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

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⊳⊲9⊳3C21

12 ⊲9⊳⊲3⊳3C⊲6⊳2⊲9⊳⊲3⊳’ mm4

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 ð106D3.6.

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

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⊳⊲4m⊳3D21.3 m4,

Iy‘DIxCIy

pD28.2°,principal moments of inertia are

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

IxD10 ft4,I

yD22 ft4,I

xy D6ft

4.

From Eq. (8.26), tan 2pD2Ixy

644

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),

D1.26 C0.655

D1.20 ð106in4

Iy0DIxCIy

Ix0y0DIxIy

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

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.

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

RD21.364

C⊲32⊳2D38.5 m4

648

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

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

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

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

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′

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

⊲I0⊳1Dr2dm D2

0

Ax2dx DA x3

3

0

⊲I0⊳1D8

3A

I0TOTAL D13

3A C8

3A D21

3A

⊲I0⊳TOTAL D7A

The total mass D3A D6kg

2

654

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 x–yplane.

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

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

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

AD4

2.

The moment of inertia about the yaxis is

Iyaxis D4

4

x2υ41

4x2dx

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

3υ41

4x23

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

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

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

⊲1C1.5Cp2⊳D5.11 kg/m.

m1D5.11 kg,

d1D0.5m,

2

l2

2

m2D7.66 kg,

d2Dp0.752C12D1.25 m

3

⊲p2⊳2

m3D7.23 kg,

d3Dp22C0.52D2.062 m.

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

D0.5⊲5.11⊳C1⊲7.66⊳C0.5⊲7.23⊳

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⊲4⊳D31.9 in.

660