III Elastic Solids12

III
Elastic Solids¹³

6. Brief recapitulation of previous work in this branch

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.

7. Strain, Stress-force, Stress-couple

12. This last however I give in my own notation. His expression in § 370 for the strain function I shall throughout denote by $χ$ . Thus

\begin{array}{l} χ ω = ω - S ω \nabla · η, & (1) \end{array}

where $η$ is the displacement that gives rise to the strain.

Let $χ$ consist of a pure strain $ψ$ followed by a rotation $q () q^{- 1}$ as explained in Tait’s Quaternions, § 365 where he obtains both $q$ and $ψ$ in terms of $χ$ . Thus

\begin{array}{l} χ ω = q ψ ω q^{- 1} . & (2) \end{array}

When the strain is small $ψ$ takes the convenient form $\bar{χ}$ where $χ$ stands for the pure part of $χ$ so that

\begin{array}{l} \bar{χ} ω = ω - \frac{1}{2} S ω \nabla · η - \frac{1}{2} \nabla_{1} S ω η_{1}, & (3) \end{array}

by equation (3) Section 2 above. Similarly $q () q^{- 1}$ becomes $V 𝜃 ()$ where $2 𝜃 = V \nabla η$ . The truth of these statements is seen by putting in equation (2) for $q$ , $1 + 𝜃 ∕ 2$ and therefore for $q^{- 1}$ , $1 - 𝜃 ∕ 2$ for $ψ$ , $\bar{χ}$ 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 $ϕ$ . Then the force on any part of the body, due to stress, is

\begin{array}{l} \iint ϕ d Σ = ∭ ϕ Δ d 𝔰, \end{array}

by equation (9) Section 4. Thus the force per unit volume $= ϕ Δ$ , for the volume considered in the equation may be taken as the element $d 𝔰$ .

Again the moment round any arbitrary origin is

\begin{array}{l} \iint V ρ ϕ d Σ = ∭ V ρ_{1} ϕ \nabla_{1} d 𝔰 + ∭ V ρ ϕ_{1} \nabla_{1} d 𝔰, \end{array}

by equation (9) Section 4. The second term on the right is that due to the force $ϕ Δ$ just considered, and the ﬁrst shews that in addition to this there is a couple per unit volume $= V ζ ϕ ζ =$ twice the “rotation” vector of $ϕ$ . Let then $\bar{ϕ}$ be the pure part and $V 𝜖 ()$ the rotatory part of $ϕ$ . Thus

Force per unit volume

\begin{array}{l} = ϕ Δ = \bar{ϕ} Δ - V \nabla 𝜖, & (4) \end{array}

Couple per unit volume

\begin{array}{l} = V ζ ϕ ζ = 2 𝜖 . & (5) \end{array}

These results are of course equivalent to those obtained by Tait, Quaternions, §§ 487–8.

The meanings just given to $η$ , $χ$ , $\bar{χ}$ , $ψ$ , $q$ , $ϕ$ , $\bar{ϕ}$ and $𝜖$ will be retained throughout this Section. In all cases of small strain as we have seen we may use $ψ$ or $\bar{χ}$ indiﬀerently and whenever we wish to indicate that we are considering the physical phenomenon of pure strain we shall use $ψ$ , $\bar{χ}$ being regarded merely as a function of $χ$ . We shall soon introduce a function $ϖ$ which will stand towards $ϕ$ somewhat as $ψ$ towards $χ$ and such that when the strain is small $ϖ = \bar{ϕ}$ .

It is to be observed that $ϕ ω$ is the force exerted on a vector area, which when strained is $ω$ , not the stress on an area which before strain is $ω$ . Similarly in equations (4) and (5) the independent variable of diﬀerentiation is $ρ + η$ so that strictly speaking in (4) we should put ${\bar{ϕ}}_{ρ + η} Δ - V_{ρ + η} \nabla 𝜖$ . In the case of small strain these distinctions need not be made.

8. Stress in terms of strain

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 $δ η$ and ﬁnd in terms of $ψ$ and $ϕ$ the increment $∭ δ w d 𝔰_{0}$ in the potential energy of the body, $w d 𝔰_{0}$ being the potential energy of any element of the body whose volume before strain was $d 𝔰_{0}$ . Thus

\begin{array}{l} ∭ δ w d 𝔰_{0} & = (work done by stresses on surface of portion considered) \\ - (work done by stresses throughout volume of same portion). \end{array}

Thus, observing that by Section 7 the rotation due to the small displacement $δ η$ is $V_{ρ + η} \nabla δ η ∕ 2$ , we have

\begin{array}{l} ∭ δ w d 𝔰_{0} = - \iint S δ η ϕ d Σ + ∭ S δ η ϕ_{1 ρ + η} \nabla_{1} d 𝔰 + ∭ S 𝜖_{ρ + η} \nabla δ η d 𝔰 . \end{array}

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 $ϕ_{ρ + η} Δ$ ); and the third is $-$ (work done by stress-couples $2 𝜖$ ). Thus converting the surface integral into a volume integral by equation (9) Section 4 above

\begin{array}{l} ∭ δ w d 𝔰_{0} & = - ∭ S δ η_{1} ϕ_{ρ + η} \nabla_{1} d 𝔰 + ∭ S δ η_{1} V 𝜖_{ρ + η} \nabla_{1} d 𝔰 \\ = - ∭ S δ η_{1} {\bar{ϕ}}_{ρ + η} \nabla_{1} d 𝔰 . \end{array}

Limiting the portion of the body considered to the element $d 𝔰$ we get

\begin{array}{l} δ w = - m S δ η_{1} {\bar{ϕ}}_{ρ + η} \nabla_{1} & (6) \end{array}

where $m$ is put for $d 𝔰 ∕ d 𝔰_{0}$ and therefore may be put by Section 2 and Section 2 above in the various forms

\begin{array}{l} 6 m & = S ζ_{1} ζ_{2} ζ_{3} S χ ζ_{1} χ ζ_{2} χ ζ_{3} & (6 a) \\ = S ζ_{1} ζ_{2} ζ_{3} S χ' ζ_{1} χ' ζ_{2} χ' ζ_{3} & (6 b) \\ = S ζ_{1} ζ_{2} ζ_{3} S ψ ζ_{1} ψ ζ_{2} ψ ζ_{3} & (6 c) \\ = S \nabla_{1} \nabla_{2} \nabla_{3} S {(ρ + η)}_{1} {(ρ + η)}_{2} {(ρ + η)}_{3} . & (6 d) \end{array}

It is to be observed that since the rotation-vector $𝜖$ of $ϕ$ does not occur in equation (6), $𝜖$ 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 $M$ be the external couple per unit volume of the unstrained solid we have always

\begin{array}{l} M + 2 m 𝜖 = 0 . & (7) \end{array}

In the particular case of equilibrium $F$ being the external force per unit volume of the unstrained solid we have

\begin{array}{l} F + m ({\bar{ϕ}}_{ρ + η} Δ - V_{ρ + η} \nabla 𝜖) = 0 . & (8) \end{array}

The mathematical problem is then the same as if for $F ∕ m$ we substituted $F ∕ m + V_{ρ + η} \nabla (M ∕ 2 m)$ and for $M$ , zero. In the case of small strains $m$ may be put $= 1$ . In this case then the mathematical problem is the same as if for $F$ we substituted $F + V \nabla M ∕ 2$ and for $M$ zero.

Returning to equation (6) observe that

\begin{array}{l} δ χ ω = - S χ ω_{ρ + η} \nabla · δ η \end{array}

[which is established just as is the equation $χ ω = - S ω \nabla · (ρ + η)$ ], so that changing $ω$ which is any vector into $χ^{- 1} ω$

\begin{array}{l} δ χ · χ^{- 1} ω = - S ω_{ρ + η} \nabla · δ η . \end{array}

Therefore by equation (6) of this section and equation (4) Section 2 above we have

\begin{array}{l} δ w = - m S δ χ χ^{- 1} ζ \bar{ϕ} ζ . \end{array}

We must express the diﬀerential on the right of this equation in terms of $δ ψ$ and $δ q$ , the latter however disappearing as we should expect. Now $χ ω = q ψ ω q^{- 1}$ [equation (2) Section 7] so that remembering that $δ · q^{- 1} = - q^{- 1} δ q q^{- 1}$

\begin{array}{l} δ χ ω & = δ q ψ ω q^{- 1} - q ψ ω q^{- 1} δ q q^{- 1} + q δ ψ ω q^{- 1} \\ = 2 V V δ q q^{- 1} · χ ω + q δ ψ ω q^{- 1}, \\ ∴ & δ w ∕ m & = - 2 S · V δ q q^{- 1} · ζ \bar{ϕ} ζ - S q δ ψ χ^{- 1} ζ q^{- 1} \bar{ϕ} ζ, \end{array}

or since $V ζ \bar{ϕ} ζ = 0$ , $\bar{ϕ}$ being self-conjugate

\begin{array}{l} δ w = - m S δ ψ χ^{- 1} ζ q^{- 1} \bar{ϕ} ζ q . \end{array}

This can be put into a more convenient shape for our present purpose. First put for $χ^{- 1}$ its value $ψ^{- 1} q^{- 1} () q$ and then apply equation (6 $a$ ) Section 2 above, putting for the $ϕ$ of that equation $q^{- 1} () q$ and therefore for the $ϕ'$ , $q () q^{- 1}$ . Thus

\begin{array}{l} δ w & = - m S δ ψ ψ^{- 1} ζ ϖ ζ, & (9) \\ where & ϖ ω & = q^{- 1} \bar{ϕ} (q ω q^{- 1}) q . & (10) \end{array}

The physical meaning of this last equation can easily be shewn. Suppose when there is no rotation that $\bar{ϕ} = ϖ'$ . Then it is natural¹⁴ to assume—in fact it seems almost axiomatic—that the superimposed rotation $q () q^{- 1}$ should merely so to speak rotate the stress along with it. Thus if $ω$ is some vector area before the rotation which becomes $ω'$ by means of the rotation

\begin{array}{l} q ϖ' ω · q^{- 1} = \bar{ϕ} ω' . \end{array}

But $ω' = q ω q^{- 1}$ , so that

\begin{array}{l} ϖ' ω = q^{- 1} \bar{ϕ} (q ω q^{- 1}) q = ϖ ω . \end{array}

Thus we see that $ϖ$ is as it were $\bar{ϕ}$ 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 $P$ and $Q$ where $P$ and $Q$ are points separated by a ﬁnite distance. This we must take as an axiom. By it we are led to the expression for $δ w$ in equation (9). This only involves the variation of the pure strain $ψ$ but not the space diﬀerential coeﬃcients of $ψ$ . 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 $w$ as a function of $ψ$ only. Therefore by equation (7) Section 3 above

\begin{array}{l} δ w = - S δ ψ ζ ψ D w ζ, \end{array}

therefore by equation (9) of this section

\begin{array}{l} S δ ψ ζ ψ D w ζ = m S δ ψ ψ^{- 1} ζ ϖ ζ \end{array}

¹⁵.

In equation (6 $a$ ) Section 2 above putting $ϕ = ψ^{- 1}$ the right-hand member of this equation becomes $m S δ ψ ζ ϖ ψ^{- 1} ζ$ . Now putting in equation (6 $a$ ) Section 2 $ϕ = ϖ ψ^{- 1}$ this member becomes $m S ζ δ ψ ψ^{- 1} ϖ ζ$ or $m S δ ψ ζ ψ^{- 1} ϖ ζ$ . Thus we have

\begin{array}{l} 2 S δ ψ ζ ψ D w ζ = m S δ ψ ζ (ϖ ψ^{- 1} + ψ^{- 1} ϖ) ζ . \end{array}

Now $ϖ ψ^{- 1} + ψ^{- 1} ϖ$ is self-conjugate. Hence by Section 2 above

\begin{array}{l} m (ϖ ψ^{- 1} + ψ^{- 1} ϖ) = 2 ψ D w . & (11) \end{array}

This equation can be looked upon as giving $ϖ$ in terms of the strain. We can obtain $ϖ$ however explicitly for $ψ^{- 1} ϖ$ is the conjugate of $ϖ ψ^{- 1}$ . Hence from equation (11) (because $(ψ^{- 1} ϖ + ϖ ψ^{- 1}) ∕ 2$ is the pure part of $ϖ ψ^{- 1}$ )

\begin{array}{l} m ϖ ψ^{- 1} ω = ψ D w ω + V 𝜃 ω \end{array}

where $𝜃$ is a vector to be found. Changing $ω$ into $ψ ω$

\begin{array}{l} m ϖ ω = ψ D w ψ ω + V 𝜃 ψ ω . \end{array}

Now $ϖ$ being self-conjugate $V ζ ϖ ζ = 0$ . Hence

\begin{array}{l} V ζ ψ D w ψ ζ & = - V ζ V 𝜃 ψ ζ = 𝜃 S ζ ψ ζ - ψ ζ S 𝜃 ζ \\ = 𝜃 S ζ ψ ζ + ψ 𝜃, \end{array}

whence

\begin{array}{l} \begin{aligned} 𝜃 & = {(ψ + S ζ_{1} ψ ζ_{1})}^{- 1} V ζ ψ D w ψ ζ \\ or & 𝜃 & = {(ψ + S ζ_{1} ψ ζ_{1})}^{- 1} V ψ ζ ψ D w ζ \end{aligned}\} & (12) \end{array}

by equation (6 $a$ ) Section 2 above. Thus ﬁnally

\begin{array}{l} \begin{aligned} m ϖ ω & = ψ D w ψ ω + V 𝜃 ψ ω \\ or & m ϖ ω & = ψ D w ψ ω - V ψ ω {(ψ + S ζ_{1} ψ ζ_{1})}^{- 1} V ψ ζ ψ D w ζ . \end{aligned}\} & (13) \end{array}

This completely solves the problem of expressing stress in terms of strain in the most general case.

15. $ϖ ω + V 𝜖' ω$ , where, as we saw in last section, $𝜖'$ is perfectly arbitrary so far as the strain is concerned, is the force on the strained area $ω$ due to the pure strain $ψ$ . And again

\begin{array}{l} \bar{ϕ} ω + V 𝜖'' ω or q (ϖ q^{- 1} ϖ q) q^{- 1} + V 𝜖'' ω \end{array}

is that due to the strain $q ψ () q^{- 1}$ or $χ$ .

To ﬁnd the force on an area which before strain was $ω_{0}$ let its strained value after $ψ$ has taken place be $ω$ . Then by equation (4), § 145 of Tait’s Quaternions, $m ω_{0} = ψ ω$ . Hence

\begin{array}{l} Required force = ϖ ω + V 𝜖' ω = ψ D w ω_{0} + V 𝜃 ω_{0} + m V 𝜖' ψ^{- 1} ω_{0} \end{array}

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 $ω_{0}$ is after the strain $χ$ or $q ψ () q^{- 1}$

\begin{array}{l} τ ω_{0} = q ψ D w ω_{0} q^{- 1} + q V 𝜃 ω_{0} · q^{- 1} + V (𝜖 q ψ^{- 1} ω_{0} · q^{- 1}) & (14) \end{array}

where $𝜖 = m q 𝜖' q^{- 1}$ and

\begin{array}{l} ∴ 𝜖 = \frac{1}{2} the stress-couple per un. vol. of unstrained body . \end{array}

This force then is a linear vector function of $ω_{0}$ , but in general even when $𝜖 = 0$ it is not self-conjugate. When both $𝜖$ and the rotation are zero we see that the rotation vector of $τ$ is $𝜃$ given by equation (12) of last section.

The stress-force can be shewn as in Section 7 to be $τ Δ$ per unit volume of the unstrained body. Thus since the corresponding stress-couple is $2 𝜖$ the moment exerted by the stresses on any portion of the body round an arbitrary origin is

\begin{array}{l} ∭ V (ρ + η) τ_{1} \nabla_{1} d 𝔰_{0} + 2 ∭ 𝜖 d 𝔰_{0} . \end{array}

But this moment may also be put in the form

\begin{array}{l} \iint V (ρ + η) τ d Σ_{0}, \\ or & ∭ V ζ τ ζ d 𝔰_{0} & + ∭ V (ρ + η) τ_{1} \nabla_{1} d 𝔰_{0} + ∭ V η_{1} τ \nabla_{1} d 𝔰_{0}, \end{array}

by equation (9) Section 4 above. Comparing these results

\begin{array}{l} or & \begin{aligned} 2 𝜖 & = V ζ τ ζ + V η_{1} τ \nabla_{1} \\ 2 𝜖 & = V {ρ'}_{1} τ \nabla_{1}, \end{aligned}\} & (14 a) \end{array}

in the notation of next section. This equation may also be deduced from equation (15 $l$ ) below but not so naturally as above.

9. The equations of equilibrium

16. We require equation (9) Section 8 above to prove the statement that no space-variations of $ψ$ or $q$ are involved in $w$ . It is also required to shew that the fact that $q$ also is not involved in $w$ 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 $ϖ$ , $τ$ , &c. and most important of all we shall obtain the equations of equilibrium by obtaining $τ$ explicitly in terms of the displacement and its space derivatives. From Section 7 above we have

\begin{matrix} χ ω = q ψ ω q^{- 1}, χ' ω = ψ (q^{- 1} ω q), \\ ∴ χ' χ = ψ^{2} = Ψ, & (15) \end{matrix}

(say) as in Tait’s Quaternions, § 365. Thus $Ψ ω = \nabla_{1} S ω \nabla_{2} S {ρ'}_{1} {ρ'}_{2}$ where $ρ'$ is put as it will be throughout this section for $ρ + η$ , the vector coordinate after displacement of the point $ρ$ . From this as we have seen in the Introduction we deduce that the coordinates of $Ψ$ are the $A$ , $B$ , $C$ , $a$ , $b$ , $c$ of Thomson and Tait’s Nat. Phil. App. C.

We may as do those authors regard $w$ as a function of $Ψ$ . Thus:—

\begin{array}{l} δ w & = - S δ Ψ ζ Ψ D w ζ, \\ = - S δ χ' χ ζ Ψ D w ζ - S χ' δ χ ζ Ψ D w ζ . \end{array}

By equation (4 $a$ ) Section 2 above, each of the terms in this last expression

\begin{array}{l} = - S δ η_{1} χ Ψ D w \nabla_{1}, \end{array}

\begin{array}{l} ∴ δ w = - 2 S δ η_{1} χ Ψ D w \nabla_{1}, & (15 a) \end{array}

Comparing this with equation (6) Section 8 and putting in both

δ η_{1} = ω' S ω ρ'

where

ω'

ω

are arbitrary constant vectors, we get

\begin{array}{l} m S ω' \bar{ϕ} ω = 2 S ω' χ Ψ D w χ' ω . \end{array}

Hence, since $ω'$ is quite arbitrary

\begin{array}{l} m \bar{ϕ} = 2 χ Ψ D w χ' . & (15 b) \end{array}

From equation (10) Section 8 above which deﬁnes $ϖ$ we see that

\begin{array}{l} m ϖ = 2 ψ Ψ D w ψ, & (15 c) \end{array}

which we should expect since we have already seen that $ϖ$ is the value of $\bar{ϕ}$ when there is no rotation and therefore $χ = χ' = ψ$ . Now since

\begin{array}{l} S δ ψ ζ ψ D ζ = S δ Ψ ζ Ψ D ζ = S (δ ψ ψ + ψ δ ψ) ζ Ψ D ζ, \end{array}

we deduce by any one of the processes already exempliﬁed that

\begin{array}{l} ψ D = Ψ D ψ + ψ Ψ D, \end{array}

where of course the diﬀerentiations of $Ψ D$ must not refer to $ψ$ . We see then from equation (15 $c$ ) that

\begin{array}{l} m (ϖ ψ^{- 1} + ψ^{- 1} ϖ) = 2 ψ D w, \end{array}

which is equation (11) Section 8.

Our present purpose however is to ﬁnd the equations of equilibrium. Let $ω_{0}$ be the vector area which by the strain $χ$ becomes $ω$ . Thus¹⁶ as in last section

\begin{array}{l} m ω_{0} = χ' ω . & (15 e) \end{array}

Further let $2 𝜖$ be the stress-couple per unit volume of the unstrained solid so that $2 𝜖 ∕ m$ is the same of the strained solid. As we know, $𝜖$ is quite independent of the strain. By Section 7 we see that the force $τ ω_{0}$ on the area which before strain was $ω_{0}$ is $\bar{ϕ} ω + V 𝜖 ω ∕ m$ . Therefore

\begin{array}{l} τ ω_{0} = 2 χ Ψ D w ω_{0} + V 𝜖 {χ'}^{- 1} ω_{0} . & (15 f) \end{array}

We saw in the last section that the force per unit volume of the unstrained solid is $τ Δ$ and the couple $2 𝜖$ . Hence

\begin{array}{l} F + τ Δ & = 0, & (15 g) \\ M + 2 𝜖 & = 0, & (15 h) \end{array}

are the equations of equilibrium, where $F$ and $M$ are the external force and couple per unit volume of the unstrained solid. All that remains to be done then is to express $τ$ in terms of $𝜖$ and the displacement. Putting

\begin{array}{l} ρ + η = ρ', & (15 i) \end{array}

as already mentioned, we have

\begin{array}{l} χ ω = - {ρ'}_{1} S ω \nabla_{1}, & (15 j) \end{array}

so that by equation (6 $h$ ) Section 2 above we have

\begin{array}{l} {χ'}^{- 1} ω = - 3 V {ρ'}_{1} {ρ'}_{2} S ω \nabla_{1} \nabla_{2} ∕ S \nabla_{1} \nabla_{2} \nabla_{3} S {ρ'}_{1} {ρ'}_{2} {ρ'}_{3} . & (15 k) \end{array}

Therefore by equation (15 $f$ )

\begin{array}{l} τ ω = - 2 {ρ'}_{1} S \nabla_{1} Ψ D w ω - 3 V 𝜖 V {ρ'}_{1} {ρ'}_{2} S ω \nabla_{1} \nabla_{2} ∕ S \nabla_{1} \nabla_{2} \nabla_{3} S {ρ'}_{1} {ρ'}_{2} {ρ'}_{3} . & (15 l) \end{array}

It is unnecessary to write down what equation (15 $g$ ) becomes when we substitute for $τ$ , changing $𝜖$ into $- M ∕ 2$ by equation (15 $h$ ). In the important case however when $M = 0$ , the equation is quite simple, viz.

\begin{array}{l} F = 2 {ρ'}_{1} S \nabla_{1} Ψ D w Δ . & (15 m) \end{array}

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 $w$ 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 $a$ , $b$ , $c$ or in accordance with Section 9 of $A$ , $B$ , $C$ where

\begin{array}{l} a & = - S ζ ψ ζ, & b & = - S ζ ψ^{2} ζ ∕ 2, & c & = - S ζ ψ^{3} ζ ∕ 3 . & (A) \\ A & = - S ζ Ψ ζ, & B & = - S ζ Ψ^{2} ζ ∕ 2, & C & = - S ζ Ψ^{3} ζ ∕ 3 . & (B) \end{array}

Let us use $x$ , $y$ , $z$ , $X$ , $Y$ , $Z$ for the diﬀerential coeﬃcients of $w$ with respect to $a$ , $b$ , $c$ , $A$ , $B$ , $C$ respectively. Thus

\begin{array}{l} d w = x d a + y d b + z d c = - S d ψ ζ (x + y ψ + z ψ^{2}) ζ \end{array}

as can be easily proved by means of equation (6 $a$ ) Section 2 above. But

\begin{array}{l} d w & = - S d ψ ζ ψ D w ζ . \\ ∴ ψ D w & = x + y ψ + z ψ^{2}, & (C) \\ by Section 2 above. Similarly we have \\ Ψ D w & = X + Y Ψ + Z Ψ^{2} . & (D) \end{array}

(Notice in passing that to pass to small strains is quite easy for $ψ D w$ is linear and homogeneous in $ψ - 1$ so that $ψ D w = x + y ψ$ where $y$ is a constant and $x + y$ a multiple of $a - 3$ .)¹⁷

From equation ( $C$ ) we see that $𝜃$ in equation (12) Section 8 is zero and therefore that from equation (13)

\begin{array}{l} m ϖ & = ψ D w ψ = x ψ + y ψ^{2} + z ψ^{3} . & (E) \\ Again from equation (14) Section 8 \\ τ ω & = x q ω q^{- 1} + y χ ω + z χ ψ ω + V 𝜖 {χ'}^{- 1} ω . & (F) \\ Similarly from equations (15 b) and (15 f) Section 9 we may prove by
equation (D) that \\ \frac{1}{2} m \bar{ϕ} & = X χ χ' + Y {(χ χ')}^{2} + Z {(χ χ')}^{3}, & (G) \end{array}

\begin{array}{l} and & τ = 2 X χ + 2 Y χ χ' χ + 2 Z χ χ' χ χ' χ + V 𝜖 {χ'}^{- 1} () . & (H) \end{array}

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

Variation of Temperature

16a. The $w$ which appears in the above sections is the same as the $w$ 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. $w$ will be a function then of the temperature as well as $ψ$ . 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 $h$ ), (15 $l$ ) and $($ instead of (15 $g$ ) $)$

\begin{array}{l} D \ddot{ρ}' = F + τ Δ, & (15 n) \end{array}

where $D$ 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 $f$ ) and (16 $d$ ) 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 $t$ be the temperature of the element which was originally $d 𝔰_{0}$ and $E d 𝔰_{0}$ its intrinsic energy. Let

\begin{array}{l} δ H = - S δ ψ ζ M ζ + N δ t, \end{array}

where $M, N$ are a linear self-conjugate function and a scalar respectively, both functions of $ψ$ and $t$ . Here $δ H d 𝔰_{0}$ is the heat required to be put into the element to raise its temperature by $δ t$ and its pure strain by $δ ψ$ . Now when $t$ is constant $δ H$ must be a perfect diﬀerential so that we may put

\begin{array}{l} M = t ψ D f, \end{array}

where $f$ is some function of $ψ$ and $t$ . Thus

\begin{array}{l} δ H = - t S δ ψ ζ ψ D f ζ + N δ t . & (16) \end{array}

Now we have seen in Section 8 that the work done on the element during the increment $δ ψ$ , divided by $d 𝔰_{0}$

\begin{array}{l} = - m S δ ψ ψ^{- 1} ζ ϖ ζ . \end{array}

Thus by the ﬁrst law of Thermo-dynamics

\begin{array}{l} δ E & = J δ H - m S δ ψ ψ^{- 1} ζ ϖ ζ \\ = J N δ t - J t S δ ψ ζ ψ D f ζ - m S δ ψ ψ^{- 1} ζ ϖ ζ, \end{array}

where $J$ is Joule’s mechanical equivalent. Thus

\begin{array}{l} J N & = d E ∕ d t, & (16 a) \\ and & \frac{1}{2} m (ϖ ψ^{- 1} & + ψ^{- 1} ϖ) = ψ D w, & (16 b) \\ where & w & = E - J t f . & (16 c) \end{array}

To apply the second law we go through exactly the same cycle as does Tait in his Thermo-dynamics, § 209, viz.

\begin{array}{l} (ψ, t) (ψ + δ ψ, t) (ψ + δ ψ, t + δ t) (ψ, t + δ t) (ψ, t) . \end{array}

We thus get¹⁸

\begin{array}{l} J ψ D f = - \frac{1}{2} m (\frac{d ϖ}{d t} ψ^{- 1} + ψ^{- 1} \frac{d ϖ}{d t}) = - \frac{d ψ D w}{d t} = - ψ D \frac{d w}{d t} \end{array}

\begin{array}{l} or & J f & = - d w ∕ d t & (16 d) \end{array}

the arbitrary function of

t

being neglected as not aﬀecting any physical phenomenon. Substituting for

w

from equation (16

c

\begin{array}{l} J t d f ∕ d t = d E ∕ d t = J N & (16 e) \end{array}

by equation (16 $a$ ). Thus from equation (16)

\begin{array}{l} δ H = t δ f, & (16 f) \end{array}

so that in elastic solids as in gases we have a convenient function which is called the “entropy”. Thus the intrinsic energy $E d 𝔰_{0}$ , the entropy $f d 𝔰_{0}$ and the stress $ϖ$ have all been determined in terms of one function $w$ of $ψ$ and $t$ which function is therefore in this general mathematical theory supposed to be known.

If instead of regarding $w$ (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 $ψ$ must also be known. For suppose $f$ the entropy be regarded as known. Then since $d w ∕ d t = - J f$

\begin{array}{l} w = W - J \int f d t, & (16 g) \end{array}

where the integral is any particular one and $W$ is a function of $ψ$ only, supposed known. Again

\begin{array}{l} E & = w + J t f \\ or & E & = W + J (t f - \int f d t) . & (16 h) \end{array}

Thus all the functions are given in terms of $f$ and $W$ . Similarly if $E$ be taken as the fundamental function

\begin{array}{l} w & = t (W' - \int E d t ∕ t^{2}), & (16 i) \\ J f & = E ∕ t + \int E d t ∕ t^{2} - W', & (16 j) \end{array}

where as before the integral is some particular one and $W'$ is a function of $ψ$ only.

10. Small strains

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 $ψ D w$ . We can deduce this case from the above more general results.

To the order considered $q = 1$ so that by equation (10) Section 8 above

\begin{array}{l} \bar{ϕ} = ϖ . \end{array}

We shall use the symbol $ϖ$ rather than $\bar{ϕ}$ for the same reasons explained in Section 7 above as induce us to use $ψ$ rather than $\bar{χ}$ .

Remembering that $ω_{0}$ and $ω$ are now the same and that $ψ$ may be put $= 1$ so that $V ζ ψ D w ψ ζ = V ζ ψ D w ζ = 0$ and therefore $𝜃$ of equation (12) $= 0$ , we have from equation (13),

\begin{array}{l} ϖ ω = ψ D w ω . & (17) \end{array}

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 $m = 1$ and $ρ + η \nabla = \nabla$ so that

\begin{array}{l} δ w = - S δ η_{1} \bar{ϕ} \nabla_{1} = - S δ η_{1} ϖ \nabla_{1}, \end{array}

and therefore by Section 2 above

\begin{array}{l} δ w = - S δ \bar{χ} ζ ϖ ζ = - S δ ψ ζ ϖ ζ . \end{array}

But by Section 3 above

\begin{array}{l} δ w = - S δ ψ ζ ψ D w ζ, \end{array}

and therefore by Section 2 we get equation (17).

18. It is convenient here to slightly change the notation. For $ψ$ , $χ$ we shall now substitute $ψ + 1$ , $χ + 1$ respectively. This leads to no confusion as will be seen.

With this notation the strain being small the stress is linear in $ψ$ i.e. $ψ D w$ is linear and therefore $w$ quadratic. Now for any such quadratic function

\begin{array}{l} w & = - S ψ ζ ψ D w ζ ∕ 2, & (18) \\ for we have by Section 3, \\ d w & = - S d ψ ζ ψ D w ζ . \end{array}

Put now $ψ = n ψ'$ . Then because $ψ D w$ is linear in $ψ$

\begin{array}{l} ψ D w = n ψ D' w, \end{array}

where $ψ D' w$ is put for the value of $ψ D w$ when the coordinates of $ψ'$ are substituted for those of $ψ$ . Thus keeping $ψ'$ constant and varying $n$ ,

\begin{array}{l} d w = - n d n S ψ' ζ ψ D' w ζ, \end{array}

whence integrating from $n = 0$ to $n = 1$ and changing $ψ'$ into $ψ$ we get equation (18).

From equation (18) we see that for small strains we have

\begin{array}{l} w = - S ψ ζ ϖ ζ ∕ 2 . & (19) \end{array}

Now $w$ is quadratic in $ψ$ and therefore also quadratic in $ϖ$ , so that regarding $w$ as a function of $ϖ$ we have as in equation (18)

\begin{array}{l} w = - S ϖ ζ ϖ D w ζ ∕ 2, \end{array}

so that by equation (19) and Section 2 above

\begin{array}{l} ψ = ϖ D w . & (20) \end{array}

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) $w$ as a function of the displacement. Now $w$ is quadratic in $ψ$ , and $ψ$ is linear and symmetrical in $\nabla_{1}$ and $η_{1}$ . In fact from equation (3) Section 7 above, remembering that the $\bar{χ}$ of that equation is our present $ψ + 1$ , we have

\begin{array}{l} 2 ψ ω = - η_{1} S ω \nabla_{1} - \nabla_{1} S ω η_{1} . \end{array}

Therefore we may put

\begin{array}{l} w = w (η_{1}, \nabla_{1}, η_{2}, \nabla_{2}), & (21) \end{array}

where $w (α, β, γ, δ)$ is linear in each of its constituents, is symmetrical in $α$ and $β$ , and again in $γ$ and $δ$ , and is also such that the pair $α$ , $β$ and the pair $γ$ , $δ$ can be interchanged. [This last statement can be made true if not so at ﬁrst, by substituting for $w (α, β, γ, δ)$ , $w (α, β, γ, δ) ∕ 2 + w (γ, δ, α, β) ∕ 2$ 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 $ψ$ , since $ψ$ involves six scalars.

Thus we have the two following expressions for $δ w$ , which we equate

\begin{array}{l} - S δ ψ ζ ϖ ζ & = w (δ η_{1}, \nabla_{1}, η_{2}, \nabla_{2}) + w (η_{1}, \nabla_{1}, δ η_{2}, \nabla_{2}) \\ = 2 w (δ η_{1}, \nabla_{1}, η_{2}, \nabla_{2}), \end{array}

or¹⁹ by Section 2 above,

\begin{array}{l} - S δ η_{1} ϖ \nabla_{1} = 2 w (δ η_{1}, \nabla_{1}, η_{2}, \nabla_{2}) . \end{array}

Now let us put $δ η = ω' S ω ρ$ where $ω'$ and $ω$ are arbitrary constant vectors. We thus get

\begin{array}{l} S ω' ϖ ω = - 2 w (ω', ω, η_{1}, \nabla_{1}) = 2 S ω' ζ w (ζ, ω, η_{1}, \nabla_{1}) . \end{array}

Whence since $ω'$ is quite arbitrary,

\begin{array}{l} ϖ ω = 2 ζ w (ζ, ω, η_{1}, \nabla_{1}) . & (22) \end{array}

The statical problem can now be easily expressed. As we saw in Section 8, equations (7) and (8), it is simply

\begin{array}{l} F + V \nabla M ∕ 2 + ϖ Δ = 0, & (22 a) \end{array}

throughout the mass; and at the surface

\begin{array}{l} F_{S} - V U d Σ M ∕ 2 - ϖ U d Σ = 0, & (22 b) \end{array}

where $F$ , $M$ are the given external force and couple per unit volume and $F_{S}$ is the given external surface traction per unit surface. Substituting for $ϖ$ from equation (22)

\begin{array}{l} \begin{aligned} F + V \nabla M ∕ 2 + 2 ζ w (ζ, Δ, η_{1}, \nabla_{1}) = 0 \\ F_{S} - V U d Σ M ∕ 2 - 2 ζ w (ζ, U d Σ, η_{1}, \nabla_{1}) = 0 . \end{aligned}\} & (23) \end{array}

11. Isotropic Bodies

20. The simplest way to treat these bodies is to consider the (linear) relations between $ϖ$ and $ψ$ .

In the ﬁrst place notice that $ψ$ can always be decomposed into three real elongations (contractions being of course considered as negative elongations). Thus $i$ being the unit vector in the direction of such an elongation,

\begin{array}{l} ψ ω = - Σ e i S i ω . \end{array}

The elongation $- e i S i ω$ will cause a stress symmetrical about the vector $i$ , i.e. a tension $A e$ in the direction of $i$ and a pressure $B e$ in all directions at right angles; $A$ and $B$ being constants (on account of the linear relation between $ϖ$ and $ψ$ ) independent of the direction of $i$ (on account of the isotropy of the solid). This stress may otherwise be described as a tension $(A + B) e$ in the direction of $i$ and a hydrostatic pressure $B e$ . Thus

\begin{array}{l} ϖ ω & = - (A + B) Σ e i S i ω - B ω Σ e \\ = (A + B) ψ ω + B ω S ζ ψ ζ . \end{array}

To obtain the values of $A$ and $B$ in terms of Thomson and Tait’s coeﬃcients $k$ and $n$ of cubical expansion and rigidity respectively; ﬁrst put

\begin{array}{l} ψ ω = e ω & and ϖ ω = 3 k e ω, \\ and then put \\ ψ ω = V λ ω μ & and ϖ ω = 2 n V λ ω μ, \end{array}

$λ$ and $μ$ being any two vectors perpendicular to each other. We thus get

\begin{array}{l} A + B = 2 n, B = - (k - 2 n ∕ 3), \end{array}

whence

\begin{array}{l} ϖ ω = 2 n ψ ω - (k - 2 n ∕ 3) ω S ζ ψ ζ . & (24) \end{array}

From this we have

\begin{array}{l} ψ ω = ϖ ω ∕ 2 n + ω S ζ ψ ζ (k - 2 n ∕ 3) ∕ 2 n, \end{array}

but from the same equation

\begin{matrix} S ζ ϖ ζ = 3 k S ζ ψ ζ; \\ ∴ ψ ω = \frac{1}{2 n} ϖ ω + (\frac{1}{6 n} - \frac{1}{9 k}) ω S ζ ϖ ζ . & (25) \end{matrix}

Equation (24) gives stress in terms of strain and (25) the converse.

21. We can now give the various useful forms of $w$ for isotropic bodies for from equation (19) Section 10,

\begin{array}{l} w = - S ψ ζ ϖ ζ ∕ 2 . \end{array}

Therefore from equations (24) and (25) respectively

\begin{array}{l} w & = - n {(ψ ζ)}^{2} + \frac{1}{2} (k - 2 n ∕ 3) S ζ_{1} ψ ζ_{1} S ζ_{2} ψ ζ_{2}, & (26) \\ w & = - \frac{1}{4 n} {(ϖ ζ)}^{2} - \frac{1}{2} (\frac{1}{6 n} - \frac{1}{9 k}) S ζ_{1} ϖ ζ_{1} S ζ_{2} ϖ ζ_{2} . & (27) \end{array}

Therefore again from Section 2 above and from equation (26),

\begin{array}{l} w = - n S η_{1} ψ \nabla_{1} + \frac{1}{2} (k - 2 n ∕ 3) {(S \nabla η)}^{2} \end{array}

\begin{array}{l} \begin{aligned} 2 w (η_{1}, \nabla_{1}, η_{2}, \nabla_{2}) = n S \nabla_{1} η_{2} S \nabla_{2} η_{1} + n S \nabla_{1} \nabla_{2} S η_{1} η_{2} \\ + (k - 2 n ∕ 3) S \nabla_{1} η_{1} S \nabla_{2} η_{2} . \end{aligned} & (28) \end{array}

Hence from equation (22)

\begin{array}{l} ϖ ω = - n S ω \nabla · η - n \nabla_{1} S η_{1} ω - (m - n) ω S \nabla η, & (29) \end{array}

where $m$ is put for $k + n ∕ 3$ . This last could have been deduced at once from equation (24) by substituting for $ψ ω$ .

Thus the equations (23) for the statical problem are

\begin{array}{l} F + V \nabla M ∕ 2 & = n \nabla^{2} η + m \nabla S \nabla η, & (30) \\ - F_{S} + V U d Σ M ∕ 2 & = n S U d Σ \nabla · η + n \nabla_{1} S η_{1} U d Σ & (31) \\ + (m - n) U d Σ S \nabla η . \end{array}

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 the equation of equilibrium

Particular integral of equation (30)²⁰

22. Since from equation (30) ( $F$ being put for simplicity instead of $F + V \nabla M ∕ 2$ ) we have

\begin{array}{l} n \nabla^{2} η = F - m \nabla S \nabla η \end{array}

we obtain as a particular case by equations (18) and (19) Section 5,

\begin{array}{l} 4 π n η = ∭ u (F - m \nabla S \nabla η) d 𝔰, \end{array}

where $u$ 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 $∭ u \nabla S \nabla η d 𝔰$ as a function of $F$ put in this term $u = - \frac{1}{2} U ρ_{1} \nabla_{1}$ where $ρ$ is taken for the $ρ - ρ'$ of Section 5, and apply equation (9) Section 4. Thus

\begin{array}{l} 4 π n η & = ∭ u F d 𝔰 + \frac{1}{2} m ∭ U ρ_{1} \nabla_{1} \nabla S \nabla η d 𝔰 \\ = ∭ u F d 𝔰 - \frac{1}{2} m ∭ U ρ \nabla^{2} S \nabla η d 𝔰 + a surf. int. \\ = ∭ u F d 𝔰 - \frac{m}{2 (m + n)} ∭ U ρ S \nabla F d 𝔰 + the surf. int. \end{array}

for by equation (30) $S \nabla F = (m + n) \nabla^{2} S \nabla η$ . Now (in order to get rid of any inﬁnite terms due to any discontinuity in $F$ ) apply equation (9) Section 4 to the second volume integral. Thus

\begin{array}{l} 4 π n η = ∭ u F d 𝔰 + \frac{m}{2 (m + n)} ∭ S F \nabla · U ρ d 𝔰 + a surf. int. \end{array}

The surface integral may be neglected as we may thus verify. Call the volume integral $4 π n η'$ . Thus

\begin{matrix} 4 π n \nabla^{2} η' = 4 π F + \frac{m}{2 (m + n)} ∭ S F \nabla · \nabla^{2} U ρ d 𝔰 \\ 4 π m \nabla S \nabla η' = \frac{m}{n} ∭ S F \nabla · \nabla u d 𝔰 + \frac{m^{2}}{2 n (m + n)} ∭ S F \nabla · \nabla S \nabla U ρ d 𝔰, \end{matrix}

so that putting $\nabla U ρ = - 2 u$ we get

\begin{array}{l} n \nabla^{2} η' + m \nabla S \nabla η' \equiv F, \end{array}

whence we have as a particular solution of equation (30) $η = η'$ or

\begin{array}{l} η = \frac{1}{4 π n} ∭ u F d 𝔰 + \frac{m}{8 π n (m + n)} ∭ S F \nabla · U ρ d 𝔰 . & (32) \end{array}

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 $F$ to make these integrals convergent. Thomson and Tait (Nat. Phil. § 730) say that this law is that $F r$ converges to zero at inﬁnity. This I think can be disproved by an example. Put²¹ $F r = r^{- a} λ$ where $λ$ is a constant vector and $a$ 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 $R$

\begin{array}{l} η = \frac{m + 3 n}{3 n (m + n)} \frac{R^{1 - a}}{1 - a} λ . \end{array}

Putting $R = \infty$ , $η$ also becomes $\infty$ . 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 $S F \nabla · U ρ$ in terms of $F$ and $r^{2} S F \nabla · \nabla u$ where $r$ is put for the reciprocal of $u$ . Noting that

\begin{array}{l} \nabla u = - u^{3} ρ, U ρ = u ρ, S F \nabla · ρ = - F, \end{array}

we have at once

\begin{array}{l} S F \nabla · U ρ & = - u F - u^{3} ρ S F ρ \\ S F \nabla · \nabla u & = u^{3} F + 3 u^{5} ρ S F ρ \end{array}

therefore eliminating $ρ S F ρ$ ,

\begin{matrix} S F \nabla · U ρ = - u F 2 ∕ 3 - r^{2} S F \nabla · \nabla u ∕ 3, \\ ∴ η = {24 π n (m + n)}^{- 1} ∭ d 𝔰 {2 (2 m + 3 n) u F - m r^{2} S F \nabla · \nabla u}, & (33) \end{matrix}

which is the required form.

23. Calling the particular solution $η'$ as before and putting

\begin{array}{l} η = η' + η'' \end{array}

the statical problem is reduced to ﬁnding $η''$ to satisfy

\begin{array}{l} n \nabla^{2} η'' + m \nabla S \nabla η'' = 0 \end{array}

and the surface equation either

\begin{array}{l} η' + η'' & = given value, \\ i.e. & η'' & = given value, \\ or & ϖ U d Σ & = given surface traction, \end{array}

i.e. by equation (29) Section 11 above,

\begin{array}{l} n S U d Σ \nabla · η'' + n \nabla_{1} S η'' U d Σ + (m - n) U d Σ S \nabla η'' = known value. \end{array}

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.

12. Orthogonal coordinates

24. It is usual to ﬁnd what equations (22 $a$ ) 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 $x$ , $y$ , $z$ be any orthogonal coordinates, i.e. let $x = const.$ , $y = const.$ , $z = const.$ , represent three families of surfaces cutting everywhere at right angles. Particular cases are of course the ordinary Cartesian coordinates, the spherical coordinates $r$ , $𝜃$ , $ϕ$ and the cylindrical coordinates $r$ , $ϕ$ , $z$ . Let $D_{x}$ , $D_{y}$ , $D_{z}$ stand for diﬀerentiations per unit length perpendicular to the three coordinate surfaces and let $λ$ , $μ$ , $ν$ be the unit vectors in the corresponding directions. Thus

\begin{array}{l} \nabla = λ D_{x} + μ D_{y} + ν D_{z} . \end{array}

Thus, using the same system of suﬃxes for the $D$ ’s as was explained in connection with $\nabla$ in Section 1,

\begin{array}{l} ϕ Δ = D_{x 1} ϕ_{1} λ + D_{y 1} ϕ_{1} μ + D_{z 1} ϕ_{1} ν, & (34) \end{array}

\begin{array}{l} ϕ Δ = D_{x} (ϕ λ) + D_{y} (ϕ μ) + D_{z} (ϕ ν) - ϕ (D_{x} λ + D_{y} μ + D_{z} ν) . & (35) \end{array}

25. Now to put equation (22 $a$ ) Section 10 into the present coordinates all that is required is to express $ϖ Δ$ in terms of those coordinates. Let the coordinates of $ϖ$ be $P Q R S T U$ . Thus from equation (35) we have

\begin{array}{l} \begin{aligned} ϖ Δ & = D_{x} (P λ + U μ + T ν) + D_{y} (U λ + Q μ + S ν) + D_{z} (T λ + S μ + R ν) \\ - ϖ (D_{x} λ + D_{y} μ + D_{z} ν) . \end{aligned} \end{array}

The ﬁrst thing then is to ﬁnd $D_{x} λ$ , $D_{x} μ$ , $D_{y} λ$ &c.

Let $p_{2}$ , $p_{3}$ be the principal curvatures normal to $x = const.$ , i.e. (by a well-known property of orthogonal surfaces) the curvatures along the lines of intersection of $x = const.$ , with $z = const.$ , and $y = const.$ $p_{2}$ , $p_{3}$ will be considered positive when²² the positive value of $d x$ is on the convex side of the corresponding curvatures. Similarly for $q_{3} q_{1} r_{1} r_{2}$ . Thus for the coordinates $r$ , $𝜃$ , $ϕ$ ;

\begin{array}{l} p_{2} = p_{3} = 1 ∕ r, q_{3} = cot 𝜃 ∕ r, q_{1} = r_{1} = r_{2} = 0 . \end{array}

Again for $r$ , $ϕ$ , $z$ ; $p_{2} = 1 ∕ r$ and the rest are each zero.

With these deﬁnitions we see geometrically that

\begin{array}{l} D_{x} λ = - μ q_{1} - ν r_{1}, D_{x} μ = λ q_{1}, D_{x} ν = λ r_{1} . & (36) \end{array}

Similarly for $D_{y} λ$ , $D_{y} μ$ , $D_{z} λ$ &c. Thus

\begin{array}{l} ϖ Δ & = λ (D_{x} P + D_{y} U + D_{z} T) + μ () + ν () \\ + λ (- Q p_{2} - R p_{3} + T r_{1} + U q_{1}) + μ () + ν () \\ + λ {(p_{2} + p_{3}) P + (q_{3} + q_{1}) U + (r_{1} + r_{2}) T} + μ {} + ν {}, \end{array}

\begin{array}{l} ϖ Δ & = λ {D_{x} P + D_{y} U + D_{z} T + P (p_{2} + p_{3}) - Q p_{2} - R p_{3} \\ + T (2 r_{1} + r_{2}) + U (q_{3} + 2 q_{1})} + μ {} + ν {} . & (37) \end{array}

26. The other chieﬂy useful thing in transformation of coordinates in the present subject is the expression for the strain function $ψ$ in terms of the coordinates of displacement. Let $u$ , $v$ , $w$ be these coordinates. Now by equation (3) Section 7, remembering (Section 10) that $ψ = \bar{χ} - 1$ we have

\begin{array}{l} - 2 ψ ω = S ω \nabla · η + \nabla_{1} S ω η_{1}, \end{array}

\begin{array}{l} whence & - 2 ψ λ = λ S λ D_{x} η + μ S λ D_{y} η + ν S λ D_{z} η - D_{x} η . \end{array}

\begin{array}{l} But & D_{x} η & = λ D_{x} u + μ D_{x} v + ν D_{x} w + u D_{x} λ + v D_{x} μ + w D_{x} ν \\ = λ (D_{x} u + v q_{1} + w r_{1}) + μ (D_{x} v - u q_{1}) + ν (D_{x} w - u r_{1}) . \end{array}

Similarly

\begin{array}{l} D_{y} η & = λ (D_{y} u - v p_{2}) + μ (D_{y} v + w r_{2} + u p_{2}) + ν (D_{y} w - v r_{2}) \\ D_{z} η & = λ (D_{z} u - w p_{3}) + μ (D_{z} v - w q_{3}) + ν (D_{z} w + u p_{3} + v q_{3}) \\ ∴ 2 ψ λ & = 2 λ (D_{x} u + v q_{1} + w r_{1}) + μ (D_{y} u + D_{x} v - u q_{1} - v p_{2}) \\ + ν (D_{z} u + D_{x} w - u r_{1} - w p_{3}) . & (38) \end{array}

But with Thomson and Tait’s notation for pure small strain

\begin{array}{l} 2 ψ λ = 2 λ e + μ c + ν b, \end{array}

\begin{array}{l} \begin{aligned} ∴ \end{aligned}\} \end{array}

²³e = Dxu + vq1 + wr1 f = Dyv + wr2 + up2 g = Dzw + up3 + vq3 a = Dyw + Dzv − wq3 − vr2 b = Dzu + Dxw − ur1 − wp3 c = Dxv + Dyu − vp2 − uq1. (39)

Thus we have $e f g a b c$ in terms of the displacement and we have already in equation (24) Section 11, which expresses $ϖ$ in terms of $ψ$ , found the values of $P Q R S T U$ in terms of $e$ , &c. Finally the expression in equation (37) for $ϖ Δ$ gives us the equations of equilibrium in terms of $P$ , &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

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 $r, ϕ, z$ as our coordinates, the axis of $z$ being parallel to the generating lines of the cylinder. Let $λ, μ, ν$ be the unit vectors in the directions of $d r, d ϕ, d z$ respectively and let

\begin{array}{l} η = u λ + v μ + w ν \end{array}

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 $z = 0$ ﬁxed let us give the cylinder a small torsion of magnitude $τ$ , i.e. let us put

\begin{array}{l} η = τ z r μ, & (40) \end{array}

for all points for which $τ z$ is small.

The practical manipulation of such expressions as this is almost always facilitated by considering the general value of $Q (\nabla_{1}, η_{1})$ where $Q$ is any function linear in each of its constituents. Thus in the present case

\begin{array}{l} Q (\nabla_{1}, η_{1}) = τ {z Q (λ, μ) - z Q (μ, λ) + r Q (ν, μ)} . \end{array}

[If $Q$ is symmetrical in its constituents, e.g. in the case of stress below this reduces to the simple form $Q (\nabla_{1}, η_{1}) = τ r Q (ν, μ)$ .] From this we at once see that $η$ satisﬁes the equation of internal equilibrium

\begin{array}{l} n \nabla^{2} η + m \nabla S \nabla η = 0, \end{array}

for putting $Q (α, β) = α β$

\begin{array}{l} \nabla η = τ (2 z ν - r λ) = τ \nabla (z^{2} - r^{2} ∕ 2) \end{array}

so that both $S \nabla η$ and $\nabla^{2} η = 0$ .

Again the value for $Q$ at once gives us the stress for

\begin{array}{l} ϖ ω = - n S ω \nabla & · η - n \nabla_{1} S ω η_{1} - (m - n) ω S \nabla η, \\ or & ϖ ω & = - n τ r (μ S ω ν + ν S ω μ), & (41) \end{array}

which is a shearing stress $n τ r$ on the interfaces perpendicular to $μ$ and $ν$ .

Putting $ω =$ the unit normal of the curved surface we have for the surface traction

\begin{array}{l} ϖ ω = - n τ r ν S ω μ . \end{array}

In the ﬁgure let the plane of the paper be $z = 0$ , $O$ the origin, $P$ a point on the curved surface and $O M$ the perpendicular from $O$ on the tangent at $P$ . Thus $- r S ω μ = O P cos O P M = P M$ , $P M$ being reckoned positive or negative according as it is in the positive or negative direction of rotation round $O z$ . Thus we see that the surface traction is parallel to $O z$ and $= n τ P M$ .

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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 $ϖ ν = n τ r μ$ by equation (41) so that its moment round the origin is $n τ \iint r^{2} d A$ , where $d A$ is an element of area and the integral extends over the whole cross-section.

28. Let us now assume a further displacement

\begin{array}{l} η = w ν, & (42) \end{array}

where $w$ is a function of $r, ϕ$ only, and let us try to determine $w$ so that there is still internal equilibrium and so that the stress on the curved surface due to $w$ shall neutralise the surface traction already considered.

In the present case

\begin{array}{l} Q (\nabla_{1}, η_{1}) = Q (\nabla w, ν) . \end{array}

Thus $S \nabla η = 0$ (since $ν$ is perpendicular to $\nabla w$ ) and therefore the equation of internal equilibrium gives

\begin{array}{l} \nabla^{2} w = 0 . \end{array}

\begin{array}{l} Again & ϖ ω & = - n S ω \nabla · η - n \nabla_{1} S ω η_{1}, \\ or & ϖ ω & = - n (ν S ω \nabla w + \nabla w S ω ν), & (43) \end{array}

a shear

= n T \nabla w

on the interfaces perpendicular to

\nabla w

and

ν

. Thus putting

ω

for the unit normal to the curved surface, the present surface traction will neutralize the former if

\begin{array}{l} S ω \nabla w & = - τ r S ω μ = - τ S ω ν (λ r) \\ = \frac{1}{2} τ S ν ω \nabla (r^{2}), \\ i.e. & d w & ∕ d n = d (τ r^{2} ∕ 2) ∕ d s, \end{array}

where $d ∕ d n$ represents diﬀerentiation along the normal outwards and $d ∕ d s$ 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 $ϖ ν = n \nabla w$ by equation (43), and therefore that the total couple is equal to $- n \iint r S μ \nabla w d A = n \iint (d w ∕ d ϕ) d A$ . This leads to the usual expression for torsional rigidity.

13. Wires

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 $s$ 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 $q$ ; the axis of $q$ being the axis of rotation, and the angle of $q$ , half the angle turned through.

$ω$ is taken so that the rate of this turning per unit length of the wire is $q ω q^{- 1}$ so that $ω$ 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 $ω$ is a function of $q$ and its derivative with reference to $s$ . This function we shall investigate later. The resolved part of $q ω q^{- 1}$ 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 $ω$ is the vector whose coordinates are the $κ, λ, τ$ of Thomson and Tait’s Nat. Phil. § 593. When we are given $q$ or $ω$ for every point we know the strain of the wire completely. $ω_{0}$ is deﬁned as the naturally normal value of $ω$ , i.e. the value of $ω$ when the wire is unstressed.

As usual we take $ρ$ as the coordinate vector of any point of the wire, $ρ$ like the rest of the functions being considered as a function of $s$ . We shall denote (after Tait, Quaternions, Chap. ix.) diﬀerentiations with regard to $s$ by dashes.

We now come to the dynamical symbols. $F$ and $M$ 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 $X, L$ 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 $δ ω$ in $ω$ and an increment $δ e$ in the elongation at any point. Then the work done on the element $d s$ by the stress-force $= - δ e S F ρ' d s$ and that done by the stress-couple $= - S q δ ω q^{- 1} M d s$ . If (as we assume, though the assumption is not justiﬁed in some useful applications of the general theory of wires) $F$ and $M$ to be of the same order of magnitude the former of these expressions can be neglected in comparison with the latter for $δ e$ is a quantity small compared with $δ ω$ . Now the work done on the element by the stress $=$ the increment of the element’s potential energy $= δ w d s$ where $w$ is some function of the strain. Hence

\begin{array}{l} δ w & = - S δ ω q^{- 1} M q . \\ Thus w is a function of ω only and \\ ∴ δ w & = - S δ ω ω \nabla w, \\ whence we see that \\ q^{- 1} M q & = ω \nabla w . & (46) \end{array}

Notice that $M$ is Thomson and Tait’s $ξ$ , $η$ , $ζ$ and $q^{- 1} M q$ their $K L M$ (Nat. Phil. §§ 594, 614).

31. Now since the strain is small, $q^{- 1} M q$ is linear in terms of the strain and therefore in terms of $ω$ . Hence we see that $w$ is quadratic in terms of $ω$ . Let us then put

\begin{array}{l} w = w_{2} (ω - ω_{0}, ω - ω_{0}) & + w_{1} (ω - ω_{0}) \\ + a function of temperature only, \end{array}

$w_{2}$ and $w_{1}$ being linear and homogeneous in each of their constituents. This is the most general quadratic function of $ω$ . Now

\begin{array}{l} ω \nabla w = ζ w_{2} (ζ, ω - ω_{0}) + ζ w_{2} (ω - ω_{0}, ζ) + ζ w_{1} ζ . \end{array}

Putting then $ω = ω_{0}$ and $ω \nabla w = 0$ we get $ζ w_{1} ζ = 0$ . Operating on the last by $S σ ()$ where $σ$ is any vector we see that $w_{1} = 0$ . Thus putting

\begin{array}{l} ω \nabla w = ϕ (ω - ω_{0}), \end{array}

where $ϕ$ has the value given by the last equation and is therefore self-conjugate we get the two following equations

\begin{array}{l} w = - S (ω - ω_{0}) ϕ (ω - ω_{0}) ∕ 2 + w_{0}, & (47) \end{array}

[as can be seen by a comparison of the last three equations $w_{0}$ being the function of the temperature] and

\begin{array}{l} q^{- 1} M q = ϕ (ω - ω_{0}) . & (48) \end{array}

When the natural shape of the wire is straight these become

\begin{matrix} w = - S ω ϕ ω ∕ 2 + w_{0}, & (49) \\ q^{- 1} M q = ϕ ω, & (50) \end{matrix}

and when further the wire is truly uniform $ϕ$ and $w_{0}$ 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 $ω$ , a moment of momentum $= ϕ ω$ where $ϕ$ has the meaning just given. If the rigid body be made to take a ﬁnite rotation $q () q^{- 1}$ and then to move with angular velocity $q ω q^{- 1}$ its moment of momentum will be $q ϕ ω q^{- 1}$ . 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 $s$ the rigid body shall have made the rotation represented by $q$ (Section 13 above). Thus by the deﬁnition of $ω$ and $q$ its angular velocity at any instant is $q ω q^{- 1}$ and its moment of momentum therefore $q ϕ ω q^{- 1}$ or $M$ .

Now consider the equilibrium of the wire when no external force or couple acts except at its ends. In this case $F$ is constant throughout and it is easy to see (what indeed is a particular case of equation (53) below) that

\begin{array}{l} M' + V ρ' F = 0 . & (51) \end{array}

Interpreting this equation for our rigid body we get as the law which governs its motion

\begin{array}{l} d (vect. mom. of mom.) ∕ d t = - V ρ' F . \end{array}

Thus the rigid body will move as if acted upon by a constant force $F$ at the end of the unit vector $ρ'$ 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 $d s$ are with the notation explained in Section 13 above

\begin{matrix} d F + X d s = 0, \\ V d ρ F + d M + L d s = 0, \end{matrix}

or dividing by $d s$

\begin{matrix} F' + X = 0, & (52) \\ V ρ' F + M' + L = 0 . & (53) \end{matrix}

Operating on the last equation by $V ρ' ()$ noting that ${ρ'}^{2} = - 1$ and putting $S ρ' F = - T$ we get

\begin{array}{l} F = ρ' T + V ρ' (M' + L), & (54) \end{array}

whence by equations (52) and (53) respectively

\begin{matrix} X + d {ρ' T + V ρ' (M' + L)} ∕ d s = 0, & (55) \\ S ρ' (M' + L) = 0 . & (56) \end{matrix}

Now by equation (48) above

\begin{array}{l} q^{- 1} M q = ϕ (ω - ω_{0}) . & (48) \end{array}

Also by the deﬁnition of $q$

\begin{array}{l} ρ' = q λ q^{- 1}, & (57) \end{array}

where $λ$ is some given constant unit vector. Finally as we are about to prove

\begin{array}{l} ω = 2 V q^{- 1} q' . & (58) \end{array}

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 $ρ$ , $T$ , $M$ , $q$ and $ω$ . 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 $T$ and $q$ in terms of which all the other unknowns are explicitly given.

To prove equation (58)²⁴ observe that

\begin{array}{l} (q + d q) σ {(q + d q)}^{- 1} = q (σ + V ω σ d s) q^{- 1}, \end{array}

where $σ$ is any vector. The truth of this is seen by noticing that $(q + d q) () {(q + d q)}^{- 1}$ is the operator that rotates any vector of the element $s + d s$ 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 $ω d s$ at the given element and then performing the strain of the wire up to the point $s$ . The ﬁrst process is represented on the left of the last equation and the second on the right. Thus we get

\begin{array}{l} q^{- 1} d q σ & + σ d (q^{- 1}) · q = V ω σ d s, \\ or ∵ & d (q^{- 1}) & = - q^{- 1} d q q^{- 1}, \\ q^{- 1} q' σ & - σ q^{- 1} q' = V ω σ, \\ whence & ω & = 2 V q^{- 1} q' . \end{array}

Returning to equations (55) to (58) observe that equations (57) and (58) give $ρ'$ and $ω$ as explicit functions of $q$ . Hence by equation (48)

\begin{array}{l} M = q ϕ (2 V q^{- 1} q' - ω_{0}) q^{- 1}, & (59) \end{array}

which gives $M$ also explicitly. Substituting for $M$ and $ρ'$ in equations (55) and (56) we have

\begin{matrix} X + \frac{d}{d s} \{q λ q^{- 1} T + V q λ q^{- 1} (L + \frac{d}{d s} [q ϕ (2 V q^{- 1} q' - ω_{0}) q^{- 1}])\} = 0, & (60) \\ S q λ q^{- 1} (L + \frac{d}{d s} [q ϕ (2 V q^{- 1} q' - ω_{0}) q^{- 1}]) = 0, & (61) \end{matrix}

which are suﬃcient equations to determine $T$ and $q$ , whereupon $M$ is given by equation (59), $ω$ by equation (58) and $ρ'$ and therefore also $ρ$ 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.

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IIIElastic Solids13