Chapter VIII: )

Any load, one at C D for example, gives to the posts a tendency to revolve on A, as a centre towards the abutment; to oppose which, there must be a force in the opposite direction. The most proper direction in which to resist such motion is the line C K, i. e., the line of the lower chord. In this bridge there is no lower chord, but in place of such are put the rods A G, A K, B H, and B C; which prevent the change of form (by the motion of the triangle) and act against the upper chord.

As an example of the estimate of strains upon this bridge take the following.

Span, 90 feet.
Rise, 18 feet.
Panel, 15 feet.
Weight per lineal foot, 2,500 lbs.
Whole weight, 225,000 lbs.
Weight on each side truss, 112,500 lbs.
Weight on each post, 18,750 lbs.

The weight borne by each system, i. e., one post and the two supporting rods, is 18,750 lbs. The strain to be resisted by any one rod depends upon its inclination.

The following figures show the elements of the truss in question:—

Section Rod. Length. Applied weight. Increased of the strain. bar in inches. A B = 90.0 C D = 18.0 A C = 23.4 (18750 – 3125) = 15625 which by 23.4/18 = 20312 1⅓ A H = 35.0 (18750 × 60)/90 = 12500 which by 35/18 = 24306 1⅔ A F = 48.5 (18750 × 45)/90 = 9375 which by 48.5/18 = 25260 1⅔ A K = 62.6 (18750 × 30)/90 = 6250 which by 62.6/18 = 21736 1⅓ A G = 77.6 (18750 × 15)/90 = 3125 which by 77.6/18 = 13472 1

Column 1, gives the name of the rod; col. 2, the calculated diagonal length; col. 4, the applied weight, (the varying weight by reason of the varying inclination) found by multiplying the whole weight upon one panel or post by the distance of that post from the abutment, and dividing the product by the span. (Thus the load applied to A G is

(W × IB)/S,

that on A K is

(W × BX)/S,

and so on.) Col. 6, shows the increase found by col. 5 on account of inclination as noticed in Chap. VIII.; and finally, col. 4 gives the necessary sectional area of the bars or rods.

The compression on the top chord is evidently the sum of the compressions of the separate systems; the compression from any one system is as follows, fig. 102.

Let _a d_, _c d_ be the rods, and _a b c_ the chord; also _b d_, the post; now if _d b_ represents the weight, _e h_ shows the tension on a lower, or the compression on an upper chord; the triangles _a c d_ and _a b e_ are similar; as also _e b h_ and _d b c_; whence

_be_ = (_ab_ × _cd_)/(_ac_);

and

_eh_ = (_cb_ × _be_)/(_dc_) compression.

Numerically we have the following figures:—

In the first system,

_be_ = (15 × 77.6)/90 = 13,

also

_eh_ = (75 × 13)/77.6 = 12.

In the second system,

_be_ = (30 × 62.6)/90 = 21,

also

_eh_ = (60 × 20)/62.6 = 20.

In the third system,

_be_ = (45 × 48.5)/90 = 24,

also

_eh_ = (45 × 24)/48.5 = 23,

that is, the compression from the system A C B, is to the weight on the post, as twelve is to the length of the post; or actually

18 to 12 as 18,750 to compression;

whence

compression = (18750 × 12)/18 = 12500

in system one, and in the second system

18 to 20 as 18750 to 20833.

In the central system,

18 to 23 as 18750 to 24000.

Doubling the sum of the first and second systems, and adding thereto the central, we have

2(12500 + 20833) + 24000 = 90666 lbs.,

as the whole compression upon one side of the bridge.

As to compression only, this would require a section of about four square inches of cast-iron, which may be obtained by a tube of four and one half inches inside, and five inches outside diameter. We may however need to increase this amount to resist flexure, or transverse strains; in which case the length of tube in one panel is to be regarded as the height of a post, or the length of a beam; and the size will be found by the table on page 138.

Each post must bear 18,750 lbs., and these being of cast-iron, to resist flexure, by the same table above referred to, should, if made as a hollow cylinder, be a little over four inches in diameter, and one half inch thick; and if of + or ¶ᕼ¶ section, should have a square of nearly five inches.

The flooring will be dimensioned by the rules given in Chapter VIII. for single beams.

There is nothing about this bridge to burn, in case of fire, except the floor; and that might easily be made of iron.

To use the words of the inventor, “The permanent principle in bridge building sustained throughout this mode of structure, and in which there is such gain in competition with any other, namely, the direct transfer of weight to the abutment, renders the calculation simple, the expense certain, and facilitates the erection of secure, economical, and durable structures.”

WHIPPLE’S IRON BRIDGES.

223. The bridges built by the above-named engineer are in all respects well proportioned, rigid, safe, and durable. Cast-iron is used as a top chord, and wrought iron is employed to resist the tensile forces. The plan put up upon the New York and Erie Railroad, consists of a hollow cast-iron top chord, circular in section. Lower chords of wrought iron rods. Posts cast cruciform in section. Diagonal tension rods, as in Pratt’s bridge, (Chapter VIII.). The whole structure is in iron exactly what the above-named bridges are in wood; and the method of calculation is the same. For spans not exceeding one hundred feet, this form answers every purpose as a railroad bridge. It is open to the same objection in larger spans as are all trusses transferring the load by a series of triangles through which the weight passes successively, namely, the effect of an enormous pressure at the feet of the second and third pairs of braces, which should be taken up by arch braces, as in fig. 69; or by rods from the top of the abutment pillars to the feet of the second and third sets of posts.

A span of this plan, upon the New York and Erie Railroad, of forty feet, and which weighed only three tons, supported a load of fifteen hundred pounds per lineal foot for two days; when the bridge had settled nearly one half inch. A load of rails weighing 1318 lbs. per foot (of bridge) was then rolled over, upon a truck without springs, thus making the whole load upwards of 2,800 lbs. per foot, when the whole deflection was three fourths of an inch. Upon removing the load the bridge returned to its original position, within one fourth of an inch.

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.

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_ = (_Y^2_)/(_b^2_) × _a_,

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

┌───────────────────────┬───────────────────────┬───────────────────────┐ │ Centre. │ Rod 1. │ Rod 2. │ ├───────────────────────┼───────────────────────┼───────────────────────┤ │ 0 │ (_E_^2)/(_b_^2) × _a_ │(4_E_^2)/(_b_^2) × _a_ │ │ 0 │ (100^2)/(500^2) × 80 │ (200^2)/(500^2) × 80 │ │ 0 │ 3.20 │ 12.80 │ └───────────────────────┴───────────────────────┴───────────────────────┘

┌───────────────────────┬───────────────────────┬───────────────────────┐ │ Rod 3. │ Rod 4. │ Rod 5. │ ├───────────────────────┼───────────────────────┼───────────────────────┤ │(9_E_^2)/(_b_^2) × _a_ │(16_E_^2)/(_b_^2) × _a_│(25_E_^2)/(_b_^2) × _a_│ │ (300^2)/(500^2) × 80 │ (400^2)/(500^2) × 80 │ (500^2)/(500^2) × 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 = (2_a_)/_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.

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_ = (_ph^2_)/(2_f_);

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 × 500^2)/160 = 9375000 lbs. or 4185 tons.

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

_T_ = ((_ph_)/(2_f_))[_h^2_ + 4_f^2_]^½:

which, in the case before us, becomes

_T_ = ((6000 × 500)/160)[500^2 + 4 × 80^2]^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.

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[_V^2_ – _d^2_]) – √(¾[_l^2_ – _d^2_]).

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[_V^2_ – _d^2_]) – √(¾[_l^2_ – _d^2_]).

_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 = (2_a_)/_b_.

Also, tang E C G = log 2_a_ – 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_)/(2_f_))√(_h^2_ + 4_f^2_),

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_ = (_Y^2_)/(_b^2_) × _a_;

and placing for _Y^2_ the distances 5, 10, 15, 20, etc., we have, commencing at the centre,

┌───────┬──────────────┬───────────────┬───────────────┐ │Centre.│ 5 │ 10 │ 15 │ ├───────┼──────────────┼───────────────┼───────────────┤ │ 0 │5^2/500^2 × 90│10^2/500^2 × 90│15^2/500^2 × 90│ │ 0 │ .009 │ .036 │ .081 │ └───────┴──────────────┴───────────────┴───────────────┘

┌───────┬───────────────┬───────────────┬───────────────┐ │Centre.│ 20 │ 25 │ 30 │ ├───────┼───────────────┼───────────────┼───────────────┤ │ 0 │20^2/500^2 × 90│25^2/500^2 × 90│30^2/500^2 × 90│ │ 0 │ .144 │ .225 │ .360 │ └───────┴───────────────┴───────────────┴───────────────┘

┌───────┬───────────────┬───────────────┬───────────────┐ │Centre.│ 35 │ 40 │ 45 │ ├───────┼───────────────┼───────────────┼───────────────┤ │ 0 │35^2/500^2 × 90│40^2/500^2 × 90│45^2/500^2 × 90│ │ 0 │ .490 │ .576 │ .729 │ └───────┴───────────────┴───────────────┴───────────────┘

┌───────┬───────────────┬─────────────────────┐ │Centre.│ 50 │ _d_ │ ├───────┼───────────────┼─────────────────────┤ │ 0 │50^2/500^2 × 90│(_d_^2)/(_b_^2) × _a_│ │ 0 │ .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.

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,

Suspension 400,000 1
Tubular 4,000,000 10

Fig. 105, shows the anchoring of the Fribourg bridge.

Fig. 107, the manner of fastening the ground stays of the Niagara bridge.

Fig. 108, connection between cable and suspender.

Figs. 109, 109 A, another method of effecting the same.

Fig. 110, floor beam attachment to suspender.

Figs. 111, 112, floor beam attachment in Niagara bridge.

Fig. 113, connection of land and water cables in Fribourg bridge.

Fig. 114.
Fig. 114 A.
]

Figs. 114, 114 A, fastening of cables at G, (fig. 105).

Fig. 115, Mr. Roebling’s pendulum connection for the cables of two adjacent spans.

BOILER PLATE BRIDGES.

Spans from 25 to 100 feet.

236. These structures fulfil every requirement of safe, durable, and rigid bridges; being open however to the contingency attendant upon all similar structures of wrought iron, namely, the becoming crystalline when exposed to vibration. Time only will show whether this is a sufficient cause for their non-adoption.

Each side truss consists as it were of a top and bottom chord connected by a vertical web. The whole being of wrought iron, requires that the section of the upper chord should be to that of the lower, as ninety to sixty-six.

The general plan of such bridges is shown in fig. 116. This is the patent wrought iron girder bridge of Mr. Fairbairn. The upper chord is formed by connecting the four plates _a a a a_, by angle irons. The web is formed either by a single or a double plate, stiffened laterally by T iron placed at the vertical plate joints, as shown generally at B, and detailed at C and D; or by a pair of plates separated by a space as at B′, thus forming a rectangular tube. The lower chord is made by bending horizontally the lower part of the web, and to the flanges thus formed riveting the plate _m m_. The suspending rod _f_ is applied to the upper chord by a washer as at E.

The central connecting web, acting as do the braces and ties in a wooden truss, should be more stiff at the ends of the span than at the centre. This is easily effected by joining the web plates towards the end by stronger T irons than at the centre. The joints for the rib, or the vertical plates, either single or double, are shown in figs. C and D.

An example of the need of such increased stiffness towards the ends, was given to the experimenters upon the Britannia model tube, which (tube) was found to yield by buckling near the ends of the span sooner than elsewhere. Thus advised, the vertical plates were made thicker as the end of the span was approached. Examination of the principles of proportioning a common wooden truss would have shown this without experiment.

The tensile and compressive strength of rolled boiler plates (by the table on page 193,) is, extension 12,740 lbs. per square inch, compression 7,500 lbs. The strength of such work depends in a very great measure upon the size and disposition of rivets. In plates exposed to compression, the strength is not so much affected by riveting as in those subjected to tensile strains; as to whatever amount the plate is cut away, by the same amount is the resistance to tension reduced.

237. Mr. Fairbairn found that to obtain the maximum strength of riveted plates, the section of the rivets should equal that of the plates, that is, in a plate four inches wide, if there are two rivets, the area of each must be one inch; or the diameter 1⅛ inches; thus leaving a section of

4 – 2¼ = 1¾ inches,

which divided by four gives seven sixteenths of an inch as the distance from the edge of the plate to the side of the first rivet; and seven eighths of an inch between rivets. If the bolt yields by shearing, the rim is destroyed by _detrusion_, or crushing across the fibres. That the rivets and plates may be equally strong, their products of area of section by the actual strength per unit of area must be equal. The detrusive strength of wrought iron (see page 193) is 12,500 lbs. per inch, whence the proportion

12,500 : 15,000 :: 1 : _d_

where 1 is the resisting length of the plate at right angles to tension, and _d_, the sum of rivet diameters. Thus suppose we have a plate 13.2 inches wide, to be fastened with nine rivets of 0.8 inch diameter; we have

9 × 0.8 = 7.2 = _d_,

and the above proportion becomes

15,000 : 12,000 :: 7.2 to 6 inches,

which is the length of plate section at right angles to tension. As there are nine rivets, there will be eight spaces between them, and one space at each edge of the plate, half as large as those between; or reducing all to the same size,

8 × 2 = 16, 16 + 2 = 18;

and as the whole plate section after punching is six inches,

6/18 = .33 or ⅓ inch

for the edge space, and two thirds inch between rivets. Proceeding thus, the result compares with the practice of Mr. Fairbairn as follows:—

Diameter of rivet. Distance between rivets.
Mr. Fairbairn ⅝ inch, or 0.625 inch, 0.8
Handbook ⅔ inch, or 0.666 inch, 0.8

The difference between the results, or 0.041 inch, less than one sixteenth inch, will be partially absorbed by the remark of Mr. Fairbairn that the area of the rivet should be _nearly_ as much as that of the plate, and partly by the difference in results showing the detrusional force of iron.

238. In experimenting to determine the resistance of rivets, Mr. Fairbairn found that by the common plan of riveting, fig. 117, the strength of plates when whole, single, and double riveted, was as follows, the section of the punched plate being in each case equal to that of the whole one.

Whole plate, 100.
Single riveted, 56.
Double riveted, 70.

This loss of strength made him fearful of the ability of the tension plates of the Britannia bridge to do their duty; and he was led to adopt what he terms “chain riveting,” which consists in placing the rivets as in fig. 118, or _in the same line of tension_. The strength of plates thus made he considers as great at the joints as elsewhere.

239. As to the diameter of rivets, we have the following results of the practice of the best English engineers.

Thickness of plate, ¼, 5/16, ⅜, 7/16, ½, 9/16, ⅝, 11/16, ¾.
Diameter of rivet, ⅝, 6/8, ⅞, 1, 1, 1⅛, 1¼, 1⅜, 1½.

240. As to the distance _in the direction of the force_ from rivet to rivet, also from the first rivet to the plate end, we gather the following from the best executed works in boiler plate. See fig. 118.

Plates exposed to compression,

_cb_ = 2 diam., _df_ = 1½ diam.

Plates exposed to extension,

_cb_ = 2½ diam., _df_ = 2 diam.

the diameter being that of the rivet.

The distance at right angles to the force has already been given.

241. If we knew the lateral adhesion of rolled plates, that is, the resistance of the fibres to sliding horizontally past each other; we should determine the distance of rivets in the direction of tension as follows:—

Let _R_, equal the resistance per unit of area for _detrusion_ or shearing, _R′_ the lateral adhesion of rolled plates, and we should have

_R_ × _a_ = _R′_ × (2 _d_ × _t_);

whence _d_ = (_R_ × _a_)/(2_R′_ × _t_);
where _a_ = area of rivet,
_d_ = distance,
_t_ = plate thickness.

also

2_d′_ + _d_ = (_R_ × _d_)/(_R′_);

and

2_d′_ = (_R_ × _d_)/(_R′_) – _d_;

whence

_d′_ = [(_R_ × _d_)/(_R′_) – _d_]/2

and finally

_d′_ = ½[(_R_ × _d_)/(_R′_) – _d_]

supposing the piece 1, 2, 3, 4, fig. 118, to split out.

The diameter of the semi-spherical head of the rivet should be three times the thickness of the plate to be riveted; that of the conical head four times; and the height of both of the heads, one and one half the plate thickness.

242. Examples of the application of the preceding remarks.

Suppose we wish to build a boiler plate bridge of one hundred feet span, twelve feet rise, weight of bridge and load 3300 lbs. per lineal foot. The tension by formula

_T_ = (_WS_)/(8_h_) (see Chap. VIII.)

becomes

33000000/96 = 343,750 lbs.

Each side truss will bear one half of this or 171,875 lbs., and as wrought iron resists eleven thousand pounds of compression per square inch, the required section of the top chord will be

171875/7500 = 22.9 square inches.

Also the lower chord resisting fifteen thousand pounds per square inch, must have

171875/12740 = 13.5 square inches

of area nearly.

If we make the tube at top of one fourth inch iron, and 8 × 10 inches; fastening the plates by one fourth inch angle iron, four inches on the side, the section becomes

One top plate 10 × ¼ = 2½ square inches.
One bottom plate 10 × ¼ = 2½ square inches.
Two side plates 8 × ¼ = 4 square inches.
Four angle irons ¼ × 8 = 8 square inches.
—
In all, 17 square inches.

In the lower chord, if we bend the web plates (of ⅜ inch) so as to form a flange of eighteen inches in width, and to that rivet a bottom plate 18 × ¼, we shall have

In the flanges, 18 × ⅜ = 6¾
Bottom plate, 18 × ¼ = 4½
———
In all, 11¼

The web acting as both ties and braces, must be able to support the following load.

Whole weight of bridge and load is, in round numbers, 344,000 lbs.
One half, 172,000 lbs.
And upon each end of the truss, 86,000 lbs.

to resist which, at eleven thousand pounds per square inch, requires eight inches nearly, regarding the plate as a brace.

Now the side of the bridge being one hundred feet long, and twelve feet wide, will contain any system of bracing that we choose to draw thereon. Suppose, for example, that we chalk a line upon the erected bridge representing an arch-brace, extending from the end to the centre. Such a brace has actual existence in the bridge; and the same idea holds good for any system of braces that may be assumed. We ought, therefore, to take the most disadvantageous system that can have place, and giving such a good bearing upon the abutment, estimate its width and thickness. Suppose that we draw a natural size representation of Howe’s bridge, the end braces must support a load of eighty-six thousand pounds, which at eleven thousand lbs. per inch, requires a section of nearly eight inches; and if the plate is one half inch thick, the brace must be sixteen inches deep. The manner, however, in which the plate would yield is by bulging laterally; which is to be checked by the before-mentioned T connecting irons at the sides. It may be thought that the above method of considering the plates as braces, would give very little thickness by assuming very wide plates. The answer to this is, that the side plates must not be so thin as to need more stiffening angle irons by weight, than a thicker plate with less stiffening. Of course the weight should be minimum.

243. As an actual example of this plan we have the following, built by Mr. Fairbairn for the Blackburn and Bolton Railroad, across the Leeds and Liverpool Canal.

Span, 60 feet,
Length, 66 feet,
End bearings, each, 3 feet,
Rise, 5 feet,
Width, 28 feet,

for a double track. Top chord of three eighths iron, web of five sixteenths, lower flange of three eighths, and vertical web plates stiffened by T irons.

This bridge was tested as follows:—

Three engines, weighing twenty tons each, running from five to twenty-five miles per hour, deflected the bridge .025 feet. Two wedges, one inch high, being placed upon the rails, and the engines being dropped from that height, the bridge was deflected at the centre .035 ft.; with wedges of one and one half inches the deflection was .045 ft. The cost of this bridge (in England) was estimated by Mr. Fairbairn at $4,500, while that of a cast-iron bridge of the same span was $7,150.

244. _Example 2._—Manchester, Sheffield, and Lincolnshire Railroad (England) Bridge, at Gainsborough, on the river Trent. Two spans, each one hundred and fifty-four feet. Rise twelve feet. Top chord, double rectangular tube, 36¾ × 16 inches, vertical web as before, and horizontal plate for the lower chord. The floor beams are wrought iron girders, cruciform in section, ten inches wide, and one foot three inches (15″) deep, placed four feet apart.

245. _Example 3._—Fifty-five feet boiler plate bridge, built by James Millholland, in 1847, for the Baltimore and Susquehanna Railroad Company. Each truss consists of two vertical plates 55 × 6 feet, formed of plates thirty-eight inches wide by six feet deep, the plates being fastened together by bolts passing through cast-iron sockets. The lower chord is formed by riveting two bars 5 × ¾ inches to each side of each truss plate; making in all eight. Top chord—one bar of the same size on each side of each plate, compression being made up by a wooden chord between the plates. Height of bridge, six feet; length, fifty-five feet; width, six feet; weight, fourteen tons; cost, $2,200, or forty dollars per foot. The inventor thinks thirty dollars per foot enough when a considerable amount of such bridging is wanted.

NOTE.—White, buff, or some light color should be used in painting
iron bridges, as such throw off, and do not absorb heat from the
sun.