through their respective centres of gravity, shew how to determine graphically the centre of gravity of the smaller part from the given centres of the whole area and the larger part.

15. A line u lies in the same plane with a circle whose centre is at O, and whose radius is equal to r. Considering the points of bisection of the polar chords, with respect to the circle, of points on the line u as having equal weight, shew that their centre of gravity lies at the point of bisection of the line OU, where U is the pole of the line u with respect to the circle.

16. Two non-intersecting circles of radii OR and O'R' respectively are described in the same plane, and the line of centres OO' intersects the second circle (R'), first in P, and secondly in P'. Considering the points of bisection of polar chords with respect to the first circle (R) of points lying on the circumference of the second (R') to be of equal weight, shew that the centre of gravity of all such points of bisection lies at a point x on the line OO' at a distance Ox k. 00' from the centre O, where k represents the ratio OP, OP', P, being the point of bisection of the polar chord of the point P with respect to circle (R).

CHAPTER V

ELLIPSES OF INERTIA AND KERNS

62. Graphic Summation.-Let polygon (1) (Fig. 87) be the polygon of forces of a series p, and (2) the corresponding funicular drawn relatively to the pole O' and the vertical lines of action of the forces, the polar distance being h. Prolong the sides of the funicular so as to cut off on the axis of y produced the series of intercepts i, i, i3, i10. Then, if x, be the abscissa of the line of action of p1, we

ני

P1

= etc. P2 h

Hence, in the sum,

1
ac=Zi==Σpx

h

(1).

Assume now that the forces p act horizontally through their points of application, the ordinates of which are Ул, Уг, Уз, ., Yo, and construct the upright funicular (3) relatively to the same series of forces and polar distance h. Then, by similar reasoning, the segment no or 2op, cut off upon the axis of a by the closing sides of the funicular, represents the sum

Moreover, if y, be the ordinate of the centre of gravity of the system p, we have

Again, if with the series of segments i, i,..., io', cut off upon no by the produced sides of the funicular, and polar

distance yo, the upright polygon (4) be drawn, we shall obtain the expression for yp = rv from the relation

Further, if with the series of segments i, i,

=

i10, cut

off upon ac, and a polar distance h' kk', a fifth polygon be

constructed relatively to the same system of forces, assumed to act in a vertical sense; the segment cut off by the closing sides upon the axis of y produced will furnish the moment of inertia of the forces, relatively to the axis of y, in the form ef=.Σpa2.

1

Lastly, if with the series of intercepts i,', i,',..., i,,', cut off upon no, and the ordinate y, as polar distance, a sixth funicular be constructed with its joints on the vertical lines through the points of application of the forces p, the segment cut off upon the axis of y produced will furnish the second moment Exyp divided by the product hy, of the two polar distances employed in the construction of the third and sixth funicular polygons.

63. Ellipses of Inertia. The following purely graphic method of constructing ellipses of inertia is for the most part taken from Culmann's Graphic Statics.

PROBLEM.-Given or found the moments of inertia Zap, Zyp, and the centrifugal moment Exyp of a series of forces p relatively to any pair of rectangular axes Ox and Oy in the plane of the forces; to find the moment of inertia Eqph. Ep of the same series of forces relatively to any axis Oq traversing the origin O, and to construct the ellipse of inertia of which O is the centre.

If, as will be proved, OC and OM (Fig. 88) be conjugate axes of the required ellipse of inertia, the moment of inertia relatively to the axis OM is

where OT is the perpendicular let fall from O upon the line through C parallel to OM (Author's Graphic and Analytic Statics, p. 198).

But OC =

от

; hence the oblique moment of inertia

relatively to the axis OM, when the distances of the points

of application of the forces of the system p are measured from OM parallel to OC, will be expressed by

Thus, it is possible to measure the distances of the points of application of the forces p in any arbitrary direction, so as to find the oblique radius of gyration in that direction. A line drawn past the end of this oblique radius parallel to the axis from which distances are measured will be a tangent to the required ellipse.