tion of pure strains.

PAD

Composi.. which (Chap. 11.) will be proved to be instantaneously, or dif

ferentially, irrotational; but which may result in leaving a
whole fluid mass merely turned round from its primitive posi-
tion, as if it had been a rigid body. The following elementary
geometrical investigation, though not bringing out a thoroughly
comprehensive view of the subject, affords a rigorous demon-
stration of the proposition, by proving it for a particular case.

Let us consider, as above (171), a simple shearing motion.
A point O being held fixed, suppose the matter of the body in
a plane, cutting that of the diagram perpendicularly in CD, to
move in this plane from right to left parallel to DC; and in
other planes parallel to it let there be motions proportional to
their distances from 0. Consider first a shear from P'to P, ;
then from P, on to P,; and let O be taken in a line through
Q:

P, perpendicular to
P A

D CD. During the shear

from P to P, a point
Q moves of course to

Q, through a distance
Q?, = PP,. Choose Q midway between P and P,, so that
P,Q = QP = {P.P. Now, as we have seen above ($ 152), the
line of the body, which is the principal axis of contraction in the
shear from Q to Q,, is 0A, bisecting the angle QOE at the be-
ginning, and 0A,, bisecting Q,0 E at the end, of the whole
motion considered. The angle between these two lines is half
the angle Q,0Q, that is to say, is equal to P,0Q. Hence, if the
plane CD is rotated through an angle equal to P,0Q, in the
plane of the diagram, in the same way as the hands of a watch,
during the shear from Q to Q,, or, which is the same thing, the
shear from P to P,, this shear will be effected without final
rotation of its principal axes. (Imagine the diagram turned
round till 0A, lies along 0A. The actual and the newly
imagined position of CD will show how this plane of the body
has moved during such non-rotational shear.)

Now, let the second step, P, to P,, be made so as to complete the whole shear, P to P,, which we have proposed to consider. Such second partial shear may be made by the common shearing process parallel to the new position (imagined in the preced

ing parenthesis) of CD, and to make itself also non-rotational, Composi

tion of pure as its predecessor has been made, we must turn further round, strains. in the same direction, through an angle equal to Q,0P,. Thus in these two steps, each made non-rotational, we have turned the plane CD round through an angle equal to Q.OQ. But now, we have a whole shear PP,; and to make this as one non-rotational shear, we must turn CD through an angle P,OP only, which is less than Q,0Q by the excess of P,0Q above QOP. Hence the resultant of the two shears, PP,, P,P,, each separately deprived of rotation, is a single shear PP,, and a rotation of its principal axes, in the direction of the hands of a watch, through an angle equal to QOP,-POQ.

185. Make the two partial shears each non-rotationally. Return from their resultant in a single non-rotational shear: we conclude with the body unstrained, but turned through the angle QOP,-POQ, in the same direction as the hands of a watch.

Axt
Y, = cx + By + az

2, = bx + ay + Cz
is (s 183) the most general possible expression for the displace-
ment of any point of a body of which one point is held fixed,
strained according to any three lines at right angles to one
another, as principal axes, which are kept fixed in direction,
relatively to the lines of reference OX, OY, OZ.

Similarly, if the body thus strained be again non-rotationally
strained, the most general possible expressions for X, Y, Z,
the co-ordinates of the position to which x, ,, , will be brought,

cy + bz

are

2, = A,x,+ 0,9, +6,22
Ya = c,, + B y,+a,,,

2,= 6,3, + a,y, +0,2.
Substituting in these, for x, ,,,, their preceding expressions,
in terms of the primitive co-ordinates, x, y, z, we have the follow-
ing expressions for the co-ordinates of the position to which the
point in question is brought by the two strains :
2. = (4,4+Cc+6,6) x + (4,c +c,B+b,a) y + (4,6 + cya +6,C) z
y,= (, A + B,c+a,b) x + (C,c +B, B+a,a) y + (0,6 + Ba +a,C);
max = (6, A +a,c +0,6) x + (b,c +a, B+C a) y +(6,6 +a a +C,C) 2.

+

Composition of pure strains.

The resultant displacement thus represented is not generally of the non-rotational character, the conditions (18) of $ 183 not being fulfilled, as we see immediately. Thus, for instance, we see that the coefficient of y in the expression for x, is not necessarily equal to the coefficient of x in the expression for yg

Cor.--If both strains are infinitely small, the resultant displacement is a pure strain without rotation. For A, B, C, A,, B, C, are each infinitely nearly unity, and a, b, etc., each infinitely small. Hence, neglecting the products of these infinitely small quantities among one another, and of any of them with the differences between the former and unity, we have a resultant displacement

2,= A Ax +(c +c) y + (6 +6,)2
Y,= (c, + c) x + B.By + (a +a,)2

mg = (6, +b) x + (a, + a) y +C,Cz,
which represents a pure strain unaccompanied by rotation,

Displacement of a curve,

186. The measurement of rotation in a strained elastic solid, or in a moving fluid, is much facilitated by considering separately the displacement of any line of the substance. We are therefore led now to a short digression on the displacement of a curve, which may either belong to a continuous solid or fluid mass, or may be an elastic cord, given in any position. The propositions at which we shall arrive are, of course, applicable to a flexible but inextensible cord ($ 14, above) as a particular case.

It must be remarked, that the displacements to be considered do not depend merely on the curves occupied by the given line in its successive positions, but on the corresponding points of these curves.

What we shall call tangential displacement is to be thus reckoned:-Divide the undisplaced curve into an infinite number of infinitely small equal parts. The sum of the tangential components of the displacements from all the points of division, multiplied by the length of each of the infinitely small parts, is the entire tangential displacement of the curve reckoned along the undisplaced curve. The same reckoning carried out in the displaced curve is the entire tangential displacement reckoned on the displaced curve.

Tangential displacement,

187. The whole tangential displacement of a curve reckoned Two reckonalong the displaced curve, exceeds the whole tangential dis- sential displacement reckoned along the undisplaced curve by half the compared. rectangle under the sum and difference of the absolute terminal displacements, taken as positive when the displacement of the end towards which the tangential components are if positive exceeds that at the other. This theorem may be proved by a geometrical demonstration which the reader may easily supply.

Analytically thus -Let x, y, z be the co-ordinates of any point, P, in the undisplaced curve; a,, Y., 77, those of P, the point to which the same point of the curve is displaced. Let dr, dy, dz be the increments of the three co-ordinates corresponding to any infinitely small arc, ds, of the first; so that

ds = (dx*+ dy + dz?)?, and let corresponding notation apply to the corresponding element of the displaced curve. Let 0 denote the angle between the line PP, and the tangent to the undisplaced curve through P; so that we have

y, - y dy, 2, - 2 dz

D ds D ds where for brevity

D={(2, -&)*+ (%, - y)* + (2, -2)}}, being the absolute space of displacement. Hence

D cos Ods = (2, - x) dx + (y, - y)dy + (z, - 2)dz. Similarly we have

D cos 6,ds, = (xc, — 2) dx, + (y, -y) dy, + (z, - ) dz,, and therefore D cos 0,ds, - D cos @ds = (2, - x)d (20, — «) + (y, - y)d(y, - y)

+ (2. – 2)d (2,-2), D cos 0,ds, - D cos 0d8 = 1d (D'). To find the difference of the tangential displacements reckoned the two ways, we have only to integrate this expression. Thus we obtain

SD cos 6,ds, – SD cos 0ds = }(D" – D')) = }(D" + D') (D" - D'), where D" and D' denote the displacements of the two ends.

+

+

or

Tangential 188. The entire tangential displacement of a closed curve displacement of a is the same whether reckoned along the undisplaced or the closcd curve.

displaced curve.

189. The entire tangential displacement from one to another of two conterminous arcs, is the same reckoned along either as along the other.

ment in a

terms of

of strann.

Rotation 190. The entire tangential displacement of a rigid closed of a rigid closed curve. curve when rotated through any angle about any axis, is equal

to twice the area of its projection on a plane perpendicular to

the axis, multiplied by the sine of the angle. Tangential (a) Prop.—The entire tangential displacement round a closed displace

curve of a homogeneously strained solid, is equal to solid, in

2(Pw + Op + Ro), components

where P, Q, R denote, for its initial position, the areas of its projections on the planes YOZ, zoX, XOY respectively, and W, P, o are as follows:

-Z{[Zy] - [Yz]} p=}{[-X2]-[2x]}

o=}{[Yx] - [Xy]}.
To prove this, let, farther,

a= {[Zy] + [Yu]}
b = }{[_X]+[2x]}

c ={[Yx] + [Xy]).
Thus we have

x, = 4x + cy + b2 + oy – pz
= cx + By + az + a2-ox

ay + Cz + på – wy.
Hence, according to the previously investigated expression, we
have, for the tangential displacement, reckoned along the undis-
placed curve,

S{(2, - x) dx +(9,- y)dy + (-2) dz} = S[]d{(4 – 1) ** + (B – 1) y* + (C – 1)2 + 2 (ayz + bzx + cxy)}

+ (ydz – zdy) +p (zdx — «dz) +0 (ædy – ydx)]. The first part, Sid{ }, vanishes for a closed curve.

[2]

bx +