176. The most common form for an arched rib is a part of a circle. If its ver. sine docs not exceed 1/4 the length of its chord the point P may be taken at 3/11 of the length of the arc B P D A (Fig. 46) from B without any sensible error. If it exceeds the above proportion the position of the point P should be corrected until the moments are equalized as described (Art. 174).

The maximum cross strain on a circular segmental rib is below the weight when the latter is placed so that the line of thrust passes through the centre of the crown, or when it is at 3/8 of the length of the half-arc from that centre. The maximum cross strain at the haunch is also when the weight is placed in the same position.

When the ver. sine does not exceed 1/4 of the length of the chord, the maximum strain at the crown is nearly double that at the haunch. But as 1/2 the length of the arc at the haunch is equal to 4/3 the length of the corresponding arc at the crown, the scantling of the rib at the haunch will require to be made for stiffness more than 1/2 the strength required at the crown. (See Example.)

Example. - Let it be required to find the proper depth, at the centre and haunches, to give to an oak rib 8" thick, 20 feet span and 5 feet rise, to sustain a weight of 5 tons at any point; the curve being the segment of a circle.

1st. Suppose the weight is placed at the crown.

To find the strength at the crown we have formula [29]

B3 = 16f Lav/T

By construction we find f = 4 1/2 tons = 10080 lbs., T = 8, v = 1.5" = 1.416, L = 3'.2 1/4" =3.208, and, from Table VI. (Art. 93), a =0119.

...B3 = 16x10080x3.208x.0119x1.416/8 = 1090.

...B = 10 1/4"

To find the strength at the haunches we have [26] B3 = 5.33xf Lav/T

By construction,f = 10080 lbs., v = .7083, L = 8.5, and a =.0119.

B3 = 5.33 x 10080 x 8.5 x.0119 x.7083/8= 481.

B = 8" nearly.

But in this case the maximum strains produced will be when the weight is situated at 3/8 of the half-are from the centre of the crown. Suppose the weight is placed in that position:

Then by construction we find f = 7168 lbs., L = 4'.4 1/8" = 4.344, v = 2'.7" = 2.5838, a = .0119.

.....B3 = 16 f Lav/T = 16 x 7168 x 4.344x.0119 x 2.5838/8 = 1915.

B = 12 1/2" at the crown nearly.

To find the strength at the haunch we have formula [26], f= 7168 lbs., L = 11.7 = 11.5833, v = 1.3 1/2 = 12916 a = .0119.

B3 =5.33xfLva/T

= 5.33 x 7168 x 11.583 x .0119 x 1.2916/8 = 850. B = 9 1/2" nearly.

In calculating the strength, therefore, of a uniform circular segmental rib draw the line of thrust through the axis at one springing, and through the centre of the crown, and let fall a perpendicular to represent the weight, cutting the axis at 3/8 of the length of the half-are from the crown. Then calculate the required strength by the formula [29].

The strength of a circular segmental rib varies imperceptibly with regard to its rise or "curvature.

177. Next let us consider a rib of any curvature acted upon by any number of forces of various magnitudes and in different directions.

Let ABCDEFG (Fig. 47) be the centre line of any curved rib, with fixed abutments at A and G, acted upon by various forces represented in magnitude and direction by the lines BCDEF.

Besolve each force into two others, one through each abutment, taking care to choose the correct segment to divide so as to equate the moments. Then we have a collection of forces at each of the points A and G. Find the resultants ff' of these collections; f and f' will represent in magnitude and direction the thrusts on the abutments at A and L. Then commencing with f or f' construct the polygon of forces AbcdefG. The thrusts along its sides multiplied into their distances from the axis of the rib will represent the moments tending to break the rib at various points, and the necessary scantlings may be found from the formulae before given.

Fig. 47.