Instructions on Modern American Bridge Building by G. B. N. Tower
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G. B. N. Tower >> Instructions on Modern American Bridge Building
INSTRUCTIONS
ON
MODERN AMERICAN
BRIDGE BUILDING.
WITH
PRACTICAL APPLICATIONS AND EXAMPLES,
ESTIMATES OF QUANTITIES, AND
VALUABLE TABLES.
Illustrated by four Plates and Thirty Figures.
BY G.B.N. TOWER,
CIVIL AND MECHANICAL ENGINEER,
_Formerly Chief Engineer U.S. Navy, and late Chandler Instructor in Civil_
_Engineering at Dartmouth College._
BOSTON:
A. WILLIAMS & COMPANY,
135 WASHINGTON STREET.
1874.
Entered according to act of Congress, in the year 1874, by
A. WILLIAMS & CO.,
in the office of the Librarian of Congress, at Washington, D.C.
PREFACE.
This little treatise was written for the purpose of supplying a want
felt by the author while giving instruction upon the subject. It was
intended for an aid to the young Engineer, and is not to be considered
as a complete substitute for the more elaborate works on the subject.
The first portion of this work mentions the various strains to which
beams are subjected, and gives the formulae 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 deg.--and let 2000 lbs., as before, be suspended from its lower end.
Now the diagonal being 20 deg.,--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:--
WL squared bd cubed x 80 E
---- = bd cubed, whence W = -----------
80 E L squared
[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
-----
/ WL squared WL squared
d = cubed/ -----, and b = ------
\/ 80 Eb 80 Ed cubed
[TeX: $d = \sqrt[3]{\frac{WL^2}{80 EB}}$, and $b = \frac{WL^2}{80 ED^3}$]
in the above formulae,
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 bd squared
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 sbd squared
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 bd squared
W = ------------,
L
[TeX: $W = \frac{4 \times 1250 bd^2}{L}$]
or
5000 bd squared
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 bd squared
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 bd squared
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 d squared
[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 bd cubed
W = --------
L squared
[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 "