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1050
SOLUTION
Moment of Inertia. The moment of inertia about the y axis for each segment can
be determined using the parallel-axis theorem,
I
y=
I
y′
+Ad
x
2.
Referring to Fig. a,
Segment A
i
(in
2
)
(dx)i
(in.)
(Iy′)i
(in
4
)(Ad
x
2
)
i
(in
4
) (Iy)i
(in
4
)
10–41.
Determine the moment of inertia for the beam’s cross-
sectional area about the y axis.
y
x
3 in. 1 in.
1 in.
4 in.
1 in.
y¿
x ¿
C
y
3 in.
Ans:
10–42.
SOLUTION
Ix=1
12 (170)(30)3+170(30)(15)2
170 mm
30 mm
30 mm
70 mm
140 mm
30 mm
30 mm
y
x¿
y¿
C
y
x
Determine the moment of inertia of the beam’s cross-
sectional area about the x axis.
1052
10–43.
Determine the moment of inertia of the beam’s cross-
sectional area about the yaxis.
SOLUTION
Iy=1
12(30)(170)3+30(170)(115)2
30 mm
70 mm
140 mm
30 mm
y
y¿
_
x
C
*10–44.
SOLUTION
Ans.
=80.68 =80.7 mm
y=170(30)(15) +170(30)(85) +100(30)(185)
170(30) +170(30) +100(30)
Determ
i
ne t
h
e
di
stance to t
h
e centro
id
Cof t
h
e
b
eam’s
cro
ss-sectional area and then compute the moment of
inertia
about the axis.x¿Ix¿
y
170 mm
30 mm
30 mm
70 mm
140 mm
30 mm
30 mm
y
x
x¿
y¿
C
y
x
1054
10–45.
Determine the distance to the centroid Cof the beam’s
cross-sectional area and then compute the moment of
inertia about the axis.y¿Iy¿
x
SOLUTION
x=170(30)(115) +170(30)(15) +100(30)(50)
170(30) +170(30) +100(30)
30 mm
70 mm
140 mm
30 mm
30 mm
y
x¿
y¿
C
y
x
1055
SOLUTION
Moment of Inertia. The moment of inertia about the x axis for each segment can
be determined using the parallel-axis theorem,
I
x=
I
x′
+Ad
y
2.
Referring to Fig. a,
Segment A
i
(in
2
)
(d
y
)
i
(in.)
(
I
x′)i
(in4)
(Ady
2
)i
(in
4
) (I
x
)
i
(in
4
)
4
Thus,
10–46.
Determine the moment of inertia for the shaded area about
the
x
axis.
x
y
3 in. 3 in.
6 in.
3 in.
3 in.
3 in.
2 in.
1056
SOLUTION
Moment of Inertia. The moment of inertia about the x axis for each segment can
be determined using the parallel-axis theorem,
I
y=
I
y′
+Ad
x
2.
Referring to Fig. a,
Segment A
i
(in
2
)
(dx)i
(in.)
(
I
x′)i
(in4)
(Ad
x
2
)
i
(in
4
) (Iy)i
(in
4
)
1 6(6) 3
1
12
(6)(63)324 432.0
y=
y
i=
10–47.
Determine the moment of inertia for the shaded area about
the y axis.
x
y
3 in. 3 in.
6 in.
3 in.
3 in.
3 in.
2 in.
1057
*10–48.
Determine the moment of inertia of the parallelogram
about the x′ axis, which passes through the centroid C of
the area.
SOLUTION
h=a sin u
y
b
x
C
a
θ
'
x
'
y
Ans:
10–49.
Determine the moment of inertia of the parallelogram
about the y′ axis, which passes through the centroid C of
the area.
SOLUTION
x=a cos u+b-a cos u
2=
1
2
(a cos u+b)
y
b
x
C
a
θ
'
x
'
y
1059
10–50.
Locate the centroid of the cross section and determine the
moment of inertia of the section about the axis.x¿
y
SOLUTION
Centroid:The area of each segment and its respective centroid are tabulated below.
0.2 m
0.05 m
0.4 m
0.2 m 0.2 m 0.2 m
0.3 m
x'
–
y
Segment Ai (m2)(dy)i (m) (Ady
2)i(m4)(Ix¿)i(m4)
(Ix¿)i (m4)
Ans:
10–51.
Determine the moment of inertia for the beam’s cross-
sectional area about the axis passing through the centroid
Cof the cross section.
x¿
SOLUTION
1061
*10–52.
SOLUTION
y
3in. 3in.
6in.
Determine the moment of inertia of the area about
the x axis.
1062
10–53.
Determine the moment of inertia of the area about the
yaxis.
SOLUTION
y
x
3 in. 3 in.
Ans:
1063
10–54.
y
l
t
Determine
the product of inertia of the thin strip of area
w
ith respect to the and axes.The strip is oriented at an
angle
from the axis. Assume that .
SOLUTION
tVlxu
yx
Ans:
10–55.
Determine the product of inertia of the shaded area with
respect to the xand yaxes.
SOLUTION
Differential Element: The area of the differential element parallel to the yaxis
shown shaded in Fig. ais .The coordinates of the centroid of
dA =y dx =1
9 x3 dx
Ans:
1065
*10–56.
Determine the product of inertia for the shaded portion of
the parabola with respect to the xand yaxes.
SOLUTION
Differential Element: Here,.The area of the differential element
parallel to the xaxis is .The coordinates of the centroid
for this element are ,.Then the product of inertia for this element is
y=yx =0
dA =2xdy =2250y1
2dy
x=250y1
2
200 mm
100 mm
x
y
yx
2
1
50
1066
10–57.
Determine the product of inertia of the shaded area with
respect to the xand yaxes, and then use the parallel-axis
theorem to find the product of inertia of the area with
respect to the centroidal and axes.
SOLUTION
Differential Element: The area of the differential element parallel to the yaxis
shown shaded in Fig. ais .The coordinates of the centroid of
this element are Thus, the product of inertia of this
element with respect to the xand yaxes is
x
'=x and y
'=y
2=1
2 x1>2
dA =y dx =x1>2 dx
y¿x¿
y2 x
2 m
y y¿
x
4 m
Cx¿
Ans:
1067
10–58.
SOLUTION
'
=x
Determ
i
ne t
h
e pro
d
uct of
i
nert
i
a for t
h
e para
b
o
li
c area w
i
t
h
respect to the xand yaxes.
y
x
a
b
yx
1/2
b
a
1/2
Ans:
10–59.
Determine the product of inertia of the shaded area with
respect to the xand yaxes.
SOLUTION
Differential Element: The area of the differential element parallel to the yaxis is
The coordinates of the centroid for this element are
Then the product of inertia for this element is
y
'=y
2=1
2
A
a1
2-x1
2
B
2.x
'=x,
dA =ydx =
A
a1
2-x1
2
B
2dx.
x
y
Oa
a
y=(a2–x
2)2
1
1
1069
*10–60.
SOLUTION
Differential Element: Here,The area of the differential element
parallel to the ydiortnec eht fo setanidrooc ehT si sixa
Product of Inertia: Performing the integration, we have
dA =ydx =24-x2dx.
y=24-x2.
Determine the product of inertia of the shaded area with
respect to the xand yaxes.
2in.
2in.
y
x
x2+y
2=4
Ans:
