Again, suppose we have a bar AB resting with its ends on two smooth inclined planes which face each other. Let G be the centre of gravity (§ 11), and let AG = a, GB = b. Let α, β be the inclinations of the planes, and θ the angle which the bar makes with the vertical. The position of equilibrium is determined by the consideration that the reactions at A and B, which are by hypothesis normal to the planes, must meet at a point J on the vertical through G. HenceJG/a = sin (θ − α) / sin α, JG/b = sin (θ + β) / sin β,whencecot θ =a cot α − b cot β.a + b(6)If the bar is uniform we have a = b, andcot θ =1⁄2(cot α − cot β).(7)The problem of a rod suspended by strings attached to two points of it is virtually identical, the tensions of the strings taking the place of the reactions of the planes.
Again, suppose we have a bar AB resting with its ends on two smooth inclined planes which face each other. Let G be the centre of gravity (§ 11), and let AG = a, GB = b. Let α, β be the inclinations of the planes, and θ the angle which the bar makes with the vertical. The position of equilibrium is determined by the consideration that the reactions at A and B, which are by hypothesis normal to the planes, must meet at a point J on the vertical through G. Hence
JG/a = sin (θ − α) / sin α, JG/b = sin (θ + β) / sin β,
whence
(6)
If the bar is uniform we have a = b, and
cot θ =1⁄2(cot α − cot β).
(7)
The problem of a rod suspended by strings attached to two points of it is virtually identical, the tensions of the strings taking the place of the reactions of the planes.
Just as a system of forces is in general equivalent to a single force, so a given force can conversely be replaced by combinations of other forces, in various ways. For instance, a given force (and consequently a system of forces) can be replaced in one and only one way by three forces acting in three assigned straight lines, provided these lines be not concurrent or parallel. Thus if the three lines form a triangle ABC, and if the given force F meet BC in H, then F can be resolved into two components acting in HA, BC, respectively. And the force in HA can be resolved into two components acting in BC, CA, respectively. A simple graphical construction is indicated in fig. 19, where the dotted lines are parallel. As an example, any system of forces acting on the lamina in fig. 9 is balanced by three determinate tensions (or thrusts) in the three links, provided the directions of the latter are not concurrent.
If P, Q, R, be any three forces acting along BC, CA, AB, respectively, the line of action of the resultant is determined by the consideration that the sum of the moments about any point on it must vanish. Hence in “trilinear” co-ordinates, with ABC as fundamental triangle, its equation is Pα + Qβ + Rγ = 0. If P : Q : R = a : b : c, where a, b, c are the lengths of the sides, this becomes the “line at infinity,” and the forces reduce to a couple.
If P, Q, R, be any three forces acting along BC, CA, AB, respectively, the line of action of the resultant is determined by the consideration that the sum of the moments about any point on it must vanish. Hence in “trilinear” co-ordinates, with ABC as fundamental triangle, its equation is Pα + Qβ + Rγ = 0. If P : Q : R = a : b : c, where a, b, c are the lengths of the sides, this becomes the “line at infinity,” and the forces reduce to a couple.
The sum of the moments of the two forces of a couple is the same about any point in the plane. Thus in the figure the sum of the moments about O is P·OA − P·OB or P·AB, which is independent of the position of O. This sum is called themoment of the couple; it must of course have the proper sign attributed to it. It easily follows that any two couples of the same moment are equivalent, and that any number of couples can be replaced by a single couple whose moment is the sum of their moments. Since a couple is for our purposes sufficiently represented by its moment, it has been proposed to substitute the nametorque(or twisting effort), as free from the suggestion of any special pair of forces.
A system of forces represented completely by the sides of a plane polygon taken in order is equivalent to a couple whosemoment 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.
Again, any coplanar system of forces can be replaced by a single force R acting at any assigned point O, together with a couple G. The force R is the geometric sum of the given forces, and the moment (G) of the couple is equal to the sum of the moments of the given forces about O. The value of G will in general vary with the position of O, and will vanish when O lies on the line of action of the single resultant.
The formal analytical reduction of a system of coplanar forces is as follows. Let (x1, y1), (x2, y2), ... be the rectangular co-ordinates of any points A1, A2, ... on the lines of action of the respective forces. The force at A1may be replaced by its components X1, Y1, parallel to the co-ordinate axes; that at A2by its components X2, Y2, and so on. Introducing at O two equal and opposite forces ±X1in Ox, we see that X1at A1may be replaced by an equal and parallel force at O together with a couple −y1X1. Similarly the force Y1at A1may be replaced by a force Y1at O together with a couple x1Y1. The forces X1, Y1, at O can thus be transferred to O provided we introduce a couple x1Y1− y1X1. Treating the remaining forces in the same way we get a force X1+ X2+ ... or Σ(X) along Ox, a force Y1+ Y2+ ... or Σ(Y) along Oy, and a couple (x1Y1− y1X1) + (x2Y2− y2X2) + ... or Σ(xY − yX). The three conditions of equilibrium are therefore
Σ(X) = 0, Σ(Y) = 0, Σ(xY − yX) = 0.
(8)
If O′ be a point whose co-ordinates are (ξ, η), the moment of the couple when the forces are transferred to O′ as a new origin will be Σ{(x − ξ) Y − (y − η) X}. This vanishes,i.e.the system reduces to a single resultant through O′, provided
−ξ·Σ(Y) + η·Σ(X) + Σ(xY − yX) = 0.
(9)
If ξ, η be regarded as current co-ordinates, this is the equation of the line of action of the single resultant to which the system is in general reducible.
If the forces are all parallel, making say an angle θ with Ox, we may write X1= P1cos θ, Y1= P1sin θ, X2= P2cos θ, Y2= P2sin θ, .... The equation (9) then becomes
{Σ(xP) − ξ·Σ(P)} sin θ − {Σ(yP) − η·Σ(P)} cos θ = 0.
(10)
If the forces P1, P2, ... be turned in the same sense through the same angle about the respective points A1, A2, ... so as to remain parallel, the value of θ is alone altered, and the resultant Σ(P) passes always through the point
(11)
which is determined solely by the configuration of the points A1, A2, ... and by the ratios P1: P2: ... of the forces acting at them respectively. This point is called thecentreof the given system of parallel forces; it is finite and determinate unless Σ(P) = 0. A geometrical proof of this theorem, which is not restricted to a two-dimensional system, is given later (§ 11). It contains the theory of thecentre of gravityas ordinarily understood. For if we have an assemblage of particles whose mutual distances are small compared with the dimensions of the earth, the forces of gravity on them constitute a system of sensibly parallel forces, sensibly proportional to the respective masses. If now the assemblage be brought into any other position relative to the earth, without alteration of the mutual distances, this is equivalent to a rotation of the directions of the forces relatively to the assemblage, the ratios of the forces remaining unaltered. Hence there is a certain point, fixed relatively to the assemblage, through which the resultant of gravitational action always passes; this resultant is moreover equal to the sum of the forces on the several particles.
The theorem that any coplanar system of forces can be reduced to a force acting through any assigned point, together with a couple, has an important illustration in the theory of the distribution of shearing stress and bending moment in a horizontal beam, or other structure, subject to vertical extraneous forces. If we consider any vertical section P, the forces exerted across the section by the portion of the structure on one side on the portion on the other may be reduced to a vertical force F at P and a couple M. The force measures theshearing stress, and the couple thebending momentat P; we will reckon these quantities positive when the senses are as indicated in the figure.If the remaining forces acting on the portion of the structure on either side of P are known, then resolving vertically we find F, and taking moments about P we find M. Again if PQ be any segment of the beam which is free from load, Q lying to the right of P, we findFP= FQ, MP− MQ= −F·PQ;(12)hence F is constant between the loads, whilst M decreases as we travel to the right, with a constant gradient −F. If PQ be a short segment containing an isolated load W, we haveFQ− FP= −W, MQ= MP;(13)Fig. 24.hence F is discontinuous at a concentrated load, diminishing by an amount equal to the load as we pass the loaded point to the right, whilst M is continuous. Accordingly the graph of F for any system of isolated loads will consist of a series of horizontal lines, whilst that of M will be a continuous polygon.To pass to the case of continuous loads, let x be measured horizontally along the beam to the right. The load on an element δx of the beam may be represented by wδx, where w is in general a function of x. The equations (12) are now replaced byδF = −wδx, δM = −Fδx,whenceFQ− FP= −∫QPw dx, MQ− MP= −∫QPF dx.(14)Fig. 25.The latter relation shows that the bending moment varies as the area cut off by the ordinate in the graph of F. In the case of uniform load we haveF = −wx + A, M =1⁄2wx2− Ax + B,(15)where the arbitrary constants A,B are to be determined by the conditions of the special problem,e.g.the conditions at the ends of the beam. The graph of F is a straight line; that of M is a parabola with vertical axis. In all cases the graphs due to different distributions of load may be superposed. The figure shows the case of a uniform heavy beam supported at its ends.
The theorem that any coplanar system of forces can be reduced to a force acting through any assigned point, together with a couple, has an important illustration in the theory of the distribution of shearing stress and bending moment in a horizontal beam, or other structure, subject to vertical extraneous forces. If we consider any vertical section P, the forces exerted across the section by the portion of the structure on one side on the portion on the other may be reduced to a vertical force F at P and a couple M. The force measures theshearing stress, and the couple thebending momentat P; we will reckon these quantities positive when the senses are as indicated in the figure.
If the remaining forces acting on the portion of the structure on either side of P are known, then resolving vertically we find F, and taking moments about P we find M. Again if PQ be any segment of the beam which is free from load, Q lying to the right of P, we find
FP= FQ, MP− MQ= −F·PQ;
(12)
hence F is constant between the loads, whilst M decreases as we travel to the right, with a constant gradient −F. If PQ be a short segment containing an isolated load W, we have
FQ− FP= −W, MQ= MP;
(13)
hence F is discontinuous at a concentrated load, diminishing by an amount equal to the load as we pass the loaded point to the right, whilst M is continuous. Accordingly the graph of F for any system of isolated loads will consist of a series of horizontal lines, whilst that of M will be a continuous polygon.
To pass to the case of continuous loads, let x be measured horizontally along the beam to the right. The load on an element δx of the beam may be represented by wδx, where w is in general a function of x. The equations (12) are now replaced by
δF = −wδx, δM = −Fδx,
whence
FQ− FP= −∫QPw dx, MQ− MP= −∫QPF dx.
(14)
The latter relation shows that the bending moment varies as the area cut off by the ordinate in the graph of F. In the case of uniform load we have
F = −wx + A, M =1⁄2wx2− Ax + B,
(15)
where the arbitrary constants A,B are to be determined by the conditions of the special problem,e.g.the conditions at the ends of the beam. The graph of F is a straight line; that of M is a parabola with vertical axis. In all cases the graphs due to different distributions of load may be superposed. The figure shows the case of a uniform heavy beam supported at its ends.
§ 5.Graphical Statics.—A graphical method of reducing a plane system of forces was introduced by C. Culmann (1864). It involves the construction of two figures, aforce-diagramand afunicular polygon. The force-diagram is constructed by placing end to end a series of vectors representing the given forces inmagnitude and direction, and joining the vertices of the polygon thus formed to an arbitrarypoleO. 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 O in the force-diagram; in particular, the two sides meeting in any vertex are respectively parallel to the lines drawn from O to the ends of that side of the force-polygon which represents the corresponding force. The relations will be understood from the annexed diagram, where corresponding lines in the force-diagram (to the right) and the funicular (to the left) are numbered similarly. 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 O; and for each force-diagram a simply infinite number of funiculars can be drawn. The two diagrams being supposed constructed, it is seen that each of the given systems of forces can be replaced by two components acting in the sides of the funicular which meet at the corresponding vertex, and that the magnitudes of these components will be given by the corresponding triangle of forces in the force-diagram; thus the force 1 in the figure is equivalent to two forces represented by 01 and 12. When this process of replacement is complete, each terminated side of the funicular is the seat of two forces which neutralize one another, and there remain only two uncompensated forces, viz., those resident in the first and last sides of the funicular. If these sides intersect, the resultant acts through the intersection, and its magnitude and direction are given by the line joining the first and last sides of the force-polygon (see fig. 26, where the resultant of the four given forces is denoted by R). As a special case it may happen that the force-polygon is closed,i.e.its first and last points coincide; the first and last sides of the funicular will then be parallel (unless they coincide), and the two uncompensated forces form a couple. If, however, the first and last sides of the funicular coincide, the two outstanding forces neutralize one another, and we have equilibrium. Hence the necessary and sufficient conditions of equilibrium are that the force-polygon and the funicular should both be closed. This is illustrated by fig. 26 if we imagine the force R, reversed, to be included in the system of given forces.
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. This is the origin of the names “link-polygon” and “funicular” (cf. § 2).
If funiculars be drawn for two positions O, O′ of the pole in the force-diagram, their corresponding sides will intersect on a straight line parallel to OO′. This is essentially a theorem of projective geometry, but the following statical proof is interesting. Let AB (fig. 27) be any side of the force-polygon, and construct the corresponding portions of the two diagrams, first with O and then with O′ as pole. The force corresponding to AB may be replaced by the two components marked x, y; and a force corresponding to BA may be represented by the two components marked x′, y′. Hence the forces x, y, x′, y′ are in equilibrium. Now x, x′ have a resultant through H, represented in magnitude and direction by OO′, whilst y, y′ have a resultant through K represented in magnitude and direction by O′O. Hence HK must be parallel to OO′. This theorem enables us, when one funicular has been drawn, to construct any other without further reference to the force-diagram.Fig. 27.The complete figures obtained by drawing first the force-diagrams of a system of forces in equilibrium with two distinct poles O, O′, and secondly the corresponding funiculars, have various interesting relations. In the first place, each of these figures may be conceived as an orthogonal projection of a closed plane-faced polyhedron. As regards the former figure this is evident at once; viz. the polyhedron consists of two pyramids with vertices represented by O, O′, and a common base whose perimeter is represented by the force-polygon (only one of these is shown in fig. 28). As regards the funicular diagram, let LM be the line on which the pairs of corresponding sides of the two polygons meet, and through it draw any two planes ω, ω′. Through the vertices A, B, C, ... and A′, B′, C′, ... of the two funiculars draw normals to the plane of the diagram, to meet ω and ω′ respectively. The points thus obtained are evidently the vertices of a polyhedron with plane faces.Fig. 28.Fig. 29.To every line in either of the original figures corresponds of course a parallel line in the other; moreover, it is seen that concurrent lines in either figure correspond to lines forming a closed polygon in the other. Two plane figures so related are calledreciprocal, since the properties of the first figure in relation to the second are the same as those of the second with respect to the first. A still simpler instance of reciprocal figures is supplied by the case of concurrent forces in equilibrium (fig. 29). The theory of these reciprocal figures was first studied by J. Clerk Maxwell, who showed amongst other things that a reciprocal can always be drawn to any figure which is the orthogonal projection of a plane-faced polyhedron. If in fact we take the pole of each face of such a polyhedron with respect to a paraboloid of revolution, these poles will be the vertices of a second polyhedron whose edges are the “conjugate lines” of those of the former. If we project both polyhedra orthogonally on a plane perpendicular to the axis of the paraboloid, we obtain two figures which are reciprocal, except that corresponding lines are orthogonal instead of parallel. Another proof will be indicated later (§ 8) in connexion with the properties of the linear complex. It isconvenient to have a notation which shall put in evidence the reciprocal character. For this purpose we may designate the points in one figure by letters A, B, C, ... and the corresponding polygons in the other figure by the same letters; a line joining two points A, B in one figure will then correspond to the side common to the two polygons A, B in the other. This notation was employed by R. H. Bow in connexion with the theory of frames (§ 6, and see alsoApplied Mechanicsbelow) where reciprocal diagrams are frequently of use (cf.Diagram).When the given forces are all parallel, the force-polygon consists of a series of segments of a straight line. This case has important practical applications; for instance we may use the method to find the pressures on the supports of a beam loaded in any given manner. Thus if 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 theclosing lineof the funicular polygon, and a parallel OE to it in the force diagram. The segments DE, EA then represent the upward pressures of the two supports on the beam, which pressures together with the given loads constitute a system of forces in equilibrium. The pressures of the beam on the supports are of course represented by ED, AE. The two diagrams are portions of reciprocal figures, so that Bow’s notation is applicable.Fig. 30.Fig. 31.A graphical method can also be applied to find the moment of a force, or of a system of forces, about any assigned point P. Let F be a force represented by AB in the force-diagram. Draw a parallel through P to meet the sides of the funicular which correspond to OA, OB in the points H, K. If R be the intersection of these sides, the triangles OAB, RHK are similar, and if the perpendiculars OM, RN be drawn we haveHK·OM = AB·RN = F·RN,which is the moment of F about P. If the given forces are all parallel (say vertical) OM is the same for all, and the moments of the several forces about P are represented on a certain scale by the lengths intercepted by the successive pairs of sides on the vertical through P. Moreover, the moments are compounded by adding (geometrically) the corresponding lengths HK. Hence if a system of vertical forces be in equilibrium, so that the funicular polygon is 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. For instance, in the case of a beam in equilibrium under any given loads and the reactions at the supports, we get a graphical representation of the distribution of bending moment over the beam. The construction in fig. 30 can easily be adjusted so that the closing line shall be horizontal; and the figure then becomes identical with the bending-moment diagram of § 4. If we wish to study the effects of a movable load, or system of loads, in different positions on the beam, it is only necessary to shift the lines of action of the pressures of the supports relatively to the funicular, keeping them at the same, distance apart; the only change is then in the position of the closing line of the funicular. It may be remarked that since this line joins homologous points of two “similar” rows it will envelope a parabola.
If funiculars be drawn for two positions O, O′ of the pole in the force-diagram, their corresponding sides will intersect on a straight line parallel to OO′. This is essentially a theorem of projective geometry, but the following statical proof is interesting. Let AB (fig. 27) be any side of the force-polygon, and construct the corresponding portions of the two diagrams, first with O and then with O′ as pole. The force corresponding to AB may be replaced by the two components marked x, y; and a force corresponding to BA may be represented by the two components marked x′, y′. Hence the forces x, y, x′, y′ are in equilibrium. Now x, x′ have a resultant through H, represented in magnitude and direction by OO′, whilst y, y′ have a resultant through K represented in magnitude and direction by O′O. Hence HK must be parallel to OO′. This theorem enables us, when one funicular has been drawn, to construct any other without further reference to the force-diagram.
The complete figures obtained by drawing first the force-diagrams of a system of forces in equilibrium with two distinct poles O, O′, and secondly the corresponding funiculars, have various interesting relations. In the first place, each of these figures may be conceived as an orthogonal projection of a closed plane-faced polyhedron. As regards the former figure this is evident at once; viz. the polyhedron consists of two pyramids with vertices represented by O, O′, and a common base whose perimeter is represented by the force-polygon (only one of these is shown in fig. 28). As regards the funicular diagram, let LM be the line on which the pairs of corresponding sides of the two polygons meet, and through it draw any two planes ω, ω′. Through the vertices A, B, C, ... and A′, B′, C′, ... of the two funiculars draw normals to the plane of the diagram, to meet ω and ω′ respectively. The points thus obtained are evidently the vertices of a polyhedron with plane faces.
To every line in either of the original figures corresponds of course a parallel line in the other; moreover, it is seen that concurrent lines in either figure correspond to lines forming a closed polygon in the other. Two plane figures so related are calledreciprocal, since the properties of the first figure in relation to the second are the same as those of the second with respect to the first. A still simpler instance of reciprocal figures is supplied by the case of concurrent forces in equilibrium (fig. 29). The theory of these reciprocal figures was first studied by J. Clerk Maxwell, who showed amongst other things that a reciprocal can always be drawn to any figure which is the orthogonal projection of a plane-faced polyhedron. If in fact we take the pole of each face of such a polyhedron with respect to a paraboloid of revolution, these poles will be the vertices of a second polyhedron whose edges are the “conjugate lines” of those of the former. If we project both polyhedra orthogonally on a plane perpendicular to the axis of the paraboloid, we obtain two figures which are reciprocal, except that corresponding lines are orthogonal instead of parallel. Another proof will be indicated later (§ 8) in connexion with the properties of the linear complex. It isconvenient to have a notation which shall put in evidence the reciprocal character. For this purpose we may designate the points in one figure by letters A, B, C, ... and the corresponding polygons in the other figure by the same letters; a line joining two points A, B in one figure will then correspond to the side common to the two polygons A, B in the other. This notation was employed by R. H. Bow in connexion with the theory of frames (§ 6, and see alsoApplied Mechanicsbelow) where reciprocal diagrams are frequently of use (cf.Diagram).
When the given forces are all parallel, the force-polygon consists of a series of segments of a straight line. This case has important practical applications; for instance we may use the method to find the pressures on the supports of a beam loaded in any given manner. Thus if 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 theclosing lineof the funicular polygon, and a parallel OE to it in the force diagram. The segments DE, EA then represent the upward pressures of the two supports on the beam, which pressures together with the given loads constitute a system of forces in equilibrium. The pressures of the beam on the supports are of course represented by ED, AE. The two diagrams are portions of reciprocal figures, so that Bow’s notation is applicable.
A graphical method can also be applied to find the moment of a force, or of a system of forces, about any assigned point P. Let F be a force represented by AB in the force-diagram. Draw a parallel through P to meet the sides of the funicular which correspond to OA, OB in the points H, K. If R be the intersection of these sides, the triangles OAB, RHK are similar, and if the perpendiculars OM, RN be drawn we have
HK·OM = AB·RN = F·RN,
which is the moment of F about P. If the given forces are all parallel (say vertical) OM is the same for all, and the moments of the several forces about P are represented on a certain scale by the lengths intercepted by the successive pairs of sides on the vertical through P. Moreover, the moments are compounded by adding (geometrically) the corresponding lengths HK. Hence if a system of vertical forces be in equilibrium, so that the funicular polygon is 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. For instance, in the case of a beam in equilibrium under any given loads and the reactions at the supports, we get a graphical representation of the distribution of bending moment over the beam. The construction in fig. 30 can easily be adjusted so that the closing line shall be horizontal; and the figure then becomes identical with the bending-moment diagram of § 4. If we wish to study the effects of a movable load, or system of loads, in different positions on the beam, it is only necessary to shift the lines of action of the pressures of the supports relatively to the funicular, keeping them at the same, distance apart; the only change is then in the position of the closing line of the funicular. It may be remarked that since this line joins homologous points of two “similar” rows it will envelope a parabola.
The “centre” (§ 4) of a system of parallel forces of given magnitudes, acting at given points, is easily determined graphically. We have only to construct the line of action of the resultant for each of two arbitrary directions of the forces; the intersection of the two lines gives the point required. The construction is neatest if the two arbitrary directions are taken at right angles to one another.
§ 6.Theory of Frames.—Aframeis a structure made up of pieces, ormembers, each of which has twojointsconnecting it with other members. In a two-dimensional frame, each joint may be conceived as consisting of a small cylindrical pin fitting accurately and smoothly into holes drilled through the members which it connects. This supposition is a somewhat ideal one, and is often only roughly approximated to in practice. We shall suppose, in the first instance, that extraneous forces act on the frame at the joints only,i.e.on the pins.
On this assumption, the reactions on any member at its two joints must be equal and opposite. This combination of equal and opposite forces is called thestressin the member; it may be atensionor athrust. For diagrammatic purposes each member is sufficiently represented by a straight line terminating at the two joints; these lines will be referred to as thebarsof the frame.
In structural applications a frame must bestiff, orrigid,i.e.it must be incapable of deformation without alteration of length in at least one of its bars. It is said to bejust rigidif it ceases to be rigid when any one of its bars is removed. A frame which has more bars than are essential for rigidity may be calledover-rigid; such a frame is in general self-stressed,i.e.it is in a state of stress independently of the action of extraneous forces. A plane frame of n joints which is just rigid (as regards deformation in its own plane) has 2n − 3 bars, for if one bar be held fixed the 2(n − 2) co-ordinates of the remaining n − 2 joints must just be determined by the lengths of the remaining bars. The total number of bars is therefore 2(n − 2) + 1. When a plane frame which is just rigid is subject to a given system of equilibrating extraneous forces (in its own plane) acting on the joints, the stresses in the bars are in general uniquely determinate. For the conditions of equilibrium of the forces on each pin furnish 2n equations, viz. two for each point, which are linear in respect of the stresses and the extraneous forces. This system of equations must involve the three conditions of equilibrium of the extraneous forces which are already identically satisfied, by hypothesis; there remain therefore 2n − 3 independent relations to determine the 2n − 3 unknown stresses. A frame of n joints and 2n − 3 bars may of course fail to be rigid owing to some parts being over-stiff whilst others are deformable; in such a case it will be found that the statical equations, apart from the three identical relations imposed by the equilibrium of the extraneous forces, are not all independent but are equivalent to less than 2n − 3 relations. Another exceptional case, known as thecritical case, will be noticed later (§ 9).
A plane frame which can be built up from a single bar by successive steps, at each of which a new joint is introduced by twonew bars meeting there, is called asimpleframe; it is obviously just rigid. The stresses produced by extraneous forces in a simple frame can be found by considering the equilibrium of the various joints in a proper succession; and if the graphical method be employed the various polygons of force can be combined into a single force-diagram. This procedure was introduced by W. J. M. Rankine and J. Clerk Maxwell (1864). It may be noticed that if we take an arbitrary pole in the force-diagram, and draw a corresponding funicular in the skeleton diagram which represents the frame together with the lines of action of the extraneous forces, we obtain two complete reciprocal figures, in Maxwell’s sense. It is accordingly convenient to use Bow’s notation (§ 5), and to distinguish the several compartments of the frame-diagram by letters. See fig. 33, where the successive triangles in the diagram of forces may be constructed in the order XYZ, ZXA, AZB. The class of “simple” frames includes many of the frameworks used in the construction of roofs, lattice girders and suspension bridges; a number of examples will be found in the articleBridges. By examining the senses in which the respective forces act at each joint we can ascertain which members are in tension and which are in thrust; in fig. 33 this is indicated by the directions of the arrowheads.
When a frame, though just rigid, is not “simple” in the above sense, the preceding method must be replaced, or supplemented, by one or other of various artifices. In some cases themethod of sectionsis sufficient for the purpose. If an ideal section be drawn across the frame, the extraneous forces on either side must be in equilibrium with the forces in the bars cut across; and if the section can be drawn so as to cut only three bars, the forces in these can be found, since the problem reduces to that of resolving a given force into three components acting in three given lines (§ 4). The “critical case” where the directions of the three bars are concurrent is of course excluded. Another method, always available, will be explained under “Work” (§ 9).
When extraneous forces act on the bars themselves the stress in each bar no longer consists of a simple longitudinal tension or thrust. To find the reactions at the joints we may proceed as follows. Each extraneous force W acting on a bar may be replaced (in an infinite number of ways) by two components P, Q in lines through the centres of the pins at the extremities. In practice the forces W are usually vertical, and the components P, Q are then conveniently taken to be vertical also. We first alter the problem by transferring the forces P, Q to the pins. The stresses in the bars, in the problem as thus modified, may be supposed found by the preceding methods; it remains to infer from the results thus obtained the reactions in the original form of the problem. To find the pressure exerted by a bar AB on the pin A we compound with the force in AB given by the diagram a force equal to P. Conversely, to find the pressure of the pin A on the bar AB we must compound with the force given by the diagram a force equal and opposite to P. This question arises in practice in the theory of “three-jointed” structures; for the purpose in hand such a structure is sufficiently represented by two bars AB, BC. The right-hand figure represents a portion of the force-diagram; in particularZX>represents the pressure of AB on B in the modified problem where the loads W1and W2on the two bars are replaced by loads P1, Q1, and P2, Q2respectively, acting on the pins. Compounding with thisXV>, which represents Q1, we get the actual pressureZV>exerted by AB on B. The directions and magnitudes of the reactions at A and C are then easily ascertained. On account of its practical importance several other graphical solutions of this problem have been devised.
When extraneous forces act on the bars themselves the stress in each bar no longer consists of a simple longitudinal tension or thrust. To find the reactions at the joints we may proceed as follows. Each extraneous force W acting on a bar may be replaced (in an infinite number of ways) by two components P, Q in lines through the centres of the pins at the extremities. In practice the forces W are usually vertical, and the components P, Q are then conveniently taken to be vertical also. We first alter the problem by transferring the forces P, Q to the pins. The stresses in the bars, in the problem as thus modified, may be supposed found by the preceding methods; it remains to infer from the results thus obtained the reactions in the original form of the problem. To find the pressure exerted by a bar AB on the pin A we compound with the force in AB given by the diagram a force equal to P. Conversely, to find the pressure of the pin A on the bar AB we must compound with the force given by the diagram a force equal and opposite to P. This question arises in practice in the theory of “three-jointed” structures; for the purpose in hand such a structure is sufficiently represented by two bars AB, BC. The right-hand figure represents a portion of the force-diagram; in particularZX>represents the pressure of AB on B in the modified problem where the loads W1and W2on the two bars are replaced by loads P1, Q1, and P2, Q2respectively, acting on the pins. Compounding with thisXV>, which represents Q1, we get the actual pressureZV>exerted by AB on B. The directions and magnitudes of the reactions at A and C are then easily ascertained. On account of its practical importance several other graphical solutions of this problem have been devised.
§ 7.Three-dimensional Kinematics of a Rigid Body.—The position of a rigid body is determined when we know the positions of three points A, B, C of it which are not collinear, for the position of any other point P is then determined by the three distances PA, PB, PC. The nine co-ordinates (Cartesian or other) of A, B, C are subject to the three relations which express the invariability of the distances BC, CA, AB, and are therefore equivalent to six independent quantities. Hence a rigid body not constrained in any way is said to have six degrees of freedom. Conversely, any six geometrical relations restrict the body in general to one or other of a series of definite positions, none of which can be departed from without violating the conditions in question. For instance, the position of a theodolite is fixed by the fact that its rounded feet rest in contact with six given plane surfaces. Again, a rigid three-dimensional frame can be rigidly fixed relatively to the earth by means of six links.
The six independent quantities, or “co-ordinates,” which serve to specify the position of a rigid body in space may of course be chosen in an endless variety of ways. We may, for instance, employ the three Cartesian co-ordinates of a particular point O of the body, and three angular co-ordinates which express the orientation of the body with respect to O. Thus in fig. 36, if OA, OB, OC be three mutually perpendicular lines in the solid, we may denote by θ the angle which OC makes with a fixed direction OZ, by ψ the azimuth of the plane ZOC measured from some fixed plane through OZ, and by φ the inclination of the plane COA to the plane ZOC. In fig. 36 these various lines and planes are represented by their intersections with a unit sphere having O as centre. This very useful, although unsymmetrical, system of angular co-ordinates was introduced by L. Euler. It is exemplified in “Cardan’s suspension,” as used in connexion with a compass-bowl or a gyroscope. Thus in the gyroscope the “flywheel” (represented by the globe in fig. 37) can turn about a diameter OC of a ring which is itself free to turn about a diametral axis OX at right angles to the former; this axis is carried by a second ring which is free to turn about a fixed diameter OZ, which is at right angles to OX.
The six independent quantities, or “co-ordinates,” which serve to specify the position of a rigid body in space may of course be chosen in an endless variety of ways. We may, for instance, employ the three Cartesian co-ordinates of a particular point O of the body, and three angular co-ordinates which express the orientation of the body with respect to O. Thus in fig. 36, if OA, OB, OC be three mutually perpendicular lines in the solid, we may denote by θ the angle which OC makes with a fixed direction OZ, by ψ the azimuth of the plane ZOC measured from some fixed plane through OZ, and by φ the inclination of the plane COA to the plane ZOC. In fig. 36 these various lines and planes are represented by their intersections with a unit sphere having O as centre. This very useful, although unsymmetrical, system of angular co-ordinates was introduced by L. Euler. It is exemplified in “Cardan’s suspension,” as used in connexion with a compass-bowl or a gyroscope. Thus in the gyroscope the “flywheel” (represented by the globe in fig. 37) can turn about a diameter OC of a ring which is itself free to turn about a diametral axis OX at right angles to the former; this axis is carried by a second ring which is free to turn about a fixed diameter OZ, which is at right angles to OX.
We proceed to sketch the theory of the finite displacements of a rigid body. It was shown by Euler (1776) that any displacementin which one point O of the body is fixed is equivalent to a purerotationabout some axis through O. Imagine two spheres of equal radius with O as their common centre, one fixed in the body and moving with it, the other fixed in space. In any displacement about O as a fixed point, the former sphere slides over the latter, as in a “ball-and-socket” joint. Suppose that as the result of the displacement a point of the moving sphere is brought from A to B, whilst the point which was at B is brought to C (cf. fig. 10). Let J be the pole of the circle ABC (usually a “small circle” of the fixed sphere), and join JA, JB, JC, AB, BC by great-circle arcs. The spherical isosceles triangles AJB, BJC are congruent, and we see that AB can be brought into the position BC by a rotation about the axis OJ through an angle AJB.
It is convenient to distinguish the two senses in which rotation may take place about an axis OA by opposite signs. We shall reckon a rotation as positive when it is related to the direction from O to A as the direction of rotation is related to that of translation in a right-handed screw. Thus a negative rotation about OA may be regarded as a positive rotation about OA′, the prolongation of AO. Now suppose that a body receives first a positive rotation α about OA, and secondly a positive rotation β about OB; and let A, B be the intersections of these axes with a sphere described about O as centre. If we construct the spherical triangles ABC, ABC′ (fig. 38), having in each case the angles at A and B equal to1⁄2α and1⁄2β respectively, it is evident that the first rotation will bring a point from C to C′ and that the second will bring it back to C; the result is therefore equivalent to a rotation about OC. We note also that if the given rotations had been effected in the inverse order, the axis of the resultant rotation would have been OC′, so that finite rotations do not obey the “commutative law.” To find the angle of the equivalent rotation, in the actual case, suppose that the second rotation (about OB) brings a point from A to A′. The spherical triangles ABC, A′BC (fig. 39) are “symmetrically equal,” and the angle of the resultant rotation, viz. ACA′, is 2π − 2C. This is equivalent to a negative rotation 2C about OC, whence the theorem that the effect of three successive positive rotations 2A, 2B, 2C about OA, OB, OC, respectively, is to leave the body in its original position, provided the circuit ABC is left-handed as seen from O. This theorem is due to O. Rodrigues (1840). The composition of finite rotations about parallel axes is a particular case of the preceding; the radius of the sphere is now infinite, and the triangles are plane.
In any continuous motion of a solid about a fixed point O, the limiting position of the axis of the rotation by which the body can be brought from any one of its positions to a consecutive one is called theinstantaneous axis. This axis traces out a certain cone in the body, and a certain cone in space, and the continuous motion in question may be represented as consisting in a rolling of the former cone on the latter. The proof is similar to that of the corresponding theorem of plane kinematics (§ 3).
It follows from Euler’s theorem that the most general displacement of a rigid body may be effected by a pure translation which brings any one point of it to its final position O, followed by a pure rotation about some axis through O. Those planes in the body which are perpendicular to this axis obviously remain parallel to their original positions. Hence, if σ, σ′ denote the initial and final positions of any figure in one of these planes, the displacement could evidently have been effected by (1) a translation perpendicular to the planes in question, bringing σ into some position σ″ in the plane of σ′, and (2) a rotation about a normal to the planes, bringing σ″ into coincidence with σ (§ 3). In other words, the most general displacement is equivalent to a translation parallel to a certain axis combined with a rotation about that axis;i.e.it may be described as atwistabout a certainscrew. In particular cases, of course, the translation, or the rotation, may vanish.
The preceding theorem, which is due to Michel Chasles (1830), may be proved in various other interesting ways. Thus if a point of the body be displaced from A to B, whilst the point which was at B is displaced to C, and that which was at C to D, the four points A, B, C, D lie on a helix whose axis is the common perpendicular to the bisectors of the angles ABC, BCD. This is the axis of the required screw; the amount of the translation is measured by the projection of AB or BC or CD on the axis; and the angle of rotation is given by the inclination of the aforesaid bisectors. This construction was given by M. W. Crofton. Again, H. Wiener and W. Burnside have employed thehalf-turn(i.e.a rotation through two right angles) as the fundamental operation. This has the advantage that it is completely specified by the axis of the rotation, the sense being immaterial. Successive half-turns about parallel axes a, b are equivalent to a translation measured by double the distance between these axes in the direction from a to b. Successive half-turns about intersecting axes a, b are equivalent to a rotation about the common perpendicular to a, b at their intersection, of amount equal to twice the acute angle between them, in the direction from a to b. Successive half-turns about two skew axes a, b are equivalent to a twist about a screw whose axis is the common perpendicular to a, b, the translation being double the shortest distance, and the angle of rotation being twice the acute angle between a, b, in the direction from a to b. It is easily shown that any displacement whatever is equivalent to two half-turns and therefore to a screw.
The preceding theorem, which is due to Michel Chasles (1830), may be proved in various other interesting ways. Thus if a point of the body be displaced from A to B, whilst the point which was at B is displaced to C, and that which was at C to D, the four points A, B, C, D lie on a helix whose axis is the common perpendicular to the bisectors of the angles ABC, BCD. This is the axis of the required screw; the amount of the translation is measured by the projection of AB or BC or CD on the axis; and the angle of rotation is given by the inclination of the aforesaid bisectors. This construction was given by M. W. Crofton. Again, H. Wiener and W. Burnside have employed thehalf-turn(i.e.a rotation through two right angles) as the fundamental operation. This has the advantage that it is completely specified by the axis of the rotation, the sense being immaterial. Successive half-turns about parallel axes a, b are equivalent to a translation measured by double the distance between these axes in the direction from a to b. Successive half-turns about intersecting axes a, b are equivalent to a rotation about the common perpendicular to a, b at their intersection, of amount equal to twice the acute angle between them, in the direction from a to b. Successive half-turns about two skew axes a, b are equivalent to a twist about a screw whose axis is the common perpendicular to a, b, the translation being double the shortest distance, and the angle of rotation being twice the acute angle between a, b, in the direction from a to b. It is easily shown that any displacement whatever is equivalent to two half-turns and therefore to a screw.
In mechanics we are specially concerned with the theory of infinitesimal displacements. This is included in the preceding, but it is simpler in that the various operations are commutative. An infinitely small rotation about any axis is conveniently represented geometrically by a length AB measures along the axis and proportional to the angle of rotation, with the convention that the direction from A to B shall be related to the rotation as is the direction of translation to that of rotation in a right-handed screw. The consequent displacement of any point P will then be at right angles to the plane PAB, its amount will be represented by double the area of the triangle PAB, and its sense will depend on the cyclical order of the letters P, A, B. If AB, AC represent infinitesimal rotations about intersecting axes, the consequent displacement of any point O in the plane BAC will be at right angles to this plane, and will be represented by twice the sum of the areas OAB, OAC, taken with proper signs. It follows by analogy with the theory of moments (§ 4) that the resultant rotation will be represented by AD, the vector-sum of AB, AC (see fig. 16). It is easily inferred as a limiting case, or proved directly, that two infinitesimal rotations α, β about parallel axes are equivalent to a rotation α + β about a parallel axis in the same plane with the two former, and dividing a common perpendicular AB in a point C so that AC/CB = β/α. If the rotations are equal and opposite, so that α + β = 0, the point C is at infinity, and the effect is a translation perpendicular to the plane of the two given axes, of amount α·AB. It thus appears that an infinitesimal rotation is of the nature of a “localized vector,” and is subject in all respects to the same mathematical laws as a force, conceived as acting on a rigid body. Moreover, that an infinitesimal translation is analogous to a couple and follows the same laws. These results are due to Poinsot.
The analytical treatment of small displacements is as follows. We first suppose that one point O of the body is fixed, and take this as the origin of a “right-handed” system of rectangularco-ordinates;i.e.the positive directions of the axes are assumed to be so arranged that a positive rotation of 90° about Ox would bring Oy into the position of Oz, and so on. The displacement will consist of an infinitesimal rotation ε about some axis through O, whose direction-cosines are, say, l, m, n. From the equivalence of a small rotation to a localized vector it follows that the rotation ε will be equivalent to rotations ξ, η, ζ about Ox, Oy, Oz, respectively, provided
ξ = lε, η = mε, ζ = nε,
(1)
and we note that
ξ2+ η2+ ζ2= ε2.
(2)
Thus in the case of fig. 36 it may be required to connect the infinitesimal rotations ξ, η, ζ about OA, OB, OC with the variations of the angular co-ordinates θ, ψ, φ. The displacement of the point C of the body is made up of δθ tangential to the meridian ZC and sin θ δψ perpendicular to the plane of this meridian. Hence, resolving along the tangents to the arcs BC, CA, respectively, we haveξ = δθ sin φ − sin θ δψ cos φ, η = δθ cos φ + sin θ δψ sin φ.(3)Fig. 40.Again, consider the point of the solid which was initially at A′ in the figure. This is displaced relatively to A′ through a space δψ perpendicular to the plane of the meridian, whilst A′ itself is displaced through a space cos θ δψ in the same direction. Henceζ = δφ + cos θ δψ.(4)
Thus in the case of fig. 36 it may be required to connect the infinitesimal rotations ξ, η, ζ about OA, OB, OC with the variations of the angular co-ordinates θ, ψ, φ. The displacement of the point C of the body is made up of δθ tangential to the meridian ZC and sin θ δψ perpendicular to the plane of this meridian. Hence, resolving along the tangents to the arcs BC, CA, respectively, we have
ξ = δθ sin φ − sin θ δψ cos φ, η = δθ cos φ + sin θ δψ sin φ.
(3)
Again, consider the point of the solid which was initially at A′ in the figure. This is displaced relatively to A′ through a space δψ perpendicular to the plane of the meridian, whilst A′ itself is displaced through a space cos θ δψ in the same direction. Hence
ζ = δφ + cos θ δψ.
(4)
To find the component displacements of a point P of the body, whose co-ordinates are x, y, z, we draw PL normal to the plane yOz, and LH, LK perpendicular to Oy, Oz, respectively. The displacement of P parallel to Ox is the same as that of L, which is made up of ηz and −ζy. In this way we obtain the formulae
δx = ηz − ζy, δy = ζx − ξz, δz = ξy − ηx.