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

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