When the load on a bridge is uniformly distributed, the curve of equilibrium is a parabola, and if the form of the rib be made to approximate to that curve, the load will have no tendency to produce derangement, the strain or pressure being transmitted in the direction of the curve.

* In Kirkaldy's experiments, a log of white Riga fir 20 feet long and 13 inches square, compressed 1 part in 930 with a load of about 53 1/2 tons (see Table XX., p. 90).

The amount of the pressure in the direction of the rib at the crown, arising from a uniformly-distributed load, is the same as the horizontal thrust, and may be determined by the Rules in Art. 40, Sect. I., or by the following method, which is more convenient for our present purpose.

Let P = the strain or pressure in lbs.; R = the rise in feet; S = the span in feet, and W the gross weight of the bridge and its load in lbs.; then, by the resolution of forces,

2R::S/2::P the pressure in the direction of the curve at the crown, or

SxW/8R = P. [A]

If 6 = the breadth of each rib; d = the depth, both in inches; and n = the number of ribs; then 1000 n x b x d = pressure, when the rib is sufficiently strong. Combining this with formula [A], we have

SxW/8x1000xRxn= b x d = sectional area in square inches. [B]

From this equation we derive the following rule: -

Rule. - Multiply the span in feet by the gross distributed load on the bridge in lbs., and divide by 8000 times the rise in feet multiplied by the number of ribs; and the result will be the sectional area in square inches of each rib at the crown capable of resisting the pressure.

Example. - To find the sectional area at the crown for each rib of a bridge of 200 feet span, and 15 feet rise, capable of supporting a gross distributed load of 1,800,000 lbs., the number of curved ribs being three,

200 x 1800000

= 1000 square niches.

8000 x 15 x 3

Such a rib may be formed of three pieces in depth and two in thickness at the crown, similar to those of the bridge near Bamberg (Plate XL.).

360. In consequence of the greater pressure, the sectional area of the ribs at the abutments must be increased to that given by the following rule: -

Rule. - Multiply half of the gross distributed load supported by each rib in lbs. by the square root of the square of the rise of the curve in feet, added to one-sixteenth part of the square of the span in feet, and the result, divided by the rise, will be the pressure in lbs. on the rib at the abutment, which, divided by 1000, will give the sectional area in square inches.

Example. - The span, rise, number of ribs, and load being as in the last example, to find the sectional area of each rib at the abutment, which, allowing 1000 lbs. to the square inch, gives 1044 square inches nearly for the area at the abutment.

361. The foregoing rules for curved ribs apply only to bridges that are uniformly loaded. When the load is variable, as in the case of a railway train in motion, there is no curve of equilibrium to which the flexible ribs of a timber bridge can be made to approximate, consequently a timber arch, as that used in the bridge at Bamberg, is not well adapted to sustain such a load, unless rendered perfectly rigid by proper bracing.

To calculate the stiffness of curved ribs without braces, the reader is referred to Mr. Guthrie's paper on the subject, given in Arts. 174 to 177, Sect. II.