SUSPENSION BRIDGES.

224. Suspension bridges of large span have been generally considered as entirely unfit for railroad purposes; but John A. Roebling has proved the contrary by erecting a wire suspension railroad bridge of eight hundred feet clear span across Niagara River; which with heavy loads and violent gales has shown itself to be both stiff and strong to any desired amount. The construction of a bridge upon any other plan would have been hardly possible at the site of Mr. Roebling’s Niagara bridge, there being no opportunity for scaffolding or for piers, pontoons or hydraulic presses.

The simple road-way supported by cables, possesses great strength with very little stiffness. It must be accompanied by stays and trusses to check vibration.

No bridge involves more simple calculations, and in none can we proceed with more absolute safety, than in the wire suspension. European suspension bridges are generally formed of cables made by linking bars of wrought iron together. This method is more expensive and more liable to failure than the American plan of forming cables of iron wire. An apparently good bar may be defective inside, while we are sure of every component fibre of the cable; indeed it is very little trouble to test each wire as it is laid into the cable.

The parts to be considered in proportioning a suspension bridge are

The anchoring masonry,

The anchor chains,

The towers and plate,

The suspension cables,

The suspending rods,

The stiffening arrangement,

The road-way.

The data given in the construction of a bridge of this description are

The span,

The load to be supported.

The assumed data

The versed sine of the cable,

The width.

And the required elements

The length of cable,

Lengths of suspending rods,

Angle of tangent of cable at point of suspension with axis of tower,

Tension upon the cables,

Section of the anchor irons,

Amount of anchoring masonry,

Size of the towers,

Dimensions of trussing and of road-way.

OF THE CABLES.

225. The curve formed by the cable of a suspension bridge lies between the parabola and the catenary. When loaded the curve is nearly the former, and when unloaded the latter.

Problem 1.

Given the horizontal distance between the points of suspension and the versed-sine, to find the length of the cable, fig. 103.

Fig. 103.

Represent C E by b, and E F by a, and the length of the semi-curve is

L = b[1 + ⅔(a
b)²].

Let the half span be five hundred feet, and the versed-sine or deflection eighty feet, the formula becomes

L = 500[1 + ⅔(80
500)²] = 500 × 1.0171 = 508.55 feet,

which is the half length of cables between towers.

Problem 2.

226. To find the length of the suspending rods. Calling E the horizontal distance between the vertical suspenders, we have the formula

X = Y²
b² × a,

in which we place E, 2E, 3E, etc., in place of Y, thus calling the rods one hundred feet apart, we have

Rod 1.	Rod 2.	Rod 3.	Rod 4.	Rod 5.
E² b² × a	4E² b² × a	9E² b² × a	16E² b² × a	25E² b² × a
100² 500² × 80	200² 500² × 80	300² 500² × 80	400² 500² × 80	500² 500² × 80
3.20	12.80	28.80	51.20	80.00

Problem 3.

227. To find the angle E C G, fig. 103. The formula for the angle between the axis of the tower, and the tangent to the curve of the cable at the point of suspension is

tang a = E C G = 2a
b.

Span being one thousand feet, b is five hundred; and a being eighty feet, we have

tang E C G = 160
500 = log 160 – log 500:

or 2.204120 – 2.698970 = tang 9.505150 = 17° 45′ = E C G.

Also, 90° – 17° 45′ = 72° 15′ = angle G C A, or A C H.

When the points of suspension are not at the same elevation, we proceed in the same manner: only using G L, G E, in place of F L, F C, in fig. 103 A.

Fig. 103 A.

That the resultant of the forces acting upon the top of the tower may be vertical, the angles G C A, and A C H, fig. 103, must be equal; if not, the masonry must be so arranged as to cause the resultant to pass through the centre of gravity. When more than one span is used, and the openings are unequal, that the intermediate pier or piers shall not be pulled over, the cable of the largest, and consequently heaviest span, must have a greater inclination from the horizontal than that of the shorter span; the product of the tensions by their respective inclinations must be equal. Mr. Roebling’s plan in connecting several spans, is to attach the cables of adjacent spans to a pendulum upon the pier, by which arrangement the difference in tension upon the different cables swings the pendulums, without racking the masonry.

Problem 4.

228. Given the weight per foot of bridge and load, to find the tension at the lowest point of the curve. The formula for the minimum tension, that at the vertex F of the curve, is

T = ph²
2f;

where p is the weight per foot of bridge and load, h the half distance between the points of suspension, and f the versed-sine. Thus the span being one thousand feet, the versed-sine eighty feet, and the load per lineal foot six thousand lbs., the formula becomes

T = 6000 × 500²
160 = 9375000 lbs. or 4185 tons.

The maximum tension is at the points of support, and is expressed by the formula

T = ph
2f[h² + 4f²]^½:

which, in the case before us, becomes

T = 6000 × 500
160[500² + 4 × 80²]^½ = 4395 tons.

229. The object of the anchoring is to connect the cable with a resistance upon the land side, which shall more than balance the weight and momentum of the bridge and load upon the opposite side. The anchoring of the Niagara bridge consists of an iron chain made of flat links, 7 feet long, 7 inches wide, and 1.4 inches thick; the chain links consist alternately of six and of seven of these bars; see fig. 104.

Fig. 104.

In the Fribourg bridge (Switzerland) the anchorage is made as in fig. 105, (see p. [220],) by a cable in place of the chain. In M. Navier’s suspension bridge at Paris, over the Seine, the anchorage depended somewhat upon the natural cohesion of the earth forming the bank of the river, and this being destroyed by the bursting of a water-pipe in the vicinity, the bridge fell. When there is no natural rock for an anchorage, the masonry of the shaft must, by its own weight, resist the tension.

230. The height of the towers must be at least as much as the versed-sine of the cable. Their duty is to support the whole bridge and load. The breadth and thickness of these columns must be determined more with a view to opposing lateral, than downward strains. The former result from the horizontal vibrations of the bridge caused by the action of the wind. Tremor and vibration caused by a passing load, tend to pull the towers into the river. The section for weight only might be very small. From the practice of the best builders, a mean section of one fifth of the height seems to give the best results; thus, if a tower is sixty feet high, the mean thickness should be twelve feet; or the top being 8 × 8 feet, the bottom should be 16 × 16 feet.

If the bridge is so little braced laterally as to swing, a dangerous momentum will be generated which would very much increase the strain, both upon the masonry and upon the cables.

231. The object of the stiffening truss is to transfer the weight applied at any one point over a considerable length, and to prevent vibration. Its dimensions should, therefore, be those of the counterbracing in an ordinary truss.

Any applied load produces a certain depression in the bridge: to use the words of Mr. Roebling, “every train that passes over the bridge causes an actual elongation of the cables, and consequently produces a depression. If the train is long, and covers nearly the whole length of the bridge, and is uniformly loaded, the depression will be uniform. If the train is short, and covers only a part of the floor, the depression will be less general and more local; and will be the joint result of an elongation of the cables, and of a disturbance of the equilibrium. Depressions will be in direct proportion to the loads, and indirectly as the length of train.” The amount of depression depends on the elongation of the cables; the elongation upon the length. The depression is shown by the formula

D = √¾[V² – d²] – √¾[l² – d²].

where D = depression,

l = half length of curve before elongation,

V = half length of curve after elongation,

d = half distance between points of suspension.”

The effect of heat, by expanding the cables, is also to depress the road-way; the amount being shown by the expression

D = √¾[V² – d²] – √¾[l² – d²].

V being the length of semi-curve as elongated by heat instead of by tension; the elongations, both by heat and tension, being found by table on page [193].

Upon the top of the towers is placed a pair of cast-iron plates separated by rollers; the upper plate (the saddle) is thus enabled to move over the lower one when pulled either way by the movement of the cables.

The length of the half cable between towers being generally greater than the distance from the top of the tower to the anchoring, expands more, when the saddle moves towards the land side. The dimensions of these castings must be sufficient to resist the whole weight of bridge and load.

232. As an example of the preceding formula, take the following:—

Assume the span as	1,000	feet.
Height of towers	100	feet.
Deflection of cables	90	feet.
Weight per foot (lineal) of bridge	2,500	lbs.
Weight per foot (lineal) of load	2,000	lbs.
Whole weight per foot	4,500	lbs.
Total weight	4,500,000	lbs.

CABLES.

The formula for the half length of cables between tops of towers is

L = b[1 + ⅔(a
b)²],

which becomes

L = 500[1 + ⅔(90
500)²] = 510.80,

which doubled, is 1021.38. To this add double the distance from the top of tower to the anchorage, (see page [206],) which is found as follows:—

tang E C G = 2a
b.

Also, tang E C G = log 2a – log b, or 2.255273 – 2.698970 = tang 9.556303 of which the angle is 19° 48′ and 90° – 19° 48′ is 70° 12′ = angle G C A or A C H.

The height of the tower being one hundred feet, and the angle at the tower 70° 12′, we have

Sin 19° 48′	9,529,864
Sin 90° 00′	10,000,000
log height (100)	2,000,000
log distance (295.2)	2,470,136

which double, and we have 590.4; finally, add twice the breadth of the tops of the towers, and the whole length of cable is, from anchorage to anchorage,

1021.38 + 590.40 + 16 = 1627.78 feet.

The formula for the maximum tension, (that at the point of suspension,) is

T = ph
2f√h² + 4f²,

which becomes

T = 4500 × 500
180√250000 + 32400 = 2966 tons.

Number 10 iron wire (20 feet per lb.) will support 1,648 lbs. per strand; this is the ultimate strength; the maximum load for safety is 400 lbs. per strand; whence 2,966 tons, or 6,642,500 lbs. will require 16,606 strands; and if we use two cables, each must have 8,303 wires; or four cables of 4,151 each. The permanent load on suspension bridges should never be more than one sixth of the ultimate strength; one eighth is a good standard. The accidental load should never exceed one fifth of the whole strength of the cables. The permanent weight supported by the Niagara bridge is only one twelfth of the ultimate strength of the cables.

ANCHOR CHAINS.

The maximum tension being 6,642,500 lbs. the whole section of the four anchorings will need to be

6642500
15000 = 443 inches,

or 111 square inches for each shaft; which is obtained by eleven links ten inches wide and one inch thick. If we so attach the anchor chains to the masonry as to reduce the tension one fourth at the first arch, (see Fribourg anchoring,) we may fasten three bars of the chain at that point, and descend from the first to the second arch with eight bars; and leaving two bars at that point, proceed to the bottom with the remaining six.

Where there is no natural rock to build the masonry into or against, enough artificial stone must be put down to balance the bridge and load.

ANCHORING MASONRY.

The entire weight of the bridge and load being 4,500,000 lbs. and the whole tension, as above found, 6,642,500 lbs., or upon each tower 3,321,250 lbs.; this is the tension tending to draw the masonry out of each shaft. This tension must be reduced on account of the inclination of the pulling force. The tower is one hundred feet high. The distance on the line of tension from the top of the tower to the anchoring, as already found, is 295.2 feet; whence the actual effort to move the anchor masonry, is thus,

295.2 to 100 as 1,660,625 to the effort or 562,542 lbs. If rock weighs 160 lbs. per cubic foot, which is resisted by a column of masonry of 3,321,250
160 = 20,758 cubic feet, or 20 × 20 × 52 feet, or by a mass 15 × 15 × 91 feet.

TOWERS.

The height of towers being one hundred feet, and the mean thickness being one fifth of the height, we have mean section 20 × 20; or top 12 × 12, and base 28 × 28.

SUSPENDING RODS.

Assuming the horizontal distances between the centres of the vertical suspenders as five feet, their lengths, then, will be found by formula

X = Y²
b² × a;

and placing for Y² the distances 5, 10, 15, 20, etc., we have, commencing at the centre,

Centre.	5	10	15	20	25	30	35	40	45	50	d
0	5² 500² × 90	10² 500² × 90	15² 500² × 90	20² 500² × 90	25² 500² × 90	30² 500² × 90	35² 500² × 90	40² 500² × 90	45² 500² × 90	50² 500² × 90	d² b² × a
0	.009	.036	.081	.144	.225	.360	.490	.576	.729	.900	h

and so on, until we arrive at the tower. Whatever distance above or below the vertex of the curve the road-way is placed, is of course constant, to be added to or taken from the above lengths.

The manner of putting in any camber is simple both in theory and practice. The strain upon the suspenders is merely the direct weight of the road-way and load. If this is 3,500 lbs. per foot, the five feet supported by two rods (one each side) will weigh 17,500 lbs.; each rod or wire rope must hold 8,750 lbs.; this can be done by a section of one half inch area. For extra strains, however, on so large a span as 1,000 feet, one inch of area is not too large.

OF THE STIFFENING TRUSSES, GIRDERS, AND STAYS.

The object of the girders supporting the rails is to diffuse the applied weight; these girders may be made of a Howe truss four or five feet deep, by trussed girders, only simply deep and stiffly framed track strings. They should be able to distribute the load applied at one point at least fifteen or twenty feet. The side trusses transfer to a still greater extent any applied load. Mr. Roebling estimates the combined effect of trusses and girders in the Niagara bridge as transferring the weight of a locomotive over a length of two hundred feet. This transferring counteracts the local depression. The Niagara truss is formed by a system of vertical posts, five feet apart, and diagonal rods passing from the top of the first post to the foot of the fifth; the inclination being 45°, spreads the weight placed upon any one pair of posts over twice the height of the truss, or about forty feet. As to the actual dimensions of the girders supporting the rails, if we intend them to spread an applied weight over forty feet, they must be as stiff as a bridge of forty feet span. And as regards the truss, if we would effectually distribute the applied weight and check vibration, the trussing should be as strong as the counterbracing in a large span upon the ordinary plans. The principle of trussing a suspension bridge may be thus explained. See fig. 106. Suppose that in place of supporting the three trusses D s w, s m m′, and m′ m d, upon piers at w and m′, we suspend these points from the cable A c B. The cable is flexible, and when we apply a load at m, the truss will assume the position D s c n d, but between D and s, s and n, n and d, the truss will be quite stiff. What we require, then, is to make the figure o p m m′, incapable of changing its form, which is done by diagonal bracing.

Fig. 106.

233. Undoubtedly the finest specimen of a bridge of large span upon the suspension principle, or indeed upon any principle, is that built by John A. Roebling, across the Niagara River, a short distance below the falls. The dimensions below of this admirable structure are from the final report of the above-named engineer.

Length of bridge from centre to centre of tower	821′	4″
Length of floor between towers	800	ft.
Number of wire cables	4
Diameter of each	10″
Solid wire section of each cable	60.40	sq. in.
Total section of four cables	241.60	sq. in.
Whole section of lower links of anchor irons	276	sq. in.
Whole section of upper links of anchor irons	372	sq. in.
Ultimate strength of chains	11,904	tons.
Whole number of wires in cables	14,560
Average strength of a wire	1,648	lbs.
Ultimate strength of four cables	12,000	tons.
Permanent weight supported by cables	1,000	tons.
Resulting tension	1,810	tons.
Length of anchor chains	66	ft.
Length of upper cables	1,261	ft.
Length of lower cables	1,193	ft.
Deflection of upper cables (mean temperature)	54	ft.
Deflection of lower cables (mean temperature)	64	ft.
Number of suspenders	624
Aggregate strength of suspenders	18,720	tons.
Number of over-floor stays	64
Aggregate strength	1,920	tons.
Number of river stays	56
Aggregate strength	1,680	tons.
Elevation of grade above mean water	245	ft.
Depth of river	200	ft.
Cost of the bridge	$400,000.

231. The following items are extracted from the report above referred to:—

“The trains of the New York Central, and Canada Great Western Railroads have crossed regularly at the rate of thirty trips per day for five months. (At present over two years.)

“A load of forty-seven tons caused a depression at the centre of five and a half inches.

“An engine of twenty-three tons weight, with four driving wheels, depressed the bridge at the centre 0.3 feet. The depression immediately under the engine was one inch; the effect of which extended one hundred feet.

“The depression caused by an engine and train of cars is so much diffused as scarcely to be noticed.

“A load of three hundred and twenty-six tons produced a deflection of 0.82 feet only. The Conway tubular bridge deflects 0.25 feet under three hundred tons; the span being only one half that of the Niagara bridge.

“The specified test for the wire was, that a strand stretched over two posts four hundred feet apart should not break at a greater deflection than nine inches; also, that it should withstand bending square and rebending over a pair of pliers without rupture. This test corresponds to a tensile strain of 90,000 lbs. per square inch, or I,300 lbs. per wire of twenty feet per pound.”

The wire is preserved from oxidation by coating with linseed oil and paint. Upon the durability of wire cables employed for suspension bridges the following fact came to light: Upon taking down the cables of the footbridge, put up in 1848, by Mr. Ellet, the wire was found so little impaired that Mr. Roebling did not hesitate to work it into the new cables; also, the original oil was found to be still soft and in good condition, having been up six years.

That iron-work lying under ground has been completely covered with cement grout, as this is found by the above-named engineer to be an effectual guard against oxidation.

Engineers wishing to study the details of the Niagara bridge, will find the final report of Mr. Roebling full of valuable matter, both as regards the making of cables, anchoring, stiffening, and the effect of passing trains.

Note.—This engineer is at present engaged upon a still greater work, namely, a railroad suspension bridge across Kentucky River, of 1,224 feet span, 300 feet above the water. There is no lower road-way in this bridge, the cross section being a triangle base upwards.

235. Note.—The Britannia tubular bridge, across the Menai Straits, is doubtless a great work, and also an enormously extravagant one. If no other structure were possible it would be admissible; but it is equalled in strength and by far surpassed in economy by Mr. Roebling’s system of trussed suspension bridges. The cost of material alone in one span of the Britannia bridge, of 460 feet, exceeds the entire cost of the Niagara bridge of 800 feet span; add to this that we are sure of the strength of wire cables, but not of tubes, and that the 800 feet span of the Niagara bridge weighs only 1,000 tons in itself against 1,400 in a 460 feet span of tube, and it will not be difficult to prove the superiority of the suspension over the tubular system; thus,

A suspension bridge of 800 feet span costs $400,000.

A tubular bridge of 460 feet span costs $500,000.

When we double the linear dimensions we increase the weight by the cube; and the cost of a tube is very nearly as the weight; whence a tubular bridge of 800 feet span will cost 2 × 2 × 2, or eight times 500,000, or $4,000,000 against $400,000. Thus,