slejejedejels

BULLETIN
UNIVERSITY OF WASHINGTON

ENGINEERING EXPERIMENT STATION

ENGINEERING EXPERIMENT STATION SERIES ButLeEtin No. 13

TENSIONS IN TRACK CABLES AND
LOGGING. SKYLINES

The Catenary Loaded at One Point

BY

SAMUEL HERBERT ANDERSON

Y
Assistant Professor of Physics

SEATTLE, WASHINGTON
PUBLISHED QUARTERLY BY THE UNIVERSITY
June, 1921

Entered as second class matter, at Seattle, under the Act of July 16, 1894.

CORRIGENDA.

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ee =

(V's’o + C? -— So)

Page treaiaton(2s ye

VSo +e x ok Bo nz@ Ax
Sa Ce a(S
2 Z

Y+V 80°+¢7| =

Page 10, equation (38) :—

io sO ee OWES
y vo — (8 eee)
2, Z

Page 15, equation (55) :—

s—y——(ys—s.).(e°—1)

LIBRARY OF CONGRESS
RECEIVED

AUG1 51922

POCUMENTS DIVISION

ns

(22)

(23)

(38)

(55)

a

TENSIONS IN TRACK CABLES AND
LOGGING SKYLINES

The Catenary Loaded at One Point

INTRODUCTION

T'wo problems of the catenary, viz., the common catenary and
the parabolic catenary, are generally treated in texts on Mechanics.
A third problem, that of a catenary loaded at one point, is equally
important, but either is not discussed at all or given a sketchy treat-
ment that is far from satisfactory. ‘This type of catenary is of con-
siderable practical importance, as a conveying cableway or a cable
tramway closely approximates in form to this curve. In the West-
ern States such devices are much used in mining and logging opera-
tions for transporting loads from higher to lower levels by gravity.
In this paper the equations of the catenary loaded at one point will
be derived and the properties of such a catenary discussed. It will
also be shown how the tensions may be computed and what the
condition is for maximum tension.

It will be assumed that (1) the cable is perfectly flexible and
(2) the load is fixed at one point, neither of which assumptions
agree with practice. The cables in use are generally steel, sometimes
two inches or more in diameter and very stiff in short lengths.
However, when they are used in lengths of 2000 to 3000 feet and
the vertical deflections are small compared to the length, the
relative stiffness is negligible, and hence, the first assumption a
legitimate one. As the load travels along, the conformation of the
cable is constantly changing and likewise the tensions. But by
assuming a fixed position of the load, it is possible to obtain the
properties of the catenary for an instantaneous position of the load,
from which the condition for maximum tension in the cable may be
determined. ‘To start with, it is assumed that the load is at any
point, so that the treatment is general.

TENSIONS IN TRACK CABLES AND

FIG: 1

LOGGING SKYLINES 5

i!

Derivation of the Equation of the Catenary Loaded at One Point.

Let OAB, figure 1, represent a heavy, flexible cord or cable,
fastened at A and B and supporting a known load W at O. Assume
B to be at a higher horizontal level than A. -At O there will be three
forces in equilibrium: (1) the weight, (2) the tension in the part
of the cable OA which will be called TI’ and (3) the tension in OB
which will be called T”. Let ¢’and ¢” be the angles that ‘IY and T”
respectively make with the horizontal. By Lame’s theorem we have

W 1” T!
sin[180°—(¢'+¢")] sin(90°-+¢”) sin(90°+¢’)

or
W Aly. ft
—— = = (1)
sin(¢’-+¢”) cos” cos¢’
Solving for W

W=T"'sind’+T’cos¢’tang” (2)
and

W=T'"tan¢’cos¢”+T’ sind” G3)

Resolving the forces vertically and horizontally, respectively,
we have
W=T'sing’-+-T’ sing” (4)
and

'T’cosd’=T"cosp’"=T, (5)

where T) is the horizontal component of the tension of the cable
and is the same at all points for a given load and position of that
load. Combining equations (2), (3), (4) and (5) we get

2W=T"sing’ + T’sing”+Titand’+Titand” (6)
W=T,tan¢’+ Titand” (7)

It is evident that in form the cable is divided into two parts
the point of division being at O, where the load is hung. From a
mathematical consideration, this is a point of discontinuity which
means that in passing from one portion of the cable to the other
the change in the properties of the catenaty with respect to the
length is very abrupt at this point. It will be necessary, then, to
consider the two parts separately and derive the equations for each
part.

6 TENSIONS IN TRACK CABLES AND

We will consider first the arc OB. Let P be any point of this
arc, and let T’ be the tension at this point making an angle y with
the horizontal. ‘The arc OP is in equilibrium under three forces:
(1) the tension at O, which with respect to P is opposite and equal
to T” of figure 1; (2) the weight of the cable between O and P
acting at the center of gravity of OP; (3) the tension T at P. If
s be the length of the cable from O to P measured along this are and
w the weight per unit length, then the second force is ws. Resolving
these three forces vertically, we have

Tsing—ws-+ Tsing” (8)
and resolving horizontally
Tcosy=T,=T" cos” (9)
Dividing equation (8) by (9) we have
dy Ws
tany = — == —-+tand¢” (10)
dx h

This is the differential equation of the curve OB which is the con-
formation the cable assumes under the action of the three forces.
The solution of this gives the desired analytical equation.

It is evident that in equation (10) w, Ty, and tang” are all con-
stants for a given cable, a given load and fixed position of that load.
We will write for these three constants

Ip
ees (LE)
WwW
and
SoC Han (12)

Fquation (11) conforms with the usual method of treatment of the
common catenary, but it will be noticed that no assumption is made
regarding the form of the curve OB. ‘That is, c is considered noth-
ing more than a constant at this stage of procedure. Introducing
these two constants equation (10) becomes

dy S+S,
— = (iS)
dx €

Since s is a variable dependent upon x and y it is necessary to
eliminate one of the three before the equation can be solved. This
may be done by the relationship

LOGGING SKYLINES 7

ds*==dy?-+dx? (14)
First eliminataing x we have

dy s+so
= Z (15)
ds V(s+s.)?+c?'
the solution of which is
yaV GFR) FEA

where A is a constant of integration. If we choose the origin at 0
so that y=0 when s—0, then

A= Vs%o+e"

and (16) becomes
y=V (s-+50)?-+o! — Vs*c?! . (17)
On eliminating y between (13) and (14) we have

ds V (s+50)? + I’

= (18)
dx e

the solution of which is
log [(s+-s)o+V (st+s.)?-+c?| =24-B ae)

where B is a constant of integration. Using the same conditions as
for evaluating the constant A,

B= — log(s.+-V+8%+¢))
Substituting this value of B (19) becomes

x ts) tVGR FE

—-_— > og (20)
c Sot V Soc?’
This may be written in the exponential form,
=z (s+%)+VGTS) Fel
@° = (21)
See Soc"
By inverting equation (21) we get
« Vets) Fel — (st5)
e = — (22)

(FEE —s.)

We may obtain an expression for y in terms of x and the con-
stant c and s, by adding (21) and (22) and combining with (17).

8 TENSIONS IN TRACK CABLES AND

Vso ec" x x ~ So x

P. x

y+Vs0?-+e| rie ks (e+ e°) mai ) (23)

By substracting (21) from (22) we get a symmetrical expres-
sion for s.

WSeaCe| a) we 55 x
c

s+ =—_——— (ee) +— (e+ &) (24)
pa 2

Equation (23) is the equation of the curve OB in rectangular
co-ordinates, and (24) gives the length of the arc measured from O
in terms of the abscissa and the constants s, and c. ‘These equations
represent a general case of the catenary of which the common
catenary is a special case. ‘This may be seen from the following
consideration. Suppose that tané” =O, which could be attained by
removal of the load W or by shifting to such a point that the are
OB became horizontal at O. Then by (12) s, =O and equations
(23) and (24) reduce to

¢ x x
YT Same gs (25)
and
c x =
es (26)

which are the well known equations of the common catenary.

In order to interpret the constants s, and \/s,?+c?| of equations
(23) and (24) let us assume that the curve OB is extended to the
left of O to a point C where the tangent is horizontal. This is rep-
resented in figure 1 by the broken line curve CO. Let the weight
per unit length of the imavinary cable be the same for the arc OB,
and let the horizontal tension be Ty. Call the co-ordinates of O,
referred to C, p and q. The are CO will be in equilibrium under
three forces, T,, the weight of the cable wr where r is the length of
the arc CO, and the tension T” at O. Resolving these forces hori-
zontally and vertically we have the following relations:

TT’ cose” : (27)
and
T’siné’—=wr (28)
Dividing equation (28) by (27) we obtain
dq a
tand” = — = — (29)

dp c

LoGcGING SKYLINES 9

This is the differential equation of the arc CO, and is of the same
form as that of the common catenary, so that the solution may be
immeditely written.

q=Vrt+e!—c (30)
eae ee
= oS oe (31)
Cc

Since the curve COB is continuous the equations for any point
referred to C as the origin will be of the same form as (30) and
(31). Let x’, y’ be the co-ordinates of any point referred to C. Then

y=vV (str)? + c?' —c . (32)

(33)

x —c log ;——_——_-

(stn +VEF +e \
c
where (s-+r) is the length of the arc measured from C.

Now let us transform the last two equations by the following
relationship,

yep
y—y+q
and substituting values of p and q from ue and (31).
y=VGHEPO —c + VP Fe +e (34)
x —— oar
c hy toe

Equations (34) and (35) must be identical with (17) and (20)
respectively, since in both cases x, y are the co-ordinates of any
point of the arc OB referred to O as an origin. Hence

T = So

That is, the constant s, of equations (17) and (18) is the length
of the fictitious arc CO. This means that the curve OB 1s a portion
of a@ common catenary, the lowest point of which is a distance So
from O measured along the curve. While OB is the only part of
this catenary that represents the conformation of a part of the
cable, nevertheless, the part of the catenary CO has a physical in-
terpretation which will be given later.

10 TENSIONS IN TRACK CABLES AND

From a consideration of equations (17), (34), and (30) it is
evident that

Vsifel=vrpl=qte | (37)
In the common catenary c is the distance from the lowest point to

a horizontal line which may be called the directrix. Hence \/s.?-c?’
is the distance of O from this same directrix. We will call this yo.
Making this substitution in equations (23) and (24) we get

Ae IE ee ee Nt

Y st Ne eg ae ae Ce ee (38)
Z Z
ame BOM yea |

S > So = z (ere) ais (eae (39)

Substituting the hyperbolic functions for exponential, these equa-
tions become

x x
y + yo= yo cosh — + s, sinh — (40)
€ fe
x x
S + so = yo sinh — + s, cosh — (41)
c €

These equations involve three constants, c, s, and yo, the last of
which is dependent upon the other two.

Yo= Vse+ C2! (42)

All three are quantities of the fictitious are CO; c being the parame-
ter, So the length of the arc, and y, the ordinate of the terminal
point O measured from the directrix. By assigning different values
to C, So, and y, (40) and (41) become the equations of a family of
curves which represent the conformation the cable will take under
various loads and horizontal tensions.

In the same manner we may obtain equations of the arc OA
which is the conformation assumed by the part of the cable to the
left of the load W. The differential equation is,

dy ws
— = —-+tan¢’ (43)
dx Ab

which differs from (10) in that x and s are taken to left of O and
¢’ replaces @”. Hence the analytical equations of the arc OA will be

LoccInG SKYLINES 11

the same as (40) and (41) except the constants s, and y, will have
different values. The constant.c is the same for the two curves,
since it depends upon the horizontal tension and the weight of the
cable per unit length, which are the same for all parts of.the cable.
Hereafter when s,,and y., are used in equations (40) and (41)
they apply to the arc OA to the left of the origin and hence the
values assigned to s and x are negative only; and when s,, and yos
are used the above equations apply to the arc OB to the right of the
origin and the values assigned to s and x are positive only. Since
the constant c is the same for both arcs it is evident that OA and OB
are arcs of the same common catenary having the parameter c. If
in figure 1 the curve DOA were shifted to the left and slightly up-
ward until the point D coincided with C, AODCOB would form a
continuous curve which is the common catenary from which OA and
OB are taken to form the discontinuous curve AOB. The length
of the arc DO is the constant s,, of the curve OA and the length of
COO is the constant s,, of the curve OB.

Ke
Determination of the Tensions.

Arbitrarily we have set
17 — we GEP)
and found that c is a perameter of both curves OB and OA.

The vertical components of the tensions T’ and T” at O may be
obtained by combining equations (5), (11) and (12).

T’sind’—=wsp, (44)
T’sind’=wso> (45)

Hence the vertical component of the tension in OA at O 1s equal to
the weight of the fictitious arc OD; and likewise, the tension in OB
at O is equal to the weight of OC. If these expressions for the
vertical components of the tensions are substituted in equation (4),
an expression is obtained giving the relation between the load W,
the weight per unit length of the cable, and the fictitious arcs.

W=w (s01+S02) (46)

This shows that the constants s,, and s,, are dependent upon the
nature of the cable and the load.

12 TENSIONS IN TRACK CABLES AND

= yWy Co Se Wan (47)

4 ha wVc? a Soo?" == Wyte . (48)

The vertical component of the tension at any point in the curve

OB is given by equation (8). Combining this with (45) we get

T.siny.—=w (s+So2) (49)
And in like manner we have for any point in the curve OA
T’siny,==w (s+s,,) (50)

Now (s+s,,) is the length of the arc COP measured from the
lowest point of the fictitious are OC, and hence the vertical compo-
nent of the tension in the cable at any point is w times the length of
the arc of the common catenary of which OB or OA is a part.

By combining (11) with (50) and (49) respectively we obtain
the total tension at any point in OA or OB respectively.

T.=w(y+yo1) (S51)
T.==w(y+Yoz) (52)

Hence the total tension at any point is w times the verticle distance
of that point above the directrix of the common catenary.

Equations (11), (44), (45), (47), (48), (49), (50), (51)
and (52) give completely the tension in the cable. In every case
the tension depends upon the weight per unit length, w, and one of
the three constants, c, S,, or yo. If these tensions are known, the
constants can then be determined and by the aid of equations (40)
and (41) the position of the cable determined. And conversely,
if the constants can be determined from the data available and

eduations (40) and (41), the tension for any point in the cable can
be found.

Equations (51) and (52) show that for a given load and given
position of that load the tension in either part of the cable is great-
est at the point where (y+y,) is a maximum, that is, at the point of
fastening at the end of the cable. So that in future considerations
of the effect of the load and of the position of the load the ten-
sions at the ends only need be taken account of. At which of the
two ends the tension is the greater depends upon the position of
the load. If in equations (51) and (52) y, is the verticle height of
the lower end above the position of the load and y, that of the
upper end, y, will be greater than y, for any position of the load

Page 12, insert before equation (47),

The tensions T’ and T” can be expressed in terms of w, c, So,
and yo. Combining (11) with (44) and (45) respectively we have,

14 TENSIONS IN TRACK CABLES AND

along the cable, but yo. will be considerably less than y,, when the
load is near the lower end and hence ‘T, will be less that 'T,. But
for the load near the middle of the span and any place between the
middle and the higher end T, will be greater than T,. (See fig-
tpe 2).

In practice, information is wanted regarding the tensions in
various parts of the cable for a certain load and different positions
of that load, or more specifically, the maximum load the cable will
carry. To answer the latter question equation (46) must be used
which involves the constants s and s,. It is desirable, then, to ex-
press s in terms of other quantities on which it depends. ‘This may
be done by eliminating y, between equations (40) and (41). Writ-
ing these in the following form

x >.<

y — s.sinh— = y,(cosh— — 1) (40)
c c
x x
s + s.(1— cosh—) = y,sinh— (41)
c €

and dividing the first by the second we get,

2 cosh x — 1 Z
Cc

which may be simplified to

y Ss
So == —coth¢ — — (53)
2 Z.
where
x
o=— (54)
2c

If x,, y, are the coordinates of one end of the cable (the
lower) referred to the position of the load as origin, s, is the length
of the portion of the cable extending from the load to the end, ie.,
OA of figure 1. But the value of s, depends upon the parameter c.
An infinite number of catenaries may be drawn thru the points A
and O, and the particular one which represents the conformation of
the cable is determined by the value of c taken. In an unloaded

LOGGING SKYLINES 15

cable the value of c depends alone upon the horizontal tension
which may be made anything desirable. In the loaded cable c may
be computed from the value of the horizontal tension, but this lat-
ter depends in part upon the load which is placed on the cable. A
relationship between x, y, s and c, which is independent of the
other constants s, and yo, may be obtained which is of considerable
use. Subtracting equation (38) from (39) we get

eee ee!) (55)
and by adding

st+y=(Yo+s) (e-—1) (56)
On multiplying (55) by (56) and substituting
=
we have
s2-y2—=c? (e°_2-+ e’)
which may also be expressed in terms of the hyperbolic function,
x
=2ce sinh— (57)
26

This may be thrown into a more useful form by substituting from
equation (54).

VEZ

sinh ¢

Vs? — y?| =x (58)

By this last equation the length of the cable between the load and
terminal point may be quickly found for any value of c. To faciil-
tate computation, it is desirable to have a curve in which sinh ¢ is

$
plotted as ordinates against ¢ as abscissas. Such a curve is given
in figure 3. In the next section it will be shown how these computa-
tions may be made.

Another equation giving s in terms of x, y, and c may be obtain-
ed by eliminating s, between equations (40) and (41).

me
s=y-t 2y, tanh—- (59)
2E ;

16 TENSIONS IN TRACK CABLES AND
Ill

Practical Solution of the Problem.

In sections I and II equations have been derived giving the
relations among all the factors involved in stresses due to loading.
But when an actual computation is undertaken, it is generally found
that not sufficient data is available. Perhaps this may be best
illustrated by taking a numerical problem, which was presented to
the writer by a logging company. It is typical of those the engi-
neers encounters.

It is required to know the maximum load a cable will carry
when used as a logging “‘sky-line” under the following conditions:

1. Breaking strength—112 tons.

2. Weight of cable—5.38 lbs. /linear ft.
3. Horizontal span—2500 ft.
4

Vertical height of upper end above lower (tail spar tree
above head spar tree)—625 ft.

5. Factor of safety—3.5.

The maximum tension allowable is

1122000
———— = 64,000 Ibs.
a5
Calling this T, and substituting in equation (52) we find
64,000
Yo + Yo = — =— ilo O mie
5.38

‘Vhis. may be used in equation (40) for finding s,., but before any
computation is possible values of x, y, s and ¢ are necessary. Since
the value of T’, used above is the maximum that the cable is to
stand, the value of x to be used should be the horizontal distance
of the load from the upper end where the tension is the greatest.
In other words, before we can proceed further it is required to know
the condition of maximum tension in the cable for a given load.
So far no general method of treating this phase of the problem has
been found, but it has been determined for a special case, which
will be taken up in detail.

ie

LoccING SKYLINES

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OEE Boome oe BESS aeeee

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18 TENSIONS IN TRACK CABLES AND

The first step necessary in this calculation is the assumption of
some value of the parameter c. A common usuage* is to take the
horizontal tension equal to one-fourth of the breaking strength of
the cable, and the “deflection” of the cable when the maximum
gross load is at the center to be one-twentieth of the span. By “de-
flection” here is meant the vertical distance of the loaded point of
the cable below the lower fixed end of the cable.

Following thse two specifications we may find values of c, x
and y. From equation (11)

Tn - 1122000

c= — = ——__ = 10,409
Ww 4X 5.38
One-twentieth of the span is 125 feet, so that
¥e==7 JOULE.
x= 1250 Fh
Ly equation (54) we find
1250
¢ = ———— = .06004
2x 10409
sinh
From tiguce.3-—— "LO 12:
p
Substituting these values of x,, y2, and sinh ¢ in equation (58)
p
we find
S456 bts
To find s, the same procedure is followed.
Kee OPE.
i= 125 it:
sinh ¢ = 1.0012 -
d

The value of sinh ¢ is the same for the two parts of the cable only

for the middle of the span where x,=—x,. This greatly simplifies
calculations. It is found that
S/—1257 th

* Mark’s Handbook of Mechanical Engineering, page 1160.

LocGcGING SKYLINES 19

The length of cable required for this span of 2500 feet is then
s, +s,==2715 ft.

We may now compute s,, and sy, using equation (53).

So—414 Et.
Seo 024 Tt.

The load which satisfies the above conditions if found by
equation (46).
W=5.38 (414+5524)—=31,946 Ibs.

To find the tensions at the two ends the constants yo, and yoz
niust be found, using equation (42).

Yo,==10,419.
Yoo=11,784.

Then by equations (51) and (52) the tensions are computed.

T=5.38(10,419-+125)=56,726 Ibs.
T—=5.38(11,784+-750)=68,203 Ibs.

Perhaps attention should be called to the fact that in this
calculation no use has been made of the factor of safety specified
above, Arbitrarily a horizontal tension was chosen equal to one-
fourth the breaking strength, and this gave a value for the -con-
stant ¢ upon which is based the calculation of the load and tensions
as just given above. The factor of safety comes out 3.29 instead
cf 3.5, It is apparent that, in using the horizontal tension recom-
mended in Mark’s Handbook, too small a factor of safety is pro-
vided for. Reference will be made to this discrepancy later.

In order that we may know where the load produces the great-
est tension, the computation must be made for several different
positions of the load in the same manner as above. In carrying out
this computation one of three proceedures must be followed:

(1). We may assume the exact position of the load measured
horizontally and vertically from one end, and the hori-
zontal tension of the cable as was used above.

(2). We may take the length of the cable fixed (say, the
length found above for the load at the middle, 2715
feet), and assume the horizontal tension the same, one-
fourth of the breaking strength.

20 TENSIONS IN TRACK CABLES AND

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LoGGING SKYLINES pai

(3). We may take the length of the cable fixed, and assume
the load the same as we found above for the middle
point of the span.

Of these three proceedures the first is the easiest, but of no
value, because the length of the cable found in this manner will be
different for each position of the load. In practice the length of
the cable is generally constant while the load is being transported.
(Although in some logging operations the cable is let down, given
a big sag, in picking up the load). The third method will give the
most useful result, but is the most difficult to carry out. In fact,
it was found impossible to make this ¢alculation until, by the second
proceedure, the tensions and positions of the load had been found
for a number of points.

The exact proceedure by the second method is as follows:

(1). Assume a value of c, say 10,409.

(2). Assume certain values of x, and y,, which seem to be
reasonable according to one’s best judgment.
(3). By equation (58) compute s,.

(4). By the same equation compute s,, using the same value of
c, and values of x, and-y,, which may be found by the
following relations,

x,+x,—horizontal span.
y2—y,—verticle height of higher end above lower.

(5). The sum of s, and s, so found should be the length of
cable 2715 feet. If it does not, new values of x, and y,
must be assumed and the same proceedure gone thru
with again.

It is thus seen that this method is one of “trial and error”
which involves a vast amount of work. ‘The computations were
carried out for seven positions or points in the span and the results
plotted in the curves of figure 4. Abscissas for all the curves are
values of x,, i.e., the horizontal distance of the load from the lower
end support. In curve L, the ordinates show the load necessary
at any point in order that the horizontal tension may be constant,
56,000 Ibs., giving a constant value of c of 10,409. ‘T, shows the
total tension in the cable at the lower end fastening, and T, the
total tension at the upper end for any position of the load.

a2 TENSIONS IN TRACK CABLES AND

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LOGGING SKYLINES 23

Curve L shows the rather surprising result that the required
load is a minimum at the center of the span. And since the load
is larger near the ends, the tensions in the cable are larger also for
the load near the ends. The curves of this figure are of little prac-
tical value except as a guide in selecting values of the coordinates
to be used in the computations made for the curves of figure 5.

In computing the data for the curves of this latter figure, we
assume :

(1). That the length of the cable is constant and equal to
27 15 ‘feet.

(2). That the load is constant and equal to 31,950 lbs., which
is the load (approximately) at the middle of the span
as first found.

The proceedure followed is one of trial and error as described
in method 2, but with this additional requisite: s,,+-s,. must equal a
certain constant value, viz., 5938, which is obtained by dividing the
load by the weight per unit length of the cable, 5.38 Ibs. /ft. This
makes the computations for these curves twice as laborious as for
figure 4.

As in the former figure, the abscissas of figure 5 are values of
x,. Curve Ty, shows how the horizontal tension varies with the
position of the load. Curves T, and T, give the tensions in the
cable at the lower and upper ends respectively. ‘The form of all
these curves might have been predicted from those of figure 4.
‘he maximum tension at the higher end occurs when the load is at
the center of the span. The maximum point in the other two curves
is slightly displaced toward the lower end.

Although these curves give the relation between the position
of the load and the tensions for a special case, they may be safely
taken as a guide in predicting in general the conditions of maximum
tensions in the cable. It is evident that if the vertical distance be-
tween the end supports of the cable are less than the value taken
above (one-fourth of the span) the tension curves will be more
nearly symmetrical, and when the two supports are at the same
level they will be perfectly symmetrical about an ordinate thru a
point corresponding to the center of the span. Hence for all cases
in which the difference in level between the ends is one-fourth the
length of the span or less, the maximum point on the tension curves
will occur at a position corresponding to the center of the span.

24 ’‘ENSIONS IN TRACK CABLES AND

For greater differences in height between the ends, the maximum
point may be displaced a little toward the lower end. But as the
curves are rather flat on top, no great error will be involved in
assuming that the maximum tension in each branch of the cable
occurs when the load is at the center of the span. ‘This conclusion
is of the greatest importance, for as it has been shown the calcula-
tion of the tensions is the simplest when the load is at the center of
the span, that is when x,;=x,; and when this computation is made
we may be certain that the maximum tension is known.

Futhermore, the conclusion stated above enables us to examine
the most favorable conditions for the use of a track cable. As was
previously stated, the use of a horizontal tension of one-fourth the
breaking strength is inconsistent with a large factor of safety.
Mark’s Handbook of Mechanical Engineering recommends a fac-
tor of safety of 4. This is absolutely impossible with a horizontal
tension of one-fourth the breaking strength. The effect of the sag
and the horizontal tension on the load and factor of safety is
shown in table 1.

Table 1.
[Sag “Horezontal| “Load | ee Fees | Factor |
Tension. | | |. fot
i | Safety _|
56,726 68,200 3.29

|
1/10th | 1/4th | 41,800 | 63,150 | 70800 | 3.17
1/20th | 1/5.25th | 30,602 | 43,300 | 56,000 | 40 |
|

|
| 1/20th | 1/4th | 31,946
:
1/10th | 1/5.25th | 33,700 | 47,900.) »56,0008 | 20rem

With a sag of one-twentieth the horizontal span and a hori-
zontal tension of one-fourth the breaking strength, the factor of
safety is 3.29, which is lower than is advisable to use. By using a
greater sag, one-tenth of the span, and the same horizontal tension
the load will be increased 30%. But this decreases the factor of
safety to 3.17. By keeping the sag one-twentieth of the span and
making the factor of safety 4 the load is reduced by 4.2 per cent
and the horizontal tension to a little less than one-fifth of the break-
ing strength. If a greater sag is used, say one-tenth of the span,
tne load is greater by 5.25 percent, and the horizontal tension is
the same.

LocciInc SKYLINES 25

Hence if it is desired to use a factor of safety of 4, the sag
should be about one-tenth of the span and the horizontal tension
one-fifth of the breaking strength. If it is necessary to keep the
sag small because of the topography of the ground, the horizontal
tension should be made as small as possible and still keep the load
at a sufficient height above the ground.

It will be observed that in the above discussion no considera-
tion is given to the bending stress which the cable undergoes in pas-
sing around. pulleys. Of course this is an important feature of
cable transportation and should be given proper weight in the de-
sign of a track cable, but it is not a part of this particular problem
which treats only of stresses due to loading.

IV
Directions for Computing the Load.

In conclusion the proceedure for finding the safe load that a
track cable will carry will be outlined. It is assumed that the fol-
lowing data is given:

(a) Breaking strength of the cable.
(b) Linear density.
(c) Horizontal span.

(d) Vertical distance between end fastenings.

(1) ‘Take the horizontal tenions to be one-fifth of the break-
ing strength and compute the parameter, c, by
Th
aa Cup)
Ww
(2) ‘Take the sag at the center of the span to be one-tenth,
(or as near that as feasible) of the span. This gives the
ordinate y, of the lower end. The ordinate y, of the
upper end is y, plus the vertical height between the twa
ends. The absicassas, x, and x,, are equal and each equal
to one-half of the span.

(3) Compute the length of each part of cable between the
load and the lower end, and between the load and the
upper end respectively, i.e., s, and s,, by

26

TENSIONS IN T'RACK CABLES AND

sinh
a ee (58)
, p
(4) The constant s,, and s,. may now be computed by
y S
So = — cothd — — | (53)
2 2
(5) The load is given by
W=w (S01+S02) 4 (40)

(6) The constants y,, and y.. may be found by
Yoo=V c-+s",) (42)

(7) The tension at each end, T, and T., respectively, may
be computed by

T=w(y-+Yo) (51) (52)

using y, and y,, for T,, and y5 and! y,, tor i. somec
y, is taken greater than y,, T, is the greater tension and
the maximum for any position of the load.

Altho the computations required by the outline above are some-

what longer than those called for by the approximate formulas

given in handbooks, they are by no means laborious, and have the

advantage of giving exactly the load which may be carried and the
maximum tension involved.

SUMMARY.

(a). When a cable is loaded at one point, the conformation
which the cable assumes is that of two arcs of the same common
catenary, with the point of intersection at point of loading.

(b). The equations of these two arcs have been derived.
"‘rhey involve three constants or parameters: (1) c, which is the
parameter of the common catenary, of which the two arcs are part,
‘and which depends implicitly upon the horizontal tension and linear
density of the cable; (2) s., which is equal to the length of an arc
ef the common catenary extending from the lowest point of the
common catenary to the point of intersection (the loaded point)
of the two arcs which represent the actual conformation of the
cable, and which may be called the “fictitious arc’? and depends
upon the load and the linear density of the cable; (3) y,, which is
dependent upon the other two parameters, and is the vertical dis-
tance of the loaded point above the directrix of the common
catenary.

(c). Expressions for the horizontal tension, vertical tension
and total tension at any point in the cable have been derived.

(d). It has been shown that in either one of the arcs the
point of maximum tension is the highest point, i.e., the point of
fastening at the end, and that in most cases the higher end of the
csble has the greater tension.

(e). It has been shown for a special case that the maximum
tension occurs in the cable when the load is at the center of the
span, and. that the point of maximum tension is the higher end of
the cable. Furthermore, it has been shown that very probably the
maximum tension a/ways occurs, in a catenary loaded at one point,
when the load is at the center of the span.

(f). A method of computing exactly the load and tensions in a
given track cable has been outlined in section IV. To the logging
engineer this will probably prove the most interesting and useful
part of the paper.

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