solutions_Chapter04
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solutions_Chapter04

Course: CIVL 3322, Spring 2011

School: U. Memphis

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4.1 A stainless steel alloy bar 25 mm wide by 16 mm thick is subjected to an axial load of P = 145 kN. Using the stress-strain diagram given in Fig. P4.1, determine: (a) the factor of safety with respect to the yield strength defined by the 0.20% offset method. (b) the factor of safety with respect to the ultimate strength. Fig. P4.1 Solution (a) From the stress-strain curve, the yield strength defined by the...

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A 4.1 stainless steel alloy bar 25 mm wide by 16 mm thick is subjected to an axial load of P = 145 kN. Using the stress-strain diagram given in Fig. P4.1, determine: (a) the factor of safety with respect to the yield strength defined by the 0.20% offset method. (b) the factor of safety with respect to the ultimate strength. Fig. P4.1 Solution (a) From the stress-strain curve, the yield strength defined by the 0.20% offset method is Y = 550 MPa The normal stress in the bar is F (145 kN)(1,000 N/kN) = = = 362.5 MPa A (25 mm)(16 mm) Therefore, the factor of safety with respect to yield is 550 MPa = 1.517 FS = Y = actual 362.5 MPa (b) From the stress-strain curve, the ultimate strength is ult = 1,100 MPa Therefore, the factor of safety with respect to the ultimate strength is 1,100 MPa = 3.03 FS = ult = actual 362.5 MPa Ans. Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 4.2 Six bolts are used in the connection between the axial member and the support, as shown in Fig. P4.2. The ultimate shear strength of the bolts is 300 MPa, and a factor of safety of 4.0 is required with respect to fracture. Determine the minimum allowable bolt diameter D required to support an applied load of P = 350 kN. Fig. P4.2 Solution The allowable shear stress for the bolts is 300 MPa = 75 MPa allow = ult = FS 4 To support a load of P = 350 kN, the total area subjected to shear stress must equal or exceed: P 350, 000 N AV = = 4, 666.67 mm 2 2 allow 75 N/mm There are six bolts, and each bolt acts in double shear; therefore, the total area subjected to shear stress is 2 Dbolt 4 The minimum bolt diameter can be found be equating the two expressions for AV: AV = (6 bolts)(2 surfaces per bolt) (12 surfaces) Dbolt 2 Dbolt 4, 666.67 mm 2 4 22.252 mm = 22.3 mm Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 4.3 A 9-kip load is supported by two bars as shown in Fig. P4.3. Bar (1) is made of cold-rolled red brass (Y = 60 ksi) and has a cross-sectional area of 0.225 in.2. Bar (2) is made of 6061-T6 aluminum (Y = 40 ksi) and has a cross-sectional area of 0.375 in.2. Determine the factor of safety with respect to yielding for each of the bars. Fig. P4.3 Solution Consider a FBD of joint B. Members (1) and (2) are two-force members; therefore, the equilibrium equations can be written as: Fx = F2 cos 55 F1 cos 40 = 0 (a) Fy = F2 sin 55 + F1 sin 40 9 kips = 0 (b) From Eq. (a): cos 40 F2 = F1 cos 55 Substitute this expression into Eq. (b) to obtain: cos 40 F1 cos 55 sin 55 + F1 sin 40 = 9 kips F1 [ cos 40 tan 55 + sin 40] = 9 kips F1 [1.736812] = 9 kips F1 = 5.181907 kips Backsubstitute to obtain F2: cos 40 F2 = F1 = (5.181907 kips)(1.335558) = 6.920735 kips cos 55 The normal stress in bar (1) is F 5.181907 kips 1 = 1 = = 23.030698 ksi 0.225 in.2 A1 Therefore, the factor of safety in bar (1) is 60 ksi = 2.605219 = 2.61 FS1 = Y ,1 = 1 23.030698 ksi The normal stress in bar (2) is F 6.920735 kips 2 = 2 = = 18.455295 ksi 0.375 in.2 A2 and thus, the factor of safety in bar (2) is 40 ksi = 2.167400 = 2.17 FS2 = Y , 2 = 2 18.455295 ksi Ans. Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 4.4 A steel plate is to be attached to a support with three bolts as shown in Fig. P4.4. The cross-sectional area of the plate is 800 mm2 and the yield strength of the steel is 250 MPa. The ultimate shear strength of the bolts is 425 MPa. A factor of safety of 1.67 with respect to yield is required for the plate. A factor of safety of 4.0 with respect to the ultimate shear strength is required for the bolts. Determine the minimum bolt diameter required to develop the full strength of the plate. (Note: consider only the gross crosssectional area of the platenot the net area.) Fig. P4.4 Solution The allowable normal stress is 250 MPa = 149.7006 MPa allow = Y = FS 1.67 The full strength of the plate (based on the gross cross-sectional area) is therefore: Pmax = allow A = (149.7006 MPa)(800 mm 2 ) = 119, 760.4790 N = 119.8 kN The allowable shear stress of the bolts is 425 MPa = 106.25 MPa allow = ult = FS 4.0 The minimum shear area required to support the maximum load P is P 119, 760.4790 N = 1,127.157449 mm 2 AV max = allow 106.25 N/mm 2 There are three bolts, each acting in single shear, which provide a shear area of (3 surfaces) 4 2 Dbolt = (3 surfaces) 2 Dbolt 4 4 Equating the two expressions for the shear area, the minimum bolt diameter can be computed: AV = (3 bolts)(1 surface per bolt) 2 Dbolt 1,127.157449 mm 2 Dbolt 21.871911 mm = 21.9 mm Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 4.5 In Fig. P4.5, member (1) is a steel bar with a crosssectional area of 1.75 in.2 and a yield strength of 50 ksi. Member (2) is a pair of 6061-T6 aluminum bars having a combined cross-sectional area of 4.50 in.2 and a yield strength of 40 ksi. A factor of safety of 1.5 with respect to yield is required for both members. Determine the maximum allowable load P that may be applied to the structure. Report the factors of safety for both members at the allowable load. Fig. P4.5 Solution Consider a FBD of joint B and write the following equilibrium equations: Fx = F2 cos 55 F1 = 0 Fy = F2 sin 55 P = 0 From Eq. (b): P F2 = sin 55 Substituting Eq. (c) into Eq. (a) gives an expression for P in terms of F1: F2 cos 55 F1 = 0 P cos 55 F1 = 0 sin 55 P = F1 tan 55 and rearranging Eq. (c) gives an expression for P in terms of F2: P = F2 sin 55 (a) (b) (c) (d) (e) The allowable normal stress in member (1) is 50 ksi allow,1 = Y ,1 = = 33.3333 ksi FS1 1.5 and the allowable force in member (1) is Fallow,1 = allow,1 A1 = (33.3333 ksi)(1.75 in.2 ) = 58.3333 kips (f) The allowable normal stress in member (2) is 40 ksi allow,2 = Y , 2 = = 26.6667 ksi FS2 1.5 and the allowable force in member (2) is Fallow,2 = allow,2 A2 = (26.6667 ksi)(4.50 in.2 ) = 120.0 kips (g) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Substitute the allowable force in member (1) from Eq. (f) into Eq. (d) to obtain the maximum load P based on the capacity of member (1): P Fallow,1 tan 55 = (58.3333 kips) tan 55 = 83.308586 kips (h) Repeat with the allowable force in member (2) from Eq. (g) substituted into Eq. (e) to obtain the maximum load P based on the capacity of member (2): (i) P Fallow,2 sin 55 = (120.0 kips) sin 55 = 98.298245 kips Compare the results in Eqs. (h) and (i) to find that the maximum load P that may be applied is controlled by the capacity of member (1): Pmax = 83.3 kips Ans. The factor of safety in member (1) is thus: FS1 = 1.5 For an applied load of P = 83.3 kips, the force in member (2) can be computed from Eq. (c): P 83.308586 kips F2 = = = 101.701005 kips sin 55 sin 55 which results in a normal stress of F 101.701005 kips 2 = 2 = = 22.600223 ksi A2 4.50 in.2 and thus, its factor of safety is: 40 ksi FS2 = Y ,2 = = 1.770 2 22.600223 ksi Ans. Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 4.6 The rigid structure ABD in Fig. P4.6 is supported at B by a 35-mm-diameter tie rod (1) and at A by a 30-mm-diameter pin used in a single shear connection. The tie rod is connected at B and C by 24-mmdiameter pins used in double shear connections. Tie rod (1) has a yield strength of 250 MPa, and each of the pins has an ultimate shear strength of 330 MPa. A concentrated load of P = 50 kN acts as shown at D. Determine: (a) the normal stress in rod (1). (b) the shearing stress in the pins at A and B. (c) the factor of safety with respect to the yield strength for tie rod (1). (d) the factor of safety with respect to the ultimate strength for the pins at A and B. Fig. P4.6 Solution Member (1) is a two-force member that is oriented at with respect to the horizontal axis: 4m = 1.052632 tan = 3.8 m = 46.4688 From a FBD of rigid structure ABD, the following equilibrium equations can be written: Fx = P cos 60 F1 cos 46.4688 + Ax = 0 Fy = P sin 60 F1 sin 46.4688 + Ay = 0 M A = ( F1 cos 46.4688)(4 m) ( P cos 60)(3 m) ( P sin 60)(5 m) = 0 From Eq. (c): (3 m) cos 60 + (5 m)sin 60 (3 m) cos 60 + (5 m)sin 60 F1 = P = (50 kN) = 105.8100 kN (4 m) cos 46.4688 (4 m) cos 46.4688 (a) (b) (c) Substitute F1 into Eq. (a) to obtain Ax: Ax = F1 cos 46.4688 P cos 60 = (105.8100 kN) cos 46.4688 (50 kN) cos 60 = 47.8766 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. and substitute F1 into Eq. (b) to obtain Ay: Ay = F1 sin 46.4688 + P sin 60 = (105.8100 kN)sin 46.4688 + (50 kN) sin 60 = 120.0135 kN The resultant pin force at A is found from Ax and Ay: 2 A = Ax2 + Ay = (47.8766 kN)2 + (120.0135 kN) 2 = 129.2107 kN (a) The cross-sectional area of the 35-mm-diameter tie rod is: A1 = (35 mm)2 = 962.112750 mm 2 4 and thus, the normal stress in rod (1) is: F 105,810 N 1 = 1 = = 109.9767 MPa = 110.0 MPa A1 962.112750 mm 2 Ans. (b) The 30-mm-diameter single shear pin at A has a shear area of (30 mm)2 = 706.858347 mm 2 4 Consequently, the shear stress in pin A is 129, 210.7 N A = = 182.7957 MPa = 182.8 MPa 706.858347 mm 2 AV , A = Ans. The 24-mm-diameter double shear pins at B and C have a shear area of (24 mm) 2 = 904.778685 mm 2 4 The shear stress in pins B and C is 105,810.0 N B = = 116.9457 MPa = 116.9 MPa 904.778685 mm 2 AV , B = (2 surfaces) (c) The factor of safety for tie rod (1) is 250 MPa FS1 = Y ,1 = = 2.27 1 109.9767 MPa (d) The factor of safety with respect to the ultimate strength for the pin at A is 330 MPa FS A = ult,A = = 1.805 A 182.7957 MPa and the factor of safety for pin B is 330 MPa FSB = ult,B = = 2.82 B 116.9457 MPa Ans. Ans. Ans. Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 4.7 Rigid bar ABD in Fig. P4.7 is supported by a pin connection at A and a tension link BC. The 8-mmdiameter pin at A is supported in a double shear connection, and the 12-mm-diameter pins at B and C are both used in single shear connections. Link BC is 30-mm wide and 6-mm thick. The ultimate shear strength of the pins is 330 MPa and the yield strength of link BC is 250 MPa. (a) Determine the factor of safety in pins A and B with respect to the ultimate shear strength. (b) Determine the factor of safety in link BC with respect to the yield strength. Fig. P4.7 Solution Consider a free-body diagram of the rigid bar. Link BC is a two-force member that is oriented at an angle of with respect to the horizontal axis: 0.4 m tan = = 48.814 0.35 m The equilibrium equations for the rigid bar can be written as: Fx = FBC cos 48.814 + (6.2 kN)cos70 + Ax = 0 Fy = FBC sin 48.814 (6.2 kN)sin70 + Ay = 0 M A = (0.6 m)FBC cos 48.814 + (0.35 m)FBC sin 48.814 (1.35 m)(6.2 kN)sin70 (0.6 m)(6.2 kN)cos70 = 0 Solving these three equations simultaneously gives: FBC = 13.876 kN Ax = 7.017 kN Ay = 4.617 kN The resultant force at pin A is: 2 A = Ax2 + Ay = (7.017 kN) 2 + (4.617 kN) 2 = 8.400 kN Before the factors of safety can be determined, we must compute the shear stresses in pins A and B as well as the normal stress in link BC. Pin shear stresses: The 8-mm-diameter pin at A is supported in a double shear connection; therefore, the shear force acting on one shear plane (which is simply equal to the cross-sectional area of the pin) is half of the resultant force at pin A: VA = 4.200 kN. The cross-sectional area of the pin at A is: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Apin A = (8 mm) 2 = 50.265 mm 2 4 and therefore, the shear stress in pin A is: V (4.200 kN)(1000 N/kN) A = A = = 83.552 N/mm 2 = 83.552 MPa AV 50.265 mm 2 The 12-mm-diameter pin at B and C is supported in a single shear connection and so the shear force acting on one shear plane is the entire force in link BC: VB = 13.876 kN. The cross-sectional area of the pin at B is: Apin B = (12 mm) 2 = 113.097 mm 2 4 and therefore, the shear stress in pin B is: V (13.876 kN)(1000 N/kN) B = B = = 122.693 N/mm 2 = 122.693 MPa 113.097 mm 2 AV Normal stress in link: The normal stress in link BC is: F (13.876 kN)(1000 N/kN) BC = BC = = 77.090 N/mm 2 = 77.090 MPa (30 mm)(6 mm) ABC Factors of safety: For pin A, the factor of safety is: 330 MPa FS A = u = = 3.95 A 83.552 MPa For pin B, the factor of safety is: 330 MPa FSB = u = = 2.69 B 122.693 MPa The factor of safety in link BC is: 250 MPa FSBC = y = = 3.24 BC 77.090 MPa Ans. Ans. Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 4.8 In Fig. P4.8, davit ABD is supported at A by a single shear pin connection and at B by a tie rod (1). The pin at A has a diameter of 1.25 in. and the pins at B and C are each 0.75-in.-diameter pins. Tie rod (1) has an area of 1.50 in.2. The ultimate shear strength in each pin is 80 ksi, and the yield strength of the tie rod is 36 ksi. A concentrated load of 25 kips is applied as shown to the davit structure at D. Determine: (a) the normal stress in rod (1). (b) the shearing stress in the pins at A and B. (c) the factor of safety with respect to the yield strength for tie rod (1). (d) the factor of safety with respect to the ultimate strength for the pins at A and B. Fig. P4.8 Solution Member (1) is a two-force member that is oriented at with respect to the horizontal axis: 9 ft tan = = 0.75 = 36.870 12 ft From a FBD of rigid structure ABD, the following equilibrium equations can be written: Fx = (25 kips) cos 60 F1 cos 36.870 + Ax = 0 (a) Fy = (25 kips) sin 60 F1 sin 36.870 + Ay = 0 (b) M A = ( F1 cos 36.870)(9 ft) (25 kips) cos 60(11 ft) (25 kips) sin 60(7 ft) = 0 (c) From Eq. (c): (11 ft) cos 60 + (7 ft)sin 60 = 40.1465 kips (9 ft) cos 36.870 Substitute F1 into Eq. (a) to obtain Ax: Ax = F1 cos 36.870 (25 kips) cos 60 = (40.1465 kips) cos 36.870 (25 kips) cos 60 = 19.6172 kips and substitute F1 into Eq. (b) to obtain Ay: Ay = F1 sin 36.870 + (25 kips) sin 60 = (40.1465 kips) sin 36.870 + (25 kips) sin 60 = 45.7386 kips The resultant pin force at A is found from Ax and Ay: F1 = (25 kips) 2 A = Ax2 + Ay = (19.6172 kips) 2 + (45.7386 kips) 2 = 49.7680 kips Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. (a) The normal stress in tie rod (1) is: F 40.1465 kips 1 = 1 = = 26.7643 ksi = 26.8 ksi A1 1.50 in.2 Ans. (b) The 1.25-in.-diameter single shear pin at A has a shear area of AV , A = (1.25 in.) 2 = 1.2272 in.2 4 Consequently, the shear stress in pin A is 49.7680 kips A = = 40.5541 ksi = 40.6 ksi 1.2272 in.2 Ans. The 0.75-in.-diameter double shear pins at B and C have a shear area of (0.75 in.) 2 = 0.8836 in.2 4 The shear stress in pins B and C is 40.1465 kips B = = 45.4352 ksi = 45.4 ksi 0.8836 in.2 AV , B = (2 surfaces) (c) The factor of safety for tie rod (1) is 36 ksi FS1 = Y ,1 = = 1.345 1 26.7643 ksi (d) The factor of safety with respect to the ultimate strength for the pin at A is 80 ksi FS A = ult,A = = 1.973 A 40.5541 ksi and the factor of safety for pin B is 80 ksi FSB = ult,B = = 1.761 B 45.4352 ksi Ans. Ans. Ans. Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 4.9 The pin-connected structure is subjected to a load P as shown in Fig. P4.9. Inclined member (1) has a crosssectional area of 250 mm2 and has a yield strength of 255 MPa. It is connected to rigid member ABC with a 12-mm-diameter pin in a double shear connection at B. The ultimate shear strength of the pin material is 300 MPa. For inclined member (1), the minimum factor of safety with respect to the yield strength is FSmin = 1.67. For the pin connections, the minimum factor of safety with respect to the ultimate strength is FSmin = 2.50. (a) Based on the capacity of member (1) and pin B, determine the maximum allowable load P that may be applied to the structure. (b) Rigid member ABC is supported by a double shear pin connection at A. Using FSmin = 2.50, determine the minimum pin diameter that may be used at support A. Fig. P4.9 Solution The allowable normal stress for inclined member (1) is 255 MPa = 152.6946 MPa allow = Y = FS 1.67 and the allowable shear stress for the pins is 300 MPa allow = ult = = 120.0 MPa FS 2.50 (a) The allowable axial force in member (1) based on normal stress is Fallow allow A1 = (152.6946 MPa)(250 mm 2 ) = 38,174 N Member (1) is connected with a 12-mm-diameter double shear pin. The shear area of this pin is: AV = (2 surfaces) (a) (12 mm) 2 = 226.1947 mm 2 4 Consequently, the shear force V (which is applied by the inclined member) that can be applied to the pin is limited to Vallow allow AV = (120.0 MPa)(226.1947 mm 2 ) = 27,143 N (b) Comparing the two values given in Eq. (a) and Eq. (b), the maximum allowable force in member (1) is F1 = 27,143 N (c) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Member (1) is a two-force member that is oriented at with respect to the horizontal axis: 1.4 m tan = = 0.6087 = 31.3287 2.3 m From a FBD of rigid structure ABC, the following equilibrium equations can be written: Fx = Ax + F1 cos 31.3287 P = 0 (d) (e) Fy = Ay F1 sin 31.3287 = 0 M A = (3.2 m)P ( F1 cos 31.3287)(1.4 m) = 0 (f) Substitute the value of F1 from Eq. (c) into Eq. (f) to obtain the maximum load P: (1.4 m) cos 31.3287 P = F1 3.2 m (1.4 m) cos 31.3287 = (27,143 N) 3.2 m = 10,144 N = 10.14 kN Ans. Substitute F1 and P into Eq. (d) to obtain Ax: Ax = P F1 cos 31.3287 = (10,144 N) (27,143 N) cos 31.3287 = 13, 042 N and substitute F1 into Eq. (e) to obtain Ay: Ay = F1 sin 31.3287 = (27,143 N) sin 31.3287 = 14,113 N The resultant pin force at A is found from Ax and Ay: 2 A = Ax2 + Ay = (13,042 N) 2 + (14,113 N) 2 = 19, 216 N The total shear area required to support the resultant pin force is A 19, 216 N AV = = 160.1333 mm 2 2 allow 120 N/mm Since the pin at A is in a double shear connection, the shear area provided by the pin can be expressed as AV = (2 surfaces) 2 Dpin 4 Equate these two expressions and solve for the required minimum pin diameter: (2 surfaces) 4 2 Dpin 160.1333 mm 2 Dpin 10.10 mm Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 4.10 After load P is applied to the pin-connected structure shown in Fig. P4.10, a normal strain of = +850 is measured in the longitudinal direction of member (1). The cross-sectional area of member (1) is A1 = 0.60 in.2, its elastic modulus is E1 = 29,000 ksi, and its yield strength is 36 ksi. (a) Determine the axial force in member (1), the applied load P, and the resultant force at pin B. (b) The ultimate shear strength of the steel pins is 54 ksi. Determine the minimum diameter for the pin at B if a factor of safety of 2.5 with respect to the ultimate shear strength is required. (c) Compute the factor of safety for member (1) with respect to its yield strength. Fig. P4.10 Solution (a) Given the strain in member (1), its stress can be computed from Hookes law: 1 = E11 = (29, 000 ksi)(0.000850 in./in.) = 24.650 ksi The force in member (1) is the product of the normal stress and the cross-sectional area: F1 = 1 A1 = (24.650 ksi)(0.60 in.2 ) = 14.79 kips Ans. Next, consider a free-body diagram of horizontal member ABC. Member (1) makes an angle of with respect to the horizontal axis: 2 ft = 33.690 tan = 3 ft Note that member (1) is a two-force member. Equilibrium equations for horizontal member ABC can be written as: Fx = Bx + F1 cos 33.690 = 0 Fy = By F1 sin 33.690 P = 0 M B = (4 ft)F1 sin 33.690 (6 ft)P = 0 Substituting the value F1 = 14.79 kips into these equations gives: Bx = 12.306 kips B y = 13.673 kips P = 5.47 kips Ans. The resultant force at pin B is: 2 B = Bx2 + By = (12.306 kips) 2 + (13.673 kips) 2 = 18.40 kips Ans. (b) The pin at B is supported in a double shear connection. Therefore, the shear force acting on one shear plane of the pin is half of the resultant force: VB = 9.20 kips. The allowable shear stress for the pin can be computed from the ultimate shear strength and the factor of safety: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. u 54 ksi = 21.6 ksi 2.5 FS Next, the cross-sectional area required for a single shear plane can be determined: V V 9.20 kips = B a AB B = = 0.4259 in.2 AB 21.6 ksi a To provide AB, the pin at B must have a diameter of at least: a = 4 = 2 DB 0.4259 in.2 DB 0.736 in. (c) The factor of safety for member (1) with respect to its yield strength is: 36 ksi FS1 = y = = 1.460 1 24.650 ksi Ans. Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 4.11 The simple pin-connected structure carries a concentrated load P as shown in Fig. P4.11. The rigid bar is supported by strut AB and by a pin support at C. The steel strut AB has a crosssectional area of 0.75 in.2 and a yield strength of 60 ksi. The diameter of the steel pin at C is 0.5 in., and the ultimate shear strength is 54 ksi. If a factor of safety of 2.0 is required in both the strut and the pin at C, determine the maximum load P that can be supported by the structure. Fig. P4.11 Solution From a FBD of the rigid bar, the following equilibrium equations can be written: Fx = F1 + Cx = 0 (a) Fy = C y P = 0 (b) M C = (8 in.)F1 (15 in.)P = 0 (c) From Eq. (c), express F1 in terms of the unknown load P: 15 in. F1 = P = 1.875P (d) 8 in. Substitute this result into Eq. (a) to express Cx in terms of P: Cx = F1 = 1.875P The resultant reaction force at pin C can now be expressed as a function of P: 2 C = Cx2 + C y = (1.875P) 2 + ( P) 2 = 2.1250 P (e) Strut AB: The allowable normal stress for strut AB is 60 ksi allow = Y = = 30 ksi FS 2.0 Therefore, the allowable axial force for strut AB is Fallow,1 = allow A1 = (30 ksi)(0.75 in.2 ) = 22.50 kips From Eq. (d), the maximum load that may be applied to the rigid bar based on the limitations on the strut normal stress is F 22.50 kips = 12.0 kips (f) Pmax allow,1 = 1.875 1.875 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Pin C: The allowable shear stress for the pin at C is 54 ksi = 27 ksi allow = ult = FS 2.0 The 0.5-in.-diameter double shear pin at C has a shear area of 2 Dpin (2 = surfaces) (0.5 in.) 2 = 0.3927 in.2 4 4 thus, the allowable reaction force at C is C allow = allow AV = (27 ksi)(0.3927 in.2 ) = 10.6029 kips AV = (2 surfaces) From Eq. (e), the maximum load that may be applied to the rigid bar based on the limitations on the pin shear stress is C allow 10.6029 kips Pmax = = 4.9896 kips (g) 2.1250 2.1250 Maximum load P: Comparing the results in Eqs. (f) and (g), the maximum load P that may be applied to the rigid bar is Pmax = 4.9896 kips = 4.99 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. 4.12 In Fig. P4.12, rigid tee-beam ABC is supported at A by a single shear pin connection and at B by a strut, which consists of two 50 mm wide by 10 mm thick steel bars. The pins at A, B, and D are each 16 mm in diameter. The yield strength of the steel bars in strut (1) is 250 MPa, and the ultimate shear strength of each pin is 500 MPa. Determine the allowable load P that may be applied to the rigid bar at C if an overall factor of safety of 3.0 is required. Use L1 = 1.4 m and L2 = 1.7 m. Fig. P4.12 Solution From a FBD of the rigid beam ABC, the following equilibrium equations can be written: Fx = Ax = 0 (a) Fy = Ay F1 P = 0 (b) M A = L1 F1 ( L1 + L2 )P = 0 (c) From Eq. (c), express F1 in terms of the unknown load P: L + L2 1.4 m + 1.7 m F1 = 1 P= P = 2.214286 P L1 1.4 m Substitute this result into Eq. (b) to express Ay in terms of P: Ay = P + F1 = P + (2.214286 P) = 1.214286 P (d) The resultant reaction force at pin A can now be expressed as a function of P: 2 A = Ax2 + Ay = (0) 2 + (1.214286 P) 2 = 1.214286 P (e) The allowable normal stress for strut (1) is 250 MPa allow = Y = = 83.333333 MPa FS 3.0 Therefore, the allowable axial force for strut (1) is Fallow,1 = allow A1 = (83.333333 MPa)(2 50 mm 10 mm) = 83,333.333333 N From Eq. (d), the maximum load that may be applied to the rigid bar based on the strut normal stress limitation is Fallow,1 83,333.333333 N Pmax = = 37,634 N (f) 2.214286 2.214286 The allowable shear stress for the pins at A, B, and D is 500 MPa = 166.666667 MPa allow = ult = FS 3.0 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. The cross-sectional area of a 16-mm-diameter pin is Apin = 4 2 Dpin = 4 (16 mm) 2 = 201.061930 mm 2 The pin at A is a single shear connection; therefore, AV,A = Apin. The allowable reaction force at A is A allow = allow AV , A = (166.666667 MPa)(201.061930 mm 2 ) = 33,510 N From Eq. (e), the maximum load that may be applied to the rigid bar based on the limitations of pin A is A allow 33,510 N Pmax = = 27,596 N (g) 1.214286 1.214286 The pins at B and D are in double shear connections; therefore, AV,B = 2Apin. The allowable force in strut (1) based on the capacity of the pins at B and D is Fallow,1 = allow AV , B = (166.666667 MPa)(2)(201.061930 mm 2 ) = 67, 021 N From Eq. (d), the maximum load that may be applied to the rigid bar based on the limitations of the pins at B and D is Fallow,1 67, 021 N Pmax = = 30, 268 N (h) 2.214286 2.214286 Comparing the results in Eqs. (f), (g), and (h), the maximum load P that may be applied to the rigid bar is Pmax = 27,596 N = 27.6 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. 4.13 In Fig. P4.13, rigid bar ABC is supported at A by a single shear pin connection and at B by a strut, which consists of two 4 in. wide by 0.5 in. thick steel bars. The pins at A, B, and D each have a diameter of 1.0 in. The yield strength of the steel bars in strut (1) is 36 ksi, and the ultimate shear strength of each pin is 72 ksi. Determine the allowable load P that may be applied to the rigid bar at C if an overall factor of safety of 3.0 is required. Use L1 = 60 in. and L2 = 40 in. Fig. P4.13 Solution From a FBD of the rigid beam ABC, the following equilibrium equations can be written: Fx = Ax = 0 (a) Fy = Ay F1 P = 0 (b) M A = L1 F1 ( L1 + L2 )P = 0 (c) From Eq. (c), express F1 in terms of the unknown load P: L + L2 60 in. + 40 in. F1 = 1 P= P = 1.666667 P L1 60 in. Substitute this result into Eq. (b) to express Ay in terms of P: Ay = P + F1 = P + (1.666667 P ) = 0.666667 P (d) The resultant reaction force at pin A can now be expressed as a function of P: 2 A = Ax2 + Ay = (0) 2 + (0.666667 P ) 2 = 0.666667 P (e) Strut (1): The allowable normal stress for strut (1) is 36 ksi allow = Y = = 12.0 ksi FS 3.0 Therefore, the allowable axial force for strut (1) is Fallow,1 = allow A1 = (12.0 ksi)(2 4 in. 0.5 in.) = 48.0 kips From Eq. (d), the maximum load that may be applied to the rigid bar based on the strut normal stress limitation is Fallow,1 48.0 kips Pmax = = 28.80 kips (f) 1.666667 1.666667 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Pins: The allowable shear stress for the pins at A, B, and D is 72 ksi = 24.0 ksi allow = ult = FS 3.0 The cross-sectional area of a 1-in.-diameter pin is Apin = 4 2 Dpin = 4 (1 in.) 2 = 0.785398 in.2 Pin A: The pin at A is a single shear connection; therefore, AV,A = Apin. The allowable reaction force at A is A allow = allow AV , A = (24.0 ksi)(0.785398 in.2 ) = 18.849552 kips From Eq. (e), the maximum load that may be applied to the rigid bar based on the limitations of pin A is A allow 18.849552 kips Pmax = = 28.274314 kips (g) 0.666667 0.666667 Pin B and D: The pins at B and D are in double shear connections; therefore, AV,B = 2Apin. The allowable force in strut (1) based on the capacity of the pins at B and D is Fallow,1 = allow AV , B = (24.0 ksi)(2)(0.785398 in.2 ) = 37.699104 kips From Eq. (d), the maximum load that may be applied to the rigid bar based on the limitations of the pins at B and D is Fallow,1 37.699104 kips Pmax = = 22.619458 kips (h) 1.666667 1.666667 Maximum load P: Comparing the results in Eqs. (f), (g), and (h), the maximum load P that may be applied to the rigid bar is Pmax = 22.619458 kips = 22.6 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. 4.14 In Fig. P4.14, davit ABD is supported at A by a single shear pin connection and at B by a tie rod (1). The 1-in.-diameter pins at A, B, and C each have an ultimate shear strength of 72 ksi. Tie rod (1) has a cross-sectional area of 1.25 in.2 and a yield strength of 50 ksi. Determine the allowable load P that may be applied to the davit at D if an overall factor of safety of 3.0 is required. Fig. P4.14 Solution Member (1) is a two-force member that is oriented at with respect to the horizontal axis: 7 ft = 0.777778 = 37.875 tan = 9 ft From a FBD of rigid structure ABD, the following equilibrium equations can be written: Fx = P cos 50 F1 cos 37.875 + Ax = 0 (a) Fy = P sin 50 F1 sin 37.875 + Ay = 0 (b) M A = ( F1 cos 37.875)(7 ft) ( P cos 50)(10 ft) ( P sin 50)(6 ft) = 0 (c) From Eq. (c): (10 ft) cos 50 + (6 ft)sin 50 F1 = P = 1.995152 P (7 ft) cos 37.875 Substitute F1 into Eq. (a) to obtain an expression for Ax in terms of the unknown load P: Ax = F1 cos 37.875 P cos 50 = (1.995152 P) cos 37.875 P cos 50 = 0.932090 P and substitute F1 into Eq. (b) to obtain an expression for Ay in terms of the unknown load P: Ay = F1 sin 37.875 + P sin 50 = (1.995152 P) sin 37.875 + P sin 50 = 1.990950 P The resultant pin force at A is found from Ax and Ay: 2 A = Ax2 + Ay = (0.932090 P) 2 + (1.990950 P) 2 = 2.198334 P (d) (e) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Tie rod (1): The allowable normal stress for tie rod (1) is 50 ksi allow = Y = = 16.666667 ksi FS 3.0 Therefore, the allowable axial force for tie rod (1) is Fallow,1 = allow A1 = (16.666667 ksi)(1.25 in.2 ) = 20.833334 kips From Eq. (d), the maximum load that may be applied to the davit based on the strut normal stress limitation is Fallow,1 20.833334 kips Pmax = = 10.441978 kips (f) 1.995152 1.995152 Pins: The allowable shear stress for the pins at A, B, and C is 72 ksi = 24.0 ksi allow = ult = FS 3.0 The cross-sectional area of a 1-in.-diameter pin is Apin = 4 2 Dpin = 4 (1 in.) 2 = 0.785398 in.2 Pin A: The pin at A is a single shear connection; therefore, AV,A = Apin. The allowable reaction force at A is A allow = allow AV , A = (24.0 ksi)(0.785398 in.2 ) = 18.849552 kips From Eq. (e), the maximum load that may be applied to the davit based on the limitations of pin A is A allow 18.849552 kips Pmax = = 8.574471 kips (g) 2.198334 2.198334 Pins B and C: The pins at B and C are in double shear connections; therefore, AV,B = 2Apin. The allowable force in strut (1) based on the capacity of the pins at B and C is Fallow,1 = allow AV , B = (24.0 ksi)(2)(0.785398 in.2 ) = 37.699104 kips From Eq. (d), the maximum load that may be applied to the rigid bar based on the limitations of the pins at B and D is Fallow,1 37.699104 kips Pmax = = 18.895354 kips (h) 1.995152 1.995152 Maximum load P: After comparing the results in Eqs. (f), (g), and (h), the maximum load P that may be applied to the rigid bar is identified as Pmax = 8.574471 kips = 8.57 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. 4.15 In Fig. P4.15, the rigid member ABDE is supported at A by a single shear pin connection and at B by a tie rod (1). The tie rod is attached at B and C with double shear pin connections. The pins at A, B, and C each have an ultimate shear strength of 80 ksi, and tie rod (1) has a yield strength of 60 ksi. A concentrated load of P = 24 kips is applied perpendicular to DE, as shown. A factor of safety of 2.0 is required for all components. Determine: (a) the minimum required diameter for the tie rod. (b) the minimum required diameter for the pin at B. (c) the minimum required diameter for the pin at A. Fig. P4.15 Solution Member (1) is a two-force member that is oriented at with respect to the horizontal axis: 9 ft tan = = 1.5 = 56.310 6 ft From a FBD of rigid structure ABDE, the following equilibrium equations can be written: Fx = P cos 70 F1 cos56.310 + Ax = 0 (a) Fy = P sin 70 F1 sin 56.310 + Ay = 0 (b) M A = ( F1 cos 56.310)(9 ft) ( P cos 70)[13 ft + (8 ft) sin 20] ( P sin 70)[(8 ft) cos 20] = 0 From Eq. (c): v (c) (d) Substitute F1 into Eq. (a) to obtain Ax: Ax = F1 cos 56.310 P cos 70 = (59.834286 kips) cos 56.310 (24 kips) cos 70 = 24.981548 kips and substitute F1 into Eq. (b) to obtain Ay: Ay = F1 sin 56.310 + P sin 70 = (59.834286 kips) sin 56.310 + (24 kips)sin 70 = 72.337797 kips Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. The resultant pin force at A is found from Ax and Ay: 2 A = Ax2 + Ay = (24.981548 kips) 2 + (72.337797 kips) 2 = 76.529959 kips (e) (a) The allowable normal stress for tie rod (1) is 60 ksi allow = Y = = 30.0 ksi FS 2.0 The minimum cross-sectional area required for the tie rod is F 59.834286 kips Amin 1 = = 1.994476 in.2 allow 30.0 ksi Therefore, the minimum tie rod diameter is 4 D12 1.994476 in.2 D1 1.593564 in. = 1.594 in. Ans. (b) The allowable shear stress for the pins at A, B, and C is 80 ksi = 40.0 ksi allow = ult = FS 2.0 The double-shear pin connection at B and C must support a load of F1 = 59.834286 kips . The shear area AV required for these pins is F 59.834286 kips AV 1 = = 1.495857 in.2 allow 40.0 ksi The pin diameter can be computed from (2 surfaces) 4 2 Dpin 1.495857 in.2 Dpin 0.975855 in. = 0.976 in. Ans. (c) The pin at A is a single shear connection; therefore, AV = Apin. The shear area AV required for this pin is A 76.529959 kips AV = = 1.913249 in.2 40.0 ksi allow The pin diameter can be computed from (1 surface) 4 2 Dpin 1.913249 in.2 Dpin 1.560777 in. = 1.561 in. Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 4.16 Rigid bar ABC is supported at B by inclined steel strut (1) and at C by a 1.25-in.-diameter pin in a single shear connection (Fig P4.16). The strut has a cross-sectional area of A1 = 2.250 in.2, and it is connected at both ends by double shear pin connections. The yield strength of the steel strut is 36 ksi. The ultimate strength of each pin is 84 ksi. A concentrated load of P = 15 kips is applied to the rigid bar at A. Determine: (a) the factor of safety in steel strut (1) with respect to the 36-ksi yield strength. (b) the factor of safety in the pin at C with respect to the 84-ksi ultimate strength. Fig. P4.16 Solution Member (1) is a two-force member that is oriented at with respect to the horizontal axis: 40 in. tan = = 0.8 = 38.660 50 in. From a FBD of rigid bar ABC, the following equilibrium equations can be written: Fx = F1 cos 38.660 + Cx = 0 (a) Fy = 15 kips F1 sin 38.660 + C y = 0 (b) M C = (80 in.)(15 kips) + ( F1 sin 38.660)(50 in.) = 0 (c) From Eq. (c): (80 in.)(15 kips) F1 = = 38.418585 kips (50 in.) sin 38.660 Substitute F1 into Eq. (a) to obtain Cx: Cx = F1 cos 38.660 = (38.418585 kips) cos 38.660 = 30.0 kips and substitute F1 into Eq. (b) to obtain Cy: C y = 15 kips + F1 sin 38.660 (d) = 15 kips + (38.418585 kips) sin 38.660 = 9.0 kips The resultant pin force at C is found from Cx and Cy: 2 C = Cx2 + C y = (30.0 kips) 2 + (9.0 kips) 2 = 31.320920 kips (e) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. (a) The normal stress magnitude in member (1) is F 38.418585 kips 1 = 1 = = 17.074927 ksi 2.250 in.2 A1 The factor of safety with respect to the 36-ksi yield strength is 36 ksi = 2.11 FS1 = Y = 1 17.074927 ksi Ans. (b) The cross-sectional area of the 1.25-in.-diameter pin at C is Apin = (1.25 in.) 2 = 1.227185 in.2 4 Since pin C is used in a single shear connection, the shear area AV is equal to the pin cross-sectional area. Therefore, the shear stress in pin C is: C 31.320920 kips C = = = 25.522582 ksi 1.227185 in.2 AV The factor of safety with respect to the 84-ksi ultimate strength is 84 ksi = 3.29 Ans. FSC = ult = C 25.522582 ksi Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 4.17 Rigid bar ABC is supported by pin-connected axial member (1) and by a pin connection at C as shown in Fig. P4.17. A 6,300-lb concentrated load is applied to the rigid bar at A. Member (1) is a 2.75 in. wide by 1.25 in. thick rectangular bar made of steel with a yield strength of Y = 36,000 psi. The pin at C has an ultimate shear strength of ult = 60,000 psi. (a) Determine the axial force in member (1). (b) Determine the factor of safety in member (1) with respect to its yield strength. (c) Determine the magnitude of the resultant reaction force acting at pin C. (d) If a minimum factor of safety of FS = 3.0 with respect to the ultimate shear strength is required, determine the minimum diameter that may be used for the pin at C. Fig. P4.17 Solution Member (1) is a two-force member that is oriented at with respect to the horizontal axis: 30 in. tan = = 0.75 = 36.870 40 in. From a FBD of rigid structure ABC, the following equilibrium equations can be written: Fx = F1 cos 36.870 + Cx = 0 (a) Fy = 6,300 lb F1 sin 36.870 + C y = 0 (b) M C = (6,300 lb)(80 in.) (c) +( F1 sin 36.870)(56 in.) = 0 (a) From Eq. (c), the axial force in member (1) is: (6,300 lb)(80 in.) F1 = = 15, 000 lb = 15, 000 lb (C) (56 in.) sin 36.870 (b) The normal stress magnitude in member (1) is F 15, 000 lb 1 = 1 = = 4,363.6364 psi A1 (2.75 in.)(1.25 in.) Therefore, its factor of safety with respect to the 36,000 psi yield stress is 36, 000 psi = 8.25 FS1 = Y = 1 4,363.6364 psi Ans. Ans. (c) Substitute F1 into Eq. (a) to obtain Cx: Cx = F1 cos 36.870 = (15, 000 lb) cos 36.870 = 12, 000 lb Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. and substitute F1 into Eq. (b) to obtain an expression for Ay in terms of the unknown load P: C y = 6,300 lb + F1 sin 36.870 = 6,300 lb + (15, 000 lb) sin 36.870 = 2, 700 lb The resultant pin force at C is found from Cx and Cy: 2 C = Cx2 + C y = (12, 000 lb) 2 + (2, 700 lb) 2 = 12,300 lb Ans. (d) The allowable shear stress for the pins at C is 60, 000 psi = 20, 000 psi allow = ult = FS 3.0 The double-shear pin connection at C must support a load of 12,300 lb. The minimum shear area AV required to support this load is C 12,300 lb AV = = 0.6150 in.2 allow 20,000 psi The pin diameter can be computed from (2 surfaces) 4 2 Dpin 0.6150 in.2 Dpin 0.625717 in. = 0.626 in. Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 4.18 A rectangular steel plate is used as an axial member to support a dead load of 70 kips and a live load of 110 kips. The yield strength of the steel is 50 ksi. (a) Use the ASD method to determine the minimum cross-sectional area required for the axial member if a factor of safety of 1.67 with respect to yielding is required. (b) Use the LRFD method to determine the minimum cross-sectional area required for the axial member based on yielding of the gross section using the LRFD method. Use a resistance factor of t = 0.9 and load factors of 1.2 and 1.6 for the dead and live loads, respectively. Solution (a) The service load on the axial member is P = D + L = 70 kips + 110 kips = 180 kips The allowable normal stress is 50 ksi allow = Y = = 29.940 ksi FS 1.67 The minimum cross-sectional area required to support the service load is P 180 kips Amin = = 6.01 in.2 allow 29.940 ksi (b) The ultimate load for LRFD is U = 1.2 D + 1.6 L = 1.2(70 kips) + 1.6(110 kips) = 260 kips The design equation for an axial member subjected to tension can be written in LRFD as t Y A U Consequently, the minimum cross-sectional area required for the tension member is U 260 kips Amin = = 5.78 in.2 t Y 0.9(50 ksi) Ans. Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 4.19 A 20-mm-thick steel plate will be used as an axial member to support a dead load of 150 kN and a live load of 220 kN. The yield strength of the steel is 250 MPa. (a) Use the ASD method to determine the minimum plate width b required for the axial member if a factor of safety of 1.67 with respect to yielding is required. (b) Use the LRFD method to determine the minimum plate width b required for the axial member based on yielding of the gross section using the LRFD method. Use a resistance factor of t = 0.9 and load factors of 1.2 and 1.6 for the dead and live loads, respectively. Solution (a) The service load on the axial member is P = D + L = 150 kN + 220 kN = 370 kN The allowable normal stress is 250 MPa allow = Y = = 149.70 MPa FS 1.67 The minimum cross-sectional area required to support the service load is P (370 kN)(1,000 N/kN) Amin = = 2, 471.61 mm 2 2 149.70 N/mm allow Since the plate is 20-mm-thick, the minimum plate width is therefore: A 2, 471.61 mm 2 bmin min = = 123.6 mm t 20 mm (b) The ultimate load for LRFD is U = 1.2 D + 1.6 L = 1.2(150 kN) + 1.6(220 kN) = 532 kN The design equation for an axial member subjected to tension can be written in LRFD as t Y A U Consequently, the minimum cross-sectional area required for the tension member is U (532 kN)(1,000 N/kN) Amin = = 2,364.44 mm 2 2 0.9(250 N/mm ) t Y Since the plate is 20-mm-thick, the minimum plate width is therefore: A 2,364.44 mm 2 bmin min = = 118.2 mm t 20 mm Ans. Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 4.20 A round steel tie rod is used as a tension member to support a dead load of 30 kips and a live load of 15 kips. The yield strength of the steel is 46 ksi. (a) Use the ASD method to determine the minimum diameter D required for the tie rod if a factor of safety of 2.0 with respect to yielding is required. (b) Use the LRFD method to determine the minimum diameter D required for the tie rod based on yielding of the gross section using the LRFD method. Use a resistance factor of t = 0.9 and load factors of 1.2 and 1.6 for the dead and live loads, respectively. Solution (a) The service load on the axial member is P = D + L = 30 kips + 15 kips = 45 kips The allowable normal stress is 46 ksi allow = Y = = 23.0 ksi FS 2.0 The minimum cross-sectional area required to support the service load is P 45 kips Amin = = 1.956522 in.2 allow 23.0 ksi and thus, the minimum tie rod diameter D is 4 2 Dmin 1.956522 in.2 Dmin 1.578 in. Ans. (b) The ultimate load for LRFD is U = 1.2 D + 1.6 L = 1.2(30 kips) + 1.6(15 kips) = 60 kips The design equation for an axial member subjected to tension can be written in LRFD as t Y A U Consequently, the minimum cross-sectional area required for the tension member is U 60 kips Amin = = 1.449275 in.2 t Y 0.9(46 ksi) and thus, the minimum tie rod diameter D is 4 2 Dmin 1.449275 in.2 Dmin 1.358 in. Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 4.21 A round steel tie rod is used as a tension member to support a dead load of 190 kN and a live load of 220 kN. The yield strength of the steel is 320 MPa. (a) Use the ASD method to determine the minimum diameter D required for the tie rod if a factor of safety of 2.0 with respect to yielding is required. (b) Use the LRFD method to determine the minimum diameter D required for the tie rod based on yielding of the gross section using the LRFD method. Use a resistance factor of t = 0.9 and load factors of 1.2 and 1.6 for the dead and live loads, respectively. Solution (a) The service load on the axial member is P = D + L = 190 kN + 220 kN = 410 kN The allowable normal stress is 320 MPa allow = Y = = 160 MPa FS 2.0 The minimum cross-sectional area required to support the service load is P (410 kN)(1,000 N/kN) Amin = = 2,562.500 mm 2 2 160 N/mm allow and thus, the minimum tie rod diameter D is 4 2 Dmin 2,562.500 mm 2 Dmin 57.1 mm Ans. (b) The ultimate load for LRFD is U = 1.2 D + 1.6 L = 1.2(190 kN) + 1.6(220 kN) = 580 kN The design equation for an axial member subjected to tension can be written in LRFD as t Y A U Consequently, the minimum cross-sectional area required for the tension member is U (580 kN)(1,000 N/kN) Amin = = 2,013.889 mm 2 2 0.9(320 N/mm ) t Y and thus, the minimum tie rod diameter D is 4 2 Dmin 2, 013.889 mm 2 Dmin 50.6 mm Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
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