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.
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
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
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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