Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

ExAmpLE A.4
40 mm
Determine the product of inertia for the zee shape shown in
Example A.3.
−18 mm
(3)
6 mm
40 mm
6 mm
(2)
y
x
40 mm
+17 mm
(1)
y
+18 mm
x
6 mm
30 mm
6 mm
6 mm
30 mm
6 mm
Plan the Solution
The zee shape can be subdivided into three rectangles. Since
rectangles are symmetric, their respective products of inertia
about their own centroidal axes are zero. Consequently, the
product of inertia for the entire zee shape will derive entirely
from the parallel-axis theorem.
SolutioN
The centroid location for the zee shape is shown in the sketch.
In computing the product of inertia using the parallel-axis theorem
[Equation (A.10)], it is essential that careful attention be
paid to the signs of x c and y c . The terms x c and y c are measured
from the centroid of the overall shape to the centroid of the individual
area. The complete calculation for I xy is summarized in
the table on the next page.
I x′y′
(mm 4 )
x c
(mm)
y c
(mm)
A i
(mm 2 )
x c y c A i
(mm 4 )
I xy
(mm 4 )
(1) 0 17.0 18.0 240 73,440 73,440
(2) 0 0 0 180 0 0
(3) 0 −17.0 −18.0 240 73,440 73,440
146,880
40 mm
−17 mm
The product of inertia for the zee shape is thus
I xy = 146,900 mm 4 .
Ans.
ExAmpLE A.5
Determine the moments of inertia and the product of inertia for the unequal-leg angle
shape shown with respect to the centroid of the area.
Plan the Solution
The unequal-leg angle is divided into two rectangles. The moments of inertia are computed
about both the x and y axes. The product of inertia calculation is performed as demonstrated
in Example A.4.
SolutioN
The centroid location for the unequal-leg angle shape is shown in the sketch. The moment
of inertia for the unequal-leg angle shape about the x centroidal axis is
I
x
3
3
(1 in.)(8 in.)
2
(5 in.)(1 in.)
2
= + (1 in.)(8 in.)(1.346 in.) + + (5 in.)(1 in.)(2.154 in.)
12
12
4
= 80.8 in.
Ans.
800