,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
Pmax allow A (200 N/mm2 )(863.938 mm2 ) 172,788 N 172.8 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.
,1.2 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
A 18 ksi
The cross-sectional area of the aluminum tube is given by
A ( D2 d 2 )
4
Set this expression equal to the minimum area and solve for the maximum inside diameter d
[(2.50 in.)2 d 2 ] 1.500 in.2
4
4
(2.50 in.)2 d 2 (1.500 in.2 )
4
(2.50 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. Ans.
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.
,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) Ans.
A1 452.3893 mm 2
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)
2 2 144.3582 MPa 144.4 MPa (C) Ans.
A2 1,385.4424 mm 2
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
,1.4 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
120 N/mm 2
The minimum rod diameter is therefore
A1,min d12 666.6667 mm 2 d1 29.1346 mm 29.1 mm Ans.
4
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
120 N/mm 2
The minimum diameter for rod (2) is therefore
A2,min d 22 1,666.6667 mm2 d 2 46.0659 mm 46.1 mm Ans.
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.
,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 cross-
sectional area that can be used for rod (1) is
F 15 kips
A1,min 1 0.375 in.2
40 ksi
The minimum rod diameter is therefore
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
A2,min 2 1.875 in.2
40 ksi
The minimum diameter for rod (2) is therefore
A2,min d 22 1.875 in.2 d 2 1.545097 in. 1.545 in. Ans.
4
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
, 1.6 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
1 1 6.23627 ksi 6.24 ksi (C) Ans.
A1 2.4053 in.2
From the given diameter of rod (2), the cross-sectional area of rod (2) is
A2 (2.50 in.) 2 4.9087 in.2
4
Accordingly, the normal stress in rod (2) is
F 75 kips
2 2 15.2789 ksi 15.28 ksi (C) Ans.
A2 2.4053 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.
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
Pmax allow A (200 N/mm2 )(863.938 mm2 ) 172,788 N 172.8 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.
,1.2 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
A 18 ksi
The cross-sectional area of the aluminum tube is given by
A ( D2 d 2 )
4
Set this expression equal to the minimum area and solve for the maximum inside diameter d
[(2.50 in.)2 d 2 ] 1.500 in.2
4
4
(2.50 in.)2 d 2 (1.500 in.2 )
4
(2.50 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. Ans.
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.
,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) Ans.
A1 452.3893 mm 2
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)
2 2 144.3582 MPa 144.4 MPa (C) Ans.
A2 1,385.4424 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.
,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
120 N/mm 2
The minimum rod diameter is therefore
A1,min d12 666.6667 mm 2 d1 29.1346 mm 29.1 mm Ans.
4
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
120 N/mm 2
The minimum diameter for rod (2) is therefore
A2,min d 22 1,666.6667 mm2 d 2 46.0659 mm 46.1 mm Ans.
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.
,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 cross-
sectional area that can be used for rod (1) is
F 15 kips
A1,min 1 0.375 in.2
40 ksi
The minimum rod diameter is therefore
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
A2,min 2 1.875 in.2
40 ksi
The minimum diameter for rod (2) is therefore
A2,min d 22 1.875 in.2 d 2 1.545097 in. 1.545 in. Ans.
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.
, 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
1 1 6.23627 ksi 6.24 ksi (C) Ans.
A1 2.4053 in.2
From the given diameter of rod (2), the cross-sectional area of rod (2) is
A2 (2.50 in.) 2 4.9087 in.2
4
Accordingly, the normal stress in rod (2) is
F 75 kips
2 2 15.2789 ksi 15.28 ksi (C) Ans.
A2 2.4053 in.2
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