E Constant. The coefficient of elasticity of concrete varies with the unit load (§ 409), but within the ordinary working stress the variation for any particular concrete is not great; and therefore E may be regarded as a constant, and may be placed before the sign of integration in the equations of condition.
Equations of Condition Restated. By placing its - 1 and 1 _ E outside of the integration sign, and using the summation sign. equations 1, 2, and 3 become, respectively, Equations of Condition in Graphic Terms. To adapt the above equations to graphic computations, it is necessary to find M in graphic terms. To do this, let GJ in Fig. 229 be a portion of the arch, ab the neutral line, ac a vertical line, and ce be the adjacent side of the equilibrium polygon, and ae a line from a perpendicular to ce. Let R represent the magnitude of the force acting in the line ce, i.e., R is the length of the ray in the force diagram parallel to the side ce of the equilibrium poly gon; and let H be the horizontal component of R, i.e., H is the true pole distance.
The force R, being eccentric, tends to bend the arch rib; and the amount of the bending, M, about a is R . ae. By similar triangles, R . ae = H. ac; that is, the bending moment at any sec tion of an arch rib acted upon by vertical loads is equal to the true pole distance multiplied by the vertical intercept between the true equilibrium polygon and the neutral line. Substituting the above value of M in equations 4, 5, and 6, and remembering that H is a constant for any particular system of loads, the equations of con dition become, respectively, in which ac is a general expression for the intercept between the true equilibrium polygon and the neutral line of the arch ring, and x is the horizontal distance of any point from the mid-span, and y is the vertical distance of any point above a horizontal line through the abutments.
c, ... c„ the true equilibrium polygon, and k, ... k, an axis of ref erence. Then at any point ac = ck — ak (e) Taking the summation of equation e, and remembering that 1 ac = o, we have 2' ac = ck — ak = o; or F ck = 1 ak . . . . (10) Multiplying equation e by x, and taking the summation, ac .x = I ck.x — Z ak.x = o; or 1 ck.x ak.x . (11)
Similarly rac.y=1ck.y—Iak.y=o; . (12) Therefore we see that the three equations of condition, equations 7, 8, and 9, will be satisfied, if equations 10, 11, and 12 are fulfilled. Furthermore, equations 10 and 11 will be fulfilled, if the axis k, .. . k, is taken so as to make • ak = 0 . . . . . . (13) Ick . . . . . . . (14) • ak.x = 0 . . . . . . . (15) • ck. y = 0 . . . . . . . (16) Hence the determination of an equilibrium polygon satisfying equations 7, 8, and 9, is accomplished by (1) dividing the arch ring into sections such that ds - I is constant, (2) finding a reference line for the arch ring such that E ak = 0 and 2' ak . x = 0, and (3) constructing thereon an equilibrium polygon such that I ck = 0, .x = 0, and Ick.y = ak y.
There are several methods of determining the successive divi sions of the arch ring, but the following graphical process is the sim plest. Divide the neutral line of the semi-arch ring into any number of equal parts, say, from 5 to 10; and measure the radial depth of the ring at each point of division. Rectify the neutral line, either by stepping around it with a pair of dividers or by computation, and lay off this distance to scale from A' to C' in Fig. 231; and divide the line A'C' into the same number of equal parts as the semi-arch ring. At each point of division of A'C' erect a vertical equal to the moment of inertia at the corresponding point on AC; or since in a plain concrete arch the moment of inertia is proportional to the cube of the depth, we may lay off the latter quantity instead of the moment of inertia. Connect the tops of these verticals by a smooth curve DF, and then it may be assumed that any ordinate to the curve DF is proportional to the moment of inertia at the corresponding point of the arch ring.