Exam (elaborations) TEST BANK FOR Mechanics of Materials An Integrated

Exam (elaborations) TEST BANK FOR Mechanics of Materials An Integrated 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 P A    (200 N/mm )(863.938 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 t   2 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) F F y 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) F F y 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) F F y 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) F F y 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, 1 1 8 kips 4 kips 4 kips 0 16 kips 16 kips (C)        F F y     F FBD through rod (1) FBD through rod (2) FBD through rod (3) Next, cut a FBD through rod (2) that includes the free end A. Again, we will assume that the internal force in rod (2) is tension. Equilibrium of this FBD reveals the internal force in rod (2): 2 2 8 kips 4 kips 4 kips 15 kips 15 kips 0 14 kips 14 kips (T)             F F y 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) F F y 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, 1 1 8 kips 4 kips 4 kips 0 16 kips 16 kips (C)        F F y     F FBD through rod (1) FBD through rod (2) FBD through rod (3) Next, cut a FBD through rod (2) that includes the free end A. Again, we will assume that the internal force in rod (2) is tension. Equilibrium of this FBD reveals the internal force in rod (2): 2 2 8 kips 4 kips 4 kips 15 kips 15 kips 0 14 kips 14 kips (T)             F F y 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) F F y 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   F F x   F   (a) 2 1 sin(35.7067 ) sin(57.9946 ) 0   F F y   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   F F x   F   (a) 2 1 sin(35.7067 ) sin(57.9946 ) 0   F F y   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