| 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.
Now substituting 3 for b, and 5 for d, we have
= 2592 lbs.
which is not enough. Using 6 for b and 8 for d, we have