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1070
10–61.
Determine the product of inertia of the shaded area with
respect to the x and y axes.
SOLUTION
Differential Element: Here, x=2y
1
2. The area of the differential element parallel
to the x axis is dA =xdy=2y
1
2
dy. The coordinates of the centroid for this element
x
y
2 in.
y = 0.25x 2
1 in.
Ixy =0.667 in
10–62.
Determine the product of inertia for the beam’s cross-
sectional area with respect to the
xand yaxes.
Ans:
1072
10–63.
SOLUTION
Moment and Product of Inertia about x and y Axes: Since the shaded area is
symmetrical about the xaxis, Ixy =0.
Moment of Inertia about the Inclined u and v Axes: Applying Eq. 10-9 with
, we have
Iu=
Ix+Iy
2+
Ix-Iy
2cos 2u-Ixy sin 2u
u=30°
Ix=1
12(1)(53)+1
12(4)(13)=10.75 in4
x
vu
0.5 in.
0.5 in.
0.5 in.
5in. 30
*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
SOLUTION
150 mm
150 mm
y
v
u
20
1074
10–65.
Determine the product of inertia for the shaded area with
respect to the xand yaxes.
SOLUTION
1075
10–66.
SOLUTION
Product of Inertia: The area for each segment, its centroid and product of inertia
with respect to xand yaxes are tabulated below.
Determine the product of inertia of the cross-sectional area
with respect to the xand yaxes.
400 mm
x
y
100 mm
20 mm
C
10–67.
SOLUTION
Centroid:
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
150 mm150 mm
y
x
u
v
50 mm
10 mm
10 mm
10 mm
y
20
C
SOLUTION
Ans.
y=300(10)(5) +2[(50)(10)(35)]
300(10) +2(50)(10) =12.5 mm
1078
10–69.
SOLUTION
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
1079
SOLUTION
Moment And Product of Inertia About x and y Axes. Since the rectangular area is
symmetrical about the x and y axes,
I
xy
=0.
Moment And Product of Inertia About The Inclined u and v Axes. With u
=30°,
Iu=
I
x
+I
y
2
+
I
x
-I
y
2
cos 2u-Ixy sin 2u
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
Ans:
1080
SOLUTION
Moment And Product of Inertia About x And y Axes. Since the rectangular Area is
symmetrical about the x and y axes,
I
xy
=0.
Construction of The Circle. The Coordinates of center O of the circle are
aI
x
+I
y
2
, 0
b
=
a0.270 +4.32
2
, 0
b
(106)=(2.295, 0)(106)
And the reference point A is
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
Ans:
1081
SOLUTION
Moment And Product of Inertia About x And y Axes. Using the parallel-axis theorem,
Principal Moments of Inertia.
I
max
min
=
Ix
+
Iy
{
Aa
Ix-Iy
2
b
2
+I2
xy
The orientation of the principal axes can now be determined
Substitute up
=-70.98°
into the equation for
Iu
,
*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.
9 in.
6 in.
O
y
x
Ans:
1082
SOLUTION
Moment And Product of Inertia About x and y Axes. Using the parallel-axis theorem,
Ix=Ix
′+
Ad
2
y
;
Ix=
1
36(9)(63)+
c1
2(9)(6)
dc1
3(6)
d2
=162 in4
Using these results of Example 10–6
And the reference point A is
Thus, the radius of the circle is
Using these results the circle shown in Fig. a can be constructed. Here, the coordinates
of points B and C represent
Imin
and
Imax
respectively. Thus
10–73.
Solve Prob. 10–72 using Mohr’s circle.
9 in.
6 in.
O
y
x
1083
The orientation of principal axes can be determined from the geometry of the
shaded triangle on the circle.
10–73. Continued
Ans:
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
Moment And Product of Inertia About x and y Axes. Using the parallel-axis
theorem by referring to Fig. a
Principal Moments of Inertia.
Imax
min
=
Ix+Iy
2
{
C
a
Ix+Iy
2b
2
+Ixy
2
x
10 mm
10 mm
100 mm
100 mm
80 mm
80 mm
y
C
1085
The orientation of the principal axes can now be determined
Substitute up
=60.05°
into the equation for
Iu
,
Thus,
10–74. Continued
Ans:
1086
10–75.
Solve Prob. 10–74 using Mohr’s circle.
SOLUTION
Moment And Product of Inertia About x and y Axes. Using the parallel-axis
theorem by referring to Fig. a
Iy
=Σ
(
Iy
′+
Adx
2
)
; Iy=
(10)(1403)+2
(100)(103)+100(10)(752)
d
Ixy
=Σ
(
Ix
′
y
′+
Adx
dy
)
; Ixy
=
0
+
3
0
+
10(100)(
-
75)(45)
4
Construction of the circle. The coordinates of center O of the circle are
And the reference point A is
Thus, the radius of the circle is
Using these results, the circle shown in Fig. b can be constructed. Here, the coordinates of
points B and C represent
Imin
and
Imax
respectively. Thus
x
10 mm
10 mm
100 mm
100 mm
80 mm
80 mm
y
C
1087
The orientation of the principal axes can be determined from the geometry of the
shaded triangle on the circle
And
10–75. Continued
Ans:
1088
*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
Moment And Product of Inertia About x and y Axes. Using the parallel-axis
theorem,
Principal of Moment of Inertia.
Imax
min
=
Ix+Iy
2
{
C
a
Ix-Iy
2b
2
+I2
xy
The orientation of principal axes can now be determined.
tan 2up=
-I
xy
(Ix-Iy)
>
2=
-81.0
(54.0 -216.0)
>
2=1.00
Substitute up
=-67.5°
into the equation for
Iu,
Ox
y
3 in.
6 in.
Ans:
1089
SOLUTION
Moment And Product of Inertia About x and y Axes. Using the parallel-axis
theorem,
I
x=
I
x′
+Ad2
y
;
Ix=
1
12
(6)(33)+6(3)(1.52)=54.0 in4
Construction of the circle. The coordinates of center O of the circle is
aI
x
+I
y
2
, 0
b
=
a54.0 +216.0
2
, 0
b
=(135, 0)
And the reference point A is
Thus, the radius of the circle is
10–77.
Solve Prob. 10–76 using Mohr’s circle.
Ox
y
3 in.
6 in.
