Problem 8.48 Determine JOand kO.
Solution: For the rectangle,
JO2D1.030 ð107mm4
JODJO1JO2
JOD6.656 ð1071.030 ð107mm4
40 mm
y
kODJO
kOD82.1mm
Problem 8.49 Determine Ixy.
Solution:
A1D⊲80⊳⊲120⊳D9600 mm2
80 mm
y
Problem 8.50 Determine Ixand kx.
40 mm
x
80
mm
120
mm
20
mm
y
622
Problem 8.51 Determine Iyand ky.
Solution: From the solution to Problem 8.50, the centroid of the
(1) Rectangle about its centroid (40, 60) mm.
IyD4.99 ð106mm4
kyD24.5mm
y
+
Problem 8.52 Determine JOand kO.
Solution: From the solutions to Problems 8.51 and 8.52,
and AD8343 mm2
y
20 mm
Problem 8.53 Determine Iyand ky.y
x
12 in
20 in
Solution: Treat the area as a circular area with a half-circular
and ⊲Iy⊳2D1
8⊲12⊳4in4,
so IyD1
4⊲20⊳41
8⊲12⊳4D1.18 ð105in4.
The area is AD⊲20⊳21
2⊲12⊳2D1030 in2
so,k
yDIy
AD1.18 ð105
1.03 ð103
D10.7in
in.
20 in. 20 in.
yy
y
Problem 8.54 Determine JOand kO.
Solution: Treating the area as a circular area with a half-circular
cutout as shown in the solution of Problem 8.53, from Appendix B,
624
Problem 8.55 Determine Iyand kyif hD3m.
y
1.2 m
Solution: Break the composite into two parts, a rectangle and a
semi-circle.
d = 4R
3
π
To get moments about the xand yaxes, the ⊲dxc,d
yc⊳for the semi-
Iy0RD1
x
b
3 m
AC
AR
1.2 m
IyD4.27 m2
Problem 8.56 Determine Ixand kxif hD3m.
Solution: Break the composite into two parts, the semi-circle and
AcD2.26 m2
dyc D3.509 m
Area: ARDbh D7.20 m2
AR
y
3 m
h
b = 2.4 m
yc′
626
Problem 8.57 If IyD5m
4, what is the dimension
of h?
For the semicircle
Iy0cDIyD⊲1.2⊳4/8D0.814 m2
Also, we know IyR CIyc D5m
4.
Hence 0.814 C1
12 ⊲h⊳⊲2.4⊳3D5
Solving, hD3.63 m
y′
R
xc′
1.2 m
Problem 8.58 Determine Iyand ky.
Solution: Let the area be divided into parts as shown. The areas
and the coordinates of their centroids are
A1D⊲40⊳⊲50⊳D2000 in2,x1D25 in, y1D20 in,
The moment of inertia of the composite area about the yaxis is
AD27.8 in.
Problem 8.59 Determine Ixand kx.
Solution: See the solution to Problem 8.58.
Let the area be divided into parts as shown. The areas and the coordi-
nates of their centroids are
A1D⊲40⊳⊲50⊳D2000 in2,x1D25 in, y1D20 in,
AD38.6 in.
628
Problem 8.60 Determine Ixy.
Solution: See the solution to Problem 8.58.
Let the area be divided into parts as shown. The areas and the coordi-
nates of their centroids are
Ixy D2.54 ð106in4.
Problem 8.61 Determine Iyand ky.
y
Solution: See the solution to Problem 8.58.
In terms of the coordinate system used in Problem 8.58, the areas and
the coordinates of their centroids are
AD23.9 in
The moment of inertia about the yaxis in terms of the coordinate
AD14.1 in.
Problem 8.62 Determine Ixand kx.
y
Solution: See the solution to Problem 8.59.
In terms of the coordinate system used in Problem 8.59, the areas and
the coordinates of their centroids are
AD33.3 in
The moment of inertia about the xaxis in terms of the coordinate
IxD4.94 ð106in4⊲33.3 in⊳2⊲3310 in2⊳D1.26 ð106in4.
The radius of gyration about the xaxis is kxDIx
Problem 8.63 Determine Ixy.
Solution: See the solution to Problem 8.60.
Ixy D[0 C⊲1.0m⊳⊲0.8m⊳⊲dx0.5m⊳⊲dy0.4m⊳]
Check using the noncentroidal product of inertia from Problem 8.60
630
Problem 8.64 Determine Iyand ky.
x
y
18 in
6 in 6 in 6 in
Solution: Divide the area into three parts:
Iyy1D1
12 18⊲183⊳D8748 in4.
2D127.23 in2,
x3D9in,
The composite area:
AD10.461 in
Problem 8.65 Determine Ixand kx.
AD15.33 in
and 8.64.
Ixy Dx1y1A1x2y2A2Cx3y3A3
Problem 8.67 Determine Iyand ky.
6 in
2 in
y
Solution: We divide the composite area into a triangle (1), rect-
angle (2), half-circle (3), and circular cutout (4):
Triangle:
632
Problem 8.68 Determine JOand kO.
Solution: Iyis determined in the solution to Problem 8.67. We
will determine Ixand use the relation JODIxCIy. Using the figures
in the solution to Problem 8.67,
Triangle:
Problem 8.69 Determine Iyand ky.
y
4 in
2 in
Solution: Divide the area into four parts: Part (1) The rectangle
8 in by 16 in. Part (2): The rectangle 4 in by 8 in. Part (3) The semi-
Iyy1D1
12 8⊲163⊳D2730.7in
4.
2D25.133 in2,
8⊲44⊳4⊲4⊳
32
Part (4): A4D⊲22⊳D12.566 in2,
x4D12 in,
The area moment of inertia:
Problem 8.70 Determine Ixand kx.
IxD8.89 ð103in4
kxDIx
634
Problem 8.71 Determine Ixy.
Solution: Use the results in the solution to Problem 8.69.
Problem 8.72 Determine Iyand ky.
y
4 in
172.567 D9.033 in,
from which
4
Problem 8.73 Determine Ixand kx.
AD5.942 in,
AD4.03 in
Problem 8.74 Determine Ixy.
Problem 8.75 Determine Iyand ky.
5 mm
15 mm
y
Solution: We divide the area into parts as shown:
⊲Iy⊳1D1
12 ⊲50 C15 C15⊳⊲30⊳3D180,000 mm4
321
2⊲15⊳2D353,274 mm4
⊲Iy⊳8D⊲Iy⊳9D⊲Iy⊳10 D1
4⊲5⊳4C⊲25⊳2⊲5⊳2D49,578 mm4.
IyD⊲Iy⊳1C3⊲Iy⊳2C3⊲Iy⊳53⊲Iy⊳8D1.46 ð106mm4.
The area is
ADA1C3A2C3A53A8
D⊲30⊳⊲80⊳C3⊲10⊳⊲30⊳C31
2⊲15⊳23⊲5⊳2
D4125 mm2
so kyDIy
AD1.46 ð106
4125 D18.8mm
15
mm
10
mm
125
8
y
Problem 8.76 Determine JOand kO.
Solution: Iyis determined in the solution to Problem 8.75. We
will determine Ixand use the relation JODIxCIy. Dividing the area
as shown in the solution to Problem 8.75, we obtain
Therefore
IxD⊲Ix⊳1C⊲Ix⊳2C2⊲Ix⊳3C⊲Ix⊳5C2⊲Ix⊳6⊲Ix⊳82⊲Ix⊳9
636
Problem 8.77 Determine Ixand Iyfor the beam’s
cross section.
y
2 in 5 in
Solution: Use the symmetry of the object
Ix
2D1
3⊲3in.⊳⊲8in⊳3C1
12 ⊲3in⊳⊲3in⊳3C⊲3in⊳2⊲11.5in⊳2
Problem 8.78 Determine Ixand Iyfor the beam’s
cross section.
y
8 in
Solution: Use Solution 8.77 and 7.39. From Problem 7.39 we
know that
yD7.48 in,AD98.987 in2
Problem 8.79 The area AD2ð104mm2. Its moment
of inertia about the yaxis is IyD3.2ð108mm4. Deter-
mine its moment of inertia about the Oyaxis.
A
yy
ˆ
x,x
ˆ
Solution: Use the parallel axis theorem. The moment of inertia
about the centroid of the figure is
The moment of inertia about the Oyaxis is
Problem 8.80 The area AD100 in2and it is
symmetric about the x0axis. The moments of inertia
Ix0D420 in4,Iy0D580 in4,JOD11000 in4, and Ixy D
4800 in4. What are Ixand Iy?
x
y
O
x’
y’
A
O’
Solution: The basic relationships:
(2) IyDx2ACIyc,
(4) JODIxCIy,
(6) Ixy DAxy CIxyc,
where the subscript capplies to the primed axes, and the others to the
unprimed axes. The x,yvalues are the displacement of the primed
axes from the unprimed axes. The steps in the demonstration are:
(i) From symmetry about the xcaxis, the product of inertia Ixyc D0.
638
Problem 8.81 Determine the moment of inertia of the
beam cross section about the xaxis. Compare your result
with the moment of inertia of a solid square cross section
of equal area. (See Example 8.5.)
x
y
20 mm
160 mm
Solution: We first need to find the location of the centroid of the
composite. Break the area into two parts. Use X,Ycoords.
y
100
XC2 = 0
A2 = 3200 mm2
20
2
1
Ix01D1
1D1
Ix1D6.20 ð106mm4
IxD1.686 ð107mm4
IxSQ D1
2.253 ð106
Problem 8.82 The area of the beam cross section is
5200 mm2. Determine the moment of inertia of the beam
cross section about the xaxis. Compare your result with
the moment of inertia of a solid square cross section of
equal area. (See Example 8.5.) x
y
2.253 ð106D1.7788 D1.78
which confirms the value given in Example 8.5.
640
c
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