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

1.654 in.
y
1.654 in.
y
1.154 in.
(1)
1.846 in.
8 in.
1 in.
1 in. 5 in.
6 in.
x
8 in.
1.346 in.
2.654 in. x
2.154 in.
2.654 in.
1 in.
(2)
1 in. 5 in.
6 in.
and the moment of inertia about the y centroidal axis is
I
y
3
3
(8 in.)(1 in.)
2
(1 in.)(5 in.)
2
= + (8 in.)(1 in.)(1.154 in.) + + (1 in.)(5 in.)(1.846 in.)
12
12
4
= 38.8 in.
Ans.
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 below.
I x′y′
(in. 4 )
x c
(in.)
y c
(in.)
A i
(in. 2 )
x c y c A i
(in. 4 )
I xy
(in. 4 )
(1) 0 −1.154 1.346 8.0 −12.426 −12.426
(2) 0 1.846 −2.154 5.0 −19.881 −19.881
−32.307
The product of inertia for the unequal-leg angle shape is thus I xy = −32.3 in. 4 .
Ans.
A.4 principal moments of Inertia
The moment of inertia of the area A in Figure A.7 with respect to the x′ axis through O will,
in general, vary with the angle θ. The x and y axes used to obtain Equation (A.6) were any
pair of orthogonal axes in the plane of the area passing through O; therefore,
J = I + I = I + I
x y x ′ y ′
where x′ and y′ are any pair of orthogonal axes through O. Since the sum of I x′ and I y′ is a
constant, I x′ will be the maximum moment of inertia and the corresponding I y′ will be the
minimum moment of inertia for one particular value of θ.
801