Mechanical Engineering Chapter 8 Problem The Beam Fig Subjected Toyy Couples Magnitude About The Axis Fig

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subject Authors Anthony M. Bedford, Wallace Fowler

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Problem 8.83 If the beam in Fig. a is subjected to
couples of magnitude Mabout the xaxis (Fig. b), the
beam’s longitudinal axis bends into a circular are whose
radius Ris given by
RDEIx
M,
where Ixis the moment of inertia of the beam’s cross
section about the xaxis. The value of the term E, which
is called the modulus of elasticity, depends on the mate-
rial of which the beam is constructed. Suppose that a
beam with the cross section shown in Fig. c is subjected
to couples of magnitude MD180 N-m. As a result,
the beam’s axis bends into a circular arc with radius
RD3 m. What is the modulus of elasticity of the beam’s
material? (See Example 8.5.)
R
MM
x
yy
z
y
3 mm
(b) Subjected to couples at the ends.
(a) Unloaded.
Solution: The moment of inertia of the beam’s cross section about
the xaxis is
IxD&1
12 3⊳⊲93C21
12 9⊳⊲33C629⊳⊲3' mm4
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Problem 8.84 Suppose that you want to design a beam
made of material whose density is 8000 kg/m3. The
beam is to be 4 m in length and have a mass of
320 kg. Design a cross section for the beam so that
8.333 ð106D3.6.
From Figure 8.6, this ratio suggests an I-beam of the form shown in
–1
–.8
.6
.8
1
f
I - beam Flange dimension
bD0.0395 m. and bD0.09 m. The lower value is an allowable value
hD39.5mm.
Problem 8.85 The area in Fig. (a) is a C230ð30
American Standard Channel beam cross section. Its cross
sectional area is AD3790 mm2and its moments of
inertia about the xand yaxes are IxD25.3ð106mm4
and IyD1ð106mm4. Suppose that two beams with
C230ð30 cross sections are riveted together to obtain a
composite beam with the cross section shown in Fig. (b).
What are the moments of inertia about the xand yaxes
of the composite beam?
14.8 mm
(a)
x
y
(b)
x
y
642
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Problem 8.86 The area in Fig. (a) is an L152ð102ð
12.7 Angle beam cross section. Its cross sectional
area is AD3060 mm2and its moments of inertia
about the xand yaxes are IxD7.24 ð106mm4and
IyD2.61 ð106mm4. Suppose that four beams with
L152ð102ð12.7 cross sections are riveted together to
obtain a composite beam with the cross section shown
in Fig. (b). What are the moments of inertia about the x
and yaxes of the composite beam?
24.9
mm
x
y
x
y
D59.8ð106mm4
Problem 8.87 In Active Example 8.6, suppose that the
vertical 3-m dimension of the triangular area is increased
to 4 m. Determine a set of principal axes and the corre-
sponding principal moments of inertia.
y
3 m
Solution: From Appendix B, the moments and products of inertia
of the area are
12 4m⊳⊲4m3D21.3 m4,
IyDIxCIy
pD28.2°,principal moments of inertia are
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Problem 8.88 In Example 8.7, suppose that the area is
reoriented as shown. Determine the moments of inertia
Ix0,I
y0and Ix0y0if D30o.
x
y
4 ft
1 ft
3 ft
1 ft
Solution: Based on Example 8.7, the moments and product of
inertia of the reoriented area are
IxD10 ft4,I
yD22 ft4,I
xy D6ft
4.
Applying Eqs. (8.23)(8.25),
Ix0DIxCIy
Problem 8.89 In Example 8.7, suppose that the area
is reoriented as shown. Determine a set of principal
axes and the corresponding principal moments of inertia.
Based on the result of Example 8.7, can you predict a
value of pwithout using Eq. (8.26)?
x
y
4 ft
1 ft
1 ft
IxD10 ft4,I
yD22 ft4,I
xy D6ft
4.
From Eq. (8.26), tan 2pD2Ixy
644
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Problem 8.90 The moment of inertia of the area are
IxD1.26 ð106in4,
IyD6.55 ð105in4,
Ixy D1.02 ð105in4
Determine the moments of inertia of the area Ix0,I
y0and
Ix0y0if D30°.x
y
x
y
u
Solution:
Applying Eqs. (8.23)(8.25),
D1.26 C0.655
D1.20 ð106in4
Iy0DIxCIy
Ix0y0DIxIy
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Problem 8.91 The moment of inertia of the area are
IxD1.26 ð106in4,
IyD6.55 ð105in4,
Ixy D1.02 ð105in4
Determine a set of principal axes and the corresponding
principal moments of inertia. x
y
x
y
u
Solution: From Eq. (8.26),
646
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Problem 8.92* Determine a set of principal axes and
the corresponding principal moments of inertia.
y
40 mm
160 mm
Solution: We divide the area into 3 rectangles as shown: In terms
C1
and (8.24) and setting D12.1°, we obtain
Ix1D80.2ð106mm4
Iy1D27.7ð106mm4.
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Problem 8.93 Solve Problem 8.87 by using Mohr’s
Circle.
y
3 m
Problem 8.87, the moments and product of inertia for the unrotated
system are
IxD1
2D42.7 m4
RD21.364
C322D38.5 m4
648
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Problem 8.94 Solve Problem 8.88 by using Mohr’s Circle.
x
y
4 ft
1 ft
3 ft
1 ft
Solution: Based on Example 8.7, the moments and product of
inertia of the reoriented area are
IxD10 ft4,I
yD22 ft4,I
xy D6ft
4.
For Mohr’s circle we have the center, radius, and angle
CD10 C22
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Problem 8.95 Solve Problem 8.89 by using Mohr’s Circle.
y
4 ft
1 ft
3 ft
1 ft
IxD10 ft4,I
yD22 ft4,I
xy D6ft
For Mohr’s circle we have the center, radius, and angle
CD10 C22
650
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Problem 8.96 Solve Problem 8.90 by using Mohr’s Circle.
x
y
x
y
u
Solution: For Mohr’s circle we have the center, radius, and angle
page-pfc
Problem 8.97 Solve Problem 8.91 by using Mohr’s Circle.
x
y
x
y
u
Solution: For Mohr’s circle we have the center, radius, and angle
652
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Problem 8.98* Solve Problem 8.92 by using Mohr’s
circle.
IxD77.91 ð106mm4
106
Problem 8.99 Derive Eq. (8.22) for the product of
inertia by using the same procedure we used to derive
Eqs. (8.20) and (8.21).
Solution: Suppose that the area moments of inertia of the area A
x
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Problem 8.100 The axis LOis perpendicular to both
segments of the L-shaped slender bar. The mass of the
bar is 6 kg and the material is homogeneous. Use the
method described in Example 8.10 to determine is moment
1 m
r
2
1
2x
0x
dy
dx
⊲I01Dr2dm D2
0
Ax2dx DA x3
3
0
⊲I01D8
3A
I0TOTAL D13
3A C8
3A D21
3A
⊲I0TOTAL D7A
The total mass D3A D6kg
2
654
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Problem 8.101 Two homogenous slender bars, each
of mass mand length l, are welded together to form
the T-shaped object. Use integration to determine the
moment of inertia of the object about the axis through
point Othat is perpendicular to the bars.
l
Ol
Problem 8.102 The slender bar lies in the xyplane.
Its mass is 6 kg and the material is homogeneous. Use
integration to determine its moment of inertia about the
zaxis.
y
2 m
50
moment of inertia of the slender bar in Problem 8.102
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Problem 8.104 The homogeneous thin plate has mass
mD12 kg and dimensions bD2 m and hD1 m. Use
the procedure described in Active Example 8.9 to
determine the moments of inertia of the plate about the
y
Solution: From Appendix B, the moments of inertia about the x
Problem 8.105 The homogenous thin plate is of
uniform thickness and mass m.
(a) Determine its moments of inertia about the xand z
axes.
(b) Let RiD0, and compare your results with the
values given in Appendix C for a thin circular
plate.
Ro
Ri
x
y
Solution:
A, thus for the plate with a circular cut,
m
RiRo
656
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Problem 8.106 The homogenous thin plate is of
uniform thickness and weighs 20 lb. Determine its
moment of inertia about the yaxis.
y
y = 4 – x2 ft
1
4
Solution:
The plate’s area is
AD4
2.
The moment of inertia about the yaxis is
Iyaxis D4
4
x2υ41
4x2dx
D1.99 slug-ft2.
xdx
y
1
4
Problem 8.107 Determine the moment of inertia of the
plate in Problem 8.106 about the xaxis.
Solution: See the solution of Problem 8.106. The mass of the strip
element is
3υ41
4x23
so the moment of inertia of the plate about the xaxis is
1
Problem 8.108 The mass of the object is 10 kg. Its
moment of inertia about L1is 10 kg-m2. What is its
moment of inertia about L2? (The three axes lie in the
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Problem 8.109 An engineer gathering data for the
design of a maneuvering unit determines that the
astronaut’s center of mass is at xD1.01 m, yD0.16 m
and that her moment of inertia about the zaxis is
105.6 kg-m2. Her mass is 81.6 kg. What is her moment
of inertia about the z0axis through her center of mass?
y
x
x
y
D1.0226 m.
Problem 8.110 Two homogenous slender bars, each of
mass mand length l, are welded together to form the T-
shaped object. Use the parallel axis theorem to determine
the moment of inertia of the object about the axis through
point Othat is perpendicular to the bars.
l
Ol
658
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Problem 8.111 Use the parallel-axis theorem to deter-
mine the moment of inertia of the T-shaped object in
Problem 8.110 about the axis through the center of mass
of the object that is perpendicular to the two bars. (See
Active Example 8.11.)
Solution: The location of the center of mass of the object is xD
Problem 8.112 The mass of the homogenous slender
bar is 20 kg. Determine its moment of inertia about the
y'
x
y
1.5 m 1 m
Dm
LD20
1C1.5Cp2D5.11 kg/m.
m1D5.11 kg,
d1D0.5m,
2
l2
2
m2D7.66 kg,
d2Dp0.752C12D1.25 m
3
p22
m3D7.23 kg,
d3Dp22C0.52D2.062 m.
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Problem 8.113 Determine the moment of inertia of the
bar in Problem 8.112 about the z0axis through its center
yD0.5m1C1m2C0.5m3
20
D0.55.11C17.66C0.57.23
Problem 8.114 The homogeneous slender bar weighs
5 lb. Determine its moment of inertia about the zaxis.
4 in
y'
y
Solution: The Bar’s mass is mD5/32.2 slugs. Its length is
LDL1CL2CL3D8Cp82C82C⊲4D31.9 in.
660

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