Chapter VIII: Wooden Bridges (1)

139. Wooden bridging, owing to its cheapness and fitness for universal application, has been and is being adopted in all parts of the country. Almost any variety of form may be seen upon our railroads, and though less durable than stone or iron, it may with proper precaution be made to last a long time.

OF THE FORCES AT WORK IN BRIDGES.

140. There are four distinct strains to which a piece of timber or a bar of metal may be exposed, each of which tends to destroy the piece in a different manner. The amount and character of these strains, depend upon the position of the bar or beam, and upon the direction of the force.

A beam may be pulled apart by stretching,—_Tension_.

It may be destroyed by crushing,—_Compression_.

It may be broken transversely,—_Cross strain_.

It may be crushed across the grain,—_Detrusion_.

TENSION.

141. If one thousand pounds were hung from the end of a suspended timber, so that the direction of the weight coincides with the axis of the timber, then will the tension upon the beam be one thousand pounds.

If the direction of the force is vertical, and the beam is inclined, then the strain is increased by as much as the diagonal of inclination exceeds the vertical; for example, let one thousand pounds be suspended from the lower end of a beam ten feet long, inclined at an angle of 45°. The diagonal being ten, the vertical will be 7.07 feet, and the strain is increased as follows:—

7.07 to 10 as 1,000 to 1,414 lbs.

As the angle of inclination, from the horizontal, increases, the strain from a given load decreases, until the beam is vertical, when a weight acts with its least power.

COMPRESSION.

142. If a vertical post is loaded with one thousand pounds, the compressive strain upon that post will also be one thousand pounds. If a post is inclined, the amount of strain is increased, as noticed in the case of tension, and to the same amount, that is, depending upon the inclination.

A piece of wood or metal acting as a post, or pillar, must not only be able to resist crushing, but also bending or bulging laterally.

143. A cylinder of which the length is only seven or eight times the diameter, will not bulge by any force that can be applied to it longitudinally, but will split. When the length exceeds this, it will be destroyed by a similar movement to that produced by a cross strain. When the length of a cast-iron pillar is thirty diameters, the fracture is produced by bending alone; when less, partly by bending and partly by fracture. When the column is cast hollow, and enlarged towards the middle, the strength is increased in a very great ratio.

144. The formula for finding the weight which any beam acting as a post, will support before bending, is, according to Barlow, who considers the weight as varying inversely as the length, as follows:—

(_WL^2_)/(80_E_) = _bd^3_,

and the value of _W_ is

(_bd^3_ × 80_E_)/(_L^2_),

and the weight being given, and the sectional dimensions assumed, we have

_d_ = ∛((_WL^2_)/(80_Eb_)),

and

_b_ = (_WL^2_)/(80_Ed^3_),

Where _W_ represents the weight in pounds,
_L_ represents the length in feet,
_E_ represents a constant,
_d_ represents the depth in inches,
_b_ represents the breadth in inches.

CROSS STRAIN.

145. The amount of strain caused by any weight applied in a transverse direction, to a beam supported at both ends, is as the breadth, as the length inversely, and as the square of the depth. Whatever depression takes place, tends to shorten the upper, and to extend the under-side; whence the fibres of the top part suffer compression, and those of the bottom extension. The amounts of compression and extension must of course be equal, and therefore if any material resists these two strains in a different degree, the number of fibres opposing each will also be different.

The top being compressed, while the bottom is extended, of course at some point within the beam there exists a line which suffers neither compression nor extension. The position of this line (the neutral axis) depends upon the relative power of the material to oppose the strains, upon its form and upon its position. Thus if wood resists two thousand pounds per square inch of extension, and one thousand pounds of compression, the axis will be twice as far from the top as from the bottom.

In some materials the neutral axis changes its place while the bar is at work; thus wrought iron, after being a little compressed, will bear a great deal more compression than when in its original state; also the lower fibres, after being extended, will resist less than at first; the effect of which two actions is to move the neutral axis up.

146. The following table shows the relative resisting powers of wood, wrought and cast-iron; with the corresponding positions of the axis, with sufficient accuracy for practice.

Distance of axis
Material. Resistance to Resistance to Ratio. from top, in
extension. compression. fractions of the
depth.

Wrought iron, 90 66 90/66 90/156 or 0.58 Cast-iron, 20 111 20/111 20/131 or 0.15 Wood, 2 1 2/1 ⅔ or 0.66

Thus in beams subjected to a cross strain, as well as to a direct extensile or compressive one, the resistance is effected by the incompressibility and inextensibility of the material.

147. The formula for dimensioning any beam to support a given weight transversely is

_S_ = (4_bd^2_)/_e_,

Where _S_ represents the ultimate strength in lbs.
_b_ represents the breadth in inches,
_d_ represents the depth in inches,
_e_ represents the length in inches,

DETRUSION.

148. Detrusion, or crushing across a fixed point, is such as occurs wherever a brace abuts against a chord, or where a bridge bears upon a bolster or wall plate; also the shearing of bolts, pins, and rivets.

GENERAL RESISTANCE OF MATERIALS.

149. The resistance to extension, to compression, (as regards simple crushing,) and to detrusion, is as the area of cross section; i. e., if we double the area, we double the strength. The resistance to a cross strain is _as the breadth_, _as the length inversely_, and _as the square of the depth_; i. e. if we double the breadth we double the strength; if we double the length, we divide the strength by two; and if we double the depth, we multiply the strength by four.

ACTUAL STRENGTH OF MATERIALS.

150. Any material will bear a much larger load for a short time than for a long one. The weight that does not so injure materials as to render them unsafe, is from one third to one fourth only of the ultimate strength. Throughout the present work one fourth will be the most that will in any case be used.

WROUGHT IRON.

151. _Extension._

lbs. per square inch.
Mean of 17 experiments by Barlow (p. 270) 62,720
Weisbach’s Mechanics (Vol. ii., p. 71) 60,500
Overman’s Mechanics, (p. 408, 409) 61,333
Brown, Rennie, and Telford, (mean) 65,251
——————
The mean 62,451
——————
Reducing by 4 for safety 15,613

Or in round numbers 15,000 lbs. per square inch, is the resistance of wrought iron to extension, to be used in practice.

152. _Compression._—Great discrepancies appear among writers on the strength of materials, as to the compressive strength of wrought iron. Though all estimate the resistance to compression, as great as to extension, yet no one in summing up the general result of experiment, places the former at more than from 50 to 75 per cent. of the latter. William Fairbairn gives, as the relative resistances to extension and compression in bars applied as girders, 2 to 1.

We have by Weisbach 56,000
We have by Rondelet 70,000
We have by Hodgkinson 65,000
——————
The mean 63,667
——————
Reducing by 4 15,917
——————
In round numbers 16,000 lbs. per square inch.

As far as practice is any guide, from 8,000 to 12,000 pounds per inch is the most to be used. The ratio of 90 to 66, seems to express very nearly the action as in the most reliable structures; which will, therefore, be adopted, or 11,000 pounds per square inch nearly. The resistance to compression is very much greater after wrought iron has been somewhat compressed.

CAST-IRON.

153. _Extension._—This material is seldom used to resist a tensile force. That the tables may be complete, however, the following is given:—

By Weisbach 20,000 pounds.
By Barlow 18,233 pounds.
By Overman 20,000 pounds.
By Rennie 18,000 pounds.
By Hodgkinson 16,577 pounds.
By the British Iron Commission 15,711 pounds.
——————
The mean 18,087 pounds.
——————
Reducing by 4 4,522 pounds.
——————
In round numbers 4,500 pounds.

154. _Compression._

By Weisbach 109,800 pounds.
By Hodgkinson 107,520 pounds.
By Iron Commission 100,000 pounds.
Stirling’s toughened 130,000 pounds.
———————
Mean of Common 105,773 pounds.
———————
Mean of Stirling’s 130,000 pounds.
———————
Reducing by 4 for safety (Common) 26,443 pounds.
———————
Reducing by 4 for safety (Stirling’s) 32,500 pounds.
———————
In round numbers (Common) 25,000 pounds.
———————
In round numbers (Stirling’s) 30,000 pounds.

155. Following are given the condensed results of the preceding figures, which may be relied upon as giving perfectly safe dimensions in practice.

Wrought Iron. Cast-Iron.
15,000 4,500 Tensile strength,
11,000 25,000 Compressive strength.

For additional remarks on iron, see chap. IX.

156. _Nature and Strength of American Woods._

Name of the Weight per Resistance Resistance Value of
wood. cubic foot. to to _S_. Elasticity.
Extension. Compression.

White Pine 26 12,000 6,000 1,229 —— Yellow Pine 31 12,000 6,000 1,185 —— Pitch Pine 46 12,000 6,000 1,727 4,900 Red Pine 35 12,000 6,000 1,527 7,359 Virginia Pine 37 12,000 6,000 1,456 —— Spruce 48 12,000 6,000 1,036 —— Larch 33 12,000 6,000 907 2,465 Tamarack 26 12,000 6,000 907 —— White Cedar 22 8,000 4,000 766 —— Canada Balsam 34 12,000 6,000 1,123 —— White Oak 48 15,000 7,500 1,743 8,595 Red Oak 41 15,000 7,600 1,687 —— Live Oak 72 15,000 7,200 1,862 —— White Beech 44 18,000 9,100 1,380 5,417 Red Beech 48 18,000 9,000 1,739 —— Birch 44 15,000 7,000 1,928 —— Black Birch 41 15,000 7,200 2,061 —— Yellow Birch 36 15,000 7,200 1,335 —— Ash 38 16,000 8,100 1,795 6,581 Black Ash 35 16,000 8,000 861 —— Swamp Ash 57 16,000 8,000 1,165 —— Hickory 51 15,000 7,200 2,129 —— Butternut 54 15,000 7,600 1,465 —— Ironwood 54 16,000 8,100 1,800 —— Rock Elm 45 16,000 8,011 1,970 2,799

The mean tensile strength of wood is 14,080 lbs. Reducing by 4 for safety 3,520 lbs. Reducing for want of seasoning 2,000 lbs. The reduced mean compressive strength 1,000 lbs. Reduced resistance to detrusion 150 lbs. Ratio of tensile to compressive strength 2 to 1. Mean value of _S_ in formula (_WL_ = 4_Sbd^2_) for the woods most used in practice 1,250.

157. The lateral adhesion of fir was found, by Barlow, to be six hundred pounds per square inch. (Lateral adhesion is the resistance which the fibres offer to sliding past each other in the direction of the grain; as, in pulling off the top of a post where it is halved on to the chord.)

158. As regards the nature of timber, seasoning, time of cutting, etc., although these are important items, still, generally, commercial considerations outbalance all else. The most complete treatise on the nature of woods, is “Du Hamel, _L′exploitation des bois_;” from which it appears that the best oaks, elms, and other large trees, are the product of good lands, rather dry than moist. They have a fine, clear bark, the sap is thinner in proportion to the diameter of the trunk, the layers are less thick, but more adherent the one to another; and more uniform than those of trees growing on moist places. The grain of the latter may look very fine and compact, but microscopic examination shows the pores to be full of gluten.

The density of the same species of timber, in the same climate, but on different soils, will vary as 7 to 5; and the strength, both before and after seasoning, as 5 to 4.

In trees not beyond their prime, the density of the butt is to that of the top, as 4 to 3; and of centre to circumference, as 7 to 5. After maturity, the reverse occurs in both cases.

Oak, in seasoning, loses from ¼ to ⅓ of its weight; but its strength is increased from 30 to 40 per cent.

GENERAL TABLE OF THE NATURE OF MATERIALS.

159. _The tensile strength of wrought iron assumed as 1,000._

Weight Sum divided
Material. Tension. Compression. Cross Sum. per by weight
Strain. cubic per cub.
ft. ft.
Cast-Iron 300 1,666 31.68 1,997.68 450 4.4
Wrought Iron 1,000 733 55.40 1,788.40 480 3.7
Wood 133 66 5.60 204.60 30 6.8

The advantage possessed by iron over wood, is in durability only. The above figures show how much more of the strength of the material is consumed by its own weight in iron than in wood. In actual practice, however, the method of making joints and other details often render iron the lightest material.

RULES FOR PRACTICE.

TENSION.

160. The tensile strength of any material, is expressed by the formula

_T_ = _Sa_,

Where _T_ represents the whole strength,
_S_ represents the strength per square inch,
_a_ represents the area of section in inches.

whence the necessary area of section of any material to resist a tensile strain, is found by the following rules:—

Wrought Iron

_a_ = _W_/15,000,

Cast-Iron

_a_ = _W_/4,500,

Wood

_a_ = _W_/2,000.

COMPRESSION.

161. Wrought Iron

_a_ = _W_/12,000,

Cast-Iron

_a_ = _W_/25,000,

Wood

_a_ = _W_/1,000.

CROSS STRAIN.

162. The power of any material to resist a cross strain, is shown by the formula

_W_ = (4_sbd_^2)/_L_,

Where _W_ represents the breaking weight in pounds,
_s_ represents the constant in the table of woods,
_b_ represents the breadth in inches,
_d_ represents the depth in inches,
and _L_ represents the length in inches,

and to reduce the load to one fourth of the breaking weight

_W_ = (4_sbd_^2)/(4_L_),

and finally, by substituting for 4_s_, 4 × 1,250, (1,250 of the table of woods,) we have

_W_ = (5000_bd_^2)/(4_L_).

Also, knowing the weight to be supported, and requiring the dimensions, we take out the values of _d_ and _b_, and have

_d_ = √((_W_ × 4_L_)/(5000_b_)) = the depth,

_b_ = (_W_ × 4_L_)/(5000_d_^2) = the breadth.

As an example of the use of the formula, take the following:—

Let the span, or length, be 20 feet,
The breadth 12 inches, and depth 18,

required the load.

The formula

_W_ = (5000_bd_^2)/(4_L_)

becomes

_W_ = (5000 × 12 × 18^2)/4 × 240 = 20,250 lbs.

Again, the weight to be supported being 15,000 lbs., length 30 feet, breadth 16 inches, the formula for the depth becomes

_d_ = √((15000 × 1440)/(5000 × 16)) = √(270) = 16 inches,

also,

_b_ = (15000 × 1440)/(5000 × 256) = 21600000/1280000 = 16 inches.

CAST-IRON.

163. The formula, expressive of the strength of a cast-iron beam, is

850_bd_^2 = _WL_,

from which we have

_b_ = (_LW_)/(850_d_^2) = the breadth,

and _d_ = √((_L_ × _W_)/(850_b_)) = the depth.

WROUGHT IRON.

164.

952_bd_^2 = _WL_,

whence

_b_ = (_WL_)/(700_d_^2) = the breadth,

and _d_ = √((_LW_)/(700_b_)) the depth.

165. Mr. Hodgekinson found, that by arranging the material in a cast-iron beam, as in fig. 60, that the resistance per unit of section was increased over that of a simple rectangular beam, in the ratio of 40 to 23. He makes the general proportion of such girders as follows:—

Length 16
Height 1
Area of top flange 1.0
Area of lower flange 6.1

In this consummate disposition of material, the areas of top and bottom flanges are made inversely proportional to the power of cast-iron to resist compression and extension.

166. Mr. Fairbairn found, that in wrought iron flanged girders, (under which come the various rails, chap. XIII.,) the top web should contain double the area of the lower one. This agrees with the conclusion adopted on page 129, as wrought iron resists more extension than compression.

167. In cast-iron girders, on no account should there be introduced webs, or openings of any kind, either from economic or ornamental motives; as the uniformity of cooling is thereby very much opposed.

168. Mr. Hodgekinson gives, as the result of his experiments, the following formula for dimensioning the cast-iron girder above referred to.

_W_ = (26_ad_)/_L_,

Where _W_ is the breaking weight in tons,
_a_ the area of the bottom flange,
_d_ the depth of the girder in inches,
_L_ the length in inches.

As it is not considered safe to load a cast-iron beam with more than one sixth of the breaking load, the formula may be expressed as follows:—

_W_ = (26_ad_)/6_L_,

for the weight in tons which may be safely borne, and transforming

_a_ = (6_WL_)/26_d_

for the area of the lower flange.

_Example._—Required the dimensions of a cast-iron beam, of Mr. Hodgekinson’s form, for a span of thirty feet, to support a load of ten tons at the centre.

Span 30 feet, Whence—
Length 34 feet, Length 34 feet,
Load 10 tons at centre. Span 30 feet,
Depth 25½ inches,
Lower flange 32.58 square inches,
Upper flange 5.34 square inches,

_a_ = (6 × 10 × 12 × 30)/(26 × (34 × 12)/16) = 32.58

and 32.58/6.1 = 5.34.

and the area of the top flange will be

36/6 = 6,

whence the following dimensions:—

Length 30 feet,
Depth 23 inches,
Lower flange 36 square inches,
Upper flange 6 square inches,

OF POSTS.

169. A post may be very well able to resist the compressive strain thrown upon it by any load, but may bulge, or bend, laterally.

The formula by which beams are dimensioned for this requirement, changes with the material, and with the form of section. For rectangular posts of wood, we have the formula below.

_W_ = (2240_bd_^3)/(_L_2),

Where _W_ represents the weight in lbs., which may be safely borne,
_b_ represents the breadth in inches,
_d_ represents the depth in inches,
and _L_ represents the length in feet.

170. The value of the formula for the strength of cast-iron posts, seems to depend more upon the authority consulted than upon the nature of iron. For example, assume the length of a post as twenty feet, and the diameter as ten inches; the load which may be safely borne is, according to six different authorities, as follows:—

_A_ 4,000,000
_B_ 181,100
_C_ 370,000
_D_ 940,000
_E_ 307,242
_F_ 300,000

and assuming the length as ten feet, and diameter as ten inches, we have

_A_ 8,007,500
_B_ 204,500
_C_ 1,442,500
_D_ 3,640,000
_E_ 1,170,000
_F_ 600,000

showing not only a great difference in the unit resistance taken, but also in the effect of the ratio between the length and diameter.

Such being the discrepancy, there will be given no formula; but in place of such, the table following, which is calculated from the rules least opposed to experimental evidence.

┌─────────────────────────────────────────────────────────────────────┐ │ TABLE SHOWING THE LOAD IN POUNDS SAFELY BORNE BY CAST-IRON COLUMNS. │ ├─────────────────────────────────────────────────────────────────────┤ │ HOLLOW CYLINDERS. │ ├──────┬──────────────────────────────────────────────────────────────┤ │Diame-│ Length or height in feet. │ │ter in│ │ │inches│ │ ├──────┼──────┬──────┬──────┬──────┬──────┬──────┬──────┬──────┬──────┤ │ │ 6 │ 8 │ 10 │ 12 │ 15 │ 18 │ 20 │ 22 │ 24 │ ├──────┼──────┼──────┼──────┼──────┼──────┼──────┼──────┼──────┼──────┤ │ 2│ 6000│ 5000│ 4000│ 3000│ 2500│ 1800│ 1500│ 1300│ 1100│ │ 3│ 16000│ 14000│ 13000│ 11000│ 9000│ 7000│ 6000│ 5000│ 5000│ │ 4│ 30000│ 29000│ 26000│ 24000│ 22000│ 18000│ 16000│ 14000│ 13000│ │ 5│ 50000│ 37000│ 45000│ 42000│ 39000│ 37000│ 31000│ 28000│ 26000│ │ 6│ 59000│ 57000│ 55000│ 52000│ 49000│ 44000│ 41000│ 38000│ 36000│ │ 7│101000│ 99000│ 96000│ 92000│ 88000│ 81000│ 76000│ 72000│ 68000│ │ 8│131000│129000│126000│122000│118000│109000│105000│100000│ 96000│ │ 9│169000│167000│164000│160000│156000│146000│141000│136000│131000│ │ 10│210000│200000│200000│200000│190000│180000│180000│170000│170000│ │ 11│250000│250000│240000│240000│240000│230000│220000│220000│210000│ │ 12│300000│300000│290000│290000│290000│270000│270000│260000│260000│ │ 14│450000│430000│410000│380000│370000│350000│330000│320000│300000│ │ 16│520000│500000│480000│460000│440000│420000│400000│370000│350000│ │ 18│650000│630000│610000│590000│560000│520000│470000│430000│400000│ │ 20│800000│760000│740000│690000│650000│590000│540000│490000│450000│ ├──────┼──────┼──────┼──────┼──────┼──────┼──────┼──────┼──────┼──────┤ │Diame-│ 6 │ 8 │ 10 │ 12 │ 15 │ 18 │ 20 │ 22 │ 24 │ │ter in│ │ │ │ │ │ │ │ │ │ │inches│ │ │ │ │ │ │ │ │ │ ├──────┼──────┴──────┴──────┴──────┴──────┴──────┴──────┴──────┴──────┤ │ │ Length or height in feet. │ └──────┴──────────────────────────────────────────────────────────────┘

┌─────────────────────────────────────────────────────────────────────┐ │ TABLE SHOWING THE LOAD IN POUNDS SAFELY BORNE BY CAST-IRON COLUMNS. │ ├─────────────────────────────────────────────────────────────────────┤ │ H AND + SECTIONS. │ ├──────┬──────────────────────────────────────────────────────────────┤ │Metal │ Length or height in feet. │ │thick-│ │ │ness. │ │ ├──────┼──────┬──────┬──────┬──────┬──────┬──────┬──────┬──────┬──────┤ │ │ 6 │ 8 │ 10 │ 12 │ 15 │ 18 │ 20 │ 22 │ 24 │ ├──────┼──────┼──────┼──────┼──────┼──────┼──────┼──────┼──────┼──────┤ │ ¼│ 4000│ 3000│ 2400│ 1800│ 1400│ 1100│ 1000│ 900│ 800│ │ ⅜│ 12000│ 11000│ 10000│ 9000│ 8000│ 7000│ 5000│ 4000│ 3000│ │ ½│ 25000│ 23000│ 21000│ 18000│ 16000│ 13000│ 12000│ 9000│ 6000│ │ ⅝│ 36000│ 34000│ 31000│ 28000│ 25000│ 23000│ 21000│ 20000│ 18000│ │ ¾│ 40000│ 38000│ 37000│ 36000│ 35000│ 34000│ 32000│ 30000│ 28000│ │ 13/16│ 60000│ 59000│ 58000│ 57000│ 56000│ 54000│ 53000│ 51000│ 49000│ │ ⅞│100000│ 98000│ 96000│ 94000│ 91000│ 88000│ 83000│ 78000│ 70000│ │ 1│140000│130000│126000│120000│114000│110000│106000│100000│ 90000│ │ 1⅛│190000│180000│170000│160000│150000│140000│130000│125000│120000│ │ 1¼│230000│220000│210000│200000│190000│180000│170000│160000│150000│ │ 1½│280000│260000│250000│240000│230000│220000│200000│190000│180000│ │ 1¾│360000│320000│310000│300000│290000│280000│270000│260000│240000│ │ 2│460000│430000│400000│370000│350000│330000│310000│300000│280000│ │ 2½│560000│530000│510000│480000│440000│410000│380000│350000│330000│ │ 3│600000│580000│550000│520000│500000│460000│430000│400000│380000│ ├──────┼──────┼──────┼──────┼──────┼──────┼──────┼──────┼──────┼──────┤ │Metal │ 6 │ 8 │ 10 │ 12 │ 15 │ 18 │ 20 │ 22 │ 24 │ │thick-│ │ │ │ │ │ │ │ │ │ │ness. │ │ │ │ │ │ │ │ │ │ ├──────┼──────┴──────┴──────┴──────┴──────┴──────┴──────┴──────┴──────┤ │ │ Length or height in feet. │ └──────┴──────────────────────────────────────────────────────────────┘

OF THE TRUSS.

171. The most simple bridge that could be built, consists of a single piece of timber placed across the opening to be spanned. This form is applicable to spans under twenty feet. The proper dimensions are found by the formula:—

_d_ = √((4_wL_)/(5000_b_)) = the depth.

_Example._—The depth of a beam of twenty feet span, and twelve inches wide, to support a load of twenty thousand two hundred and fifty lbs. is

_d_ = √((4 × 20,250 × 20 × 12)/(5000 × 12)) = 18 inches.

A beam 12 × 18, and of 20 feet span, will therefore bear safely a load of 20,250 lbs., applied at the centre.

In this manner is formed the following table, giving the scantling of sticks for railroad stringer bridges, of twenty feet span and under.

Span. Breadth. Depth.
5 12 12
10 12 13
12 12 15
15 12 18
18 12 20
20 12 21 inches.

The first scantlings exceed the requirement of the rule, but are none too large to resist the shocks to which such sticks are exposed.

Cross-ties of plank, 2 or 3 by 6 or 8 inches, and plank braces underneath, (as shown in the fig. at the end of chapter VIII.,) should be bolted to the main timbers; the same bolt passing through the tie beam and plank. The longitudinal pieces should be firmly notched and bolted to the wall-plates, and these latter either built in or scribed on to the masonry.

172. For a span of from 20 to 50 feet, we may use the combination shown in fig. 61. The piece A B, must be so strong as not to yield between A and D, or D and B. The pieces C E must be stiff enough to resist the load coming upon them which is as follows. A locomotive engine of the heaviest class will not exceed fifty tons weight, each pair of driving wheels will support ten tons, and on each side five tons, 2240 × 5 = 11,200 lbs.; or to allow for shocks and extra strains, 15,000 lbs. Each brace, then, must support seven thousand five hundred pounds, which for compression simply would require only seven and one half square inches of sectional area; but the brace being inclined, the strain is increased as follows:—

A E to E C as 7,500 to X.

And A E being ten feet, and A D fifteen feet, E C becomes eighteen feet, whence

10 to 18 as 7,500 to 13,500 lbs.;

which would require only thirteen and one half inches for compression, or a piece 4 × 3½. But is this enough for flexure?

On page 124 the load which may be safely borne, by a rectangular post of wood, is shown by the formula

_W_ = (2240_bd^3_)/(_L^2_).

Substituting for _b_ and _d_, the dimensions 4 × 4, we have

_W_ = (2240 × 4 × 4^3)/(18^2), or 573,440/324, = 1,770,

which is evidently too small.

Placing 6 × 7, for _b_ × _d_, we have

_W_ = (2240 × 6 × 7^3)/(18^2) = 14,227;

exceeding by a small amount the requirement.

173. It is evidently immaterial whether we _support_ the point D _upon_ C, or _suspend_ it as in fig. 62, provided we prevent any motion in the feet of the inclines A _c_ B _c_. Abutting them against A B, throws a tension against A B, found as follows:—

Representing by _c_ D, the applied weight, draw D E parallel to _c_ B; also E _f_ parallel to A B; E _f_ is the tension. The graphic construction gives results near enough for practice. Rigorously we have

A _c_ D, similar to E _c_ _f_;

also,

A _c_, to E _c_, as A D, to E _f_;

and

E_f_ = (E_c_ × AD)/(A_c_).

When _a d_ and _c d_ are differently inclined, proceed as follows. See fig. 102, p. 200 inverted.

Let _d b_ represent the weight; _e h_ shows the tension. The triangles _a c d_, and _a b e_, are similar; as also _e b h_ and _d b c_; whence

_b e_ = (_a b_, _c d_)/(_a c_), and _e h_ = (_c b_, _b e_)/(_d c_) =
tension.

In practice place _w_ for _b d_; i. e. the actual weight.

In this plan, if the chord is able to resist the cross strain between A and D, it will also resist the tension. This cross strain is found by the formula already given and illustrated.

174. From what precedes, we have the following dimensions for bridges such as are shown in figs. 61 and 62. The details of 62, at _f_ and _c_, and at E, 61, are shown in figs. 62 A, 62 B, and 61 C.

Span. Rise. A B. C E. Rod _b_.
20 8 12 × 12 (5 × 8)—2 1¼ inches.
25 10 12 × 15 (5 × 9)—2 1⅜ inches.
30 12 12 × 18 (5 × 10)—2 1½ inches.
35 13 12 × 20 (5 × 10)—2 1⅝ inches.
40 14 14 × 21 (5 × 12)—2 1⅝ inches.
45 15 14 × 22 (6 × 12)—2 1¾ inches.
50 16 14 × 24 (6 × 12)—2 1¾ inches.

The braces, (column 4,) being in pairs and blocked together. In spans exceeding twenty-five feet, the braces _d f_, and the rods _f g_, should never be omitted. The size of the rod _g f_, is found by considering A, _d_, _f_, as a small bridge.

175. In all light bridges, like the one under consideration, all parts should be _fastened_ by bolts, to prevent springing by reaction. A bridge with but little inertia, or dead weight, tends to jump up when the engine has passed over it. _Fastening_ takes the place of _weight_ in a large span.

As soon as the rise admits, the points C, on each side of the bridge, should be connected to resist lateral motion. When the height is not enough for this, the same points may be joined to a floor beam extended out beyond the truss.

Though the dimensions are given for this plan up to fifty feet span, it is very seldom advisable to go beyond twenty-five or thirty feet; as frames consisting of a few long timbers are not so rigid, and free from vibration, as those made of a greater number of short pieces.

176. In extending this system one hundred or two hundred feet, we see at once that the pieces A _c_, B _c_, would become very long and would need to be made large and heavy. We should always so proportion any beam in a bridge that it is at once able to resist all of the several strains to which it may be exposed, without being unnecessarily large.

As to compression, the above system might be extended to almost any amount; but the braces would yield by flexure.

Instead of producing the braces A _c_, A′ _c′_, fig. 64, to their intersection, we stop at _c_ and _c′_, insert _c c′_; to prevent the approach of these points, suspend the points B and B′ from _c_ and _c′_, and commence again with the braces B D, B′ D; and so on as far as necessary.

To prevent the backward motion of the points B, and B′, either the chord A A′, or the counter-braces _m_, _m_, are necessary.

The pieces A _c_, A′_c′_ must support all of the load, including the weight of the bridge, lying within the rectangle B _c_, B′ _c′_. The next set of braces must sustain that part of the load only which comes over the centre of the bridge. Thus the braces should decrease in size as the centre is approached. The rods _c_ B, _c′_ B′, must resist a tension equal in amount to the pressure on the braces, only being vertical they do not need the increase given to the braces on account of their inclination.

177. There is another method of stiffening a beam, as shown in figs. 65 and 66, by trussing rods, and a post. The dimensions being the same, the forces in both cases will be equal. The second, fig. 66, leaves the passage beneath the bridge clear.

The tension on the rods A _c_, B _c_, fig. 66, tends to draw the points A and B together, an effort which is resisted by the top chord A B.

In extending this system, as in art. 176, the rods become either very long, or very large, from the small angle of inclination; evils which remedied as before, by supporting the post _c_ B, fig. 64, from the foot of the first rod, fig. 64, and commencing again from _c_.

To prevent the motion of the triangle _c_ B G, fig. 64, about the angle B, we must introduce either the upper chord _c c′_, or the counter rod _c_ A. If the lower chord is omitted the rod D B must be of the same size as E B. In this truss, either the top or the lower chord simply may theoretically be omitted, due allowance being made in the size of the rods. In practice it is never advisable to omit either, as both are required for lateral bracing, and for support of the road-way.

Having said thus much of the general ideas that apply to all bridges, let us now look at some of the plans most in use; and to become familiar with the subject, work out the dimensions of an example of each kind.

178. As rods, nuts, and washers are used in all bridges, the following table may not be out of place:—

Column 1 gives the diameter of rod.
Column 2 strength at 15,000 lbs. per square inch.
Column 3 the weight per lineal foot.
Column 4 side of the square nut.
Column 5 the thickness of the same.
Column 6 the dimensions of washers.
Column 7 the thickness of washers.
Column 8 breadth (side to side) six-sided nut.
Column 9 breadth (across angles) six-sided nut.
Column 10 thickness of six-sided nut.
Column 11 number of screw threads per inch.
Column 12 gives the diameter of rod.

┌─────────┬────────┬──────┬──────┬──────────┬───────┐ │ 1 │ 2 │ 3 │ 4 │ 5 │ 6 │ │Diameter.│Strength│Weight│Square│Thickness.│Square │ │ │of Rod. │ per │ Nut. │ │ of │ │ │ │Foot. │ │ │Washer.│ ├─────────┼────────┼──────┼──────┼──────────┼───────┤ │ ½│ 2,940│ 0.66│ 1¼│ ¾│ 2½│ │ ¾│ 6,630│ 1.49│ 1½│ ⅞│ 3│ │ 1│ 11,775│ 2.65│ 2│ 1│ 4│ │ 1⅛│ 14,910│ 3.36│ 2│ 1⅛│ 4½│ │ 1¼│ 18,405│ 4.17│ 2¼│ 1¼│ 5│ │ 1⅜│ 22,260│ 5.02│ 2½│ 1⅜│ 5½│ │ 1½│ 26,505│ 5.97│ 2¾│ 1½│ 6│ │ 1⅝│ 31,095│ 7.01│ 2⅞│ 1⅝│ 6½│ │ 1¾│ 36,075│ 8.13│ 3│ 1¾│ 7│ │ 1⅞│ 41,415│ 9.33│ 3¼│ 1⅞│ 7½│ │ 2│ 47,130│ 10.62│ 3½│ 1⅞│ 8│ │ 2⅛│ 53,190│ 12.00│ 3¾│ 2│ 8½│ │ 2¼│ 59,640│ 13.40│ 4│ 2⅛│ 9│ │ 2⅜│ 66,450│ 15.00│ 4⅛│ 2¼│ 9½│ │ 2½│ 73,620│ 16.70│ 4¼│ 2½│ 10│ ├─────────┼────────┼──────┼──────┼──────────┼───────┤ │ 1 │ 2 │ 3 │ 4 │ 5 │ 6 │ └─────────┴────────┴──────┴──────┴──────────┴───────┘

┌─────────┬───────┬─────┬──────┬──────┬──────┬─────────┐ │ 1 │ 7 │ 8 │ 9 │ 10 │ 11 │ 12 │ │Diameter.│ Top │Six- │ Six- │ Six- │Screw.│Diameter.│ │ │Washer.│Sided│Sided │Sided │ │ │ │ │ │Nut. │ Nut. │ Nut. │ │ │ ├─────────┼───────┼─────┼──────┼──────┼──────┼─────────┤ │ ½│ ¼│ 1⅜│ 1½│ 9/16│ 12│ ½│ │ ¾│ ¼│ 1¾│ 2│ ⅞│ 10│ ¾│ │ 1│ ⅜│ 1⅞│ 2¼│ 1⅛│ 8│ 1│ │ 1⅛│ ⅜│ 2⅛│ 27/16│ 1¼│ 7│ 1⅛│ │ 1¼│ ½│ 2¼│211/16│ 17/12│ 7│ 1¼│ │ 1⅜│ ½│ 2½│ 2⅞│ 17/16│ 6│ 1⅜│ │ 1½│ ½│ 2⅝│ 3⅛│111/16│ 6│ 1½│ │ 1⅝│ ⅝│ 2⅞│ 35/16│113/16│ 5│ 1⅝│ │ 1¾│ ⅝│ 3│ 3½│ 2│ 5│ 1¾│ │ 1⅞│ ⅝│ 3¼│ 3¾│ 2⅛│ 4½│ 1⅞│ │ 2│ ⅝│ 3½│ 4│ 2¼│ 4½│ 2│ │ 2⅛│ ¾│ 3⅝│ 43/16│ 2⅜│ 4│ 2⅛│ │ 2¼│ ¾│ 3¾│ 46/16│ 2½│ 4│ 2¼│ │ 2⅜│ ¾│ 4│ 4⅝│ 2⅝│ 4│ 2⅜│ │ 2½│ ¾│ 4¼│ 4⅞│ 2¾│ 3½│ 2½│ ├─────────┼───────┼─────┼──────┼──────┼──────┼─────────┤ │ 1 │ 7 │ 8 │ 9 │ 10 │ 11 │ 12 │ └─────────┴───────┴─────┴──────┴──────┴──────┴─────────┘

179. Let us now assume the following data:—

Span 200 feet,
Rise (centre to centre of chords) 25 feet,
Width 20 feet,
Length of panel 15 feet,
Weight (bridge and load) per lineal ft. 4,000 lbs.

HOWE’S BRIDGE.

Fig. 67.

_Lower Chords._—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

_T_ = (_W_ × _S_)/(8_h_),

which becomes, with the above data,

_T_ = (800000 × 200)/200 = 800,000 lbs.

Here the tension and the total weight are equal, a result which can occur only when the rise is one eighth of the span. This is the best ration between these dimensions, as then the horizontal and vertical forces are equal.

As to the proportion of the _panel_, (or the rectangle inclosed by the chords and any two adjacent posts,) the ratio of base to height should be such as to make the inclination of diagonal about 50° from the horizontal; if much less, the timbers become large and heavy; and if more, the number of pieces is unnecessarily increased.

The braces at the end of a long span, may be nearer to 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.

To determine the size of the lower chords, to resist the above 800,000 pounds of tension, proceed as follows: Each side truss will support one half of the whole load, or 400,000 pounds; which, at 2,000 pounds per inch, will require 200 square inches of section. Four sticks of 8 × 12 inches, give an area of 384 square inches, which must be reduced as follows: Deduct 72 square inches for the area cut out by the splicing blocks, 40 inches for the bolts connecting the pieces, 28 inches for inserting the foot blocks, and 10 inches for inserting the washer, and we have remaining 234 square inches; which exceeds by a little the exact demand. This excess (about one seventh) is a necessary allowance for accidental strain, to which all bridges are subjected.

The splices used in bridge framing are shown in fig. 67 A and fig. 67 B. For the first, the depth of insertion and length of the block depend upon the tension upon the chord. The following dimensions have been much used and are perfectly reliable:—

Span of Bridge. A C B C C D
Feet. Feet. Inches. Feet.
50 1.00 1½ 1.50
100 1.25 2 2.00
150 1.75 2½ 2.25
200 2.00 3 2.75

There is no need of cutting more than one notch, as in the figure; the resistance of the triangles is thereby lessened, and the work increased.

In fig. 67 B, the rods must of course be able to resist the tension upon the one piece which is cut.

_Upper Chord._—The upper chords of a bridge suffer compression, to the same amount numerically, as the tension on the lower chord; as whatever tension is thrown by any brace upon the lower chord, reacts as just so much compression upon the upper. In the case at hand, 800,000 lbs. in all, or 400,000 on each chord.

The resistance to compression being one thousand pounds per inch, renders necessary four hundred inches of section to each chord; four pieces 8 × 12 give in all three hundred and eighty-four inches of section, which requires no reduction, as the whole chord pressing together and being properly framed is not weakened by splicing. The splicing blocks in the upper are merely plain pieces, inserted one half inch, the only duty being to keep the sticks at the proper horizontal distance.

The spaces between the pieces should be large enough to allow the rods to pass without cutting the chords; (two inches answers every purpose). The bolts for splicing, have no very great strain to bear. In small spans from ½ to ⅝, and in large bridges from ⅝ to an inch is enough.

The object in framing a built beam for a bridge chord, is to make a stick which shall be uniformly strong. This is done by cutting the pieces in the centre of the panel, and by having no two joints in either chord in one panel; though in long spans this cannot always be done. Figs. 67 D and 67 E (page 153)

BRACES.

The whole load being 800,000 pounds,
Each truss supports 400,000 pounds,
Each set of braces 200,000 pounds,
Each brace (there being 4) 50,000 pounds,

which must be increased for inclination as follows: The length of diagonal is twenty-nine feet, (the height being twenty-five and length 15,) whence

25 to 29 as 50,000 to 58,000 lbs.;

which would need fifty-eight square inches, or 7 × 8 for compression; which, however, is quite too small for flexure. 12 × 12 placed in the formula gives

_W_ = (2240 × _bd^3_)/(_L^2_),

or

_W_ = (2240 × 12 × 1728)/841 = 55,296 lbs.

In practice, smaller braces than 12 × 12 would answer, because the four braces in a set may be fastened together, making a post of four pieces 8 × 12, or in all a built post of 44 × 12 inches; twelve being the depth, whence

_W_ = (2240 × 44 × 1728)/841 = 202,511 lbs.;

the forty-four inches being made by blocking the braces four inches apart. The second set of braces are to be treated in the same manner, the weight to be supported being only the rectangle included by those braces; i. e. the whole bridge and load less the two end panels.

As the centre of the span is approached, the pressure on the braces becomes very small; and the scantling of the braces will be reduced to about 6 × 7 inches.

RODS.

The weight upon the first set of rods is the same as that upon the end sets of braces; in the present case 800000 ÷ 2 = 400000 on each side truss, and 400000 ÷ 2 = 200000 on each end; and if there are five rods in each set, each rod bears 40,000 lbs. Referring to the table on p. 146, opposite to 41,415 lbs., is the diameter 1⅞ inches; whence the first set must contain five rods, of 1⅞ inches diameter. The second set decrease in size as the weight is lessened by the two end panels. The nut and washer for the rod are also found in the same table.

COUNTERBRACING.

180. When a load is placed on the point C′, fig. 64, the truss tends to sink at that point, and a corresponding rise takes place at C. This motion changes the figure A B C E, from a rectangle to an oblique angled figure; the diagonal E B being shortened, and A C lengthened. This motion is easily checked by the introduction of the counter brace E B.

The action which this timber is called upon to resist, being caused by the moving or variable load on one panel, the brace must resist the load coming thereon, (say fifteen feet,) and is thus the same size as the brace at the centre of the span.

The counter braces may be so confined between the braces, at the intersection, as not to move laterally or vertically, but must not be fastened to the braces; because the action of the separate timbers is thus trammelled.

The manner of adjusting the braces and counter braces to the chord is shown in fig. 67 C. It was formerly the custom to abut the braces against a block on one side of the chord, and to screw the rod against a block on the opposite side; the whole strain acting to crush the chord crosswise. This has been remedied by the arrangement shown in the figure, the two blocks being cast in one piece and connected by a small hollow cylinder passing between the chord sticks.

This system is known as Howe’s bridge, and may be seen in almost any section of the country; and though in many cases badly proportioned, and of bad material, if properly made answers a very good end.

The following table has been formed for the use of engineers and builders, giving, together with the table of nuts and washers, all dimensions required.

┌───────┬───────┬───────┬────────┬───────┬───────┬───────┬───────┐
│ Span. │ Rise. │Panel. │Chords. │ End │Centre │ End │Centre │
│ │ │ │ │braces.│braces.│ rod. │ rod. │
├───────┼───────┼───────┼────────┼───────┼───────┼───────┼───────┤
│ 50│ 10 │ 7│2—8 × 10│ 7^2│ 5 × 5 │2—1⅛ │ 2—1 │
│ 75│ 12 │ 9│2—8 × 10│ 8^2│ 5 × 5 │2—1½ │ 2—1 │
│ 100│ 15 │ 11│3—8 × 10│ 9^2│ 6 × 6 │2—1¾ │ 2—1 │
│ 150│ 20 │ 13│4—8 × 12│ 10^2│ 6 × 7 │3—2 │ 3—1 │
│ 200│ 25 │ 15│4—8 × 16│ 12^2│ 7 × 7 │5—2 │ 5—1 │
└───────┴───────┴───────┴────────┴───────┴───────┴───────┴───────┘

PRATT’S BRIDGE.

181. Assume the following data for an example:—

Span 100 feet.
Rise 12 feet.
Panel 10 feet.
Weight per lineal ft. 2,500 lbs.

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

(250000 × 100)/96 = 260,417 lbs.

The manner of dimensioning the chord, and of splicing, is the same as already described for Howe’s.

SUSPENSION RODS.

The first sets of rods, A B, A′ B′, must sustain the whole weight of the bridge and load; which is 250,000 lbs. Each side 125,000 lbs.; and each end set of rods 62,500 lbs.; and if each set has four rods, each rod must support 15,625 lbs.

The rod being inclined, this amount is increased by the following proportion:—

12 (height) to 15.8 (diagonal) as 15,625 to 20,573 lbs.

This is half-way between the tabular numbers for rods of 1¼ and 1⅜ inches in diameter; 1⅜ will therefore answer. The next set of rods must be considered as supporting the whole load, less the two end panels, and so on as already explained for Howe’s bridge. The manner of applying the rods to the chords is shown in fig. 68 A. The bevel block should be connected with the block at the foot of the post, so as to prevent crushing the chord.

COUNTER RODS.

As both top and bottom chords are always used in this bridge, the counter rods have only the variable load on one panel to resist. The action is, in amount, the same as that on the counter braces in Howe’s bridge; but acts in a different direction, and in the other diagonal.

The weight of a passing load cannot be more than two thousand pounds per lineal foot. The panel being ten feet long, the whole weight coming on two sets of counter rods, (one set in each side truss,) is twenty thousand pounds; or ten thousand pounds on each set; and if there are put three rods in each set, we have 3,333 pounds per rod, which increase for inclination as follows:—

12 : 15.8 :: 3333 : 4389 lbs.,

requiring a rod of three fourths inch diameter.

The posts in this structure, correspond to the braces in the Howe bridge; only being vertical, they need not be so large.

182. The following table gives all the dimensions necessary for proportioning this truss.

┌───────┬───────┬────────┬───────┬───────┬───────┬───────┬───────┐
│ Span. │ Rise. │Chords. │ End │C post.│ End │C rod. │Counter│
│ │ │ │ post. │ │ rod. │ │ rod. │
├───────┼───────┼────────┼───────┼───────┼───────┼───────┼───────┤
│ 50│ 10│2—8 × 10│ 5 × 5│ 4^2│ 2—1⅜│ 2—1│ 1—1½│
│ 75│ 12│2—8 × 10│ 6 × 6│ 5^2│ 2—1⅝│ 2—1│ 1—1½│
│ 100│ 15│3—8 × 10│ 7 × 7│ 6^2│ 2—1¾│ 2—1│ 2—1⅛│
│ 125│ 18│3—8 × 10│ 8 × 8│ 6^2│ 3—1⅞│ 3—1│ 2—1⅜│
│ 150│ 21│4—8 × 12│ 9 × 9│ 6^2│ 3—2⅛│ 3—1│ 3—1⅛│
│ 200│ 24│4—8 × 16│10 × 10│ 6^2│ 5—1⅞│ 5—1│ 3—1⅛│
└───────┴───────┴────────┴───────┴───────┴───────┴───────┴───────┘

And the following, the sizes of counter rods, for different panels.

Length of Height of Approximate
panel. panel. diagonal of Diameter of the rod.
panel.
One in a Two per Three per
set. set. set.
10 12 16 1⅝ 1⅛ ⅞
11 13 17 1⅝ 1¼ 1⅛
12 14 18 1¾ 1¼ 1⅛
13 15 20 1¾ 1¼ 1⅛
14 16 21 1⅞ 1⅜ 1⅛
15 18 23 1⅞ 1⅜ 1⅛
16 21 26 2 1⅜ 1⅛
18 25 27 2 1⅜ 1⅛

The advantage possessed by this bridge, over Howe’s plan, is that the panel diagonals may be adjusted by the screws; by which control is had over the form of the truss, and of the duty done by the several parts. Change of form cannot be had by working upon verticals. Howe’s bridge must be adjusted by wedging the braces and the counter braces.

183. The manner of drawing the bevel block in this bridge, is shown in fig. 68 b. The proportions of the block depend upon the proportions of the panel; and the dimensions, upon the size of washer used.

Let C C be the centre line of the post, and A B the chord. Let _o m_, and _o n_, be the panel diagonals, and H and _y_, the length of the washers.

The depth of insertion of the block into the chord, depends upon the horizontal strain upon it. In a span of one hundred and fifty feet, with the rods at an angle of 50°, two inches have been found ample at the end of the truss, and one half inch at centre.

From D, perpendicular to _m m_, lay off D E; equal to H, also from E, at right angles to _n n_, make E E′ = _y_. From E′ draw the vertical E′ L.

The strain upon the rod _o m_, being represented by _o m_; and that upon _o n_, by _o n_, the resultant is shown, both in direction and amount, by _o_ V. It is not necessary that this should pass through the centre of the post, as the excess of tension on _o m_, over that on _o n_, is absorbed by the lower chord.

NOTE.—Screwing up truss bridges, is a more scientific operation than
is generally supposed. Many builders commence at each end, and lift
the bridge from the scaffolding. By this method the greater part of
the load is often borne by a few of the end sets of rods. The better
method is to begin at the centre and work both ways towards the
ends, being sure that each set of rods does its duty before the next
is touched. The lift to be made by each set of rods, should first be
calculated, and tested while screwing up, with the level.

LATTICE BRIDGES.

184. _Town’s lattice_, consists of a simple lattice-work of plank, 3 × 12 inches, treenailed together at an angle of forty, forty-five, or fifty degrees. It possesses great stiffness, without by any means having the material disposed in the best manner. Such bridges might well be made by the mile, and cut off to order according to the span.

The improved lattice, by Hermann Haupt, Esq., C. E., avoids all of the evils attendant upon the common lattice, and gives a very cheap, strong, and rigid bridge. In this plan the braces are placed in pairs, with vertical tie planks between them; by which the twisting seen in the common lattice, is removed. The braces are also brought to the vertical, as the point of support is reached, by which a good bearing is given to the end sets of timbers.

To vary the size of the braces, as the strain upon them decreases, would be both inconvenient and expensive; but the same effect may be produced by varying the distance between them, making it greater as the centre is approached.

S. W. HALL’S WOODEN TRUSS AND ARCH BRIDGE.

Inverting Mr. Haupt’s design for a lattice of improved construction, (which consists of vertical ties and inclined braces,) we have the base of the above-named bridge; where the inclined timbers are used to resist tension, as below.

This being a very good plan, and the arrangement for building being such as to secure the thorough execution of the work in its most minute detail; it is thought best to extract at some length from a letter from the inventor, dated July 31st, 1856, not however being confined to the matter therein.

The first claim, is for a _new form of truss_, formed of posts vertical, or nearly so, and _tension_ pieces, inclining downwards toward the centre; thus differing from nearly all other plans. Timber resists double as much extension as compression; and when large enough to resist the simple tension, does not have to be increased as in resisting compression for flexure; but requires a larger allowance for joints, as tension tends to pull the joints apart, while compression forces them together.

The following result was obtained, showing the superior strength of timber work in resisting by _tension_. Two models, containing the same amount of timber, were tested. The one built with vertical ties and _braces_, broke by crippling the brace, under 2,400 lbs.; while that constructed with verticals and _suspenders_, inclining towards the centre, sustained 4,200 lbs. with no visible change of form.

The second claim is for more efficient bearings and connections than common, and this with less cutting away of timber. The arch and arch braces have a full, fair bearing at top and bottom. The first sets of tension braces, (those extending from the top of the arch braces towards the centre,) are sustained by two pins at each joint; which gives six pin bearings, or twelve for one set of braces, of six inches each, (the pin being two inches in diameter, and plank three inches thick,) equal in all, to seventy-two inches of bearing surface at least, for each five feet lineal of bridge, or one hundred and forty-four inches for ten feet.

The third claim, is that the bearings, at joints, are _central_, and that the shrinkage of the timber is _towards_ and not _from_ them as in many plans.

The pin holes are bored by machinery smooth and true; the treenails when of wood are of seasoned oak or locust, turned to a perfect fit, and when of iron are made hollow.

These bridges, after three years, stand within an inch of their shape as framed without exception. One indeed supporting an aqueduct, which throws upon the truss a constant load of 2¾ tons per foot, not including the weight of the bridge, without any apparent settling.

The connections being fast, prevent reaction and vibration from variable loads, the strains in this case are reversed, the bridge tending to spring up instead of settling.

The fourth claim, is for the small brace connecting the lower with the intermediate chord; by which additional connections are obtained, and smaller timbers rendered available.

The fifth claim, is the formation of a stronger chord than by the plan of using a few large sticks. The chord being made of a great number of small pieces, the strength is of course less affected at any one point, by a joint, than when only a few pieces are used.

In the bridges built by the above engineer, are to be seen some of the most perfect built beams in the country. The following conditions being observed, the most uniform, and highest average strength possible is obtained.

First. To cut but one stick in any one panel.

Second. To cut no stick at the centre of the bridge.

Third. To place every joint in the middle of the collateral piece.