SOLUTION
10–41.
Determine the moment of inertia for the beam’s cross-
sectional area about the y axis.
y
x
3 in. 1 in.
4 in.
1 in.
y¿
x ¿
C
y
3 in.
10–42.
SOLUTION
Ans.Ix=154(106)mm4
+1
12 (100)(30)3+100(30)(185)2
+1
12 (30)(170)3+30(170)(85)2
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
x¿
y¿
C
x
Determine the moment of inertia of the beam’s cross-
sectional area about the x axis.
10–43.
Determine the moment of inertia of the beam’s cross-
sectional area about the yaxis.
SOLUTION
30 mm
70 mm
30 mm
y
y¿
*10–44.
SOLUTION
Ans.
Ans.Ix¿=67.6(106)mm4
+1
12 (100)(30)3+100(30)(185 – 80.68)2
+c1
12 (30)(170)3+30(170)(85 – 80.68)2 d
Ix¿=c1
12 (170)(30)3+170(30)(80.68 –15)2d
=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
30 mm
70 mm
140 mm
30 mm
y
x
x¿
y¿
C
x
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
Ans.
Ans.Iy¿=41.2(106)mm4
+1
12 (30)(100)3+100(30)(50 – 61.59)2
+c1
12 (170)(30)3+30(170)(15 – 61.59)2 d
Iy¿=c1
12 (30)(170)3+170(30)(115 – 61.59)2 d
=61.59 =61.6 mm
x=170(30)(115) +170(30)(15) +100(30)(50)
170(30) +170(30) +100(30)
170 mm
30 mm
30 mm
70 mm
140 mm
30 mm
y
x
x¿
y¿
C
x
SOLUTION
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.
1057
*10–48.
Determine the moment of inertia of the parallelogram
about the x′ axis, which passes through the centroid C of
the area.
12
12
12
y
b
x
C
a
θ
‘
x
‘
y
12
10–49.
Determine the moment of inertia of the parallelogram
about the y′ axis, which passes through the centroid C of
the area.
SOLUTION
1
12
y
b
x
C
a
θ
‘
x
‘
y
10–50.
Locate the centroid of the cross section and determine the
moment of inertia of the section about the axis.x¿
y
SOLUTION
0.2 m
0.05 m
0.4 m
0.2 m 0.2 m 0.2 m
0.3 m
x’
–
y
Th
us,
Ans.
Moment
of Inertia:The moment of inertia about the axis for each segment can be
determined
using the parallel-axis theorem Ix¿=I
x
¿+Ady
2.
x¿
y=
©yA
©A=0.046042
0.255 =0.1806 m =0.181 m
Thus,
Ans.Ix¿=©Ix¿i=4.233 10–3m4=4.23 10–3m4
Segment A (m2)y (m) yA (m3)
1 0.3(0.4) 0.25 0.03
21
210.4210.420.1833 0.014667
3 1.1(0.05) 0.025 0.001375
©0.255 0.046042
Segment Ai (m2)(dy)i (m) (Ady
2)i(m4)(Ix¿)i(m4)
1 0.3(0.4) 0.06944 1
12 10.3210.4320.5787110–322.1787110–32
21
210.4210.420.002778 1
36 10.4210.4320.6173110–620.7117110–32
31.1(0.05) 0.1556 1
12 11.1210.05321.3309110–321.3423110–32
(Ix¿)i (m4)
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
–2c1
4(200)4ap
4–1
2sin90°bd
Ix¿=1
12(200)(332.8)3+4c1
36(141.4)(141.4)3+a1
2(141.4)(141.4)ba141.4
3b2d
x¿
100mm100mm
200mm
200mm
C
25 mm
25 mm
45
4545
45
*10–52.
SOLUTION
Ans.–
B
1
R
=1.19(103) in4
Ix=
B
1
12 (6)(10)3+6(10)(5)2
R
–
B
1
36 (3)(6)3+a1
2b(3)(6)(8)2
R
y
3in. 3in.
6in.
4in.
2in.
Determine the moment of inertia of the area about
the x axis.
10–53.
Determine the moment of inertia of the area about the
yaxis.
y
3 in. 3 in.
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
Ans.=1
6 l3t sin 2u
lxy =L
A
xydA =Ll
0
(s cos u)(s sin u)tds =sin u cos utL1
0
s2ds
tVlxu
yx
3
10–55.
y x
3
1
9
3 in.
y
Determine the product of inertia of the shaded area with
respect to the xand yaxes.
SOLUTION
1065
*10–56.
Determine the product of inertia for the shaded portion of
the parabola with respect to the xand yaxes.
SOLUTION
2
200 mm
100 mm
y
yx
2
1
50
L
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
this element are Thus, the product of inertia of this
element with respect to the xand yaxes is
Product of Inertia: Performing the integration, we have
Ans.
Using the information provided on the inside back cover of this book, the location of
the centroid of the parabolic area is at and
and its area is given by Thus,
Ans. Ix¿y¿=1.07 m4
10.67 =Ix¿y¿+5.333(2.4)(0.75)
Ixy =Ix¿y¿+Adxdy
A=2
3 (4)(2) =5.333 m2.
y=3
8 (2) =0.75 mx=4–2
5 (4) =2.4 m
Ixy =LdI
xy =L4 m
0
1
2 x2dx =a1
6x3b20
4 m
=10.67 m4=10.7 m4
=1
2 x2dx
=0+
A
x1>2 dx
B
(x)a1
2 x1>2b
dIxy =dIx¿y¿+dA~
x~
y
x
‘=x and y
‘=y
2=1
2 x1>2
y¿x¿
y2 x
2 m
y y¿
x
Cx¿
1067
10–58.
SOLUTION
1
=
1
6
B
ab2
abx3
R
a
0
=La
0
1
2ab2
abx2dx
Ixy =LdI
xy
dIxy =
xy2
2dx
dA =ydx
y
‘
=
y
2
x
‘
=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
6
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
Product of Inertia: Performing the integration, we have
Ans.=a4
280
=1
2
x4
4
+a2
2x2+2ax3–8
5a3
2x5
2–8
7a1
2x7
2
`
0
a
Ixy =LdIxy =1
2La
0
¢
x3+a2x+6ax2–4a3
2x3
2–4a1
2x5
2bdx
=1
2
A
x3+a2x+6ax2–4a3
2x3
2–4a1
2x5
2
B
dx
=0+c
A
a1
2–x1
2
B
2dx d1x2c1
2aa1
2–x1
2b2d
dIxy =dIx¿y¿+dAx
‘y
‘
y
‘=y
2=1
2
A
a1
2–x1
2
B
2.x
‘=x,
dA =ydx =
A
a1
2–x1
2
B
x
y
a
y=(a2–x
2)2
1
1
*10–60.
SOLUTION
parallel to the ydiortnec eht fo setanidrooc ehT si sixa
siht rof aitreni fo tcudorp eht nehT era tnemele siht rof
element is
Product of Inertia: Performing the integration, we have
Ans.=1
22x2–x4
40
2in.
=2.00 in4
Ixy =LdIxy =1
2L2in.
014x–x32dx
=1
214x–x32dx
=0+
A
24–x2dx
B
1x2a1
224–x2b
dIxy =dIx¿y¿+dAx
‘y
‘
y
‘=y
2=1
224–x2.x
‘=x,
dA =ydx =24–x2dx.
Determine the product of inertia of the shaded area with
respect to the xand yaxes.
2in.
y
x2+y
2=4