Solution manual for mechanics of materials 3rd edition by philpot

Solution Manual for Mechanics of Materials 3rd Edition by Philpot Full file P1.1at A https://TestbankDirect.eu/ stainless steel tube with an outside diameter of 60 mm and a wall thickness of 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   A  ( D  d )  [(60 mm)  (50 mm) ]  863.938 mm 4 The normal stress in the tube can be expressed as P  A The maximum normal stress in the tube must be limited to 200 MPa Using 200 MPa as the allowable normal stress, rearrange this expression to solve for the maximum load P Ans Pmax   allow A  (200 N/mm2 )(863.938 mm2 )  172,788 N  172.8 kN Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Full file at https://TestbankDirect.eu/ Solution Manual for Mechanics of Materials 3rd Edition by Philpot Full file P1.2atA https://TestbankDirect.eu/ 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 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 ) Set this expression equal to the minimum area and solve for the maximum inside diameter d  [(2.50 in.)  d ]  1.500 in.2 (2.50 in.)  d  (2.50 in.)    (1.500 in.2 ) (1.500 in.2 )  d  d max  2.08330 in The outside diameter D, the inside diameter d, and the wall thickness t are related by D  d  2t Therefore, the minimum wall thickness required for the aluminum tube is D  d 2.50 in  2.08330 in tmin    0.20835 in  0.208 in 2 Ans Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Full file at https://TestbankDirect.eu/ Solution Manual for Mechanics of Materials 3rd Edition by Philpot Full file https://TestbankDirect.eu/ P1.3atTwo 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, Fy   F1  15 kips   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   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    0.375 in.2  40 ksi The minimum rod diameter is therefore  Ans A1,min  d12  0.375 in.2  d1  0.69099 in  0.691 in Similarly, the normal stress in rod (2) must be limited to 40 ksi, which requires a minimum area of F 75 kips A2,min    1.875 in.2  40 ksi The minimum diameter for rod (2) is therefore  Ans A2,min  d 22  1.875 in.2  d  1.545097 in  1.545 in Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Full file at https://TestbankDirect.eu/ Solution Manual for Mechanics of Materials 3rd Edition by Philpot Full file https://TestbankDirect.eu/ P1.4atTwo 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, Fy   F1  15 kips   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   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.)  2.4053 in.2 and thus, the normal stress in rod (1) is F 15 kips 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.)  4.9087 in.2 Accordingly, the normal stress in rod (2) is F 75 kips 2    15.2789 ksi  15.28 ksi (C) A2 2.4053 in.2 Ans Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Full file at https://TestbankDirect.eu/ Solution Manual for Mechanics of Materials 3rd Edition by Philpot Full file https://TestbankDirect.eu/ P1.5atAxial 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  kips  kips  kips   F1  16 kips  16 kips (C) FBD through rod (1) FBD through rod (2) FBD through rod (3) Next, cut a FBD through rod (2) that includes the free end A Again, we will assume that the internal force in rod (2) is tension Equilibrium of this FBD reveals the internal force in rod (2): Fy   F2  kips  kips  kips  15 kips  15 kips   F2  14 kips  14 kips (T) Similarly, cut a FBD through rod (3) that includes the free end A From this FBD, the internal force in rod (3) is: Fy   F3  kips  kips  kips  15 kips  15 kips  20 kips  20 kips   F3  26 kips  26 kips (C) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Full file at https://TestbankDirect.eu/ Solution Manual for Mechanics of Materials 3rd Edition by Philpot Full file at https://TestbankDirect.eu/ From the given diameter of rod (1), the cross-sectional area of rod (1) is  A1  (2.00 in.)  3.1416 in.2 and thus, the normal stress in aluminum rod (1) is F 16 kips 1    5.0930 ksi  5.09 ksi (C) A1 3.1416 in.2 From the given diameter of rod (2), the cross-sectional area of rod (2) is  A2  (1.50 in.)  1.7671 in.2 Accordingly, the normal stress in brass rod (2) is F 14 kips 2    7.9224 ksi  7.92 ksi (T) A2 1.7671 in.2 Finally, the cross-sectional area of rod (3) is  A3  (3.00 in.)  7.0686 in.2 and the normal stress in the steel rod is F 26 kips 3    3.6782 ksi  3.68 ksi (C) A3 7.0686 in.2 Ans Ans Ans Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Full file at https://TestbankDirect.eu/ Solution Manual for Mechanics of Materials 3rd Edition by Philpot Full file https://TestbankDirect.eu/ P1.6at 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  kips  kips  kips   F1  16 kips  16 kips (C) FBD through rod (1) FBD through rod (2) FBD through rod (3) Next, cut a FBD through rod (2) that includes the free end A Again, we will assume that the internal force in rod (2) is tension Equilibrium of this FBD reveals the internal force in rod (2): Fy   F2  kips  kips  kips  15 kips  15 kips   F2  14 kips  14 kips (T) Similarly, cut a FBD through rod (3) that includes the free end A From this FBD, the internal force in rod (3) is: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Full file at https://TestbankDirect.eu/ Solution Manual for Mechanics of Materials 3rd Edition by Philpot Full file atF https://TestbankDirect.eu/   F  kips  kips  kips  15 kips  15 kips  20 kips  20 kips  y  F3  26 kips  26 kips (C) Notice that two of the three rods are in compression In these situations, we are concerned only with the stress magnitude; therefore, we will use the force magnitudes to determine the minimum required crosssectional areas, and in turn, the minimum rod diameters The normal stress in aluminum rod (1) must be limited to 18 ksi; therefore, the minimum cross-sectional area required for rod (1) is F 16 kips A1,min    0.8889 in.2  18 ksi The minimum rod diameter is therefore  Ans A1,min  d12  0.8889 in.2  d1  1.0638 in  1.064 in The normal stress in brass rod (2) must be limited to 25 ksi, which requires a minimum area of F 14 kips A2,min    0.5600 in.2  25 ksi which requires a minimum diameter for rod (2) of  A2,min  d 22  0.5600 in.2  d  0.8444 in  0.844 in Ans The normal stress in steel rod (3) must be limited to 15 ksi The minimum cross-sectional area required for this rod is: F 26 kips A3,min    1.7333 in.2  15 ksi which requires a minimum diameter for rod (3) of  Ans A3,min  d32  1.7333 in.2  d3  1.4856 in  1.486 in Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Full file at https://TestbankDirect.eu/ Solution Manual for Mechanics of Materials 3rd Edition by Philpot Full file https://TestbankDirect.eu/ P1.7atTwo 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 Fx  F2 cos(35.7067)  F1 cos(57.9946)  Fy  F2 sin(35.7067)  F1 sin(57.9946)  P  (a) (b) Unknown forces F1 and F2 can be found from the simultaneous solution of Eqs (a) and (b) Using the substitution method, Eq (b) can be solved for F2 in terms of F1: cos(57.9946) F2  F1 (c) cos(35.7067) Substituting Eq (c) into Eq (b) gives cos(57.9946) F1 sin(35.7067)  F1 sin(57.9946)  P cos(35.6553) F1  cos(57.9946) tan(35.7067)  sin(57.9946)   P  F1  P P  cos(57.9946) tan(35.7067)  sin(57.9946) 1.2289 For the given load of P = 50 kN, the internal force in rod (1) is therefore: 50 kN F1   40.6856 kN 1.2289 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Full file at https://TestbankDirect.eu/ Solution Manual for Mechanics of Materials 3rd Edition by Philpot Full file at https://TestbankDirect.eu/ Backsubstituting this result into Eq (c) gives force F2: F2  F1 cos(57.9946) cos(57.9946)  (40.6856 kN)  26.5553 kN cos(35.7067) cos(35.7067) The normal stress in rod (1) must be limited to 130 MPa; therefore, the minimum cross-sectional area required for rod (1) is F (40.6856 kN)(1,000 N/kN) A1,min    312.9664 mm 2 1 130 N/mm The minimum rod diameter is therefore  Ans A1,min  d12  312.9664 mm  d1  19.9620 mm  19.96 mm The minimum area required for rod (2) is F (26.5553 kN)(1,000 N/kN) A2,min    204.2718 mm 2 2 130 N/mm which requires a minimum diameter for rod (2) of  A2,min  d 22  204.2718 mm2  d  16.1272 mm  16.13 mm Ans Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Full file at https://TestbankDirect.eu/ Solution Manual for Mechanics of Materials 3rd Edition by Philpot Full file at The https://TestbankDirect.eu/ P1.32 beam shown in Figure P1.32 is supported by a pin at C and by a short link AB If w = 30 kN/m, determine the average shear stress in the pins at A and C Each pin has a diameter of 25 mm Assume L = 1.8 m and  = 35° FIGURE P1.32 Solution Equilibrium: Using the FBD shown, calculate the reaction forces that act on the beam  1.8 m  M C   F1 sin(35)(1.8 m)  (30 kN/m)(1.8 m)  0    F1  47.0731 kN Fx  Cx  (47.0731 kN)cos(35)   Cx  38.5600 kN  1.8 m  M B  C y (1.8 m)  (30 kN/m)(1.8 m)  0    C y  27.0000 kN The resultant force at C is C  (38.5600 kN)2  (27.0000 kN)2  47.0731 kN Shear stress in pin A The cross-sectional area of a 25-mm-diameter pin is: Apin   (25 mm)  490.8739 mm Pin A is a single shear connection; therefore, its average shear stress is 47,073.1 N  pin A   95.9 MPa 490.8739 mm Shear stress in pin C Pin C is a double shear connection; therefore, its average shear stress is 47,073.1 N  pin C   47.9 MPa 2(490.8739 mm ) Ans Ans Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Full file at https://TestbankDirect.eu/ Solution Manual for Mechanics of Materials 3rd Edition by Philpot Full file at The https://TestbankDirect.eu/ P1.33 bell-crank mechanism shown in Figure P1.33 is in equilibrium for an applied load of P = kN applied at A Assume a = 200 mm, b = 150 mm, and  = 65° Determine the minimum diameter d required for pin B for each of the following conditions: (a) The average shear stress in the pin may not exceed 40 MPa (b) The bearing stress in the bell crank may not exceed 100 MPa (c) The bearing stress in the support bracket may not exceed 165 MPa FIGURE P1.33 Solution Equilibrium: Using the FBD shown, calculate the reaction forces that act on the bell crank M B  (7,000 N)sin(65)(200 mm)  F2 (150 mm)   F2  8, 458.873 N Fx  Bx  (7,000 N)cos(65) 8,458.873 N   Bx  11,417.201 N Fy  By  (7,000 N)sin(65)   By  6,344.155 N The resultant force at B is B  (11,417.201 N)2  (6,344.155 N)2  13,061.423 N (a) The average shear stress in the pin may not exceed 40 MPa The shear area required for the pin at B is 13,061.423 N AV   326.536 mm2 40 N/mm Since the pin at B is supported in a double shear connection, the required cross-sectional area for the pin is A Apin  V  163.268 mm 2 and therefore, the pin must have a diameter of d (163.268 mm )  14.42 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 Full file at https://TestbankDirect.eu/ Solution Manual for Mechanics of Materials 3rd Edition by Philpot Full file at https://TestbankDirect.eu/ (b) The bearing stress in the bell crank may not exceed 100 MPa The projected area of pin B on the bell crank must equal or exceed 13,061.423 N Ab   130.614 mm 2 100 N/mm The bell crank thickness is mm; therefore, the projected area of the pin is Ab = (8 mm)d Calculate the required pin diameter d: 130.614 mm Ans d  16.33 mm mm (c) The bearing stress in the support bracket may not exceed 165 MPa The pin at B bears on two 6mm-thick support brackets Thus, the minimum pin diameter required to satisfy the bearing stress limit on the support bracket is 13,061.423 N Ab   79.160 mm2 165 N/mm2 d 79.160 mm  6.60 mm 2(6 mm) Ans Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Full file at https://TestbankDirect.eu/ Solution Manual for Mechanics of Materials 3rd Edition by Philpot Full file atAhttps://TestbankDirect.eu/ P1.34 structural steel bar with a 25 mm × 75 mm rectangular cross section is subjected to an axial load of 150 kN Determine the maximum normal and shear stresses in the bar Solution The maximum normal stress in the steel bar is F (150 kN)(1,000 N/kN)  max    80 MPa A (25 mm)(75 mm) The maximum shear stress is one-half of the maximum normal stress   max  max  40 MPa Ans Ans Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Full file at https://TestbankDirect.eu/ Solution Manual for Mechanics of Materials 3rd Edition by Philpot Full file at Ahttps://TestbankDirect.eu/ P1.35 steel rod of circular cross section will be used to carry an axial load of 92 kips The maximum stresses in the rod must be limited to 30 ksi in tension and 12 ksi in shear Determine the required diameter for the rod Solution Based on the allowable 30 ksi tension stress limit, the minimum cross-sectional area of the rod is F 92 kips Amin    3.0667 in.2  max 30 ksi For the 12-ksi shear stress limit, the minimum cross-sectional area of the rod must be F 92 kips Amin    3.8333 in.2 2 max 2(12 ksi) Therefore, the rod must have a cross-sectional area of at least 3.8333 in.2 in order to satisfy both the normal and shear stress limits The minimum rod diameter D is therefore  d  3.8333 in.2  d  2.2092 in  2.21 in Ans Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Full file at https://TestbankDirect.eu/ Solution Manual for Mechanics of Materials 3rd Edition by Philpot Full file at An https://TestbankDirect.eu/ P1.36 axial load P is applied to the rectangular bar shown in Figure P1.36 The cross-sectional area of the bar is 400 mm2 Determine the normal stress perpendicular to plane AB and the shear stress parallel to plane AB if the bar is subjected to an axial load of P = 70 kN FIGURE P1.36 Solution The angle θ for the inclined plane is 35° The normal force N perpendicular to plane AB is found from N = P cos θ = (40 kN)cos35° = 57.3406 kN and the shear force V parallel to plane AB is V = − P sin θ = −(70 kN) sin 35° = −40.1504 kN The cross-sectional area of the bar is 400 mm2, but the area along inclined plane AB is A 400 mm An = = = 488.3098 mm cos θ cos35° The normal stress σn perpendicular to plane AB is N (57.3406 kN)(1,000 N/kN) σn = = = 117.4268 MPa = 117.4 MPa An 488.3098 mm The shear stress τnt parallel to plane AB is V (−40.1504 kN)(1,000 N/kN) τ nt = = = −82.2231 MPa = −82.2 MPa An 488.3098 mm Ans Ans Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Full file at https://TestbankDirect.eu/ Solution Manual for Mechanics of Materials 3rd Edition by Philpot Full file atAn https://TestbankDirect.eu/ P1.37 axial load P is applied to the 1.75 in by 0.75 in rectangular bar shown in Figure P1.37 Determine the normal stress perpendicular to plane AB and the shear stress parallel to plane AB if the bar is subjected to an axial load of P = 18 kips FIGURE P1.37 Solution The angle  for the inclined plane is 60° The normal force N perpendicular to plane AB is found from N  P cos  (18 kips)cos60  9.0 kips and the shear force V parallel to plane AB is V  P sin   (18 kips)sin 60  15.5885 kips The cross-sectional area of the bar is (1.75 in.)(0.75 in.) = 1.3125 in.2, but the area along inclined plane AB is 1.3125 in.2 An  A / cos    2.6250 in.2 cos 60 The normal stress n perpendicular to plane AB is N 9.0 kips n    3.4286 ksi  3.43 ksi An 2.6250 in.2 The shear stress nt parallel to plane AB is V 15.5885 kips  nt    5.9385 ksi  5.94 ksi An 2.6250 in.2 Ans Ans  Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Full file at https://TestbankDirect.eu/ Solution Manual for Mechanics of Materials 3rd Edition by Philpot Full file atAhttps://TestbankDirect.eu/ P1.38 compression load of P = 80 kips is applied to a in by in square post, as shown in Figure P1.38/39 Determine the normal stress perpendicular to plane AB and the shear stress parallel to plane AB FIGURE P1.38/39 Solution The angle  for the inclined plane is 55° The normal force N perpendicular to plane AB is found from N  P cos  (80 kips)cos55  45.8861 kips and the shear force V parallel to plane AB is V  P sin   (80 kips)sin55  65.5322 kips The cross-sectional area of the post is (4 in.)(4 in.) = 16 in.2, but the area along inclined plane AB is 16 in.2 An  A / cos    27.8951 in.2 cos55 The normal stress n perpendicular to plane AB is N 45.8861 kips n    1.6449 ksi  1.645 ksi An 27.8951 in.2 The shear stress nt parallel to plane AB is V 65.5322 kips  nt    2.3492 ksi  2.35 ksi An 27.8951 in.2 Ans Ans Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Full file at https://TestbankDirect.eu/ Solution Manual for Mechanics of Materials 3rd Edition by Philpot Full file at Specifications https://TestbankDirect.eu/ P1.39 for the 50 mm × 50 mm square bar shown in Figure P1.38/39 require that the normal and shear stresses on plane AB not exceed 120 MPa and 90 MPa, respectively Determine the maximum load P that can be applied without exceeding the specifications FIGURE P1.38/39 Solution The general equations for normal and shear stresses on an inclined plane in terms of the angle  are P (a) n  (1  cos 2 ) 2A and P (b)  nt  sin 2 2A The cross-sectional area of the square bar is A = (50 mm)2 = 2,500 mm2, and the angle  for plane AB is 55° The normal stress on plane AB is limited to 120 MPa; therefore, the maximum load P that can be supported by the square bar is found from Eq (a): A n 2(2,500 mm )(120 N/mm ) P   911,882 N  cos 2  cos 2(55) The shear stress on plane AB is limited to 90 MPa From Eq (b), the maximum load P based the shear stress limit is A nt 2(2,500 mm )(90 N/mm ) P   478,880 N sin 2 sin 2(55) Thus, the maximum load that can be supported by the bar is Pmax  479 kN Ans Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Full file at https://TestbankDirect.eu/ Solution Manual for Mechanics of Materials 3rd Edition by Philpot Full file atSpecifications https://TestbankDirect.eu/ P1.40 for the in × in square post shown in Figure P1.40 require that the normal and shear stresses on plane AB not exceed 800 psi and 400 psi, respectively Determine the maximum load P that can be applied without exceeding the specifications FIGURE P1.40 Solution The general equations for normal and shear stresses on an inclined plane in terms of the angle  are P (a) n  (1  cos 2 ) 2A and P (b)  nt  sin 2 2A The cross-sectional area of the square post is A = (6 in.)2 = 36 in.2, and the angle  for plane AB is 40° The normal stress on plane AB is limited to 800 psi; therefore, the maximum load P that can be supported by the square post is found from Eq (a): A n 2(36 in.2 )(800 psi) P   49,078 lb  cos 2  cos 2(40) The shear stress on plane AB is limited to 400 psi From Eq (b), the maximum load P based the shear stress limit is A nt 2(36 in.2 )(400 psi) P   29, 244 lb sin 2 sin 2(40) Thus, the maximum load that can be supported by the post is Pmax  29,200 lb  29.2 kips Ans Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Full file at https://TestbankDirect.eu/ Solution Manual for Mechanics of Materials 3rd Edition by Philpot Full file atAhttps://TestbankDirect.eu/ P1.41 90 mm wide bar will be used to carry an axial tension load of 280 kN as shown in Figure P1.41 The normal and shear stresses on plane AB must be limited to 150 MPa and 100 MPa, respectively Determine the minimum thickness t required for the bar FIGURE P1.41 Solution The general equations for normal and shear stresses on an inclined plane in terms of the angle  are P (a) n  (1  cos 2 ) 2A and P (b)  nt  sin 2 2A The angle  for plane AB is 50° The normal stress on plane AB is limited to 150 MPa; therefore, the minimum cross-sectional area A required to support P = 280 kN can be found from Eq (a): P (280 kN)(1,000 N/kN) A (1  cos 2 )  (1  cos 2(50))  771.2617 mm 2 n 2(150 N/mm ) The shear stress on plane AB is limited to 100 MPa; therefore, the minimum cross-sectional area A required to support P = 280 kN can be found from Eq (b): P (280 kN)(1,000 N/kN) A sin 2  sin 2(50)  1,378.7309 mm 2 nt 2(100 N/mm ) To satisfy both the normal and shear stress requirements, the cross-sectional area must be at least Amin = 1,379.7309 mm2 Since the bar width is 90 mm, the minimum bar thickness t must be 1,378.7309 mm tmin   15.3192 mm  15.32 mm Ans 90 mm Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Full file at https://TestbankDirect.eu/ Solution Manual for Mechanics of Materials 3rd Edition by Philpot Full file at A https://TestbankDirect.eu/ P1.42 rectangular bar having width w = 6.00 in and thickness t = 1.50 in is subjected to a tension load P as shown in Figure P1.42/43 The normal and shear stresses on plane AB must not exceed 16 ksi and ksi, respectively Determine the maximum load P that can be applied without exceeding either stress limit FIGURE P1.42/43 Solution The general equations for normal and shear stresses on an inclined plane in terms of the angle  are P (a) n  (1  cos 2 ) 2A and P (b)  nt  sin 2 2A The angle  for inclined plane AB is calculated from tan      71.5651 The cross-sectional area of the bar is A = w×t = (6.00 in.)(1.50 in.) = 9.0 in.2 The normal stress on plane AB is limited to 16 ksi; therefore, the maximum load P can be found from Eq (a): A n 2(9.0 in.2 )(16 ksi) P   1, 440 ksi  cos 2  cos 2(71.5651) The shear stress on plane AB is limited to ksi From Eq (b), the maximum load P based the shear stress limit is A nt 2(9.0 in.2 )(8 ksi) P   240 kips sin 2 sin 2(71.5651) Thus, the maximum load that can be supported by the bar is Pmax  240 kips Ans Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Full file at https://TestbankDirect.eu/ Solution Manual for Mechanics of Materials 3rd Edition by Philpot Full file at https://TestbankDirect.eu/ P1.43 In Figure P1.42/43, a rectangular bar having width w = 1.25 in and thickness t is subjected to a tension load of P = 30 kips The normal and shear stresses on plane AB must not exceed 12 ksi and ksi, respectively Determine the minimum bar thickness t required for the bar FIGURE P1.42/43 Solution The general equations for normal and shear stresses on an inclined plane in terms of the angle  are P (a) n  (1  cos 2 ) 2A and P (b)  nt  sin 2 2A The angle  for inclined plane AB is calculated from tan      71.5651 The normal stress on plane AB is limited to 12 ksi; therefore, the minimum cross-sectional area A required to support P = 30 kips can be found from Eq (a): P 30 kips A (1  cos 2 )  (1  cos 2(71.5651))  0.2500 in.2 2 n 2(12 ksi) The shear stress on plane AB is limited to ksi; therefore, the minimum cross-sectional area A required to support P = 30 kips can be found from Eq (b): P 30 kips A sin 2  sin 2(71.5651)  1.1250 in.2 2 nt 2(8 ksi) To satisfy both the normal and shear stress requirements, the cross-sectional area must be at least Amin = 1.1250 in.2 Since the bar width is 1.25 in., the minimum bar thickness t must be 1.1250 in.2 tmin   0.900 in  0.900 in Ans 1.25 in Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Full file at https://TestbankDirect.eu/ Solution Manual for Mechanics of Materials 3rd Edition by Philpot Full file at The https://TestbankDirect.eu/ P1.44 rectangular bar has a width of w = 3.00 in and a thickness of t = 2.00 in The normal stress on plane AB of the rectangular block shown in Figure P1.44/45 is ksi (C) when the load P is applied Determine: (a) the magnitude of load P (b) the shear stress on plane AB (c) the maximum normal and shear stresses in the block at any possible orientation FIGURE P1.44/45 Solution The general equation for normal stress on an inclined plane in terms of the angle  is P n  (1  cos 2 ) 2A and the angle  for inclined plane AB is tan    0.75   36.8699 The cross-sectional area of the rectangular bar is A = (3.00 in.)(2.00 in.) = 6.00 in.2 (a) (a) Since the normal stress on plane AB is given as ksi, the magnitude of load P can be calculated from Eq (a): A n 2(6.0 in.2 )(6 ksi) P   56.25 kips  56.3 kips Ans  cos 2  cos 2(36.8699) (b) The general equation for shear stress on an inclined plane in terms of the angle  is P  nt  sin 2 2A therefore, the shear stress on plane AB is 56.25 kips  nt  sin 2(36.8699)  4.50 ksi 2(6.00 in.2 ) (c) The maximum normal stress at any possible orientation is P 56.25 kips  max    9.3750 ksi  9.38 ksi A 6.00 in.2 and the maximum shear stress at any possible orientation in the block is P 56.25 kips  max    4.6875 ksi  4.69 ksi A 2(6.00 in.2 ) Ans Ans Ans Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Full file at https://TestbankDirect.eu/ Solution Manual for Mechanics of Materials 3rd Edition by Philpot Full file atThe https://TestbankDirect.eu/ P1.45 rectangular bar has a width of w = 100 mm and a thickness of t = 75 mm The shear stress on plane AB of the rectangular block shown in Figure P1.44/45 is 12 MPa when the load P is applied Determine: (a) the magnitude of load P (b) the normal stress on plane AB (c) the maximum normal and shear stresses in the block at any possible orientation FIGURE P1.44/45 Solution The general equation for shear stress on an inclined plane in terms of the angle  is P  nt  sin 2 2A and the angle  for inclined plane AB is tan    0.75   36.8699 The cross-sectional area of the rectangular bar is A = (100 mm)(75 mm) = 7,500 mm2 (a) (a) Since the shear stress on plane AB is given as 12 MPa, the magnitude of load P can be calculated from Eq (a): A nt 2(7,500 mm )(12 N/mm ) P   187,500 N  187.5 kN Ans sin 2 sin 2(36.8699) (b) The general equation for normal stress on an inclined plane in terms of the angle  is P n  (1  cos 2 ) 2A therefore, the normal stress on plane AB is 187,500 N n  (1  cos 2(36.8699))  16.00 MPa 2(7,500 mm ) (c) The maximum normal stress at any possible orientation is P 187,500 N  max    25.0 MPa A 7,500 mm and the maximum shear stress at any possible orientation in the block is P 187,500 N  max    12.50 MPa A 2(7,500 mm ) Ans Ans Ans Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Full file at https://TestbankDirect.eu/ ... https://TestbankDirect.eu/ Solution Manual for Mechanics of Materials 3rd Edition by Philpot Full file JointatA:https://TestbankDirect.eu/ Begin the solution process by considering a FBD of joint A Consider... https://TestbankDirect.eu/  Solution Manual for Mechanics of Materials 3rd Edition by Philpot Full file JointatA:https://TestbankDirect.eu/ Begin the solution process by considering a FBD of joint A Consider... https://TestbankDirect.eu/ Solution Manual for Mechanics of Materials 3rd Edition by Philpot Full file JointatA:https://TestbankDirect.eu/ Begin the solution process by considering a FBD of joint A Consider