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:  $W = \frac{2240 \times 6 \times 512}{324} = 21238$]