Solution_Manual_for_Mechanics_of_Materia (1)

Prévia do material em texto

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.1 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 
 2 2 2 2 2( ) [(60 mm) (50 mm) ] 863.938 mm
4 4
A D d
 
     
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 
 2 2
max allow (200 N/mm )(863.938 mm ) 172,788 172.8 kN NP A    Ans. 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.2 A 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 
2
min
27 kips
1.500 in.
18 ksi
P P
A
A


     
The cross-sectional area of the aluminum tube is given by 
 2 2( )
4
A D d

  
Set this expression equal to the minimum area and solve for the maximum inside diameter d 
 
2 2 2
2 2 2
2 2 2
max
[(2.50 in.) ] 1.500 in.
4
4
(2.50 in.) (1.500 in. )
4
(2.50 in.) (1.500 in. )
2.08330 in.
d
d
d
d



 
 
 
 
 
 
The outside diameter D, the inside diameter d, and the wall thickness t are related by 
 2D d t  
Therefore, the minimum wall thickness required for the aluminum tube is 
 
min
2.50 in. 2.08330 in.
0.20835 in. 0.208 in.
2 2
D d
t
 
    Ans. 
 
 
 
 
 
 
 
 
 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.3 Two 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, 
 
1
1
15 kips 0
15 kips 15 kips (C)
yF F
F
    
    
 
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): 
 
2
2
30 kips 30 kips 15 kips 0
75 kips 75 kips (C)
yF F
F
      
    
 
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 
 21
1,min
15 kips
0.375 in.
40 ksi
F
A

   
The minimum rod diameter is therefore 
 2 21,min 1 10.375 in. 0.6909 0.691 9 i
4
inn. .A d d

     Ans. 
Similarly, the normal stress in rod (2) must be limited to 40 ksi, which requires a minimum area of 
 222,min
75 kips
1.875 in.
40 ksi
F
A

   
The minimum diameter for rod (2) is therefore 
 2 22,min 2 21.875 in. 1.54509 1.545 in.7 in.
4
A d d

     Ans. 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.4 Two 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, 
 
1
1
15 kips 0
15 kips 15 kips (C)
yF F
F
    
    
 
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): 
 
 
2
2
30 kips 30 kips 15 kips 0
75 kips 75 kips (C)
yF F
F
      
    
 
From the given diameter of rod (1), the cross-sectional area of rod (1) is 
 2 2
1 (1.75 in.) 2.4053 in.
4
A

  
and thus, the normal stress in rod (1) is 
 11 2
1
15 kips
6.23627 ksi
2.4053 in
6.24 ksi )
.
(C
F
A


     Ans. 
 
From the given diameter of rod (2), the cross-sectional area of rod (2) is 
 2 2
2 (2.50 in.) 4.9087 in.
4
A

  
Accordingly, the normal stress in rod (2) is 
 22 2
2
75 kips
15.2789 ksi
2.4053 in.
15.28 ksi (C)
F
A


     Ans. 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.5 Axial 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, 
 1 18 kips 4 kips 4 kips 0 16 kips 16 kips (C)yF F F           
 
 
 
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): 
 2 28 kips 4 kips 4 kips 15 kips 15 kips 0 14 kips 14 kips (T)yF F F            
 
Similarly,cut a FBD through rod (3) that includes the free end A. From this FBD, the internal force in 
rod (3) is: 
 
3
3
8 kips 4 kips 4 kips 15 kips 15 kips 20 kips 20 kips 0
26 kips 26 kips (C)
yF F
F
          
   
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
 
From the given diameter of rod (1), the cross-sectional area of rod (1) is 
 2 2
1 (2.00 in.) 3.1416 in.
4
A

  
and thus, the normal stress in aluminum rod (1) is 
 11 2
1
16 kips
5.0930 ksi
3.1416 in
5.09 ksi (C)
.
F
A


     Ans. 
 
From the given diameter of rod (2), the cross-sectional area of rod (2) is 
 2 2
2 (1.50 in.) 1.7671 in.
4
A

  
Accordingly, the normal stress in brass rod (2) is 
 22 2
2
14 kips
7.9224 ksi
1.7671 in.
7.92 ksi (T)
F
A
     Ans. 
 
Finally, the cross-sectional area of rod (3) is 
 2 2
3 (3.00 in.) 7.0686 in.
4
A

  
and the normal stress in the steel rod is 
 33 2
3
26 kips
3.6782 ksi
7.0686 in
3.68 ksi (C)
.
F
A


     Ans. 
 
 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.6 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, 
 1 18 kips 4 kips 4 kips 0 16 kips 16 kips (C)yF F F           
 
 
 
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): 
 2 28 kips 4 kips 4 kips 15 kips 15 kips 0 14 kips 14 kips (T)yF F F            
 
Similarly, cut a FBD through rod (3) that includes the free end A. From this FBD, the internal force in 
rod (3) is: 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
 
3
3
8 kips 4 kips 4 kips 15 kips 15 kips 20 kips 20 kips 0
26 kips 26 kips (C)
yF F
F
          
   
 
 
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 cross-
sectional 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 
 
21
1,min
1
16 kips
0.8889 in.
18 ksi
F
A

   
The minimum rod diameter is therefore 
 2 2
1,min 1 10.8889 in. 1.0638 in 1.064 in..
4
A d d

     Ans. 
 
The normal stress in brass rod (2) must be limited to 25 ksi, which requires a minimum area of 
 
22
2,min
2
14 kips
0.5600 in.
25 ksi
F
A

   
which requires a minimum diameter for rod (2) of 
 2 2
2,min 2 20.5600 in. 0.8444 in 0.844 in..
4
A d d

     Ans. 
 
The normal stress in steel rod (3) must be limited to 15 ksi. The minimum cross-sectional area required 
for this rod is: 
 
23
3,min
3
26 kips
1.7333 in.
15 ksi
F
A

   
which requires a minimum diameter for rod (3) of 
 2 23,min 3 31.7333 in. 1.4856 in 1.486 in..
4
A d d

     Ans. 
 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.7 Two 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. 
 
 2 1cos(35.7067 ) cos(57.9946 ) 0xF F F      (a) 
 2 1sin(35.7067 ) sin(57.9946 ) 0yF F F P       (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: 
 2 1
cos(57.9946 )
cos(35.7067 )
F F



 (c) 
Substituting Eq. (c) into Eq. (b) gives 
  
1 1
1
1
cos(57.9946 )
sin(35.7067 ) sin(57.9946 )
cos(35.6553 )
cos(57.9946 ) tan(35.7067 ) sin(57.9946 )
cos(57.9946 ) tan(35.7067 ) sin(57.9946 ) 1.2289
F F P
F P
P P
F

   

    
  
   
 
 
For the given load of P = 50 kN, the internal force in rod (1) is therefore: 
 1
50 kN
40.6856 kN
1.2289
F   
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
Backsubstituting this result into Eq. (c) gives force F2: 
2 1
cos(57.9946 ) cos(57.9946 )
(40.6856 kN) 26.5553 kN
cos(35.7067 ) cos(35.7067 )
F F
 
  
 
 
 
The normal stress in rod (1) must be limited to 130 MPa; therefore, the minimum cross-sectional area 
required for rod (1) is 
 
21
1,min 2
1
(40.6856 kN)(1,000 N/kN)
312.9664 mm
130 N/mm
F
A

   
The minimum rod diameter is therefore 
 2 2
1,min 1 1312.9664 mm 19.9620 19.
4
96 mmmmA d d

     Ans. 
 
The minimum area required for rod (2) is 
 
22
2,min 2
2
(26.5553 kN)(1,000 N/kN)
204.2718 mm
130 N/mm
F
A

   
which requires a minimum diameter for rod (2) of 
 2 2
2,min 2 2204.2718 mm 16.1272 16.
4
13 mmmmA d d

     Ans. 
 
 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted.Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.8 Two 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. 
 
 2 1cos(35.7067 ) cos(57.9946 ) 0xF F F      (a) 
 2 1sin(35.7067 ) sin(57.9946 ) 0yF F F P       (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: 
 2 1
cos(57.9946 )
cos(35.7067 )
F F



 (c) 
Substituting Eq. (c) into Eq. (b) gives 
  
1 1
1
1
cos(57.9946 )
sin(35.7067 ) sin(57.9946 )
cos(35.6553 )
cos(57.9946 ) tan(35.7067 ) sin(57.9946 )
cos(57.9946 ) tan(35.7067 ) sin(57.9946 ) 1.2289
F F P
F P
P P
F

   

    
  
   
 
 
For the given load of P = 27 kN, the internal force in rod (1) is therefore: 
 1
27 kN
21.9702 kN
1.2289
F   
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
Backsubstituting this result into Eq. (c) gives force F2: 
2 1
cos(57.9946 ) cos(57.9946 )
(21.9702 kN) 14.3399 kN
cos(35.7067 ) cos(35.7067 )
F F
 
  
 
 
 
The diameter of rod (1) is 16 mm; therefore, its cross-sectional area is: 
 2 2
1 (16 mm) 201.0619 mm
4
A

  
and the normal stress in rod (1) is: 
 
21
1 2
1
(21.9702 kN)(1,000 N/kN)
109.2710 N/mm
201.0
109.3 MPa (T)
619 mm
F
A
     Ans. 
 
The diameter of rod (2) is 12 mm; therefore, its cross-sectional area is: 
 2 2
2 (12 mm) 113.0973 mm
4
A

  
and the normal stress in rod (2) is: 
 
22
2 2
2
(14.3399 kN)(1,000 N/kN)
126.7924 N/mm
113.0
126.8 MPa (T)
973 mm
F
A
     Ans. 
 
 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.9 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: 

 
2 kips
2 ki
0
ps
x
x
xF A
A
   
   
 
(6 ft) (5 kips)(14 ft) (2 kips)(7 ft)
14 kips
0
y
A yB
B
M   
 

 
 
5 kips 0
9 kips
y y y
y
F A B
A  
    
 
 
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 0.50 26.565
14 ft
7 ft
tan 0.875 41.186
8 ft
AC AC
BC BC
 
 
    
    
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
Joint A: 
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. 
 cos(26.565 ) 0x AC AB xF F F A      (a) 
 sin(26.565 ) 0y AC yF F A     (b) 
 
 
Solve Eq. (b) for FAC: 
 
9 kips
sin(26.565 ) sin(26.5
20.125 kip
65 )
s
y
AC
A
F

    
 
 
 
and then compute FAB using Eq. (a): 
 
cos(26.565 )
(20.125 kips)cos(26.5 16.000 kips65 ) ( 2 kips)
AB AC xF F A   
      
 
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. 
 cos(41.186 ) 0x AB BCF F F      (c) 
 sin(41.186 ) 0y BC yF F B     (d) 
 
Solve Eq. (d) for FBC: 
 
14 kips
sin(41.186 ) sin(41.18
21.260 kip
6
s
)
y
BC
B
F      
 
 
 
Eq. (c) can be used as a check on our calculations: 
 
cos(41.186 )
( 16.000 kips) ( 21.260 kips)cos(41.186 ) 0
x AB BCF F F    
       Checks! 
 
Section properties: 
For each of the three truss members: 
 2 2 24.00 in. 2(0.226 in.) 3.548 in. (4.00 in.) (3.548 in.) 2.67954 in.
4
d A

        
Normal stress in each truss member: 
 
2
16.000 kips
5.971 ksi
2.67954
5.97 ksi (C)
 in.
AB
AB
AB
F
A


     Ans. 
 
2
20.125 kips
7.510 ksi
2.67954
7.51 ksi (T)
 in.
AC
AC
AC
F
A
     Ans. 
 
2
21.260 kips
7.934 ksi
2.67954
7.93 ksi (C)
 in.
BC
BC
BC
F
A


     Ans. 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.10 A 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 free-
body diagram (FBD) of the entire truss is 
shown. The followingequilibrium equations 
can be written for this structure: 
 
12 k
12
N 0
 kNx
x xF A
A
   
   
 
(1 m) (15 kN)(4.3 m) 0
64.5 kNy
A yB
B
M   
  
 
15 kN
49.5 kN
0
y
y y yF
A
A B
  
    
 
 
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 1.50 56.310
1.0 m
1.5 m
tan 0.454545 24.444
3.3 m
AB AB
BC BC
 
 
    
    
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
Joint A: 
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. 
 cos(56.310 ) 0x AC AB xF F F A      (a) 
 sin(56.310 ) 0y y ABF A F     (b) 
 
 
Solve Eq. (b) for FAB: 
 
49.5 kN
sin(56.310 ) sin(56.310 )
59.492 kN
y
AB
A
F

   
 
 
 
and then compute FAC using Eq. (a): 
 
cos(56.310 )
( 59.492 kN)cos(56.3 45.000 10 ) ( 12 kN) kN
AC AB xF F A   
       
 
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. 
 cos(24.444 ) 12 kN 0x AC BCF F F       (c) 
 sin(24.444 ) 15 kN 0y BCF F      (d) 
 
 
Solve Eq. (d) for FBC: 
 
15 kN
sin(24.444 )
36.249 kNBCF

  

 
Eq. (c) can be used as a check on our calculations: 
 
cos(24.444 ) 12 kN 0
(45.000 kN) ( 36.249 kN)cos(24.444 ) 12 kN 0
x AC BCF F F      
       Checks! 
 
Section properties: 
For each of the three truss members: 
 2 2 260 mm 2(4 mm) 52 mm (60 mm) (52 mm) 703.7168 mm
4
d A

        
 
Normal stress in each truss member: 
 
2
( 59.492 kN)(1,000 N/kN)
84.539 MPa
70
84.5 MPa (C)
3.7168 mm
AB
AB
AB
F
A


     Ans. 
 
2
(45.000 kN)(1,000 N/kN)
63.946 MPa
70
63.9 MPa 
3.7168
)
 mm
(TACAC
AC
F
A
     Ans. 
 
2
( 36.249 kN)(1,000 N/kN)
51.511 MPa
70
51.5 MPa (C)
3.7168 mm
BC
BC
BC
F
A


     Ans. 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.11 A 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: 
 
30 kN
30 kN
0y
y
yF A
A 
   
 

 
(30 kN)(4.5 m) (15 kN)(1.6 m) (5.6 m)
19.821 kN
0
x
A x
B
M B   
 

 
 
15 kN 0
15 kN 15 kN ( 19.821 kN 34.821 ) kN
x x
x x
x
x
F A B
A AB
    
       
 
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 0.355556 19.573
4.5 m
4 m
tan 0.888889 41.634
4.5 m
AC AC
BC BC
 
 
    
    
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
Joint A: 
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. 
 cos(19.573 ) 0x x ACF A F     (a) 
 sin(19.573 ) 0y y AC ABF A F F      (b) 
 
 
Solve Eq. (a) for FAC: 
 
34.821 kN
cos(19.573 ) cos(19.573 )
36.957 kNxAC
A
F   
 
 
 
and then compute FAB using Eq. (b): 
 
sin(19.573 )
(30.000 kN) (36.957 kN)sin(19. 17.619 573 ) kN
AB y ACF A F  
    
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. 
 cos(41.634 ) 0x x BCF B F     (c) 
 sin(41.634 ) 0y BC ABF F F     (d) 
 
Solve Eq. (c) for FBC: 
 
( 19.821 kN)
cos(41.634 ) cos(41.634 )
26.520 kNxBC
B
F

   
 
 
Eq. (d) can be used as a check on our calculations: 
 
sin(41.634 )
( 26.520 kN)sin(41.634 ) (17.619 kN) 0
y BC ABF F F   
     Checks! 
 
Section properties: 
For each of the three truss members: 
 2 2 242 mm 2(3.5 mm) 35 mm (42 mm) (35 mm) 423.3296 mm
4
d A

        
 
Normal stress in each truss member: 
 
2
(17.619 kN)(1,000 N/kN)
41.620 MPa
42
41.6 MPa 
3.3296
)
 mm
(TABAB
AB
F
A
     Ans. 
 
2
(36.957 kN)(1,000 N/kN)
87.301 MPa
42
87.3 MPa 
3.3296
)
 mm
(TACAC
AC
F
A
     Ans. 
 
2
( 26.520 kN)(1,000 N/kN)
62.647 MPa
42
62.6 MPa (C)
3.3296 mm
BC
BC
BC
F
A


     Ans. 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.12 The rigid beam BC shown in Figure P1.12 is 
supported by rods (1) and (2) that have cross-sectional 
areas of 175 mm
2
 and 300 mm
2
, 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
2
1
2
1.8 m
(3 m) (15 kN/m)(1.8 m) 0
2
1.8 m
(3 m) (15 kN/m)(1.8 m) 3 m 0
8.100 kN
18.900
2
 kN
C
B
M F
M
F
F
F
 
      
 
     
 
 
 
 
 
Stresses: 
 
21
1 2
1
(8.100kN)(1,000 N/kN)
46.286 N/mm
175 m
46.3 MPa
m
F
A
     Ans. 
 
 
22
2 2
2
(18.900 kN)(1,000 N/kN)
63.000 N/mm
300 m
63.0 MPa
m
F
A
     Ans. 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.13 Bar (1) in Figure P1.15 has a cross-
sectional 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 
 2
1,max 1 1 (30 ksi)(0.75 in. ) 22.5 kipsF A   
 
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: 
 
1
1
(6 ft) (10 ft) 0
6 ft
10 ft
AM F P
P F
   
 
 
 
 
Substitute the maximum force F1,max = 22.5 kips into this relationship to obtain the maximum load that 
may be applied to the structure: 
 1
6 ft 6 ft
(22.5 kips)
10 ft 10 ft
13.50 kipsP F   Ans. 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.14 The 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:
 (13 kN/m)(1.2 m ) (9 kN) 0
(13 kN/m)(1.2 m ) (9 kN)
xF x F
F x

     
   
 
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: 
 
(13 kN/m)(1.2 m ) 0
(13 kN/m)(1.2 m )
xF x F
F x

    
  
 
The largest force in this interval occurs at x = a where F = 9.1 
kN. 
 
 
Maximum Normal Stress: 
 max
(9.1 kN)(1,000 N/kN)
(15 mm
15.17 MPa 
)(40 mm)
at 0.5 mx   Ans. 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.15 The 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:
 
(750 lb/ft)(1 ft/12 in.)(48 in. )
(2,000 lb) (1,000 lb) 0
(62.5 lb/in.)(48 in. ) 3,000 lb
xF x
F
F x

  
   
   
 
At x = 10 in., F = 5,375 lb. 
 
 
Stress: The normal stress at this location can be calculated as follows. 
 
2 2
2
(1.25 in.) 1.227185 in.
4
5,375 lb
4,379 4,380 p.944 psi
1
si
.227185 in.
A


 
   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:
 
(750 lb/ft)(1 ft/12 in.)(48 in. )
(1,000 lb) 0
(62.5 lb/in.)(48 in. ) 1,000 lb
xF x
F
F x

  
  
   
 
At x = 30 in., F = 2,125 lb. 
 
 
Stress: The normal stress at this location can be calculated as follows. 
 
2
1,730 ps
2,125 lb
1,731.606 psi
1.227185 i
i
n.
    Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.16 Two 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 
 
0
10,000 lb
5,000 lb
2
xF P V V
V
    
   
 
 
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 
 2min
5,000 lb
41.6667 in.
120 psi
A   
The boards are 6-in. wide; therefore, glue must be spread along board (2) for a length of at least 
 
2
glue joint
41.6667 in.
6.9444 in.
6 in.
L   
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 
 
min 6.9444 in. 6.9444 in. 0.5 in 14.39 in..L     Ans. 
 
 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.17 For 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: 
 0 2xF P V V P V       
 
 
The area of the pin surfaceexposed by the FBD is simply the cross-sectional area of the pin: 
 2 2 2
pin pin (10 mm) 78.539816 mm
4 4
A d
 
   
 
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 
 2 2bolt (95 N/mm )(78.539816 mm ) 7,461.283 NV A   
 
Therefore, the maximum load P that may be applied to the connection is 
 2 2(7,461.283 N) 14,922.565 14.92 kNNP V    Ans. 
 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.18 For the connection shown in Figure P1.18, 
determine the average shear stress produced in the 3/8-
in. 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
625 lb
4 bolts
V   
The bolts in this connection act in single shear. The cross-sectional area of a single bolt is 
 2 2 2 2
bolt bolt (3 / 8 in.) (0.375 in.) 0.110447 in.
4 4 4
A d
  
    
Therefore, the average shear stress in each bolt is 
 
2
bolt
625 lb
5,658.8427 psi
0.110447 in.
5,660 psi
V
A
     Ans. 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.19 The five-bolt connection shown in Figure P1.19 must 
support an applied load of P = 265 kN. If the average shear stress 
in the bolts must be limited to 120 MPa, determine the minimum 
bolt diameter that may be used for this connection. 
 
 FIGURE P1.19 
Solution 
There are five 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 
 
265 kN
53 kN 53,000 N
5 bolts
V    
Since the average shear stress must be limited to 120 MPa, each bolt must provide a shear area of at 
least: 
 2
2
53,000 N
441.6667 mm
120 N/mm
VA   
Each bolt in this connection acts in double shear; therefore, two cross-sectional bolt surfaces are 
available to transmit shear stress in each bolt. 
 
2
2
bolt
441.6667 mm
220.8333 mm per surface
2 surfaces per bolt 2 surfaces
VAA    
The minimum bolt diameter must be 
 2 2
bolt bolt 16.77 m220.8333 mm 16.7682 mm m
4
d d

    Ans. 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.20 A coupling is used to connect a 2 in. diameter 
plastic pipe (1) to a 1.5 in. diameter pipe (2), as 
shown in Figure P1.20. If the average shear stress in 
the adhesive must be limited to 400 psi, determine 
the minimum lengths L1 and L2 required for the joint 
if the applied load P is 5,000 lb. 
 
 FIGURE P1.24 
Solution 
To resist a shear force of 5,000 lb, the area of adhesive required on each pipe is 
 
2
adhesive
5,000 lb
12.5 in.
400 psi
V
V
A

   
 
Consider the coupling on pipe (1). The adhesive is applied to the circumference of the pipe, and the 
circumference C1 of pipe (1) is 
 1 1 (2.0 in.) 6.2832 in.C D    
The minimum length L1 is therefore 
 
2
1
1
12.5 in.
1.9894 in.
6.2832 i
1.989 i
n
 n.
.
VAL
C
    Ans. 
 
Consider the coupling on pipe (2). The circumference C2 of pipe (2) is 
 2 2 (1.5 in.) 4.7124 in.C D    
The minimum length L2 is therefore 
 
2
2
2
12.5 in.
2.6526 in.
4.7124 
2.65 in.
in.
VAL
C
    Ans. 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
1.21 A hydraulic punch press is used to 
punch a slot in a 0.50-in. thick plate, as 
illustrated in Fig. P1.21. If the plate shears 
at a stress of 30 ksi, determine the 
minimum force P required to punch the 
slot. 
 
 FIGURE P1.21 
Solution 
The shear stress associated with removal of the slug exists on its perimeter. The perimeter of the slug is 
given by 
 perimeter 2(3.00 in.) + (0.75 in.) 8.35619 in.  
Thus, the area subjected to shear stress is 
 2perimeter plate thickness (8.35619 in.)(0.50 in.) 4.17810 in.VA     
Given that the plate shears at  = 30 ksi, the force required to remove the slug is therefore 
 2
min (30 ksi)(4.17810 in. ) 125.343 kips 125.3 kipsVP A    Ans.
 
 
 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.22 The handle shown in Figure P1.22 is 
attached to a 40-mm-diameter shaft with a 
square shear key. The forces applied to the 
lever are P = 1,300 N. If the average shear 
stress in the key must not exceed 150 MPa, 
determine the minimum dimension a that must 
be used if the key is 25 mm long. The overall 
length of the handle is L = 0.70 m. 
 FIGURE P1.22 
Solution 
To determine the shear force V that must be resisted by the shear key, sum moments about the center of 
the shaft (which will be denoted O): 
 
700 mm 700 mm 40 mm
(1,300 N) (1,300 N) 0
2 2 2
45,500 N
OM V
V
     
              
 
 
Since the average shear stress in the key must not exceed 150 MPa, the shear area required is 
 2
2
45,500 N
303.3333 mm
150 N/mm
V
V
A

   
The shear area in the key is given by the product of its length L (i.e., 25 mm) and its width a. Therefore, 
the minimum key width a is 
 
2303.3333 mm
12.1333 mm
25 m
12.1 
m
3 mmV
A
a
L
    Ans. 
 
 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.23 An axial load P is supported by the short steel 
column shown in Figure P1.23. The column has a cross-
sectional area of 14,500 mm
2
. If the average normal stress 
in the steel column must not exceed 75 MPa, determine the 
minimum required dimension a so that the bearing stress 
between the base plate and the concrete slab does not 
exceed 8 MPa. Assume b = 420 mm. 
 
 FIGURE P1.23 
Solution 
Since the normal stress in the steel column must not exceed 75 MPa, the maximum column load is 
 2 2max (75 N/mm)(14,500 mm ) 1,087,500 NP A   
The maximum column load must be distributed over a large enough area so that the bearing stress 
between the base plate and the concrete slab does not exceed 8 MPa; therefore, the minimum plate area 
is 
 
2
min 2
1,087,500 N
135,937.5 mm
8 N/mmb
P
A

   
The area of the plate is a ×b. Since b = 420, the minimum length of a must be 
 
2
min
2
135,937.5 mm
135,937.5 mm
420 mm
324 mm
A a b
a
  
   Ans. 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.24 The two wooden boards shown in Figure P1.24 are 
connected by a 0.5-in.-diameter bolt. Washers are installed 
under the head of the bolt and under the nut. The washer 
dimensions are D = 2 in. and d = 5/8 in. The nut is tightened 
to cause a tensile stress of 9,000 psi in the bolt. Determine 
the bearing stress between the washer and the wood. 
 
 FIGURE P1.24 
Solution 
The tensile stress in the bolt is 9,000 psi; therefore, the tension force that acts in the bolt is 
 2 2
bolt bolt bolt (9,000 psi) (0.5 in.) (9,000 psi)(0.196350 in. ) 1,767.146 lb
4
F A

    
The contact area between the washer and the wood is 
 2 2 2
washer (2 in.) (0.625 in.) 2.834796 in.
4
A

     
Thus, the bearing stress between the washer and the wood is 
 
2
1,767.146 lb
2.834796 in.
623 psib   Ans. 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.25 For the beam shown in Figure P1.25, the 
allowable bearing stress for the material under 
the supports at A and B is b = 800 psi. 
Assume w = 2,100 lb/ft, P = 4,600 lb, a = 20 ft, 
and b = 8 ft. Determine the size of square 
bearing plates required to support the loading 
shown. Dimension the plates to the nearest ½ 
in. 
 
 FIGURE P1.25 
Solution 
Equilibrium: Using the FBD shown, 
calculate the beam reaction forces. 
 
20 ft
(20 ft) (2,100 lb/ft)(20 ft) (4,600 lb)(28 ft) 0
2
27,440 lb
A y
y
M B
B
 
      
  
 
 
20 ft
(20 ft) (2,100 lb/ft)(20 ft) (4,600 lb)(8 ft) 0
2
19,160 lb
B y
y
M A
A
 
       
  
 
Bearing plate at A: The area of the bearing plate required for support A is 
 2
19,160 lb
23.950 in.
800 psi
AA   
Since the plate is to be square, its dimensions must be 
 
2 use 5 in. 23.95 5 in. bearing plate at0 in. 4.894 in . Awidth   Ans. 
 
Bearing plate at B: The area of the bearing plate required for support B is 
 2
27,440 lb
34.300 in.
800 psi
BA   
Since the plate is to be square, its dimensions must be 
 
2 use 6 in. 34.30 6 in. bearing plate at0 in. 5.857 in . Bwidth   Ans. 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.26 The d = 15-mm-diameter solid rod shown in 
Figure P1.26 passes through a D = 20-mm-diameter 
hole in the support plate. When a load P is applied 
to the rod, the rod head rests on the support plate. 
The support plate has a thickness of b = 12 mm. 
The rod head has a diameter of a = 30 mm and the 
head has a thickness of t = 10 mm. If the normal 
stress produced in the rod by load P is 225 MPa, 
determine: 
(a) the bearing stress acting between the support 
plate and the rod head. 
(b) the average shear stress produced in the rod 
head. 
(c) the punching shear stress produced in the 
support plate by the rod head. 
 
 FIGURE P1.26 
Solution 
The cross-sectional area of the rod is: 
 
2 2
rod (15 mm) 176.715 mm
4
A

  
The tensile stress in the rod is 225 MPa; therefore, the tension force in the rod is 
 2 2rod rod rod (225 N/mm )(176.715 mm ) 39,760.782 NF A   
 
(a) The contact area between the support plate and the rod head is 
 2 2 2
contact (30 mm) (20 mm) 392.699 mm
4
A

     
Thus, the bearing stress between the support plate and the rod head is 
 
2
39,760.782 N
392.699 mm
101.3 MPab   Ans. 
 
(b) In the rod head, the area subjected to shear stress is equal to the perimeter of the rod times the 
thickness of the head. 
 
2(15 mm)(10 mm) 471.239 mmVA  
 and therefore, the average shear stress in the rod head is 
 
2
39,760.782 N
471.
84.4 MP
m
a
239 m
   Ans. 
 
(c) In the support plate, the area subjected to shear stress is equal to the product of the rod head 
perimeter and the thickness of the plate. 
 
2(30 mm)(12 mm) 1,130.973 mmVA  
 and therefore, the average punching shear stress in the support plate is 
 
2
39,760.782 N
1,130
35.2 MPa
.973 mm
   Ans. 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.27 The rectangular bar is connected to the support bracket with 
a circular pin, as shown in Figure P1.27. The bar width is w = 1.75 
in. and the bar thickness is 0.375 in. For an applied load of P = 
5,600 lb, determine the average bearing stress produced in the bar 
by the 0.625-in.-diameter pin. 
 
 FIGURE P1.27 
Solution 
The average bearing stress produced in the bar by the pin is based on the projected area of the pin. The 
projected area is equal to the pin diameter times the bar thickness. Therefore, the average bearing stress 
in the bar is 
 
5,600 lb
23,893.33 psi
(0.625 in.)(0.375 i
23,
n.
900
)
 psib    Ans. 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.28 The steel pipe column shown in Figure P1.28 has an 
outside diameter of 8.625 in. and a wall thickness of 0.25 in. 
The timber beam is 10.75 in wide, and the upper plate has the 
same width. The load imposed on the column by the timber 
beam is 80 kips. Determine 
 (a) The average bearing stress at the surfaces between the pipe 
column and the upper and lower steel bearing plates. 
(b) The length L of the rectangular upper bearing plate if its 
width is 10.75 in. and the average bearing stress between the 
steel plate and the wood beam is not to exceed 500 psi. 
(c) The dimension “a” of the square lower bearing plate if the 
average bearing stress between the lower bearing plate and 
the concrete slab is not to exceed 900 psi. 
 
 Figure P1.28 
Solution 
(a) The area of contact between the pipe column and one of the bearing plates is simply the cross-
sectional area of the pipe. To calculate the pipe area, wemust first calculate the pipe inside diameter d: 
 2 2 8.625 in. 2(0.25 in.) 8.125 in.D d t d D t        
The pipe cross-sectional area is 
 2 2 2 2 2
pipe (8.625 in.) (8.125 in.) 6.5777 in.
4 4
A D d
 
           
Therefore, the bearing stress between the pipe and one of the bearing plates is 
 
2
80 kips
12.1623 ksi
6.5777 in.
12.16 ksib
b
P
A
     Ans. 
 
(b) The bearing stress between the timber beam and the upper bearing plate must not exceed 500 psi 
(i.e., 0.5 ksi). To support a load of 80 kips, the contact area must be at least 
 
280 kips 160 in.
0.5 ksi
b
b
P
A

   
If the width of the timber beam is 10.75 in., then the length L of the upper bearing plate must be 
 
2160 in.
14.8837 in.
beam width 10.75
14.88 in
 
.
in.
bAL     Ans. 
 
(c) The bearing stress between the concrete slab and the lower bearing plate must not exceed 900 psi 
(i.e., 0.9 ksi). To support the 80-kip pipe load, the contact area must be at least 
 
280 kips 88.8889 in.
0.9 ksi
b
b
P
A

   
Since the lower bearing plate is square, its dimension a must be 
 288.8889 in. 9.4 9.43 in281 n. .ibA a a a      Ans. 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.29 A clevis-type pipe hanger supports an 8-in-
diameter pipe as shown in Figure P1.29. The hanger 
rod has a diameter of 1/2 in. The bolt connecting the 
top yoke and the bottom strap has a diameter of 5/8 in. 
The bottom strap is 3/16-in.-thick by 1.75-in.-wide by 
36-in.-long. The weight of the pipe is 2,000 lb. 
Determine the following: 
(a) the normal stress in the hanger rod 
(b) the shear stress in the bolt 
(c) the bearing stress in the bottom strap 
 
 FIGURE P1.29 
Solution 
(a) The normal stress in the hanger rod is 
 
2 2
rod
rod 2
(0.5 in.) 0.196350 in.
4
2,000 lb
10,185.917 psi
0.196350 in.
10,190 psi
A


 
   Ans. 
 
(b) The cross-sectional area of the bolt is: 
 
2 2
bolt (0.625 in.) 0.306796 in.
4
A

  
The bolt acts in double shear; therefore, its average shear stress is 
 bolt 2
2,000 lb
3,259.493 psi
2(0.306796 in.
3,260 i
)
 ps    Ans. 
 
(c) The bearing stress in the bottom strap is based on the projected area of the bolt in contact with the 
strap. Also, keep in mind that there are two ends of the strap that contact the bolt. The bearing stress is 
thus 
 
2,000 lb
8,533.334 psi
2(0.625 in.)(3/16 
8,530 psi
in.)
b    Ans. 
 
 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.30 Rigid bar ABC shown in Figure P1.30 is 
supported by a pin at bracket A and by tie rod (1). 
Tie rod (1) has a diameter of 5 mm, and it is 
supported by double-shear pin connections at B and 
D. The pin at bracket A is a single-shear 
connection. All pins are 7 mm in diameter. 
Assume a = 600 mm, b = 300 mm, h = 450 mm, P 
= 900 N, and  = 55°. Determine the following: 
(a) the normal stress in rod (1) 
(b) the shear stress in pin B 
(c) the shear stress in pin A 
 
 FIGURE P1.30 
Solution 
Equilibrium: Using the FBD shown, calculate 
the reaction forces that act on rigid bar ABC. 
 
1
1
sin(36.87 )(600 mm)
(900 N)sin (55 )(900 mm) 0
1,843.092 N
AM F
F
  
  
  
 
 
(1,843.092 N)cos(36.87 ) (900 N)cos(55 ) 0
958.255 N
x x
x
F A
A
      
  
 
 
(1,843.092 N)sin (36.87 ) (900 N)sin (55 ) 0
368.618 N
y y
y
F A
A
      
  
 
The resultant force at A is 
 2 2(958.255 N) ( 368.618 N) 1,026.709 NA    
 
 
(a) Normal stress in rod (1). 
 
2 2
rod
rod 2
(5 mm) 19.635 mm
4
1,843.092 N
19.635 mm
93.9 MPa
A


 
  Ans. 
 
(b) Shear stress in pin B. The cross-sectional area of a 7-mm-diameter pin is: 
 
2 2
pin (7 mm) 38.485 mm
4
A

  
Pin B is a double shear connection; therefore, its average shear stress is 
 pin 2
1,843.092 N
2(38.485 m
23.9 M a
)
P
m
B   Ans. 
 
(c) Shear stress in pin A. 
Pin A is a single shear connection; therefore, its average shear stress is 
 pin 2
1,026.709 N
38.485 m
26.7 MPa
m
A   Ans. 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.31 The bell crank shown in Figure P1.31 is in 
equilibrium for the forces acting in rods (1) and (2). 
The bell crank is supported by a 10-mm-diameter 
pin at B that acts in single shear. The thickness of 
the bell crank is 5 mm. Assume a = 65 mm, b = 
150 mm, F1 = 1,100 N, and  = 50°. Determine the 
following: 
(a) the shear stress in pin B 
(b) the bearing stress in the bell crank at B 
 
 FIGURE P1.31 
Solution 
Equilibrium: Using the FBD shown, calculate the 
reaction forces that act on the bell crank. 
 
2
2
(1,100 N)sin(50 )(65 mm)
(150 mm) 0
365.148 N
BM
F
F
   
 
 
 
 
 
(1,100 N)cos(50 )
365.148 N 0
341.919 N
x x
x
F B
B
   
 
 
 
 
 
(1,100 N)sin(50 ) 0
842.649 N
y y
y
F B
B
    
   
 
 
 
The resultant force at B is 
 2 2(341.919 N) ( 842.649 N) 909.376 NB    
 
 
(a) Shear stress in pin B. The cross-sectional area of the 10-mm-diameter pin is: 
 
2 2
pin (10 mm) 78.540 mm
4
A

  
Pin B is a single shear connection; therefore, its average shear stress is 
 pin 2
909.376 N
78.540 mm
11.58 MPaB   Ans. 
 
(b) Bearing stress in the bell crank at B. The average bearing stress produced in the bell crank by the 
pin is based on the projected area of the pin. The projected area is equal to the pin diameter times the 
bell crank thickness. Therefore, the average bearing stress in the bell crank is 
 
909.376 N
(10 mm)(5 mm)
18.19 MPab   Ans. 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.32 The beam shown in Figure P1.32 is 
supported by a pin at C and by a short link 
AB. If w = 30 kN/m, determine the average 
shear stress in the pins at A and C. Each 
pin has a diameter of 25 mm. Assume L = 
1.8 m and  = 35°. 
 
 FIGURE P1.32 
Solution 
Equilibrium: Using the FBD shown, 
calculate the reaction forces that act on the 
beam. 
 
 
 
1
1
1.8 m
sin(35 )(1.8 m) (30 kN/m)(1.8 m) 0
2
47.0731 kN
CM F
F
 
       
 
 
 
 
(47.0731 kN)cos(35 ) 0
38.5600 kN
x x
x
F C
C
    
 
 
 
 
1.8 m
(1.8 m) (30 kN/m)(1.8 m) 0
2
27.0000 kN
B y
y
M C
C
 
     
  
 
The resultant force at C is 
 2 2(38.5600 kN) (27.0000 kN) 47.0731 kNC   
 
 
Shear stress in pin A. The cross-sectional area of a 25-mm-diameter pin is: 
 
2 2
pin (25 mm) 490.8739mm
4
A

  
Pin A is a single shear connection; therefore, its average shear stress is 
 pin 2
47,073.1 N
490.8739 m
95.9 MPa
m
A   Ans. 
 
Shear stress in pin C. 
Pin C is a double shear connection; therefore, its average shear stress is 
 pin 2
47,073.1 N
2(490.8739 m
47.9 M a
)
P
m
C   Ans. 
 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.33 The bell-crank mechanism shown 
in Figure P1.33 is in equilibrium for an 
applied load of P = 7 kN applied at A. 
Assume a = 200 mm, b = 150 mm, and  = 
65°. Determine the minimum diameter d 
required for pin B for each of the following 
conditions: 
(a) The average shear stress in the pin may 
not exceed 40 MPa. 
(b) The bearing stress in the bell crank 
may not exceed 100 MPa. 
(c) The bearing stress in the support 
bracket may not exceed 165 MPa. 
 
 
 FIGURE P1.33 
Solution 
Equilibrium: Using the FBD shown, calculate 
the reaction forces that act on the bell crank. 
 
2
2
(7,000 N)sin(65 )(200 mm)
(150 mm) 0
8,458.873 N
BM
F
F
  
 
 
 
 
 
(7,000 N)cos(65 )
8,458.873 N 0
11,417.201 N
x x
x
F B
B
   
 
  
 
 
 
(7,000 N)sin(65 ) 0
6,344.155 N
y y
y
F B
B
    
  
 
 
 
The resultant force at B is 
 2 2( 11,417.201 N) (6,344.155 N) 13,061.423 NB    
 
 
(a) The average shear stress in the pin may not exceed 40 MPa. The shear area required for the pin 
at B is 
 2
2
13,061.423 N
326.536 mm
40 N/mm
VA   
Since the pin at B is supported in a double shear connection, the required cross-sectional area for the pin 
is 
 2pin 163.268 mm
2
VAA   
and therefore, the pin must have a diameter of 
 
24 (163.268 m 14.m 4) 2 mmd

  Ans. 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
(b) The bearing stress in the bell crank may not exceed 100 MPa. The projected area of pin B on the 
bell crank must equal or exceed 
 2
2
13,061.423 N
130.614 mm
100 N/mm
bA   
The bell crank thickness is 8 mm; therefore, the projected area of the pin is Ab = (8 mm)d. Calculate the 
required pin diameter d: 
 
2130.614 mm
8
16.
 
3
mm
3 mmd   Ans. 
 
(c) The bearing stress in the support bracket may not exceed 165 MPa. The pin at B bears on two 6-
mm-thick support brackets. Thus, the minimum pin diameter required to satisfy the bearing stress limit 
on the support bracket is 
 2
2
13,061.423 N
79.160 mm
165 N/mm
bA   
 
 
279.160 mm
2(6 mm)
6.60 mmd   Ans. 
 
 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.34 A structural steel bar with a 25 mm × 75 mm rectangular cross section is subjected to an axial 
load of 150 kN. Determine the maximum normal and shear stresses in the bar. 
 
Solution 
The maximum normal stress in the steel bar is 
 max
(150 kN)(1,000 N/kN)
(25 mm)(75 mm)
80 MPa
F
A
    Ans. 
The maximum shear stress is one-half of the maximum normal stress 
 max
max
2
40 MPa

   Ans. 
 
 
 
 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.35 A steel rod of circular cross section will be used to carry an axial load of 92 kips. The 
maximum stresses in the rod must be limited to 30 ksi in tension and 12 ksi in shear. Determine the 
required diameter for the rod. 
 
Solution 
Based on the allowable 30 ksi tension stress limit, the minimum cross-sectional area of the rod is 
 
2
min
max
92 kips
3.0667 in.
30 ksi
F
A

   
For the 12-ksi shear stress limit, the minimum cross-sectional area of the rod must be 
 
2
min
max
92 kips
3.8333 in.
2 2(12 ksi)
F
A

   
Therefore, the rod must have a cross-sectional area of at least 3.8333 in.
2
 in order to satisfy both the 
normal and shear stress limits. 
 
The minimum rod diameter D is therefore 
 2 2
min min3.8333 in. 2.2092 in. 2.21 in.
4
d d

    Ans. 
 
 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.36 An axial load P is applied to the
rectangular bar shown in Figure P1.36. The 
cross-sectional area of the bar is 400 mm2. 
Determine the normal stress perpendicular to
plane AB and the shear stress parallel to
plane AB if the bar is subjected to an axial
load of P = 70 kN. 
 FIGURE P1.36 
Solution 
The angle θ for the inclined plane is 35°. The 
normal force N perpendicular to plane AB is found 
from 
 cos (40 kN)cos35 57.3406 kNN P θ= = ° = 
 
and the shear force V parallel to plane AB is 
 sin (70 kN)sin 35 40.1504 kNV P θ= − = − ° = −
 
The cross-sectional area of the bar is 400 mm2, but the area along inclined plane AB is 
 
2
2400 mm 488.3098 mm
cos cos35n
AA
θ
= = =
°
 
 
The normal stress σn perpendicular to plane AB is 
 2
(57.3406 kN)(1,000 N/kN) 117.4268 MPa
488
117.4 MP
.3098 mm
an
n
N
A
σ = = = = Ans. 
 
The shear stress τnt parallel to plane AB is 
 2
( 40.1504 kN)(1,000 N/kN) 82.2231 MPa
488
82.2 M
.3098 m
Pa
mnt n
V
A
τ −−= = = − = Ans. 
 
 
 
 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.37 An axial load P is applied to the 1.75 in. 
by 0.75 in. rectangular bar shown in Figure 
P1.37. Determine the normal stress 
perpendicular to plane AB and the shear stress 
parallel to plane AB if the bar is subjected to 
an axial load of P = 18 kips. 
 
 FIGURE P1.37 
Solution 
The angle  for the inclined plane is 60°. The 
normal force N perpendicular to plane AB is 
found from 
 cos (18 kips)cos60 9.0 kipsN P     
 
and the shear force V parallel to plane AB is 
 sin (18 kips)sin60 15.5885 kipsV P     
 
 
The cross-sectional area of the bar is (1.75 in.)(0.75 in.) = 1.3125 in.
2
, but the area along inclined plane 
AB is 
 
2
21.3125 in./ cos 2.6250 in.
cos60
nA A   

 
 
The normal stress n perpendicular to plane AB is 
 
2
9.0 kips
3.4286 ksi
2.6250 in
3.43 ks
.
inn
N
A
     Ans. 
 
The shear stress nt parallel to plane AB is 
 
2
15.5885 kips
5.9385 ksi
2.6250
5.94 ks
 i
i
n.
nt
n
V
A
     Ans. 

 
 
 
 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.38 A compression load of P = 80 kips is applied to a 4 in. by 4 
in. square post, as shown in Figure P1.38/39. Determine the normal 
stress perpendicular to plane AB and the shear stress parallel to 
plane AB. 
 
 FIGURE P1.38/39 
Solution 
The angle  for the inclined plane is 55°. The normal force N 
perpendicular to plane AB is found from 
 cos (80 kips)cos55 45.8861 kipsN P     
 
and the shear force V parallel to plane AB is 
 sin (80 kips)sin55 65.5322 kipsV P     
 
The cross-sectional area of the post is (4 in.)(4 in.) = 16 in.
2
, but the area 
along inclined plane AB is 
 
2
216 in./ cos 27.8951 in.
cos55
nA A   

 
 
 
 
The normal stress n perpendicular to plane AB is 
 
2
45.8861 kips
1.6449 ksi
27.8951 
1.645 ksi
in.
n
n
N
A
     Ans. 
 
The shear stress nt parallel to plane AB is 
 
2
65.5322 kips
2.3492 ksi
27.8951
2.35 ks
 i
i
n.
nt
n
V
A
     Ans. 
 
 
 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.39 Specifications for the 50 mm × 50 mm square bar 
shown in Figure P1.38/39 require that the normal and shear 
stresses on plane AB not exceed 120 MPa and 90 MPa, 
respectively. Determine the maximum load P that can be 
applied without exceeding the specifications. 
 
 FIGURE P1.38/39 
Solution 
The general equations for normal and shear stresses on an inclined plane in terms of the angle  are 
 (1 cos2 )
2
n
P
A
   (a) 
and 
 sin 2
2
nt
P
A
  (b) 
The cross-sectional area of the square bar is A = (50 mm)
2
 = 2,500 mm
2
, and the angle  for plane AB is 
55°. 
 
The normal stress on plane AB is limited to 120 MPa; therefore, the maximum load P that can be 
supported by the square bar is found from Eq. (a): 
 
2 22 2(2,500 mm )(120 N/mm )
911,882 N
1 cos2 1 cos2(55 )
nAP


  
  
 
 
The shear stress on plane AB is limited to 90 MPa. From Eq. (b), the maximum load P based the shear 
stress limit is 
 
2 22 2(2,500 mm )(90 N/mm )
478,880 N
sin 2 sin 2(55 )
ntAP


  

 
 
Thus, the maximum load that can be supported by the bar is 
 
max 479 kNP  Ans. 
 
 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.40 Specifications for the 6 in. × 6 in. square post shown in 
Figure P1.40 require that the normal and shear stresses on 
plane AB not exceed 800 psi and 400 psi, respectively. 
Determine the maximum load P that can be applied without 
exceeding the specifications. 
 
 FIGURE P1.40 
Solution 
The general equations for normal and shear stresses on an inclined plane in terms of the angle  are 
 (1 cos2 )
2
n
P
A
   (a) 
and 
 sin 2
2
nt
P
A
  (b) 
The cross-sectional area of the square post is A = (6 in.)
2
 = 36 in.
2
, and the angle  for plane AB is 40°. 
 
The normal stress on plane AB is limited to 800 psi; therefore, the maximum load P that can be 
supported by the square post is found from Eq. (a): 
 
22 2(36 in. )(800 psi)
49,078 lb
1 cos2 1 cos2(40 )
nAP


  
  
 
 
The shear stress on plane AB is limited to 400 psi. From Eq. (b), the maximum load P based the shear 
stress limit is 
 
22 2(36 in. )(400 psi)
29,244 lb
sin 2 sin 2(40 )
ntAP


  

 
 
Thus, the maximum load that can be supported by the post is 
 
max 29,200 l 29.2 ipb k sP   Ans. 
 
 
 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.41 A 90 mm wide bar will be used to carry an axial 
tension load of 280 kN as shown in Figure P1.41. The 
normal and shear stresses on plane AB must be limited 
to 150 MPa and 100 MPa, respectively. Determine the 
minimum thickness t required for the bar. 
 
 FIGURE P1.41 
Solution 
The general equations for normal and shear stresses on an inclined plane in terms of the angle  are 
 (1 cos2 )
2
n
P
A
   (a) 
and 
 sin 2
2
nt
P
A
  (b) 
The angle  for plane AB is 50°. 
 
The normal stress on plane AB is limited to 150 MPa; therefore, the minimum cross-sectional area A 
required to support P = 280 kN can be found from Eq. (a): 
 
2
2
(280 kN)(1,000 N/kN)
(1 cos2 ) (1 cos2(50 )) 771.2617 mm
2 2(150 N/mm )n
P
A 

      
 
The shear stress on plane AB is limited to 100 MPa; therefore, the minimum cross-sectional area A 
required to support P = 280 kN can be found from Eq. (b): 
 
2
2
(280 kN)(1,000 N/kN)
sin 2 sin 2(50 ) 1,378.7309 mm
2 2(100 N/mm )nt
P
A 

    
 
To satisfy both the normal and shear stress requirements, the cross-sectional area must be at least Amin = 
1,379.7309 mm
2
. Since the bar width is 90 mm, the minimum bar thickness t must be 
 
2
min
1,378.7309 mm
15.3192 mm
90 m
15.3 
m
2 mmt    Ans. 
 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
P1.42 A rectangular bar having width w = 
6.00 in. and thickness t = 1.50 in. is subjected 
to a tension load P as shown in Figure 
P1.42/43. The normal and shear stresses on 
plane AB must not exceed 16 ksi and 8 ksi, 
respectively. Determine the maximum load P 
that can be applied without exceeding either 
stress limit. 
 
 FIGURE P1.42/43 
Solution 
The general equations for normal and shear stresses on an inclined plane in terms of the angle  are 
 (1 cos2 )
2
n
P
A
   (a) 
and 
 sin 2
2
nt
P
A
  (b) 
The angle  for inclined plane AB is calculated from 
 
3
tan 3 71.5651
1
      
The cross-sectional area of the bar is A = w×t = (6.00 in.)(1.50 in.) = 9.0 in.
2
. 
 
The normal stress on plane AB is limited to 16 ksi; therefore, the maximum load P can be found from 
Eq. (a): 
 
22 2(9.0 in. )(16 ksi)
1,440 ksi
1 cos2 1 cos2(71.5651 )
nAP


  
  
 
 
The shear stress on plane AB is limited to 8 ksi. From Eq. (b), the maximum load P based the shear 
stress limit is 
 
22 2(9.0 in. )(8 ksi)
240 kips
sin 2 sin 2(71.5651 )
ntAP


  

 
 
Thus, the maximum load that can be supported