1090
10–78.
SOLUTION
Ans.
Ans.
Ans.
I
min =435 in4
I
max =1.74(103)in4
=450 +1730
2;Aa450 –1730
2b2
+1382
I
max/min =
Ix+Iy
2;AaIx–Iy
2b2
+I2
xy
u
=6.08°
ta
n 2u=
–2Ixy
Ix–Iy
=
–2(138)
450–1730
The area of the cross section of an airplane wing has the
f
ollowing properties about the xand yaxes passing through
t
he centroid C:
D
etermine the orientation of the principal axes and the
prin
cipal moments of inertia.
Ixy =138 in4.Iy=1730in4,Ix=450 in4,
y
x
C
10–79.
Ans.
Center of circle:
Ans.
Ans.Imin =1090 –654.71 =435 in4
Imax =1090 +654.71 =1.74(103)in4
R=2(1730 –1090)2+(138)2=654.71 in4
Ix+Iy
2=450 +1730
2=1090 in4
u=6.08°
y
x
C
Solve Prob. 10–78 using Mohr’s circle.
*10–80.
Determine the moments and product of inertia for the
shaded area with respect to the u and v axes.
SOLUTION
y
x
v
u
10 mm
10 mm
10 mm
120 mm
60
10–81.
Solve Prob. 10–80 using Mohr’s circle.
SOLUTION
y
x
v
u
10 mm
10 mm
10 mm
120 mm
60
10–82.
Determine the directions of the principal axes with origin
located at point
O, and the principal moments of inertia for
the area about these axes
.
SOLUTION
T
hus,
Ans.
I
max
min
=
Ix+Iy
2;
C
aIx–Iy
2b2
+I2
xy
u
P1=-31.4°;
uP2=58.6°
uP=-31.388°;
58.612°
tan
2uP=
–Ixy
Ix–Iy
2
=
–118.87
(236.95 –114.65)
2
I
xy =[0 +(4)(6)(2)(3)] –[0 +p(1)(2)(4)] =118.87 in4
=114.65 in4
I
y=c1
12(6)(4)3+(4)(6)(2)2d–c1
4p(1)4+p(1)2(2)2d
=236.95 in4
I
x=c1
12(4)(6)3+(4)(6)(3)2d–c1
4p(1)4+p(1)2(4)2d
Ox
2 in.
y
2in.
2in.
1 in.
4in.
10–83.
Solve Prob. 10–82 using Mohr’s circle.
SOLUTION
2 in.
y
2in.
2in.
1 in.
*10–84.
SOLUTION
T
hus,
Ans.Iz=mR
2
m=L2p
0
rARdu=2prAR
Iz=L2p
0
rA(Rdu)R2=2prAR
3
Determine
the moment of inertia of the thin ring about the
z
axis.The ring has a mass m.
x
y
R
1098
10–85.
Determine the moment of inertia of the ellipsoid with respect
to the xaxis and express the result in terms of the mass mof
the ellipsoid.Thematerial has aconstant density r.
SOLUTION
Thus,
Ans.Ix=
2
5mb2
Ix=La
–a
1
2rpb4a1–
x2
a2b2
dx =
8
15 prab4
m=L
V
rdV =La
–a
rp b2a1–
x2
a2bdx =
4
3prab2
dI
x=
y2dm
dm=Lpy2dx
2
y
x
b
1
b
2
y
2
a
2
x
2
5
10–86.
SOLUTION
Determ
i
ne t
h
e ra
di
us of gyrat
i
on of t
h
e para
b
o
l
o
id
.T
h
e
density of the material is r=5Mg>m3.
kx
y
x
100mm
y
2
50 x
200mm
1100
10–87.
SOLUTION
The paraboloid is formed by revolving the shaded area
around the xaxis. Determine the moment of inertia about
the xaxis and express the result in terms of the total mass m
of the paraboloid. The material has a constant density .r
y
x
a
a2
–
hxy2=
3
1101
*10–88.
Determine the moment of inertia of the homogenous
triangular prism with respect to the yaxis. Express the
result in terms of the mass mof the prism. Hint: For
integration, use thin plate elements parallel to the x-y plane
having a thickness of dz.
SOLUTION
z
–h
a(x–a)z=
h
6
1102
10–89.
Determ
i
ne t
h
e moment of
i
nert
i
a of t
h
e sem
i
–e
lli
pso
id
w
i
t
h
respect to the xaxis and express the result in terms of the
mass mof the semiellipsoid. The material has a constant
density r.
SOLUTION
.The mass moment of inertia of this element is
.
Total Mass: Performing the integration,we have
Mass Moment of Inertia: Performing the integration,we have
The mass moment of inertia expressed in terms of the total mass is.
Ans.Ix=
2
5a2
3rp ab2bb2=
2
5mb2
=
4
15 rp ab4
=rp b4
2ax5
5a4–
2x3
3a2+xb
`
a
0
Ix=LdIx=La
0
rp b4
2ax4
a4–
2x2
a2+1bdx
=
2
3rpab2
m=Lm
dm =La
0
rp b2a1–x2
a2bdx =rp b2ax–x3
3a2b
`
a
0
dIx=1
2dmy2=1
2crp b2a1–x2
a2bdx dcb2a1–x2
a2bd=rp b4
2ax4
a4–
2x2
a2+1bdx
dm =rdV =rp y2dx =rp b2a1–x2
a2bdx
y
x
b
a
1
b
2
y
2
a
2
x
2
5
1103
10–90.
Determine the radius of gyration kx of the solid formed by
revolving the shaded area about x axis. The density of the
material is
r.
c
a
n+4b
rp ah2
y
x
h
a
yn x
a
hn
An+2
2(n
+
4) h
10–91.
The concrete shape is formed by rotating the shaded area
about the yaxis.Determine the moment of inertia The
specific weight of concrete is g=150 lb>ft3.
Iy.
SOLUTION
Ans.Iy=2.25 slug#ft2
=324.1 slug#in2
=
1
2p(150)
32.2(12)3
B
(10)4(8)–a9
2b2a1
3b(8)3
R
Iy=
1
2pr
B
L8
0
(10)4dy –L8
0a9
2b2
y2dy
R
=
1
2[pr (10)2dy](10)2–
1
2prx2dyx2
dI
y=
1
2(dm)(10)2–
1
2(dm)x2
y
x
8in.
6in. 4in.
2
9x
2
y
1105
*10–92.
Determine the moment of inertia of the sphere and
express the result in terms of the total mass mof the sphere.
The sphere has a constant density r.
Ix
SOLUTION
Thus,
Ans.Ix=
2
5mr2
=
4
3rpr3
m=Lr
–r
rp(r2–x2)dx
=
8
15 prr5
Ix=Lr
–r
1
2rp(r2–x2)2dx
dI
x=
1
2rp(r2–x2)2dx
dm =rdV =r(py2dx)=rp(r2–x2)dx
dI
x=
y2dm
2
x
y
r
x
2
y
2
r
2
5
1106
10–93.
The right circular cone is formed by revolving the shaded
area around the xaxis. Determine the moment of inertia
and express the result in terms of the total mass mof the
cone.The cone has a constant density .r
Ix
SOLUTION
dm =rdV =r(py2dx)
y
x
r
r
–
hxy
10