up | next | prev | ptail | tail |
11. As far as I am aware the only author who has applied Quaternions to Elasticity is Prof. Tait. In the chapter on Kinematics of his treatise on Quaternions, §§ 360–371, he has considered the mathematics of strain with some elaboration and again in the chapter on Physical Applications, §§ 487–491, he has done the same with reference to stress and also its expression in terms of the displacement at every point of an elastic body.
In the former he has very successfully considered various useful decompositions of strain into pure and rotational parts and so far as strain alone is considered, i.e. without reference to what stress brings it about he has left little or nothing to be done. In the latter he has worked out the expressions for stress by means of certain vector functions at each point, which express the elastic properties of the body at that point.
But as far as I can see his method will not easily adapt itself to the solution of problems which have already been considered by other methods, or prepare the way for the solution of fresh problems. To put Quaternions in this position is our present object. I limit myself to the statical aspect of Elasticity, but I believe that Quaternions can be as readily, or nearly so, applied to the Kinetics of the subject.
For the sake of completeness I shall repeat in my own notation a small part of the work that Tait has given.
Tait shews (§ 370 of his Quaternions) that in any small portion of a strained medium the strain is homogeneous and (§ 360) that a homogeneous strain function is a linear vector one. He also shews (§ 487) that the stress function is a linear vector one and he obtains expressions (§§ 487–8) for the force per unit volume due to the stress, in terms of the space-variation of the stress.
12. This last however I give in my own notation. His expression in § 370 for the strain function I shall throughout denote by . Thus
where is the displacement that gives rise to the strain.
Let consist of a pure strain followed by a rotation as explained in Tait’s Quaternions, § 365 where he obtains both and in terms of . Thus
When the strain is small takes the convenient form where stands for the pure part of so that
by equation (3) Section 2 above. Similarly becomes where . The truth of these statements is seen by putting in equation (2) for , and therefore for , for , and then neglecting all small quantities of an order higher than the first.
13. Next let us find the force and couple per unit volume due to a stress which varies from point to point. Let the stress function be . Then the force on any part of the body, due to stress, is
by equation (9) Section 4. Thus the force per unit volume , for the volume considered in the equation may be taken as the element .
Again the moment round any arbitrary origin is
by equation (9) Section 4. The second term on the right is that due to the force just considered, and the first shews that in addition to this there is a couple per unit volume twice the “rotation” vector of . Let then be the pure part and the rotatory part of . Thus
Force per unit volume
Couple per unit volume
These results are of course equivalent to those obtained by Tait, Quaternions, §§ 487–8.
The meanings just given to , , , , , , and will be retained throughout this Section. In all cases of small strain as we have seen we may use or indifferently and whenever we wish to indicate that we are considering the physical phenomenon of pure strain we shall use , being regarded merely as a function of . We shall soon introduce a function which will stand towards somewhat as towards and such that when the strain is small .
It is to be observed that is the force exerted on a vector area, which when strained is , not the stress on an area which before strain is . Similarly in equations (4) and (5) the independent variable of differentiation is so that strictly speaking in (4) we should put . In the case of small strain these distinctions need not be made.
14. To express stress in terms of strain we assume any displacement and consequent strain at every point of the body and then give to every point a small additional displacement and find in terms of and the increment in the potential energy of the body, being the potential energy of any element of the body whose volume before strain was . Thus
Thus, observing that by Section 7 the rotation due to the small displacement is , we have
The first of the terms on the right is the work done on the surface of the portion of the body considered; the second is (work done by stress-forces ); and the third is (work done by stress-couples ). Thus converting the surface integral into a volume integral by equation (9) Section 4 above
Limiting the portion of the body considered to the element we get
where is put for and therefore may be put by Section 2 and Section 2 above in the various forms
It is to be observed that since the rotation-vector of does not occur in equation (6), and therefore the stress-couple are quite arbitrary so far as the strain and potential energy are concerned. I do not know whether this has been pointed out before. Of course other data in the problem give the stress-couple. In fact it can be easily shewn that in all cases whether there be equilibrium or not the external couple per unit volume balances the stress-couple. [Otherwise the angular acceleration of the element would be infinite.] Thus if be the external couple per unit volume of the unstrained solid we have always
In the particular case of equilibrium being the external force per unit volume of the unstrained solid we have
The mathematical problem is then the same as if for we substituted and for , zero. In the case of small strains may be put . In this case then the mathematical problem is the same as if for we substituted and for zero.
Returning to equation (6) observe that
[which is established just as is the equation ], so that changing which is any vector into
Therefore by equation (6) of this section and equation (4) Section 2 above we have
We must express the differential on the right of this equation in terms of and , the latter however disappearing as we should expect. Now [equation (2) Section 7] so that remembering that
or since , being self-conjugate
This can be put into a more convenient shape for our present purpose. First put for its value and then apply equation (6) Section 2 above, putting for the of that equation and therefore for the , . Thus
The physical meaning of this last equation can easily be shewn. Suppose when there is no rotation that . Then it is natural14 to assume—in fact it seems almost axiomatic—that the superimposed rotation should merely so to speak rotate the stress along with it. Thus if is some vector area before the rotation which becomes by means of the rotation
But , so that
Thus we see that is as it were with the rotation undone.
Before proceeding further with the calculation let us see what we have assumed and what equation (9) teaches us. The one thing we have assumed is that the potential energy of the body can be taken as the sum of the potential energies of its elements, in other words that no part of the potential energy depends conjointly on the strains at and where and are points separated by a finite distance. This we must take as an axiom. By it we are led to the expression for in equation (9). This only involves the variation of the pure strain but not the space differential coefficients of . This is not an obvious result as far as I can see but it is I believe always assumed without proof.
We may now regard as a function of only. Therefore by equation (7) Section 3 above
therefore by equation (9) of this section
In equation (6) Section 2 above putting the right-hand member of this equation becomes . Now putting in equation (6) Section 2 this member becomes or . Thus we have
Now is self-conjugate. Hence by Section 2 above
This equation can be looked upon as giving in terms of the strain. We can obtain however explicitly for is the conjugate of . Hence from equation (11) (because is the pure part of )
where is a vector to be found. Changing into
Now being self-conjugate . Hence
whence
by equation (6) Section 2 above. Thus finally
This completely solves the problem of expressing stress in terms of strain in the most general case.
15. , where, as we saw in last section, is perfectly arbitrary so far as the strain is concerned, is the force on the strained area due to the pure strain . And again
is that due to the strain or .
To find the force on an area which before strain was let its strained value after has taken place be . Then by equation (4), § 145 of Tait’s Quaternions, . Hence
by equation (13) of last section. If the rotation now take place this force rotates with it so that the force on the area which was originally is after the strain or
where and
This force then is a linear vector function of , but in general even when it is not self-conjugate. When both and the rotation are zero we see that the rotation vector of is given by equation (12) of last section.
The stress-force can be shewn as in Section 7 to be per unit volume of the unstrained body. Thus since the corresponding stress-couple is the moment exerted by the stresses on any portion of the body round an arbitrary origin is
But this moment may also be put in the form
by equation (9) Section 4 above. Comparing these results
in the notation of next section. This equation may also be deduced from equation (15) below but not so naturally as above.
16. We require equation (9) Section 8 above to prove the statement that no space-variations of or are involved in . It is also required to shew that the fact that also is not involved in is a mathematical sequence of the assumption that the potential energy of a solid is the sum of the potential energies of its elements. Assuming these facts however we can arrive at the equation of stress (11) Section 8 in a different way from the above. We shall also obtain quite different expressions for , , &c. and most important of all we shall obtain the equations of equilibrium by obtaining explicitly in terms of the displacement and its space derivatives. From Section 7 above we have
(say) as in Tait’s Quaternions, § 365. Thus where is put as it will be throughout this section for , the vector coordinate after displacement of the point . From this as we have seen in the Introduction we deduce that the coordinates of are the , , , , , of Thomson and Tait’s Nat. Phil. App. C.
We may as do those authors regard as a function of . Thus:—
By equation (4) Section 2 above, each of the terms in this last expression
Comparing this with equation (6) Section 8 and putting in both where , are arbitrary constant vectors, we get
Hence, since is quite arbitrary
From equation (10) Section 8 above which defines we see that
which we should expect since we have already seen that is the value of when there is no rotation and therefore . Now since
we deduce by any one of the processes already exemplified that
where of course the differentiations of must not refer to . We see then from equation (15) that
which is equation (11) Section 8.
Our present purpose however is to find the equations of equilibrium. Let be the vector area which by the strain becomes . Thus16 as in last section
Further let be the stress-couple per unit volume of the unstrained solid so that is the same of the strained solid. As we know, is quite independent of the strain. By Section 7 we see that the force on the area which before strain was is . Therefore
We saw in the last section that the force per unit volume of the unstrained solid is and the couple . Hence
are the equations of equilibrium, where and are the external force and couple per unit volume of the unstrained solid. All that remains to be done then is to express in terms of and the displacement. Putting
as already mentioned, we have
so that by equation (6) Section 2 above we have
Therefore by equation (15)
It is unnecessary to write down what equation (15) becomes when we substitute for , changing into by equation (15). In the important case however when , the equation is quite simple, viz.
Addition to Section 9, Dec., 1887 (sent in with the Essay). [The following considerations occurred just before I was obliged to send the essay in, so that though I thought them worth giving I had not time to incorporate them in the text.
It is interesting to consider the case of an isotropic body. Here is a function of the three principal elongations only and therefore we may in accordance with Section 8 and Section 8 suppose it a function of , , or in accordance with Section 9 of , , where
Let us use , , , , , for the differential coefficients of with respect to , , , , , respectively. Thus
as can be easily proved by means of equation (6) Section 2 above. But
(Notice in passing that to pass to small strains is quite easy for is linear and homogeneous in so that where is a constant and a multiple of .)17
From equation () we see that in equation (12) Section 8 is zero and therefore that from equation (13)
If we wish to neglect all small quantities above a certain order the present equations pave the way for suitably treating the subject. I do not however propose to consider the problem here as I have not considered it sufficiently to do it justice.]
Variation of Temperature
16a. The which appears in the above sections is the same as the which occurs in Tait’s Thermo-dynamics, § 209, and therefore all the above work is true whether the solid experience change of temperature or not. will be a function then of the temperature as well as . To express the complete mathematical problem of the physical behaviour of a solid we ought of course instead of the above equations of equilibrium, to have corresponding equations of motion, viz. equations (15), (15) and instead of (15)
where is the original density of the solid at the point considered. Further we ought to put down the equations of conduction of heat and lastly equations (16) and (16) below.
We do not propose to consider the conduction of heat, but it will be well to shew how the thermo-dynamics of the present question are treated by Quaternions.
Let be the temperature of the element which was originally and its intrinsic energy. Let
where are a linear self-conjugate function and a scalar respectively, both functions of and . Here is the heat required to be put into the element to raise its temperature by and its pure strain by . Now when is constant must be a perfect differential so that we may put
where is some function of and . Thus
Now we have seen in Section 8 that the work done on the element during the increment , divided by
Thus by the first law of Thermo-dynamics
where is Joule’s mechanical equivalent. Thus
To apply the second law we go through exactly the same cycle as does Tait in his Thermo-dynamics, § 209, viz.
We thus get18
the arbitrary function of being neglected as not affecting any physical phenomenon. Substituting for from equation (16),
by equation (16). Thus from equation (16)
so that in elastic solids as in gases we have a convenient function which is called the “entropy”. Thus the intrinsic energy , the entropy and the stress have all been determined in terms of one function of and which function is therefore in this general mathematical theory supposed to be known.
If instead of regarding (which with the generalised meaning it now bears may still conveniently be called the “potential energy” per unit volume) as the fundamental function of the substance we regard the intrinsic energy or the entropy as such it will be seen that one other function of must also be known. For suppose the entropy be regarded as known. Then since
where the integral is any particular one and is a function of only, supposed known. Again
Thus all the functions are given in terms of and . Similarly if be taken as the fundamental function
where as before the integral is some particular one and is a function of only.
17. We now make the usual assumption that the strains are so small that their coordinates can be neglected in comparison with ordinary quantities such as the coefficients of the linear vector function . We can deduce this case from the above more general results.
To the order considered so that by equation (10) Section 8 above
We shall use the symbol rather than for the same reasons explained in Section 7 above as induce us to use rather than .
Remembering that and are now the same and that may be put so that and therefore of equation (12) , we have from equation (13),
Of course we do not require to go through the somewhat complicated process of Section 8, Section 8 to arrive at this result. In fact in equation (6) Section 8, we may put and so that
and therefore by Section 2 above
But by Section 3 above
and therefore by Section 2 we get equation (17).
18. It is convenient here to slightly change the notation. For , we shall now substitute , respectively. This leads to no confusion as will be seen.
With this notation the strain being small the stress is linear in i.e. is linear and therefore quadratic. Now for any such quadratic function
Put now . Then because is linear in
where is put for the value of when the coordinates of are substituted for those of . Thus keeping constant and varying ,
whence integrating from to and changing into we get equation (18).
From equation (18) we see that for small strains we have
Now is quadratic in and therefore also quadratic in , so that regarding as a function of we have as in equation (18)
so that by equation (19) and Section 2 above
All these results for small strains are well-known in their Cartesian form, but it cannot be bias that makes these quaternion proofs appear so much more natural and therefore more simple and beautiful than the ordinary ones.
19. Let us now consider (as in Section 9 is really done) as a function of the displacement. Now is quadratic in , and is linear and symmetrical in and . In fact from equation (3) Section 7 above, remembering that the of that equation is our present , we have
Therefore we may put
where is linear in each of its constituents, is symmetrical in and , and again in and , and is also such that the pair , and the pair , can be interchanged. [This last statement can be made true if not so at first, by substituting for , as this does not affect equation (21).] Such a function can be proved to involve 21 independent scalars, which is the number also required to determine an arbitrary quadratic function of , since involves six scalars.
Thus we have the two following expressions for , which we equate
Now let us put where and are arbitrary constant vectors. We thus get
Whence since is quite arbitrary,
The statical problem can now be easily expressed. As we saw in Section 8, equations (7) and (8), it is simply
throughout the mass; and at the surface
where , are the given external force and couple per unit volume and is the given external surface traction per unit surface. Substituting for from equation (22)
20. The simplest way to treat these bodies is to consider the (linear) relations between and .
In the first place notice that can always be decomposed into three real elongations (contractions being of course considered as negative elongations). Thus being the unit vector in the direction of such an elongation,
The elongation will cause a stress symmetrical about the vector , i.e. a tension in the direction of and a pressure in all directions at right angles; and being constants (on account of the linear relation between and ) independent of the direction of (on account of the isotropy of the solid). This stress may otherwise be described as a tension in the direction of and a hydrostatic pressure . Thus
To obtain the values of and in terms of Thomson and Tait’s coefficients and of cubical expansion and rigidity respectively; first put
and being any two vectors perpendicular to each other. We thus get
whence
From this we have
but from the same equation
Equation (24) gives stress in terms of strain and (25) the converse.
21. We can now give the various useful forms of for isotropic bodies for from equation (19) Section 10,
Therefore from equations (24) and (25) respectively
Therefore again from Section 2 above and from equation (26),
or
Hence from equation (22)
where is put for . This last could have been deduced at once from equation (24) by substituting for .
Thus the equations (23) for the statical problem are
We now proceed to apply these results for small strains in isotropic bodies to particular cases. These particular cases have all been worked out by the aid of Cartesian Geometry and they are given to illustrate the truth of the assertion made in the Introduction that the consideration of general problems is made simpler by the use of Quaternions instead of the ordinary methods.
Particular integral of equation (30)20
22. Since from equation (30) ( being put for simplicity instead of ) we have
we obtain as a particular case by equations (18) and (19) Section 5,
where has the meaning explained in Section 5, and the volume integral extends over any portion (say the whole) of the body we may choose to consider. To express as a function of put in this term where is taken for the of Section 5, and apply equation (9) Section 4. Thus
for by equation (30) . Now (in order to get rid of any infinite terms due to any discontinuity in ) apply equation (9) Section 4 to the second volume integral. Thus
The surface integral may be neglected as we may thus verify. Call the volume integral . Thus
so that putting we get
whence we have as a particular solution of equation (30) or
This is generally regarded as a solution of the statical problem for an infinite isotropic body. In this case some law of convergence must apply to to make these integrals convergent. Thomson and Tait (Nat. Phil. § 730) say that this law is that converges to zero at infinity. This I think can be disproved by an example. Put21 where is a constant vector and a positive constant less than unity. Equation (32) then gives for the displacement at the origin due to the part of the integral extending throughout a sphere whose centre is the origin and radius
Putting , also becomes . The real law of convergence does not seem to me to be worth seeking as the practical utility of equation (32) is owing to the fact that it is a particular integral.
The present solution of the problem has only to be compared with the one in Thomson and Tait’s Nat. Phil. §§ 730–1 to see the immense advantage to be derived from Quaternions.
It is easy to put our result in the form given by them. We have merely to express in terms of and where is put for the reciprocal of . Noting that
we have at once
therefore eliminating ,
which is the required form.
23. Calling the particular solution as before and putting
the statical problem is reduced to finding to satisfy
and the surface equation either
i.e. by equation (29) Section 11 above,
This general problem for the spherical shell, the only case hitherto solved, I do not propose to work out by Quaternions, as the methods adopted are the same as those used by Thomson and Tait in the same problem. But though each step of the Cartesian proof would be represented in the Quaternion, the saving in mental labour which is effected by using the peculiarly happy notation of Quaternions can only be appreciated by him who has worked the whole problem in both notations. The only remark necessary to make is that we may just as easily use vector, surface or solid, harmonics or indeed quaternion harmonics as ordinary scalar harmonics.
24. It is usual to find what equations (22) of Section 10 and (3) of Section 7 become when expressed in terms of any orthogonal coordinates. This can be done much more easily by Quaternions than Cartesian Geometry. Compare the following investigation with the corresponding one in Ibbetson’s Math. Theory of Elasticity, Chap. V.
Let , , be any orthogonal coordinates, i.e. let , , , represent three families of surfaces cutting everywhere at right angles. Particular cases are of course the ordinary Cartesian coordinates, the spherical coordinates , , and the cylindrical coordinates , , . Let , , stand for differentiations per unit length perpendicular to the three coordinate surfaces and let , , be the unit vectors in the corresponding directions. Thus
Thus, using the same system of suffixes for the ’s as was explained in connection with in Section 1,
or
25. Now to put equation (22) Section 10 into the present coordinates all that is required is to express in terms of those coordinates. Let the coordinates of be . Thus from equation (35) we have
The first thing then is to find , , &c.
Let , be the principal curvatures normal to , i.e. (by a well-known property of orthogonal surfaces) the curvatures along the lines of intersection of , with , and , will be considered positive when22 the positive value of is on the convex side of the corresponding curvatures. Similarly for . Thus for the coordinates , , ;
Again for , , ; and the rest are each zero.
With these definitions we see geometrically that
Similarly for , , &c. Thus
or
26. The other chiefly useful thing in transformation of coordinates in the present subject is the expression for the strain function in terms of the coordinates of displacement. Let , , be these coordinates. Now by equation (3) Section 7, remembering (Section 10) that we have
Similarly
But with Thomson and Tait’s notation for pure small strain
Thus we have in terms of the displacement and we have already in equation (24) Section 11, which expresses in terms of , found the values of in terms of , &c. Finally the expression in equation (37) for gives us the equations of equilibrium in terms of , &c. Thus we have all the materials for considering any problem with the coordinates we have chosen.
All these results can be at once applied to spherical and cylindrical coordinates, but as this has nothing to do with our present purpose—the exemplification of Quaternion methods—we leave the matter here.
Let us as an example of particular coordinates to which this section forms a suitable introduction consider St Venant’s Torsion Problem by means of cylindrical coordinates.
Saint-Venant’s Torsion Problem
27. In this problem we consider the equilibrium of a cylinder with any given cross-section, subjected to end-couples, but to no bodily forces and no stress on the curved surface.
We shall take as our coordinates, the axis of being parallel to the generating lines of the cylinder. Let be the unit vectors in the directions of respectively and let
as before.
We shall follow Thomson and Tait’s lines of proof—i.e. we shall first find the effect of a simple torsion and then add another displacement and so try to get rid of stress on the curved surface.
Holding the section fixed let us give the cylinder a small torsion of magnitude , i.e. let us put
for all points for which is small.
The practical manipulation of such expressions as this is almost always facilitated by considering the general value of where is any function linear in each of its constituents. Thus in the present case
[If is symmetrical in its constituents, e.g. in the case of stress below this reduces to the simple form .] From this we at once see that satisfies the equation of internal equilibrium
for putting
so that both and .
Again the value for at once gives us the stress for
which is a shearing stress on the interfaces perpendicular to and .
Putting the unit normal of the curved surface we have for the surface traction
In the figure let the plane of the paper be , the origin, a point on the curved surface and the perpendicular from on the tangent at . Thus , being reckoned positive or negative according as it is in the positive or negative direction of rotation round . Thus we see that the surface traction is parallel to and .
Hence in the case of a circular cylinder a torsion round the axis satisfies all the conditions of our original problem, but this is true in no other case.
The surface traction at any point on the plane ends necessary to produce this strain is by equation (41) so that its moment round the origin is , where is an element of area and the integral extends over the whole cross-section.
28. Let us now assume a further displacement
where is a function of only, and let us try to determine so that there is still internal equilibrium and so that the stress on the curved surface due to shall neutralise the surface traction already considered.
In the present case
Thus (since is perpendicular to ) and therefore the equation of internal equilibrium gives
a shear on the interfaces perpendicular to and . Thus putting for the unit normal to the curved surface, the present surface traction will neutralize the former if
where represents differentiation along the normal outwards and differentiation along the positive direction of the bounding curve.
We leave the problem here to the theory of complex variables and Fourier’s Theorem.
Observe however that the surface traction at any point on the plane end is equal to by equation (43), and therefore that the total couple is equal to . This leads to the usual expression for torsional rigidity.
29. In the following general treatment of Wires some of the processes are merely Thomson and Tait’s translated into their shorter Quaternion forms; others are quite different. The two will be easily distinguished by such as are acquainted with Thomson and Tait’s Nat. Phil.
The one thing to be specially careful about is the notation and its exact meaning. This meaning we give at the outset.
The wires we consider are not necessarily naturally straight but we assume some definite straight condition of the wire as the “geometrically normal” condition.
The variable in terms of which we wish to express everything is the distance along the wire from some definite point on it.
Any element of the wire, since it is only slightly strained, may be assumed to have turned as a rigid body from its geometrically normal position. This rotation is expressed as usual (Tait’s Quaternions, § 354) by the quaternion ; the axis of being the axis of rotation, and the angle of , half the angle turned through.
is taken so that the rate of this turning per unit length of the wire is so that is the rate of turning per unit of length when the whole wire is moved as a rigid body so as to bring the element under consideration back to its geometrically normal position. Of course is a function of and its derivative with reference to . This function we shall investigate later. The resolved part of parallel to the wire is the vector twist and the resolved part perpendicular to the wire is in the direction of the binormal and equal to the curvature. In fact is the vector whose coordinates are the of Thomson and Tait’s Nat. Phil. § 593. When we are given or for every point we know the strain of the wire completely. is defined as the naturally normal value of , i.e. the value of when the wire is unstressed.
As usual we take as the coordinate vector of any point of the wire, like the rest of the functions being considered as a function of . We shall denote (after Tait, Quaternions, Chap. ix.) differentiations with regard to by dashes.
We now come to the dynamical symbols. and are the force and couple respectively exerted across any normal section of the wire on the part of the wire which is on the negative side of the section by the part on the positive side.
Finally let be the external force and couple per unit length exerted upon the wire.
30. When the wire is strained in any way let us impose a small additional strain represented by an increment in and an increment in the elongation at any point. Then the work done on the element by the stress-force and that done by the stress-couple . If (as we assume, though the assumption is not justified in some useful applications of the general theory of wires) and to be of the same order of magnitude the former of these expressions can be neglected in comparison with the latter for is a quantity small compared with . Now the work done on the element by the stress the increment of the element’s potential energy where is some function of the strain. Hence
Notice that is Thomson and Tait’s , , and their (Nat. Phil. §§ 594, 614).
31. Now since the strain is small, is linear in terms of the strain and therefore in terms of . Hence we see that is quadratic in terms of . Let us then put
and being linear and homogeneous in each of their constituents. This is the most general quadratic function of . Now
Putting then and we get . Operating on the last by where is any vector we see that . Thus putting
where has the value given by the last equation and is therefore self-conjugate we get the two following equations
[as can be seen by a comparison of the last three equations being the function of the temperature] and
When the natural shape of the wire is straight these become
and when further the wire is truly uniform and are constant along the wire.
32. Assuming the truth of these restrictions let us conceive a rigid body moving about a fixed point which, when placed in a certain position which we shall call the normal position, has, if then rotating with any vector angular velocity , a moment of momentum where has the meaning just given. If the rigid body be made to take a finite rotation and then to move with angular velocity its moment of momentum will be . Now let a point move along the wire with unit velocity and let the rigid body so move in unison with it that when the moving point reaches the point the rigid body shall have made the rotation represented by (Section 13 above). Thus by the definition of and its angular velocity at any instant is and its moment of momentum therefore or .
Now consider the equilibrium of the wire when no external force or couple acts except at its ends. In this case is constant throughout and it is easy to see (what indeed is a particular case of equation (53) below) that
Interpreting this equation for our rigid body we get as the law which governs its motion
Thus the rigid body will move as if acted upon by a constant force at the end of the unit vector or—since this vector is fixed in the body—as if acted upon by a constant force acting through a point fixed in the body. From this kinetic analogue of Kirchhoff’s the mathematical problem of the shape of such a wire as we are now considering, under the given circumstances, is shewn to be identical with the general problem of the pendulum of which the top is a variety.
33. We will now give the general equations for any wire under any external actions. The comparison of the Quaternion treatment of this with the Cartesian as given in Thomson and Tait’s Nat. Phil. § 614 seems to me to be all in favour of the former.
The equations of equilibrium of an element are with the notation explained in Section 13 above
or dividing by
Operating on the last equation by noting that and putting we get
whence by equations (52) and (53) respectively
Now by equation (48) above
Also by the definition of
where is some given constant unit vector. Finally as we are about to prove
It is usual in the Cartesian treatment to leave the problem in the form of equations equivalent to the above (55) to (58), 13 scalar equations for the 13 unknown scalars of , , , and . We can however as we shall directly in the Quaternion treatment quite easily reduce the general problem to one vector and one scalar equation involving the four unknown scalars of and in terms of which all the other unknowns are explicitly given.
To prove equation (58)24 observe that
where is any vector. The truth of this is seen by noticing that is the operator that rotates any vector of the element from its geometrically normal position to its strained position. But we can also get to this final position by first in the geometrically normal wire making the small strain at the given element and then performing the strain of the wire up to the point . The first process is represented on the left of the last equation and the second on the right. Thus we get
Returning to equations (55) to (58) observe that equations (57) and (58) give and as explicit functions of . Hence by equation (48)
which gives also explicitly. Substituting for and in equations (55) and (56) we have
which are sufficient equations to determine and , whereupon is given by equation (59), by equation (58) and and therefore also by equation (57).
Spiral springs can be treated very simply by means of the above equations, but we have already devoted sufficient space to this subject.
up | next | prev | ptail | top |