Chapter 2

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.

Span. Rise. Stringer. Post. Rod. Rods. Feet. In Feet. (single.) (double.)

15 1-7/8 12 x 12 6 x 8 2-1/8 diam. or 1-1/2 diam. 20 2-1/2 12 x 14 7 x 8 2-1/2 " 1-3/4 " 25 3-1/8 12 x 16 8 x 8 2-3/4 " 2 " 30 3-3/4 13 x 18 9 x 9 3 " 2-1/8 "

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 A F and F C, 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 A C until they meet in a point, as before we stop them at a, and c, and place the straining beam, a c, between them to prevent those points from approaching, suspend the points B and D from them, and start the braces B b and D b--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' b C', and A E C, by any passing load, we insert the counter braces marked R.

The braces A a and C c, must support all of the weight of the bridge and its load within the parallelogram B a c D--and the next set of braces, B b and D b, 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 sight 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

W x S T = ----- --. 8 h

[TeX: $T = \frac{W \times S}{8 h}$]

In this case we have taken the rise at 1/8 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 1/3 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,

W x S T = ----- and supplying values, we have 8 h

[TeX: $T = \frac{W \times S}{8 h}$]

240 x 120 T = --------- = 240 tons, or 537600 lbs., 8 x 15

[TeX: $T = \frac{240 \times 120}{8 \times 15 = 240$}

for the two Lower Chords, and 1/2 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 ------ = 134.4 sq. inches. 2000

[TeX: $\frac{268800}{2000} = 134.4$]

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 l2" = 288 square inches area of section, and

allowing for splicing 72 square inches, " " foot blocks, 24 " " " " bolts, 24 " " " " washers, 8 " "

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

268800 area required, ------ = 268.8 1000

[TeX: $\frac{268800}{1000} = 268.8$]

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

2240 bd³ W = --------, and we have L²

[TeX: $W = \frac{2240 \times bd^3}{L^2}$]

2240 x 8 x 1000 W = ---------------, or reducing, W=55308 lbs. 324

[TeX: $W = \frac{2240 \times 8 \times 1000}{324} = 55308$]

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

2240 x 24 x 1000 of 24" x 10"--whence W = ---------------- = 165925 lbs., 324

[TeX: $W = \frac{2240 \times 24 \times 1000}{324} = 165925$]

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 2-1/2 inches. The sizes of the rods in each set will decrease towards the centre of the bridge as the weight becomes less.

=Counterbraces.= Now, as to the necessity of Counterbracing, there are various opinions. The object of it is to stiffen the truss and check vibrations. If a load be placed over any panel point, it causes that portion of the truss to sink, and produces an elevation of the corresponding panel point at the other end of the truss--thus producing a distortion, which change of form is resisted by proper counterbraces. The strain to which this timber is subjected is caused by the moving load on one panel only--and requires only scantling of the size of the middle braces. These counterbraces should not be pinned or bolted to the braces where the cross--as their action is thereby entirely altered--but it is well to so confine them as to prevent vertical or lateral motion.

=Shoes.= Formerly it was the custom to foot the braces and counters on hard wood blocks on one side of the chord, the vertical rods passing through and screwing against a block on the other side--thus the whole strain tended to crush the chord across its fibres. This is now remedied by the use of cast iron blocks, bearing on one side of the chord, but having tubes extending through to the other side, where the washer plate for the bolts fits firmly on their ends, forming a complete protection, as all the crushing strain is received on the block itself.

=Width.= It now becomes necessary to determine upon the width between the two trusses. For a single track bridge for a railroad, 14 ft. is the usual width adopted, and for a highway bridge, from 12 to 16 ft. When a double track is required, three trusses are usually employed, with a width for each roadway of 14 ft. for railroads.

=Bolsters.= Large timbers 12 x 12, or thereabouts, are laid on the bridge seats of the abutments to support the ends of the trusses, one of these should be directly under each of the extreme panel points. A panel point is the intersection of the centre line of a brace produced, with the centre line of a chord. The rise of a truss is the vertical distance between the centre lines of the upper and lower chords.

=Camber.= Were a bridge to be framed with its chords perfectly horizontal, it would be found to fall below the horizontal line on being placed in its proper position, owing to the closing up of the joints in the upper parts of the structure, and opening of joints in the lower parts, as well as to the compression of the parts. To obviate this defect, it is usual to curve the chords slightly in a vertical direction, by elongating the upper chord, so that the bays or panels are no longer rectangular but of a trapezoidal form--and, as a consequence, the inclined web members are slightly lengthened, and the verticals become radii of the curve. The amount of deviation from a horizontal line is called the Camber.

A table of Cambers for different spans will be found further on, as also a table of multipliers, by which to multiply the camber in order to find the elongation of the upper chord. Part of the Camber table is taken from Trautwine's Engineer's Pocket-Book, (which should be the inseparable companion of every engineer,) and part was calculated for this pamphlet, according to Trautwine's rules. The table of multipliers is Trautwine's.

=Diagonal Bracing.= In order to stiffen a bridge, it should have the two Trusses braced together at the Lower Chords always, at the Upper Chords when practicable--and in case of a deck bridge, where the roadway is supported on the upper chords, it is as well to have rods for vertical diagonal braces, their planes being perpendicular to the axis of the bridge. The more usual form is similar to the web members of the Howe Truss--the rods from 3/4" to 1" in diameter, and the braces of 6" x 7" scantling, footed on wooden blocks, usually. It is more usual to have the tie rods of the horizontal diagonal bracing, and the braces themselves, meet in a point about midway of a Truss panel on the centre line, nearly, of the chord. This will generally give a half panel of diagonal bracing near each end of the truss--and it is very usual to have the diagonals foot at their intersection there against a cross timber interposed between the trusses, while the tie rod prevents any spreading.

=Floor Timbers.= The general dimensions of the transverse floor beams, when about 3 feet apart, from centre te centre, are 8" x 14", the largest dimension being the depth. The stringers should be notched to the floor beams about 1" or 2", and should be about 10" or 12" x 14". The cross ties should be 18" to 24" apart, from centre to centre, and be 3-1/2" x 6".

Large, heavy bridges require no fastening to connect them with their seats, but light bridges should be fastened, as the spring on the sudden removal of a load, (as when the last car of a train has passed,) may move it from its proper position.

=Splices.= As the upper and lower chords have to be made in several lengths, securely fastened to each other, and, in order to weaken the built beam as little as possible, it is necessary to adopt some form of splicing whereby the greatest amount of tensional strength may be retained in the chord with the least amount of cutting, and yet have a secure joint. Such a splice is shown in Pl. II, Fig. 4, and below is a table from Vose's Hand-book, giving reliable dimensions.

Span. A C B B C D Feet. Feet. Inches. Feet. 50 1.00 1-1/2 1.50 100 1.25 2 2.00 150 1.75 2-1/2 2.25 200 2.00 3 2.75

This manner of splicing requires the back of the splice block to be let into the chord stick, against which it lies, about 3/4 of an inch. To show how the various Engineers differ, as to their estimates of the sizes of the several parts of bridges, I subjoin two Tables--one by Prof. G.L. Vose, a well known Engineer, and one by Jno. C. Trautwine, an Engineer of note also--and I would premise that a bridge built according to either would be amply strong.

TABLE FOR DIMENSIONING A HOWE TRUSS BRIDGE. G.L. VOSE.

End Centre Centre Span. Rise. Panel. Chords. Braces. Braces. End Rods. Rods. 50 10 7 2--8 x 10 7 x 7 5 x 5 1--1-1/8 2--1 75 12 9 2--8 x 10 8 x 8 5 x 5 2--1-1/2 2--1 100 15 11 2--8 x 10 8 x 9 6 x 6 2--1-3/4 2--1 150 20 13 4--8 x 12 10 x 10 6 x 7 3--2 3--1 200 25 15 4--8 x 16 12 x 12 7 x 7 5--2 5--1

TABLE FOR DIMENSIONING A HOWE TRUSS BRIDGE. JNO. C. TRAUTWINE, C.E.

| | |An Upper | A Lower | An End |A Centre| | End | Centre Clear| | No.| Chord. | Chord. | Brace. | Brace.|Counter.| Rod. | Rod. Span |Rise| of |---------|---------|---------|--------|--------|-----------|----------- in | in |Pan-| No| | No| | No.| | No| | No| | No.| | No.| feet.|feet|els.|Pcs|Size.|Pcs|Size.|Pcs.|Size|Pcs|Size|Pcs|Size|Rods|Size. |Rods|Size. -----|----|----|---|-----|---|-----|----|----|---|----|---|----|----|------|----|----- 25 | 6 | 8 | 3 | 4x5 | 3 | 4x10| 2 |4x6 | 2 |5x5 | 1 |4x5 | 2 |1-5/16| 2 | 7/8 50 | 9 | 9 | 3 | 6x7 | 3 | 6x10| 2 |6x7 | 2 |5x6 | 1 |5x6 | 2 |1-5/8 | 2 |1-1/16 75 | 12 | 10 | 3 | 6x9 | 3 | 6x11| 2 |6x8 | 2 |6x6 | 1 |6x6 | 2 |1-7/8 | 2 |1-3/16 100 | 15 | 11 | 3 | 6x10| 3 | 6x12| 2 |8x9 | 2 |6x8 | 1 |6x8 | 2 |2-3/16| 2 |1-5/16 125 | 18 | 12 | 4 | 6x10| 4 | 6x13| 2 |8x10| 2 |6x9 | 1 |6x9 | 2 |2-5/8 | 2 |1-3/8 150 | 21 | 13 | 4 | 8x10| 4 | 8x14| 3 |9x10| 3 |6x9 | 2 |6x9 | 3 |2-3/8 | 3 |1-3/16 175 | 24 | 14 | 4 |10x12| 4 |10x15| 3 |9x11| 3 |8x8 | 2 |8x8 | 3 |2-5/8 | 3 |1-1/4 200 | 27 | 15 | 4 |12x12| 4 |12x16| 3 |9x12| 3 |8x10| 2 |8x10| 3 |2-7/8 | 3 |1-3/8

Both of these tables were calculated for a single Railroad track, and would answer equally well for a double Highway Bridge. In the bridge according to Trautwine's Table, each lower chord is supposed to have a piece of plank, half as thick as one of the chord pieces, and as long as three panels, firmly bolted on each of its sides, in the middle of its length.

* * * * *

=PRATT'S BRIDGE.=

This is opposite in arrangement of parts to a Howe Bridge, as the diagonals are rods, and sustain tension, and the verticals are posts, and suffer compression:

_Example._--Span = 100 feet. Rise = 12 " Panel = 10 " Weight per lineal ft. = 3000 lbs.

The tension on the lower, or compression on the upper chord, will be

300000 x 100 ------------ = 3333333 lbs. 96

[TeX: $\frac{300000 \times 100}{96} = 3333333$]

The dimensions of the chord and splicing would be found in the same manner as for a Howe Truss.

=Suspension Rods.= Fig. 1, Pl. III., represents an elevation of a Pratt Bridge. Now, it is evident that the first sets of rods must support the weight of the whole bridge and its load, which we have found to be 300000 lbs. Each truss will have to sustain 150,000 lbs., and each end set of rods 75,000 lbs. Now, if there are two rods in each set,--each rod will have to bear a strain of 37500 lbs., and this will have an increase due to its inclination, so that the strain on it must be found by the following proportion:

Height : diagonal :: W : W' or

12 : 15.8 :: 37500 : 49375 lbs.

Referring to the Table for bolts, we find that 2-1/8 gives a strength a little in excess, and will be the proper size. The next set of rods bear the weight of the whole load, less that due to the two end panels, and so on. Fig. 2, Pl. III, shows the manner of applying the rods. The bevel block should be so fitted to the chord that it will not have a crushing action.

=Counters.= Top and bottom chords are always used in this bridge, and consequently the counter rods have only to sustain the movable load on one panel. The weight of the moving load cannot be more than 2000 lbs. per lineal foot which, for a panel of 10 ft., gives 20000 lbs., or 10,000 lbs. for each set, and if we have two rods in a set, the strain on each rod will be 5000 lbs., increasing this for inclination, we shall have,

12 : 15.8 :: 5000 : 6585 lbs.,

requiring a rod of 3/4 of an inch diameter. The posts in this bridge correspond to the braces of the Howe Truss, but being vertical, are not so large.

Subjoined are two Tables, one by Prof. G.L. Vose, and one by Mr. Trautwine, giving principal dimensions for bridges of different spans of the Pratt type of Truss.

TABLE OF DIMENSIONS OF A PRATT TRUSS.

PROF. G. L. VOSE.

End Centre End Centre Counter Span. Rise. Chords. Post. Post. Rod. Rod. Rod.

50 10 2--8x10 5 x 5 4 x 4 2--1-3/8 2--1 1--1-1/2 75 12 2--8x10 6 x 6 5 x 5 2--1-5/8 2--1 1--1-1/2 100 15 3--8x10 7 x 7 6 x 6 2--1-3/4 2--1 2--1-1/8 125 18 3--8x10 8 x 8 6 x 6 3--1-7/8 3--1 2--1-1/3 150 21 4--8x12 9 x 9 6 x 6 3--2-1/8 3--1 8--1-1/8 200 24 4--8x16 10 x 10 6 x 6 5--1-7/8 5--1 3--1-1/8

TABLE OF DIMENSIONS OF A PRATT'S TRUSS.