Mechanical Engineering Chapter 7 Problem The Crosssectional Area The Beam Shown Problem And The Coordinate The

Problem 7.38 If the cross-sectional area of the beam

shown in Problem 7.37 is 8400 mm2and the ycoordi-

nate of the centroid of the area is yD90 mm, what are

the dimensions band h?

Solution: From the solution to Problem 7.37

A1D120 b, A2D200 h

and yDy1A1Cy2A2

bD39.7mm

200 mm

120 mm

A2h

b

Problem 7.39 Determine the ycoordinate of the cen-

troid of the beam’s cross section.

x

y

2 in 5 in

8 in

Problem 7.40 Determine the coordinates of the cen-

troid of the airplane’s vertical stabilizer.

70°

48°

11 m

12.5 m

y

x

dD12.5mC11 m cot 70°D16.50 m

eD11 m tan 48°D12.22 m

11 m

48°

532

Problem 7.41 The area has elliptical boundaries. If

aD30 mm, bD15 mm, and εD6 mm, what is the x

coordinate of the centroid of the area?

y

b

Solution: The equation of the outer ellipse is

x2

⊲a Cε⊳2Cy2

⊲b Cε⊳2D1

and for the inner ellipse

x2

a2Cy2

b2D1

˛˛2

2

2D˛ˇ/4

(The area of a full ellipse is ˛ˇ so this checks.

Now for the composite area.

A1D2375 mm2A2D1414 mm2

Problem 7.42 By determining the xcoordinate of the

centroid of the area shown in Problem 7.41 in terms of

a,b, and ε, and evaluating its limit as ε!0, show that

the xcoordinate of the centroid of a quarter-elliptical

line is

3A2Dab

4

so x1A1D⊲a Cε⊳2⊲b Cε⊳

3

x2A2Da2b

3

C⊲2aCb⊳ε2Cε3⊳

Finally xDx1A1x2A2

A1A2

1

3⊲2ab Ca2⊳C⊲2aCb⊳ε Cε2ε

Problem 7.43 Three sails of a New York pilot

schooner are shown. The coordinates of the points are

(a)

Solution: Divide the object into three areas: (1) The triangle with

altitude 21 ft and base 20 ft. (2) The triangle with altitude 21 ft

(2) A2D42 ft2,

x2D16 C2

534

Problem 7.44 Determine the centroid of sail 2 in

Problem 7.43.

by 9.5 ft, (3) a triangle at the top with base of 9.5 ft and altitude of

(1) A1D3⊲20⊳

2D30 ft2,

y3D20 C2

33D22 ft

AD6.472 ft ,

yDA1y1CA2y2A3y3A4y4

Problem 7.45 Determine the centroid of sail 3 in

Problem 7.43.

Solution: Divide the object into six areas: (1) The triangle Oef,

(1) A1D36.75 ft2,

(4) A4D130.5ft

x4D17 ft,

y4D9.67 ft.

bga

AD10.877 ft

AD11.23 ft

Problem 7.46 In Active Example 7.5, suppose that the

distributed load is modiﬁed as shown. Determine the

reactions on the beam at Aand B.

Solution: We can treat the distributed load as two triangular distri-

buted loads. Using the area analogy, the magnitude of the left one

is 1

2⊲8m⊳⊲60 N/m⊳D240 N, and the magnitude of the right one is

Problem 7.47 Determine the reactions at Aand B.

200 lb/ft

4 ft

6 ft

B

A

200 lb/ft

536

Problem 7.48 In Example 7.6, suppose that the

distributed loads are modiﬁed as shown. Determine the

reactions on the beam at Aand B.

600 N/m

400 N/m

6 m 6 m

6 m

AB

400 N/m

Problem 7.49 In Example 7.7, suppose that the

distributed load acting on the beam from xD0toxD

10 ft is given by wD350 C0.3x3lb/ft. (a) Determine

the downward force and the clockwise moment about

Aexerted by the distributed load. (b) Determine the

reactions at the ﬁxed support.

10,000 ft-lb

2000 lb

w

y

A

x

10 ft 10 ft

Solution:

(a) The force and moment are

Problem 7.50 Determine the reactions at the ﬁxed

support A.

5 m

= 3(1 – x2/25) kN/m

x

A

y

Solution: The free-body diagram of the beam is: The downward

From the equilibrium equations

FxDAxD0,

538

Problem 7.51 An engineer measures the forces

exerted by the soil on a 10-m section of a building

y

x

A

10 m2 m

Solution:

(a) The total force is

jFjD333.3kN

(b) The moment about the origin is

3x31

4x4C0.2

5x510

0

jMjD1833.33 kN.

FD5.5m,

Problem 7.52 Determine the reactions on the beam at

Aand B.

3 kN/m

2 kN/m

Solution: Replace the distributed load with three equivalent single

forces.

The equilibrium equations

Fx:AxD0

2 kN

Problem 7.53 The aerodynamic lift of the wing is

described by the distributed load

wD300p10.04x2N/m.

The mass of the wing is 27 kg, and its center of mass is

located 2 m from the wing root R.

y

x

R

jFjD1178.1N.

The moment about the root due to the lift is

MD300 5

0

⊲10.04x2⊳1/2xdx,

540

Problem 7.54 Determine the reactions on the bar at A

and B.

x

2 ft

y

400 lb/ft

4 ft 4 ft

600 lb/ft

B

A

400 lb/ft

Solution: First replace the distributed loads with three equiva-

lent forces.

The equilibrium equations

Solving:

Problem 7.55 Determine the reactions on member AB

at Aand B.

6 ft

6 ft 6 ft

300 lb/ft

AB

C

542

Problem 7.56 Determine the axial forces in members

BD, CD, and CE of the truss and indicate whether they

are in tension (T) or compression (C). ADFH

G

E

C

2 m

2 m 2 m 2 m 2 m

8 kN/m

4 kN/m

B

Problem 7.57 Determine the reactions on member

ABC at Aand B.

160

mm

160

mm

160

mm

160 mm

240 mm

400 N/m

200 N/m

A

B

C

D

E

Solution: Work on the entire structure ﬁrst to ﬁnd the reactions

at A. Replace the distributed forces with equivalent concentrated forces

Fx:AxC160 N D0

MB:Ax⊲0.24 m⊳C⊲160 N⊳⊲0.04 m⊳

⊲48 N⊳⊲0.16 m⊳⊲96 N⊳⊲0.24 m⊳

7

p65 CD⊲0.32 m⊳C4

p65 CD⊲0.16 m⊳D0

Fx:AxCBxC160 N 4

p65 CD D0

Fy:AyCBy48 N 96 N 7

p65 CD D0

Solving:

AxD160 N,A

yD52 N

E

Ax

Ay

48 N

96 N

48 N

544

Problem 7.58 Determine the forces on member ABC

of the frame.

2 m

A

C

2 m 1 m

1 m

1 m 3 kN/m

B

Solution: The free body diagram of the member on which the

distributed load acts is

From the equilibrium equations

we obtain CyD16 kN. Then from the middle free body diagram,

we write the equilibrium equations

BX

(4 m)(3 kN/m) = 12 kN

DY

2 m 2 m

Problem 7.59 Use the method described in Active

Example 7.8 to determine the centroid of the truncated

cone.

y

Solution: Just as in Active Example 7.8, the volume of the disk

Problem 7.60 A grain storage tank has the form of a

surface of revolution with the proﬁle shown. The height

of the tank is 7 m and its diameter at ground level is

10 m. Determine the volume of the tank and the height

above ground level of the centroid of its volume.

y

x

10 m

7 m

y = ax1/2

Solution:

0

xDx3/37

0

x2/27

The volume is

VD7

a2xdx Da249

546

Problem 7.61 The object shown, designed to serve

as a pedestal for a speaker, has a proﬁle obtained by

revolving the curve yD0.167x2about the xaxis. What

is the xcoordinate of the centroid of the object?

x

z

y

0.75 m

Solution:

y = 0.167 x2

dV = y2dx

π

Problem 7.62 The volume of a nose cone is generated

by rotating the function yDx0.2x2about the xaxis.

(a) What is the volume of the nose cone?

(b) What is the xcoordinate of the centroid of

the volume?

y

x

z

Problem 7.63 Determine the centroid of the hemisphe-

rical volume.

zR

y

x

Solution: The equation of the surface of a sphere is x2Cy2C

z2DR2.

4R4.

Divide by the volume:

y

548

Problem 7.64 The volume consists of a segment of a

sphere of radius R. Determine its centroid.

R

2

x

y

z

Solution: The volume: The element of volume is a disk of radius

and thickness dx. The area of the disk is 2, and the element of

x

R

Problem 7.65 A volume of revolution is obtained

by revolving the curve x2

a2Cy2

b2D1 about the xaxis.

Determine its centroid.

x2

y2

y

550