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