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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.  Compression.  Detrusion.

Wood,          2000       1000          150

Wro't Iron,   15000      11000

Cast Iron,     4500      25000


=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:--

  WL²                    bd³ x 80 E
  ---- = bd³, whence W = -----------
  80 E                       L²

[TeX: $\frac{WL^2}{80 E} = {bd^3}$, whence $W = \frac{{bd^3} x 80 E}{L^2}$]

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

          -----
         / WL²              WL²
  d =  ³/ -----,  and b = ------
      \/  80 Eb           80 Ed³

[TeX: $d = \sqrt[3]{\frac{WL^2}{80 EB}}$, and $b = \frac{WL^2}{80 ED^3}$]

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.

                                                 Dist. of axis
               Resistance  Resistance            from top in
                   to          to                frac's of
  Material.    Extension.  Compression.  Ratio.  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    2/3, or 0.66.


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

      4 bd²
  S = ----
       l

[TeX: $S = \frac{4 bd^2}{l}$]

  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 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:

  Weight                Ult.    Ult.   Working Strengths Factor  Safety.
  in lbs.  Materials.   Ext.    Comp.  Exten.    Comp.   Tension  Comp.

    30    Wood.        14,000   7,000   2,000    1,000      7      7
   480    Wrou't Iron. 60,000  64,000  15,000   12,000      4      5.33
   450    Cast Iron.   18,000 100,000   4,500   25,000      4      4


=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.=

  Length      Safety       Length    Safety     Length    Safety
  given in    Weig't in   given in   Wt. in    given in   Wt. in
  Diameters.  Pounds.    Diameters.  Pounds.  Diameters.  Pounds.

     6         1000         24         440       42         203
     8          960         26         394       44         185
    10          910         28         358       46         169
    12          860         30         328       48         155
    14          810         32         299       50         143
    16          760         34         276       52         132
    18          710         36         258       54         122
    20          660         38         239       56         114
    22          570         40         224       58         106
                                                 60          99

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:

       T
  a = ---
       S

[TeX: $a = \frac{T}{S}$]

or, by substituting the working strengths 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,

      4 sbd²
  W = ------
        L

[TeX: $W = \frac{4 sbd^2}{L}$]

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

      4 x 1250 bd²
  W = ------------,
           L

[TeX: $W = \frac{4 \times 1250 bd^2}{L}$]

or

      5000 bd²
  W = --------.
          L


[TeX: $W = \frac{5000 bd^2}{L}$]

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

      5000 bd²
  W = --------.
         4L

[TeX: $W = \frac{5000 bd^2}{4 L}$]

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.

                  5000 bd²
  The formula W = --------
                     4L

[TeX: $W = \frac{5000 bd^2}{4 L}$]

becomes, by substitution,

      5000 x 8 x 196
  W = -------------- = 11.666 lbs.
          4 x 8

[TeX: $W = \frac{5000 \times 8 \times 196}{4 \times 168} = 11,666$ 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

         -------
        / W X 4L
  d =  /  ------
     \/   5000 b


[TeX: $d = \sqrt{\frac{w \times 4L}{5000 b}}$]

substituting

         ----------------
        / 18000 x 168 x 4      ------
  d =  /  --------------- =   / 268.8 = 16, inches nearly.
     \/      5000 x 9       \/


[TeX: $d = \sqrt{\frac{1800 \times 168 \times 4}{5000 \times 9}}
 = \sqrt{268.8} = 16$]

  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,

        W x 4L
  b = ----------
      5000 x d²

[TeX: $b = \frac{W \times 4L}{5000 \times d^2}$]

substituting

      22400 x 4 x 168
  b = --------------- = 9.3 inches nearly.
        5000 x 324

[TeX: $b = \frac{22400 \times 4 \times 168}{5000 \times 324} = 9.3$]

For a cast iron beam or girder--Mr. Hodgkinson found 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

       26 ad
  W = ------
         L

[TeX: $W = \frac{26 ad}{L}$]

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

       26 ad
  W = ------
        6 L

[TeX: $W = \frac{26 ad}{6 L}$]

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

      6 WL
  a = ----
      26 d

[TeX: $a = \frac{6 WL}{26 d}$]

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.

      6 x 10 x 60 x 12
  a = ---------------- = 37 inches nearly.
            60 x 12
       26 x -------
              16

[TeX: $a = \frac{6 \times 10 \times 60 \times 12}{26 \times \frac{60
\times 12}{16}} = 37$]

and the area of the top flange will be,

  37
  -- = 6.16 inches--
  6

[TeX: $\frac{37}{6} = 6.16$]

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.

[Illustration: Pl. 1.]

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,     4 1/4 x 1 1/2 inches nearly.
  Bottom Flange,  6 x 2 1/2 inches nearly.
  Web,            1 1/2 inches thick.

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

          ------
         /  4WL
  d =   / ------ --
      \/  5000 b


[TeX: $d = \sqrt{\frac{4 WL}{5000 b}}$]

_Example._--The depth of a beam, of 12 feet span and 12
feet wide, to support a load of 22400 lbs. will be

          ------       --------------------
         / 4WL        / 4 X 22400 x 12 x 12      -------
  d =   / ------ =   /  ------------------- =   / 215.04 = 15 in. nearly
      \/  5000 b   \/        5000 x 12        \/


[TeX: $d = \sqrt\frac{4 WL}{5000 b}} = \sqrt\frac{4 \times 22400
\times 12 \times 12}{5000 \times 12} = \sqrt{215.04} = 15$]

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

  Span.  Breadth.  Depth.

    4       10       12
    6       10       12
    8       12       12
   10       12       13
   12       12       15
   16       12       18
   18       12       20
   20       12       22

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.

  Span.  Bolsters.  Stringers.   Ties.   Braces.   Diameter of Bolts.

    4     12 x 12    10 x 12     6 x 8    2 x 8         1 inch.
   10     12 x 12    12 x 13     6 x 8    2 x 8         1  "
   16     14 x 14    12 x 18     6 x 8    2 x 8         1  "
   20     14 x 14    12 x 22     6 x 8    2 x 8         1  "


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 D F and E F 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 2-1/2 tons = 2-1/2 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.

      2240 bd³
  W = --------
         L²

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

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

      2240 x 3 x 125   840000
  W = -------------- = ------ = 2592 lbs.
           324           324

[TeX: $W=\frac{2240 \times 3 \times 125}{324}=\frac{840000}{324}=2592$]

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

      2240 x 6 x 512
  W = -------------- = 21238 lbs.
            324

[TeX: $W = \frac{2240 \times 6 \times 512}{324} = 21238$]

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 F B, Pl. II, Fig. 2, draw
B D parallel to F C, also D H parallel to A C--D H is the tension.
This is the graphical construction, and is near enough for practice.
Geometrically we have the two similar triangles A F B and D F H,
whence

  A F : D F :: A B : D H

            D F x A B
  and D H = ---------
               A F

[TeX: $DH = \frac{DF \times AB}{AF}$]

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:

  Span.    Rise.    Bolster.    Stringer.     Braces.     Rod.
                                                         No. Size.

   15        6      12 x 12     12 x 12       2--5 x 6    1-1/8
   20        7      14 x 14     12 x 13       2--5 x 8    1-3/8
   25        8      14 x 14     12 x 15       2--6 x 8    1-1/2
   30       10      14 x 14     12 x 18       2--6 x 9    1-5/8

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.

[Illustration: Pl. II. with Fig. 1., Fig. 2., Fig. 3., Fig. 4.]


=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.

       |    |    | Upper   |  Lower  |  Main Brace Rods.  |  Counter  |   |     |          |
  Clear|    | No.| Chord.  |  Chord. |                    |   Rods.   |   |     |  Posts.  |
  Span |Rise| of |---------|---------|--------------------|-----------|   |     |----------|
   in  | in |Pan-|No.|     |No.|     |No.| Size |No.|Size.|Num|       |No.| Size|No.|Size. |
  feet.|feet|els.|Pcs|Size.|Pcs|Size.|Ctr|Centre|End| End.|ber| Size. |End| End.|Ctr|Centre|
  -----|----|----|---|-----|---|-----|---|------|---|-----|---|-------|---|-----|---|------|
    25 |  6 |  8 | 3 | 4x5 | 3 | 4x10| 2 |1     | 2 |1-3/8| 1 |1-7/16 | 3 | 4x5 | 3 |  4x4 |
    50 |  9 |  9 | 3 | 6x7 | 3 | 6x10| 2 |1-3/16| 2 |1-1/8| 1 |1-5/8  | 3 | 6x6 | 3 |  6x5 |
    75 | 12 | 10 | 3 | 6x9 | 3 | 6x11| 2 |1-5/16| 2 |2-1/2| 1 |1-7/8  | 3 | 6x7 | 3 |  6x5 |
   100 | 15 | 11 | 3 | 6x10| 3 | 6x12| 2 |1-7/16| 2 |2-7/8| 1 |2      | 3 | 6x9 | 3 |  6x7 |
   125 | 18 | 12 | 4 | 6x10| 4 | 6x13| 2 |1-1/2 | 2 |2-3/8| 1 |2-1/8  | 4 | 6x9 | 4 |  6x7 |
   150 | 21 | 13 | 4 | 8x10| 4 | 8x14| 3 |1-5/16| 3 |2-1/2| 2 |1-5/8  | 4 | 8x8 | 4 |  8x7 |
   175 | 24 | 14 | 4 |10x12| 4 |10x15| 3 |1-5/8 | 3 |2-3/4| 2 |1-11/16| 4 |10x10| 4 | 10x8 |
   200 | 27 | 15 | 4 |12x12| 4 |12x16| 3 |1-1/2 | 3 |3-1/8| 2 |1-13/16| 4 |12x10| 4 | 10x8 |

This table is partly given in Trautwine's Engineer's Pocket Book, and
partly made up from directions therein given.


TABLE OF DIMENSIONS FOR SMALL SINGLE TRACK PRATT TRUSSES.

                                 At centre   At end    Centre      End
  Clear Chords  Centre    End    of truss,  of truss,  Counter,  Counter,
  Span, each,    Post,   Posts,  Diam. of   Diam. of   Diameter, Diameter,
  Ft.   Ins.     Ins.     Ins.     Rods.     Rods.      Ins.      Ins.

   30    9 x 11  4 x 9    7 x 9    1         1-5/8      1-3/8      1
   40   10 x 12  4 x 10   8 x 10   1-1/8     1-7/8      1-5/8      1
   50   10 x 14  5 x 10   9 x 10   1-1/4     2-1/8      1-3/4      1
   60   12 x 15  5 x 12   9 x 12   1-3/8     2-3/8      2          1
   70   12 x 17  6 x 12  11 x 12   1-1/2     2-1/2      2-1/8      1

This bridge possesses an advantage over the Howe Truss, for the panel
diagonals can be tightened up by screws, so that every part of the
truss can be forced to perform its work. In Howe's bridge the
adjustments must be made by wedging the braces and counters.

Below are given the dimensions of a Howe bridge on the Vermont Central
R.R., at South Royalton, (single track, deck.)

              No. of     Upper        Lower
  Span. Rise. Panels.    Chord.       Chord.      Braces.   Counters.
  150    20    12     4--6-1/2 x 13 4--6-1/2 x 13 2--8 x 9  1--8 x 9

    Rods.    Transverse Bracing.
              Braces.    Rods.
  3--1-1/4"   6 x 8      7/8

The bridge over the White River, on the Passumpsic R.R., is a Howe
Truss, strengthened by an arch. The verticals are of wood, and the
diagonals foot on steps formed by enlarging the ends of the verticals.
The counters are in two lengths, and are adjusted by wedges at the
points where they intersect the braces. The bridge is in two spans,
and has a double track, and consequently three trusses. There are two
timber arches to each truss, and the truss is supported on them by
connecting them to the verticals by short cross pieces notched into
the posts, and resting on the upper surface of the arches. It is a
very stiff bridge, and similar to the one at Bellows Falls, both
having their axis oblique to the channel of the stream they cross. The
timbers could hardly be procured now, except at great expense.

       No.
       of        Upper     Lower
  Span Pan- Rods Chord     Chord     Braces   Counters  Uprights Arches
       els
  182   14   21  2--8 x16  2--8 x17, 1--21 x8 1--8 x10  21 x11   2--8 x9
                 1--5 x16  2--4 x17,
                           1--5 x17,

Diagonals 6 x 8, Rods 7/8. Floor timbers suspended both from
arches and truss, 9 x 13; stringers 10 x 14.

In the Cheshire Bridge, the braces are only 20x8, and the span is only
175 feet, the number of Panels being 14, as in the W.R. Bridge--the
other dimensions are the same. Below are given the dimensions of a
Howe Truss of 108 ft. span, weight to be borne on upper chord.

               No.
               of    Upper    Lower                        E.        Floor
  Rise  Camber Pan-  Chord    Chord    Braces    Counters  Rods     Timbers
   Ft.   Ins.  els    Ins.     Ins.     Ins.       Ins.    Ins.      Ins.

  13-1/2  3     12   8--3 x12 8--3 x12 2--8 x10  1--7 x10  2--2-1/8  9 x16

As plank is used for the chords, the pieces must be bolted thoroughly
with 5/8 bolts.


A form of bridge that has been used to some extent on the Baltimore
and Ohio Railroad, by Mr. Latrobe, is the Arch Brace Truss. In this
form of Truss the braces lead directly from the abutments to the head
of each vertical; thus the load is transferred at once to the
abutments, without passing through a series of web members. The
counterbracing is effected by means of a light lattice,--and is
applied to both sides of the chords, and the intersections of the
diagonals are fastened while the bridge is strained by a load--thus
preventing recoil--so that the effect of a moving load is to lighten
the strain on the lattice--without otherwise affecting the Truss.

[Illustration: Pl. III. with Fig. 1., Fig. 2., Fig. 3., Fig. 4., Fig. 5.]

There are two models of this style of bridge, to my knowledge; one
built by Prof. G.L. Vose, on a scale of 1/2 an inch to the foot,
and representing a span of 150 feet, which supported 2,500 lbs. at
the centre, and a movable load of 150 lbs., proving itself to be
strong and rigid enough for any thing. The other, on a scale of 1
inch to the foot, and representing a span of 76 feet, was built by
the Class of '73, of the Thayer Engineering School, under the
writer's direction, and though bearing very heavy weights, has never
been thoroughly tested--it has, however, been subjected to the
sudden shock of 1040 lbs. falling 20 inches, without injury, several
times. Subjoined are the dimensions of the models mentioned.

DIMENSIONS OF A MODEL OF AN ARCH BRACE TRUSS.

                  G.L. VOSE.

        Length,  7 feet.
        Height,  1 foot.
        Width,   1 foot.
        Chords,  4--1/4 x 1/2  inch.
        Braces   4--1/4 x 1/8   "
        Lattice,    1/4 x 1/16  "

This represented a span of 150 ft, a rise of 20 feet, and a panel
of 15 ft. Weight, per running foot of bridge and load, was taken
at 3000 lbs.

The method of calculating the dimensions of this truss, from the
foregoing data, is as follows. The half number of panels is 5, and the
lengths of the corresponding diagonals (neglecting fractions) are

     ---------
    /20² + 15² = 25 feet. [TeX: $\root{20^2 + 15^2} = 25$]
  \/

     ---------
    /20² + 30² = 37  "    [TeX: $\root{20^2 + 30^2} = 37$]
  \/

     ---------
    /20² + 45² = 49  "    [TeX: $\root{20^2 + 45^2} = 49$]
  \/

     ---------
    /20² + 60² = 64  "    [TeX: $\root{20^2 + 60^2} = 64$]
  \/

     ---------
    /20² + 75² = 78  "    [TeX: $\root{20^2 + 75^2} = 78$]
  \/

The weight upon each set of braces is that due to one panel, or 3000
x 15 = 45000 lbs., half of this, or 22500 lbs., is the weight for one
truss only--and, as there is a brace under each of the 4 chord sticks,
we divide by 4, and have 5625 lbs. per stick of the brace;--now,
correcting for inclination, we shall have

  20 : 25 :: 5625 :  7031 lbs.
  20 : 37 :: 5625 : 10406 lbs.
  20 : 49 :: 5625 : 13781 lbs.
  20 : 64 :: 5625 : 18000 lbs.
  20 : 78 :: 5625 : 21937 lbs.

The weights fouud show the compressional strains on the several
braces;--and, were the pieces to be proportioned for compression
only, their Scantling would be quite small--but on account
of their elasticity, they require larger dimensions.

These braces should not be fastened to the verticals,--but
should be confined both laterally and vertically, where they pass
them. The length of beam, for which we have to guard agains
flexure, is the length between verticals in any panel.

  In panel No. 1, it will be 25 feet,
       "    "  2,    "   "   18  "
       "    "  3,    "   "   17  "
       "    "  4,    "   "   16  "
       "    "  5,    "   "   16  "

Now, using the formula

  2240 b d³
  --------- = W,
      L²

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

we shall have, in round numbers, the following dimensions:

  For the 1st panel, 25 feet long, 8 x 10
     "    2d   "     37   "    "   8 x 10
     "    3d   "     49   "    "   8 x 10
     "    4th  "     64   "    "   8 x 10
     "    5th  "     78   "    "   8 x 10

For the lattice work, a double course on each side of each truss, in
long spans; and a single course, in shorter spans, of 3 x 6, or 2 x 9
plank, bolted at intersections, is sufficient.


GENERAL TABLE OF DIMENSIONS FOR ARCH BRACE TRUSS.

                    G.L. VOSE.

  Span.  Rise.   Chords.     Ties.      Braces.     Lattice.
    50    10    2--8 x 10   1--8 x 10   2--6 x 6
    75    12    2--8 x 10   1--8 x 10   2--6 x 6     2 x 9
   100    15    3--8 x 10   2--8 x 10   3--6 x 6      or
   150    20    4--8 x l2   3--8 x 10   4--6 x 8     3 x 6
   200    25    4--8 x 16   3--8 x 10   4--6 x 9

The arch braces must all foot on an iron thrust block, of which a view
is given in Fig. 4, Pl. III; and the centre of pressure of the braces
must be directly over a bolster, to prevent crippling.

The several sticks forming a brace must be blocked together at
intervals, and When they are spliced,--a butt joint Should be
used--and it should come in the centre of a panel. Below are given the
dimensions of the Thayer Engineering School model.

  Height Ins.    12
  No. Panels      8
  Chords Ins.    2--1 x 1/2
  Posts Ins.     1--2/3 x 5/6
  Braces Ins.    2--1/2 x 1/2
  Lattice Ins.   1/4 x 1/2
  Width Ins.     13

There are several other forms of Bridge, the most notable among which
are the Whipple, McCallum's, Post's, Towne's, Haupt's, and Burr's. But
enough has been said to give the student an idea of the general
arrangement of the different parts of a Truss, and to enable him to
determine the strains to which the various members are subjected.
Nothing will be said in regard to Wooden Arches, as our space is too
limited.


=Pile Bridging.= A bridge of this description is useful in crossing
marshes, or in shallow water. Fig. 5, Pl. III, gives a good example of
this kind of bridge, under 20 feet in height. If on a curve, there
must be extra bracing on the convex side.


=Trestle Work.= This is a combination of posts, caps, and braces; and
is used for both temporary and permanent works. Plate IV, Figs. 1, 2,
3 and 4, give some of the best varieties in use. Figs. 1 and 2, may be
used up to 15 feet in height; Fig. 4, up to 20 feet; and Fig. 3, to 30
ft. The distance apart of the various bents should not exceed 10 or 12
ft., unless bracing is introduced between them, and the bents should
always be raised above the ground a few feet on a solid masonry
foundation. Want of space forbids any mention of abutments and piers,
which really come more properly under the head of masonry.

Iron Bridging is gradually working its way into favor, and Will
probably eventually supersede wooden trusses;--but in many cases wood
is the only material at hand--and therefore some knowledge of Wooden
Bridging is desirable. It is intended to follow this pamphlet with a
portfolio of sheets containing working drawings of several kinds of
Wooden Bridges, taken from actual measurements of some of the best
specimens of the different styles of Truss in use.

       *       *       *       *       *


=PRACTICAL NOTES.=


When putting a truss together in its proper position, on the
abutments, 'false works' must first be erected to support the parts
until they are so joined together as to form a complete
self-sustaining truss. The bottom chords are first laid as level as
possible on the false works, then the top chords are raised on
temporary supports, sustained by those of the lower chord, and are
placed a few inches higher at first than their proper position, in
order that the web members may be slipped into place. When this is
done the top chords are gradually lowered into place. The screws are
then gradually tightened, (beginning at the centre and working towards
both ends,) to bring the surfaces of the joints into proper contact,
and by this method, the camber forms itself, and lifts the lower
chords clear of the false works, leaving the truss resting only upon
its proper supports. The subjoined Table will be found useful in
estimating the strains on a truss when proportioning a bridge for any
moving load.

Table of weights per running foot of a bridge, (either of wood or
iron,) including weights of floor, lateral bracing, &c., complete, for
a single track.

  Clear      Weight of
  Span.       Bridge.
            Tons.  lbs.

   25       .266    596
   30       .281    629
   40       .313    701
   50       .343    768
   60       .374    838
   70       .404    905
   80       .434    972
   90       .464   1039
  100       .494   1106
  120       .554   1241
  140       .614   1375
  150       .643   1440
  160       .673   1507
  170       .703   1575
  180       .733   1642
  200       .792   1774
  225       .867   1942
  250       .940   2105
  275      1.013   2269
  300      1.087   2435


The weight of a single track railway bridge may be taken as equal to
that of a double track highway bridge,--and the trusses that will be
large enough for one will be large enough for the other.

The greatest load that a highway bridge can be subjected to is 120
lbs. to the square foot of surface.


TABLE OF CAMBERS FOR BRIDGE TRUSSES.

  Span.   Camber.   Span.   Camber.   Span.   Camber.   Span.   Camber.
  feet.   Inches.   Feet.   Inches.   Feet.   Inches.   Feet.   Inches.

   25       0.8      75      2.5      175       5.8     275      9.2
   30       1.0     100      3.3      200       6.7     300      10.0
   50       1.7     120      4.0      225       7.5     325      10.8
   60       2.0     150      5.0      250       8.3     350      11.7


TRAUTWINE'S TABLE FOR FINDING INCREASE IN
LENGTH OF UPPER CHORD BEYOND THE
LOWER CHORD ON ACCOUNT OF THE CAMBER.

             Multiply               Multiply
  Depth of   Camber      Depth of   Camber
   Truss.      by         Truss.      by

  1-4 span    2.00      1-12 span    .666
  1-5  "      1.60      1-13  "      .614
  1-6  "      1.33      1-14  "      .571
  1-7  "      1.15      1-15  "      .533
  1-8  "      1.00      1-16  "      .500
  1-9  "      .888      1-17  "      .470
  1-10 "      .800      1-18  "      .444
  1-11 "      .727      1-20  "      .400



TABLE OF AMERICAN WOODS.

                   Weight per   Resistance in lbs. per
     Kind.         cubic foot        square inch.         Value of s.
                   in pounds.   Extension  Compression.

  White Pine.          26         12,000       6000          1229
  Yellow Pine.         31         12,000       6000          1185
  Pitch Pine.          46         12,000       6000          1727
  Red Pine.            35         12,000       6000          1527
  Virginia Pine.       37         12,000       6000          1456
  Spruce.              48         12,000       6000          1036
  Tamarack.            26         12,000       6000           907
  Canada Balsam.       34         12,000       6000          1123
  White Oak.           48         15,000       7500          1743
  Red Oak.             41         15,000       7600          1687
  Birch.               44         15,000       7000          1928
  Ash.                 38         16,000       8100          1795
  Hickory.             51         15,000       7200          2129
  Elm.                 45         16,000       8011          1970


The above table is compiled from a much fuller one in Vose's Treatise
on R.R. Construction.


TABLE OF BOLTS AND NUTS CALCULATED FOR A
WORKING STRAIN OF 15,000 LBS. PER
SQUARE INCH OF SECTION.

  Diameter.  Area.    Strength in Weight per           Thick's  No. thr's
  Inches.  Sq. inches.  Pounds      Foot.  Square nut.  of nut. per inch.

    1/2      .19635        2940     0.66     1-1/4 in    3/4 in    12
    5/8      .30680        4602     1.03     1-3/8       3/4       10
    3/4      .44179        6630     1.49     1-1/2       7/8       10
    7/8      .60132        9019     2.03     1-3/4     1            9
  1          .78540       11775     2.65     2         1            8
  1-1/8      .99402       14910     3.36     2         1-1/8        7
  1-1/4     1.2272        18405     4.17     2-1/4     1-1/4        7
  1-3/8     1.4849        22260     5.02     2-1/2     1-3/8        6
  1-1/2     1.7671        25505     5.97     2-3/4     1-1/2        6
  1-5/8     2.0739        31095     7.01     2-7/8     1-5/8        5
  1-3/4     2.4053        36075     8.13     3         1-3/4        5
  1-7/8     2.7612        41415     9.33     3-1/4     1-7/8        4-1/2
  2         3.1416        47130    10.62     3-1/2     2            4-1/2
  2-1/8     3.5166        53190    12.00     3-3/4     2-1/8        4
  2-1/4     3.9761        59640    13.40     4         2-1/4        4
  2-3/8     4.4301        66450    15.00     4-1/8     2-3/8        4
  2-1/2     4.9087        73620    16.70     4-1/4     2-1/2        3-1/2
  2-5/8     5.4119        81178    18.20     4-1/2     2-5/8        3-1/2
  2-3/4     5.9396        89094    20.00     4-3/4     2-3/4        3-1/2
  2-7/8     6.4918        97377    21.90     5         2-7/8        3
  3         7.0686       106029    23.80     5-1/4     3            3
  3-1/4     8.2958       124437    27.90     5-3/4     3-1/4        3
  3-1/2     9.6211       144316    32.40     6         3-1/2        2-1/2


TABLE OF SAFE WORKING LOAD IN LBS., FOR HOLLOW CAST-IRON COLUMNS.

[_G.L. Vose._]

   Outside             Length or height in Feet              Metal
   Diameter                                                 Thickness
  in inches.  6      8      10     12     15     18     20  in inches.

       3    16000  14000  13000  11000   9000   7000   6000    3/8
       4    30000  29000  26000  24000  22000  18000  16000    1/2
       5    50000  37000  45000  42000  39000  37000  31000    5/8
       6    59000  57000  55000  52000  49000  44000  41000    3/4
       7   101000  99000  96000  92000  88000  81000  76000   13/16
       8   131000 129000 126000 122000 118000 109000 105000    7/8
       9   169000 167000 164000 160000 156000 146000 141000   1
      10   210000 200000 200000 200000 190000 180000 180000   1-1/8
      11   250000 250000 240000 240000 240000 230000 220000   1-1/4
      12   300000 300000 290000 290000 290000 270000 270000   1-1/2
      14   450000 430000 410000 380000 370000 350000 330000   1-3/4
      16   520000 500000 480000 460000 440000 420000 400000   2
      18   650000 630000 610000 590000 560000 520000 470000   2-1/2
      20   800000 760000 740000 690009 650000 590000 540000   3


[Illustration: Pl. IV. with Fig. 1., Fig. 2., Fig. 3., Fig. 4.]




Transcriber's Notes:

DISCLAIMER: This document should NOT be used to engineer any bridge
projects! Many typesetting errors were found, and it is possible that
there are further errors in the information that were not caught.

Formulas have been provided both as ASCII and TeX following in brackets.
Bold headings are handled with equal signs before and after the bold text.
Italicised text uses the standard underlines before and after the text.
Fractions are expressed in the format: 2-1/4 means two and one quarter.

The page numbers listed below are project page numbers.
(The original book used Roman numerals to number the pages.)

Note that the book uses the "long" ton equal to 2,240 pounds.

CORRECTIONS MADE:

   1. Page 8--the formula for "d" must use a cube root, which is how it
      is shown here, but the '3' to indicate a cube root is not found in
      the original document.
   2. Page 8--typo in word 'sectien'--changed to 'section'.
   3. Page 10--Value for working comp. strength of cast iron in the
      table had a typo (25,v00). Since other values use round numbers,
      it is assumed the value should be 25,000.
   4. Page 10--Two other typos. Changed 'the the' to 'the', and in
      table heading, original word was 'detrution', changed to correct
      spelling of 'detrusion'.
   5. Page 12--changed 'woooden' to 'wooden'.
   6. Page 13--Example II--In the calculations, the intermediate value
      in the book was printed as the square root of 67.2. The left part
      is correct, but reduces to the square root of 268.8, and that is
      ~16.395. So I have corrected the intermediate value.
   7. Page 13--Because the original page scan cut off the text on the
      right edge, I have made assumptions on what text was missing.
      Because the scans came from an outside source, I could not get
      the missing information, which was the words at the end of Example
      II, and words in the last paragraph of the page.
   8. Page 14--Three typos found: 'dimensiens' for 'dimensions', 'betng'
      for 'being', and 'ars' for 'are'.
   9. Page 17--a value in a formula was printed as 6000, but in the
      context of the other information, particularly the example
      immediately following, the value was believed to be incorrect, and
      was changed to 5000.
  10. Page 23--The value of 388 sq. inches at the top of the page in the
      book is incorrect; 3 x 8 x 12 = 288, so has been corrected.
  11. Page 24--Rods section, numerical value appeared to be 15.000 in
      the book, but from context, must be 15,000 instead.
  12. Page 28--Typos: changed 'Trautwine's Edgineer's Pocket-Bood' to
      Trautwine's Engineer's Pocket-Book'; corrected  'af' to 'as',
      'bracas' to 'braces'.
  13. Page 30--Apparent typo in the table at the bottom of page.  Value
      for Center Brace size for 200' span was shown as '8 x 1', believed
      from context of table to be '8 x 10'.
  14. Page 33--table of dimensions of a Pratt Truss, last column, row
      starting with 150, the original says 8--1-1/8, this is believed to
      be, and has been changed to, 3--1-1/8.
  15. Page 37--The five formulas with square roots were incorrectly
      printed in the book, multiplying the terms inside the square root
      instead of adding them, which is obviously incorrect per the
      Pythagorean theorem of right triangles.
  16. Page 38: The fifth ratio in the group of five near the top of the
      page must start with 20, not 10 as in the book
  17. Page 38--The equation for W as printed on this page is not
      consistent with that found on pages 18 to 24, so has been corrected
      from 'bd^2' to 'bd^3'.
  18. Page 39--arch brace truss table heading typo-changed 'FOE' to 'FOR'.