Instructions on Modern American Bridge Building

INSTRUCTIONS

ON

MODERN AMERICAN

BRIDGE BUILDING.

WITH

PRACTICAL APPLICATIONS AND EXAMPLES,

ESTIMATES OF QUANTITIES, AND
VALUABLE TABLES.

Illustrated by four Plates and Thirty figures.

BY G.B.N. TOWER,

CIVIL AND MECHANICAL ENGINEER,

Formerly Chief Engineer U.S. Navy, and late Chandler Instructor in Civil
Engineering at Dartmouth College.

BOSTON:

A. WILLIAMS & COMPANY,

135 WASHINGTON STREET.

1874.

Entered according to act of Congress, in the year 1874, by

A. WILLIAMS & CO.,

in the office of the Librarian of Congress, at Washington, D.C.

PREFACE.

This little treatise was written for the purpose of supplying a want felt by the author while giving instruction upon the subject. It was intended for an aid to the young Engineer, and is not to be considered as a complete substitute for the more elaborate works on the subject.

The first portion of this work mentions the various strains to which beams are subjected, and gives the formulæ used in determining the amount of those strains, together with a few examples to illustrate their application, and also the method of calculating a simple truss.

The second portion names and explains the various members of a Bridge Truss, and, by means of examples, shows the method of calculating the strains upon the various timbers, bolts, etc., as well as their proper dimensions; and gives, in addition, several useful tables.

The explanatory plates, which are referred to freely throughout the work, are believed to be amply sufficient for the purpose intended.

So much has been written on this subject that it is next to impossible to be wholly original, and no claim of that nature is preferred. It is simply an arrangement of ideas, gleaned from the various works of standard authorities, and modified by the author's practice, embodied in book form.

To give a correct list of all the books consulted would be simply impossible;—but it is well to state that the Hand-book of Railroad Construction, by Prof. G.L. Vose, under whom the author served as an Engineer, has been used as authority in many cases where there has been a difference of opinions among other authors. Some parts have been quoted entirely; but due credit has been given, it is believed, wherever such is the case.

It is not claimed that this little work covers the whole ground, but it is intended to describe, and explain thoroughly, three or four of the more prominent styles of Truss, leaving the other forms of Wooden Bridges to a subsequent volume.

Abutments and Piers, as well as Box and Arch Culverts, belonging more properly to masonry, will be treated of hereafter under that head.

Iron Bridges form a distinct class, and may be mentioned separately at some future period.

If this small volume should lead the student of Engineering to examine carefully the best Bridges of modern practice, and study the larger scientific works on this art, the author will feel satisfied that his efforts have not been entirely in vain.

Cambridge, February 23, 1874.

TOWER'S

Modern American Bridge Building.

BRIDGE BUILDING.

The simplest bridge that can be built, is a single beam, or stick of timber, spanning the opening between the abutments—but this is only of very limited application—(only for spans of 20 feet and less) owing to the rapid increase in sectional dimensions which is required as the span becomes greater.

Next comes the single beam supported by an inclined piece from each abutment meeting each other at the middle point of the under side of the beam—or, another arrangement, of two braces footing securely on the beam and meeting at a point above the middle point of the beam, which is suspended from the apex of the triangle formed by them, by means of an iron rod—These arrangements may be used up to 50 feet. For any span beyond 50 feet, modifications of this arrangement are used which will be described hereafter. Now let us investigate shortly the different strains that the various parts of a bridge have to bear—and the strength of the materials used. The theory of strains in bridge trusses is merely that of the Composition and Resolution of Forces. The various strains, to which the materials of a bridge are subjected—are compression, extension and detrusion.

Wood and Iron are the materials more generally employed in bridge construction—and in this pamphlet we shall take the following as the working strength of the materials—per square inch of section.

Tension. If a weight of 2000 lbs. were hung to the lowest end of a vertical beam, so that the line of action of the weight and axis of the beam formed one and the same straight line—the tension on the beam would be 2000 lbs. But, if the beam were inclined, and the force acted in a vertical direction, then the strain would be increased in the ratio of the increase of the diagonal of inclination over the vertical;—suppose the beam is 20 ft. long and inclined at an angle of 45°—and let 2000 lbs., as before, be suspended from its lower end. Now the diagonal being 20,—the vertical will be 14.014 ft.—and the strain will be found as follows,—

14.014 : 20 :: 2000 : 2854—lbs.

The greater the angle of inclination from the horizontal, the less the strain from a given load—and when the beam is vertical the weight causes the least strain.
 

Compression. If we load a vertical post with a weight of 2000 lbs., the strain of compression exerted upon the post will be 2000 lbs. Now, if we incline the post—the strain will be increased, as we have shown above under the head of tension, and in like manner, dependent upon the inclination.

But when wood, iron, or any other material is used for a pillar or strut, it has not only to resist a crushing force, but also a force tending to bend or bulge it laterally.

A post of circular section with a length of 7 or 8 diameters will not bulge with any force applied longitudinally, but will split. But if the length exceeds this limit—it will be destroyed by an action similar to that of a transverse strain.

A cast iron column of thirty diameters in length, is fractured by bending; when the length is less than this ratio—by bending and splitting off of wedge shaped pieces. But by casting the column hollow, and swelling it in the middle, its strength is greatly increased.

Barlow's formula for finding the weight that can be sustained by any beam, acting as a pillar or strut, before bending, is:—

, whence

now, having the weight given, and assuming the dimensions of the cross-section—we shall have

and

in the above formulæ,

W = weight in pounds.
L = length in feet.
E = a constant.
b = breadth in inches.
d = depth in inches.
 

Transverse Strains. The strain caused by any weight, applied transversely, to a beam supported at both ends, is directly as the breadth, and square of the depth, and inversely as the length. It causes the beam to be depressed towards the middle of its length, forming a curve, concave to the horizontal and below it. In assuming this form—the fibres of the upper part of the beam are compressed, and those of the lower part are extended—consequently there must be some line situated between the upper and lower surfaces of the beam where the fibers are subjected to neither of these two forces, this line is called the neutral axis.

These two strains of compression and extension must be equal in amount—and upon the relative strength of the material to resist these strains, as well as its form and position, the situation of this axis depends. If wood resists a compression of 1000 lbs. per square inch of section, and a tension of 2000 lbs. the axis will be twice as far from the top as from the bottom in a rectangular beam.

The following table by Mr. G.L. Vose gives, with sufficient accuracy for practice, the relative resisting powers of wood, wrought, and cast iron, with the corresponding positions of the axis.

Thus we see that the resistance of a beam to a cross strain, as well as to tension and compression, is affected by the incompressibility and inextensibility of the material.

The formula for the dimensions of any beam to support a strain transversely is

S = the ultimate strength in lbs.
b = the breadth in inches.
d = the depth in inches.
l = the length in inches.

Detrusion. Detrusion is the crushing against some fixed point, such as obtains where a brace abuts against a chord, or where a bridge rests on a bolster; and the shearing of pins, bolts and rivets, also comes under this head.
 

General Abstract. The resistance to the above mentioned strains varies as the the area of the cross section; so that by doubling the area we double the strength. Any material will bear a much greater strain for a short time than for a long one. The working strength of materials, or the weight which does not injure them enough to render them unsafe, is a mooted point, and varies, according to the authority, from 1-3 to 1-10 of the ultimate strength. The ratio of the ultimate strength to the working strength is called the factor of safety.

The following is a table of ultimate and working strengths of materials, and factors of safety:

Lateral Adhesion. Lateral adhesion is the resistance offered by the fibres to sliding past each other in the direction of the grain, as when a brace is notched into a chord, or tie beam, at its foot, it is prevented by the lateral adhesion of the fibres from crowding off the piece, to the depth of the notch, against which it toes. Barlow's experiments give the lateral adhesion of fir as 600 lbs. per square inch, and the factor of safety employed varies in practice from 4 to 6, giving a working strength of from 150 to 100 lbs. per square inch.

TABLE OF COMPRESSIVE RESISTANCE OF TIMBER.

In tensional strains, the length of the beam does not affect the strength; but in the beams submitted to compression, the length is a most important element, and in the table given above, the safety strains to which beams may be subjected, without crushing or bending, has been given for lengths, varying from 6 to 60 diameters.

PRACTICAL RULES.

Tensional Strain.

Let T = whole tensional strain.
  "   S = strength per square inch.
  "   a = sectional area in inches.
Then we have T = Sa.

Now to find the necessary sectional area for resisting any strain, we have the following general formula:

a = T ÷ S

or, by substituting the working strenths for the various materials in the formula, we have for wood,

a = T ÷ 2000

Wrought Iron,

a = T ÷ 1500

Cast Iron,

a = T ÷ 4500

But, in practice, cast iron is seldom used except to resist compression.

Strains of Compression. Allowing the same letters to denote the same things as above, we have for

Wood,

a = T ÷1000

Wrought Iron,

a = T ÷ 12000

Cast Iron,

a = T ÷ 25000

As this pamphlet has to do with wooden bridges only, nothing will be said of the proper relative dimensions of cast-iron columns to sustain the strains to which they may be subjected, but a table of the strength of columns will be found further on.

Transverse Strains.

Let W = breaking weight in lbs.
  "   s = constant in table.
  "   b = breadth in inches.
  "   d = depth in inches.
  "   L = length in inches.

Then, for the power of a beam to resist a transverse strain, we shall have,

This formula has been derived from experiments made by the most reliable authorities.

The constant, 1250, adopted for wood in the following formula, is an average constant, derived from the table, of those woods more commonly used.

Now to reduce the formula to the most convenient shape for use, we substitute the value of s, and we have

or

But, to reduce the load to the proper working strain, we must divide this equivalent by 4, the factor of safety, and we shall have

.

Let us apply the formula—

Case I. Given a span of 14 feet,
a breadth of 8 inches,
a depth of 14 inches.

Required the safe load.

The formula becomes, by substitution,

= 11,666 lbs.

Case II. Given the safety load 18000 lbs.,
the breadth 9 inches,
the length 14 feet.

Required the depth.

From the above formula we have

substituting

= 16 inches nearly.

Case III. Given the safety load 22,400 lbs.,
the depth 18 inches,
the length 14 feet.

Required the breadth.

Deriving b from the foregoing, we have,

substituting

= 9.3 inches nearly.

For a cast iron beam or girder—Mr. Hodgkinson from numerous carefully conducted experiments that, by arranging the material in the form of an inverted T—thus creating a small top flange as well as the larger bottom one, the resistance was increased, per unit of section, over that of a rectangular beam, in the ratio of 40 to 23.

In this beam the areas of the top and bottom flanges are inversely proportional to the power of the iron to resist compression and extension. Mr. Hodgkinson's formula for the dimensions of his girder, is

The factor of safety being 6 for cast iron beams—the formula for the working load will be,

and, to find area of lower flange, we shall have

The general proportions of his girders are as follows:

Length, 16
Height, 1
Area Top Flange, 1.0
Area Bottom Flange, 6.1
In the above formula for cast iron beams,
  W = weight in tons.
  a = area in square inches of bottom flange.
  d = depth in inches.
  h = length in inches.

The web uniting the two flanges must be made solid—as any opening, by causing irregularity in cooling, would seriously affect the strength of the beam.

Example.—Required the dimensions of a Hodgkinson girder—for a span of 60 feet—with a load of 10 tons in the centre.

= 37 inches nearly.

and the area of the top flange will be,

37 ÷ 6 = 6.16 inches—

so that our dimensions will be as follows:

Length, 30 feet.
Depth, 45 inches.
Area Top Flange, 6.16 inches.
Area Bottom Flange, 37 inches.

The thickness of web is usually a little greater at the bottom than at the top, and varies from 1/14 to 1/24 of the depth of the girder. The bottom rib is usually made from six to eight times as wide as it is thick, and the top rib from three to six times as wide as thick, so that, in the example above given, we could have as dimensions for the parts

Top Flange, 4¼ X 1½ inches nearly.
Bottom Flange, 6 X 2½ inches nearly.
Web, 1½ inches thick.

The simplest bridge, consisting of a single stick, to span openings of 20 feet and under, is calculated according to the formula

Example.—The depth of a beam, of 12 feet span and 12 feet wide, to support a load of 22400 lbs. will be

= 15 in. nearly.

The following Table was calculated by the above rule—and the dimensions altered according to the actual practice of the writer.

 
These dimensions will give ample strength and stiffness. Fig. 1, Plate I. gives an illustration of this kind of bridge—in which a, a, are the bolsters or wall plates, shown in section, to which the bridge beams are notched and bolted. Fig. 1, A, Plate I, shows the method of diagonally bracing these beams by planks, dimensions of which in general use are 6 to 8 by 2 to 3 inches. The track should rest on ties, about 6 inches by 8 or 10 inches—the same bolt confining the ends of the ties and diagonal braces when practicable. These ties should be notched on the string pieces 2 or 3 inches—without cutting the stringers. Below is a table giving general dimensions, in inches, of the several parts of a bridge of this description.

 
Each bolt must have a washer under the head, and also under the nut. For a span of from 15 to 30 feet, we can use the combination shown in Plate II, Fig. 3. The piece A F must have the same dimensions as a simple string piece of a length A B—so that it may not yield between B and either of the points A or D. The two braces DF and EF must be stiff enough to support the load coming upon them. Suppose the weight on a pair of drivers of a Locomotive to be 10 tons, then each side must bear 5 tons, and each brace 2½ tons = 2½ x 2240 = 5600 lbs.

Now, to allow for sudden or extra strains, call 8000 lbs. the strain to be supported by each brace, and, accordingly, 8 square inches of sectional area would be sufficient for compression only; but, as the brace is inclined, the strain is increased. Let the vertical distance from A to D be 10 ft., and, calling the span 30 ft.—A B will be 15 ft.—from whence D F must be 18 ft., then we shall have the proportion

10:18::8000:14400 lbs.

which would require an area of about 15 square inches of section to resist compression, or a piece 3x5 inches. Now, as this stick is more than 6 or 8 diameters in length, it will yield by bending—and consequently its area must be increased. The load, which a piece of wood acting as a post or strut will safely sustain, is found by the formula already given.

Now substituting 3 for b, and 5 for d, we have

= 2592 lbs.

which is not enough. Using 6 for b and 8 for d, we have

= 21238 lbs.

which is something larger than is actually required, but it is no harm to have an excess of strength. Now in many cases this arrangement would be objectionable, as not affording sufficient head room on account of the braces—and we can as well use the form of structure given in Pl. I. Fig. 3, since it is evidently immaterial whether the point B be supported on F or suspended from it, provided we can prevent motion in the feet of the braces, which is done by notching them into the stringer at that point. This of course creates a tensional strain along the stringer, which is found as follows:—Representing the applied weight by FB, Pl. II, Fig. 2, draw BD parallel to FC, also DH parallel to AC—DH is the tension. This is the graphical construction, and is near enough for practice. Geometrically we have the two similar triangles AFB and DFH, whence

AF: DF:: AB: DH

and

This style of structure may be used up to 50 feet, but it is not employed for spans exceeding 30 feet in length. It is very customary to make the braces in pairs so as to use smaller scantling, and gain in lateral stiffness—the two pieces forming one brace by being properly blocked and bolted together. Below is given a table of dimensions for the various parts of this style of structure:

 
Single Beams under each rail firmly braced laterally, and trussed by an iron rod, (or preferably by two iron rods,) and a post on the under side of the beam. The deflection of the rod is usually taken at 1/8 of the span. Pl. II., Fig. 1, represents this style of trussing a beam—which is generally used for spans of from 15 to 30 ft. Below is a table of dimensions for this truss with single and double rods; if double rods are used only half the given section will be necessary for each one of the pair.

 
It is as well to tenon the post into the beam, and also strap it firmly with iron plates—and the end should be shod with iron to form a saddle for the rods to bear upon.

Now if we should make a bridge, on the plan of Fig. 3, Pl. I., 75 or 100 feet, or perhaps more, in length, the braces AF and FC, would not only be very long but very large and heavy, and one chief requisite in a good bridge is, to have all the beams so proportioned that they will resist all the strains acting upon them, without being unnecessarily large. It now becomes necessary to have a different arrangement of the parts of the truss in order to obtain increased length of span.

Suppose we have a span, of 40 feet, as represented in Fig 2, Pl. I. Now instead of running the braces from AC until they meet in a point, as before we stop them at a, and c, and place the straining beam, ac, between them to prevent those points from approaching, suspend the points B and D from them, and start the braces Bb and Db—and, if the truss were longer, would continue on in the same manner as far as needful.

To prevent the truss from altering its form, as shown by the dotted lines A'bC', and AEC, by any passing load, we insert the counter braces marked R.

The braces Aa and Cc, must support all of tho weight of the bridge and its load within the parallelogram BacD—and the next set of braces, Bb and Db, sustain that part of the load which comes over the centre of the bridge. Consequently the braces must increase in size from the centre towards the abutments. The rods resist the same pressure in amount as their braces—but being vertical, do not need the increase, given to the braces on account of their inclination—but increase simply with the strain upon them, from the centre to the ends of the truss.

There are many forms of small bridges differing from those enumerated, in various minor details, but sufficient has been said to give the reader a fair idea of the strains upon the different parts, and how to arrange and proportion the materials to resist them.

PRACTICAL RULES AND EXAMPLES IN WOODEN BRIDGE BUILDING.

In any case that may arise, we must determine approximately the gross weight of the bridge and its load—as a basis, and then we can proceed as follows—in case of a Howe, Pratt, or Arch Brace Truss.
 

To find the dimensions of the Lower Chord.

The tension at the centre of the Lower Chord is found by dividing the product of the weight of the whole bridge and load by the span, by eight times the height—or letting T=tension in lbs., W=weight of bridge and load in lbs., S=span in feet, and h=rise or height—we have —. In this case we have taken the rise at ⅛ of the span, which is evidently the best ratio between those dimensions, as it equalizes the vertical and horizontal forces. As to the proportions of the bays or panels, (or that portion of the truss bounded by two adjacent verticals, as struts or ties, and the chords,) the ratio of the rise (or the vertical distance between the centre lines of the two chords,) and the length on the chord should be such, that the diagonal truss members may make an angle of about 50° with the chords; as the size of the timbers is increased by decreasing the angle, and, if the angle is increased, there are more timbers required.

Mr. G.L. Vose, in his admirable work on R.R. Construction, observes very truly that "The braces, at the end of a long span, may be nearer the vertical than those near the centre, as they have more work to do. If the end panel be made twice as high as long, and the centre panel square, the intermediates varying as their distance from the end, a good architectural effect is produced."

Now it is necessary for us to have some data from which to determine the approximate weight of the bridge, and also its load. These can be found by comparing weights of bridges in common use, as obtained from reports. In a small bridge of short span, the weight of the structure itself may be entirely neglected, because of the very small proportion the strains caused by it bear to those due to the load;—but, in long spans, the weight becomes a very important element in the calculations for strength and safety—inasmuch as it may exceed the weight of the load.

In all Bridges of 120 ft. span, about ⅓ of a ton, per foot run, will be the weight of each truss for a single track, including floor timbers—transverse bracing, &c. If the bridge were loaded with Locomotives only, the greatest load would be, on the whole bridge—160 tons = 1.33 tons per ft. run of the bridge or .666 tons per ft. run of each truss. Now if we make the rise of the bridge 15 ft., and divide the span into 12 panels of 10 ft. each, we shall have for total weight of bridge and load 240 tons, or for a single truss 10 tons to each panel.

Lower Chords. Now to find the tension on the Lower Chords, and supplying values, we have = 240 tons, or 537600 lbs., for the two Lower Chords, and ½ of this, or 268800 lbs. for one chord. The Tensional Strength of timber for safety may be taken at 2000 lbs. per square inch of section, and hence the area of timber required to sustain the above strain will be 268800 ÷ 2000 = 134.4 sq. inches. But this chord has also to sustain the transverse strains arising from the weights passing over it, and, as in the case of a Locomotive, the weight of 20 tons on 2 pair of drivers, (or 10 tons for one truss,) may be concentrated on the middle point of a panel—the chord must be so proportioned as to safely bear, as a horizontal beam, this weight. Suppose we take three sticks of 8" x 12", to form the chord (the greater dimension being the depth,) we shall have 3 x 8" x 12" = 288 square inches area of section, and

we shall have after deducting allowances (288-128) 160 square inches area, giving an excess over 134.4, the area demanded, sufficient to cover allowances for any accidental strain.
 

Upper Chords. The upper chords are compressed as forcibly as the lower ones suffer tension—owing to the action and reaction of the diagonals. In this case the compression is 268800 lbs., and as 1 square inch of section will safely bear 1000 lbs., we have for the area required, 268800 ÷ 1000 = 268.8 square inches,—three pieces 8" x 11" will give 264 square inches and this area will require no reduction, as the whole chord presses together when properly framed and is not weakened by splicing. So far, the calculations made would apply to either of the three Bridges mentioned, as well as to a Warren Truss. But now, to obtain the dimensions of the web members, so called, of the Truss, it is necessary to decide upon the specific variety. The form of Bridge in more general use in the United States is called the Howe Truss, from its inventor, and in spans of 150 feet, and under, is very reliable; for spans exceeding 150 ft. it should be strengthened either by Arch Braces or by the addition of Arches, as the heavy strains from the weight of bridge and load bearing on the feet of the braces near the abutments, tend to cripple and distort the truss by sagging, although the Baltimore Bridge Co. have built a Wooden Howe Bridge of two Trusses of 300 ft. span, 30 ft. rise, and 26 ft. wide, without any arch, but it has a wrought iron lower chord, and is only proportioned for a moving load of 1000 lbs. per ft. run. [Vide Vose on R.R. construction.]

In order to ensure uniformity in strength in the chords—but one joint should be allowed in a panel—and that should come at the centre of the panel length—but in long spans this cannot always be done.
 

Web Members. We will now proceed to calculate the web members of a Howe Truss of the foregoing dimensions, when subjected to the strains above mentioned.
 

Braces. The end braces must evidently support the whole weight of the bridge and load, which for one end of one truss will be 134400 lbs., and as these braces are in pairs,—67200 lbs. will be the strain vertically on the stick—but as this stick is a diagonal—whose vertical is 15 ft., and horizontal 10 ft., we shall have for its length 18 ft. in round numbers, whence the strain along the diagonal will be found from the proportion 15 : 18 :: 67200 : 80640 lbs., whence we have an area of 80 inches required for compression, or a stick of 8" x 10". Now, to ascertain if this is stiff enough for flexure, we will substitute these values in the equation , and we have , or reducing, W = 55308 lbs. Now, these proportions will give ample strength for both flexure and compression, for if we block the two sticks composing the end brace together, and firmly connect them by bolts, we shall have a built beam of 24" x 10"—whence = 165925 lbs., and as 134400 lbs. was all that the conditions demand, we really have an excess of strength. The next set of braces supports the weight of the rectangle included between the upper ends of the braces and the two chords, and the dimensions of the sticks are calculated in the same manner. We find, as we approach the centre of the bridge, that the strains on the braces become less, and consequently their scantling should be reduced, but in ordinary practice this is seldom done.
 

Rods. The next thing is to ascertain the dimensions of the various tie rods. It is evident that the same weight comes upon the first set of rods, as on the first set of braces—which will give for the rods at one end of one truss, 134400 lbs.; and as there are two of these rods, each will sustain a strain of 67200 lbs.—and, at 15,000 lbs. per square inch, will have an area of 4.48 sq. inches, and, by Vose's Tables, must have a diameter of 2½ inches. The sizes of the rods in each set will decrease towards the centre of the bridge as the weight becomes less.