PRODUCT OF INERTIA OF AN AREA
Calculate the product of inertia of the isosceles trapezoid in Fig. 14 with respect to the
rectangular axes u and v.
Calculation Procedure:- Locate the centroid of
the trapezoid
Using the AISC Manual or another
suitable reference, we find h = cen-
troid distance from the axis (Fig. 14)
= (9/3)[(2 x 5 + 10)/(5 + 1O)] = 4 in
(101.6mm). - Compute the area and
product of inertia Pxy
The area of the trapezoid is A =
FIGURE 14 l/'(9)(5 + 10) = 67.5 in^2 (435.5 cm^2 ).
Since the area is symmetrically dis-
posed with respect to the y axis, the
product of inertia with respect to the
x andy axes is P^ = O. - Compute the product of inertia by applying the transfer
equation
The transfer equation for the product of inertia is Puv = P^ + Axmym, where xm and ym are
the coordinates of O' with respect to the centroidal x and y axes, respectively. Thus Puv =•
O + 67.5(-5)(-4) = 1350 in^4 (5.6 dm^4 ).
PROPERTIES OF AN AREA WITH RESPECT
TO ROTATED AXESIn Fig 15, x and y are rectangular axes through the centroid of the isosceles triangle; x'
and y' are axes parallel to x and y, respectively; jc"and>>"are axes making an angle of 30°
with jc' and y', respectively. Compute the moments of inertia and the product of inertia of
the triangle with respect to the x" and y" axes.
Calculation Procedure:
- Compute the area of the figure
The area of this triangle = 0.5(base)(altirude) = 0.5(8)(9) = 36 in^2 (232.3 cm^2 ). - Compute the properties of the area with respect to the
x and y axes
Using conventional moment-of-inertia relations, we find Ix = bcP/3>6 = 8(9)3/36 = 162 in^4
(0.67 dm^4 ); Iy = b^3 d/4$ = (8)3(9)748 = 96 in^4 (0.39 dm^4 ). By symmetry, the product of in-
ertia with respect to the x and y axes is P^ = O.