PROBLEM 9.103
The moments and product of inertia of an L4 3
1
4
-in. angle cross section with respect to two
rectangular axes x and y through C are, respectively,
1.33 in
4
,
y
I
2.75 in
4
, and
0,
xy
I
with the
minimum value of the moment of inertia of the area with respect to any axis through C being
min
0.692
in
4
. Using Mohr’s circle, determine (a) the product of inertia
y
of the area, (b) the orientation of the
principal axes, (c) the value of
max
.I
when the 4-in. leg of the angle is parallel to the x axis. Further, for
0,
xy
I
the angle must be oriented as
shown.)
Now
4
ave
11
()(1.332.75)2.040 in
22
xy
III
and
min ave
or 2.040 0.692 IIR R
4
1.348 in
Using
ave
I
and R, the Mohr’s circle is then drawn as shown; note that for the diameter XY,
(1.33, )
xy
XI
and
(2.75, | |).
xy
YI
(a) We have
2
22
1()
2
xy xy
RIII
or
2
22
1
1.348 (1.33 2.75)
2
xy
I
Solving for
xy
I
and taking the negative root (since
0)
xy
I
yields 4
1.14586 in .
xy
I
4
1.146 in
xy
I
(b) We have
22( 1.14586)
tan 2 1.33 2.75
1.61389
xy
mxy
I
II
or
2 58.217 29.1
mm
The principal axes are obtained by rotating the xy axes through
29.1° clockwise
about C.
(c) We have
max ave
2.040 1.348 IIR
or
4
max
3.39 inI