Solution Manual for Mechanics of Materials 7th Edition by Beer
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CHAPTER 1

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Solution Manual for Mechanics of Materials 7th Edition by Beer
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Solution Manual for Mechanics of Materials 7th Edition by Beer
Full file at />d1

PROBLEM 1.1

d2

125 kN B

C

A

Two solid cylindrical rods AB and BC are
welded together at B and loaded as shown.
Knowing that d1  30 mm and d 2  50 mm,
find the average normal stress at the
midsection of (a) rod AB, (b) rod BC.

60 kN
125 kN

0.9 m

1.2 m

SOLUTION
(a)

Rod AB:
Force:

P  60  103 N tension

Area:

A

Normal stress:
(b)

 AB 


4

d12 


4

(30  103 ) 2  706.86  106 m 2


P
60  103

 84.882  106 Pa
A 706.86  106

 AB  84.9 MPa 

Rod BC:
Force:

P  60  103  (2)(125  103 )  190  103 N

Area:

A

Normal stress:

 BC 


4

d 22 


4


(50  103 )2  1.96350  103 m 2

P
190  103

 96.766  106 Pa
A 1.96350  103

 BC  96.8 MPa 

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Solution Manual for Mechanics of Materials 7th Edition by Beer
Full file at />d1

PROBLEM 1.2

d2

125 kN B

C

A


Two solid cylindrical rods AB and BC are
welded together at B and loaded as shown.
Knowing that the average normal stress must
not exceed 150 MPa in either rod, determine
the smallest allowable values of the diameters
d1 and d2.

60 kN
125 kN
0.9 m

1.2 m

SOLUTION
(a)

Rod AB:
Force:

P  60  103 N

Stress:

 AB  150  106 Pa
 2
A

Area:

 AB


4

d1
4
P
P

 A
A
 AB

d12 
d12 

P

 AB
4P

 AB



(4)(60  103 )
 509.30  106 m 2
 (150  106 )

d1  22.568  103 m


(b)

d1  22.6 mm 

Rod BC:
Force:
Stress:
Area:

P  60  103  (2)(125  103 )  190  103 N

 BC  150  106 Pa
 2
A

 BC

d2
4
P
4P


A  d 22

d 22 

4P

 BC




(4)(190  103 )
 1.61277  103 m 2
 (150  106 )

d 2  40.159  103 m

d 2  40.2 mm 

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Solution Manual for Mechanics of Materials 7th Edition by Beer
Full file at />PROBLEM 1.3

A

Two solid cylindrical rods AB and BC are welded together at B and loaded as
shown. Knowing that P = 10 kips, find the average normal stress at the
midsection of (a) rod AB, (b) rod BC.

30 in.
1.25 in.
B

12 kips
25 in.
0.75 in.
C
P

SOLUTION
(a)

Rod AB:
P  12  10  22 kips
A

 AB
(b)



d12 



(1.25) 2  1.22718 in 2

4
4
P
22



 17.927 ksi
A 1.22718

 AB  17.93 ksi 

Rod BC:
P  10 kips

 AB



d 22 



(0.75)2  0.44179 in 2
4
4
P
10


 22.635 ksi
A 0.44179

A

 AB  22.6 ksi 


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Solution Manual for Mechanics of Materials 7th Edition by Beer
Full file at />PROBLEM 1.4

A

Two solid cylindrical rods AB and BC are welded together at B and loaded as
shown. Determine the magnitude of the force P for which the tensile stresses in
rods AB and BC are equal.

30 in.
1.25 in.
B
12 kips
25 in.
0.75 in.
C
P

SOLUTION
(a)

Rod AB:
P  P  12 kips

A

d2
4




4

(1.25 in.)2

A  1.22718 in 2

 AB 
(b)

P  12 kips
1.22718 in 2

Rod BC:
P P
A


4

d2 



4

(0.75 in.)2

A  0.44179 in 2

 BC 

P
0.44179 in 2

 AB   BC
P  12 kips
P

2
1.22718 in
0.44179 in 2
5.3015  0.78539 P

P  6.75 kips 

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Solution Manual for Mechanics of Materials 7th Edition by Beer

Full file at />1200 N

PROBLEM 1.5
A strain gage located at C on the surface of bone AB indicates that the average normal stress
in the bone is 3.80 MPa when the bone is subjected to two 1200-N forces as shown.
Assuming the cross section of the bone at C to be annular and knowing that its outer diameter
is 25 mm, determine the inner diameter of the bone’s cross section at C.

A

C

B

1200 N

SOLUTION

 
Geometry:

A


4

P
P
 A
A



(d12  d 22 )
d 22  d12 

4A



 d12 

d 22  (25  103 )2 

4P



(4)(1200)
 (3.80  106 )

 222.92  106 m 2
d 2  14.93  103 m

d 2  14.93 mm 

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Solution Manual for Mechanics of Materials 7th Edition by Beer
Full file at />PROBLEM 1.6

A
a

15 mm
B

100 m
b

Two brass rods AB and BC, each of uniform diameter, will be brazed together
at B to form a nonuniform rod of total length 100 m, which will be suspended
from a support at A as shown. Knowing that the density of brass is 8470 kg/m3,
determine (a) the length of rod AB for which the maximum normal stress in
ABC is minimum, (b) the corresponding value of the maximum normal stress.

10 mm

C

SOLUTION
Areas:

AAB 
ABC 


4


4

(15 mm) 2  176.715 mm 2  176.715  106 m 2
(10 mm)2  78.54 mm 2  78.54  106 m 2

b  100  a

From geometry,
Weights:



WAB   g AAB AB  (8470)(9.81)(176.715  106 ) a  14.683 a
WBC   g ABC  BC  (8470)(9.81)(78.54  106 )(100  a)  652.59  6.526 a

Normal stresses:
At A,

PA  WAB  WBC  652.59  8.157a

A 
At B,

(a)

PA
 3.6930  106  46.160  103a

AAB

PB  WBC  652.59  6.526a

B 

(1)

(2)

PB
 8.3090  106  83.090  103a
ABC

Length of rod AB. The maximum stress in ABC is minimum when  A   B or
4.6160  106  129.25  103a  0

a  35.71 m
(b)

 AB  a  35.7 m 

Maximum normal stress.

 A  3.6930  106  (46.160  103 )(35.71)
 B  8.3090  106  (83.090  103 )(35.71)
 A   B  5.34  106 Pa

  5.34 MPa 


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Solution Manual for Mechanics of Materials 7th Edition by Beer
Full file at />PROBLEM 1.7
0.4 m
C
0.25 m

0.2 m

B

Each of the four vertical links has an 8  36-mm uniform rectangular
cross section and each of the four pins has a 16-mm diameter. Determine
the maximum value of the average normal stress in the links connecting
(a) points B and D, (b) points C and E.

E

20 kN
D
A

SOLUTION
Use bar ABC as a free body.


M C  0 :

(0.040) FBD  (0.025  0.040)(20  103 )  0
FBD  32.5  103 N

Link BD is in tension.
3

M B  0 :  (0.040) FCE  (0.025)(20  10 )  0
FCE  12.5  103 N

Link CE is in compression.

Net area of one link for tension  (0.008)(0.036  0.016)  160  106 m 2
For two parallel links,
(a)

 BD 

A net  320  106 m 2

FBD
32.5  103

 101.563  106
6
Anet
320  10


 BD  101.6 MPa 

Area for one link in compression  (0.008)(0.036)  288  106 m 2
For two parallel links,
(b)

 CE 

A  576  106 m 2

FCE
12.5  103

 21.701  106
6
A
576  10

 CE  21.7 MPa 

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Solution Manual for Mechanics of Materials 7th Edition by Beer
Full file at />PROBLEM 1.8
B


2 in.

Link AC has a uniform rectangular cross section

12 in.

in. thick and 1 in. wide.

Determine the normal stress in the central portion of the link.

120 lb
4 in.
30Њ

1
8

120 lb

A
C

10 in.

8 in.

SOLUTION
Use the plate together with two pulleys as a free body. Note that the cable tension causes at 1200 lb-in.
clockwise couple to act on the body.


M B  0:  (12  4)( FAC cos 30)  (10)( FAC sin 30)  1200 lb  0
FAC  

1200 lb
 135.500 lb
16 cos 30  10 sin 30

Area of link AC:
Stress in link AC:

1
in.  0.125 in 2
8
F
135.50
 AC  
 1084 psi  1.084 ksi
A
0.125

A  1 in. 

 AC



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Solution Manual for Mechanics of Materials 7th Edition by Beer
Full file at />PROBLEM 1.9

0.100 m
E
P

P

P
D

A

0.150 m

B

Three forces, each of magnitude P  4 kN, are applied to the mechanism
shown. Determine the cross-sectional area of the uniform portion of rod
BE for which the normal stress in that portion is 100 MPa.

C

0.300 m


0.250 m

SOLUTION
Draw free body diagrams of AC and CD.

Free Body CD:

M D  0: 0.150P  0.250C  0
C  0.6 P

Free Body AC:

Required area of BE:

M A  0: 0.150 FBE  0.350 P  0.450 P  0.450C  0
FBE 

1.07
P  7.1333 P  (7.133)(4 kN)  28.533 kN
0.150

 BE 

FBE
ABE

ABE 

FBE


 BE



28.533  103
 285.33  106 m 2
100  106
ABE  285 mm 2 

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Solution Manual for Mechanics of Materials 7th Edition by Beer
Full file at />4 kips

6

C

in.

u

B

1


308

Link BD consists of a single bar 1 in. wide and
1
in. thick. Knowing that each pin has a 83 -in.
2
diameter, determine the maximum value of the
average normal stress in link BD if (a)  = 0,
(b)  = 90.

.
2 in

A

PROBLEM 1.10

D

SOLUTION
Use bar ABC as a free body.

(a)

  0.
M A  0: (18 sin 30)(4)  (12 cos30) FBD  0
FBD  3.4641 kips (tension)

Area for tension loading:

Stress:
(b)

3  1 

A  (b  d )t  1     0.31250 in 2
8  2 

F
3.4641 kips
  BD 
A
0.31250 in 2

  11.09 ksi 

  90.
M A  0:  (18 cos30)(4)  (12 cos 30) FBD  0
FBD  6 kips i.e. compression.

Area for compression loading:
Stress:

1
A  bt  (1)    0.5 in 2
2
F
6 kips
  BD 
A

0.5 in 2

  12.00 ksi 

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Solution Manual for Mechanics of Materials 7th Edition by Beer
Full file at />B

D

PROBLEM 1.11

F

12 ft
H

A

C
9 ft

E
9 ft


80 kips

For the Pratt bridge truss and loading shown, determine the
average normal stress in member BE, knowing that the crosssectional area of that member is 5.87 in2.

G
9 ft

80 kips

9 ft
80 kips

SOLUTION


Use entire truss as free body.
M H  0: (9)(80)  (18)(80)  (27)(80)  36 Ay  0
Ay  120 kips
Use portion of truss to the left of a section cutting members
BD, BE, and CE.



   Fy  0: 120  80 

 BE 

12

FBE  0
15

 FBE  50 kips

FBE
50 kips

A
5.87 in 2

 BE  8.52 ksi 



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Solution Manual for Mechanics of Materials 7th Edition by Beer
Full file at />45 in.

A

B

PROBLEM 1.12


30 in.
C

480 lb
4 in.

4 in.

40 in.

D

15 in.

E

30 in.

The frame shown consists of four wooden members, ABC,
DEF, BE, and CF. Knowing that each member has a 2  4-in.
rectangular cross section and that each pin has a 12 -in.
diameter, determine the maximum value of the average
normal stress (a) in member BE, (b) in member CF.

F

SOLUTION



Add support reactions to figure as shown.


Using entire frame as free body,
M A  0: 40Dx  (45  30)(480)  0
Dx  900 lb



Use member DEF as free body.
Reaction at D must be parallel to FBE and FCF .
Dy 

4
Dx  1200 lb
3

4

M F  0:  (30)  FBE   (30  15) DY  0
5


FBE  2250 lb
4

M E  0: (30)  FCE   (15) DY  0
5

FCE  750 lb

Stress in compression member BE:
A  2 in.  4 in.  8 in 2

Area:
(a)

 BE 

FBE
2250

A
8

 BE  281 psi 

Minimum section area occurs at pin.
Amin  (2)(4.0  0.5)  7.0 in 2
Stress in tension member CF:

(b)

 CF 

FCF
750

Amin
7.0


 CF 107.1 psi 

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Solution Manual for Mechanics of Materials 7th Edition by Beer
Full file at />PROBLEM 1.13

Dimensions in mm
1150

D

100
C

G

A

F

850

B


250

E
500

450

675

825

An aircraft tow bar is positioned by means of a single
hydraulic cylinder connected by a 25-mm-diameter steel
rod to two identical arm-and-wheel units DEF. The mass
of the entire tow bar is 200 kg, and its center of gravity
is located at G. For the position shown, determine the
normal stress in the rod.

SOLUTION


FREE BODY – ENTIRE TOW BAR:
W  (200 kg)(9.81 m/s 2 )  1962.00 N
M A  0: 850R  1150(1962.00 N)  0
R  2654.5 N




FREE BODY – BOTH ARM & WHEEL UNITS:


tan  

100
675

  8.4270

M E  0: ( FCD cos  )(550)  R(500)  0
FCD 

500
(2654.5 N)
550 cos 8.4270

 2439.5 N (comp.)



 CD  

2439.5 N
FCD

ACD
 (0.0125 m)2

 4.9697  106 Pa

 CD  4.97 MPa 


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Solution Manual for Mechanics of Materials 7th Edition by Beer
Full file at />150 mm
300 mm
A

D
F

150 mm

PROBLEM 1.14
Two hydraulic cylinders are used to control the position
of the robotic arm ABC. Knowing that the control rods
attached at A and D each have a 20-mm diameter and
happen to be parallel in the position shown, determine the
average normal stress in (a) member AE, (b) member DG.

C

B

400 mm

E

800 N

600 mm

G

200 mm

SOLUTION
Use member ABC as free body.

M B  0: (0.150)

4
FAE  (0.600)(800)  0
5

FAE  4  103 N

Area of rod in member AE is
Stress in rod AE:

A

 AE 


4


d2 


4

(20  103 ) 2  314.16  106 m 2

FAE
4  103

 12.7324  106 Pa
6
A
314.16  10

(a)

 AE  12.73 MPa 

Use combined members ABC and BFD as free body.
4

4

M F  0: (0.150)  FAE   (0.200)  FDG   (1.050  0.350)(800)  0
5

5


FDG  1500 N

Area of rod DG:
Stress in rod DG:

A


4

d2 

 DG 


4

(20  103 ) 2  314.16  106 m 2

FDG
1500

 4.7746  106 Pa
A
3.1416  106
(b)

 DG  4.77 MPa 

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Solution Manual for Mechanics of Materials 7th Edition by Beer
Full file at />PROBLEM 1.15
Determine the diameter of the largest circular hole that can be punched into a sheet of polystyrene 6 mm
thick, knowing that the force exerted by the punch is 45 kN and that a 55-MPa average shearing stress is
required to cause the material to fail.

SOLUTION
For cylindrical failure surface:

A   dt

Shearing stress:

 

Therefore,
Finally,

P



P
or

A

A

P



  dt

d 


P

 t
45  103 N
 (0.006 m)(55  106 Pa)

 43.406  103 m
d  43.4 mm 

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Solution Manual for Mechanics of Materials 7th Edition by Beer

Full file at />5
8

P'

1 in.
2 in.

PROBLEM 1.16

in.
5
8

in.

2 in.
1 in.

9 in.

Two wooden planks, each 12 in. thick and 9 in.
wide, are joined by the dry mortise joint shown.
Knowing that the wood used shears off along its
grain when the average shearing stress reaches
1.20 ksi, determine the magnitude P of the axial
load that will cause the joint to fail.

P


SOLUTION
Six areas must be sheared off when the joint fails. Each of these areas has dimensions

5
8

in. 

1
2

in., its area

being
A

5 1
5 2
 
in  0.3125 in 2
8 2 16

At failure, the force carried by each area is
F   A  (1.20 ksi)(0.3125 in 2 )  0.375 kips

Since there are six failure areas,
P  6 F  (6)(0.375)

P  2.25 kips 


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Solution Manual for Mechanics of Materials 7th Edition by Beer
Full file at />PROBLEM 1.17

0.6 in.

P

P'
Steel

3 in.

Wood

When the force P reached 1600 lb, the wooden specimen shown failed
in shear along the surface indicated by the dashed line. Determine the
average shearing stress along that surface at the time of failure.

SOLUTION
Area being sheared:

A  3 in.  0.6 in.  1.8 in 2


Force:

P  1600 lb

Shearing stress:

 

P 1600 lb

 8.8889  102 psi
2
A 1.8 in

  889 psi 

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Solution Manual for Mechanics of Materials 7th Edition by Beer
Full file at />PROBLEM 1.18

40 mm
10 mm
8 mm
12 mm


A load P is applied to a steel rod supported as shown by an aluminum
plate into which a 12-mm-diameter hole has been drilled. Knowing that
the shearing stress must not exceed 180 MPa in the steel rod and 70 MPa
in the aluminum plate, determine the largest load P that can be applied to
the rod.

P

SOLUTION
A1   dt   (0.012 m)(0.010 m)

For steel:

 376.99  106 m 2

1 

P
 P  A11  (376.99  106 m 2 )(180  106 Pa)
A
 67.858  103 N

A2   dt   (0.040 m)(0.008 m)  1.00531  103 m 2

For aluminum:

2 

P

 P  A2 2  (1.00531  103 m 2 )(70  106 Pa)  70.372  103 N
A2
P  67.9 kN 

Limiting value of P is the smaller value, so

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Solution Manual for Mechanics of Materials 7th Edition by Beer
Full file at />PROBLEM 1.19
The axial force in the column supporting the timber beam shown is
P  20 kips. Determine the smallest allowable length L of the bearing
plate if the bearing stress in the timber is not to exceed 400 psi.

L

6 in.

P

SOLUTION
Bearing area: Ab  Lw

b 
L


P
P

Ab
Lw
20  103 lb
P

 8.33 in.
 b w (400 psi)(6 in.)

L  8.33 in. 

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Solution Manual for Mechanics of Materials 7th Edition by Beer
Full file at />
d

PROBLEM 1.20
Three wooden planks are fastened together by a series of bolts to form
a column. The diameter of each bolt is 12 mm and the inner diameter
of each washer is 16 mm, which is slightly larger than the diameter of
the holes in the planks. Determine the smallest allowable outer

diameter d of the washers, knowing that the average normal stress in
the bolts is 36 MPa and that the bearing stress between the washers
and the planks must not exceed 8.5 MPa.

12 mm

SOLUTION
Bolt:

ABolt 

Tensile force in bolt:

 

d2
4



 (0.012 m)2
4

 1.13097  104 m 2

P
 P  A
A

 (36  106 Pa)(1.13097  104 m 2 )

 4.0715  103 N
Bearing area for washer:

Aw 

and

Aw 



d

4

2
o

 di2



P

 BRG

Therefore, equating the two expressions for Aw gives


4


d

2
o



 di2 
d o2 
d o2 

P

 BRG
4P

 BRG

 di2

4 (4.0715  103 N)
 (0.016 m) 2
 (8.5  106 Pa)

d o2  8.6588  104 m 2
d o  29.426  103 m
d o  29.4 mm 

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Solution Manual for Mechanics of Materials 7th Edition by Beer
Full file at />PROBLEM 1.21

P 5 40 kN

120 mm

b

A 40-kN axial load is applied to a short wooden post that is
supported by a concrete footing resting on undisturbed soil.
Determine (a) the maximum bearing stress on the concrete
footing, (b) the size of the footing for which the average bearing
stress in the soil is 145 kPa.

100 mm

b

SOLUTION
(a)

Bearing stress on concrete footing.
P  40 kN  40  103 N

A  (100)(120)  12  103 mm 2  12  103 m 2

 
(b)

P
40  103

 3.3333  106 Pa
A 12  103

Footing area. P  40  103 N

 

P
A

3.33 MPa 

  145 kPa  45  103 Pa
A

P





40  103

 0.27586 m 2
3
145  10

Since the area is square, A  b 2
b

A 

0.27586  0.525 m

b  525 mm 

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Solution Manual for Mechanics of Materials 7th Edition by Beer
Full file at />PROBLEM 1.22
a

P

a

An axial load P is supported by a short W8  40 column of crosssectional area A  11.7 in 2 and is distributed to a concrete foundation
by a square plate as shown. Knowing that the average normal stress in

the column must not exceed 30 ksi and that the bearing stress on the
concrete foundation must not exceed 3.0 ksi, determine the side a of
the plate that will provide the most economical and safe design.

SOLUTION
For the column,  

P
or
A
P   A  (30)(11.7)  351 kips

For the a  a plate,   3.0 ksi

A

P





351
 117 in 2
3.0

Since the plate is square, A  a 2
a

A  117


a  10.82 in. 

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Solution Manual for Mechanics of Materials 7th Edition by Beer
Full file at />PROBLEM 1.23
Link AB, of width b = 2 in. and thickness t = 14 in., is used to support the end of a
horizontal beam. Knowing that the average normal stress in the link is 20 ksi and
that the average shearing stress in each of the two pins is 12 ksi, determine (a) the
diameter d of the pins, (b) the average bearing stress in the link.

A
d

b

t

B
d

SOLUTION
Rod AB is in compression.
A  bt


1
in.
4

1
P   A  (20)(2)    10 kips
4
P
P 
AP

Pin:

AP 

and
(a)

where b  2 in. and t 

d 

4 AP





4P


 P




4

d2
(4)(10)
 1.03006 in.
 (12)

d  1.030 in. 

(b)

b 

P
10

 38.833 ksi
(1.03006)(0.25)
dt

 b  38.8 ksi 

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