Namas ChandraIntroduction to Mechanical engineering

HibblerChapter 2-1

EML 3004C

Problem 6-5 (page 256)

Determine the location (x’,y’) of the centroid of the shaded area.

Solution:Area of the differential element dA = ydx = (1/x)dx and xc = x, yc= y/2 = 1/(2x)

xc

Axc

d

A1

d

xc0.5

2

xx1x

d

0.5

2

x1x

d

xc 1.08in

yc

Ayc

d

A1

d

yc0.5

2

x12x

1x

d

0.5

2

x1x

d

in yc 0.541in

Namas ChandraIntroduction to Mechanical engineering

HibblerChapter 2-2

EML 3004C

Problem 6-20 (page 259)

Determine the distance y’ to the center of gravity of the volume. The material is homogeneous.

Solution:

The volume of the differential thin disk element dV = y^2dz = (a^2-z^2)dz and zc = z

zc

Vz

d

V1

d

zc0

a

zz a2 z2

d

0

a

z a2 z2

d

zc38

a

Namas ChandraIntroduction to Mechanical engineering

HibblerChapter 2-3

EML 3004C

Problem 6-25 (page 263)

Determine the location (x’,y’) of the centroid of the area.

Solution:

yc

yc A

A yc

1 4 2( )[ ] 4 8 2( )[ ] 7 4 2( )[ ]2 4( ) 2( ) 8 2( )

in yc 4 in

xc

xc A

A xc

1 8 2( )[ ] 4 2 4( ) 2( )[ ]2 4( ) 2( ) 8 2( )

in xc 2.5in

Namas ChandraIntroduction to Mechanical engineering

HibblerChapter 2-4

EML 3004C

Problem 6-35 (page 265)

Determine the distance x’ to the center of gravity of the generator assembly. The weight and the center of gravity of each of the various components are indicated below. What are the vertical relations at blocks A and B needed to support the assembly?

Solution:W 200 1000 250 W 1450lb

xc W MA 1450xc 200 1( ) 10004( ) 250 8( )

xc 4.276ftEquilibrium :

Namas ChandraIntroduction to Mechanical engineering

HibblerChapter 2-5

EML 3004C

Problem 6-35 (continued)

xc 4.276ftEquilibrium :

MA 0 By 10( ) 14504.276( ) 0 By 620 lb

Fy 0 Ay 620 1450 0 Ay 830 lb

Namas ChandraIntroduction to Mechanical engineering

HibblerChapter 2-6

EML 3004C

Problem 6-76 (page 289)

Determine the moment of inertia of the shaded area about the x axis.

Solution:

Area of the differential element (shaded) dA = 2x dy where x = a/b(b^2-y^2)^1/2, hencedA = 2x dy = 2a/b(b^2-y^2)^1/2 dy

Ix Ax2

d Ix

2

2

xx2 114

x2

d Ix

2

2

xx2 14

x4

d

Ixb3

4

Namas ChandraIntroduction to Mechanical engineering

HibblerChapter 2-7

EML 3004C

Problem 6-82 (page 293)

The composite beam consists of a wide-flange beam and cover plates welded together as shown. Determine the moment of inertia of the cross-sectional area with respect to a horizontal axis passing through the beam’s centroid.

Solution:

Ixcxc112

175( ) 200( )3 112

160 1703 2112

275153 2 15 275 107.52

mm4

Ixcxc 1.467 108 mm4

Namas ChandraIntroduction to Mechanical engineering

HibblerChapter 2-8

EML 3004C

Problem 6-95 (page 295)

Determine the distance y’ to the centroid of the plate area.

Solution:

Area of the differential element dA = y dx = (1 - 1/4(x^2)) dx and yc = y/2 = 1/2(1-(1/4)x^2)

yc

Ayc

d

A1

d

yc2

2

x12

114

x2

114

x2

d

2

2

x114

x2

d

m yc 0.4m