Michell has discussed also the hollow vortex stationary inside a polygon (Phil.
Trans., 1890); the solution is given by ch nS2=sn w, shnS2=i cn w (II) so that, round the boundary of the polygon, ik = K', sin n8 =o; and on the surface of the vortex 1P= o, q = Q, and cos n8=sn4p,nB= Zit -am sic, (12) the intrinsic equation of the curve.
For figures of more than four sides this method is not usually convenient, except for such special cases as that of a regular polygon, which can be divided into triangles C by radii drawn from its centre.
The area of the polygon in fig.
By considering the circle as the limit of a polygon, it follows that the formulae (iii) and (v) of § 26 hold for a right circular cylinder and a right circular cone; i.e.
For longer bridges the funicular polygon affords a method of determining maximum bending moments which is perhaps more convenient.
The following definition of reciprocal figures: - " Two plane figures are reciprocal when they consist of an equal number of lines so that corresponding lines in the two figures are parallel, and corresponding lines which converge to a point in one figure form a closed polygon in the other."
The sides of the polygon may be arranged in any order, provided care is taken so to draw them that in passing round the polygon in C D _E FIG.
This polygon of forces may, by a slight extension of the above definition, be called the reciprocal figure of the external forces, if the sides are arranged in the same order as that of the joints on which they act, so that if the joints and forces be numbered I, 2, 3, 4, &c., passing round the outside of the frame in one direction, and returning at last to joint 1, then in the polygon the side representing the force 2 will be next the side representing the force I, and will be followed by the side representing the force 3, and so forth.
This polygon falls under the definition of a reciprocal figure given by Clerk Maxwell, if we consider the frame as a point in equilibrium under the external forces.
When all the forces are vertical, as will be the case in girders, the polygon of external forces will be reduced to two straight lines, fig.
If there are no redundant members in the frame there will be only two members abutting at the point of support, for these two members will be sufficient to balance the reaction, whatever its direction may be; we can therefore draw two triangles, each having as one side the reaction YX, and having the two other sides parallel to these two members; each of these triangles will represent a polygon of forces in equilibrium at the point of support.
67 a) in a fitting position to represent part of the polygon of forces at Xefa; beginning with the upward thrust EX, continuing down XA, and drawing AF parallel to AF in the frame we complete the polygon by drawing EF parallel to EF in the frame.
67 d as the complete reciprocal figure of the frame and forces upon it, and we see that each line in the reciprocal figure measures the stress on the corresponding member in the frame, and that the polygon of forces acting at any point, as Ijky, in the frame is represented by a polygon of the same name in the reciprocal figure.
69 b is the polygon of external forces, and 69 c is half the reciprocal figure.
The latter, as we know, calculated the perimeters of successive polygons, passing from one polygon to another of double the number of sides; in a similar manner Gregory calculated the areas.
It is a regular polygon with five bastions, founded by Frederick III.
This is the proposition known as the polygon of forces.
The case of the funicular polygon will be of use to us later.
This diagram consists of a polygon whose successive sides represent /\p9
The polygon of forces is then made up of segments of a vertical line.
This is equivalent to imagining the polygon Jii Jar J14..., supposed fixed in the l~mina, to roll on the polygon J12 Jar which is supposed fixed in space.
This result is easily extended to the case of a polygon of any number of sides; it has an important application in hydrostatics.
A system of forces represented completely by the sides of I plane polygon taken in order is equivalent to a couple whosc moment is represented by twice the area of the polygon; this is proved by taking moments about any point.
If the polygon intersects itself, care must be taken to attribute to the different parts of the area their proper signs.
In all cases the magnitude and direction, and joining the vertices of the polygon thus formed to an arbitrary pole 0.
The funicular or link polygon has its vertices on the lines of action of the given forces, and its sides respectively parallel to the lines drawn from 0 in the force-diagram; in particular, the two sides meeting in any vertex are respectively parallel to the lines drawn from 0 to the ends of that side of the force-polygon which represents the corresponding force.
The sides of the force-polygon may in the first instance be arranged in any order; the force-diagram can then be completed in a doubly infinite number of ways, owing to the arbitrary position of 0; and for each force-diagram a simply infinite number of funiculars can be drawn.
As a special case it may happen that the force-polygon is closed, i.e.
Hence the necessary and sufficient conditions of equilibrium are that the force-polygon and the funicular should both be closed.
It is evident that a system of jointed bars having the shape of the funicular polygon would be in equilibrium under the action of the given forces, supposed applied to the joints; moreover any bar in which the stress is of the nature of a tension (as distinguished from a thrust) might be replaced by a string.
27) he any side of the force-polygon, and construct the corresponding portions of the two diagrams, first with 0 and then with 0 as pole.
When the given forces are all parallel, the force-polygon consists of a series of segments of a straight line.
Thus ii AB, BC, CD represent the given loads, in the force-diagram, we construct the sides corresponding to OA, OB, OC, OD in the funicular; we then draw the closing line of the funicular polygon, and a parallel OE to it in the force diagram.
Hence if a system of vertical forces be in equilibrium, so that the funicular polygon ii closed, the length which this polygon intercepts on the vertical through any point P gives the sum of the moments about P of all the forces on one side of this vertical.
Considering, in the first place, the case in which the load and the two resistances by which each piece is balanced meet in one point, which may be called the centre of load, there will be as many such points of intersection, or centres of load, as there are pieces in the structure; and the directions and positions of the resistances or mutual pressures exerted between the pieces will be represented by the sides of a polygon joining Pi h2 ~, ~ those points, as in fig.
P4 represent the centres of load in a structure of four pieces, and the sides of the ~ polygon of resistances P1 P2 P2 P4 represent respectively the direc~ I~~ tions and positions FIG.
Further, at any one of the centres of load let PL represent the magnitude and direction of the gross load, and Pa, Pb the two resistances by which the piece to which that load is applied is supported; then wifl those three lines be respectively the diagonal and sides of a parallelogram; or, what is the same thing, they will be equal to the three sides of a triangleS and they must be in the same plane, although the sides of the polygon of resistances may be in different planes.
According to a well-known principle of statics, because the loads or external pressures P1L~, &c., balance each other, they must be proportional to the sides of a closed polygon drawn respectively parallel to their directions.
87 construct such a polygon of loads by L2 tional to, and joined end to end in the order o R ~s of, the gross loads on the pieces of the structure.
Then from the proportionality and parallelism sides of a triangle, there results the following of the load and the two resistances applied to each piece of the structure to the three theorem (originally due to Rankine): If from the angles of the polygon of loads there be drawn lines (Ri, R2, &c.), each of which is parallel to the resistance (as Pi F2, &c.) exerted FIG.
At the joint between the pieces to which the two loads reprfsented by the contiguous sides of the polygon of loads (such as L1, L2, &c.) are applied; then will all those lines meet in one point (0), and their lengths, measured from that point to the angles of Ike polygon, will represent the magnitudes of the resistances to which they are respectively parallel.
When the load on one of the pieces is parallel to the resistances which balance it, the polygon of resistances ceases to be closed, two of the sides becoming parallel to each other and to the load in question, and extending indefinitely.
In the polygon of loads the direction of a load sustained by parallel resistances traverses the point O-i i Since the relation discussed in 7 was enunciated by Rankine, an enormous development has taken place in the subject of Graphic Statics, the first comprehensive textbook on the subject being Die Graphische Statik by K.
Partial Polygons of Resistance.In a structure in which there are pieces supported at more than two joints, let a polygon be con-.
This may be called a partial polygon of resistances.
In considering its properties, the load at each centre of load is to be held to include the resistances of those joints which are not comprehended in the partial polygon of resistances, to which the theorem of 7 will then apply in every respect.
Line of PressuresCentres and Line of Resistance.The line of pressures is a line to which the directions of all the resistances in one polygon are tangents.
This polygon is the partial polygon of resistance.
A curve tangential to all the sides of the polygon is the line of pressures.