## Y

Fig. 1.7. The momental ellipse.

Fig. 1.7. The momental ellipse.

u2 sin26dA

But UU and VV are the principal axes so that Iuv = f uvdA is zero.

Similarly,

= J u2cos20dA- J 2uv sin 9 cos 9dA + J v2sin 29dA and with J uvdA = 0

Also

Ixy — J xydA — J(u cos 9 — v sin 9)(v cos 9 + u sin 9) dA

= J [uv(cos2 9 - sin2 9) + (u2 - v2) sin 9 cos 9] dA

— /««cos 29 + | (/„—/„) sin 29 and /„„ = 0 Ixy = \(h-Iu) sin 29 (1.12)

IXX — Iyy = Iu cos2 9 + Iv sin2 9 — Iv cos2 9 — /„ sin2 9

= (/„-/„) cos2 9 -(/„-/„) sin2 9 Ixx - Iyy = (Iu - Iv) COS 29 (1.13)

*yy *xx and combining eqns. (1.10) and (1.11) gives

Substitution into eqns. (1.10) and (1.11) then yields

Iu = \\(hx +Iyy) + (Ixx - Iyy)sec29] (1.16) as (1.6)

Iv = \l(Ixx +Iyy) - (Ixx - Iyy)sec26] (1.17) as (1.7)

1.6. The ellipse of second moments of area

The above relationships can be used as the basis for construction of the moment of area ellipse proceeding as follows:

(1) Plot the values of /„ and Iv on two mutually perpendicular axes and draw concentric circles with centres at the origin, and radii equal to /„ and Iv (Fig. 1.8).

(2) Plot the point with coordinates x = Iu cos 9 and y = /„ sin 0, the value of 9 being given by eqn. (1.14).

This equation is the locus of the point P and represents the equation of an ellipse - the ellipse of second moments of area. (3) Draw OQ at an angle 6 to the /„ axis, cutting the circle through Iv in point S and join SP which is then parallel to the /„ axis. Construct a perpendicular to OQ through P to meet OQ in R. Then

— lu — (Iu sin 6 — Iv sin 6) sin 0 = /„-(/„-/„) sin2 6 = /„ cos2 6 + Iv sin2 0

0 0