To construct the trial equilibrium polygon proceed as follows: 1. Lay off a load line 1.... 19 to represent the loads. 2. Choose a trial pole, which may be at any point but which should preferably be at a point such that the pole distance will be some round number as 10,000, 20,000, etc., and also approximately opposite the point on the load line that divides the load line into the two reactions. The true pole distance can be approximately determined by applying Navier's principle (§ 1214) to the crown of the arch. In the case in hand, the crown thrust as computed by Navier's principle is: T = p p = (2 X 150 + 2 X 100 + 100) 36.25 = 21,749 lb. Hence the trial pole distance was taken at 20,000 lb., and the trial pole was located a little below the center of the load line at P. 3. Draw the several rays. 4. Construct the equilibrium polygon b, .... 5. Draw a line from v, (= to (= and drop vertical from L„ b;, etc., upon v„ and mark the points v,. . the arch were either hinged or simply supported at A and B, there no moment at these points, and hence the clos ing line of. the equilibrium polygon would be parallel to the line AB, and the true equilibrium polygon could readily be found; but as the arch under consideration has fixed ends, there is a moment at each abutment, and therefore the position of the closing line is not known, and hence the true equilibrium polygon can not be found by the usual method.
The first step toward finding the true equilibrium polygon is to find the position of the closing line, of the trial equilibrium polygon, b, .... .... Since the summation of the moments at the various points of an arch ring having fixed ends is zero, and since the ordinates of an equilibrium polygon are proportional to the moments, the closing line should have such a position that the sum mation of the intercepts between it and the equilibrium polygon will be zero, i.e., the closing line should satisfy the condition 2' M = 0, or its equivalent But the above condition is not enough to fix the position of the clos ing line, since any number of lines can be drawn which will make the sum of the intercepts equal to zero. The other condition which may be employed to fix the closing line is that stated in equation 5, page 674, viz.: 2' M . x = 0, or its equivalent To show how to utilize the above conditions in finding the closing line, the problem may be restated as follows: If the trial equilibrium polygon be considered without reference to the arch, the intercepts .... .... may be regarded as forces; and then the prob lem of finding the closing line may be regarded as that to find what system of minus forces must be added to the positive forces b,v, ... . to satisfy the conditions 2' M = 0 and 2' M . x = 6, or their equivalents r bm = 0 and S bm ..r = 0. Then, since the summation of the moments is to be made equal to zero, Le.. since we are to have I bm = 0, the total minus forces must be equal to the total positive forces; and since we arc also to have I bm . z = 0, the resultant of the minus forces must lie in the same line as the re sultant of the positive forces.* The preceding principles make it possible to find (1) the resultant of the positive forces and (2) the closing line of the trial equilibrium polygon.
Table 94 gives the values of the coordinates z and y to the points of intersection of the lines of action and neutral line of the arch ring. and also various intercepts and products employed in the solution to follow. (Taking the origin of coordinates at the middle of the span gives smaller values of x and otherwise materially shortens the sub sequent work.)
To find the magnitude of R', the resultant of the forces repre sented by the intercepts by, take the sum of 6,r, which is shown in Table 94 to be 164.71 ft.t To find the position of the resultant, compute the successive prod ucts by . z, and divide their sum by the sum of the intercepts by. The several products by . x are given in Table 94; and their sum is — 19.4, which divided by 164.71 gives — 0.12 ft. Hence, R' acts 0.12 ft. to the left of C, i.e., on the side toward the abutment that has the heavier load; or z = — 0.12 ft. Or, since the position of the inter cepts by is symmetrical about the crown, the equation of moments may be stated thus: R . z = (b,v, — bnvu) x, + (b9v9 — In solving the problem on a drawing board, this formula affords a method of finding the position of the resultant which is a little shorter than the preceding one.
To find the Closing Line of the Trial Equilibrium Polygon. The next step is to find a closing line such that if the ordinates from it to v18 are treated as forces, their resultant will be equal to R' in magnitude and coincide with it in position. We will assume a trial closing line parallel to such that v,n, is equal to the average of the by ordinates, i.e., v,n, = uienis = R' _ (16 + 2) = + 9.15 ft. (The line could be drawn in any position, but drawing it parallel to v,v18 and making R _ (16 + 2) simplifies the subsequent work).
We have said that the ordinates v,n, ... may be regarded as representing the negative forces which must be added to the given forces to give the closing line and similarly, if lines n,v„, and be drawn, the total negative load may be regarded as being represented by the ordinates to the two triang!°s and n,n18v13. Designate the resultant of the forces represented by the triangle as trial and the resultant for the triangle as trial (The subscript of T indicates whether the resultant lies to the right or to the left of the center line C'D'). It is required to find the magnitude and position of T1 and _ _ The magnitude of trial is the sum of the ordinates of .the triangle Since the line was drawn parallel to the triangles and are equal; and therefore trial — trial T = 1 R' _ (164.71) = 82.35 ft.; and trial is as far to the right of C' as trial Ti is to the left. Let represent the distance of trial from C'. As just stated The position of trial Ti is most conveniently found by taking moments of the intercepts about C', and dividing by the sum of the intercepts. If a line be drawn from v, to F' (the point where crosses the vertical through C'), then the moment of the tri angle is equal to that of and consequently the moment of about F' is equal to the moment of about the same point. The intercepts of the triangle are given in Table 94, page 683, as also the moments of these intercepts about F', the former being designated / and the latter f . x. Forming the equation of moments as above and solving, we get The magnitude of trial Tl is increased, if the point n, is moved vertically upward; and is decreased, if is moved down. Further, the position of trial is not changed if is moved vertically, since all ordinates will be increased proportional to their lengths and hence the sum of the momenta divided by the sum of the forces will remain constant; i.e., the distance from trial to C' remains un changed if n, is moved vertically either up or down. The movement of n, does not affect either the position or magnitude of trial since neither the magnitudes of the several ordinates nor their position horizontally is altered. Similarly a movement of n„ alters the value of trial but not its position.