As an example of the estimate of strains upon this bridge take the following.
| Span, | 90 | feet. |
| Rise, | 18 | feet. |
| Panel, | 15 | feet. |
| Weight per lineal foot, | 2,500 | lbs. |
| Whole weight, | 225,000 | lbs. |
| Weight on each side truss, | 112,500 | lbs. |
| Weight on each post, | 18,750 | lbs. |
The weight borne by each system, i. e., one post and the two supporting rods, is 18,750 lbs. The strain to be resisted by any one rod depends upon its inclination.
The following figures show the elements of the truss in question:—
| Rod. | Length. | Applied weight. | Increased strain. | Section of the bar in inches. | |
|---|---|---|---|---|---|
| A B = | 90.0 | ||||
| C D = | 18.0 | ||||
| A C = | 23.4 | (18750 – 3125) = | 15625 which by | 23.4 18 = 20312 | 1⅓ |
| A H = | 35.0 | 18750 × 60 90 = | 12500 which by | 35 18 = 24306 | 1⅔ |
| A F = | 48.5 | 18750 × 45 90 = | 9375 which by | 48.5 18 = 25260 | 1⅔ |
| A K = | 62.6 | 18750 × 30 90 = | 6250 which by | 62.6 18 = 21736 | 1⅓ |
| A G = | 77.6 | 18750 × 15 90 = | 3125 which by | 77.6 18 = 13472 | 1 |
Column 1, gives the name of the rod; col. 2, the calculated diagonal length; col. 4, the applied weight, (the varying weight by reason of the varying inclination) found by multiplying the whole weight upon one panel or post by the distance of that post from the abutment, and dividing the product by the span. (Thus the load applied to A G is
W × IB
S,
that on A K is
W × BX
S,
and so on.) Col. 6, shows the increase found by col. 5 on account of inclination as noticed in Chap. VIII.; and finally, col. 4 gives the necessary sectional area of the bars or rods.