CHAPTER IX.
IRON BRIDGES.

“A little knowledge is a dangerous thing.”

213. Within the past ten years iron has been brought extensively into use for railroad bridging; and when employed by those who understand its chemical and mechanical nature is unequalled for strength, durability, and elegance of appearance; but when, as is too often the case in America, it is intrusted to men who neither know nor care for any thing but the price they get for it, nothing can be more unsafe. No material requires so complete a knowledge of its properties, to be safely used, as cast-iron.

NATURE AND STRENGTH OF IRON.

214. The table below shows the properties of the several descriptions of iron used in engineering.

Wrought Iron. Cast-Iron. Iron Wire. Boiler Plate. Designation of the quality.
480 450
480 Weight per cubic foot in lbs.
15000 4500 25000 12740 Resistance to extension in lbs. per sq. inch.
11000 25000
7500 Resistance to compression in lbs. per sq. in.
.0000066 .00000608 .00000685 .0000066 Expansion per degree Fahrenheit in lengths.
.0000000424 .000000106 .0000000446 .0000000524 Extension per lb. per square inch.
.000000149 .000000083
.000000189 Compression per lb. per square inch.
90 to 66 20 to 111
127 to 75 Ratio of extensive to compressive strength.
12500 17500

Resistance to detrusion, or shearing.
55 31

Relative transverse strength.

Column four refers to boiler plate when built into tubes.

After wrought iron has become a little compressed, its power to resist a crushing force is very much increased.

215. The tenacity of wrought iron is increased by heating. Experiments upon thirty varieties gave the following mean result, the temperature ranging from 500° to 700° Fahrenheit.

Strength when

Cold. Hot. Cooled.
60,000 64,000 70,000

216. Stirling’s process of toughening cast-iron, by the addition of malleable scrap, increases the strength in the following ratio:—

The mean tensile strength of cast-iron being 18,000 lbs.
And the compressive strength being 105,000 lbs.
When Stirling-toughened the tensile strength is 23,000 lbs.
And the compressive strength 130,000 lbs.

The strength of cast-iron increases rapidly up to the twelfth or fifteenth recasting, when it is nearly doubled; after the fifteenth melting the strength decreases.

217. Wrought iron exposed for some time to vibration, as in the case of railroad axles, or iron which has been wrought with light hammers, loses its toughness and becomes “short,” (crystalline). The fibre may be restored in such cases by reheating and cooling slowly.

218. GENERAL RATIOS OF THE STRENGTH OF IRON.

Tension. Compression. Cross Strain.
Cast, 300 1,666 31.68
Wrought, 1,000 733 55.40

OF THE STRENGTH OF BOILER PLATES.

219. The strength of rolled boiler plates is no greater in the direction of the fibres than crosswise, but is more regular; whence the length of the fibre must be placed as nearly as possible with the direction of the force.

A mean of twelve experiments, by Mr. Fairbairn, gives the tensile strength of wrought iron plates as 50,960 lbs. per square inch; and the compressive strength of plates, when built into tubes, as 30,464 lbs., or for safe use in practice, for extension, 12,740 lbs., and for compression, 7,500 lbs. In the remarks upon girder bridges the matter of riveting will be considered.

CLASSIFICATION OF IRON BRIDGES.

220. Iron bridges may be classified as follows:—

Those entirely of cast-iron, or Arch and Girder bridge.
Those of wrought iron alone, or Tubular and Girder.
Those of iron wire, or Suspension bridges.
Those of cast and wrought iron, or Trussed bridges.

The order in which these bridges may be placed as regards cost of construction, and extent of application, is as follows:—

Number. Span. Description of bridge.
1 10 to 50 feet Cast-iron girder.
2 50 to 200 feet Cast and wrought combinations.
3 200 to 2000 feet Suspension.
4 200 to 500 feet Cast arch.
5 25 to 100 feet Boiler plate girder.
6 100 to 500 feet Tubular.

Numbers 2, 3, and 5, are the forms which are in use upon American roads. No. 1, is very liable to failure, requires much more knowledge and care in building, and is far more expensive than a wooden truss, or trussed girder. No. 4, is very expensive, and causes a greater obstruction to the water-way than any other. The enormous expense of No. 6, should, and will prevent its adoption in the United States. Let us look at the principles of construction of numbers 2, 3, and 5.

COMBINATIONS OF CAST AND WROUGHT IRON.

221. Under this head come all of the iron trussed frames used in this country.

As before observed, skill in bridge construction consists in using always that material which with the least expense is the best able to resist the particular strain to which it may be exposed. Thus wrought iron must always be used to resist tension, and cast-iron compression. Posts, braces, and upper chords should always be cast, while ties and lower chords should be made of wrought iron.

The strength of a railroad bridge must be such as to resist all extra shocks and strains, such as are produced by derailment of engines, and breakage of axles; also incidental strains arising from change of form by expansion and contraction of the metal, and from high winds and gales.

Fig. 101.

Every part of a bridge not resisting some force is worse than useless, as it adds to the weight. Lightness not only increases the economy directly, but indirectly by removing a part of the permanent load.

222. Foremost in class number two stands Wendel Bollman’s Iron Suspension and trussed bridge. For simplicity of construction and directness of action, this bridge is unsurpassed. The weight at each post is transferred at once to the abutment or pier. The upper chord is of cast iron, hollow, octagonal without, and circular within. The posts consist of an casting, the central web cast open and the flanges whole. The top is adjusted to the chord, and the bottom to the tension or suspending rods. These latter are of wrought iron, rectangular in section, joined when the length requires it by an eye bolt. Each set after leaving the foot of the post, passes through the chair at A B, fig. 101, and is secured by a nut. The junction of the tension rod A C, and the counter rod B C, is attached indirectly to the foot of the post by a pendulum or link; which serves to equalize the effect of expansion upon the rods. Vibration and reaction are prevented by the panel diagonal ties D H, and C E. The floor is supported by flanges at the foot of each post. The lateral bracing consists of a system of hollow cast-iron posts, and of wrought diagonal tie rods. A lower chord is plainly unnecessary, its place being taken by the rods C B, F B, F A, G A.

A bridge of this description upon the Baltimore and Ohio Railroad of the following dimensions,

Clear span, 124 feet
Length of top chord, 128 feet
Length of panel, 15 feet
Height of truss, 17 feet
Width, 16 feet
Lbs. of cast-iron, 65,137
Lbs. of wrought iron, 33,527
Whole weight, 98,664
Weight per lineal foot, 796

was subjected to the following tests.

Three locomotives with tenders attached, and weighing in all one hundred and twenty-two tons, (nearly one ton per foot,) were run over the bridge at eight miles per hour, when the deflection at centre was one and three eighths inches, and at the first post nine sixteenths of an inch. The following tests were applied to a bridge of seventy-six feet span upon the Washington branch of the same road:

An engine and tender weighing forty tons, caused a deflection of five eighths of an inch. A fast passenger train deflected the bridge nine sixteenths of an inch.

Two engines and tenders, back to back, at rest, and weighing in all 77½ tons, caused a deflection of 11
16
inch,
The same at ten miles per hour, 13
16
inch,
Engines head to head at four miles per hour, 13
16
inch,
Engines head to head at eight miles per hour, 13
16
inch,
Engines head to head at twenty miles per hour, 14
16
inch.

The extreme expansion of the one hundred and twenty-eight feet chord from heat, was five sixteenths of an inch at each end, or five eighths of an inch in all, or 1
2457
th of the length; and that without the slightest derangement of masonry. The rod C B, being five times as long as C A, expands five times as much, but at the same time the lengths D A, D B, being so nearly proportional to C A, and C B, expand also in the ratio of one to five; and thus no bad result is experienced.

The estimate of strains upon this bridge is extremely simple; the whole consisting of as many separate systems as there are posts. Each set of rods sustain a rectangle equal to one panel, i. e., the two adjacent half panels. Thus A C, and C B, support the rectangle m m, m m, the rods A F, F B, the rectangle n n, n n. Allowance must of course be made for the inclination of the rods. The dimensions of the central pair will of course be the same; but those of the other sets will vary. The diagonals D H, and H L, prevent reaction; and must be able to resist the action produced by the variable load upon one panel (as noticed in 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:—

Rod. Length. Applied weight. Increased strain. Section of the 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.

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 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.

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

Centre. Rod 1. Rod 2. Rod 3. Rod 4. Rod 5.
0 E2
b2
× a
4E2
b2
× a
9E2
b2
× a
16E2
b2
× a
25E2
b2
× a
0 1002
5002
× 80
2002
5002
× 80
3002
5002
× 80
4002
5002
× 80
5002
5002
× 80
0 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[V2d2] – √¾[l2d2].
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[V2d2] – √¾[l2d2].

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:—