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1030
10–21.
Determine the moment of inertia for the shaded area about
the x axis.
SOLUTION
Differential Element. Here
x2=y
and x1=
1
2
y2. The area of the differential element
y
x
2 m
2 m
y2 2x
y x
Ans:
10–22.
Determine the moment of inertia for the shaded area about
the y axis.
SOLUTION
Differential Element. Here, y
2x
1
2 and
y1=x
. The area of the differential
Moment of Inertia. Perform the integration,
Iy=
x2dA =
x2
a2
2x1
2-x
b
dx
y
x
2 m
2 m
y2 2x
y x
Ans:
10–23.
Determine the moment of inertia for the shaded area about
the x axis.
SOLUTION
Differential Element. Here x2=
a
b
1
2
y
1
2 and x1=
a
b
2y2. Thus, the area of the
differential element parallel to the x axis shown shaded in Fig. a is
dA =(x2-x1) dy
=
aa
b
1
2
y1
2-
a
b2y2
b
dy.
b
x
y
a
y2 —x
b2
a
y — x2
b
a2
1033
*10–24.
Determine the moment of inertia for the shaded area about
the y axis.
SOLUTION
Differential Element. Here, y2=
b
a
1
2
x1
2 and y1=
b
a
2
x2. Thus, the area of the
differential element parallel to the y axis shown shaded in Fig. a is
dA =(y2-y1)dx
=
a
b
a
1
2
x1
2-b
a2x2
b
dx
Moment of Inertia. Perform the integration,
b
x
y
a
y2 —x
b2
a
y — x2
b
a2
10–25.
Determine the moment of inertia of the composite area
about the xaxis.
SOLUTION
Composite Parts: The composite area can be subdivided into three segments as
shown in Fig. a.The perpendicular distance measured from the centroid of each
segment to the xaxis is also indicated.
y
x
6 in.
3 in.
3 in.
3 in.
10–26.
Determine
the moment of inertia of the composite area
about the
yaxis.
SOLUTION
Composite
Parts: The composite area can be subdivided into three segments as
shown
in Fig. a.The perpendicular distance measured from the centroid of each
segment to the
xaxis is also indicated.
y
x
6 in.
3 in.
3 in.
3 in.
1036
10–27.
The polar moment of inertia for the area is
JC
= 642 (106) mm4, about the z
′
axis passing through the
centroid C. The moment of inertia about the y
′
axis is
264 (106) mm4, and the moment of inertia about the x axis is
938 (106) mm4. Determine the area A.
SOLUTION
Applying the parallel-axis theorem with
d
y
=200 mm
and I
x=
938
(
10
6
)
mm
4
,
I
x=
I
x′
+Ad2
y
y
200 mm
Cx
¿
x
¿
1037
*10–28.
50 mm
50 mm
x
–
y
50 mm
350 mm
250 mm
Determine the location of the centroid of the channel’s
cross-sectional area and then calculate the moment of
inertia of the area about this axis.
y
SOLUTION
Centroid: The area of each segment and its respective centroid are tabulated below.
Segment A(mm2)y
'(mm) y
'A(mm3)
1 100(250) 125 3.12511062
Ans:
1038
10–29.
Determine
y
, which locates the centroidal axis
x′
for the
cross-sectional area of the T-beam, and then find the
moments of inertia
Ix′
and
I
y
′
.
SOLUTION
Centroid. Referring to Fig. a, the areas of the segments and their respective centroids
are tabulated below.
Segment
A(mm2)
y
∼(mm)
y
∼A(mm3)
1 150(20) 10 30
(103)
Moment of Inertia. The moment of inertia about the
x′
axis for each segment can
be determined using the parallel axis theorem, Ix′=Ix′+Ad
2
y. Referring to Fig. b,
Segment
Ai(mm2)
(d
y
)
i
(mm)
(Ix′)i
(mm4)
(Ad
y
2)
i
(mm4)
(Ix′)i
(mm4)
Thus
75 mm
x¿
y¿
C
75 mm
150 mm
20 mm
20 mm
y
10–30.
Determine the moment of inertia for the beam’s cross-
sectional area about the x axis.
SOLUTION
Segment A
i
(in
2
)
(d
y
)
i
(in.)
(
I
x′)i
(in4)
(Ady
2
)i
(in
4
) (I
x
)
i
(in
4
)
1 1(8) 4
1
12
(1)(83)128 170.67
8 in.
y
x
10 in.
3 in.
1 in.
1 in.
1 in.
SOLUTION
Moment of Inertia. The moments of inertia about the y axis for each segment can
be determined using the parallel axis theorem,
I
x=
I
x′
+Ad2
y
.
Referring to Fig. a,
Segment A
i
(in
2
)
(dx)i
(in.)
(Iy')i
(in
4
)(Adx
2
)
i
(in
4
) (Iy)i
(in
4
)
1 8(1) 9.5
1
12
(8)(1
3)722 722.67
10–31.
Determine the moment of inertia for the beam’s cross-
sectional area about the y axis.
8 in.
y
x
10 in.
3 in.
1 in.
1 in.
1 in.
SOLUTION
Moment of Inertia. The moment of inertia about the x axis for each segment can
be determined using the parallel axis theorem,
Ix=Ix′+Ad2y.
Referring to Fig. a
Segment A
i
(mm
2
)
(d
y
)
i
(mm)
(I
x′
)
i
(mm
4
) (Ady)
2
i
(mm
4
) (I
x
)
i
(mm
4
)
1 200(300) 150
(200)(3003)1.35(10
9
) 1.80(10
9
)
*10–32.
Determine the moment of inertia Ix of the shaded area
about the x axis.
Ox
150 mm
150 mm
100 mm 100 mm
75 mm
150 mm
y
Ans:
1042
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′
+Ad2
x
.
Referring to Fig. a
Segment A
i
(mm
2
)
(dx)i
(mm)
Iy
′
(mm
4
) (Ad
2
x
)
i
(mm
4
) (Iy)i
(mm
4
)
10–33.
Determine the moment of inertia Ix of the shaded area
about the x axis.
Ox
150 mm
150 mm
100 mm 100 mm
75 mm
150 mm
y
10–34.
C
y
x¿
250 mm
50 mm
150 mm
150 mm
Determine
the moment of inertia of the beam’s cross-
sectional area about the
yaxis.
SOLUTION
M
oment of Inertia: The dimensions and location of centroid of each segment are
shown
in Fig. a.Since the yaxis passes through the centroid of both segments, the
moment of inertia about
yaxis for each segment is simply
(Iy)i=(Iy¿)i.
10–35.
Determine which locates the centroidal axis for the
cross-sectional area of the T-beam, and then find the
moment of inertia about the x¿ axis.
x
¿
y
,
SOLUTION
y=
©yA
©A=125(250)(50) +(275)(50)(300)
250(50) +50(300)
C
y
x¿
x¿
250 mm
50 mm
150 mm
150 mm
SOLUTION
Moment of Inertia. Since the x axis passes through the centroids of the two segments,
Fig. a,
*10–36.
Determine the moment of inertia about the x axis.
150 mm
150 mm
y
x
C
200 mm
200 mm
20 mm
20 mm
20 mm
10–37.
Determine the moment of inertia about the y axis.
SOLUTION
Moment of Inertia. Since the y axis passes through the centroid of the two segments,
Fig. a,
150 mm
150 mm
y
x
C
200 mm
200 mm
20 mm
20 mm
20 mm
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
)
1 6(6) 3
1
12
(6)(63)324 432.0
10–38.
Determine the moment of inertia of the shaded area about
the x axis.
x
6 in.
3 in.
6 in.
y
6 in.
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
)
1 6(6) 3
(6)(63)324 432.0
Thus,
Iy=Σ(Iy)i=2376 in
4
Ans.
10–39.
Determine the moment of inertia of the shaded area about
the y axis.
x
6 in.
3 in.
6 in.
y
6 in.
1049
SOLUTION
Centroid. Referring to Fig. a, the areas of the segments and their respective centroids
are tabulated below.
Segment
A(in2)
y
∼(in.)
y
∼A(in3)
Σ
14.0 19.00
Segment A
i(in2)
(d
y
)
i
(in.)
(
I
x′)i
(in4)
(
Ady
2)
i
(in4)
(
I
x′)i
(in4)
*10–40.
Determine the distance
y
to the centroid of the beam’s
cross-sectional area; then find the moment of inertia about
the centroidal
x′
axis.
y
x
3 in. 1 in.
1 in.
4 in.
1 in.
y¿
x ¿
C
y
3 in.
Ans:
