Exam (elaborations) TEST BANK FOR Mechanics of Materials An Integrated Learning System 3rd Edition By Timothy A. Philpot (Instructors Solution Manual)

Exam (elaborations) TEST BANK FOR Mechanics of Materials An Integrated Learning System 3rd Edition By Timothy A. Philpot (Instructors Solution Manual) CONTENTS Philpot MoM 3rd 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 2. Strain 2.1 Displacement, Deformation, and the Concept of Strain 2.2 Normal Strain 2.3 Shear Strain 2.4 Thermal Strain 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 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 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 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 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 8. Bending 8.1 Introduction 8.2 Flexural Strains 8.3 Normal Stresses 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 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 9.9 Shear Stress and Shear Flow in Thin-Walled Members 9.10 Shear Centers of Thin-Walled Open Sections 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 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 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 Plane Stress 12.5 Generating the Stress Element 12.6 Equilibrium Method for Plane Stress Transformations 12.7 General 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 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 14. Thin-Walled Pressure Vessels 14.1 Introduction 14.2 Spherical Pressure Vessels 14.3 Cylindrical Pressure Vessels 14.4 Strains in Pressure Vessels 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 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 17. Energy Methods 17.1 Introduction 17.2 Work and Strain Energy 17.3 Elastic Strain Energy for Axial Deformation 17.4 Elastic Strain Energy for Torsional Deformation 17.5 Elastic Strain Energy for Flexural Deformation 17.6 Impact Loading 17.7 Work-Energy Method for Single Loads 17.8 Method of Virtual Work 17.9 Deflections of Trusses by the Virtual-Work Method 17.10 Deflections of Beams by the Virtual-Work Method 17.11 Castigliano’s Second Theorem 17.12 Calculating Deflections of Trusses by Castigliano’s Theorem 17.13 Calculating Deflections of Beams by Castigliano’s Theorem Appendix A: Geometric Properties of an Area A.1 Centroid of an Area A.2 Moment of Inertia for an Area A.3 Product of Inertia for an Area A.4 Principal Moments of Inertia A.5 Mohr’s Circle for Principal Moments of Inertia Appendix B: Geometric Properties of Structural Steel Shapes Appendix C: Table of Beam Slopes and Deflections Appendix D: Average Properties of Selected Materials Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. P1.1 A stainless steel tube with an outside diameter of 60 mm and a wall thickness of 5 mm is used as a compression member. If the axial normal stress in the member must be limited to 200 MPa, determine the maximum load P that the member can support. Solution The cross-sectional area of the stainless steel tube is 2 2 2 2 2 ( ) [(60 mm) (50 mm) ] 863.938 mm 4 4 A D d        The normal stress in the tube can be expressed as P A   The maximum normal stress in the tube must be limited to 200 MPa. Using 200 MPa as the allowable normal stress, rearrange this expression to solve for the maximum load P 2 2 max allow (200 N/mm )(863.938 P  A mm ) 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. P1.2 A 2024-T4 aluminum tube with an outside diameter of 2.50 in. will be used to support a 27-kip load. If the axial normal stress in the member must be limited to 18 ksi, determine the wall thickness required for the tube. Solution From the definition of normal stress, solve for the minimum area required to support a 27-kip load without exceeding a stress of 18 ksi 2 min 27 kips 1.500 in. 18 ksi P P A A        The cross-sectional area of the aluminum tube is given by 2 2 ( ) 4 A D d    Set this expression equal to the minimum area and solve for the maximum inside diameter d 2 2 2 2 2 2 2 2 2 max [(2.50 in.) ] 1.500 in. 4 4 (2.50 in.) (1.500 in. ) 4 (2.50 in.) (1.500 in. ) 2.08330 in. d d d d            The outside diameter D, the inside diameter d, and the wall thickness t are related by D  d  2t Therefore, the minimum wall thickness required for the aluminum tube is min 2.50 in. 2.08330 in. 0.20835 in. 0.208 in. 2 2 D d t       Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. P1.3 Two solid cylindrical rods (1) and (2) are joined together at flange B and loaded, as shown in Figure P1.3/4. If the normal stress in each rod must be limited to 40 ksi, determine the minimum diameter required for each rod. FIGURE P1.3/4 Solution Cut a FBD through rod (1). The FBD should include the free end of the rod at A. As a matter of course, we will assume that the internal force in rod (1) is tension (even though it obviously will be in compression). From equilibrium, 1 1 15 kips 0 15 kips 15 kips (C) yF F F          Next, cut a FBD through rod (2) that includes the free end of the rod at A. Again, we will assume that the internal force in rod (2) is tension. Equilibrium of this FBD reveals the internal force in rod (2): 2 2 30 kips 30 kips 15 kips 0 75 kips 75 kips (C) yF F F            Notice that rods (1) and (2) are in compression. In this situation, we are concerned only with the stress magnitude; therefore, we will use the force magnitudes to determine the minimum required cross-sectional areas. If the normal stress in rod (1) must be limited to 40 ksi, then the minimum cross-sectional area that can be used for rod (1) is 1 2 1,min 15 kips 0.375 in. 40 ksi F A     The minimum rod diameter is therefore 2 2 1,min 1 1 0.375 in. 0.69099 i 0.691 4 A d d n. in.       Ans. Similarly, the normal stress in rod (2) must be limited to 40 ksi, which requires a minimum area of 2 2 2,min 75 kips 1.875 in. 40 ksi F A     The minimum diameter for rod (2) is therefore 2 2 2,min 2 2 1.875 in. 1. in. 1.545 in. 4 A d d       Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. P1.4 Two solid cylindrical rods (1) and (2) are joined together at flange B and loaded, as shown in Figure P1.3/4. The diameter of rod (1) is 1.75 in. and the diameter of rod (2) is 2.50 in. Determine the normal stresses in rods (1) and (2). FIGURE P1.3/4 Solution Cut a FBD through rod (1). The FBD should include the free end of the rod at A. We will assume that the internal force in rod (1) is tension (even though it obviously will be in compression). From equilibrium, 1 1 15 kips 0 15 kips 15 kips (C) yF F F          Next, cut a FBD through rod (2) that includes the free end of the rod at A. Again, we will assume that the internal force in rod (2) is tension. Equilibrium of this FBD reveals the internal force in rod (2): 2 2 30 kips 30 kips 15 kips 0 75 kips 75 kips (C) yF F F            From the given diameter of rod (1), the cross-sectional area of rod (1) is 2 2 1 (1.75 in.) 2.4053 in. 4 A    and thus, the normal stress in rod (1) is 1 1 2 1 15 kips 6.23627 ksi 2.4053 in 6.24 ksi ) . (C F A        Ans. From the given diameter of rod (2), the cross-sectional area of rod (2) is 2 2 2 (2.50 in.) 4.9087 in. 4 A    Accordingly, the normal stress in rod (2) is 2 2 2 2 75 kips 15.2789 ksi 2.4053 in. 15.28 ksi (C) F A        Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. P1.5 Axial loads are applied with rigid bearing plates to the solid cylindrical rods shown in Figure P1.5/6. The diameter of aluminum rod (1) is 2.00 in., the diameter of brass rod (2) is 1.50 in., and the diameter of steel rod (3) is 3.00 in. Determine the axial normal stress in each of the three rods. FIGURE P1.5/6 Solution Cut a FBD through rod (1). The FBD should include the free end A. We will assume that the internal force in rod (1) is tension (even though it obviously will be in compression). From equilibrium, 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): 2 2 8 kips 4 kips 4 kips 15 kips 15 kips 0 14 kips 14 kips (T) y F  F       F   Similarly, cut a FBD through rod (3) that includes the free end A. From this FBD, the internal force in rod (3) is: 3 3 8 kips 4 kips 4 kips 15 kips 15 kips 20 kips 20 kips 0 26 kips 26 kips (C) yF F F                Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. From the given diameter of rod (1), the cross-sectional area of rod (1) is 2 2 1 (2.00 in.) 3.1416 in. 4 A    and thus, the normal stress in aluminum rod (1) is 1 1 2 1 16 kips 5.0930 ksi 3.1416 in 5.09 ksi (C) . F A        Ans. From the given diameter of rod (2), the cross-sectional area of rod (2) is 2 2 2 (1.50 in.) 1.7671 in. 4 A    Accordingly, the normal stress in brass rod (2) is 2 2 2 2 14 kips 7.9224 ksi 1.7671 in. 7.92 ksi (T) F A      Ans. Finally, the cross-sectional area of rod (3) is 2 2 3 (3.00 in.) 7.0686 in. 4 A    and the normal stress in the steel rod is 3 3 2 3 26 kips 3.6782 ksi 7.0686 in 3.68 ksi (C) . F A        Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. P1.6 Axial loads are applied with rigid bearing plates to the solid cylindrical rods shown in Figure P1.5/6. The normal stress in aluminum rod (1) must be limited to 18 ksi, the normal stress in brass rod (2) must be limited to 25 ksi, and the normal stress in steel rod (3) must be limited to 15 ksi. Determine the minimum diameter required for each of the three rods. FIGURE P1.5/6 Solution The internal forces in the three rods must be determined. Begin with a FBD cut through rod (1) that includes the free end A. We will assume that the internal force in rod (1) is tension (even though it obviously will be in compression). From equilibrium, 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): 2 2 8 kips 4 kips 4 kips 15 kips 15 kips 0 14 kips 14 kips (T) y F  F       F   Similarly, cut a FBD through rod (3) that includes the free end A. From this FBD, the internal force in rod (3) is: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3 3 8 kips 4 kips 4 kips 15 kips 15 kips 20 kips 20 kips 0 26 kips 26 kips (C) yF F F                Notice that two of the three rods are in compression. In these situations, we are concerned only with the stress magnitude; therefore, we will use the force magnitudes to determine the minimum required 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 1 2 1,min 1 16 kips 0.8889 in. 18 ksi F A     The minimum rod diameter is therefore 2 2 1,min 1 1 0.8889 in. 1.0638 in. 1.064 in. 4 A d d       Ans. The normal stress in brass rod (2) must be limited to 25 ksi, which requires a minimum area of 2 2 2,min 2 14 kips 0.5600 in. 25 ksi F A     which requires a minimum diameter for rod (2) of 2 2 2,min 2 2 0.5600 in. 0.8444 in. 0.844 in. 4 A d d       Ans. The normal stress in steel rod (3) must be limited to 15 ksi. The minimum cross-sectional area required for this rod is: 3 2 3,min 3 26 kips 1.7333 in. 15 ksi F A     which requires a minimum diameter for rod (3) of 2 2 3,min 3 3 1.7333 in. 1.4856 in. 1.486 in. 4 A d d       Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. P1.7 Two solid cylindrical rods support a load of P = 50 kN, as shown in Figure P1.7/8. If the normal stress in each rod must be limited to 130 MPa, determine the minimum diameter required for each rod. FIGURE P1.7/8 Solution Consider a FBD of joint B. Determine the angle  between rod (1) and the horizontal axis: 4.0 m tan 1.600 57.9946 2.5 m       and the angle  between rod (2) and the horizontal axis: 2.3 m tan 0.7188 35.7067 3.2 m       Write equilibrium equations for the sum of forces in the horizontal and vertical directions. Note: Rods (1) and (2) are two-force members. 2 1 cos(35.7067 ) cos(57.9946 ) 0 x F  F   F   (a) 2 1 sin(35.7067 ) sin(57.9946 ) 0 y F  F   F   P  (b) Unknown forces F1 and F2 can be found from the simultaneous solution of Eqs. (a) and (b). Using the substitution method, Eq. (b) can be solved for F2 in terms of F1: 2 1 cos(57.9946 ) cos(35.7067 ) F F    (c) Substituting Eq. (c) into Eq. (b) gives   1 1 1 1 cos(57.9946 ) sin(35.7067 ) sin(57.9946 ) cos(35.6553 ) cos(57.9946 ) tan(35.7067 ) sin(57.9946 ) cos(57.9946 ) tan(35.7067 ) sin(57.9946 ) 1.2289 F F P F P P P F                   For the given load of P = 50 kN, the internal force in rod (1) is therefore: 1 50 kN 40.6856 kN 1.2289 F   Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Backsubstituting this result into Eq. (c) gives force F2: 2 1 cos(57.9946 ) cos(57.9946 ) (40.6856 kN) 26.5553 kN cos(35.7067 ) cos(35.7067 ) F F        The normal stress in rod (1) must be limited to 130 MPa; therefore, the minimum cross-sectional area required for rod (1) is 1 2 1,min 2 1 (40.6856 kN)(1,000 N/kN) 312.9664 mm 130 N/mm F A     The minimum rod diameter is therefore 2 2 1,min 1 1 312.9664 mm 19.9620 19. 4 A d d mm 96 mm       Ans. The minimum area required for rod (2) is 2 2 2,min 2 2 (26.5553 kN)(1,000 N/kN) 204.2718 mm 130 N/mm F A     which requires a minimum diameter for rod (2) of 2 2 2,min 2 2 204.2718 mm 16.1272 16. 4 A d d mm 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. P1.8 Two solid cylindrical rods support a load of P = 27 kN, as shown in Figure P1.7/8. Rod (1) has a diameter of 16 mm and the diameter of rod (2) is 12 mm. Determine the axial normal stress in each rod. FIGURE P1.7/8 Solution Consider a FBD of joint B. Determine the angle  between rod (1) and the horizontal axis: 4.0 m tan 1.600 57.9946 2.5 m       and the angle  between rod (2) and the horizontal axis: 2.3 m tan 0.7188 35.7067 3.2 m       Write equilibrium equations for the sum of forces in the horizontal and vertical directions. Note: Rods (1) and (2) are two-force members. 2 1 cos(35.7067 ) cos(57.9946 ) 0 x F  F   F   (a) 2 1 sin(35.7067 ) sin(57.9946 ) 0 y F  F   F   P  (b) Unknown forces F1 and F2 can be found from the simultaneous solution of Eqs. (a) and (b). Using the substitution method, Eq. (b) can be solved for F2 in terms of F1: 2 1 cos(57.9946 ) cos(35.7067 ) F F    (c) Substituting Eq. (c) into Eq. (b) gives   1 1 1 1 cos(57.9946 ) sin(35.7067 ) sin(57.9946 ) cos(35.6553 ) cos(57.9946 ) tan(35.7067 ) sin(57.9946 ) cos(57.9946 ) tan(35.7067 ) sin(57.9946 ) 1.2289 F F P F P P P F                   For the given load of P = 27 kN, the internal force in rod (1) is therefore: 1 27 kN 21.9702 kN 1.2289 F   Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Backsubstituting this result into Eq. (c) gives force F2: 2 1 cos(57.9946 ) cos(57.9946 ) (21.9702 kN) 14.3399 kN cos(35.7067 ) cos(35.7067 ) F F        The diameter of rod (1) is 16 mm; therefore, its cross-sectional area is: 2 2 1 (16 mm) 201.0619 mm 4 A    and the normal stress in rod (1) is: 1 2 1 2 1 (21.9702 kN)(1,000 N/kN) 109.2710 N/mm 201.0 109.3 MPa (T) 619 mm F A      Ans. The diameter of rod (2) is 12 mm; therefore, its cross-sectional area is: 2 2 2 (12 mm) 113.0973 mm 4 A    and the normal stress in rod (2) is: 2 2 2 2 2 (14.3399 kN)(1,000 N/kN) 126.7924 N/mm 113.0 126.8 MPa (T) 973 mm F A      Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. P1.9 A simple pin-connected truss is loaded and supported as shown in Figure P1.9. All members of the truss are aluminum pipes that have an outside diameter of 4.00 in. and a wall thickness of 0.226 in. Determine the normal stress in each truss member. FIGURE P1.9 Solution Overall equilibrium: Begin the solution by determining the external reaction forces acting on the truss at supports A and B. Write equilibrium equations that include all external forces. Note that only the external forces (i.e., loads and reaction forces) are considered at this time. The internal forces acting in the truss members will be considered after the external reactions have been computed. The free-body diagram (FBD) of the entire truss is shown. The following equilibrium equations can be written for this structure:  2 kips 2 ki 0 ps x x x F A A        (6 ft) (5 kips)(14 ft) (2 kips)(7 ft) 14 kips 0 y A y B B M       5 kips 0 9 kips y y y y F A B A        Method of joints: Before beginning the process of determining the internal forces in the axial members, the geometry of the truss will be used to determine the magnitude of the inclination angles of members AC and BC. Use the definition of the tangent function to determine AC and BC: 7 ft tan 0.50 26.565 14 ft 7 ft tan 0.875 41.186 8 ft AC AC BC BC               Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Joint A: Begin the solution process by considering a FBD of joint A. Consider only those forces acting directly on joint A. In this instance, two axial members, AB and AC, are connected at joint A. Additionally, two reaction forces, Ax and Ay, act at joint A. Tension forces will be assumed in each truss member. cos(Fx  FAC 26.565)  FAB  Ax  0 (a) sin(26.565 ) 0 y AC y F  F   A  (b) Solve Eq. (b) for FAC: 9 kips sin(26.565 ) sin(26.5 20.125 kip 65 ) s y AC A F         and then compute FAB using Eq. (a): cos(26.565 ) (20.125 kips)cos(26.565 ) ( 2 kips) 16.000 kips AB AC x F  F   A        Joint B: Next, consider a FBD of joint B. In this instance, the equilibrium equations associated with joint B seem easier to solve than those that would pertain to joint C. As before, tension forces will be assumed in each truss member. cos(41.186 ) 0 x AB BC F  F  F   (c) sin(41.186 ) 0 y BC y F  F   B  (d) Solve Eq. (d) for FBC: 14 kips sin(41.186 ) sin(41.18 21.260 kip 6 s ) y BC B F         Eq. (c) can be used as a check on our calculations: cos(41.186 ) ( 16.000 kips) ( 21.260 kips)cos(41.186 ) 0 x AB BC F  F  F         Checks! Section properties: For each of the three truss members: 2 2 2 4.00 in. 2(0.226 in.) 3.548 in. (4.00 in.) (3.548 in.) 2.67954 in. 4 d A          Normal stress in each truss member: 2 16.000 kips 5.971 ksi 2.67954 5.97 ksi (C) in. AB AB AB F A        Ans. 2 20.125 kips 7.510 ksi 2.67954 7.51 ksi (T) in. AC AC AC F A      Ans. 2 21.260 kips 7.934 ksi 2.67954 7.93 ksi (C) in. BC BC BC F A        Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. P1.10 A simple pin-connected truss is loaded and supported as shown in Figure P1.10. All members of the truss are aluminum pipes that have an outside diameter of 60 mm and a wall thickness of 4 mm. Determine the normal stress in each truss member. FIGURE P1.10 Solution Overall equilibrium: Begin the solution by determining the external reaction forces acting on the truss at supports A and B. Write equilibrium equations that include all external forces. Note that only the external forces (i.e., loads and reaction forces) are considered at this time. The internal forces acting in the truss members will be considered after the external reactions have been computed. The freebody diagram (FBD) of the entire truss is shown. The following equilibrium equations can be written for this structure:  12 k 12 N 0 kN x x x F A A        (1 m) (15 kN)(4.3 m) 0 64.5 kN y A y B B M      15 kN 49.5 kN 0 y y y y F A A B         Method of joints: Before beginning the process of determining the internal forces in the axial members, the geometry of the truss will be used to determine the magnitude of the inclination angles of members AB and BC. Use the definition of the tangent function to determine AB and BC: 1.5 m tan 1.50 56.310 1.0 m 1.5 m tan 0. 24.444 3.3 m AB AB BC BC               Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Joint A: Begin the solution process by considering a FBD of joint A. Consider only those forces acting directly on joint A. In this instance, two axial members, AB and AC, are connected at joint A. Additionally, two reaction forces, Ax and Ay, act at joint A. Tension forces will be assumed in each truss member. cos(Fx  FAC  FAB 56.310)  Ax  0 (a) sin(56.310 ) 0 y y AB F  A  F   (b) Solve Eq. (b) for FAB: 49.5 kN sin(56.310 ) sin(56.310 ) 59.492 kN y AB A F       and then compute FAC using Eq. (a): cos(56.310 ) ( 59.492 kN)cos(56.310 ) ( 12 kN) 45.000 kN AC AB x F  F   A        Joint C: Next, consider a FBD of joint C. In this instance, the equilibrium equations associated with joint C seem easier to solve than those that would pertain to joint B. As before, tension forces will be assumed in each truss member. cos(24.444 ) 12 kN 0 x AC BC F  F  F    (c) sin(24.444 ) 15 kN 0 y BC F  F    (d) Solve Eq. (d) for FBC: 15 kN sin(24.444 ) 36.249 kN BC F     Eq. (c) can be used as a check on our calculations: cos(24.444 ) 12 kN 0 (45.000 kN) ( 36.249 kN)cos(24.444 ) 12 kN 0 x AC BC F  F  F           Checks! Section properties: For each of the three truss members: 2 2 2 60 mm 2(4 mm) 52 mm (60 mm) (52 mm) 703.7168 mm 4 d A          Normal stress in each truss member: 2 ( 59.492 kN)(1,000 N/kN) 84.539 MPa 70 84.5 MPa (C) 3.7168 mm AB AB AB F A        Ans. 2 (45.000 kN)(1,000 N/kN) 63.946 MPa 70 63.9 MPa 3.7168 ) mm (T AC AC AC F A      Ans. 2 ( 36.249 kN)(1,000 N/kN) 51.511 MPa 70 51.5 MPa (C) 3.7168 mm BC BC BC F A        Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. P1.11 A simple pin-connected truss is loaded and supported as shown in Figure P1.11. All members of the truss are aluminum pipes that have an outside diameter of 42 mm and a wall thickness of 3.5 mm. Determine the normal stress in each truss member. FIGURE P1.11 Solution Overall equilibrium: Begin the solution by determining the external reaction forces acting on the truss at supports A and B. Write equilibrium equations that include all external forces. Note that only the external forces (i.e., loads and reaction forces) are considered at this time. The internal forces acting in the truss members will be considered after the external reactions have been computed. The free-body diagram (FBD) of the entire truss is shown. The following equilibrium equations can be written for this structure: 30 kN 30 kN 0 y y y F A A       (30 kN)(4.5 m) (15 kN)(1.6 m) (5.6 m) 19.821 kN 0 x A x B M   B      15 kN 0 15 kN 15 kN ( 19.821 kN) 34.821 kN x x x x x x F A B A B A             Method of joints: Before beginning the process of determining the internal forces in the axial members, the geometry of the truss will be used to determine the magnitude of the inclination angles of members AC and BC. Use the definition of the tangent function to determine AC and BC: 1.6 m tan 0. 19.573 4.5 m 4 m tan 0. 41.634 4.5 m AC AC BC BC               Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Joint A: Begin the solution process by considering a FBD of joint A. Consider only those forces acting directly on joint A. In this instance, two axial members, AB and AC, are connected at joint A. Additionally, two reaction forces, Ax and Ay, act at joint A. Tension forces will be assumed in each truss member. Fx  Ax  FAC cos(19.573)  0 (a) sin(19.573 ) 0 y y AC AB F  A  F   F  (b) Solve Eq. (a) for FAC: 34.821 kN cos(19.573 ) cos(19.573 ) 36.957 kN x AC A F      and then compute FAB using Eq. (b): sin(19.573 ) (30.000 kN) (36.957 kN)sin(19.573 ) 17.619 kN AB y AC F  A  F      Joint B: Next, consider a FBD of joint B. In this instance, the equilibrium equations associated with joint B seem easier to solve than those that would pertain to joint C. As before, tension forces will be assumed in each truss member. cos(41.634 ) 0 x x BC F  B  F   (c) sin(41.634 ) 0 y BC AB F  F   F  (d) Solve Eq. (c) for FBC: ( 19.821 kN) cos(41.634 ) cos(41.634 ) 26.520 kN x BC B F        Eq. (d) can be used as a check on our calculations: sin(41.634 ) ( 26.520 kN)sin(41.634 ) (17.619 kN) 0 y BC AB F  F   F      Checks! Section properties: For each of the three truss members: 2 2 2 42 mm 2(3.5 mm) 35 mm (42 mm) (35 mm) 423.3296 mm 4 d A          Normal stress in each truss member: 2 (17.619 kN)(1,000 N/kN) 41.620 MPa 42 41.6 MPa 3.3296 ) mm (T AB AB AB F A      Ans. 2 (36.957 kN)(1,000 N/kN) 87.301 MPa 42 87.3 MPa 3.3296 ) mm (T AC AC AC F A      Ans. 2 ( 26.520 kN)(1,000 N/kN) 62.647 MPa 42 62.6 MPa (C) 3.3296 mm BC BC BC F A        Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. P1.12 The rigid beam BC shown in Figure P1.12 is supported by rods (1) and (2) that have cross-sectional areas of 175 mm2 and 300 mm2, respectively. For a uniformly distributed load of w = 15 kN/m, determine the normal stress in each rod. Assume L = 3 m and a = 1.8 m. FIGURE P1.12 Solution Equilibrium: Calculate the internal forces in members (1) and (2). 1 2 1 2 1.8 m (3 m) (15 kN/m)(1.8 m) 0 2 1.8 m (3 m) (15 kN/m)(1.8 m) 3 m 0 8.100 kN 18.900 2 kN C B M F M F F F                           Stresses: 1 2 1 2 1 (8.100 kN)(1,000 N/kN) 46.286 N/mm 175 m 46.3 MPa m F A      Ans. 2 2 2 2 2 (18.900 kN)(1,000 N/kN) 63.000 N/mm 300 m 63.0 MPa m F A      Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. P1.13 Bar (1) in Figure P1.15 has a crosssectional area of 0.75 in.2. If the stress in bar (1) must be limited to 30 ksi, determine the maximum load P that may be supported by the structure. FIGURE P1.13 Solution Given that the cross-sectional area of bar (1) is 0.75 in.2 and its normal stress must be limited to 30 ksi, the maximum force that may be carried by bar (1) is 2 1,max 1 1 (30 ksi)(F  A  0.75 in. )  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: 1 1 (6 ft) (10 ft) 0 6 ft 10 ft A M F P P F       Substitute the maximum force F1,max = 22.5 kips into this relationship to obtain the maximum load that may be applied to the structure: 1 6 ft 6 ft (22.5 kips) 10 ft 10 ft P  F   13.50 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. P1.14 The rectangular bar shown in Figure P1.14 is subjected to a uniformly distributed axial loading of w = 13 kN/m and a concentrated force of P = 9 kN at B. Determine the magnitude of the maximum normal stress in the bar and its location x. Assume a = 0.5 m, b = 0.7 m, c = 15 mm, and d = 40 mm. FIGURE P1.14 Solution Equilibrium: Draw a FBD for the interval between A and B where 0  x  a . Write the following equilibrium equation: (13 kN/m)(1.2 m ) (9 kN) 0 (13 kN/m)(1.2 m ) (9 kN) x F x F F x            The largest force in this interval occurs at x = 0 where F = 6.6 kN. In the interval between B and C where a  x  a  b , and write the following equilibrium equation: (13 kN/m)(1.2 m ) 0 (13 kN/m)(1.2 m ) x F x F F x          The largest force in this interval occurs at x = a where F = 9.1 kN. Maximum Normal Stress: max (9.1 kN)(1,000 N/kN) (15 mm 15.17 MPa )(40 mm)    at x  0.5 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. P1.15 The solid 1.25-in.-diameter rod shown in Figure P1.15 is subjected to a uniform axial distributed loading along its length of w = 750 lb/ft. Two concentrated loads also act on the rod: P = 2,000 lb and Q = 1,000 lb. Assume a = 16 in. and b = 32 in. Determine the normal stress in the rod at the following locations: (a) x = 10 in. (b) x = 30 in. FIGURE P1.15 Solution (a) x = 10 in. Equilibrium: Draw a FBD for the interval between A and B where 0  x  a , and write the following equilibrium equation: (750 lb/ft)(1 ft/12 in.)(48 in. ) (2,000 lb) (1,000 lb) 0 (62.5 lb/in.)(48 in. ) 3,000 lb xF x F F x             At x = 10 in., F = 5,375 lb. Stress: The normal stress at this location can be calculated as follows. 2 2 2 (1.25 in.) 1. in. 4 5,375 lb 4,379.944 psi 4,380 p 1 si . in. A        Ans. (b) x = 30 in. Equilibrium: Draw a FBD for the interval between B and C where a  x  a  b , and write the following equilibrium equation: (750 lb/ft)(1 ft/12 in.)(48 in. ) (1,000 lb) 0 (62.5 lb/in.)(48 in. ) 1,000 lb xF x F F x            At x = 30 in., F = 2,125 lb. Stress: The normal stress at this location can be calculated as follows. 2 1,730 ps 2,125 lb 1,731.606 psi 1. i i n.     Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. P1.16 Two 6 in. wide wooden boards are to be joined by splice plates that will be fully glued on the contact surfaces. The glue to be used can safely provide a shear strength of 120 psi. Determine the smallest allowable length L that can be used for the splice plates for an applied load of P = 10,000 lb. Note that a gap of 0.5 in. is required between boards (1) and (2). FIGURE P1.16 Solution Consider a FBD of board (2). The glue on the splice plates provides resistance to the 10,000 lb applied load on both the top and bottom surfaces of board (2). Denoting the shear resistance on a glue surface as V, equilibrium in the horizontal direction requires 0 10,000 lb 5,000 lb 2 x F P V V V         In other words, each glue surface must be large enough so that 5,000 lb of shear resistance can be provided to board (2). Since the glue has a shear strength of 120 psi, the area of each glue surface on board (2) must be at least 2 min 5,000 lb 41.6667 in. 120 psi A   The boards are 6-in. wide; therefore, glue must be spread along board (2) for a length of at least 2 glue joint 41.6667 in. 6.9444 in. 6 in. L   Although we’ve discussed only board (2), the same rationale applies to board (1). For both boards (1) and (2), the glue must be applied along a length of at least 6.9444 in. on both the top and bottom of the boards in order to resist the 10,000 lb applied load. The glue applied to boards (1) and (2) must be matched by glue applied to the splice plates. Therefore, the splice plates must be at least 6.9444 in. + 6.9444 in. = 13.8889 in. long. However, we are told that a 0.5-in. gap is required between boards (1) and (2); therefore, the splice plates must be 0.5-in. longer. Altogether, the length of the splice plates must be at least min 6.9444 in. 6.9444 L   in. 0.5 in.  14.39 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. P1.17 For the clevis connection shown in Figure P1.17, determine the maximum applied load P that can be supported by the 10-mm-diameter pin if the average shear stress in the pin must not exceed 95 MPa. FIGURE P1.17 Solution Consider a FBD of the bar that is connected by the clevis, including a portion of the pin. If the shear force acting on each exposed surface of the pin is denoted by V, then the shear force on each pin surface is related to the load P by: Fx  P V V  0 P  2V The area of the pin surface exposed by the FBD is simply the cross-sectional area of the pin: 2 2 2 pin pin (10 mm) 78. mm 4 4 A d      If the average shear stress in the pin must be limited to 95 MPa, the maximum shear force V on a single cross-sectional surface must be limited to 2 2 bolt V  A  (95 N/mm )(78. mm )  7,461.283 N Therefore, the maximum load P that may be applied to the connection is P  2V  2(7,461.283 N)  14,922.565 N  14.92 kN Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. P1.18 For the connection shown in Figure P1.18, determine the average shear stress produced in the 3/8- in. diameter bolts if the applied load is P = 2,500 lb. FIGURE P1.18 Solution There are four bolts, and it is assumed that each bolt supports an equal portion of the external load P. Therefore, the shear force carried by each bolt is 2,500 lb 625 lb 4 bolts V   The bolts in this connection act in single shear. The cross-sectional area of a single bolt is 2 2 2 2 bolt bolt (3 / 8 in.) (0.375 in.) 0. in. 4 4 4 A d        Therefore, the average shear stress in each bolt is 2 bolt 625 lb 5,658.8427 psi 0. in. 5,660 psi V A      Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. P1.19 The five-bolt connection shown in Figure P1.19 must support an applied load of P = 265 kN. If the average shear stress in the bolts must be limited to 120 MPa, determine the minimum bolt diameter that may be used for this connection. FIGURE P1.19 Solution There are five bolts, and it is assumed that each bolt supports an equal portion of the external load P. Therefore, the shear force carried by each bolt is 265 kN 53 kN 53,000 N 5 bolts V    Since the average shear stress must be limited to 120 MPa, each bolt must provide a shear area of at least: 2 2 53,000 N 441.6667 mm 120 N/mm V A   Each bolt in this connection acts in double shear; therefore, two cross-sectional bolt surfaces are available to transmit shear stress in each bolt. 2 2 bolt 441.6667 mm 220.8333 mm per surface 2 surfaces per bolt 2 surfaces V A A    The minimum bolt diameter must be 2 2 bolt 220.8333 mm bolt 16.7682 mm 16.77 mm 4 d d  