Solution manual for mechanics of materials 3rd edition by philpot

Solution Manual for Mechanics of Materials 3rd Edition by Philpot
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P1.1at
A />stainless steel tube with an outside diameter of 60 mm and a wall thickness of 5 mm is used as
a compression member. If the axial normal stress in the member must be limited to 200 MPa,
determine the maximum load P that the member can support.

Solution
The cross-sectional area of the stainless steel tube is


A  ( D 2  d 2 )  [(60 mm) 2  (50 mm) 2 ]  863.938 mm 2
4
4
The normal stress in the tube can be expressed as
P

A
The maximum normal stress in the tube must be limited to 200 MPa. Using 200 MPa as the allowable
normal stress, rearrange this expression to solve for the maximum load P
Ans.
Pmax   allow A  (200 N/mm2 )(863.938 mm2 )  172,788 N  172.8 kN

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Solution Manual for Mechanics of Materials 3rd Edition by Philpot
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P1.2atA />2024-T4 aluminum tube with an outside diameter of 2.50 in. will be used to support a 27-kip
load. If the axial normal stress in the member must be limited to 18 ksi, determine the wall thickness
required for the tube.

Solution
From the definition of normal stress, solve for the minimum area required to support a 27-kip load
without exceeding a stress of 18 ksi
P
P 27 kips

 Amin  
 1.500 in.2
A
 18 ksi
The cross-sectional area of the aluminum tube is given by

A  (D2  d 2 )
4
Set this expression equal to the minimum area and solve for the maximum inside diameter d



4

[(2.50 in.) 2  d 2 ]  1.500 in.2

(2.50 in.) 2  d 2 
(2.50 in.) 2 

4



4



(1.500 in.2 )

(1.500 in.2 )  d 2

 d max  2.08330 in.
The outside diameter D, the inside diameter d, and the wall thickness t are related by
D  d  2t
Therefore, the minimum wall thickness required for the aluminum tube is
D  d 2.50 in.  2.08330 in.
tmin 

 0.20835 in.  0.208 in.
2
2

Ans.

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Solution Manual for Mechanics of Materials 3rd Edition by Philpot
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/>P1.3atTwo
solid cylindrical rods (1) and

(2)
are joined together at flange B and loaded, as
shown in Figure P1.3/4. If the normal stress
in each rod must be limited to 40 ksi,
determine the minimum diameter required
for each rod.

FIGURE P1.3/4

Solution
Cut a FBD through rod (1). The FBD should include the free end of the rod at A.
As a matter of course, we will assume that the internal force in rod (1) is tension
(even though it obviously will be in compression). From equilibrium,
Fy   F1  15 kips  0

 F1  15 kips  15 kips (C)
Next, cut a FBD through rod (2) that includes the free end of the rod at A. Again,
we will assume that the internal force in rod (2) is tension. Equilibrium of this
FBD reveals the internal force in rod (2):
Fy   F2  30 kips  30 kips  15 kips  0

 F2  75 kips  75 kips (C)
Notice that rods (1) and (2) are in compression. In this situation, we are
concerned only with the stress magnitude; therefore, we will use the force
magnitudes to determine the minimum required cross-sectional areas. If

the normal stress in rod (1) must be limited to 40 ksi, then the minimum
cross-sectional area that can be used for rod (1) is
F 15 kips
A1,min  1 
 0.375 in.2

40 ksi
The minimum rod diameter is therefore

Ans.
A1,min  d12  0.375 in.2
 d1  0.69099 in.  0.691 in.
4
Similarly, the normal stress in rod (2) must be limited to 40 ksi, which requires a minimum area of
F
75 kips
A2,min  2 
 1.875 in.2

40 ksi
The minimum diameter for rod (2) is therefore

Ans.
A2,min  d 22  1.875 in.2
 d 2  1.545097 in.  1.545 in.
4

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/>P1.4atTwo
solid cylindrical rods (1) and

(2) are
joined together at flange B and loaded, as shown in
Figure P1.3/4. The diameter of rod (1) is 1.75 in.
and the diameter of rod (2) is 2.50 in. Determine the
normal stresses in rods (1) and (2).

FIGURE P1.3/4

Solution
Cut a FBD through rod (1). The FBD should include the free end of the rod at A. We
will assume that the internal force in rod (1) is tension (even though it obviously will
be in compression). From equilibrium,
Fy   F1  15 kips  0

 F1  15 kips  15 kips (C)
Next, cut a FBD through rod (2) that includes the free end of the rod at A. Again, we
will assume that the internal force in rod (2) is tension. Equilibrium of this FBD
reveals the internal force in rod (2):

Fy   F2  30 kips  30 kips  15 kips  0
 F2  75 kips  75 kips (C)
From the given diameter of rod (1), the cross-sectional area of rod (1) is


A1  (1.75 in.) 2  2.4053 in.2
4
and thus, the normal stress in rod (1) is
F
15 kips
1  1 
 6.23627 ksi  6.24 ksi (C)
Ans.
A1 2.4053 in.2
From the given diameter of rod (2), the cross-sectional area of rod (2) is

A2  (2.50 in.) 2  4.9087 in.2
4
Accordingly, the normal stress in rod (2) is
F
75 kips
2  2 
 15.2789 ksi  15.28 ksi (C)
A2 2.4053 in.2

Ans.

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Solution Manual for Mechanics of Materials 3rd Edition by Philpot

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/>P1.5atAxial
loads are applied with rigid bearing plates to the
solid cylindrical rods shown in Figure P1.5/6. The diameter
of aluminum rod (1) is 2.00 in., the diameter of brass rod (2)
is 1.50 in., and the diameter of steel rod (3) is 3.00 in.
Determine the axial normal stress in each of the three rods.

FIGURE P1.5/6

Solution
Cut a FBD through rod (1). The FBD should include the free end A. We will assume that the internal
force in rod (1) is tension (even though it obviously will be in compression). From equilibrium,
Fy   F1  8 kips  4 kips  4 kips  0
 F1  16 kips  16 kips (C)

FBD through rod (1)
FBD through rod (2)

FBD through rod (3)
Next, cut a FBD through rod (2) that includes the free end A. Again, we will assume that the internal
force in rod (2) is tension. Equilibrium of this FBD reveals the internal force in rod (2):
Fy   F2  8 kips  4 kips  4 kips  15 kips  15 kips  0  F2  14 kips  14 kips (T)
Similarly, cut a FBD through rod (3) that includes the free end A. From this FBD, the internal force in
rod (3) is:
Fy   F3  8 kips  4 kips  4 kips  15 kips  15 kips  20 kips  20 kips  0

 F3  26 kips  26 kips (C)
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Solution Manual for Mechanics of Materials 3rd Edition by Philpot
Full file at />From the given diameter of rod (1), the cross-sectional area of rod (1) is

A1  (2.00 in.) 2  3.1416 in.2
4
and thus, the normal stress in aluminum rod (1) is
F
16 kips
1  1 
 5.0930 ksi  5.09 ksi (C)
A1 3.1416 in.2
From the given diameter of rod (2), the cross-sectional area of rod (2) is

A2  (1.50 in.) 2  1.7671 in.2
4
Accordingly, the normal stress in brass rod (2) is
F
14 kips
2  2 
 7.9224 ksi  7.92 ksi (T)
A2 1.7671 in.2
Finally, the cross-sectional area of rod (3) is

A3  (3.00 in.) 2  7.0686 in.2
4
and the normal stress in the steel rod is

F
26 kips
3  3 
 3.6782 ksi  3.68 ksi (C)
A3 7.0686 in.2

Ans.

Ans.

Ans.

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Solution Manual for Mechanics of Materials 3rd Edition by Philpot
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/>P1.6at
Axial
loads are applied with rigid bearing plates to the
solid cylindrical rods shown in Figure P1.5/6. The normal
stress in aluminum rod (1) must be limited to 18 ksi, the
normal stress in brass rod (2) must be limited to 25 ksi, and
the normal stress in steel rod (3) must be limited to 15 ksi.
Determine the minimum diameter required for each of the
three rods.

FIGURE P1.5/6

Solution
The internal forces in the three rods must be determined. Begin with a FBD cut through rod (1) that
includes the free end A. We will assume that the internal force in rod (1) is tension (even though it
obviously will be in compression). From equilibrium,
Fy   F1  8 kips  4 kips  4 kips  0
 F1  16 kips  16 kips (C)

FBD through rod (1)
FBD through rod (2)

FBD through rod (3)
Next, cut a FBD through rod (2) that includes the free end A. Again, we will assume that the internal
force in rod (2) is tension. Equilibrium of this FBD reveals the internal force in rod (2):
Fy   F2  8 kips  4 kips  4 kips  15 kips  15 kips  0  F2  14 kips  14 kips (T)
Similarly, cut a FBD through rod (3) that includes the free end A. From this FBD, the internal force in
rod (3) is:
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Solution Manual for Mechanics of Materials 3rd Edition by Philpot
Full file atF />  F  8 kips  4 kips  4 kips  15 kips  15 kips  20 kips  20 kips  0
y

3

 F3  26 kips  26 kips (C)
Notice that two of the three rods are in compression. In these situations, we are concerned only with the
stress magnitude; therefore, we will use the force magnitudes to determine the minimum required crosssectional areas, and in turn, the minimum rod diameters. The normal stress in aluminum rod (1) must be
limited to 18 ksi; therefore, the minimum cross-sectional area required for rod (1) is
F 16 kips
A1,min  1 
 0.8889 in.2
 1 18 ksi
The minimum rod diameter is therefore

Ans.
A1,min  d12  0.8889 in.2
 d1  1.0638 in.  1.064 in.
4
The normal stress in brass rod (2) must be limited to 25 ksi, which requires a minimum area of
F 14 kips
A2,min  2 
 0.5600 in.2
 2 25 ksi
which requires a minimum diameter for rod (2) of

A2,min  d 22  0.5600 in.2
 d 2  0.8444 in.  0.844 in.
4

Ans.

The normal stress in steel rod (3) must be limited to 15 ksi. The minimum cross-sectional area required
for this rod is:
F

26 kips
A3,min  3 
 1.7333 in.2
 3 15 ksi
which requires a minimum diameter for rod (3) of

Ans.
A3,min  d32  1.7333 in.2
 d3  1.4856 in.  1.486 in.
4

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Solution Manual for Mechanics of Materials 3rd Edition by Philpot
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/>P1.7atTwo
solid cylindrical rods support a load

of
P = 50 kN, as shown in Figure P1.7/8. If the
normal stress in each rod must be limited to 130
MPa, determine the minimum diameter required
for each rod.

FIGURE P1.7/8

Solution
Consider a FBD of joint B. Determine the angle  between
rod (1) and the horizontal axis:
4.0 m
tan  
 1.600
   57.9946
2.5 m
and the angle  between rod (2) and the horizontal axis:
2.3 m
tan  
 0.7188
   35.7067
3.2 m
Write equilibrium equations for the sum of forces in the
horizontal and vertical directions. Note: Rods (1) and (2)
are two-force members.
Fx  F2 cos(35.7067)  F1 cos(57.9946)  0
Fy  F2 sin(35.7067)  F1 sin(57.9946)  P  0

(a)
(b)

Unknown forces F1 and F2 can be found from the simultaneous solution of Eqs. (a) and (b). Using the
substitution method, Eq. (b) can be solved for F2 in terms of F1:
cos(57.9946)
F2  F1
(c)
cos(35.7067)
Substituting Eq. (c) into Eq. (b) gives

cos(57.9946)
F1
sin(35.7067)  F1 sin(57.9946)  P
cos(35.6553)

F1  cos(57.9946) tan(35.7067)  sin(57.9946)   P
 F1 

P
P

cos(57.9946) tan(35.7067)  sin(57.9946) 1.2289

For the given load of P = 50 kN, the internal force in rod (1) is therefore:
50 kN
F1 
 40.6856 kN
1.2289

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Solution Manual for Mechanics of Materials 3rd Edition by Philpot
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at />Backsubstituting
this result into Eq. (c) gives force F2:
F2  F1

cos(57.9946)
cos(57.9946)
 (40.6856 kN)
 26.5553 kN
cos(35.7067)
cos(35.7067)

The normal stress in rod (1) must be limited to 130 MPa; therefore, the minimum cross-sectional area
required for rod (1) is
F (40.6856 kN)(1,000 N/kN)
A1,min  1 
 312.9664 mm 2
2
1
130 N/mm
The minimum rod diameter is therefore

Ans.
A1,min  d12  312.9664 mm 2
 d1  19.9620 mm  19.96 mm
4
The minimum area required for rod (2) is
F
(26.5553 kN)(1,000 N/kN)
A2,min  2 
 204.2718 mm 2
2
2
130 N/mm

which requires a minimum diameter for rod (2) of

A2,min  d 22  204.2718 mm2
 d 2  16.1272 mm  16.13 mm
4

Ans.

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Solution Manual for Mechanics of Materials 3rd Edition by Philpot
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/>P1.8atTwo
solid cylindrical rods support a load
of P = 27 kN, as shown in Figure P1.7/8. Rod
(1) has a diameter of 16 mm and the diameter
of rod (2) is 12 mm. Determine the axial
normal stress in each rod.

FIGURE P1.7/8

Solution
Consider a FBD of joint B. Determine the angle  between rod (1)
and the horizontal axis:
4.0 m
tan  

 1.600
   57.9946
2.5 m
and the angle  between rod (2) and the horizontal axis:
2.3 m
tan  
 0.7188
   35.7067
3.2 m
Write equilibrium equations for the sum of forces in the horizontal
and vertical directions. Note: Rods (1) and (2) are two-force
members.
Fx  F2 cos(35.7067)  F1 cos(57.9946)  0
Fy  F2 sin(35.7067)  F1 sin(57.9946)  P  0

(a)
(b)

Unknown forces F1 and F2 can be found from the simultaneous solution of Eqs. (a) and (b). Using the
substitution method, Eq. (b) can be solved for F2 in terms of F1:
cos(57.9946)
F2  F1
(c)
cos(35.7067)
Substituting Eq. (c) into Eq. (b) gives
cos(57.9946)
F1
sin(35.7067)  F1 sin(57.9946)  P
cos(35.6553)

F1  cos(57.9946) tan(35.7067)  sin(57.9946)   P
 F1 

P
P

cos(57.9946) tan(35.7067)  sin(57.9946) 1.2289

For the given load of P = 27 kN, the internal force in rod (1) is therefore:
27 kN
F1 
 21.9702 kN
1.2289

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Solution Manual for Mechanics of Materials 3rd Edition by Philpot
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at />Backsubstituting
this result into Eq. (c) gives force F2:
F2  F1

cos(57.9946)
cos(57.9946)
 (21.9702 kN)
 14.3399 kN

cos(35.7067)
cos(35.7067)

The diameter of rod (1) is 16 mm; therefore, its cross-sectional area is:

A1  (16 mm) 2  201.0619 mm 2
4
and the normal stress in rod (1) is:
F (21.9702 kN)(1,000 N/kN)
1  1 
 109.2710 N/mm 2  109.3 MPa (T)
2
A1
201.0619 mm
The diameter of rod (2) is 12 mm; therefore, its cross-sectional area is:

A2  (12 mm)2  113.0973 mm2
4
and the normal stress in rod (2) is:
F
(14.3399 kN)(1,000 N/kN)
2  2 
 126.7924 N/mm 2  126.8 MPa (T)
2
A2
113.0973 mm

Ans.

Ans.

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Solution Manual for Mechanics of Materials 3rd Edition by Philpot
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P1.9at
A />simple pin-connected truss is loaded
and supported as shown in Figure P1.9. All
members of the truss are aluminum pipes that
have an outside diameter of 4.00 in. and a wall
thickness of 0.226 in. Determine the normal
stress in each truss member.

FIGURE P1.9

Solution
Overall equilibrium:
Begin the solution by determining the
external reaction forces acting on the
truss at supports A and B. Write
equilibrium equations that include all
external forces. Note that only the
external forces (i.e., loads and
reaction forces) are considered at this
time. The internal forces acting in the
truss members will be considered

after the external reactions have been
computed. The free-body diagram
(FBD) of the entire truss is shown.
The following equilibrium equations
can be written for this structure:
Fx  Ax  2 kips  0



 Ax  2 kips
M A  By (6 ft)  (5 kips)(14 ft)  (2 kips)(7 ft)  0
 By  14 kips
Fy  Ay  By  5 kips  0
 Ay  9 kips
Method of joints:
Before beginning the process of determining the internal forces in the axial members, the geometry of
the truss will be used to determine the magnitude of the inclination angles of members AC and BC. Use
the definition of the tangent function to determine AC and BC:
7 ft
tan  AC 
 0.50
  AC  26.565
14 ft
7 ft
tan  BC 
 0.875
  BC  41.186
8 ft

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Solution Manual for Mechanics of Materials 3rd Edition by Philpot
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JointatA: />Begin the solution process by considering a FBD of joint A. Consider
only those forces acting directly on joint A. In this instance, two axial
members, AB and AC, are connected at joint A. Additionally, two
reaction forces, Ax and Ay, act at joint A. Tension forces will be
assumed in each truss member.
Fx  FAC cos(26.565)  FAB  Ax  0
(a)
Fy  FAC sin(26.565)  Ay  0
(b)
Solve Eq. (b) for FAC:
Ay
9 kips
FAC  

 20.125 kips
sin(26.565)
sin(26.565)
and then compute FAB using Eq. (a):
FAB   FAC cos(26.565)  Ax
  (20.125 kips) cos(26.565)  ( 2 kips)  16.000 kips

Joint B:
Next, consider a FBD of joint B. In this instance, the equilibrium

equations associated with joint B seem easier to solve than those that
would pertain to joint C. As before, tension forces will be assumed in
each truss member.
Fx   FAB  FBC cos(41.186)  0
(c)
Fy  FBC sin(41.186)  By  0
(d)
Solve Eq. (d) for FBC:
By
14 kips
FBC  

 21.260 kips
sin(41.186)
sin(41.186)
Eq. (c) can be used as a check on our calculations:
Fx   FAB  FBC cos(41.186)
  ( 16.000 kips)  ( 21.260 kips) cos(41.186)  0

Section properties:
For each of the three truss members:
d  4.00 in.  2(0.226 in.)  3.548 in.

A

Checks!



(4.00 in.) 2  (3.548 in.) 2   2.67954 in.2

4

Normal stress in each truss member:
F
16.000 kips
 AB  AB 
 5.971 ksi  5.97 ksi (C)
AAB
2.67954 in.2

Ans.

 AC 

FAC
20.125 kips

 7.510 ksi  7.51 ksi (T)
AAC 2.67954 in.2

Ans.

 BC 

FBC 21.260 kips

 7.934 ksi  7.93 ksi (C)
ABC
2.67954 in.2

Ans.

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atA />P1.10
simple pin-connected truss is loaded
and supported as shown in Figure P1.10. All
members of the truss are aluminum pipes that
have an outside diameter of 60 mm and a wall
thickness of 4 mm. Determine the normal
stress in each truss member.

FIGURE P1.10

Solution
Overall equilibrium:
Begin the solution by determining the
external reaction forces acting on the truss at
supports A and B. Write equilibrium
equations that include all external forces.
Note that only the external forces (i.e., loads
and reaction forces) are considered at this
time. The internal forces acting in the truss
members will be considered after the external

reactions have been computed. The freebody diagram (FBD) of the entire truss is
shown. The following equilibrium equations
can be written for this structure:
Fx  Ax  12 kN  0

 Ax  12 kN
M A  By (1 m)  (15 kN)(4.3 m)  0
 By  64.5 kN
Fy  Ay  By  15 kN  0
 Ay  49.5 kN
Method of joints:
Before beginning the process of determining the internal forces in the axial members, the geometry of
the truss will be used to determine the magnitude of the inclination angles of members AB and BC. Use
the definition of the tangent function to determine AB and BC:
1.5 m
tan  AB 
 1.50
  AB  56.310
1.0 m
1.5 m
tan  BC 
 0.454545
  BC  24.444
3.3 m

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


Solution Manual for Mechanics of Materials 3rd Edition by Philpot
Full file
JointatA: />Begin the solution process by considering a FBD of joint A. Consider
only those forces acting directly on joint A. In this instance, two axial
members, AB and AC, are connected at joint A. Additionally, two
reaction forces, Ax and Ay, act at joint A. Tension forces will be assumed
in each truss member.
Fx  FAC  FAB cos(56.310)  Ax  0
(a)
Fy  Ay  FAB sin(56.310)  0
(b)
Solve Eq. (b) for FAB:
Ay
49.5 kN
FAB 

 59.492 kN
sin(56.310) sin(56.310)
and then compute FAC using Eq. (a):
FAC   FAB cos(56.310)  Ax
  ( 59.492 kN)cos(56.310)  ( 12 kN)  45.000 kN

Joint C:
Next, consider a FBD of joint C. In this instance, the equilibrium
equations associated with joint C seem easier to solve than those that
would pertain to joint B. As before, tension forces will be assumed in
each truss member.

Fx   FAC  FBC cos(24.444)  12 kN  0
(c)
Fy   FBC sin(24.444)  15 kN  0
(d)
Solve Eq. (d) for FBC:
15 kN
FBC 
 36.249 kN
sin(24.444)
Eq. (c) can be used as a check on our calculations:
Fx   FAC  FBC cos(24.444)  12 kN  0
  (45.000 kN)  ( 36.249 kN) cos(24.444)  12 kN  0

Section properties:
For each of the three truss members:
d  60 mm  2(4 mm)  52 mm

A

Checks!



(60 mm) 2  (52 mm) 2   703.7168 mm2
4

Normal stress in each truss member:
F
( 59.492 kN)(1,000 N/kN)
 AB  AB 

 84.539 MPa  84.5 MPa (C)
AAB
703.7168 mm 2
F
(45.000 kN)(1,000 N/kN)
 AC  AC 
 63.946 MPa  63.9 MPa (T)
AAC
703.7168 mm 2
F
( 36.249 kN)(1,000 N/kN)
 BC  BC 
 51.511 MPa  51.5 MPa (C)
ABC
703.7168 mm 2

Ans.
Ans.
Ans.

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atA />P1.11
simple pin-connected truss is loaded

and supported as shown in Figure P1.11. All
members of the truss are aluminum pipes that
have an outside diameter of 42 mm and a wall
thickness of 3.5 mm. Determine the normal
stress in each truss member.

FIGURE P1.11

Solution

Overall equilibrium:
Begin the solution by determining the external
reaction forces acting on the truss at supports A
and B. Write equilibrium equations that include all
external forces. Note that only the external forces
(i.e., loads and reaction forces) are considered at
this time. The internal forces acting in the truss
members will be considered after the external
reactions have been computed. The free-body
diagram (FBD) of the entire truss is shown. The
following equilibrium equations can be written for
this structure:
Fy  Ay  30 kN  0

 Ay  30 kN

M A  (30 kN)(4.5 m)  (15 kN)(1.6 m)  Bx (5.6 m)  0



 Bx  19.821 kN
Fx  Ax  Bx  15 kN  0
Ax  15 kN  Bx  15 kN  (19.821 kN)

 Ax  34.821 kN

Method of joints:
Before beginning the process of determining the internal forces in the axial members, the geometry of
the truss will be used to determine the magnitude of the inclination angles of members AC and BC. Use
the definition of the tangent function to determine AC and BC:
1.6 m
tan  AC 
 0.355556
  AC  19.573
4.5 m
4m
tan  BC 
 0.888889
  BC  41.634
4.5 m
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Solution Manual for Mechanics of Materials 3rd Edition by Philpot
Full file
JointatA: />Begin the solution process by considering a FBD of joint A. Consider
only those forces acting directly on joint A. In this instance, two axial

members, AB and AC, are connected at joint A. Additionally, two
reaction forces, Ax and Ay, act at joint A. Tension forces will be
assumed in each truss member.
Fx  Ax  FAC cos(19.573)  0
(a)
Fy  Ay  FAC sin(19.573)  FAB  0
(b)
Solve Eq. (a) for FAC:
Ax
34.821 kN
FAC 

 36.957 kN
cos(19.573) cos(19.573)
and then compute FAB using Eq. (b):
FAB  Ay  FAC sin(19.573)

 (30.000 kN)  (36.957 kN)sin(19.573)  17.619 kN
Joint B:
Next, consider a FBD of joint B. In this instance, the equilibrium
equations associated with joint B seem easier to solve than those that
would pertain to joint C. As before, tension forces will be assumed in
each truss member.
Fx  Bx  FBC cos(41.634)  0
(c)
Fy  FBC sin(41.634)  FAB  0
(d)
Solve Eq. (c) for FBC:
Bx
( 19.821 kN)

FBC 

 26.520 kN
cos(41.634) cos(41.634)
Eq. (d) can be used as a check on our calculations:
Fy  FBC sin(41.634)  FAB

 (26.520 kN)sin(41.634)  (17.619 kN)  0
Section properties:
For each of the three truss members:
d  42 mm  2(3.5 mm)  35 mm

A

Checks!



(42 mm) 2  (35 mm) 2   423.3296 mm2
4

Normal stress in each truss member:
F
(17.619 kN)(1,000 N/kN)
 AB  AB 
 41.620 MPa  41.6 MPa (T)
AAB
423.3296 mm 2

Ans.

 AC 

FAC (36.957 kN)(1,000 N/kN)

 87.301 MPa  87.3 MPa (T)
AAC
423.3296 mm 2

Ans.

 BC 

FBC ( 26.520 kN)(1,000 N/kN)

 62.647 MPa  62.6 MPa (C)
ABC
423.3296 mm 2

Ans.

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Solution Manual for Mechanics of Materials 3rd Edition by Philpot
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atThe

/>P1.12
rigid beam BC shown in Figure P1.12 is
supported by rods (1) and (2) that have cross-sectional
areas of 175 mm2 and 300 mm2, respectively. For a
uniformly distributed load of w = 15 kN/m, determine
the normal stress in each rod. Assume L = 3 m and a =
1.8 m.

FIGURE P1.12

Solution
Equilibrium: Calculate the internal forces in members (1) and (2).
 1.8 m 
M C   F1 (3 m)  (15 kN/m)(1.8 m) 
0
 2 

 F1  8.100 kN
1.8 m 

M B  F2 (3 m)  (15 kN/m)(1.8 m)  3 m 
 0

2 
 F2  18.900 kN
Stresses:

1 

F1 (8.100 kN)(1,000 N/kN)


 46.286 N/mm 2  46.3 MPa
2
A1
175 mm

Ans.

2 

F2 (18.900 kN)(1,000 N/kN)

 63.000 N/mm 2  63.0 MPa
2
A2
300 mm

Ans.

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Solution Manual for Mechanics of Materials 3rd Edition by Philpot
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at Bar
/>P1.13
(1) in Figure P1.15 has a crosssectional area of 0.75 in.2. If the stress in bar

(1) must be limited to 30 ksi, determine the
maximum load P that may be supported by
the structure.

FIGURE P1.13

Solution
Given that the cross-sectional area of bar (1) is 0.75 in.2 and its normal stress must be limited to 30 ksi,
the maximum force that may be carried by bar (1) is
F1,max  1 A1  (30 ksi)(0.75 in.2 )  22.5 kips
Consider a FBD of ABC. From the moment equilibrium
equation about joint A, the relationship between the force in
bar (1) and the load P is:
M A  (6 ft)F1  (10 ft)P  0
P 

6 ft
F1
10 ft

Substitute the maximum force F1,max = 22.5 kips into this relationship to obtain the maximum load that
may be applied to the structure:
6 ft
6 ft
Ans.
P
F1 
(22.5 kips)  13.50 kips
10 ft
10 ft

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Solution Manual for Mechanics of Materials 3rd Edition by Philpot
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at The
/>P1.14
rectangular bar shown in Figure
P1.14 is subjected to a uniformly distributed
axial loading of w = 13 kN/m and a
concentrated force of P = 9 kN at B.
Determine the magnitude of the maximum
normal stress in the bar and its location x.
Assume a = 0.5 m, b = 0.7 m, c = 15 mm, and
d = 40 mm.
FIGURE P1.14

Solution
Equilibrium:
Draw a FBD for the interval between A and B where
0  x  a . Write the following equilibrium equation:



 Fx  (13 kN/m)(1.2 m  x)  (9 kN)  F  0
 F  (13 kN/m)(1.2 m  x)  (9 kN)

The largest force in this interval occurs at x = 0 where F = 6.6
kN.

In the interval between B and C where a  x  a  b , and write
the following equilibrium equation:



 Fx  (13 kN/m)(1.2 m  x)  F  0
 F  (13 kN/m)(1.2 m  x)
The largest force in this interval occurs at x = a where F = 9.1
kN.

Maximum Normal Stress:
(9.1 kN)(1,000 N/kN)
 max 
 15.17 MPa at x  0.5 m
(15 mm)(40 mm)

Ans.

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Solution Manual for Mechanics of Materials 3rd Edition by Philpot
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at The

/>P1.15
solid 1.25-in.-diameter rod shown in
Figure P1.15 is subjected to a uniform axial
distributed loading along its length of w = 750 lb/ft.
Two concentrated loads also act on the rod: P =
2,000 lb and Q = 1,000 lb. Assume a = 16 in. and b
= 32 in. Determine the normal stress in the rod at
the following locations:
(a) x = 10 in.
(b) x = 30 in.
FIGURE P1.15

Solution
(a) x = 10 in.
Equilibrium: Draw a FBD for the interval between A and B
where 0  x  a , and write the following equilibrium equation:



Fx  (750 lb/ft)(1 ft/12 in.)(48 in.  x)
 (2,000 lb)  (1,000 lb)  F  0
 F  (62.5 lb/in.)(48 in.  x)  3,000 lb
At x = 10 in., F = 5,375 lb.
Stress: The normal stress at this location can be calculated as follows.
A





4

(1.25 in.) 2  1.227185 in.2

5,375 lb
 4,379.944 psi  4,380 psi
1.227185 in.2

Ans.

(b) x = 30 in.
Equilibrium: Draw a FBD for the interval between B and C
where a  x  a  b , and write the following equilibrium
equation:



Fx  (750 lb/ft)(1 ft/12 in.)(48 in.  x)
 (1,000 lb)  F  0
 F  (62.5 lb/in.)(48 in.  x)  1,000 lb
At x = 30 in., F = 2,125 lb.
Stress: The normal stress at this location can be calculated as follows.
2,125 lb

 1,731.606 psi  1,730 psi
1.227185 in.2

Ans.

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Solution Manual for Mechanics of Materials 3rd Edition by Philpot
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at Two
/>P1.16
6 in. wide wooden boards are
to be joined by splice plates that will be
fully glued on the contact surfaces. The
glue to be used can safely provide a shear
strength of 120 psi. Determine the smallest
allowable length L that can be used for the
splice plates for an applied load of P =
10,000 lb. Note that a gap of 0.5 in. is
required between boards (1) and (2).

FIGURE P1.16

Solution
Consider a FBD of board (2). The glue on the splice plates provides resistance to the 10,000 lb applied
load on both the top and bottom surfaces of board (2). Denoting the shear resistance on a glue surface as
V, equilibrium in the horizontal direction requires
Fx  P  V  V  0
V 

10,000 lb
 5,000 lb

2

In other words, each glue surface must be large enough so that 5,000 lb of shear resistance can be
provided to board (2). Since the glue has a shear strength of 120 psi, the area of each glue surface on
board (2) must be at least
5,000 lb
Amin 
 41.6667 in.2
120 psi
The boards are 6-in. wide; therefore, glue must be spread along board (2) for a length of at least
41.6667 in.2
Lglue joint 
 6.9444 in.
6 in.
Although we’ve discussed only board (2), the same rationale applies to board (1). For both boards (1)
and (2), the glue must be applied along a length of at least 6.9444 in. on both the top and bottom of the
boards in order to resist the 10,000 lb applied load.
The glue applied to boards (1) and (2) must be matched by glue applied to the splice plates. Therefore,
the splice plates must be at least 6.9444 in. + 6.9444 in. = 13.8889 in. long. However, we are told that a
0.5-in. gap is required between boards (1) and (2); therefore, the splice plates must be 0.5-in. longer.
Altogether, the length of the splice plates must be at least
Ans.
Lmin  6.9444 in.  6.9444 in.  0.5 in.  14.39 in.

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Solution Manual for Mechanics of Materials 3rd Edition by Philpot
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at For
/>P1.17
the clevis connection shown

in Figure
P1.17, determine the maximum applied load P that
can be supported by the 10-mm-diameter pin if the
average shear stress in the pin must not exceed 95
MPa.

FIGURE P1.17

Solution
Consider a FBD of the bar that is connected by the clevis,
including a portion of the pin. If the shear force acting on each
exposed surface of the pin is denoted by V, then the shear force
on each pin surface is related to the load P by:
Fx  P  V  V  0
 P  2V
The area of the pin surface exposed by the FBD is simply the cross-sectional area of the pin:
 2

Apin  d pin
 (10 mm) 2  78.539816 mm 2
4
4
If the average shear stress in the pin must be limited to 95 MPa, the maximum shear force V on a single
cross-sectional surface must be limited to

V   Abolt  (95 N/mm 2 )(78.539816 mm 2 )  7, 461.283 N
Therefore, the maximum load P that may be applied to the connection is
P  2V  2(7, 461.283 N)  14,922.565 N  14.92 kN

Ans.

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Solution Manual for Mechanics of Materials 3rd Edition by Philpot
Full file
at />P1.18
For the connection shown in

Figure P1.18,
determine the average shear stress produced in the 3/8in. diameter bolts if the applied load is P = 2,500 lb.

FIGURE P1.18

Solution
There are four bolts, and it is assumed that each bolt supports an equal portion of the external load P.
Therefore, the shear force carried by each bolt is
2,500 lb
V
 625 lb
4 bolts
The bolts in this connection act in single shear. The cross-sectional area of a single bolt is

 2


Abolt  d bolt
 (3 / 8 in.) 2  (0.375 in.) 2  0.110447 in.2
4
4
4
Therefore, the average shear stress in each bolt is
V
625 lb


 5,658.8427 psi  5,660 psi
Ans.
Abolt 0.110447 in.2

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