ISM for
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
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
A
( D2  d 2 ) 
[(60 mm)2  (50 mm)2 ]  863.938 mm2
4
4
The normal stress in the tube can be expressed as
P

A
The maximum normal stress in the tube must be limited to 200 MPa. Using 200 MPa as the allowable
normal stress, rearrange this expression to solve for the maximum load P
Ans.
Pmax   allow A  (200 N/mm2 )(863.938 mm2 )  172,788 N  172.8 kN
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.
1.2 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
P
P 27 kips

 Amin  
 1.500 in.2
 18 ksi
A
The cross-sectional area of the aluminum tube is given by
A

( D2  d 2 )
4
Set this expression equal to the minimum area and solve for the maximum inside diameter d

4
[(2.50 in.)2  d 2 ]  1.500 in.2
(2.50 in.)2  d 2 
(2.50 in.)2 

4

4
(1.500 in.2 )
(1.500 in.2 )  d 2
 d max  2.08330 in.
The outside diameter D, the inside diameter d, and the wall thickness t are related by
D  d  2t
Therefore, the minimum wall thickness required for the aluminum tube is
D  d 2.50 in.  2.08330 in.
tmin 

 0.20835 in.  0.208 in.
2
2
Ans.
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.
1.3 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,
Fx  F1  80 kN  0
 F1  80 kN (T)
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):
Fx  F2  140 kN  140 kN  80 kN  0

 F2  200 kN  200 kN (C)
From the given diameter of rod (1), the cross-sectional area of rod (1) is
A1 
(24 mm) 2  452.3893 mm 2
4
and thus, the normal stress in rod (1) is
F (80 kN)(1,000 N/kN)
1  1 
 176.8388 MPa  176.8 MPa (T)
A1
452.3893 mm 2

Ans.
From the given diameter of rod (2), the cross-sectional area of rod (2) is
A2 
(42 mm) 2  1,385.4424 mm2
4
Accordingly, the normal stress in rod (2) is
F (200 kN)(1,000 N/kN)
 144.3582 MPa  144.4 MPa (C)
2  2 
1,385.4424 mm 2
A2
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.
1.4 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,
Fx  F1  80 kN  0
 F1  80 kN (T)
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):
Fx  F2  140 kN  140 kN  80 kN  0
 F2  200 kN  200 kN (C)
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
F (80 kN)(1,000 N/kN)
A1,min  1 
 666.6667 mm2
2
120 N/mm

The minimum rod diameter is therefore
A1,min 

4
d12  666.6667 mm 2
 d1  29.1346 mm  29.1 mm
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.
F (200 kN)(1,000 N/kN)
A2,min  2 
 1, 666.6667 mm 2
2
120 N/mm

The minimum diameter for rod (2) is therefore
A2,min 

4
d 22  1,666.6667 mm2
 d 2  46.0659 mm  46.1 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.
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,
Fy   F1  15 kips  0
 F1  15 kips  15 kips (C)
Next, cut a FBD through rod (2) that includes the free end of the rod at A. Again, we
will assume that the internal force in rod (2) is tension. Equilibrium of this FBD reveals
the internal force in rod (2):
Fy   F2  30 kips  30 kips  15 kips  0
 F2  75 kips  75 kips (C)
Notice that rods (1) and (2) are in compression. In this situation, we are
concerned only with the stress magnitude; therefore, we will use the force
magnitudes to determine the minimum required cross-sectional areas. If the
normal stress in rod (1) must be limited to 40 ksi, then the minimum crosssectional area that can be used for rod (1) is
F 15 kips
A1,min  1 
 0.375 in.2
40 ksi

The minimum rod diameter is therefore
A1,min 

d12  0.375 in.2
 d1  0.69099 in.  0.691 in.
Ans.
4
Similarly, the normal stress in rod (2) must be limited to 40 ksi, which requires a minimum area of
F 75 kips
 1.875 in.2
A2,min  2 

40 ksi
The minimum diameter for rod (2) is therefore
A2,min 

4
d 22  1.875 in.2
 d 2  1.545097 in.  1.545 in.
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.
1.6 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,
Fy   F1  15 kips  0
 F1  15 kips  15 kips (C)
Next, cut a FBD through rod (2) that includes the free end of the rod at A. Again, we
will assume that the internal force in rod (2) is tension. Equilibrium of this FBD
reveals the internal force in rod (2):
Fy   F2  30 kips  30 kips  15 kips  0
 F2  75 kips  75 kips (C)

From the given diameter of rod (1), the cross-sectional area of rod (1) is
A1 
(1.75 in.)2  2.4053 in.2
4
and thus, the normal stress in rod (1) is
F
15 kips
 6.23627 ksi  6.24 ksi (C)
1  1 
A1 2.4053 in.2

Ans.
From the given diameter of rod (2), the cross-sectional area of rod (2) is
A2 
(2.50 in.) 2  4.9087 in.2
4
Accordingly, the normal stress in rod (2) is
F
75 kips
 15.2789 ksi  15.28 ksi (C)
2  2 
A2 2.4053 in.2
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.
1.7 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,
Fy   F1  8 kips  4 kips  4 kips  0
 F1  16 kips  16 kips (C)
FBD through rod (1)
FBD through rod (2)
FBD through rod (3)
Next, cut a FBD through rod (2) that includes the free end A. Again, we will assume that the internal
force in rod (2) is tension. Equilibrium of this FBD reveals the internal force in rod (2):
Fy   F2  8 kips  4 kips  4 kips  15 kips  15 kips  0  F2  14 kips  14 kips (T)
Similarly, cut a FBD through rod (3) that includes the free end A. From this FBD, the internal force in
rod (3) is:
Fy   F3  8 kips  4 kips  4 kips  15 kips  15 kips  20 kips  20 kips  0
 F3  26 kips  26 kips (C)
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
A1 
(2.00 in.) 2  3.1416 in.2
4
and thus, the normal stress in aluminum rod (1) is
F
16 kips
1  1 
 5.0930 ksi  5.09 ksi (C)
A1 3.1416 in.2

Ans.
From the given diameter of rod (2), the cross-sectional area of rod (2) is
A2 
(1.50 in.)2  1.7671 in.2
4
Accordingly, the normal stress in brass rod (2) is
F
14 kips
2  2 
 7.9224 ksi  7.92 ksi (T)
A2 1.7671 in.2

Ans.
Finally, the cross-sectional area of rod (3) is
A3 
(3.00 in.) 2  7.0686 in.2
4
and the normal stress in the steel rod is
F
26 kips
 3.6782 ksi  3.68 ksi (C)
3  3 
A3 7.0686 in.2
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.
1.8 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,
Fy   F1  8 kips  4 kips  4 kips  0
 F1  16 kips  16 kips (C)
FBD through rod (1)
FBD through rod (2)
FBD through rod (3)
Next, cut a FBD through rod (2) that includes the free end A. Again, we will assume that the internal
force in rod (2) is tension. Equilibrium of this FBD reveals the internal force in rod (2):
Fy   F2  8 kips  4 kips  4 kips  15 kips  15 kips  0  F2  14 kips  14 kips (T)
Similarly, cut a FBD through rod (3) that includes the free end A. From this FBD, the internal force in
rod (3) is:
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.
Fy   F3  8 kips  4 kips  4 kips  15 kips  15 kips  20 kips  20 kips  0
 F3  26 kips  26 kips (C)
Notice that two of the three rods are in compression. In these situations, we are concerned only with the
stress magnitude; therefore, we will use the force magnitudes to determine the minimum required crosssectional areas, and in turn, the minimum rod diameters. The normal stress in aluminum rod (1) must be
limited to 18 ksi; therefore, the minimum cross-sectional area required for rod (1) is
F 16 kips
A1,min  1 
 0.8889 in.2
 1 18 ksi
The minimum rod diameter is therefore
A1,min 

4
d12  0.8889 in.2
 d1  1.0638 in.  1.064 in.
Ans.
The normal stress in brass rod (2) must be limited to 25 ksi, which requires a minimum area of
F 14 kips
A2,min  2 
 0.5600 in.2
 2 25 ksi
which requires a minimum diameter for rod (2) of
A2,min 

4
d 22  0.5600 in.2
 d 2  0.8444 in.  0.844 in.
Ans.
The normal stress in steel rod (3) must be limited to 15 ksi. The minimum cross-sectional area required
for this rod is:
26 kips
F
A3,min  3 
 1.7333 in.2
 3 15 ksi
which requires a minimum diameter for rod (3) of
A3,min 

4
d32  1.7333 in.2
 d3  1.4856 in.  1.486 in.
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.
1.9 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.
Fx  F2 cos(35.7067)  F1 cos(57.9946)  0
Fy  F2 sin(35.7067)  F1 sin(57.9946)  P  0
(a)
(b)
Unknown forces F1 and F2 can be found from the simultaneous solution of Eqs. (a) and (b). Using the
substitution method, Eq. (b) can be solved for F2 in terms of F1:
cos(57.9946)
F2  F1
(c)
cos(35.7067)
Substituting Eq. (c) into Eq. (b) gives
cos(57.9946)
sin(35.7067)  F1 sin(57.9946)  P
F1
cos(35.6553)
F1  cos(57.9946) tan(35.7067)  sin(57.9946)   P
 F1 
P
P

cos(57.9946) tan(35.7067)  sin(57.9946) 1.2289
For the given load of P = 50 kN, the internal force in rod (1) is therefore:
50 kN
F1 
 40.6856 kN
1.2289
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:
cos(57.9946)
cos(57.9946)
F2  F1
 (40.6856 kN)
 26.5553 kN
cos(35.7067)
cos(35.7067)
The normal stress in rod (1) must be limited to 130 MPa; therefore, the minimum cross-sectional area
required for rod (1) is
F (40.6856 kN)(1,000 N/kN)
A1,min  1 
 312.9664 mm 2
2
130 N/mm
1
The minimum rod diameter is therefore
A1,min 

4
d12  312.9664 mm 2
 d1  19.9620 mm  19.96 mm
Ans.
The minimum area required for rod (2) is
F
(26.5553 kN)(1,000 N/kN)
A2,min  2 
 204.2718 mm 2
2
130 N/mm
2
which requires a minimum diameter for rod (2) of
A2,min 

4
d 22  204.2718 mm2
 d 2  16.1272 mm  16.13 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.
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.
Fx  F2 cos(35.7067)  F1 cos(57.9946)  0
Fy  F2 sin(35.7067)  F1 sin(57.9946)  P  0
(a)
(b)
Unknown forces F1 and F2 can be found from the simultaneous solution of Eqs. (a) and (b). Using the
substitution method, Eq. (b) can be solved for F2 in terms of F1:
cos(57.9946)
F2  F1
(c)
cos(35.7067)
Substituting Eq. (c) into Eq. (b) gives
cos(57.9946)
sin(35.7067)  F1 sin(57.9946)  P
F1
cos(35.6553)
F1  cos(57.9946) tan(35.7067)  sin(57.9946)   P
 F1 
P
P

cos(57.9946) tan(35.7067)  sin(57.9946) 1.2289
For the given load of P = 27 kN, the internal force in rod (1) is therefore:
27 kN
F1 
 21.9702 kN
1.2289
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:
cos(57.9946)
cos(57.9946)
F2  F1
 (21.9702 kN)
 14.3399 kN
cos(35.7067)
cos(35.7067)

The diameter of rod (1) is 16 mm; therefore, its cross-sectional area is:
A1 
(16 mm)2  201.0619 mm 2
4
and the normal stress in rod (1) is:
F (21.9702 kN)(1,000 N/kN)
1  1 
 109.2710 N/mm 2  109.3 MPa (T)
2
201.0619 mm
A1

Ans.
The diameter of rod (2) is 12 mm; therefore, its cross-sectional area is:
A2 
(12 mm)2  113.0973 mm 2
4
and the normal stress in rod (2) is:
(14.3399 kN)(1,000 N/kN)
F
2  2 
 126.7924 N/mm 2  126.8 MPa (T)
113.0973 mm 2
A2
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.
1.11 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:
Fx  Ax  2 kips  0

 Ax  2 kips
M A  By (6 ft)  (5 kips)(14 ft)  (2 kips)(7 ft)  0
 By  14 kips
Fy  Ay  By  5 kips  0
 Ay  9 kips
Method of joints:
Before beginning the process of determining the internal forces in the axial members, the geometry of
the truss will be used to determine the magnitude of the inclination angles of members AC and BC. Use
the definition of the tangent function to determine AC and BC:
7 ft
tan  AC 
 0.50
 AC  26.565
14 ft
7 ft
tan  BC 
 0.875
 BC  41.186
8 ft
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.
Fx  FAC cos(26.565)  FAB  Ax  0
(a)
Fy  FAC sin(26.565)  Ay  0
(b)
Solve Eq. (b) for FAC:
Ay
9 kips
FAC  

 20.125 kips
sin(26.565)
sin(26.565)
and then compute FAB using Eq. (a):
FAB   FAC cos(26.565)  Ax
  (20.125 kips) cos(26.565)  ( 2 kips)  16.000 kips
Joint B:
Next, consider a FBD of joint B. In this instance, the equilibrium
equations associated with joint B seem easier to solve than those that
would pertain to joint C. As before, tension forces will be assumed in
each truss member.
Fx   FAB  FBC cos(41.186)  0
(c)
Fy  FBC sin(41.186)  By  0
(d)
Solve Eq. (d) for FBC:
By
14 kips
FBC  

 21.260 kips
sin(41.186)
sin(41.186)
Eq. (c) can be used as a check on our calculations:
Fx   FAB  FBC cos(41.186)
  ( 16.000 kips)  ( 21.260 kips) cos(41.186)  0
Section properties:
For each of the three truss members:
d  4.00 in.  2(0.226 in.)  3.548 in.
A

(4.00 in.) 2  (3.548 in.) 2   2.67954 in.2
4
Normal stress in each truss member:
F
16.000 kips
 5.971 ksi  5.97 ksi (C)
 AB  AB 
AAB
2.67954 in.2
 AC 
 BC 
Checks!
20.125 kips
FAC

 7.510 ksi  7.51 ksi (T)
AAC 2.67954 in.2
FBC 21.260 kips

 7.934 ksi  7.93 ksi (C)
ABC
2.67954 in.2
Ans.
Ans.
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.
1.12 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 freebody diagram (FBD) of the entire truss is
shown. The following equilibrium equations
can be written for this structure:
Fx  Ax  12 kN  0
 Ax  12 kN
M A  By (1 m)  (15 kN)(4.3 m)  0
 By  64.5 kN
Fy  Ay  By  15 kN  0
 Ay  49.5 kN
Method of joints:
Before beginning the process of determining the internal forces in the axial members, the geometry of
the truss will be used to determine the magnitude of the inclination angles of members AB and BC. Use
the definition of the tangent function to determine AB and BC:
1.5 m
 1.50
  AB  56.310
tan  AB 
1.0 m
1.5 m
 0.454545
  BC  24.444
tan  BC 
3.3 m
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.
Fx  FAC  FAB cos(56.310)  Ax  0
(a)
Fy  Ay  FAB sin(56.310)  0
(b)
Solve Eq. (b) for FAB:
Ay
49.5 kN
FAB 

 59.492 kN
sin(56.310) sin(56.310)
and then compute FAC using Eq. (a):
FAC   FAB cos(56.310)  Ax
  ( 59.492 kN)cos(56.310)  ( 12 kN)  45.000 kN
Joint C:
Next, consider a FBD of joint C. In this instance, the equilibrium
equations associated with joint C seem easier to solve than those that
would pertain to joint B. As before, tension forces will be assumed in
each truss member.
Fx   FAC  FBC cos(24.444)  12 kN  0
(c)
Fy   FBC sin(24.444)  15 kN  0
(d)
Solve Eq. (d) for FBC:
15 kN
FBC 
 36.249 kN
sin(24.444)
Eq. (c) can be used as a check on our calculations:
Fx   FAC  FBC cos(24.444)  12 kN  0
  (45.000 kN)  ( 36.249 kN) cos(24.444)  12 kN  0
Section properties:
For each of the three truss members:
d  60 mm  2(4 mm)  52 mm
A
Checks!

(60 mm) 2  (52 mm) 2   703.7168 mm2
4
Normal stress in each truss member:
F
( 59.492 kN)(1,000 N/kN)
 84.539 MPa  84.5 MPa (C)
 AB  AB 
703.7168 mm 2
AAB
F
(45.000 kN)(1,000 N/kN)
 AC  AC 
 63.946 MPa  63.9 MPa (T)
703.7168 mm 2
AAC
 BC 
FBC ( 36.249 kN)(1,000 N/kN)

 51.511 MPa  51.5 MPa (C)
ABC
703.7168 mm 2
Ans.
Ans.
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.
1.13 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:
Fy  Ay  30 kN  0
 Ay  30 kN
M A  (30 kN)(4.5 m)  (15 kN)(1.6 m)  Bx (5.6 m)  0
 Bx  19.821 kN
Fx  Ax  Bx  15 kN  0
Ax  15 kN  Bx  15 kN  (19.821 kN)

 Ax  34.821 kN
Method of joints:
Before beginning the process of determining the internal forces in the axial members, the geometry of
the truss will be used to determine the magnitude of the inclination angles of members AC and BC. Use
the definition of the tangent function to determine AC and BC:
1.6 m
 0.355556
  AC  19.573
tan  AC 
4.5 m
4m
 0.888889
  BC  41.634
tan  BC 
4.5 m
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.
Fx  Ax  FAC cos(19.573)  0
(a)
Fy  Ay  FAC sin(19.573)  FAB  0
(b)
Solve Eq. (a) for FAC:
Ax
34.821 kN
FAC 

 36.957 kN
cos(19.573) cos(19.573)
and then compute FAB using Eq. (b):
FAB  Ay  FAC sin(19.573)
 (30.000 kN)  (36.957 kN)sin(19.573)  17.619 kN
Joint B:
Next, consider a FBD of joint B. In this instance, the equilibrium
equations associated with joint B seem easier to solve than those that
would pertain to joint C. As before, tension forces will be assumed in
each truss member.
Fx  Bx  FBC cos(41.634)  0
(c)
Fy  FBC sin(41.634)  FAB  0
(d)
Solve Eq. (c) for FBC:
Bx
( 19.821 kN)
FBC 

 26.520 kN
cos(41.634) cos(41.634)
Eq. (d) can be used as a check on our calculations:
Fy  FBC sin(41.634)  FAB
 (26.520 kN)sin(41.634)  (17.619 kN)  0
Section properties:
For each of the three truss members:
d  42 mm  2(3.5 mm)  35 mm
A
Checks!

(42 mm) 2  (35 mm) 2   423.3296 mm2
4
Normal stress in each truss member:
F
(17.619 kN)(1,000 N/kN)
 AB  AB 
 41.620 MPa  41.6 MPa (T)
AAB
423.3296 mm 2
 AC 
 BC 
FAC (36.957 kN)(1,000 N/kN)

 87.301 MPa  87.3 MPa (T)
423.3296 mm 2
AAC
FBC ( 26.520 kN)(1,000 N/kN)

 62.647 MPa  62.6 MPa (C)
423.3296 mm 2
ABC
Ans.
Ans.
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.
1.14 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:
Fx  Ax  (15 kips)cos50  0
 Ax  9.642 kips
M A  By (4 ft)  (15 kips)(4 ft)cos50  (15 kips)(18 ft)sin50  0
 By  61.350 kips
Fy  Ay  By  (15 kips)sin 50  0
 Ay  49.859 kips
Method of joints:
Before beginning the process of determining the internal forces in the axial members, the geometry of
the truss will be used to determine the magnitude of the inclination angles of members AB, AC, and BC.
Use the definition of the tangent function to determine AB, AC, and BC:
6 ft
tan  AB 
 1.5
  AB  56.3099
4 ft
4 ft
tan  AC 
 0.222222
  AC  12.5288
18 ft
10 ft
tan  BC 
 0.714286
  BC  35.5377
14 ft
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.
Fx  FAC cos(12.5288)  FAB cos(56.3099)  Ax  0
(a)
Fy  FAC sin(12.5288)  FAB sin(56.3099)  Ay  0
(b)
Solve Eqs. (a) and (b) simultaneously to obtain:
FAB  49.948 kips
FAC  38.259 kips
Joint B:
Next, consider a FBD of joint B. In this instance, the equilibrium
equations associated with joint B seem easier to solve than those that
would pertain to joint C. As before, tension forces will be assumed in
each truss member.
Fx  FBC cos(35.5377)  FAB cos(56.3099)  0
(c)
Fy  FBC sin(35.5377)  FAB sin(56.3099)  By  0
(d)
Solve Eq. (c) for FBC:
cos(56.3099)
cos(56.3099)
FBC  FAB
 ( 49.9484)
 34.048 kips
cos(35.5377)
cos(35.5377)
Eq. (d) can be used as a check on our calculations:
Fy  FBC sin(35.5377)  FAB sin(56.3099)  By
 (34.0485 kips)sin(35.5377)  ( 49.9484 kips)sin(56.3099)  61.350 kips  0
Section properties:
For each of the three truss members:
d  4.50 in.  2(0.237 in.)  4.026 in.
A

(4.50 in.) 2  (4.026 in.) 2   3.17405 in.2
4
Normal stress in each truss member:
F
49.948 kips
 15.736 ksi  15.74 ksi (C)
 AB  AB 
3.17405 in.2
AAB
F
38.259 kips
 AC  AC 
 12.054 ksi  12.05 ksi (T)
AAC 3.17405 in.2
 BC 
Checks!
FBC 34.048 kips

 10.727 ksi  10.73 ksi (C)
3.17405 in.2
ABC
Ans.
Ans.
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.
1.15 Bar (1) in Fig. P1.15 has a crosssectional area of 0.75 in.2. If the stress in bar
(1) must be limited to 30 ksi, determine the
maximum load P that may be supported by
the structure.
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
F1,max  1 A1  (30 ksi)(0.75 in.2 )  22.5 kips
Consider a FBD of ABC. From the moment equilibrium
equation about joint A, the relationship between the force in
bar (1) and the load P is:
M A  (6 ft)F1  (10 ft)P  0
P 
6 ft
F1
10 ft
Substitute the maximum force F1,max = 22.5 kips into this relationship to obtain the maximum load that
may be applied to the structure:
6 ft
6 ft
P
F1 
(22.5 kips)  13.50 kips
Ans.
10 ft
10 ft
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.
1.16 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
Fx  P  V  V  0
V 
10, 000 lb
 5, 000 lb
2
In other words, each glue surface must be large enough so that 5,000 lb of shear resistance can be
provided to board (2). Since the glue has a shear strength of 120 psi, the area of each glue surface on
board (2) must be at least
5, 000 lb
Amin 
 41.6667 in.2
120 psi
The boards are 6-in. wide; therefore, glue must be spread along board (2) for a length of at least
41.6667 in.2
Lglue joint 
 6.9444 in.
6 in.
Although we’ve discussed only board (2), the same rationale applies to board (1). For both boards (1)
and (2), the glue must be applied along a length of at least 6.9444 in. on both the top and bottom of the
boards in order to resist the 10,000 lb applied load.
The glue applied to boards (1) and (2) must be matched by glue applied to the splice plates. Therefore,
the splice plates must be at least 6.9444 in. + 6.9444 in. = 13.8889 in. long. However, we are told that a
0.5-in. gap is required between boards (1) and (2); therefore, the splice plates must be 0.5-in. longer.
Altogether, the length of the splice plates must be at least
Ans.
Lmin  6.9444 in.  6.9444 in.  0.5 in.  14.39 in.
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.
1.17 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
Fx  P  V  V  0
V 
 45 kN
2


The area of the bolt surface exposed by the FBD is simply the cross-sectional area of the bolt:
Abolt 
4
2
d bolt

4
(22 mm)2  380.1327 mm 2
Therefore, the shear stress in the bolt is
V
(45 kN)(1,000 N/kN)


 118.3797 N/mm 2  118.4 MPa
Abolt
380.1327 mm 2
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.
1.18 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:
Fx  P  V  V  0
 P  2V



The area of the bolt surface exposed by the FBD is simply the cross-sectional area of the bolt:
Abolt 
4
2
d bolt

4
(3 / 8 in.) 2 
4
(0.3750 in.) 2  0.1104466 in.2
If the shear stress in the bolt must be limited to 36 ksi, the maximum shear force V on a single crosssectional surface must be limited to
V   Abolt  (36 ksi)(0.1104466 in.2 )  3.976078 kips
Therefore, the maximum load P that may be applied to the connection is
P  2V  2(3.976078 kips)  7.952156 kips  7.95 kips
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.
1.19 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
Abolt 
2
d bolt

(7 / 8 in.)2 
(0.875 in.) 2  0.6013205 in.2
4
4
4
Since there are five bolts, the total area that carries shear stress is
AV  5 Abolt  5(0.6013205 in.2 )  3.006602 in.2
Therefore, the shear stress in each bolt is
P
45 kips


 14.96706 ksi  14.97 ksi
AV 3.006602 in.2
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.
1.20 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
P (300 kN)(1,000 N/kN)
AV  
 1,333.3333 mm 2
2
225 N/mm

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
AV 1,333.3333 mm 2
Abolt 

 266.6667 mm 2
5
5
The minimum bolt diameter is therefore
Abolt  266.6667 mm2 

4
2
d bolt
 d bolt  18.4264 mm  18.43 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.
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
V 40 kips
 1.66667 in.2
AV  
24 ksi

The three bolts in this connection act in double shear; therefore, six cross-sectional bolt surfaces are
available to transmit shear stress.
AV
1.66667 in.2
Abolt 

 0.27778 in.2 per surface
(2 surfaces per bolt)(3 bolts)
6 surfaces
The minimum bolt diameter must be

4
2
d bolt
 0.27778 in.2
 d bolt  0.59471 in.  0.595 in.
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.
1.22 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
Abolt 
2
d bolt

(12 mm)2  113.097355 mm2
4
4
This is a double-shear connection. Therefore, the three bolts provide a total shear area of
AV  2(3 bolts)Abolt  2(3 bolts)(113.097355 mm 2 )  678.58401 mm 2
Since the shear stress must be limited to 160 MPa, the total shear force that can be resisted by the three
bolts is
Vmax   AV  (160 N/mm 2 )(678.58401 mm 2 )  108,573.442 N
In this connection, the shear force in the bolts is equal to the applied load P; therefore,
Ans.
Pmax  108.6 kN
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.
1.23 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
AV  perimeter  plate thickness  (8.35619 in.)(0.50 in.)  4.17810 in.2
Given that the plate shears at  = 30 ksi, the force required to remove the slug is therefore
Pmin   AV  (30 ksi)(4.17810 in.2 )  125.343 kips  125.3 kips
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.
1.24 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
V
5, 000 lb
AV 

 12.5 in.2
 adhesive 400 psi
Consider the coupling on pipe (1). The adhesive is applied to the circumference of the pipe, and the
circumference C1 of pipe (1) is
C1   D1   (2.0 in.)  6.2832 in.
The minimum length L1 is therefore
A
12.5 in.2
Ans.
L1  V 
 1.9894 in.  1.989 in.
C1 6.2832 in.
Consider the coupling on pipe (2). The circumference C2 of pipe (2) is
C2   D2   (1.5 in.)  4.7124 in.
The minimum length L2 is therefore
A
12.5 in.2
L2  V 
 2.6526 in.  2.65 in.
C2 4.7124 in.
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.
1.25 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 
M O  (400 N)(750 mm)  
V 0
V  12,000 N
 2 
Since the shear stress in the key must not exceed 90 MPa, the shear area required is
V 12,000 N
AV  
 133.3333 mm2
2
 90 N/mm
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
A
133.3333 mm 2
a V 
 8.8889 mm  8.89 mm
Ans.
L
15 mm
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.
1.26 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
V
4, 000 lb
AV  
 0.1333 in.2
 30, 000 psi
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
AV 0.1333 in.2
AV  2 Apin
 Apin 

 0.0667 in.2
2
2
and the minimum pin diameter is

4
2
d pin
 0.0667 in.2
 d pin  0.2913 in.  0.291 in.
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.
1.27 An axial load P is supported by a short steel
column, which has a cross-sectional area of
11,400 mm2. 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
Pmax   A  (110 N/mm 2 )(11, 400 mm 2 )  1, 254, 000 N
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
P 1, 254, 000 N
Amin 

 156, 750 mm 2
2
b
8 N/mm
Since the plate is square, the minimum plate dimension a must be
Amin  156, 750 mm2  a  a
 a  395.9167 mm  396 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.
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 crosssectional area of the pipe. To calculate the pipe area, we must first calculate the pipe inside diameter d:
D  d  2t
 d  D  2t  8.625 in.  2(0.25 in.)  8.125 in.
The pipe cross-sectional area is


 D 2  d 2   (8.625 in.)2  (8.125 in.) 2   6.5777 in.2
4
4
Therefore, the bearing stress between the pipe and one of the bearing plates is
P
80 kips
b 

 12.1623 ksi  12.16 ksi
Ab 6.5777 in.2
Apipe 
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
P 80 kips

 160 in.2
Ab 
 b 0.5 ksi
If the width of the timber beam is 10.75 in., then the length L of the upper bearing plate must be
Ab
160 in.2
L

 14.8837 in.  14.88 in.
Ans.
beam width 10.75 in.
(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
P 80 kips
Ab 

 88.8889 in.2
 b 0.9 ksi
Since the lower bearing plate is square, its dimension a must be
Ans.
Ab  a  a  88.8889 in.2
 a  9.4281 in.  9.43 in.
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.
1.29 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,
AV  shaft circumference  collar thickness   (1.0 in.)(0.5 in.)  1.5708 in.2
If the shear stress must not exceed 18 ksi, the maximum load that can be supported by the vertical shaft
is:
P   AV  (18 ksi)(1.5708 in.2 )  28.2743 kips
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
Acollar 

(1.5 in.) 2  1.7671 in.2
4
is reduced by the area taken up by the shaft
Ashaft 

(1.0 in.) 2  0.7854 in.2
4
Therefore, the area of the collar that actually contacts the plate is
Ab  Acollar  Ashaft  1.7671 in.2  0.7854 in.2  0.9817 in.2
If the bearing stress must not exceed 24 ksi, the maximum load that can be supported by the vertical
shaft is:
P   b Ab  (24 ksi)(0.9817 in.2 )  23.5619 kips
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
Ans.
Pmax  23.6 kips
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.
1.30 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
F (150 kN)(1,000 N/kN)
 max  
 80 MPa
A
(25 mm)(75 mm)
The maximum shear stress is one-half of the maximum normal stress
 max 
 max
2
 40 MPa
Ans.
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.
1.31 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-sectional area of the rod is
92 kips
F
Amin 

 3.0667 in.2
 max 30 ksi
For the 12-ksi shear stress limit, the minimum cross-sectional area of the rod must be
92 kips
F
Amin 

 3.8333 in.2
2 max 2(12 ksi)
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
4
2
 3.8333 in.2
d min
 d min  2.2092 in.  2.21 in.
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.
1.32 An axial load P is applied to the
rectangular bar shown in Fig. P1.32. The
cross-sectional area of the bar is 400 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.
Fig. P1.32
Solution
The angle  for the inclined plane is 35°. The
normal force N perpendicular to plane AB is
found from
N  P cos  (40 kN)cos35  57.3406 kN
and the shear force V parallel to plane AB is
V  P sin   (70 kN)sin35  40.1504 kN
The cross-sectional area of the bar is 400 mm2, but the area along inclined plane AB is
A
400 mm 2
An 

 488.3098 mm 2
cos 
cos35
The normal stress n perpendicular to plane AB is
N (57.3406 kN)(1,000 N/kN)
n 

 117.4268 MPa  117.4 MPa
488.3098 mm 2
An
The shear stress nt parallel to plane AB is
V
(40.1504 kN)(1,000 N/kN)
 nt 

 82.2231 MPa  82.2 MPa
488.3098 mm 2
An

Ans.
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.
1.33 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
N  P cos  (18 kips)cos60  9.0 kips
and the shear force V parallel to plane AB is
V  P sin   (18 kips)sin 60  15.5885 kips
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
1.3125 in.2
An  A / cos  
 2.6250 in.2
cos 60
The normal stress n perpendicular to plane AB is
N
9.0 kips
n 

 3.4286 ksi  3.43 ksi
An 2.6250 in.2
The shear stress nt parallel to plane AB is
V 15.5885 kips
 nt 

 5.9385 ksi  5.94 ksi
2.6250 in.2
An

Ans.
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.
1.34 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
N  P cos  (80 kips) cos55  45.8861 kips
and the shear force V parallel to plane AB is
V  P sin   (80 kips)sin 55  65.5322 kips
The cross-sectional area of the post is (4 in.)(4 in.) = 16 in.2, but the area
along inclined plane AB is
16 in.2
An  A / cos  
 27.8951 in.2
cos 55
The normal stress n perpendicular to plane AB is
N 45.8861 kips
n 

 1.6449 ksi  1.645 ksi
An 27.8951 in.2
The shear stress nt parallel to plane AB is
V
65.5322 kips
 nt 

 2.3492 ksi  2.35 ksi
27.8951 in.2
An
Ans.
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.
1.35 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
P
(a)
n 
(1  cos 2 )
2A
and
P
(b)
 nt 
sin 2
2A
The cross-sectional area of the square bar is A = (50 mm)2 = 2,500 mm2, 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 A n
2(2,500 mm 2 )(120 N/mm 2 )
P

 911,882 N
1  cos 2
1  cos 2(55)
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 A nt 2(2,500 mm 2 )(90 N/mm 2 )

 478,880 N
P
sin 2
sin 2(55)
Thus, the maximum load that can be supported by the bar is
Pmax  479 kN
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.
1.36 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
P
(a)
n 
(1  cos 2 )
2A
and
P
(b)
 nt 
sin 2
2A
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):
2 A n
2(36 in.2 )(800 psi)
P

 49,078 lb
1  cos 2
1  cos 2(40)
The shear stress on plane AB is limited to 400 psi. From Eq. (b), the maximum load P based the shear
stress limit is
2 A nt 2(36 in.2 )(400 psi)
P

 29, 244 lb
sin 2
sin 2(40)
Thus, the maximum load that can be supported by the post is
Pmax  29,200 lb  29.2 kips
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.
1.37 A 90 mm wide bar will be used to carry an axial
tension load of 280 kN. 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.
Fig. P1.37
Solution
The general equations for normal and shear stresses on an inclined plane in terms of the angle  are
P
(a)
n 
(1  cos 2 )
2A
and
P
(b)
 nt 
sin 2
2A
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):
(280 kN)(1,000 N/kN)
P
(1  cos 2 ) 
(1  cos 2(50))  771.2617 mm 2
A
2
2 n
2(150 N/mm )
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):
(280 kN)(1,000 N/kN)
P
sin 2 
sin 2(50)  1,378.7309 mm 2
A
2 nt
2(100 N/mm 2 )
To satisfy both the normal and shear stress requirements, the cross-sectional area must be at least Amin =
1,379.7309 mm2. Since the bar width is 90 mm, the minimum bar thickness t must be
1,378.7309 mm 2
tmin 
 15.3192 mm  15.32 mm
Ans.
90 mm
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.
1.38 A rectangular bar having width w = 6.00
in. and thickness t = 1.50 in. is subjected to a
tension load P. 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.
Fig. P1.38
Solution
The general equations for normal and shear stresses on an inclined plane in terms of the angle  are
P
(a)
n 
(1  cos 2 )
2A
and
P
(b)
 nt 
sin 2
2A
The angle  for inclined plane AB is calculated from
3
  71.5651
tan    3
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):
2 A n
2(9.0 in.2 )(16 ksi)
P

 1, 440 ksi
1  cos 2 1  cos 2(71.5651)
The shear stress on plane AB is limited to 8 ksi. From Eq. (b), the maximum load P based the shear
stress limit is
2 A nt 2(9.0 in.2 )(8 ksi)

 240 kips
P
sin 2
sin 2(71.5651)
Thus, the maximum load that can be supported by the bar is
Pmax  240 kips
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.
1.39 In Fig. P1.39, a rectangular bar having width
w = 1.25 in. and thickness t is subjected to a
tension load of P = 30 kips. The normal and shear
stresses on plane AB must not exceed 12 ksi and 8
ksi, respectively. Determine the minimum bar
thickness t required for the bar.
Fig. P1.39
Solution
The general equations for normal and shear stresses on an inclined plane in terms of the angle  are
P
(a)
n 
(1  cos 2 )
2A
and
P
(b)
 nt 
sin 2
2A
The angle  for inclined plane AB is calculated from
3
tan    3
  71.5651
1
The normal stress on plane AB is limited to 12 ksi; therefore, the minimum cross-sectional area A
required to support P = 30 kips can be found from Eq. (a):
P
30 kips
A
(1  cos 2 ) 
(1  cos 2(71.5651))  0.2500 in.2
2 n
2(12 ksi)
The shear stress on plane AB is limited to 8 ksi; therefore, the minimum cross-sectional area A required
to support P = 30 kips can be found from Eq. (b):
P
30 kips
A
sin 2 
sin 2(71.5651)  1.1250 in.2
2 nt
2(8 ksi)
To satisfy both the normal and shear stress requirements, the cross-sectional area must be at least Amin =
1.1250 in.2. Since the bar width is 1.25 in., the minimum bar thickness t must be
1.1250 in.2
tmin 
 0.900 in.  0.900 in.
Ans.
1.25 in.
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.
1.40 The rectangular bar has a width of w = 3.00
in. and a thickness of t = 2.00 in. The normal
stress on plane AB of the rectangular block
shown in Fig. P1.40 is 6 ksi (C) when the load P
is applied. Determine:
(a) the magnitude of load P.
(b) the shear stress on plane AB.
(c) the maximum normal and shear stresses in
the block at any possible orientation.
Fig. P1.40
Solution
The general equation for normal stress on an inclined plane in terms of the angle  is
P
n 
(1  cos 2 )
2A
and the angle  for inclined plane AB is
3
tan    0.75
  36.8699
4
The cross-sectional area of the rectangular bar is A = (3.00 in.)(2.00 in.) = 6.00 in.2.
(a)
(a) Since the normal stress on plane AB is given as 6 ksi, the magnitude of load P can be calculated from
Eq. (a):
2 A n
2(6.0 in.2 )(6 ksi)
P

 56.25 kips  56.3 kips
Ans.
1  cos 2 1  cos 2(36.8699)
(b) The general equation for shear stress on an inclined plane in terms of the angle  is
P
 nt 
sin 2
2A
therefore, the shear stress on plane AB is
56.25 kips
 nt 
sin 2(36.8699)  4.50 ksi
2(6.00 in.2 )
(c) The maximum normal stress at any possible orientation is
P 56.25 kips
 max  
 9.3750 ksi  9.38 ksi
6.00 in.2
A
and the maximum shear stress at any possible orientation in the block is
56.25 kips
P
 max 

 4.6875 ksi  4.69 ksi
2 A 2(6.00 in.2 )
Ans.
Ans.
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.
1.41 The rectangular bar has a width of w = 100
mm and a thickness of t = 75 mm. The shear stress
on plane AB of the rectangular block shown in
Fig. P1.41 is 12 MPa when the load P is applied.
Determine:
(a) the magnitude of load P.
(b) the normal stress on plane AB.
(c) the maximum normal and shear stresses in the
block at any possible orientation.
Fig. P1.41
Solution
The general equation for shear stress on an inclined plane in terms of the angle  is
P
 nt 
sin 2
2A
and the angle  for inclined plane AB is
3
tan    0.75
  36.8699
4
The cross-sectional area of the rectangular bar is A = (100 mm)(75 mm) = 7,500 mm2.
(a)
(a) Since the shear stress on plane AB is given as 12 MPa, the magnitude of load P can be calculated
from Eq. (a):
2 A nt 2(7,500 mm 2 )(12 N/mm 2 )
P

 187,500 N  187.5 kN
Ans.
sin 2
sin 2(36.8699)
(b) The general equation for normal stress on an inclined plane in terms of the angle  is
P
n 
(1  cos 2 )
2A
therefore, the normal stress on plane AB is
187,500 N
n 
(1  cos 2(36.8699))  16.00 MPa
2(7,500 mm 2 )
(c) The maximum normal stress at any possible orientation is
P 187,500 N
 max  
 25.0 MPa
A 7,500 mm 2
and the maximum shear stress at any possible orientation in the block is
P
187,500 N
 max 

 12.50 MPa
2 A 2(7,500 mm 2 )
Ans.
Ans.
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.
2.1 When an axial load is applied to the
ends of the bar shown in Fig. P2.1, the total
elongation of the bar between joints A and C
is 0.15 in. In segment (2), the normal strain
is measured as 1,300 in./in. Determine:
(a) the elongation of segment (2).
(b) the normal strain in segment (1) of
the bar.
Fig. P2.1
Solution
(a) From the definition of normal strain, the elongation in segment (2) can be computed as
(1,300 10 6 )(90 in.) 0.1170 in.
2
2 L2
Ans.
(b) The combined elongations of segments (1) and (2) is given as 0.15 in. Therefore, the elongation that
occurs in segment (1) must be
0.15 in. 0.1170 in. 0.0330 in.
1
total
2
The strain in segment (1) can now be computed:
0.0330 in.
1
0.000825 in./in. 825 μin./in.
1
40 in.
L1
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.
2.2 A rigid steel bar is supported by three rods, as
shown in Fig. P2.2. There is no strain in the rods before
the load P is applied. After load P is applied, the
normal strain in rod (2) is 1,080 in./in. Assume initial
rod lengths of L1 = 130 in. and L2 = 75 in. Determine:
(a) the normal strain in rods (1).
(b) the normal strain in rods (1) if there is a 1/32-in. gap
in the connections between the rigid bar and rods
(1) at joints A and C before the load is applied.
(c) the normal strain in rods (1) if there is a 1/32-in. gap
in the connection between the rigid bar and rod (2)
at joint B before the load is applied.
Fig. P2.2
Solution
(a) From the normal strain in rod (2) and its length, the deformation of rod (2) can be calculated:
(1,080 10 6 )(75 in.) 0.0810 in.
2
2 L2
Since rod (2) is assumed to be connected to the rigid bar with a perfect connection, the rigid bar must
move downward by an amount equal to the deformation of rod (2); therefore,
vB
0.0810 in. (downward)
2
By symmetry, the rigid bar must remain horizontal as it moves downward, and thus, vB = vA = vC. Rods
(1) are connected to the rigid bar at A and C, and again, perfect connections are assumed. The
deformation of rod (1) must be equal to the deflection of joint A (or C); thus, 1 = 0.0810 in. The normal
strain in rods (1) can now be calculated as:
0.0810 in.
1
0.0006231 in./in. 623 μin./in.
Ans.
1
130 in.
L1
(b) We can assume that the bolted connection at B is
perfect; therefore, vB = 0.0810 in. (downward).
Further, the rigid bar must remain horizontal as it
deflects downward by virtue of symmetry.
Therefore, the deflection downward of joints A and C
is still equal to 0.0810 in.
What effect is caused by the gap at A and C? When
joint A (or C) moves downward by 0.0810 in., the
first 1/32-in. of this downward movement does not
stretch rod (1)—it just closes the gap. Therefore, rod
(1) only gets elongated by the amount
vA 0.03125 in.
1
0.0810 in. 0.03125 in.
0.04975 in.
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.
This deformation creates a strain in rod (1) of:
0.04975 in.
1
0.0003827 in./in. 383 μin./in.
1
L1
130 in.
Ans.
(c) The gap is now at joint B. We know the strain
in rod (2); hence, we know its deformation must
be 0.0810 in. However, the first 1/32-in. of
downward movement by the rigid bar does not
elongate the rod—it simply closes the gap. To
elongate rod (2) by 0.0810 in., joint B must move
down:
vB
0.03125 in.
2
0.0810 in. 0.03125 in.
0.11225 in.
Again, since the rigid bar remains horizontal, vB =
vA = vC. The joints at A and C are assumed to be
perfect; thus,
vA
0.11225 in.
1
and the normal strain in rod (1) is:
0.11225 in.
1
0.00086346 in./in. 863 μin./in.
1
130 in.
L1
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.
2.3 A rigid steel bar is supported by three rods, as
shown in Fig. P2.3. There is no strain in the rods
before the load P is applied. After load P is
applied, the normal strain in rods (1) is 860 m/m.
Assume initial rod lengths of L1 = 2,400 mm and L2
= 1,800 mm. Determine:
(a) the normal strain in rod (2).
(b) the normal strain in rod (2) if there is a 2mm gap in the connections between the rigid
bar and rods (1) at joints A and C before the
load is applied.
(c) the normal strain in rod (2) if there is a 2mm gap in the connection between the rigid bar
and rod (2) at joint B before the load is applied.
Fig. P2.3
Solution
(a) From the normal strain in rod (1) and its length, the deformation of rod (1) can be calculated:
(860 10 6 )(2, 400 mm) 2.064 mm
1
1 L1
Since rod (1) is assumed to be connected to the rigid bar with a perfect connection, the rigid bar must
move downward by an amount equal to the deformation of rod (1); therefore,
vA vC
2.064 mm (downward)
1
By symmetry, the rigid bar must remain horizontal as it moves downward, and thus, vB = vA = vC. Rod
(2) is connected to the rigid bar at B, and again, a perfect connection is assumed. The deformation of
rod (2) must be equal to the deflection of joint B; thus, 2 = 2.064 mm. The normal strain in rod (2) can
now be calculated as:
2.064 mm
2
0.0011467 mm/mm 1,147 μmm/mm
Ans.
2
L2 1,800 mm
(b) We know the strain in rod (1); hence, we know its
deformation must be 2.064 mm. However, the joints
at A and C are not perfect connections. The first 2
mm of downward movement by the rigid bar does not
elongate rod (1)—it simply closes the gap. To
elongate rod (1) by 2.064 mm, joint A must move
down:
vA
2 mm 2.064 mm 2 mm 4.064 mm
1
By symmetry, the rigid bar remains horizontal;
therefore, vB = vA = vC. The joint at B is assumed to be
perfect; thus, any downward movement of the rigid
bar also elongates rod (2) by the same amount:
vB
4.064 mm
2
and the normal strain in rod (2) 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.
2
2
L2
4.064 mm
1,800 mm
0.0022578 mm/mm
2, 260 μmm/mm
Ans.
(c) We now assume that the bolted connection at A
(or C) is perfect; therefore, the rigid bar deflection
as calculated in part (a) must be vA = 2.064 mm
(downward). Further, the rigid bar must remain
horizontal as it deflects downward by virtue of
symmetry; thus, vB = vA = vC. Therefore, the
deflection downward of joint B is vB = 2.064 mm
What effect is caused by the gap at B? When joint
B moves downward by 2.064 mm, the first 2 mm
of this downward movement does not stretch rod
(2)—it just closes the gap. Therefore, rod (2) only
gets elongated by the amount
vB 2 mm
2
2.064 mm 2 mm
0.064 mm
This deformation creates a strain in rod (2) of:
0.064 mm
2
0.0000356 mm/mm
2
L2 1,800 mm
35.6 μmm/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.
2.4 A rigid bar ABCD is supported by two bars as
shown in Fig. P2.4. There is no strain in the
vertical bars before load P is applied. After load P
is applied, the normal strain in rod (1) is −570
m/m. Determine:
(a) the normal strain in rod (2).
(b) the normal strain in rod (2) if there is a 1-mm
gap in the connection at pin C before the load is
applied.
(c) the normal strain in rod (2) if there is a 1-mm
gap in the connection at pin B before the load is
applied.
Fig. P2.4
Solution
(a) From the strain given for rod (1)
1
1 L1
( 570 10 6 )(900 mm)
0.5130 mm
Therefore, vB = 0.5130 mm (downward).
From the deformation diagram of rigid bar ABCD
vB
vC
240 mm (240 mm 360 mm)
600 mm
(0.5130 mm) 1.2825 mm
vC
240 mm
Therefore,
2
2
2
L2
= 1.2825 mm (elongation), and thus, from the definition of strain:
1.2825 mm
855 10 6 mm/mm 855 με
1,500 mm
Ans.
(b) The 1-mm gap at C doesn’t affect rod (1);
therefore, 1 = −0.5130 mm. The rigid bar
deformation diagram is unaffected; thus, vB =
0.5130 mm (downward) and vC = 1.2825 mm
(downward).
The rigid bar must move downward 1 mm at C
before it begins to elongate member (2).
Therefore, the elongation of member (2) is
vC 1 mm
2
1.2825 mm 1 mm
0.2825 mm
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.
and so the strain in member (2) is
0.2825 mm
2
188.3 10
2
L2
1,500 mm
6
mm/mm
188.3 με
Ans.
(c) From the strain given for rod (1), 1 =
−0.5130 mm. In order to contract rod (1) by this
amount, the rigid bar must move downward at B
by
vB 0.5130 mm 1 mm 1.5130 mm
From deformation diagram of rigid bar ABCD
vB
vC
240 mm (240 mm 360 mm)
600 mm
(1.5130 mm) 3.7825 mm
vC
240 mm
and so
3.7825 mm
2
2,521.7 10 6 mm/mm
2
L2
1,500 mm
2,520 με
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.
2.5 In Fig. P2.5, rigid bar ABC is supported by a
pin connection at B and two axial members. A slot
in member (1) allows the pin at A to slide 0.25-in.
before it contacts the axial member. If the load P
produces a compression normal strain in member
(1) of −1,300 in./in., determine the normal strain
in member (2).
Fig. P2.5
Solution
From the strain given for member (1)
1
1 L1
( 1,300 10 6 )(32 in.)
0.0416 in.
Pin A has to move 0.25 in. before it
contacts member (1); therefore, vA =
1.080 mm (downward).
vA
0.0416 in. 0.25 in.
0.2916 in.
0.2916 in. (to the left)
From the deformation diagram of
rigid bar ABC
vA
vC
12 in. 20 in.
20 in.
(0.2916 in.) 0.4860 in. (downward)
vC
12 in.
Therefore, 2 = 0.4860 in. (elongation). From the definition of strain,
0.4860 in.
2
3,037.5 10 6 in./in. 3,040 με
2
L2
160 in.
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.
2.6 The sanding-drum mandrel shown in Fig.
P2.6 is made for use with a hand drill. The
mandrel is made from a rubber-like material that
expands when the nut is tightened to secure the
sanding sleeve placed over the outside surface. If
the diameter D of the mandrel increases from 2.00
in. to 2.15 in. as the nut is tightened, determine
(a) the average normal strain along a diameter of
the mandrel.
(b) the circumferential strain at the outside surface
of the mandrel.
Fig. P2.6
Solution
(a) The change in strain along a diameter is found from
D 2.15 in. 2.00 in.
0.075 in./in.
D
D
2.00 in.
(b) Note that the circumference of a circle is given by D. The change in strain around the
circumference of the mandrel is found from
(2.15 in.)
(2.00 in.)
C
0.075 in./in.
C
(2.00 in.)
C
Ans.
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.
2.7 The normal strain in a suspended bar of material of varying cross section due to its own weight is
given by the expression y/3E where is the specific weight of the material, y is the distance from the
free (i.e., bottom) end of the bar, and E is a material constant. Determine, in terms of , L, and E,
(a) the change in length of the bar due to its own weight.
(b) the average normal strain over the length L of the bar.
(c) the maximum normal strain in the bar.
Solution
(a) The strain of the suspended bar due to its own weight is given as
y
3E
Consider a slice of the bar having length dy. In general, = L.
Applying this definition to the bar slice, the deformation of slice dy is
given by
y
d
dy
dy
3E
Since this strain expression varies with y, the total deformation of the bar must be found by integrating
d over the bar length:
L
y
dy
E
3
0
y2
3E 2
L
0
L2
6E
(b) The average normal strain is found by dividing the expression above for
L2 / 6 E
L
avg
L
6E
Ans.
by the bar length L
Ans.
(c) Since the given strain expression varies with y, the maximum normal strain occurs at the maximum
value of y, that is, at y = L:
y
L
Ans.
max
y L
3E y L 3E
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.
2.8 A steel cable is used to support an elevator cage at the bottom of a 2,000-ft deep mineshaft. A
uniform normal strain of 250 in./in. is produced in the cable by the weight of the cage. At each point,
the weight of the cable produces an additional normal strain that is proportional to the length of the cable
below the point. If the total normal strain in the cable at the cable drum (upper end of the cable) is 700
in./in., determine
(a) the strain in the cable at a depth of 500 ft.
(b) the total elongation of the cable.
Solution
Call the vertical coordinate y and establish the origin of the y
axis at the lower end of the steel cable. The strain in the cable
has a constant term (i.e., = 250 ) and a term (we will call it
k) that varies with the vertical coordinate y.
250 10 6 k y
The problem states that the normal strain at the cable drum (i.e.,
y = 2,000 ft) is 700 in./in. Knowing this value, the constant k
can be determined
700 10 6 250 10 6 k (2, 000 ft)
700 10 6 250 10 6
0.225 10 6 /ft
2, 000 ft
Substituting this value for k in the strain expression gives
250 10 6 (0.225 10 6 /ft) y
k
(a) At a depth of 500 ft, the y coordinate is y = 1,500 ft. Therefore, the cable strain at a depth of 500 ft
is
Ans.
250 10 6 (0.225 10 6 /ft)(1,500 ft) 587.5 10 6 588 με
(b) The total elongation is found by integrating the strain expression over the cable length; thus,
2000
250 10
dy
6
(0.225 10 6 /ft) y dy
0
6
(250 10 ) y
0.225 10
2
0.950 ft 11.40 in.
2000
6
y
2
0
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.
2.9 The 16 × 22 × 25-mm rubber blocks shown
in Fig. P2.9 are used in a double U shear mount
to isolate the vibration of a machine from its
supports. An applied load of P = 690 N causes
the upper frame to be deflected downward by 7
mm. Determine the average shear strain and the
shear stress in the rubber blocks.
Fig. P2.9
Solution
Consider the deformation of one block. After a downward
deflection of 7 mm:
7 mm
tan
0.4375
0.412410 rad
16 mm
and thus, the shear strain in the block is
Ans.
0.412 rad 412,000 μrad
Note that the small angle approximation most definitely does not
apply here!
The applied load of 690 N is
carried by two blocks; therefore,
the shear force applied to one
block is V = 345 N. The area
subjected to shear stress is the area
that is parallel to the direction of
the shear force; that is, the 22 mm
by 25 mm surface of the block.
The shear stress is
345 N
(22 mm)(25 mm)
0.627 MPa
627 kPa
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.
2.10
A thin polymer plate PQR is
deformed such that corner Q is displaced
downward 1/16-in. to new position Q’ as
shown in Fig. P2.10. Determine the shear
strain at Q’ associated with the two edges
(PQ and QR).
Fig. P2.10
Solution
Before deformation, the angle of the plate at
Q was 90° or /2 radians. We must now
determine the plate angle at Q′ after
deformation. The difference between these
angles is the shear strain.
After point Q displaces downward by 1/16in., the angle ′ is
25 in.
25 in.
tan
(10 in. 1 / 16-in.) 10.0625 in.
68.075215
and the angle ′ is
4 in.
4 in.
tan
(10 in. 1 / 16-in.) 10.0625 in.
21.678589
After deformation, the angle of the plate at Q′ is
68.075215 21.678589 89.753804 1.566499 rad
The difference in the plate angle at Q before and after deformation is the shear strain:
2
1.566499 rad
0.004297 rad
4,300 μrad
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.
2.11 A thin polymer plate PQR is
deformed so that corner Q is displaced
downward 1.0 mm to new position Q’ as
shown in Fig. P2.11. Determine the
shear strain at Q’ associated with the two
edges (PQ and QR).
Fig. P2.11
Solution
Before deformation, the angle of the plate
at Q was 90° or /2 radians. We must
now determine the plate angle at Q′ after
deformation. The difference between
these angles is the shear strain.
After deformation, the angle ′ is
120 mm
120 mm
tan
(300 mm 1 mm) 301 mm
21.735741
and the angle ′ is
750 mm
750 mm
tan
(300 mm 1 mm) 301 mm
68.132764
Therefore, after deformation, the angle of the plate at Q′ is
21.735741 68.132764 89.868505 1.568501 rad
The difference in the angle at Q before and after deformation is the shear strain:
2
1.568501 rad
0.002295 rad
2,300 μrad
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.
2.12 A thin rectangular plate is uniformly
deformed as shown in Fig. P2.12.
Determine the shear strain xy at P.
Fig. P2.12
Solution
Before deformation, the angle of the plate at P
was 90° or /2 radians. We must now
determine the plate angle at P after deformation.
The difference between these angles is the shear
strain.
After deformation, the angle that side PQ makes
with the horizontal axis is
1.4 mm
tan
1,100 mm
0.072922
and the angle that side PR makes with the
vertical axis is
0.7 mm
tan
600 mm
0.066845
Therefore, the angle of the plate at P that was initially 90° has been reduced by the sum of and :
0.072922 0.066845 0.139767 0.002439 rad
Since shear strain is defined as the change in angle between two initially perpendicular lines, the shear
strain in the plate at P is
Ans.
0.002439 rad 2,440 μrad
P
Note: Since these angles are small, we could have just as well used tan ≈ and tan
the extra steps involved in using the inverse tangent function. Thus,
1.4 mm
tan
0.001273 rad
1,100 mm
0.7 mm
tan
0.001167 rad
600 mm
0.001273 rad 0.001167 rad 0.002439 rad 2,440 μrad
P
≈
and saved
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.
2.13 A thin rectangular plate is uniformly
deformed as shown in Fig. P2.13.
Determine the shear strain xy at P.
Fig. P2.13
Solution
Before deformation, the angle of the plate at P was 90° or /2
radians. We must now determine the plate angle at P after
deformation. The difference between these angles is the shear strain.
After deformation, the angle that side PQ makes with the horizontal
axis is
0.125 in.
tan
0.397881
18 in.
and the angle that side PR makes with the vertical axis is
0.125 in.
tan
0.286477
25 in.
By inspection, the angle of the plate at P has been increased by
and decreased by ; thus, the angle at P after deformation is:
90
90 0.397881 0.286477 90.111404
The angle in the plate at P that was initially 90° has been increased to 90.111404° = 1.572741 rad. The
shear strain in the plate at P is thus
P
2
1.572741 rad
0.001944 rad
1,944 μrad
Ans.
Note: Since these angles are small, we could have just as well used tan ≈ and tan ≈ and saved
the extra steps involved in using the inverse tangent functions. Thus,
0.125 in.
tan
0.006944 rad
18 in.
0.125 in.
tan
0.005000 rad
25 in.
By inspection, the angle of the plate at P has been increased by and decreased by ; thus, the angle at
P after deformation is:
0.006944 rad 0.005000 rad
0.001944 rad
2
2
2
Therefore, the shear strain in the plate at P is P = −0.001944 rad = −1,944 rad.
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.
2.14 A thin polymer plate PQR is deformed
so that corner Q is displaced downward 1.0
mm to new position Q’ as shown in Fig.
P2.14. Determine the shear strain at Q’
associated with the two edges (PQ and QR).
Fig. P2.14
Solution
Before deformation, the angle of the plate at Q was 90° or
/2 radians. We must now determine the plate angle at Q′
after deformation. The difference between these angles is
the shear strain.
After deformation, the angle ′ is
180 mm 1.0 mm 181 mm
tan
300 mm
300 mm
31.103981
and the angle ′ is
500 mm 1.0 mm 499 mm
tan
300 mm
300 mm
58.985614
Therefore, after deformation, the angle of the plate at Q′ is
31.103981 58.985614 90.089595 1.572360 rad
The difference in plate angle before and after deformation is the shear strain:
2
1.572360 rad
0.001564 rad
1,564 μrad
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.
2.15 An airplane has a half-wingspan of 33 m. Determine the change in length of the aluminum
alloy [ A = 22.5×10-6/°C] wing spar if the plane leaves the ground at a temperature of 15°C and
climbs to an altitude where the temperature is –55°C.
Solution
The change in temperature between the ground and the altitude in flight is
T Tfinal Tinitial
55°C 15°C
70°C
The thermal strain is given by
T (22.5 10 6 /°C)( 70°C)
T
0.001575 mm/mm
and thus the deformation in the 33-m wing is
( 0.001575 mm/mm)(33 m)
0.0520 m
TL
52.0 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.
2.16 A large cement kiln has a length of 400 ft and a diameter of 20 ft. Determine the change in
length and diameter of the structural steel [ S = 6.5×10-6/°F] shell caused by an increase in
temperature of 350°F.
Solution
The thermal strain is given by
T (6.5 10 6 /°F)( 350°F)
T
0.002275 in./in.
The change in length of the 225-ft kiln length is
L T L ( 0.002275 in./in.)(400 ft) 0.9100 ft
The change in diameter of the 20-ft diameter is
D T L ( 0.002275 in./in.)(20 ft) 0.004550 ft
10.92 in.
0.546 in.
Ans.
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.
2.17 A cast iron pipe has an inside diameter of d = 208 mm and an outside diameter of D = 236 mm. The
length of the pipe is L = 3.0 m. The coefficient of thermal expansion for cast iron is I = 12.1×10-6/°C.
Determine the dimension changes caused by an increase in temperature of 70°C.
Solution
The thermal strain caused by a temperature increase of 70°C is given by
T (12.1 10 6 /°C)(70°C) 0.000847 mm/mm
T
The dimension changes caused by a temperature increase of 70°C are
D T D (0.000847 mm/mm)(236 mm) 0.1999 mm
d
T
d
(0.000847 mm/mm)(208 mm)
L
T
L (0.000847 mm/mm)(3.0 m) 0.002541 m
Ans.
Ans.
0.1762 mm
2.54 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.
2.18 At a temperature of 40°F, a 0.08-in.
gap exists between the ends of the two
bars shown in Fig. P2.18. Bar (1) is an
aluminum alloy [ = 12.5 × 10−6/°F] and
bar (2) is stainless steel [ = 9.6 ×
10−6/°F]. The supports at A and C are
rigid. Determine the lowest temperature
at which the two bars contact each other.
Fig. P2.18
Solution
Write expressions for the temperature-induced deformations and set this equal to the 0.08-in. gap:
0.08 in.
1 T L1
2 T L2
T
L
1 1
2
L2
0.08 in.
0.08 in.
0.08 in.
77.821°F
-6
(12.5 10 / °F)(40 in.) (9.6 10-6 / °F)(55 in.)
1 L1
2 L2
Since the initial temperature is 40°F, the temperature at which the gap is closed is 117.8°F.
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.
2.19 At a temperature of 5°C, a 3-mm gap
exists between two polymer bars and a rigid
support, as shown in Fig. P2.19. Bars (1) and
(2) have coefficients of thermal expansion of
= 140 × 10−6/°C and
= 67 × 10−6/°C,
respectively. The supports at A and C are rigid.
Determine the lowest temperature at which the
3-mm gap is closed.
Fig. P2.19
Solution
Write expressions for the temperature-induced deformations and set this equal to the 3-mm gap:
3 mm
1 T L1
2 T L2
T
L
1 1
2
L2
3 mm
3 mm
3 mm
30.084°C
-6
(140 10 / °C)(540 mm) (67 10-6 / °C)(360 mm)
1 L1
2 L2
Since the initial temperature is 5°C, the temperature at which the gap is closed is 35.084°C = 35.1°C. Ans.
T
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.
2.20 An aluminum pipe has a length of 60 m at a temperature of 10°C. An adjacent steel pipe at
the same temperature is 5 mm longer. At what temperature will the aluminum pipe be 15 mm
longer than the steel pipe? Assume that the coefficient of thermal expansion for the aluminum is
22.5×10-6/°C and that the coefficient of thermal expansion for the steel is 12.5×10-6/°C.
Solution
The length of the aluminum pipe after a change in temperature can be expressed as
Lfinal A
Linitial A
L A
Linitial A
A T Linitial A
Similarly, the length of the steel pipe after the same change in temperature is given by
Lfinal S
Linitial S
L S
(a)
Linitial S
(b)
S T Linitial S
From the problem statement, we are trying to determine the temperature change that will cause the final
length of the aluminum pipe to be 15 mm longer than the steel pipe. This requirement can be expressed
as
(c)
Lfinal A Lfinal S 15 mm
Substitute Eqs. (a) and (b) into Eq. (c) to obtain the following relationship
Linitial A
Linitial S
15 mm
A T Linitial A
S T Linitial S
Collect the terms with T on the left-hand side of the equation
Linitial S Linitial A 15 mm
A T Linitial A
S T Linitial S
Factor out T
T A Linitial
A
S
Linitial
S
Linitial
S
Linitial
A
15 mm
and thus, T can be expressed as
Linitial S Linitial A 15 mm
T
A Linitial A
S Linitial S
Convert all length dimensions to units of millimeters and solve
60, 005 mm 60, 000 mm 15 mm
T
6
(22.5 10 /°C)(60, 000 mm) (12.5 10 6 /°C)(60, 005 mm)
20 mm
20 mm
1.35 mm/°C 0.75 mm/°C 0.60 mm/°C
33.3°C
Initially, the pipes were at a temperature of 10°C. With the temperature change determined above, the
temperature at which the aluminum pipe is 15 mm longer than the steel pipe is
Ans.
Tfinal Tinitial
T 10°C 33.3°C 43.3°C
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.
2.21 Determine the movement of the pointer
of Fig. P2.21 with respect to the scale zero in
response to a temperature increase of 60°F.
The coefficients of thermal expansion are
6.6×10-6/°F for the steel and 12.5×10-6/°F for
the aluminum.
Fig. P2.21
Solution
In response to the 60°F temperature increase, the steel pieces elongate by the amount
(6.6 10 6 /°F)(60°F)(12 in.) 0.004752 in.
Steel
Steel T LSteel
and the aluminum piece elongates by
(12.5 10 6 /°F)(60°F)(12 in.) 0.009000 in.
Alum
Alum T LAlum
From the deformation diagram of the pointer, the scale reading can be determined from similar triangles
vpointer
Alum
Steel
1.5 in.
vpointer
8.5 in.
8.5 in.
Alum
1.5 in.
Steel
5.6667 0.009000 in. 0.004752 in.
0.0241 in. (upward)
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.
2.22 Determine the horizontal movement of
point A of Fig. P2.22 due to a temperature
increase of 75°C. Assume that member AE
has a negligible coefficient of thermal
expansion.
The coefficients of thermal
expansion are 11.9×10-6/°C for the steel and
22.5×10-6/°C for the aluminum alloy.
Fig. P2.22
Solution
In response to the 75°C temperature increase, the steel piece elongates by the amount
(11.9 10 6 /°C)(75°C)(300 mm) 0.267750 mm
Steel
Steel T LSteel
Thus, joint C moves to the right by 0.267750 mm.
Next, calculate the deformation of the aluminum piece:
(22.5 10 6 /°C)(75°C)(300 mm)
Alum
Alum T LAlum
Joint E moves to the right by 0.50625 mm.
0.50625 mm
Take the initial position of E as the origin. The coordinates of E in the deflected position are (0.50625
mm, 0) and the coordinates of C are (0.267750 mm, 25 mm). Use the deflected-position coordinates of
E and C to determine the slope of the pointer.
yC yE
25 mm 0
25 mm
slope
104.821803
xC xE 0.267750 mm 0.50625 mm
0.2385 mm
A general equation for the deflected pointer can be expressed as
y mx b
104.821803x b
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.
Use the known coordinates of point E to determine b:
0
104.821803(0.50625 mm) b
b 53.066038 mm
Thus, the deflected pointer can be described by the line
y
104.821803x 53.066038 mm
or rearranging to solve for x
(53.066038 mm) y
x
104.821803
At pointer tip A, y = 275 mm; therefore, the x coordinate of the pointer tip is
(53.066038 mm 275 mm)
221.933962 mm
Ans.
x
2.12 mm
2.11725 mm
104.821803
104.821803
The x coordinate is the same as the horizontal movement since we took the initial position of joint E as
the origin.
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.
2.23 At a temperature of 25°C, a cold-rolled red brass [ B = 17.6×10-6/°C] sleeve has an inside
diameter of dB = 299.75 mm and an outside diameter of DB = 310 mm. The sleeve is to be placed
on a steel [ S = 11.9×10-6/°C] shaft with an outside diameter of DS = 300 mm. If the temperatures
of the sleeve and the shaft remain the same, determine the temperature at which the sleeve will slip
over the shaft with a gap of 0.05 mm.
Solution
The inside diameter of the brass sleeve after a change in temperature can be expressed as
d final B
dinitial B
d B
dinitial B
B T d initial B
Similarly, the outside diameter of the steel shaft after the same change in temperature is given by
Dfinal S
Dinitial S
D S
(a)
Dinitial S
(b)
S T Dinitial S
From the problem statement, we are trying to determine the temperature change that will cause the
inside diameter of the brass sleeve to be 0.05 mm greater than the outside diameter of the steel shaft.
This requirement can be expressed as
(c)
dfinal B Dfinal S 0.05 mm
Substitute Eqs. (a) and (b) into Eq. (c) to obtain the following relationship
dinitial B
Dinitial S
0.05 mm
B T dinitial B
S T Dinitial S
Collect the terms with T on the left-hand side of the equation
Dinitial S dinitial B 0.05 mm
B T dinitial B
S T Dinitial S
Factor out T
T B dinitial
B
S
Dinitial
S
Dinitial
S
dinitial
B
0.05 mm
and thus, T can be expressed as
Dinitial S dinitial B 0.05 mm
T
B d initial B
S Dinitial S
Solve for the temperature change
300 mm 299.75 mm 0.05 mm
T
6
(17.6 10 /°C)(299.75 mm) (11.9 10 6 /°C)(300 mm)
0.30 mm
0.30 mm
0.005276 mm/°C 0.003570 mm/°C 0.001706 mm/°C
175.9°C
Initially, the shaft and sleeve were at a temperature of 25°C. With the temperature change determined
above, the temperature at which the sleeve fits over the shaft with a gap of 0.05 mm is
Ans.
Tfinal Tinitial
T 25°C 175.9°C 201°C
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.
3.1 At the proportional limit, a 2-inch gage length of a 0.375-in.-diameter alloy bar has elongated 0.0083
in. and the diameter has been reduced 0.0005 in. The total tension force on the bar was 4.75 kips.
Determine the following properties of the material:
(a) the modulus of elasticity.
(b) Poisson’s ratio.
(c) the proportional limit.
Solution


(a) The bar cross-sectional area is
A
D2 
(0.375 in.)2  0.110447 in.2
4
4
and thus, the normal stress corresponding to the 4.75-kip force is
4.75 kips

 43.007204 ksi
0.110447 in.2
The longitudinal strain in the bar is
 0.0083 in.
 0.004150 in./in.
 long  
L
2 in.
The modulus of elasticity is therefore

43.007204 ksi

 10,363.2 ksi  10,360 ksi
E
 long 0.004150 in./in.
(b) The longitudinal strain in the bar was calculated previously as
 long  0.004150 in./in.
The lateral strain can be determined from the reduction of the diameter:
D 0.0005 in.
 lat 

 0.001333 in./in.
D
0.375 in.
Poisson’s ratio for this specimen is therefore

0.001333 in./in.
 0.321285  0.321
   lat  
 long
0.004150 in./in.
Ans.
Ans.
(c) Based on the problem statement, the stress in the bar is equal to the proportional limit; therefore,
Ans.
 PL  43.0 ksi
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.
3.2 At the proportional limit, a 30 mm wide × 12 mm thick bar elongates 2.0 mm under an axial load of
41.5 kN. The bar is 1.5-m long. If Poisson’s ratio is 0.33 for the material, determine:
(a) the modulus of elasticity.
(b) the proportional limit.
(c) the change in each lateral dimension.
Solution
(a) The bar cross-sectional area is
A  (30 mm)(12 mm)  360 mm 2
and thus, the normal stress corresponding to the 41.5-kN axial load is
(41.5 kN)(1,000 N/kN)

 115.2778 MPa
360 mm2
The longitudinal strain in the bar is

2.0 mm
 long  
 0.001333 mm/mm
L (1.5 m)(1,000 mm/m)
The modulus of elasticity is therefore

115.2778 MPa
E

 86, 458 MPa  86.5 GPa
 long 0.001333 mm/mm
Ans.
(b) Based on the problem statement, the stress in the bar is equal to the proportional limit; therefore,
Ans.
 PL  115.3 MPa
(c) Poisson’s ratio is given as  = 0.33. The longitudinal strain in the bar was calculated previously as
 long  0.001333 mm/mm
The corresponding lateral strain can be determined from Poisson’s ratio:
 lat   long  (0.33)(0.001333 mm/mm)  0.000440 mm/mm
Using this lateral strain, the change in bar width is
width  lat (width)  (0.000440 mm/mm)(30 mm)  0.01320 mm
and the change in bar thickness is
thickness  lat (thickness)  (0.000440 mm/mm)(12 mm)  0.00528 mm
Ans.
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.
3.3 At an axial load of 22 kN, a 45-mm-wide × 15-mm-thick polyimide polymer bar elongates 3.0 mm
while the bar width contracts 0.25 mm. The bar is 200 mm long. At the 22-kN load, the stress in the
polymer bar is less than its proportional limit. Determine:
(a) the modulus of elasticity.
(b) Poisson’s ratio.
(c) the change in the bar thickness.
Solution
(a) The bar cross-sectional area is
A  (15 mm)(45 mm)  675 mm 2
and thus, the normal stress corresponding to the 22-kN axial load is
(22 kN)(1,000 N/kN)

 32.592593 MPa
675 mm2
The longitudinal strain in the bar is
 3.0 mm
 long  
 0.0150 mm/mm
L 200 mm
The modulus of elasticity is therefore

32.592593 MPa

 2,173 MPa  2.17 GPa
E
 long 0.0150 mm/mm
Ans.
(b) The longitudinal strain in the bar was calculated previously as
 long  0.0150 mm/mm
The lateral strain can be determined from the reduction of the bar width:
width 0.25 mm
 lat 

 0.005556 mm/mm
width
45 mm
Poisson’s ratio for this specimen is therefore

0.005556 mm/mm
   lat  
 0.370370  0.370
 long
0.0150 mm/mm
(c) The change in bar thickness can be found from the lateral strain:
thickness  lat (thickness)  (0.005556 mm/mm)(15 mm)  0.0833 mm
Ans.
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.
3.4 A 0.75-in.-thick rectangular alloy bar is subjected to a tensile load P by pins at A and B. The width
of the bar is w = 3.0 in. Strain gages bonded to the specimen measure the following strains in the
longitudinal (x) and transverse (y) directions: x = 840  and y = −250 .
(a) Determine Poisson’s ratio for this
specimen.
(b) If the measured strains were produced
by an axial load of P = 32 kips, what is the
modulus of elasticity for this specimen?
Fig. P3.4
Solution
(a) Poisson’s ratio for this specimen is

250 με

 0.298
   lat   y  
840 με
 long
x
(b) The bar cross-sectional area is
A  (3.0 in.)(0.75 in.)  2.25 in.2
and so the normal stress for an axial load of P = 32 kips is
32 kips

 14.222222 ksi
2.25 in.2
The modulus of elasticity is thus

14.222222 ksi
E

 16,931.2 ksi  16,930 ksi
 long
 1 in./in. 
(840 με) 

 1,000,000 με 
Ans.
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.
3.5 A 6-mm-thick rectangular alloy bar is subjected to a tensile load P by pins at A and B. The width
of the bar is w = 30 mm. Strain gages bonded to the specimen measure the following strains in the
longitudinal (x) and transverse (y) directions: x = 900  and y = −275 .
(a) Determine Poisson’s ratio for this
specimen.
(b) If the measured strains were produced
by an axial load of P = 19 kN, what is the
modulus of elasticity for this specimen?
Fig. P3.5
Solution
(a) Poisson’s ratio for this specimen is

275 με

 0.306
   lat   y  
900 με
 long
x
(b) The bar cross-sectional area is
A  (30 mm)(6 mm)  180 mm 2
and so the normal stress for an axial load of P = 19 kN is
(19 kN)(1,000 N/kN)

 105.556 MPa
180 mm2
The modulus of elasticity is thus

105.556 MPa
E

 117,284.0 MPa  117.3 GPa
 long
 1 mm/mm 
(900 με) 
 1,000,000 με 
Ans.
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.
3.6 A copper rod [E = 110 GPa] originally 350-mm long is pulled in tension with a normal stress of 180
MPa. If the deformation is entirely elastic, what is the resulting elongation?
Solution
Since the deformation is elastic, the strain in the rod can be determined from Hooke’s Law,

180 MPa
 0.0016364 mm/mm
 
E 110,000 MPa
The elongation in the rod is thus
   L  (0.0016364 mm/mm)(350 mm)  0.573 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.
3.7 A 6061-T6 aluminum tube [E = 10,000 ksi;  = 0.33] has an outside diameter of 4.000 in. and a wall
thickness of 0.065 in.
(a) Determine the tension force that must be applied to the tube to cause its outside diameter to contract
by 0.005 in.
(b) If the tube is 84-in. long, what is the overall elongation?
Solution
(a) The lateral strain associated with the given diameter contraction is
0.005 in.
 0.001250 in./in.
 lat 
4.000 in.
From the given Poisson’s ratio, the longitudinal strain in the tube must be

0.001250 in./in.
 long   lat  
 0.003788 in./in.

0.33
and from Hooke’s Law, the normal stress can be calculated as
  E long  (10, 000 ksi)(0.003788 in./in.)  37.878788 ksi
The cross-sectional area of the tube is needed to determine the tension force. Given that the outside
diameter of the tube is 4.000 in. and the wall thickness is 0.065 in., the inside diameter of the tube is
3.870 in. The tube cross-sectional area is thus

(4.000 in.)2  (3.870 in.)2   0.803541 in.2
4
and the force applied to the tube is
F   A  (37.878788 ksi)(0.803541 in.2 )  30.437154 kips  30.4 kips
A
Ans.
(b) The longitudinal strain was calculated previously. Use the longitudinal strain to determine the
overall elongation:
Ans.
   long L  (0.003788 in./in.)(84 in.)  0.318 in.
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.
3.8 A metal specimen with an original diameter of 0.500 in. and a gage length of 2.000 in. is tested in
tension until fracture occurs. At the point of fracture, the diameter of the specimen is 0.260 in. and the
fractured gage length is 3.08 in. Calculate the ductility in terms of percent elongation and percent
reduction in area.
Solution
Percent elongation is simply the longitudinal strain at fracture:
 (3.08 in.  2.000 in.) 1.08 in.

 0.54 in./in.
 
L
2.000 in.
2.000 in.
 percent elongation  54%
Ans.

The initial cross-sectional area of the specimen is
A0 
(0.500 in.) 2  0.196350 in.2
4
The final cross-sectional area at the fracture location is
Af 

(0.260 in.) 2  0.053093 in.2
4
The percent reduction in area is
percent reduction of area 
A0  Af
A0
(100%)
(0.196350 in.2  0.053093 in.2 )
(100%)  73.0%

0.196350 in.2
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.
3.9 A portion of the stress-strain curve for a
stainless steel alloy is shown in Fig. P3.9. A 350mm-long bar is loaded in tension until it elongates
2.0 mm and then the load is removed.
(a) What is the permanent set in the bar?
(b) What is the length of the unloaded bar?
(c) If the bar is reloaded, what will be the
proportional limit?
Fig. P3.9
Solution
(a) The normal strain in the specimen is
 2.0 mm
 0.005714 mm/mm
 
L 350 mm
Construct a line parallel to the elastic modulus line that passes through the data curve at a strain of  =
0.005714 mm/mm. The strain value at which this modulus line intersects the strain axis is the
permanent set:
permanent set  0.0035 mm/mm
Ans.
(b) The length of the unloaded bar is therefore:
   L  (0.0035 mm/mm)(350 mm)  1.225 mm
L f  350 mm  1.225 mm  351.225 mm
(c) From the stress-strain curve, the reload proportional limit is 444 MPa .
Ans.
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.
3.10 The 16 by 22 by 25-mm rubber blocks
shown in Fig. P3.10 are used in a double U
shear mount to isolate the vibration of a
machine from its supports. An applied load of
P = 285 N causes the upper frame to be
deflected downward by 5 mm. Determine the
shear modulus G of the rubber blocks.
Fig. P3.10
Solution
Consider a FBD of the upper U frame. The
downward force P is resisted by two upward shear
forces V; therefore, V = 285 N / 2 = 142.5 N.
Next, consider a FBD of one of the rubber blocks.
The shear force acting on one rubber block is V =
142.5 N. The area of the rubber block that is
parallel to the direction of V is
AV  (22 mm)(25 mm)  550 mm 2
Consequently, the shear stress in one rubber block is
142.5 N
V


 0.259091 MPa
AV 550 mm 2
The shear strain associated with the 5-mm downward displacement of the rubber blocks is given by:
5 mm
 0.312500
   0.302885 rad
tan  
16 mm
From Hooke’s Law for shear stress and shear strain, the shear modulus G can be computed:
 0.259091 MPa
G  
 0.855411 MPa  0.855 MPa
  G
0.302885 rad

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.
3.11 Two hard rubber blocks are used in an anti-vibration
mount to support a small machine, as shown in Fig. P3.11.
An applied load of P = 150 lb causes a downward deflection
of 0.25 in. Determine the shear modulus of the rubber blocks.
Assume a = 0.5 in., b = 1.0 in., and c = 2.5 in.
Fig. P3.11
Solution
Determine the shear strain from the angle formed by the
downward deflection and the block thickness a:
0.250 in.
 0.500
   0.463648 rad
tan  
0.50 in.
Note: The small angle approximation tan  is not applicable in this instance.
Determine the shear stress from the applied load P and the block area.
Note that this is a double shear configuration; therefore, the shear force V acting on a single rubber block
is half of the total load: V = P/2 = 75 lb.
To determine the area needed here, consider the surface that is bonded to the plate. This area has
dimensions of bc. The shear stress acting on a single block is therefore:
V
75 lb
 
 30 psi
A (1.0 in.)(2.5 in.)
The shear modulus G can be calculated from Hooke’s law for shear stress and strain:

30 psi
G  
 64.704 psi  64.7 psi
  G
 0.463648 rad
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.
3.12 Two hard rubber blocks [G = 350 kPa] are used in an antivibration mount to support a small machine, as shown in Fig.
P3.12. Determine the downward deflection that will occur for an
applied load of P = 900 N. Assume a = 20 mm, b = 50 mm, and
c = 80 mm.
Fig. P3.12
Solution
Note that this is a double shear configuration; therefore, the shear force V acting on a single rubber block
is half of the total load: V = P/2 = 450 N.
Determine the shear stress from the shear force V and the block area. To determine the area needed
here, consider the surface that is bonded to the plate. This area has dimensions of bc. The shear stress
acting on a single block is therefore:
450 N
V
 
 0.112500 MPa
A (50 mm)(80 mm)
Since the shear modulus G is given, the shear strain can be calculated from Hooke’s law for shear stress
and shear strain:
 0.112500 MPa
  G
  
 0.321429 rad
G
0.350 MPa
From the angle  and the block thickness a, the downward deflection  of the block can be determined
from:
tan  

a
   a tan   (20 mm) tan(0.321429 rad)  6.659512 mm  6.66 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.
3.13 A load test on a 6 mm diameter by 225 mm long aluminum alloy rod found that a tension load of
4,800 N caused an elastic elongation of 0.52 mm in the rod. Using this result, determine the elastic
elongation that would be expected for a 24-mm-diameter rod of the same material if the rod were 1.2 m
long and subjected to a tension force of 37 kN.
Solution

The cross-sectional area of the 6-mm-diameter rod is
A
(6 mm) 2  28.274334 mm 2
4
Thus, the normal stress in the rod due to a 4,800-N load is
F
4,800 N
 
 169.76527 MPa
A 28.274334 mm2
The strain in the 225-mm long rod associated with a 0.52-mm elongation is
 0.52 mm
 0.0023111 mm/mm
 
L 225 mm
Therefore, the elastic modulus of the aluminum alloy is

169.76527 MPa
 73,456.128 MPa
E 
 0.0023111 mm/mm
The cross-sectional area of the 24-mm-diameter rod is

A  (24 mm) 2  252.389342 mm 2
4
Thus, the normal stress in the 24-mm-diameter rod due to a 37-kN load is
F (37 kN)(1,000 N/kN)
 
 81.787957 MPa
A
452.389342 mm 2
From Hooke’s Law, the strain that would be expected is
 81.787957 MPa
 0.0011134 mm/mm
 
E 73, 456.128 MPa
Since the 42-mm-diameter rod is 1.2-m long, the expected elongation is
   L  (0.0011134 mm/mm)(1.2 m)(1,000 mm/m)  1.336111 mm  1.336 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.
3.14 The stress-strain diagram for a
particular stainless steel alloy is shown in
Fig. P3.14. A rod made from this material is
initially 800 mm long at a temperature of
20°C. After a tension force is applied to the
rod and the temperature is increased by
200°C, the length of the rod is 804 mm.
Determine the stress in the rod and state
whether the elongation in the rod is elastic
or inelastic. Assume the coefficient of
thermal expansion for this material is 18 ×
10−6/°C.
Fig. P3.14
Solution
The 4-mm total elongation of the rod is due to a combination of load and temperature increase. The
200°C temperature increase causes a normal strain of:
 T   T  (18 106 / C )(200C )  0.003600 mm/mm
which means that the rod elongates
 T  T L  (0.003600 mm/mm)(800 mm)  2.8800 mm
The portion of the 4-mm total elongation due to load is therefore
      T  4 mm  2.8800 mm  1.1200 mm
The strain corresponding to this elongation is

1.1200 mm
   
 0.001400 mm/mm
L
800 mm
By inspection of the stress-strain curve, this strain is clearly in the linear region. Therefore, the rod is
elastic in this instance.
For the linear region, the elastic modulus can be determined from the lower scale plot:

(400 MPa  0)
E

 200, 000 MPa
 (0.002 mm/mm  0)
Using Hooke’s Law (or directly from the - diagram), the stress corresponding to the 0.001400
mm/mm strain is
  E  (200,000 MPa)(0.001400 mm/mm)  280 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.
3.15 In Fig. P3.15, rigid bar ABC is supported by
axial member (1), which has a cross-sectional area
of 400 mm2, an elastic modulus of E = 70 GPa, and
a coefficient of thermal expansion of  = 22.5 ×
10−6 /°C. After load P is applied to the rigid bar
and the temperature rises 40°C, a strain gage
affixed to member (1) measures a strain increase of
2,150 . Determine:
(a) the normal stress in member (1).
(b) the magnitude of applied load P.
(c) the deflection of the rigid bar at C.
Fig. P3.15
Solution
(a) The strain measured in member (1) is due to both the internal force in the member and the
temperature change. The strain caused by the temperature change is
 T   T  (22.5  106 / °C)(40°C)  0.000900 mm/mm
Since the total strain is  = 2,150  = 0.002150 mm/mm, the strain caused by the internal force in
member (1) must be
    T  0.002150 mm/mm  0.000900 mm/mm  0.001250 mm/mm
The elastic modulus of member (1) is E = 70 GPa; thus, from Hooke’s Law, the stress in the member is:
Ans.
1  E  (70,000 MPa)(0.001250 mm/mm)  87.5 MPa
(b) If the normal stress in member (1) is 87.5 MPa, the axial force in the member is
F1   1 A1  (87.5 N/mm 2 )(400 mm 2 )  35,000 N
Consider moment equilibrium of rigid bar ABC about joint
A to determine the magnitude of P:
M A  (1.75 m)(35,000 N)  (3.0 m)P  0
 P  20,417 N  20.4 kN
Ans.
(c) The strain in member (1) was measured as  = 2,150  = 0.002150 mm/mm; therefore, the total
elongation of member (1) is
1  1L1  (0.002150 mm/mm)(3,750 mm)  8.0625 mm
The deflection of the rigid bar at B is equal to this elongation; therefore, vB = 1 = 8.0625 mm
(downward). By similar triangles, the deflection of the rigid bar at C is given by:
vB
v
 C
1.75 m 3.0 m
3.0 m
3.0
 vC  vB
 (8.0625 mm)
 13.821429 mm  13.82 mm 
Ans.
1.75 m
1.75
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.
3.16 A tensile test specimen of 1045 hot-rolled
steel having a diameter of 0.505 in. and a gage
length of 2.00 in. was tested to fracture. Stress
and strain data obtained during the test are
shown in Fig. P3.16. Determine
(a) the modulus of elasticity.
(b) the proportional limit.
(c) the ultimate strength.
(d) the yield strength (0.20% offset).
(e) the fracture stress.
(f) the true fracture stress if the final diameter of
the specimen at the location of the fracture was
0.392 in.
Fig. P3.16
Solution
From the stress-strain curve, the proportional limit will be taken as  = 60 ksi at a strain of  = 0.0019.
(Obviously, there can be quite a bit of leeway in pulling numbers from such a limited plot.)
(a) The modulus of elasticity is

60 ksi
E 
 31,600 ksi
 0.0019 in./in.
(b) From the diagram, the proportional limit is taken as
 PL  60 ksi
(c) The ultimate strength is
U  105 ksi
(d) The yield strength is
Y  68 ksi
(e) The fracture stress is
 fracture  98 ksi

Ans.
Ans.
Ans.
Ans.
Ans.
(f) The original cross-sectional area of the specimen is
A0 
(0.505 in.)2  0.200296 in.2
4
The cross-sectional area of the specimen at the fracture location is
Af 

(0.392 in.) 2  0.120687 in.2
4
The true fracture stress is therefore
0.200296 in.2
true  fracture  (98 ksi)
 162.6 ksi
0.120687 in.2
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.
3.17 A tensile test specimen of stainless
steel alloy having a diameter of 0.495 in.
and a gage length of 2.00 in. was tested to
fracture. Stress and strain data obtained
during the test are shown in Fig. P3.17.
Determine:
(a) the modulus of elasticity.
(b) the proportional limit.
(c) the ultimate strength.
(d) the yield strength (0.20% offset).
(e) the fracture stress.
(f) the true fracture stress if the final
diameter of the specimen at the location of
the fracture was 0.350 in.
Fig. P3.17
Solution
From the stress-strain curve, the proportional limit will be taken as  = 60 ksi at a strain of  = 0.002.
(Obviously, there can be quite a bit of leeway in pulling numbers from such a limited plot.)
(a) The modulus of elasticity is

60 ksi
E 
 30,000 ksi
 0.002 in./in.
(b) From the diagram, the proportional limit is taken as
 PL  60 ksi
(c) The ultimate strength is
U  159 ksi
(d) The yield strength is
Y  80 ksi
(e) The fracture stress is
 fracture  135 ksi

Ans.
Ans.
Ans.
Ans.
Ans.
(f) The original cross-sectional area of the specimen is
A0 
(0.495 in.)2  0.192442 in.2
4
The cross-sectional area of the specimen at the fracture location is
Af 

(0.350 in.) 2  0.096211 in.2
4
The true fracture stress is therefore
0.192442 in.2
 270 ksi
true  fracture  (135 ksi)
0.096211 in.2
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.
3.18 A bronze alloy specimen having a
diameter of 12.8 mm and a gage length
of 50 mm was tested to fracture. Stress
and strain data obtained during the test
are shown in Fig. P3.18. Determine:
(a) the modulus of elasticity.
(b) the proportional limit.
(c) the ultimate strength.
(d) the yield strength (0.20% offset).
(e) the fracture stress.
(f) the true fracture stress if the final
diameter of the specimen at the location
of the fracture was 10.5 mm.
Fig. P3.18
Solution
From the stress-strain curve, the proportional limit will be taken as  = 210 MPa at a strain of  = 0.002.
(Obviously, there can be quite a bit of leeway in pulling numbers from such a limited plot.)
(a) The modulus of elasticity is

210 MPa
E 
 105,000 MPa
 0.002 in./in.
(b) From the diagram, the proportional limit is taken as
 PL  210 MPa
(c) The ultimate strength is
U  380 MPa
(d) The yield strength is
Y  290 MPa
(e) The fracture stress is
 fracture  320 MPa

Ans.
Ans.
Ans.
Ans.
Ans.
(f) The original cross-sectional area of the specimen is
A0 
(12.8 mm) 2  128.679635 mm 2
4
The cross-sectional area of the specimen at the fracture location is

Af  (10.5 mm)2  86.590148 mm 2
4
The true fracture stress is therefore
128.679635 mm 2
true  fracture  (320 MPa)
 476 MPa
86.590148 mm 2
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.
3.19 An alloy specimen having a diameter of 12.8
mm and a gage length of 50 mm was tested to
fracture. Load and deformation data obtained
during the test are given. Determine:
(a) the modulus of elasticity.
(b) the proportional limit.
(c) the ultimate strength.
(d) the yield strength (0.05% offset).
(e) the yield strength (0.20% offset).
(f) the fracture stress.
(g) the true fracture stress if the final diameter of
the specimen at the location of the fracture was
11.3 mm.
(kN)
Change in
Length
(mm)
0
7.6
14.9
22.2
28.5
29.9
30.6
32.0
33.0
33.3
36.8
41.0
0
0.02
0.04
0.06
0.08
0.10
0.12
0.16
0.20
0.24
0.50
1.00
Load
(kN)
Change in
Length
(mm)
43.8
45.8
48.3
49.7
50.4
50.7
50.4
50.0
49.7
47.9
45.1
1.50
2.00
3.00
4.00
5.00
6.00
7.00
8.00
9.00
10.00
fracture
Load
Solution
The plot of the stress-strain data is shown below.
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.
(a) Using the data point for 28.5 kN load and 0.08 mm elongation, the modulus of elasticity can be
calculated as

221.48 MPa
Ans.
E 
 138, 400 MPa
 0.0016 mm/mm
(b) From the diagram, the proportional limit is taken as
 PL  234 MPa
(c) The ultimate strength is
(50.7 kN)(1,000 N/kN)
U 
 394 MPa

(12.8 mm)2
4
Ans.
Ans.
(d) The yield strength by the 0.05% offset method is
Y  239 MPa
Ans.
(e) The yield strength by the 0.2% offset method is
Y  259 MPa
Ans.
(f) The fracture stress is
(45.1 kN)(1,000 N/kN)
 fracture 
 350.48 MPa  350 MPa

2
(12.8 mm)
4

Ans.
(f) The original cross-sectional area of the specimen is
A0 
(12.8 mm) 2  128.679635 mm 2
4
The cross-sectional area of the specimen at the fracture location is
Af 

(11.3 mm)2  100.287492 mm 2
4
The true fracture stress is therefore
128.679635 mm 2
true  fracture  (356 MPa)
 457 MPa
100.287492 mm 2
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.
3.20 A 1035 hot-rolled steel specimen with a
diameter of 0.500 in. and a 2.0-in. gage length was
tested to fracture. Load and deformation data
obtained during the test are given. Determine:
(a) the modulus of elasticity.
(b) the proportional limit.
(c) the ultimate strength.
(d) the yield strength (0.05% offset).
(e) the yield strength (0.20% offset).
(f) the fracture stress.
(g) the true fracture stress if the final diameter of
the specimen at the location of the fracture was
0.387 in.
(lb)
Change in
Length
(in.)
0
2,690
5,670
8,360
11,050
12,540
13,150
13,140
12,530
12,540
12,840
12,840
0
0.0009
0.0018
0.0028
0.0037
0.0042
0.0046
0.0060
0.0079
0.0098
0.0121
0.0139
Load
(lb)
Change
in Length
(in.)
12,540
12,540
14,930
17,020
18,220
18,820
19,110
19,110
18,520
17,620
16,730
16,130
15,900
0.0209
0.0255
0.0487
0.0835
0.1252
0.1809
0.2551
0.2968
0.3107
0.3246
0.3339
0.3385
fracture
Load
Solution
The plot of the stress-strain data is shown below.
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.
(a) Using the data point for 12,540 lb load and 0.0042 in. elongation, the modulus of elasticity can be
calculated as

63.866 ksi
Ans.
E 
 30, 400 ksi
 0.0021 in./in.
(b) From the diagram, the proportional limit is taken as
 PL  63.8 ksi
Ans.
(c) The ultimate strength is
U  97.3 ksi
Ans.
(d) The yield strength using the 0.05% offset method is
Y  65.4 ksi
Ans.
(e) The yield strength using the 0.2% offset method is
Y  63.8 ksi
Ans.
(f) The fracture stress is
15,900 lb
 fracture 
 80,978 psi  81.0 ksi

2
(0.500 in.)
4
Ans.

(g) The original cross-sectional area of the specimen is
A0 
(0.500 in.) 2  0.196350 in.2
4
The cross-sectional area of the specimen at the fracture location is
Af 

(0.387 in.) 2  0.117628 in.2
4
The true fracture stress is therefore
0.196350 in.2
 135,172 psi  135.2 ksi
true  fracture  (80,978 psi)
0.117628 in.2
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.
3.21 A 2024-T4 aluminum test specimen with a
diameter of 0.505 in. and a 2.0-in. gage length was
tested to fracture. Load and deformation data
obtained during the test are given. Determine:
(a) the modulus of elasticity.
(b) the proportional limit.
(c) the ultimate strength.
(d) the yield strength (0.05% offset).
(e) the yield strength (0.20% offset).
(f) the fracture stress.
(g) the true fracture stress if the final diameter of
the specimen at the location of the fracture was
0.452 in.
Load
(lb)
0
1,300
2,390
3,470
4,560
5,640
6,720
7,380
8,240
8,890
9,330
9,980
10,200
10,630
Change in
Length
(in.)
0.0000
0.0014
0.0023
0.0032
0.0042
0.0051
0.0060
0.0070
0.0079
0.0088
0.0097
0.0107
0.0116
0.0125
Load
(lb)
11,060
11,500
12,360
12,580
12,800
13,020
13,230
13,450
13,670
13,880
14,100
14,100
14,100
14,100
14,100
Change in
Length
(in.)
0.0139
0.0162
0.0278
0.0394
0.0603
0.0788
0.0974
0.1159
0.1391
0.1623
0.1994
0.2551
0.3200
0.3246
fracture
Solution
The plot of the stress-strain data is shown below.
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.
(a) Using the data point for 6,720 lb load and 0.0060 in. elongation, the modulus of elasticity can be
calculated as

33.55 ksi
Ans.
E 
 11,180 ksi
 0.003 in./in.
(b) From the diagram, the proportional limit is taken as
 PL  33.6 ksi
Ans.
(c) The ultimate strength is
U  70.4 ksi
Ans.
(d) The yield strength using the 0.05% offset method is
Y  44.4 ksi
Ans.
(e) The yield strength using the 0.2% offset method is
Y  54.5 ksi
Ans.
(f) The fracture stress is
 fracture  70.4 ksi
Ans.

(g) The original cross-sectional area of the specimen is
A0 
(0.505 in.)2  0.200296 in.2
4
The cross-sectional area of the specimen at the fracture location is
Af 

(0.452 in.) 2  0.160460 in.2
4
The true fracture stress is therefore
0.200296 in.2
true  fracture  (70.4 ksi)
 87.9 ksi
0.160460 in.2
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.
3.22 A 1045 hot-rolled steel tension test specimen
has a diameter of 6.00 mm and a gage length of 25
mm. In a test to fracture, the stress and strain data
below were obtained. Determine:
(a) the modulus of elasticity.
(b) the proportional limit.
(c) the ultimate strength.
(d) the yield strength (0.05% offset).
(e) the yield strength (0.20% offset).
(f) the fracture stress.
(g) the true fracture stress if the final diameter of
the specimen at the location of the fracture was
4.65 mm.
(kN)
Change in
Length
(mm)
0.00
2.94
5.58
8.52
11.16
12.63
13.02
13.16
13.22
13.22
13.25
13.22
0.00
0.01
0.02
0.03
0.05
0.05
0.06
0.08
0.08
0.10
0.14
0.17
Load
(kN)
Change in
Length
(mm)
13.22
16.15
18.50
20.27
20.56
20.67
20.72
20.61
20.27
19.97
19.68
19.09
18.72
0.29
0.61
1.04
1.80
2.26
2.78
3.36
3.83
3.94
4.00
4.06
4.12
fracture
Load
Solution
The plot of the stress-strain data is shown below.
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.
(a) Using the data point for 8.52 kN load and 0.03 mm elongation, the modulus of elasticity can be
calculated as

301.3 MPa
Ans.
E 
 251, 000 MPa
 0.0012 mm/mm
(b) From the diagram, the proportional limit is taken as
 PL  400 MPa
(c) The ultimate strength is
(20.72 kN)(1,000 N/kN)
U 
 732.82 MPa  733 MPa

(6.00 mm)2
4
Ans.
Ans.
(d) The yield strength by the 0.05% offset method is
Y  465 MPa
Ans.
(e) The yield strength by the 0.2% offset method is
Y  465 MPa
Ans.
(f) The fracture stress is
(18.72 kN)(1,000 N/kN)
 fracture 
 662.08 MPa  662 MPa

2
(6.00 mm)
4
Ans.
(f) The original cross-sectional area of the specimen is

A0  (6 mm) 2  28.274334 mm 2
4
The cross-sectional area of the specimen at the fracture location is
Af 

(4.65 mm) 2  16.982272 mm2
4
The true fracture stress is therefore
28.274334 mm 2
true  fracture  (662.08 MPa)
 1,102 MPa
16.982272 mm 2
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.
3.23 Rigid bar BCD in Fig. P3.23 is
supported by a pin at C and by aluminum
rod (1). A concentrated load P is applied to
the lower end of aluminum rod (2), which is
attached to the rigid bar at D. The crosssectional area of each rod is A = 0.20 in.2
and the elastic modulus of the aluminum
material is E = 10,000 ksi. After the load P
is applied at E, the strain in rod (1) is
measured as 1,350  (tension).
(a) Determine the magnitude of load P.
(b) Determine the total deflection of point E
relative to its initial position.
Fig. P3.23
Solution
(a) From the measured strain, the stress in rod (1) is
 1  E11  (10,000 ksi)(1,350  10 6 in./in.)  13.5 ksi
and thus, the force in rod (1) is
F1   1 A1  (13.5 ksi)(0.20 in.2 )  2.70 kips (T)
Consider the equilibrium of the rigid bar, and write a
moment equilibrium equation about C to determine the
magnitude of load P:
M C  (20 in.)(2.70 kips)  (30 in.)P  0
 P  1.8 kips
Ans.
(b) From the measured strain, the elongation of rod (1) is
1  1L1  (1,350  106 in./in.)(50 in.)  0.0675 in.
From similar triangles, the deflection of the rigid bar at D can
be expressed in terms of the deflection at B:
vB
v
 D
20 in. 30 in.
30 in.
30 in.
 vD  vB
 (0.0675 in.)
 0.10125 in.
20 in.
20 in.
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.
The elongation of rod (2) due to the 1.8-kip load must be determined, also. The stress in rod (2) is
F 1.8 kips
2  2 
 9 ksi
A2 0.2 in.2
and consequently, the strain in rod (2) is

9 ksi
 0.000900 in./in.
2  2 
E2 10,000 ksi
From the strain, the elongation in rod (2) can be computed:
 2   2 L2  (0.000900 in./in.)(100 in.)  0.0900 in.
The deflection of joint E is the sum of the rigid bar deflection at D and the elongation in rod (2):
vE  vD   2  0.10125 in.  0.09 in.  0.19125 in.  0.1913 in. 
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.
3.24 Rigid bar ABC is supported by pin-connected axial member (1) and by a 0.75-in.-diameter double
shear pin connection at C as shown in Fig. P3.24. Member (1) is a 2.75 in. wide by 1.25 in. thick
rectangular bar made of aluminum with an elastic modulus of 10,000 ksi. When a concentrated load P is
applied to the rigid bar at A, the normal strain in member (1) is measured as –880 .
(a) Determine the magnitude of
applied load P.
(b) Determine the average shear
stress in the pin at C.
Fig. P3.24
Solution
Member (1) is a two-force member that is oriented at 
with respect to the horizontal axis:
30 in.
tan  
 0.75
  36.870
40 in.
From a FBD of rigid structure ABC, the following
equilibrium equations can be written:
Fx   F1 cos 36.870  Cx  0
(a)
Fy   P  F1 sin 36.870  C y  0
(b)
M C  P(80 in.)  ( F1 sin 36.870)(56 in.)  0
(c)
(a) The force in member (1) can be determined from the measured strain of –880 . From Hooke’s
law, the stress in member (1) is:
1 in./in.
 1  E1  (10,000 ksi)(  880 με)
 8.800 ksi
1  106 με
and thus, the force in member (1) is:
F1  1 A1  ( 8.800 ksi)(2.75 in.)(1.25 in.)  30.250 kips
Substitute this value into Eq. (c) to compute P:
( F sin 36.870)(56 in.)
( 30.250 kips)(sin 36.870)(56 in.)
P 1

80 in.
80 in.
 12.705 kips  12.71 kips
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) Substitute the values for P and F1 into Eqs. (a) and (b) to obtain Cx and Cy:
Cx  F1 cos36.870  ( 30.250 kips)cos36.870  24.200 kips
C y  P  F1 sin 36.870  12.705 kips  ( 30.250 kips)sin 36.870  5.445 kips
The resultant pin force at C is found from Cx and Cy:
C  Cx2  C y2  (24.200 kips)2  ( 5.445 kips) 2  24.805 kips
The pin at C is supported in a double-shear connection. The cross-sectional area of the 0.75-in.diameter pin is:
Apin 

4
(0.75 in.)2  0.4421786 in.2
The average shear stress in the pin is thus:
C
24.805 kips


 28.074 ksi  28.1 ksi
AV (2 surfaces)(0.441786 in.2 /surface)
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.
3.25 A concentrated load P is supported by two
bars as shown in Fig. P3.25. Bar (1) is made of
cold-rolled red brass [E = 16,700 ksi;  = 10.4 ×
10−6 /°F] and has a cross-sectional area of 0.225
in.2. Bar (2) is made of 6061-T6 aluminum [E =
10,000 ksi;  = 13.1 × 10−6 /°F] and has a crosssectional area of 0.375 in.2. After load P has been
applied and the temperature of the entire assembly
has increased by 50°F, the total strain in bar (1) is
measured as 1,400  (elongation). Determine:
(a) the magnitude of load P.
(b) the total strain in bar (2).
Fig. P3.25
Solution
Consider a FBD of joint B. Members (1) and (2) are two-force members; therefore,
the equilibrium equations can be written as:
Fx  F2 cos 55  F1 cos 40  0
(a)
Fy  F2 sin 55  F1 sin 40  P  0
(b)
From Eq. (a):
cos 40
F2  F1
cos55
Substitute this expression into Eq. (b) to obtain:
 cos 40 
 F1 cos55  sin 55  F1 sin 40  P
F1  cos 40 tan 55  sin 40  P
 P  F1 1.736812
(c)
(d)
(a) The total strain in bar (1) is measured as 1,400  (elongation). Part of this strain is due to the stress
acting in the bar and part of this strain is due to the temperature increase. The strain caused by the
temperature change is
T   T  (10.4  106 / °F)(50°F)  520  106 in./in.
Since the total strain is  = 1,400  = 0.001400 in./in., the strain caused by the stress in bar (1) must be:
    T  0.001400 in./in.  0.000520 in./in.  0.000880 in./in.
From Hooke’s law, the stress in bar (1) is therefore:
1  E1  (16,700 ksi)(0.000880 in./in.)  14.696 ksi
and thus, the force in bar (1) is:
F1   1 A1  (14.696 ksi)(0.225 in.2 )  3.3066 kips
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 Eq. (d), the applied load P is:
P  F1 1.736812  (3.3066 kips)(1.736812)  5.7429 kips  5.74 kips
Ans.
(b) From Eq. (c), the force in bar (2) is:
cos 40
F2  F1
 (3.3066 kips)(1.335558)  4.4162 kips
cos55
The normal stress in bar (2) is therefore:
4.4162 kips
F
2  2 
 11.7764 ksi
A2
0.375 in.2
The normal strain due to this stress is:

11.7764 ksi
 1,177.64  10 6 in./in.
2  2 
E2 10,000 ksi
The strain caused in bar (2) by the temperature change is:
T   T  (13.1  106 / °F)(50°F)  655  10 6 in./in.
and thus, the total strain in bar (2) is:
    T  1,177.64  106 in./in.  655  10 6 in./in.
 1,832.64  10 6 in./in.  1,833 με
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.
3.26 The rigid bar AC in Fig. P3.26 is
supported by two axial bars (1) and (2).
Both axial bars are made of bronze [E =
100 GPa;  = 18 × 10−6 /°C]. The crosssectional area of bar (1) is A1 = 240 mm2
and the cross-sectional area of bar (2) is A2
= 360 mm2. After load P has been applied
and the temperature of the entire assembly
has increased by 30°C, the total strain in
bar (2) is measured as 1,220 
(elongation). Determine:
(a) the magnitude of load P.
(b) the vertical displacement of pin A.
Fig. P3.26
Solution
(a) The total strain in bar (2) is caused partly by the axial force in the bar and partly by the increase in
temperature. The strain caused by the 30°C temperature increase is:
T   T  (18  10 6 / C)(30C)  0.000540 mm/mm
The strain caused by the axial force in the bar is thus:
 2,   2   2,T  0.001220 mm/mm  0.000540 mm/mm  0.000680 mm/mm
The stress in bar (2) is
 2  E2 2,  (100,000 MPa)(0.000680 mm/mm)  68 MPa
and the force in bar (2) is
F2   2 A2  (68 N/mm 2 )(360 mm 2 )  24, 480 N
Next, consider a FBD of the rigid bar AC. Equilibrium
equations for this FBD are:
Fy  F1  F2  P  0
M A  (1, 400 mm)F2  (500 mm)P  0
which can be solve simultaneously to give:
500 mm
F2 
P  0.357143P
1,400 mm
and
F1  0.642857 P
The applied load P can be expressed in terms of F2 as
1
P
F2  2.8F2
0.357143
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.
and so the magnitude of load P is
P  2.8F2  2.8(24,480 N)  68,544 N  68.5 kN
Ans.
(b) The force in bar (1) is
F1  0.642857 P  (0.642857)(68,544 N)  44,064 N
Thus, the stress in bar (1) is
F 44,064 N
1  1 
 183.60 MPa
A1 240 mm 2
The normal strain due to the axial force in bar (1) is

183.60 MPa
1,  1 
 0.001836 mm/mm
E1 100,000 MPa
The normal strain caused by the 30°C temperature increase is:
1,T   T  (18  106 /C)(30C)  0.000540 mm/mm
Therefore, the total strain in bar (1) is
1  1,  1,T  0.001836 mm/mm  0.000540 mm/mm  0.002376 mm/mm
and the elongation in bar (1) is
1  1, L1  (0.002376 mm/mm)(1,300 mm)  3.0888 mm
Since rigid bar ABC is connected to bar (1) (with a perfect connection), joint A displaces downward by
an amount equal to the elongation of bar (1); therefore,
vA  1  3.09 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.
3.27 The rigid bar in Fig. P3.27 is supported
by axial bar (1) and by a pin connection at
C. Axial bar (1) has a cross-sectional area of
A1 = 275 mm2, an elastic modulus of E =
200 GPa, and a coefficient of thermal
expansion of  = 11.9 × 10−6 /°C. The pin at
C has a diameter of 25 mm. After load P has
been applied and the temperature of the
entire assembly has been increased by 20°C,
the total strain in bar (1) is measured as 925
 (elongation). Determine:
(a) the magnitude of load P.
(b) the average shear stress in pin C.
Fig. P3.27
Solution
The total strain in bar (1) consists of thermal strain as well as normal strain caused by normal stress:
    T
The normal strain due to the increase in temperature is:
T   T  (11.9  106 /C)(20C)  0.000238 mm/mm
Therefore, the normal stress in bar (1) causes a normal strain of:
    T  0.000925 mm/mm  0.000238 mm/mm  0.000687 mm/mm
From Hooke’s law, the normal stress in bar (1) can be calculated as:
1  E  (200,000 MPa)(0.000687 mm/mm)  137.4 MPa
and thus the axial force in bar (1) must be:
F1   1 A1  (137.4 N/mm 2 )(275 mm 2 )  37,785 N
Next, consider a free-body diagram of the rigid bar. Write
a moment equilibrium equation about pin C:
M C  (260 mm)F1  (540 mm)P
 (260 mm)(37,785 N)  (540 mm)P  0
 P  18,192.8 N  18.19 kN
Ans.
Now that P is known, the horizontal and vertical reactions
at C can be calculated:
Fx  Cx  F1  0
 Cx  F1  37,785 N
Fy  C y  P  0
 C y  P  19,192.8 N
The resultant force acting on pin C is:
C  Cx2  C y2  (37,785 N)2  (18,192.8 N)2  41,936.659 N
Since the pin at C is a double shear connection, the shear force acting on one shear plane is half of the
resultant force: V = 20,968.330 N. The area of one shear plane of the 25-mm-diameter pin at C (in
other words, the cross-sectional area of the pin) 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.
Apin 

(25 mm)2  490.874 mm2
4
and thus the average shear stress in pin C is:
V
20,968.330 N
C 

 42.716 N/mm 2  42.7 MPa
2
AV 490.874 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.
3.28 The rigid bar in Fig. P3.28 is
supported by axial bar (1) and by a pin
connection at C. Axial bar (1) has a crosssectional area of A1 = 275 mm2, an elastic
modulus of E = 200 GPa, and a coefficient
of thermal expansion of  = 11.9 × 10−6
/°C. The pin at C has a diameter of 25 mm.
After load P has been applied and the
temperature of the entire assembly has
been decreased by 30°C, the total strain in
bar (1) is measured as 925  (elongation).
Determine:
(a) the magnitude of load P.
(b) the average shear stress in pin C.
Fig. P3.28
Solution
The total strain in bar (1) consists of thermal strain as well as normal strain caused by normal stress:
    T
The normal strain due to the decrease in temperature is:
T   T  (11.9  106 /C)(  30C)  0.000357 mm/mm
Therefore, the normal stress in bar (1) causes a normal strain of:
    T  0.000925 mm/mm  ( 0.000357 mm/mm)  0.001282 mm/mm
From Hooke’s law, the normal stress in bar (1) can be calculated as:
1  E  (200,000 MPa)(0.001282 mm/mm)  256.4 MPa
and thus the axial force in bar (1) must be:
F1   1 A1  (256.4 N/mm 2 )(275 mm 2 )  70,510 N
Next, consider a free-body diagram of the rigid bar. Write
a moment equilibrium equation about pin C:
M C  (260 mm)F1  (540 mm)P
 (260 mm)(70,510 N)  (540 mm)P  0
 P  33,949.3 N  33.9 kN
Ans.
Now that P is known, the horizontal and vertical reactions
at C can be calculated:
Fx  Cx  F1  0
 Cx  F1  70,510 N
Fy  C y  P  0
 C y  P  33,949.3 N
The resultant force acting on pin C is:
C  Cx2  C y2  (70,510 N)2  (33,949.3 N) 2  78,257.3 N
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.
Since the pin at C is a double shear connection, the shear force acting on one shear plane is half of the
resultant force: V = 39,128.7 N. The area of one shear plane of the 25-mm-diameter pin at C (in other
words, the cross-sectional area of the pin) is:
Apin 

(25 mm)2  490.874 mm2
4
and thus the average shear stress in pin C is:
V
39,128.7 N
C 

 79.712 N/mm 2  79.7 MPa
AV 490.874 mm 2
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.
4.1 A stainless steel alloy bar 25 mm wide by
16 mm thick is subjected to an axial load of
P = 145 kN. Using the stress-strain diagram
given in Fig. P4.1, determine:
(a) the factor of safety with respect to the
yield strength defined by the 0.20% offset
method.
(b) the factor of safety with respect to the
ultimate strength.
Fig. P4.1
Solution
(a) From the stress-strain curve, the yield strength defined by the 0.20% offset method is
 Y  550 MPa
The normal stress in the bar is
F (145 kN)(1,000 N/kN)
 
 362.5 MPa
A
(25 mm)(16 mm)
Therefore, the factor of safety with respect to yield is
550 MPa

FS  Y 
 1.517
 actual 362.5 MPa
(b) From the stress-strain curve, the ultimate strength is
 U  1,100 MPa
Therefore, the factor of safety with respect to the ultimate strength is

1,100 MPa
FS  U 
 3.03
 actual 362.5 MPa
Ans.
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.
4.2 Six bolts are used in the connection between the
axial member and the support, as shown in Fig.
P4.2. The ultimate shear strength of the bolts is 300
MPa, and a factor of safety of 4.0 is required with
respect to fracture.
Determine the minimum
allowable bolt diameter required to support an
applied load of P = 475 kN.
Fig. P4.2
Solution
The allowable shear stress for the bolts is

300 MPa
 allow  U 
 75 MPa
FS
4
To support a load of P = 350 kN, the total area subjected to shear stress must equal or exceed:
475,000 N
P
AV 

 6,333.333 mm 2
2
 allow 75 N/mm
There are six bolts, and each bolt acts in double shear; therefore, the total area subjected to shear stress
is

2
AV  (6 bolts)(2 surfaces per bolt) d bolt
4
The minimum bolt diameter can be found be equating the two expressions for AV:

2
 6,333.333 mm2
(12 surfaces) d bolt
4
 d bolt  25.923 mm  25.9 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.
4.3 A 14-kip load is supported by two bars as
shown in Fig. P4.3. Bar (1) is made of cold-rolled
red brass (Y = 60 ksi) and has a cross-sectional
area of 0.225 in.2. Bar (2) is made of 6061-T6
aluminum (Y = 40 ksi) and has a cross-sectional
area of 0.375 in.2. Determine the factor of safety
with respect to yielding for each of the bars.
Fig. P4.3
Solution
Consider a FBD of joint B. Members (1) and (2) are two-force members; therefore,
the equilibrium equations can be written as:
Fx  F2 cos50  F1 cos35  0
(a)
Fy  F2 sin 50  F1 sin 35  14 kips  0
(b)
From Eq. (a):
cos35
F2  F1
cos50
Substitute this expression into Eq. (b) to obtain:
 cos35 
 F1 cos50  sin 50  F1 sin 35  14 kips
F1  cos35 tan 50  sin 35  14 kips
F1 1.549804  14 kips
F1  9.033401 kips
Backsubstitute to obtain F2:
cos35
F2  F1
 (9.033401 kips)(1.274374)  11.511935 kips
cos50
The normal stress in bar (1) is
F 9.033401 kips
1  1 
 40.148452 ksi
A1
0.225 in.2
Therefore, the factor of safety in bar (1) is

60 ksi
FS1  Y 
 1.494454  1.494
 1 40.148452 ksi
The normal stress in bar (2) is
F 11.511935 kips
2  2 
 30.698495 ksi
A2
0.375 in.2
and thus, the factor of safety in bar (2) is

40 ksi
 1.302995  1.303
FS2  Y 
 2 30.698495 ksi
Ans.
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.
4.4 A steel plate is to be attached to a support with three bolts as
shown in Fig. P4.4. The cross-sectional area of the plate is 560
mm2 and the yield strength of the steel is 250 MPa. The ultimate
shear strength of the bolts is 425 MPa. A factor of safety of 1.67
with respect to yield is required for the plate. A factor of safety of
4.0 with respect to the ultimate shear strength is required for the
bolts. Determine the minimum bolt diameter required to develop
the full strength of the plate. (Note: consider only the gross crosssectional area of the plate—not the net area.)
Fig. P4.4
Solution
The allowable normal stress is

250 MPa
 149.7006 MPa
 allow  Y 
FS
1.67
The full strength of the plate (based on the gross cross-sectional area) is therefore:
Pmax   allow A  (149.7006 MPa)(560 mm 2 )  83,832.3353 N
The allowable shear stress of the bolts is

425 MPa
 106.25 MPa
 allow  U 
FS
4.0
The minimum shear area required to support the maximum load P is
83,832.3353 N
P
AV  max 
 789.0102 mm 2
2
 allow 106.25 N/mm
There are three bolts, each acting in single shear, which provide a shear area of


2
2
AV  (3 bolts)(1 surface per bolt) d bolt
 (3 surfaces) d bolt
4
4
Equating the two expressions for the shear area, the minimum bolt diameter can be computed:

2
(3 surfaces) d bolt
 789.0102 mm2
4
 d bolt  18.2994 mm  18.30 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.
4.5 In Fig. P4.5, member (1) is a steel bar with a crosssectional area of 1.35 in.2 and a yield strength of 50 ksi.
Member (2) is a pair of 6061-T6 aluminum bars having a
combined cross-sectional area of 3.50 in.2 and a yield
strength of 40 ksi. A factor of safety of 1.6 with respect to
yield is required for both members. Determine the
maximum allowable load P that may be applied to the
structure. Report the factors of safety for both members at
the allowable load.
Fig. P4.5
Solution
Consider a FBD of joint B and write the following equilibrium equations:
Fx  F2 cos 65  F1  0
Fy  F2 sin 65  P  0
From Eq. (b):
P
F2 
sin 65
Substituting Eq. (c) into Eq. (a) gives an expression for P in terms of F1:
F2 cos 65  F1  0
(a)
(b)
(c)
 P 

 cos 65  F1  0
sin 65 
P  F1 tan 65
and rearranging Eq. (c) gives an expression for P in terms of F2:
P  F2 sin 65
(d)
(e)
The allowable normal stress in member (1) is

50 ksi
 31.25 ksi
 allow,1  Y ,1 
FS1
1.6
and the allowable force in member (1) is
Fallow,1   allow,1 A1  (31.25 ksi)(1.35 in.2 )  42.1875 kips
(f)
The allowable normal stress in member (2) is

40 ksi
 25.00 ksi
 allow,2  Y ,2 
FS2
1.6
and the allowable force in member (2) is
Fallow,2   allow,2 A2  (25.00 ksi)(3.50 in.2 )  87.50 kips
(g)
Substitute the allowable force in member (1) from Eq. (f) into Eq. (d) to obtain the maximum load P
based on the capacity of member (1):
P  Fallow,1 tan 65  (42.1875 kips) tan 65  90.471386 kips
(h)
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.
Repeat with the allowable force in member (2) from Eq. (g) substituted into Eq. (e) to obtain the
maximum load P based on the capacity of member (2):
P  Fallow,2 sin 65  (87.50 kips)sin 65  79.301931 kips
(i)
Compare the results in Eqs. (h) and (i) to find that the maximum load P that may be applied is controlled
by the capacity of member (2):
Ans.
Pmax  79.3 kips
The factor of safety in member (2) is thus:
FS2  1.6
For an applied load of P = 79.3 kips, the force in member (1) can be computed from Eq. (d):
79.301931 kips
P
F1 

 36.979096 kips
tan 65
tan 65
which results in a normal stress of
F 36.979096 kips
1  1 
 27.391924 ksi
A1
1.350 in.2
and thus, its factor of safety is:

50 ksi
FS1  Y ,1 
 1.825
1 27.391924 ksi
Ans.
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.
4.6 The rigid structure ABD in Fig. P4.6 is supported at B by a 35-mm-diameter tie rod (1) and at A by a
30-mm-diameter pin used in a single shear connection. The tie rod is connected at B and C by 24-mmdiameter pins used in double shear connections. Tie rod (1) has a yield strength of 250 MPa, and each of
the pins has an ultimate shear strength of 330 MPa. A concentrated load of P = 50 kN acts as shown at
D. Determine:
(a) the normal stress in rod (1).
(b) the shearing stress in the pins at A and B.
(c) the factor of safety with respect to the yield
strength for tie rod (1).
(d) the factor of safety with respect to the
ultimate strength for the pins at A and B.
Fig. P4.6
Solution
Member (1) is a two-force member that is
oriented at  with respect to the horizontal axis:
8m
 1.509434
tan  
5.3 m
  56.4755
From a FBD of rigid structure ABD, the following equilibrium equations can be written:
Fx  P cos 60  F1 cos56.4755  Ax  0
Fy   P sin 60  F1 sin 56.4755  Ay  0
(a)
(b)
M A  ( F1 cos56.4755)(8 m)  ( P cos 60)(3.4 m)  ( P sin 60)(7.5 m)  0
(c)
From Eq. (c):
(3.4 m)cos 60  (7.5 m)sin 60
(3.4 m)cos 60  (7.5 m)sin 60
F1  P
 (50 kN)
 92.7405 kN
(8 m)cos56.4755
(8 m)cos56.4755
Substitute F1 into Eq. (a) to obtain Ax:
Ax  F1 cos56.4755  P cos 60  (92.7405 kN)cos56.4755  (50 kN)cos 60  26.2199 kN
and substitute F1 into Eq. (b) to obtain Ay:
Ay  F1 sin 56.4755  P sin 60  (92.7405 kN)sin 56.4755  (50 kN)sin 60  120.6144 kN
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.
The resultant pin force at A is found from Ax and Ay:
A
Ax2  Ay2  (26.2199 kN)2  (120.6144 kN)2  123.4314 kN

(a) The cross-sectional area of the 35-mm-diameter tie rod is:
A1 
(35 mm)2  962.112750 mm2
4
and thus, the normal stress in rod (1) is:
F
92,740.5 N
1  1 
 96.3925 MPa  96.4 MPa
A1 962.112750 mm 2

Ans.
(b) The 30-mm-diameter single shear pin at A has a shear area of
AV , A 
(30 mm)2  706.858347 mm2
4
Consequently, the shear stress in pin A is
123,431.4 N
A 
 174.6197 MPa  174.6 MPa
706.858347 mm2

Ans.
The 24-mm-diameter double shear pins at B and C have a shear area of
AV , B  (2 surfaces)
(24 mm) 2  904.778685 mm 2
4
The shear stress in pins B and C is
92,740.5 N
B 
 102.5008 MPa  102.5 MPa
904.778685 mm 2
(c) The factor of safety for tie rod (1) is

250 MPa
FS1  Y 
 2.59
 1 96.3925 MPa
(d) The factor of safety with respect to the ultimate strength for the pin at A is

330 MPa
FS A  U 
 1.890
 A 174.6197 MPa
and the factor of safety for pin B is

330 MPa
FSB  U 
 3.22
 B 102.5008 MPa
Ans.
Ans.
Ans.
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.
4.7 Rigid bar ABD in Fig. P4.7 is supported by a pin
connection at A and a tension link BC. The 8-mmdiameter pin at A is supported in a double shear
connection, and the 12-mm-diameter pins at B and C are
both used in single shear connections. Link BC is 30-mm
wide and 6-mm thick. The ultimate shear strength of the
pins is 330 MPa and the yield strength of link BC is 250
MPa.
(a) Determine the factor of safety in pins A and B with
respect to the ultimate shear strength.
(b) Determine the factor of safety in link BC with respect
to the yield strength.
Fig. P4.7
Solution
Consider a free-body diagram of the rigid bar. Link BC is a
two-force member that is oriented at an angle of  with
respect to the horizontal axis:
0.6 m
  53.1301
tan  
0.45 m
The equilibrium equations for the rigid bar can be written as:
Fx   FBC cos53.1301  (8.2 kN)cos70  Ax  0
Fy  FBC sin 53.1301  (8.2 kN)sin70  Ay  0
M A  (0.75 m)FBC cos53.1301  (0.45 m)FBC sin 53.1301
(1.85 m)(8.2 kN)sin70  (0.75 m)(8.2 kN)cos70  0
Solving these three equations simultaneously gives:
FBC  20.1958 kN
Ax  9.3129 kN
Ay  8.4511 kN
The resultant force at pin A is:
A
Ax2  Ay2  (9.3129 kN)2  ( 8.4511 kN) 2  12.5758 kN
Before the factors of safety can be determined, we must compute the shear stresses in pins A and B as
well as the normal stress in link BC.
Pin shear stresses:
The 8-mm-diameter pin at A is supported in a double shear connection; therefore, the shear force acting
on one shear plane (which is simply equal to the cross-sectional area of the pin) is half of the resultant
force at pin A: VA = 6.2879 kN. The cross-sectional area of the pin at A is:
Apin A 

(8 mm)2  50.265 mm2
4
and therefore, the shear stress in pin A is:
V
(6.2879 kN)(1000 N/kN)
A  A 
 125.0940 N/mm 2  125.0940 MPa
2
AV
50.265 mm
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.
The 12-mm-diameter pin at B and C is supported in a single shear connection and so the shear force
acting on one shear plane is the entire force in link BC: VB = 20.1958 kN. The cross-sectional area of
the pin at B is:
Apin B 

(12 mm)2  113.097 mm2
4
and therefore, the shear stress in pin B is:
(20.1958 kN)(1000 N/kN)
V
B  B 
 178.5697 N/mm 2  178.5697 MPa
113.097 mm 2
AV
Normal stress in link:
The normal stress in link BC is:
(20.1958 kN)(1000 N/kN)
F
 BC  BC 
 112.1986 N/mm 2  112.1986 MPa
(30 mm)(6 mm)
ABC
Factors of safety:
For pin A, the factor of safety is:

330 MPa
FS A  U 
 2.64
 A 125.0940 MPa
For pin B, the factor of safety is:

330 MPa
 1.848
FSB  U 
 B 178.5697 MPa
The factor of safety in link BC is:

250 MPa
 2.23
FSBC  Y 
 BC 112.1986 MPa
Ans.
Ans.
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.
4.8 In Fig. P4.8, davit ABD is supported at A by a single
shear pin connection and at B by a tie rod (1). The pin at
A has a diameter of 1.25 in. and the pins at B and C are
each 0.75-in.-diameter pins. Tie rod (1) has an area of
1.50 in.2. The ultimate shear strength in each pin is 80 ksi,
and the yield strength of the tie rod is 36 ksi. A
concentrated load of 25 kips is applied as shown to the
davit structure at D. Determine:
(a) the normal stress in rod (1).
(b) the shearing stress in the pins at A and B.
(c) the factor of safety with respect to the yield strength
for tie rod (1).
(d) the factor of safety with respect to the ultimate
strength for the pins at A and B.
Fig. P4.8
Solution
Member (1) is a two-force member that is oriented at  with
respect to the horizontal axis:
9 ft
tan  
 0.75   36.870
12 ft
From a FBD of rigid structure ABD, the following equilibrium
equations can be written:
Fx  (25 kips) cos 60  F1 cos 36.870  Ax  0
(a)
Fy  (25 kips)sin 60  F1 sin 36.870  Ay  0
(b)
M A  ( F1 cos 36.870)(9 ft)  (25 kips) cos 60(11 ft)
(25 kips) sin 60(7 ft)  0
(c)
(11 ft) cos 60  (7 ft)sin 60
 40.1465 kips
(9 ft) cos 36.870
Substitute F1 into Eq. (a) to obtain Ax:
Ax  F1 cos 36.870  (25 kips) cos 60  (40.1465 kips) cos36.870  (25 kips) cos 60  19.6172 kips
and substitute F1 into Eq. (b) to obtain Ay:
Ay  F1 sin 36.870  (25 kips)sin 60  (40.1465 kips)sin 36.870  (25 kips)sin 60  45.7386 kips
From Eq. (c):
F1  (25 kips)
The resultant pin force at A is found from Ax and Ay:
A  Ax2  Ay2  (19.6172 kips)2  (45.7386 kips)2  49.7680 kips
(a) The normal stress in tie rod (1) is:
F 40.1465 kips
1  1 
 26.7643 ksi  26.8 ksi
1.50 in.2
A1
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 1.25-in.-diameter single shear pin at A has a shear area of
AV , A 
(1.25 in.)2  1.2272 in.2
4
Consequently, the shear stress in pin A is
49.7680 kips
A 
 40.5541 ksi  40.6 ksi
1.2272 in.2

Ans.
The 0.75-in.-diameter double shear pins at B and C have a shear area of
AV , B  (2 surfaces)
(0.75 in.) 2  0.8836 in.2
4
The shear stress in pins B and C is
40.1465 kips
B 
 45.4352 ksi  45.4 ksi
0.8836 in.2
(c) The factor of safety for tie rod (1) is

36 ksi
FS1  Y 
 1.345
 1 26.7643 ksi
(d) The factor of safety with respect to the ultimate strength for the pin at A is

80 ksi
 1.973
FS A  U 
 A 40.5541 ksi
and the factor of safety for pin B is

80 ksi
 1.761
FSB  U 
 B 45.4352 ksi
Ans.
Ans.
Ans.
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.
4.9 The pin-connected structure is subjected to a load P
as shown in Fig. P4.9. Inclined member (1) has a crosssectional area of 250 mm2 and has a yield strength of
255 MPa. It is connected to rigid member ABC with a
16-mm-diameter pin in a double shear connection at B.
The ultimate shear strength of the pin material is 300
MPa. For inclined member (1), the minimum factor of
safety with respect to the yield strength is FSmin = 1.5.
For the pin connections, the minimum factor of safety
with respect to the ultimate strength is FSmin = 3.0.
(a) Based on the capacity of member (1) and pin B,
determine the maximum allowable load P that may be
applied to the structure.
(b) Rigid member ABC is supported by a double shear
pin connection at A. Using FSmin = 3.0, determine the
minimum pin diameter that may be used at support A.
Fig. P4.9
Solution
The allowable normal stress for inclined member (1) is

255 MPa
 170.00 MPa
 allow  Y 
FS
1.50
and the allowable shear stress for the pins is

300 MPa
 allow  U 
 100.0 MPa
FS
3.00
(a) The allowable axial force in member (1) based on normal stress is
Fallow   allow A1  (170.00 MPa)(250 mm 2 )  42,500 N
Member (1) is connected with a 16-mm-diameter double shear pin. The shear area of this pin is:
(a)

AV  (2 surfaces) (16 mm) 2  402.1239 mm 2
4
Consequently, the shear force V (which is applied by the inclined member) that can be applied to the pin
is limited to
Vallow   allow AV  (100.0 MPa)(402.1239 mm 2 )  40, 212.4 N
(b)
Comparing the two values given in Eq. (a) and Eq. (b), the maximum allowable force in member (1) is
F1  40, 212.4 N
(c)
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.
Member (1) is a two-force member that is oriented at 
with respect to the horizontal axis:
1.4 m
tan  
 0.6087
  31.3287
2.3 m
From a FBD of rigid structure ABC, the following
equilibrium equations can be written:
Fx  Ax  F1 cos 31.3287  P  0
(d)
Fy  Ay  F1 sin 31.3287  0
(e)
M A  (3.2 m)P  ( F1 cos 31.3287)(1.4 m)  0
(f)
Substitute the value of F1 from Eq. (c) into Eq. (f) to obtain
the maximum load P:
(1.4 m)cos31.3287
P  F1
3.2 m
(1.4 m)cos31.3287
 (40,212.4 N)
3.2 m
Ans.
 15,027.8 N  15.03 kN
Substitute F1 and P into Eq. (d) to obtain Ax:
Ax  P  F1 cos31.3287  (15,027.8 N)  (40, 212.4 N)cos31.3287  19,321.5 N
and substitute F1 into Eq. (e) to obtain Ay:
Ay  F1 sin 31.3287  (40, 212.4 N)sin 31.3287  20,908.3 N
The resultant pin force at A is found from Ax and Ay:
A
Ax2  Ay2  ( 19,321.5 N)2  (20,908.3 N) 2  28,468.9 N
The total shear area required to support the resultant pin force is
A
28,468.9 N

 284.689 mm2
AV 
 allow 100 N/mm2
Since the pin at A is in a double shear connection, the shear area provided by the pin is
 2
AV  (2 surfaces) d pin
4
Equate these two expressions and solve for the required minimum pin diameter:

2
(2 surfaces) d pin
 284.689 mm 2
4
 d pin  13.46 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.
4.10 After load P is applied to the pin-connected structure shown in Fig. P4.10, a normal strain of
 = +550  is measured in the longitudinal direction of member (1). The cross-sectional area of
member (1) is A1 = 0.60 in.2, its elastic modulus is E1 = 29,000 ksi, and its yield strength is 36 ksi.
(a) Determine the axial force in
member (1), the applied load P, and the
resultant force at pin B.
(b) The ultimate shear strength of the
steel pins is 54 ksi. Determine the
minimum diameter for the pin at B if a
factor of safety of 2.5 with respect to
the ultimate shear strength is required.
(c) Compute the factor of safety for
member (1) with respect to its yield
strength.
Fig. P4.10
Solution
(a) Given the strain in member (1), its stress can be computed from Hooke’s law:
1  E11  (29,000 ksi)(0.000550 in./in.)  15.9500 ksi
The force in member (1) is the product of the normal stress and the cross-sectional area:
F1  1 A1  (15.9500 ksi)(0.60 in.2 )  9.57 kips
Ans.
Next, consider a free-body diagram of
horizontal member ABC. Member (1) makes
an angle of  with respect to the horizontal
axis:
3.50 ft
tan  
  36.3844
4.75 ft
Note that member (1) is a two-force member.
Equilibrium equations for horizontal member
ABC can be written as:
Fx  Bx  F1 cos36.3844  0
Fy  By  F1 sin 36.3844  P  0
M B  (6.50 ft)F1 sin 36.3844  (9.50 ft)P  0
Substituting the value F1 = 14.79 kips into these equations gives:
Bx  7.7044 kips
By  9.5611 kips
P  3.8842 kips  3.88 kips
Ans.
The resultant force at pin B is:
B  Bx2  By2  (7.7044 kips)2  (9.5611 kips) 2  12.2789 kips  12.28 kips
Ans.
(b) The pin at B is supported in a double shear connection. Therefore, the shear force acting on one
shear plane of the pin is half of the resultant force: VB = 6.1395 kips. The allowable shear stress for the
pin is computed from the ultimate shear strength and the factor of safety:

54 ksi
a  U 
 21.6 ksi
FS
2.5
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.
Next, the cross-sectional area required for a single shear plane can be determined:
V
V
6.1395 kips
  B  a
 AB  B 
 0.2842 in.2
AB
21.6 ksi
a
To provide AB, the pin at B must have a diameter of at least:
 2
d B  0.2842 in.2
 d B  0.602 in.
4
(c) The factor of safety for member (1) with respect to its yield strength is:

36 ksi
FS1  Y 
 2.26
 1 15.9500 ksi
Ans.
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.
4.11 The simple pin-connected structure carries a
concentrated load P as shown in Fig. P4.11. The
rigid bar is supported by strut AB and by a pin
support at C. The steel strut AB has a crosssectional area of 0.25 in.2 and a yield strength of
60 ksi. The diameter of the steel pin at C is 0.375
in., and the ultimate shear strength is 54 ksi. If a
factor of safety of 2.0 is required in both the strut
and the pin at C, determine the maximum load P
that can be supported by the structure.
Fig. P4.11
Solution
From a FBD of the rigid bar, the following equilibrium
equations can be written:
Fx   F1  Cx  0
(a)
Fy  C y  P  0
(b)
M C  (10 in.)F1  (22 in.)P  0
(c)
From Eq. (c), express F1 in terms of the unknown load P:
22 in.
F1 
P  2.2 P
(d)
10 in.
Substitute this result into Eq. (a) to express Cx in terms of P:
Cx  F1  2.2 P
The resultant reaction force at pin C can now be expressed as a function of P:
C  Cx2  C y2  (2.2P)2  ( P)2  2.41661P
(e)
Strut AB:
The allowable normal stress for strut AB is

60 ksi
 allow  Y 
 30 ksi
FS
2.0
Therefore, the allowable axial force for strut AB is
Fallow,1   allow A1  (30 ksi)(0.25 in.2 )  7.50 kips
From Eq. (d), the maximum load that may be applied to the rigid bar based on the limitations on the strut
normal stress is
F
7.50 kips
Pmax  allow,1 
 3.4091 kips
(f)
2.2
2.2
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.
Pin C:
The allowable shear stress for the pin at C is

54 ksi
 27 ksi
 allow  U 
FS
2.0
The 0.375-in.-diameter double shear pin at C has a shear area of

AV  (2 surfaces) (0.375 in.) 2  0.2209 in.2
4
thus, the allowable reaction force at C is
C allow   allow AV  (27 ksi)(0.2209 in.2 )  5.9641 kips
From Eq. (e), the maximum load that may be applied to the rigid bar based on the limitations on the pin
shear stress is
C allow
5.9641 kips
Pmax 

 2.4680 kips
(g)
2.41661
2.41661
Maximum load P:
Comparing the results in Eqs. (f) and (g), the maximum load P that may be applied to the rigid bar is
Ans.
Pmax  2.4680 kips  2.47 kips
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.
4.12 In Fig. P4.12, rigid tee-beam ABC is supported at
A by a single shear pin connection and at B by a strut,
which consists of two 30 mm wide by 8 mm thick steel
bars. The pins at A, B, and D are each 12 mm in
diameter. The yield strength of the steel bars in strut (1)
is 250 MPa, and the ultimate shear strength of each pin
is 500 MPa. Determine the allowable load P that may
be applied to the rigid bar at C if an overall factor of
safety of 3.0 is required. Use L1 = 1.1 m and L2 = 1.3 m.
Fig. P4.12
Solution
From a FBD of the rigid beam ABC, the following
equilibrium equations can be written:
Fx  Ax  0
(a)
Fy  Ay  F1  P  0
(b)
M A   L1 F1  ( L1  L2 )P  0
(c)
From Eq. (c), express F1 in terms of the unknown load P:
L  L2
1.1 m  1.3 m
F1   1
P
P  2.181818P
L1
1.1 m
Substitute this result into Eq. (b) to express Ay in terms of P:
Ay  P  F1  P  ( 2.181818P )  1.181818 P
(d)
The resultant reaction force at pin A can now be expressed as a function of P:
A
Ax2  Ay2  (0)2  ( 1.181818P) 2  1.181818P
(e)
The allowable normal stress for strut (1) is

250 MPa
 allow  Y 
 83.333333 MPa
FS
3.0
Therefore, the allowable axial force for strut (1) is
Fallow,1   allow A1  (83.333333 MPa)(2  30 mm  8 mm)  40,000 N
From Eq. (d), the maximum load that may be applied to the rigid bar based on the strut normal stress
limitation is
Fallow,1
40,000 N
Pmax 

 18,333 N
(f)
2.181818 2.181818
The allowable shear stress for the pins at A, B, and D is
500 MPa

 166.666667 MPa
 allow  U 
FS
3.0
The cross-sectional area of a 12-mm-diameter pin is

Apin  (12 mm)2  113.097336 mm 2
4
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.
The pin at A is a single shear connection; therefore, AV,A = Apin. The allowable reaction force at A is
A allow   allow AV , A  (166.666667 MPa)(113.097336 mm2 )  18,849.5560 N
From Eq. (e), the maximum load that may be applied to the rigid bar based on the limitations of pin A is
A allow
18,849.5560 N
Pmax 

 15,949.627 N
(g)
1.181818
1.181818
The pins at B and D are in double shear connections; therefore, AV,B = 2Apin. The allowable force in strut
(1) based on the capacity of the pins at B and D is
Fallow,1   allow AV , B  (166.666667 MPa)(2)(113.097336 mm2 )  37,699.1121 N
From Eq. (d), the maximum load that may be applied to the rigid bar based on the limitations of the pins
at B and D is
Fallow,1
37,699.1121 N
Pmax 

 17, 278.7611 N
(h)
2.181818
2.181818
Comparing the results in Eqs. (f), (g), and (h), the maximum load P that may be applied to the rigid bar
is
Ans.
Pmax  15,949.627 N  15.95 kN
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.
4.13 In Fig. P4.13, rigid bar ABC is supported at A by a
single shear pin connection and at B by a strut, which
consists of two 2 in. wide by 0.25 in. thick steel bars.
The pins at A, B, and D each have a diameter of 0.5 in.
The yield strength of the steel bars in strut (1) is 36 ksi,
and the ultimate shear strength of each pin is 72 ksi.
Determine the allowable load P that may be applied to
the rigid bar at C if an overall factor of safety of 3.0 is
required. Use L1 = 36 in. and L2 = 24 in.
Fig. P4.13
Solution
From a FBD of the rigid beam ABC, the following
equilibrium equations can be written:
Fx  Ax  0
(a)
Fy  Ay  F1  P  0
(b)
M A   L1 F1  ( L1  L2 )P  0
(c)
From Eq. (c), express F1 in terms of the unknown load P:
L  L2
36 in.  24 in.
F1   1
P
P  1.666667 P
L1
36 in.
Substitute this result into Eq. (b) to express Ay in terms of P:
Ay  P  F1  P  (1.666667 P )  0.666667 P
(d)
The resultant reaction force at pin A can now be expressed as a function of P:
A  Ax2  Ay2  (0)2  (0.666667 P) 2  0.666667 P
(e)
Strut (1):
The allowable normal stress for strut (1) is

36 ksi
 allow  Y 
 12.0 ksi
FS
3.0
Therefore, the allowable axial force for strut (1) is
Fallow,1   allow A1  (12.0 ksi)(2  2 in.  0.25 in.)  12.0 kips
From Eq. (d), the maximum load that may be applied to the rigid bar based on the strut normal stress
limitation is
Fallow,1
12.0 kips
Pmax 

 7.20 kips
(f)
1.666667 1.666667
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.
Pins:
The allowable shear stress for the pins at A, B, and D is

72 ksi
 24.0 ksi
 allow  U 
FS
3.0
The cross-sectional area of a 0.5-in.-diameter pin is
Apin 

4
(0.5 in.)2  0.196350 in.2
Pin A:
The pin at A is a single shear connection; therefore, AV,A = Apin. The allowable reaction force at A is
A allow   allow AV , A  (24.0 ksi)(0.196350 in.2 )  4.7124 kips
From Eq. (e), the maximum load that may be applied to the rigid bar based on the limitations of pin A is
A allow
4.7124 kips
Pmax 

 7.0686 kips
(g)
0.666667
0.666667
Pin B and D:
The pins at B and D are in double shear connections; therefore, AV,B = 2Apin. The allowable force in strut
(1) based on the capacity of the pins at B and D is
Fallow,1   allow AV , B  (24.0 ksi)(2)(0.196350 in.2 )  9.4248 kips
From Eq. (d), the maximum load that may be applied to the rigid bar based on the limitations of the pins
at B and D is
Fallow,1
9.4248 kips
Pmax 

 5.6549 kips
(h)
1.666667
1.666667
Maximum load P:
Comparing the results in Eqs. (f), (g), and (h), the maximum load P that may be applied to the rigid bar
is
Ans.
Pmax  5.6549 kips  5.65 kips
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.
4.14 A concentrated load of P = 70 kips is applied to
beam AB as shown in Fig. P4.14. Rod (1) has a
diameter of 1.50 in., and its yield strength is 60 ksi. Pin
A is supported in a double shear connection, and the
ultimate shear strength of pin A is 80 ksi.
(a) Determine the normal stress in rod (1).
(b) Determine the factor of safety with respect to the
yield strength for rod (1).
(c) If a factor of safety of 3.0 with respect to the
ultimate strength is specified for pin A, determine the
minimum required pin diameter.
Fig. P4.14
Solution
Tie rod (1) is a two-force member that is oriented at 
with respect to the horizontal axis:
8.5 ft
 0.7083
  35.3112
tan  
12 ft
From a FBD of beam AB, the following equilibrium
equations can be written:
Fx  Ax  F1 cos35.3112  0
(a)
Fy  Ay  F1 sin 35.3112  P  0
(b)
 10 in. 
M A  ( F1 cos35.3112) 
 12 in./ft 
( F1 sin 35.3112)(12 ft)  P(8 ft)  0
(c)
From Eq. (c):
F1 
(70 kips)(8 ft)
 73.5272 kips
(cos35.3112)  0.833333 ft   (sin35.3112)(12 ft)
Substitute F1 into Eq. (a) to obtain Ax:
Ax   F1 cos35.3112  (73.5272 kips)cos35.3112  60.0 kips
and substitute F1 into Eq. (b) to obtain Ay:
Ay  P  F1 sin 35.3112  70 kips  (73.5272 kips)sin 35.3112  27.50 kips
The resultant pin force at A is found from Ax and Ay:
A
Ax2  Ay2  (60.0 kips)2  (27.5 kips)2  66.0019 kips

(a) The cross-sectional area of the 1.5-in.-diameter tie rod (1) is
A1 
4
(1.5 in.) 2  1.7671 in.2
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.
The normal stress in tie rod (1) is:
F 73.5272 kips
1  1 
 41.6079 ksi  41.6 ksi
A1
1.7671 in.2
(b) The factor of safety for tie rod (1) is

60 ksi
 1.442
FS1  Y 
 1 41.6079 ksi
Ans.
Ans.
(c) The allowable shear stress is:
80 ksi

 26.6667 ksi
 allow  U 
FS
3.0
The shear area required for pin A is thus
A 66.0019 kips

 2.4751 in.2
AV 
 ult 26.6667 ksi
Since pin A is in a double shear connection, the minimum required pin area is
AV
2.4751 in.2
Apin 

 1.2375 in.2
2 shear surfaces
2
Thus, the diameter of pin A must be

4
d A2  1.2375 in.2
 d A  1.255 in.
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.
4.15 Beam AB is supported as shown in Fig. P4.15. Tie
rod (1) is attached at B and C with double shear pin
connections while the pin at A is attached with a single
shear connection. The pins at A, B, and C each have an
ultimate shear strength of 54 ksi, and tie rod (1) has a
yield strength of 36 ksi. A concentrated load of P = 16
kips is applied to the beam as shown. A factor of safety
of 3.0 is required for all components. Determine:
(a) the minimum required diameter for tie rod (1).
(b) the minimum required diameter for the double shear
pins at B and C.
(c) the minimum required diameter for the single shear
pin at A.
Fig. P4.15
Solution
Tie rod (1) is a two-force member that is oriented at 
with respect to the horizontal axis:
8.5 ft
 0.7083
  35.3112
tan  
12 ft
From a FBD of beam AB, the following equilibrium
equations can be written:
Fx  Ax  F1 cos35.3112  0
(a)
Fy  Ay  F1 sin 35.3112  P  0
(b)
 10 in. 
M A  ( F1 cos35.3112) 
 12 in./ft 
( F1 sin 35.3112)(12 ft)  P(8 ft)  0
(c)
From Eq. (c):
F1 
(16 kips)(8 ft)
 16.8062 kips
(cos35.3112)  0.833333 ft   (sin35.3112)(12 ft)
Substitute F1 into Eq. (a) to obtain Ax:
Ax   F1 cos35.3112  (16.8062 kips)cos35.3112  13.7143 kips
and substitute F1 into Eq. (b) to obtain Ay:
Ay  P  F1 sin 35.3112  16 kips  (16.8062 kips)sin 35.3112  6.2857 kips
The resultant pin force at A is found from Ax and Ay:
A
Ax2  Ay2  (13.7143 kips)2  (6.2857 kips)2  15.0861 kips
(a) The allowable normal stress for tie rod (1) is

36 ksi
 12.0 ksi
 allow  Y 
FS
3.0
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.
The minimum cross-sectional area required for the tie rod is
F
16.8062 kips
 1.400518 in.2
Amin  1 
 allow
12 ksi
Therefore, the minimum tie rod diameter is

4
d12  1.400518 in.2
 d1  1.335363 in.  1.335 in.
Ans.
(b) The allowable shear stress for the pins at A, B, and C is

54 ksi
 allow  U 
 18 ksi
FS
3.0
The double-shear pin connection at B and C must support a load of F1 =16.8062 kips . The shear area
AV required for these pins is
16.8062 kips
F
AV  1 
 0.933679 in.2
18 ksi
 allow
The pin diameter can be computed from

2
(2 surfaces) d pin
 0.933679 in.2
4
 d pin  0.7710 in.  0.771 in.
Ans.
(c) The pin at A is a single shear connection; therefore, AV = Apin. The shear area AV required for this pin
is
A
15.0861 kips
AV 

 0.838119 in.2
18 ksi
 allow
The pin diameter can be computed from

2
(1 surface) d pin
 0.838119 in.2
4
 d pin  1.0330 in.  1.033 in.
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.
4.16 In Fig. P4.16, the rigid member ABDE is supported at A by a
single shear pin connection and at B by a tie rod (1). The tie rod is
attached at B and C with double shear pin connections. The pins at
A, B, and C each have an ultimate shear strength of 80 ksi, and tie
rod (1) has a yield strength of 60 ksi. A concentrated load of
P = 24 kips is applied perpendicular to DE, as shown. A factor of
safety of 2.0 is required for all components. Determine:
(a) the minimum required diameter for the tie rod.
(b) the minimum required diameter for the pin at B.
(c) the minimum required diameter for the pin at A.
Fig. P4.16
Solution
Member (1) is a two-force member that is oriented at  with respect to
the horizontal axis:
9 ft
tan  
 1.5
  56.310
6 ft
From a FBD of rigid structure ABDE, the following equilibrium
equations can be written:
Fx  P cos 70  F1 cos 56.310  Ax  0
(a)
Fy   P sin 70  F1 sin 56.310  Ay  0
(b)
M A  ( F1 cos 56.310)(9 ft)  ( P cos 70)[13 ft  (8 ft) sin 20]
( P sin 70)[(8 ft) cos 20]  0
(c)
From Eq. (c):
(cos70)[13 ft  (8 ft)sin 20]  (sin 70)[(8 ft)cos 20]
F1  P
(cos56.310)(9 ft)
 (24 kips)
5.382084 ft  7.064178 ft
 59.834286 kips
4.992293 ft
(d)
Substitute F1 into Eq. (a) to obtain Ax:
Ax  F1 cos56.310  P cos 70
 (59.834286 kips) cos56.310  (24 kips) cos 70
 24.981548 kips
and substitute F1 into Eq. (b) to obtain Ay:
Ay  F1 sin 56.310  P sin 70
 (59.834286 kips)sin 56.310  (24 kips)sin 70
 72.337797 kips
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.
The resultant pin force at A is found from Ax and Ay:
A  Ax2  Ay2  (24.981548 kips)2  (72.337797 kips)2  76.529959 kips
(e)
(a) The allowable normal stress for tie rod (1) is

60 ksi
 30.0 ksi
 allow  Y 
FS
2.0
The minimum cross-sectional area required for the tie rod is
F
59.834286 kips
Amin  1 
 1.994476 in.2
30.0 ksi
 allow
Therefore, the minimum tie rod diameter is

4
d12  1.994476 in.2
 d1  1.593564 in.  1.594 in.
Ans.
(b) The allowable shear stress for the pins at A, B, and C is

80 ksi
 40.0 ksi
 allow  U 
FS
2.0
The double-shear pin connection at B and C must support a load of F1 = 59.834286 kips . The shear
area AV required for these pins is
F
59.834286 kips
AV  1 
 1.495857 in.2
 allow
40.0 ksi
The pin diameter can be computed from

2
(2 surfaces) d pin
 1.495857 in.2
4
 d pin  0.975855 in.  0.976 in.
Ans.
(c) The pin at A is a single shear connection; therefore, AV = Apin. The shear area AV required for this pin
is
A
76.529959 kips

 1.913249 in.2
AV 
 allow
40.0 ksi
The pin diameter can be computed from

2
(1 surface) d pin
 1.913249 in.2
4
 d pin  1.560777 in.  1.561 in.
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.
4.17 Rigid bar ABC is subjected to a concentrated load P as
shown in Fig. P4.17. Inclined member (1) has a cross
sectional area of A1 = 2.250 in.2 and is connected at ends B
and D by 1.00-in.-diameter pins in double shear
connections. The rigid bar is supported at C by a 1.00-in.diameter pin in a single shear connection. The yield
strength of inclined member (1) is 36 ksi, and the ultimate
strength of each pin is 60 ksi. For inclined member (1), the
minimum factor of safety with respect to the yield strength
is FSmin = 1.5. For the pin connections, the minimum factor
of safety with respect to the ultimate strength is FSmin = 2.0.
Determine the maximum load P that can be supported by
the structure.
Fig. P4.17
Solution
Member (1) is a two-force member that is oriented at  with
respect to the horizontal axis:
40 in.
 0.8   38.6598
tan  
50 in.
From a FBD of rigid bar ABC, the following equilibrium
equations can be written:
Fx  F1 cos38.6598  Cx  0
(a)
Fy   P  F1 sin 38.6598  C y  0
(b)
M C  (80 in.)P  ( F1 sin 38.6598)(50 in.)  0
(c)
From Eq. (c), express F1 in terms of the unknown load P:
(80 in.)P
F1  
 2.561250 P
(50 in.)sin 38.6598
Substitute this result into Eq. (a) to express Cx in terms of P:
Cx  F1 cos38.6598  ( 2.561250 P)cos38.6598  2.0 P
And similarly, express Cy in terms of P from Eq. (b):
C y  P  F1 sin 38.6598  P  ( 2.561250 P)sin 38.6598  0.60 P
(d)
The resultant reaction force at pin C can now be expressed as a function of P:
C  Cx2  C y2  (2.0 P)2  (0.60 P)2  2.088061P
(e)
Inclined member (1):
The allowable normal stress for inclined member (1) is

36 ksi
 24 ksi
 allow  Y 
FS
1.5
Therefore, the allowable axial force for inclined member (1) is
Fallow,1   allow A1  (24 ksi)(2.25 in.2 )  54 kips
From Eq. (d), the maximum load that may be applied to the rigid bar based on the limitations on the
inclined member normal stress is
Fallow,1
54 kips
Pmax 

 21.083455 kips
(f)
2.561250 2.561250
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.
Pin B:
The allowable shear stress for the pins is

60 ksi
 30 ksi
 allow  U 
FS
2.0
A 1.0-in.-diameter pin has a cross-sectional area of
Apin 

(1.0 in.) 2  0.785398 in.2
4
For a double shear connection, the corresponding shear area is
AV  2 Apin  2(0.785398 in.2 )  1.570796 in.2
Thus, the maximum allowable force in member (1) based on the pin at B is:
Fallow,1   allow AV  (30 ksi)(1.570796 in.2 )  47.123880 kips
From Eq. (d), the maximum load that may be applied to the rigid bar based on the limitations on the pin
at B is
Fallow,1
47.123880 kips
Pmax 

 18.398782 kips
(g)
2.561250
2.561250
Pin C:
The pin at C is in a single shear connection; therefore, the shear area of pin C is equal to the crosssectional area of the 1.0-in.-diameter pin. The maximum allowable shear force in this pin is thus
Fallow,C   allow AV  (30 ksi)(0.785398 in.2 )  23.561940 kips
From Eq. (e), the maximum load that may be applied to the rigid bar based on the limitations on the pin
shear stress is
C allow
23.561940 kips
Pmax 

 11.284125 kips
(h)
2.088061
2.088061
Maximum load P:
Comparing the results in Eqs. (f), (g), and (h), the maximum load P that may be applied to the rigid bar
is
Ans.
Pmax  11.284125 kips  11.28 kips
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.
4.18 Rigid bar ABC is supported by pin-connected axial member (1) and by a pin connection at C as
shown in Fig. P4.18. A 6,300-lb concentrated load is applied to the rigid bar at A. Member (1) is a 2.75
in. wide by 1.25 in. thick rectangular bar made of steel with a yield strength of Y = 36,000 psi. The pin
at C has an ultimate shear strength of U = 60,000 psi.
(a) Determine the axial force in
member (1).
(b) Determine the factor of safety in
member (1) with respect to its yield
strength.
(c) Determine the magnitude of the
resultant reaction force acting at pin C.
(d) If a minimum factor of safety of FS
= 3.0 with respect to the ultimate shear
strength is required, determine the
minimum diameter that may be used
for the pin at C.
Fig. P4.18
Solution
Member (1) is a two-force member that is oriented at 
with respect to the horizontal axis:
30 in.
tan  
 0.75
  36.870
40 in.
From a FBD of rigid structure ABC, the following
equilibrium equations can be written:
Fx   F1 cos 36.870  Cx  0
(a)
Fy  6,300 lb  F1 sin 36.870  C y  0
(b)
M C  (6,300 lb)(80 in.)
(c)
( F1 sin 36.870)(56 in.)  0
(a) From Eq. (c), the axial force in member (1) is:
(6,300 lb)(80 in.)
F1  
 15,000 lb  15,000 lb (C)
(56 in.)sin 36.870
(b) The normal stress magnitude in member (1) is
F
15,000 lb
1  1 
 4,363.6364 psi
A1 (2.75 in.)(1.25 in.)
Therefore, its factor of safety with respect to the 36,000 psi yield stress is

36,000 psi
FS1  Y 
 8.25
 1 4,363.6364 psi
Ans.
Ans.
(c) Substitute F1 into Eq. (a) to obtain Cx:
Cx  F1 cos36.870  (15, 000 lb) cos36.870  12, 000 lb
and substitute F1 into Eq. (b) to obtain an expression for Ay in terms of the unknown load P:
C y  6,300 lb  F1 sin 36.870  6,300 lb  (15, 000 lb)sin 36.870  2, 700 lb
The resultant pin force at C is found from Cx and Cy:
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.
C  Cx2  C y2  ( 12,000 lb)2  ( 2,700 lb)2  12,300 lb
Ans.
(d) The allowable shear stress for the pins at C is

60,000 psi
 20,000 psi
 allow  U 
FS
3.0
The double-shear pin connection at C must support a load of 12,300 lb. The minimum shear area AV
required to support this load is
C
12,300 lb
AV 

 0.6150 in.2
 allow 20,000 psi
The pin diameter can be computed from

2
(2 surfaces) d pin
 0.6150 in.2
4
 d pin  0.625717 in.  0.626 in.
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.
4.19 A rectangular steel plate is used as an axial member to support a dead load of 70 kips and a live
load of 110 kips. The yield strength of the steel is 50 ksi.
(a) Use the ASD method to determine the minimum cross-sectional area required for the axial member if
a factor of safety of 1.67 with respect to yielding is required.
(b) Use the LRFD method to determine the minimum cross-sectional area required for the axial member
based on yielding of the gross section using the LRFD method. Use a resistance factor of t = 0.9 and
load factors of 1.2 and 1.6 for the dead and live loads, respectively.
Solution
(a) The service load on the axial member is
P  D  L  70 kips  110 kips  180 kips
The allowable normal stress is

50 ksi
 29.940 ksi
 allow  Y 
FS 1.67
The minimum cross-sectional area required to support the service load is
P
180 kips

 6.01 in.2
Amin 
 allow 29.940 ksi
(b) The ultimate load for LRFD is
U  1.2D  1.6L  1.2(70 kips)  1.6(110 kips)  260 kips
The design equation for an axial member subjected to tension can be written in LRFD as
t Y A  U
Consequently, the minimum cross-sectional area required for the tension member is
U
260 kips

 5.78 in.2
Amin 
t Y 0.9(50 ksi)
Ans.
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.
4.20 A 20-mm-thick steel plate will be used as an axial member to support a dead load of 150 kN and a
live load of 220 kN. The yield strength of the steel is 250 MPa.
(a) Use the ASD method to determine the minimum plate width b required for the axial member if a
factor of safety of 1.67 with respect to yielding is required.
(b) Use the LRFD method to determine the minimum plate width b required for the axial member based
on yielding of the gross section using the LRFD method. Use a resistance factor of t = 0.9 and load
factors of 1.2 and 1.6 for the dead and live loads, respectively.
Solution
(a) The service load on the axial member is
P  D  L  150 kN  220 kN  370 kN
The allowable normal stress is

250 MPa
 allow  Y 
 149.70 MPa
FS
1.67
The minimum cross-sectional area required to support the service load is
(370 kN)(1,000 N/kN)
P
Amin 

 2, 471.61 mm 2
 allow
149.70 N/mm 2
Since the plate is 20-mm-thick, the minimum plate width is therefore:
A
2, 471.61 mm 2
bmin  min 
 123.6 mm
20 mm
t
(b) The ultimate load for LRFD is
U  1.2D  1.6L  1.2(150 kN)  1.6(220 kN)  532 kN
The design equation for an axial member subjected to tension can be written in LRFD as
t Y A  U
Consequently, the minimum cross-sectional area required for the tension member is
(532 kN)(1,000 N/kN)
U
Amin 

 2,364.44 mm 2
2
0.9(250 N/mm )
t Y
Since the plate is 20-mm-thick, the minimum plate width is therefore:
A
2,364.44 mm 2
bmin  min 
 118.2 mm
t
20 mm
Ans.
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.
4.21 A round steel tie rod is used as a tension member to support a dead load of 30 kips and a live load
of 15 kips. The yield strength of the steel is 46 ksi.
(a) Use the ASD method to determine the minimum diameter required for the tie rod if a factor of safety
of 2.0 with respect to yielding is required.
(b) Use the LRFD method to determine the minimum diameter required for the tie rod based on yielding
of the gross section using the LRFD method. Use a resistance factor of t = 0.9 and load factors of 1.2
and 1.6 for the dead and live loads, respectively.
Solution
(a) The service load on the axial member is
P  D  L  30 kips  15 kips  45 kips
The allowable normal stress is

46 ksi
 23.0 ksi
 allow  Y 
FS
2.0
The minimum cross-sectional area required to support the service load is
45 kips
P

 1.956522 in.2
Amin 
 allow 23.0 ksi
and thus, the minimum tie rod diameter is

4
2
d min
 1.956522 in.2
 d min  1.578 in.
Ans.
(b) The ultimate load for LRFD is
U  1.2D  1.6L  1.2(30 kips)  1.6(15 kips)  60 kips
The design equation for an axial member subjected to tension can be written in LRFD as
t Y A  U
Consequently, the minimum cross-sectional area required for the tension member is
60 kips
U
Amin 

 1.449275 in.2
t Y 0.9(46 ksi)
and thus, the minimum tie rod diameter is

4
2
d min
 1.449275 in.2
 d min  1.358 in.
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.
4.22 A round steel tie rod is used as a tension member to support a dead load of 190 kN and a live load
of 220 kN. The yield strength of the steel is 320 MPa.
(a) Use the ASD method to determine the minimum diameter D required for the tie rod if a factor of
safety of 2.0 with respect to yielding is required.
(b) Use the LRFD method to determine the minimum diameter D required for the tie rod based on
yielding of the gross section using the LRFD method. Use a resistance factor of t = 0.9 and load factors
of 1.2 and 1.6 for the dead and live loads, respectively.
Solution
(a) The service load on the axial member is
P  D  L  190 kN  220 kN  410 kN
The allowable normal stress is

320 MPa
 allow  Y 
 160 MPa
FS
2.0
The minimum cross-sectional area required to support the service load is
(410 kN)(1,000 N/kN)
P
Amin 

 2,562.500 mm 2
 allow
160 N/mm 2
and thus, the minimum tie rod diameter is

4
2
d min
 2,562.500 mm 2
 d min  57.1 mm
Ans.
(b) The ultimate load for LRFD is
U  1.2D  1.6L  1.2(190 kN)  1.6(220 kN)  580 kN
The design equation for an axial member subjected to tension can be written in LRFD as
t Y A  U
Consequently, the minimum cross-sectional area required for the tension member is
U
(580 kN)(1,000 N/kN)
Amin 

 2, 013.889 mm 2
t Y
0.9(320 N/mm 2 )
and thus, the minimum tie rod diameter is

4
2
d min
 2,013.889 mm2
 d min  50.6 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.
5.1 A steel [E = 200 GPa] rod with a circular cross section is 7.5-m long. Determine the minimum
diameter required if the rod must transmit a tensile force of 50 kN without exceeding an allowable stress
of 180 MPa or stretching more than 5 mm.
Solution
If the normal stress in the rod cannot exceed 180 MPa, the cross-sectional area must equal or exceed
P (50 kN)(1,000 N/kN)
 277.7778 mm 2
A 
2
180 N/mm

If the elongation must not exceed 5 mm, the cross-sectional area must equal or exceed
PL (50 kN)(1,000 N/kN)(7,500 mm)
A

 375.0000 mm 2
2
E
(200,000 N/mm )(5 mm)
Therefore, the minimum cross-sectional area that may be used for the rod is Amin = 375 mm2. The
corresponding rod diameter is

4
2
d rod
 375 mm2
 d rod  21.8510 mm  21.9 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.
5.2 An aluminum [E = 10,000 ksi] control rod with a circular cross section must not stretch more than
0.25 in. when the tension in the rod is 2,200 lb. If the maximum allowable normal stress in the rod is 12
ksi, determine:
(a) the smallest diameter that can be used for the rod.
(b) the corresponding maximum length of the rod.
Solution
(a) If the normal stress in the rod cannot exceed 20 ksi, the cross-sectional area must equal or exceed
P 2.20 kips
 0.1833 in.2
A 
12 ksi


The corresponding rod diameter is
4
2
d rod
 0.1833 in.2
 d rod  0.483 in.
Ans.
(b) If the elongation must not exceed 0.25 in., the aluminum control rod having a cross-sectional area of
A = 0.1833 in.2 can have a length no greater than
AE (0.1833 in.2 )(10,000 ksi)(0.25 in.)

 208.3333 in.  208 in.
L
Ans.
2.20 kips
P
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.
5.3 A 12-mm-diameter steel [E = 200 GPa] rod (2)
is connected to a 30-mm-wide by 8-mm-thick
rectangular aluminum [E = 70 GPa] bar (1), as
shown in Fig. P5.3. Determine the force P required
to stretch the assembly 10.0 mm.
Fig. P5.3
Solution
The elongations in the two axial members are expressed by
FL
FL
1  1 1
and
2  2 2
A1 E1
A2 E2
The total elongation of the assembly is thus
FL F L
uC  1   2  1 1  2 2
A1 E1 A2 E2
Since the internal forces F1 and F2 are equal to external load P, this expression can be simplified to
 L
L 
uC  P  1  2 
 A1 E1 A2 E2 
For rectangular aluminum bar (1), the cross-sectional area is
A1  (30 mm)(8 mm)  240 mm 2
and the cross-sectional area of steel rod (2) is
A2 

4
(12 mm)2  113.0973 mm2
The force P required to stretch the assembly 10.0 mm is thus
uC
P
L1
L
 2
A1E1 A2 E2

10.0 mm


450 mm
1,300 mm
 (240 mm 2 )(70,000 N/mm 2 )  (113.09739 mm 2 )(200, 000 N/mm 2 ) 


 118,682.6 N
= 118.7 kN
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.
5.4 Two polymer bars are connected to
a rigid plate at B, as shown in Fig. P5.4.
Bar (1) has a cross-sectional area of
1.65 in.2 and an elastic modulus of
2,400 ksi. Bar (2) has a cross-sectional
area of 0.975 in.2 and an elastic modulus
of 4,000 ksi. Determine the total
deformation of the bar.
Fig. P5.4
Solution
Draw a FBD that cuts through member (1) to find
that the internal axial force in member (1) is
F1  8 kips (T)
Similarly, draw a FBD that cuts through member (2)
to find that the internal axial force in member (2) is
F2  14 kips (T)
The elongation in bar (1) can be computed as
FL
(8 kips)(20 in.)
1  1 1 
 0.040404 in.
A1 E1 (1.65 in.2 )(2, 400 ksi)
and the elongation in bar (2) can be computed as
FL
(14 kips)(30 in.)
2  2 2 
 0.107692 in.
A2 E2 (0.975 in.2 )(4, 000 ksi)
The total elongation of the bar is thus
1  2  0.040404 in.  0.107692 in.  0.1481 in.
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.
5.5 An axial member consisting of two polymer
bars is supported at C as shown in Fig. P5.5. Bar
(1) has a cross-sectional area of 540 mm2 and an
elastic modulus of 28 GPa. Bar (2) has a crosssectional area of 880 mm2 and an elastic
modulus of 16.5 GPa. Determine the deflection
of point A relative to support C.
Fig. P5.5
Solution
Draw a FBD that cuts through member (1) to find that the internal
axial force in member (1) is
F1  35 kN (T)
Similarly, draw a FBD that cuts through member
(2) and includes the free end of the axial member.
From this FBD, the equilibrium equation is
Fx  35 kN  50 kN  50 kN  F2  0
Therefore, the internal axial force in member (2) is
F2  65 kN  65 kN (C)
The deformation in bar (1) can be computed as
F L (35 kN)(1,000 N/kN)(850 mm)
1  1 1 
 1.9676 mm
A1 E1
(540 mm 2 )(28, 000 N/mm 2 )
and the deformation in bar (2) can be computed as
FL
(65 kN)(1,000 N/kN)(1,150 mm)
 5.1481 mm
2  2 2 
A2 E2
(880 mm 2 )(16,500 N/mm 2 )
The deflection of point A relative to the support at C is the sum of these two deformations:
uA  1  2  1.9676 mm  (5.1481 mm)  3.18 mm  3.18 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.
5.6 The roof and second floor of a building are supported
by the column shown in Fig. P5.6. The column is a
structural steel W10 × 60 wide-flange section [E = 29,000
ksi; A = 17.6 in2]. The roof and floor subject the column to
the axial forces shown. Determine:
(a) the amount that the first floor will settle.
(b) the amount that the roof will settle.
Fig. P5.6
Solution
(a) Draw a FBD that cuts through member (1) and includes the free
end of the column. From this FBD, the sum of forces in the
vertical direction can be written as
Fy  110 kips  155 kips  F1  0
Therefore, the internal axial force in member (1) is
F1  270 kips  270 kips (C)
The settlement of the first floor is determined by the deformation
(i.e., contraction in this case) that occurs in member (1).
FL
u B  1  1 1
A1 E1

(270 kips)(16 ft)(12 in./ft)
(17.6 in.2 )(29, 000 ksi)
 0.101567 in.  0.1016 in. 
Ans.
(b) Draw a FBD that cuts through member (2) and includes the free
end of the column. From this FBD, the equilibrium equation is
Fx  115 kips  F2  0
Therefore, the internal axial force in member (2) is
F2  115 kips  115 kips (C)
The deformation in member (2) (which will be contraction in this
instance) can be computed as
F2 L2 (115 kips)(14 ft)(12 in./ft)

 0.037853 in.
A2 E2
(17.6 in.2 )(29, 000 ksi)
The settlement of the roof is found from the sum of the contractions in the two members:
2 
uC  1   2  0.101567 in.  (0.037853 in.)  0.139420 in.  0.1394 in. 
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.
5.7 Aluminum [E = 70 GPa] member ABC supports a
load of 28 kN, as shown in Fig. P5.7. Determine:
(a) the value of load P such that the deflection of joint C
is zero.
(b) the corresponding deflection of joint B.
Fig. P5.7
Solution
Cut a FBD that exposes the internal axial force in member (2):
Fy  28 kN  F2  0
 F2  28 kN
Similarly, cut a FBD that exposes the internal axial force in
member (1):
Fy  28 kN  P  F1  0
 F1  28 kN  P
The deflection at joint C, which must ultimately equal zero, can be
expressed in terms of the member deformations 1 and 2:
FL F L
uC  1   2  1 1  2 2  0
A1 E1 A2 E2
or
FL
FL
uC  1 1  2 2
A1E1 A2 E2

(28,000 N  P)(1,000 mm)

4
2
2
(50 mm) (70,000 N/mm )


4
(28,000 N)(1,300 mm)
2
2
0
(32 mm) (70,000 N/mm )
The only unknown in this equation is P, which can be computed as
P  80.5841 kN  80.6 kN
The corresponding deflection at joint B can be found from the deformation in member (1):
F L (28, 000 N  80,584 N)(1,300 mm)
uB  1  1 1 
 0.497 mm  0.497 mm 

A1 E1
2
2
(50 mm) (70, 000 N/mm )
4
Ans.
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.
5.8 A solid brass [E = 100 GPa] axial member is loaded and
supported as shown in Fig. P5.8. Segments (1) and (2) each have a
diameter of 25 mm and segment (3) has a diameter of 14 mm.
Determine:
(a) the deformation of segment (2).
(b) the deflection of joint D with respect to the fixed support at A.
(c) the maximum normal stress in the entire axial member.
Fig. P5.8
Solution
(a) Draw a FBD that cuts through segment (2) and includes the free
end of the axial member. From this FBD, the sum of forces in the
vertical direction reveals the internal force in the segment:
Fy  F2  14 kN  28 kN  0
 F2  42 kN (T)
The cross-sectional area of the 25-mm-diameter segment is

A2  (25 mm) 2  490.8739 mm 2
4
The deformation in segment (2) is thus
FL
(42,000 N)(1,200 mm)
2  2 2 
A2 E2 (490.8739 mm 2 )(100, 000 N/mm 2 )
 1.026740 mm  1.027 mm
Ans.
(b) The internal forces in segments (1) and (3) must be determined at
the outset. From a FBD that cuts through segment (1) and includes
the free end of the axial member:
Fy  F1  40 kN  14 kN  28 kN  0
 F1  82 kN (T)
The deformation in segment (1) can be computed as
FL
(82,000 N)(1,800 mm)
1  1 1 
 3.006883 mm
A1 E1 (490.8739 mm 2 )(100, 000 N/mm 2 )
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.
Similarly, consider a FBD that cuts through segment (3) and includes the free end of
the axial member:
Fy  F3  28 kN  0
 F3  28 kN (T)
The cross-sectional area of the 14-mm-diameter segment is
A3 

(164 mm)2  153.9380 mm 2
4
The deformation in segment (3) can be computed as
FL
(28,000 N)(1,600 mm)
3  3 3 
 2.810262 mm
A3 E3 (153.9380 mm 2 )(100, 000 N/mm 2 )
The deflection of joint D with respect to the fixed support at A is found from the sum of the three
segment deformations:
u D  1   2   3
 3.006883 mm  1.026740 mm  2.810262 mm
 6.943885 mm  6.94 mm
Ans.
(c) Since segments (1) and (2) have the same cross-sectional area, the maximum normal stress in these
two segments occurs where the axial force is greater; that is, in segment (1):
F
82,000 N
1  1 
 167.0490 MPa (T)
A1 490.8739 mm 2
The normal stress in segment (3) is
F
28,000 N
3  3 
 181.8914 MPa (T)
A3 153.9380 mm 2
Therefore, the maximum normal stress in the axial member occurs in segment (3):
Ans.
 max  181.9 MPa (T)
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.
5.9 A hollow steel [E = 30,000 ksi] tube (1) with
an outside diameter of 2.75 in. and a wall
thickness of 0.25 in. is fastened to a solid
aluminum [E = 10,000 ksi] rod (2) that has a 2in.-diameter and a solid 1.375-in.-diameter
aluminum rod (3). The bar is loaded as shown in
Fig. P5.9. Determine:
(a) the change in length of steel tube (1).
(b) the deflection of joint D with respect to the
fixed support at A.
(c) the maximum normal stress in the entire axial
assembly.
Fig. P5.9
Solution
Before proceeding, it is convenient to determine the internal forces in each of the three axial segments.
Segment (1): Draw a FBD that cuts through
segment (1) and includes the free end of the axial
member. From this FBD, the sum of forces in the
horizontal direction gives the force in segment (1):
Fx   F1  2(34 kips)  2(18 kips)  25 kips  0
 F1  57 kips
Segment (2): Draw a FBD that cuts through
segment (2) and includes the free end of the axial
member. From this FBD, the sum of forces in the
horizontal direction gives the force in segment (2):
Fx   F2  2(18 kips)  25 kips  0
 F2  11 kips
Segment (3): Draw a FBD that cuts through
segment (3) and includes the free end of the axial
member. From this FBD, the sum of forces in the
horizontal direction gives the force in segment (3):
Fx   F3  25 kips  0
 F3  25 kips

(a) The cross-sectional area of the hollow steel tube (inside diameter = 2.25 in.) is
(2.75 in.)2  (2.25 in.) 2   1.963495 in.2
4
The deformation in segment (1) is thus
FL
(57 kips)(60 in.)
1  1 1 
 0.058060 in.  0.0581 in.
A1 E1 (1.963495 in.2 )(30, 000 ksi)
A1 

Ans.
(b) For segment (2), the cross-sectional area of the 2-in.-diameter aluminum rod is
A2 
(2 in.)2  3.141593 in.2
4
The deformation in segment (2) can be computed as
FL
(11 kips)(40 in.)
2  2 2 
 0.014006 in.
A2 E2 (3.141593 in.2 )(10, 000 ksi)
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.

For segment (3), the cross-sectional area of the 1.375-in.-diameter aluminum rod is
A3 
(1.375 in.) 2  1.484893 in.2
4
The deformation in segment (3) can be computed as
FL
(25 kips)(30 in.)
3  3 3 
 0.050509 in.
A3 E3 (1.484893 in.2 )(10, 000 ksi)
The deflection of joint D with respect to the fixed support at A is found from the sum of the three
segment deformations:
uD  1   2   3  0.058060 in.  0.014006 in.  0.050509 in.
 0.094563 in.  0.0946 in.  0.0946 in. 
Ans.
(c) Compute the normal stress in each of the three segments:
57 kips
F
1  1 
 29.0299 ksi (C)
A1 1.963495 in.2
2 
3 
11 kips
F2

 3.5014 ksi (T)
A2 3.141593 in.2
25 kips
F3

 16.8362 ksi (C)
A3 1.484893 in.2
Therefore, the maximum normal stress in the axial member occurs in segment (1):
 max  29.0 ksi (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.
5.10 A solid 5/8-in. steel [E = 29,000 ksi] rod
(1) supports beam AB, as shown in Fig. P5.10.
If the stress in the rod must not exceed 30 ksi
and the maximum deformation in the rod must
not exceed 0.25 in., determine the maximum
load P that may be supported.
Fig. P5.10
Solution

The area of the 5/8-in.-diameter rod is
A1 
(0.625 in.)2  0.306796 in.2
4
If the stress in the rod must not exceed 30 ksi, then the maximum force that can be applied to rod (1) is
F1   allow,1 A1  (30 ksi)(0.306796 in.2 )  9.203880 kips
The length of rod (1) is
L1  (20 ft)2  (16 ft)2  25.612497 ft  307.35 in.
If the maximum deformation in the rod must not exceed 0.25 in., the maximum internal force in the rod
must be limited to
1 A1E1 (0.25 in.)(0.306796 in.2 )(29,000 ksi)

 7.236932 kips
F1 
307.35 in.
L1
Therefore, the maximum internal force in rod (1) must not exceed 7.237 kips.
Consider a FBD of the rigid beam. Rod (1) is a twoforce member at an orientation  defined by:
16 ft
tan  
 0.8
  38.660
20 ft
Write the equilibrium equation for the sum of moments
about A to find the relationship between F1 and P:
M A  (20 ft)(F1 sin 38.660)  (8 ft)P  0
P 
(20 ft)(7.236932 kips)sin 38.660
 11.302236 kips  11.30 kips
8 ft
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.
5.11 A 1-in.-diameter by 16-ft-long cold-rolled bronze bar [E = 15,000 ksi and  = 0.320 lb/in3] hangs
vertically while suspended from one end. Determine the change in length of the bar due to its own
weight.
Solution
An incremental length dy of the bar has an incremental deformation given by
F ( y)
d 
dy
AE
The force in the bar can be expressed as the product of the unit density of the bronze (bronze) and the
volume of the bar below the incremental slice dy:
F  y    bronze A y
Therefore, the incremental deformation can be expressed as


Ay
y
d  bronze
dy  bronze dy
AE
E
Integrate this expression over the entire length L of the bar:
L
L
L
y



L2
1
   bronze dy  bronze  y dy  bronze  y 2  0  bronze
0
E
E 0
E 2
2E
The change in length of the bar due to its own weight is therefore:
L2 (0.320 lb/in.3 )(16 ft 12 in./ft) 2

  bronze 
 0.000393216 in.  0.000393 in.
2E
2(15, 000, 000 psi)
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.
5.12 A homogenous rod of length L and elastic modulus E is a truncated cone with diameter that varies
linearly from d0 at one end to 2d0 at the other end. A concentrated axial load P is applied to the ends of
the rod, as shown in Fig. P5.12. Assume that the taper of the cone is slight enough for the assumption of
a uniform axial stress distribution over a cross section to be valid.
(a) Determine an expression for the stress distribution on an arbitrary cross section at x.
(b) Determine an expression for the elongation of the rod.
P5.12
Solution
(a) Determine an expression for the diameter of the
truncated cone through the use of the similar triangles
concept:
2d 0  d 0 d ( x )  d 0

L
x
x

(a)
 d ( x)  d 0 1  

L
Next, consider equilibrium of the truncated cone:
 F ( x)  P
 Fx  F ( x)  P  0
F ( x)    dA   ( x) A( x)
The internal force F(x) can be expressed in terms of stress and area:
A
Equating this expression to the equilibrium equation gives:
F ( x)   ( x) A( x)  P
  ( x) 
P
A( x)
From Eq. (a), the area of the cone at any location x is expressed as
x

 d 1  
2

 d ( x)
L
A( x) 

4
4
Therefore, the stress can be written as
P
4P
 ( x) 

2
A( x)
x

 d 02 1  

L
2
2
0
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 elongation in the axial member is found from Eq. (5.5):
L
L F ( x)
dx
   d  
0
0 A( x ) E
For the truncated cone axial member:
L
4P
4 P L dx


dx
2
2
0
 Ed 02 0 
x
x

 Ed 02 1  
1


L
L
4P   L 
2 PL


2 

x
 Ed 0 1 
 Ed 02



L 0
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.
5.13 Determine the extension, due to its own weight, of
the conical bar shown in Fig. P5.13. The bar is made of
aluminum alloy [E = 10,600 ksi and  = 0.100 lb/in.3].
The bar has a 2-in. radius at its upper end and a length of
L = 20 ft. Assume the taper of the bar is slight enough for
the assumption of a uniform axial stress distribution over
a cross section to be valid.
Fig. P5.13
Solution
An incremental length dy of the bar has an incremental deformation given by
F ( y)
d 
dy
AE
The force in the bar can be expressed as the product of the unit density of the aluminum (alum) and the
volume of the bar below the incremental slice dy. The volume below the slice dy is a cone. At y, the
cross-sectional area of the base of the cone is:
y 
Ay    r 
L 
and the volume of a cone is given by:
1
V  (area of base)(altitude)
3
The internal force at y can be expressed as
1
F ( y )   alum  Ay y
3
The incremental deformation can be expressed as
 alum Ay y
 y
d 
dy  alum dy
3 Ay E
3E
Integrate this expression over the entire length L of the bar:
L
L
L
y


 L2
1
   alum dy  alum  y dx  alum  y 2  0  alum
0
3E
3E 0
3E 2
6E
2
The change in length of the bar due to its own weight is therefore:
 L2 (0.100 lb/in.3 )(20 ft 12 in./ft) 2
 9.0566  105 in.  90.6 106 in.
  alum 
6E
6(10, 600, 000 psi)
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.
5.14 Rigid bar ABCD is loaded and supported as
shown in Fig. P5.14. Steel [E = 30,000 ksi] bars
(1) and (2) are unstressed before the load P is
applied. Bar (1) has a cross-sectional area of
0.75 in.2 and bar (2) has a cross-sectional area of
0.425 in.2. After load P is applied, the strain in
bar (1) is found to be 780 . Determine:
(a) the stresses in bars (1) and (2).
(b) the vertical deflection of point D.
(c) the load P.
Fig. P5.14
Solution
From the strain in bar (1), the deformation in bar (1) is
(780 10 6 in./in.)(60 in.) 0.0468 in.
1
1 L1
Since the joint at B is a perfect connection, the rigid bar deflection at B must equal the deformation of
bar (1):
vB
0.0468 in.
1
From a deformation diagram of the rigid bar,
the vertical deflection of joint C is related to B
by similar triangles:
vB
vC
40 in. 80 in.
80 in.
vC vB
2vB
40 in.
2(0.0468 in.) 0.0936 in.
The joint at C is also a perfect connection; therefore, the downward displacement of C also causes an
equal deformation in bar (2):
vC 0.0936 in.
2
(a) Now that the deformations in both bars are known, the normal stresses in each can be computed.
The normal stress in bar (1) is
F1L1
(0.0468 in.)(30,000 ksi)
1 L1
1 E1
23.4 ksi (T)
Ans.
1
1
A1E1
E1
L1
60 in.
and the normal stress in bar (2) is
F2 L2
(0.0936 in.)(30,000 ksi)
2 L2
2 E2
31.2 ksi (T)
Ans.
2
2
A2 E2
E2
L2
90 in.
Alternatively, the stress in (1) could be calculated from Hooke’s Law:
E1 1 (30,000 ksi)(780 10 6 in./in.) 23.4 ksi (T)
1
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) From a deformation diagram of the rigid bar, the vertical deflection of joint D is related to B and C
by similar triangles:
vB
vC
vD
40 in. 80 in. 110 in.
110 in.
vD vB
2.75(0.0468 in.) 0.1287 in.
Ans.
40 in.
(c) The force in each bar can be determined from the stresses computed previously:
F1
(23.4 ksi)(0.75 in.2 ) 17.55 kips
1 A1
F2
2
A2
(31.2 ksi)(0.425 in.2 ) 13.26 kips
Consider a FBD of the rigid bar. The
equilibrium equation for the sum of moments
about A can be used to determine the load P:
M A (40 in.)F1 (80 in.)F2 (110 in.)P
P
0
(40 in.)(17.55 kips) (80 in.)(13.26 kips)
110 in.
16.0255 kips
16.03 kips
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.
5.15 Rigid bar ABCD is loaded and supported as
shown in Fig. P5.15. Bars (1) and (2) are
unstressed before the load P is applied. Bar (1) is
made of bronze [E = 100 GPa] and has a crosssectional area of 520 mm2. Bar (2) is made of
aluminum [E = 70 GPa] and has a cross-sectional
area of 960 mm2. After the load P is applied, the
force in bar (2) is found to be 25 kN (in tension).
Determine:
(a) the stresses in bars (1) and (2).
(b) the vertical deflection of point A.
(c) the load P.
Fig. P5.15
Solution
Given that the axial force in bar (2) is 25 kN (in tension), the deformation can be computed as:
F2 L2
(25, 000 N)(800 mm)
0.297619 mm
2
A2 E2 (960 mm 2 )(70, 000 N/mm 2 )
Since the pin at C is a perfect connection, the deflection of the rigid bar at C is equal to the deformation
of bar (2):
vC
0.297619 mm
2
From a deformation diagram of the rigid bar, the
vertical deflection of joint C is related to B by
similar triangles:
vB
vC
1.6 m 0.5 m
1.6 m
vB vC
3.2vC
0.5 m
3.2(0.297619 mm) 1.116071 mm
The joint at B is also a perfect connection; therefore, the downward displacement of B also causes an
equal contraction in bar (1):
vB 1.116071 mm
1
Note that a downward displacement at B causes contraction (and hence, compression) in bar (1).
(a) The normal stress in aluminum bar (2) can be computed from the known force in the bar:
F2 25,000 N
26.041667 MPa 26.0 MPa (T)
2
A2 960 mm 2
The normal stress in bronze bar (1) can be computed from its contraction:
F1L1
1 L1
1
A1E1
E1
E
L1
1 1
1
( 1.116071 mm)(100,000 N/mm 2 )
600 mm
186.011905 MPa
186.0 MPa (C)
Ans.
Ans.
(b) From a deformation diagram of the rigid bar, the vertical deflection of joint A is related to joint B by
similar triangles:
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.
vA
2.0 m
vC
0.5 m
vA
vC
2.0 m
0.5 m
4vC
4(0.297619 mm) 1.488095 mm
1.488 mm
Ans.
(c) The force in bar (2) is given as F2 = 25 kN. The force in bar (1) can be determined from the stress
computed previously:
F1
( 186.011905 N/mm 2 )(520 mm 2 )
96,726.1905 N
96.7262 kN
1 A1
Consider a FBD of the rigid bar. The
equilibrium equation for the sum of moments
about D can be used to determine the load P:
M D (1.6 m)F1 (0.5 m)F2 (2.0 m)P 0
P
(0.5 m)(25 kN) (1.6 m)( 96.7262 kN)
2.0 m
77.5446 kN 77.5 kN
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.
5.16 In Fig. P5.16, aluminum [E = 70
GPa] links (1) and (2) support rigid beam
ABC. Link (1) has a cross-sectional area
of 300 mm2 and link (2) has a crosssectional area of 450 mm2. For an
applied load of P = 55 kN, determine the
rigid beam deflection at point B.
Fig. P5.16
Solution
From a FBD of the rigid beam, write two equilibrium
equations:
Fy F1 F2 55 kN 0
(a)
M A (2, 200 mm)F2 (1, 400 mm)(55 kN)
Solve Eq. (b) for F2:
(1, 400 mm)(55 kN)
F2
35 kN
2, 200 mm
and backsubstitute into Eq. (a) to obtain F1:
F1 55 kN F2 55 kN 35 kN 20 kN
0 (b)
Next, determine the deformations in links (1) and (2):
F1L1
(20,000 N)(2,500 mm)
2.380952 mm
1
A1E1 (300 mm 2 )(70,000 N/mm 2 )
F2 L2
(35,000 N)(4,000 mm)
4.444444 mm
2
A2 E2 (450 mm 2 )(70,000 N/mm 2 )
Since the connections at A and C are perfect, the rigid beam deflections at these joints are equal to the
deformations of links (1) and (2), respectively:
vA
2.380952 mm
and
vC
4.444444 mm
1
2
The deflection at B can be determined
from similar triangles:
vC v A
vB v A
2, 200 mm 1, 400 mm
Solve this expression for vB:
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.
vB
1, 400 mm
(vC v A ) v A
2,200 mm
1, 400 mm
(4.444444 mm 2.380952 mm) 2.380952 mm
2,200 mm
(0.636364)(2.053492 mm) 2.380952 mm
= 3.69 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.
5.17 Rigid bar ABC is supported by bronze rod (1) and aluminum
rod (2), as shown in Fig P5-17. A concentrated load P is applied
to the free end of aluminum rod (3). Bronze rod (1) has an elastic
modulus of E1 = 15,000 ksi and a diameter of d1 = 0.50 in.
Aluminum rod (2) has an elastic modulus of E2 = 10,000 ksi and
a diameter of d2 = 0.75 in. Aluminum rod (3) has a diameter of d3
= 1.0 in. The yield strength of the bronze is 48 ksi and the yield
strength of the aluminum is 40 ksi.
(a) Determine the magnitude of load P that can safely be applied
to the structure if a minimum factor of safety of 1.67 is required.
(b) Determine the deflection of point D for the load determined
in part (a).
(c) The pin used at B has an ultimate shear strength of 54 ksi. If a
factor of safety of 3.0 is required for this double shear pin
connection, determine the minimum pin diameter that can be
used at B.
Fig. P5.17
Solution
Before beginning, the cross-sectional areas of the three rods can be calculated from the specified
diameters:
A1 0.196350 in.2
A2 0.441786 in.2
A3 0.785398 in.2
The allowable stress of the bronze is
48 ksi
Y ,bronze
28.742515 ksi
allow,1
FS
1.67
and the allowable stress of the aluminum is
40 ksi
Y ,alum
23.952096 ksi
allow,2
allow,3
FS
1.67
(a) From a FBD cut through rod (3), equilibrium requires that the internal force in rod (3) is F3 = P.
From a FBD of the rigid bar, write two equilibrium equations:
Fy F1 F2 F3 F1 F2 P 0
(a)
M A (4 ft)F2 (2.5 ft)F3 (4 ft)F2 (2.5 ft)P 0
(b)
Solve Eq. (b) for F2 in terms of the unknown load P:
2.5 ft
(c)
F2
P 0.6250 P
4 ft
Backsubstitute this expression into Eq. (a) to obtain F1 in terms of
the unknown load P:
F1 P F2 P 0.6250 P 0.3750 P
(d)
Rearrange Eqs. (c) and (d) to express P in terms of F2 and F1, respectively:
P 1.6000 F2
P 2.6667 F1
(e)
(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.
Based on the allowable stress for the bronze, the maximum allowable force that can be supported by rod
(1) is:
F1
(28.742515 ksi)(0.196350 in.2 ) 5.643579 kips
allow,1 A1
Thus, from Eq. (f), the corresponding maximum load P is:
P 2.6667 F1 2.6667(5.643579 kips) 15.049545 kips
(g)
Next, the maximum allowable force that can be supported by rod (2) based on the allowable stress for
the aluminum is:
F2
(23.952096 ksi)(0.441786 in.2 ) 10.581711 kips
allow,2 A2
which leads to a corresponding maximum load P from Eq. (e):
P 1.6000 F2 1.6000(10.581711 kips) 16.930738 kips
(h)
Finally, we should also check the capacity of rod (3):
F3
(23.952096 ksi)(0.785398 in.2 ) 18.811931 kips
allow,3 A3
which means that
P 18.811931 kips
(i)
From a comparison of Eqs. (g), (h), and (i), the maximum load that may be applied to the structure is
Ans.
Pmax 15.049545 kips 15.05 kips
Based on this maximum load P, the internal forces in the three rods are:
F1 0.3750 P 0.3750(15.049545 kips) 5.643579 kips
F2
0.6250 P
0.6250(15.049545 kips)
F3
P 15.049545 kips
9.405966 kips
(b) Next, determine the deformations in rods (1), (2), and (3):
F1L1 (5.643579 kips)(6 ft)(12 in./ft)
0.137964 in.
1
A1E1
(0.196350 in.2 )(15,000 ksi)
2
F2 L2
A2 E2
(9.405966 kips)(8 ft)(12 in./ft)
(0.441786 in.2 )(10,000 ksi)
0.204391 in.
3
F3 L3
A3 E3
(15.049545 kips)(3 ft)(12 in./ft)
(0.785398 in.2 )(10,000 ksi)
0.068982 in.
Since the connections at A and C are perfect, the rigid bar deflections at these joints are equal to the
deformations of rods (1) and (2), respectively:
vA
0.137964 in.
and
vC
0.204391 in.
1
2
The rigid bar deflection at B can be
determined from similar triangles:
vC vA vB vA
4 ft
2.5 ft
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.
Solve this expression for vB:
2.5 ft
(vC vA ) vA
vB
4 ft
2.5 ft
(0.204391 in. 0.137964 in.) 0.137964 in.
4 ft
(0.6250)(0.066427 in.) 0.137964 in.
0.179481 in.
The deflection of joint D is equal to the deflection of the rigid bar at B plus the deformation in rod (3):
vD
vB
0.179481 in. 0.068982 in. 0.248463 in.
3
0.248 in.
Ans.
(c) The pin at B has an allowable shear stress of
54 ksi
U
18 ksi
allow
FS
3.0
The force tending to shear this pin is equal to the load P = 15.049545 kips. The shear area AV required
for this pin is thus
V
15.049545 kips
AV
0.836086 in.2
18
ksi
allow
Since the pin is used in a double shear connection, the shear area is equal to twice the cross-sectional
area of the pin:
AV
2 Apin
2
2
d pin
2
2
d pin
4
and so the minimum pin diameter is
4
0.836086 in.2
d pin
0.729568 in.
0.730 in.
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.
5.18 Solve problem 5.14 when there is a
clearance of 0.05 in. in the pin connection at C.
5.14 (Repeated here for convenience). Rigid
bar ABCD is loaded and supported as shown in
Fig. P5.14. Steel [E = 30,000 ksi] bars (1) and (2)
are unstressed before the load P is applied. Bar
(1) has a cross-sectional area of 0.75 in.2 and bar
(2) has a cross-sectional area of 0.425 in.2. After
load P is applied, the strain in bar (1) is found to
be 780 . Determine:
(a) the stresses in bars (1) and (2).
(b) the vertical deflection of point D.
(c) the load P.
Fig. P5.14 (repeated)
Solution
From the strain in bar (1), the deformation in bar (1) is
(780 10 6 in./in.)(60 in.) 0.0468 in.
1
1 L1
Since the joint at B is a perfect connection, the rigid bar deflection at B must equal the deformation of
bar (1):
vB
0.0468 in.
1
From a deformation diagram of the rigid bar,
the vertical deflection of joint C is related to B
by similar triangles:
vB
vC
40 in. 80 in.
80 in.
vC vB
2vB
40 in.
2(0.0468 in.) 0.0936 in.
There is a 0.05-in. clearance in the connection at C; therefore,
vC
0.05 in.
2
2
vC
0.05 in. 0.0936 in. 0.05 in. 0.0436 in.
(a) Now that the deformations in both bars are known, the normal stresses in each can be computed.
The normal stress in bar (1) is
F1L1
(0.0468 in.)(30,000 ksi)
1 L1
1 E1
23.4 ksi (T)
Ans.
1
1
A1E1
E1
L1
60 in.
and the normal stress in bar (2) is
F2 L2
(0.0436 in.)(30,000 ksi)
2 L2
2 E2
14.5333 ksi 14.53 ksi (T)
Ans.
2
2
90 in.
A2 E2
E2
L2
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) From a deformation diagram of the rigid bar, the vertical deflection of joint D is related to B and C
by similar triangles:
vB
vC
vD
40 in. 80 in. 110 in.
110 in.
vD vB
2.75(0.0468 in.) 0.1287 in.
Ans.
40 in.
(c) The force in each bar can be determined from the stresses computed previously:
F1
(23.4 ksi)(0.75 in.2 ) 17.55 kips
1 A1
F2
2
A2
(14.5333 ksi)(0.425 in.2 )
6.1767 kips
Consider a FBD of the rigid bar. The
equilibrium equation for the sum of moments
about A can be used to determine the load P:
M A (40 in.)F1 (80 in.)F2 (110 in.)P
P
0
(40 in.)(17.55 kips) (80 in.)(6.1767 kips)
110 in.
10.8739 kips
10.87 kips
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.
5.19 The rigid beam in Fig. P5.19 is supported by
links (1) and (2), which are made from a polymer
material [E = 16 GPa]. Link (1) has a crosssectional area of 400 mm2 and link (2) has a crosssectional area of 800 mm2. Determine the
maximum load P that may by applied if the
deflection of the rigid beam is not to exceed 20
mm at point C.
Fig. P5.19
Solution
Equilibrium: Consider a FBD of the rigid beam
and assume tension in each link.
Fy
F1 F2 P 0
(a)
M A (600 mm)F2 (900 mm)P
From Eq. (b):
900 mm
F2
P 1.5P
600 mm
Backsubstituting into Eq. (a):
F1 F2 P 1.5P P 0.5P
0
(b)
(c)
(d)
Force-deformation relationships: The relationship between internal force and member deformation
for links (1) and (2) can be expressed as:
F1L1
F2 L2
and 2
(e)
1
A1E1
A2 E2
Geometry of deformations:
Consider a deformation diagram of
the rigid beam. Using similar
triangles, one way to express the
relationships between vA, vB, and vC
is:
v A vC
vC vB
(f)
900 mm 300 mm
Note: Here, vA, vB, and vC are
treated as unsigned magnitudes in
the directions shown on the
deformation diagram.
The rigid beam deflection at A will equal the deformation that occurs in link (1):
vA
1
and similarly at B:
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.
vB
2
Equation (f) can now be rewritten in terms of e1 and e2 as:
vC
vC
1
2
900 mm 300 mm
or
900 mm
vC
(vC
3vC 3 2
1
2)
300 mm
Solving for vC gives:
2vC
3 2
vC 0.5 1 1.5 2
1
Substitute the force-deformation relationships from Eq. (e) to obtain:
0.5 F1L1 1.5 F2 L2
vC 0.5 1 1.5 2
A1E1
A2 E2
and then substitute Eqs. (c) and (d) to derive an expression for vC in terms of the unknown load P:
0.5(0.5P) L1 1.5(1.5P) L2
0.25L1 2.25L2
vC
P
A1 E1
A2 E2
A1 E1
A2 E2
The unknown load P is thus related to the rigid beam deflection at C by:
vC
P
0.25L1 2.25 L2
A1 E1
A2 E2
Substituting the appropriate values into this relationship gives the maximum load P that may be applied
to the rigid beam at C without causing more than 20 mm deflection:
20 mm
P
(0.25)(1,000 mm)
(2.25)(1, 250 mm)
2
2
(400 mm )(16,000 N/mm ) (800 mm 2 )(16,000 N/mm 2 )
77, 283 N
77.3 kN
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.
5.20 Three aluminum [E = 10,000 ksi] bars are
used to support the loads shown in Fig. P5.20. The
elongation in each bar must be limited to 0.25 in.
Determine the minimum cross-sectional area
required for each bar.
Fig. P5.20
Solution
Determine the inclination angles for bars (1), (2), and (3):
7 ft
tan 1
34.992
1
10 ft
3 ft
tan 2
18.435
2
9 ft
8 ft
tan 3
57.995
1
5 ft
The bar lengths are:
L1
(7 ft)2 (10 ft) 2
12.206556 ft 146.4787 in.
L2
(3 ft)2 (9 ft) 2
9.486833 ft 113.8420 in.
L3
(8 ft) 2 (5 ft) 2
9.433981 ft 113.2078 in.
Equilibrium: Consider a FBD of joint B and write two equilibrium
equations:
Fx
F1 cos34.992 F2 cos18.435 0
Fy
F1 sin 34.992
F2 sin18.435
31 kips
0
Solve these two equations simultaneously to obtain F1 = 36.61967 kips
and F2 = 31.62278 kips.
Next, consider a FBD of joint C and write two additional equilibrium
equations:
Fx
F2 cos18.435 F3 cos57.995 0
Fy
F2 sin18.435
F3 sin 57.995
38 kips
0
From this, the internal force in bar (3) is F3 = 56.60389 kips.
The minimum cross-sectional area required for each bar is:
F1L1 (36.61967 kips)(146.4787 in.)
A1
2.15 in.2
(0.25 in.)(10,000 ksi)
e1E1
A2
F2 L2
e2 E2
(31.62278 kips)(113.8420 in.)
(0.25 in.)(10,000 ksi)
1.440 in.2
Ans.
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.
A3
F3 L3
e3 E3
(56.60389 kips)(113.2078 in.)
(0.25 in.)(10,000 ksi)
2.56 in.2
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.
5.21 A tie rod (1) and a pipe strut (2) are used to
support an 80-kip load as shown in Fig. P5.21. Pipe
strut (2) has an outside diameter of 6.625 in. and a
wall thickness of 0.280 in. Both the tie rod and the
pipe strut are made of structural steel with a modulus
of elasticity of E = 29,000 ksi and a yield strength of
Y = 36 ksi. For the tie rod, the minimum factor of
safety with respect to yield is 1.5 and the maximum
allowable axial elongation is 0.20 in.
(a) Determine the minimum diameter required to
satisfy both constraints for tie rod (1).
(b) Draw a deformation diagram showing the final
position of joint B.
Fig. P5.21
Solution
The angles of inclination for members (1) and (2) are:
12 ft
tan 1
26.565
1
24 ft
30 ft
tan 2
51.340
2
24 ft
The member lengths are:
L1
(12 ft)2 (24 ft) 2
26.8328 ft
321.9938 in.
L2
(30 ft)2 (24 ft) 2
38.4187 ft
461.0249 in.
(a) Equilibrium: Consider a FBD of joint B and write two equilibrium
equations:
Fx
F1 cos 26.565 F2 cos51.340 0
Fy
F1 sin 26.565
F2 sin 51.340
80 kips
0
Solve these two equations simultaneously to obtain F1 = 51.1103 kips and
F2 = −73.1786 kips.
The allowable normal stress for tie rod (1) is
36 ksi
Y
24 ksi
allow
FS
1.5
The minimum cross-sectional area required to satisfy the normal stress requirement is
F1
51.1103 kips
A
2.1296 in.2
24 ksi
allow
The maximum allowable axial deformation for the tie rod is 0.20 in. The minimum cross-sectional area
required to satisfy the deformation requirement is
F1L1 (51.1103 kips)(321.9938 in.)
A
2.8374 in.2
(0.20 in.)(29,000 ksi)
1 E1
To satisfy both the stress and deformation requirements, the tie rod must have a minimum crosssectional area of A = 2.8374 in.2. The corresponding rod diameter 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.
4
2
d rod
2.8374 in.2
d rod
1.901 in.
Ans.
(b) The pipe has a cross-sectional area of A2 = 5.5814 in.2. The pipe deformation is
F2 L2 ( 73.1786 kips)(461.0249 in.)
0.2084 in.
2
A2 E2
(5.5814 in.2 )(29,000 ksi)
Rod (1) elongates 0.20 in. and pipe (2) contracts
0.2084 in. Therefore, the deformation diagram
showing the final position of joint B is shown.
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.
5.22 Two axial members are used to support a load of P = 72
kips as shown in Fig. P5.22. Member (1) is 12-ft long, it has a
cross-sectional area of A1 = 1.75 in.2, and it is made of structural
steel [E = 29,000 ksi]. Member (2) is 16-ft long, it has a crosssectional area of A2 = 4.50 in.2, and it is made of an aluminum
alloy [E = 10,000 ksi].
(a) Compute the normal stress in each axial member.
(b) Compute the deformation of each axial member.
(c) Draw a deformation diagram showing the final position of
joint B.
(d) Compute the horizontal and vertical displacements of joint B.
Fig. P5.22
Solution
(a) Consider a FBD of joint B and write two equilibrium equations:
Fx
F1 F2 cos55 0
Fy
F2 sin 55
72 kips
0
Solve these two equations simultaneously to obtain F1 = 50.4149 kips
and F2 = 87.8958 kips.
The normal stress in member (1) is:
F1 50.4149 kips
28.8 ksi
1
1.75 in.2
A1
and the normal stress in member (2) is:
F2 87.8958 kips
19.53 ksi
2
4.50 in.2
A2
(b) The member deformations are
F1L1 (50.4149 kips)(12 ft)(12 in./ft)
1
A1E1
(1.75 in.2 )(29,000 ksi)
2
F2 L2
A2 E2
(87.8958 kips)(16 ft)(12 in./ft)
(4.50 in.2 )(10,000 ksi)
Ans.
Ans.
0.1430 in.
Ans.
0.375 in.
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.
(c) Member (1) elongates 0.1430 in.
and member (2) elongates 0.375 in.
Therefore, the deformation diagram
showing the final position of joint B is
shown.
(d) The horizontal displacement of B is
x 0.1430 in.
The vertical displacement is found from
0.375 in. 0.1430cos55
sin 55
y
0.375 in. 0.1430cos55
y
sin 55
Ans.
0.375 in. 0.0820 in.
sin 55
0.558 in.
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.
5.23 The 200 × 200 × 1,200-mm oak [E = 12 GPa] block (2)
shown in Fig. P5.23 was reinforced by bolting two 6 × 200 × 1,200
mm steel [E = 200 GPa] plates (1) to opposite sides of the block. A
concentrated load of 360 kN is applied to a rigid cap. Determine:
(a) the normal stresses in the steel plates (1) and the oak block (2).
(b) the shortening of the block when the load is applied.
Fig. P5.23
Solution
Equilibrium
Consider a FBD of the rigid cap. Sum forces in the vertical direction to obtain:
Fy
2 F1 F2 360 kN 0
(a)
Geometry of Deformations Relationship
For this configuration, the deformations of both members will be equal;
therefore,
1
2
(b)
Force-Deformation Relationships
The relationship between the internal force and the deformation of an axial member can be stated for
members (1) and (2):
F1L1
F2 L2
(c)
1
2
A1E1
A2 E2
Compatibility Equation
Substitute the force-deformation relationships (c) into the geometry of deformation relationship (b) to
derive the compatibility equation:
F1 L1 F2 L2
(d)
A1 E1 A2 E2
Solve the Equations
Solve Eq. (d) for F2, recognizing that steel plates (1) and oak block (2) have the same length:
L AE
L A E
A E
F2 F1 1 2 2 F1 1 2 2 F1 2 2
A1 E1 L2
L2 A1 E1
A1 E1
Substitute this expression into equilibrium equation (a) and solve for F1:
A E
A2 E2
2 F1 F2
2 F1 F1 2 2
360 kN
Fy
F1 2
A1 E1
A1 E1
(e)
(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.
For this structure, the areas and elastic moduli are given below:
A1 (6 mm)(200 mm) 1, 200 mm2
A2 (200 mm)2
40,000 mm2
E1 200 GPa
E2 12 GPa
Substitute these values into Eq. (f) and calculate F1 = −90 kN. Backsubstitute into Eq. (e) to calculate
F2 = −180 kN.
Normal Stresses
The normal stresses in each axial member can now be calculated:
F1 ( 90 kN)(1,000 N/kN)
75 MPa 75 MPa (C)
1
1,200 mm 2
A1
2
F2
A2
( 180 kN)(1,000 N/kN)
40,000 mm 2
4.50 MPa
4.50 MPa (C)
Ans.
Ans.
(b) The shortening of the block is equal to the deformation (i.e., contraction in this instance) of member
(2):
( 180,000 N)(1,200 mm)
F2 L2
0.450 mm 0.450 mm
uB
Ans.
2
A2 E2 (40,000 mm 2 )(12,000 N/mm 2 )
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.
5.24 Two identical steel [E = 200 GPa] pipes, each with a
cross-sectional area of 1,475 mm2, are attached to
unyielding supports at the top and bottom, as shown in
Fig. P5.24. At flange B, a concentrated downward load of
120 kN is applied. Determine:
(a) the normal stresses in the upper and lower pipes.
(b) the deflection of flange B.
Fig. P5.24
Solution
(a) Equilibrium: Consider a FBD of flange B. Sum forces in the vertical
direction to obtain:
Fy F1 F2 120 kN 0
(a)
Geometry of Deformations:
0
1
2
(b)
Force-Deformation Relationships:
F1L1
F2 L2
1
2
A1E1
A2 E2
(c)
Compatibility Equation: Substitute Eqs. (c) into Eq. (b) to derive the compatibility equation:
F1 L1 F2 L2
0
A1 E1 A2 E2
Solve the Equations: Solve Eq. (d) for F2:
L AE
L A E
F2
F1 1 2 2
F1 1 2 2
A1 E1 L2
L2 A1 E1
and substitute this expression into Eq. (a) to determine F1:
L A E
L A E
Fy F1 F2 F1
F1 1 2 2
F1 1 1 2 2
L2 A1 E1
L2 A1 E1
(d)
(e)
120 kN
(f)
Notice that A1 = A2 and E1 = E2 for this structure. Therefore, F1 can be computed as:
120 kN
120 kN
120 kN
66.268657 kN
F1
3.0 m 1.810811
L1 A2 E2
1
1+
L2 A1 E1
3.7 m
and from Eq. (e), F2 has a value of
L A E
3.0 m
F2
F1 1 2 2
(66.268657 kN)
L2 A1 E1
3.7 m
53.731343 kN
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.
Normal Stresses: The normal stresses in each axial member can now be calculated:
F1 66.268657 N
44.927903 MPa 44.9 MPa (T)
1
1,475 mm 2
A1
2
F2
A2
53,731.343 N
1,475 mm 2
36.428029 MPa
36.4 MPa (C)
Ans.
Ans.
(b) The deflection of flange B is equal to the deformation (i.e., contraction in this instance) of member
(2):
F2 L2
( 53.731343 N)(3,700 mm)
uB
0.673919 mm 0.674 mm
Ans.
2
A2 E2 (1,475 mm 2 )(200,000 N/mm 2 )
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.
5.25 Solve Problem 5.24 if the lower support in Fig. P5.24 yields
and displaces downward 1.0 mm as the load P is applied.
5.24 (Repeated here for convenience) Two identical steel [E =
200 GPa] pipes, each with a cross-sectional area of 1,475 mm2,
are attached to unyielding supports at the top and bottom, as
shown in Fig. P5.24. At flange B, a concentrated downward load
of 120 kN is applied. Determine:
(a) the normal stresses in the upper and lower pipes.
(b) the deflection of flange B.
Fig. P5.24 (repeated)
Solution
(a) Equilibrium: Consider a FBD of flange B. Sum forces in the vertical
direction to obtain:
Fy F1 F2 120 kN 0
(a)
Geometry of Deformations:
1.0 mm
1
2
(b)
Force-Deformation Relationships:
F1L1
F2 L2
1
2
A1E1
A2 E2
(c)
Compatibility Equation: Substitute Eqs. (c) into Eq. (b) to derive the compatibility equation:
F1L1 F2 L2
1.0 mm
A1E1 A2 E2
Solve the Equations: Solve Eq. (d) for F2:
F1L1 A2 E2 (1.0 mm)A2 E2
F2 1.0 mm
A1E1 L2
L2
F1
L1 A2 E2
L2 A1 E1
and substitute this expression into Eq. (a) to determine F1:
(1.0 mm)A2 E2
L A E
120 kN
Fy F1 F2 F1
F1 1 2 2
L2
L2 A1 E1
(d)
(e)
(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.
Simplify this expression, gathering terms with F1 on the left-hand side of the equation:
(1.0 mm)A2 E2
L A E
120 kN
F1
F1 1 2 2
L2
L2 A1 E1
F1 1
L1 A2 E2
L2 A1 E1
120 kN
(1.0 mm)A2 E2
L2
Notice that A1 = A2 and E1 = E2 for this structure. Therefore, F1 can be computed as:
(1.0 mm)(1,475 mm 2 )(200,000 N/mm 2 )
120,000 N
199,729.7297 N
3,700 mm
110, 298.5075 N
F1
3,000 mm
1.810811
1
3,700 mm
and from Eq. (a), F2 has a value of
F2 F1 120 kN 110, 298.507 N 120,000 N
9,701.493 N
Normal Stresses: The normal stresses in each axial member can now be calculated:
F1 110, 298.507 N
74.778649 MPa 74.8 MPa (T)
1
1,475 mm 2
A1
2
F2
A2
9,701.493 N
1,475 mm 2
6.577283 MPa
6.58 MPa (C)
(b) The downward deflection of flange B is equal to the deformation of member (1):
F1L1
(110, 298.507 N)(3,000 mm)
uB
1.121680 mm 1.122 mm
1
A1E1 (1,475 mm 2 )(200,000 N/mm 2 )
Ans.
Ans.
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.
5.26 A load P is supported by a structure consisting of
rigid bar ABC, two identical solid bronze [E = 15,000
ksi] rods, and a solid steel [E = 30,000 ksi] rod, as
shown in Fig. P5.26. The bronze rods (1) each have a
diameter of 0.75 in. and they are symmetrically
positioned relative to the center rod (2) and the
applied load P. Steel rod (2) has a diameter of 0.50 in.
If all bars are unstressed before the load P is applied,
determine the normal stresses in the bronze and steel
rods after a load of P = 20 kips is applied.
Fig. P5.26
Solution
Equilibrium
By virtue of symmetry, the forces in the two bronze rods (1) are
identical. Consider a FBD of the rigid bar. Sum forces in the
vertical direction to obtain:
Fy 2 F1 F2 P 0
(a)
Geometry of Deformations Relationship
For this configuration, the deformations of all rods will be equal;
therefore,
(b)
1
2
Force-Deformation Relationships
The relationship between the internal force and the deformation of an axial member can be stated for
members (1) and (2):
F1L1
F2 L2
(c)
1
2
A1E1
A2 E2
Compatibility Equation
Substitute the force-deformation relationships (c) into the geometry of deformation relationship (b) to
derive the compatibility equation:
F1 L1 F2 L2
(d)
A1 E1 A2 E2
Solve the Equations
Solve Eq. (d) for F2:
L AE
F2 F1 1 2 2
A1 E1 L2
F1
L1 A2 E2
L2 A1 E1
Substitute this expression into equilibrium equation (a) and solve for F1:
L A E
L A E
Fy 2 F1 F2 2 F1 F1 1 2 2 F1 2 1 2 2
P
L2 A1 E1
L2 A1 E1
(e)
(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.
For this structure, P = 20 kips, and the lengths, areas, and elastic moduli are given below:
L1 50 in.
L2 90 in.
(0.75 in.) 2 0.441786 in.2
(0.50 in.) 2 0.196350 in.2
A2
4
4
E1 15,000 ksi
E2 30,000 ksi
Substitute these values into Eq. (f) and calculate F1 = 8.019802 kips. Backsubstitute into Eq. (e) to
calculate F2 = 3.960396 kips.
A1
Normal Stresses
The normal stresses in each axial member can now be calculated:
F1 8.019802 kips
18.15 ksi
1
A1 0.441786 in.2
2
F2
A2
3.960396 kips
0.196350 in.2
20.2 ksi
Ans.
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.
5.27 An aluminum alloy [E = 10,000 ksi] pipe
with a cross-sectional area of A1 = 4.50 in.2 is
connected at flange B to a steel [E = 30,000 ksi]
pipe with a cross-sectional area of A2 = 3.20 in.2.
The assembly (shown in Fig. P5.27) is connected
to rigid supports at A and C. For the loading
shown, determine:
(a) the normal stresses in aluminum pipe (1) and
steel pipe (2).
(b) the deflection of flange B.
Fig. P5.27
Solution
(a) Equilibrium: Consider a FBD of flange B. Sum forces
in the horizontal direction to obtain:
Fx
F1 F2 90 kips 0
(a)
Geometry of Deformations:
0
1
2
(b)
Force-Deformation Relationships:
F1L1
F2 L2
1
2
A1E1
A2 E2
(c)
Compatibility Equation: Substitute Eqs. (c) into Eq. (b) to derive the compatibility equation:
F1 L1 F2 L2
0
A1 E1 A2 E2
(d)
Solve the Equations: Solve Eq. (d) for F2:
L AE
L A E
F2
F1 1 2 2
F1 1 2 2
A1 E1 L2
L2 A1 E1
and substitute this expression into Eq. (a) to determine F1:
L A E
L A E
Fx
F1 F2
F1
F1 1 2 2
F1 1 1 2 2
90 kips
L2 A1 E1
L2 A1 E1
F1 can be computed as:
90 kips
90 kips
90 kips
35.273159 kips
F1
2
L1 A2 E2
2.551515
160
in.
3.20
in.
30,000
ksi
1
1+
L2 A1 E1
220 in. 4.50 in.2 10,000 ksi
and from Eq. (a), F2 has a value of
F2 F1 90 kips 35.273159 kips 90 kips
54.726841 kips
Normal Stresses: The normal stresses in each axial member can now be calculated:
F1 35.273159 kips
7.838480 ksi 7.84 ksi (T)
1
4.50 in.2
A1
(e)
(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.
2
F2
A2
54.726841 kips
3.20 in.2
17.102138 ksi
17.10 ksi (C)
(b) The deflection of flange B is equal to the deformation of member (1):
F1L1 (35.273159 kips)(160 in.)
uB
0.125416 in. 0.1254 in.
1
A1E1
(4.50 in.2 )(10,000 ksi)
Ans.
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.
5.28 The concrete [E = 29 GPa] pier shown in
Fig. P5.28 is reinforced using four steel [E =
200 GPa] reinforcing rods, each having a
diameter of 19 mm. If the pier is subjected to an
axial load of 670 kN, determine:
(a) the normal stress in the concrete and in the
steel reinforcing rods.
(b) the shortening of the pier.
Fig. P5.28
Solution
(a) Equilibrium: Consider a FBD of the pier, cut around the upper end.
The concrete will be designated member (1) and the reinforcing steel bars
will be designated member (2). Sum forces in the vertical direction to
obtain:
Fy
F1 F2 670 kN 0
(a)
Geometry of Deformations:
1
2
Force-Deformation Relationships:
F1L1
F2 L2
1
2
A1E1
A2 E2
(b)
(c)
Compatibility Equation: Substitute Eqs. (c) into Eq. (b) to derive the compatibility equation:
F1 L1 F2 L2
A1 E1 A2 E2
Solve the Equations: Solve Eq. (d) for F2:
L AE
L A E
F2 F1 1 2 2 F1 1 2 2
A1 E1 L2
L2 A1 E1
and substitute this expression into Eq. (a) to determine F1:
L A E
L A E
Fy
F1 F2
F1 F1 1 2 2
F1 1 1 2 2
L2 A1 E1
L2 A1 E1
(d)
(e)
670 kN
(f)
Since L1 = L2, the term L1/L2 = 1 can be eliminated and Eq. (f) simplified to
A E
F1 1 2 2
670 kN
(g)
A1 E1
The gross cross-sectional area of the pier is (250 mm)2 = 62,500 mm2; however, the reinforcing bars
take up a portion of this area. The cross-sectional area of four 19-mm-diameter reinforcing bars 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.
A2
(4 bars)
(19 mm) 2
1,134.115 mm 2
4
and thus the area of the concrete is
A1 62,500 mm 2 1,134.115 mm 2
F1 can now be computed as:
1,134.115 mm 2 200 GPa
F1 1
61,365.885 mm 2 29 GPa
61,365.885 mm 2
670 kN
F1
670 kN
1.127457
and from Eq. (a), F2 has a value of
F2
F1 670 kN
( 594.248 kN) 670 kN
594.258 kN
75.742 kN
Normal Stresses: The normal stresses in each axial member can now be calculated. In the concrete, the
normal stress is:
594, 258 N
F1
9.68 MPa 9.68 MPa (C)
Ans.
1
A1 61,365.885 mm 2
and the normal stress in the reinforcing bars is
F2
75,742 N
66.8 MPa 66.8 MPa (C)
Ans.
2
A2 1,134.115 mm 2
(b) The shortening of the 1.5-m-long pier is equal to the deformation (i.e., contraction in this instance) of
member (1) or member (2):
F1L1
( 594, 258 N)(1,500 mm)
u
0.501 mm 0.501 mm
Ans.
1
A1E1 (61,365.885 mm 2 )(29,000 N/mm 2 )
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.
5.29 The concrete [E = 29 GPa] pier
shown in Fig. P5.29 is reinforced using
four steel [E = 200 GPa] reinforcing rods.
If the pier is subjected to an axial force of
670 kN, determine the required diameter D
of each rod so that 20% of the total load is
carried by the steel.
Fig. P5.29
Solution
(a) Equilibrium: Consider a FBD of the pier, cut around the upper end. The
concrete will be designated member (1) and the reinforcing steel bars will be
designated member (2). Sum forces in the vertical direction to obtain:
Fy
F1 F2 670 kN 0
(a)
Geometry of Deformations:
1
(b)
2
Force-Deformation Relationships:
F1L1
F2 L2
1
2
A1E1
A2 E2
(c)
Compatibility Equation: Substitute Eqs. (c) into Eq. (b) to derive the compatibility equation:
F1 L1 F2 L2
A1 E1 A2 E2
Solve the Equations: Solve Eq. (d) for F1:
L AE
L A E
F1 F2 2 1 1 F2 2 1 1
A2 E2 L1
L1 A2 E2
and substitute this expression into Eq. (a) to determine F2:
L A E
L A E
Fy
F1 F2
F2 2 1 1
F2
F2 2 1 2 1
L1 A2 E2
L1 A2 E2
(d)
(e)
670 kN
Since L1 = L2, the term L1/L2 = 1 can be eliminated and Eq. (f) simplified to
A E
F2 1 1 1
670 kN
A2 E2
(f)
(g)
From the problem statement, 20% of the total load must be carried by the reinforcing steel; therefore,
F2 0.20( 670 kN)
134 kN
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.
The gross cross-sectional area of the pier is Agross = (250 mm)2 = 62,500 mm2; however, the reinforcing
bars take up a portion of this area. Therefore, the areas of the concrete (1) and steel (2) are related by:
Agross A1 A2
and thus, the area of the concrete can be expressed as A1 = Agross − A2. Equation (g) can now be restated
and solved for the area of the reinforcing steel bars:
62,500 mm 2 A2 29 GPa
( 134 kN)
1
670 kN
A2
200 GPa
62,500 mm 2
A2
A2
27.586207
A2
2,186.369 mm 2
Knowing the area required from four bars, the diameter of each bar can be computed:
2
(4 bars) d bar
4
2,186.369 mm 2
d bar
26.4 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.
5.30 A load of P = 100 kN is supported by a structure
consisting of rigid bar ABC, two identical solid bronze
[E = 100 GPa] rods, and a solid steel [E = 200 GPa]
rod, as shown in Fig. P5.30. The bronze rods (1) each
have a diameter of 20 mm and they are symmetrically
positioned relative to the center rod (2) and the applied
load P. Steel rod (2) has a diameter of 24 mm. All bars
are unstressed before the load P is applied; however,
there is a 3-mm clearance in the bolted connection at
B. Determine:
(a) the normal stresses in the bronze and steel rods.
(b) the downward deflection of rigid bar ABC.
Fig. P5.30
Solution
(a) Equilibrium: By virtue of symmetry, the forces in the
two bronze rods (1) are identical. Consider a FBD of the rigid
bar. Sum forces in the vertical direction to obtain:
Fy 2 F1 F2 P 0
(a)
Geometry of Deformations: For this configuration, the
deflections of joints A, B, and C are equal:
v A vB vC
(b)
The pin connections at A and C are ideal; therefore, the deflection of joints A and C will cause an
identical deformation of rods (1):
vA
(c)
1
The pin at B has a 3-mm clearance; thus, the deformation of rod (2) is related to rigid bar deflection vB
by:
vB
3 mm
(d)
2
Substitute Eqs. (c) and (d) into Eq. (b) to obtain the geometry of deformation equation:
3 mm
(e)
1
2
Force-Deformation Relationships: The relationship between the internal force and the deformation of
an axial member can be stated for members (1) and (2):
F1L1
F2 L2
(f)
1
2
A1E1
A2 E2
Compatibility Equation: Substitute the force-deformation relationships (f) into the geometry of
deformation relationship (e) to derive the compatibility equation:
F1 L1 F2 L2
3 mm
A1 E1 A2 E2
(g)
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.
Solve the Equations: Solve Eq. (g) for F1:
F2 L2
AE
L A E (3 mm)A1 E1
F1
3 mm 1 1 F2 2 1 1
A2 E2
L1
L1 A2 E2
L1
Substitute this expression into Eq. (a)
L A E (3 mm)A1 E1
Fy 2 F1 F2 2 F2 2 1 1
F2 P
L1 A2 E2
L1
and derive an expression for F2:
L A E
(3 mm)A1 E1
F2 2 2 1 1 1 P 2
L1 A2 E2
L1
(3 mm)A1 E1
L1
L A E
2 2 1 1 1
L1 A2 E2
P 2
F2
(h)
For this structure, P = 100 kN = 100,000 N, and the lengths, areas, and elastic moduli are given below:
L1 3,000 mm
L2 1,500 mm
A1
(20 mm)2
314.159265 mm2
A2
(24 mm)2
452.389342 mm2
4
4
E1 100,000 MPa
E2 200,000 MPa
Substitute these values into Eq. (h) and calculate F2 = 27,588.728 N. Backsubstitute into Eq. (a) to
calculate F1 = 36,205.636 N.
Normal Stresses: The normal stresses in each rod can now be calculated:
F1 36, 205.636 N
115.2 MPa
1
A1 314.1593 mm 2
2
F2
A2
27,588.728 N
452.3893 mm 2
61.0 MPa
(b) The downward deflection of the rigid bar can be determined from the deformation of rods (1):
F1L1
(36, 205.636 N)(3,000 mm)
v A vB vC
3.46 mm
1
A1E1 (314.1593 mm 2 )(100,000 MPa)
Ans.
Ans.
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.
5.31 Two steel [E = 30,000 ksi] pipes (1) and (2) are
connected at flange B, as shown in Fig. P5.31. Pipe (1)
has an outside diameter of 6.625 in. and a wall
thickness of 0.28 in. Pipe (2) has an outside diameter of
4.00 in. and a wall thickness of 0.226 in. If the normal
stress in each steel pipe must be limited to 18 ksi,
determine:
(a) the maximum downward load P that may be applied
at flange B.
(b) the deflection of flange B at the load determined in
part (a).
Fig. P5.31
Solution
(a) Pipe section properties: The pipe cross-sectional areas are:
D1 6.625 in.
d1 6.625 in. 2(0.28 in.) 6.0650 in.
A1
D2
A2
4
(6.625 in.) 2 (6.0650 in.) 2
4.00 in.
4
d2
5.5814 in.2
4.00 in. 2(0.226 in.)
(4.00 in.) 2 (3.5480 in.) 2
3.5480 in.
2.6795 in.2
Equilibrium: Consider a FBD of flange B. Sum forces in the vertical
direction to obtain:
Fy F2 F1 P 0
(a)
Geometry of Deformations:
0
1
2
Force-Deformation Relationships:
F1L1
F2 L2
1
2
A1E1
A2 E2
(b)
(c)
Compatibility Equation: Substitute Eqs. (c) into Eq. (b) to derive the compatibility equation:
F1 L1 F2 L2
0
A1 E1 A2 E2
(d)
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.
For this problem, it is convenient to express the compatibility equation in terms of the normal stress 1
and 2:
1 L1
2 L2
0
(e)
E1
E2
Solve the Equations: Solve Eq. (e) for 1:
L2 E1
(f)
1
2
L1 E2
The elastic moduli of the two pipes are equal; therefore, E1/E2 = 1. The normal stresses in pipes (1) and
(2) are limited to 18 ksi. Assume that the normal stress in pipe (2) controls, meaning that 2 = 18 ksi.
Calculate the corresponding stress in pipe (1):
L2 E1
16 ft
(18 ksi)
28.80 ksi
1
2
L1 E2
10 ft
Since this stress magnitude is greater than 18 ksi, our assumption is proven incorrect. We now know
that the normal stress in pipe (1) actually controls; thus, 1 = −18 ksi (negative by inspection). The
normal stress in pipe (1) can be found from Eq. (f)
L1 E2
10 ft
( 18 ksi)
11.250 ksi
(f)
2
1
L2 E1
16 ft
Now that the stresses are known, the allowable forces F1 and F2 can be computed:
F1
( 18 ksi)(5.5814 in.2 )
100.4652 kips
1 A1
F2
2
A2
(11.250 ksi)(2.6795 in.2 ) 30.1444 kips
Substitute these values into Eq. (a) to obtain the allowable load P:
Pmax F2 F1 30.1444 kips ( 100.4652 kips) 130.6 kips
(b) The deflection of flange B is equal to the deformation of pipe (2):
(11.250 ksi)(16 ft)(12 in./ft)
2 L2
uB
0.0720 in.
2
30,000 ksi
E2
Ans.
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.
5.32 A solid aluminum [E = 70 GPa] rod (1) is connected to a solid
bronze [E = 100 GPa] rod at flange B, as shown in Fig P5.32. Aluminum
rod (1) has an outside diameter of 35 mm and bronze rod (2) has an
outside diameter of 20 mm. The normal stress in the aluminum rod must
be limited to 160 MPa, and the normal stress in the bronze rod must be
limited to 110 MPa. Determine:
(a) the maximum downward load P that may be applied at flange B.
(b) the deflection of flange B at the load determined in part (a).
Fig. P5.32
Solution
(a) Rod section properties: The rod cross-sectional areas are:
A1
4
(35 mm)2
962.1128 mm2
A2
4
(20 mm)2
314.1593 mm2
Equilibrium: Consider a FBD of flange B. Sum forces in the vertical direction to
obtain:
Fy F2 F1 P 0
(a)
Geometry of Deformations:
0
1
2
Force-Deformation Relationships:
F1L1
F2 L2
1
2
A1E1
A2 E2
(b)
(c)
Compatibility Equation: Substitute Eqs. (c) into Eq. (b) to derive the compatibility equation:
F1 L1 F2 L2
0
(d)
A1 E1 A2 E2
For this problem, it is convenient to express the compatibility equation in terms of the normal stress 1
and 2:
1 L1
2 L2
0
(e)
E1
E2
Solve the Equations: Solve Eq. (e) for 1:
L2 E1
(f)
1
2
L1 E2
The normal stress in aluminum rod (1) is limited to 160 MPa. We will assume that the aluminum rod
controls, and then, we will calculate the corresponding stress in bronze rod (2):
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.
L1 E2
175 mm 100 GPa
(160 MPa)
117.647 MPa
L2 E1
340 mm 70 GPa
This stress magnitude is greater than the 110 MPa allowable stress for the bronze rod; therefore, this
calculation shows that the stress in the bronze rod actually controls. If the stress in the bronze rod is
limited to 110 MPa, the normal stress in aluminum rod (1) can be found from Eq. (f)
L2 E1
340 mm 70 GPa
(110 MPa)
149.600 MPa
(f)
1
2
L1 E2
175 mm 100 GPa
2
1
Now that the stresses are known, the allowable forces F1 and F2 can be computed:
F1
( 149.600 MPa)(962.1128 mm 2 )
143,932.1 N
143.9321 kN
1 A1
F2
2
A2
(110 MPa)(314.1593 mm2 )
34,557.5 N
34.5575 kN
Substitute these values into Eq. (a) to obtain the allowable load P:
Pmax F2 F1 34.5575 kN ( 143.9321 kN) 178.5 kN
(b) The deflection of flange B is equal to the deformation of rod (2):
(110 MPa)(340 mm)
2 L2
uB
0.374 mm
2
E2
100,000 MPa
Ans.
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.
5.33 A pin-connected structure is supported and
loaded as shown in Fig. P5.33. Member ABCD
is rigid and is horizontal before the load P is
applied. Bars (1) and (2) are both made from
steel [E = 30,000 ksi] and both have a crosssectional area of 1.25 in.2. A concentrated load
of P = 25 kips acts on the structure at D.
Determine:
(a) the normal stresses in both bars (1) and (2).
(b) the downward deflection of point D on the
rigid bar.
Fig. P5.33
Solution
Equilibrium
Consider a FBD of the rigid bar. Assume tension forces in
members (1) and (2). A moment equation about pin A gives
the best information for this situation:
M A (66 in.)F1 (168 in.)F2 (212 in.)P 0
(a)
Geometry of Deformations Relationship
Draw a deformation diagram of the rigid bar. The deflections
of the rigid bar are related by similar triangles:
vB
vC
(b)
66 in. 168 in.
Since there are no gaps, clearances, or other misfits at pins B
and C, the deformation of member (1) will equal the
deflection of the rigid bar at B and the deformation of member
(2) will equal the deflection of the rigid bar at C. Therefore,
Eq. (b) can be rewritten in terms of the member deformations
as:
1
66 in.
2
168 in.
(c)
Force-Deformation Relationships
The relationship between the internal force and the deformation of an axial member can be stated for
members (1) and (2):
F1L1
F2 L2
(d)
1
2
A1E1
A2 E2
Compatibility Equation
Substitute the force-deformation relationships (d) into the geometry of deformation relationship (c) to
derive the compatibility equation:
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 F1L1
66 in. A1E1
1 F2 L2
168 in. A2 E2
(e)
Solve the Equations
Solve Eq. (e) for F1:
66 in. L2 A1E1
66 in. L2 A1 E1
F1 F2
F2
168 in. A2 E2 L1
168 in. L1 A2 E2
Substitute this expression into equilibrium equation (a) and solve for F2:
M A (66 in.)F1 (168 in.)F2 (212 in.)P 0
(66 in.)F2
66 in. L2 A1 E1
168 in. L1 A2 E2
F2 (66 in.)
66 in. L2 A1 E1
168 in. L1 A2 E2
(168 in.)F2
(212 in.)P
(168 in.)
(212 in.)P
(f)
(g)
For this structure, P = 20 kips, and the lengths, areas, and elastic moduli are given below:
L1 80 in.
L2 120 in.
A1 1.25 in.2
A2 1.25 in.2
E1 30, 000 ksi
E2 30, 000 ksi
Substitute these values into Eq. (g) and calculate F2 = 25.617124 kips. Backsubstitute into Eq. (f) to
calculate F1 = 15.095805 kips.
Normal Stresses
The normal stresses in each axial member can now be calculated:
F1 15.095805 kips
12.08 ksi
1
1.25 in.2
A1
F2 25.617124 kips
20.5 ksi
2
1.25 in.2
A2
Ans.
Ans.
Deflections of the rigid bar
Calculate the deformation of one of the axial members, say member (1):
F1L1 (15.095805 kips)(80 in.)
0.032204 in.
(h)
1
A1E1
(1.25 in.2 )(30,000 ksi)
Since there are no gaps at pin B, the rigid bar deflection at B is equal to the deformation of member (1);
therefore, vB = 1 = 0.032204 in. (downward). From similar triangles, the deflection of the rigid bar at D
is related to vB by:
vB
vD
(i)
66 in. 212 in.
From Eq. (i), the deflection of the rigid bar at D is:
212 in.
212 in.
vD
vB
(0.032204 in.) 0.1034 in.
66 in.
66 in.
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.
5.34 A pin-connected structure is supported and
loaded as shown in Fig. P5.34. Member ABCD
is rigid and is horizontal before the load P is
applied. Bars (1) and (2) are both made from
steel [E = 30,000 ksi] and both have a crosssectional area of 1.25 in.2. If the normal stress in
each steel bar must be limited to 18 ksi,
determine the maximum load P that may be
applied to the rigid bar.
Fig. P5.34
Solution
Equilibrium
Consider a FBD of the rigid bar. Assume tension forces in
members (1) and (2). A moment equation about pin A
gives the best information for this situation:
M A (66 in.)F1 (168 in.)F2 (212 in.)P 0
(a)
Geometry of Deformations Relationship
Draw a deformation diagram of the rigid bar. The
deflections of the rigid bar are related by similar triangles:
vB
vC
(b)
66 in. 168 in.
Since there are no gaps, clearances, or other misfits at pins
B and C, the deformation of member (1) will equal the
deflection of the rigid bar at B and the deformation of
member (2) will equal the deflection of the rigid bar at C.
Therefore, Eq. (b) can be rewritten in terms of the member
deformations as:
1
66 in.
2
168 in.
(c)
Force-Deformation Relationships
The relationship between the internal force and the deformation of an axial member can be stated for
members (1) and (2):
F1L1
F2 L2
(d)
1
2
A1E1
A2 E2
Compatibility Equation
Substitute the force-deformation relationships (d) into the geometry of deformation relationship (c) to
derive the compatibility equation:
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 F1L1
66 in. A1E1
1 F2 L2
168 in. A2 E2
(e)
Solve the Equations
Since allowable stresses are specified, it is convenient to express Eq. (e) in terms of stress:
66 in. 2 L2
1 L1
E1
168 in. E2
and solve for 2:
168 in. L1 E2
2
1
66 in. L2 E1
Since both bars have the same elastic modulus, E1/E2 = 1. Substitute 1 = 18 ksi and solve for
168 in. L1 E2
168 in. 80 in.
(18 ksi)
30.54546 ksi 18 ksi
N.G.
2
1
66 in. L2 E1
66 in. 120 in.
It is now evident that the stress in bar (2) controls. Substitute 2 = 18 ksi and solve for
66 in. L2 E1
66 in. 120 in.
(18 ksi)
10.60714 ksi 18 ksi
OK
1
2
168 in. L1 E2
168 in. 80 in.
(f)
2:
1:
Now that the stresses are known, the allowable forces F1 and F2 can be computed:
F1
(10.60714 ksi)(1.25 in.2 ) 13.25893 kips
1 A1
F2
2
A2
(18 ksi)(1.25 in.2 )
22.5 kips
Substitute these values into Eq. (a) to obtain the allowable load P:
M A (66 in.)(16.875 kips) (168 in.)(22.5 kips) (212 in.)P
Pmax
21.95797 kips
22.0 kips
0
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.
5.35 The pin-connected structure shown in Fig.
P5.35 consists of a rigid beam ABCD and two
supporting bars. Bar (1) is an aluminum alloy [E =
70 GPa] with a cross-sectional area of A1 = 800
mm2. Bar (2) is a bronze alloy [E = 100 GPa] with
a cross-sectional area of A2 = 500 mm2. The
normal stress in the aluminum bar must be limited
to 70 MPa, and the normal stress in the bronze rod
must be limited to 90 MPa. Determine:
(a) the maximum downward load P that may be
applied at B.
(b) the deflection of the rigid beam at B.
Fig. P5.35
Solution
Equilibrium
Consider a FBD of the rigid bar. Assume tension forces
in members (1) and (2). A moment equation about pin D
gives the best information for this situation:
MD
(3.5 m)F1 (1.2 m)F2 (2.7 m)P 0
(a)
Geometry of Deformations Relationship
Draw a deformation diagram of the rigid bar. The
deflections of the rigid bar are related by similar
triangles:
vC
vA
3.5 m 1.2 m
(b)
Since there are no gaps, clearances, or other misfits at
pins A and C, the deformation of member (1) will equal
the deflection of the rigid bar at A and the deformation of
member (2) will equal the deflection of the rigid bar at C.
Therefore, Eq. (b) can be rewritten in terms of the
member deformations as:
1
3.5 m
2
1.2 m
(c)
Force-Deformation Relationships
The relationship between the internal force and the deformation of an axial member can be stated for
members (1) and (2):
F1L1
F2 L2
(d)
1
2
A1E1
A2 E2
Compatibility Equation
Substitute the force-deformation relationships (d) into the geometry of deformation relationship (c) to
derive the compatibility equation:
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 F1 L1
3.5 m A1 E1
1 F2 L2
1.2 m A2 E2
(e)
Solve the Equations
Since allowable stresses are specified, it is convenient to express Eq. (e) in terms of stress:
L
3.5 m 2 L2
1 L1
2.9167 2 2
E1
1.2 m E2
E2
and solve for 1:
L E
2.9167 2 2 1
1
L1 E2
Substitute 2 = 90 MPa and solve for 1:
L E
3.3 m 70 GPa
2.9167 2 2 1 2.9167(90 MPa)
209.1 MPa 70 MPa
N.G.
1
L1 E2
2.9 m 100 GPa
It is now evident that the stress in bar (1) controls. Substitute 1 = 70 MPa and solve for 2:
L1 E2
1
1
2.9 m 100 GPa
(70 MPa)
30.13 MPa 90 MPa
O.K.
2
1
2.9167 L2 E1 2.9167
3.3 m 70 GPa
(f)
Now that the stresses are known, the allowable forces F1 and F2 can be computed:
F1
(70 MPa)(800 mm2 ) 56, 000 N 56.00 kN
1 A1
F2
(30.13 MPa)(500 mm2 ) 15, 065 N 15.065 kN
2 A2
Substitute these values into Eq. (a) to obtain the allowable load P:
MD
(3.5 m)(56.00 kN) (1.2 m)(15.065 kN) (2.7 m)P 0
Pmax
79.288 kN
79.3 kN
(b) Deflections of the rigid bar
Calculate the deformation of one of the axial members, say member (1):
F1L1
(56,000 N)(2,900 mm)
2.9000 mm
1
A1E1 (800 mm 2 )(70,000 MPa)
Since there are no gaps at pin A, the rigid bar deflection at
A is equal to the deformation of member (1); therefore, vA
= 1 = 2.9000 mm (downward). From similar triangles, the
deflection of the rigid bar at B is related to vA by:
vA
vB
(h)
3.5 m 2.7 m
From Eq. (h), the deflection of the rigid bar at B is:
2.7 m
2.7 m
(2.9000 mm) 2.24 mm
vB
vA
3.5 m
3.5 m
Ans.
(g)
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.
5.36 The pin-connected structure shown in Fig.
P5.36 consists of a rigid beam ABCD and two
supporting bars. Bar (1) is an aluminum alloy [E =
70 GPa] with a cross-sectional area of A1 = 800
mm2. Bar (2) is a bronze alloy [E = 100 GPa] with
a cross-sectional area of A2 = 500 mm2. All bars
are unstressed before the load P is applied;
however, there is a 3-mm clearance in the pin
connection at A. If a load of P = 54 kN is applied
at B, determine:
(a) the normal stresses in both bars (1) and (2).
(b) the normal strains in bars (1) and (2).
(c) determine the downward deflection of point A
on the rigid bar.
Fig. P5.36
Solution
Equilibrium
Consider a FBD of the rigid bar. Assume tension forces in
members (1) and (2). A moment equation about pin D gives
the best information for this situation:
MD
(3.5 m)F1 (1.2 m)F2 (2.7 m)P 0
(a)
Geometry of Deformations Relationship
Draw a deformation diagram of the rigid bar. The deflections
of the rigid bar are related by similar triangles:
vC
vA
(b)
3.5 m 1.2 m
Since there is no clearance at pin C, the deformation of
member (2) will equal the deflection of the rigid bar at C.
However, there is a 3-mm clearance at A. Consequently, not
all of the rigid beam deflection at A will go toward elongating
bar (1). The relationship between rigid beam deflection and
axial member deformation can be expressed:
vA
3 mm
1
Equation (c) can be rewritten in terms of the member deformations as:
3 mm
1
2
3.5 m
1.2 m
(c)
(d)
Force-Deformation Relationships
The relationship between the internal force and the deformation of an axial member can be stated for
members (1) and (2):
F1L1
F2 L2
(e)
1
2
A1E1
A2 E2
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.
Compatibility Equation
Substitute the force-deformation relationships (e) into the geometry of deformation relationship (d) to
derive the compatibility equation:
F1 L1
FL
3.5 m F2 L2
3 mm
2.9167 2 2
(f)
A1 E1
1.2 m A2 E2
A2 E2
Solve the Equations
Solve Eq. (f) for F2:
A2 E2
F1 L1
F2
2.9167 L2 A1 E1
F1
3 mm
L1
A2 E2
2.9167 L2 A1 E1
(3 mm)A2 E2
2.9167 L2
and substitute into Eq. (a) to solve for F1:
(3.5 m)F1 (1.2 m)F2
(3.5 m)F1 (1.2 m) F1
L1
A2 E2
2.9167 L2 A1 E1
(3.5 m)F1 (1.2 m)F1
(3 mm)A2 E2
2.9167 L2
L1
A2 E2
2.9167 L2 A1 E1
(2.7 m)(54 kN)
(2.7 m)(54 kN)
(2.7 m)(54 kN) (1.2 m)
(3 mm)A2 E2
2.9167 L2
(3 mm)A2 E2
2.9167 L2
L1
A2 E2
(3.5 m) (1.2 m)
2.9167 L2 A1 E1
(2.7 m)(54 kN) (1.2 m)
F1
The value of F1 is thus calculated as:
(3 mm)(500 mm 2 )(100, 000 N/mm 2 )
2.9167(3,300 mm)
2.9 m
500 mm 2 100 GPa
3.5 m (1.2 m)
2.9167(3.3 m) 800 mm 2 70 GPa
33, 247.4 N 33.2474 kN
(2.7 m)(54, 000 N) (1.2 m)
F1
and backsubstituting into Eq. (a) gives F2:
(2.7 m)(54 kN) (3.5 m)F1
F2
1.2 m
(2.7 m)(54 kN) (3.5 m)(33.2474 kN)
1.2 m
24.5285 kN
(a) Normal stresses:
F1 33,247.4 N
1
A1
800 mm 2
2
F2
A2
24,528.5 N
500 mm 2
41.6 MPa (T)
Ans.
49.1 MPa (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.
(b) Normal strains:
41.5592 MPa
1
593.7 10 6 mm/mm
1
E1
70,000 MPa
49.0570 MPa
2
490.6 10 6 mm/mm
2
E2 100,000 MPa
594 με
Ans.
491 με
Ans.
(c) Deflections of the rigid bar
Calculate the deformation of member (1):
F1L1 (33, 247.4 N)(2,900 mm)
1.721739 mm
1
A1E1 (800 mm 2 )(70,000 MPa)
The relationship between the rigid bar deflection at A and
the deformation of member (1) was expressed in Eq. (c).
Therefore, the rigid bar deflection at A is:
vA
3 mm
1
1.721739 mm 3 mm
4.72 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.
5.37 The pin-connected structure shown in Fig.
P5.37 consists of a rigid bar ABCD and two axial
members. Bar (1) is bronze [E = 100 GPa] with a
cross-sectional area of A1 = 620 mm2. Bar (2) is
an aluminum alloy [E = 70 GPa] with a crosssectional area of A2 = 340 mm2. A concentrated
load of P = 75 kN acts on the structure at D.
Determine:
(a) the normal stresses in both bars (1) and (2).
(b) the downward deflection of point D on the
rigid bar.
Fig. P5.37
Solution
Equilibrium
Consider a FBD of the rigid bar. Assume tension
forces in members (1) and (2). A moment equation
about pin C gives the best information for this situation:
M C (825 mm)F1 (325 mm)F2
(1,100 mm)P
0
Geometry of Deformations Relationship
Draw a deformation diagram of the rigid bar. The
deflections of the rigid bar are related by similar
triangles:
vA
vB
vD
825 mm 325 mm 1,100 mm
(a)
(b)
Since there are no gaps or clearances at either pin A or
pin B, the deformations of members (1) and (2) will
equal the deflections of the rigid bar at A and B,
respectively.
1
825 mm
2
325 mm
(c)
Force-Deformation Relationships
The relationship between the internal force and the deformation of an axial member can be stated for
members (1) and (2):
F1L1
F2 L2
(d)
1
2
A1E1
A2 E2
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.
Compatibility Equation
Substitute the force-deformation relationships (d) into the geometry of deformation relationship (c) to
derive the compatibility equation:
F1L1 825 mm F2 L2
FL
2.538462 2 2
(e)
A1E1 325 mm A2 E2
A2 E2
Solve the Equations
Solve Eq. (e) for F1:
L2 A1 E1
L1 A2 E2
and substitute into Eq. (a) to solve for F2:
F1
2.538462 F2
(f)
(825 mm)F1
(325 mm)F2
(1,100 mm)P
L2 A1 E1
L1 A2 E2
(325 mm)F2
(1,100 mm)P
(825 mm)(2.538462) F2
540 mm 620 mm 2 100 GPa
F2 (825 mm)(2.538462)
900 mm 340 mm 2 70 GPa
325 mm
(1,100 mm)(75 kN)
F2
22.9273 kN
Backsubstitute this result into Eq. (f) to compute F1:
540 mm 620 mm 2 100 GPa
(2.538462)
(22.9273 kN)
F1
900 mm 340 mm 2 70 GPa
90.9680 kN
(a) Normal stresses:
F1 90,968.0 N
1
620 mm 2
A1
2
F2
A2
22,927.3 N
340 mm 2
146.7 MPa (T)
Ans.
67.4 MPa (T)
Ans.
(b) Deflections of the rigid bar
Calculate the deformation of member (1):
F1L1
(90,968.0 N)(900 mm)
1
A1E1 (620 mm 2 )(100,000 MPa)
1.3205 mm
Since the pin at A is assumed to have a perfect connection, vA =
1,100 mm
vD
vA 1.333333(1.3205 mm) 1.761 mm
825 mm
1
= 1.3205 mm. From Eq. (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.
5.38 The pin-connected structure shown in Fig.
P5.38 consists of a rigid bar ABCD and two axial
members. Bar (1) is bronze [E = 100 GPa] with a
cross-sectional area of A1 = 620 mm2. Bar (2) is
an aluminum alloy [E = 70 GPa] with a crosssectional area of A2 = 340 mm2. All bars are
unstressed before the load P is applied; however,
there is a 3-mm clearance in the pin connection at
A. If a concentrated load of P = 90 kN acts on the
structure at D, determine:
(a) the normal stresses in both bars (1) and (2).
(b) the normal strains in bars (1) and (2).
(c) the downward deflection of point D on the
rigid bar.
Fig. P5.38
Solution
Equilibrium
Consider a FBD of the rigid bar. Assume tension
forces in members (1) and (2). A moment equation
about pin C gives the best information for this situation:
M C (825 mm)F1 (325 mm)F2
(1,100 mm)P
0
Geometry of Deformations Relationship
Draw a deformation diagram of the rigid bar. The
deflections of the rigid bar are related by similar
triangles:
vA
vB
vD
825 mm 325 mm 1,100 mm
(a)
(b)
Since there is no clearance at pin B, the deformation of
member (2) will equal the deflection of the rigid bar at
B. However, there is a 3-mm clearance in the
connection at A. Consequently, not all of the rigid bar
deflection at A will go toward elongating bar (1). The
relationship between rigid bar deflection and axial
member deformation can be expressed:
vA
3 mm
1
Equation (b) can be rewritten in terms of the member deformations as:
3 mm
1
2
825 mm
325 mm
(c)
(d)
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.
Force-Deformation Relationships
The relationship between the internal force and the deformation of an axial member can be stated for
members (1) and (2):
F1L1
F2 L2
(e)
1
2
A1E1
A2 E2
Compatibility Equation
Substitute the force-deformation relationships (e) into the geometry of deformation relationship (d) to
derive the compatibility equation:
F1L1
FL
825 mm F2 L2
3 mm
2.538462 2 2
(f)
A1E1
325 mm A2 E2
A2 E2
Solve the Equations
Solve Eq. (f) for F2:
A2 E2
F1L1
L1
A2 E2 (3 mm)A2 E2
F2
F1
3 mm
2.538462 L2 A1E1
2.538462 L2 A1 E1 2.538462 L2
and substitute into Eq. (a) to solve for F1:
(825 mm)F1 (325 mm)F2 (1,100 mm)P
(825 mm)F1
(325 mm) F1
L1
A2 E2
2.538462 L2 A1 E1
(825 mm)F1
(325 mm)F1
(3 mm)A2 E2
2.538462 L2
L1
A2 E2
2.538462 L2 A1 E1
(1,100 mm)P
(1,100 mm)P (325 mm)
(3 mm)A2 E2
2.538462 L2
(3 mm)A2 E2
2.538462 L2
L1
A2 E2
825 mm (325 mm)
2.538462 L2 A1 E1
(1,100 mm)P (325 mm)
F1
The value of F1 is thus calculated as:
(3 mm)(340 mm 2 )(70,000 N/mm 2 )
2.538462(540 mm)
900 mm
340 mm 2 70,000 N/mm 2
825 mm (325 mm)
2.538462(540 mm) 620 mm 2 100,000 N/mm 2
90, 495.6 N 90.4956 kN
(1,100 mm)(90,000 N) (325 mm)
F1
and backsubstituting into Eq. (a) gives F2:
(1,100 mm)(90 kN) (825 mm)F1
F2
325 mm
(1,100 mm)(90 kN) (825 mm)(90.4956 kN)
325 mm
74.8957 kN
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.
(a) Normal stresses:
F1 90, 495.6 N
1
620 mm 2
A1
F2 74,895.7 N
2
A2
340 mm 2
146.0 MPa (T)
Ans.
220 MPa (T)
Ans.
(b) Normal strains:
F1
90, 495.6 N
1
A1E1 (620 mm 2 )(100,000 N/mm 2 )
2
F2
A2 E2
74,895.7 N
(340 mm 2 )(70,000 N/mm 2 )
1, 459.6 10
6
mm/mm
1, 460 με
Ans.
3,146.9 10
6
mm/mm
3,150 με
Ans.
(c) Deflections of the rigid bar
Calculate the deformation of member (2):
F2 L2
(74,895.7 N)(540 mm)
1.6993 mm
2
A2 E2 (340 mm 2 )(70,000 MPa)
Since the pin at B is assumed to have a perfect connection, vB =
1,100 mm
vD
vB 3.384615(1.6993 mm) 5.75 mm
325 mm
2
= 1.6993 mm. From Eq. (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.
5.39 The assembly shown in Fig. P5.39 consists of
a solid aluminum alloy [E = 70 GPa] post (2)
surrounded by a bronze [E = 100 GPa] tube (1).
Before the load P is applied, there is a clearance of
2 mm between the post and the tube. The yield
stress for the aluminum post is 260 MPa and the
yield stress for the bronze tube is 340 MPa.
Determine:
(a) the maximum load P that may be applied to the
assembly without causing yielding of either the
post or the tube.
(b) the downward displacement of rigid cap B.
(c) the normal strain in the bronze tube.
Fig. P5.39
Solution
Section properties:
R1 35 mm r1
29 mm
R2 15 mm
A1
A2
(35 mm)2 (29 mm)2
(15 mm) 2
1, 206.372 mm2
706.858 mm 2
Equilibrium: Consider a FBD around rigid cap B after the gap has been closed.
Sum forces in the vertical direction to obtain:
Fy
F1 F2 P 0
(a)
Geometry of Deformations:
2 mm
2
1
Force-Deformation Relationships:
F1L1
F2 L2
1
2
A1E1
A2 E2
(b)
(c)
Compatibility Equation
Substitute the force-deformation relationships (c) into the geometry of deformation relationship (b) to
derive the compatibility equation:
F2 L2 F1 L1
2 mm
(d)
A2 E2 A1 E1
Solve the Equations
Since allowable stresses are specified, it is convenient to express Eq. (d) in terms of stress:
2 L2
1 L1
2 mm
E2
E1
and solve for 2:
(e)
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.
2
1
L1 E2
L2 E1
(2 mm)
E2
L2
600 mm 70, 000 MPa
70, 000 MPa
(2 mm)
602 mm 100, 000 MPa
602 mm
Substitute 1 = 340 MPa and solve for 2:
70, 000 MPa
0.697674(340 MPa) (2 mm)
469.8 MPa 260 MPa
N.G.
2
602 mm
It is now evident that the stress in post (2) controls. Substitute 2 = 260 MPa and solve for 1:
70, 000 MPa L2 E1
(2 mm)
1
2
602 mm
L1 E2
1
70, 000 MPa 602 mm 100, 000 MPa
602 mm
600 mm 70, 000 MPa
340 MPa
OK
260 MPa (2 mm)
39.333 MPa
(a) Maximum load: Now that the stresses are known, the allowable forces F1 and F2 can be computed:
F1
(39.333 MPa)(1,206.372 mm2 ) 47, 450 N 47.450 kN
1 A1
F2
(260 MPa)(706.858 mm2 ) 183, 783 N 183.783 kN
2 A2
Substitute these values into Eq. (a) to obtain the allowable load P. By inspection, the forces in the post
and the tube must be compression; therefore:
Fy
F1 F2 P 0
Pmax
( 47.450 kN) ( 183.783 kN)
231 kN
Ans.
(b) Displacement of cap B: The contraction of tube (1) is:
F1L1
( 47, 451 N)(600 mm)
0.2360 mm
1
A1E1 (1,206.372 mm 2 )(100,000 N/mm 2 )
The displacement of cap B is 2 mm greater:
uB
0.2360 mm 2 mm
2.2360 mm
2.24 mm
(c) Normal strain in tube (1): The strain in bronze tube (1) is:
39.333 MPa
1
393 με
1
E1 100,000 MPa
Ans.
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.
5.40 A 4.5-m-long aluminum tube (1) is to be connected to a 2.4-mlong bronze pipe (2) at B. When put in place, however, a gap of 8
mm exists between the two members, as shown in Fig P5.40.
Aluminum tube (1) has an elastic modulus of 70 GPa and a crosssectional area of 2,000 mm2. Bronze pipe (2) has an elastic modulus
of 100 GPa and a cross-sectional area of 3,600 mm2. If bolts are
inserted in the flanges and tightened so that the gap at B is closed,
determine:
(a) the normal stresses produced in each of the members.
(b) the final position of flange B with respect to support A.
Fig. P5.40
Solution
Equilibrium: Consider a FBD of flange B after the bolts have been tightened
and the gap at B has been closed. Sum forces in the vertical direction to obtain:
Fy F1 F2 0
F1 F2
(a)
Geometry of Deformations:
8 mm
1
2
(b)
Force-Deformation Relationships:
F1L1
F2 L2
1
2
A1E1
A2 E2
(c)
Compatibility Equation: Substitute Eqs. (c) into Eq. (b) to derive the compatibility equation:
F1L1 F2 L2
8 mm
A1E1 A2 E2
(d)
Solve the Equations: Substitute Eq. (a) into Eq. (d) and solve for F1:
L1
L2
F1
8 mm
A1E1 A2 E2
F1
8 mm
4,500 mm
(2,000 mm 2 )(70,000 N/mm 2 )
206,135.0 N
(a) Normal stresses:
F1 206,135.0 N
1
2,000 mm 2
A1
2
F2
A2
206,135.0 N
3,600 mm 2
2,400 mm
(3,600 mm 2 )(100,000 N/mm 2 )
(f)
103.0675 MPa
103.1 MPa (T)
Ans.
57.2597 MPa
57.3 MPa (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.
(b) The deflection of flange B is equal to the deformation of bronze pipe (2):
(206,135.0 N)(2,400 mm)
F2 L2
1.374 mm
2
A2 E2 (3,600 mm 2 )(100,000 N/mm 2 )
Ans.
In its final position, flange B is located 2,401.374 mm away from support A.
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.
5.41 The assembly shown in Fig. P5.41 consists of a steel [E1 =
30,000 ksi; A1 = 1.25 in.2] rod (1), a rigid bearing plate B that is
securely fastened to rod (1), and a bronze [E2 = 15,000 ksi; A2 =
3.75 in.2] post (2). The yield strengths of the steel and bronze are
62 ksi and 75 ksi, respectively. A clearance of 0.125 in. exists
between the bearing plate B and bronze post (2) before the
assembly is loaded. After a load of P = 65 kips is applied to the
bearing plate, determine:
(a) the normal stresses in bars (1) and (2).
(b) the factors of safety with respect to yield for each of the
members.
(c) the vertical displacement of bearing plate B.
Fig. P5.41
Solution
Deformation in rod (1) alone: Check to see if the bearing plate attached to rod (1) will contact the
bronze post for the 65-kip load.
F1L1 (65 kips)(14 ft)(12 in./ft)
0.2912 in. 0.125 in. gap
contact will occur
1
A1E1
(1.25 in.2 )(30,000 ksi)
This calculation proves that the bearing plate attached to rod (1) will contact the bronze post when the
65-kip load is applied; therefore, this structure must be analyzed as a statically indeterminate structure.
Equilibrium: Consider a FBD of flange B after the gap at B has been closed.
Sum forces in the vertical direction to obtain:
Fy F1 F2 P 0
(a)
Geometry of Deformations:
0.125 in.
1
2
Force-Deformation Relationships:
F1L1
F2 L2
1
2
A1E1
A2 E2
(b)
(c)
Compatibility Equation: Substitute Eqs. (c) into Eq. (b) to derive the compatibility equation:
F1L1 F2 L2
0.125 in.
A1E1 A2 E2
(d)
Solve the Equations: Solve Eq. (d) for F1:
F2 L2 A1E1
AE
L A E
F1
0.125 in.
(0.125 in.) 1 1 F2 2 1 1
A2 E2 L1
L1
L1 A2 E2
and substitute the resulting expression into Eq. (a) to determine an expression for F2:
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.
(0.125 in.)
A1E1
L1
F2
L2 A1 E1
L1 A2 E2
F2
F2 1
65 kips
L2 A1 E1
L1 A2 E2
0
(0.125 in.)
A1E1
L1
65 kips
A1E1
65 kips
L1
L2 A1 E1
1
L1 A2 E2
(0.125 in.)
F2
The axial force in post (2) is thus:
(1.25 in.2 )(30,000 ksi)
(0.125 in.)
65 kips
(168 in.)
F2
32.4609 kips
36 in. 1.25 in.2 30,000 ksi
1
168 in. 3.75 in.2 15,000 ksi
and the axial force in rod (1) is:
F1 F2 P
32.4609 kips 65 kips 32.5391 kips
(a) Normal stresses:
F1 32.5391 kips
1
1.25 in.2
A1
2
F2
A2
32.4609 kips
3.75 in.2
(b) Factors of safety:
62 ksi
FS1
26.0313 ksi
2.38
26.0313 ksi
26.0 ksi (T)
8.6563 ksi
FS2
Ans.
8.66 ksi (C)
75 ksi
8.6563 ksi
Ans.
8.66
(c) Displacement of plate B: The displacement of plate B is equal to the deformation of rod (1).
F1L1 (32.5391 kips)(168 in.)
0.145775 in.
1
A1E1 (1.25 in.2 )(30,000 ksi)
uB
Ans.
Ans.
0.1458 in.
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.
5.42 A hollow steel [E = 30,000 ksi] tube (1) with
an outside diameter of 3.50 in. and a wall
thickness of 0.216 in. is fastened to a solid 2-in.diameter aluminum [E = 10,000 ksi] rod. The
assembly is attached to unyielding supports at the
left and right ends and is loaded as shown in Fig.
P5.42. Determine:
(a) the stresses in all parts of the axial structure.
(b) the deflections of joints B and C.
Fig. P5.42
Solution
Section properties: The steel tube cross-sectional area is:
D1 3.50 in.
d1 3.50 in. 2(0.216 in.) 3.068 in.
(3.50 in.) 2 (3.068 in.) 2
2.2285 in.2
4
and the aluminum rod has a cross-sectional area of:
D2 D3 2.00 in.
A1
A2
A3
4
(2.00 in.) 2
3.1416 in.2
Equilibrium: Consider a FBD of flange B. Sum forces in
the horizontal direction to obtain:
Fx
F1 F2 34 kips 0
(a)
Consider a FBD of flange C. Sum forces in the horizontal
direction to obtain:
Fx
F2 F3 26 kips 0
(b)
Geometry of Deformations:
0
1
2
3
Force-Deformation Relationships:
F1L1
F2 L2
1
2
3
A1E1
A2 E2
(c)
F3 L3
A3 E3
(d)
Compatibility Equation: Substitute Eqs. (d) into Eq. (c) to derive the compatibility equation:
F1 L1 F2 L2 F3 L3
0
A1 E1 A2 E2 A3 E3
(e)
Solve the Equations: Solve Eq. (a) for F1:
F1 F2 34 kips
(f)
Solve Eq. (b) for F3:
F3 F2 26 kips
(g)
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.
Substitute Eqs. (f) and (g) into compatibility equation (e):
( F2 34 kips)L1 F2 L2 ( F2 26 kips)L3
0
A1 E1
A2 E2
A3 E3
and expand terms:
F2 L1 (34 kips)L1 F2 L2 F2 L3 (26 kips)L3
0
A1 E1
A1 E1
A2 E2 A3 E3
A3 E3
Regroup terms:
L3
L1
L2
(34 kips)L1
F2
A1E1 A2 E2 A3 E3
A1E1
and solve for F2:
(34 kips)L1 (26 kips)L3
A1 E1
A3 E3
F2
L3
L1
L2
A1 E1 A2 E2 A3 E3
(26 kips)L3
A3 E3
(34 kips)(48 in.)
(26 kips)(60 in.)
2
(2.2285 in. )(30, 000 ksi) (3.1416 in.2 )(10, 000 ksi)
48 in.
60 in.
60 in.
2
2
(2.2285 in. )(30, 000 ksi) (3.1416 in. )(10, 000 ksi) (3.1416 in.2 )(10, 000 ksi)
16.3227 kips
Backsubstitute into Eqs. (f) and (g) to obtain F1 and F3:
F1 F2 34 kips 16.3227 kips 34 kips
17.6773 kips
F3
F2 26 kips 16.3227 kips 26 kips
9.6773 kips
(a) Normal stresses: The normal stresses in each axial member can now be calculated:
17.6773 kips
F1
7.9325 ksi 7.93 ksi (C)
1
2.2285 in.2
A1
2
F2
A2
16.3227 kips
3.1416 in.2
3
F3
A3
9.6773 kips
3.1416 in.2
5.1957 ksi
3.0834 ksi
5.20 ksi (T)
3.08 ksi (C)
Ans.
Ans.
Ans.
(b) Joint deflections: The deflection of flange B is equal to the deformation (i.e., contraction in this
instance) of member (1):
( 17.6773 kips)(48 in.)
F1L1
0.012692 in. 0.01269 in.
uB
Ans.
1
A1E1 (2.2285 in.2 )(30,000 ksi)
The deflection of flange C is equal to the sum of the deformations in members (1) and (2). The
deformation of member (2) 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.
2
F2 L2
A2 E2
(16.3227 kips)(60 in.)
(3.1416 in.2 )(10,000 ksi)
0.031174 in.
And thus, the deflection of flange C is:
uC
0.012692 in. 0.031174 in. 0.018482 in.
1
2
0.01848 in.
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.
5.43 Rigid bar ABCD in Fig. P5.43 is supported by a
pin connection at A and by two axial bars (1) and (2).
Bar (1) is a 30-in.-long bronze [E = 15,000 ksi] bar
with a cross-sectional area of 1.25 in.2. Bar (2) is a
40-in.-long aluminum alloy [E = 10,000 ksi] bar with
a cross-sectional area of 2.00 in.2. Both bars are
unstressed before the load P is applied. If a
concentrated load of P = 27 kips is applied to the rigid
bar at D, determine:
(a) the normal stresses in bars (1) and (2).
(b) the deflection of the rigid bar at point D.
Fig. P5.43
Solution
Equilibrium
Consider a FBD of the rigid bar. Assume tension forces
in members (1) and (2). A moment equation about pin
A gives the best information for this situation:
MA
(36 in.)F1 (84 in.)F2 (98 in.)P 0
(a)
Geometry of Deformations Relationship
Draw a deformation diagram of the rigid bar. The
deflections of the rigid bar are related by similar
triangles:
vC
vB
vD
(b)
36 in. 84 in. 98 in.
There are no gaps, clearances, or other misfits at the
pins in this structure. The vertical deflection of the
rigid bar at C will produce deformation in member (2);
however, and the vertical deflection of the rigid bar at B
will create contraction in member (1) (see the
discussion under the heading Structures with a
rotating rigid bar in the text and examine Fig. 5.11).
Therefore, Eq. (b) can be rewritten in terms of the
member deformations as:
1
36 in.
2
84 in.
(c)
Force-Deformation Relationships
The relationship between the internal force and the deformation of an axial member can be stated for
members (1) and (2):
F1L1
F2 L2
(d)
1
2
A1E1
A2 E2
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.
Compatibility Equation
Substitute the force-deformation relationships (d) into the geometry of deformation relationship (c) to
derive the compatibility equation:
1 F1 L1
1 F2 L2
(e)
36 in. A1 E1 84 in. A2 E2
Solve the Equations
Solve Eq. (e) for F1:
36 in. L2 A1 E1
36 in. L2 A1 E1
F1
F2
F2
84 in. A2 E2 L1
84 in. L1 A2 E2
Substitute this expression into equilibrium equation (a) and solve for F2:
MA
(36 in.)F1 (84 in.)F2 (98 in.)P 0
(36 in.)F2
36 in. L2 A1 E1
84 in. L1 A2 E2
F2 (36 in.)
3 L2 A1 E1
7 L1 A2 E2
(84 in.)F2
(84 in.)
(f)
(98 in.)P
(98 in.)P
(g)
For this structure, P = 27 kips, and the lengths, areas, and elastic moduli are given below:
L1 30 in.
L2 40 in.
A1 1.25 in.2
A2
2.00 in.2
E1 15, 000 ksi
E2 10, 000 ksi
Substitute these values into Eq. (g) and calculate F2 = 25.6183 kips. Backsubstitute into Eq. (f) to
calculate F1 = −13.7241 kips.
(a) Normal Stresses
The normal stresses in each axial member can now be calculated:
F1
13.7241 kips
10.98 ksi (C)
1
A1
1.25 in.2
2
F2
A2
25.6183 kips
2.00 in.2
12.81 ksi (T)
Ans.
Ans.
(b) Deflections of the rigid bar
Calculate the deformation of one of the axial members, say member (2):
F2 L2 (25.6183 kips)(40 in.)
0.051237 in.
(h)
2
A2 E2 (2.00 in.2 )(10,000 ksi)
Since there are no gaps at pin C, the rigid bar deflection at C is equal to the deformation of member (2);
therefore, vC = 2 = 0.051237 in. (downward). From similar triangles [Eq. (b)], the deflection of the
rigid bar at D is therefore:
98 in.
98 in.
vD
vC
(0.051237 in.) 0.0598 in.
Ans.
84 in.
84 in.
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.
5.44 A 25-mm-diameter by 3.5-m-long steel rod (1)
is stress free after being attached to rigid supports,
as shown in Fig. P5.44. At A, a 16-mm-diameter
bolt is used to connect the rod to the support.
Determine the normal stress in steel rod (1) and the
shear stress in bolt A after the temperature drops
60°C. Use E = 200 GPa and  = 11.9 × 10−6/°C.
Fig. P5.44
Solution
Section properties:
For the 25-mm-diameter rod, the cross-sectional area is:
A1 

4
(25 mm) 2  490.873852 mm2
Force-Temperature-Deformation Relationship
The relationship between internal force, temperature change, and deformation of an axial member is:
FL
1  1 1  1T L1
A1E1
Since the rod is attached to rigid supports,  1 = 0.
F1 L1
 1T L1  0
A1 E1
Thus, the force produced in the rod by the temperature drop is:
AE
F1  1T L1 1 1  1T A1E1
L1
 (11.9  106 / C)(  60C)(490.873852 mm 2 )(200,000 N/mm 2 )
 70,096.786 N
The normal stress in the steel rod is:
70,096.786 N
1 
 142.8 MPa (T)
490.873852 mm 2

Ans.
The 16-mm-diameter bolt has a cross-sectional area of:
Abolt 
(16 mm) 2  201.061930 mm 2
4
Since the bolt is loaded in double shear, the shear stress in the bolt is
70,096.786 N
 bolt 
 174.3 MPa
2(201.061930 mm 2 )
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.
5.45 A 0.875-in.-diameter by 15-ft-long steel rod
(1) is stress free after being attached to rigid
supports. A clevis-and-bolt connection, as shown in
Fig. P5.45, connects the rod with the support at A.
The normal stress in the steel rod must be limited to
18 ksi, and the shear stress in the bolt must be
limited to 42 ksi. Assume E = 29,000 ksi and
 = 6.6 × 10−6/°F and determine:
(a) the temperature decrease that can be safely
accommodated by rod (1) based on the allowable
normal stress.
(b) the minimum required diameter for the bolt at A
using the temperature decrease found in part (a).
Fig. P5.45
Solution
Section properties:
For the 0.875-in.-diameter rod, the cross-sectional area is:

A1  (0.875 in.) 2  0.601320 in.2
4
Force-Temperature-Deformation Relationship
The relationship between internal force, temperature change, and deformation of an axial member is:
FL
1  1 1  1T L1
A1E1
Since the rod is attached to rigid supports, 1 = 0.
F1 L1
 1T L1  0
A1 E1
which can also be expressed in terms of the rod normal stress:
L
 1 1  1T L1  0
E1
Solve for T corresponding to a 24 ksi normal stress in the steel rod:
L 1

T   1 1
 1
1E1
E1 1L1

18 ksi
 94.0F
(6.6  10 / F)(29,000 ksi)
6
Ans.
The normal force in the steel rod is:
F1  (18 ksi)(0.601320 in.2 )  10.823768 kips
If the allowable shear stress in the bolt is 42 ksi, the minimum diameter required for the double shear
bolt is
  2  10.823768 kips

 0.257709 in.2
2  d bolt

42 ksi
4

 d bolt  0.405 in.
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.
5.46 A steel [E = 29,000 ksi and  = 6.6×10-6/°F] rod containing a turnbuckle has its ends attached
to rigid walls. During the summer when the temperature is 82°F, the turnbuckle is tightened to
produce a stress in the rod of 5 ksi. Determine the stress in the rod in the winter when the
temperature is 10°F.
Solution

The normal strain in the rod can be expressed as:

  T
E
Since the rod is attached to rigid walls, the rod strain after the temperature change is  = 0.


  T  0
E
The change in temperature between the summer and winter is
T  Twinter  Tsummer  10F  82F  72F
Solve for the stress increase created by the 72°F drop in temperature.
   T E

  (6.6  10 6 / F)(  72F)(29,000 ksi)  13.78 ksi (T)
In the summer, the rod had a tension normal stress of 5 ksi; therefore, the rod stress in the winter is:
Ans.
 winter  5 ksi  13.78 ksi  18.78 ksi (T)
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.
5.47 A high-density polyethylene [E = 120 ksi and  = 78 ×
10−6/°F] block (1) is positioned in a fixture as shown in Fig.
P5.47. The block is 2-in. by 2-in. by 32-in.-long. At room
temperature, a gap of 0.10 in. exists between the block and
the rigid support at B. Determine:
(a) the normal stress in the block caused by a temperature
increase of 100°F.
(b) the normal strain in block (1) at the increased
temperature.
Fig. P5.47
Solution
If the polyethylene block were completely free to elongate, a temperature change of 100°F would cause
an elongation of
1  1T L1  (78  10 6 / F)(100F)(32 in.)  0.2496 in.
Since this elongation is greater than the 0.10-in. gap, the temperature change will cause the polyethylene
block to contact the support at B, which will create normal stress in the block.
Force-Temperature-Deformation Relationship
The relationship between the internal force, temperature change, and the deformation of an axial
member is:
FL
1  1 1  1T1L1
A1E1
In this situation, the deformation of member (1) equals the 0.10-in. gap:
F1 L1
 1T1 L1  0.10 in.
A1 E1
This relationship can be stated in terms of normal stress as
 1 L1
 1T1 L1  0.10 in.
E1
(a) Normal stress:
The normal stress in the block due to the 100°F temperature increase is:
E
1   0.10 in.  1T1L1  1
L1
 0.10 in.  (78  106 / F)(100F)(32 in.) 
120 ksi
 0.561 ksi  561 psi (C)
32 in.
(b) Normal strain:
The normal strain in the polyethylene block is:
 0.10 in.
1  1 
 0.003125 in./in.  3,125 με
L1
32 in.
Ans.
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.
5.48 The assembly shown in Fig. P5.48 consists of a brass shell
(1) fully bonded to a solid ceramic core (2). The brass shell [E
= 115 GPa,  = 18.7 × 10−6/°C] has an outside diameter of 50
mm and an inside diameter of 35 mm. The ceramic core [E =
290 GPa,  = 3.1 × 10−6/°C] has a diameter of 35 mm. At a
temperature of 15°C, the assembly is unstressed. Determine the
largest temperature increase that is acceptable for the assembly
if the normal stress in the longitudinal direction of the brass
shell must not exceed 80 MPa.
Fig. P5.48
Solution


Section properties: The cross-sectional areas of brass shell (1) and ceramic core (2) are:
A1 
(50 mm) 2  (35 mm) 2   1,001.3827 mm 2
4
A2 
4
(35 mm) 2  962.1128 mm2
Equilibrium
Consider a FBD cut through the assembly. Sum forces in the
horizontal direction to obtain:
Fx   F1  F2  0
 F2   F1
(a)
Geometry of Deformations Relationship
For this configuration, the deformations of both members will be equal; therefore,
1   2
(b)
Force-Temperature-Deformation Relationships
The relationship between internal force, temperature change, and deformation of an axial member can be
stated for members (1) and (2):
FL
FL
1  1 1  1T L1
 2  2 2   2 T L2 (c)
A1E1
A2 E2
Compatibility Equation
Substitute the force-deformation relationships (c) into the geometry of deformation relationship (b) to
derive the compatibility equation:
F1 L1
FL
 1T L1  2 2   2 T L2
(d)
A1 E1
A2 E2
Solve the Equations
Since a limiting stress is specified for brass shell (1), express Eq. (d) in terms of normal stress:
L
L
 1 1  1T L1   2 2   2 T L2
E1
E2
Based on Eq. (a), the normal stress 2 can be expressed in terms of 1 as:
F
F
F A
A
 2  2   1   1 1   1 1
A2
A2
A2 A1
A2
Substitute this expression into Eq. (e) to obtain
L
A L
 1 1  1T L1   1 1 2   2 T L2
E1
A2 E2
(e)
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.
Rearrange terms
L A L 
 1  1  1 2    2 T L2  1T L1  ( 2 L2  1L1 )T
 E1 A2 E2 
and solve for T, recognizing that both the shell and the core have the same length:
L A L 
1 A 1
1  1  1 2  1   1 
E A2 E2 
E A2 E2 
T   1
  1
 2 L2  1 L1
 2  1
(f)
Substitute the problem data into Eq. (f) to compute T that will produce a normal stress of 80 MPa in
the brass shell:


1
1,001.3827 mm 2
1
(80 MPa) 


2
115,000 MPa 962.1128 mm 290,000 MPa 
T 
3.1  106 / C  18.7  106 / C
 62.9983C
Since the problem asks for the largest temperature increase,
Tmax  63.0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.
5.49 At a temperature of 60°F, a 0.04-in. gap exists between the ends of the two bars shown in Fig.
P5.49. Bar (1) is an aluminum alloy [E = 10,000 ksi,  = 0.32,  = 12.5 × 10−6/°F] bar with a width of 3
in. and a thickness of 0.75 in. Bar (2) is a stainless steel [E = 28,000 ksi,  = 0.12,  = 9.6 × 10−6/°F] bar
with a width of 2 in. and a thickness of 0.75 in. The supports at A and C are rigid. Determine:
(a) the lowest temperature at which the two bars
contact each other.
(b) the normal stress in the two bars at a
temperature of 250°F.
(c) the normal strain in the two bars at 250°F.
(d) the change in width of the aluminum bar at a
temperature of 250°F.
Fig. P5.49
Solution
(a) Lowest Contact Temperature
Before the gap is closed, only thermal strains and the associated axial elongations exist. Write
expressions for the temperature-induced elongations and set this equal to the 0.04-in. gap:
1T L1   2 T L2  0.04 in.
T 1 L1   2 L2   0.04 in.
T 
0.04 in.
0.04 in.

 48.6381°F
-6
1 L1   2 L2 (12.5 10 / °F)(32 in.)  (9.6 10-6 / °F)(44 in.)
Since the initial temperature is 60°F, the temperature at which the gap is closed is 108.6°F.
Ans.
(b) Equilibrium
From the results obtained for part (a), we know that the gap will be closed at 250°F, making this a
statically indeterminate axial configuration. Knowing this, consider a FBD at joint B.
Assume that both internal axial forces will be tension, even
though we know intuitively that both F1 and F2 will turn out
to be compression.
Fx   F1  F2  0
(a)
Geometry of Deformations Relationship
For this configuration, the sum of the deformations of members (1) and (2) must equal the initial gap:
1   2  0.04 in.
(b)
Force-Temperature-Deformation Relationships
The relationship between the internal force, temperature change, and the deformation of an axial
member can be stated for members (1) and (2):
FL
FL
1  1 1  1T1L1
 2  2 2   2 T2 L2
A1E1
A2 E2
(c)
Compatibility Equation
Substitute the force-temperature-deformation relationships (c) into the geometry of deformation
relationship (b) to derive the compatibility equation:
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.
F1 L1
FL
 1T1 L1  2 2   2 T2 L2  0.04 in.
A1 E1
A2 E2
(d)
Solve the Equations
This part is no fun, but it must be done. From Eq. (a), F1 = F2. Substituting this into Eq. (d) gives:
F1 L1
FL
 1T1 L1  1 2   2 T2 L2  0.04 in.
A1 E1
A2 E2
F1 L1 F1 L2

 0.04 in.  1T1 L1   2 T2 L2
A1 E1 A2 E2
 L
L 
F1  1  2   0.04 in.  1T1 L1   2 T2 L2
 A1 E1 A2 E2 
0.04 in.  1T1 L1   2 T2 L2
F1 
 L1
L2 
AE  A E 
 1 1
2 2 
(e)
For this structure, the lengths, areas, and elastic moduli are given below. The temperature change is the
same for both members; therefore, T1 = T2 = T = 190°F:
L1  32 in.
L2  44 in.
A1  (3 in.)(0.75 in.)  2.25 in.2
E1  10,000 ksi
1  12.5 10 / °F
6
A2  (2 in.)(0.75 in.)  1.50 in.2
E2  28,000 ksi
 2  9.6 106 / °F
Substitute these values into Eq. (e) and calculate F1 = −47.0702 kips. Backsubstitute into Eq. (a) to
calculate F2 = −47.0702 kips. Note that the internal forces are compression, as expected.
Normal Stresses
The normal stresses in each axial member can now be calculated:
F 47.0702 kips
 20.9201 ksi  20.9 ksi (C)
1  1 
2.25 in.2
A1
2 
F2 47.0702 kips

 31.3801 ksi  31.4 ksi (C)
1.50 in.2
A2
Ans.
(c) Normal Strains
The force-temperature-deformation relationships were expressed in Eq. (b). By definition,  = /L.
Therefore, the normal strain for each axial member can be determined by dividing the relationships in
Eq. (c) by the respective member lengths:
F
F
1  1  1T1
 2  2   2 T2
(f)
A1 E1
A2 E2
Substitute the appropriate values to calculate the normal strains in each member:
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 

F1
 1T1
A1E1
47.0702 kips
 (12.5  106 / °F)(190F)
2
(2.25 in. )(10,000 ksi)
 0.000283 in./in.  283 με
2 

Ans.
F2
  2 T2
A2 E2
47.0702 kips
 (9.6  106 / °F)(190F)
(1.50 in.2 )(28,000 ksi)
 0.000703 in./in.  703 με
Ans.
(d) The change in width of the aluminum bar (1) is caused partly by the Poisson effect and partly by the
temperature change. The longitudinal strain in the aluminum bar caused by the internal force F1 is:
F
47.0702 kips
 2, 092.0110 6 in./in.
 long,  1 
2
A1 E1 (2.25 in. )(10, 000 ksi)
The accompanying lateral strain due to the Poisson effect is thus
 lat,   long,  (0.32)(2,092.01106 in./in.)  669.44 106 in./in.
The lateral strain caused by the temperature change is
 lat,T  1T1  (12.5 106 / °F)(190F)  2,375 106 in./in.
Therefore, the total lateral strain in aluminum bar (1) is
 lat   lat,   lat,T
 669.44 106 in./in.  2,375 106 in./in.  3,044.44 106 in./in.
The change is width of the aluminum bar is thus
width  lat (width)  (3,044.44  106 in./in.)(3 in.)  0.00913 in.
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.
5.50 At a temperature of 5°C, a 3-mm gap exists
between two polymer bars and a rigid support, as
shown in Fig. P5.50. Bar (1) is 50-mm wide and
20-mm thick [E = 800 MPa,  = 140 × 10−6/°C].
Bar (2) is 75-mm wide and 25-mm thick [E = 2.7
GPa,  = 67 × 10−6/°C] bar. The supports at A and
C are rigid. Determine:
(a) the lowest temperature at which the 3-mm gap
is closed.
(b) the normal stress in the two bars at a
temperature of 60°C.
(c) the normal strain in the two bars at 60°C.
Fig. P5.50
Solution
(a) Before the gap is closed, only thermal strains and the associated axial elongations exist. Write
expressions for the temperature-induced elongations and set this equal to the 3-mm gap:
1T L1   2 T L2  3 mm
T 1 L1   2 L2   3 mm
T 
3 mm
3 mm

 24.038°C
-6
1 L1   2 L2 (140 10 / °C)(700 mm)  (67 10-6 / °C)(400 mm)
Since the initial temperature is 5°C, the temperature at which the gap is closed is 29.038°C = 29.0°C. Ans.
Equilibrium
(b) From the results obtained for part (a), we know that the gap will be closed at 60°C, making this a
statically indeterminate axial configuration. Knowing this, consider a FBD at joint B.
Assume that both internal axial forces will be tension, even
though we know intuitively that both F1 and F2 will turn out
to be compression.
Fx   F1  F2  0
(a)
Geometry of Deformations Relationship
For this configuration, the sum of the deformations of members (1) and (2) must equal the initial gap:
1   2  3 mm
(b)
Force-Temperature-Deformation Relationships
The relationship between the internal force, temperature change, and the deformation of an axial
member can be stated for members (1) and (2):
FL
FL
1  1 1  1T1L1
 2  2 2   2 T2 L2
A1E1
A2 E2
Compatibility Equation
Substitute the force-temperature-deformation relationships (c) into the geometry of deformation
relationship (b) to derive the compatibility equation:
F1 L1
FL
 1T1 L1  2 2   2 T2 L2  3 mm
A1 E1
A2 E2
(c)
(d)
Solve the Equations
This part is no fun, but it must be done. From Eq. (a), F1 = F2. Substituting this into Eq. (d) gives:
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.
F1 L1
FL
 1T1 L1  1 2   2 T2 L2  3 mm
A1 E1
A2 E2
F1 L1 F1 L2

 3 mm  1T1 L1   2 T2 L2
A1 E1 A2 E2
 L
L 
F1  1  2   3 mm  1T1 L1   2 T2 L2
 A1 E1 A2 E2 
3 mm  1T1 L1   2 T2 L2
F1 
 L1
L2 
AE  A E 
 1 1
2 2 
(e)
For this structure, the lengths, areas, and elastic moduli are given below. The temperature change is the
same for both members; therefore, T1 = T2 = T = 55°C:
L1  700 mm
L2  400 mm
A1  (50 mm)(20 mm)  1,000 mm2
E1  800 MPa
1  140  10 / °C
6
A2  (75 mm)(25 mm)  1,875 mm 2
E2  2,700 MPa
 2  67  106 / °C
Substitute these values into Eq. (e) and calculate F1 = −4.050262 kN. Backsubstitute into Eq. (a) to
calculate F2 = −4.050262 kN. Note that the internal forces are compression, as expected.
Normal Stresses
The normal stresses in each axial member can now be calculated:
F ( 4.050262 kN)(1,000 N/kN)
 4.050262 MPa  4.05 MPa (C)
1  1 
1,000 mm 2
A1
2 
F2 ( 4.050262 kN)(1,000 N/kN)

 2.160140 MPa  2.16 MPa (C)
1,875 mm 2
A2
Ans.
Normal Strains
The force-temperature-deformation relationships were expressed in Eq. (c). By definition,  = /L.
Therefore, the normal strain for each axial member can be determined by dividing the relationships in
Eq. (c) by the respective member lengths:
F
F
1  1  1T1
 2  2   2 T2
(f)
A1 E1
A2 E2
Substitute the appropriate values to calculate the normal strains in each member:
F
1  1  1T1
A1E1

( 4.050262 kN)(1,000 N/kN)
 (140  106 / °C)(55C)
2
2
(1,000 mm )(800 N/mm )
 0.00263717 mm/mm  2,640 με
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.
2 

F2
  2 T2
A2 E2
( 4.050262 kN)(1,000 N/kN)
 (67  10 6 / °C)(55C)
2
2
(1,875 mm )(2,700 N/mm )
 0.00288495 mm/mm  2,880 με
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.
5.51 The axial assembly shown in Fig. P5.51
consists of a solid 1-in.-diameter aluminum alloy
[E = 10,000 ksi,  = 0.32,  = 12.5 × 10−6/°F] rod
(1) and a solid 1.5-in.-diameter bronze [E =
15,000 ksi,  = 0.15,  = 9.4 × 10−-6/°F] rod (2). If
the supports at A and C are rigid and the assembly
is stress free at 0°F, determine:
(a) the normal stress in both rods at 160°F.
(b) the displacement of flange B.
(c) the change in diameter of the aluminum rod.
Fig. P5.51
Solution
Equilibrium
Consider a FBD at joint B. Assume that both internal
axial forces will be tension.
Fx   F1  F2  0
 F1  F2
(a)
Geometry of Deformations Relationship
1   2  0
(b)
Force-Temperature-Deformation Relationships
FL
FL
1  1 1  1T1L1
 2  2 2   2 T2 L2
A1E1
A2 E2
(c)
Compatibility Equation
F1 L1
FL
 1T1 L1  2 2   2 T2 L2  0
A1 E1
A2 E2
(d)
Solve the Equations
From Eq. (a), F1 = F2. The temperature change is the same for both members; therefore, T1 = T2 =
T. Eq. (d) then can be written as:
F1 L1
FL
 1T L1  1 2   2 T L2  0
A1 E1
A2 E2
Solving for F1:
F1 L1 F1 L2

 1T L1   2 T L2
A1 E1 A2 E2
 L
L 
F1  1  2   T 1 L1   2 L2 
 A1 E1 A2 E2 
T 1 L1   2 L2 
F1  
 L1
L2 
AE  A E 
 1 1
2 2
(e)
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.
For this structure, the lengths, areas, coefficients of thermal expansion, and elastic moduli are given
below.
L1  15 in.
L2  22 in.
A1 

(1 in.) 2  0.7854 in.2
4
E1  10, 000 ksi
A2 

(1.5 in.) 2  1.7671 in.2
4
E2  15, 000 ksi
1  12.5 106 / °F
 2  9.4 106 / °F
Substitute these values along with T = +160°F into Eq. (e) and calculate F1 = −23.0263 kips. From Eq.
(a), F2 = F1 = −23.0263 kips.
(a) Normal Stresses
The normal stresses in each rod can now be calculated:
F 23.0263 kips
1  1 
 29.318 ksi  29.3 ksi (C)
0.7854 in.2
A1
2 
F2 23.0263 kips

 13.030 ksi  13.03 ksi (C)
1.7671 in.2
A2
Ans.
Ans.
(b) Displacement of Flange B
The displacement of flange B is equal to the deformation (i.e., contraction in this instance) of rod (1).
The deformation of rod (1) is given by:
FL
1  1 1  1T1L1
A1E1
( 23.0263 kips)(15 in.)
 (12.5  106 / °F)(160°F)(15 in.)  0.013977 in.
(0.7854 in.2 )(10,000 ksi)
The displacement of flange B is thus:
uB  1  0.01398 in. 

Ans.
(c) Change in diameter of the aluminum rod
The change in diameter of aluminum rod (1) is caused partly by the Poisson effect and partly by the
temperature change. The longitudinal strain in the aluminum rod caused by the internal force F1 is:
F
23.0263 kips
 2,931.78 106 in./in.
 long,  1 
2
A1 E1 (0.7854 in. )(10, 000 ksi)
The accompanying lateral strain due to the Poisson effect is thus
 lat,   long,  (0.32)(2,931.78106 in./in.)  938.17 106 in./in.
The lateral strain caused by the temperature change is
 lat,T  1T1  (12.5 106 / °F)(160F)  2,000 106 in./in.
Therefore, the total lateral strain in aluminum rod (1) is
 lat   lat,   lat,T
 938.17 106 in./in.  2, 000 106 in./in.  2,938.17 106 in./in.
The change in diameter of the aluminum rod is thus
d1  lat d1  (2,938.17  106 in./in.)(1 in.)  0.00294 in.
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.
5.52 The pin-connected structure shown in Fig.
P5.52 consists of a rigid bar ABC, a solid bronze
[E = 100 GPa,  = 16.9 × 10−6/°C] rod (1), and a
solid aluminum alloy [E = 70 GPa,  =
22.5×10−6/°C] rod (2). Bronze rod (1) has a
diameter of 24 and aluminum rod (2) has a
diameter of 16 mm. The bars are unstressed
when the structure is assembled at 25°C. After
assembly, the temperature of rod (2) is
decreased by 40°C while the temperature of rod
(1) remains constant at 25°C. Determine the
normal stresses in both rods for this condition.
Fig. P5.52
Solution
Equilibrium
Consider a FBD of the rigid bar. Assume tension forces in
members (1) and (2). A moment equation about pin B gives
the best information for this situation:
M B  (200 mm)F1  (350 mm)F2  0
(a)
Geometry of Deformations Relationship
Draw a deformation diagram of the rigid bar. The deflections
of the rigid bar are related by similar triangles:
vA
vC

(b)
200 mm 350 mm
There are no gaps, clearances, or other misfits at pins A and
C; therefore, Eq. (c) can be rewritten in terms of the member
deformations as:
1
200 mm

2
350 mm
(c)
Force-Temperature-Deformation Relationships
FL
FL
1  1 1  1T1L1
 2  2 2   2 T2 L2
A1E1
A2 E2
(d)
Compatibility Equation
 F1L1

 F2 L2

1
1
 1T1L1  
  2 T2 L2 


200 mm  A1E1
 350 mm  A2 E2

(e)
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.
Solve the Equations
Solve Eq. (a) for F1:
350 mm
F1  
F2  1.75F2
200 mm
(f)
Substitute this result into Eq. (e):
 ( 1.75F2 ) L1

 F2 L2

1
1
 1T1L1  
  2 T2 L2 


A1E1
200 mm 
 350 mm  A2 E2

Note that T1 = 0°C; therefore, the equation simplifies to:

1.75F2 L1 200 mm  F2 L2


  2 T2 L2 

A1E1
350 mm  A2 E2

Solve this equation for F2:

1.75 F2 L1 200 mm  F2 L2


  2 T2 L2 

350 mm  A2 E2
A1E1

1.75 L1 200 L2 
200

F2 
 2 T2 L2

350
350
A
E
A
E
 1 1
2 2
200
 2 T2 L2
350
(g)
F2  
1.75 L1 200 L2

350 A2 E2
A1E1
For this structure, the lengths, areas, coefficients of thermal expansion, and elastic moduli are given
below.
L1  250 mm
L2  500 mm
A1 

(24 mm) 2  452.3893 mm 2
4
E1  100,000 MPa
A2 

(16 mm) 2  201.0619 mm 2
4
E2  70,000 MPa
1  16.9  10 6 / °C
 2  22.5  10 6 / °C
Substitute these values along with T2 = −40°C into Eq. (g) and calculate F2 = −8,579.65 N. From Eq.
(f), calculate F1 = −15,014.39 N.
Normal Stresses
The normal stresses in each axial member can now be calculated:
F
15,014.39 N
 33.2 MPa (C)
1  1 
A1 452.3893 mm 2
F
8,579.65 N
2  2 
 42.7 MPa (T)
A2 201.0619 mm 2
Ans.
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.
5.53 Rigid bar ABC is supported by two identical solid bronze
[E = 100 GPa,  = 16.9 × 10−6/°C] rods, and a solid steel [E =
200 GPa,  = 11.9×10−6/°C] rod, as shown in Fig. P5.53. The
bronze rods (1) each have a diameter of 16 mm and they are
symmetrically positioned relative to the center rod (2) and the
applied load P. Steel rod (2) has a diameter of 20 mm. The
bars are unstressed when the structure is assembled at 30°C.
When the temperature decreases to –20°C, determine:
(a) the normal stresses in the bronze and steel rods.
(b) the normal strains in the bronze and steel rods.
Fig. P5.53
Solution
Equilibrium: By virtue of symmetry, the forces in the two
bronze rods (1) are identical. Consider a FBD of the rigid
bar. Sum forces in the vertical direction to obtain:
Fy  2 F1  F2  0
(a)
Geometry of Deformations: For this configuration, the deflections of joints A, B, and C are equal:
v A  vB  vC
All pin connections are ideal; therefore, the deflection of joints A and C will cause an identical
deformation of rods (1):
v A  1
and the rigid bar deflection vB will cause an identical deformation of rod (2):
vB   2
Substitute Eqs. (c) and (d) into Eq. (b) to obtain the geometry of deformation equation:
1   2
Force-Temperature-Deformation Relationships: The relationship between internal force,
temperature change, and deformation can be stated for members (1) and (2):
FL
FL
1  1 1  1T L1
 2  2 2   2 T L2
A1E1
A2 E2
Compatibility Equation: Substitute the force-deformation relationships (f) into the geometry of
deformation relationship (e) to derive the compatibility equation:
F1 L1
FL
 1T1 L1  2 2   2 T2 L2
A1 E1
A2 E2
(b)
(c)
(d)
(e)
(f)
(g)
Solve the Equations: For this situation, T1 = T2 = T. Solve Eq. (g) for F2:
FL
AE
F2   1 1  1T1 L1   2 T2 L2  2 2
 A1 E1
 L2
 F1
L1 A2 E2
AE
 T 1 L1   2 L2  2 2
L2 A1 E1
L2
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.
Substitute this expression into Eq. (a)
L A E
AE
2 F1  F1 1 2 2  T 1L1   2 L2  2 2  0
L2 A1 E1
L2
and derive an expression for F1, noting that L1 = L2:

A E 
F1  2  2 2   T 1   2  A2 E2
A1 E1 

T 1   2  A2 E2
F1  
A E
2 2 2
A1 E1
For this structure, the areas, coefficients of thermal expansion, and elastic moduli are given below:

A1 
(16 mm) 2  201.0619 mm 2
4
E1  100,000 MPa
1  16.9  106 / °C
A2 

(h)
(20 mm)2  314.1593 mm2
4
E2  200,000 MPa
 2  11.9  106 / °C
Substitute these values along with T = −50°C into Eq. (h) and calculate F1:
T 1   2  A2 E2
F1  
A E
2 2 2
A1 E1

( 50C) 16.9  10 6  11.9  10 6  (452.3893 mm 2 )(200,000 N/mm 2 )
 314.1593 mm 2   200,000 MPa 
2
 201.0619 mm 2   100,000 MPa 
 3,064.97 N
Backsubstitute this result into Eq. (a) to calculate F2 = –6,129.94 N.
(a) Normal Stresses: The normal stresses in each rod can now be calculated:
3,064.97 N
F
1  1 
 15.24 MPa (T)
A1 201.0619 mm 2
2 
F2
6,129.94 N

 19.51 MPa (C)
A2 314.1593 mm 2
Ans.
Ans.
(b) Normal Strains: The force-temperature-deformation relationships were expressed in Eq. (f). By
definition,  = /L. Therefore, the normal strain for each axial member can be determined by dividing
the relationships in Eq. (f) by the respective member lengths:
F

F

1  1  1T1  1  1T1
 2  2   2 T2  2   2 T2
(f)
A1E1
E1
A2 E2
E2
Substitute the appropriate values to calculate the normal strains in each member:
1 

1
E1
 1T1
15.2439 MPa
 (16.9  106 / °C)(  50C)
100,000 MPa
 0.00069256 mm/mm  693 με
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.
2 

F2
  2 T2
A2 E2
19.5122 MPa
 (11.9  106 / °C)(  50C)
200,000 MPa
 0.00069256 mm/mm  693 με
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.
5.54 A load of P = 85 kN is supported by a structure
consisting of rigid bar ABC, two identical solid bronze
[E = 100 GPa,  = 16.9 × 10−6/°C] rods, and a solid
steel [E = 200 GPa,  = 11.9×10−6/°C] rod, as shown
in Fig. P5.54. The bronze rods (1) each have a
diameter of 20 mm and they are symmetrically
positioned relative to the center rod (2) and the applied
load P. Steel rod (2) has a diameter of 14 mm. The
bars are unstressed when the structure is assembled.
After the temperature of the structure has been
increased by 45°C, determine:
(a) the normal stresses in the bronze and steel rods.
(b) the normal strains in the bronze and steel rods.
(c) the downward deflection of rigid bar ABC.
Fig. P5.54
Solution
Equilibrium: By virtue of symmetry, the forces in the two
bronze rods (1) are identical. Consider a FBD of the rigid
bar. Sum forces in the vertical direction to obtain:
Fy  2 F1  F2  P  0
(a)
Geometry of Deformations: For this configuration, the deflections of joints A, B, and C are equal:
v A  vB  vC
All pin connections are ideal; therefore, the deflection of joints A and C will cause an identical
deformation of rods (1):
v A  1
and the rigid bar deflection vB will cause an identical deformation of rod (2):
vB   2
Substitute Eqs. (c) and (d) into Eq. (b) to obtain the geometry of deformation equation:
1   2
Force-Temperature-Deformation Relationships: The relationship between internal force,
temperature change, and deformation can be stated for members (1) and (2):
FL
FL
1  1 1  1T L1
 2  2 2   2 T L2
A1E1
A2 E2
Compatibility Equation: Substitute the force-deformation relationships (f) into the geometry of
deformation relationship (e) to derive the compatibility equation:
F1 L1
FL
 1T1 L1  2 2   2 T2 L2
A1 E1
A2 E2
(b)
(c)
(d)
(e)
(f)
(g)
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.
Solve the Equations: For this situation, T1 = T2 = T. Solve Eq. (g) for F2:
FL
AE
F2   1 1  1T1 L1   2 T2 L2  2 2
 A1 E1
 L2
L1 A2 E2
AE
 T 1 L1   2 L2  2 2
L2 A1 E1
L2
Substitute this expression into Eq. (a)
L A E
AE
2 F1  F1 1 2 2  T 1 L1   2 L2  2 2  P
L2 A1 E1
L2
and derive an expression for F1:

L A E 
AE
F1  2  1 2 2   P  T 1L1   2 L2  2 2
L2 A1 E1 
L2

 F1
F1 
P  T 1L1   2 L2 
2
A2 E2
L2
L1 A2 E2
L2 A1 E1
For this structure, P = 85 kN = 85,000 N, and the lengths, areas, and elastic moduli are given below:
L1  1, 200 mm
L2  2, 200 mm
A1 

(20 mm) 2  314.1593 mm 2
4
E1  100,000 MPa
A2 
(h)

(14 mm) 2  153.9380 mm 2
4
E2  200,000 MPa
1  16.9  10 6 / °C
 2  11.9  10 6 / °C
Substitute these values along with T = +45°C into Eq. (h) and calculate F1:
AE
P  T 1L1   2 L2  2 2
L2
F1 
L A E
2 1 2 2
L2 A1 E1

(85,000 N)  (45C) (16.9  10 6 )(1, 200)  (11.9  10 6 )(2, 200) 
(153.9380 mm 2 )(200,000 N/mm 2 )
2, 200 mm
2
 1, 200 mm   153.9380 mm   200,000 MPa 
2
 2, 200 mm   314.1593 mm 2   100,000 MPa 
 35,002.53 N
Backsubstitute this result into Eq. (a) to calculate F2 = 14,994.94 N.
(a) Normal Stresses: The normal stresses in each rod can now be calculated:
F
35,002.53 N
1  1 
 111.4 MPa
A1 314.1593 mm 2
2 
F2
14,994.94 N

 97.4 MPa
A2 153.9380 mm 2
Ans.
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) Normal Strains: The force-temperature-deformation relationships were expressed in Eq. (f). By
definition,  = /L. Therefore, the normal strain for each axial member can be determined by dividing
the relationships in Eq. (f) by the respective member lengths:
F
F
1  1  1T1
 2  2   2 T2
(f)
A1 E1
A2 E2
Substitute the appropriate values to calculate the normal strains in each member:
F
1  1  1T1
A1E1

35,002.53 N
 (16.9  106 / °C)(45C)
(314.1593 mm 2 )(100,000 N/mm2 )
 0.00187467 mm/mm  1,875 με
2 

Ans.
F2
  2 T2
A2 E2
14,994.94 N
 (11.9  10 6 / °C)(45C)
2
2
(153.9380 mm )(200,000 N/mm )
 0.0010225 mm/mm  1,023 με
Ans.
(c) Rigid bar deflection: The downward deflection of the rigid bar can be determined from the
deformation of rods (1):
FL
vB  1  1 1  1T1L1
A1E1

(35,002.53N)(1,200 mm)
 (16.9  106 / °C)(45°C)(1,200 mm)
(314.1593 mm2 )(100,000 MPa )
 2.25 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.
5.55 A solid aluminum [E = 70 GPa,  = 22.5 ×
10−6/°C] rod (1) is connected to a solid bronze
[E = 100 GPa,  = 16.9 × 10−6/°C] rod at flange
B, as shown in Fig. P5.55. Aluminum rod (1)
has a diameter of 40 mm and bronze rod (2) has
a diameter of 120 mm. The bars are unstressed
when the structure is assembled at 30°C. After
the 300-kN load is applied to flange B, the
temperature increases to 45°C. Determine:
(a) the normal stresses in rods (1) and (2).
(b) the deflection of flange B.
Fig. P5.55
Solution
Equilibrium: Consider a FBD of flange B. Sum forces in
the horizontal direction to obtain:
Fx   F1  F2  300 kN  0
(a)
Geometry of Deformations:
1   2  0
(b)
Force-Deformation Relationships:
FL
FL
1  1 1  1T1L1
 2  2 2   2 T2 L2
A1E1
A2 E2
(c)
Compatibility Equation: Substitute Eqs. (c) into Eq. (b) to derive the compatibility equation:
F1 L1
FL
 1T1 L1  2 2   2 T2 L2  0
A1 E1
A2 E2
(d)
Solve the Equations: For this situation, T1 = T2 = T. Solve Eq. (d) for F1:
L A E
AE
F1   F2 2 1 1  T 1 L1   2 L2  1 1
L1 A2 E2
L1
Substitute this expression into Eq. (a):
L A E
AE
F2 2 1 1  T 1 L1   2 L2  1 1  F2  300 kN
L1 A2 E2
L1
and solve for F2:
L A E

AE
F2  2 1 1  1  300 kN  T 1 L1   2 L2  1 1
L1
 L1 A2 E2 
AE
300 kN  T 1 L1   2 L2  1 1
L1
F2  
L2 A1 E1
1
L1 A2 E2
(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.
Member lengths, areas, elastic moduli, and coefficients of thermal expansion are given below:
L1  2, 600 mm
L2  1, 000 mm
A1 

(40 mm) 2  1, 256.637 mm 2
4
E1  70, 000 MPa
A2 

(120 mm) 2  11,309.734 mm 2
4
E2  100, 000 MPa
1  22.5 106 / °C
 2  16.9 10 6 / °C
Substitute these values along with T = +15°C into Eq. (f) and compute F2 = −328.4395 kN.
Backsubstitute this result into Eq. (a) to find F1 = −28.4395 kN.
(a) Normal Stresses:
The normal stresses in each rod can now be calculated:
F
28, 439.5 N
1  1 
 22.6 MPa (C)
A1 1,256.637 mm 2
2 
F2
328,439.5 N

 29.0 MPa (C)
A2 11,309.734 mm 2
Ans.
Ans.
(b) Deflection of Flange B:
The deflection of flange B can be determined from the deformation of rod (1):
FL
uB  1  1 1  1T1L1
A1E1

( 28,439.5 N)(2,600 mm)
 (22.5  106 / °C)(15°C)(2,600 mm)
(1,256.637 mm 2 )(70,000 MPa )
 0.0369 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.
5.56 A steel [E = 30,000 ksi,  = 6.6 ×
10−6/°F] pipe column (1) with a crosssectional area of A1 = 5.60 in.2 is connected at
flange B to an aluminum alloy [E = 10,000
ksi,  = 12.5 × 10−6/°F] pipe (2) with a crosssectional area of A2 = 4.40 in.2. The assembly
(shown in Fig. P5.56) is connected to rigid
supports at A and C. It is initially unstressed at
a temperature of 90°F.
(a) At what temperature will the normal stress
in steel pipe (1) be reduced to zero?
(b) Determine the normal stresses in steel pipe
(1) and aluminum pipe (2) when the
temperature reaches –10°F.
Fig. P5.56
Solution
Equilibrium: Consider a FBD of flange B. Sum forces in
the horizontal direction to obtain:
Fx   F1  F2  60 kips  0
(a)
Geometry of Deformations:
1   2  0
Force-Temperature-Deformation Relationships:
FL
FL
1  1 1  1T1L1  2  2 2   2 T2 L2
A1E1
A2 E2
Compatibility Equation: Substitute Eqs. (c) into Eq. (b) to derive the compatibility equation:
F1 L1
FL
 1T1 L1  2 2   2 T2 L2  0
A1 E1
A2 E2
(b)
(c)
(d)
Solve the Equations: Set F1 = 0 and solve Eq. (a) to find F2 = 60 kips. Substitute these values for F1
and F2 into Eq. (d) along with the observation that the temperature change for both axial members is the
same (i.e., T1 = T2 = T) and solve for T:
FL F L
(60 kips)(144 in.)
 1 1 2 2
0
A1 E1 A2 E2
(4.40 in.2 )(10,000 ksi)
T 

 75.758F
(6.6 106 / F)(120 in.)  (12.5 106 / F)(144 in.)
1 L1   2 L2
Since the pipes are initially at a temperature 90°F, the temperature at which the normal stress in steel
pipe (1) is reduced to zero is
T  90F  75.748F  14.24F
Ans.
(b) Solve Eq. (a) for F2 to obtain
F2  F1  60 kips
(e)
When the temperature reaches −10°F, the total change in temperature is T = −100°F. Substitute this
value along with Eq. (e) into the compatibility equation [Eq. (d)] and derive an expression for F1:
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.
F1 L1 ( F1  60 kips)L2

 1T L1   2 T L2
A1 E1
A2 E2

 L
L  (60 kips)L2
F1  1  2  
 T 1 L1   2 L2 
A2 E2
 A1 E1 A2 E2 
 L
L 
(60 kips)L2 
F1  1  2   T 1 L1   2 L2  
A2 E2
 A1 E1 A2 E2 
(60 kips)L2
T 1 L1   2 L2  
A2 E2
F1 
L1
L
 2
A1 E1 A2 E2
and compute F1:
F1 
(100F) (6.6 106 )(120 in.)  (12.5 10 6 )(144 in.)  
(60 kips)(144 in.)
(4.40 in.2 )(10, 000 ksi)
120 in.
144 in.

2
(5.60 in. )(30, 000 ksi) (4.40 in.2 )(10, 000 ksi)
0.062836 in.
0.259200 in.  0.196364 in.


6
6
714.2857 10 in./kip  3, 272.7273 10 in./kip 3,987.0130 10 6 in./kip
 15.7602 kips
From Eq. (a), F2 has a value of
F2  F1  60 kips  15.7602 kips  60 kips  75.7602 kips
(a) Normal Stresses: The normal stresses in each axial member can now be calculated:
F 15.7602 kips
1  1 
 2.8143 ksi  2.81 ksi (T)
5.60 in.2
A1
2 
F2 75.7602 kips

 17.2182 ksi  17.22 ksi (T)
4.40 in.2
A2
Ans.
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.
5.57 A load P will be supported by a structure
consisting of a rigid bar ABCD, a polymer [E =
2,300 ksi,  = 2.9 × 10−6/°F] bar (1) and an
aluminum alloy [E = 10,000 ksi,  = 12.5 ×
10−6/°F] bar (2), as shown in Fig. P5.57. Each bar
has a cross-sectional area of 2.00 in.2. The bars are
unstressed when the structure is assembled at
30°F. After a concentrated load of P = 26 kips is
applied and the temperature is increased to 100°F,
determine:
(a) the normal stresses in bars (1) and (2).
(b) the vertical deflection of joint D.
Fig. P5.57
Solution
Equilibrium
Consider a FBD of the rigid bar. Assume tension
forces in members (1) and (2). A moment equation
about pin A gives the best information for this situation:
M A  (30 in.)F1  (84 in.)F2  (66 in.)P  0
(a)
Geometry of Deformations Relationship
Draw a deformation diagram of the rigid bar. The
deflections of the rigid bar are related by similar
triangles:
vB
v
(b)
 D
30 in. 84 in.
Since there are no gaps, clearances, or other misfits
at pins B and D, the deformation of member (1) will
equal the deflection of the rigid bar at B and the
deformation of member (2) will equal the deflection
of the rigid bar at D. Therefore, Eq. (b) can be
rewritten in terms of the member deformations as:
1
30 in.

2
84 in.
(c)
Force-Temperature-Deformation Relationships
The relationship between the internal force, temperature change, and deformation of an axial member
can be stated for members (1) and (2):
FL
FL
1  1 1  1T1L1
 2  2 2   2 T2 L2
(d)
A1E1
A2 E2
Compatibility Equation
Substitute the force-deformation relationships (d) into the geometry of deformation relationship (c) to
derive the compatibility equation:
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  F1L1
1  F2 L2
 1T1L1  
  2 T2 L2 


30 in.  A1E1
 84 in.  A2 E2

(e)
Solve the Equations
Note that the temperature change for both axial members is the same (i.e., T1 = T2 = T). Solve Eq.
(e) for F1:

F1 L1 30 in.  F2 L2

  2 T L2   1T L1

A1 E1 84 in.  A2 E2

AE
30 in. L2 A1 E1 30 in.

 2 T L2 1 1  1T A1 E1
84 in. L1 A2 E2 84 in.
L1
Substitute this expression into equilibrium equation (a):
 30 in. L2 A1 E1 30 in.

AE

 2 T L2 1 1  1T A1E1   (84 in.)F2  (66 in.)P  0
(30 in.)  F2
L1
 84 in. L1 A2 E2 84 in.

and solve for F2:
AE
(30 in.)2
(66 in.)P 
 2 T L2 1 1  (30 in.)1T A1E1
84 in.
L1
F2 
2
(30 in.) L2 A1 E1
 84 in.
84 in. L1 A2 E2
For this structure, P = 26 kips, and the lengths, areas, elastic moduli, and coefficients of thermal
expansion are listed below:
L1  72 in.
L2  96 in.
F1  F2
A1  2.00 in.2
E1  2,300 ksi
1  2.9 106 / °F
(f)
(g)
A2  2.00 in.2
E2  10, 000 ksi
 2  12.5 106 / °F
Substitute these values along with T = 70°F into Eq. (g) and calculate F2 = 19.3218 kips.
Backsubstitute into Eq. (f) to calculate F1 = 3.0991 kips.
Normal Stresses
The normal stresses in each axial member can now be calculated:
F 3.0991 kips
1  1 
 1.550 ksi (T)
2.00 in.2
A1
2 
F2 19.3218 kips

 9.66 ksi (T)
A2
2.00 in.2
Ans.
Ans.
Deflections of the rigid bar
Calculate the deformation of member (2):
FL
(19.3218 kips)(96 in.)
 2  2 2   2 T2 L2 
 (12.5  10 6 / °F)(70°F)(96 in.)  0.1767 in.
(h)
2
A2 E2
(2.00 in. )(10,000 ksi)
Since there are no gaps at pin D, the rigid bar deflection at D is equal to the deformation of member (2);
therefore:
vD   2  0.1767 in. 
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.
5.58 A load P will be supported by a structure
consisting of a rigid bar ABCD, an aluminum alloy
[E = 10,000 ksi,  = 12.5 × 10−6/°F] bar (1) and a
steel [E = 29,000 ksi,  = 6.5 × 10−6/°F] bar (2), as
shown in Fig. P5.58. Each bar has a crosssectional area of 2.00 in.2. The bars are unstressed
when the structure is assembled at 80°F. After a
concentrated load of P = 42 kips is applied and the
temperature is decreased to –10°F, determine:
(a) the normal stresses in bars (1) and (2).
(b) the vertical deflection of joint D.
Fig. P5.58
Solution
Equilibrium
Consider a FBD of the rigid bar. Assume tension
forces in members (1) and (2). A moment equation
about pin A gives the best information for this situation:
M A  (30 in.)F1  (84 in.)F2  (66 in.)P  0 (a)
Geometry of Deformations Relationship
Draw a deformation diagram of the rigid bar. The
deflections of the rigid bar are related by similar
triangles:
vB
v
(b)
 D
30 in. 84 in.
Since there are no gaps, clearances, or other misfits
at pins B and D, the deformation of member (1) will
equal the deflection of the rigid bar at B and the
deformation of member (2) will equal the deflection
of the rigid bar at D. Therefore, Eq. (b) can be
rewritten in terms of the member deformations as:
1
30 in.

2
84 in.
(c)
Force-Temperature-Deformation Relationships
The relationship between the internal force, temperature change, and deformation of an axial member
can be stated for members (1) and (2):
FL
FL
1  1 1  1T1L1
 2  2 2   2 T2 L2
(d)
A1E1
A2 E2
Compatibility Equation
Substitute the force-deformation relationships (d) into the geometry of deformation relationship (c) to
derive the compatibility equation:


1  F1L1
1  F2 L2
 1T1L1  
  2 T2 L2 
(e)


30 in.  A1E1
 84 in.  A2 E2

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.
Solve the Equations
Note that the temperature change for both axial members is the same (i.e., T1 = T2 = T). Solve Eq.
(e) for F1:

F1 L1 30 in.  F2 L2

  2 T L2   1T L1

A1 E1 84 in.  A2 E2

AE
30 in. L2 A1 E1 30 in.

 2 T L2 1 1  1T A1 E1
84 in. L1 A2 E2 84 in.
L1
Substitute this expression into equilibrium equation (a):
 30 in. L2 A1 E1 30 in.

AE

 2 T L2 1 1  1T A1E1   (84 in.)F2  (66 in.)P  0
(30 in.)  F2
L1
 84 in. L1 A2 E2 84 in.

and solve for F2:
AE
(30 in.)2
 2 T L2 1 1  (30 in.)1T A1E1
(66 in.)P 
L1
84 in.
F2 
2
(30 in.) L2 A1 E1
 84 in.
84 in. L1 A2 E2
For this structure, P = 42 kips, and the lengths, areas, elastic moduli, and coefficients of thermal
expansion are listed below:
L1  72 in.
L2  96 in.
F1  F2
A1  2.00 in.2
E1  10,000 ksi
1  12.5  10 / °F
6
(f)
(g)
A2  2.00 in.2
E2  29,000 ksi
 2  6.5  10 6 / °F
Substitute these values along with T = –90°F into Eq. (g) and calculate F2 = 25.4609 kips.
Backsubstitute into Eq. (f) to calculate F1 = 21.1094 kips.
Normal Stresses
The normal stresses in each axial member can now be calculated:
F 21.1094 kips
1  1 
 10.55 ksi (T)
2.00 in.2
A1
2 
F2 25.4609 kips

 12.73 ksi (T)
A2
2.00 in.2
Ans.
Ans.
Deflections of the rigid bar
Calculate the deformation of member (2):
FL
 2  2 2   2 T2 L2
A2 E2

(25.4609 kips)(96 in.)
(h)
 (6.5  106 / °F)( 90°F)(96 in.)  0.01402 in.
(2.00 in.2 )(29,000 ksi)
Since there are no gaps at pin D, the rigid bar deflection at D is equal to the deformation of member (2);
therefore:
vD   2  0.01402 in. 
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.
5.59 The pin-connected structure shown in Fig.
P5.59 consists of a rigid bar ABCD and two
axial members. Bar (1) is steel [E = 200 GPa, 
= 11.7 × 10−6/°C ] with a cross-sectional area of
A1 = 400 mm2. Bar (2) is an aluminum alloy [E
= 70 GPa,  = 22.5 × 10−6/°C] with a crosssectional area of A2 = 400 mm2. The bars are
unstressed when the structure is assembled.
After a concentrated load of P = 36 kN is
applied and the temperature is increased by
25°C, determine:
(a) the normal stresses in bars (1) and (2).
(b) the deflection of point D on the rigid bar.
Fig. P5.59
Solution
Equilibrium
Consider a FBD of the rigid bar. Assume tension
forces in members (1) and (2). A moment equation
about pin C gives the best information for this situation:
M C  (950 mm)F1  (600 mm)F2
 (720 mm)(36 kN)  0
Geometry of Deformations Relationship
Draw a deformation diagram of the rigid bar. The
deflections of the rigid bar are related by similar
triangles:
vA
vB
vD


950 mm 600 mm 720 mm
(a)
(b)
Since there are no gaps or clearances at either pin A or
pin B, the deformations of members (1) and (2) will
equal the deflections of the rigid bar at A and B,
respectively.
1
950 mm

2
600 mm
(c)
Force-Temperature-Deformation Relationships
The relationship between the internal force, temperature change, and deformation of an axial member
can be stated for members (1) and (2)
FL
FL
1  1 1  1T1L1
 2  2 2   2 T2 L2
(d)
A1E1
A2 E2
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.
Compatibility Equation
Substitute the force-deformation relationships (d) into the geometry of deformation relationship (c) to
derive the compatibility equation:
 F1L1

 F2 L2

1
1
(e)
 1T1L1  
  2 T2 L2 


A
E
950 mm  A1E1
600
mm

 2 2

Solve the Equations
Solve Eq. (e) for F1:



 950  F2 L2
 AE
  2 T2 L2   1T1L1  1 1
F1  




 600  A2 E2
 L1
(f)
and substitute this expression into Eq. (a):
 950  F2 L2
 AE



(950 mm) 
  2 T2 L2   1T1L1  1 1  (600 mm)F2  (720 mm)(36 kN)

 600  A2 E2
 L1



(g)
For this structure, the lengths, areas, elastic moduli, and coefficients of thermal expansion are listed
below:
L1  900 mm
L2  900 mm
A1  400 mm 2
E1  200,000 MPa
1  11.7  10 / °C
6
A2  400 mm 2
E2  70,000 MPa
 2  22.5  10 6 / °C
Substitute these values along with T1 = T2 = +25°C into Eq. (g) and calculate F2 = –3,989.18 N.
Backsubstitute into Eq. (f) to calculate F1 = 29,803.69 N.
(a) Normal stresses:
F 29,803.69 N
1  1 
 74.5 MPa (T)
400 mm 2
A1
F
3,989.18 N
2  2 
 9.97 MPa (C)
A2
400 mm 2
Ans.
Ans.
(b) Deflections of the rigid bar
Calculate the deformation of member (1):
FL
1  1 1  1T1L1
A1E1

(29,803.69 N)(900 mm)
 (11.7  106 / C)(25C)(900 mm)
2
(400 mm )(200,000 MPa)
 0.5985 mm
Since the pin at A is assumed to have a perfect connection, vA = 1 = 0.5985 mm. From Eq. (b),
720 mm
vD 
vA  0.757895(0.5985 mm)  0.454 mm 
950 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.
5.60 The pin-connected structure shown in Fig.
P5.60 consists of a rigid bar ABCD and two
axial members. Bar (1) is steel [E = 200 GPa, 
= 11.7 × 10−6/°C ] with a cross-sectional area of
A1 = 400 mm2. Bar (2) is an aluminum alloy [E
= 70 GPa,  = 22.5 × 10−6/°C] with a crosssectional area of A2 = 400 mm2. The bars are
unstressed when the structure is assembled.
After a concentrated load of P = 36 kN is
applied and the temperature is decreased by
50°C, determine:
(a) the normal stresses in bars (1) and (2).
(b) the deflection of point D on the rigid bar.
Fig. P5.60
Solution
Equilibrium
Consider a FBD of the rigid bar. Assume tension
forces in members (1) and (2). A moment equation
about pin C gives the best information for this situation:
M C  (950 mm)F1  (600 mm)F2
 (720 mm)(36 kN)  0
Geometry of Deformations Relationship
Draw a deformation diagram of the rigid bar. The
deflections of the rigid bar are related by similar
triangles:
vA
vB
vD


950 mm 600 mm 720 mm
(a)
(b)
Since there are no gaps or clearances at either pin A or
pin B, the deformations of members (1) and (2) will
equal the deflections of the rigid bar at A and B,
respectively.
1
950 mm

2
600 mm
(c)
Force-Temperature-Deformation Relationships
The relationship between the internal force, temperature change, and deformation of an axial member
can be stated for members (1) and (2)
FL
FL
1  1 1  1T1L1
 2  2 2   2 T2 L2
(d)
A1E1
A2 E2
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.
Compatibility Equation
Substitute the force-deformation relationships (d) into the geometry of deformation relationship (c) to
derive the compatibility equation:
 F1L1

 F2 L2

1
1
(e)
 1T1L1  
  2 T2 L2 


A
E
950 mm  A1E1
600
mm

 2 2

Solve the Equations
Solve Eq. (e) for F1:



 950  F2 L2
 AE
  2 T2 L2   1T1L1  1 1
F1  




 600  A2 E2
 L1
(f)
and substitute this expression into Eq. (a):
 950  F2 L2
 AE



(950 mm) 
  2 T2 L2   1T1L1  1 1  (600 mm)F2  (720 mm)(36 kN)

 600  A2 E2
 L1



(g)
For this structure, the lengths, areas, elastic moduli, and coefficients of thermal expansion are listed
below:
L1  900 mm
L2  900 mm
A1  400 mm 2
E1  200,000 MPa
1  11.7  10 / °C
6
A2  400 mm 2
E2  70,000 MPa
 2  22.5  10 6 / °C
Substitute these values along with T1 = T2 = –50°C into Eq. (g) and calculate F2 = 23,855.47 N.
Backsubstitute into Eq. (f) to calculate F1 = 12,217.60 N.
(a) Normal stresses:
F 12, 217.60 N
1  1 
 30.5 MPa (T)
400 mm 2
A1
23,855.47 N
F
2  2 
 59.6 MPa (T)
400 mm 2
A2
Ans.
Ans.
(b) Deflections of the rigid bar
Calculate the deformation of member (1):
FL
1  1 1  1T1L1
A1E1

(12, 217.60 N)(900 mm)
 (11.7  10 6 / C)( 50C)(900 mm)
2
(400 mm )(200,000 MPa)
 0.3891 mm
Since the pin at A is assumed to have a perfect connection, vA = 1 = –0.3891 mm (i.e., joint A moves to
the left). From Eq. (b),
720 mm
vD 
vA  0.757895( 0.3891 mm)  0.295 mm 
Ans.
950 mm
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.
5.61 Rigid bar ABCD is loaded and supported as
shown in Fig. P5.61. Bar (1) is made of bronze [E
= 100 GPa,  = 16.9 × 10−6/°C] and has a crosssectional area of 400 mm2. Bar (2) is made of
aluminum [E = 70 GPa,  = 22.5 × 10−6/°C] and
has a cross-sectional area of 600 mm2. Bars (1)
and (2) are initially unstressed. After the
temperature has increased by 40°C, determine:
(a) the stresses in bars (1) and (2).
(b) the vertical deflection of point A.
Fig. P5.61
Solution
Equilibrium
Consider a FBD of the rigid bar. Assume tension forces
in members (1) and (2). A moment equation about pin
D gives the best information for this situation:
M D  (3 m)F1  (1 m)F2  0
 F2  3F1
(a)
Geometry of Deformations Relationship
Draw a deformation diagram of the rigid bar. The
deflections of the rigid bar are related by similar
triangles:
v
vA
v
 B  C
4 m 3m 1m
(b)
There are no gaps, clearances, or other misfits at pins B
and C; therefore, Eq. (b) can be rewritten in terms of the
member deformations as:
1

 2
 1  3 2
(c)
3m 1m
Note: To understand the negative sign associated with 1, see Section 5.5 for discussion of statically
indeterminate rigid bar configurations with opposing members.
Force-Temperature-Deformation Relationships
FL
FL
1  1 1  1T1L1  2  2 2   2 T2 L2
A1E1
A2 E2
(d)
Compatibility Equation
FL

F1L1
 1T1L1  3  2 2   2 T2 L2 
A1E1
 A2 E2

(e)
Solve the Equations
For this situation, T1 = T2 = T = 40°C. Substitute Eq. (a) into Eq. (e):
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.
  3F  L

F1 L1
 1T L1  3  1 2   2 T L2 
A1E1
 A2 E2

and solve for F1:
T  3 2 L2  1 L1 
F1  
L1
9 L2

A1 E1 A2 E2
(40C) 3(22.5 106 / C)(920 mm)  (16.9 106 / C)(840 mm) 
840 mm
9(920 mm)

2
2
(400 mm )(100, 000 N/mm ) (600 mm 2 )(70, 000 N/mm 2 )
 13,990 N  13.990 kN
Backsubstitute into Eq. (a) to find F2 = −41.970 kN.

(a) Normal Stresses
The normal stresses in each axial member can now be calculated:
F 13,990 N
 35.0 MPa (C)
1  1 
A1
400 mm 2
2 
F2 41,970 N

 70.0 MPa (C)
A2
600 mm 2
Ans.
Ans.
(b) Deflection of the rigid bar at A
Calculate the deformation of one of the axial members, say member (1):
FL
1  1 1  1T1L1
A1E1
( 13,990 N)(840 mm)
 (16.9  106 / °C)(40C)(840 mm)
2
2
(400 mm )(100,000 N/mm )
 0.27405 mm
Since there are no gaps at pin B, the rigid bar deflection at B is equal to the deformation of member (1);
therefore, vB = 1 = 0.27405 mm (upward). From similar triangles, the deflection of the rigid bar at A is
related to vB by:
vA
v
 B
4m 3m
The deflection of the rigid bar at A is thus:
4m
4m
vA 
vB 
(0.27405 mm)  0.365 mm 
Ans.
3m
3m

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.
5.62 Rigid bar ABCD in Fig. P5.62 is supported by a
pin connection at A and by two axial bars (1) and (2).
Bar (1) is a 30-in.-long bronze [E = 15,000 ksi,  =
9.4 × 10−6/°F] bar with a cross-sectional area of 1.25
in.2. Bar (2) is a 40-in.-long aluminum alloy [E =
10,000 ksi,  = 12.5 × 10−6/°F] bar with a crosssectional area of 2.00 in.2. Both bars are unstressed
before the load P is applied. If a concentrated load of
P = 27 kips is applied to the rigid bar at D and the
temperature is decreased by 100°F, determine:
(a) the normal stresses in bars (1) and (2).
(b) the normal strains in bars (1) and (2).
(c) the deflection of the rigid bar at point D.
Fig. P5.62
Solution
Equilibrium
Consider a FBD of the rigid bar. Assume tension forces in
members (1) and (2). A moment equation about pin A gives
the best information for this situation:
M A  (36 in.)F1  (84 in.)F2  (98 in.)(27 kips)  0 (a)
Geometry of Deformations Relationship
Draw a deformation diagram of the rigid bar. The deflections
of the rigid bar are related by similar triangles:
v
vB
 C
(b)
36 in. 84 in.
There are no gaps, clearances, or other misfits at pins B and
C; therefore, Eq. (b) can be rewritten in terms of the member
deformations as:
1
2

(c)
36 in. 84 in.
Note: To understand the negative sign associated with 1, see Section 5.5 for discussion of statically
indeterminate rigid bar configurations with opposing members.
Force-Deformation Relationships
FL
FL
1  1 1  1T1L1
 2  2 2   2 T2 L2
A1E1
A2 E2
(d)
Compatibility Equation


1  F1 L1
1  F2 L2

 1T1 L1  
  2 T2 L2 


36 in.  A1 E1
 84 in.  A2 E2

(e)
Solve the Equations
Solve Eq. (e) for F1:
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.
 36 in.  F2 L2
 AE

F1   
  2 T2 L2   1T1 L1  1 1


 84 in.  A2 E2
 L1
 3 F2 L2 3
 AE
 
  2 T2 L2  1T1 L1  1 1
 7 A2 E2 7
 L1
Substitute this expression into equilibrium equation (a) and solve for F2 using T1 = T2 = −100°F:
(36 in.)F1  (84 in.)F2  (98 in.)(27 kips)
  3 F2 L2 3
 AE 
(36 in.)   
  2 T2 L2  1T1 L1  1 1   (84 in.)F2  (98 in.)(27 kips)
 L1 
  7 A2 E2 7
 3 L2 A1 E1

(36 in.)  84 in. F2  (98 in.)(27 kips)

 7 L1 A2 E2

AE
AE
3
(36 in.)  2 T2 L2 1 1  (36 in.)1T1 L1 1 1
7
L1
L1
 3 40 in. 1.25 in.2 15, 000 ksi

(36 in.)  84 in. F2  (98 in.)(27 kips)

2
 7 30 in. 2.00 in. 10, 000 ksi

3
(1.25 in.2 )(15, 000 ksi)
(36 in.) (12.5 106 )(100F)(40 in.)
7
30 in.
2
(1.25 in. )(15, 000 ksi)
(36 in.)(9.4 106 )(100F)(30 in.)
30 in.
2, 646 kip-in.  482.1429 kip-in.  634.5000 kip-in. 3,762.6429 kip-in.

19.2857 in.  84 in.
103.2857 in.
 36.4295 kips
Backsubstitute into Eq. (a) to find F1:
(84 in.)(36.4295 kips)  (98 in.)(27 kips)
F1 
 11.5022 kips
36 in.
F2 
Normal Stresses
The normal stresses in each axial member can now be calculated:
F 11.5022 kips
1  1 
 9.20 ksi (T)
A1
1.25 in.2
2 
F2 36.4295 kips

 18.21 ksi (T)
A2
2.00 in.2
Ans.
Ans.
(b) Normal Strains
The force-temperature-deformation relationships were expressed in Eq. (d). By definition,  = /L.
Therefore, the normal strain for each axial member can be determined by dividing the relationships in
Eq. (d) by their respective member lengths:
F
F
1  1  1T1
 2  2   2 T2
A1 E1
A2 E2
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.
Substitute the appropriate values to calculate the normal strains in each member:
F
1  1  1T1
A1E1

11.5022 kips
 (9.4  106 / °F)(  100F)
2
(1.25 in. )(15,000 ksi)
 326.55  106 in./in.  327 με
2 

Ans.
F2
  2 T2
A2 E2
36.4295 kips
 (12.5  106 / °F)(  100F)
2
(2.00 in. )(10,000 ksi)
 571.48  10 6 in./in.  571 με
Ans.
(c) Deflection of the rigid bar at D
Calculate the deformation of one of the axial members, say member (2):
(36.4295 kips)(40 in.)
FL
 2  2 2   2 T2 L2 
 (12.5  10 6 / °F)(  100F)(40 in.)
(2.00 in.2 )(10,000 ksi)
A2 E2
 0.022859 in.
This deformation can also be determined from the strain in member (2):
 2   2 L2  (571.48  106 in./in.)(40 in.)  0.022859 in.
Since there are no gaps at pin C, the rigid bar deflection at C is equal to the deformation of member (2);
therefore, vC = 2 = 0.022859 in. (downward). From similar triangles, the deflection of the rigid bar at D
is related to vC by:
vC
v
 D
84 in. 98 in.
The deflection of the rigid bar at D is thus:
98 in.
98 in.
vD 
vC 
(0.022859 in.)  0.0267 in. 
84 in.
84 in.
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.
5.63 The pin-connected structure shown in Fig.
P5.63 consists of a rigid bar ABC, a steel bar
(1), and a steel rod (2). The cross-sectional area
of bar (1) is 1.5 in.2 and the diameter of rod (2)
is 0.75 in. Assume E = 30,000 ksi and  = 6.6
× 10−6/°F for both axial members. The bars are
unstressed when the structure is assembled at
70°F. After application of a concentrated force
of P = 20 kips, the temperature is decreased to
30°F. Determine:
(a) the normal stresses in bar (1) and rod (2).
(b) the normal strains in bar (1) and rod (2).
(c) the deflection of pin C from its original
position.
Fig. P5.63
Solution
Equilibrium
Consider a FBD of the rigid bar. Assume tension
forces in members (1) and (2). A moment
equation about pin B gives the best information
for this situation:
M B  (12 in.)F1  (20 in.)F2
(15 in.)(20 kips)  0
(a)
Geometry of Deformations Relationship
Draw a deformation diagram of the rigid bar. The
deflections of the rigid bar are related by similar
triangles:
v
vA
(b)
 C
12 in. 20 in.
There are no gaps, clearances, or other misfits at
pins A and C; therefore, Eq. (b) can be rewritten
in terms of the member deformations as:
1
2

(c)
12 in. 20 in.
Note: To understand the negative sign associated with 1, see Section 5.5 for a discussion of statically
indeterminate rigid bar configurations with opposing members.
Force-Temperature-Deformation Relationships
FL
FL
1  1 1  1T1L1
 2  2 2   2 T2 L2
A1E1
A2 E2
(d)
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.
Compatibility Equation


1  F1L1
1  F2 L2

 1T1L1  
  2 T2 L2 


12 in.  A1E1
 20 in.  A2 E2

(e)
Solve the Equations
Solve Eq. (e) for F1:
 12 in.  F2 L2
 AE

F1   
  2 T2 L2   1T1 L1  1 1


 20 in.  A2 E2
 L1
 12 F2 L2 12
 AE
 
  2 T2 L2  1T1 L1  1 1
 20 A2 E2 20
 L1
Substitute this expression into equilibrium equation (a) and solve for F2 using T1 = T2 = −40°F:
(12 in.)F1  (20 in.)F2  (15 in.)(20 kips)
  12 F2 L2 12
 AE 
(12 in.)   
  2 T2 L2  1T1 L1  1 1   (20 in.)F2  (15 in.)(20 kips)
 L1 
  20 A2 E2 20
Note that the area of rod (2) is A2 = /4(0.75 in.)2 = 0.4418 in.2.

 12 L2 A1 E1

(12 in.)  20 in. F2  (15 in.)(20 kips)

 20 L1 A2 E2

(12 in.)
AE
AE
12
 2 T2 L2 1 1  (12 in.)1T1 L1 1 1
L1
L1
20
 12 80 in. 1.5 in.2 30, 000 ksi

 20 32 in. 0.4418 in.2 30, 000 ksi (12 in.)  20 in. F2  (15 in.)(20 kips)


12
(1.5 in.2 )(30, 000 ksi)
(12 in.) (6.6 106 )( 40F)(80 in.)
20
32 in.
2
(1.5 in. )(30, 000 ksi)
(12 in.)(6.6 106 )( 40F)(32 in.)
32 in.
300 kip-in.  213.84 kip-in.  142.56 kip-in. 656.40 kip-in.

61.113626 in.  20 in.
81.1136267 in.
 8.092 kips
F2 
Backsubstitute into Eq. (a) to find F1:
(20 in.)(8.092 kips)  (15 in.)(20 kips)
F1 
 11.513 kips
12 in.
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.
(a) Normal Stresses
The normal stresses in each axial member can now be calculated:
F 11.513 kips
1  1 
 7.68 ksi (C)
A1
1.5 in.2
8.092 kips
F
2  2 
 18.32 ksi (T)
A2 0.4418 in.2
Ans.
Ans.
(b) Normal Strains
The force-temperature-deformation relationships were expressed in Eq. (d). By definition,  = /L.
Therefore, the normal strain for each axial member can be determined by dividing the relationships in
Eq. (d) by the respective member lengths:
F
F
1  1  1T1
 2  2   2 T2
A1 E1
A2 E2
Substitute the appropriate values to calculate the normal strains in each member:
F
1  1  1T1
A1E1

11.513 kips
 (6.6  10 6 / °F)(  40F)
(1.5 in.2 )(30,000 ksi)
 519.8  106 in./in.  520 με
2 

Ans.
F2
  2 T2
A2 E2
8.092 kips
 (6.6  106 / °F)(  40F)
(0.4418 in.2 )(30,000 ksi)
 346.5  10 6 in./in.  347 με
Ans.
(c) Deflection of the rigid bar at C
Calculate the deformation of one of the axial members, say member (2):
FL
 2  2 2   2 T2 L2
A2 E2

(8.092 kips)(80 in.)
 (6.6  106 / °F)(  40F)(80 in.)
2
(0.4418 in. )(30,000 ksi)
 0.027723 in.
This deformation can also be determined from the strain in member (2):
 2   2 L2  (346.5  10 6 in./in.)(80 in.)  0.027723 in.
Since there are no gaps at pin C, the rigid bar deflection at C is equal to the deformation of member (2);
therefore:
vC   2  0.0277 in. 
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.
5.64 Three rods of different materials are connected
and placed between rigid supports at A and D, as
shown in Fig. P5.64. Properties for each of the three
rods are given below. The bars are initially
unstressed when the structure is assembled at 70°F.
After the temperature has been increased to 250°F,
determine:
(a) the normal stresses in the three rods.
(b) the force exerted on the rigid supports.
(c) the deflections of joints B and C relative to rigid
support A.
Aluminum (1)
L1 = 10 in.
A1 = 0.8 in.2
E1 = 10,000 ksi
 = 12.5×10-6/°F
Fig. P5.64
Cast Iron (2)
L2 = 5 in.
A2 = 1.8 in.2
E2 = 22,500 ksi
 = 7.5×10-6/°F
Bronze (3)
L3 = 7 in.
A3 = 0.6 in.2
E3 = 15,000 ksi
 = 9.4×10-6/°F
Solution
Equilibrium
Consider a FBD at joint B. Assume that both internal axial
forces will be tension.
Fx   F1  F2  0
 F1  F2
(a)
Similarly, consider a FBD at joint C. Assume that both internal
axial forces will be tension.
Fx   F2  F3  0
 F3  F2
(b)
Geometry of Deformations Relationship
1   2   3  0
Force-Temperature-Deformation Relationships
FL
FL
1  1 1  1T1L1
 2  2 2   2 T2 L2
A1E1
A2 E2
Compatibility Equation
FL
F1 L1
FL
 1T1 L1  2 2   2 T2 L2  3 3   3 T3 L3  0
A1 E1
A2 E2
A3 E3
(c)
3 
F3 L3
  3 T3 L3
A3 E3
(d)
(e)
Solve the Equations
From Eq. (a), F1 = F2, and from Eq. (b), F3 = F2. The temperature change is the same for all members;
therefore, T1 = T2 = T3 = T. Eq. (e) then can be written as:
 F2  L1   T L  F2 L2   T L   F2  L3   T L  0
1
1
2
2
3
3
A1E1
A2 E2
A3 E3
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.
Solving for F2:
F2 L1 F2 L2 F3 L3


 1T L1   2 T L2   3T L3
A1 E1 A2 E2 A3 E3
 L
L 
L
F2  1  2  3   T 1 L1   2 L2   3 L3 
 A1 E1 A2 E2 A3 E3 
T 1 L1   2 L2   3 L3 
F2  
L
L1
L
 2  3
A1 E1 A2 E2 A3 E3
(f)
Substitute the problem data along with T = +180°F into Eq. (f) and calculate F1 = −19.1025 kips.
From Eq. (a), F1 = −19.1025 kips and from Eq. (b), F3 = −19.1025 kips.
(a) Normal Stresses
The normal stresses in each rod can now be calculated:
F 19.1025 kips
1  1 
 23.878 ksi  23.9 ksi (C)
0.8 in.2
A1
2 
3 
F2 19.1025 kips

 10.613 ksi  10.61 ksi (C)
A2
1.8 in.2
F3 19.1025 kips

 31.838 ksi  31.8 ksi (C)
0.6 in.2
A3
(b) Force on Rigid Supports
The force exerted on the rigid supports is equal to the internal axial force:
RA  RD  19.10 kips
Ans.
Ans.
Ans.
Ans.
(c) Deflection of Joints B and C
The deflection of joint B is equal to the deformation (i.e., contraction in this instance) of rod (1). The
deformation of rod (1) is given by:
FL
1  1 1  1T1L1
A1E1
( 19.1025 kips)(10 in.)
 (12.5  106 / °F)(180°F)(10 in.)  0.001378 in.
(0.8 in.2 )(10,000 ksi)
The deflection of joint B is thus:
uB  1  0.001378 in. 
The deformation of rod (2) is given by:
FL
 2  2 2   2 T2 L2
A2 E2

( 19.1025 kips)(5 in.)
 (7.5  106 / °F)(180°F)(5 in.)  0.004392 in.
2
(1.8 in. )(22,500 ksi)
The deflection of joint C is:
uC  uB   2  0.001378 in.  0.004392 in.  0.00301 in. 
Ans.

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.
5.65 Three rods of different materials are connected
and placed between rigid supports at A and D, as
shown in Fig. P5.65. Properties for each of the three
rods are given below. The bars are initially
unstressed when the structure is assembled at 20°C.
After the temperature has been increased to 100°C,
determine:
(a) the normal stresses in the three rods.
(b) the force exerted on the rigid supports.
(c) the deflections of joints B and C relative to rigid
support A.
Aluminum (1)
L1 = 440 mm
A1 = 1,200 mm2
E1 = 70 GPa
 = 22.5×10-6/°C
Fig. P5.65
Cast Iron (2)
L2 = 200 mm
A2 = 2,800 mm2
E2 = 155 GPa
 = 13.5×10-6/°C
Bronze (3)
L3 = 320 mm
A3 = 800 mm2
E3 = 100 GPa
 = 17.0×10-6/°C
Solution
Equilibrium
Consider a FBD at joint B. Assume that both internal axial
forces will be tension.
Fx   F1  F2  0
 F1  F2
(a)
Similarly, consider a FBD at joint C. Assume that both internal
axial forces will be tension.
Fx   F2  F3  0
 F3  F2
(b)
Geometry of Deformations Relationship
1   2   3  0
Force-Temperature-Deformation Relationships
FL
FL
1  1 1  1T1L1
 2  2 2   2 T2 L2
A1E1
A2 E2
Compatibility Equation
FL
F1 L1
FL
 1T1 L1  2 2   2 T2 L2  3 3   3 T3 L3  0
A1 E1
A2 E2
A3 E3
(c)
3 
F3 L3
  3 T3 L3
A3 E3
(d)
(e)
Solve the Equations
From Eq. (a), F1 = F2, and from Eq. (b), F3 = F2. The temperature change is the same for all members;
therefore, T1 = T2 = T3 = T. Eq. (e) then can be written as:
 F2  L1   T L  F2 L2   T L   F2  L3   T L  0
1
1
2
2
3
3
A1E1
A2 E2
A3 E3
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.
Solving for F2:
F2 L1 F2 L2 F3 L3


 1T L1   2 T L2   3T L3
A1 E1 A2 E2 A3 E3
 L
L 
L
F2  1  2  3   T 1 L1   2 L2   3 L3 
 A1 E1 A2 E2 A3 E3 
T 1 L1   2 L2   3 L3 
F2  
(f)
L3
L1
L2


A1 E1 A2 E2 A3 E3
Substitute the problem data along with T = +80°C into Eq. (f) and calculate F1 = −148.80 kN. From
Eq. (a), F1 = −148.80 kN and from Eq. (b), F3 = −148.80 kN.
(a) Normal Stresses
The normal stresses in each rod can now be calculated:
F 148,800 N
1  1 
 124.00 MPa  124.0 MPa (C)
A1 1,200 mm 2
2 
3 
F2 148,800 N

 53.143 MPa  53.1 MPa (C)
A2 2,800 mm 2
F3 148,800 N

 186.0 MPa  186.0 MPa (C)
800 mm 2
A3
(b) Force on Rigid Supports
The force exerted on the rigid supports is equal to the internal axial force:
RA  RD  148.8 kN
Ans.
Ans.
Ans.
Ans.
(c) Deflection of Joints B and C
The deflection of joint B is equal to the deformation (i.e., contraction in this instance) of rod (1). The
deformation of rod (1) is given by:
FL
1  1 1  1T1L1
A1E1
( 148,800 N)(440 mm)
 (22.5  106 / °C)(80°C)(440 mm)  0.01257 mm
(1,200 mm 2 )(70,000 N/mm 2 )
The deflection of joint B is thus:
uB  1  0.01257 mm 

Ans.
The deformation of rod (2) is given by:
FL
 2  2 2   2 T2 L2
A2 E2

( 148,800 N)(200 mm)
 (13.5  106 / °C)(80°C)(200 mm)  0.14743 mm
2
2
(2,800 mm )(155,000 N/mm )
The deflection of joint C is:
uC  uB   2  0.01257 mm  0.14743 mm  0.1600 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.
5.66 The machine part shown in Fig. P5.66
is 3/8-in. thick and is made of cold-rolled
18-8 stainless steel (see Appendix D for
properties). Determine the maximum safe
load P if a factor of safety of 2.5 with
respect to failure by yield is specified.
Fig. P5.66
Solution
 allow 
Y
FS

165 ksi
 66 ksi
2.5
Fillet:
D 4.0 in.

 2.0
d 2.0 in.
Pallow 
 allow Amin
K

r 0.5 in.

 0.25
d 2.0 in.
(66 ksi)(2.0 in.)(0.375 in.)
 27.2 kips
1.82
Hole:
d 1.25 in.

 0.3125
D 4.0 in.
Pallow 
 allow Anet
K

 K  1.82
(a)
 K  2.34
(66 ksi)(4.0 in.  1.25 in.)(0.375 in.)
 29.1 kips
2.34
Controlling Load:
The fillet controls in this case; therefore,
Pallow  27.2 kips
(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.
5.67 The machine part shown in Fig. P5.67
is 12 mm thick and is made of SAE 4340
heat-treated steel (see Appendix D for
properties). The holes are centered in the
bar. Determine the maximum safe load P
if a factor of safety of 3.0 with respect to
failure by yield is specified. 
Fig. P5.67
Solution
 allow 
Y
FS

910 MPa
 303.33 MPa
3.0
Small Hole:
d
15 mm

 0.1
D 100 mm
Pallow 
 allow Anet
K

(303.33 N/mm 2 )(100 mm  10 mm)(12 mm)
 120,000 N  120.0 kN
2.73
Large Hole:
d
35 mm

 0.35
D 100 mm
Pallow 
 allow Anet
K

 K  2.73
(a)
 K  2.29
(303.33 N/mm 2 )(100 mm  35 mm)(12 mm)
 103,319 N  103.3 kN
2.29
Controlling Load:
The large hole controls in this case; therefore,
Pallow  103.3 kN
(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.
5.68 A 100-mm-wide by 8-mm-thick steel
bar is transmitting an axial tensile load of
3,000 N. After the load is applied, a 4mm-diameter hole is drilled through the
bar, as shown in Fig. P5.68. The hole is
centered in the bar.
(a) Determine the stress at point A (on the
edge of the hole) in the bar before and
after the hole is drilled.
(b) Does the axial stress at point B on the
edge of the bar increase or decrease as the
hole is drilled? Explain.
Fig. P5.68
Solution
(a) Stress at point A:
Before hole is drilled:
3,000 N
P
A  
 3.75 MPa
A (100 mm)(8 mm)
After hole is drilled:
d
4 mm

 0.04
D 100 mm
A 
Ans.
 K  2.89
PK
(3,000 N)(2.89)

 11.29 MPa
Anet (100 mm  4 mm)(8 mm)
Ans.
(b) Stress at point B:
The axial stress at point B decreases. Since the average stress changes very little with the introduction
of the small hole and since the stress at A is larger than the average stress, the axial stress far from the
hole must be less than the average stress.
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.
5.69 The machine part shown in Fig. P5.69
is 90-mm-wide by 12-mm-thick and is
made of 2014-T4 aluminum (see Appendix
D for properties). The hole is centered in
the bar. Determine the maximum safe load
P if a factor of safety of 1.50 with respect
to failure by yield is specified.
Fig. P5.69
Solution
 allow 
Y
FS

290 MPa
 193.33 MPa
1.50
Hole:
d 30 mm

 0.33
D 90 mm
Pallow 
 allow Anet
K
(193.33 N/mm 2 )(90 mm  30 mm)(12 mm)

 60, 260 N  60.3 kN
2.31
Notches:
r 10 mm

 0.20
d 50 mm
Pallow 
 allow Amin
K
 K  2.31

D 90 mm

 1.8
d 50 mm
 K  2.27
(193.33 N/mm 2 )(50 mm)(12 mm)
 51,101 N  51.1 kN
2.27
Controlling Load:
The notches control in this case; therefore,
Pallow  51.1 kN
(a)
(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.
5.70 The machine part shown in Fig.
P5.70 is 8-mm thick and is made of
AISI 1020 cold-rolled steel (see
Appendix D for properties). Determine
the maximum safe load P if a factor of
safety of 3 with respect to failure by
yield is specified.
Fig. P5.70
Solution
 allow 
Y
FS

427 MPa
 142.33 MPa
3
Fillets:
D 120 mm

 1.5
d
80 mm
Pallow 
 allow Amin
K

r 16 mm

 0.2
d 80 mm
(142.33 MPa)(80 mm)(8 mm)
 49.8 kN
1.83
Hole:
d
15 mm

 0.125
D 120 mm
Pallow 
 allow Anet
K

 allow Amin
K

(a)
 K  2.67
(142.33 MPa)(120 mm  15 mm)(8 mm)
 44.8 kN
2.67
Notches:
r 10 mm

 0.167
d 60 mm
Pallow 
 K  1.83
D 800 mm

 1.333
d
60 mm
 K  2.28
(142.33 MPa)(60 mm)(8 mm)
 30.0 kN
2.28
Controlling Load:
The notches control in this case; therefore,
Pallow  30.0 kN
(b)
(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.
5.71 The machine part shown in Fig. P5.71
is 10 mm thick, is made of AISI 1020 coldrolled steel (see Appendix D for
properties), and is subjected to a tensile
load of P = 45 kN. Determine the minimum
radius r that can be used between the two
sections if a factor of safety of 2 with
respect to failure by yield is specified.
Round the minimum fillet radius up to the
nearest 1-mm multiple.
Fig. P5.71
Solution
 allow 
K
Y
FS

 allow Amin
Pallow
427 MPa
 213.5 MPa
2
(213.5 N/mm2 )(40 mm)(10 mm)

 1.90
45,000 N
D 80 mm

2
d 40 mm
Then, from Fig. 5.15
r
 0.214
d
 rmin  (0.214)d  (0.214)(40 mm)  8.56 mm
 say rmin  9 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.
5.72 The 0.25-in.-thick bar shown in Fig.
P5.72 is made of 2014-T4 aluminum (see
Appendix D for properties) and will be
subjected to an axial tensile load of P =
1,500 lbs. A 0.5625-in.-diameter hole is
located on the centerline of the bar.
Determine the minimum safe width D for
the bar if a factor of safety of 2.5 with
respect to failure by yield must be
maintained.
Fig. P5.72
Solution
 allow 
Y
FS

42 ksi
 16.8 ksi  16,800 psi
2.5
By definition:

K  max
 nom
In this instance, the maximum stress equals the allowable stress:
16,800 psi

K  max 
 K  nom  16,800 psi
 nom
 nom
If we denote the net width of the bar at the hole location as w and the bar thickness at t, then w must
satisfy:
P
P
KP
K  nom  K
K
 16,800 psi
w 
(a)
(16,800 psi) t
Anet
wt
And the bar width D is the sum of the net width w and the hole diameter d:
(b)
D  w d
For a bar with a hole, the value of K is dependent on the ratio of d/D. Since K cannot be determined
until D is known, and since the calculation of D from Eqs. (a) and (b) depends on K, a trial-and-error
process can be established. One possible sequence is summarized in the table below.
d
D
d
D
K
[from Fig. 5.14]
w (in.)
[from Eq. (a)]
D (in.)
[from Eq. (b)]
0.25
2.43
0.868
1.430
0.393
0.39
2.25
0.804
1.366
0.412
0.41
2.23
0.796
1.359
0.414
(agrees with trial value)
Trial value of
Therefore, the minimum bar width is:
Dmin  1.359 in.
Resulting value of
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.
5.73 The stepped bar with a circular hole,
shown in Fig. P5.73, is made of annealed
18-8 stainless steel. The bar is 12 mm thick
and will be subjected to an axial tensile
load of P = 70 kN. The normal stress in the
bar is not to exceed 150 MPa. To the
nearest millimeter, determine:
(a) the maximum allowable hole diameter
d.
(b) the minimum allowable fillet radius r.
Fig. P5.73
Solution
(a) Maximum hole diameter d:
By definition:

K  max
 nom
In this instance, the maximum stress equals the 150-MPa allowable stress:
150 MPa

K  max 
 K  nom  150 MPa
 nom
 nom
If we denote the net width of the bar at the hole location as w and the bar thickness at t, then w can be
expressed as w = D – d. The nominal stress at the hole location depends on Anet, which in turn can be
expressed in terms of w:
P
P
P
KP
K  nom  K
K
K
 150 N/mm 2
 (D  d ) 
Anet
wt
(D  d ) t
(150 N/mm 2 ) t
From this relationship, the hole diameter d is:
KP
d D
(a)
(150 N/mm 2 ) t
For a bar with a hole, the value of K is dependent on the ratio of d/D. Since K cannot be determined
until d is known, and since the calculation of d from Eq. (a) depends on K, a trial-and-error process can
be established. One possible sequence is summarized in the table below.
d
D
d
D
K
[from Fig. 5.14]
d (mm)
[from Eq. (a)]
0.10
0.18
2.73
2.56
23.83
30.44
0.183
0.234
0.23
0.26
0.28
2.47
2.42
2.39
33.94
35.89
37.06
0.29
2.37
37.83
0.261
0.276
0.285
0.291
(agrees with trial value)
Trial value of
Therefore, the maximum hole diameter is:
dmax  37 mm
Resulting value of
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) Minimum fillet radius r:
D 130 mm

 1.3
d 100 mm
Therefore, the curve for D/d = 1.30 on Fig. 5.15 will be used to determine the stress concentration
factor.
 max
 allow
150 N/mm2
K


 2.571
 nom P / Amin (70,000 N) / (100 mm)(12 mm)
From Fig. 5.15,
r
 0.047
d
 rmin  0.047(100 mm)  4.7 mm
say rmin  5 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.
6.1 A solid circular steel shaft having an outside diameter of d = 0.75 in. is subjected to a pure torque of
T = 650 lb-in. Determine the maximum shear stress in the shaft.
Solution

The polar moment of inertia for the shaft is
J
(0.75 in.) 4  0.031063 in.4
32
The maximum shear stress in the steel shaft is found from the elastic torsion formula:
Tc (650 lb-in.)(0.75 in. / 2)
 max 

 7,846.93 psi  7,850 psi
J
0.031063 in.4
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.
6.2 A hollow aluminum shaft with an outside diameter of 80 mm and a wall thickness of 5 mm has an
allowable shear stress of 75 MPa. Determine the maximum torque T that may be applied to the shaft.
Solution


The polar moment of inertia for the shaft is
 D 4  d 4  
(80 mm)4  (70 mm) 4   1,664,062 mm 4
32
32
Rearrange the elastic torsion formula to determine the maximum torque T:
 J (75 N/mm 2 )(1,664,062 mm 4 )
T  allow 
 3,120,117 N-mm  3,120 N-m
c
80 mm / 2
J
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.
6.3 A hollow steel shaft with an outside diameter of 100 mm and a wall thickness of 10 mm is subjected
to a pure torque of T = 5,500 N-m.
(a) Determine the maximum shear stress in the hollow shaft.
(b) Determine the minimum diameter of a solid steel shaft for which the maximum shear stress is the
same as in part (a) for the same torque T.
Solution


The polar moment of inertia for the shaft is
 D 4  d 4  
(100 mm)4  (80 mm) 4   5,796, 238 mm 4
32
32
(a) The maximum shear stress in the hollow steel shaft is found from the elastic torsion formula:
Tc (5,500 N-m)(100 mm / 2)(1,000 mm/m)
 max 

 47.445 MPa  47.4 MPa
J
5,796, 238 mm 4
J
Ans.
(b) The polar moment of inertia for a solid shaft can be expressed as

J  d4
32
Rearrange the elastic torsion formula to group terms with d on the left-hand side:
 d4
T

32 (d / 2) 
and simplify to
d 3 T

16

From this equation, the unknown diameter of the solid shaft can be expressed as
16T
d3

To support a torque of T = 5,500 N-m without exceeding the maximum shear stress determined in part
(a), a solid shaft must have a diameter of
16T
16(5,500 N-m)(1,000 mm/m)
Ans.
d3
3
 83.891 mm  83.9 mm

 (47.445 N/mm2 )
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.
6.4 A compound shaft consists of two pipe
segments. Segment (1) has an outside diameter of
200 mm and a wall thickness of 10 mm. Segment
(2) has an outside diameter of 150 mm and a wall
thickness of 10 mm. The shaft is subjected to
torques TB = 42 kN-m and TC = 18 kN-m, which act
in the directions shown in Fig. P6.4. Determine the
maximum shear stress magnitude in each shaft
segment.
Fig. P6.4
Solution
Equilibrium:
M x  T1  42 kN-m  18 kN-m  0
M x  T2  18 kN-m  0
Section properties:




J1 
J2 
 T1  24 kN-m
 T2  18 kN-m
 D14  d14  
(200 mm) 4  (180 mm) 4   54,019,686 mm 4
32 
32 
 D24  d 24  
(150 mm)4  (130 mm)4   21,551,281 mm 4
32 
32 
Shear stress magnitudes:
(24 kN-m)(200 mm / 2)(1,000 N/kN)(1,000 mm/m)
1 
 44.4 MPa
54,019,686 mm 4
(18 kN-m)(150 mm / 2)(1,000 N/kN)(1,000 mm/m)
2 
 62.3 MPa
21,661,281 mm 4
Ans.
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.
6.5 A compound shaft consists of two pipe
segments. Segment (1) has an outside diameter of
10.75 in. and a wall thickness of 0.365 in. Segment
(2) has an outside diameter of 6.625 in. and a wall
thickness of 0.280 in. The shaft is subjected to
torques TB = 60 kip-ft and TC = 24 kip-ft, which act
in the directions shown in Fig. P6.5. Determine the
maximum shear stress magnitude in each shaft
segment.
Fig. P6.5
Solution
Equilibrium:
M x  T1  60 kip-ft  24 kip-ft  0
M x  T2  24 kip-ft  0
Section properties:




J1 
J2 
 T1  36 kip-ft
 T2  24 kip-ft
 D14  d14  
(10.75 in.) 4  (10.02 in.) 4   321.4685 in.4
32 
32 
 D24  d 24  
(6.625 in.)4  (6.0650 in.) 4   56.2844 in.4
32 
32 
Shear stress magnitudes:
(36 kip-ft)(10.75 in. / 2)(12 in./ft)
1 
 7.22 ksi
321.4685 in.4
(24 kip-ft)(6.625 in. / 2)(12 in./ft)
2 
 16.95 ksi
56.2844 in.4
Ans.
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.
6.6 A compound shaft (Fig. P6.6) consists of brass segment (1) and
aluminum segment (2). Segment (1) is a solid brass shaft with an outside
diameter of 0.625 in. and an allowable shear stress of 6,000 psi. Segment
(2) is a solid aluminum shaft with an outside diameter of 0.50 in. and an
allowable shear stress of 9,000 psi. Determine the magnitude of the largest
torque TC that may be applied at C.
Fig. P6.6
Solution

Section properties:
J1 
J2 

32
(0.625 in.) 4  0.014980 in.4
32
(0.50 in.) 4  0.006136 in.4
Allowable internal torques:
 J (6,000 psi)(0.014980 in.4 )
T1  1 1 
 287.621 lb-in.
c1
0.625 in./2
T2 
 2 J2
c2

(9,000 psi)(0.006136 in.4 )
 220.893 lb-in.
0.50 in./2
 controls
Equilibrium:
The internal torque magnitude in each segment equals the external torque; therefore, T1 = T2 = TC. The
controlling internal torque is T2 = 220.893 lb-in.; therefore, the maximum external torque TC that may be
applied to the compound shaft is
Ans.
TC  220.893 lb-in.  221 lb-in.
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.
6.7 A compound shaft (Fig. P6.7) consists of brass segment (1) and
aluminum segment (2). Segment (1) is a solid brass shaft with an
allowable shear stress of 60 MPa. Segment (2) is a solid aluminum shaft
with an allowable shear stress of 90 MPa. If a torque of TC = 23,000 N-m
is applied at C, determine the minimum required diameter of (a) the brass
shaft and (b) the aluminum shaft.
Fig. P6.7
Solution
The polar moment of inertia for a solid shaft can be expressed as

J  d4
32
Rearrange the elastic torsion formula to group terms with d on the left-hand side:
 d4
T

32 (d / 2) 
and simplify to
d 3 T

16

From this equation, the unknown diameter of the solid shaft can be expressed as
16T
d3

Equilibrium:
For this shaft, the internal torque magnitude in each segment equals the external torque; therefore, T1 =
T2 = TC = 23,000 N-m.
Minimum shaft diameters:
(a) Brass shaft (1)
16T1
16(23,000 N-m)(1,000 mm/m)
d1  3
3
 125.0 mm
 allow,1
 (60 N/mm2 )
(b) Aluminum shaft (2)
16T2
16(23,000 N-m)(1,000 mm/m)
d2  3
3
 109.2 mm
 allow,2
 (90 N/mm 2 )
Ans.
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.
6.8 A solid 0.75-in.-diameter shaft is subjected to
the torques shown in Fig. P6.8. The bearings shown
allow the shaft to turn freely.
(a) Plot a torque diagram showing the internal
torque in segments (1), (2), and (3) of the shaft. Use
the sign convention presented in Section 6-6.
(b) Determine the maximum shear stress magnitude
in the shaft.
Fig. P6.8
Solution
Equilibrium:
M x  T1  10 lb-ft  0
M x  T2  10 lb-ft  50 lb-ft  0
M x  T3  30 lb-ft  0

Section properties:
J1  J 2  J 3 
32
 T1  10 lb-ft
T2  40 lb-ft
 T3  30 lb-ft
(0.75 in.) 4  0.031063 in.4
Shear stress magnitudes:
T c (10 lb-ft)(0.75 in. / 2)(12 in./ft)
1  1 1 
 1, 448.7 psi
J1
0.031063 in.4
2 
3 
T2c2 (40 lb-ft)(0.75 in. / 2)(12 in./ft)

 5,794.7 psi
J2
0.031063 in.4
T3c3 (30 lb-ft)(0.75 in. / 2)(12 in./ft)

 4,346.0 psi
J3
0.031063 in.4
The maximum shear stress in the shaft occurs in segment (2):
 max  5,790 psi
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.
6.9 A solid constant-diameter shaft is subjected to
the torques shown in Fig. P6.9. The bearings shown
allow the shaft to turn freely.
(a) Plot a torque diagram showing the internal
torque in segments (1), (2), and (3) of the shaft. Use
the sign convention presented in Section 6-6.
(b) If the allowable shear stress in the shaft is 80
MPa, determine the minimum acceptable diameter
for the shaft.
Fig. P6.9
Solution
M x  160 N-m  T3  0
M x  160 N-m  380 N-m  T2  0
M x  160 N-m  380 N-m  330 N-m  T1  0
T3  160 N-m
T2  220 N-m
T1  110 N-m
The maximum torque magnitude in the shaft occurs in segment (2): Tmax = 220 N-m.
(b) The elastic torsion formula gives the relationship between shear stress and torque in a shaft.
Tc

J
In this instance, the torque and the allowable shear stress are known for the shaft. Rearrange the elastic
torsion formula, putting the known terms on the right-hand side of the equation:
J T

c 
Express the left-hand side of this equation in terms of the shaft diameter D:

32   d 3  T

d / 2 16
d4
and solve for the minimum acceptable diameter:
16 T 16(220 N-m)(1,000 mm/m)

 14,005.635 mm3
d3 
 
 (80 N/mm 2 )
 d  24.1 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.
6.10 A solid circular steel shaft having an outside diameter of 1.25 in. is subjected to a pure torque of T
= 2,200 lb-in. The shear modulus of the steel is G = 12,000 ksi. Determine:
(a) the maximum shear stress in the shaft.
(b) the magnitude of the angle of twist in a 6-ft length of shaft.
Solution

The polar moment of inertia for the shaft is
J
32
(1.25 in.) 4  0.239684 in.4
(a) The maximum shear stress in the steel shaft is found from the elastic torsion formula:
Tc (2,200 lb-in.)(1.25 in. / 2)
 max 

 5, 736.7 psi  5.74 ksi
0.239684 in.4
J
(b) The magnitude of the angle of twist in a 6-ft length of shaft is
TL
(2, 200 lb-in.)(6 ft)(12 in./ft)


 0.055072 rad  0.0551 rad  3.16
JG (0.239684 in.4 )(12,000,000 psi)
Ans.
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.
6.11 A solid circular steel shaft having an outside diameter of 35 mm is subjected to a pure torque of T =
640 N-m. The shear modulus of the steel is G = 80 GPa. Determine:
(a) the maximum shear stress in the shaft.
(b) the magnitude of the angle of twist in a 1.5-m length of shaft.
Solution

The polar moment of inertia for the shaft is
J
32
(35 mm)4  147,323.515 mm4
(a) The maximum shear stress in the steel shaft is found from the elastic torsion formula:
Tc (640 N-m)(35 mm / 2)(1,000 mm/m)
 max 

 76.023 MPa  76.0 MPa
147,323.515 mm 4
J
(b) The magnitude of the angle of twist in a 1.5-m length of shaft is
(640 N-m)(1.5 m)(1,000 mm/m) 2
TL


 0.081453 rad  0.0815 rad  4.67
JG (147,323.515 mm 4 )(80,000 N/mm 2 )
Ans.
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.
6.12 A hollow steel shaft with an outside diameter of 85 mm and a wall thickness of 10 mm is subjected
to a pure torque of T = 7,000 N-m. The shear modulus of the steel is G = 80 GPa. Determine:
(a) the maximum shear stress in the shaft.
(b) the magnitude of the angle of twist in a 2.5-m length of shaft.
Solution


The polar moment of inertia for the shaft is
J
 D 4  d 4  
(85 mm)4  (65 mm) 4   3,372,303 mm 4
32
32
(a) The maximum shear stress in the steel shaft is found from the elastic torsion formula:
Tc (7,000 N-m)(85 mm / 2)(1,000 mm/m)
 max 

 88.219 MPa  88.2 MPa
J
3,372,303 mm 4
(b) The magnitude of the angle of twist in a 2.5-m length of shaft is
TL (7,000 N-m)(2.5 m)(1,000 mm/m) 2


 0.064867 rad  0.0649 rad  3.72
(3,372,303 mm 4 )(80,000 N/mm 2 )
JG
Ans.
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.
6.13 A solid stainless steel [G = 12,500 ksi] shaft that is 72-in. long will be subjected to a pure torque of
T = 900 lb-in. Determine the minimum diameter required if the shear stress must not exceed 8,000 psi
and the angle of twist must not exceed 5°. Report both the maximum shear stress  and the angle of twist
 at this minimum diameter.
Solution
Consider shear stress:
The polar moment of inertia for a solid shaft can be expressed as

J  d4
32
The elastic torsion formula can be rearranged to gather terms with d:
 d4
d 3 T


32 ( d / 2) 16

From this equation, the unknown diameter of the solid shaft can be expressed as
16T
d3

For the solid stainless steel shaft, the minimum diameter that will satisfy the allowable shear stress is:
16(900 lb-in.)
d3
 0.831 in.
 (8,000 psi)
Consider angle of twist:
Rearrange the angle of twist equation:
TL
 4 TL
J 
d 

JG
32
G
and solve for the minimum diameter that will satisfy the angle of twist limitation:
32TL
32(900 lb-in.)(72 in.)
d4
4
 0.882 in.
G
 (5)( rad/180)(12,500,000 psi)
Therefore, the minimum diameter that could be used for the shaft is
dmin  0.882 in.
Ans.
The angle of twist for this shaft is  = 5° = 0.087266 rad. To compute the shear stress in a 0.882-in.diameter shaft, first compute the polar moment of inertia:
J

(0.882 in.)4  0.059404 in.4
32
The shear stress in the shaft is thus:
(900 lb-in.)(0.882 in. / 2)
 max 
 6,681.12 psi  6,680 psi
0.059404 in.4
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.
6.14 A solid stainless steel [G = 86 GPa] shaft that is 2.0 m long will be subjected to a pure torque of T =
75 N-m. Determine the minimum diameter required if the shear stress must not exceed 50 MPa and the
angle of twist must not exceed 4°. Report both the maximum shear stress  and the angle of twist  at
this minimum diameter.
Solution
Consider shear stress:
The polar moment of inertia for a solid shaft can be expressed as

J  d4
32
The elastic torsion formula can be rearranged to gather terms with d:
 d4
d 3 T



32 ( d / 2) 16
From this equation, the unknown diameter of the solid shaft can be expressed as
16T
d3

For the solid stainless steel shaft, the minimum diameter that will satisfy the allowable shear stress is:
16(75 N-m)(1,000 mm/m)
d3
 19.70 mm
 (50 N/mm2 )
Consider angle of twist:
Rearrange the angle of twist equation:
TL
 4 TL
d 
J 

JG
32
G
and solve for the minimum diameter that will satisfy the angle of twist limitation:
32TL
32(75 N-m)(2,000 mm)(1,000 mm/m)
d4
4
 22.46 mm
G
 (0.069813 rad)(86,000 N/mm 2 )
Therefore, the minimum diameter that could be used for the shaft is
dmin  22.5 mm
The angle of twist for this shaft is  = 0.069813 rad. To compute the shear stress in a 22.46-mmdiameter shaft, first compute the polar moment of inertia:

J  (22.46 mm) 4  24,983.625 mm 4
32
The shear stress in the shaft is thus:
(75 N-m)(22.46 mm / 2)(1,000 mm/m)
 max 
 33.7 MPa
24,983.625 mm 4
Ans.
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.
6.15 A hollow steel [G = 12,000 ksi] shaft with an outside diameter of 3.50 in. will be subjected to a
pure torque of T = 3,750 lb-ft. Determine the maximum inside diameter d that can be used if the shear
stress must not exceed 8,000 psi and the angle of twist must not exceed 3° in an 8-ft length of shaft.
Report both the maximum shear stress  and the angle of twist  for this maximum inside diameter.
Solution
Consider shear stress: The polar moment of inertia for a hollow shaft can be expressed as

 D 4  d 4 
J
32
The elastic torsion formula can be rearranged to gather terms with D:
4
4
  D  d  T

32 D / 2

Rearrange this equation to isolate the inside diameter d term:
32 T D
32 T D
D4  d 4 
 d 4  D4 
2
2
From this equation, the unknown inside diameter of the hollow shaft can be expressed as
32 T D
d  4 D4 
2
For the hollow steel shaft, the maximum diameter that will satisfy the allowable shear stress is:
32(3,750 lb-ft)(3.50 in.)(12 in./ft)
d  4 (3.50 in.)4 
 2.6564 in.
2 (8,000 psi)
Consider angle of twist: Rearrange the angle of twist equation:
TL
TL

 D 4  d 4  
J 

JG
32
G
Rearrange this equation to isolate the inside diameter d term:
32TL
d 4  D4 
 G
and solve for the maximum inside diameter that will satisfy the angle of twist limitation:
d  4 D4 
32 TL
32(3,750 lb-ft)(8 ft)(12 in./ft) 2
 4 (3.50 in.) 4 
 2.991 in.
 G
 (3)( /180)(12,000,000 psi)
Therefore, the maximum inside diameter that could be used for the shaft is
dmax  2.66 in.
Ans.
The shear stress for this shaft is  = 8,000 psi. To compute the angle of twist in an 8-ft length of the
hollow shaft, first compute the polar moment of inertia:


 D 4  d 4  
(3.50 in.)4  (2.66 in.) 4   9.817318 in.4
32
32
The angle of twist in the shaft is thus:
TL
(3,750 lb-ft)(8 ft)(12 in./ft) 2


 0.0367 rad  2.10
JG (9.817318 in.4 )(12,000,000 psi)
J
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.
6.16 A compound steel [G = 80 GPa] shaft
(Fig. P6.16) consists of a solid 55-mmdiameter segment (1) and a solid 40-mmdiameter segment (2). The allowable shear
stress of the steel is 70 MPa, and the
maximum rotation angle at the free end of
the compound shaft must be limited to C ≤
3°. Determine the magnitude of the largest
torque TC that may be applied at C.
Fig. P6.16
Solution

Section properties:
J1 
J2 

32
32
(55 mm)4  898,360.5 mm4
(40 mm) 4  251,327.4 mm 4
Consider shear stress: The larger shear stress will occur in the smaller diameter segment; that is,
segment (2).
 J (70 N/mm 2 )(251,327.4 mm 4 )
Tmax 

 879,646 N-mm
40 mm/2
c
Consider angle of twist: The rotation angle at C is the sum of the angles of twist in segments (1) and
(2):
TL
TL
C  1  2  1 1  2 2
J1G1 J 2G2
Since T1 = T2 = TC and G1 = G2, this equation can be simplified to
Tmax  L1 L2 
    3
G  J1 J 2 
and solved for the maximum torque:
(3)( / 180)(80,000 N/mm 2 )
 739,395 N-mm
Tmax 
1,200 mm 
 800 mm
 898,360.5 mm4  251,327.4 mm4 


Therefore, the magnitude of the largest torque TC that may be applied at C is
TC ,max  739,395 N-mm  739 N-m
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.
6.17 A compound shaft (Fig. P6.17) consists of brass segment (1) and
aluminum segment (2). Segment (1) is a solid brass [G = 5,600 ksi]
shaft with an outside diameter of 1.75 in. and an allowable shear stress
of 9,000 psi. Segment (2) is a solid aluminum [G = 4,000 ksi] shaft
with an outside diameter of 1.25 in. and an allowable shear stress of
12,000 psi. The maximum rotation angle at the upper end of the
compound shaft must be limited to C ≤ 4°. Determine the magnitude
of the largest torque TC that may be applied at C.
Fig. P6.17
Solution
From equilibrium, the internal torques in segments (1) and (2) are equal to the external torque TC. The
elastic torsion formula gives the relationship between shear stress and torque in a shaft.
Tc

J
Consider shear stress: In this compound shaft, the diameters and allowable shear stresses in segments
(1) and (2) are known. The elastic torsion formula can be rearranged to solve for the unknown torque.
An expression can be written for each shaft segment:
J
 J
T1  1 1
T2  2 2
c1
c2
For shaft segment (1), the polar moment of inertia is:
J1 

(1.75 in.)4  0.920772 in.4
32
Use this value along with the 9,000 psi allowable shear stress to determine the allowable torque T1:
 J (9,000 psi)(0.920772 in.4 )
(a)
T1  1 1 
 9,470.8 lb-in.
c1
(1.75 in./2)

For shaft segment (2), the polar moment of inertia is:
J2 
(1.25 in.) 4  0.239684 in.4
32
Use this value along with the 12,000 psi allowable shear stress to determine the allowable torque T2:
 2 J 2 (12,000 psi)(0.239684 in.4 )
(b)
T2 

 4,601.9 lb-in.
c2
(1.25 in./2)
Consider angle of twist: The angles of twists in segments (1) and (2) can be expressed as:
TL
TL
1  1 1
2  2 2
J1G1
J 2G2
The rotation angle at C is the sum of these two angles of twist:
TL
TL
C  1  2  1 1  2 2
J1G1 J 2G2
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.
and since T1 = T2 = TC:
 L
L 
C  TC  1  2 
 J1G1 J 2G2 
C
Solving for TC gives:
TC 

L1
L
 2
J1G1 J 2G2
  rad 
(4) 
 180 
12 in.
18 in.

4
(0.920772 in. )(5,600,000 psi) (0.239684 in.4 )(4,000,000 psi)
 3,308.4 lb-in.
(c)
Compare the torque magnitudes in Eqs. (a), (b), and (c). The smallest torque controls; therefore, the
maximum torque that can be applied to the compound shaft at C is TC = 3,308.4 lb-in. = 276 lb-ft. 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.
6.18 A compound steel [G = 80 GPa] shaft
consists of solid 30-mm-diameter segments
(1) and (3) and tube segment (2), which has
an outside diameter of 60 mm and an inside
diameter of 50 mm (Fig. P6.18). Determine:
(a) the maximum shear stress in tube
segment (2).
(b) the angle of twist in tube segment (2).
(c) the rotation angle of gear D relative to
gear A.
Fig. P6.18
Solution
The torques in all shaft segments are equal; therefore, T1 = T2 = T3 = 240 N-m.
Polar moments of inertia in the shaft segments will be needed for this calculation. For segments (1) and
(3), which are solid 30-mm-diameter shafts, the polar moment of inertia is:
J1 

(30 mm) 4  79,521.56 mm 4  J 3
32
For segment (2), which has a tube cross section, the polar moment of inertia is:
J2 

(60 mm) 4  (50 mm) 4   658,752.71 mm 4
32 
Shear stress in tube segment (2): Use the elastic torsion formula to calculate the shear stress caused by
an internal torque of T2 = 240 N-m. Note that the radius term used in the elastic torsion formula for the
tube is the outside radius; that is, c2 = 60 mm/2 = 30 mm.
Tc
(240 N-m)(30 mm)(1,000 mm/m)
2  2 2 
 10.93 MPa
Ans.
J2
658,752.71 mm 4
Angle of twist in tube segment (2): Apply the angle of twist equation to segment (2).
TL
(240 N-m)(2,000 mm)(1,000 mm/m)
2  2 2 
 0.009108 rad  0.00911 rad
(658,752.71 mm 4 )(80,000 N/mm 2 )
J 2G2
Ans.
Rotation angle of gear D relative to gear A: The angles of twist in segments (1) and (3) must be
calculated. Since both segments have the same shear modulus, polar moment of inertia, and length, they
will both have the same angle of twist:
TL
(240 N-m)(300 mm)(1,000 mm/m)
1  1 1 
 0.011318 rad  3
J1G1 (79,521.56 mm 4 )(80,000 N/mm 2 )
Since gear A is the origin of the coordinate system for this problem, we will arbitrarily define the
rotation angle at gear A to be zero; that is, A = 0. The rotation angle of gear D relative to gear A is
found by adding the angles of twist for the three segments to A:
 D   A  1  2  3
 0  0.011318 rad  0.009108 rad  0.011318 rad
 0.031743 rad  0.0317 rad
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.
6.19 A compound steel [G = 80 GPa] shaft
consists of solid 40-mm-diameter segments
(1) and (3) and tube segment (2), which has
an outside diameter of 75 mm (Fig. P6.19).
If the rotation of gear D relative to gear A
must not exceed 0.01 rad, determine the
maximum inside diameter that may be used
for tube segment (2).
Fig. P6.19
Solution

The polar moment of inertia for shaft segments (1) and (3) is:
J1  J 3 
(40 mm) 4  251,327 mm 4
32
The rotation of gear D relative to gear A is equal to the sum of the angles of twist in segments (1), (2),
and (3):
D  1  2  3
The angles of twist in segments (1) and (3) is
(240 N-m)(300 mm)(1,000 mm/m)
1  3 
 0.003581 rad
(251,327 mm 4 )(80,000 N/mm 2 )
Therefore, the angle of twist that can allowed for segment (2) is
D  1  2  3  0.01 rad
2  0.01 rad  2(0.003581 rad)  0.002838 rad
The minimum polar moment of inertia required for segment (2) is thus:
T2 L2
(240 N-m)(2 m)(1,000 mm/m)2
 0.002838 rad
 J2 
 2,114,165 mm4
2
J 2G2
(80,000 N/mm )(0.002838 rad)
Since the outside diameter of segment (2) is D2 = 75 mm, the maximum inside diameter d2 is:


 D24  d 24  
(75 mm)4  d 24   2,114,165 mm4
32 
32 
 d 2  56.4 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.
6.20 The compound shaft shown in Fig. P6.20 consists of aluminum segment (1) and steel segment (2).
Aluminum segment (1) is a tube with an outside diameter of D1 = 4.00 in., a wall thickness of t1 = 0.25
in., and a shear modulus of G1 = 4,000 ksi. Steel segment (2) is a tube with an outside diameter of D2 =
2.50 in., a wall thickness of t2 = 0.125 in., and a shear modulus of G2 = 12,000 ksi. The compound shaft
is subjected to torques applied at B and C, as shown in Fig. P6.20.
(a) Prepare a diagram that shows the internal
torque and the maximum shear stress in
segments (1), and (2) of the shaft. Use the sign
convention presented in Section 6-6.
(b) Determine the rotation angle of B with
respect to the support at A.
(c) Determine the rotation angle of C with
respect to the support at A.
Fig. P6.20
Solution




Section properties: Polar moments of inertia in the shaft segments will be needed for this calculation.
J1 
J2 
 D14  d14  
(4.00 in.) 4  (3.50 in.) 4   10.4004 in.4
32 
32 
 D24  d 24  
(2.50 in.) 4  (2.25 in.) 4   1.3188 in.4
32 
32 
(a) Equilibrium
M x  950 lb-ft  2,100 lb-ft  T1  0
T1  1,150 lb-ft
M x  950 lb-ft  T2  0
T2  950 lb-ft
Shear stress:
T c (1,150 lb-ft)(4.00 in./2)(12 in./ft)
1  1 1 
 2,653.75 psi  2,650 psi
10.4004 in.4
J1
2 
T2c2 ( 950 lb-ft)(2.50 in./2)(12 in./ft)

 10,804.95 psi  10,800 psi
J2
1.3188 in.4
Ans.
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) Rotation angle of B with respect to A:
1  B   A  B  0  B
Therefore,
(1,150 lb-ft)(9 ft)(12 in./ft)2
TL
B  1  1 1 
 0.035826 rad  0.0358 rad
J1G1 (10.4004 in.4 )(4,000,000 psi)
(c) Rotation angle of C with respect to A:
2  C  B
C  B  2
The angle of twist in shaft (2) is
T2 L2
(950 lb-ft)(6 ft)(12 in./ft)2

 0.051864 rad
2 
J 2G2 (1.3188 in.4 )(12,000,000 psi)
and thus, the rotation angle at C is
C  B  2  0.035826 rad  (0.051864 rad)  0.016038 rad  0.01604 rad
Ans.
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.
6.21 A solid 1.00-in.-diameter steel [G =
12,000 ksi] shaft is subjected to the torques
shown in Fig. P6.21.
(a) Prepare a diagram that shows the internal
torque and the maximum shear stress in
segments (1), (2), and (3) of the shaft. Use
the sign convention presented in Section 6-6.
(b) Determine the rotation angle of pulley C
with respect to the support at A.
(c) Determine the rotation angle of pulley D
with respect to the support at A.
Fig. P6.21
Solution
Section properties: The polar moment of inertia for the solid 1.00-in.-diameter steel shaft segments
will be needed for this calculation.
J1 

32
d14 

32
(1.00 in.)4  0.098175 in.4  J 2  J 3
(a) Equilibrium
M x  240 lb-ft  280 lb-ft  130 lb-ft  T1  0
T1  90 lb-ft
M x  280 lb-ft  130 lb-ft  T2  0
T2  150 lb-ft
M x  130 lb-ft  T3  0
T3  130 lb-ft
Shear stress:
T c ( 90 lb-ft)(1.00 in./2)(12 in./ft)
 5,500.4 psi  5,500 psi
1  1 1 
J1
0.098175 in.4
2 
3 
T2c2 (150 lb-ft)(1.00 in./2)(12 in./ft)

 9,167.3 psi  9,170 psi
J2
0.098175 in.4
T3c3 ( 130 lb-ft)(1.00 in./2)(12 in./ft)

 7,945.0 psi  7,950 psi
J3
0.098175 in.4
The maximum shear stress in the entire shaft is max = 9,170 psi.
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) Rotation angle of C with respect to A:
The angles of twist in the three shaft segments are:
TL
( 90 lb-ft)(36 in.)(12 in./ft)
1  1 1 
 0.033002 rad
J1G1 (0.098175 in.4 )(12,000,000 psi)
(150 lb-ft)(36 in.)(12 in./ft)
TL
2  2 2 
 0.055004 rad
J 2G2 (0.098175 in.4 )(12,000,000 psi)
T3 L3
( 130 lb-ft)(36 in.)(12 in./ft)

 0.047670 rad
J 3G3 (0.098175 in.4 )(12,000,000 psi)
The rotation angle of C with respect to A is found from the sum of the angles of twist in segments (1)
and (2):
Ans.
C  1  2  0.033002 rad  0.055004 rad  0.022002 rad  0.0220 rad
3 
(c) Rotation angle of D with respect to A:
The rotation angle of D with respect to A is found from the sum of the angles of twist in segments (1),
(2), and (3):
 D  1  2  3
 0.033002 rad  0.055004 rad  ( 0.047670 rad)
 0.025669 rad  0.0257 rad
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.
6.22 A compound shaft supports several
pulleys as shown in Fig. P6.22. Segments
(1) and (4) are solid 25-mm-diameter steel
[G = 80 GPa] shafts. Segments (2) and (3)
are solid 50-mm-diameter steel shafts. The
bearings shown allow the shaft to turn
freely.
(a) Determine the maximum shear stress in
the compound shaft.
(b) Determine the rotation angle of pulley D
with respect to pulley B.
(c) Determine the rotation angle of pulley E
with respect to pulley A.
Fig. P6.22
Solution
Section properties: The polar moments of inertia for the shaft segments will be needed for this
calculation.
J1 
J2 


32
32
d14 
d 24 


32
32
(25 mm) 4  38,349.52 mm 4  J 4
(50 mm)4  613,592.32 mm 4  J 3
(a) Equilibrium
M x  T1  175 N-m  0
T1  175 N-m
M x  T2  2,100 N-m  175 N-m  0
T2  1,925 N-m
M x  650 N-m  225 N-m  T3  0
T3  425 N-m
M x  225 N-m  T4  0
T4  225 N-m
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.
Shear stress:
T c ( 175 N-m)(25 mm/2)(1,000 mm/m)
1  1 1 
 57.041 MPa  57.0 MPa
J1
38,349.52 mm 4
Tc
(1,925 N-m)(50 mm/2)(1,000 mm/m)
2  2 2 
 78.432 MPa  78.4 MPa
J2
613,592.32 mm 4
3 
4 
T3c3 (425 N-m)(50 mm/2)(1,000 mm/m)

 17.316 MPa  17.32 MPa
J3
613,592.32 mm 4
T4c4 ( 225 N-m)(25 mm/2)(1,000 mm/m)

 73.339 MPa  73.3 MPa
J4
38,349.52 mm 4
The maximum shear stress in the entire shaft is max = 78.4 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.
(b) Angles of twist:
The angles of twist in the four shaft segments are:
TL
( 175 N-m)(750 mm)(1,000 mm/m)
1  1 1 
 0.042781 rad
J1G1
(38,349.52 mm 4 )(80,000 N/mm 2 )
2 
3 
4 
T2 L2 (1,925 N-m)(500 mm)(1,000 mm/m)

 0.019608 rad
J 2G2
(613,592.32 mm 4 )(80,000 N/mm 2 )
T3 L3
(425 N-m)(625 mm)(1,000 mm/m)

 0.005411 rad
J 3G3 (613,592.32 mm 4 )(80,000 N/mm 2 )
T4 L4 ( 225 N-m)(550 mm)(1,000 mm/m)

 0.040336 rad
J 4G4
(38,349.52 mm 4 )(80,000 N/mm 2 )
Rotation angle of pulley D with respect to pulley B: The rotation angle of D with respect to B is
found from the sum of the angles of twist in segments (2) and (3):
Ans.
D / B  2  3  0.019608 rad  0.005411 rad  0.025019 rad  0.0250 rad
(c) Rotation angle of pulley E with respect to pulley A: The rotation angle of E with respect to A is
found from the sum of the angles of twist in all four segments:
 E  1  2  3  4
 0.042781 rad  0.019608 rad  0.005411 rad  ( 0.040336 rad)
 0.058098 rad  0.0581 rad
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.
6.23 A solid steel [G = 80 GPa] shaft of variable
diameter is subjected to the torques shown in Fig.
P6.23. The diameter of the shaft in segments (1) and
(3) is 50 mm, and the diameter of the shaft in
segment (2) is 80 mm. The bearings shown allow
the shaft to turn freely.
(a) Determine the maximum shear stress in the
compound shaft.
(b) Determine the rotation angle of pulley D with
respect to pulley A.
Fig. P6.23
Solution
Section properties: The polar moments of inertia for the shaft segments will be needed for this
calculation.
J1 
J2 


32
32
(50 mm) 4  613,592.32 mm 4  J 3
(80 mm) 4  4,021, 238.60 mm 4
(a) Equilibrium
M x  1, 200 N-m  T1  0
T1  1, 200 N-m
M x  1, 200 N-m  4,500 N-m  T2  0
T2  3,300 N-m
M x  T3  500 N-m  0
T3  500 N-m
Shear stress:
T c (1, 200 N-m)(50 mm/2)(1,000 mm/m)
1  1 1 
 48.892 MPa  48.9 MPa
J1
613,592.32 mm 4
2 
3 
T2c2 ( 3,300 N-m)(80 mm/2)(1,000 mm/m)

 32.826 MPa  32.8 MPa
J2
4,021,238.60 mm 4
T3c3 ( 500 N-m)(50 mm/2)(1,000 mm/m)

 20.372 MPa  20.4 MPa
J3
613,592.32 mm 4
The maximum shear stress in the compound shaft is  max  1  48.9 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.
Angles of twist:
The angles of twist in the three shaft segments are:
T1L1 (1,200 N-m)(0.7 m)(1,000 mm/m)2
1 

 0.017112 rad
J1G1 (613,592.32 mm4 )(80,000 N/mm2 )
T2 L2 (3,300 N-m)(1.8 m)(1,000 mm/m) 2

 0.018464 rad
2 
J 2G2 (4,021,238.60 mm4 )(80,000 N/mm2 )
T3 L3 ( 500 N-m)(0.7 m)(1,000 mm/m)2

 0.007130 rad
3 
J 3G3 (613,592.32 mm4 )(80,000 N/mm2 )
(b) Rotation angles:
 A  0 rad
 B   A  1  0 rad  0.017112 rad  0.017112 rad
C   B  2  0.017112 rad  (0.018464 rad)  0.001352 rad
 D  C  3  0.001352 rad  ( 0.007130 rad)  0.008482 rad
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.
6.24 A compound shaft drives three gears,
as shown in Fig. P6.24. Segments (1) and
(2) of the compound shaft are hollow
aluminum [G = 4,000 ksi] tubes, which
have an outside diameter of 3.00 in. and a
wall thickness of 0.25 in. Segments (3) and
(4) are solid 2.00-in.-diameter steel [G =
12,000 ksi] shafts. The bearings shown
allow the shaft to turn freely.
(a) Determine the maximum shear stress in
the compound shaft.
(b) Determine the rotation angle of flange
C with respect to flange A.
(c) Determine the rotation angle of gear E
with respect to flange A.
Fig. P6.24
Solution
Section properties: The polar moment of inertia for the solid 1.00-in.-diameter steel shaft segments
will be needed for this calculation.
J1 
J3 


 D14  d14  
(3.00 in.)4  (2.50 in.)4   4.117204 in.4  J 2
32 
32 

32
d34 

32
(2.00 in.) 4  1.570796 in.4  J 4
(a) Equilibrium
M x  14 kip-in.  42 kip-in.
35 kip-in.  T1  0
 T1  7 kip-in.
M x  14 kip-in.  42 kip-in.  T2  0
T2  28 kip-in.
M x  14 kip-in.  42 kip-in.  T3  0
T3  28 kip-in.
M x  14 kip-in.  T4  0
T4  14 kip-in.
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.
Shear stress:
T c ( 7 kip-in.)(3.00 in./2)
1  1 1 
 2.55 ksi
J1
4.117204 in.4
Tc
( 28 kip-in.)(3.00 in./2)
2  2 2 
 10.20 ksi
J2
4.117204 in.4
3 
4 
T3c3 ( 28 kip-in.)(2.00 in./2)

 17.83 ksi
J3
1.570796 in.4
T4c4 (14 kip-in.)(2.00 in./2)

 8.91 ksi
J4
1.570796 in.4
The maximum shear stress in the compound shaft is  max   3  17.83 ksi
Ans.
(b) Angles of Twist: The angles of twist in the four shaft segments are:
TL
(7 kip-in.)(60 in.)
1  1 1 
 0.025503 rad
J1G1 (4.117204 in.4 )(4,000 ksi)
2 
3 
4 
T2 L2
( 28 kip-in.)(6 in.)

 0.010201 rad
J 2G2 (4.117204 in.4 )(4,000 ksi)
T3 L3
( 28 kip-in.)(18 in.)

 0.026738 rad
J 3G3 (1.570796 in.4 )(12,000 ksi)
T4 L4
(14 kip-in.)(12 in.)

 0.008913 rad
J 4G4 (1.570796 in.4 )(12,000 ksi)
 A  0 rad
 B   A  1  0 rad  0.025503 rad  0.0255 rad
C   B  2  0.025503 rad  ( 0.010201 rad)  0.01530 rad
 D  C  3  0.015302 rad  ( 0.026738 rad)  0.01144 rad
 E   D  4  0.011436 rad  0.008913 rad  0.00252 rad
The rotation angle of flange C with respect to A is
C  0.01530 rad
Ans.
(c) Rotation angle of E with respect to A:
E  0.00252 rad
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.
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.
6.25 A compound shaft drives several
pulleys, as shown in Fig. P6.25. Segments (1)
and (2) of the compound shaft are hollow
aluminum [G = 4,000 ksi] tubes, which have
an outside diameter of 3.00 in. and a wall
thickness of 0.125 in. Segments (3) and (4)
are solid 1.50-in.-diameter steel [G = 12,000
ksi] shafts. The bearings shown allow the
shaft to turn freely.
(a) Determine the maximum shear stress in
the compound shaft.
(b) Determine the rotation angle of flange C
with respect to pulley A.
(c) Determine the rotation angle of pulley E
with respect to pulley A.
Fig. P6.25
Solution
(a) Equilibrium
M x  95 lb-ft  T4  0
 T4  95 lb-ft
M x  95 lb-ft  270 lb-ft  T3  0
 T3  175 lb-ft
Note: The internal torque in shaft segment (2) is the same as
in segment (3); therefore, T2 = –175 lb-ft.
M x  T1  650 lb-ft  0
 T1  650 lb-ft
Polar moments of inertia in the shaft segments will be needed for this calculation. Segments (1) and (2)
are hollow tubes with an outside diameter of 3.00 in. and an inside diameter of 3.00 in. – 2(0.125 in.) =
2.75 in. The polar moment of inertia of segments (1) and (2) is:


 D14  d14  
(3.00 in.) 4  (2.75 in.) 4   2.33740 in.4  J 2
32
32
Segments (3) and (4) are solid 1.50-in.-diameter shafts, which have a polar moment of inertia of:
J1 
J3 

32
d34 

32
(1.50 in.)4  0.49701 in.4  J 4
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.
Shear stresses: The shear stresses in each segment can be calculated by the elastic torsion formula:
T c (650 lb-ft)(1.50 in.)(12 in./ft)
1  1 1 
 5,005.56 psi  5,010 psi
J1
2.33740 in.4
Tc
( 175 lb-ft)(1.50 in.)(12 in./ft)
2  2 2 
 1,347.6 psi  1,348 psi
J2
2.33740 in.4
3 
4 
T3c3 ( 175 lb-ft)(0.75 in.)(12 in./ft)

 3,169.0 psi  3,170 psi
J3
0.49701 in.4
T4c4 (95 lb-ft)(0.75 in.)(12 in./ft)

 1,720.3 psi  1,720 psi
J4
0.49701 in.4
The maximum shear stress in the compound shaft is  max  1  5,010 psi
Ans.
Angles of twist:
TL
(650 lb-ft)(72 in.)(12 in./ft)
1  1 1 
 0.060067 rad
J1G1 (2.33740 in.4 )(4,000,000 lb/in.2 )
TL
( 175 lb-ft)(36 in.)(12 in./ft)
2  2 2 
 0.008086 rad
J 2G2 (2.33740 in.4 )(4,000,000 lb/in.2 )
3 
4 
T3 L3
( 175 lb-ft)(54 in.)(12 in./ft)

 0.019014 rad
J 3G3 (0.49701 in.4 )(12,000,000 lb/in.2 )
T4 L4
(95 lb-ft)(54 in.)(12 in./ft)

 0.010322 rad
J 4G4 (0.49701 in.4 )(12,000,000 lb/in.2 )
Rotation angles: The angles of twist can be defined in terms of the rotation angles at the ends of each
segment:
1  B  A
2  C  B
3  D  C
4  E  D
The origin of the coordinate system is located at pulley A. We will arbitrarily define the rotation angle
at pulley A to be zero (A = 0). The rotation angle at B can be calculated from the angle of twist in
segment (1):
1  B   A
 B   A  1  0  0.060067 rad  0.0601 rad
Similarly, the rotation angle at C is determined from the angle of twist in segment (2) and the rotation
angle of pulley B:
2  C  B
 C  B  2  0.060067 rad  (0.008086 rad)  0.051981 rad  0.0520 rad
The rotation angle at D is:
3  D  C
 D  C  3  0.051981 rad  (0.019014 rad)  0.032967 rad  0.0330 rad
and the rotation angle at E is:
4   E   D
 E  D  4  0.032967 rad  0.010322 rad  0.043289 rad  0.0433 rad
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) Rotation angle of flange C with respect to pulley A: Using the rotation angles determined for the
system, the rotation angle of flange C with respect to pulley A is simply:
Ans.
C  0.051981 rad  0.0520 rad
(c) Rotation angle of pulley E with respect to pulley A: Using the rotation angles determined for the
system, the rotation angle of flange C with respect to pulley A is simply:
Ans.
E  0.043289 rad  0.0433 rad
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.
6.26 A torque of TA = 460 lb-ft is applied to gear
A of the gear train shown in Fig. P6.26. The
bearings shown allow the shafts to rotate freely.
(a) Determine the torque TD required for
equilibrium of the system.
(b) Assume shafts (1) and (2) are solid 1.5-in.diameter steel shafts. Determine the magnitude
of the maximum shear stresses acting in each
shaft.
(c) Assume shafts (1) and (2) are solid steel
shafts, which have an allowable shear stress of
6,000 psi. Determine the minimum diameter
required for each shaft.
Fig. P6.26
Solution
(a) Torque TD required for equilibrium:
7 in.
7 in.
805 lb-ft
(460 lb-ft)
TD TA
4 in.
4 in.
Ans.
(b) Shear stress magnitudes if shafts are solid 1.5-in.-diameter:
d4
(1.5 in.) 4 0.497010 in.4
32
32
T1c1 (460 lb-ft)(1.5 in./2)(12 in./ft)
J1
0.497010 in.4
J
1
(805 lb-ft)(1.5 in./2)(12 in./ft)
0.497010 in.4
T2c2
J2
2
8,329.82 psi
14,577.18 psi
8,330 psi
14,580 psi
Ans.
Ans.
(c) Determine minimum diameters: The elastic torsion formula can be rearranged and rewritten in
terms of the unknown diameter d:
Tc
J
J
c
d4
32
d /2
16
d3
T
Using this expression, solve for the minimum acceptable diameter of each shaft:
16 T1
16(460 lb-ft)(12 in./ft)
d13
4.685522 in.3
(6,000
psi)
allow
d1 1.673347 in.
d 23
16 T2
allow
d2
1.673 in.
16(805 lb-ft)(12 in./ft)
(6,000 psi)
2.016502 in.
2.02 in.
Ans.
8.199663 in.3
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.
6.27 A torque of TD = 450 N-m is applied to gear
D of the gear train shown in Fig. P6.27. The
bearings shown allow the shafts to rotate freely.
(a) Determine the torque TA required for
equilibrium of the system.
(b) Assume shafts (1) and (2) are solid 30-mmdiameter steel shafts. Determine the magnitude
of the maximum shear stresses acting in each
shaft.
(c) Assume shafts (1) and (2) are solid steel
shafts, which have an allowable shear stress of
60 MPa. Determine the minimum diameter
required for each shaft.
Fig. P6.27
Solution
(a) Torque TA required for equilibrium:
90 mm
90 mm
TA TD
(450 N-m)
150 mm
150 mm
Ans.
270 N-m
(b) Shear stress magnitudes if shafts are solid 30-mm-diameter:
d4
(30 mm)4 79,521.564 mm4
32
32
T1c1 (270 N-m)(30 mm/2)(1,000 mm/m)
79,521.564 mm 4
J1
J
1
T2c2
J2
2
(450 N-m)(30 mm/2)(1,000 mm/m)
79,521.564 mm 4
50.930 MPa
84.883 MPa
50.9 MPa
Ans.
84.9 MPa
Ans.
(c) Determine minimum diameters: The elastic torsion formula can be rearranged and rewritten in
terms of the unknown diameter d:
Tc
J
J
c
d4
32
d /2
16
d3
T
Using this expression, solve for the minimum acceptable diameter of each shaft:
16 T1
16(270 N-m)(1,000 mm/m)
d13
22,918.312 mm3
2
(60
N/mm
)
allow
d1
d 23
28.405 mm
16 T2
allow
d2
28.4 mm
16(450 N-m)(1,000 mm/m)
(60 N/mm 2 )
33.678 mm
33.7 mm
Ans.
38,197.186 mm3
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.
6.28 The gear train system shown in Fig. P6.28
includes shafts (1) and (2), which are solid 1.375in.-diameter steel shafts. The allowable shear
stress of each shaft is 8,000 psi. The bearings
shown allow the shafts to rotate freely.
Determine the maximum torque TD that can be
applied to the system without exceeding the
allowable shear stress in either shaft.
Fig. P6.28
Solution
Section properties:
J1
32
(1.375 in.) 4
0.350922 in.4
J2
Maximum torque in either shaft:
(8,000 psi)(0.350922 in.4 )
allow J1
Tmax
c1
1.375 in./2
Torque relationship:
T1
T2
or T1
10 in. 6 in.
T2
10 in.
1.667 T2
6 in.
4,083.457 lb-in.
T1 controls
The torque in shaft (1) controls; therefore, T1 = 4,083.457 lb-in. Consequently, the maximum torque in
shaft (2) must be limited to:
6 in.
T2 T1
0.6(4,083.457 lb-in.) 2,450.074 lb-in. 204 lb-ft
Ans.
10 in.
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.
6.29 The gear train system shown in Fig. P6.29
includes shafts (1) and (2), which are solid 20-mmdiameter steel shafts. The allowable shear stress of
each shaft is 50 MPa. The bearings shown allow the
shafts to rotate freely. Determine the maximum
torque TD that can be applied to the system without
exceeding the allowable shear stress in either shaft.
Fig. P6.29
Solution
Section properties:
J1
32
(20 mm)4
15,707.963 mm 4
J2
Maximum torque in either shaft:
(50 N/mm2 )(15,707.963 mm4 )
allow J1
Tmax
c1
20 mm/2
Torque relationship:
T1
T2
200 mm 80 mm
or T1
T2
200 mm
80 mm
78,539.816 N-mm
2.50 T2
T1 controls
The torque in shaft (1) controls; therefore, T1 = 78,539.816 N-mm. Consequently, the maximum torque
in shaft (2) must be limited to:
80 mm
0.4(78,539.816 N-mm) 31,415.927 N-mm 31.4 N-m
T2 T1
Ans.
200 mm
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.
6.30 In the gear system shown in Fig. P6.30, the
motor applies a 210 lb-ft torque to the gear at A. A
torque of TC = 300 lb-ft is removed from the shaft
at gear C, and the remaining torque is removed at
gear D. Segments (1) and (2) are solid 1.5-in.diameter steel [G = 12,000 ksi] shafts, and the
bearings shown allow free rotation of the shaft.
Determine:
(a) the maximum shear stress in segments (1) and
(2) of the shaft.
(b) the rotation angle of gear D relative to gear B.
Fig. P6.30
Solution
Section properties:
J1
32
(1.5 in.) 4
0.497010 in.4
Torque relationship:
TA
TB
hence TB
4 in. 10 in.
TA
J2
10 in.
4 in.
2.50 TA
2.50(210 lb-ft)
525 lb-ft
Shaft (1) has the same torque as gear B; therefore, T1 = 525 lb-ft. A torque of 300 lb-ft is removed from
the shaft at gear C; therefore, TC = 300 lb-ft. The torque in shaft (2) must satisfy equilibrium:
Mx
T1 TC T2 0
T2 T1 TC
To summarize:
T1 525 lb-ft
525 lb-ft 300 lb-ft
T2
225 lb-ft
225 lb-ft
(a) Maximum shear stress in shafts (1) and (2):
T1c1 (525 lb-ft)(1.5 in./2)(12 in./ft)
9,506.856 psi
1
0.497010 in.4
J1
9,510 psi
T2c2 (225 lb-ft)(1.5 in./2)(12 in./ft)
4,074.367 psi 4,070 psi
J2
0.497010 in.4
(b) Angles of twist in shafts (1) and (2):
T1L1
(525 lb-ft)(60 in.)(12 in./ft)
0.063379 rad
1
J1G1 (0.497010 in.4 )(12,000,000 psi)
2
Ans.
Ans.
T2 L2
(225 lb-ft)(40 in.)(12 in./ft)
0.018108 rad
J 2G2 (0.497010 in.4 )(12,000,000 psi)
Note: There is not enough information given to say whether these twist angles are positive or negative.
However, both twist angles will have the same sign.
say B 0 rad
(since this is our reference)
2
D
B
1
2
0 rad 0.063379 rad 0.018108 rad
0.081487 rad
0.0815 rad
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.
6.31 In the gear system shown in Fig. P6.31, the motor applies a 360 lb-ft torque to the gear at A. A
torque of TC = 500 lb-ft is removed from the shaft at gear C, and the remaining torque is removed at gear
D. Segments (1) and (2) are solid steel [G = 12,000 ksi] shafts, and the bearings shown allow free
rotation of the shaft.
(a) Determine the minimum permissible
diameters for segments (1) and (2) of the
shaft if the maximum shear stress must not
exceed 6,000 psi.
(b) If the same diameter is to be used for
both segments (1) and (2), determine the
minimum permissible diameter that can be
used for the shaft if the maximum shear
stress must not exceed 6,000 psi and the
rotation angle of gear D relative to gear B
must not exceed 0.10 rad.
Fig. P6.31
Solution
Torque relationship: The motor applies a torque of 360 lb-ft to gear A. Since gear B is bigger than
gear A, the torque on gear B will be increased in proportion to the gear ratio:
TA
TB
10 in.
hence TB TA
2.50 TA 2.50(360 lb-ft) 900 lb-ft
4 in. 10 in.
4 in.
Let the positive x axis for shaft BCD extend from gear B toward gear D. In order for shaft BCD to be in
equilibrium with the torques as shown on gears C and D, the torque at B must act in the negative x
direction. Consequently, TB = −900 lb-ft.
Draw a free-body diagram that cuts through shaft (1) and includes gear B. From this FBD, the torque in
shaft (1) is:
Mx
900 lb-ft T1 0
T1 900 lb-ft
From the problem statement, we are told that a torque
of 500 lb-ft is removed from the shaft at gear C. In
other words, TC = 500 lb-ft. Draw a FBD that cuts
through shaft (2) and includes both gears B and C.
From this FBD, the torque in shaft (2) is:
Mx
900 lb-ft 500 lb-ft T2 0
T2
400 lb-ft
To summarize:
T1 900 lb-ft
T2
400 lb-ft
(a) Determine minimum diameters: The elastic torsion formula can be rearranged and rewritten in
terms of the unknown diameter d:
Tc
J
J
c
d4
32
d /2
16
d3
T
Using this expression, solve for the minimum acceptable diameter of each shaft:
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.
d13
16 T1
allow
2.092900 in.
d1
d 23
16(900 lb-ft)(12 in./ft)
(6,000 psi)
16 T2
allow
2.09 in.
16(400 lb-ft)(12 in./ft)
(6,000 psi)
1.597178 in.
d2
9.167325 in.3
1.597 in.
Ans.
4.074367 in.3
Ans.
(b) The rotation angle of gear D with respect to gear B is equal to the sum of the angles of twist in shaft
segments (1) and (2):
0.10 rad
D
B
1
2
Since the same solid steel shaft is to be used for both segments (1) and (2):
T1L1 T2 L2
1
0.01 rad
T1L1 T2 L2
J1G1 J 2G2 JG
or rearranging:
(900 lb-ft)(60 in.)(12 in./ft) (400 lb-ft)(40 in.)(12 in./ft)
1.883606 in.4
J
d4
32
(12,000,000 psi)(0.1 rad)
d
1.634 in.
Since the shaft has already been designed for stresses and since the diameter that is required to satisfy
the rotation angle requirement is smaller than d1 determined in part (a), the minimum diameter required
for a constant diameter shaft between gears B and D is:
Ans.
dmin 2.09 in.
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.
6.32 In the gear system shown in Fig. P6.32, the
motor applies a torque of 220 N-m to the gear at
A. A torque of TC = 400 N-m is removed from the
shaft at gear C, and the remaining torque is
removed at gear D. Segments (1) and (2) are solid
40-mm-diameter steel [G = 80 GPa] shafts, and
the bearings shown allow free rotation of the
shaft.
(a) Determine the maximum shear stress in
segments (1) and (2) of the shaft.
(b) Determine the rotation angle of gear D
relative to gear B.
Fig. P6.32
Solution
Section properties:
J1
32
(40 mm)4
Torque relationship:
TA
TB
100 mm 300 mm
251,327.41 mm 4
hence TB
TA
J2
300 mm
100 mm
3TA
3(220 N-m)
660 N-m
Shaft (1) has the same torque as gear B; therefore, T1 = 660 N-m. A torque of TC = 400 N-m is removed
from the shaft at gear C. The torque in shaft (2) must satisfy equilibrium:
Mx
T1 TC T2 0
T2
T1 TC
660 N-m 400 N-m 260 N-m
To summarize:
T1 660 N-m
T2
260 N-m
(a) Maximum shear stress in shafts (1) and (2):
T1c1 (660 N-m)(40 mm/2)(1,000 mm/m)
1
251,327.41 mm 4
J1
2
T2c2
J2
(260 N-m)(40 mm/2)(1,000 mm/m)
251,327.41 mm 4
52.521 MPa
20.690 MPa
52.5 MPa
20.7 MPa
Ans.
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) Angles of twist in shafts (1) and (2):
T1L1
(660 N-m)(1.5 m)(1,000 mm/m) 2
1
J1G1 (251,327.41 mm4 )(80,000 N/mm2 )
0.049239 rad
T2 L2
(260 N-m)(1.0 m)(1,000 mm/m) 2
0.012931 rad
2
J 2G2 (251,327.41 mm4 )(80,000 N/mm 2 )
Note: There is not enough information given to say whether these twist angles are positive or negative.
However, both twist angles will have the same sign.
say
B
D
0 rad
B
(since this is our reference)
1
2
0 rad 0.049239 rad 0.012931 rad
0.062170 rad
0.0622 rad
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.
6.33 In the gear system shown in Fig. P6.33, the
motor applies a torque of 400 N-m to the gear at
A. A torque of TC = 700 N-m is removed from the
shaft at gear C, and the remaining torque is
removed at gear D. Segments (1) and (2) are solid
steel [G = 80 GPa] shafts, and the bearings shown
allow free rotation of the shaft.
(a) Determine the minimum permissible
diameters for segments (1) and (2) of the shaft if
the maximum shear stress must not exceed 40
MPa.
(b) If the same diameter is to be used for
segments (1) and (2), determine the minimum
permissible diameter that can be used for the
shaft if the maximum shear stress must not
exceed 40 MPa and the rotation angle of gear D
relative to gear B must not exceed 3.0°.
Fig. P6.33
Solution
Torque relationship:
TA
TB
100 mm 300 mm
hence TB
TA
300 mm
100 mm
3TA
3(400 N-m) 1, 200 N-m
Shaft (1) has the same torque as gear B; therefore, T1 = 1,200 N-m. A torque of TC = 700 N-m is
removed from the shaft at gear C. The torque in shaft (2) must satisfy equilibrium:
Mx
T1 TC T2 0
T2 T1 TC
To summarize:
T1 1, 200 N-m
1, 200 N-m 700 N-m 500 N-m
T2
500 N-m
(a) Determine minimum diameters: The elastic torsion formula can be rearranged and rewritten in
terms of the unknown diameter d:
Tc
J
J
c
d4
32
d /2
16
d3
T
Using this expression, solve for the minimum acceptable diameter of each shaft:
16 T1
16(1, 200 N-m)(1,000 mm/m)
152,788.745 mm3
d13
2
(40
N/mm
)
allow
d1
d 23
53.460 mm
16 T2
allow
d2
53.5 mm
16(500 N-m)(1,000 mm/m)
(40 N/mm 2 )
39.929 mm
39.9 mm
Ans.
63,661.977 mm3
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 rotation angle of gear D with respect to gear B is equal to the sum of the angles of twist in shaft
segments (1) and (2):
(3 )( /180 ) 0.052360 rad
D
B
1
2
Since the same solid steel shaft is to be used for both segments (1) and (2):
T1L1 T2 L2
1
0.052360 rad
T1L1 T2 L2
J1G1 J 2G2 JG
or rearranging:
J
32
d
d4
(1, 200 N-m)(1.5 m)(1,000 mm/m) 2 (500 N-m)(1.0 m)(1,000 mm/m) 2
(80,000 N/mm 2 )(0.052360 rad)
549,083.270 mm 4
48.6 mm
Since the shaft has already been designed for stresses and since the diameter that is required to satisfy
the rotation angle requirement is smaller than d1 determined in part (a), the minimum diameter required
for a constant diameter shaft between gears B and D is:
Ans.
dmin 53.5 mm
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.
6.34 In the gear system shown in Fig. P6.34, the
motor applies a torque of 250 N-m to the gear at A.
Shaft (1) is a solid 35-mm-diameter shaft, and
shaft (2) is a solid 50-mm-diameter shaft. The
bearings shown allow free rotation of the shafts.
(a) Determine the torque TE provided by the gear
system at gear E.
(b) Determine the maximum shear stresses in
shafts (1) and (2).
Fig. P6.34
Solution
(a) Torque relationship: The torque on gear B is:
72
TB TA
3TA 3(250 N-m) 750 N-m
24
Shaft (1) has the same torque as gear B; therefore, T1 = 750 N-m and TC = 750 N-m. The torque on gear
D is found from:
60
TD TC
2 TC 2(750 N-m) 1,500 N-m
30
Shaft (2) has the same torque as gear D; therefore, T2 = 1,500 N-m and
Ans.
TE 1,500 N-m
To summarize:
T1 750 N-m
T2
1,500 N-m
(b) Section properties:
J1
J2
32
32
(35 mm) 4
147,323.51 mm 4
(50 mm) 4
613,592.32 mm 4
Maximum shear stress in shafts (1) and (2):
T1c1 (750 N-m)(35 mm/2)(1,000 mm/m)
89.090 MPa 89.1 MPa
1
147,323.51 mm 4
J1
T2c2 (1,500 N-m)(50 mm/2)(1,000 mm/m)
61.115 MPa 61.1 MPa
2
J2
613,592.32 mm 4
Ans.
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.
6.35 In the gear system shown in Fig. P6.35, the
motor applies a torque of 600 N-m to the gear at A.
Shafts (1) and (2) are solid shafts, and the bearings
shown allow free rotation of the shafts.
(a) Determine the torque TE provided by the gear
system at gear E.
(b) If the allowable shear stress in each shaft must
be limited to 70 MPa, determine the minimum
permissible diameter for each shaft.
Fig. P6.35
Solution
(a) Torque relationship: The torque on gear B is:
72
3TA 3(600 N-m) 1,800 N-m
TB TA
24
Shaft (1) has the same torque as gear B; therefore, T1 = 1,800 N-m and TC = 1,800 N-m. The torque on
gear D is found from:
60
2 TC 2(1,800 N-m) 3,600 N-m
TD TC
30
Shaft (2) has the same torque as gear D; therefore, T2 = 3,600 N-m and
Ans.
TE 3,600 N-m
To summarize:
T1 1,800 N-m
T2
3,600 N-m
(b) Determine minimum diameters: The elastic torsion formula can be rearranged and rewritten in
terms of the unknown diameter d:
Tc
J
J
c
d4
32
d /2
16
d3
T
Using this expression, solve for the minimum acceptable diameter of each shaft:
16 T1
16(1,800 N-m)(1,000 mm/m)
130,961.782 mm3
d13
2
(70
N/mm
)
allow
d1
d 23
50.783 mm
16 T2
allow
d2
50.8 mm
16(3,600 N-m)(1,000 mm/m)
(70 N/mm 2 )
63.982 mm
64.0 mm
Ans.
261,923.563 mm3
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.
6.36 In the gear system shown in Fig. P6.36, a
torque of TE = 1,500 lb-ft is delivered at gear E.
Shafts (1) and (2) are solid shafts, and the bearings
shown allow free rotation of the shafts.
(a) Determine the torque provided by the motor to
gear A.
(b) If the allowable shear stress in each shaft must
be limited to 4,000 psi, determine the minimum
permissible diameter for each shaft.
Fig. P6.36
Solution
(a) Torque relationship: The torque on gear E creates a torque in shaft (2) of T2 = 1,500 lb-ft.
Accordingly, the torque applied to gear D is also equal to 1,500 lb-ft. The torque on gear C is found
with the gear ratio:
30
TC TD
0.5TD 0.5(1,500 lb-ft) 750 lb-ft
60
Shaft (1) has the same torque as gear C; therefore, T1 = 750 lb-ft and TB = 750 N-m. The torque on gear
A is found from:
24 TC 750 lb-ft
250 lb-ft
TA TB
Ans.
72 3
3
To summarize:
T1 750 lb-ft
T2
1,500 lb-ft
(b) Determine minimum diameters: The elastic torsion formula can be rearranged and rewritten in
terms of the unknown diameter d:
Tc
J
J
c
d4
32
d /2
16
d3
T
Using this expression, solve for the minimum acceptable diameter of each shaft:
16 T1
16(750 lb-ft)(12 in./ft)
11.459156 in.3
d13
(4,000 psi)
allow
d1
d 23
2.255 in.
16 T2
allow
d2
2.26 in.
16(1,500 lb-ft)(12 in./ft)
(4, 000 psi)
2.840 in.
2.84 in.
Ans.
22.918313 in.3
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.
6.37 In the gear system shown in Fig. P6.37, a
torque of TE = 720 lb-ft is delivered at gear E.
Shaft (1) is a solid 1.50-in.-diameter shaft, and
shaft (2) is a solid 2.00-in.-diameter shaft. The
bearings shown allow free rotation of the shafts.
(a) Determine the torque provided by the motor to
gear A.
(b) Determine the maximum shear stresses in
shafts (1) and (2).
Fig. P6.37
Solution
(a) Torque relationship: The torque on gear E creates a torque in shaft (2) of T2 = 720 lb-ft.
Accordingly, the torque applied to gear D is also equal to 720 lb-ft. The torque on gear C is found with
the gear ratio:
30
TC TD
0.5TD 0.5(720 lb-ft) 360 lb-ft
60
Shaft (1) has the same torque as gear C; therefore, T1 = 360 lb-ft and TB = 360 N-m. The torque on gear
A is found from:
24 TC 360 lb-ft
120 lb-ft
TA TB
Ans.
72 3
3
To summarize:
T1 360 lb-ft
T2
720 lb-ft
(b) Section properties:
J1
J2
32
32
(1.50 in.) 4
0.497010 in.4
(2.00 in.) 4
1.570796 in.4
Maximum shear stress in shafts (1) and (2):
T1c1 (360 lb-ft)(1.50 in./2)(12 in./ft)
1
J1
0.497010 in.4
2
T2c2
J2
(720 lb-ft)(2.00 in./2)(12 in./ft)
1.570796 in.4
6,519.98 psi
5,500.396 psi
6,520 psi
5,500 psi
Ans.
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.
6.38 Two solid 30-mm-diameter steel shafts are
connected by the gears shown in Fig. P6.38. The
shaft lengths are L1 = 300 mm and L2 = 500 mm.
Assume that the shear modulus of both shafts is
G = 80 GPa and that the bearings shown allow
free rotation of the shafts. If the torque applied at
gear D is TD = 160 N-m,
(a) determine the internal torques T1 and T2 in
the two shafts.
(b) determine the angles of twist 1 and 2.
(c) determine the rotation angles B and C of
gears B and C.
(d) determine the rotation angle of gear D.
Fig. P6.38
Solution
(a) Torque relationship: The torque on gear D creates a torque in shaft (2)
of T2 = 160 N-m.
M x TD T2 0
T2 TD 160 N-m
Accordingly, the torque applied to gear C is also equal to 160 N-m. The
torque on gear B is found with the gear ratio:
54 teeth
TB
TC
1.8 TC
1.8(160 N-m)
288 N-m
30 teeth
Shaft (1) has the same torque as gear B; therefore, T1 = −288 N-m. To summarize:
T1
288 N-m
T2 160 N-m
Ans.
Section properties:
(30 mm) 4 79,521.56 mm 4 J 2
32
(b) Angles of twist in shafts (1) and (2):
T1L1 ( 288 N-m)(300 mm)(1,000 mm/m)
1
J1G1
(79,521.56 mm 4 )(80,000 N/mm 2 )
J1
2
T2 L2
J 2G2
(160 N-m)(500 mm)(1,000 mm/m)
(79,521.56 mm 4 )(80,000 N/mm 2 )
0.013581 rad
0.012575 rad
0.01358 rad
0.01258 rad
Ans.
Ans.
(c) Rotation angles of gears B and C: The rotation of gear B is found from the angle of twist in shaft
(1):
Ans.
0 rad ( 0.013581 rad)
0.01358 rad
1
B
A
B
A
1
As gear B rotates, gear C also rotates but it rotates in the opposite direction. The magnitude of the
rotation is dictated by the gear ratio:
54 teeth
54 teeth
( 0.013581 rad)
0.024446 rad 0.0244 rad
Ans.
C
B
30 teeth
30 teeth
The rotation of gear D is found from:
Ans.
0.024446 rad 0.012575 rad 0.037021 rad 0.0370 rad
2
D
C
D
C
2
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.
6.39 Two solid 1.75-in.-diameter steel shafts are
connected by the gears shown in Fig. P6.39. The
shaft lengths are L1 = 6 ft and L2 = 10 ft. Assume
that the shear modulus of both shafts is G =
12,000 ksi and that the bearings shown allow
free rotation of the shafts. If the torque applied at
gear D is TD = 225 lb-ft,
(a) determine the internal torques T1 and T2 in
the two shafts.
(b) determine the angles of twist 1 and 2.
(c) determine the rotation angles B and C of
gears B and C.
(d) determine the rotation angle of gear D.
Fig. P6.39
Solution
(a) Torque relationship: The torque on gear D creates a torque in shaft (2)
of T2 = 225 lb-ft.
M x TD T2 0
T2 TD 225 lb-ft
Accordingly, the torque applied to gear C is also equal to 225 lb-ft. The
torque on gear B is found with the gear ratio:
54 teeth
TB
TC
1.8 TC
1.8(225 lb-ft)
405 lb-ft
30 teeth
Shaft (1) has the same torque as gear B; therefore, T1 = −405 lb-ft. To summarize:
T1
405 lb-ft
T2 225 lb-ft
Ans.
Section properties:
(1.75 in.)4 0.920772 in.4 J 2
32
(b) Angles of twist in shafts (1) and (2):
( 405 lb-ft)(6 ft)(12 in./ft)2
T1L1
1
J1G1 (0.920772 in.4 )(12,000,000 psi)
J1
2
T2 L2
J 2G2
(225 lb-ft)(10 ft)(12 in./ft)2
(0.920772 in.4 )(12,000,000 psi)
0.031669 rad
0.029323 rad
Ans.
0.0317 rad
Ans.
0.0293 rad
(c) Rotation angles of gears B and C: The rotation of gear B is found from the twist angle in shaft (1):
Ans.
0.0317 rad
0 rad ( 0.031669 rad)
B
A
B
A
1
1
As gear B rotates, gear C also rotates but it rotates in the opposite direction. The magnitude of the
rotation is dictated by the gear ratio:
54 teeth
54 teeth
( 0.031669 rad)
0.057004 rad 0.0570 rad
Ans.
C
B
30 teeth
30 teeth
(d) Rotation angle of gear D:The rotation of gear D is found from:
0.057004 rad 0.029323 rad 0.086327 rad
2
D
C
D
C
2
0.0863 rad
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.
6.40 Two solid steel shafts are connected by the
gears shown in Fig. P6.40. The design
requirements for the system require (1) that both
shafts must have the same diameter, (2) that the
maximum shear stress in each shaft must be less
than 6,000 psi, and (3) that the rotation angle of
gear D must not exceed 3°. Determine the
minimum required diameter of the shafts if the
torque applied at gear D is TD = 345 lb-ft. The
shaft lengths are L1 = 10 ft and L2 = 8 ft.
Assume that the shear modulus of both shafts is
G = 12,000 ksi and that the bearings shown
allow free rotation of the shafts.
Fig. P6.40
Solution
Torque relationship: The torque on gear D creates a torque in shaft (2) of
T2 = 1,000 lb-ft.
M x TD T2 0
T2 TD 345 lb-ft 4,140 lb-in.
Accordingly, the torque applied to gear C is also 4,140 lb-in. The torque on
gear B is found with the gear ratio:
48 teeth
TB
TC
0.6667 TC
0.6667(4,140 lb-in.)
2,760 lb-in.
72 teeth
Shaft (1) has the same torque as gear B; therefore, T1 = −2,760 lb-in.
Diameters based on allowable shear stresses: The elastic torsion formula gives the relationship
between shear stress and torque in a shaft.
Tc
J
The torques in both shafts have been determined, and the allowable shear stress is specified. Rearrange
the elastic torsion formula, putting the known terms on the right-hand side of the equation:
J T
c
Express the left-hand side of this equation in terms of the shaft diameter D:
d4
32
d /2
16
d3
T
Solve for the minimum acceptable diameter in shaft (1):
16 T1 16(2,760 lb-in.)
d13
2.342761 in.3
2
(6,000 lb/in. )
d1 1.328 in.
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.
Solve for the minimum acceptable diameter in shaft (2):
16 T2 16(4,140 lb-in.)
3.514141 in.3
d 23
(6,000 lb/in.2 )
d 2 1.520 in.
Of these two values, d2 controls. Therefore, both shafts could have a diameter of 1.520 in. or more and
the shear stress constraint would be satisfied.
The angles of twists in shafts (1) and (2) can be expressed as:
T1L1
T2 L2
1
2
J1G1
J 2G2
The rotations of gears B and C are related since the arclengths turned by the two gears have the same
magnitude. The gears turn in opposite directions; therefore, a negative sign is introduced.
RC C
RB B
The rotation angle of gear B is equal to the angle of twist in shaft (1): B = 1. Therefore, the rotation
angle of gear C can be expressed in terms of T1.
RB
RB
RB T1L1
N B T1L1
C
B
1
RC
RC
RC J1G1
N C J1G1
The rotation angle of gear D is equal to the rotation angle of gear C plus the twist that occurs in shaft
(2):
D
C
2
and so the rotation angle of gear D can be expressed in terms of the torques T1 and T2:
N B T1L1 T2 L2
3
D
N C J1G1 J 2G2
Shafts having the same diameters and the same shear moduli are required for this system; therefore, J1 =
J2 = J and G1 = G2 = G. Factor these terms out to obtain:
1
1
NB
NB
3
T1L1 T2 L2
J
T1L1 T2 L2
JG NC
G (3 ) NC
Express the polar moment of inertia in terms of diameter to obtain the following relationship:
NB
32
d4
T1L1 T2 L2
G (3 ) NC
32
(12,000,000 psi)(3 )
rad
180
48 teeth
( 2,760 lb-in.)(120 in.) (4,140 lb-in.)(96 in.)
72 teeth
4
10.022529 in.
d 1.779 in.
Since this minimum diameter is greater than the diameter needed to satisfy the shear stress requirements,
the rotation angle constraint controls.
Therefore, the minimum shaft diameter that satisfies all requirements is d ≥ 1.779 in.
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.
6.41 Two solid steel shafts are connected by the
gears shown in Fig. P6.41. The design
requirements for the system require (1) that both
shafts must have the same diameter, (2) that the
maximum shear stress in each shaft must be less
than 50 MPa, and (3) that the rotation angle of
gear D must not exceed 3°. Determine the
minimum required diameter of the shafts if the
torque applied at gear D is TD = 750 N-m. The
shaft lengths are L1 = 2.5 m and L2 = 2.0 m.
Assume that the shear modulus of both shafts is
G = 80 GPa and that the bearings shown allow
free rotation of the shafts.
Fig. P6.41
Solution
Torque relationship: The torque on gear D creates a torque in shaft (2) of
T2 = 1,200 N-m.
M x TD T2 0
T2 TD 750 N-m
Accordingly, the torque applied to gear C is also 750 N-m. The torque on
gear B is found with the gear ratio:
48 teeth
TB
TC
0.6667 TC
0.6667(750 N-m)
500 N-m
72 teeth
Shaft (1) has the same torque as gear B; therefore, T1 = −500 N-m.
Diameters based on allowable shear stresses: The elastic torsion formula gives the relationship
between shear stress and torque in a shaft. This equation can be rearranged in terms of the outside
diameter:
Tc
T
d3
J
16
Solve for the minimum acceptable diameter in shaft (1):
16 T1 16(500 N-m)(1,000 mm/m)
d13
50,929.582 mm3
2
(50 N/mm )
d1 37.067 mm
Solve for the minimum acceptable diameter in shaft (2):
16 T2 16(750 N-m)(1,000 mm/m)
d 23
76,394.373 mm3
2
(50 N/mm )
d 2 42.431 mm
Of these two values, d2 controls. Therefore, both shafts could have a diameter of 42.431 mm or more
and the shear stress constraint would be satisfied.
The angles of twists in shafts (1) and (2) can be expressed as:
T1L1
T2 L2
1
2
J1G1
J 2G2
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.
The rotations of gears B and C are related since the arclengths turned by the two gears have the same
magnitude. The gears turn in opposite directions; therefore, a negative sign is introduced.
RC C
RB B
The rotation angle of gear B is equal to the angle of twist in shaft (1): B = 1. Therefore, the rotation
angle of gear C can be expressed in terms of T1.
RB
RB
RB T1L1
N B T1L1
C
B
1
RC
RC
RC J1G1
N C J1G1
The rotation angle of gear D is equal to the rotation angle of gear C plus the twist that occurs in shaft
(2):
D
C
2
and so the rotation angle of gear D can be expressed in terms of the torques T1 and T2:
N B T1L1 T2 L2
3
D
N C J1G1 J 2G2
Shafts having the same diameters and the same shear moduli are required for this system; therefore, J1 =
J2 = J and G1 = G2 = G. Factor these terms out to obtain:
1
1
NB
NB
3
T1L1 T2 L2
J
T1L1 T2 L2
JG NC
G (3 ) NC
Express the polar moment of inertia in terms of diameter to obtain the following relationship:
NB
32
d4
T1L1 T2 L2
G (3 ) N C
32
48 teeth
( 500 N-m)(2.5 m) (750 N-m)(2.0 m) (1,000 mm/m) 2
72 teeth
(80,000 N/mm 2 )(3 )( / 180 )
5,673,986.284 mm 4
d
48.806 mm
Since this minimum diameter is greater than the diameter needed to satisfy the shear stress requirements,
the rotation angle constraint controls.
Therefore, the minimum shaft diameter that satisfies all requirements is d ≥ 48.8 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.
6.42 The driveshaft of an automobile is being designed to transmit 180 hp at 3,500 rpm. Determine the
minimum diameter required for a solid steel shaft if the allowable shear stress in the shaft is not to
exceed 6,000 psi.
Solution
The torque in the driveshaft is:
550 lb-ft/s
180 hp
1 hp
P
T
3,500 rev 2 rad 1 min
min
1 rev
60 s
270.109 lb-ft
The minimum diameter required for the shaft can be found from:
T
(270.109 lb-ft)(12 in./ft)
d3
0.540217 in.3
16
6,000
psi
allow
d
1.401241 in.
1.401 in.
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.
6.43 A tubular steel shaft transmits 225 hp at 4,000 rpm. Determine the maximum shear stress produced
in the shaft if the outside diameter is D = 3.000 in. and the wall thickness is t = 0.125 in.
Solution
The torque in the tubular steel shaft is:
550 lb-ft/s
225 hp
1 hp
P
T
4,000 rev 2 rad 1 min
min
1 rev
60 s
295.431 lb-ft
The polar moment of inertia of the shaft is:
J
32
D4
d4
32
(3.000 in.)4
(2.750 in.)4
and the maximum shear stress in the shaft is
Tc (295.431 lb-ft)(3.000 in./2)(12 in./ft)
2.337403 in.4
J
2.337403 in.4
2, 275.074 psi
2, 280 psi
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.
6.44 A tubular steel shaft is being designed to transmit 225 kW at 1,700 rpm. The maximum shear stress
in the shaft must not exceed 30 MPa. If the outside diameter of the shaft is D = 75 mm, determine the
minimum wall thickness for the shaft.
Solution
The torque in the tubular steel shaft is:
1,000 N-m/s
225 kW
P
1 kW
T
1,700 rev 2 rad 1 min
min
1 rev
60 s
1, 263.877 N-m
The maximum shear stress in the shaft will be determined from the elastic torsion formula:
Tc
D4 d 4
where J
J
32
Rearrange this equation, grouping the diameter terms on the left-hand side of the equation:
D4 d 4
T
32 D / 2
Substitute the known values for T, , and D and solve for the inside diameter d:
(75 mm) 4 d 4
(1, 263.877 N-m)(1,000 mm/m)
32
75 mm / 2
(75 mm) 4 d 4
d4
30 N/mm 2
16,092,181.76 mm 4
(75 mm)4 16,092,181.76 mm 4
d 62.7945 mm
15,548, 443.24 mm 4
The minimum wall thickness for the tubular steel shaft is thus
D d 2t
D d 75 mm 62.7945 mm
t
6.10275 mm 6.10 mm
2
2
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.
6.45 A solid 20-mm-diameter bronze shaft transmits 11 kW at 25 Hz to the propeller of a small sailboat.
Determine the maximum shear stress produced in the shaft.
Solution
The torque in the tubular steel shaft is:
1,000 N-m/s
11 kW
P
1 kW
70.028175 N-m
T
25 rev 2 rad
s
1 rev
The polar moment of inertia of the shaft is:
J
32
(20 mm)4
15,707.963 mm4
and the maximum shear stress in the shaft is
Tc (70.028175 N-m)(20 mm/2)(1,000 mm/m)
J
15,707.963 mm 4
44.581 MPa
44.6 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.
6.46 A steel propeller for a windmill transmits 5.5 kW at 65 rpm. If the allowable shear stress in the
shaft must be limited to 60 MPa, determine the minimum diameter required for a solid shaft.
Solution
The torque in the tubular steel shaft is:
1,000 N-m/s
5.5 kW
P
1 kW
T
65 rev 2 rad 1 min
min
1 rev
60 s
808.017 N-m
The minimum diameter required for the shaft can be found from:
(808.017 N-m)(1,000 mm/m)
T
13, 466.957 mm3
d3
2
16
60 N/mm
allow
d
40.934 mm
40.9 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.
6.47 A solid steel [G = 12,000 ksi] shaft with a 3-in. diameter must not twist more than 0.06 rad in a 16ft length. Determine the maximum horsepower that the shaft can transmit at 4 Hz.
Solution
Section properties: The polar moment of inertia of the shaft is:
J
32
(3 in.)4
7.952156 in.4
Angle of twist relationship: From the angle of twist relationship, compute the allowable torque:
JG (0.06 rad)(7.952156 in.4 )(12,000 ksi)
T
29.820585 kip-in. 2, 485.05 lb-ft
L
(16 ft)(12 in./ft)
Power transmission: The maximum power that can be transmitted at 4 Hz is:
4 rev 2 rad
P T
(2, 485.05 lb-ft)
62, 456.088 lb-ft/s
s
1 rev
or in units of horsepower,
62, 456.088 lb-ft/s
P
113.6 hp
550 lb-ft/s
1 hp
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.
6.48 A tubular steel [G = 80 GPa] shaft with an outside diameter of D = 100 mm and a wall thickness of
t = 6 mm must not twist more than 0.05 rad in a 7-m length. Determine the maximum power that the
shaft can transmit at 375 rpm.
Solution
Section properties: The polar moment of inertia of the shaft is:
J
32
D4
d4
32
(100 mm) 4
(88 mm) 4
3,929,981.613 mm 4
Angle of twist relationship: From the angle of twist relationship, compute the allowable torque:
JG (0.05 rad)(3,929,981.613 mm 4 )(80,000 N/mm 2 )
T
L
(7 m)(1,000 mm/m)
2,245,703.779 N-mm 2,245.704 N-m
Power transmission: The maximum power that can be transmitted at 375 rpm is:
375 rev 2 rad 1 min
P T
(2, 245.704 N-m)
88,188.581 N-m/s 88.2 kW
60 s
min
1 rev
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.
6.49 A solid 3-in.-diameter bronze [G = 6,000 ksi] shaft is 7-ft long. The allowable shear stress in the
shaft is 8 ksi and the angle of twist must not exceed 0.03 rad. Determine the maximum horsepower that
this shaft can deliver
(a) when rotating at 150 rpm.
(b) when rotating at 540 rpm.
Solution
Section properties: The polar moment of inertia of the shaft is:
J
32
(3 in.)4
7.952156 in.4
Torque based on allowable shear stress: Compute the allowable torque if the shear stress must not
exceed 8 ksi:
J (8 ksi)(7.952156 in.4 )
42.4115 kip-in.
T
(a)
1.5 in.
c
Torque based on angle of twist: Compute the allowable torque if the angle of twist must not exceed
0.03 rad:
JG (0.03 rad)(7.952156 in.4 )(6,000 ksi)
T
17.0403 kip-in.
(b)
L
(7 ft)(12 in./ft)
Controlling torque: From comparison of Eqs. (a) and (b), the maximum torque allowed for this shaft
is:
Tmax 17.0403 kip-in. 1, 420.028 lb-ft
(a) Power transmission at 150 rpm: The maximum power that can be transmitted at 150 rpm is:
150 rev 2 rad 1 min
P T
(1, 420.028 lb-ft)
22,305.746 lb-ft/s
min
1 rev
60 s
or in units of horsepower,
22,305.746 lb-ft/s
P
40.6 hp
Ans.
550 lb-ft/s
1 hp
(b) Power transmission at 540 rpm: The maximum power that can be transmitted at 540 rpm is:
540 rev 2 rad 1 min
P T
(1, 420.028 lb-ft)
80,300.684 lb-ft/s
min
1 rev
60 s
or in units of horsepower,
80,300.684 lb-ft/s
P
146.0 hp
Ans.
550 lb-ft/s
1 hp
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.
6.50 A hollow titanium [G = 43 GPa] shaft has an outside diameter of D = 50 mm and a wall thickness of t
= 1.25 mm. The maximum shear stress in the shaft must be limited to 150 MPa. Determine:
(a) the maximum power that can be transmitted by the shaft if the rotation speed must be limited to 20 Hz.
(b) the magnitude of the angle of twist in a 700-mm length of the shaft when 30 kW is being transmitted at
8 Hz.
Solution
Section properties: The polar moment of inertia of the shaft is:
J
32
D4
d4
32
(50 mm)4
(47.5 mm)4
113,817.540 mm4
Torque based on allowable shear stress: Compute the allowable torque if the shear stress must not
exceed 100 MPa:
J (150 N/mm 2 )(113,817.540 mm 4 )
682,905.237 N-mm 682.905 N-m
T
50 mm/2
c
(a) Power transmission: The maximum power that can be transmitted at 20 Hz is:
20 rev 2 rad
P T
(682.905 N-m)
85,816.403 N-m/s 85.8 kW
s
1 rev
Ans.
(b) Angle of twist: The torque in the shaft when 30 kW is being transmitted at 8 Hz is:
P
30,000 N-m/s
T
596.831 N-m
8 rev 2 rad
s
1 rev
The corresponding angle of twist is:
T L (596.831 N-m)(700 mm)(1,000 mm/m)
JG
(113,817.540 mm 4 )(43,000 N/mm 2 )
0.0854 rad
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.
6.51 A tubular aluminum alloy [G = 4,000 ksi] shaft is being designed to transmit 400 hp at 1,500 rpm.
The maximum shear stress in the shaft must not exceed 6 ksi and the angle of twist is not to exceed 5° in
an 8-ft length. Determine the minimum permissible outside diameter if the inside diameter is to be threefourths of the outside diameter.
Solution
Torque from power transmission equation: The torque in the shaft when 400 hp is being transmitted
at 1,500 rpm is:
550 lb-ft/s
(400 hp)
1 hp
P
1, 400.56 lb-ft
T
1,500 rev 2 rad 1 min
min
1 rev
60 s
Polar moment of inertia: The inside diameter of the shaft is to be three-fourths of the outside
diameter; therefore, d = 0.75D. From this, the polar moment of inertia can be expressed as:
D4 4
D4
4
4
4
4
4
1 (0.75)
0.683594
0.067112 D 4
J
D d
D (0.75D)
(a)
32
32
32
32
Diameter based on shear stress: The maximum shear stress must not exceed 6 ksi; thus, from the
elastic torsion formula:
Tc
J T
J
c
Use the results of Eq. (a) to simplify the left-hand side of this equation, giving an expression in terms of
the outside diameter D:
J 0.067112 D 4
0.134224 D3
c
D/2
Solve for the outside diameter D:
(1, 400.56 lb-ft)(12 in./ft)
0.134224 D3
2.801120 in.3
D 2.753175 in.
(b)
6, 000 psi
Diameter based on twist angle: The angle of twist is not to exceed 5° in an 8-ft length; thus, from the
angle of twist formula:
TL
TL
J
JG
G
Use the results of Eq. (a) to simplify this equation and solve for the outside diameter D:
(1, 400.56 lb-ft)(8 ft)(12 in./ft) 2
(c)
0.067112 D4
4.622180 in.4
D 2.881972 in.
(5 )( /180 )(4,000,000 psi)
From the results of Eqs. (b) and (c), the minimum outside diameter D acceptable in this instance is:
Ans.
Dmin 2.88 in.
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.
6.52 A tubular steel [G = 80 GPa] shaft is being designed to transmit 150 kW at 30 Hz. The maximum
shear stress in the shaft must not exceed 80 MPa and the angle of twist is not to exceed 6° in a 4-m
length. Determine the minimum permissible outside diameter if the ratio of the inside diameter to the
outside diameter is 0.80.
Solution
Torque from power transmission equation: The torque in the shaft when 150 kW is being transmitted
at 1,500 rpm is:
P
150, 000 N-m/s
T
795.7747 N-m
30 rev 2 rad
s
1 rev
Polar moment of inertia: The ratio of inside diameter to outside diameter for the shaft is to be d/D =
0.80 or in other words d = 0.80D. From this, the polar moment of inertia can be expressed as:
D4 4
D4
4
4
4
4
4
1 (0.80)
0.590400
0.057962 D 4
J
D d
D (0.80 D)
(a)
32
32
32
32
Diameter based on shear stress: The maximum shear stress must not exceed 80 MPa; thus, from the
elastic torsion formula:
Tc
J T
J
c
Use the results of Eq. (a) to simplify the left-hand side of this equation, giving an expression in terms of
the outside diameter D:
J 0.057962 D 4
0.115925D3
c
D/2
Solve for the outside diameter D:
(795.7747 N-m)(1,000 mm/m)
0.115925D3
80 N/mm2
9,947.18375 mm3
D 44.1070 mm
(b)
Diameter based on twist angle: The angle of twist is not to exceed 6° in a 4-m length; thus, from the
angle of twist formula:
TL
TL
J
JG
G
Use the results of Eq. (a) to simplify this equation and solve for the outside diameter D:
(795.7747 N-m)(4 m)(1,000 mm/m) 2
0.057962 D4
379,954.43 mm 4
D 50.5996 mm
(6 )( /180 )(80,000 N/mm2 )
(c)
From the results of Eqs. (b) and (c), the minimum outside diameter D acceptable in this instance is:
Ans.
Dmin 50.6 mm
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.
6.53 An automobile engine supplies 180 hp at 4,200 rpm to a driveshaft. If the allowable shear stress in
the driveshaft must be limited to 5 ksi, determine:
(a) the minimum diameter required for a solid driveshaft.
(b) the maximum inside diameter permitted for a hollow driveshaft if the outside diameter is 2.00 in.
(c) the percent savings in weight realized if the hollow shaft is used instead of the solid shaft. (Hint: The
weight of a shaft is proportional to its cross-sectional area.)
Solution
Power transmission equation: The torque in the shaft is
550 lb-ft/s
180 hp
1 hp
P
225.0906 lb-ft
T
4,200 rev 2 rad 1 min
min
1 rev
60 s
(a) Solid driveshaft: The minimum diameter required for a solid shaft can be found from:
T
(225.0906 lb-ft)(12 in./ft)
d3
0.540217 in.3
16
5,000 psi
allow
d
1.401241 in.
Ans.
1.401 in.
(b) Hollow driveshaft: From the elastic torsion formula:
D4 d 4
Tc
J
T
J
c 32 D / 2
Solve this equation for the maximum inside diameter d:
(2.00 in.)4 d 4
(225.0906 lb-ft)(12 in./ft)
32
2.00 in. / 2
(2.00 in.)
4
5,000 psi
d
4
d4
0.540217 in.3
5.502605 in.4
16 in.4
d
5.502605 in.4
10.497395 in.4
Ans.
1.800 in.
(c) Weight savings: The weights of the solid and hollow shafts are proportional to their respective
cross-sectional areas. The cross-sectional area of the solid shaft is
Asolid
(1.401241 in.)2
1.5421 in.2
4
and the cross-sectional area of the hollow shaft is
Ahollow
(2.00 in.)2 (1.80 in.) 2
0.5969 in.2
4
The weight savings can be determined from
( Asolid Ahollow )
weight savings (in percent)
Asolid
(1.5421 in.2 0.5929 in.2 )
1.5421 in.2
61.3%
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.
6.54 The impeller shaft of a fluid agitator transmits 28 kW at 440 rpm. If the allowable shear stress in
the impeller shaft must be limited to 80 MPa, determine:
(a) the minimum diameter required for a solid impeller shaft.
(b) the maximum inside diameter permitted for a hollow impeller shaft if the outside diameter is 40 mm.
(c) the percent savings in weight realized if the hollow shaft is used instead of the solid shaft. (Hint: The
weight of a shaft is proportional to its cross-sectional area.)
Solution
Power transmission equation: The torque in the shaft is
P
28, 000 N-m/s
T
607.6825 N-m
440 rev 2 rad 1 min
min
1 rev
60 s
(a) Solid impeller shaft: The minimum diameter required for a solid shaft can be found from:
T
(607.6825 N-m)(1,000 mm/m)
d3
7,596.0313 mm3
2
16
80 N/mm
allow
d
33.8209 mm
33.8 mm
Ans.
(b) Hollow driveshaft: From the elastic torsion formula:
D4 d 4
Tc
J
T
J
c 32 D / 2
Solve this equation for the maximum inside diameter:
(40 mm) 4 d 4
(607.6825 N-m)(1,000 mm/m)
32
40 mm / 2
(40 mm) 4 d 4
d4
80 N/mm
1,547, 450.779 mm 4
2
7,596.0313 mm3
2,560,000 mm 4 1,547, 450.779 mm 4
d
31.7215 mm
1,012,549.221 mm 4
31.7 mm
Ans.
(c) Weight savings: The weights of the solid and hollow shafts are proportional to their respective
cross-sectional areas. The cross-sectional area of the solid shaft is
Asolid
(33.8209 mm)2
898.3803 mm 2
4
and the cross-sectional area of the hollow shaft is
(40 mm)2 (31.7215 mm) 2
466.3262 mm 2
4
The weight savings can be determined from
( Asolid Ahollow ) (898.3803 mm2 466.3262 mm2 )
weight savings (in percent)
898.3803 mm2
Asolid
Ahollow
48.1%
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.
6.55 A motor supplies 200 kW at 6 Hz to flange A of the shaft shown in Fig. P6.55. Gear B transfers 125
kW of power to operating machinery in the factory, and the remaining power in the shaft is transferred
by gear D. Shafts (1) and (2) are solid aluminum [G = 28 GPa] shafts that have the same diameter and an
allowable shear stress of = 40 MPa. Shaft (3) is a solid steel [G = 80 GPa] shaft with an allowable
shear stress of = 55 MPa. Determine:
(a) the minimum permissible diameter for
aluminum shafts (1) and (2).
(b) the minimum permissible diameter for
steel shaft (3).
(c) the rotation angle of gear D with respect
to flange A if the shafts have the minimum
permissible diameters as determined in (a)
and (b).
Fig. P6.55
Solution
Power transmission: The torque on flange A is
P
200, 000 N-m/s
TA
5,305.165 N-m
6 rev 2 rad
s
1 rev
thus, the torque in shaft segment (1) is T1 = 5,305.165 N-m. The torque transferred by gear B is
P
125, 000 N-m/s
TB
3,315.728 N-m
6 rev 2 rad
s
1 rev
therefore, the remaining torque in segments (2) and (3) of the shaft is
T2 TA TB 5,305.165 N-m 3,315.728 N-m 1,989.437 N-m
(a) Minimum diameter for solid aluminum shaft: The minimum diameter required for the solid
aluminum shaft can be found from:
(5,305.165 N-m)(1,000 mm/m)
T1
132,629.125 mm3
d13
2
16
40 N/mm
1,allow
d1
Ans.
87.7 mm
(b) Minimum diameter for solid steel shaft: The minimum diameter required for the solid steel shaft
can be found from:
T3
(1,989.437 N-m)(1,000 mm/m)
d33
36,171.582 mm3
2
16
55 N/mm
3,allow
d3
Ans.
56.9 mm
(c) Angles of twist: The polar moments of inertia of the aluminum and steel shafts are:
J1
J3
J2
32
32
(87.7 mm) 4
(56.9 mm) 4
5,807,621 mm 4
1,029,080 mm 4
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.
The angles of twist in the three shaft segments are:
T1L1 (5,305.165 N-m)(0.8 m)(1,000 mm/m) 2
1
J1G1
(5,807,621 mm 4 )(28,000 N/mm 2 )
0.026100 rad
2
T2 L2
J 2G2
(1,989.437 N-m)(0.4 m)(1,000 mm/m) 2
(5,807,621 mm 4 )(28,000 N/mm 2 )
0.004894 rad
3
T3 L3
J 3G3
(1,989.437 N-m)(0.8 m)(1,000 mm/m)2
(1,029,080 mm 4 )(80,000 N/mm 2 )
0.019332 rad
Rotation angle at D: The rotation angle at D is equal to the sum of the angles of twist in the three shaft
segments:
D
1
2
3
0.026100 rad 0.004894 rad 0.019332 rad
0.050326 rad
0.0503 rad
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.
6.56 A motor supplies 150 hp at 520 rpm to flange A of the shaft shown in Fig. P6.56. Gear B transfers
90 hp of power to operating machinery in the factory, and the remaining power in the shaft is transferred
by gear D. Shafts (1) and (2) are solid aluminum [G = 4,000 ksi] shafts that have the same diameter and
an allowable shear stress of = 6 ksi. Shaft (3) is a solid steel [G = 12,000 ksi] shaft with an allowable
shear stress of = 8 ksi. Determine:
(a) the minimum permissible diameter for
aluminum shafts (1) and (2).
(b) the minimum permissible diameter for
steel shaft (3).
(c) the rotation angle of gear D with respect
to flange A if the shafts have the minimum
permissible diameters as determined in (a)
and (b).
Fig. P6.56
Solution
Power transmission: The torque on flange A is
550 lb-ft/s
150 hp
1 hp
P
1,515.033 lb-ft
TA
520 rev 2 rad 1 min
min
1 rev
60 s
thus, the torque in shaft segment (1) is T1 = 1,515.033 lb-ft. The torque transferred by gear B is
550 lb-ft/s
90 hp
1 hp
P
TB
909.020 lb-ft
520 rev 2 rad 1 min
min
1 rev
60 s
therefore, the remaining torque in segments (2) and (3) of the shaft is
T2 TA TB 1,515.033 lb-ft 909.020 lb-ft 606.013 lb-ft
(a) Minimum diameter for solid aluminum shaft: The minimum diameter required for the solid
aluminum shaft can be found from:
T1
(1,515.033 lb-ft)(12 in./ft)
d13
3.030066 in.3
16
6,000 psi
1,allow
d1
2.49 in.
Ans.
(b) Minimum diameter for solid steel shaft: The minimum diameter required for the solid steel shaft
can be found from:
T3
(606.013 lb-ft)(12 in./ft)
d33
0.909020 in.3
16
8,000 psi
3,allow
d3 1.667 in.
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.
(c) Angles of twist: The polar moments of inertia of the aluminum and steel shafts are:
J1
J3
J2
32
32
(2.49 in.) 4
(1.667 in.) 4
3.773960 in.4
0.758128 in.4
The angles of twist in the three shaft segments are:
T1L1 (1,515.033 lb-ft)(2 ft)(12 in./ft)2
0.028904 rad
1
J1G1
(3.773960 in.4 )(4,000,000 psi)
2
T2 L2
J 2G2
(606.013 lb-ft)(1 ft)(12 in./ft)2
(3.773960 in.4 )(4,000,000 psi)
3
T3 L3
J 3G3
(606.013 lb-ft)(2 ft)(12 in./ft)2
(0.758128 in.4 )(12,000,000 psi)
0.005781 rad
0.019185 rad
Rotation angle at D: The rotation angle at D is equal to the sum of the angles of twist in the three shaft
segments:
D
1
2
3
0.028904 rad 0.005781 rad 0.019185 rad
0.053870 rad
0.0539 rad
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.
6.57 A motor supplies sufficient power to the
system shown in Fig. P6.57 so that gears C and
D provide torques of TC = 800 N-m and TD =
550 N-m, respectively, to machinery in a
factory. Power shaft segments (1) and (2) are
hollow steel tubes with an outside diameter of D
= 60 mm and an inside diameter of d = 50 mm.
If the power shaft [i.e., segments (1) and (2)]
rotates at 40 rpm, determine:
(a) the maximum shear stress in power shaft
segments (1) and (2).
(b) the power (in kW) that must be provided by
the motor as well as the rotation speed (in rpm).
(c) the torque applied to gear A by the motor.
Fig. P6.57
Solution
Equilibrium:
M x TD T2
T2
Mx
TC
0
550 N-m
TD
TD
T1
T1
TC
0
TD
800 N-m 550 N-m
1,350 N-m
Section properties:
J1
32
(60 mm) 4
(50 mm) 4
658,752.71 mm 4
(a) Shear stresses:
T1c1 (1,350 N-m)(60 mm/2)(1,000 mm/m)
1
658,752.71 mm 4
J1
2
T2c2
J2
(550 N-m)(60 mm/2)(1,000 mm/m)
658,752.71 mm 4
J2
61.480 MPa
61.5 MPa
Ans.
25.047 MPa
25.0 MPa
Ans.
(b) Power transmission: Compute the power in shaft segment (1):
40 rev 2 rad 1 min
P T1
(1,350 N-m)
5,654.867 N-m/s
min
1 rev
60 s
This power in the shaft must be provided by the motor. Therefore:
Pmotor 5.65 kW
5.65 kW
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.
The rotation speed of the motor is found with the use of the gear ratio:
72 teeth
3 B 3(40 rpm) 120 rpm
A
B
motor
24 teeth
(c) Motor torque: The motor must provide a torque of
N
24 teeth
450 N-m
TA TB A (1,350 N-m)
72 teeth
NB
Ans.
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.
6.58 A motor supplies sufficient power to the
system shown in Fig. P6.58 so that gears C and
D provide torques of TC = 100 lb-ft and TD = 80
lb-ft, respectively, to machinery in a factory.
Power shaft segments (1) and (2) are hollow
steel tubes with an outside diameter of D = 1.75
in. and an inside diameter of d = 1.50 in. If the
power shaft [i.e., segments (1) and (2)] rotates
at 540 rpm, determine:
(a) the maximum shear stress in power shaft
segments (1) and (2).
(b) the power (in hp) that must be provided by
the motor as well as the rotation speed (in rpm).
(c) the torque applied to gear A by the motor.
Fig. P6.58
Solution
Equilibrium:
M x TD T2
T2
Mx
TC
0
TD
TD
T1
80 lb-ft
T1
TC
0
TD
100 lb-ft 80 lb-ft
180 lb-ft
Section properties:
J1
32
(1.75 in.)4
(1.50 in.)4
0.423762 in.4
(a) Shear stresses:
T1c1 (180 lb-ft)(1.75 in./2)(12 in./ft)
1
J1
0.423762 in.4
2
T2c2
J2
(80 lb-ft)(1.75 in./2)(12 in./ft)
0.423762 in.4
J2
4, 460.049 psi
1,982.244 psi
4, 460 psi
1,982 psi
Ans.
Ans.
(b) Power transmission: Compute the power in shaft segment (1):
540 rev 2 rad 1 min
P T1
(180 lb-ft)
10,178.760 lb-ft/s 18.507 hp
min
1 rev
60 s
This power in the shaft must be provided by the motor. Therefore:
Pmotor 18.51 hp
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.
The rotation speed of the motor is found with the use of the gear ratio:
72 teeth
3 B 3(540 rpm) 1,620 rpm
A
B
motor
24 teeth
(c) Motor torque: The motor must provide a torque of
N
24 teeth
TA TB A (180 lb-ft)
60 lb-ft 720 lb-in.
NB
72 teeth
Ans.
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.
6.59 A motor supplies 25 hp at 6 Hz to gear A
of the drive system shown in Fig. P6.59. Shaft
(1) is a solid 2.25-in.-diameter aluminum [G =
4,000 ksi] shaft with a length of L1 = 16 in.
Shaft (2) is a solid 1.5-in.-diameter steel [G =
12,000 ksi] shaft with a length of L2 = 12 in.
Shafts (1) and (2) are connected at flange C, and
the bearings shown permit free rotation of the
shaft. Determine:
(a) the maximum shear stress in shafts (1) and
(2).
(b) the rotation angle of gear D with respect to
gear B.
Fig. P6.59
Solution
Power transmission: The torque acting on gear A is:
550 lb-ft/s
25 hp
1 hp
P
364.730 lb-ft
TA
6 rev 2 rad
s
1 rev
Equilibrium: The torque applied to gear B is:
N
60 teeth
TB TA B (364.730 lb-ft)
911.825 lb-ft
NA
24 teeth
The torque on shaft segments (1) and (2) as well as gear D is equal to the torque applied to gear B.
T1 T2 TD TB 911.825 lb-ft
Section properties: The polar moments of inertia for aluminum shaft (1) and steel shaft (2) are:
J1
32
d14
32
(2.25 in.)4
2.516112 in.4
J2
(a) Maximum shear stresses:
T1c1 (911.825 lb-ft)(2.25 in./2)(12 in./ft)
1
2.516112 in.4
J1
2
T2c2
J2
(911.825 lb-ft)(1.5 in./2)(12 in./ft)
0.497010 in.4
32
d 24
32
(1.5 in.)4
0.497010 in.4
4,892.326 psi
4,890 psi
Ans.
16,511.601 psi
16,510 psi
Ans.
(b) Angles of twist: The angles of twist in the two shaft segments are:
T1L1 (911.825 lb-ft)(16 in.)(12 in./ft)
0.017395 rad
1
J1G1 (2.516112 in.4 )(4,000,000 psi)
2
T2 L2
J 2G2
(911.825 lb-ft)(12 in.)(12 in./ft)
(0.497010 in.4 )(12,000,000 psi)
Rotation angle of D relative to B:
0.017395 rad 0.022015 rad
D
B
1
2
0.022015 rad
0.039410 rad
0.0394 rad
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.
6.60 A motor supplies 20 kW at 400 rpm to
gear A of the drive system shown in Fig. P6.60.
Shaft (1) is a solid 50-mm-diameter aluminum
[G = 28 GPa] shaft with a length of L1 = 1,200
mm. Shaft (2) is a solid 40-mm-diameter steel
[G = 80 GPa] shaft with a length of L2 = 750
mm. Shafts (1) and (2) are connected at flange
C, and the bearings shown permit free rotation
of the shaft. Determine:
(a) the maximum shear stress in shafts (1) and
(2).
(b) the rotation angle of gear D with respect to
gear B.
Fig. P6.60
Solution
Power transmission: The torque acting on gear A is:
P
20,000 N-m/s
TA
477.465 N-m
400 rev 2 rad 1 min
min
1 rev
60 s
Equilibrium: The torque applied to gear B is:
N
60 teeth
TB TA B (477.465 N-m)
1,193.662 N-m
NA
24 teeth
The torque on shaft segments (1) and (2) as well as gear D is equal to the torque applied to gear B.
T1 T2 TD TB 1,193.662 N-m
Section properties: The polar moments of inertia for aluminum shaft (1) and steel shaft (2) are:
J1
32
d14
32
(50 mm)4
613,592.32 mm 4
J2
(a) Maximum shear stresses:
T1c1 (1,193.662 N-m)(50 mm/2)(1,000 mm/m)
1
613,592.32 mm 4
J1
2
T2c2
J2
(1,193.662 N-m)(40 mm/2)(1,000 mm/m)
251,327.41 mm 4
32
d 24
32
(40 mm) 4
251,327.41 mm4
48.634 MPa
48.6 MPa
Ans.
94.989 MPa
95.0 MPa
Ans.
(b) Angles of twist: The angles of twist in the two shaft segments are:
T1L1 (1,193.662 N-m)(1,200 mm)(1,000 mm/m)
0.083373 rad
1
(613,592.32 mm 4 )(28,000 N/mm 2 )
J1G1
2
T2 L2
J 2G2
(1,193.662 N-m)(750 mm)(1,000 mm/m)
(251,327.41 mm 4 )(80,000 N/mm 2 )
Rotation angle of D relative to B:
0.083373 rad 0.044526 rad
D
B
1
2
0.044526 rad
0.127899 rad
0.1279 rad
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.
6.61 The motor shown in Fig. P6.61 supplies 10 hp
at 1,500 rpm at A. The bearings shown permit free
rotation of the shafts.
(a) Shaft (1) is a solid 0.875-in.-diameter steel
shaft. Determine the maximum shear stress
produced in shaft (1).
(b) If the shear stress in shaft (2) must be limited to
6,000 psi, determine the minimum acceptable
diameter for shaft (2) if a solid shaft is used.
Fig. P6.61
Solution
Power transmission: The torque acting in shaft (1) and on gear B is:
550 lb-ft/s
10 hp
1 hp
P
35.014 lb-ft
T1 TB
1,500 rev 2 rad 1 min
min
1 rev
60 s
Section properties: The polar moment of inertia for shaft (1) is:
J1
32
d14
32
(0.875 in.)4
0.057548 in.4
(a) Maximum shear stress in shaft (1):
T1c1 (35.014 lb-ft)(0.875 in./2)(12 in./ft)
1
0.057548 in.4
J1
3,194.258 psi
3,194 psi
Ans.
(b) Equilibrium: The torque applied to gear C is:
30 teeth
N
21.884 lb-ft
TC TB C (35.014 lb-ft)
48 teeth
NB
The torque on shaft segment (2) as well as gear D is equal to the torque applied to gear C.
T2 TD TC 21.884 lb-ft
The minimum diameter required for solid shaft (2) can be found from:
(21.884 lb-ft)(12 in./ft)
T2
0.043768 in.3
d 23
16
6,000
psi
2,allow
d2
0.606328 in.
0.606 in.
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.
6.62 The motor shown in Fig. P6.62 supplies 12
kW at 15 Hz at A. The bearings shown permit free
rotation of the shafts.
(a) Shaft (2) is a solid 35-mm-diameter steel shaft.
Determine the maximum shear stress produced in
shaft (2).
(b) If the shear stress in shaft (1) must be limited to
40 MPa, determine the minimum acceptable
diameter for shaft (1) if a solid shaft is used.
Fig. P6.62
Solution
Power transmission: The torque acting in shaft (1) and on gear B is:
12,000 N-m/s
P
127.324 N-m
T1 TB
15 rev 2 rad
s
1 rev
The torque applied to gear C is:
30 teeth
N
TC TB C (127.324 N-m)
48 teeth
NB
79.577 N-m
The torque on shaft segment (2) as well as gear D is equal to the torque applied to gear C.
T2 TD TC 79.577 N-m
Section properties: The polar moment of inertia for shaft (2) is:
J2
32
d 24
32
(35 mm) 4
147,323.515 mm 4
(a) Maximum shear stress in shaft (2):
T2c2 (79.577 N-m)(35 mm/2)(1,000 mm/m)
2
147,323.515 mm 4
J2
(b) Minimum diameter for solid shaft (1):
(127.324 N-m)(1,000 mm/m)
T1
d13
16
40 N/mm 2
1,allow
d1
25.309 mm
25.3 mm
9.453 MPa
9.45 MPa
Ans.
3,183.099 mm3
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.
6.63 The motor shown in Fig. P6.63 supplies 9
kW at 15 Hz at A. Shafts (1) and (2) are each
solid 25-mm-diameter steel [G = 80 GPa] shafts
with lengths of L1 = 900 mm and L2 = 1,200 mm,
respectively. The bearings shown permit free
rotation of the shafts. Determine:
(a) the maximum shear stress produced in shafts
(1) and (2).
(b) the rotation angle of gear D with respect to
flange A.
Fig. P6.63
Solution
Power transmission: The magnitude of the torque acting in shaft (1) and on gear B is:
P
9,000 N-m/s
T1 TB
95.493 N-m
15 rev 2 rad
s
1 rev
The magnitude of the torque applied to gear C is:
54 teeth
N
143.239 N-m
TC TB C (95.493 N-m)
36 teeth
NB
The torque magnitude in shaft segment (2) as well as the magnitude of the torque acting on gear D is
equal to the torque applied to gear C.
T2 TD TC 143.239 N-m
To determine the proper signs for T1 and T2, consider equilibrium. First,
consider a FBD that cuts through shaft (2) and includes gear D.
M x TD T2 0
T2 TD
143.239 N-m
Since T2 is positive, the torque in shaft (1) must be negative; therefore,
T1
95.493 N-m
Section properties: The polar moments of inertia for shafts (1) and (2) are equal:
J1
32
d14
32
(25 mm) 4
38,349.520 mm 4
J2
(a) Maximum shear stress magnitudes:
T1c1 (95.493 N-m)(25 mm/2)(1,000 mm/m)
1
38,349.520 mm 4
J1
2
T2c2
J2
(143.239 N-m)(25 mm/2)(1,000 mm/m)
38,349.520 mm 4
31.126 MPa
46.689 MPa
31.1 MPa
46.7 MPa
Ans.
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.
Angles of twist in shafts (1) and (2):
T1L1 ( 95.493 N-m)(900 mm)(1,000 mm/m)
0.028013 rad
1
J1G1
(38,349.520 mm 4 )(80,000 N/mm 2 )
T2 L2 (143.239 N-m)(1,200 mm)(1,000 mm/m)
0.056027 rad
2
J 2G2
(38,349.520 mm 4 )(80,000 N/mm 2 )
Rotation angles of gears B and C: The rotation of gear B is found from the angle of twist in shaft (1):
0 rad ( 0.028013 rad)
0.028013 rad
1
B
A
B
A
1
As gear B rotates, gear C also rotates but it rotates in the opposite direction. The magnitude of the
rotation is dictated by the gear ratio:
36 teeth
NB
( 0.028013 rad)
0.018676 rad
C
B
54 teeth
NC
(b) Rotation angle of gear D: The rotation of gear D with respect to flange A is found from:
0.018676 rad 0.056027 rad 0.074702 rad 0.0747 rad
2
D
C
D
C
2
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.
6.64 The motor shown in Fig. P6.64 supplies 150
hp at 1,800 rpm at A. Shafts (1) and (2) are each
solid 2-in.-diameter steel [G = 12,000 ksi] shafts
with lengths of L1 = 75 in. and L2 = 90 in.,
respectively. The bearings shown permit free
rotation of the shafts.
(a) Determine the maximum shear stress
produced in shafts (1) and (2).
(b) Determine the rotation angle of gear D with
respect to flange A.
Fig. P6.64
Solution
Power transmission: The magnitude of the torque acting in shaft (1) and on gear B is:
550 lb-ft/s
150 hp
1 hp
P
437.676 lb-ft
T1 TB
1,800 rev 2 rad 1 min
min
1 rev
60 s
The magnitude of the torque applied to gear C is:
N
54 teeth
TC TB C (437.676 lb-ft)
656.514 lb-ft
NB
36 teeth
The torque magnitude in shaft segment (2) as well as the magnitude of the torque acting on gear D is
equal to the torque applied to gear C.
T2 TD TC 656.514 lb-ft
To determine the proper signs for T1 and T2, consider equilibrium. First,
consider a FBD that cuts through shaft (2) and includes gear D.
M x TD T2 0
T2 TD
656.514 lb-ft
Since T2 is positive, the torque in shaft (1) must be negative; therefore,
T1
437.676 lb-ft
Section properties: The polar moments of inertia for shafts (1) and (2) are equal:
J1
32
d14
32
(2 in.) 4
1.570796 in.4
J2
(a) Maximum shear stress magnitudes:
T1c1 (437.676 lb-ft)(2 in./2)(12 in./ft)
1
J1
1.570796 in.4
2
T2c2
J2
(656.514 lb-ft)(2 in./2)(12 in./ft)
1.570796 in.4
3,343.599 psi
5,015.399 psi
3,340 psi
5,020 psi
Ans.
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.
Angles of twist in shafts (1) and (2):
T1L1 ( 437.676 lb-ft)(75 in.)(12 in./ft)
0.020897 rad
1
J1G1
(1.570796 in.4 )(12,000,000 psi)
T2 L2 (656.514 lb-ft)(90 in.)(12 in./ft)
0.037615 rad
2
J 2G2 (1.570796 in.4 )(12,000,000 psi)
Rotation angles of gears B and C: The rotation of gear B is found from the angle of twist in shaft (1):
0 rad ( 0.020897 rad)
0.020897 rad
1
B
A
B
A
1
As gear B rotates, gear C also rotates but it rotates in the opposite direction. The magnitude of the
rotation is dictated by the gear ratio:
36 teeth
NB
( 0.020897 rad)
0.013932 rad
C
B
54 teeth
NC
(b) Rotation angle of gear D: The rotation of gear D with respect to flange A is found from:
0.013932 rad 0.037615 rad 0.051547 rad 0.0515 rad
2
D
C
D
C
2
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.
6.65 The gear train shown in Fig. P6.65
transmits power from a motor to a machine at E.
The motor turns at a frequency of 50 Hz. The
diameter of solid shaft (1) is 25 mm, the
diameter of solid shaft (2) is 32 mm, and the
allowable shear stress for each shaft is 60 MPa.
Determine:
(a) the maximum power that can be transmitted
by the gear train.
(b) the torque provided at gear E.
(c) the rotation speed of gear E (in Hz).
Fig. P6.65
Solution
Section properties:
J1
J2
32
32
d14
32
d 24
32
(25 mm) 4
38,349.52 mm 4
(32 mm) 4
102,943.71 mm 4
Allowable torques in the shafts:
(60 N/mm2 )(38,349.52 mm4 )
allow J1
T1,allow
c1
25 mm/2
T2,allow
allow
c2
J2
(60 N/mm2 )(102,943.71 mm4 )
32 mm/2
184,078 N-mm
386,039 N-mm
Torque relationships at the gears: The magnitudes of the torques acting on the gears are related by:
N
N
TB TA B
TD TC D
(a)
NA
NC
Equilibrium: The magnitudes of the torques in the gears and shafts are related by:
TB T1 TC
TD T2 TE
(b)
Equations (a) and (b) express the relationship between the torque in shafts (1) and (2):
N
T2 T1 D
NC
(c)
Controlling shaft torque: In Eq. (c), let T1 = T1,allow and compare the result with the allowable torque
for shaft (2):
N
check T1,allow D T2,allow
NC
(184, 078 N-mm)
60 teeth
30 teeth
368,156 N-mm 386, 039 N-m
O.K.
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.
This calculation shows that the torque in shaft (1) controls the capacity of the gear train. Using this fact,
the torques throughout the gear train can be determined:
T1 184,078 N-mm
T2 368,156 N-mm
TB
TC
TA
TB
184,078 N-mm
NA
NB
(184,078 N-mm)
TD
24 teeth
72 teeth
TE
368,156 N-mm
61,359 N-mm
(a) Power transmission: The maximum power that can be transmitted by the gear train can be found
from the torque and rotation speed at A:
50 rev 2 rad
19, 276.571 N-m/s 19.28 kW
P TA A (61.359 N-m)
Ans.
s
1 rev
(b) Torque at gear E: Equations (a) and (b) can be combined to express the relationship between the
torque supplied by the motor at A and the torque available at gear E:
N
N
N ND
NB ND
(d)
TE TD TC D TB D
TA B
TA
NC
NC
N A NC
N A NC
The torque available at gear E is thus:
NB ND
72 teeth
TE TA
(61.359 N-m)
N A NC
24 teeth
60 teeth
30 teeth
368.154 N-m
368 N-m
Ans.
This result could also be found from the torque in shaft (2), as calculated previously.
(c) Rotation speed of gear E: There are a couple of ways to compute this speed. Using the gear ratios,
the speed of gear E can be found from:
N A NC
24 teeth 30 teeth
(50 Hz)
8.33 Hz
Ans.
E
A
72 teeth 60 teeth
NB ND
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.
6.66 The gear train shown in Fig. P6.66
transmits power from a motor to a machine at
E. The motor turns at 2,250 rpm. The diameter
of solid shaft (1) is 1.50 in., the diameter of
solid shaft (2) is 2.00 in., and the allowable
shear stress for each shaft is 6 ksi. Determine:
(a) the maximum power that can be
transmitted by the gear train.
(b) the torque provided at gear E.
(c) the rotation speed of gear E (in Hz).
Fig. P6.66
Solution
Section properties:
J1
J2
32
32
d14
32
d 24
32
(1.50 in.) 4
0.497010 in.4
(2.00 in.) 4
1.570796 in.4
Allowable torques in the shafts:
(6,000 psi)(0.497010 in.4 )
allow J1
T1,allow
c1
1.50 in./2
T2,allow
allow
c2
J2
(6,000 psi)(1.570796 in.4 )
2.00 in./2
3,976.1 lb-in.
9,424.8 lb-in.
Torque relationships at the gears: The magnitudes of the torques acting on the gears are related by:
N
N
TB TA B
TD TC D
(a)
NA
NC
Equilibrium: The magnitudes of the torques in the gears and shafts are related by:
TB T1 TC
TD T2 TE
Equations (a) and (b) express the relationship between the torque in shafts (1) and (2):
N
T2 T1 D
NC
(b)
(c)
Controlling shaft torque: In Eq. (c), let T1 = T1,allow and compare the result with the allowable torque
for shaft (2):
N
check T1,allow D T2,allow
NC
(3,976.1 lb-in.)
60 teeth
30 teeth
7,952.2 lb-in. 9, 424.8 lb-in.
O.K.
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.
This calculation shows that the torque in shaft (1) controls the capacity of the gear train. Using this fact,
the torques throughout the gear train can be determined:
T1 3,976.1 lb-in.
T2 7,952.2 lb-in.
TB
TC
TA
TB
3,976.1 lb-in.
NA
NB
(3,976.1 lb-in.)
TD
24 teeth
72 teeth
TE
7,952.2 lb-in.
1,325.4 lb-in. 110.45 lb-ft
(a) Power transmission: The maximum power that can be transmitted by the gear train can be found
from the torque and rotation speed at A:
2,250 rev 2 rad 1 min
P TA A (110.45 lb-ft)
26,023.381 lb-ft/s 47.3 hp
Ans.
min
1 rev
60 s
(b) Torque at gear E: Equations (a) and (b) can be combined to express the relationship between the
torque supplied by the motor at A and the torque available at gear E:
N
N
N ND
NB ND
(d)
TE TD TC D TB D
TA B
TA
NC
NC
N A NC
N A NC
The torque available at gear E is thus:
NB ND
72 teeth
TE TA
(110.45 lb-ft)
N A NC
24 teeth
60 teeth
30 teeth
662.7 lb-ft
663 lb-ft
Ans.
This result could also be found from the torque in shaft (2), as calculated previously.
(c) Rotation speed of gear E: There are a couple of ways to compute this speed. Using the gear ratios,
the speed of gear E can be found from:
N A NC
24 teeth 30 teeth
(2,250 rpm)
375 rpm
Ans.
E
A
72 teeth 60 teeth
NB ND
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.
6.67 A hollow circular cold-rolled bronze [G1 = 6,500 ksi] tube (1) with an outside diameter of 1.75 in.
and an inside diameter of 1.25 in. is securely bonded to a solid 1.25-in.-diameter cold-rolled stainless
steel [G2 = 12,500 ksi] core (2) as shown in Fig. P6.67. The allowable shear stress of tube (1) is 27 ksi,
and the allowable shear stress of core (2) is 60 ksi. Determine:
(a) the allowable torque T that can be applied to
the tube-and-core assembly.
(b) the corresponding torques produced in tube (1)
and core (2).
(c) the angle of twist produced in a 10-in. length
of the assembly by the allowable torque T.
Fig. P6.67
Solution

Section Properties: For tube (1) and core (2), the polar moments of inertia are:
J1 
J2 
(1.75 in.) 4  (1.25 in.) 4   0.681087 in.4
32

32
(1.25 in.) 4  0.239684 in.4
Equilibrium:
M x  T1  T2  T  0
(a)
Geometry of Deformation Relationship: Since the tube and core are securely bonded together, the
angles of twist in both members must be equal; therefore,
1  2
(b)
Torque-Twist Relationships:
TL
TL
1  1 1
2  2 2
J1G1
J 2G2
Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of
deformation relationship [Eq. (b)] to obtain the compatibility equation:
T1L1 T2 L2

J1G1 J 2G2
(c)
(d)
Solve the Equations: Since the problem is expressed in terms of allowable stresses, it is convenient to
rewrite Eq. (d) in terms of stresses. In general, the elastic torsion formula can be rearranged as:
Tc
T 
 

J
J c
which allows Eq. (d) to be rewritten as:
 1L1  2 L2

(e)
G1c1 G2c2
Solve Eq. (e) for  1:
L G c
 6,500 ksi   1.75 in./2 
(f)
1   2 2 1 1   2 
 0.728 2
 12,500 ksi   1.25 in./2 
L1 G2 c2
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.
Assume the core controls: If the shear stress in core (2) reaches its allowable value of 60 ksi (in other
words, if the core controls), then the shear stress in tube (1) will be:
1  0.728(60 ksi)  43.68 ksi  27 ksi
N.G.
which is larger than the 27-ksi allowable shear stress for the tube. Therefore, the shear stress in the tube
controls.
Tube shear stress actually controls: Rearrange Eq. (f) to solve for the shear stress in core (2) given
that the shear stress in tube (1) is at its 27-ksi allowable value:
27 ksi
2 
 37.088 ksi  60 ksi
O.K.
0.728
Now that the maximum shear stresses in the tube and the core are known, the torques in each component
can be computed:
 J (27 ksi)(0.681087 in.4 )
T1  1 1 
 21.016 kip-in.
c1
1.75 in./2
T2 
 2J2
c2
(37.088 ksi)(0.239684 in.4 )

 14.223 kip-in.
1.25 in./2
(a) Total Torque: From Eq. (a), the total torque acting on the assembly must not exceed:
Tmax  T1  T2  21.016 kip-in.  14.223 kip-in.  35.239 kip-in.  35.2 kip-in.
Ans.
(b) Torques in each component: As computed previously, the torque in tube (1) is:
T1  21.0 kip-in.
Ans.
and the torque in core (2) is:
T2  14.22 kip-in.
Ans.
(c) Angle of Twist produced in a 10-in. length: Since both the tube and the core twist exactly the
same amount [i.e., Eq. (b)], either torque-twist relationship can be used to compute the angle of twist of
the entire assembly.
TL
(21.016 kip-in.)(10 in.)
1  1 1 
 0.047473 rad  0.0475 rad
Ans.
J1G1 (0.681087 in.4 )(6,500 ksi)
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.
6.68 A hollow circular cold-rolled bronze [G1 = 6,500 ksi] tube (1) with an outside diameter of 1.75 in.
and an inside diameter of 1.25 in. is securely bonded to a solid 1.25-in.-diameter cold-rolled stainless
steel [G2 = 12,500 ksi] core (2) as shown in Fig. P6.68. If a torque of T = 20 kip-in. is applied to the
tube-and-core assembly, determine:
(a) the torques produced in tube (1) and core (2).
(b) the maximum shear stress in the bronze tube
and in the stainless steel core.
(c) the angle of twist produced in a 10-in. length
of the assembly.
Fig. P6.68
Solution

Section Properties: For tube (1) and core (2), the polar moments of inertia are:
J1 
J2 
(1.75 in.) 4  (1.25 in.) 4   0.681087 in.4
32

32
(1.25 in.) 4  0.239684 in.4
Equilibrium:
M x  T1  T2  20 kip-in.  0
(a)
Geometry of Deformation Relationship: Since the tube and core are securely bonded together, the
angles of twist in both members must be equal; therefore,
1  2
(b)
Torque-Twist Relationships:
TL
TL
1  1 1
2  2 2
J1G1
J 2G2
Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of
deformation relationship [Eq. (b)] to obtain the compatibility equation:
T1L1 T2 L2

J1G1 J 2G2
(c)
(d)
Solve the Equations: Solve Eq. (d) for T1:
 0.681087 in.4   6,500 ksi 
L2 J1 G1
 T2 
 1.477632T2
T1  T2
L1 J 2 G2
 0.239684 in.4   12,500 ksi 
and substitute this result into Eq. (a) to compute the torque T2 in core (2):
T1  T2  1.477632 T2  T2  2.477632 T2  20 kip-in.
 T2  8.072 kip-in.  8.07 kip-in.
The torque in tube (1) is therefore:
T1  20 kip-in.  T2  20 kip-in.  8.072 kip-in.  11.928 kip-in.  11.93 kip-in.
Ans.
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) Maximum Shear Stress: The maximum shear stress in tube (1) is:
T c (11.928 kip-in.)(1.75 in. / 2)
1  1 1 
 15.32 ksi
0.681087 in.4
J1
The maximum shear stress in core (2) is:
(8.072 kip-in.)(1.25 in. / 2)
Tc
2  2 2 
 21.0 ksi
0.239684 in.4
J2
Ans.
Ans.
(c) Angle of Twist produced in a 10-in. length: Since both the tube and the core twist exactly the
same amount [i.e., Eq. (b)], either torque-twist relationship can be used to compute the angle of twist of
the entire assembly.
TL
(11.928 kip-in.)(10 in.)
1  1 1 
 0.026943 rad  0.0269 rad
Ans.
J1G1 (0.681087 in.4 )(6,500 ksi)
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.
6.69 A composite assembly consisting of a steel [G = 80 GPa] core (2) connected by rigid plates at the
ends of an aluminum [G = 28 GPa] tube (1) is shown in Fig. P6.69a. The cross-sectional dimensions of
the assembly are shown in Fig. P6.69b. If a torque of T = 1,100 N-m is applied to the composite
assembly, determine:
(a) the maximum shear stress in the aluminum tube and in the steel core.
(b) the rotation angle of end B relative to end A.
Fig. P6.69a Tube-and-Core Composite Shaft
Fig. P6.69b Cross-Sectional Dimensions
Solution

Section Properties: For aluminum tube (1) and steel core (2), the polar moments of inertia are:
J1 
J2 
(50 mm) 4  (40 mm) 4   362, 265 mm 4
32 

32
(20 mm) 4  15,708.0 mm 4
Equilibrium:
M x  T1  T2  1,100 N-m  0
(a)
Geometry of Deformation Relationship: Since the tube and core are connected by rigid plates, the
angles of twist in both members must be equal; therefore,
1  2
(b)
Torque-Twist Relationships:
TL
TL
1  1 1
2  2 2
J1G1
J 2G2
Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of
deformation relationship [Eq. (b)] to obtain the compatibility equation:
T1L1 T2 L2

J1G1 J 2G2
(c)
(d)
Solve the Equations: Solve Eq. (d) for T1:
 362,265 mm4   28 GPa 
L J G
T1  T2 2 1 1  T2 

  8.071858T2
L1 J 2 G2
 15,708.0 mm4   80 GPa 
and substitute this result into Eq. (a) to compute the torque T2 in steel core (2):
T1  T2  8.071858T2  T2  9.071858T2  1,100 N-m
T2  121.254 N-m
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.
The torque in aluminum tube (1) is therefore:
T1  1,100 N-m  T2  1,100 N-m  121.254 N-m  978.746 N-m
(a) Maximum Shear Stress: The maximum shear stress in aluminum tube (1) is:
T c (978.746 N-m)(50 mm / 2)(1,000 mm/m)
1  1 1 
 67.5 MPa
362, 265 mm 4
J1
The maximum shear stress in steel core (2) is:
(121.254 N-m)(20 mm / 2)(1,000 mm/m)
Tc
2  2 2 
 77.2 MPa
15,708.0 mm 4
J2
Ans.
Ans.
(b) Rotation Angle of B relative to A: Since both the tube and the core twist exactly the same amount
[i.e., Eq. (b)], either torque-twist relationship can be used to compute the angle of twist of the entire
assembly.
TL
(978.746 N-m)(300 mm)(1,000 mm/m)
1  1 1 
 0.028947 rad  0.0289 rad
Ans.
J1G1
(362,265 mm 4 )( 28,000 N/mm 2 )
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.
6.70 A composite assembly consisting of a steel [G = 80 GPa] core (2) connected by rigid plates at the
ends of an aluminum [G = 28 GPa] tube (1) is shown in Fig. P6.70a. The cross-sectional dimensions of
the assembly are shown in Fig. P6.70b. The allowable shear stress of aluminum tube (1) is 90 MPa, and
the allowable shear stress of steel core (2) is 130 MPa. Determine:
(a) the allowable torque T that can be applied to the composite shaft.
(b) the corresponding torques produced in tube (1) and core (2).
(c) the angle of twist produced by the allowable torque T.
Fig. P6.70a Tube-and-Core Composite Shaft
Fig. P6.70b Cross-Sectional Dimensions
Solution

Section Properties: For aluminum tube (1) and steel core (2), the polar moments of inertia are:
J1 
J2 
(50 mm) 4  (40 mm) 4   362, 265 mm 4
32

32
(20 mm) 4  15,708.0 mm 4
Equilibrium:
M x  T1  T2  T  0
(a)
Geometry of Deformation Relationship: Since the tube and core are connected by rigid plates, the
angles of twist in both members must be equal; therefore,
1  2
(b)
Torque-Twist Relationships:
TL
TL
1  1 1
2  2 2
J1G1
J 2G2
Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of
deformation relationship [Eq. (b)] to obtain the compatibility equation:
T1L1 T2 L2

J1G1 J 2G2
(c)
(d)
Solve the Equations: Since the problem is expressed in terms of allowable stresses, it is convenient to
rewrite Eq. (d) in terms of stresses. In general, the elastic torsion formula can be rearranged as:
Tc
T 
 

J
J c
which allows Eq. (d) to be rewritten as:
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.
 1L1

 2 L2
(e)
G1c1 G2c2
Solve Eq. (e) for  1:
L G c
 28 GPa   50 mm/2 
 0.875 2
1   2 2 1 1   2 
(f)
 80 GPa   20 mm/2 
L1 G2 c2
Assume the core controls: If the shear stress in core (2) reaches its allowable value of 130 MPa (in
other words, if the core controls), then the shear stress in tube (1) will be:
1  0.875(130 MPa)  113.75 MPa  90 MPa
N.G.
which is larger than the 90 MPa allowable shear stress for the tube. Therefore, the shear stress in the
tube controls.
Tube shear stress actually controls: Rearrange Eq. (f) to solve for the shear stress in core (2) given
that the shear stress in tube (1) is at its 90 MPa allowable value:
90 MPa
2 
 102.857 MPa  130 MPa
O.K.
0.875
Now that the maximum shear stresses in the tube and the core are known, the torques in each component
can be computed:
1 J1 (90 N/mm 2 )(362,265 mm 4 )
T1 

 1,304,154 N-mm
c1
50 mm/2
T2 
2J2
c2
(102.857 MPa)(15,708.0 mm 4 )

 161,568 N-mm
20 mm/2
(a) Total Torque: From Eq. (a), the total torque acting on the assembly must not exceed:
Tmax  T1  T2  1,304,154 N-mm  161,568 N-mm  1,465,721 N-mm  1,466 N-m
(b) Torques in each component: As computed previously, the torque in tube (1) is:
T1  1,304 N-m
and the torque in core (2) is:
T2  161.6 N-m
Ans.
Ans.
Ans.
(c) Angle of Twist: Since both the tube and the core twist exactly the same amount [i.e., Eq. (b)], either
torque-twist relationship can be used to compute the angle of twist of the entire assembly.
TL
(1,304,154 N-mm)(300 mm)
1  1 1 
 0.038571 rad  0.0386 rad
Ans.
J1G1 (362,265 mm 4 )(28, 000 N/mm 2 )
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.
6.71 The composite shaft shown in Fig. P6.71 consists of a bronze sleeve (1) securely bonded to an inner
steel core (2). The bronze sleeve has an outside diameter of 35 mm, an inside diameter of 25 mm, and a
shear modulus of G1 = 45 GPa. The solid steel core has a diameter of 25 mm and a shear modulus of G2
= 80 GPa. The allowable shear stress of sleeve (1) is 180 MPa, and the allowable shear stress of core (2)
is 150 MPa. Determine:
(a) the allowable torque T that can be applied to the
composite shaft.
(b) the corresponding torques produced in sleeve (1) and
core (2).
(c) the rotation angle of end B relative to end A that is
produced by the allowable torque T.
Fig. P6.71
Solution

Section Properties: For bronze sleeve (1) and steel core (2), the polar moments of inertia are:
J1 
J2 
(35 mm) 4  (25 mm) 4   108,974 mm 4
32

32
(25 mm) 4  38,349.5 mm 4
Equilibrium:
M x  T1  T2  T  0
(a)
Geometry of Deformation Relationship: Since the sleeve and core are securely bonded together, the
angles of twist in both members must be equal; therefore,
1  2
(b)
Torque-Twist Relationships:
TL
TL
1  1 1
2  2 2
J1G1
J 2G2
Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of
deformation relationship [Eq. (b)] to obtain the compatibility equation:
T1L1 T2 L2

J1G1 J 2G2
(c)
(d)
Solve the Equations: Since the problem is expressed in terms of allowable stresses, it is convenient to
rewrite Eq. (d) in terms of stresses. In general, the elastic torsion formula can be rearranged as:
Tc
T 
 

J
J c
which allows Eq. (d) to be rewritten as:
 1L1  2 L2

(e)
G1c1 G2c2
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.
Solve Eq. (e) for  1:
L G c
 45 GPa   (35 mm/2) 
 0.7875 2
1   2 2 1 1   2 
 80 GPa   (25 mm/2) 
L1 G2 c2
(f)
Assume the core controls: If the shear stress in core (2) reaches its allowable value of 150 MPa (in
other words, if the core controls), then the shear stress in sleeve (1) will be:
 1  0.7875(150 MPa)  118.125 MPa  180 MPa O.K.
This calculation shows that the shear stress in the core does in fact controls.
Now that the maximum shear stresses in the sleeve and the core are known, the torques in each
component can be computed:
 J (118.125 N/mm 2 )(108,974 mm 4 )
T1  1 1 
 735,574 N-mm
c1
35 mm/2
T2 
 2J2
c2
(150 MPa)(38,349.5 mm 4 )

 460,194 N-mm
25 mm/2
(a) Total Torque: From Eq. (a), the total torque acting on the assembly must not exceed:
Tmax  T1  T2  735,574 N-mm  460,194 N-mm  1,195,769 N-mm  1,196 N-m
(b) Torques in each component: As computed previously, the torque in sleeve (1) is:
T1  736 N-m
and the torque in core (2) is:
T2  460 N-m
Ans.
Ans.
Ans.
(c) Angle of Twist: Since both the sleeve and the core twist exactly the same amount [i.e., Eq. (b)],
either torque-twist relationship can be used to compute the angle of twist of the entire assembly.
TL
(735,574 N-mm)(360 mm)
1  1 1 
 0.054000 rad  0.0540 rad
Ans.
J1G1 (108,974 mm 4 )(45,000 N/mm 2 )
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.
6.72 The composite shaft shown in Fig. P6.72 consists
of a bronze sleeve (1) securely bonded to an inner steel
core (2). The bronze sleeve has an outside diameter of
35 mm, an inside diameter of 25 mm, and a shear
modulus of G1 = 45 GPa. The solid steel core has a
diameter of 25 mm and a shear modulus of G2 = 80 GPa.
The composite shaft is subjected to a torque of T = 900
N-m. Determine:
(a) the maximum shear stresses in the bronze sleeve and
the steel core.
(b) the rotation angle of end B relative to end A.
Fig. P6.72
Solution

Section Properties: For bronze sleeve (1) and steel core (2), the polar moments of inertia are:
J1 
J2 
(35 mm) 4  (25 mm) 4   108,974 mm 4
32

32
(25 mm) 4  38,349.5 mm 4
Equilibrium:
M x  T1  T2  900 N-m  0
(a)
Geometry of Deformation Relationship: Since the sleeve and core are securely bonded together, the
angles of twist in both members must be equal; therefore,
1  2
(b)
Torque-Twist Relationships:
TL
TL
1  1 1
2  2 2
J1G1
J 2G2
Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of
deformation relationship [Eq. (b)] to obtain the compatibility equation:
T1L1 T2 L2

J1G1 J 2G2
(c)
(d)
Solve the Equations: Solve Eq. (d) for T1:
 108,974 mm4   45 GPa 
L2 J1 G1
 T2 
T1  T2

  1.598401T2
L1 J 2 G2
 38,349.5 mm4   80 GPa 
and substitute this result into Eq. (a) to compute the torque T2 in steel core (2):
T1  T2  1.598401T2  T2  2.598401T2  900 N-m
T2  346.367 N-m
The torque in bronze sleeve (1) is therefore:
T1  900 N-m  T2  900 N-m  346.367 N-m  553.633 N-m
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.
(a) Maximum Shear Stress: The maximum shear stress in bronze sleeve (1) is:
T c (553.633 N-m)(35 mm / 2)(1,000 mm/m)
1  1 1 
 88.9 MPa
108,974 mm 4
J1
The maximum shear stress in steel core (2) is:
Tc
(346.367 N-m)(25 mm / 2)(1,000 mm/m)
2  2 2 
 112.9 MPa
J2
38,349.5 mm 4
Ans.
Ans.
(b) Rotation Angle of B relative to A: Since both the sleeve and the core twist exactly the same
amount [i.e., Eq. (b)], either torque-twist relationship can be used to compute the angle of twist of the
entire assembly.
(553.633 N-m)(360 mm)(1,000 mm/m)
TL
1  1 1 
 0.040643 rad  0.0406 rad
Ans.
(108,974 mm 4 )( 45,000 N/mm 2 )
J1G1
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.
6.73 The composite shaft shown in Fig. P6.73
consists of two steel pipes that are connected at
flange B and securely attached to rigid walls at A
and C. Steel pipe (1) has an outside diameter of
168 mm and a wall thickness of 7 mm. Steel pipe
(2) has an outside diameter of 114 mm and a wall
thickness of 6 mm. Both pipes are 3-m long and
have a shear modulus of 80 GPa. If a concentrated
torque of 20 kN-m is applied to flange B,
determine:
(a) the maximum shear stress magnitudes in pipes
(1) and (2).
(b) the rotation angle of flange B relative to
support A.
Fig. P6.73
Solution

Section Properties: The polar moments of inertia for the two steel pipes are:
J1 
J2 
(168 mm) 4  (154 mm) 4   22,987,183 mm 4
32

(114 mm) 4  (102 mm) 4   5,954,575 mm 4
32
Equilibrium:
M x  T1  T2  20 kN-m  0
(a)
Geometry of Deformation Relationship: Since the two steel pipes are securely attached to fixed
supports at A and C, the sum of the angles of twist in the two pipes must equal zero:
1  2  0
Torque-Twist Relationships:
TL
TL
1  1 1
2  2 2
J1G1
J 2G2
Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of
deformation relationship [Eq. (b)] to obtain the compatibility equation:
T1 L1 T2 L2

0
J1G1 J 2G2
(b)
(c)
(d)
Solve the Equations: Solve Eq. (d) for T1:
L J G
T1  T2 2 1 1
L1 J 2 G2
4
 3 m   22,987,183 mm   80 GPa 
 T2 
 3 m   5,954,575 mm 4   80 GPa 
 22,987,183 mm 4 
 T2 
 3.860424 T2
 5,954,575 mm 4 
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.
and substitute this result into Eq. (a) to compute the torque T2:
T1  T2    3.860424 T2   T2  4.860424 T2  20 kN-m
T2  4.1149 kN-m
The torque in member (1) is therefore:
T1  T2  20 kN-m  4.1149 kN-m  20 kN-m  15.8851 kN-m
(a) Maximum Shear Stress: The maximum shear stress magnitude in member (1) is:
T1c1 (15.8851 kN-m)(168 mm / 2)(1,000)2
1 

 58.0 MPa
J1
22,987,183 mm4
The maximum shear stress magnitude in member (2) is:
Tc
(4.1149 kN-m)(114 mm / 2)(1,000)2
2  2 2 
 39.4 MPa
J2
5,954,575 mm4
Ans.
Ans.
(b) Rotation Angle of Flange B Relative to Support A: The angle of twist in member (1) can be
defined by the difference in rotation angles at the two ends; hence,
1  B   A
Since joint A is restrained from rotating, A = 0 and thus
1  B
The rotation angle at B can be determined by computing the angle of twist in member (1):
TL
1  1 1
J1G1

(15.8851 kN-m)(3,000 mm)(1,000 mm/m)(1,000 N/kN)
(22,987,183 mm 4 )(80,000 N/mm 2 )
 0.025914 rad  0.0259 rad
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.
6.74 The composite shaft shown in Fig. P6.74 consists of two steel pipes that are connected at flange B
and securely attached to rigid walls at A and C. Steel pipe (1) has an outside diameter of 8.625 in., a wall
thickness of 0.322 in., and a length of L1 = 15 ft. Steel pipe (2) has an outside diameter of 6.625 in., a
wall thickness of 0.280 in., and a length of L2 = 25 ft. Both pipes have a shear modulus of 12,000 ksi.
If a concentrated torque of 36 kip-ft is
applied to flange B, determine:
(a) the maximum shear stress magnitudes
in pipes (1) and (2).
(b) the rotation angle of flange B relative
to support A.
Fig. P6.74
Solution

Section Properties: The polar moments of inertia for the two steel pipes are:
J1 
J2 
(8.625 in.) 4  (7.981 in.) 4   144.978 in.4
32

(6.625 in.) 4  (6.065 in.) 4   56.284 in.4
32
Equilibrium:
M x  T1  T2  36 kip-ft  0
(a)
Geometry of Deformation Relationship: Since the two steel pipes are securely attached to fixed
supports at A and C, the sum of the angles of twist in the two pipes must equal zero:
1  2  0
Torque-Twist Relationships:
TL
TL
1  1 1
2  2 2
J1G1
J 2G2
Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of
deformation relationship [Eq. (b)] to obtain the compatibility equation:
T1 L1 T2 L2

0
J1G1 J 2G2
(b)
(c)
(d)
Solve the Equations: Solve Eq. (d) for T1:
L J G
T1  T2 2 1 1
L1 J 2 G2
4
 25 ft   144.978 in.   12,000 ksi 
 T2 
 4.293050 T2
 15 ft   56.284 in.4   12,000 ksi 
and substitute this result into Eq. (a) to compute the torque T2:
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.
T1  T2    4.293050T2   T2  5.293050T2  36 kip-ft
 T2  6.801 kip-ft
The torque in member (1) is therefore:
T1  T2  36 kip-ft  6.801 kip-ft  36 kip-ft  29.199 kip-ft
(a) Maximum Shear Stress: The maximum shear stress magnitude in member (1) is:
T c (29.199 kip-ft)(8.625 in. / 2)(12 in./ft)
1  1 1 
 10.42 ksi
144.978 in.4
J1
The maximum shear stress magnitude in member (2) is:
Tc
(6.801 kip-ft)(6.625 in. / 2)(12 in./ft)
2  2 2 
 4.80 ksi
J2
56.284 in.4
Ans.
Ans.
(b) Rotation Angle of Flange B Relative to Support A: The angle of twist in member (1) can be
defined by the difference in rotation angles at the two ends; hence,
1  B   A
Since joint A is restrained from rotating, A = 0 and thus
1  B
The rotation angle at B can be determined by computing the angle of twist in member (1):
TL
1  1 1
J1G1

(29.199 kip-ft)(15 ft)(12 in./ft) 2
(144.978 in.4 )(12,000 ksi)
 0.036252 rad  0.0363 rad
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.
6.75 The composite shaft shown in Fig. P6.75 consists of a solid brass segment (1) and a solid aluminum
segment (2) that are connected at flange B and securely attached to rigid supports at A and C. Brass
segment (1) has a diameter of 1.00 in., a length of L1 = 15 in., a shear modulus of 5,600 ksi, and an
allowable shear stress of 8 ksi. Aluminum segment (2) has a diameter of 0.75 in., a length of L2 = 20 in.,
a shear modulus of 4,000 ksi, and an allowable shear stress of 6 ksi. Determine:
(a) the allowable torque TB that can be
applied to the composite shaft at flange B.
(b) the magnitudes of the internal torques in
segments (1) and (2).
(c) the rotation angle of flange B that is
produced by the allowable torque TB.
Fig. P6.75
Solution
Section Properties: The polar moments of inertia for the two shafts are:

J1  (1.00 in.) 4  0.098175 in.4
32

J 2  (0.75 in.) 4  0.031063 in.4
32
Equilibrium:
M x  T1  T2  TB  0
(a)
Geometry of Deformation Relationship: Since the two shafts are securely attached to fixed supports
at A and C, the sum of the angles of twist in the two members must equal zero:
1  2  0
(b)
Torque-Twist Relationships:
TL
TL
1  1 1
2  2 2
J1G1
J 2G2
Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of
deformation relationship [Eq. (b)] to obtain the compatibility equation:
T1 L1 T2 L2

0
J1G1 J 2G2
(c)
(d)
Solve the Equations: Since the problem is expressed in terms of allowable stresses, it is convenient to
rewrite Eq. (d) in terms of stresses. In general, the elastic torsion formula can be rearranged as:
Tc
T 
 

J
J c
which allows Eq. (d) to be rewritten as:
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.
 1L1

 2 L2
G1c1
G2c2
Solve Eq. (e) for  1:
L G c
 20 in.  5,600 ksi   1.00 in./2 
1   2 2 1 1   2 
 2.488889 2
 15 in.   4,000 ksi   0.75 in./2 
L1 G2 c2
(e)
(f)
Assume the shaft (2) controls: If the shear stress in shaft (2) reaches its allowable magnitude of 6 ksi,
then the shear stress magnitude in shaft (1) will be:
1  2.488889(6 ksi)  14.9333 ksi  8 ksi
N.G.
which is larger than the 8 ksi allowable shear stress magnitude for shaft (1). Therefore, the shear stress
magnitude in shaft (1) must control.
Shaft (1) actually controls: Rearrange Eq. (f) to solve for the shear stress magnitude in shaft (2) given
that the shear stress magnitude in shaft (1) is at its 8 ksi allowable magnitude:
8 ksi
2 
 3.214 ksi  6 ksi
O.K.
2.488889
Now that the maximum shear stress magnitudes in the two shafts are known, the torque magnitudes in
each component can be computed:
 J (8 ksi)(0.098175 in.4 )
T1  1 1 
 1.571 kip-in.
c1
1.00 in./2
T2 
2J2

(3.214 ksi)(0.031063 in.4 )
 0.266 kip-in.
c2
0.75 in./2
Note that these are torque magnitudes. From inspection of the FBD of flange B and the associated
equilibrium equation, it is apparent that T2 must act opposite to the direction assumed in the FBD, giving
it a negative value. Therefore, by inspection
T2  0.266 kip-in.
(a) Total Torque: From Eq. (a), the total torque acting at flange B must not exceed:
TB,max  T1  T2  1.571 kip-in.  (0.266 kip-in.)  1.837 kip-in.
Ans.
(b) Torques Magnitudes: As computed previously, the torque magnitude in shaft (1) is:
T1  1.571 kip-in.
Ans.
and the torque magnitude in shaft (2) is:
T2  0.266 kip-in.
Ans.
(c) Rotation Angle of Flange B Relative to Support A: The angle of twist in member (1) can be
defined by the difference in rotation angles at the two ends; hence,
1  B   A
Since joint A is restrained from rotating, A = 0 and thus
1  B
The rotation angle at B can be determined by computing the angle of twist in member (1):
TL
(1.571 kip-in.)(15 in.)
1  1 1 
 0.042857 rad  0.0429 rad
Ans.
J1G1 (0.098175 in.4 )(5,600 ksi)
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.
6.76 The composite shaft shown in Fig. P6.76
consists of a solid brass segment (1) and a solid
aluminum segment (2) that are connected at flange
B and securely attached to rigid walls at A and C.
Brass segment (1) has a diameter of 18 mm, a
length of L1 = 235 mm, and a shear modulus of 39
GPa. Aluminum segment (2) has a diameter of 24
mm, a length of L2 = 165 mm, and a shear modulus
of 28 GPa. If a concentrated torque of 270 N-m is
applied to flange B, determine:
(a) the maximum shear stress magnitudes in
segments (1) and (2).
(b) the rotation angle of flange B relative to
support A.
Fig. P6.76
Solution

Section Properties: The polar moments of inertia for the two steel pipes are:
J1 
J2 

32
32
(18 mm) 4  10,306.0 mm 4
(24 mm) 4  32,572.0 mm 4
Equilibrium:
M x  T1  T2  270 N-m  0
(a)
Geometry of Deformation Relationship: Since the two steel pipes are securely attached to fixed
supports at A and C, the sum of the angles of twist in the two pipes must equal zero:
1  2  0
Torque-Twist Relationships:
TL
TL
1  1 1
2  2 2
J1G1
J 2G2
Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of
deformation relationship [Eq. (b)] to obtain the compatibility equation:
T1 L1 T2 L2

0
J1G1 J 2G2
(b)
(c)
(d)
Solve the Equations: Solve Eq. (d) for T1:
L J G
T1  T2 2 1 1
L1 J 2 G2
4
 165 mm   10,306.0 mm   39 GPa 
 T2 
 0.309434 T2
 235 mm   32,572.0 mm 4   28 GPa 
and substitute this result into Eq. (a) to compute the torque T2:
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.
T1  T2    0.309434 T2   T2  1.309434 T2  270 N-m
T2  206.1959 N-m
The torque in member (1) is therefore:
T1  T2  270 N-m  206.1959 N-m  270 N-m  63.8041 N-m
(a) Maximum Shear Stress: The maximum shear stress magnitude in member (1) is:
T c (63.8041 N-m)(18 mm / 2)(1,000 mm/m)
1  1 1 
 55.7 MPa
10,306.0 mm 4
J1
The maximum shear stress magnitude in member (2) is:
(206.1959 N-m)(24 mm / 2)(1,000 mm/m)
Tc
2  2 2 
 76.0 MPa
32,572.0 mm 4
J2
Ans.
Ans.
(b) Rotation Angle of Flange B Relative to Support A: The angle of twist in member (1) can be
defined by the difference in rotation angles at the two ends; hence,
1  B   A
Since joint A is restrained from rotating, A = 0 and thus
1  B
The rotation angle at B can be determined by computing the angle of twist in member (1):
TL
1  1 1
J1G1

(63.8041 N-m)(235 mm)(1,000 mm/m)
(10,306.0 mm 4 )(39,000 N/mm 2 )
 0.037305 rad  0.0373 rad
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.
6.77 The composite shaft shown in Fig. P6.77
consists of a stainless steel tube (1) and a brass
tube (2) that are connected at flange B and securely
attached to rigid supports at A and C. Stainless
steel tube (1) has an outside diameter of 2.25 in., a
wall thickness of 0.250 in., a length of L1 = 40 in.,
and a shear modulus of 12,500 ksi. Brass tube (2)
has an outside diameter of 3.500 in., a wall
thickness of 0.219 in., a length of L2 = 20 in., and a
shear modulus of 5,600 ksi. If a concentrated
torque of TB = 42 kip-in. is applied to flange B,
determine:
(a) the maximum shear stress magnitudes in tubes
(1) and (2).
(b) the rotation angle of flange B relative to
support A.
Fig. P6.77
Solution

Section Properties: The polar moments of inertia for the two tubes are:
J1 
J2 
(2.250 in.) 4  (1.750 in.) 4   1.595340 in.4
32

(3.500 in.) 4  (3.062 in.) 4   6.102156 in.4
32
Equilibrium:
M x  T1  T2  42 kip-in.  0
(a)
Geometry of Deformation Relationship: Since the two tubes are securely attached to fixed supports at
A and C, the sum of the angles of twist in the two tubes must equal zero:
1  2  0
(b)
Torque-Twist Relationships:
TL
TL
1  1 1
2  2 2
J1G1
J 2G2
Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of
deformation relationship [Eq. (b)] to obtain the compatibility equation:
T1 L1 T2 L2

0
J1G1 J 2G2
(c)
(d)
Solve the Equations: Solve Eq. (d) for T1:
L J G
T1  T2 2 1 1
L1 J 2 G2
4
 20 in.  1.595340 in.   12,500 ksi 
 T2 
 0.291784 T2
 40 in.   6.102156 in.4   5,600 ksi 
and substitute this result into Eq. (a) to compute the torque T2:
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.
T1  T2    0.291784 T2   T2  1.291784 T2  42 kip-in.
T2  32.5132 kip-in.
The torque in member (1) is therefore:
T1  T2  42 kip-in.  32.5132 kip-in.  42 kip-in.  9.4868 kip-in.
(a) Maximum Shear Stress: The maximum shear stress magnitude in member (1) is:
T c (9.4868 kip-in.)(2.250 in. / 2)
1  1 1 
 6.69 ksi
1.595340 in.4
J1
The maximum shear stress magnitude in member (2) is:
Tc
(32.5132 kip-in.)(3.500 in. / 2)
2  2 2 
 9.32 ksi
J2
6.102156 in.4
Ans.
Ans.
(b) Rotation Angle of Flange B Relative to Support A: The angle of twist in member (1) can be
defined by the difference in rotation angles at the two ends; hence,
1  B   A
Since joint A is restrained from rotating, A = 0 and thus
1  B
The rotation angle at B can be determined by computing the angle of twist in member (1):
TL
1  1 1
J1G1

(9.4868 kip-in.)(40 in.)
(1.595340 in.4 )(12,500 ksi)
 0.019029 rad  0.01903 rad
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.
6.78 The composite shaft shown in Fig. P6.78
consists of a stainless steel tube (1) and a brass
tube (2) that are connected at flange B and securely
attached to rigid supports at A and C. Stainless
steel tube (1) has an outside diameter of 2.25 in., a
wall thickness of 0.250 in., a length of L1 = 40 in.,
and a shear modulus of 12,500 ksi. Brass tube (2)
has an outside diameter of 3.500 in., a wall
thickness of 0.219 in., a length of L2 = 20 in., and a
shear modulus of 5,600 ksi. The allowable shear
stress in the stainless steel is 50 ksi, and the
allowable shear stress in the brass is 18 ksi.
Determine:
(a) the allowable torque TB that can be applied to
the composite shaft at flange B.
(b) the rotation angle of flange B that is produced
by the allowable torque TB.
Fig. P6.78
Solution

Section Properties: The polar moments of inertia for the two tubes are:
J1 
J2 
(2.250 in.) 4  (1.750 in.) 4   1.595340 in.4
32 

(3.500 in.) 4  (3.062 in.) 4   6.102156 in.4
32 
Equilibrium:
M x  T1  T2  TB  0
(a)
Geometry of Deformation Relationship: Since the two tubes are securely attached to fixed supports at
A and C, the sum of the angles of twist in the two tubes must equal zero:
1  2  0
(b)
Torque-Twist Relationships:
TL
TL
1  1 1
2  2 2
J1G1
J 2G2
Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of
deformation relationship [Eq. (b)] to obtain the compatibility equation:
T1 L1 T2 L2

0
J1G1 J 2G2
(c)
(d)
Solve the Equations: Since the problem is expressed in terms of allowable stresses, it is convenient to
rewrite Eq. (d) in terms of stresses. In general, the elastic torsion formula can be rearranged as:
Tc
T 
 

J
J c
which allows Eq. (d) to be rewritten as:
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.
 1L1

 2 L2
(e)
G1c1
G2c2
Solve Eq. (e) for  1:
L G c
 20 in.  12,500 ksi   2.250 in./2 
1   2 2 1 1   2 
 0.717474 2
 40 in.  5,600 ksi   3.500 in./2 
L1 G2 c2
(f)
Assume the shaft (2) controls: If the shear stress in shaft (2) reaches its allowable magnitude of 18 ksi,
then the shear stress magnitude in shaft (1) will be:
1  0.717474(18 ksi)  12.914541 ksi  50 ksi
O.K.
This calculation shows that the shear stress in shaft (2) does in fact control.
Now that the maximum shear stress magnitudes in the two shafts are known, the torque magnitudes in
each component can be computed:
 J (12.914541 ksi)(1.595340 in.4 )
T1  1 1 
 18.3139 kip-in.
c1
2.250 in./2
2J2
(18 ksi)(6.102156 in.4 )
T2 

 62.7650 kip-in.
c2
3.500 in./2
Note that these are torque magnitudes. From inspection of the FBD of flange B and the associated
equilibrium equation, it is apparent that T2 must act opposite to the direction assumed in the FBD, giving
it a negative value. Therefore, by inspection
T2  62.7650 kip-in.
(a) Allowable Torque TB: From Eq. (a), the total torque acting at flange B must not exceed:
TB,max  T1  T2  18.3138 kip-in.  (62.7650 kip-in.)  81.0788 kip-in.  81.1 kip-in.
Ans.
(b) Rotation Angle of Flange B Relative to Support A: The angle of twist in member (1) can be
defined by the difference in rotation angles at the two ends; hence,
1  B   A
Since joint A is restrained from rotating, A = 0 and thus
1  B
The rotation angle at B can be determined by computing the angle of twist in member (1):
TL
(18.3139 kip-in.)(40 in.)
1  1 1 
 0.036735 rad  0.0367 rad
Ans.
J1G1 (1.595340 in.4 )(12,500 ksi)
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.
6.79 The torsional assembly of Fig. P6.79
consists of a cold-rolled stainless steel tube
connected to a solid cold-rolled brass
segment at flange C. The assembly is
securely fastened to rigid supports at A and
D. Stainless steel tube (1) and (2) has an
outside diameter of 3.50 in., a wall thickness
of 0.120 in., and a shear modulus of G =
12,500 ksi. The solid brass segment (3) has a
diameter of 2.00 in. and a shear modulus of
G = 5,600 ksi. A concentrated torque of TB =
6 kip-ft is applied to the stainless steel pipe
at B. Determine:
(a) the maximum shear stress magnitude in
the stainless steel tube.
(b) the maximum shear stress magnitude in
brass segment (3).
(c) the rotation angle of flange C.
Fig. P6.79
Solution
Section Properties: The polar moments of inertia for the stainless steel tube and the solid brass
segment are:
J1 
J3 

(3.500 in.) 4  (3.260 in.) 4   3.643915 in.4  J 2
32

32
(2.0 in.) 4  1.570796 in.4
Equilibrium: Consider a free-body diagram cut around joint
B, where the external torque TB is applied:
M x  T1  T2  TB  0
(a)
and also consider a FBD cut around joint C. Although there is
not an external torque applied at joint C, the section properties
of the torsion structure change at C.
M x  T2  T3  0 T2  T3
(b)
Geometry of Deformation Relationship: Since the two ends of the torsion structure are securely
attached to fixed supports at A and D, the sum of the angles of twist in the three shafts must equal zero:
1  2  3  0
(c)
Torque-Twist Relationships:
TL
TL
1  1 1
2  2 2
J1G1
J 2G2
3 
T3 L3
J 3G3
Compatibility Equation: Substitute the torque-twist relationships [Eqs. (d)] into the geometry of
deformation relationship [Eq. (c)] to obtain the compatibility equation:
T1L1 T2 L2 T3 L3


0
J1G1 J 2G2 J 3G3
(d)
(e)
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.
Solve the Equations: The approach used here will be to reduce the variables in Eq. (e) by replacing T1
and T3 with equivalent expressions involving T2. From Eq. (a):
T1  T2  TB
(f)
Substitute Eqs. (b) and (f) into Eq. (e) and simplify to derive an expression for T2:
(T2  TB ) L1 T2 L2 T2 L3


0
J1G1
J 2G2 J 3G3
 L
L
L 
TL
T2  1  2  3    B 1
J1G1
 J1G1 J 2G2 J 3G3 
TB L1
J1G1
T2  
L1
L
L
 2  3
J1G1 J 2G2 J 3G3
Note that G1 = G2 and J1 = J2. Compute T2:
(6 kip-ft)(42 in.)(12 in./ft)
(3.643915 in.4 )(12,500 ksi)
T2  
42 in.  30 in.
24 in.

4
(3.643915 in. )(12,500 ksi) (1.570796 in.4 )(5,600 ksi)
 15.4070 kip-in.
Substitute this result into Eq. (f) to compute internal torque T1:
T1  T2  TB  15.4070 kip-in.  (6 kip-ft)(12 in./ft)  56.5930 kip-in.
and from Eq. (b):
T3  T2  15.4070 kip-in.
(e)
Maximum Shear Stress: The maximum shear stress magnitudes in the three members are:
T c (56.5930 kip-in.)(3.500 in. / 2)
1  1 1 
 27.2 ksi
J1
3.643915 in.4
2 
3 
T2c2 (15.4070 kip-in.)(3.500 in. / 2)

 7.40 ksi
J2
3.643915 in.4
T3c3 (15.4070 kip-in.)(2.000 in. / 2)

 9.81 ksi
J3
1.570796 in.4
(a) Maximum Shear Stress Magnitude in Stainless Steel Tube:
1  27.2 ksi
Ans.
(b) Maximum Shear Stress Magnitude in Brass Shaft:
 3  9.81 ksi
Ans.
(c) Rotation Angle of Flange C Relative to Support A: The angle of twist in member (1) can be
defined by the difference in rotation angles at the two ends:
1  B   A
B   A  1
Similarly, the angle of twist in member (2) can be defined by:
2  C  B
C  B  2
(f)
(g)
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.
To derive an expression for C, substitute Eq. (f) into Eq. (g), and note that joint A is restrained from
rotating; therefore, A = 0. 
C  1  2
The angle of twist in member (1) is:
TL
(56.5930 kip-in.)(42 in.)
1  1 1 
 0.052184 rad
J1G1 (3.643915 in.4 )(12,500 ksi)
The angle of twist in member (2) is:
TL
( 15.4070 kip-in.)(30 in.)
2  2 2 
 0.010148 rad
J 2G2 (3.643915 in.4 )(12,500 ksi)
The rotation angle of flange C is thus:
C  1  2  0.052184 rad  (0.010148 rad)  0.042036 rad  0.0420 rad
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.
6.80 The torsional assembly of Fig. P6.80
consists of a cold-rolled stainless steel tube
connected to a solid cold-rolled brass
segment at flange C. The assembly is
securely fastened to rigid supports at A and
D. Stainless steel tubes (1) and (2) have an
outside diameter of 3.50 in., a wall thickness
of 0.120 in., a shear modulus of G = 12,500
ksi, and an allowable shear stress of 30 ksi.
The solid brass segment (3) has a diameter
of 2.00 in., a shear modulus of G = 5,600
ksi, and an allowable shear stress of 18 ksi.
Determine the maximum permissible
magnitude for the concentrated torque TB.
Fig. P6.80
Solution
Section Properties: The polar moments of inertia for the stainless steel tube and the solid brass segment
are:
J1 
J3 

(3.500 in.) 4  (3.260 in.) 4   3.643915 in.4  J 2
32

32
(2.0 in.) 4  1.570796 in.4
Equilibrium: Consider a free-body diagram cut around joint
B, where the external torque TB is applied:
M x  T1  T2  TB  0
(a)
and also consider a FBD cut around joint C. Although there is
not an external torque applied at joint C, the section properties
of the torsion structure change at C.
M x  T2  T3  0 T2  T3
(b)
Geometry of Deformation Relationship: Since the two ends of the torsion structure are securely
attached to fixed supports at A and D, the sum of the angles of twist in the three shafts must equal zero:
1  2  3  0
(c)
Torque-Twist Relationships:
TL
TL
1  1 1
2  2 2
J1G1
J 2G2
3 
T3 L3
J 3G3
Compatibility Equation: Substitute the torque-twist relationships [Eqs. (d)] into the geometry of
deformation relationship [Eq. (c)] to obtain the compatibility equation:
T1L1 T2 L2 T3 L3


0
J1G1 J 2G2 J 3G3
(d)
(e)
Solve the Equations: Since the problem is expressed in terms of allowable stresses, it is convenient to
rewrite Eq. (e) in terms of stresses. In general, the elastic torsion formula can be rearranged as:
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.
Tc
T 
 
J
J c
which allows Eq. (e) to be rewritten as:
 1L1  2 L2  3 L3


0
(f)
G1c1 G2c2 G3c3
Also using the elastic torsion formula, Eq. (b) can be expressed in terms of stress:
 2 J 2  3J3

 J

 3 2 2
(g)
c2
c3
c3 c2 J 3
Substitute Eq. (g) into Eq. (f) to obtain
1L1  2 L2 L3  2 J 2


0
G1c1 G2c2 G3 c2 J 3
Simplify:
L J 
 L2
 3 2

 L
L
L J 
G c G3c2 J 3

1 1   2  2  3 2 
 1   2  2 2
L1
G1c1


 G2c2 G3 c2 J 3 


G1c1
Substitute values:

 3.643915 in.4  
30 in.
24 in.



(12,500 ksi)(3.50 in./2) (5,600 ksi)(3.50 in./2)  1.570796 in.4  

1   2
42 in.




(12,500 ksi)(3.50 in./2)


 3.6732  2
This calculation demonstrates that the shear stress in segment (1) of the stainless steel tube is much
larger than the shear stress in segment (2). If the shear stress magnitude in segment (1) is 30 ksi, then
the shear stress magnitude in segment (2) will be:

2  
1
 8.1673 ksi
3.6732
Next, we need to check the corresponding shear stress in brass shaft (3). From Eq. (g):
c J
3  2 3 2
c2 J 3
(h)
Substitute the magnitude obtained for 2 in Eq. (h) into this expression and calculate the corresponding
shear stress magnitude in shaft (3):
4
 2.00 in./2   3.643915 in. 
 3  (8.1673 ksi) 
O.K. 
 10.8265 ksi  18 ksi
 3.50 in./2   1.570796 in.4 
Now that the maximum shear stresses in the three shaft segments are known, the torques in each
component can be computed:
 J (30 ksi)(3.643915 in.4 )
 62.4671 kip-in.
T1  1 1 
c1
3.50 in./2
T2 
T3 
2J2
c2
 3J3
c3

( 8.1673 ksi)(3.643915 in.4 )
 17.0062 kip-in.
3.50 in./2
( 10.8265 ksi)(1.570796 in.4 )

 17.0062 kip-in.
2.00 in./2
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.
Maximum Permissible Torque TB: From Eq. (a), the torque TB acting on the assembly must not
exceed:
Ans.
TB,max  T1  T2  62.4671 kip-in.  (17.062 kip-in.)  79.4733 kip-in.  79.5 kip-in.
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.
6.81 The torsional assembly of Fig. P6.81a consists of a solid 75-mm-diameter bronze [G = 45 GPa]
segment (1) securely connected at flange B to solid 75-mm-diameter stainless steel [G = 86 GPa]
segments (2) and (3). The flange at B is secured by four 14-mm-diameter bolts, which are each located
on a 120-mm-diameter bolt circle (Fig. P6.81b). The allowable shear stress of the bolts is 90 MPa, and
friction effects in the flange can be neglected for this analysis. Determine:
(a) the allowable torque TC that can be applied to the assembly at C without exceeding the capacity of
the bolted flange connection.
(b) the maximum shear stress magnitude in bronze segment (1).
(c) the maximum shear stress magnitude in stainless steel segments (2) and (3).
Fig. P6.81a
Fig. P6.81b Flange B Bolts
Solution
Section Properties: For bronze segment (1) and stainless steel segments (2) and (3), the polar moments
of inertia are identical:
J1  J 2  J 3 

32
(75 mm)4  3,106,311 mm4
Equilibrium: Consider a free-body diagram cut around
flange B:
M x  T1  T2  0
T1  T2
(a)
and also consider a FBD cut around joint C, where the
external torque TC is applied:
M x  T2  T3  TC  0
(b)
Capacity of Flange B: The flange at B is secured by four 14-mm-diameter bolts, which are each
located on a 120-mm-diameter bolt circle. The allowable shear stress of the bolts is 90 MPa, and each
bolt acts in single shear. The cross-sectional area of a single bolt is:
Abolt 

(14 mm)2  153.9380 mm2
4
The shear force that can be provided by each bolt is:
Vbolt  (90 N/mm 2 )(153.9380 mm 2 )  13,854.4236 N
The total torque that can resisted by the bolts is thus:
TB ,max  (4 bolts)(13,954.4236 N/bolt)(120 mm/2)  3,325, 062 N-mm
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.
The internal torques in segments (1) and (2) cannot exceed this magnitude; therefore, based on Eq. (a):
T1  T2  3,325, 062 N-mm
(c)
From Eq. (b), we observe that the external torque applied to the shaft is related to internal torques T2 and
T3. We must determine the relationship between these two internal torques by developing a
compatibility equation, which is based on the geometry of deformations for this configuration.
Geometry of Deformation Relationship: Since the two ends of the torsion structure are securely
attached to fixed supports at A and D, the sum of the angles of twist in the three segments must equal
zero:
1  2  3  0
(d)
Torque-Twist Relationships:
TL
TL
1  1 1
2  2 2
J1G1
J 2G2
3 
T3 L3
J 3G3
Compatibility Equation: Substitute the torque-twist relationships [Eqs. (e)] into the geometry of
deformation relationship [Eq. (d)] to obtain the compatibility equation:
T1L1 T2 L2 T3 L3


0
J1G1 J 2G2 J 3G3
Solve the Equations: Solve Eq. (f) for T3, noting that J1 = J2 = J3, G2 = G3, and T1 = T2:
L G L 
TL
T L G J
T3    1 1  2 2  3 3  T1  1 3  2 
 J1G1 J 2G2  L3
 L3 G1 L3 
Substitute the values determined for T1 and T2 in Eq. (c) into Eq. (g) to compute T3:
 400 mm  86 GPa  750 mm 
T3  (3,325, 062 N-mm) 



 400 mm  45 GPa  400 mm 
(e)
(f)
(g)
 (3,325, 062 N-mm)(3.786111)
 12,589, 054 N-mm
(a) Allowable Torque TC: From equilibrium equation (b),
TC  T2  T3
 3,325,062 N-mm  ( 12,589,054 N-mm)
 15,914,116 N-mm  15.91 kN-m
Ans.
(b) Shear Stress Magnitude in Bronze Segment (1):
T c (3,325,062 N-mm)(75 mm / 2)
1  1 1 
 40.1 MPa
J1
3,106,311 mm 4
Ans.
(c) Shear Stress Magnitude in Stainless Steel Segments (2) and (3):
Tc
(12,589,054 N-mm)(75 mm / 2)
3  3 3 
 152.0 MPa
J3
3,106,311 mm 4
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.
6.82 The torsional assembly shown in Fig. P6.82
consists of solid 2.50-in.-diameter aluminum [G =
4,000 ksi] segments (1) and (3) and a central solid
3.00-in.-diameter bronze [G = 6,500 ksi] segment
(2). Concentrated torques of TB = T0 and TC = 2T0
are applied to the assembly at B and C,
respectively. If T0 = 20 kip-in., determine:
(a) the maximum shear stress magnitude in
aluminum segments (1) and (3).
(b) the maximum shear stress magnitude in
bronze segment (2).
(c) the rotation angle of joint C.
Fig. P6.82
Solution

Section Properties: The polar moments of inertia for the aluminum and bronze segments are:
J1  J 3 
J2 

32
32
(2.50 in.) 4  3.834952 in.4
(3.00 in.) 4  7.952156 in.4
Equilibrium: Consider a free-body diagram cut around
joint B, where the external torque TB is applied:
M x  T1  T2  T0  0
(a)
and also consider a FBD cut around joint C, where the
external torque TC is applied:
M x  T2  T3  2T0  0
(b)
Geometry of Deformation Relationship: Since the two ends of the torsion structure are securely
attached to fixed supports at A and D, the sum of the angles of twist in the three shafts must equal zero:
1  2  3  0
(c)
Torque-Twist Relationships:
TL
TL
1  1 1
2  2 2
J1G1
J 2G2
3 
T3 L3
J 3G3
Compatibility Equation: Substitute the torque-twist relationships [Eqs. (d)] into the geometry of
deformation relationship [Eq. (c)] to obtain the compatibility equation:
T1L1 T2 L2 T3 L3


0
J1G1 J 2G2 J 3G3
(d)
(e)
Solve the Equations: The approach used here will be to reduce the variables in Eq. (e) by replacing T1
and T3 with equivalent expressions involving T2. From Eq. (a):
T1  T2  T0
(f)
and from Eq. (b):
T3  T2  2T0
(g)
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.
Substitute Eqs. (f) and (g) into Eq. (e) and simplify to derive an expression for T2:
(T2  T0 ) L1 T2 L2 (T2  2T0 ) L3


0
J1G1
J 2G2
J 3G3
 L
L
L  2T L T L
T2  1  2  3   0 3  0 1
 J1G1 J 2G2 J 3G3  J 3G3 J1G1
2T0 L3 T0 L1

J 3G3 J1G1
T2 
L1
L
L
 2  3
J1G1 J 2G2 J 3G3
Since G1 = G3 and J1 = J3, Eq. (h) can be further simplified and T2 can be computed as:
T   2 L3  L1  
T2  0 

J1G1  L1  L3  L2 
J 2G2 
 J1G1
(h)
20 kip-in.
2(12 in.)  12 in.


4



12
in.
12
in.
24
in.
(3.834952 in. )(4,000 ksi)



4
4
 (3.834952 in. )(4,000 ksi) (7.952156 in. )(6,500 ksi) 
 7.711461 kip-in.
Backsubstitute this result into Eqs. (f) and (g) to obtain T1 and T3:
T1  7.711461 kip-in.  20 kip-in.  27.711461 kip-in.
T3  7.711461 kip-in.  2(20 kip-in.)  32.288539 kip-in.

Maximum Shear Stress: The maximum shear stress magnitudes in the three segments are:
T c (27.711461 kip-in.)(2.500 in. / 2)
1  1 1 
 9.03 ksi
J1
3.834952 in.4
Tc
(7.711461 kip-in.)(3.000 in. / 2)
2  2 2 
 1.455 ksi
J2
7.952156 in.4
3 
T3c3 (32.288539 kip-in.)(2.500 in. / 2)

 10.52 ksi
J3
3.834952 in.4
(a) Maximum Shear Stress Magnitude in Aluminum Segments:
 3  10.52 ksi
Ans.
(b) Maximum Shear Stress Magnitude in Bronze Segment:
 2  1.455 ksi
Ans.
(c) Rotation Angle of Flange C Relative to Support A: The angle of twist in member (1) can be
defined by the difference in rotation angles at the two ends:
1  B   A
B   A  1
Similarly, the angle of twist in member (2) can be defined by:
2  C  B
C  B  2
To derive an expression for C, substitute Eq. (i) into Eq. (j), and note that joint A is restrained from
rotating; therefore, A = 0. 
(i)
(j)
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.
C  1  2
The angle of twist in member (1) is:
TL
(27.711461 kip-in.)(12 in.)
1  1 1 
 0.021678 rad
J1G1 (3.834952 in.4 )(4,000 ksi)
The angle of twist in member (2) is:
TL
(7.711461 kip-in.)(24 in.)
2  2 2 
 0.003581 rad
J 2G2 (7.952156 in.4 )(6,500 ksi)
The rotation angle of flange C is thus:
C  1  2  0.021678 rad  0.003581 rad  0.025259 rad  0.0253 rad
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.
6.83 The torsional assembly shown in Fig. P6.83
consists of solid 2.50-in.-diameter aluminum [G =
4,000 ksi] segments (1) and (3) and a central solid
3.00-in.-diameter bronze [G = 6,500 ksi] segment
(2). Concentrated torques of TB = T0 and TC = 2T0
are applied to the assembly at B and C,
respectively. If the rotation angle at joint C must
not exceed 3°, determine:
(a) the maximum magnitude of T0 that may be
applied to the assembly.
(b) the maximum shear stress magnitude in
aluminum segments (1) and (3).
(c) the maximum shear stress magnitude in
bronze segment (2).
Fig. P6.83
Solution

Section Properties: The polar moments of inertia for the aluminum and bronze segments are:
J1  J 3 
J2 

32
32
(2.50 in.) 4  3.834952 in.4
(3.00 in.) 4  7.952156 in.4
Equilibrium: Consider a free-body diagram cut around
joint B, where the external torque TB is applied:
M x  T1  T2  T0  0
(a)
and also consider a FBD cut around joint C, where the
external torque TC is applied:
M x  T2  T3  2T0  0
Torque-Twist Relationships:
TL
TL
1  1 1
2  2 2
J1G1
J 2G2
3 
(b)
T3 L3
J 3G3
Torque in Segment (3): From the problem statement, the rotation angle at C must not exceed 3°;
consequently, the angle of twist in segment (3) must be the opposite of this value:
3  3  0.052360 rad
From the torque-twist relationships, the internal torque in segment (3) can be computed:
TL
 J G (0.052360 rad)(3.834952 in.4 )(4,000 ksi)
T3  3 3 3 
 66.9325 kip-in.
3  3 3
J 3G3
L3
12 in.
Consider Rotation Angle of Flange C Relative to Support A: The rotation angle at C is given by the
sum of the angles of twist in segments (1) and (2). This rotation angle must not exceed 3°. 
TL
TL
C  1  2  1 1  2 2  3  0.052360 rad
(c)
J1G1 J 2G2
from Eq. (b):
T2  T3  2T0
(d)
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.
and from Eq. (a):
T1  T2  T0  (T3  2T0 )  T0  T3  3T0
(e)
Substitute the expressions derived for T1 and T2 in Eqs. (d) and (e) into Eq. (c):
(T3  3T0 ) L1 (T3  2T0 ) L2

 0.052360 rad
J1G1
J 2G2
Simplify
 L
 3L
L 
2L2 
T3  1  2   T0  1 
  0.052360 rad
 J1G1 J 2G2 
 J1G1 J 2G2 
Solve for T0:
 L
L 
0.052360 rad  T3  1  2 
 J1G1 J 2G2 
T0 
3L1
2 L2

J1G1 J 2G2
and compute:


12 in.
24 in.

0.052360 rad  ( 66.9325 kip-in.) 
4
4

 (3.834952 in. )(4,000 ksi) (7.952156 in. )(6,500 ksi) 
T0 
3(12 in.)
2(24 in.)

4
(3.834952 in. )(4,000 ksi) (7.952156 in.4 )(6,500 ksi)
 41.4590 kip-in.
Backsubstitute this value into Eq. (d) to compute T2:
T2  T3  2T0  66.9325 kip-in.  2(41.4590 kip-in.)  15.9855 kip-in.
and backsubstitute this value into Eq. (e) to compute T1:
T1  T3  3T0  66.9325 kip-in.  3(41.4590 kip-in.)  57.4445 kip-in.
(a) Maximum magnitude of T0:
T0,max  41.5 kip-in.
Ans.
Maximum Shear Stress Magnitudes: The maximum shear stress magnitudes are:
T c (57.4445 kip-in.)(2.500 in. / 2)
1  1 1 
 18.72 ksi
J1
3.834952 in.4
Tc
(15.9855 kip-in.)(3.000 in. / 2)
2  2 2 
 3.02 ksi
J2
7.952156 in.4
3 
T3c3 (66.9325 kip-in.)(2.500 in. / 2)

 21.8 ksi
J3
3.834952 in.4
(b) Maximum Shear Stress Magnitude in Aluminum Segments:
 3  21.8 ksi
Ans.
(c) Maximum Shear Stress Magnitude in Bronze Segment:
 2  3.02 ksi
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.
6.84 The torsional assembly shown in Fig. P6.84
consists of a solid 60-mm-diameter aluminum [G
= 28 GPa] segment (2) and two bronze [G = 45
GPa] tube segments (1) and (3), which have an
outside diameter of 75 mm and a wall thickness
of 5 mm. If concentrated torques of TB = 9 kN-m
and TC = 9 kN-m are applied in the directions
shown, determine:
(a) the maximum shear stress magnitude in
bronze tube segments (1) and (3).
(b) the maximum shear stress magnitude in
aluminum segment (2).
(c) the rotation angle of joint C.
Fig. P6.84
Solution

Section Properties: The polar moments of inertia for the aluminum and bronze segments are:
J1  J 3 
J2 

32
(75 mm) 4  (65 mm) 4   1,353,830 mm 4
32
(60 mm) 4  1, 272,345 mm 4
Equilibrium: Consider a free-body diagram cut around
joint B, where the external torque TB is applied:
M x  T1  T2  TB  0
(a)
and also consider a FBD cut around joint C, where the
external torque TC is applied:
M x  T2  T3  TC  0
(b)
Geometry of Deformation Relationship: Since the two ends of the torsion structure are securely
attached to fixed supports at A and D, the sum of the angles of twist in the three shafts must equal zero:
1  2  3  0
(c)
Torque-Twist Relationships:
TL
TL
1  1 1
2  2 2
J1G1
J 2G2
3 
T3 L3
J 3G3
Compatibility Equation: Substitute the torque-twist relationships [Eqs. (d)] into the geometry of
deformation relationship [Eq. (c)] to obtain the compatibility equation:
T1L1 T2 L2 T3 L3


0
J1G1 J 2G2 J 3G3
(d)
(e)
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.
Solve the Equations: The approach used here will be to reduce the variables in Eq. (e) by replacing T1
and T3 with equivalent expressions involving T2. From Eq. (a):
T1  T2  TB
(f)
and from Eq. (b):
T3  T2  TC
(g)
Substitute Eqs. (f) and (g) into Eq. (e) and simplify to derive an expression for T2:
(T2  TB ) L1 T2 L2 (T2  TC ) L3


0
J1G1
J 2G2
J 3G3
 L
L
L 
TL T L
T2  1  2  3    B 1  C 3
J1G1 J 3G3
 J1G1 J 2G2 J 3G3 
TB L1 TC L3

J1G1 J 3G3
T2  
L1
L
L
 2  3
J1G1 J 2G2 J 3G3
Since G1 = G3 and J1 = J3, Eq. (h) can be simplified further and T2 can be computed as:
1  TB L1  TC L3  
T2  


J1G1  L1  L3  L2 
J 2G2 
 J1G1
(h)
1
 (9,000,000 N-mm)(325 mm)  (9,000,000 N-mm)(325 mm) 
2 

325 mm  325 mm
400 mm
(1,353,830 mm )(45,000 N/mm )



4
4
 (1,353,830 mm )(45,000) (1,272,345 mm )(28,000) 
 4,385, 216 N-mm
Backsubstitute this result into Eqs. (f) and (g) to obtain T1 and T3:
T1  4,385, 216 N-mm  9, 000, 000 N-mm  4, 614, 784 N-mm
T3  4,385, 216 N-mm  9, 000, 000 N-mm  4, 614, 784 N-mm

4
Maximum Shear Stress: The maximum shear stress magnitudes in the three segments are:
T c (4,614,784 N-mm)(75 mm / 2)
1  1 1 
 127.8 MPa
J1
1,353,830 mm 4
Tc
(4,385, 216 N-mm)(60 mm / 2)
2  2 2 
 103.4 MPa
J2
1, 272,345 mm 4
3 
T3c3 (4,614,784 N-mm)(75 mm / 2)

 127.8 MPa
J3
1,353,830 mm 4
(a) Maximum Shear Stress Magnitude in Bronze Segments:
1  127.8 MPa
Ans.
(b) Maximum Shear Stress Magnitude in Aluminum Segment:
 2  103.4 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.
(c) Rotation Angle of Joint C Relative to Support A: The angle of twist in member (1) can be
defined by the difference in rotation angles at the two ends:
1  B   A
B   A  1
Similarly, the angle of twist in member (2) can be defined by:
2  C  B
C  B  2
To derive an expression for C, substitute Eq. (i) into Eq. (j), and note that joint A is restrained from
rotating; therefore, A = 0. 
C  1  2
The angle of twist in member (1) is:
TL
(4,614,784 N-mm)(325 mm)
1  1 1 
 0.024618 rad
J1G1 (1,353,830 mm 4 )(45,000 N/mm 2 )
The angle of twist in member (2) is:
TL
( 4,385, 216 N-mm)(400 mm)
2  2 2 
 0.049237 rad
J 2G2 (1, 272,345 mm 4 )(28,000 N/mm 2 )
The rotation angle of flange C is thus:
C  1  2  0.024618 rad  (0.049237 rad)  0.024618 rad  0.0246 rad
(i)
(j)
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.
6.85 The torsional assembly shown in Fig. P6.85
consists of a solid 60-mm-diameter aluminum [G
= 28 GPa] segment (2) and two bronze [G = 45
GPa] tube segments (1) and (3), which have an
outside diameter of 75 mm and a wall thickness
of 5 mm. If concentrated torques of TB = 6 kN-m
and TC = 10 kN-m are applied in the directions
shown, determine:
(a) the maximum shear stress magnitude in
bronze tube segments (1) and (3).
(b) the maximum shear stress magnitude in
aluminum segment (2).
(c) the rotation angle of joint C.
Fig. P6.85
Solution

Section Properties: The polar moments of inertia for the aluminum and bronze segments are:
J1  J 3 
J2 

32
(75 mm) 4  (65 mm) 4   1,353,830 mm 4
32
(60 mm) 4  1, 272,345 mm 4
Equilibrium: Consider a free-body diagram cut around
joint B, where the external torque TB is applied:
M x  T1  T2  TB  0
(a)
and also consider a FBD cut around joint C, where the
external torque TC is applied:
M x  T2  T3  TC  0
(b)
Geometry of Deformation Relationship: Since the two ends of the torsion structure are securely
attached to fixed supports at A and D, the sum of the angles of twist in the three shafts must equal zero:
1  2  3  0
(c)
Torque-Twist Relationships:
TL
TL
1  1 1
2  2 2
J1G1
J 2G2
3 
T3 L3
J 3G3
Compatibility Equation: Substitute the torque-twist relationships [Eqs. (d)] into the geometry of
deformation relationship [Eq. (c)] to obtain the compatibility equation:
T1L1 T2 L2 T3 L3


0
J1G1 J 2G2 J 3G3
(d)
(e)
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.
Solve the Equations: The approach used here will be to reduce the variables in Eq. (e) by replacing T1
and T3 with equivalent expressions involving T2. From Eq. (a):
T1  T2  TB
(f)
and from Eq. (b):
T3  T2  TC
(g)
Substitute Eqs. (f) and (g) into Eq. (e) and simplify to derive an expression for T2:
(T2  TB ) L1 T2 L2 (T2  TC ) L3


0
J1G1
J 2G2
J 3G3
 L
L
L 
TL T L
T2  1  2  3    B 1  C 3
J1G1 J 3G3
 J1G1 J 2G2 J 3G3 
TB L1 TC L3

J1G1 J 3G3
T2  
L1
L
L
 2  3
J1G1 J 2G2 J 3G3
Since G1 = G3 and J1 = J3, Eq. (h) can be simplified further and T2 can be computed as:
1  TB L1  TC L3  
T2  


J1G1  L1  L3  L2 
J 2G2 
 J1G1
(h)
1
 (6,000,000 N-mm)(325 mm)  (10,000,000 N-mm)(325 mm) 
2 

325 mm  325 mm
400 mm
(1,353,830 mm )(45,000 N/mm )



(1,353,830 mm 4 )(45,000) (1,272,345 mm 4 )(28,000)


 3,897,970 N-mm

4
Backsubstitute this result into Eqs. (f) and (g) to obtain T1 and T3:
T1  3,897,970 N-mm  6, 000, 000 N-mm  2,102, 030 N-mm
T3  3,897,970 N-mm  10, 000, 000 N-mm  6,102, 030 N-mm
Maximum Shear Stress: The maximum shear stress magnitudes in the three segments are:
T c (2,102,030 N-mm)(75 mm / 2)
1  1 1 
 58.2 MPa
J1
1,353,830 mm 4
Tc
(3,897,970 N-mm)(60 mm / 2)
2  2 2 
 91.9 MPa
J2
1, 272,345 mm 4
3 
T3c3 (6,102,030 N-mm)(75 mm / 2)

 169.0 MPa
J3
1,353,830 mm 4
(a) Maximum Shear Stress Magnitude in Bronze Segments:
 3  169.0 MPa
Ans.
(b) Maximum Shear Stress Magnitude in Aluminum Segment:
 2  91.9 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.
(c) Rotation Angle of Joint C Relative to Support A: The angle of twist in member (1) can be
defined by the difference in rotation angles at the two ends:
1  B   A
B   A  1
Similarly, the angle of twist in member (2) can be defined by:
2  C  B
C  B  2
To derive an expression for C, substitute Eq. (i) into Eq. (j), and note that joint A is restrained from
rotating; therefore, A = 0. 
C  1  2
The angle of twist in member (1) is:
TL
(2,102,030 N-mm)(325 mm)
1  1 1 
 0.011214 rad
J1G1 (1,353,830 mm 4 )(45,000 N/mm 2 )
The angle of twist in member (2) is:
TL
( 3,897,970 N-mm)(400 mm)
2  2 2 
 0.043766 rad
J 2G2 (1, 272,345 mm 4 )(28,000 N/mm 2 )
The rotation angle of flange C is thus:
C  1  2  0.011214 rad  (0.043766 rad)  0.032552 rad  0.0326 rad
(i)
(j)
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.
6.86 A solid 1.50-in.-diameter brass [G = 5,600 ksi] shaft [segments (1), (2), and (3)] has been stiffened
between B and C by the addition of a cold-rolled stainless steel tube (4) (Fig. P6.86a). The tube (Fig.
P6.86b) has an outside diameter of 3.50 in., a wall thickness of 0.12 in., and a shear modulus of G =
12,500 ksi. The tube is attached to the brass shaft by means of rigid flanges welded to the tube and to the
shaft. (The thickness of the flanges can be neglected for this analysis.) If a torque of 400 lb-ft is applied
to the shaft as shown in Fig. P6.86a, determine:
(a) the maximum shear stress magnitude in segment (1) of the brass shaft.
(b) the maximum shear stress magnitude in segment (2) of the brass shaft (i.e., between flanges B and C)
(c) the maximum shear stress magnitude in the stainless steel tube (4).
(d) the rotation angle of end D relative to end A.
Fig. P6.86a
Fig. P6.86b Cross Section Through Tube
Solution

Section Properties: The polar moments of inertia for the brass shaft and the stainless steel tube are:
J1  J 2  J 3 
J4 

32
(1.50 in.) 4  0.497010 in.4
(3.50 in.) 4  (3.26 in.) 4   3.643915 in.4
32
(a) Shear Stress Magnitude in Brass Segment (1):
T c (400 lb-ft)(1.500 in. / 2)(12 in./ft)
1  1 1 
 7, 243.3 psi  7, 240 psi
J1
0.497010 in.4
Ans.
Statically Indeterminate Portion of the Shaft: The portion of the shaft between B and C is statically
indeterminate.
Equilibrium: Consider a free-body diagram that cuts
through the shaft between B and C and includes the
free end of the shaft at A:
M x  T2  T4  400 lb-ft  0
Geometry of Deformation Relationship: Since the tube and shaft are securely connected by the rigid
flanges, the angles of twist in both members must be equal; therefore,
2  4
(b)
Torque-Twist Relationships:
TL
TL
2  2 2
4  4 4
J 2G2
J 4G4
(c)
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.
Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of
deformation relationship [Eq. (b)] to obtain the compatibility equation:
T2 L2
TL
 4 4
J 2G2 J 4G4
(d)
Solve the Equations: Solve Eq. (d) for T2:
4
L J G
 20 in.  0.497010 in.   5,600 ksi 
 0.0611047 T4
T2  T4 4 2 2  T4 
 20 in.  3.643915 in.4   12,500 ksi 
L2 J 4 G4
and substitute this result into Eq. (a) to compute the torque T4 in the stainless steel tube (4):
T2  T4  0.0611047 T4  T4  1.0611047 T4  400 lb-ft
T4  376.9656 lb-ft
The torque in brass shaft segment (2) is therefore:
T2  400 lb-ft  T4  400 lb-ft  ( 376.9656 lb-ft)  23.0344 lb-ft
(b) Maximum Shear Stress in Brass Segment (2): The maximum shear stress in segment (2) is:
(23.0344 lb-ft)(1.50 in. / 2)(12 in./ft)
Tc
2  2 2 
 417.1 psi  417 psi
Ans.
0.497010 in.4
J2
(c) Maximum Shear Stress in Stainless Steel Tube (4): The maximum shear stress in tube (4) is:
(376.9656 lb-ft)(3.50 in. / 2)(12 in./ft)
Tc
4  4 4 
 2,172.5 psi  2,170 psi
Ans.
3.643915 in.4
J4
(d) Rotation Angle of End D Relative to End A: The angle of twist of segment (1) is:
TL
( 400 lb-ft)(12 in.)(12 in./ft)
1  1 1 
 0.020695 rad
J1G1 (0.497010 in.4 )(5,600,000 psi)
The angle of twist in brass segment (3) has the same value:
3  0.020695 rad
The angle of twist in brass segment (2) is:
TL
( 23.0344 lb-ft)(20 in.)(12 in./ft)
 0.0019863 rad
2  2 2 
J 2G2
(0.497010 in.4 )(5,600,000 psi)
The rotation angle of end D relative to end A is the sum of these three angles of twist:
 D  1  2  3
 0.020695 rad  ( 0.0019863 rad)  ( 0.020695 rad)
 0.0433763 rad  0.0434 rad
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.
6.87 A solid 60-mm-diameter cold-rolled brass [G = 39 GPa] shaft that is 1.25-m long extends through
and is completely bonded to a hollow aluminum [G = 28 GPa] tube, as shown in Fig. P6.87. Aluminum
tube (1) has an outside diameter of 90 mm, an inside diameter of 60 mm, and a length of 0.75 m. Both
the brass shaft and the aluminum tube are securely attached to the wall support at A. When the two
torques shown are applied to the composite shaft, determine:
(a) the maximum shear stress magnitude in
aluminum tube (1).
(b) the maximum shear stress magnitude in brass
shaft segment (2).
(c) the maximum shear stress magnitude in brass
shaft segment (3).
(d) the rotation angle of joint B.
(e) the rotation angle of end C.
Fig. P6.87
Solution

Section Properties: The polar moments of inertia for tube (1) and brass shafts (2) and (3) are:
J1 
(90 mm) 4  (60 mm) 4   5,168,902 mm 4
32 
J 2  J3 

32
(60 mm) 4  1, 272,345 mm 4
Equilibrium: Consider a free-body diagram cut
around end C through segment (3):
M x  T3  8 kN-m  0
T3  8 kN-m (a)
and also consider a FBD cut around joint B:
M x  T1  T2  T3  20 kN-m  0
T1  T2  T3  20 kN-m  12 kN-m
(b)
Eq. (b) reveals that the portion of the shaft between A and B is statically indeterminate. The five-step
solution procedure will be used to determine T1 and T2 in this portion of the shaft.
Geometry of Deformation Relationship: Since the aluminum tube and the brass shaft are securely
bonded together, the angles of twist in both members must be equal; therefore,
1  2
(c)
Torque-Twist Relationships:
TL
TL
1  1 1
2  2 2
J1G1
J 2G2
(d)
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.
Compatibility Equation: Substitute the torque-twist relationships [Eqs. (d)] into the geometry of
deformation relationship [Eq. (c)] to obtain the compatibility equation:
T1L1 T2 L2

J1G1 J 2G2
(e)
Solve the Equations: Solve Eq. (e) for T1:
 5,168,902 mm4   28 GPa 
L J G
T1  T2 2 1 1  T2 

  2.916667 T2
L1 J 2 G2
 1,272,345 mm4   39 GPa 
and substitute this result into Eq. (b) to compute the torque T2 in brass shaft (2):
T1  T2  2.916667 T2  T2  3.916667 T2  12,000 N-m
T2  3,063.830 N-m
The torque in aluminum tube (1) is therefore:
T1  12, 000 N-m  T2  12, 000 N-m  3, 063.830 N-m  8,936.170 N-m
(a) Maximum Shear Stress in Aluminum Tube: The maximum shear stress in aluminum tube (1) is:
T c (8,936.170 N-m)(90 mm / 2)(1,000 mm/m)
1  1 1 
 77.8 MPa
Ans.
J1
5,168,902 mm 4
(b) Maximum Shear Stress Magnitude in Brass Shaft Segment (2):
Tc
(3,063.830 N-m)(60 mm / 2)(1,000 mm/m)
2  2 2 
 72.2 MPa
J2
1, 272,345 mm 4
Ans.
(c) Maximum Shear Stress Magnitude in Brass Shaft Segment (3):
(8,000 N-m)(60 mm / 2)(1,000 mm/m)
Tc
3  3 3 
 188.6 MPa
1, 272,345 mm 4
J3
Ans.
(d) Rotation Angle of Joint B:
(8,936.170 N-m)(0.75 m)(1,000 mm/m) 2
TL
1  1 1 
 0.046308 rad
(5,168,902 mm 4 )(28,000 N/mm 2 )
J1G1
or
2 
T2 L2 (3,063.830 N-m)(0.75 m)(1,000 mm/m) 2

 0.046308 rad
(1,272,345 mm 4 )(39,000 N/mm 2 )
J 2G2
  B  0.0463 rad
Ans.
(e) Rotation Angle of End C:
TL
( 8,000 N-m)(0.5 m)(1,000 mm/m) 2
 0.080610 rad
3  3 3 
(1,272,345 mm 4 )(39,000 N/mm 2 )
J 3G3
 C   B  3  0.046308 rad  ( 0.080610 rad)  0.0343 rad
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.
6.88 The gear assembly shown in Fig. P6.88 is
subjected to a torque of TC = 140 N-m. Shafts (1)
and (2) are solid 20-mm-diameter steel shafts, and
shaft (3) is a solid 25-mm-diameter steel shaft.
Assume L = 400 mm and G = 80 GPa. Determine:
(a) the maximum shear stress magnitude in shaft
(1).
(b) the maximum shear stress magnitude in shaft
segment (3).
(c) the rotation angle of gear E.
(d) the rotation angle of gear C.
Fig. P6.88
Solution

Section Properties: The polar moments of inertia for the shafts are:
J1  J 2 
J3 

32
32
(20 mm) 4  15,708.0 mm 4
(25 mm) 4  38,349.5 mm 4
Equilibrium: Consider a free-body diagram cut through
shaft (2) and around gear C:
M x  T2  TC  0
 T2  TC  140 N-m (a)
Next, consider a FBD cut around gear B through shafts (1)
and (2). The teeth of gear C exert a force F on the teeth of
gear B. This force F opposes the rotation of gear B. The
radius of gear B will be denoted by RB for now (even though
the gear radius is not given explicitly).
M x  T2  T1  F  RB  0
(b)
Finally, consider a FBD cut around gear E through shaft (3).
The teeth of gear B exert an equal magnitude force F on the
teeth of gear C, acting opposite to the direction assumed in
the previous FBD. The radius of gear E will be denoted by
RE for now.
T
M x  T3  F  RE  0
F   3
(c)
RE
The results of Eqs. (a) and (c) can be substituted into Eq. (b) to give
R
T1  140 N-m  T3 B
RE
The ratio RB/RE is simply the gear ratio between gears B and E, which can also be expressed in terms of
gear teeth NB and NE:
N
 24 teeth 
 140 N-m  0.4T3
T1  140 N-m  T3 B  140 N-m  T3 
(d)
 60 teeth 
NE
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.
Equation (d) summarizes the results of the equilibrium considerations, but there are still two unknowns
in this equation: T1 and T3. Consequently, this problem is statically indeterminate. To solve the
problem, an additional equation must be developed. This second equation will be derived from the
relationship between the angles of twist in shafts (1) and (3).
Geometry of Deformation Relationship:
The rotation of gear B is equal to the angle of twist in shaft (1):
B   1
and the rotation of gear E is equal to the angle of twist in shaft (3):
E  3
However, since the gear teeth mesh, the rotation angles for gears B and E are not independent. The
arclengths associated with the respective rotations must be equal, but the gears turn in opposite
directions. The relationship between gear rotations can be stated as:
RBB   REE
where RB and RE are the radii of gears B and E, respectively. Since the gear rotation angles are related to
the shaft angles of twist, this relationship can be expressed as:
RB1   RE3
(e)
Torque-Twist Relationships:
TL
TL
1  1 1
3  3 3
J1G1
J 3G3
(f)
Compatibility Equation:
Substitute the torque-twist relationships [Eqs. (f)] into the geometry of deformation relationship [Eq. (e)]
to obtain:
TL
TL
RB 1 1   RE 3 3
J1G1
J 3G3
which can be rearranged and expressed in terms of the gear ratio NB /NE:
N B T1 L1
TL
 3 3
(g)
N E J1G1
J 3G3
Note that the compatibility equation has two unknowns: T1 and T3. This equation can be solved
simultaneously with Eq. (d) to calculate the internal torques in shafts (1) and (3).
Solve the Equations: Solve for internal torque T3 in Eq. (g):
 N  L  J G 
T3  T1  B   1   3   3 
 N E   L3   J1   G1 
and substitute this result into Eq. (d):
T1  140 N-m  0.4T3
  N   L   J  G 
 140 N-m  0.4  T1  B   1   3   3  
  N E   L3   J1   G1  
  24 teeth   38,349.5 mm 4  
 140 N-m  0.4 T1 

4
  60 teeth   15,708.0 mm  
 140 N-m  0.390624 T1
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.
Group the T1 terms to obtain:
140 N-m
T1 
 100.6742 N-m
1.390624
Backsubstitute this result into Eq. (d) to find the internal torque in shaft (3):
T1  140 N-m  0.4T3
T1  140 N-m 100.6742 N-m  140 N-m

 98.3146 N-m
0.4
0.4
(a) Maximum Shear Stress Magnitude in Shaft (1):
T c (100.6742 N-m)(20 mm / 2)(1,000 mm/m)
1  1 1 
 64.1 MPa
J1
15,708.0 mm 4
 T3 
(b) Maximum Shear Stress Magnitude in Shaft (3):
Tc
(98.3146 N-m)(25 mm / 2)(1,000 mm/m)
3  3 3 
 32.0 MPa
J3
38,349.5 mm 4
Ans.
Ans.
(c) Rotation Angle of Gear E:
TL
( 98.3146 N-m)(1.25)(400 mm)(1,000 mm/m)
3  3 3 
 0.016023 rad
J 3G3
(38,349.5 mm 4 )(80,000 N/mm 2 )
  E  3  0.01602 rad
Ans.
(d) Rotation Angle of Gear C:
(100.6742 N-m)(1.25)(400 mm)(1,000 mm/m)
TL
1  1 1 
 0.040057 rad
J1G1
(15,708.0 mm 4 )(80,000 N/mm 2 )
2 
T2 L2 (140 N-m)(400 mm)(1,000 mm/m)

 0.044563 rad
J 2G2
(15,708.0 mm 4 )(80,000 N/mm 2 )
 C   B  2  1  2  0.040057 rad  0.044563 rad  0.0846 rad
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.
6.89 The gear assembly shown in Fig. P6.89 is
subjected to a torque of TC = 1,100 lb-ft. Shafts
(1) and (2) are solid 1.625-in.-diameter aluminum
shafts and shaft (3) is a solid 2.00-in.-diameter
aluminum shaft. Assume L = 20 in. and G = 4,000
ksi. Determine:
(a) the maximum shear stress magnitude in shaft
(1).
(b) the maximum shear stress magnitude in shaft
segment (3).
(c) the rotation angle of gear E.
(d) the rotation angle of gear C.
Fig. P6.89
Solution

Section Properties: The polar moments of inertia for the shafts are:
J1  J 2 
J3 

32
32
(1.625 in.) 4  0.684563 in.4
(2.00 in.) 4  1.570796 in.4
Equilibrium: Consider a free-body diagram cut through
shaft (2) and around gear C:
M x  T2  TC  0
 T2  TC  1,100 lb-ft (a)
Next, consider a FBD cut around gear B through shafts (1)
and (2). The teeth of gear C exert a force F on the teeth of
gear B. This force F opposes the rotation of gear B. The
radius of gear B will be denoted by RB for now (even though
the gear radius is not given explicitly).
M x  T2  T1  F  RB  0
(b)
Finally, consider a FBD cut around gear E through shaft (3).
The teeth of gear B exert an equal magnitude force F on the
teeth of gear C, acting opposite to the direction assumed in
the previous FBD. The radius of gear E will be denoted by
RE for now.
T
M x  T3  F  RE  0
F   3
(c)
RE
The results of Eqs. (a) and (c) can be substituted into Eq. (b) to give
R
T1  1,100 lb-ft  T3 B
RE
The ratio RB/RE is simply the gear ratio between gears B and E, which can also be expressed in terms of
gear teeth NB and NE:
N
 24 teeth 
 1,100 lb-ft  0.4T3
T1  1,100 lb-ft  T3 B  1,100 lb-ft  T3 
(d)
 60 teeth 
NE
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.
Equation (d) summarizes the results of the equilibrium considerations, but there are still two unknowns
in this equation: T1 and T3. Consequently, this problem is statically indeterminate. To solve the
problem, an additional equation must be developed. This second equation will be derived from the
relationship between the angles of twist in shafts (1) and (3).
Geometry of Deformation Relationship:
The rotation of gear B is equal to the angle of twist in shaft (1):
B   1
and the rotation of gear E is equal to the angle of twist in shaft (3):
E  3
However, since the gear teeth mesh, the rotation angles for gears B and E are not independent. The
arclengths associated with the respective rotations must be equal, but the gears turn in opposite
directions. The relationship between gear rotations can be stated as:
RBB   REE
where RB and RE are the radii of gears B and E, respectively. Since the gear rotation angles are related to
the shaft angles of twist, this relationship can be expressed as:
RB1   RE3
(e)
Torque-Twist Relationships:
TL
TL
1  1 1
3  3 3
J1G1
J 3G3
(f)
Compatibility Equation:
Substitute the torque-twist relationships [Eqs. (f)] into the geometry of deformation relationship [Eq. (e)]
to obtain:
TL
TL
RB 1 1   RE 3 3
J1G1
J 3G3
which can be rearranged and expressed in terms of the gear ratio NB /NE:
N B T1 L1
TL
 3 3
(g)
N E J1G1
J 3G3
Note that the compatibility equation has two unknowns: T1 and T3. This equation can be solved
simultaneously with Eq. (d) to calculate the internal torques in shafts (1) and (3).
Solve the Equations: Solve for internal torque T3 in Eq. (g):
 N  L  J G 
T3  T1  B   1   3   3 
 N E   L3   J1   G1 
and substitute this result into Eq. (d):
T1  1,100 lb-ft  0.4T3
  N   L   J  G 
 1,100 lb-ft  0.4  T1  B   1   3   3  
  N E   L3   J1   G1  
  24 teeth   1.570796 in.4  
 1,100 lb-ft  0.4 T1 

4
  60 teeth   0.684563 in.  
 1,100 lb-ft  0.367135T1
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.
Group the T1 terms to obtain:
1,100 lb-ft
T1 
 804.602 lb-ft
1.367135
Backsubstitute this result into Eq. (d) to find the internal torque in shaft (3):
T1  1,100 lb-ft  0.4T3
T1  1,100 lb-ft 804.602 lb-ft  1,100 lb-ft

 738.495 lb-ft
0.4
0.4
(a) Maximum Shear Stress Magnitude in Shaft (1):
T c (804.602 lb-ft)(1.625 in. / 2)(12 in./ft)
1  1 1 
 11, 459.677 psi  11.46 ksi
0.684563 in.4
J1
 T3 
(b) Maximum Shear Stress Magnitude in Shaft (3):
(738.495 lb-ft)(2.00 in. / 2)(12 in./ft)
Tc
3  3 3 
 5,641.687 psi  5.64 ksi
1.570796 in.4
J3
Ans.
Ans.
(c) Rotation Angle of Gear E:
TL
( 738.495 lb-ft)(1.25)(20 in.)(12 in./ft)
3  3 3 
 0.035261 rad
J 3G3
(1.570796 in.4 )(4,000,000 psi)
  E  3  0.0353 rad
Ans.
(d) Rotation Angle of Gear C:
(804.602 lb-ft)(1.25)(20 in.)(12 in./ft)
TL
1  1 1 
 0.088151 rad
J1G1
(0.684563 in.4 )(4,000,000 psi)
2 
T2 L2
(1,100 lb-ft)(20 in.)(12 in./ft)

 0.096412 rad
J 2G2 (0.684563 in.4 )(4,000,000 psi)
 C   B  2  1  2  0.088151 rad  0.096412 rad  0.1846 rad
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.
6.90 A torque of TC = 460 N-m acts on gear C of
the assembly shown in Fig. P6.90. Shafts (1)
and (2) are solid 35-mm-diameter aluminum
shafts and shaft (3) is a solid 25-mm-diameter
aluminum shaft. Assume L = 200 mm and G =
28 GPa. Determine:
(a) the maximum shear stress magnitude in
shaft (1).
(b) the maximum shear stress magnitude in
shaft segment (3).
(c) the rotation angle of gear E.
(d) the rotation angle of gear C.
Fig. P6.90
Solution
Section Properties: The polar moments of inertia for the shafts are:

J1  J 2 
(35 mm) 4  147,323.5 mm 4
32

J 3  (25 mm) 4  38,349.5 mm 4
32
Equilibrium: Consider a free-body diagram cut through
shaft (2) and around gear C:
M x  T2  TC  0
 T2  TC  460 N-m (a)
Next, consider a FBD cut around gear B through shafts (1)
and (2). The teeth of gear E exert a force F on the teeth of
gear B. This force F opposes the rotation of gear B. The
radius of gear B will be denoted by RB for now (even though
the gear radius is not given explicitly).
M x  T2  T1  F  RB  0
(b)
Finally, consider a FBD cut around gear E through shaft (3).
The teeth of gear B exert an equal magnitude force F on the
teeth of gear C, acting opposite to the direction assumed in
the previous FBD. The radius of gear E will be denoted by
RE for now.
T
M x  T3  F  RE  0
F   3
(c)
RE
The results of Eqs. (a) and (c) can be substituted into Eq. (b) to give
R
T1  460 N-m  T3 B
RE
The ratio RB/RE is simply the gear ratio between gears B and E, which can also be expressed in terms of
gear teeth NB and NE:
N
 54 teeth 
T1  460 N-m  T3 B  460 N-m  T3 
 460 N-m  1.285714 T3
(d)
 42 teeth 
NE
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.
Equation (d) summarizes the results of the equilibrium considerations, but there are still two unknowns
in this equation: T1 and T3. Consequently, this problem is statically indeterminate. To solve the
problem, an additional equation must be developed. This second equation will be derived from the
relationship between the angles of twist in shafts (1) and (3).
Geometry of Deformation Relationship:
The rotation of gear B is equal to the angle of twist in shaft (1):
B   1
and the rotation of gear E is equal to the angle of twist in shaft (3):
E  3
However, since the gear teeth mesh, the rotation angles for gears B and E are not independent. The
arclengths associated with the respective rotations must be equal, but the gears turn in opposite
directions. The relationship between gear rotations can be stated as:
RBB   REE
where RB and RE are the radii of gears B and E, respectively. Since the gear rotation angles are related to
the shaft angles of twist, this relationship can be expressed as:
RB1   RE3
(e)
Torque-Twist Relationships:
TL
TL
1  1 1
3  3 3
J1G1
J 3G3
(f)
Compatibility Equation:
Substitute the torque-twist relationships [Eqs. (f)] into the geometry of deformation relationship [Eq. (e)]
to obtain:
TL
TL
RB 1 1   RE 3 3
J1G1
J 3G3
which can be rearranged and expressed in terms of the gear ratio NB /NE:
N B T1 L1
TL
 3 3
(g)
N E J1G1
J 3G3
Note that the compatibility equation has two unknowns: T1 and T3. This equation can be solved
simultaneously with Eq. (d) to calculate the internal torques in shafts (1) and (3).
Solve the Equations: Solve for internal torque T3 in Eq. (g):
 N  L  J G 
T3  T1  B   1   3   3 
 N E   L3   J1   G1 
and substitute this result into Eq. (d):
T1  460 N-m  1.285714 T3
  N   L   J  G 
 460 N-m  1.285714  T1  B   1   3   3  
  N E   L3   J1   G1  
  54 teeth   38,349.5 mm 4  
 460 N-m  1.285714 T1 

4
  42 teeth   147,323.5 mm  
 460 N-m  0.430305 T1
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.
Group the T1 terms to obtain:
460 N-m
T1 
 321.610 N-m
1.430305
Backsubstitute this result into Eq. (d) to find the internal torque in shaft (3):
T1  460 N-m  1.285714 T3
 T3 
T1  460 N-m 321.610 N-m  460 N-m

 107.637 N-m
1.285714
1.285714
(a) Maximum Shear Stress Magnitude in Shaft (1):
T c (321.610 N-m)(35 mm / 2)(1,000 mm/m)
1  1 1 
 38.2 MPa
147,323.5 mm 4
J1
Ans.
(b) Maximum Shear Stress Magnitude in Shaft (3):
(107.637 N-m)(25 mm / 2)(1,000 mm/m)
Tc
3  3 3 
 35.1 MPa
38,349.5 mm 4
J3
Ans.
(c) Rotation Angle of Gear E:
TL
( 107.637 N-m)(2)(200 mm)(1,000 mm/m)
3  3 3 
 0.040096 rad
J 3G3
(38,349.5 mm 4 )(28, 000 N/mm 2 )
  E  3  0.0401 rad
Ans.
(d) Rotation Angle of Gear C:
(321.610 N-m)(2)(200 mm)(1,000 mm/m)
TL
1  1 1 
 0.031186 rad
J1G1
(147,323.5 mm 4 )(28,000 N/mm 2 )
2 
T2 L2 (460 N-m)(200 mm)(1,000 mm/m)

 0.022303 rad
J 2G2 (147,323.5 mm 4 )(28,000 N/mm 2 )
 C   B  2  1  2  0.031186 rad  0.022303 rad  0.0535 rad
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.
6.91 A torque of TC = 40 kip-in. acts on gear C
of the assembly shown in Fig. P6.91. Shafts
(1) and (2) are solid 2.00-in.-diameter stainless
steel shafts and shaft (3) is a solid 1.75-in.diameter stainless steel shaft. Assume L = 8 in.
and G = 12,500 ksi. Determine:
(a) the maximum shear stress magnitude in
shaft (1).
(b) the maximum shear stress magnitude in
shaft segment (3).
(c) the rotation angle of gear E.
(d) the rotation angle of gear C.
Fig. P6.91
Solution

Section Properties: The polar moments of inertia for the shafts are:
J1  J 2 
J3 

32
32
(2.00 in.) 4  1.570796 in.4
(1.75 in.) 4  0.920772 in.4
Equilibrium: Consider a free-body diagram cut through
shaft (2) and around gear C:
M x  T2  TC  0
 T2  TC  40 kip-in. (a)
Next, consider a FBD cut around gear B through shafts (1)
and (2). The teeth of gear E exert a force F on the teeth of
gear B. This force F opposes the rotation of gear B. The
radius of gear B will be denoted by RB for now (even though
the gear radius is not given explicitly).
M x  T2  T1  F  RB  0
(b)
Finally, consider a FBD cut around gear E through shaft (3).
The teeth of gear B exert an equal magnitude force F on the
teeth of gear C, acting opposite to the direction assumed in
the previous FBD. The radius of gear E will be denoted by
RE for now.
T
M x  T3  F  RE  0
F   3
(c)
RE
The results of Eqs. (a) and (c) can be substituted into Eq. (b) to give
R
T1  40 kip-in.  T3 B
RE
The ratio RB/RE is simply the gear ratio between gears B and E, which can also be expressed in terms of
gear teeth NB and NE:
N
 54 teeth 
 40 kip-in.  1.285714 T3
T1  40 kip-in.  T3 B  40 kip-in.  T3 
(d)
 42 teeth 
NE
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.
Equation (d) summarizes the results of the equilibrium considerations, but there are still two unknowns
in this equation: T1 and T3. Consequently, this problem is statically indeterminate. To solve the
problem, an additional equation must be developed. This second equation will be derived from the
relationship between the angles of twist in shafts (1) and (3).
Geometry of Deformation Relationship:
The rotation of gear B is equal to the angle of twist in shaft (1):
B   1
and the rotation of gear E is equal to the angle of twist in shaft (3):
E  3
However, since the gear teeth mesh, the rotation angles for gears B and E are not independent. The
arclengths associated with the respective rotations must be equal, but the gears turn in opposite
directions. The relationship between gear rotations can be stated as:
RBB   REE
where RB and RE are the radii of gears B and E, respectively. Since the gear rotation angles are related to
the shaft angles of twist, this relationship can be expressed as:
RB1   RE3
(e)
Torque-Twist Relationships:
TL
TL
1  1 1
3  3 3
J1G1
J 3G3
(f)
Compatibility Equation:
Substitute the torque-twist relationships [Eqs. (f)] into the geometry of deformation relationship [Eq. (e)]
to obtain:
TL
TL
RB 1 1   RE 3 3
J1G1
J 3G3
which can be rearranged and expressed in terms of the gear ratio NB /NE:
N B T1 L1
TL
 3 3
(g)
N E J1G1
J 3G3
Note that the compatibility equation has two unknowns: T1 and T3. This equation can be solved
simultaneously with Eq. (d) to calculate the internal torques in shafts (1) and (3).
Solve the Equations: Solve for internal torque T3 in Eq. (g):
 N  L  J G 
T3  T1  B   1   3   3 
 N E   L3   J1   G1 
and substitute this result into Eq. (d):
T1  40 kip-in.  1.285714 T3
  N   L   J  G 
 40 kip-in.  1.285714  T1  B   1   3   3  
  N E   L3   J1   G1  
  54 teeth   0.920772 in.4  
 40 kip-in.  1.285714 T1 

4 
  42 teeth   1.570796 in.  
 40 kip-in.  0.968994 T1
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.
Group the T1 terms to obtain:
40 kip-in.
T1 
 20.3149 kip-in.
1.968994
Backsubstitute this result into Eq. (d) to find the internal torque in shaft (3):
T1  40 kip-in.  1.285714 T3
 T3 
T1  40 kip-in. 20.3149 kip-in.  40 kip-in.

 15.3106 kip-in.
1.285714
1.285714
(a) Maximum Shear Stress Magnitude in Shaft (1):
T c (20.3149 kip-in.)(2.00 in. / 2)
1  1 1 
 12.93 ksi
1.570796 in.4
J1
Ans.
(b) Maximum Shear Stress Magnitude in Shaft (3):
(15.3106 kip-in.)(1.75 in. / 2)
Tc
3  3 3 
 14.55 ksi
0.920772 in.4
J3
Ans.
(c) Rotation Angle of Gear E:
TL
( 15.3106 kip-in.)(2)(8 in.)
 0.021284 rad
3  3 3 
J 3G3 (0.920772 in.4 )(12, 500 ksi)
  E  3  0.0213 rad
Ans.
(d) Rotation Angle of Gear C:
TL
(20.3149 kip-in.)(2)(8 in.)
1  1 1 
 0.016554 rad
J1G1 (1.570796 in.4 )(12,500 ksi)
2 
T2 L2
(40 kip-in.)(8 in.)

 0.016297 rad
J 2G2 (1.570796 in.4 )(12,500 ksi)
 C   B  2  1  2  0.016554 rad  0.016297 rad  0.0329 rad
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.
6.92 The steel [G = 12,000 ksi] pipe shown in Fig. P6.92 is fixed to the wall support at C. The bolt holes
in the flange at A were supposed to align with mating holes in the wall support; however, an angular
misalignment of 4° was found to exist. To connect the pipe to its supports, a temporary installation
torque T′B must be applied at B to align flange A with the mating holes in the wall support. The outside
diameter of the pipe is 3.50 in. and its wall thickness is 0.216 in.
(a) Determine the temporary installation
torque T′B that must be applied at B to align
the bolt holes at A.
(b) Determine the maximum shear stress
initial in the pipe after the bolts are
connected and the temporary installation
torque at B is removed.
(c) If the maximum shear stress in the pipe
shaft must not exceed 12 ksi, determine the
maximum external torque TB that can be
applied at B after the bolts are connected.
Fig. P6.92
Solution

Section Properties: The polar moment of inertia for the pipe is:
J1  J 2 
(3.50 in.)4  (3.068 in.)4   6.034313 in.4
32
(a) Temporary Installation Torque T′B: The shaft, which is fixed to the support at C, must be rotated
4° so that the connection at A can be completed. Consequently, segment (2) must be twisted 4° (or
0.069813 rad) by the temporary torque T′B. From the torque-twist relationship for shaft (2)
TL
2  2 2
J 2 G2
the temporary torque T′B is:
2 J 2G2 (0.069813 rad)(6.034313 in.4 )(12,000 ksi)
Ans.
TB 

 28.1 kip-in.
L2
(15 ft)(12 in./ft)
(b) Maximum Shear Stress after Temporary Torque is Removed: After T′B is removed, the 4° angle
of twist now applies to the total pipe length, i.e., 25 ft. The internal torque magnitude in the pipe due to
the 4° misfit is thus:
 JG (0.069813 rad)(6.034313 in.4 )(12,000 ksi)
Tinitial  misfit

 16.851 kip-in.
( L1  L2 )
(25 ft)(12 in./ft)
and the maximum shear stress magnitude in the pipe due to this internal torque is:
T c (16.851 kip-in.)(3.50 in. / 2)
 initial  initial 
 4.89 ksi
Ans.
J
6.034313 in.4
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.
After Bolts Installed: After the bolts are installed, the pipe is analyzed as a statically indeterminate
torsion structure.
Equilibrium:
M x  T1  T2  TB  0
(a)
Note: By inspection, the internal torque T2 will end up having
a negative value.
Geometry of Deformation Relationship: The two shafts are securely attached to fixed supports at A
and C; however, in this instance, the sum of the angles of twist in the two members must equal the misfit
angle. Note: If the rotation angle at A is a positive 4° (as can be inferred from the problem sketch), then
the pipe must twist in a negative sense to reach a zero rotation angle at C.
1  2  0.069813 rad
(b)
Torque-Twist Relationships:
TL
TL
1  1 1
2  2 2
J1G1
J 2G2
Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of
deformation relationship [Eq. (b)] to obtain the compatibility equation:
T1L1 T2 L2

 0.069813 rad
J1G1 J 2G2
(c)
(d)
Solve the Equations: Since the problem is expressed in terms of allowable stresses, it is convenient to
rewrite Eq. (d) in terms of stresses. In general, the elastic torsion formula can be rearranged as:
Tc
T 
 

J
J c
which allows Eq. (d) to be rewritten as:
1L1
 L
 0.069813 rad  2 2
(e)
G1c1
G2c2
Solve Eq. (e) for  1:
GR
L G c
 1  ( 0.069813 rad) 1 1   2 2 1 1
L1
L1 G2 c2
(12,000 ksi)(3.50 in./2)
 15 ft 
 2 
 10 ft 
(10 ft)(12 in./ft)
 12.217305 ksi  1.5  2
 ( 0.069813 rad)
(f)
Assume the shaft (2) controls: By inspection of the FBD, the internal torque T2 should have a negative
value. Therefore, we can assume that the maximum shear stress in shaft (2) will be −12 ksi. If the shear
stress in shaft (2) reaches this value, then the shear stress in shaft (1) will be:
1  12.217305 ksi  1.5( 12 ksi)  5.7827 ksi  12 ksi
O.K.
This calculation demonstrates that the shear stress magnitude in shaft (2) must control.
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.
Now that the maximum shear stress magnitudes in the two shafts are known, the torque magnitudes in
each component can be computed:
1 J1 (5.7827 ksi)(6.034313 in.4 )
T1 

 19.9398 kip-in.
c1
3.50 in./2
T2 
2J2
c2

( 12 ksi)(6.034313 in.4 )
 41.3781 kip-in.
3.50 in./2
(c) Maximum Allowable Torque: From Eq. (a), the total torque acting at B must not exceed:
TB,max  T1  T2  19.9398 kip-in.  (41.3781 kip-in.)  61.3179 kip-in.  61.3 kip-in.
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.
6.93 The steel [G = 12,000 ksi] pipe shown in Fig. P6.93 is fixed to the wall support at C. The bolt holes
in the flange at A were supposed to align with mating holes in the wall support; however, an angular
misalignment of 4° was found to exist. To connect the pipe to its supports, a temporary installation
torque T′B must be applied at B to align flange A with the mating holes in the wall support. The outside
diameter of the pipe is 2.875 in. and its wall thickness is 0.203 in.
(a) Determine the temporary installation
torque T′B that must be applied at B to align
the bolt holes at A.
(b) Determine the maximum shear stress
initial in the pipe after the bolts are
connected and the temporary installation
torque at B is removed.
(c) Determine the magnitude of the
maximum shear stress in segments (1) and
(2) if an external torque of TB = 80 kip-in. is
applied at B after the bolts are connected.
Fig. P6.93
Solution

Section Properties: The polar moment of inertia for the pipe is:
J1  J 2 
(2.875 in.) 4  (2.469 in.) 4   3.059108 in.4
32
(a) Temporary Installation Torque T′B: The shaft, which is fixed to the support at C, must be rotated
4° so that the connection at A can be completed. Consequently, segment (2) must be twisted 4° (or
0.069813 rad) by the temporary torque T′B. From the torque-twist relationship for shaft (2)
TL
2  2 2
J 2 G2
the temporary torque T′B is:
2 J 2G2 (0.069813 rad)(3.059108 in.4 )(12,000 ksi)
Ans.

 14.24 kip-in.
TB 
(15 ft)(12 in./ft)
L2
(b) Maximum Shear Stress after Temporary Torque is Removed: After T′B is removed, the 4° angle
of twist now applies to the total pipe length, i.e., 25 ft. The internal torque magnitude in the pipe due to
the 4° misfit is thus:
 JG (0.069813 rad)(3.059108 in.4 )(12,000 ksi)
Tinitial  misfit

 8.5426 kip-in.
( L1  L2 )
(25 ft)(12 in./ft)
and the maximum shear stress magnitude in the pipe due to this internal torque is:
T c (8.5426 kip-in.)(2.875 in. / 2)
 initial  initial 
 4.01 ksi
Ans.
J
3.059108 in.4
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.
After Bolts Installed: After the bolts are installed, the pipe is analyzed as a statically indeterminate
torsion structure.
Equilibrium:
M x  T1  T2  TB  0
(a)
Note: By inspection, the internal torque T2 will end up having
a negative value.
Geometry of Deformation Relationship: The two shafts are securely attached to fixed supports at A
and C; however, in this instance, the sum of the angles of twist in the two members must equal the misfit
angle. Note: If the rotation angle at A is a positive 4° (as can be inferred from the problem sketch), then
the pipe must twist in a negative sense to reach a zero rotation angle at C.
1  2  0.069813 rad
(b)
Torque-Twist Relationships:
TL
TL
1  1 1
2  2 2
J1G1
J 2G2
Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of
deformation relationship [Eq. (b)] to obtain the compatibility equation:
T1L1 T2 L2

 0.069813 rad
J1G1 J 2G2
(c)
(d)
Solve the Equations: Solve Eq. (d) for T1:
JG
L J G
T1  ( 0.069813 rad) 1 1  T2 2 1 1
L1
L1 J 2 G2
(3.059108 in.4 )(12,000 ksi)
 15 ft 
 T2 
 10 ft 
(10 ft)(12 in./ft)
 21.356600 kip-in.  1.5 T2
and substitute this result into Eq. (a) to compute the torque T2:
T1  T2    21.356600 kip-in.  1.5T2   T2  21.356600 kip-in.  2.5T2  80 kip-in.
 ( 0.069813 rad)
101.3566 kip-in.
 40.5426 kip-in.
2.5
The torque in member (1) is therefore:
T1  T2  80 kip-in.  40.5426 kip-in.  80 kip-in.  39.4574 kip-in.
 T2 
(c) Maximum Shear Stress: The maximum shear stress magnitude in member (1) is:
T c (39.4574 kip-in.)(2.875 in. / 2)
1  1 1 
 18.54 ksi
J1
3.059108 in.4
The maximum shear stress magnitude in member (2) is:
Tc
(40.5426 kip-in.)(2.875 in. / 2)
2  2 2 
 19.05 ksi
J2
3.059108 in.4
Ans.
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.
6.94 A stepped shaft with a major diameter of D = 1.375 in. and a minor diameter of d = 1.00 in. is
subjected to a torque of 500 lb-in. A fillet with a radius of r = 3/16 in. is used to transition from the
major diameter to the minor diameter. Determine the maximum shear stress in the shaft.
Solution
Fillet: From Fig. 6.17b:
D 1.375 in.
1.375
d
1.0 in.
r
d
0.1875 in.
1.0 in.
0.1875
K
1.22
Section Properties of Minor Diameter Section:
J
32
(1.00 in.) 4
0.098175 in.4
Maximum Shear Stress:
Tc
(500 lb-in.)(1.00 in./2)
K
(1.22)
max
J
0.098175 in.4
3,106.7 psi
3,110 psi
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.
6.95 A stepped shaft with a major diameter of D = 20 mm and a minor diameter of d = 16 mm is
subjected to a torque of 25 N-m. A full quarter-circular fillet having a radius of r = 2 mm is used to
transition from the major diameter to the minor diameter. Determine the maximum shear stress in the
shaft.
Solution
Fillet: From Fig. 6.17b:
D 20 mm
1.25
d 16 mm
r
d
2 mm
16 mm
0.125
K
1.30
Section Properties of Minor Diameter Section:
J
32
(16 mm) 4
6,434.0 mm 4
Maximum Shear Stress:
Tc
(25 N-m)(16 mm/2)(1,000 mm/m)
K
(1.30)
J
6, 434.0 mm 4
40.4 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.
6.96 A fillet with a radius of 1/2 in. is used at the junction in a stepped shaft where the diameter is
reduced from 8.00 in. to 6.00 in. Determine the maximum torque that the shaft can transmit if the
maximum shear stress in the fillet must be limited to 5 ksi.
Solution
Fillet: From Fig. 6.17b:
D 8.0 in.
r
1.33
d 6.0 in.
d
0.50 in.
6.0 in.
0.0833
K
1.41
Section Properties of Minor Diameter Section:
J
32
(6.00 in.) 4
Maximum Torque:
Tc
K
J
allow J
Tmax
Kc
127.2345 in.4
(5 ksi)(127.2345 in.4 )
(1.41)(6.00 in./2)
150.4 kip-in.
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.
6.97 A stepped shaft with a major diameter of D = 2.50 in. and a minor diameter of d = 1.25 in. is
subjected to a torque of 1,200 lb-in. If the maximum shear stress must not exceed 4,000 psi, determine
the minimum radius r that may be used for a fillet at the junction of the two shaft segments. The fillet
radius must be chosen as a multiple of 0.05 in.
Solution
Section Properties of Minor Diameter Section:
J
32
(1.25 in.) 4
0.239684 in.4
Fillet Requirement: From Fig. 6.17b:
Tc
J (4,000 psi)(0.239684 in.4 )
K
K
J
Tc (1, 200 lb-in.)(1.25 in. / 2)
D 2.5 in.
2.00
d 1.25 in.
r
0.17
d
Minimum Fillet Radius:
r
rmin
d (0.17)(1.25 in.)
d
0.213 in.
say rmin
1.28
0.25 in.
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.
6.98 A fillet with a radius of 16 mm is used at the junction in a stepped shaft where the diameter is
reduced from 200 mm to 150 mm. Determine the maximum torque that the shaft can transmit if the
maximum shear stress in the fillet must be limited to 55 MPa.
Solution
Fillet: From Fig. 6.17b:
D 200 mm
1.33
d 150 mm
r
d
16 mm
150 mm
0.107
K
1.34
Section Properties of Minor Diameter Section:
J
32
(150 mm)4
Maximum Torque:
Tc
K
J
allow J
Tmax
Kc
49,700,978 mm 4
(55 N/mm 2 )(49,700,978 mm 4 )
(1.34)(150 mm/2)
27.1995 106 N-mm
27.2 kN-m
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.
6.99 A stepped shaft with a major diameter of D = 50 mm and a minor diameter of d = 32 mm is
subjected to a torque of 210 N-m. If the maximum shear stress must not exceed 40 MPa, determine the
minimum radius r that may be used for a fillet at the junction of the two shaft segments. The fillet
radius must be chosen as a multiple of 1 mm.
Solution
Section Properties of Minor Diameter Section:
J
32
(32 mm) 4
102,943.7 mm 4
Fillet Requirement: From Fig. 6.17b:
Tc
J
(40 N/mm 2 )(102,943.7 mm 4 )
K
K
J
Tc (210 N-m)(32 mm / 2)(1,000 mm/m)
D 50 mm
1.56
d 32 mm
r
0.19
d
Minimum Fillet Radius:
r
rmin
d (0.19)(32 mm)
d
6.08 mm
say rmin
7 mm
1.23
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.
6.100 A stepped shaft has a major diameter of D = 2.00 in. and a minor diameter of d = 1.50 in. A fillet
with a 0.25-in. radius is used to transition between the two shaft segments. The maximum shear stress
in the shaft must be limited to 9,000 psi. If the shaft rotates at a constant angular speed of 800 rpm,
determine the maximum power that may be delivered by the shaft.
Solution
Fillet: From Fig. 6.17b:
D 2.00 in.
1.33
d 1.50 in.
r
d
0.25 in.
1.50 in.
0.167
K
1.24
Section Properties of Minor Diameter Section:
J
32
(1.50 in.) 4
0.497010 in.4
Maximum Torque:
Tc
K
allow
J
Tmax
allow
Kc
J
(9,000 lb/in.2 )(0.497010 in.4 )
(1.24)(1.50 in./2)
4,809.77 lb-in. 400.8143 lb-ft
Power transmission: The maximum power that can be transmitted at 800 rpm is:
800 rev 1 min 2 rad
Pmax Tmax
(400.8143 lb-ft)
33,578.54 lb-ft/s
min
60 s
1 rev
or in units of horsepower,
33,578.54 lb-ft/s
Pmax
61.1 hp
550 lb-ft/s
1 hp
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.
6.101 A stepped shaft has a major diameter of D = 100 mm and a minor diameter of d = 75 mm. A fillet
with a 10-mm radius is used to transition between the two shaft segments. The maximum shear stress in
the shaft must be limited to 60 MPa. If the shaft rotates at a constant angular speed of 500 rpm,
determine the maximum power that may be delivered by the shaft.
Solution
Fillet: From Fig. 6.17b:
D 100 mm
1.33
d
75 mm
r
d
10 mm
75 mm
0.133
K
1.29
Section Properties of Minor Diameter Section:
J
32
(75 mm) 4
3,106,311.1 mm4
Maximum Torque:
Tc
K
allow
J
Tmax
allow
Kc
J
(60 N/mm 2 )(3,106,311.1 mm4 )
(1.29)(75 mm/2)
3,852,789 N-mm
3,852.789 N-m
Power transmission: The maximum power that can be transmitted at 500 rpm is:
500 rev 1 min 2 rad
(3,852.789 N-m)
Pmax Tmax
min
60 s
1 rev
201,731.6 N-m/s
202 kW
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.
6.102 A 2-in.-diameter shaft contains a ½-in.-deep U-shaped groove that has a ¼-in. radius at the bottom
of the groove. The shaft must transmit a torque of T = 720 lb-in. Determine the maximum shear stress in
the shaft.
Solution
Groove: From Fig. 6.17a:
d 2.0 in. 2(0.5 in.) 1.0 in.
D 2.0 in.
r 0.25 in.
2.00
d 1.0 in.
d
1.0 in.
K
0.25
1.29
Section Properties at Minimum Diameter Section:
J
32
(1.00 in.) 4
0.098175 in.4
Shaft Shear Stress:
Tc
(720 lb-in.)(1.00 in./2)
K
(1.29)
max
J
0.098175 in.4
4,730.34 psi
4,730 psi
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.
6.103 A semicircular groove with a 6-mm radius is required in a 50-mm-diameter shaft. If the maximum
allowable shear stress in the shaft must be limited to 40 MPa, determine the maximum torque that can be
transmitted by the shaft.
Solution
Groove: From Fig. 6.17a:
d 50 mm 2(6 mm) 38 mm
D 50 mm
r
1.32
d 38 mm
d
6 mm
38 mm
0.158
K
1.39
Section Properties at Minimum Diameter Section:
J
32
(38 mm) 4
Maximum Torque:
Tc
K
J
allow J
Tmax
Kc
204,707.75 mm 4
(40 N/mm 2 )(204,707.75 mm 4 )
(1.39)(38 mm/2)
310,045.8 N-mm
310 N-m
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.
6.104 A 40-mm-diameter shaft contains a 10-mm-deep U-shaped groove that has a 6-mm radius at the
bottom of the groove. The maximum shear stress in the shaft must be limited to 60 MPa. If the shaft
rotates at a constant angular speed of 22 Hz, determine the maximum power that may be delivered by
the shaft.
Solution
Groove: From Fig. 6.17a:
d 40 mm 2(10 mm)
D 40 mm
2.00
d 20 mm
20 mm
r
d
6 mm
20 mm
0.30
K
1.25
Section Properties at Minimum Diameter Section:
J
32
(20 mm) 4
Maximum Torque:
Tc
K
J
allow J
Tmax
Kc
15,707.96 mm 4
(60 N/mm 2 )(15,707.96 mm 4 )
(1.25)(20 mm/2)
75.3982 N-m
Power transmission: The maximum power that can be transmitted at 22 Hz is:
22 rev 2 rad
Pmax Tmax
(75.3982 N-m)
s
1 rev
10, 422.3 N-m/s
10.42 kW
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.
6.105 A 1.25-in. -diameter shaft contains a 0.25-in.-deep U-shaped groove that has a 1/8-in. radius at the
bottom of the groove. The maximum shear stress in the shaft must be limited to 12,000 psi. If the shaft
rotates at a constant angular speed of 6 Hz, determine the maximum power that may be delivered by the
shaft.
Solution
Groove: From Fig. 6.17a:
d 1.25 in. 2(0.25 in.)
D 1.25 in.
1.67
d 0.75 in.
0.75 in.
r
d
0.125 in.
0.75 in.
0.167
K
1.39
Section Properties at Minimum Diameter Section:
J
32
(0.75 in.) 4
Maximum Torque:
Tc
K
J
allow J
Tmax
Kc
0.031063 in.4
(12,000 psi)(0.031063 in.4 )
(1.39)(0.75 in./2)
715.1219 lb-in. 59.5935 lb-ft
Power transmission: The maximum power that can be transmitted at 6 Hz is:
6 rev 2 rad
Pmax Tmax
(59.5935 lb-ft)
s
1 rev
2,246.6219 lb-ft/s
or in units of horsepower,
2, 246.6219 lb-ft/s
Pmax
4.08 hp
550 lb-ft/s
1 hp
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.
6.106 A torque of magnitude T = 1.5
kip-in. is applied to each of the bars
shown in Fig. P6.106. If the allowable
shear stress is specified as allow = 8 ksi,
determine the minimum required
dimension b for each bar.
Fig. P6.106
Solution
(a) Circular Section
Rearrange the elastic torsion formula to group terms with d on the left-hand side:
 d4
d 3 T
T

which can be simplified to


32 ( d / 2) 
16
From this equation, the unknown diameter of the solid shaft can be expressed as
16T
d3

To support a torque of T = 1.5 kip-in. without exceeding the maximum shear stress of 8 ksi, the solid
shaft must have a diameter (i.e., dimension b shown in the problem statement) of
16T
16(1.5 kip-in.)
Ans.
bmin  3
3
 0.985 in.

 (8 ksi)
(b) Square Section
From Table 6.1,
b
1

  0.208
a
The maximum shear stress in a rectangular section is given by Eq. (6.22):
T
 max  2
a b
For a square section where a = b,
T
1.5 kip-in.
b3 

 0.901442 in.3
 bmin  0.966 in.
 max (0.208)(8 ksi)
ab
aspect ratio
(c) Rectangular Section
From Table 6.1,
2b
2 
  0.246
aspect ratio
b
For a rectangular section where a = 2b,
1.5 kip-in.
T
b3 

 0.381098 in.3
2 max 2(0.246)(8 ksi)
 bmin  0.725 in.
Ans.
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.
6.107 A torque of magnitude T = 270 Nm is applied to each of the bars shown in
Fig. P6.107. If the allowable shear stress
is specified as allow = 70 MPa, determine
the minimum required dimension b for
each bar.
Fig. P6.107
Solution
(a) Circular Section
Rearrange the elastic torsion formula to group terms with d on the left-hand side:
 d4
d 3 T
T

which can be simplified to


32 ( d / 2) 
16
From this equation, the unknown diameter of the solid shaft can be expressed as
16T
d3

To support a torque of T = 270 N-m without exceeding the maximum shear stress of 70 MPa, the solid
shaft must have a diameter (i.e., dimension b shown in the problem statement) of
16T
16(270 N-m)(1,000 mm/m)
Ans.
bmin  3
3
 27.0 mm

 (70 N/mm2 )
(b) Square Section
From Table 6.1,
b
1

  0.208
a
The maximum shear stress in a rectangular section is given by Eq. (6.22):
T
 max  2
a b
For a square section where a = b,
T
(270 N-m)(1,000 mm/m)
b3 

 18,543.946 mm3
 bmin  26.5 mm
2
 max
(0.208)(70 N/mm )
ab
aspect ratio
(c) Rectangular Section
From Table 6.1,
2b
2 
  0.246
aspect ratio
b
For a rectangular section where a = 2b,
(270 N-m)(1,000 mm/m)
T
b3 

 7,839.721 mm3
2
2 max
2(0.246)(70 N/mm )
 bmin  19.87 mm
Ans.
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.
6.108 The bars shown in Fig. P6.108
have equal cross-sectional areas and they
are each subjected to a torque of T = 160
N-m. Determine:
(a) the maximum shear stress in each
bar.
(b) the rotation angle at the free end if
each bar has a length of 300 mm.
Assume G = 28 GPa.
Fig. P6.108
Solution
Theoretical Basis: The maximum shear stress in a rectangular section is given by Eq. (6.22):
T
 max  2
a b
and the angle of twist is given by Eq. (6.23)
TL
 3
 a bG
Rectangular Section 15 mm by 50 mm
From linear interpolation of the values given in Table 6.1,
50 mm
aspect ratio
 3.333 
  0.272
  0.269
15 mm
Shear stress:
T
(160 N-m)(1,000 mm/m)
 max  2 
 52.3 MPa
 a b (0.272)(15 mm) 2 (50 mm)
Angle of twist:
TL
(160 N-m)(300 mm)(1,000 mm/m)
 3 
 0.0378 rad
 a bG (0.269)(15 mm)3 (50 mm)(28,000 N/mm 2 )
Rectangular Section 25 mm by 30 mm
From Table 6.1,
30 mm
 1.2

  0.219
  0.166
aspect ratio
25 mm
Shear stress:
T
(160 N-m)(1,000 mm/m)
 max  2 
 39.0 MPa
 a b (0.219)(25 mm) 2 (30 mm)
Angle of twist:
TL
(160 N-m)(300 mm)(1,000 mm/m)
 3 
 0.0220 rad
 a bG (0.166)(25 mm)3 (30 mm)(28,000 N/mm 2 )
Ans.
Ans.
Ans.
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.
6.109 The allowable shear stress for each
bar shown in Fig. P6.109 is 75 MPa.
Determine:
(a) the largest torque T that may be
applied to each bar.
(b) the corresponding rotation angle at the
free end if each bar has a length of 300
mm. Assume G = 28 GPa.
Fig. P6.109
Solution
Theoretical Basis: The maximum shear stress in a rectangular section is given by Eq. (6.22):
T
 max  2
a b
and the angle of twist is given by Eq. (6.23)
TL
 3
 a bG
Rectangular Section 15 mm by 50 mm
From linear interpolation of the values given in Table 6.1,
50 mm
aspect ratio
 3.333 
  0.272
  0.269
15 mm
Maximum torque:
T   maxa2b  (75 N/mm2 )(0.272)(15 mm)2 (50 mm)  229,500 N-mm  230 N-m
Angle of twist:
TL
(229,500 N-mm)(300 mm)
 3 
 0.0542 rad
 a bG (0.269)(15 mm)3 (50 mm)(28,000 N/mm 2 )
Rectangular Section 25 mm by 30 mm
From Table 6.1,
30 mm
 1.2

  0.219
  0.166
aspect ratio
25 mm
Maximum torque:
T   maxa2b  (75 N/mm2 )(0.219)(25 mm)2 (30 mm)  307,979 N-mm  308 N-m
Angle of twist:
TL
(307,979 N-mm)(300 mm)
 3 
 0.0424 rad
 a bG (0.166)(25 mm)3 (30 mm)(28,000 N/mm 2 )
Ans.
Ans.
Ans.
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.
6.110 A solid circular rod having
diameter D is to be replaced by a
rectangular tube having cross-sectional
dimensions D × 2D (which are
measured to the wall centerlines of the
cross section shown in Fig. P6.110).
Determine the required minimum
thickness tmin of the tube so that the
maximum shear stress in the tube will
not exceed the maximum shear stress
in the solid bar.
Fig. P6.110
Solution
For the solid circular rod, the maximum shear stress is given by
Tc T ( D / 2) 16T
 max 


 4
 D3
J
D
32
For the rectangular tube, the area enclosed by the median line is
Am  ( D)(2 D)  2 D 2
The maximum shear stress for the thin-walled section is given by Eq. (6.25)
T
T
T
 max 


2
2 Amt 2(2 D )t 4 D 2t
Equate Eqs. (a) and (b)
16T
T

3
 D 4 D 2t
and solve for t:
D
tmin 
64
(a)
(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.
6.111 A 24 in. wide by 0.100 in. thick by 100 in. long steel sheet is to be formed into a hollow section
by bending through 360° and welding (i.e., butt-welding) the long edges together. Assume a crosssectional medial length of 24 in. (no stretching of the sheet due to bending). If the maximum shear stress
must be limited to 12 ksi, determine the maximum torque that can be carried by the hollow section if
(a) the shape of the section is a circle.
(b) the shape of the section is an equilateral triangle.
(c) the shape of the section is a square.
(d) the shape of the section is an 8 × 4 in. rectangle.
Solution
The maximum shear stress for a thin-walled section is given by Eq. (6.25)
T
 max 
2 Am t
and thus, the maximum torque that can be carried by the hollow section is
Tmax  2 max Amt
(a) Circle:
 d m  24 in.
Am 
Tmax

 d m  7.639437 in.
(7.639437 in.) 2  45.8366 in.2
4
 2 max Amt  2(12 ksi)(45.8366 in.2 )(0.100 in.)  110.0 kip-in.
(b) Equilateral triangle:
triangle sides are each 24 in./3  8 in.
1
1
Am  bh  (8 in.)(8 in.)sin 60  27.7128 in.2
2
2
Tmax  2 max Amt  2(12 ksi)(27.7128 in.2 )(0.100 in.)  66.5 kip-in.
Ans.
Ans.
(c) Square:
sides of the square are each 24 in./4  6 in.
Am  bh  (6 in.)(6 in.)  36 in.2
Tmax  2 max Amt  2(12 ksi)(36 in.2 )(0.100 in.)  86.4 kip-in.
Ans.
(d) 8 × 4 in. rectangle:
Am  bh  (8 in.)(4 in.)  32 in.2
Tmax  2 max Amt  2(12 ksi)(32 in.2 )(0.100 in.)  76.8 kip-in.
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.
6.112 A 500 mm wide by 3 mm thick by 2 m long aluminum sheet is to be formed into a hollow section
by bending through 360° and welding (i.e., butt-welding) the long edges together. Assume a crosssectional medial length of 500 mm (no stretching of the sheet due to bending). If the maximum shear
stress must be limited to 75 MPa, determine the maximum torque that can be carried by the hollow
section if
(a) the shape of the section is a circle.
(b) the shape of the section is an equilateral triangle.
(c) the shape of the section is a square.
(d) the shape of the section is a 150 × 100 mm rectangle.
Solution
The maximum shear stress for a thin-walled section is given by Eq. (6.25)
T
 max 
2 Am t
and thus, the maximum torque that can be carried by the hollow section is
Tmax  2 max Amt
(a) Circle:
 d m  500 mm
Am 

 d m  159.155 mm
(159.155 mm) 2  19,894.382 mm 2
4
 2 max Amt  2(75 N/mm 2 )(19,894.382 mm 2 )(3 mm)  8,952, 472 N-mm  8.95 kN-m
Ans.
(b) Equilateral triangle:
triangle sides are each 500 mm/3  166.667 mm
1
1
Am  bh  (166.667 mm)(166.667 mm)sin 60  12,028.131 mm 2
2
2
Tmax  2 max Amt  2(75 N/mm 2 )(12,028.131 mm 2 )(3 mm)  5, 412,659 N-mm  5.41 kN-m
Ans.
Tmax
(c) Square:
sides of the square are each 500 mm/4  125 mm
Am  bh  (125 mm)(125 mm)  15,625 mm 2
Tmax  2 max Amt  2(75 N/mm 2 )(15,625 mm 2 )(3 mm)  7,031, 250 N-mm  7.03 kN-m
Ans.
(d) 150 × 100 mm rectangle:
Am  bh  (150 mm)(100 mm)  15,000 mm 2
Tmax  2 max Amt  2(75 N/mm 2 )(15,000 mm 2 )(3 mm)  6,750,000 N-mm  6.75 kN-m
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.
6.113 A torque of T = 150 kip-in. will be applied to
the hollow, thin-walled aluminum alloy section
shown in Fig. P6.113. If the maximum shear stress
must be limited to 10 ksi, determine the minimum
thickness required for the section. (Note: The
dimensions shown are measured to the wall
centerline.)
Fig. P6.113
Solution
The maximum shear stress for a thin-walled section is given by Eq. (6.25)
T
 max 
2 Am t

For the aluminum alloy section,
Am  (6 in.)(8 in.) 
4
(6 in.)2  76.274 in.2
The minimum thickness required for the section if the maximum shear stress must be limited to 10 ksi is
thus:
T
150 kip-in.

 0.0983 in.
tmin 
Ans.
2 max Am 2(10 ksi)(76.274 in.2 )
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.
6.114 A torque of T = 2.5 kN-m will be applied to the hollow,
thin-walled aluminum alloy section shown in Fig. P6.114. If
the maximum shear stress must be limited to 50 MPa,
determine the minimum thickness required for the section.
(Note: The dimensions shown are measured to the wall
centerline.)
Fig. P6.114
Solution
The maximum shear stress for a thin-walled section is given by Eq. (6.25)
T
 max 
2 Am t
For the aluminum alloy section,
1
Am   (100 mm)2  7,853.982 mm 2
4
The minimum thickness required for the section if the maximum shear stress must be limited to 50 MPa
is thus:
T
(2.5 kN-m)(1,000 N/kN)(1,000 mm/m)
tmin 

 3.18 mm
Ans.
2 max Am
2(50 N/mm 2 )(7,853.982 mm 2 )
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.
6.115 A torque of T = 100 kip-in. will be applied to
the hollow, thin-walled aluminum alloy section
shown in Fig. P6.115. If the section has a uniform
thickness of 0.100 in., determine the magnitude of
the maximum shear stress developed in the section.
(Note: The dimensions shown are measured to the
wall centerline.)
Fig. P6.115
Solution
The maximum shear stress for a thin-walled section is given by Eq. (6.25)
T
 max 
2 Am t

For the aluminum alloy section,
Am 
8
(10 in.)2  39.2699 in.2
The maximum shear stress developed in the section is:
100 kip-in.
T
 max 

 12.73 ksi
2 Amt 2(39.2699 in.2 )(0.100 in.)
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.
6.116 A torque of T = 2.75 kN-m will be applied to
the hollow, thin-walled aluminum alloy section
shown in Fig. P6.116. If the section has a uniform
thickness of 4 mm, determine the magnitude of the
maximum shear stress developed in the section.
(Note: The dimensions shown are measured to the
wall centerline.)
Fig. P6.116
Solution
The maximum shear stress for a thin-walled section is given by Eq. (6.25)
T
 max 
2 Am t

For the aluminum alloy section,
Am  (150 mm)(50 mm) 
2
(25 mm) 2  8, 481.748 mm 2
The maximum shear stress developed in the section is:
T
(2.75 kN-m)(1,000 N/kN)(1,000 mm/m)
 max 

 40.5 MPa
2 Amt
2(8, 481.748 mm 2 )(4 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.
6.117 A cross section of the leading edge of an
airplane wing is shown in Fig. P6.117. The enclosed
area is 82 in.2. Sheet thicknesses are shown on the
diagram. For an applied torque of T = 100 kip-in.,
determine the magnitude of the maximum shear
stress developed in the section. (Note: The
dimensions shown are measured to the wall
centerline.)
Fig. P6.117
Solution
The maximum shear stress for a thin-walled section is given by Eq. (6.25)
T
 max 
2 Am t
The maximum shear stress developed in the section is:
100 kip-in.
T
 max 

 12.20 ksi
2 Amt 2(82 in.2 )(0.050 in.)
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.
6.118 A cross section of an airplane fuselage made
of aluminum alloy is shown in Fig. P6.118. For an
applied torque of T = 1,250 kip-in. and an allowable
shear stress of  = 7.5 ksi, determine the minimum
thickness of the sheet (which must be constant for
the entire periphery) required to resist the torque.
(Note: The dimensions shown are measured to the
wall centerline.)
Fig. P6.118
Solution
The maximum shear stress for a thin-walled section is given by Eq. (6.25)
T
 max 
2 Am t
For the fuselage,
Am  (30 in.)(20 in.)   (15 in.) 2  1,306.858 in.2
The minimum thickness required for the sheet if the maximum shear stress must be limited to 7.5 ksi is
thus:
T
1, 250 kip-in.

 0.0638 in.
tmin 
Ans.
2 max Am 2(7.5 ksi)(1,306.858 in.2 )
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.