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10–61.
Determine the product of inertia of the shaded area with
respect to the x and y axes.
SOLUTION
1
0
3
x
y
2 in.
y = 0.25x 2
1 in.
10–62.
SOLUTION
Ans.=97.8 in
4
I
xy =0.5(4)(8)(1) +6(0.5)(10)(1) +11.5(1.5)(3)(1)
Determine the product of inertia for the beam’s cross-
sectional area with respect to the
xand yaxes.
8in.
y
x
12 in.
3 in.
1in.
1in.
1in.
10–63.
Determine the moments of inertia for the shaded area with
respect to the uand axes.v
y
vu
0.5 in.
1073
*10–64.
Determ
i
ne t
h
e pro
d
uct of
i
nert
i
a for t
h
e
b
eam’s cross-
sectional area with respect to the uand axes.v
The section is symmetric about both xand yaxes; therefore .
Ans.
=
135
(
10
)
6mm4
=a511.36 -90.24
2sin 40° +0 cos 40°b106
Iuv =
Ix-Iy
2sin 2u+Ixy cos 2u
Ixy =0
Iy=2c1
12(20)(300)3d+1
12(360)(20)3=90.24(10)6mm4
150 mm
150 mm
y
v
u
20
10–65.
Determine the product of inertia for the shaded area with
respect to the xand yaxes.
SOLUTION
Ans.=119 in4
Ixy =©(Ix¿y¿+x
'y
'A)=[0 +2(3)(4)(6)] -
C
0+2(4)(p)(1)2
D
y
x
4in.
2in.
1 in.
2in.2 in.
10–66.
SOLUTION
Determine the product of inertia of the cross-sectional area
with respect to the xand yaxes.
400 mm
y
100 mm
20 mm
10–67.
SOLUTION
Determine the location (x, y) of the centroid C of the
angle’s cross-sectional area, and then compute the product
of inertia with respect to the x′ and y′ axes.
150 mm 18 mm
150 mm
18 mm
yy
¿
Cx¿
x
y
x
1077
*10–68.
Determ
i
ne t
h
e
di
stance to t
h
e centro
id
of t
h
e area an
d
then
calculate the moments of inertia and for the
channel’
s cross-sectional area. The uand axes have their
origin
at the centroid C.For the calculation, assume all
corners to be square
.
v
Iv
Iu
y
y
x
u
v
50 mm
10 mm
10 mm
10 mm
y
20
C
SOLUTION
Ans.
Ans.
=
38.5
(
106
)
mm4
=0.9083(106)+43.53(106)
2-0.9083(106)-43.53(106)
2cos 40°+0
Iv=
Ix+Iy
2-
Ix-Iy
2cos 2u+Ixy sin 2u
=5.89(106)mm4
=0.9083(106)+43.53(106)
2+0.9083(106)-43.53(106)
2cos 40° -0
Iu=
Ix+Iy
2+
Ix-Iy
2cos 2u-Ixy sin 2u
Ixy =0 (By symmetry)
=43.53(106)mm4
Iy=1
12 (10)(300)3+2c1
12 (50)(10)3+50(10)(150 -5)2d
=0.9083(106)mm4
+2c1
12 (10)(50)3+10(50)(35 -12.5)2d
Ix=c1
12 (300)(10)3+300(10)(12.5 -5)2d
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10–69.
SOLUTION
Ans.
Ans.
=5.515 -1.419
2(103) sin 90° +0
Iuv =
Ix-Iy
2sin 2u+Ixy cos 2u
Iv=3.47(103)in4
=3.47(103)in4
=5.515 +1.419
2(103)+5.515 -1.419
2(103) cos 90° -0
Iu=
Ix+Iy
2+
Ix-Iy
2cos 2u-Ixy sin 2u
Ixy =0
=1.419(103)in4
Iy=1
12(2)(20)3+1
12(16)(4)3
=5.515(103)in4
Ix=1
12 (20)(2)3+20(2)(1)2+1
12(4)(16)3+4(16)(8)2
Determine the moments of inertia and the product of
inertia for the beam’s cross-sectional area.Take u=45°.Iuv
I
v
I
u
,
y
x
vu
16 in.
O
8in.
2in. 2in.
8in. 2in.
u
SOLUTION
10–70.
Determine the moments of inertia Iu, Iv and the product of
inertia Iuv for the rectangular area. The u and v axes pass
through the centroid C.
x
u
v
y
30 mm
120 mm
C
30
SOLUTION
10–71.
Solve Prob. 10–70 using Mohr’s circle. Hint: To solve, find
the coordinates of the point P(Iu, Iuv) on the circle, measured
counterclockwise from the radial line OA. (See Fig. 10–19.)
The point Q(Iv, -Iuv) is on the opposite side of the circle.
x
u
v
y
30 mm
120 mm
C
30
SOLUTION
*10–72.
Determine the directions of the principal axes having an
origin at point O, and the principal moments of inertia for
the triangular area about the axes.
6 in.
O
y
x
1082
SOLUTION
10–73.
Solve Prob. 10–72 using Mohr’s circle.
6 in.
y
x
1083
1084
10–74.
Determine the orientation of the principal axes having an
origin at point C, and the principal moments of inertia of
the cross section about these axes.
SOLUTION
x
10 mm
100 mm
100 mm
80 mm
80 mm
y
C
1086
10–75.
Solve Prob. 10–74 using Mohr’s circle.
SOLUTION
x
10 mm
100 mm
100 mm
80 mm
80 mm
y
C
1087
*10–76.
Determine the orientation of the principal axes having an
origin at point O, and the principal moments of inertia for
the rectangular area about these axes.
SOLUTION
Ox
y
3 in.
6 in.
1089
