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164. When the position of the principal axes, and the magni- Excate tudes of the principal elongations of a strain are given, the main elongation of any line of the body, and the alteration of angle of the boss. between any two lines, may be obviously determined by a siraple geometrical construction,

Analytically thus :—let a-1, B-1, y-1 denote the principal
elongations, so that a, b, y may be now the ratios of alteration
along the three principal axes, as we used thein formerly for the
ratios for any three oblique or rectangular lines. Let l, m, n
be the direction cosines of any line, with reference to the three
principal axes. Thus,

lr, mr, nr
being the three initial co-ordinates of a point P, at a distance
OP=r, from the origin in the direction l, m, n; the co-ordinates
of the same point of the body, with reference to the same rect-
angular axes, become, in the altered state,

alr, Bmr, ynr. Hence the altered length of OP is

(a^2 + B^ + Y) ,
and therefore the "elongation" of the body in that direction is

(a’l® + ß'm + ynyt-1.
For brevity, let this be denoted by 6-1, i.e.

= (a’l* + Bomo +ying?
The direction cosines of OP in its altered position are

ζ' ζ ' ζ Ś ;
and therefore the angles XOP, YOP, ZOP are altered to having
their cosines of these values respectively, from having them of
the values l, m, n.

The cosine of the angle between any two lines OP and OP,
specified in the initial condition of the body by the direction
cosines l', m, n, is

Il + mm' + nn',
in the initial condition of the body, and becomes

aʼll' + ß'mm' +y'nn'
(a*l + B*m* +y'no)} (a?!? +Bm+

in the altered condition.

al ßm gn


= 0.

ιξ, η

Change of 165. With the same data the alteration of angle between plane in the

any two planes of the body may also be easily determined, either geometrically or analytically.

Let l, m, n be the cosines of the angles which a plane makes with the planes YOZ, ZoXY, XOY, respectively, in the initial condition of the body. The effects of the change being the same on all parallel planes, we may suppose the plane in question to pass through 0; and therefore its equation will be

lx + my + nz = In the altered condition of the body we shall have, as before,

&= ax, n= By, &= yz, for the altered co-ordinates of any point initially x, y, z. Hence the equation of the altered plane is



B Y But the planes of reference are still rectangular, according to our present supposition. Hence the cosines of the inclinations of the plane in question, to YOZ, ZOX, XOY, in the altered condition of the body, are altered from l, m, n to


a' B9' ys' respectively, where for brevity


If we have a second plane similarly specified by l', m', n', in the
initial condition of the body, the cosine of the angle between the
two planes, which is

Il + mm' + nn'
in the initial condition, becomes altered to

IV mm'







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Conical sur.

166. Returning to elongations, and considering that these are elongation. generally different in different directions, we perceive that all

lines through any point, in which the elongations have any one

or, since

value intermediate between the greatest and least, must lie on Conical sura determinate conical surface. This is easily proved to be in elongation. general a cone of the second degree.

For, in a direction denoted by direction cosines l, m, n, we

a'l + B'm' + y'n' = %,
where & denotes the ratio of elongation, intermediate between a
the greatest and y the least. This is the equation of a cone of
the second degree, l, m, n being the direction cosines of a gene-

rating line. 167. In one particular case this cone becomes two planes, Two planes

of no dis. the planes of the circular sections of the strain ellipsoid. tortion, Let S = B. The preceding equation becomes

alo +Yono - 8* (1 – m*) = 0,


(a' - B9) 29 (BP - 7°) no = 0.
The first member being the product of two factors, the equation
is satisfied by putting either = 0, and therefore the equation re-
presents the two planes whose equations are

1 (a' B2)} + n (B* – °)* = 0,

l(a® B9)}- n (B? – y)} = 0, respectively. This is the case in which the given elongation is equal being the to that along the mean principal axis of the strain ellipsoid. sections of The two planes are planes through the mean principal axis of ellipsoid. the ellipsoid, equally inclined on the two sides of either of the other axes. The lines along which the elongation is equal to the mean principal elongation, all lie in, or parallel to, either of these two planes. This is easily proved as follows, without any analytical investigation.

168. Let the ellipse of the annexed diagram represent the section of the strain ellipsoid through the greatest and least principal axes. Let S'OS, T OT be the two diameters of this ellipse, which are

TE equal to the mean principal axis of the


X ellipsoid. Every plane through 0, perpendicular to the plane of the diagram, cuts the ellipsoid in an ellipse of which


of no distortion, being the circular sections of the strain ellipsoid.

in parallel


Two planes one principal axis is the diameter in which it cuts the ellipse of

the diagram, and the other, the mean principal diameter of the ellipsoid. Hence a plane through either SS', or TT", perpendicular to the plane of the diagram, cuts the ellipsoid in an ellipse of which the two principal axes are equal, that is to say, in a circle. Hence the elongations along all lines in either of these planes are equal to the elongation along the mean princi

pal axis of the strain ellipsoid. Distortion

169. The consideration of the circular sections of the strain planes with. ellipsoid is highly instructive, and leads to important views of volume. with reference to the analysis of the most general character of

a strain. First, let us suppose there to be no alteration of volume on the whole, and neither elongation nor contraction along the mean principal axis. That is to say, let B=1,

1 and y= = ($ 162). Let OX and OZ be the directions of elongation a-1 and

1 contraction 1 respectively. Let A be any point of the z

body in its primitive condition, IC

and A, the same point of the altered body, so that 0 A,= a0 A.

Now, if we take OC= 0A,,

and if C, be the position of that X


point of the body which was in
the position C initially, we shall

have OC, = 10C, and therefore Z'

OC,=0A. Hence the two triangles COA and C,0A, are equal and similar.

Hence CA experiences no alteration of length, but takes altered position of lines the altered position C, A, in the altered position of the body.

Similarly, if we measure on XO produced, OA' and 04; equal respectively to OA and 0A,, we find that the line C A' experiences no alteration in length, but takes the altered position C,A,'.

Consider now a plane of the body initially through CA perpendicular to the plane of the diagram, which will be altered into a plane through C,A,, also perpendicular to the plane of


Initial and


the diagram. All lines initially perpendicular to the plane of Initial and the diagram remain so, and remain unaltered in length. A tion of lines has just been proved to remain unaltered in length. Hence gation. ($ 158) all lines in the plane we have just drawn remain unaltered in length and in mutual inclination. Similarly we see that all lines in a plane through CA', perpendicular to the plane of the diagram, altering to a plane through C,A', perpendicular to the plane of the diagram, remain unaltered in length and in mutual inclination.

170. The precise character of the strain we have now under consideration will be elucidated by the following :-Produce CO, and take OC and OC) respectively equal to OC and OC, Join CA, CA', CA,, and CA, by plain and dotted lines as in the diagram. Then we see that the rhombus CAC' A' (plain lines) of the body in its initial state becomes the rhombus C, A,C A (dotted) in the altered condition. Now imagine the body thus strained to be moved as a rigid body (i.e., with its state of strain kept unchanged) till A, coincides with A, and C with C", keeping all the lines of the diagram still in the same plane. A' C, will take a position in CA' produced, as shown in the new diagram, and the original and the altered parallelogram will be on the same base A C', and between the same parallels AC and CA', and their other sides will be equally inclined on the two sides of a perpendicular to these parallels. Hence, irrespectively of any rotation, or other absolute motion of the body not involving change of form or dimensions, the strain under consideration may be produced by holding fast and unaltered the plane of the body through AC' perpendicular to the plane of the diagram, and making every plane parallel to it slide, keeping the same distance, through a space proportional to this distance (i. e., different planes parallel to the fixed plane slide through spaces proportional to their distances).

171. This kind of strain is called a simple sheur. The simple plane of a shear is a plane perpendicular to the undistorted planes, and parallel to the lines of their relative motion. It


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