assumption which shall be more suitable, it being necessary to make the form of the arch conform more closely to that of the equilibrium polygon for the given loading. The question may be regarded as one of economy of material, and ease of construction, analogous to that of the truss bridge. In this latter case, constructors have long since abandoned any idea of making bridges in which the inclination of the ties and posts should be such as to require theoretically the minimum amount of material. Indeed, the amount of material in the case of a theoretic minimum, differs by such an inconsiderable quantity from that in cases in which the ties and posts have a very different inclination, that the attainment of the minimum is of no practical consequence. Similar considerations applied to the arch, lead us to the conclusion that the form adopted can in every case be composed of segments of one or more circles, and that for the purpose of construction every requirement will then be met as fully as by the more complicated transcendental curves found by the writers previously mentioned. If considerations of an artistic nature render it desirable to adopt segments of parabolas, ellipses or other ovals, it will be a matter of no more consequence than is the particular style of truss adopted by rival bridge builders. The It arc of a circle; having a chord or span 8 Let bab in Fig. 2, be the neutral axis of the arch of which the rise is onetenth the span. Let a x y z be the area representing the load on the left half of the arch, and a x'y' z' that on the right, so that yp=a. P=xy on the left, and ypx'y' on the right. Divide the span into sixteen equal parts bb, bb', etc, and consider that the load, which is really uniformly distributed, is applied to the arch at the points a, a,, a etc., in the verticals through b, b1, b', etc.; so that the equal weights P are applied at each of the points on the left of a and the equal weights Pat each point on the right of a, while 3 Pis applied at a. We can also readily treat the problem in an inverse manner, viz:-find the system of loading, of which the assumed curve of the arch is the equilibrium polygon. From this it will be known how to load a given arch so that there Take b as the pole of a force polygon shall be no bending moments in it. for these weights, and lay off the weights This, as may be seen, is often a very which are applied at the left of a on the useful item of information; for, by leav- vertical through b, viz., bw, P=the ing open spaces in the masonry of the weight coming to a from the left; spandrils, or by properly loading the w, w,=P=the weight applied at a,; crown to a small extent, we may fre-, P=the weight applied at a, etc. quently render a desirable form entirely Using b still as the pole, lay off bw,' = stable and practicable. P=the weight coming to a from the right; w,' w, P=the weight applied at a,', etc. This amounts to the same thing as if all the weights were laid off in the same vertical. Part are put at the left and part at the right for convenience of construction. Now draw bw, until it intersects the vertical 1 at c1; then draw c, c, || bw,; and c, c, || bw,, CHAPTER II. THE ARCH RIB WITH FIXED ENDS. LET us take, as the particular case to be treated, that of the St. Louis Bridge, which is a steel arch in the form of the 3 2 8 8 etc. In the same manner draw bw,' to negative loading must be equal numericc'; then cc' || bw', etc. Then the ally to the whole positive loading, if we broken line bc,...c. is the equilibrium are to have (M)=0. Next, as the polygon due to the weights on the left closing line is to be straight, the negaof a, and be,... c.' is that due to the tive load c. c,' h, h' may be considered weights on the right. Had the polygon in two parts, viz., the two triangles, been constructed for the uniformly dis- c. c'h, and c'h, h.'. Let the whole tributed load (not considered as concen- span be trisected at t and t', then the trated), on the left we should have a total negative loading may be considered parabola passing through the points to be applied in the verticals through be,... c, and another parabola on the t and t', since the centers of gravity of right through bc'...c'. From the the triangles fall in these verticals. properties of this parabola it is easily Again, the positive loading we shall find seen that c, must bisect w, w, as c' must it convenient to distribute in this manalso bisect ww; which fact serves to ner: viz., the triangle c, bc,' applied in test the accuracy of our construction. the vertical through b, the parabolic area This test is not so simple in cases of bc,... c. in the vertical 4 which conmore irregular loading. tains its center of gravity, and the parabolic area be'. . . c' in 4'. 5 4 The equilibrium polygon cb c' is that due to the applied weights, but if these Now these areas must be reduced to weights act on a straight girder with equivalent triangles or rectangles, with fixed ends, this manner of support re- a common base, in order that we may quires that the total bending be zero, compare the loads they represent. Let when the sum is taken of the bending the common base be half the span : then at the various points along the entire bb=pp' is the positive load due to the girder; for, the position of the ends triangle cbc; and c, c=pp, and does not change under the action of the c, c'p'p,' are the positive loads due weights, hence the positive must cancel to the parabolic areas. the negative bending. To express this by our equations : yb=e. Σ(M)=0 .. Σ(M)=0. This is one of two conditions which are to enable us to fix the position of the true closing line hh' in this case. The other condition results from the fact that the algebraic sum of all the deflections of this straight girder must be zero if the ends are fixed horizontally. This is evident from the fact that when one end of a girder is built in, if a tangent be drawn to its neutral axis at that end, the tangent is unmoved whatever deflections may be given to the girder; and if the other end be also fixed, its position with reference to this tangent is likewise unchanged by any deflections which may be given to the girder. To express this by our equations: yd=f. Σ (Mx)=0 .. ≥ (Mx)=0 Now assume any point g as a pole for the load line p, p,' and find the center of gravity of the positive loading by drawing the equilibrium polygon, whose sides are parallel to the lines of this force polygon viz., use qp, and qp as the 1st and 2nd sides, and make pq' || qp', and q'q, || qp,'. The first and last sides intersect at 7,; therefore the center of gravity of the positive loads must lie in the vertical through q Now the negative loading must have its center of gravity in the same vertical, in order that the condition (Mx)=0 may be satisfied, for it is the numerator of the general expression for finding the center of gravity of the loading. The question then assumes this form: what negative loads must be applied in the verticals through t and t' that their sum may be p, p,', and that they may have their center of gravity in the vertical through q The method of introducing these con- The shortest way to obtain these two ditions is due to Mohr. Consider the segments of p, p,' is to join r and r' area included between the straight line which are in the horizontals through cc and the polygon c. b c as some p, and p', and draw an horizontal species of plus loading; we wish to find through, which is the intersection of what minus loading will fulfill the above rr' with the vertical through q,; then two conditions. Evidently the whole rr, and r'r,' are the required segments 8 2 0 0 2 = of the negative load. For, let rr=p,'p, we intend to make between the polyand take as the pole of the load r; gons c and d (as we may briefly desigthen, since, 。 || r2r' and 9. rrr we nate the polygons c. b c and d d d'), let have the equilibrium polygon r, qr' ful- us notice the significance of certain operfilling the required conditions. lations which are of use in the construcNow these two negative loads rr2 tion before us. One of these is the r'r' and r1, are the required heights of multiplication of the ordinates of the the triangles c, h. c and c. c'h'; there- polygon or curve a to obtain those of d. fore lay off c, h=r'r,' and c'h'=rr. If a was inverted, certain weights might The closing line hh can then be be hung at the points a, a, etc., such drawn, and the moments bending the that the curve would be in stable equistraight girder will then be proportional librium, even though there are flexible to h, c,, hc, etc., the points of inflexion joints at these points. Equilibrium being where the closing line intersects would still exist in the present upright the polygon. If the construction has position under these same applied been correctly made, the area above the weights, though it would be unstable. closing line is equal to that below, a test easy to apply. 8 If now, radiating from any point, we draw lines, one parallel to each of the sides aa1, a, a, aa', etc., of the polygon, then any vertical line intersecting this pencil of radiating lines will be cut by it in segments, which represent the relative weights needed to make a their equilibri Let us now turn to the consideration of the curve of the arch itself, and treat it as an equilibrium polygon. Since the rise of the arch is such a small fraction of the span, the curve itself is rather flat for our purposes, and we shall therefore um polygon. By drawing the vertical line multiply its ordinates ab, a, b,, etc., by at a proper distance from the pole, its any number convenient for our purpose: total length, i. e., the total load on the in this case, say, by 3. We thereby get arch can be made of any amount we a polygon ddd' such that d b=3 ab, please. The horizontal line from the db=3d, b1, etc. If a curve be de- pole to this vertical will be the actual scribed through d ̧...d... d.' it will be horizontal thrust of the arch measured the arc of an ellipse, of which d is the on the same scale as the load. If a like extremity of the major axis. pencil of radiating lines be drawn paralIf we wish to find the closing line kk'lel to the sides of the polygon d and the of this curve, such that it shall make load be the same as that we had sup(Ma)=0_and (Mdx)=0, the same posed upon the polygon a, it is at once process we have just used is here appli- seen that the pole distance for d is onecable; but since the curve is symmetri- third of that for a; for, every line in d cal, the object can be effected more has three times the rise of the correeasily. By reason of the symmetry sponding one in a, and hence with the about the vertical through b, the center same rise, only one-third the horizontal of gravity of the positive area above the span. The increase of ordinates, then, horizontal through b lies in the vertical means a decrease of pole distance in the through b. The center of gravity of the same ratio, and vice versa. As is well negative area lies there also; hence the known, the product of the pole distance negative area is symmetrical about the by the ordinate of the equilibrium polycenter vertical; the closing line must then gon is the bending moment. This probe horizontal. It only remains then to find duct is not changed by changing the the height of a rectangle having the same pole distance. area as the elliptical segment, and having the span for its base. This is done very approximately by taking (in this case where the span is divided into 16 equal segments) the sum of the ordinates b, d,, etc. We thus find the height bk and the horizontal through k is the required closing line. Before effecting the comparison which Again, suppose the vertical load-line of a force polygon to remain in a given position, and the pole to be moved vertically to a new position. No vertical or horizontal dimension of the force polygon is affected by this change, neither will any such dimension of the equilibrium polygon corresponding to the new position of the pole be different from that in the polygon corre sponding to the first position of the pole; determining the pole distance of the the direction of the closing line, how- polygon e, which is one-third of the ever, is changed. Thus we see that the closing line of any equilibrium polygon can be made to coincide with any line not vertical, and that its ordinates will be unchanged by the operation. It is unnecessary to draw the force polygon to effect this change. Now to make clear the relationship between the polygons c and d, let us suppose, for the instant, that the polygone has been drawn by some means as yet unknown, so that its ordinates from d, viz., e,d=y1, e,d=y, etc., are proportional to the actual moments Me which tend to bend the arch. The conditions which then hold respecting these moments Me, are three: (Me)=0, (Mex)=0, Σ (Mey)=0. 7 6 2 actual thrust of the arch measured on the scale of the weights w, w,, etc. The physical significance of this condition may be stated according to Prop. V, thus: if the moments Md are applied to a uniform vertical girder bd at the points b, b.", b b", etc., at the same height total deflection æde. (May) as will with b, d, etc., they will cause the same the moments Me when applied at the same points. Hence if Ma are used as a species of loading, we can obtain the deflection by an equilibrium polygon. Suppose the load at d, is d, k, and that at de is dek, etc., then that at b is b. This approximation is sufficiently accurate for our purposes. 8 8 6 4 67 Now lay off on as a load line | dm=i̟bk, m ̧m=dk, m ̧m=d ̧ k etc. The direction of these loads must The first condition exists because the side of the line k; e. g., m ̧ m ̧=k ̧ d ̧. be changed when they fall on the other total bending from end to end is zero when the ends are fixed. The second entire arch m' (not drawn) will fall as If this process be continued through the and third are true, because the total de far to the right of d as m, does to the flection is zero both vertically and hori- left, and the last load will just reach zontally, since the span is unvariable as to d again. This is a test of the corwell as the position of the tangents at the ends. These results are in accord-line kk' has been found. Now using rectness with which the position of the ance with Prop. V. Now by Prop. III these moments Me are the differences of the moments of a straight girder and of the arch itself; hence the polygon e is simply the polygon e in a new position new pole distance. As moments are unchanged by such transformations, let us denote these moments by Mc. We have before seen that and with a Again, (Mey)=(Me - Ma) y=0 ... ≥ (Mey)=Σ (May). 7 6 any point as b for a pole, draw bm, to f The curve bf is then the exaggerated then draw fƒ¢ \| bm, ƒ¢Ï's || bm, etc. shape of a vertical girder bd, fixed at b, under the action of that part of moments Ma which are in the left half of the arch. The moments Ma on the right may act on another equal girder, having the same initial position bd, and it will then be equally deflected to the right of bd. This is not drawn. Again, suppose these vertical girders. fixed at bare bent instead by the moments Mc. We do not know just how much these moments are, though we do know that they are proportional to the ordinates of the polygon c. Therefore make dn, h, c, n n h1 C n, n ̧ = h ̧ c ̧, etc. When all these loads are laid off, the last one n' d = i̟h.' c.' must just return to d. This tests the accuracy of the work in determining the position of hh.. 6 69 8 89 = Now using b as a pole as before, construct the deflection curves by and bg'. Since these two deflections, viz., 2 f This last condition we shall use for and gg' ought to be the same, this fact informs us that each of the ordinates multiply the ordinates of the arch by h, c, h, c, must be increased in the ratio some number greater than 3. 2 2 4 of gg' to df, in order that when they | As a final test of the accuracy of the are considered as loads, they may pro- work, let us see whether (Mey) is acduce a total deflection equal to 2 df. tually zero, as should be. At d, for exTo effect this, lay off bj=df and bi ample, yd, and Me is proportional gg', and draw the horizontals through to d, e,. Then d, s, is proportional to i and j. At any convenient distance Mey at that point if e, s, is the arc of draw the vertical ij, and draw bi, and a circle, of which e 7 is the diameter. bj. These last two lines enable us to Similarly find d's, etc. 7 When e for effect the required proportions for any example falls above d, the circle must ordinates on the left, and these or two be described on the sum of 1, d, and de̟ lines of the same slope on the right to as a diameter, and as is proportional do the same thing on the right. E. 9. to a moment of different sign from that lay off the ordinate bi=h' c', then the required new ordinate is b. Then of the moments at the different points at d. We have distinguished the sign lay off ke bj'. In the same manner find ke from hb, and ke from he signs before the letter s. along the arch, by putting different It would have been slightly more accurate to have used only one-half the ordinates be, and be, but as they nearly equal in this case and of opposite sign, we have introduced no appreciable error. In the same manner can the other ordinates k, e, etc., be found; but this is not the best way to determine the rest of them, for we can now find the pole and pole distance of the polygon e. As we have previously seen, the pole distance is decreased in the same ratio as the ordinates of the moment curve 2 ds 4 4 and at right angles to it 8, 8, 8, then Now at any point s lay off ss,=d. 84, are increased, therefore prolong bi, to make 8, s=d's', etc. at right angles to the hypothenuse ss. Then the sum and draw a horizontal line through of the positive squares is ss,', and simi̟intersecting bj, at v, and the middle vertical at v; then is v, v, the pole dis- larly the sum of the negative squares is tance decreased in the required ratio. 8. If these are equal, then (Mey) Hence we move up the weight-line w, wtion is correctly made. vanishes as it should, and the constructo the position u, u vertically through v; and for convenience, lay off the It would have been equally correct to weights w, w, at u,' u,', etc. suppose the two vertical girders fixed at d, and bent by the moments acting. We could have determined the required ratio equally well from this construction. Further, in proving the correctness of the construction by taking the algebraic of the squares, we could have reckoned the ordinates, y, from any other horizontal line as well as from li̟'. on 1 sum 8 8 stress Furthermore, we know that the new closing-line is horizontal. To find the position of the pole o so that this shall occur, draw by parallel to hh, and from v the horizontal vo. As is well known, divides the total weight into the two segments, which are the vertical resistances of the abutments, and if the pole o is the same horizontal with To find the resultant the v, in closing line will be horizontal. the different portions of the arch, Now having determined the positions we must prolong v'o to o', say, of the points e, e, e', starting from one (not drawn) so that the pole distance of them, say e, draw e e, || ou, e, e̟ || ou, v'o'=3 v'o; then if we join o' and u etc.; then if the work be accurate, the o'u will be the resultant stress in the polygon will pass through the other two segment b. a,; o'u will be the stress in points e and e'. The bending moments a, a, etc., measured in the same scale as of the arch d or the arch a at a,, a,, etc., the weights w, w, etc. This resultant is the product of the pole distance stress is not directly along the neutral vv=vo by the ordinates d, e,, d, e, axis of the arch. etc., respectively, and between these points a similar product gives the moment with sufficient accuracy. It would be useful for the sake of accuracy to 6 8 The vertical shearing stress is constructed in the same manner as for a girder, by drawing one horizontal through w between the verticals 7 and 8, another |