ds so that ds _ I shall be constant, draw a line C'a at any slope and then a line ab at the same slope, and continue the construction by drawing other isosceles triangles as shown, using always the same slope. This divides the rectified arch ring into a number of parts, Cb, bd, df, etc., such that the length of each part divided by the moment of inertia at its center is constant, i.e., 48 _ I = 2 tan a, in which a is the angle between the sides of the isosceles triangles and the vertical. It is not important that a point of division shall fall exactly at A', since many arches join the abutment by a gradually increasing section, and hence there is really no springing line, and also since most arches are so thick at the springing that the position of the line of resistance is mainly determined by the portion of th" 3.rch ring over the central half of the span.
of parts only by successive approximations. To make the first approximation, find the average value of the moment of inertia at several points, say, the equidistance points used in constructing DF, Fig. 231, i.e., find the average of the ordinates used in constructing Fig. 231, and designate the result 1„. Then if n = the number of parts into which AC or A'C' is to be divided, The I, found as above is not the average of the ordinates at a, c,
e, etc.; and consequently a solution depending upon it will be only an approximation. To make a first approximation, lay off tan a a little smaller than the value computed above, and construct a series of isosceles triangles. By using a Brown and Sharp protractor these
triangles can be constructed very quickly. To make a second approx imation, find a new value of 1, by taking the mean of the ordinates at a, c, e, etc., and construct a new series of triangles. For a third approximation, determine a third value of I, by measuring the or dinates at a, c, e, etc., of the second series of triangles. The third approximation is usually practically exact. In making the exact division, it is better to begin at A in constructing the isosceles triangles, since then the first side of each triangle intersects the curve DF more nearly at right angles, and hence the solution is more accurate.
The second step (§ 1310) is to locate the line kk, Fig. 230, page 675, so as to satisfy the conditions ak = 0 and I ak .x = 0. Since the line joining the abutments is horizontal, if a horizontal line k,k, is drawn at a distance above AB equal to the average of the ordinates to the neutral line, then I ak = 0; that is, if a, k, = r ad - (n + 2) in which n is the num ber of sections in the arch ring (§ 1312), then equation 13, page 675, i.e., ak = 0, is satisfied.
Since the arch is symmetrical with reference to a vertical line through the crown, and since the points of divisions of the arch ring are also symmetrical with reference to the crown, the line k,
drawn as above, also satisfies the equation 15, page 675, which is r ak .x = 0 .
Equations 14 and 16, page 675, and also equation 9, page 674, involve the equilibrium polygon, and hence they can not be satisfied until that is determined.