P : concentrated load -
SECTION 3
i4
Suspended length between supports, L:
вн where‘“d” is 5% of “s”or less (approx) -(7)
8? 4d? ee 3 7 è
L=2 я += where “d” is more than 3% of “s”
r a
(approx) — —— — — — — === == (8)
_ 4d/ 16%. (ad 16d
7 sd “a 3 EE ке)!
(Ехасё) — — — —— —— (9)
_5 16d* 8 4d 16d*
L 5 1:18 tag” SCH = ) -(9a)
E 21/17, ody) 4@
Liss | 7 y В + log, 7 +1 5) — — -(8b)
В | ss
L= = Vian: a * SeC* a+ log, (tan q + seca) | - —(9c)
Based on the exact formula (9), the “Span Factors”
below will be of use:
w — uniform dead load weight per foot s/d Span
p —uniform live load per foot Е т
P — concentrated load | % Fraction actor an. Sec.
m ==distance from left support to “P” 20 1/5 1.098042 8 1.2800
Г, = suspended rope length, whole span 15, о 3/20 1.057039 8 1.1662
œ — angle of maximum rope stress -1/ ‘| 1.11803
Log — Napierian or Hyperbolic Logarithms 11 1/9.09 1.03 1239 44 1.0925
A — metallic area 10 1/10 1.026055 ‚А 1.0770
à = modulus of elasticity 833 | 1/13 | гово | 3938 | 106%
Arm : . . . .
elongation 8 1/125 | 1.016867 | .32 1.0500
Equation of curve with origin at left support and xx
: | , 7.7 1/13 1.015600 3077 1.0462
axis through supports: 71/7 | 1/14 11.013430 | 2857 | 10400
5 | 6-2/3 | 1/15 1.011722 2667 1.0349
nu ; A dx 6-1/4 1/16 1.010316 25 1.0308
+ y= §—x)—— |
7 Ir y= (8-2) —- (1) 5.838 1/17 [1009139 | 2353 | 1.0273
a 4 5.55 1/18 1.008154 2222 1.0244
| а a _ wWæ(s—x) 5-1/4 1/19 11.007323 | .2105 | 1.0219
| Хде, Aw OT Y= — (a) 5 1/20 | 1.006626 | 2 1.0198
4 1/25 1.004228 ‚16 1.0127
rue 19 | 38% |199 | 329 | 9
Horizontal cable tension, H= nn — 3 1.00
© vension, 2d 8d (2) 2.88 1/35 [1.002152 | 1143 | 10065
2d Ad 2-1/2 1/40 1.001615 . 1.0050
tang = ——=—— — тонн он к. (3) 2.22 1/45 1.001264 0888 1.0039
Los - 2 1/50 |1.001008 | .08 | 1.0032
та . .
sec à 1: о a ООО ООО (4) (Fixed Ends) Inclined Span
| a) > ° iD
ax N— a= 5-55 — -——-(10)
16а? + ` —
. T=H-sec x -н)/1+ a---------- (5) * ] | Piz а 20
- q x
vg r : 7 à _ md _ md (11)
M=d- = — (6) o |“ 1+ да):
FIG 2
wm?
2 a 12
H=5 i (12)
tan e a (13)
m
2
sec 21:38 AA - (14)
tang = 22 — dm) mm ee = == — = (15)
secf = V1 + tan*s Ee == = == — — — (16)
Т= Н sec a, at high support — —~ — — — — — — (17)
То== Н + весВ, at low support — — — — nn. (18)
Equation of curve, origin at left support, xx axis
thru origin:
4dr ÇÎ
y= 5 e-m)+e-
=
—
Lowest point x, where m =
tan 6 = -(19)
РН. Ва
8 8 a
51+ aa - tan 8) - - -(20)
Approximate formulas for cableways on inclined
spans:
== = о === не = = = — — —› (21)
Т = Н «вес а 8 ws TT TT TT TT TTT (22)
dee Esa) mmm mmm (23)
de "WMA aa (94)
os 2H
m= x : р) __. (25)
Concentrated Loads on Suspended Spans. (Fixed
Ends) Horizontal Span.
o s | pulls PH -
<> > TT — A (26)
1 Р
: X | H= 5d + (26a)
FIG, 3
Load “P” lbs. Cable at “wo” 108°
Ibs. per ft. ú= 32H (27)
а, = Fe неее (28)
tan a = a EE (29)
| ‘ В
Approx. cable length: Б = 2 | rt d+ ae (30)
sp£ +d
For cableway work, the supporting rope is hung
so that the deflection with “P” at the center is a
predetermined amount, “d”. The problem (for fixed
ends) is to determine at what deflection, “d” to erect
the rope to result in the desired “d” with “P” at
midspan.
Find “H” from Eq. 26, then
| 5
Se: _ se
; — — do + = —= — — — — (31)
Za dE
- 20 bk—2a) — — —(32)
FIG, 4 a
2H
= НН 33
ws + 2P (33)
Б= Аа? + 1 — — = — = = = = = — — — — = —- (34)
pal BY aa (35)
6 4H
ws ws” 10d”
To=7 sseof=Hoseof= gr 1+ ао (36)
Average tension in rope,
T= He Heoseep + 1) ——— (7
> Length on Ground, Z, — Lo —
SECTION 3
Another more accurate method of figuring “d,” is
from:
do=dt NDd——— — + — — — — — — — — — — — — (38)
ISAL
d= НН
A 16(5r —24r) (39)
where r =d/s
167? , 16d?
Арал 5 (1+ hy =e чо) ——— (40)
H
A=" — === 40
b= dns (40a)
Ps
rte (41)
(for Concentrated Load “P”)
ог И, = неее (42)
(For a uniform superimposed load “p” Ibs. per foot
over. entire span. );
or H= BE (1-4) == === == === (43)
(For a partially loaded section of “p” lbs. per foot
centrally located, similar to Figure 15, Page 299.)
Me
Suspended length, unloaded (dotted curve),
8d?
Lo 3s
. Talo
Hlongation under stress of T4 = A = ЧЕ о (45)
where “A” is metallic area of rope, square inches;
and В, is Modulus of Elasticity of rope, Section 8,
page 11.
— — — — (46)
Further correction for temperature from Equation 86.
fr
A
в -1
a”
x
©
— F —
FIG. 5
(ws + 2P)Y'mn
ms ns se нее ен (47)
2H ws + 4P mn |
tana MOD — — —]—l48)
a s ;
a т
< =
a a
o he § —
FIG 6
tan Balad —— ————— — (49)
Peg ws”
Le sin( a + a-+6) 8dcoss TT (50)
15
SECTION 3
16
nu + —_—
я о | J |
a ON — # Ton
DO |
TE =
FIG 7 +
P cosa ws
T= sin{ « + f) + 8d cos 6 (51)
P cosg wm
7 sin(a + 8) 2d o (52)
P cos « wmn
== +
* sin(a +8) 2d cos
a” 18 + or — a9 vertex 10 halsw 9 or
т wri ue RT
above the lower support.
The suspended length of rope for a single load
“P” at any position, “m” from left support and with
a difference in elevation of supports of “D”, Fig. VII,
as given by F. C. Carstarphen* 1s,
Pm | ws D |
=g+ „еее Во —
L=s+ os + ws) + 473" 2s (54)
- which for “P” at midspan, or where m=
D |
Lm =s+ 17,P+00 + gat == — — — — — — (54a)
and the pt with empty span (“P” removed):
+
Im S $+ 57 XT at TT 54D)
**" Acrial ae by F. C- Carstarphen
presented before A.S.C.E. November 2, 1927.
Counterweighted ropes
(Constant Tension)
The rope supporting a load does not present a
continuous curve, but consists of two parabolas. The
curve of the path of a moving load is continuous and
may be expressed by:
la $ ——
ro
; pet} 5
7 гор
| | © ee
À; Te 7 >
FIG. 8
y= Um PI tan ————— (35)
if a point on the curve to the left of “P” is con-
sidered, the equation of the left parabola is:
We, , Pnx
n= AT + + atan = == тнт = (56)
and for the right parabola:
WIR 4 Pmz, | dan A e
у *— + tan 8 (57)
with “P” at midspan, m=n= 5 then,
_ws+2Ps, D
BT + a TT = (58)
I the supports are the same elevation, the last
term in equations 55-58 becomes zero.
The following equations are derived from Egs.
55-58:
я y f f я
но 7 108
7 а, = - —— — (59
EN | ° ST cos« (59)
x rf
w omy FIG. 9
^^ °
“5 | a f
| a д
7 7 E % of, #
x я = wre 1
5 * OT cosa (60)
x {| Fa
FIG. De
AL
К: О f HT
ut A + 2P
r ws‘ + 2Ps
yon dp=—m— —. —(61)
, (2) ST cos a
—
i A pe ra —
FIG. 70
—(62)
Hy —— ee. 7
Fig Ma
k—————— $ — |
a T Za e
“7 2 5 я”
| de dm & ( .. Pp,
F 1 ‚ mn (ws + LU
® de ens 69
|| 2sT cos o
m ñ
i Хх — A 7—
FIG. 1
NU 5
HH 7
A д %
f WEszas + 2 Pm
д, = Léa FETE (64)
25 T cosa
_ | (lett side)
— X mm Zn wxes + 2Pxn
© FIG. Ha de = ут -(65)
2s T cos «
(right aide)
In the formulas, 55-65, on preceding pages, it is
not always easy to obtain the value of H or cosa,
(H = T+ cos a) so that the value of “T” is used.
This substitution of “T” for “H” would reduce Cd”
by a maximum of about 5%.
Tiger Brand Bridge Cables — The San Francisco-Ookland Bay Bridge
Cable spans may be divided into two general classes, An-
chored Spans, and Counterweighted Spans. In each of
these divisions, we find it necessary to solve for stresses and
deflections of uniformly loaded spans and also of spans
supporting one or more individual concentrated loads. Ît
is, therefore, necessary to analyze the conditions of each
problem carefully and the following points must be con-
sidered:
1. Horizontal distance between supports.
2. Difference in elevation between supports.
3. Maximum ollowable deflection, measured vertically
from chord to cable. |
4. length of cable between supports.
5. Weight per foot of cable, to which must bé added in
certain cases the additional weight imposed by snow
and ice,
ó. Maximum load to be supported by the cable.
a, Load uniformly distributed over the length of the
span.
b. A single load supported at any point inthe span,
c. Multiple individual loads.
7. Is the cable anchored at both ends or is it anchored at
one end and counterweighted at the other end?
8. Modulus of elasticity in tension.
9. Wind loads on the cable and on the suspended load.
10. Changes in length of cable due to changes in tem-
perature.
Since our purpose is to present means for obtaining results
quickly, we will not give derivations of the following for-
mulas. Computations are simplified by the assumption that
176
-
uniform loading is distributed horizontally, and that the
cable assumes a parabolic arc, For the great majority of
cases encountered in practice, the results thus obtained are
sufficiently accurate. If special cases occur where the ratio
of deflection to span is very large, then the catenary equa-
tions should be applied. These are available in several
textbooks.
The following nomenclature will be used:
A = == Net cross sectional area of cable.
a — Horizontal spacing of loads.
min-1) ulu-1)
b =, с = — _——
2 | 2
e - == Васе of Naperian system of logarithms
— 2.7182818.
—E — Modulus of elasticity in tension.
6 = Weight of an individual concenirated load.
.h == Vertical difference in elevation of supports.
k = Ratio of deflection to span == — for level spans
and Ws еее for inclined spans.
L; == Length along cable when the cable only is sup-
ported in a span, |
1, — Hypothetical length along cable at zero tension.
| = Length along cable when either a uniformly dis-
tributed load or one or more concentrated loads
are suspended.
m == Horizontal distance from left support to the first
load.
51
à
— —
=
Li
—— maja
~
=
+
nr
I
Number of concentrated loads.
Change in total length of cable per pound of ten-
sion == ——
AE
Horizontal distance between supports,
Chord length of sub-span between load and sup-
port or between two loads,
— Horizontal component of cable tension.
|
Maximum cable tension af left support,
Maximum cable tension at right support.
Erection tension of empty cable in an anchored
span,
== Number of loads to left of xy in a multiple loaded
I
|
span,
Weight per foot of horizontal length of span for
a uniformly distributed load, w == w' sec a.
Weight per foot of uniformiy distributed load
along the cable, which is assumed for purposes of
parabolic curve calculations, as equivalent to uni-
formiy distributed load along the chord.
Weight per foot of uniformly distributed load
along the cable for purposes of catenary curve
calculations.
Horizontal distance from support to xy.
Vertical deflection from support to xy.
Vertical deflection from support at center of span.
A term in the general formula for multiple loaded
counterweighted spans. |
Albha — Angle between the horizontal and a
chord between supports.
Beta; — Angle between the horizontal and a tan-
gent to a cable curve af the left support.
Betas — Angle between the horizontal and a tan-
gent to a cable curve at the right support.
Beta, — Angle between the horizontal and a tan- :
gent to a cable curve at any point in. a span.
Beta, — Angle between the horizontal and a tan-
gent to a cable curve at a load,
Lambda — Change in length of cable per foot of
length, per pound of tension.
Delta — Total change in length of cable == A + Lo,
SECTION
—
© = Theta — Angle between the horizontal and the
chord of a half span,
sec — Secant of an angle —
— cosine
ANCHORED SPANS are principally employed for suppori-
ing electrical cables, for guy lines, for suspension bridges,
and usually for track cables of cableways and reversible
aerial tramways where a single moving load is supported
in a clear span.
When a cable span is erected, anchored af hoth ends, and
a load of any kind supported from the cable, the deflection
increases because of the elastic properties of the cable.
The tension also increases when the load is applied.
It is necessary to select the size, construction, and grade of
the cable, with a proper factor of safety, after having de-
termined the maximum tension in the cable due to dead
and live loads, It is then necessary fo erect the cable at
such a deflection that the maximum safe working tension
will not be exceeded when the [oad is applied,
In the case of cableways with high self-supporting towers,
the cable tension and deflection may be affected by yield.
ing of the supports. A complete study of such a span in-
cludes the application of the theory of deflection in framed
structures, but such a special condition does not come
within the scope of this handbook. In all cases we will as-
sume that cables are anchored to rigid supports or immov-
able ground anchorages.
The determination of the proper erection deflection and
tension involves the use of the modulus of elasticity in ten-
sion for the particular construction of cable which is being
used,
lt is well known that the modulus of elasticity ranges be-
tween 28,000,000 and 30,600,000 for structural steel, but
the modulus of elasticity of a wire cable, considering the
cable as a whole, has various values depending on its con-
struction, and also on the work that has been put into it.
The modulus can be appreciably increased by a prestress-
ing operation. This is frequently done to bridge cables. In
the case of track cables carrying rolling loads somewhat
the same effect is secured after a period of operation, as
most of the structural stretch is removed. See Moduli of
Elasticity page 167, and Prestressed Strands and Ropes
page 16.
LEVEL SPAN — UNIFORMLY LOADED — ANCHORED
ao
S i
Figure 1
177
When the tension is known, the center deflection is found
from:
or
WS“
Ye == at (1)
and the deflection at any point in the span is:
wx[s-x) .
= | (2)
When the center deflection is known, the horizontal com-
ponent of tension is found from:
2
= (3)
When the deflection at some point other than the center of
span is known:
— WXls-x)
+ — y (4)
Р == | 5ес В, (5)
The cable slope at any point in the span is:
tan == (4 — x) (6)
At either support the cable slope is:
Ау, ;
tan 8; or 8g —— (7)
also tan 5, or Ps — м (8)
When the tension is Known, the length of cable i is:
LLorL—s 4 2 “approx. ) (9)
517
When the deflection is known: L or L —
т ,, 01 |
s En + 16k? — log. (4k + + 16k? ) no)
An easier formula, giving “ont approximate results is:
Lorl=s(1 +55 e ZB 20 es) (11)
— Sufficient accuracy can be od for many of the cases
encountered in practice, by contracting formula (11)4o:
8.
Lorl=s (1 +51)
In determining the erection tension for a uniformly loaded
span, the values of Ly and t. must satisfy the equation:
|= НЕ 113)
——— == РО(+ — 4
AE |
By substitution of (9) for L, in (13) and using corresponding
values of w and t.,
was?
Pt L— (Pt + 5) — ,
fe + Pe = 54 te”
This equation can be solved for t., using the trial and error
method.
(14)
EXAMPLE:
178
A 750,000 C.M. bare, hard drawn, stranded copper cable
is to be supported across a river. Supports will be af the
same elevation and 1350 feet apart. The copper cable is
9981 inches in diameter, and weighs 2.325 pounds рег
foot. Conditions require consideration of a coating of ice
1%” thick on both conductor and messenger, plus a hori-
zontal wind load of eight pounds per square foot on the
(12)
projected area of the ice coated cables. We are limited to
a maximum center deflection of 75 feet.
(a) What are the specifications of the necessary messen-
ger cable, assuming the same ice and wind loads?
(b} What is the cable slope at supports and at the quarter
poinfs of the span?
(c) What is the erection tension and deflection for the
messenger strand only, assuming there are no ice or wind
conditions at time of erection?
№ is necessary to assume the diameter of messenger strand
to figure the loading on the span. It may then be necessary
to revise the figures if the first selection does not prove
suitable. We will assume a 7%" diameter strand weighing
1.581 pounds per foot.
Copper cable + ice = 3.240 pounds per foot
Messenger strand -}- ice — 2.421 pounds per foot
Total vertical load — 5.661 pounds per foot
Horizontal wind load on both cables — 2.582 pounds
per foot
Total resultant load — 6.222 pounds
per foot .
; 6.222 X 1350”
Th f 3 t — — 1 ,
en from {3) | 87 75 8,200 pounds
6.222 X 1350
Th f == -
en from (8) tan fi 2 18900
== ‚2222, В, == 127-372
Then from (5) tr — 18900 X sec 127.32
— 19365 pounds
Then from {6} when x — 337.5 feet
6.222
tan Ba —— 18900 (6735 — 337.5) = Л 1 i i,
Ba — 6°-20°
With à factor of safety of 4, the required breaking strength
will be 4 X 19365 — 77460 pounds. Page 117 shows % ”
diameter, 19 wire, Extra Galvanized Extra High Strength
Strand has a breaking strength of 79,700 pounds, and will
be satisfactory for the purpose intended. From Page 117
we find W to be 1.581 pounds per foot. From Page 166
we find A to be 4444 square inches.
6. 222° X 13307
24 x 18900?
In order to set up (14) in convenient form, first calculate
the following:
1381110
74444 x 21,000,000
Pt 4 s = (.0001458 < 18900} + 1350—1352.756 ft.
= 1361.110 ft.
(9) L= 1350 +
— 0001458 feet per pound
1.5812 X 1350
wis?
or = 57 — 256,240,000 |
Substituting these values in {14),
0001458 t, + 8.354 — 236,249,000
Î
E
The following shows the results of a series of tentative
computations for assumed values of t. until the above
equation is satisfied {the values in the last two columns are
equal).
to .0001458t, + 8354 + _226,740,000
A
5100 ‚744 9.098 9.852
5200 758 2.112 9.476
5290 77 9.125 9,157
5299 773 9,127 7.126
f. — 5299 pounds.
From [1] y, — 1.381 X 13507 — 67,97 ft.
’ 8 x 5299
| 2 3
From (9) L; = 1350 4281 > 1350 1350 126 61
24 X 5299:
Therefore: |
la) One piece %” diam, 19 wire Extra Galvanized Extra
High Strength Strand with sockets attached so as to
give a length of 1359.13 feet center to center of
SECTION
(b) Maximum cable slope at supports — 12°-32".
Maximum cable slope at quarter points of span —
6-20.
{c} Erection tension = 5299 pounds
Erection deflection = 67.97 ft.
LENGTH AND MAXIMUM TENSION
The following table gives factors for obtaining maximum
tension Y at the supports of a uniformly loaded level span
when w, the weight per horizontal foot and s, the hori-
zontal length of span, are known. See column 2. The close
relation between the parabola and the catenary is shown
by a comparison of the values in columns 2 and 3. Column
3 gives the factor for obtaining t when w”, the weight per
foot along the cable, and s is known. The length of a uni-
formly loaded level span, based on a parabolic curve, can
be obtained from the factors in column 4. If the span is
inclined see formulas (24) and (25).
The factors in column 4 can also be used for the catenary
for k ratios up to 0.12 with an error less than 0.02%, and
supports. for k ratios as high as 0.20 with an error of only 0.1%.
CABLE LENGTH AND MAXIMUM TENSION
FACTORS FOR MAXIMUM TENSION CABLE LENGTH FACTORS
When weight of load per foot When weight of load per foot To get length of cable, muitiply
Ratio of center deflection to of span, w, is known of cable, w”, is known total span by factor below
chord length of span k Y = ws X factor U = ws X factor LL = 5 X factor
COLUMN 1 COLUMN 2 COLUMN 3 COLUMN 4
01 12.510 12.51 1.00027
012 10.429 10.43 1.60038
014 8.942 8.94 1.00052 -
016 7.828 7.83 1.06068
018 6.962 6.96 1.00085
02 6.270 6.27 1.06107
022 5.704 5.70 1.00129
‚024 5.232 5.23 1.00153
026 4.834 4,83 1.00180
‚028 4.492 4.49 1.00209
03 4.196 4.20 1.00240
032 3.938 3.94 1.00272
‚034 3.710 3.71 1.00307
036 3.508 3.51 1.00344
038. 3.327 333. 1.00384
04 3.165 3.17 1.00425
042 3.018 3.02 1.00468
‚044 2.884 2.88 1.00514
‚046 2.763 2.76 1.00561
‚048 2.652 2.65 1.00611
‚05 2.545 2.55 1.00663
055 2.327 2.33 1.00801
.06 2.142 2.14 1.60952
‚065 1.987 1.99 101115 —
07 1.854 1.86 1.01291
075 1.740 1.74 1.01480
08 1.640 1.64 1.01681
‚085 1.553 1.56 1.01894
09 1.476 1.49 1.02119
095 1.408 1.42 1.02356
179
CABLE LENGTH AND MAXIMUM TENSION (Cont.)
Ratio of center deflection to
chord length of span k
FACTORS FOR MAXIMUM TENSION
CABLE LENGTH FACTORS
When weight of load per foot
of span, w, Is known
t = ws X factor
When weight of load per foot
of cable, w”, is known
t = ws X factor
To get length of cable, muitiply
total span by factor below
Li = 5 X factor
COLUMN 1 COLUMN 2 COLUMN 3 COLUMN 4
10 1.346 1.36 1.02604
105 1.291 1.31 1.02865
11 1.242 1.26 1.03136
115 1.197 1.22 1.03419
17 1.155 1.18 1,03713
125 1.118 1.14 1.04021
13 1.084 1.11 1.04333
135 1.052 1.08 1.04659
14 1.023 1.05 1.04995
145 0.997 1.03 1.05341
15 0.972 1.005 1.05711
16 0.928 0.964 1.06455
17 0.890 0.930 1.07236
18 0.856 0.900 - 1.08063
19 0.826 0.874 . | 1.08919
20 0.800 0.853 1.09822
21 0.777 0.834 1107899
22 0.757 0.820 1.11706
23 0.738 0.807 1.12701
‚24 0.722 0.796 1.13724
25 0.707 0.788 1.14778
180
doen
=
SECTION
INCLINED SPAN — UNIFORMLY LOADED — ANCHORED
Figure 2
The following formulas give the increments of deflection
and slope due to inclination of the chord. "Down" slopes
are usually considered as plus values and “up” slopes os
minus values. SE,
ws” h
tan a — uE : (16)
5
At any point—
wxls-x]
— +
oF x tana | (17)
tan В, == о Мапа (18)
tan Вы == — tan a (19)
ВИ ws
tan 2; (at any point) — 7 —X + tan a (20)
A
=
y
When center deflection is known:
1 (21)
Low point of an inclined span occurs when tan Ba — 0
5 t
e X== —-—— fane
2
When deflection af any ofher point is known:
wx{s-x}
'= Sx tam a) 22
Ко = { 5ес В) © | (23)
1" —t sec Ga *
To find the lengths of cable in an inclined span formulas
(9) and (11) are modified:
o Same |
L, or e + h* + wa {approx.) {24)
L; or Le + е( +516 e + (25)
lt will be seen that the solutions for inclined spans are quite
similar to those for level spans,
18}
LEVEL SPAN — SINGLE LOAD AT CENTER — ANCHORED
No»
{'
i {= tv
Are tan y
Fi gure 3
The deflection produced by a concenirated load suspended
midway between two fixed points A and B forms two equal
sub-chords AC and CB. The cable assumes two catenary
arcs which intersect at С. The following formulas are, how-
ever, based on the parabola, as the difference in results is
negligible,
The center deflection is found from:
— Gs ws” __s(2G- ws)
Yo == "4 + 8t 7 8t (26)
and + — s(2G — ws} 127)
Bye
f'—tsec B; =t sec B,=!" | (28)
tan 8; = e == tan fs (29)
Example: A rolling load weighing 2000 pounds is to be
supported in a level span 2000 ft, long by a cable an-
chored at both ends. The deflection must not exceed 83
feet. No wind or ice conditions.
(a) What are the specifications of the cable?
_ (b} What is the maximum tension in the cable?
(с) What is the slope at the supports with the load at cen-
ter of span? |
td) What is the cable length between supports, with no
rolling load on the cable?
tel What is the erection tension and erection deflection of |
the cable?
lt is necessary to assume a size and grade ef cable for the
calculations. If the first selection does not prove suitable,
the calculations must be revised. We shall assume that a
1%” diameter Standard Grade Locked Coil Cable will be
suitable.
Since this is alevel span, a —o and w == м”
w = 3.16 pounds per foot (from page 111), A— .8567
square inches (from page 166).
2000 (2 X 2000 | 3.16 X 2000)
From {27) + — 8 x 83
= 31,084 pounds
2000 + {3.16 X 2000)
2 x 31084
‚1338 В) == 7°-37'
31084 X 1.0089
31360 pounds
The maximum cable length occurs when load is af center
of span,
From (29) tan 8, —
|
From (28) r
182
3 ——
$“ = (=) + y? =a] 000% — 832 = 1003.439 4.
83 |
tan © = ооо == 083 9—4AÀ 45
5 3
> L==2 (a + lero Y
3.1621 000 cord 45)
— 2 C 003.439 — .
24 x 310847
— 2007.730
In order to set up {14} in convenient form, first calculate
the following:
2007.730
P= 8567 Xx 19,000,000 — — 0001233
Pt As — 0001233 X 31,084 + 2000 — 2003. 833
wèss _ 3.162 x 20003
24 7 24
Substituting these values in (14)
0001233, - 2007.730 — 2003.833
_ 3 328,533,333
+2
== 3,328,533,333
The following shows the results of a series of slide rule
— computations for assumed values of t, until the above
equation is satisfied {the values in the last two columns are
equal}.
3,328,533,333
te ‚0001233,
3.897 AI
22,000 2713 6.610 6.877
22,300 2.750 6.647 6.693
22,360 2.757 6.654 6.657
te == 22,360 pounds :
3.16 « 20002
from (1) y. — 3 22360 — 70.66 ft.
2 2
from (9} Li == 2000 + 3.167 x 2000° __ 2006.66 ft.
24 x 22360:
(a) With a factor of safety of 3.2, the required breaking
strength will be 3.2 x 31360 = 100352 pounds — 50.18
tans, The breaking strength of a 1%” diameter Standard
Grade Lotked Coil Cable is 54 tons [from page 111}.
Therefore, this size cable is satisfactory, and our Locked
Coil Cable is the most suitable construction where rolling
loads are to be handled. If the proposed installation is
temporary, or If first cost of the cable is a prime consider-
184
INCLINED SPAN — SINGLE LOAD AT ANY POINT-— ANCHORED
Ls $
>
pla (5X) ———
N
Figure 6
Formula (32) can be applied to inclined spans by adding
r
hx which becomes when X =. Then, for inclined
S
spans:
(5-х) hx
x {ws + 2G)? |
ws t+ + = (34)
у 2t (ws? + 46 Vx 5-х) 5
MULTIPLE LOADS
IN ANCHORED SPANS
Multiple loads in anchored spans are seldom encountered
in practice. However, the subject is important enough to
merit some attention. When speaking of multiple loads, it
will be assumed loads are equal in amount and spaced
uniformly. |
The loads should be placed symmetrically about the center
line of the span to compute the maximum tension or de-
flection in the span. Use formula {52}, page 188, to deter-
mine the deflection and formula (54) to determine the
maximum fension. To determine the length along the cable
at maximum tension, consider the loads as stationary in
the position stated above and treat the lengths of cable
between supports and the first load, and the lengths be-
tween loads, as separate spans. After this length, L, has
been determined, the erection tension, deflection of empty
cable, etc., aré calculated by ihe trial method ina similar
manner to that for a single load in an anchored span.
wT TTF ЦРО “Y Ow Ww YT
wow Wr My PO CTO YO wr WOW OW О Бр о Y "TT Y VW TF ЦР
SECTION
LEVEL SPAN — UNIFORMLY LOADED — COUNTERWEIGHTED
Figure 7
The tension and deflection of either an anchored or a
counterweighted span are thé same, under the same con-
ditions of loading, when the cable supports a uniformly
distributed load. However, an important difference occurs
when the live load is removed. In the case of an anchored
span, the deflection and length of the cable remain con-
stant, except as they are affected by the elastic properties
of the cable, backstays, and supporis. The tension, how-
ever, decreases when the live load is removed. Comparing
this performance with a counterweighted span, we find
that the tension remains constant when the live load is
removed, while the deflection and length of the cable de-
crease in proportion to the change in loading. These are
the effects due to equalizing the moment-sum of all forces
for any origin of moments.
o'w't,
The same comparison holds true of spans supporting one
or more individual concentrated loads, when the loads are
so placed as fo produce the maximum deflection.
The use of a counterweighted track cable for rolling loads
results in a constant angle under the load, the angle whose
‚ ©
tangent is — at all points of a span, Álso, it produces a
smaller angle at each support than would be the case with
an anchored span. These two factors are of definite advan-
tage in the design of aerial tramways having intermediate
supports,
Apply formulas (1) to (12) inclusive, page 178, under “An-
chored Spans.”
INCLINED SPAN — UNIFORMLY LOADED — COUNTERWEIGHTED
< 8 =
E 5
2
4+— X ——
A a
Bs у- Ус h
В:
B32
Figure 8 Pow
Apply formulas (15) to (25) inclusive,
page 181, under "Anchored Spans.”
185
LEVEL SPAN — SINGLE LOAD
AT CENTER — COUNTERWEIGHTED
ton
+
Arc tan e
Figure 9
Apply formulas (26) to (29) inclusive, page 182, under “Anchored Spans."
INCLINED SPANS— SINGLE LOAD AT CENTER — COUNTERWEIGHTED
Apply formulas {30} and (31) inclusive, page 183, under “Anchored Spans.”
LEVEL SPAN — SINGLE LOAD AT ANY POINT — COUNTERWEIGHTED
Figure 10
In à constant tension span the deflection af the load may
be determined from:
__ Gxis- x} wx{s-x)
+
si 24
Also the deflection of the cable may be determined for any
point In the span, with the load at any point, x; yı being
coordinates to points to the left of G and x» y being coor-
186
dinates of points to the "a of G.
y, (points left of 6) =— (s-m) + (sx) (36)
Y» (points right of G) — {s- xo) 4. * (s-xo) (37)
The cable slope at left support when x; == E 15;
tan 8: = Ss — m) + (38)
The cable slope at right support, when x; == s, is:
Gm WS
tan 62 — ot or (39)
The cable slope at any point between the load and either
support is:
tan Os (points to left of G} =
к) | (40)
G w/s
Fm MG =
(35)
C'w't,
tan Ss (points to right of GC) =="
Sm + (e — +)
When x == m, the slope at and to the left of the load is:
{41}
tan By (sloping to left of G} —
G Gx S
+ TG T x)
Da st
The slope at and to the right of the load is:
tan By {sloping to the right of G} —
Gx w 5 |
The tangent of the angle under the load is equal fo
(42) + (43) == =
(42)
If we take half the difference between the angles obtained
from (42) and (43); the tangent of the resulting angle will
be the slope which a moving load must climb. The maxi-
mum slope thus obtained will determine the maximum pull
on a haulage rope.
SECTION
INCLINED SPAN — SINGLE LOAD AT ANY POINT — COUNTERWEIGHTED
8 . >
<
<
E —— M-———m ——J
E Y Bi
4 к —ы
X2 ————
y! y
А a A 4
4 -h
—]————]]
Figure 11
G
In these formulas, as in all others, we have placed the
higher support at the left-hand end of span, and have
made this point the origin of moments.
For y — at the load — add to formula (35)
x tan a (44}
For y; — points left of G — add to formula {36)
x; tan a (45)
For yo — points right of G — add to formula (37)
хо tan a (46)
The cable slopes are determined by taking the chord into
account as an additional term in the above equations,
tan 2; — at left support — formula (38)
+ tana (47)
tan Py — at right support — formula (39)
— tana (48)
tan $; — points to left of G — formula {40)
— tan « [49]
tan Ba — points to right of G — formula (41)
— tan а (50)
tan 54 — at the load — formula (42) and (43)
— tana (51)
EXAMPLE: A 2,000 pound rolling load is to be supported
on an inclined span 800 ft. long with difference in eleva-
tion of 67 ft. The cable is 134” diameter Standard Grade
Locked Coil; w' == 4.73 pounds per foot {from Page 111).
A == 1.280 sq. in. {from Page 166). The center deflection
must not exceed 18 ft, from the chord.
(a) What is the horizontal component of cable tension with
load at center of span?
—(b) What is the slope of the cable at the higher support
y?
c'w't
(1) with the load at center of span, (2) with the load 100
ft. horizontally away from the upper support and (3} with
the cable unloaded?
(c} What is the center deflection of the unloaded cable?
08375, a —= 4° —47', sec a — 1.0035
tan в —=
7 — rrr
800
w==4.73 X 1.0035 == 4.75
800 (2 x 2000 + 4.75 x 800)
8х 51.5 — 4 х 67
— 43,333 pounds
From {31}, f=
2000 x 400 4.75 x 800
800 x 43333 ! 2 x 43333
— .1507, @, — 8°-34’
From (47), tan В, == =
2000 x 700 A475 x 800 67
From (471, tan PL == 500 x 43333 T2 x 43333 ! 800
= 1680, 8, = 9°-32'
4, 75 300 67
From (18), tan 81 — 3243333 + 300 800
== 1276, В; == 7°-16'
4.75 x 800% 67
8 x 43333 7 9 4227.
From (15), y. ==
(a) 43,333 pounds.
(bt) Slope 8°-34' with load at center of span.
(b2} Slope 9°-32" with m = 100 feet,
(b3) Slope 7°-16" with cable unloaded.
(c} Center deflection 42.27 feet with cable unloaded.
187
LEVEL SPAN — MULTIPLE LOADS — COUNTERWEIGHTED
Figure 12
A cable supporting multiple loads forms a series of para-
bolic arcs between the loads. For many cases encountered
in practice, it will be sufficiently accurate to calculate spans
carrying more than five loads as uniformly loaded spans,
If this is done, the load per foot equals weight of cable
G
plus —.
a
However, the general formula for deflection y, at any point
xy, of a span supporting n loads of uniform spacing and
equal weight, the cable tension being constant, is:
al (2-1) E]
гу ===——| ох (п — у} — т (| — — и — dl — — С
+ $ 5
wx [s — x}
tr (52)
#2 xin—ul—m —u )—a EE
$ $ ;
Then y = 2Gz + wx {s — x} (53)
2%
and t — 2Gz + wx {s — x} (54)
2y
The cable slope af any point may be found from the gen-
eral formula:
(55)
Example: A 114" diameter Standard Grade Locked Coil
Cable is to be used to support 5 loads, each weighing
2000 pounds, and spaced uniformly 400 feet apart. Length
of span 2000 feet. Horizontal component of working ten-
sion t — 45,964 pounds.
м == у’ == 5.63 роипаз (тот Раде 111).
188
{a} What is maximum deflection?
(b) What is the slope of the cable at a point 500 feet from
the support? |
Maximum deflection occurs with one load at center of span,
x — 1000 ft,
m=— 200 #.
U = 2
5x4
b = 2 == 10
с — 2x] — 1
2000 Pen
From 152) Ус == 45964 | 1000 (5—2) — 200
1000 x 5 10 x 1000
( 2000 —2 )-— 400 ( 2000 -1)|
+ 5.63 X 1000 X 1000
2 х 45964
— 56.566 +- 61.244 = 117.81 feet
From (55) with x = 500 feet, u == 1
2000 5 x 200 + 400 X 10
ton Bs = T5964] BTN 2000
+23 1000-500
= 0653 1 .0612—.1265, Bs ==7°-13'
{a} == 117.81 feet
b) = 7-19
Meur wr
Ng a
SECTION
INCLINED SPAN — MULTIPLE LOADS — COUNTERWEIGHTED
< $ a
G С: G G
CWt.
Figure 13
For y add to formula {52} and (53) x tan a (56)
Formula (54) becomes t —=——C7 = WXIs—x) 157
2 [y—x tan a
tan 3 is found by completing
formula (55) with + tan a — (58) WIND AND ICE LOADS
The change in length of cables due to change in tempera-
ture has not been taken into account in the examples given
in this section. In counterweighted spans such a change in
length results in a small movement of the counterweight,
the tension and deflection remaining constant. However, in
anchored spans fhe change in length due to temperature
changes results in changes in cable tension, and frequently
the effect of such changes must be carefully considered.
To find the change in length, multiply the length of the
cable by the number of degrees (F.) variation in tempera-
ture and the product by the coefficient .00000689 for steel
rope wire,
Wind loads on cylindrical surfaces, such as wire cables,
are determined from maximum wind velocities. If P equals
wind pressure in pounds per square foot of projected area
and Y — actual wind velocity in miles per hour, then. P —
0.0025 V2. This gives 4.0 pounds per square foot for 40
miles per hour, 12.2 pounds per square foot for 70 miles
per hour, and 20.2 pounds per square foot for 90 miles
per hour,
Where exceptionally severe sleet conditions occur, the co-
bles are assumed to be covered with a coating of ice %
inch thick, or a diametric total of 1% inches of ice plus the
diameter of cable. Where the sleet conditions are less se-
vere, the ice coating is assumed to be % inch thick, or a
total of | inch plus the diameter of cable, Then wind load
is based on the total diameter of ice plus cable, and the
resultant cable load is determined from the horizontal
wind load and vertical load of cable and ice. The weight
of ice is approximately 56 pounds per cubic foot, or .0324
pounds per cubic inch,
189
wT ALT aT Fo on EE TET
a FIL =n =. T1
RTE PE RE, PA re RE
ASE ERA PER AE ERA RAI RES PR RD EE ASE EE AVE EEE
1
TRAMWAY AND CABLEWAY DATA
A tramway can be described as a means of
aerial or overhead transportation, wherein the
material transported is carried in cars or buck-
ets which are both supported and transported
by means of wire rope.
Continuous Tramways
The continuous tramway consists of a series
of carriers traveling around an endless cireuit
which connects the two terminals of the
tramway.
1. The Single Rope Fixed Clip Tramway—In
this case a single rope supports and trans-
ports the carriers. Carriers are fixed to the
rope and do not leave it at the terminals.
Although this type of tramway is obsolete
for many industrial purposes, it is used very
effectively for the modern ski lift and chair.
lift.
2. The Bicabie or Double Rope Tramway—The
carriers are supported by a stationary track
cable and transported by a separate hauling
rope. They are connected to the hauling rope
by detachable grips and are removed from
the track cable for loading and unloading at
the terminals. Many industrial tramways
now in use are of this type.
3. The Modern Monocable or Single Rope
Saddle Clip Tramweay— The carriers are sup-
ported and transported by a single endless
rope. They Test on the rope and specially-
designed saddle clips prevent slippage along
the rope. The carriers are removed at the
terminals for loading and unloading. This
type of tramway is extremely simple and
rugged in design and has an exceptionally
low operating cost. It is in use for many
industrial operations throughout the world.
Reversible Tramways
In these tramways, which are usually of the
bicable type; one or two carriers shuttle be-
tween the terminals.
1. The To-and-Fro or Single Reversible Tram-
way--As its name implies, this type of
tramway involves the use of a single carrier.
It is frequently used for the disposal of slate
and other refuse in mining operations.
TEE
ETS Fri pt т E E a < SAE
Cet Ee Trond SAR a RTE EEE я WTC
era e ESA o FER FE TA a аи)
CES due О Te e ers E eE AL a
A ea ET en aT EE er ren
re a
à. The Jigback or Double Reversible Tram-
way This type involves the use of two
carriers. It is sometimes used for waste dis-
posal but more often for transportation be-
tween two fixed points, such as transporta-
tion of ore from mine to mill where relatively
short distances are involved. This type of
tramway is often employed on passenger
tramways to gain access to mountain-top
resorts. |
Cableways
The cableway 1s a special type of tramway
which hoists as well as transports its loads. It is
usually of the to-and-fro type. It is very often
used in construction work and is a common,
sight during the installation of most large
dams where it is used for placing concrete.
TRAMWAY CABLE CALCULATIONS
On continuous tramways the weights of the
carriers and their loads are usually divided by
their spacing and the quotient added to the
rope weight in order to obtain an equivalent
uniform weight (w) per unit of length along the
tramway. The cable tensions and sags can
then be found by the common parabolic
formulas, such as equations 1 to 8, page 146.
In the case of tramways, however, w in these
formulas should be replaced by w sec «. The
following equation is another useful tool for
the tramway designer:
Ty=Ti+wB................. 28°
where Ty = Tension in rope at any _
point (1).
T, = Tension in rope at a higher
point (2).
B = Vertical rise from (1) to (2). a
w = Equivalent uniform weight
of rope and loads.
Note: For a bicable track cable, w is the weight of the
track cable alone. For the hauling rope of the icable or
the rope of a monocable, w is the uniform weight of rope
and buckets taken together, |
Reversible tramways require special treat-
ment, as the loads are concentrated rather
than uniform. The following are methods of
solution for a few of the problems involved:
Ti To sr ra, TRE Rp a es A Ee ela TCR Err Li
Ri a Sh NX uu. ay
Une En E El Fo a
PEL AMET
TET mt SHEE EEE
A Typical Loaded Span.
de E ”
i и”
N KL = 4
|
В
2
B
+
4 L Ï ta.
H = Р | =
Tu Ï й
Vi IL Ur= Wer./ FT. OF CABLE
Wo= W SEC-X
r= We K+ 2P)K
H = oF 3 A | aa 29
RNA
few) 30
Je =} 8H
B,—4
Tan PL == ye fz aaa aaa vaa 31
Tan de = 22M 32
Ку
И у, = H tan Bile ove tee ener 33
Tr = H sec Ф1;, 212 1 1 2 1 1 4 Ene nria 34
Var = H tan PUT. aaa es 35
Toy = H sec pay... Le 36
Short-Cut Method of Computing Equations
29 and 36
For convenience equation 29 is broken down
into two sections, one representing fH, for the
load and the other H, for the rope weight:
PK
H = рено 37
wK* -
H, = BF 38
Then equation 29 may be written:
H = H, + (H» x multiplier) naaa 39
Also equation 36 may be written:
Т = Н X multiplier............... 40
On Charts 2a, 2b and 2¢, values of H,, H.,
and the multiplier are plotted for various con-
ditions. On all charts, each curve represents a
different value of sag ratio <>, stated as a per-
K
centage.
Example for Horizontal Spans
K |
nj
К = 2,000 ft.
f = 60 ft. = 3%
Load = 10,000 Ibs.
= 10.44 Ibs./ft. for two 115” diam.
Locked Smooth Coil Tramway
Track Strands.
wK = 2,000 x 10.44 = 20,880
B
T= 0
From Chart 2a, H; = 83,000 (for load = 10,000)
« «9h H, = 87,000 (for wK = 20,880)
“ “дс, Multiplier = 1.002 (for B_
о)
Hs X Multiplier = 87,174
H, = 83,000
H = 170,174
T = 170,174 x 1.002 = 170,500
_ Track Cable Strength
T
Factor of Safety
_ 2 x 255,000
170,500 900
E UA RA pe TAZA ZZ
ea AT
EEE EN ro Ta Tn TT Sr To IE NoE Cr A PAS A SET ETE EN TE TES уд
AEREA ESE EA = те REESE EAT TASA DA TR TDA TAT Fa
Example for Inclined Spans
- K |] From Chart 2a, H; = 83,000 (forload = 10,0001bs.)
о || « «3h H, = 100,500
“ “2e, Multiplier = 1.048 (mg = 25)
со] =
|
105,300
H, X Multiplier = 100,500 x 1.048
83,000
B H:
f H 188,300
T = 1.048 x 188,300 197,330.
= "ce Track Cable Strength
Factor of Safety = - F ©
|
K = 2,000 ft. 2 x 292,000 _ 3 06
f= 60 ft. = 39, 197,330 7 “
“ur Load = 10,000 Ibs. | | |
ER | _ 1 at . . hy Note: The recommended Factor of Safety for Locked
pile w= 1 10 ibs 7 ft. for two 154” diam. -.. Smooth Coil Tramway Track Strand is 3.0 and for Smooth
E Locked “Smooth Coil Tramway + Coil Tramway Track Strand it is 4.0,
Hh Track Strands. | о |
Te B = 500 ft.
Ih B
DE K — 25
wk = 24,200
EL 35000 AY ROPE
Ef: a {LOAD CURVES)
23000
20000
LOAD
nd 15000
10900
HORIZONTAL COMPONENT OF CABLE STRESS DUE TO LOAD ONLY = Hi
10000 20009 30000 40000 50000 60000 70000 80000 20000 100000
Chart 2a
157
se
$
CABLEWAY ROPE STRESS CURVES
CABLE WEIGHT CURYES;
wk
—
«=
Effective Cable Weight
м
.
Ee as a . —— Ae EN TUNA a ann nr ты
i
| -
!
| OF CABLE STRESS QUE TO
| 50000 0000
| Chart 2b | | 7
- CABLEWAY ROPE STRESS CURVES
88 |& (MULTIPLIER
i
| =
i
| 4
tH
| =
[=
ОП
Ratio of Difference in Tower Elevation to Hori
1,25 1.3 1.05 1.4 1.45 1.5
| Chart 2e
de EE ER En A a то
ов
ео EE es a pea
RUCCI me ЕН a nO E
= a e ten S ВЕ se HE ur are ET i
Te
Path of Load
The designer is frequently interested in the
path of travel of an individual concenirated
load as it crosses a single span. The method of
determining this so-called “Path of Load” de-
pends on whether the track cables are main-
tained at a constant tension by means of
counterweighis or whether they are solidly
anchored and, therefore, subject to changes
in tension as the load traverses the span.
Track Cables Counterweighted
In this case the path of load can be com-
puted as a parabola for all practical purposes.
The formulas to be used are:
y-4R(1-%) aaa 41
and = (1-5) +702
and the meanings of the terms of these form-
ulas are shown in Figure 3. The sag from the
chord to the cable at mid-span when the load
is at mid-span 1s represented by f. (See “A
Typical Loaded Span,” page 155.) /
K
A N
a | 2 7
5 | = |
| Ï c
i
LL
When x = у =
0 0
AK and 9K 36f
2K and .8K 64f
3K and ЛК 84f
AK and 6K —.96f
SK f
Figure 3
frack Cables Anchored -
1. Single Span Tramway or Cableway with
Relatively Short Backstays to the Anchorages
—The path of load may be found with suffi-
cient accuracy by the use of equations 43 and
44 shown in Figure 4.
. К №
a К.
2 æ
8 A |
E | J
7
> Pa Total weight of moving z load (carriage,
car, payload and moving ropes.)
= Weight per foot of track cable.
В Horizontal component of track cable
tension with load at mid-span. |
f, K, B, P, w given as design conditions.
= Effective weight per horizontal foot of
track cable (= w sec œ).
Then:
_ (w, K + 2P)K
H = ay PN BU 29
— 2
- =( E) que X-+ 27) TZ aaa 43
2H Hiw, +% EY 31 — 51 — #) |
_ (w.K+2P)? (K—2x) o K*L2PV ZA B
Tan $ +=
2H [0 KE HAPY Ko —x7 Kx—x? K
e. 44
Fi igure 4
2. Multiple Span Tramway —In this case
the spans adjacent to the one in which the
path of load is-desired have the effect of par-
tial counterweights; thus the solution is more
complicated. It is somewhere between the
results obtained by Figures 3 and 4 and there
is no satisfactory simple approximate method
for this case. However, any tramway problem
can be solved with the aid of the parabolic or
catenary formulas mentioned above (equa-
tions 1-14, 18-25, 26, 27) together with the
method outlined for a suspension bridge free
cable (see Figure 2, page 150). _
Ap ==: e NEE AN a i =
RIE bree fr eT, a Aa A The Terry Far i
Не AAA ar ee MES EE PB o
E E E = E E
HA da AT Ta Lo
Was this manual useful for you? yes no
Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Download PDF

advertising