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.