Chapter XIII: Part I: Statics (1)

§ 1. _Statics of a Particle._--By a _particle_ is meant a body whose position can for the purpose in hand be sufficiently specified by a mathematical point. It need not be "infinitely small," or even small compared with ordinary standards; thus in astronomy such vast bodies as the sun, the earth, and the other planets can for many purposes be treated merely as points endowed with mass.

A _force_ is conceived as an effort having a certain direction and a certain magnitude. It is therefore adequately represented, for mathematical purposes, by a straight line AB drawn in the direction in question, of length proportional (on any convenient scale) to the magnitude of the force. In other words, a force is mathematically of the nature of a "vector" (see VECTOR ANALYSIS, QUATERNIONS). In most questions of pure statics we are concerned only with the _ratios_ of the various forces which enter into the problem, so that it is indifferent what _unit_ of force is adopted. For many purposes a gravitational system of measurement is most natural; thus we speak of a force of so many pounds or so many kilogrammes. The "absolute" system of measurement will be referred to below in PART II., KINETICS. It is to be remembered that all "force" is of the nature of a push or a pull, and that according to the accepted terminology of modern mechanics such phrases as "force of inertia," "accelerating force," "moving force," once classical, are proscribed. This rigorous limitation of the meaning of the word is of comparatively recent origin, and it is perhaps to be regretted that some more technical term has not been devised, but the convention must now be regarded as established.

The fundamental postulate of this part of our subject is that the two forces acting on a particle may be compounded by the "parallelogram rule." Thus, if the two forces P,Q be represented by the lines OA, OB, they can be replaced by a single force R represented by the diagonal OC of the parallelogram determined by OA, OB. This is of course a physical assumption whose propriety is justified solely by experience. We shall see later that it is implied in Newton's statement of his Second Law of motion. In modern language, forces are compounded by "vector-addition"; thus, if we draw in succession vectors [->HK], [->KL] to represent P, Q, the force R is represented by the vector [->HL] which is the "geometric sum" of [->HK], [->KL].

By successive applications of the above rule any number of forces acting on a particle may be replaced by a single force which is the vector-sum of the given forces: this single force is called the _resultant_. Thus if [->AB], [->BC], [->CD] ..., [->HK] be vectors representing the given forces, the resultant will be given by [->AK]. It will be understood that the figure ABCD ... K need not be confined to one plane.

If, in particular, the point K coincides with A, so that the resultant vanishes, the given system of forces is said to be in _equilibrium_--i.e. the particle could remain permanently at rest under its action. This is the proposition known as the _polygon of forces_. In the particular case of three forces it reduces to the _triangle of forces_, viz. "If three forces acting on a particle are represented as to magnitude and direction by the sides of a triangle taken in order, they are in equilibrium."

A sort of converse proposition is frequently useful, viz. if three forces acting on a particle be in equilibrium, and any triangle be constructed whose sides are respectively parallel to the forces, the magnitudes of the forces will be to one another as the corresponding sides of the triangle. This follows from the fact that all such triangles are necessarily similar.

As a simple example of the geometrical method of treating statical
problems we may consider the equilibrium of a particle on a "rough"
inclined plane. The usual empirical law of sliding friction is that
the mutual action between two plane surfaces in contact, or between a
particle and a curve or surface, cannot make with the normal an angle
exceeding a certain limit [lambda] called the _angle of friction_. If
the conditions of equilibrium require an obliquity greater than this,
sliding will take place. The precise value of [lambda] will vary with
the nature and condition of the surfaces in contact. In the case of a
body simply resting on an inclined plane, the reaction must of course
be vertical, for equilibrium, and the slope [alpha] of the plane must
therefore not exceed [lambda]. For this reason [lambda] is also known
as the _angle of repose_. If [alpha] > [lambda], a force P must be
applied in order to maintain equilibrium; let [theta] be the
inclination of P to the plane, as shown in the left-hand diagram. The
relations between this force P, the gravity W of the body, and the
reaction S of the plane are then determined by a triangle of forces
HKL. Since the inclination of S to the normal cannot exceed [lambda]
on either side, the value of P must lie between two limits which are
represented by L1H, L2H, in the right-hand diagram. Denoting these
limits by P1, P2, we have

P1/W = L1H/HK = sin ([alpha] - [lambda])/cos ([theta] + [lambda]),
P2/W = L2H/HK = sin ([alpha] + [lambda])/cos ([theta] - [lambda]).

It appears, moreover, that if [theta] be varied P will be least when
L1H is at right angles to KL1, in which case P1 = W sin ([alpha] -
[lambda]), corresponding to [theta] = -[lambda].

Just as two or more forces can be combined into a single resultant, so a single force may be _resolved_ into _components_ acting in assigned directions. Thus a force can be uniquely resolved into two components acting in two assigned directions in the same plane with it by an inversion of the parallelogram construction of fig. 1. If, as is usually most convenient, the two assigned directions are at right angles, the two components of a force P will be P cos [theta], P sin [theta], where [theta] is the inclination of P to the direction of the former component. This leads to formulae for the analytical reduction of a system of coplanar forces acting on a particle. Adopting rectangular axes Ox, Oy, in the plane of the forces, and distinguishing the various forces of the system by suffixes, we can replace the system by two forces X, Y, in the direction of co-ordinate axes; viz.--

X = P1 cos [theta]1 + P2 cos [theta]2 + ... = [Sigma](P cos [theta]), }
Y = P1 sin [theta]1 + P2 sin [theta]2 + ... = [Sigma](P sin [theta]). } (1)

These two forces X, Y, may be combined into a single resultant R making an angle [phi] with Ox, provided

X = R cos [phi], Y = R sin [phi], (2)

whence

R² = X² + Y², tan [phi] = Y/X. (3)

For equilibrium we must have R = 0, which requires X = 0, Y = 0; in words, the sum of the components of the system must be zero for each of two perpendicular directions in the plane.

A similar procedure applies to a three-dimensional system. Thus if, O being the origin, [->OH] represent any force P of the system, the planes drawn through H parallel to the co-ordinate planes will enclose with the latter a parallelepiped, and it is evident that [->OH] is the geometric sum of [->OA], [->AN], [->NH], or [->OA], [->OB], [->OC], in the figure. Hence P is equivalent to three forces Pl, Pm, Pn acting along Ox, Oy, Oz, respectively, where l, m, n, are the "direction-ratios" of [->OH]. The whole system can be reduced in this way to three forces

X = [Sigma] (Pl), Y = [Sigma] (Pm), Z = [Sigma] (Pn), (4)

acting along the co-ordinate axes. These can again be combined into a single resultant R acting in the direction ([lambda], [mu], [nu]), provided

X = R[lambda], Y = R[mu], Z = R[nu]. (5)

If the axes are rectangular, the direction-ratios become direction-cosines, so that [lambda]² + [mu]² + [nu]² = 1, whence

R² = X² + Y² + Z². (6)

The conditions of equilibrium are X = 0, Y = 0, Z = 0.

§ 2. _Statics of a System of Particles._--We assume that the mutual forces between the pairs of particles, whatever their nature, are subject to the "Law of Action and Reaction" (Newton's Third Law); i.e. the force exerted by a particle A on a particle B, and the force exerted by B on A, are equal and opposite in the line AB. The problem of determining the possible configurations of equilibrium of a system of particles subject to extraneous forces which are known functions of the positions of the particles, and to internal forces which are known functions of the distances of the pairs of particles between which they act, is in general determinate. For if n be the number of particles, the 3n conditions of equilibrium (three for each particle) are equal in number to the 3n Cartesian (or other) co-ordinates of the particles, which are to be found. If the system be subject to frictionless constraints, e.g. if some of the particles be constrained to lie on smooth surfaces, or if pairs of particles be connected by inextensible strings, then for each geometrical relation thus introduced we have an unknown reaction (e.g. the pressure of the smooth surface, or the tension of the string), so that the problem is still determinate.

The case of the _funicular polygon_ will be of use to us later. A
number of particles attached at various points of a string are acted
on by given extraneous forces P1, P2, P3 ... respectively. The
relation between the three forces acting on any particle, viz. the
extraneous force and the tensions in the two adjacent portions of the
string can be exhibited by means of a triangle of forces; and if the
successive triangles be drawn to the same scale they can be fitted
together so as to constitute a single _force-diagram_, as shown in
fig. 6. This diagram consists of a polygon whose successive sides
represent the given forces P1, P2, P3 ..., and of a series of lines
connecting the vertices with a point O. These latter lines measure the
tensions in the successive portions of string. As a special, but very
important case, the forces P1, P2, P3 ... may be parallel, e.g. they
may be the weights of the several particles. The polygon of forces is
then made up of segments of a vertical line. We note that the tensions
have now the same horizontal projection (represented by the dotted
line in fig. 7). It is further of interest to note that if the weights
be all equal, and at equal horizontal intervals, the vertices of the
funicular will lie on a parabola whose axis is vertical. To prove this
statement, let A, B, C, D ... be successive vertices, and let H, K ...
be the middle points of AC, BD ...; then BH, CK ... will be vertical
by the hypothesis, and since the geometric sum of [->BA], [->BC] is
represented by 2[->BH], the tension in BA: tension in BC: weight at B

as BA: BC: 2BH.

The tensions in the successive portions of the string are therefore
proportional to the respective lengths, and the lines BH, CK ... are
all equal. Hence AD, BC are parallel and are bisected by the same
vertical line; and a parabola with vertical axis can therefore be
described through A, B, C, D. The same holds for the four points B, C,
D, E and so on; but since a parabola is uniquely determined by the
direction of its axis and by three points on the curve, the successive
parabolas ABCD, BCDE, CDEF ... must be coincident.

§ 3. _Plane Kinematics of a Rigid Body._--The ideal _rigid body_ is one in which the distance between any two points is invariable. For the present we confine ourselves to the consideration of displacements in two dimensions, so that the body is adequately represented by a thin lamina or plate.

The position of a lamina movable in its own plane is determinate when we know the positions of any two points A, B of it. Since the four co-ordinates (Cartesian or other) of these two points are connected by the relation which expresses the invariability of the length AB, it is plain that virtually three independent elements are required and suffice to specify the position of the lamina. For instance, the lamina may in general be fixed by connecting any three points of it by rigid links to three fixed points in its plane. The three independent elements may be chosen in a variety of ways (e.g. they may be the lengths of the three links in the above example). They may be called (in a generalized sense) the _co-ordinates_ of the lamina. The lamina when perfectly free to move in its own plane is said to have _three degrees of freedom_.

By a theorem due to M. Chasles any displacement whatever of the lamina in its own plane is equivalent to a rotation about some finite or infinitely distant point J. For suppose that in consequence of the displacement a point of the lamina is brought from A to B, whilst the point of the lamina which was originally at B is brought to C. Since AB, BC, are two different positions of the same line in the lamina they are equal, and it is evident that the rotation could have been effected by a rotation about J, the centre of the circle ABC, through an angle AJB. As a special case the three points A, B, C may be in a straight line; J is then at infinity and the displacement is equivalent to a pure _translation_, since every point of the lamina is now displaced parallel to AB through a space equal to AB.

Next, consider any continuous motion of the lamina. The latter may be brought from any one of its positions to a neighbouring one by a rotation about the proper centre. The limiting position J of this centre, when the two positions are taken infinitely close to one another, is called the _instantaneous centre_. If P, P´ be consecutive positions of the same point, and [delta][theta] the corresponding angle of rotation, then ultimately PP´ is at right angles to JP and equal to JP·[delta][theta]. The instantaneous centre will have a certain locus in space, and a certain locus in the lamina. These two loci are called _pole-curves_ or _centrodes_, and are sometimes distinguished as the _space-centrode_ and the _body-centrode_, respectively. In the continuous motion in question the latter curve rolls without slipping on the former (M. Chasles). Consider in fact any series of successive positions 1, 2, 3... of the lamina (fig. 11); and let J12, J23, J34... be the positions in space of the centres of the rotations by which the lamina can be brought from the first position to the second, from the second to the third, and so on. Further, in the position 1, let J12, J´23, J´34 ... be the points of the lamina which have become the successive centres of rotation. The given series of positions will be assumed in succession if we imagine the lamina to rotate first about J12 until J´23 comes into coincidence with J23, then about J23 until J´34 comes into coincidence with J34, and so on. This is equivalent to imagining the polygon J12 J´23 J´34 ..., supposed fixed in the lamina, to roll on the polygon J12 J23 J34 ..., which is supposed fixed in space. By imagining the successive positions to be taken infinitely close to one another we derive the theorem stated. The particular case where both centrodes are circles is specially important in mechanism.

The theory may be illustrated by the case of "three-bar motion." Let
ABCD be any quadrilateral formed of jointed links. If, AB being held
fixed, the quadrilateral be slightly deformed, it is obvious that the
instantaneous centre J will be at the intersection of the straight
lines AD, BC, since the displacements of the points D, C are
necessarily at right angles to AD, BC, respectively. Hence these
displacements are proportional to JD, JC, and therefore to DD´ CC´,
where C´D´ is any line drawn parallel to CD, meeting BC, AD in C´, D´,
respectively. The determination of the centrodes in three-bar motion
is in general complicated, but in one case, that of the "crossed
parallelogram" (fig. 13), they assume simple forms. We then have AB =
DC and AD = BC, and from the symmetries of the figure it is plain that

AJ + JB = CJ + JD = AD.

Hence the locus of J relative to AB, and the locus relative to CD are
equal ellipses of which A, B and C, D are respectively the foci. It
may be noticed that the lamina in fig. 9 is not, strictly speaking,
fixed, but admits of infinitesimal displacement, whenever the
directions of the three links are concurrent (or parallel).

The matter may of course be treated analytically, but we shall only require the formula for infinitely small displacements. If the origin of rectangular axes fixed in the lamina be shifted through a space whose projections on the original directions of the axes are [lambda], [mu], and if the axes are simultaneously turned through an angle [epsilon], the co-ordinates of a point of the lamina, relative to the original axes, are changed from x, y to [lambda] + x cos [epsilon] - y sin [epsilon], [mu] + x sin [epsilon] + y cos [epsilon], or [lambda] + x - y[epsilon], [mu] + x[epsilon] + y, ultimately. Hence the component displacements are ultimately

[delta]x = [lambda] - y[epsilon], [delta]y = [mu] + x[epsilon] (1)

If we equate these to zero we get the co-ordinates of the instantaneous centre.

§ 4. _Plane Statics._--The statics of a rigid body rests on the following two assumptions:--

(i) A force may be supposed to be applied indifferently at any point in its line of action. In other words, a force is of the nature of a "bound" or "localized" vector; it is regarded as resident in a certain line, but has no special reference to any particular point of the line.

(ii) Two forces in intersecting lines may be replaced by a force which is their geometric sum, acting through the intersection. The theory of parallel forces is included as a limiting case. For if O, A, B be any three points, and m, n any scalar quantities, we have in vectors

m · [->OA] + n·[->OB] = (m + n) [->OC], (1)

provided

m · [->CA] + n·[->CB] = 0. (2)

Hence if forces P, Q act in OA, OB, the resultant R will pass through C, provided

m = P/OA, n = Q/OB;

also

R = P·OC/OA + Q·OC/OB, (3)

and

P·AC : Q·CB = OA : OB. (4)

These formulae give a means of constructing the resultant by means of any transversal AB cutting the lines of action. If we now imagine the point O to recede to infinity, the forces P, Q and the resultant R are parallel, and we have

R = P + Q, P·AC = Q·CB. (5)

When P, Q have opposite signs the point C divides AB externally on the side of the greater force. The investigation fails when P + Q = 0, since it leads to an infinitely small resultant acting in an infinitely distant line. A combination of two equal, parallel, but oppositely directed forces cannot in fact be replaced by anything simpler, and must therefore be recognized as an independent entity in statics. It was called by L. Poinsot, who first systematically investigated its properties, a _couple_.

We now restrict ourselves for the present to the systems of forces in one plane. By successive applications of (ii) any such coplanar system can in general be reduced to a _single resultant_ acting in a definite line. As exceptional cases the system may reduce to a couple, or it may be in equilibrium.

The _moment_ of a force about a point O is the product of the force into the perpendicular drawn to its line of action from O, this perpendicular being reckoned positive or negative according as O lies on one side or other of the line of action. If we mark off a segment AB along the line of action so as to represent the force completely, the moment is represented as to magnitude by twice the area of the triangle OAB, and the usual convention as to sign is that the area is to be reckoned positive or negative according as the letters O, A, B, occur in "counter-clockwise" or "clockwise" order.

The sum of the moments of two forces about any point O is equal to the moment of their resultant (P. Varignon, 1687). Let AB, AC (fig. 16) represent the two forces, AD their resultant; we have to prove that the sum of the triangles OAB, OAC is equal to the triangle OAD, regard being had to signs. Since the side OA is common, we have to prove that the sum of the perpendiculars from B and C on OA is equal to the perpendicular from D on OA, these perpendiculars being reckoned positive or negative according as they lie to the right or left of AO. Regarded as a statement concerning the orthogonal projections of the vectors [->AB] and [->AC] (or BD), and of their sum [->AD], on a line perpendicular to AO, this is obvious.

It is now evident that in the process of reduction of a coplanar system no change is made at any stage either in the sum of the projections of the forces on any line or in the sum of their moments about any point. It follows that the single resultant to which the system in general reduces is uniquely determinate, i.e. it acts in a definite line and has a definite magnitude and sense. Again it is necessary and sufficient for equilibrium that the sum of the projections of the forces on each of two perpendicular directions should vanish, and (moreover) that the sum of the moments about some one point should be zero. The fact that three independent conditions must hold for equilibrium is important. The conditions may of course be expressed in different (but equivalent) forms; e.g. the sum of the moments of the forces about each of the three points which are not collinear must be zero.

The particular case of three forces is of interest. If they are not all parallel they must be concurrent, and their vector-sum must be zero. Thus three forces acting perpendicular to the sides of a triangle at the middle points will be in equilibrium provided they are proportional to the respective sides, and act all inwards or all outwards. This result is easily extended to the case of a polygon of any number of sides; it has an important application in hydrostatics.

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 [alpha], [beta] be the
inclinations of the planes, and [theta] 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 ([theta] - [alpha])/sin [alpha], JG/b = sin ([theta] + [beta])/sin [beta],

whence

a cot [alpha] - b cot [beta]
cot [theta] = ----------------------------. (6)
a + b

If the bar is uniform we have a = b, and

cot [theta] = ½ (cot [alpha] - cot [beta]). (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 P[alpha] + Q[beta] + R[gamma] = 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 the _moment 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 name _torque_ (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 whose 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.

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 A1 may be replaced by its components X1, Y1, parallel to the co-ordinate axes; that at A2 by its components X2, Y2, and so on. Introducing at O two equal and opposite forces ±X1 in Ox, we see that X1 at A1 may be replaced by an equal and parallel force at O together with a couple -y1X1. Similarly the force Y1 at A1 may be replaced by a force Y1 at 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 [Sigma](X) along Ox, a force Y1 + Y2 + ... or [Sigma](Y) along Oy, and a couple (x1Y1 - y1X1) + (x2Y2 - y2X2) + ... or [Sigma](xY - yX). The three conditions of equilibrium are therefore

[Sigma](X) = 0, [Sigma](Y) = 0, [Sigma](xY - yX) = 0. (8)

If O´ be a point whose co-ordinates are ([xi], [eta]), the moment of the couple when the forces are transferred to O´ as a new origin will be [Sigma]{(x - [xi]) Y - (y - [eta]) X}. This vanishes, i.e. the system reduces to a single resultant through O´, provided

-[xi]·[Sigma](Y) + [eta]·[Sigma](X) + [Sigma](xY - yX) = 0. (9)

If [xi], [eta] 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 [theta] with Ox, we may write X1 = P1 cos [theta], Y1 = P1 sin [theta], X2 = P2 cos [theta], Y2 = P2 sin [theta], .... The equation (9) then becomes

{[Sigma](xP) - [xi]·[Sigma](P)} sin [theta] - {[Sigma](yP) - [eta]·[Sigma](P)} cos [theta] = 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 [theta] is alone altered, and the resultant [Sigma](P) passes always through the point

[Sigma](xP) [Sigma](yP)
[|x] = -----------, [|y] = -----------, (11)
[Sigma](P) [Sigma](P)

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 the _centre_ of the given system of parallel forces; it is finite and determinate unless [Sigma](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 the _centre of gravity_ as 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 the _shearing stress_, and the couple the _bending moment_ at
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

F_P = F_Q, M_P - M_Q = -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

F_Q - F_P = -W, M_Q = M_P; (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
[delta]x of the beam may be represented by w[delta]x, where w is in
general a function of x. The equations (12) are now replaced by

[delta]F = -w[delta]x, [delta]M = -F[delta]x,

whence
_ _
/ Q / Q
F_Q - F_P = - | w dx, M_Q - M_P = - | F dx. (14)
_/P _/P

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 = ½wx² - 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, a _force-diagram_ and a _funicular polygon_. The force-diagram is constructed by placing end to end a series of vectors representing the given forces in magnitude and direction, and joining the vertices of the polygon thus formed to an arbitrary _pole_ O. 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.

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 [omega],
[omega]´. Through the vertices A, B, C, ... and A´, B´, C´, ... of the
two funiculars draw normals to the plane of the diagram, to meet
[omega] and [omega]´ 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 called _reciprocal_, 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 is convenient
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 also APPLIED MECHANICS below)
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 the _closing line_ of 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._--A _frame_ is a structure made up of pieces, or _members_, each of which has two _joints_ connecting 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 the _stress_ in the member; it may be a _tension_ or a _thrust_. For diagrammatic purposes each member is sufficiently represented by a straight line terminating at the two joints; these lines will be referred to as the _bars_ of the frame.

In structural applications a frame must be _stiff_, or _rigid_, i.e. it must be incapable of deformation without alteration of length in at least one of its bars. It is said to be _just rigid_ if 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 called _over-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 the _critical 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 two new bars meeting there, is called a _simple_ frame; 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 article BRIDGES. 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 the _method of sections_ is 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 particular [->ZX]
represents the pressure of AB on B in the modified problem where the
loads W1 and W2 on the two bars are replaced by loads P1, Q1, and P2,
Q2 respectively, acting on the pins. Compounding with this [->XV],
which represents Q1, we get the actual pressure [->ZV] 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.