Saint V enant1 ingeniously side-stepped the major difficulty of the problem by substituting a simplified problem. Of the six com ponents of stress (the unknowns of
Section 2 gives the general m athem atical statem ent of the problem of Saint Venant, and of the relaxed problems associated with it. Section 3 contains an old
friend in new clothes— the Saint Venant solution in tensor notation. I believe that is worth while to give such a presentation, avoiding as far as possible the customary special choices of axes and particular integrals, in order to reveal the true m athe
matical structure of the solution. In Section 4 we pass beyond the Saint Venant solu
tion, and set up the basic eigen-value problem associated with the exponential type of solution. In Section 5 we link this formulation with the solution given by Dougall for a cylinder of circular section. Section 6 contains some questions.
2. T he problem of Saint Venant and the relaxed problems. Latin suffixes will have the range 0, 1, 2, and Greek suffixes the range 1, 2, with the usual sum m ation conven
tion for repeated suffixes. Let Xi be rectangular Cartesian coordinates, with the axis of x 0 in the direction of the generators of the cylinder. T he position of the origin and the directions of the axes of x x and x2 remain arbitrary. L et u ( be the displacem ent and E ij ( = E n ) the reduced stress, i.e., the stress divided by Y oung’s modulus. W e have the basic stress-strain relations
K u i,i + « • ,; ) = (1 + ° ) E i j — a S ijE k k , ( 2 . 1 )
or equivalently
2 ( 1 + tr)(l — 2cr)E{j = 2crSi,Uk,k + ( 1 — 2cr)(w/,i -f- (2.2) Here a is Poisson’s ratio (a constant which may, in theory, take any value in the range
— 1< (7< j ) , and the comma denotes partial differentiation ( / , i = d //d x ,) ; 5,-,- is the Kronecker delta.
In any problem in elasticity, we have a choice between two m ethods: we can work w ith the displacem ent «,• or with the stress E ij. W hen the boundary conditions are given in terms of stress (as they are in the problem of Saint V enant), the relative ad
vantages m ay be set down as follows:
Displacem ent m ethod: Simple p.d.e. and complicated b.c.
Stress m ethod: Complicated p.d.e. and simple b.c.
Here, and throughout, “p .d .e.” means “partial differential equations,” and “b .c.”
means “boundary conditions.”
T he two rival formulations of the problem of Saint Venant are set ou t below in (2.3) and (2.4). In each case, (a) contains the p.d.e., (b) contains the b.c. on the free sides (na are the direction cosines of the normal), and (c) contains the b.c. on the ends. To, T aare the com ponents of the assigned stress. T he symbol A 3 is the 3 - d i m e n
sional Laplacian differential operator (A 3 = d 2/ d x j d x j ) . S a i n t V enant problem i n terms o f displacem ent:
(1 — 2cr)Ajw,’ -f- tij,ji — 0; (2.3a) («3.0 + uo,p)n$ — 0, 2 out.ktia + (1 — 2 o - ) ( t t i3 , „ + u a,f)n$ = 0; (2.3b) auk.k + (1 — 2a)Mo.o = (1 "k <r)(l — 2a)To, «o.a T «a.o— 2(1 -f- a)T a. (2.3c)
S a i n t V enant problem i n terms o f stress: solutions m ay be superimposed. We m ay hope that, by superimposing solutions of the relaxed problem, we m ay succeed in satisfying the b.c. (c), either accurately or ap
proximately.
In spite of the underdetermination, the relaxed problem is too com plicated to yield to mere guessing, except for one very simple solution: £ 0o = const., E0a =0, -Ea 0 = O. problem w ith (2.5) imposed, and transitory free modes for solutions of the relaxed prob
lem with (2.6) imposed. (This terminology is suggested by the theory of vibrations,
mental. In the following treatm ent, stress is used and the axes OxiXi remain com pletely general.
W e substitute the auxiliary conditions (2.5) in (2.4a, b), and obtain
jEoo.o + Eop,p = 0 , (1 + <t) A 3E 00 -+- Eoo.oo — 0 , ' Ea 0,0 — 0, (1 + Cr)A3£ao + Eoo.aO = 0,
Eoo.ap — 0 ;
(3.1a) (3.1b) Eopnp = 0.
It follows at once th at Ea0 are independent of *0, and that
Eoo = — Xo(ApXp + A ) + Bpxp + B, (3.2) where Ap, A , Bp, B are six constants, at present arbitrary. T he remaining equations in (3.1a) are equivalent to
A E ao = (1 + cr)~lAa, Epo.p = Apxp + A , (3.3) where A is the plane Laplacian- (A = d-/dxpdxp). T hese three equations are to be solved for the two unknowns E a0, with the b.c. (3.1b). T he problem is a plane problem, the domain being the section of the cylinder.
Given the vector E a0, there exist invariants <f>, \p, such that
E ao = <f>,a — tat'f'.l (€11 = 622 = 0| e12 = ~ «21 = 1). (3.4) Substitution in (3.3) leads at once to
A <p = ApXp + A , (1 + <r)Ap — trepyA p x 7 + C, (3.5a) where C is another arbitrary constant. (N ote that eapeay — Spy.) The b.c. is
4>,pUp — epynpp,y = 0 or d f / d n — d f / d s = 0, (3.5b) where d n is a elem ent of the normal n a, drawn out of the material and to the right of the elem ent ds of the bounding curve (Fig. 2). On integrating (3.5b) around the bounding curve, and using (3.5a), we find that
A = — Apxp, (3.6) where xp are the coordinates of the centroid of the section. T hus there are only six arbitrary constants, Ap, Bp, B , C.
Remembering the Riemann-Cau- chy relations, we note th at <j>, rp are indeterm inate to the extent of adding to (p+iip an arbitrary analytic func
tion of X i+ ixo.
Let <£(1>, ypa) be any single-valued particular solutions of (3.5a). L et us define <f>, Sk by
$ = ^ - ¿U), <fr = p ~ (3.7)
so th at 3>, Sk are harmonic. L et 12 be the harmonic conjugate of 4 r, so that
H .a ~ “ ^ay^P ,y ^a't}p \y '' ( 3 * 8 )
If P is defined by
P = 4 - Q, (3.9)
then (3.4) reads
E ao = P.a + 4 $ - t a j f f , (3.10)
and the b.c. (3.5b) reads
d P / d n = - d<j>w / d n + # < » /ch . (3.11) Thus the relaxed problem of Saint Venant reduces to a problem of Neumann— to find a harmonic function P under the b.c. (3.11). T he solution m ust be single-valued in order that the displacem ent m ay be single-valued.
This presentation of the problem has a fictitious sim plicity; the constants Ap, C are wrapped up in the Neum ann problem. W e m ay split up the problem into problems involving only the geom etry of the section and Poisson’s ratio. T o do this, we write
(3.12) P = A p P p + (1 + <r) lCQ,
<j>w = A ^ \ V l) = + (1 + * )-'C x k>, Here <j>p\ \ p p \ x (1) are single-valued particular solutions of
A4>P = x fi — X p , (1 - f <r)A\ p p = a t p y X y , A x 0 ) = 1, ( 3 . 1 3 )
and Pp, Q are single-valued harmonic functions satisfying the b.c.
d P p /d n = - d<t>P/dn - f d f P / d s , dQ /dn = f c (l)/ ^ - ( 3 - 1 4 )
The determination of the vector P p constitutes the f l e x u r e p r o b l e m , and the determ i
nation of Q the t o r s i o n p r o b l e m .
Since A ( r 2x „ ) = 8 x „ , A ( r 2) = 4 , where r 2 = x p x p , particular solutions of ( 3 . 1 3 ) are given by
<j>P = r 2x p / 8 — r 2x p / 4 ,
w ( 3 . 1 5 )
(1 + c ) p P = c rr2e p y X y / 8 , x (1) = r2/ 4 .
T he determination of the displacem ent in terms of ( 3 .2 ) and ( 3 . 1 0 ) in this general notation is interesting, because it reveals w hy the com ponents u a are independent of the solution of the Neum ann problem ( 3 . 1 1 ) . Using ( 2 . 5 ) in ( 2 . 1 ) , we have
«0 , 0 = P oo, « a, 0 + Mo,a = 2 ( 1 + tr)PaO,
( 3 . 1 6 ) up. a + «0,3 — — 2<rdapEoo.
B y ( 3 . 2 ) , the first two equations give at once
«o = ~ i%o(ApXp + A ) + x 0(Bpxp + B ) + fo,
« „ = $ x l A a - i x 0B a — X ofo.a + 2 ( 1 + <r)x0E a0 + / „ ,
where f o , f a are unknown functions of X i, x t . When we substitute this expression for u a
in the last of ( 3 . 1 6 ) , w e get
fo .a p — (1 + tr)(P oO ,3 + P/SO,a) — c d a p ^ A y X y -f- A ) , ( 3 . 1 8 )
fp .a Ar fa ,P = — 2 trd a p (B yX y + B ) . ( 3 . 1 9 )
N o w , b y ( 3 . 1 0 )
E a0,ß — Eß0,a = — tay^ß + « 0 7 ^ 1 = ~ Caß&'P™, (3.20) and so, since ip a) satisfies (3.5a),
(1 “1” E )(^E a 0,ß E ß 0,a) £aß(.G£pvA ^Xy "T C ) . (3.21) W e now use this equation to sbustitute for Eß0,ain (3.18). When we do so, the linear • expression on the right m ust be a partial derivative with respect to Xß. N ow if Faß . . . is a single-valued homogeneous function of degree n in x u x it it follows from Euler’s theorem that
where K is a constant. Hence we deduce from (3.18)
fo,a — 2(1 -f- <r)Eao + eapXp^ae^A^x, + C) — <rxa(§AyXy + A ) -f- ha, (3.23) where h a is constant. N ow substitute for E a0 from (3.10) and integrate; this gives
where h is constant. B y virtue of the p.d.e. (3.5a) satisfied by i¡/w , this integral has the same value for reconcilable paths. B ut we have no guarantee th at it has the same value for irreconcilable paths in the case of a m ultiply connected section. To secure a single valued displacement, we should choose for \ a function which satisfies the p.d.e. (3.5a) throughout the whole interior of the outer boundary. W e may, for ex
ample, use the expression given in (3.12), (3.15).
T he equations (3.19) determine f a to within a plane rigid body displacem ent, and so
where k a, k are constants.
T o find the displacement, we substitute for / 0 from (3.24) in the first of (3.17), in (3.24). T o find we substitute in the second of (3.17). There are two substitutions, one for fq,a — 2 (l+ < r )£ Q0 from (3.23), and the other for/„ from (3.25). W e see that P does not occur, and thus we verify the well known fact that the lateral displacem ent is independent of the solution of the Neum ann problem.
4. T he exponential type of solution. L et us now investigate the solution of the relaxed problem with the auxiliary condition (2.6). Since stress determines displace
m ent to within a rigid body displacem ent (which we shall om it), we m ay write for the corresponding displacem ent
(3.22)
/o = 2(1 + o ) P + 2(1 + + f [ — 2(1 -T <r)e«7^ + eaßXß(j<refl^A)1x, + C)
(3.24)
— crxa(%AyXy + A ) \d x a + haxa — h.
f a — <r{hBar* — x aByXy — B x a) + keayx 7 + k a, (3.25)
and so obtain u a. It involves the solution of the Neumann problem, since P appears
Ui = ekzi>Vi(xi, xf). (4.1)
Once more we have a choice between two methods— displacem ent and stress—
and we shall choose displacement.
When we substitute from (4.1) in (2.3a, b), the equations reduce to
(A + V ) V = 0, (A + k*)va = - V , a, (4.2a) 2 <rVna + (vp,a + Va,e)ng = 0, (1 — 2<r)V,pnp + khptip — v p jyn y = 0. (4.2b) Here V is an auxiliary variable, given by
(1 - 2 a )V = h 0 + v m; (4.3)
it is the dilatation, to within a factor.
Inspection of (4.2) shows that we have before us an eig en va lu e problem of con
siderable com plexity. T he system will (presumably) be consistent only for certain values of k. There is no objection to complex eigen-values, with complex solutions for V, v„, and the corresponding stress. D enoting complex conjugates by bars, we should take in such cases for the real displacement and stress
+ eixov ,■), i ( e kX0F<i + ekziFi,). (4.4) If k is an eigen-value of (4.2), so also are — k and ± k . In fact, the eigen-values occur in sets of two if they are real or purely imaginary, and in sets of four if complex.
B y a simple and ingenious argument, D ougall2 has shown that the system (4.2) has no purely imaginary eigen-values. A purely imaginary k implies a periodic dis
tribution of displacem ent and stress. Consider the energy in a length of cylinder equal to this period. It is equal to the work done by the terminal stress in passing from the natural state to the strained state. But, from the periodicity, this is zero. Hence, the energy of a strained state is zero, which is contrary to a basic postulate of elasticity.
Hence there can he no p u re ly i m a g in a r y eigen-value k. It should be added that we can
not assert this if a is arbitrary. It is necessarily true only if strain-energy is positive- definite, i.e., if — 1 < a < J.
In m any problems in applied m athematics, harmonic functions play a funda
mental role. In our problem (4.2) that role is taken over by plane wave functions, where by a w ave fu n ctio n /(x i, x^) we mean a solution of (A + ^ 2) / = 0 . We note th at V is a w ave function, but va is not.
We can, however, easily reduce the unknowns to wave functions by writing
va = w a + w*a, (4.5)
where w* is any particular solution of
(A + k 2)w* = — V , a. (4.6)
Let us not tie ourselves down to any definite particular solution. Two, however, seem to be particularly interesting:
w t = - \ x aV, (4.7)
w*a = - W , „ , (A + k*)W = V. (4.8)
Our problem (4.2) m ay now be stated as follows:
(A + ¿2) F = 0, (A + k*)wa = 0, (4.9a) 2 <xVna + (v/fa + + (wp,a + W a ^ n ? = 0,
( 4 .9b)
(1 — 2 a ) V t$np - f k 2w p i p — W ^ y n y + k 2Wpn$ — w ^,g y n y = 0 .
In sim plifying the p.d.e., we have com plicated the b.c.
In dealing with Saint V enant’s solution, we found it advisable to pass in (3.4)
* / 32 d \ ( d d d \
"v + - W + * + • * ) ♦ - ( * s - * w * - *■
* * / d a a \ / a 2 a \
,
+ Wp.cdtanp + ( K — )<£ — ( —- + I* 2 + K— = 0, (4.17b)
\35 3» ds/ \ 3 s 2 3?«/
3F * * d<f> d ip
(1 — 2 < r) H k 2w p n p — Wpjytt-, + 2&2 k 2 — = 0 .
3» 3» 3i
Here is any particular solution of (4.6).
If we choose w * as in (4.7), we m ay substitute in (4.1 7b ):
* 1 . 1 3 3F
aV+u>a,0nan p = (1 —2<r)F--- (r2) --->
2 4 3« 3«
1 * * 1 3 3F 1 3 3F ,
— (w?,a+ w a,p)tan p = - — — (r2) - — — (r3) —— , (4.18)
2 8 3s dn 8 an ds
3F * * 3 F 1 3 1 3
(1 — 2 c r) (}7«t = 2 ( 1 — < r)---¿ 2F — (r2)4--- (x jsF .g ).
3« 3« 4 dn 2 dn
On the other hand, if we choose w * as in (4.8), we m ay substitute for these three quantities, respectively,
/ 3 3 3 \ 3F 3 IF
trF - d H V / d n 2, - ( ---k — ) W , 2(1 - <r) 2 k 2 — ■ (4.19)
\3 s 3w 35/ dn dn
If w e choose w * as in (4.7), we have the following expressions for the displacem ent in terms of F, <p, i/', k:
« 0 = e t l ! l » 0 , W a = e kzl>V a ,
kv0 = 2(1 - <r)F + fx « F ,CT + ¿V , (4.20) Va = — iX a V + 4>.a ~
It takes only a mom ent to verify th at these expressions satisfy (4.2a). T hese formulas (in a slightly more general form) are the key to D ougall’s treatm ent of the circular cylinder.2 H owever, he gives no indication as to how to he obtained them , or whether th ey are a general representation of the exponential typ e of displacement. I have shown that th ey are in fact a general representation. Given and k, then V, cf>, \J/
are uniquely determined.
T he stress corresponding to (4.20) is
E i , - = e kxoF i j , F a o ~ ~ k ~ lF a p j , F o o = k ~2F a $ , a f>,
2(lT<r)2?a/3= — ( 1 -2<j)5ap V — %(xaV ,0+ XpV ,a) + 2<j> ,ap— (iaW',70 + eS7'A.Ta)>
2 k ( 1 + tr)Fo0 = 2 (1 — <r) V, „ + K ^ F .i) .« — § k2x aV + 2 k 24>,« — k 2e ,B, 2(l+ < r)i,oo = 2 (2 -< r )F + x aF ,a+ 2 A ^ .
5. T he case of the circular section. T he circular section has been dealt w ith so thoroughly by D ougall2 th at there is little more to be said about it. However, it is
( 5 . 4 )
interesting to see how his method of solution connects up with the preceding ap
proach.
Let the boundary be r = a, and let w be the polar angle. Then the following are wave functions
(F, <*>, <A) = ( A , B, C ) J m( k r ) e (5.1) where A , B , C are any constants. Each of these functions satisfies on r = a the equa
tions
d f _ K f d f _ i m f d d f _ i m K f
dn a ds a ds dn a2 (5.2)
k = $ J U Q / M Q , £ = ka.
T hey also satisfy for r = a ,
— (xfsf.ii) = ( m2 - ? ) f / a . ( 5 . 3 ) a n
Substitution in (4.18) and (4.17b) gives
L i o V a 2 Ln<j) -T L a f = 0 ,
Too = §(1 — 2<r) -(- \ K , Lai — — ?w2 T i2 T K , Z.02 = i m ( K — 1), L01 = — m /4 , L n = i m ( K — 1), L u = m2 — — K ,
L i0 = (1 - c ) K + W ~ = K e , L a = - |m ? 2.
The characteristic equation is
I ¿«1 = 0, (5.5)
which is essentially the determinantal equation of Dougall for the determination of £, and hence k.
For modes independent of co, we put m = 0. Then (5.5) reduces to
(£2 + 2K ) [£2(t f2 + ? ) - 2 K \ l - a) ] = 0, (5 .6) and we get the two characteristic equations
/*(€) = 0, » ( 4 + To2) = 2(1 - <r)Ji2. (5.7) T he first equation has an infinite sequence of real roots; the second has an infinite sequence of complex roots.
Dougall states (p. 902) that for every m , (5.5) has an infinite number of real roots and an infinite number of complex roots (but no purely imaginary roots, as pointed out earlier).
For other work bearing on the circular section, reference m ay be made to. Poch- hammer,7 T hom ae,8 Schiff,9 Chree,10 T edone,11,12,13'14'15 F ilon,16 Purser,17’18 T im p e,19
7 L . P o c h h a m m e r, J o u r n a l f. M a th ., 8 1 , 33-6 1 (1876).
• J . T h o m a e , B e ric h te V erh. K . S ac h s. G es. d . W iss. zu L eipzig, M a th . P h y s . C l., 3 7 , 3 9 9 -4 1 8 (1885).
* Schiff, J o u rn a l d e M a th ., 9 , 4 0 7 -4 2 4 (1883).
10 C. C h ree, C a m b rid g e P h ilo so p h ical T ra n s a c tio n s , 14, 2 5 0 -3 6 9 (1887).
11 O. T e d o n e , A tti R . A cc. L incei R e n d . C l. sci. fis. m a t. n a t., 10, 131-137 (1901).
17 O. T ed o n e , A tti R . A cc. L incei R e n d . C l. sci. fis. m a t. n a t., 13i, 232-2 4 0 (1904).
11 O. T ed o n e , A tti R . A cc. L incei R e n d . C l. sci. fis. m a t. n a t., 20j, 617-6 2 2 (1911).
W olf,20 B arton.21 Barton makes use of the general Papcovitch22-Neuber23,24 solution of (2.3a), viz.,
1 d
U. = yf,. _ --- (Xrf. _J_ ( 5 . g )
4(1 — <r)
da;,-where A3t / \ =0, A z4> = 0. A ny solution of (2.3a) m ay be so expressed, but it seems that for present purposes D ougall’s expressions (4.20) are more convenient.
6. Conclusion. T he method of Dougall can be extended to a pipe, i.e., a section bounded by two concentric circles. Here are some questions:
1) Are there any other sections which can be solved by simple extensions of the method used for the circle?
2) Do eigen-values exist under reasonably general assumptions regarding the sm oothness of the boundary curve?
3) Is there always a set of real eigen-values, or is that a peculiarity of the circle?
4) Write k = p + i q , and let m be the least value of \ p \ in the sequence of eigen
values for a given section. Then m 2S , where S is the area of the section, depends only on the shape of the section. For arbitrary sections, m 2S forms a positive se
quence. Is it bounded below, and, if so, w hat is the lower bound?
T his last question is very interesting from a practical point of view, because \ p \
represents the rate at which end effects decay as we pass along the cylinder. The greater \ p \ , the more rapid the decay. Engineers are worried by end effects, because the Saint Venant solution gives no information about them. The assignm ent of such a lower bound might be more valuable than the description of a com plicated process for the evaluation of eigen-values.
14 O. T ed o n e , A tti R . A cc. L incei R e n d . Cl. sci. fis. m a t. n a t., 21i, 384-3 8 9 (1912).
15 O. T ed o n e , E n c y k . d . m a th . W iss., 4(, p. 150.
u L. N . G . F ilo n , P h il. T ra n s . R o y . Soc. L o n d o n , A 198, 147-233 (1902).
17 F . P u rs e r, T ra n s . R o y . Iris h A cad em y , 32 A, 3 1 -6 0 (1902).
18 F . P u rs e r, P ro c. R o y . Irish A c ad e m y , 26 A, 5 4 -6 0 (1906).
17 A . T im p e , M a th e m a tis c h e A n n a len , 71, 4 8 0 -5 0 9 (1912).
70 K . W olf, K . A k a d . d . W iss. W ien , M a th .-n a tu rw is s . K l., A b t. H a , 125, 1149-1166 (1916).
21 M . V. B a rto n , J . A p p . M ech an ics, 8 , A -97-A -104 (1941).
22 P . F . P a p c o v itc h , C o m p te s re n d u s A cad , d es Sci. P a ris, 195, 513-515 (1932).
23 H . N e u b er, Z eit. an g ew . M a th . u. M ech ., 14, 203-212 (1934).
24 H . N e u b er, K erbspannungslehre, B erlin , 1937.