
pressure. For the depth of flange a we have
a 1.5* = 0.07/4- 0.2
From which the other dimensions can be determined as in the
illustrations
TABLES AND GENERAL DATA.
Nf
Fig. 37 b
Fig-37 c
- 37
Hooks.
Formulas prepared by the YALE & TOWNE MANUFACTURING Co.
A = capacity of hook in tons of 2000 Ibs.
D = .$4-\-i.2$ G.^D-,
= .644+ i. 60 O= .363/4 + .66;
F= .33J + .85; Q =
H = i.oSA; L= u
/= 1.33^4 ; M .
J=i.2oA\ N=.%$B 16
Fig. 38
Capacity of hook iiiii234568io tons.
Dimension A I H I *& T i J $ J t 2 2 i 2 \ 2 s 3i '"
OF THB
UNIVEBSITY
66
DESIGNING OF DRA W-SPANS.
Fig 39
Fig. 39 a -
SHAFT-BEARING.
TABLES AND GENERAL DATA.
6 7
DRILL HOLES FOR
%'TURNED BOLTS
CUP OUT FOfU
BABBITT $1"DEEP
%STUD BOLTS 3& LONG
BABBITTED FOR 3*ie'8HAFT
[CORED
Pig. 40.
SHAFT-BEARING.
68
DESIGNING OF DRA W-SPANS.
At middle of
At point of 3
JF c
support = >. '3
At middle of
WL
beam = .
8
At centre of
WL*
span = .
31.9^7
At middle
support = \WY. 2.
At middle of
WL
support = -.
o
..StsZ
O 4)
II II
o o
B S
II II
6 .
II
Q
At end of
WL
beam= rr
At end of
WL*
At middle of
WL*
K.S
At point of
support = W.
At point of
support = W.
At point of
W
At point of
support = WZ-.
At point of
WL
support =
At middle of
WL
beam = .
TABLES AND GENERAL DATA.
DRAW-SPAN MOMENTS AND SHEARS.
(See Fig. n.)
COEFFICIENTS C' FOR LOADS IN FIRST ARM AND COEFFICIENTS
AND Z> FOR LOADS IN SECOND ARM.
Number
or Panels
in Half-
J
B'
C
C'
D
D'
E
E'
f
F'
G
G'
H
H'
7
I'
Totals.
span.
4
.0586
.0938
.0820
2344
5
.048
.084
.096
.072
.300
6
7
.0406
.0350
.0740
.0656
0937
.0875
.0925
.0962
.0637
.0875
.0568
3645
.4285
8
.0308
.0586
.0806
.0938
.0952
.0820
0513
4923
9
.0274
.0527
.0740
.0891
0960
.0925
.0767
.0466
5550
COEFFICIENTS D' FOR LOADS IN FIRST ARM.
4
.691
.406
.168
1.265
5
752
.516
304
.128
1.700
6
.792
S9 2
.406
.241
.103
2-134
7
.822
.649
.484
332
.198
.086
2-57 1
8
-844
.691
544
.406
.280
.168
.074
3.007
9
.861
725
592
.466
.348
.241
.146
.065
3-444
VALUES OF E' FOR LOADS IN FIRST ARM.
4
.810
.842
.goo
5
.807
.827
.862
.916
6
-805
.818
.842
.879
.929
1
-803
.803
:SII
.830
.824
.856
.842
893
.868
943
.900
943
9
.801
.809
.820
.832
.852
.879
.910
95
LOADS FOR MAXIMUM NEGATIVE MOMENTS-FIRST ARM.
4
I
B
C
For maximum at F
7
B
C
D
E
" G
8
B
C
D
E
F
" " H
9
B
c
D
E
F
G
u /
All loads on second arm in each case. All loads cause negative moments over pier.
LOADS FOR MAXIMUM POSITIVE MOMENTS-FIRST ARM.
Max. at
4
B
C
D
B to D
B
C
D
E
B to E
6
B
C
D
E
F
B to E
6
D
E
F
F
7
B
C
D
E
F
G
B to F
7
F
G
G
8
B
C
D
E
F
G
H
B toG
8
G
H
H
9
B
C
D
E
F
G
H
I
Bto H
H
I
I
SHEARS: All loads on second arm cause negative shear in first arm.
Loads moving A towards Z cause negative shear in first arm
Loads moving Z towards A cause positive shear in first arm.
PI = any load in first arm.
PI = any load in second arm.
St = reaction at .<4 from PI or /> a .
7J/2 = moment at pier from P or />.
X = distance from A to point of zero moment in first arm.
L = length of half-span.
WEB-STRESSES: Max. stress in any -j m A ?^'' J-
Max. stress in any \ ;' [ tQ piece jn
M- =
L or C*PL.
or Z>/V
.
E'L.
When ' ad CXtendS tromA
DESIGNING OF DRA W-SPANS.
I
I
1
GENERAL DATA DETAILS, ETC.
Fig. 44-
Fig. 45-
END MACHINERY.
72
DESIGNING OF DRA W-SPANS.
Fig. 46.
CENTRE MACHINERY.
Fig. 47-
CENTRE MACHINERY
GENERAL DATA DETAILS, ETC.
73
Fig. 48. Fig. 49. Fig. 50.
BALL-BEARING CENTRE. PIVOT CENTRE. ADJUSTABLE END WEDGE.
Fig- 5'-
SHAFT BALL-BEARING.
Fig- 52.
CENTRE ON CONICAL ROLLERS.
74
DESIGNING OF DRA W-SPANS.
G2- *a. SAILS IV 'All.
25.000 IB $ W/GffT 01VAlS.
Fig. 53-
/3-STEEL
20-3AUS
MYALLS.
54-
Fig. 55-
BALL-BEARINGS.
GENERAL DATA DETAILS, ETC.
75
Fig, 56.
CENTRE PIVOT.
Pivot 33" diam. to be forged in steel. Friction disks turned and ground
spherically to a 36" radius. Upper part steel. Lower part phosphor-bronze.
Base of cast iron, to be faced top and bottom, turned inside.
WRCIRCLe.
&/-AIUSMtt V-D/AM.
WX.WM 125 GROSS 7WS. \ 23 25 ZJ'S PR ffALL '.
Fig. 57-
BALL-BEARING.
DESIGNING OF DRA W-SPANS.
Fig. 5 8.
CENTRE FOR SMALL DRAW.
Fig- 59
SELLERS COUPLING FOR SHAFTING.
GENERAL DATA DETAILS, ETC.
77
End Elevation
Sectional Plan
Details of End Lifting Machinery
Fig. 60.
Section- Shovi
Screw;
Section
Kail Lift
teA Segment
;ftack Pinion
Details of Turn Table.
Fig. 6z.
DESIGNING OF DRA W-SPANS.
Fig. 62.
END SUPPORTS, LATCH, ETC.
Fig. 63.
END LIFT, LATCH MACHINERY, ETC,
GENERAL DATA DETAILS, ETC.
79
Fig. 6 5 b.
Fig. 65.
END LIFT.
Fig. 6Sa.
When draw is closed and ends are raised the middle pins of toggle-joint
Stand \" inside of vertical line through top and bottom pins to prevent the
toggle from opening In above position castings bear against each other.
8o
DESIGNING OF DRA W- SPANS.
JSLEMTION
Fig. 66.
WEDGING GEAR.
'ELEVATION
Fig. 67.
TURNING GEAR.
0W
16
Fig. 68.
PIVOT CENTRE.
GENERAL DATA DETAILS, ETC.
81
Fig. 70.
LATCHING DEVICES.
Fig. 71.
SLEEVE FOR CLAMPING RAILS.
OF THB
UNIVERSITY
82
DESIGNING OF DRA IV-SPANS.
Fig. 72.
Fig. 73-
MACHINERY FOR OPERATING SAFETY-SIGNALS.
GENERAL DATA DETAILS, ETC. 83
VIEWS SHOWING PLATE-GIRDER DRAW IN PROCESS OF CON-
STRUCTION.
Balance-wheel and Centre Wedge.
End Wedges and Portion of Machinery in Position.
84 DESIGNING OP DRAW-SPANS.
VIEWS SHOWING PLATE-GIRDER DRAW IN PROCESS OF CON-
STRUCTION.
End Wedging Arrangement.
Portion of Machinery at Centre.
EXPLANATORY NOTES.
Where the term " moment of resistance" and the letter
R designating the same have been employed in this work, they
are used as indicating the moment of resistance for a fibre-
stress of i ; or the term indicates the " section modulus" as
given by some authors.
2. In Case 5, page 68, for continuous beams on three
supports, note that the moments are obtained by scaling the
ordinates between the curve and the inclined line, and not
by scaling between the curve and the horizontal line as in the
other cases.
3. On page 16 it will be noticed t-hat the centre moments
have been given for the loads on one arm only. The moments
for the loads on the other arm are the same, and have been
included in obtaining the total moment.
85
86
DESIGNING OF DRA W- SPANS.
Friction of Worm-thread. (See page 38.) The efficiency
of the worm is very much reduced by the friction. In many
cases a coefficient as high as o. 15 would be nearer correct than
o. 10. The formula for the available vertical force is
W -
, where W = vertical force, r = radius of
6.28
turning lever, F force at end of turning lever to overcome
the vertical force W, F 1 = force at end of turning lever to
overcome the friction produced by W, P= pitch of the worm-
thread, D = the distance from centre of shaft to the centre of
the worm-thread, c = the coefficient of friction. A force of
i Ib. at the end of a 6-ft. lever gives an available vertical force
on the worm-nut, after deducting the friction of the thread
and of the guides, as follows :
Diameter of Shaft.
Pitch.
Size of Thread.
W, in Pounds.
f
*K"
J*"
iH"
k"
M irrsq.
So
161
182
207
242
WORKING VALUES FOR WORM-SHAFTS.
SQUARE THREAD
EMERY THREAD
A
,5
'
C
Area
of A.
Area
of C.
Safe Tensile
Strain Iron
at 10.000.
Safe Tensile
Strain Steel
at 12,500.
D
D'
W
W
i^
6
.167
1.284
1.767
i .'230
12,300
1 5-375
.696
.708
91.7
90.5
i6
5%
.182
1.389
2.073
1.496
14,960
18,700
754
767
84.5
83-2
i%
5
.2OO
i 491
2.405
1 750
1 7*, 5o
21,875
.810
.825
78.3
77.1
*%
5
.200
1.616
2.761
2.OOO
20,000
25,000
-8 73
.888
73-7
72.7
2
4^*2
222
1.712
3 MI
2.300
23.000
28,750
.928
944
68.8
67.8
2*4
4^
222
1.962
3-976
2 990
29,900
37,375
1-053
1.070
62.1
61.3
2 %2
4
.250
2 . 176
4.908
3 640
36,400
45,5oo
i ..169
1. 188
55-8
55-o
2%
4
.250
2.426
5-939
4.806
48,060
60,070
1.294
1 3 T 3
50.7
\y
Ig
.286
.286
2 629
2.879
7 068
8.295
5-4II
6.491
54,"o
64.910
67.638
81,137
1.407
1.420
I -554
4 6 - 8
43 6
46.2
42.6
3%
3/4
.308
3-100
9.621
7.080
70,800
88.500
i 650
1-673
40 5
40.0
3%
3
333
3.317
11.044
8.395
83,950
104,930
1.767
1.792
37 7
37-3
4
3
333
3-S67
12.566
9.970
99,700
124.620
i 892
1.917
35-6
35-2
4?4
2%
348
3 798
14.186
11.144
111,440
139,300
2.012
2.038
33 S
33-2
4^
2%
364
4.028
15.904
12.567
125,670
157,080
2.132
2 I5Q
3i 8
Number ot threads per inch on above bolts is the number given in the Sellers System.
A = external diameter ; B number ol threads per inch, C = diameter at root of thread;
D, D' = radius of centre of thread ; W (for v thread), W (for square thread; = the weight
which can be raised by a force of i Ib. with a leverage of i foot. Coefficient of friction = .15.
INDEX.
PAGE
Anchor-bolts 22
Areas of flanges 17
Arms of gears 49, 52
Beams, deflection of 68
moments in . . . . 68
Bearings of shafts 40, 60
Bevel-gears 48
Bracing, lateral 22
Camber 28
Care of draw-spans 45
Centre-post 22
Collar-friction 55
Conditions of loading 2
Cover-plates, length of 17
Cross-girder 22
Deflection, formulae for 23
upward 4> 35
Diagram for moments and shear 14
Elbow- joint 42
example of 43
Eccentrics 64
Explanatory notes 85
Force to overcome friction of centre. 29
shafts 31, 40
trailing-wheel 30
worm-nut 38
inertia 28
total 31
unbalanced condition of draw 30
87
88 INDEX.
PAGE
Formulae for strength of shafts 56
Friction of collars 38, 55
shafts 55
Gears, breadth of face 52
mitre 48
proportions of , 46, 47
strength of 50
arms 52
table of strength 51
weight of 53
Hammer at ends of draw 4
Horse-power 58
Hooks 65
Keys 4 2
for shafting 59
Key ways 42
Latch 44
Lateral bracing 22
Length of cover-plates 17
Levers, strength of 41
Load on rollers 61
wheels , 61
Loading, conditions of 2
Machinery 28
for turning 28
latching 44
wedging 36
Mitre-gears 48
Moment of inertia 27
Moments, bending, in beams 68, 69
maximum ... 9, 1 1
signs of 9
Parabola, to draw 6
Pivot 34
load on 62
table of safe load on 62
wind-pressure on 29
Polygon, equilibrium 8
force 8
INDEX. 89
PAGE
Rack 47
Rail-lift ?i, 79
-splice 44
Reactions 10, 14, 19
Rollers 33, 61
Set-screws . . 45
Shaft, horizontal 36
moments in 39
worm 35
Shafting, bending and twisting 58
deflection of 58
friction of 55
general formulae 56
horse-power of 58
keys 59
strength of 54
table of 57
Signals 44
Shear at pier, dead load 19
in web 19
Shearing forces, table of 69
Shears at end from live load 20
centre from live load 20
combination of 21
Springs. 63
Steel rollers 33
Stiff eners 21
Strains, combination of 17
dead load continuous 16
swinging 4
live load as single span 7
continuous 9
uniform 15
position of load for maximum 15
Stresses, unit 17
Strength of levers 41
teeth 33
Type of draw most satisfactory 2
Twisting force in draw 2
how best resisted 2
Teeth, strength of 33, 50
table, strength of 51
Time for turning 31
QO INDEX.
PACK
Trailing-wheels 33
Turning-shaft, size of . 32
Web-stresses 19
Wedging arrangement 35
Wedges 70
Weight of gears 53
Wheels, load on 61
Worm-shaft , 35
Worms 36
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Weisbach's Hydraulics. (Du Bois.) Svo, 5 00
Wilson's Irrigation Engineering, ... Svo, 4 00
Wolff's Windmill as a Prime Mover Svo, 3 CO
Wood's Theory of Turbines Svo, 250
8
MANUFACTURES.
ANILINE BOILERS EXPLOSIVES IRON SUGAR WATCHES
WOOLLENS, ETC.
Allen's Tables for Iron Analysis 8vo, $3 00
Beaumont's Woollen and Worsted Manufacture 12rno, 1 50
Bollaud's Encyclopaedia of Founding Terms .... 12mo, 3 00
" The Iron Founder 12mo, 250
Supplement 12mo, 250
Booth's Clock and Watch Maker's Manual 12mo, 2 00
Bouviers Handbook on Oil Painting 12mo, 2 00
Eissler's Explosives, Nitroglycerine and Dynamite 8vo, 4 00
Ford's Boiler Making for Boiler Makers 18mo, 1 00
Metcalfe's Cost of Manufactures ' 8vo, 5 00
Metcalf 's Steel A Manual for Steel Users 12mo, 2 00
Reimann's Aniline Colors. (Crookes.). ... 8vo, 2 50
* Reisig's Guide to Piece Dyeing 8vo, 25 00
Spencer's Sugar Manufacturer's Handbook 12mo, inor. flap, 2 00
Svedelius's Handbook for Charcoal Burners 12mo, 1 50
The Lathe and Its Uses 8vo, 600
Thurston's Manual of Steam Boilers 8vo, 5 00
West's American Foundry Practice 12mo, 2 50
" Moulder's Text-book 12mo, 2 50
Wiechrnarm's Sugar Analysis 8vo, 2 50
Woodbury's Fire Protection of Mills 8vo, 2 50
MATERIALS OF ENGINEERING.
STRENGTH ELASTICITY RESISTANCE, ETC.
(See also ENGINEERING, p. 6.)
Baker's Masonry Construction 8vo, 5 00
T3eardslee and Kent's Strength of Wrought Iron 8vo, 1 50
Bovey's Strength of Materials 8vo, 7 50
Burr's Elasticity and Resistance of Materials 8vo, 5 00
Byrne's Highway Construction 8vo, 5 00
Carpenter's Testing Machines and Methods of Testing Materials
Church's Mechanic's of Engineering Solids and Fluids 8vo, 6 00
Du Bois's Stresses in Framed Structures 4to, 10 00
Hatfield's Transverse Strains 8vo, 5 00
Johnson's Materials of Construction 8vo, 6 00
9
Lanza's Applied Mechanics 8vo, $7 50
" Strength of Wooden Columns 8vo, paper, 50
Merrill's Stones for Building and Decoration 8vo, 5 00
Merriman's Mechanics of Materials 8vo, 4 00
Pattou's Treatise on Foundations 8vo, 5 00
Rockwell's Roads and Pavements in France 12mo, 1 25
Spaldiug's Roads and Pavements 12mo, 2 00
" Hydraulic Cement 12mo,
Thurston's Materials of Construction , 8vo, 5 00
Thurston's Materials of Engineering 3 vols., 8vo, 8 00
Vol. I., Non-metallic 8vo, 2 00
Vol. II., Iron and Steel 8vo, 3 50
Vol. III., Alloys, Brasses, and Bronzes , 8vo, 2 50
Weyrauch's Strength of Iron and Steel. (Du Bois.) 8vo, 1 50
Wood's Resistance of Materials 8vo, 2 00
MATHEMATICS.
CALCULUS GEOMETRY TRIGONOMETRY, ETC.
Baker's Elliptic Functions 8vo, 1 50
Ballard's Pyramid Problem 8vo, 1 50
Barnard's Pyramid Problem 8vo, 1 50
Bass's Differential Calculus 12mo, 4 00
Brigg's Plane Analytical Geometry 12mo, 1 00
Chapman's Theory of Equations 12mo, 1 50
Chessin's Elements of the Theory of Functions
Compton's Logarithmic Computations 12mo, 1 50
Craig's Linear Differential Equations 8vo, 5 00
Davis's Introduction to the Logic of Algebra 8vo, 1 50
Halsted's Elements of Geometry , . .8vo, 1 75
" Synthetic Geometry 8vo, 150
Johnson's Curve Tracing 12mo, 1 00
" Differential Equations Ordinary and Partial 8vo, 350
Integral Calculus 12mo, 150
" Least Squares , 12mo, 150
Ludlow's Logarithmic and Other Tables. (Bass.) 8vo, 2 00
Trigonometry with Tables. (Bass.) 8vo, 300
Mahan's Descriptive Geometry (Stone Cutting). 8vo, 1 50
Merriman and Woodward's Higher Mathematics 8vo, 5 00
Merriman's Method of Least Squares 8vo, 2 00>
10
Parker's Quadrature of the Circle 8vo, $2 50
Rice and Johnson's Differential and Integral Calculus,
2 vols. inl, 12mo, 2 50
" Differential Calculus 8vo, 350
" Abridgment of Differential Calculus 8vo, 1 50
Searles's Elements of Geometry 8vo, 1 50
Totten's Metrology 8vo, 2 50
Warren's Descriptive Geometry 2 vols., 8vo, 3 50
' * Drafting Instruments 12mo, 1 25
" Free-hand Drawing 12mo, 1 00
" Higher Linear Perspective 8vo, 3 50
" Linear Perspective 12mo, 1 00
" Primary Geometry 12mo, 75
Plane Problems 12mo, 1 25
Plane Problems 12mo, 125
" Problems and Theorems 8vo, 2 50
" Projection Drawing 12mo, 1 50
Wood's Co-ordinate Geometry 8vo, 2 00
" Trigonometry 12mo, 1 00
Woolf's Descriptive Geometry Royal 8vo, 3 00
MECHANICS-MACHINERY.
TEXT-BOOKS AND PRACTICAL WORKS.
(See also ENGINEERING, p. 6.)
Baldwin's Steam Heating for Buildings 12mo, 2 50-
Benjamin's Wrinkles and Recipes 12mo, 2 00
Carpenter's Testing Machines and Methods of Testing
Materials 8vo,
Chordal's Letters to Mechanics 12mo, 2 Oft
Church's Mechanics of Engineering. ... 8vo, 6 00
" Notes and Examples in Mechanics 8vo, 2 00 ;
Crehore's Mechanics of the Girder 8vo, 5 00
Cromwell's Belts and Pulleys -12mo, 1 50
Toothed Gearing 12mo, 1 50>
Compton's First Lessons in Metal Working 12mo, 1 50
Dana's Elementary Mechanics 12mo, 1 50
Dingey's Machinery Pattern Making 12mo, 2 OO
11
Dredge's Trans. Exhibits Building, World Exposition,
4to, half morocco,$15 00
Du Bois's Mechanics. Vol. I., Kinematics , 8vo, 3 50
" Vol.11., Statics 8vo, 400
Vol. III., Kinetics 8vo, 350
Fitzgerald's Boston Machinist 1 8m o, 1 00
Flather's Dynamometers 12mo, 2 00
" Rope Driving 12mo, 200
Hall's Car Lubrication 12mo, 1 00
Holly's Saw Filing 18mo, 75
Lanza's Applied Mechanics 8vo, 7 50
MacCord's Kinematics 8vo, 5 00
Merriman's Mechanics of Materials 8vo, 4 00
Metcalfe's Cost of Manufactures 8vo, 5 00
Michie's Analytical Mechanics 8vo, 4 00
Mosely's Mechanical Engineering. (Mahau.) 8vo, 5 00
Richards's Compressed Air 12mo, 1 50
Robinson's Principles of Mechanism 8vo, 3 00
Smith's Press- working of Metals 8vo, 8 00
The Lathe and Its Uses . 8vo, 6 00
Thurston's Friction and Lost Work 8vo, 3 00
" The Animal as a Machine 12mo, 1 00
Warren's Machine Construction 2 vols., 8vo, 7 50
Weisbach's Hydraulics and Hydraulic Motors. (Du Bois.)..8vo, 5 00
Mechanics of Engineering. Vol. III., Part I.,
Sec. I. (Klein.) 8vo, 500
Weisbach's Mechanics of Engineering. Vol. III., Part I.,
Sec. II. (Klein.). 8vo, 5 00
Weisbach's Steam Engines. (Du Bois.). < Svo, 5 00
"Wood's Analytical Mechanics Svo, 3 00
" Elementary Mechanics 12mo, 125
Supplement and Key 1 25
METALLURGY.
IKON GOLD SILVER ALLOYS, ETS.
Allen's Tables for Iron Analysis , 8vo, 3 00
Egleston's Gold and Mercury Svo, 7 50
12
Eglestou's Metallurgy of Silver 8vo, $7 50
* Kerl's Metallurgy Copper and Iron 8vo, 15 00
* ' Steel, Fuel, etc 8vo, 1500
Kunbardt's Ore Dressing in Europe 8vo, 1 50
Metcalf Steel A Manual for Steel Users 12mo, 2 00
O'Driscoll's Treatment of Gold Ores 8vo, 2 00
Thurston's Iron and Steel 8vo, 3 50
Alloys 8vo, 250
Wilson's Cyanide Processes 12mo, 1 50
MINERALOGY AND MINING.
MINE ACCIDENTS VENTILATION ORE DRESSING, ETC.
Beard's Ventilation of Mines 12mo, 2 50
Boyd's Resources of Soutb Western Virginia Svo, 3 00
Map of South Western Virginia Pocket-book form, 2 00
Brush and Penfield's Determinative Mineralogy Svo, 3 50
Chester's Catalogue of Minerals 8vo, 1 25
Dictionary of the Names of Minerals.. Svo, 3 00
Dana's American Localities of Minerals Svo, 1 00
" Descriptive Mineralogy. (E. S.) Svo, half morocco, 1250
" Mineralogy and Petrography. (J. D.) 12mo, 200
" Minerals aud How to Study Them. (E. S.)., 12uio, 1 50
" Text-book of Mineralogy. (E. S.) Svo, 3 50
*Drinker's Tunnelling, Explosives, Compounds, and Rock Drills.
4to, half morocco, 25 00
Eglestou's Catalogue of Minerals and Synonyms Svo, 2 50
Eissler's Explosives Nitroglycerine and Dynamite Svo, 4 00
Goodyear 's Coal Mines of the Western Coast. 12mo, 2 50
Hussak's Rock- forming Minerals. (Smith.) Svo, 2 00
Ihlseng's Manual of Mining . . Svo, 4 00
Kunbardt's Ore Dressing in Europe , Svo, 1 50
O'Driscoll's Treatment of Gold Ores Svo, 2 00
Rosenbusch's Microscopical Physiography of Minerals aud
Rocks. (Iddings.) 8vo, 500
V 5 )
Sawyer's Accidents in Mines 8vo, 7 00
StDokbridge's Rocks and Soils 8vo, 2 50
13
Williams's Lithology 8vo, $3 00
Wilson's Mine Ventilation 16mo, 1 25
STEAM AND ELECTRICAL ENGINES, BOILERS, Etc.
STATIONARY MARINE LOCOMOTIVE GAS ENGINES, ETC.
(See also ENGINEERING, p. 6.)
Baldwin's Steam Heating for Buildings 12mo, 2 50
Clerk's Gas Engine t 12mo, 400
Ford's Boiler Making for Boiler Makers 18mo, 1 00
Hemenway 's Indicator Practice 12mo, 2 00
Hoadley's Warm-blast Furnace 8vo, 1 50
Kneass's Practice and Theory of the Injector 8vo, 1 50
MacCord's Slide Valve 8vo,
* Maw's Marine Engines Folio, half morocco, 18 00
Meyer's Modern Locomotive Construction 4to, 10 00
Peabody and Miller's Steam Boilers Svo,
Peabody's Tables of Saturated Steam Svo, 1 00
" Thermodynamics of the Steam Engine 8vo, 5 00
Valve Gears for the Steam-Engine 8vo, 2 50
Pray's Twenty Years with the Indicator Royal Svo, 2 50
Pupin and Osterberg's Thermodynamics 12mo, 1 25
Reagan's Steam and Electrical Locomotives. 12rno, 2 00
RSntgen's Thermodynamics. (Du Bois.) Svo, 5 00
Sinclair's Locomotive Running 12mo, 2 00
Thurston's Boiler Explosion 12mo, 1 50
" Engine and Boiler Trials Svo, 500
" Manual of the Steam Engine. Part I., Structure
and Theory, Svo, 7 50
" Manual of the Steam Engine. Part II., Design,
Construction, and Operation Svo, 7 50
2 parts, 12 00
" Philosophy of the Steam Engine 12mo, 75
" Reflection on the Motive Power of Heat. (Caruot.)
12mo, 2 00
" Stationary Steam Engines 12mo, 1 50
" Steam-boiler Construction and Operation Svo, 500
14
Spaugler's Valve Gears 8vo, $2 50
Trowbridge's Stationary Steam Engines 4to, boards, 2 50
Weisbach's Steam Engine. (Du Bois.) 8vo, 5 00
Whitham's Constructive Steam Engineering 8vo, 10 00
' ' Steam-engine Design ... 8vo, 6 00
Wilson's Steam Boilers. (Fluther.) 12mo, 2 50
Wood's Thermodynamics, Heat Motors, etc 8vo, 4 00
TABLES, WEIGHTS, AND MEASURES.
FOR ACTUARIES, CHEMISTS, ENGINEERS, MECHANICS METRIC
TABLES, ETC.
Adriance's Laboratory Calculations 12mo, 1 25
Allen's Tables for Iron Analysis 8vo, 3 00
Bixby's Graphical Computing Tables Sheet, 25
Compton's Logarithms 12mo, 1 50
Crandall's Railway and Earthwork Tables 8vo, 1 50
Egleston's Weights and Measures 18mo, 75
Fisher's Table of Cubic Yards Cardboard, 25
Hudson's Excavation Tables. Vol. II 8vo, 1 00
Johnson's Stadia and Earthwork Tables , .8vo, 1 25
Ludlow's Logarithmic and Other Tables. (Bass.) 12mo, 2 00
Thurston's Conversion Tables 8vo, 1 00
Totten's Metrology 8vo, 2 50
VENTILATION.
STEAM HEATING HOUSE INSPECTION MINE VENTILATION.
Baldwin's Steam Heating 12mo, . 2 50
Beard's Ventilation of Mines 12mo, 2 50
Carpenter's Heating and Ventilating of Buildings 8vo, 3 00
Gerhard's Sanitary House Inspection Square 16mo, 1 00
Mott's The Air We Breathe, and Ventilation 16mo, 1 00
Reid's Ventilation of American Dwellings 12mo, 1 50
Wilson's Mine Ventilation 16mo, 1 25
niSCELLANEOUS PUBLICATIONS.
Alcott's Gems, Sentiment, Language Gilt edges, 5 00
Bailey's The New Tale of a Tub , 8vo, 75
15
Ballard's Solution of the Pyramid Problem 8vo, $1 50
Barnard's The Metrological System of the Great Pyramid. .8vo, 1 50
Enimou's Geological Guide-book of the Rocky Mountains. .8vo, 1 50
Ferrel' s Treatise on the Winds 8vo, 4 00
Mott's The Fallacy of the Present Theory of Sound. .Sq. IGnio, 1 00
Perkins's Cornell University Oblong 4to, 1 50
Ricketts's History of Rensselaer Polytechnic Institute 8vo, 3 00
Rotherham's The New Testament Critical^ Emphathized.
12ino, 1 50
Totteu's An Important Question in Metrology 8vo, 2 50
Whitehouse's Lake Moeris Paper, 25
* Wiley's Yosemite, Alaska, and Yellowstone 4to, 3 00
HEBREW AND CHALDEE TEXT=BOOKS.
FOR SCHOOLS AND THEOLOGICAL, SEMINARIES.
Gesenius's Hebrew and Chaldee Lexicon to Old Testament.
(Tregelles.) Small 4to, half morocco, 5 00
Green's Elementary Hebrew Grammar 12mo, 1 25
" Grammar of the Hebrew Language (New Edition). 8 vo, 3 00
" Hebrew Chrestomathy 8vo, 2 00
Letteris's Hebrew Bible (Massoretic Notes in English).
8vo, arabesque, 2 25
Luzzato's Grammar of the Biblical Chaldaic Language and the
Talmud Babli Idioms 12mo, 1 50
MEDICAL.
Bull's Maternal Management in Health and Disease 12mo, 1 00
Hammarsten's Physiological Chemistry. (Maudel.) 8vo, 4 00
Mott's Composition, Digestibility, and Nutritive Value of Food.
Large mounted chart, 1 25
Steel's Treatise on the Diseases of the Ox 8vo, 6 00
" Treatise on the Diseases of the Dog 8vo, 3 50
Worcester's Small Hospitals Establishment and Maintenance,
including Atkinson's Suggestions for Hospital Archi-
tecture 12mo, 1 25