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قراءة كتاب Instructions on Modern American Bridge Building

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Instructions on Modern American Bridge Building

Instructions on Modern American Bridge Building

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دار النشر: Project Gutenberg
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class="center">37 ÷ 6 = 6.16 inches—

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.

 

Plate 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, 4¼ X 1½ inches nearly.
Bottom Flange, 6 X 2½ inches nearly.
Web, 1½ inches thick.

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

Equation: d = (sqrt)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

Equation: d = (sqrt)4 WL / 5000 b = (sqrt) 4 x 22400 x 12 / 500 x 12 = (sqrt)215.04 = 15 in. nearly.

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. Ties. 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 DF and EF 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 2½ tons = 2½ 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.

Equation: W = 2240 bd^3 / L^2

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

Equation: W = 2240 x 3 x 125 / 324 = 840000 / 324 = 2592 lbs.

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

Equation: W = 2240 x 6 x 512 / 324 = 21238 lbs.

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 FB, Pl. II, Fig. 2, draw BD parallel to FC, also DH parallel to AC—DH is the tension. This is the graphical construction, and is near enough for practice. Geometrically we have the two similar triangles AFB and DFH, whence

AF: DF:: AB: DH

and Equation: DH = DF x 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

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