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Chapter 1  Stress 
  1.1 Introduction 
  1.2 Normal Stress Under Axial Loading 
  1.3 Direct Shear Stress 
  1.4 Bearing Stress 
  1.5 Stresses on Inclined Sections 
  1.6 Equality of Shear Stresses on Perpendicular Planes 
 
Chapter 2  Strain 
  2.1 Displacement, Deformation, and the Concept of Strain 
  2.2 Normal Strain 
  2.3 Shear Strain 
  2.4 Thermal Strain 
 
Chapter 3  Mechanical Properties of Materials 
  3.1 The Tension Test 
  3.2 The Stress‐Strain Diagram 
  3.3 Hooke’s Law 
  3.4 Poisson’s Ratio 
 
Chapter 4  Design Concepts 
  4.1 Introduction 
  4.2 Types of Loads 
  4.3 Safety 
  4.4 Allowable Stress Design 
  4.5 Load and Resistance Factor Design 
 
Chapter 5  Axial Deformation 
  5.1 Introduction 
  5.2 Saint‐Venant’s Principle 
  5.3 Deformations in Axially Loaded Bars 
  5.4 Deformations in a System of Axially Loaded Bars 
  5.5 Statically Indeterminate Axially Loaded Members 
  5.6 Thermal Effects on Axial Deformation 
  5.7 Stress Concentrations 
 
Chapter 6  Torsion 
  6.1 Introduction 
  6.2 Torsional Shear Strain 
  6.3 Torsional Shear Stress 
  6.4 Stresses on Oblique Planes 
  6.5 Torsional Deformations 
  6.6 Torsion Sign Conventions 
  6.7 Gears in Torsion Assemblies 
  6.8 Power Transmission 
  6.9 Statically Indeterminate Torsion Members 
  6.10 Stress Concentrations in Circular Shafts Under Torsional Loadings 
  6.11 Torsion of Noncircular Sections 
  6.12 Torsion of Thin‐Walled Tubes: Shear Flow 
 
Chapter 7  Equilibrium of Beams 
  7.1 Introduction 
  7.2 Shear and Moment in Beams 
  7.3 Graphical Method for Constructing Shear and Moment Diagrams 
  7.4 Discontinuity Functions to Represent Load, Shear, and Moment 
 
Chapter 8  Bending 
  8.1 Introduction 
  8.2 Flexural Strains 
  8.3 Normal Strains in Beams 
  8.4 Analysis of Bending Stresses in Beams 
  8.5 Introductory Beam Design for Strength 
  8.6 Flexural Stresses in Beams of Two Materials 
  8.7 Bending Due to Eccentric Axial Load 
  8.8 Unsymmetric Bending 
  8.9 Stress Concentrations Under Flexural Loadings 
 
Chapter 9  Shear Stress in Beams 
  9.1 Introduction 
  9.2 Resultant Forces Produced by Bending Stresses 
  9.3 The Shear Stress Formula 
  9 .4 The First Moment of Area Q 
  9.5 Shear Stresses in Beams of Rectangular Cross Section 
  9.6 Shear Stresses in Beams of Circular Cross Section 
  9.7 Shear Stresses in Webs of Flanged Beams 
  9.8 Shear Flow in Built‐Up Members 
 
Chapter 10  Beam Deflections 
  10.1 Introduction 
  10.2 Moment‐Curvature Relationship 
  10.3 The Differential Equation of the Elastic Curve 
  10.4 Deflections by Integration of a Moment Equation 
  10.5 Deflections by Integration of Shear‐Force or Load Equations 
  10.6 Deflections Using Discontinuity Functions 
  10.7 Method of Superposition 
 
Chapter 11  Statically Indeterminate Beams 
  11.1 Introduction 
  11.2 Types of Statically Indeterminate Beams 
  11.3 The Integration Method 
  11.4 Use of Discontinuity Functions for Statically Indeterminate Beams 
  11.5 The Superposition Method 
 
 
 
 
 
 
Chapter 12  Stress Transformations 
  12.1 Introduction 
  12.2 Stress at a General Point in an Arbitrarily Loaded Body 
  12.3 Equilibrium of the Stress Element 
  12.4 Two‐Dimensional or Plane Stress 
  12.5 Generating the Stress Element 
  12.6 Equilibrium Method for Plane Stress Transformation 
  12.7General Equations of Plane Stress Transformation 
  12.8 Principal Stresses and Maximum Shear Stress 
  12.9 Presentation of Stress Transformation Results 
  12.10 Mohr’s Circle for Plane Stress 
  12.11 General State of Stress at a Point 
 
Chapter 13  Strain Transformations 
  13.1 Introduction 
  13.2 Two‐Dimensional or Plane Strain 
  13.3 Transformation Equations for Plane Strain 
  13.4 Principal Strains and Maximum Shearing Strain 
  13.5 Presentation of Strain Transformation Results 
  13.6 Mohr’s Circle for Plane Strain 
  13.7 Strain Measurement and Strain Rosettes 
  13.8 Generalized Hooke’s Law for Isotropic Materials 
 
Chapter 14  Thin‐Walled Pressure Vessels 
  14.1 Introduction 
  14.2 Spherical Pressure Vessels 
  14.3 Cylindrical Pressure Vessels 
  14.4 Strains in Pressure Vessels 
 
Chapter 15  Combined Loads 
  15.1 Introduction 
  15.2 Combined Axial and Torsional Loads 
  15.3 Principal Stresses in a Flexural Member 
  15.4 General Combined Loadings 
  15.5 Theories of Failure 
 
Chapter 16  Columns 
  16.1 Introduction 
  16.2 Buckling of Pin‐Ended Columns 
  16.3 The Effect of End Conditions on Column Buckling 
  16.4 The Secant Formula 
  16.5 Empirical Column Formulas & Centric Loading 
  16.6 Eccentrically Loaded Columns 
 
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1.1 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 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. 
 
 
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1.2 A 2024-T4 aluminum tube with an outside diameter of 2.50 in. will be used to support a 27-kip 
load. If the 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. 
 
 
 
 
 
 
 
 
 
 
 
 
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1.3 Two solid cylindrical rods (1) and (2) are 
joined together at flange B and loaded, as shown in 
Fig. P1.3. The diameter of rod (1) is d1 = 24 mm 
and the diameter of rod (2) is d2 = 42 mm. 
Determine the normal stresses in rods (1) and (2). 
 
 Fig. P1.3 
Solution 
Cut a FBD through rod (1) that includes the free end of the rod at A. 
Assume that the internal force in rod (1) is tension. From equilibrium, 
 
1 180 kN 0 80 kN (T)xF F F     
 
 
 
Next, cut a FBD through rod (2) that includes the free 
end of the rod A. Assume that the internal force in rod 
(2) is tension. Equilibrium of this FBD reveals the 
internal force in rod (2): 
 
 
2 2140 kN 140 kN 80 kN 0 200 kN 200 kN (C)xF F F         
 
 
From the given diameter of rod (1), the cross-sectional area of rod (1) is 
 
2 2
1 (24 mm) 452.3893 mm
4
A

 
 
and thus, the normal stress in rod (1) is 
 
1
1 2
1
(80 kN)(1,000 N/kN)
176.8388 MPa
452.389
176
3 m
.8 M )
m
Pa (T
F
A
     Ans. 
 
From the given diameter of rod (2), the cross-sectional area of rod (2) is 
 
2 2
2 (42 mm) 1,385.4424 mm
4
A

 
 
Accordingly, the normal stress in rod (2) is 
 
2
2 2
2
( 200 kN)(1,000 N/kN)
144.3582 MPa
1,385.4
144.4 MPa (
424
C)
 mm
F
A
      Ans. 
 
 
 
 
 
 
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1.4 Two solid cylindrical rods (1) and (2) are 
joined together at flange B and loaded, as shown in 
Fig. P1.4. If the normal stress in each rod must be 
limited to 120 MPa, determine the minimum 
diameter required for each rod. 
 
 Fig. P1.4 
Solution 
Cut a FBD through rod (1) that includes the free end of the rod at A. 
Assume that the internal force in rod (1) is tension. From equilibrium, 
 
1 180 kN 0 80 kN (T)xF F F     
 
 
 
Next, cut a FBD through rod (2) that includes the free 
end of the rod A. Assume that the internal force in rod 
(2) is tension. Equilibrium of this FBD reveals the 
internal force in rod (2): 
 
 
2 2140 kN 140 kN 80 kN 0 200 kN 200 kN (C)xF F F         
 
 
If the normal stress in rod (1) must be limited to 120 MPa, then the minimum cross-sectional area that 
can be used for rod (1) is 
 
21
1,min 2
(80 kN)(1,000 N/kN)
666.6667 mm
120 N/mm
F
A   
 
The minimum rod diameter is therefore 
 
2 2
1,min 1 1666.6667 mm 29.1346 mm 29. mm
4
1 A d d

    
 Ans. 
 
Similarly, the normal stress in rod (2) must be limited to 120 MPa. Notice that rod (2) is in 
compression. In this situation, we are concerned only with the magnitude of the stress; therefore, we 
will use the magnitude of F2 in the calculations for the minimum required cross-sectional area. 
 
22
2,min 2
(200 kN)(1,000 N/kN)
1,666.6667 mm
120 N/mm
F
A   
 
The minimum diameter for rod (2) is therefore 
 
2 2
2,min 2 21,666.6667 mm 46.0659 mm 46. mm
4
1 A d d

    
 Ans. 
 
 
 
 
 
 
 
 
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1.5 Two solid cylindrical rods (1) and (2) are 
joined together at flange B and loaded, as 
shown in Fig. P1.5. If the normal stress in 
each rod must be limited to 40 ksi, 
determine the minimum diameter required 
for each rod. 
Fig. P1.5 
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 2
1,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 
 
22
2,min
75 kips
1.875 in.
40 ksi
F
A   
 
The minimum diameter for rod (2) is therefore 
 
2 2
2,min 2 21.875 in. 1.54509 1.545 in.7 in.
4
A d d

    
 Ans. 
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1.6 Two solid cylindrical rods (1) and (2) are joined 
together at flange B and loaded, as shown in Fig. 
P1.6. 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). 
Fig. P1.6 
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 
 
1
1 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 
 
2
2 2
2
75 kips
15.2789 ksi
2.4053 in.
15.28 ksi (C)
F
A
     
 Ans. 
 
 
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1.7 Axial loads are applied with rigid bearing plates to the 
solid cylindrical rods shown in Fig. P1.7. 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 normal stress in each of the three rods. 
Fig. P1.7 
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
          
   
 
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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 
 
1
1 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 
 
2
2 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 
 
3
3 2
3
26 kips
3.6782 ksi
7.0686 in
3.68 ksi (C)
.
F
A
     
 Ans. 
 
 
 
 
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1.8 Axial loads are applied with rigid bearing plates to the solid 
cylindrical rods shown in Fig. P1.8. 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. 
Fig. P1.8 
 
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: 
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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 2
3,min 3 31.7333 in. 1.4856 in 1.486 in..
4
A d d

    
 Ans. 
 
 
 
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1.9 Two solid cylindrical rods support 
a load of P = 50 kN, as shown in Fig. 
P1.9. If the normal stress in each rod 
must be limited to 130 MPa, determine 
the minimum diameter required for 
each rod. 
Fig. P1.10 
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  
 
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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. 
 
 
 
 
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1.10 Two solid cylindrical rods 
support a load of P = 27 kN, as 
shown in Fig. P1.10. Rod (1) has a 
diameter of 16 mm and the diameter 
of rod (2) is 12 mm. Determine the 
normal stress in each rod. 
Fig. P1.10 
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  
 
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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. 
 
 
 
 
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1.11 A simple pin-connected truss is loaded 
and supported as shown in Fig. P1.11. 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. 
 
 
Fig. 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: 

 
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
 
 
    
    
 
 
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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 computeFAB 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. 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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1.12 A simple pin-connected truss is loaded 
and supported as shown in Fig. P1.12. 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. 
 
 
Fig. P1.12 
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: 
 
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
 
 
    
    
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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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. 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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1.13 A simple pin-connected truss is loaded 
and supported as shown in Fig. P1.13. 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. 
 
 
Fig. P1.13 
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
 
 
    
    
 
 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
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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. 
 
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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. 
1.14 The members of the truss shown in Fig. 
P1.14 are aluminum pipes that have an outside 
diameter of 4.50 in. and a wall thickness of 
0.237 in. Determine the normal stress in each 
truss member. 
 
 
Fig. P1.14 
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: 
 (15 kips)cos5
9.642 k
0
ip
0
sx
x xF A
A
    
  
 
 
 
(4 ft) (15 kips)(4 ft)cos50 (15 kips)(18 ft)sin50 0
61.350 kips
(15 kips)sin50
49.859 kips
0
A y
y y y
y
y
M B
F A B
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 AB, AC, and BC. 
Use the definition of the tangent function to determine AB, AC, and BC: 
 
6 ft
tan 1.5 56.3099
4 ft
4 ft
tan 0.222222 12.5288
18 ft
10 ft
tan 0.714286 35.5377
14 ft
AB AB
AC AC
BC BC
 
 
 
    
    
    
 
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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(12.5288 ) cos(56.3099 ) 0x AC AB xF F F A      
 (a) 
 
sin(12.5288 ) sin(56.3099 ) 0y AC AB yF F F A      
 (b) 
Solve Eqs. (a) and (b) simultaneously to obtain: 
 49.948 kips
38.259 kips
AB
AC
F
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(35.5377 ) cos(56.3099 ) 0x BC ABF F F     
 (c) 
 
sin(35.5377 ) sin(56.3099 ) 0y BC AB yF F F B      
 (d) 
 
Solve Eq. (c) for FBC: 
 
cos(56.3099 ) cos(56.3099 )
( 49.9484)
cos(35.5377 ) cos(35.5377
34.048 k s
)
ipBC ABF F
 
   
 

 
Eq. (d) can be used as a check on our calculations: 
 
sin(35.5377 ) sin(56.3099 )
( 34.0485 kips)sin(35.5377 ) ( 49.9484 kips)sin(56.3099 ) 61.350 kips 0
y BC AB yF F F B     
        Checks! 
 
Section properties: 
For each of the three truss members: 
 
2 2 24.50 in. 2(0.237 in.) 4.026 in. (4.50 in.) (4.026 in.) 3.17405 in.
4
d A

       
 
 
Normal stress in each truss member: 
 
2
49.948 kips
15.736 ksi
3.17405 
15.74 ksi (C
in
)
.
AB
AB
AB
F
A
      Ans. 
 
2
38.259 kips
12.054 ksi
3.17405 
12.05 ksi (T
n
)
i .
AC
AC
AC
F
A
     Ans. 
 
2
34.048 kips
10.727 ksi
3.17405 
10.73 ksi (C
in
)
.
BC
BC
BC
F
A
      Ans. 
 
 
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1.15 Bar (1) in Fig. 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. 
 
 Fig. P1.15 
 
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 relationshipto 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. 
 
 
 
 
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1.16 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). 
 
 
Fig. 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 
 
2
min
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. 
 
 
 
 
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1.17 For the clevis connection shown in Fig. P1.17, 
determine the shear stress in the 22-mm diameter bolt 
for an applied load of P = 90 kN. 
 
 Fig. P1.17 
Solution 
Consider a FBD of the bar that is connected by the clevis, 
including a portion of the bolt. If the shear force acting on each 
exposed surface of the bolt is denoted by V, then the shear force 
on each bolt surface is 
 
90 kN
0 45 kN
2
xF P V V V       
 
 
 
The area of the bolt surface exposed by the FBD is simply the cross-sectional area of the bolt: 
 
2 2 2
bolt bolt (22 mm) 380.1327 mm
4 4
A d
 
  
 
 
Therefore, the shear stress in the bolt is 
 
2
2
bolt
(45 kN)(1,000 N/kN)
118.3797 N/mm
380.1
118.4 MPa
327 mm
V
A
     Ans. 
 
 
 
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1.18 For the clevis connection shown in Fig. 
P1.18, the shear stress in the 3/8 in. diameter 
bolt must be limited to 36 ksi. Determine the 
maximum load P that may be applied to the 
connection. 
 Fig. P1.18 
Solution 
Consider a FBD of the bar that is connected by the clevis, 
including a portion of the bolt. If the shear force acting on each 
exposed surface of the bolt is denoted by V, then the shear force 
on each bolt surface is related to the load P by: 
 
0 2xF P V V P V      
 
 
 
The area of the bolt surface exposed by the FBD is simply the cross-sectional area of the bolt: 
 
2 2 2 2
bolt bolt (3 / 8 in.) (0.3750 in.) 0.1104466 in.
4 4 4
A d
  
   
 
 
If the shear stress in the bolt must be limited to 36 ksi, the maximum shear force V on a single cross-
sectional surface must be limited to 
 
2
bolt (36 ksi)(0.1104466 in. ) 3.976078 kipsV A   
 
Therefore, the maximum load P that may be applied to the connection is 
 
2 2(3.976078 kips) 7.952156 k 7ips .95 kipsP V   
 Ans. 
 
 
 
 
 
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1.19 For the connection shown in Fig. P1.19, 
determine the average shear stress in the 7/8-in. 
diameter bolts if the load is P = 45 kips. 
 
 Fig. P1.19 
Solution 
The bolts in this connection act in single shear. The cross-sectional area of a single bolt is 
 
2 2 2 2
bolt bolt (7 / 8 in.) (0.875 in.) 0.6013205 in.
4 4 4
A d
  
   
 
Since there are five bolts, the total area that carries shear stress is 
 
2 2
bolt5 5(0.6013205 in. ) 3.006602 in.VA A  
 
Therefore, the shear stress in each bolt is 
 
2
45 kips
14.96706 ksi
3.006602 in.
14.97 ksi
V
P
A
    
 Ans. 
 
 
 
 
 
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1.20 The five-bolt connection shown in Fig. 
P1.20 must support an applied load of P = 300 
kN. If the average shear stress in the bolts must 
be limited to 225 MPa, determine the minimum 
bolt diameter that may be used in the connection. 
 
 Fig P1.20 
Solution 
To support a load of 300 kN while not exceeding an average shear stress of 225 MPa, the total shear 
area provided by the bolts must be at least 
 
2
2
(300 kN)(1,000 N/kN)
1,333.3333 mm
225 N/mm
V
P
A   
 
Since there are five single-shear bolts in this connection, five cross-sectional surfaces carry shear stress. 
Consequently, each bolt must provide a minimum area of 
 2
2
bolt
1,333.3333 mm
266.6667 mm
5 5
VAA   
 
The minimum bolt diameter is therefore 
 
2 2
bolt bolt bolt 18.43 266.6667 mm 18.42 mm64 mm
4
A d d

    
 Ans. 
 
 
 
 
 
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1.21 The three-bolt connection shown in Fig. P1.21 
must support an applied load of P = 40 kips. If the 
average shear stress in the bolts must be limited to 
24 ksi, determine the minimum bolt diameter that 
may be used in the connection. 
 
 Fig. P1.21 
Solution 
The shear force V that must be provided by the bolts equals the applied load of P = 40 kips. The total 
shear area required is thus 
 
240 kips 1.66667 in.
24 ksi
V
V
A   
 
The three bolts in this connection act in double shear; therefore, six cross-sectional bolt surfaces are 
available to transmit shear stress. 
 2
2
bolt
1.66667 in.
0.27778 in. per surface
(2 surfaces per bolt)(3 bolts) 6 surfaces
VAA   
 
The minimum bolt diameter must be 
 
2 2
bolt bolt0.27778 in. 0.59471 in. 0.595 i .
4
nd d

   
 Ans. 
 
 
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1.22 For the connection shown in Fig. P1.22, 
the average shear stress in the 12-mm-diameter 
bolts must be limited to 160 MPa. Determine 
the maximum load P that may be applied to the 
connection. 
 
 Fig. P1.22 
Solution 
The cross-sectional area of a 12-mm-diameter bolt is 
 
2 2 2
bolt bolt (12 mm) 113.097355 mm
4 4
A d
 
  
 
This is a double-shear connection. Therefore, the three bolts provide a total shear area of 
 
2 2
bolt2(3 bolts) 2(3 bolts)(113.097355 mm ) 678.58401 mmVA A  
 
Since the shear stress must be limited to 160 MPa, the total shear force that can be resisted by the three 
bolts is 
 
2 2
max (160 N/mm )(678.58401 mm ) 108,573.442 NVV A   
In this connection, the shear force in the bolts is equal to the applied load P; therefore, 
 
max 108.6 kNP 
 Ans. 
 
 
 
 
 
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1.23 A hydraulic punch press is used to 
punch a slot in a 0.50-in. thick plate, as 
illustrated in Fig. P1.23. If the plate shears 
at a stress of 30 ksi, determine the 
minimum force P required to punch the 
slot. 
 
 Fig. P1.23 
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.
 
 
 
 
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1.24 A coupling is used to connect a 2 in. diameter 
plastic pipe (1) to a 1.5 in. diameter pipe (2), as 
shown in Fig. P1.24. 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. 
 
 Fig. 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. 
 
 
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1.25 A lever is attached to a shaft with a 
square shear key, as shown in Fig. P1.25. 
The force applied to the lever is P = 400 N. 
If the shear stress in the key must not exceed 
90 MPa, determine the minimum dimension 
“a” that must be used if the key is 15 mm 
long. 
 
 Fig. P1.25 
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): 
 
50 mm
(400 N)(750 mm) 0 12,000 N
2
OM V V
 
        
 
Since the shear stress in the key must not exceed 90 MPa, the shear area required is 
 
2
2
12,000 N
133.3333 mm
90 N/mm
V
V
A   
 
The shear area in the key is given by the product of its length L (i.e., 15 mm) and its width a. Therefore, 
the minimum key width a is 
 2133.3333 mm
8.8889 mm
15 
8.89 mm
mm
VAa
L
   
 Ans. 
 
 
 
 
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1.26 A common trailer hitch connection is shown in 
Fig. P1.26. The shear stress in the pin must be limited 
to 30,000 psi. If the applied load is P = 4,000 lb, 
determine the minimum diameter that must be used 
for the pin. 
 
 Fig. P1.26 
Solution 
The shear force V acting in the hitch pin is equal to the applied load; therefore, V = P = 4,000 lb. The 
shear area required to support a 4,000 lb shear force is 
 
24,000 lb 0.1333 in.
30,000 psi
V
V
A   
 
The hitch pin is used in a double-shear connection; therefore, two cross-sectional areas of the pin are 
subjected to shear stress. Thus, the cross-sectional area of the pin is given by 
 2
2
pin pin
0.1333 in.
2 0.0667 in.
2 2
V
V
A
A A A    
 
and the minimum pin diameter is 
 
2 2
pin pin0.0667 in. 0.2913 in. 0.291 n.
4
id d
   
 Ans. 
 
 
 
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1.27 An axial load P is supported by a short steel 
column, which has a cross-sectional area of 
11,400 mm
2
. If the average normal stress in the steel 
column must not exceed 110 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. 
 
 Fig. P1.27 
Solution 
Since the normal stress in the steel column must not exceed 110 MPa, the maximum column load is 
 
2 2
max (110 N/mm )(11,400 mm ) 1,254,000 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
1,254,000 N
156,750 mm
8 N/mmb
P
A   
 
Since the plate is square, the minimum plate dimension a must be 
 
2
min
396 
156,
mm
750 mm
395.9167 mm
A a a
a
  
   Ans. 
 
 
 
 
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1.28 The steel pipe column shown in Fig. 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. 
 
 Fig. 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, we must 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. 
 
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1.29 A vertical shaft is supported by a thrust 
collar and bearing plate, as shown in Fig. 
P1.29. The average shear stress in the collar 
must be limited to 18 ksi. The average bearing 
stress between the collar and the plate must be 
limited to 24 ksi. Based on these limits, 
determine the maximum axial load P that can 
be applied to the shaft. 
 
 Fig. P1.29 
Solution 
Consider collar shear stress: The area subjected to shear stress in the collar is equal to the product of 
the shaft circumference and the collar thickness; therefore, 
 
2shaft circumference collar thickness (1.0 in.)(0.5 in.) 1.5708 in.VA     
If the shear stress must not exceed 18 ksi, the maximum load that can be supported by the vertical shaft 
is: 
 
2(18 ksi)(1.5708 in. ) 28.2743 kipsVP A   
 
Consider collar bearing stress: We must determine the area of contact between the collar and the 
plate. The overall cross-sectional area of the collar is 
 
2 2
collar (1.5 in.) 1.7671 in.
4
A

 
 
is reduced by the area taken up by the shaft 
 
2 2
shaft (1.0 in.) 0.7854 in.
4
A

 
 
Therefore, the area of the collar that actually contacts the plate is 
 
2 2 2
collar shaft 1.7671 in. 0.7854 in. 0.9817 in.bA A A    
 
If the bearing stress must not exceed 24 ksi, the maximum load that can be supported by the vertical 
shaft is: 
 
2(24 ksi)(0.9817 in. ) 23.5619 kipsb bP A   
 
Controlling P: Considering both shear stress in the collar and bearing stress between the collar and the 
plate, the maximum load that can be supported by the shaft is 
 
max 23.6 kipsP 
 Ans. 
 
 
 
 
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1.30 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. 
 
 
 
 
 
 
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1.31 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-sectionalarea 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. 
 
 
 
 
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1.32 An axial load P is applied to the 
rectangular bar shown in Fig. P1.32. The 
cross-sectional area of the bar is 400 mm
2
. 
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. 
 
 Fig. P1.32 
Solution 
The angle  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     
 
and the shear force V parallel to plane AB is 
 
sin (70 kN)sin35 40.1504 kNV P     
 
 
The cross-sectional area of the bar is 400 mm
2
, but the area along inclined plane AB is 
 2
2400 mm 488.3098 mm
cos cos35
n
A
A   
 
 
The normal stress n 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
     Ans. 
 
The shear stress nt parallel to plane AB is 
 
2
(40.1504 kN)(1,000 N/kN)
82.2231 MPa
4
82.2 MP
88.309 
a
8 mm
nt
n
V
A
     Ans. 

 
 
 
 
 
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1.33 An axial load P is applied to the 1.75 in. 
by 0.75 in. rectangular bar shown in Fig. 
P1.33. 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. 
 
 Fig. P1.33 
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
.
in
n
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. 

 
 
 
 
 
 
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1.34 A compression load of P = 80 kips is applied to a 4 in. 
by 4 in. square post, as shown in Fig. P1.34. Determine the 
normal stress perpendicular to plane AB and the shear stress 
parallel to plane AB. 
 Fig. P1.34 
 
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. 
 
 
 
 
 
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1.35 Specifications for the 50 mm × 50 mm square bar 
shown in Fig. P1.35 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. 
 Fig. P1.35 
Solution 
The general equations for normal and shear stresses on an inclined plane in terms of the angle  are 
 
(1 cos 2 )
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. 
 
 
 
 
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1.36 Specifications for the 6 in. × 6 in. square post shown in 
Fig. P1.36 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. 
 Fig. P1.36 
Solution 
The general equations for normal and shear stresses on an inclined plane in terms of the angle  are 
 
(1 cos 2 )
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