CHAPTER 6 PROBLEM 6.1 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Free body: Entire truss: ΣFy = 0: By = 0 By = 0 ΣM C = 0: − Bx (3.2 m) − (48 kN)(7.2 m) = 0 Bx = −108 kN B x = 108 kN ΣFx = 0: C − 108 kN + 48 kN = 0 C = 60 kN C = 60 kN Free body: Joint B: FAB FBC 108 kN = = 5 4 3 FAB = 180.0 kN T FBC = 144.0 kN T Free body: Joint C: FAC FBC 60 kN = = 13 12 5 FAC = 156.0 kN C FBC = 144 kN (checks) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 743 PROBLEM 6.2 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Reactions: ΣM A = 0: C = 1260 lb ΣFx = 0: A x = 0 ΣFy = 0: A y = 960 lb Joint B: FAB FBC 300 lb = = 12 13 5 FAB = 720 lb T FBC = 780 lb C Joint A: ΣFy = 0: − 960 lb − 4 FAC = 0 5 FAC = 1200 lb FAC = 1200 lb C 3 ΣFx = 0: 720 lb − (1200 lb) = 0 (checks) 5 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 744 PROBLEM 6.3 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION AB = 32 + 1.252 = 3.25 m BC = 32 + 42 = 5 m Reactions: ΣM A = 0: (84 kN)(3 m) − C (5.25 m) = 0 C = 48 kN ΣFx = 0: Ax − C = 0 A x = 48 kN ΣFy = 0: Ay = 84 kN = 0 A y = 84 kN Joint A: ΣFx = 0: 48 kN − 12 FAB = 0 13 FAB = +52 kN ΣFy = 0: 84 kN − 5 (52 kN) − FAC = 0 13 FAC = +64.0 kN FAB = 52.0 kN T FAC = 64.0 kN T Joint C: FBC 48 kN = 5 3 FBC = 80.0 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 745 PROBLEM 6.4 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: C = D = 600 lb From the symmetry of the truss and loading, we find Free body: Joint B: FAB 5 = FBC 300 lb = 2 1 FAB = 671 lb T FBC = 600 lb C Free body: Joint C: ΣFy = 0: 3 FAC + 600 lb = 0 5 FAC = −1000 lb ΣFx = 0: From symmetry: 4 ( −1000 lb) + 600 lb + FCD = 0 5 FAC = 1000 lb C FCD = 200 lb T FAD = FAC = 1000 lb C , FAE = FAB = 671 lb T , FDE = FBC = 600 lb C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 746 PROBLEM 6.5 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Reactions: ΣM D = 0: Fy (24) − (4 + 2.4)(12) − (1)(24) = 0 Fy = 4.2 kips ΣFx = 0: Fx = 0 ΣFy = 0: D − (1 + 4 + 1 + 2.4) + 4.2 = 0 D = 4.2 kips Joint A: FAB = 0 ΣFx = 0: FAB = 0 ΣFy = 0 : −1 − FAD = 0 FAD = −1 kip Joint D: ΣFy = 0: − 1 + 4.2 + ΣFx = 0: 8 FBD = 0 17 FBD = −6.8 kips 15 (−6.8) + FDE = 0 17 FDE = +6 kips FAD = 1.000 kip C FBD = 6.80 kips C FDE = 6.00 kips T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 747 PROBLEM 6.5 (Continued) Joint E: ΣFy = 0 : FBE − 2.4 = 0 FBE = +2.4 kips FBE = 2.40 kips T Truss and loading symmetrical about cL. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 748 PROBLEM 6.6 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION AD = 52 + 122 = 13 ft BCD = 122 + 162 = 20 ft Reactions: ΣFx = 0: Dx = 0 ΣM E = 0: D y (21 ft) − (693 lb)(5 ft) = 0 ΣFy = 0: 165 lb − 693 lb + E = 0 Joint D: D y = 165 lb E = 528 lb ΣFx = 0: 5 4 FAD + FDC = 0 13 5 (1) ΣFy = 0: 12 3 FAD + FDC + 165 lb = 0 13 5 (2) Solving Eqs. (1) and (2) simultaneously, FAD = −260 lb FDC = +125 lb FAD = 260 lb C FDC = 125 lb T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 749 PROBLEM 6.6 (Continued) Joint E: ΣFx = 0: 5 4 FBE + FCE = 0 13 5 (3) ΣFy = 0: 12 3 FBE + FCE + 528 lb = 0 13 5 (4) Solving Eqs. (3) and (4) simultaneously, FBE = 832 lb C FBE = −832 lb FCE = 400 lb T FCE = +400 lb Joint C: Force polygon is a parallelogram (see Fig. 6.11, p. 209). FAC = 400 lb T FBC = 125.0 lb T Joint A: ΣFx = 0: 5 4 (260 lb) + (400 lb) + FAB = 0 13 5 FAB = −420 lb ΣFy = 0: FAB = 420 lb C 12 3 (260 lb) − (400 lb) = 0 13 5 0 = 0 (Checks) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 750 PROBLEM 6.7 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Free body: Entire truss: ΣFx = 0: Cx + 2(5 kN) = 0 Cx = −10 kN C x = 10 kN ΣM C = 0: D(2 m) − (5 kN)(8 m) − (5 kN)(4 m) = 0 D = +30 kN D = 30 kN ΣFy = 0: C y + 30 kN = 0 C y = −30 kN C y = 30 kN Free body: Joint A: FAB FAD 5 kN = = 4 1 17 FAB = 20.0 kN T FAD = 20.6 kN C Free body: Joint B: ΣFx = 0: 5 kN + 1 5 FBD = 0 FBD = −5 5 kN ΣFy = 0: 20 kN − FBC − 2 5 (−5 5 kN) = 0 FBC = +30 kN FBD = 11.18 kN C FBC = 30.0 kN T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 751 PROBLEM 6.7 (Continued) Free body: Joint C: ΣFx = 0: FCD − 10 kN = 0 FCD = +10 kN FCD = 10.00 kN T ΣFy = 0: 30 kN − 30 kN = 0 (checks) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 752 PROBLEM 6.8 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Reactions: ΣM C = 0: A x = 16 kN ΣFy = 0: A y = 9 kN ΣFx = 0: C = 16 kN Joint E: FBE FDE 3 kN = = 5 4 3 FBE = 5.00 kN T FDE = 4.00 kN C Joint B: ΣFx = 0: 4 (5 kN) − FAB = 0 5 FAB = +4 kN 3 ΣFy = 0: − 6 kN − (5 kN) − FBD = 0 5 FBD = −9 kN FAB = 4.00 kN T FBD = 9.00 kN C Joint D: ΣFy = 0: − 9 kN + 3 FAD = 0 5 FAD = +15 kN 4 ΣFx = 0: − 4 kN − (15 kN) − FCD = 0 5 FCD = −16 kN FAD = 15.00 kN T FCD = 16.00 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 753 PROBLEM 6.9 Determine the force in each member of the Gambrel roof truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: ΣFx = 0: H x = 0 Because of the symmetry of the truss and loading, A = Hy = 1 total load 2 A = H y = 1200 lb Free body: Joint A: FAB = 1500 lb C FAB FAC 900 lb = = 5 4 3 FAC = 1200 lb T Free body: Joint C: BC is a zero-force member. FCE = 1200 lb T FBC = 0 Free body: Joint B: ΣFx = 0: 24 4 4 FBD + FBE + (1500 lb) = 0 25 5 5 24 FBD + 20 FBE = −30, 000 lb or ΣFy = 0: (1) 7 3 3 FBD − FBE + (1500) − 600 = 0 25 5 5 7 FBD − 15FBE = −7,500 lb or (2) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 754 PROBLEM 6.9 (Continued) Multiply Eq. (1) by 3, Eq. (2) by 4, and add: FBD = 1200 lb C 100 FBD = −120, 000 lb Multiply Eq. (1) by 7, Eq. (2) by –24, and add: FBE = 60.0 lb C 500 FBE = −30,000 lb Free body: Joint D: 24 24 FDF = 0 (1200 lb) + 25 25 ΣFx = 0: FDF = −1200 lb FDF = 1200 lb C 7 7 (1200 lb) − (−1200 lb) − 600 lb − FDE = 0 25 25 ΣFy = 0: FDE = 72.0 lb FDE = 72.0 lb T Because of the symmetry of the truss and loading, we deduce that FEF = FBE FEF = 60.0 lb C FEG = FCE FEG = 1200 lb T FFG = FBC FFG = 0 FFH = FAB FFH = 1500 lb C FGH = FAC FGH = 1200 lb T Note: Compare results with those of Problem 6.11. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 755 PROBLEM 6.10 Determine the force in each member of the Howe roof truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: ΣFx = 0: H x = 0 Because of the symmetry of the truss and loading, A = Hy = 1 total load 2 A = H y = 1200 lb Free body: Joint A: FAB FAC 900 lb = = 5 4 3 FAB = 1500 lb C FAC = 1200 lb T Free body: Joint C: BC is a zero-force member. FCE = 1200 lb T FBC = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 756 PROBLEM 6.10 (Continued) Free body: Joint B: ΣFx = 0: or FBD + FBE = −1500 lb ΣFy = 0: or 4 4 4 FBD + FBC + (1500 lb) = 0 5 5 5 (1) 3 3 3 FBD − FBE + (1500 lb) − 600 lb = 0 5 5 5 FBD − FBE = −500 lb Add Eqs. (1) and (2): 2 FBD = −2000 lb Subtract Eq. (2) from Eq. (1): 2 FBE = −1000 lb (2) FBD = 1000 lb C FBE = 500 lb C Free Body: Joint D: 4 4 (1000 lb) + FDF = 0 5 5 ΣFx = 0: FDF = −1000 lb FDF = 1000 lb C 3 3 (1000 lb) − ( −1000 lb) − 600 lb − FDE = 0 5 5 ΣFy = 0: FDE = +600 lb Because of the symmetry of the truss and loading, we deduce that FEF = FBE FEG = FCE FFG = FBC FFH = FAB FGH = FAC FDE = 600 lb T FEF = 500 lb C FEG = 1200 lb T FFG = 0 FFH = 1500 lb C FGH = 1200 lb T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 757 PROBLEM 6.11 Determine the force in each member of the Pratt roof truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: ΣFx = 0: Ax = 0 Due to symmetry of truss and load, Ay = H = 1 total load = 21 kN 2 Free body: Joint A: FAB FAC 15.3 kN = = 37 35 12 FAB = 47.175 kN FAC = 44.625 kN FAB = 47.2 kN C FBD = 47.175 kN, FBC = 10.5 kN FBC = 10.50 kN C FAC = 44.6 kN T Free body: Joint B: From force polygon: FBD = 47.2 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 758 PROBLEM 6.11 (Continued) Free body: Joint C: ΣFy = 0: FCD = 17.50 kN T 3 FCD − 10.5 = 0 5 ΣFx = 0: FCE + 4 (17.50) − 44.625 = 0 5 FCE = 30.625 kN Free body: Joint E: DE is a zero-force member. FCE = 30.6 kN T FDE = 0 Truss and loading symmetrical about cL . PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 759 PROBLEM 6.12 Determine the force in each member of the Fink roof truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: ΣFx = 0: A x = 0 Because of the symmetry of the truss and loading, Ay = G = 1 total load 2 A y = G = 6.00 kN Free body: Joint A: F FAB 4.50 kN = AC = 2.462 2.25 1 FAC = 10.125 kN FAB = 11.08 kN C FAC = 10.13 kN T Free body: Joint B: ΣFx = 0: 3 2.25 2.25 FBC + FBD + (11.08 kN) = 0 5 2.462 2.462 (1) F 4 11.08 kN FBC + BD + − 3 kN = 0 5 2.462 2.462 (2) ΣFy = 0: − Multiply Eq. (2) by –2.25 and add to Eq. (1): 12 FBC + 6.75 kN = 0 5 FBC = −2.8125 FBC = 2.81 kN C Multiply Eq. (1) by 4, Eq. (2) by 3, and add: 12 12 FBD + (11.08 kN) − 9 kN = 0 2.462 2.462 FBD = −9.2335 kN FBD = 9.23 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 760 PROBLEM 6.12 (Continued) Free body: Joint C: ΣFy = 0: 4 4 FCD − (2.8125 kN) = 0 5 5 FCD = 2.8125 kN, FCD = 2.81 kN T 3 3 ΣFx = 0: FCE − 10.125 kN + (2.8125 kN) + (2.8125 kN) = 0 5 5 FCE = +6.7500 kN Because of the symmetry of the truss and loading, we deduce that FDE = FCD FDF = FBD FEF = FBC FEG = FAC FFG = FAB FCE = 6.75 kN T FCD = 2.81 kN T FDF = 9.23 kN C FEF = 2.81 kN C FEG = 10.13 kN T FFG = 11.08 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 761 PROBLEM 6.13 Using the method of joints, determine the force in each member of the double-pitch roof truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: ΣM A = 0: H (18 m) − (2 kN)(4 m) − (2 kN)(8 m) − (1.75 kN)(12 m) − (1.5 kN)(15 m) − (0.75 kN)(18 m) = 0 H = 4.50 kN ΣFx = 0: Ax = 0 ΣFy = 0: Ay + H − 9 = 0 Ay = 9 − 4.50 A y = 4.50 kN Free body: Joint A: FAB FAC 3.50 kN = 2 1 5 FAB = 7.8262 kN C = FAB = 7.83 kN C FAC = 7.00 kN T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 762 PROBLEM 6.13 (Continued) Free body: Joint B: ΣFx = 0: 2 5 FBD + 2 5 (7.8262 kN) + 1 2 FBC = 0 FBD + 0.79057 FBC = −7.8262 kN or ΣFy = 0: 1 5 FBD + 1 5 (7.8262 kN) − 1 2 (1) FBC − 2 kN = 0 FBD − 1.58114 FBC = −3.3541 or (2) Multiply Eq. (1) by 2 and add Eq. (2): 3FBD = −19.0065 FBD = 6.34 kN C FBD = −6.3355 kN Subtract Eq. (2) from Eq. (1): 2.37111FBC = −4.4721 FBC = 1.886 kN C FBC = −1.8861 kN Free body: Joint C: 2 ΣFy = 0: 5 FCD − 1 2 (1.8861 kN) = 0 FCD = 1.491 kN T FCD = +1.4911 kN ΣFx = 0: FCE − 7.00 kN + FCE 1 (1.8861 kN) + 2 = +5.000 kN 1 5 (1.4911 kN) = 0 FCE = 5.00 kN T Free body: Joint D: ΣFx = 0: 2 5 FDF + 1 2 FDE + 2 5 (6.3355 kN) − 1 5 (1.4911 kN) = 0 FDF + 0.79057 FDE = −5.5900 kN or ΣFy = 0: or 1 5 FDF − 1 2 FDE + 1 5 (6.3355 kN) − (1) 2 5 (1.4911 kN) − 2 kN = 0 FDF − 0.79057 FDE = −1.1188 kN (2) 2 FDF = −6.7088 kN FDF = 3.35 kN C Add Eqs. (1) and (2): FDF = −3.3544 kN Subtract Eq. (2) from Eq. (1): 1.58114 FDE = −4.4712 kN FDE = −2.8278 kN FDE = 2.83 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 763 PROBLEM 6.13 (Continued) Free body: Joint F: ΣFx = 0: 1 FFG + 2 2 (3.3544 kN) = 0 5 FFG = 4.24 kN C FFG = −4.243 kN ΣFy = 0: −FEF − 1.75 kN + 1 5 (3.3544 kN) − 1 2 FEF = 2.750 kN (−4.243 kN) = 0 FEF = 2.75 kN T Free body: Joint G: ΣFx = 0: 1 FGH − 2 ΣFy = 0: − 1 2 1 2 (4.243 kN) = 0 FGH − 1 2 FEG − 1 2 (1) (4.243 kN) − 1.5 kN = 0 FGH + FEG = −6.364 kN or 2 FGH = −10.607 (2) FGH = 5.30 kN C FGH = −5.303 Subtract Eq. (1) from Eq. (2): 2 FEG + FGH − FEG = −4.243 kN or Add Eqs. (1) and (2): 1 2 FEG = −2.121 kN FEG = −1.0605 kN FEG = 1.061 kN C Free body: Joint H: FEH = 3.75 kN T FEH 3.75 kN = 1 1 We can also write FGH 2 = 3.75 kN 1 FGH = 5.30 kN C (Checks) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 764 PROBLEM 6.14 The truss shown is one of several supporting an advertising panel. Determine the force in each member of the truss for a wind load equivalent to the two forces shown. State whether each member is in tension or compression. SOLUTION Free body: Entire truss: ΣM F = 0: (800 N)(7.5 m) + (800 N)(3.75 m) − A(2 m) = 0 A = +2250 N A = 2250 N ΣFy = 0: 2250 N + Fy = 0 Fy = −2250 N Fy = 2250 N ΣFx = 0: −800 N − 800 N + Fx = 0 Fx = +1600 N Fx = 1600 N Joint D: 800 N FDE FBD = = 8 15 17 FBD = 1700 N C FDE = 1500 N T Joint A: 2250 N FAB FAC = = 15 17 8 FAB = 2250 N C FAC = 1200 N T Joint F: ΣFx = 0: 1600 N − FCF = 0 FCF = +1600 N ΣFy = 0: FEF − 2250 N = 0 FEF = +2250 N FCF = 1600 N T FEF = 2250 N T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 765 PROBLEM 6.14 (Continued) Joint C: ΣFx = 0: 8 FCE − 1200 N + 1600 N = 0 17 FCE = −850 N ΣFy = 0: FBC + 15 FCE = 0 17 FBC = − 15 15 FCE = − ( −850 N) 17 17 FBC = +750 N Joint E: FCE = 850 N C ΣFx = 0: − FBE − 800 N + 8 (850 N) = 0 17 FBE = −400 N ΣFy = 0: 1500 N − 2250 N + FBC = 750 N T FBE = 400 N C 15 (850 N) = 0 17 0 = 0 (checks) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 766 PROBLEM 6.15 Determine the force in each of the members located to the left of line FGH for the studio roof truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: ΣFx = 0: A x = 0 Because of symmetry of loading, Ay = L = 1 total load 2 A y = L = 1200 lb Zero-Force Members: Examining joints C and H, we conclude that BC, EH, and GH are zero-force members. Thus, FBC = FEH = 0 Also, FCE = FAC (1) Free body: Joint A: FAB FAC 1000 lb = 2 1 5 FAB = 2236 lb C = FAB = 2240 lb C FAC = 2000 lb T FCE = 2000 lb T From Eq. (1): Free body: Joint B: ΣFx = 0: 2 5 FBD + 2 5 FBE + 2 5 (2236 lb) = 0 FBD + FBE = −2236 lb or ΣFy = 0: 1 5 FBD − 1 5 FBE + 1 FBD − FBE = −1342 lb or (2) 5 (2236 lb) − 400 lb = 0 (3) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 767 PROBLEM 6.15 (Continued) FBD = 1789 lb C 2 FBD = −3578 lb Add Eqs. (2) and (3): FBE = 447 lb C Subtract Eq. (3) from Eq. (1): 2 FBE = − 894 lb Free body: Joint E: ΣFx = 0: 2 5 FEG + ΣFy = 0: FDE + 1 5 2 5 (447 lb) − 2000 lb = 0 FEG = 1789 lb T (1789 lb) − 1 5 FDE = − 600 lb (447 lb) = 0 FDE = 600 lb C Free body: Joint D: ΣFx = 0: 2 5 FDF + 2 5 FDG + 2 5 (1789 lb) = 0 FDF + FDG = −1789 lb or ΣFy = 0: 1 FDF − 1 FDG + (4) 1 5 5 5 + 600 lb − 400 lb = 0 FDF − FDG = −2236 lb or Add Eqs. (4) and (5): (1789 lb) (5) 2 FDF = − 4025 lb FDF = 2010 lb C Subtract Eq. (5) from Eq. (4): 2 FDG = 447 lb FDG = 224 lb T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 768 PROBLEM 6.16 Determine the force in member FG and in each of the members located to the right of FG for the studio roof truss shown. State whether each member is in tension or compression. SOLUTION Reaction at L: Because of the symmetry of the loading, L= 1 total load, 2 L = 1200 lb (See F.B. diagram to the left for more details.) Free body: Joint L: 9 = 26.57° 18 3 β = tan −1 = 9.46° 18 FJL FKL 1000 lb = = sin 63.43° sin 99.46° sin17.11° α = tan −1 FKL = 3352.7 lb C FJL = 3040 lb T FKL = 3350 lb C Free body: Joint K: ΣFx = 0: − or 2 5 FIK − 2 5 2 FJK − 5 (3352.7 lb) = 0 FIK + FJK = − 3352.7 lb ΣFy = 0: or 1 5 FIK − 1 5 FJK + (1) 1 5 (3352.7) − 400 = 0 FIK − FJK = − 2458.3 lb 2 FIK = − 5811.0 Add Eqs. (1) and (2): FIK = − 2905.5 lb Subtract Eq. (2) from Eq. (1): 2 FJK = − 894.4 FJK = − 447.2 lb (2) FIK = 2910 lb C FJK = 447 lb C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 769 PROBLEM 6.16 (Continued) Free body: Joint J: 2 ΣFx = 0: − 13 3 ΣFy = 0: FIJ + 13 6 FIJ − 37 1 37 6 FGJ + FGJ − 37 1 37 2 (3040 lb) − (3040 lb) − 5 1 5 (447.2) = 0 (447.2) = 0 (3) (4) Multiply Eq. (4) by 6 and add to Eq. (3): 16 13 FIJ − 8 (447.2) = 0 5 FIJ = 361 lb T FIJ = 360.54 lb Multiply Eq. (3) by 3, Eq. (4) by 2, and add: − 16 37 ( FGJ − 3040) − 8 (447.2) = 0 5 FGJ = 2430 lb T FGJ = 2431.7 lb Free body: Joint I: 2 ΣFx = 0: − 5 2 FFI − 5 2 FGI − 5 2 (2905.5) + 13 (360.54) = 0 FFI + FGI = − 2681.9 lb or ΣFy = 0: 1 5 FFI − 1 5 FGI + 1 5 (5) (2905.5) − 3 13 (360.54) − 400 = 0 FFI − FGI = −1340.3 lb or 2 FFI = −4022.2 Add Eqs. (5) and (6): FFI = −2011.1 lb Subtract Eq. (6) from Eq. (5): 2 FGI = −1341.6 lb (6) FFI = 2010 lb C FGI = 671 lb C Free body: Joint F: From ΣFx = 0: FDF = FFI = 2011.1 lb C 1 2011.1 lb = 0 ΣFy = 0: FFG − 400 lb + 2 5 FFG = + 1400 lb FFG = 1400 lb T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 770 PROBLEM 6.17 Determine the force in each of the members located to the left of FG for the scissors roof truss shown. State whether each member is in tension or compression. SOLUTION Free Body: Truss: ΣFx = 0: A x = 0 ΣM L = 0: (1 kN)(12 m) + (2 kN)(10 m) + (2 kN)(8 m) + (1 kN)(6 m) − Ay (12 m) = 0 A y = 4.50 kN FBC = 0 We note that BC is a zero-force member: Also, FCE = FAC Free body: Joint A: ΣFx = 0: ΣFy = 0: (1) 1 2 1 2 FAB + FAB + 2 5 1 5 FAC = 0 (2) FAC + 3.50 kN = 0 (3) Multiply Eq. (3) by –2 and add Eq. (2): − 1 2 FAB − 7 kN = 0 FAB = 9.90 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 771 PROBLEM 6.17 (Continued) Subtract Eq. (3) from Eq. (2): 1 5 FAC − 3.50 kN = 0 FAC = 7.826 kN FCE = 7.83 kN T FCE = FAC = 7.826 kN From Eq. (1): FAC = 7.83 kN T Free body: Joint B: ΣFy = 0: 1 2 FBD + 1 2 (9.90 kN) − 2 kN = 0 FBD = −7.071 kN 1 ΣFx = 0: FBE + 2 (9.90 − 7.071) kN = 0 FBE = −2.000 kN Free body: Joint E: ΣFx = 0: 2 5 FBD = 7.07 kN C FBE = 2.00 kN C ( FEG − 7.826 kN) + 2.00 kN = 0 FEG = 5.590 kN ΣFy = 0: FDE − 1 5 FEG = 5.59 kN T (7.826 − 5.590) kN = 0 FDE = 1.000 kN FDE = 1.000 kN T Free body: Joint D: ΣFx = 0: 2 5 ( FDF + FDG ) + 1 2 (7.071 kN) FDF + FDG = −5.590 kN or ΣFy = 0: or 1 5 ( FDF − FDG ) + 1 2 (4) (7.071 kN) = 2 kN − 1 kN = 0 FDE − FDG = −4.472 (5) Add Eqs. (4) and (5): 2 FDF = −10.062 kN FDF = 5.03 kN C Subtract Eq. (5) from Eq. (4): 2 FDG = −1.1180 kN FDG = 0.559 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 772 PROBLEM 6.18 Determine the force in member FG and in each of the members located to the right of FG for the scissors roof truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: ΣM A = 0: L(12 m) − (2 kN)(2 m) − (2 kN)(4 m) − (1 kN)(6 m) = 0 L = 1.500 kN Angles: tan α = 1 tan β = 1 2 α = 45° β = 26.57° Zero-force members: Examining successively joints K, J, and I, we note that the following members to the right of FG are zeroforce members: JK, IJ, and HI. FHI = FIJ = FJK = 0 Thus, We also note that and FGI = FIK = FKL (1) FHJ = FJL (2) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 773 PROBLEM 6.18 (Continued) Free body: Joint L: FJL F 1.500 kN = KL = sin116.57° sin 45° sin18.43° FJL = 4.24 kN C FJL = 4.2436 kN From Eq. (1): FGI = FIK = FKL From Eq. (2): FHJ = FJL = 4.2436 kN FKL = 3.35 kN T FGI = FIK = 3.35 kN T FHJ = 4.24 kN C Free body: Joint H: FGH FFH 4.2436 = = sin108.43° sin18.43° sin 53.14° FFH = 5.03 kN C FGH = 1.677 kN T Free body: Joint F: ΣFx = 0: − FDF cos 26.57° − (5.03 kN) cos 26.57° = 0 FDF = −5.03 kN ΣFy = 0: − FFG − 1 kN + (5.03 kN) sin 26.57° − ( −5.03 kN)sin 26.57° = 0 FFG = 3.500 kN FFG = 3.50 kN T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 774 PROBLEM 6.19 Determine the force in each member of the Warren bridge truss shown. State whether each member is in tension or compression. SOLUTION ΣFx = 0: Ax = 0 Free body: Truss: Due to symmetry of truss and loading, Ay = G = Free body: Joint A: 1 total load = 6 kips 2 FAB = 7.50 kips C FAB FAC 6 kips = = 5 3 4 Free body: Joint B: FBC FBD 7.5 kips = = 5 6 5 FAC = 4.50 kips T FBC = 7.50 kips T FBD = 9.00 kips C Free body: Joint C: ΣFy = 0: 4 4 (7.5) + FCD − 6 = 0 5 5 3 ΣFx = 0: FCE − 4.5 − (7.5) = 0 5 FCE = +9 kips FCD = 0 FCE = 9.00 kips T Truss and loading is symmetrical about cL. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 775 PROBLEM 6.20 Solve Problem 6.19 assuming that the load applied at E has been removed. PROBLEM 6.19 Determine the force in each member of the Warren bridge truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: ΣFx = 0: Ax = 0 ΣM G = 0: 6(36) − Ay (54) = 0 A y = 4 kips ΣFy = 0: 4 − 6 + G = 0 G = 2 kips Free body: Joint A: Free body Joint B: FAB FAC 4 kips = = 5 3 4 FBC FBD 5 kips = = 5 6 5 FAB = 5.00 kips C FAC = 3.00 kips T FBC = 5.00 kips T FBD = 6.00 kips C Free body Joint C: ΣM y = 0: 4 4 (5) + FCD − 6 = 0 5 5 3 3 ΣFx = 0: FCE + (2.5) − (5) − 3 = 0 5 5 FCD = 2.50 kips T FCE = 4.50 kips T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 776 PROBLEM 6.20 (Continued) Free body: Joint D: 4 4 ΣFy = 0: − (2.5) − FDE = 0 5 5 FDE = −2.5 kips 3 3 ΣFx = 0: FDF + 6 − (2.5) − (2.5) = 0 5 5 FDF = −3 kips Free body: Joint F: FEF FFG 3 kips = = 5 5 6 Free body: Joint G: FEG 2 kips = 3 4 Also, FFG 2 kips = 5 4 FDE = 2.50 kips C FDF = 3.00 kips C FEF = 2.50 kips T FFG = 2.50 kips C FEG = 1.500 kips T FFG = 2.50 kips C (Checks) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 777 PROBLEM 6.21 Determine the force in each member of the Pratt bridge truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: ΣFz = 0: A x = 0 ΣM A = 0 : H (36 ft) − (4 kips)(9 ft) − (4 kips)(18 ft) − (4 kips)(27 ft) = 0 H = 6 kips ΣFy = 0: Ay + 6 kips − 12 kips = 0 A y = 6 kips Free body: Joint A: FAB FAC 6 kips = = 5 3 4 FAB = 7.50 kips C FAC = 4.50 kips T Free body: Joint C: ΣFx = 0: ΣFy = 0: FCE = 4.50 kips T FBC = 4.00 kips T Free body: Joint B: ΣFy = 0: − ΣFx = 0: 4 4 FBE + (7.50 kips) − 4.00 kips = 0 5 5 8 3 (7.50 kips) + (2.50 kips) + FBD = 0 5 5 FBD = −6.00 kips FBE = 2.50 kips T FBD = 6.00 kips C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 778 PROBLEM 6.21 (Continued) Free body: Joint D: We note that DE is a zero-force member: Also, From symmetry: FFE = FBE FEG = FCE FFG = FBC FFH = FAB FGH = FAC FDE = 0 FDF = 6.00 kips C FEF = 2.50 kips T FEG = 4.50 kips T FFG = 4.00 kips T FFH = 7.50 kips C FGH = 4.50 kips T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 779 PROBLEM 6.22 Solve Problem 6.21 assuming that the load applied at G has been removed. PROBLEM 6.21 Determine the force in each member of the Pratt bridge truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: ΣFx = 0: A x = 0 ΣM A = 0: H (36 ft) − (4 kips)(9 ft) − (4 kips)(18 ft) = 0 H = 3.00 kips ΣFy = 0: A y + 5.00 kips We note that DE and FG are zero-force members. FDE = 0, Therefore, FFG = 0 Also, FBD = FDF (1) and FEG = FGH (2) Free body: Joint A: FAB FAC 5 kips = = 5 3 4 FAB = 6.25 kips C FAC = 3.75 kips T Free body: Joint C: ΣFx = 0: ΣFy = 0: FCE = 3.75 kips T FBC = 4.00 kips T Free body: Joint B: ΣFx = 0: 4 4 (6.25 kips) − 4.00 kips − FBE = 0 5 5 FBE = 1.250 kips T 3 3 ΣFx = 0: FBD + (6.25 kips) + (1.250 kips) = 0 5 5 FBD = −4.50 kips FBD = 4.50 kips C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 780 PROBLEM 6.22 (Continued) Free body: Joint F: We recall that FFG = 0, and from Eq. (1) that FDF = FBD FEF FFH 4.50 kips = = 5 5 6 FDF = 4.50 kips C FEF = 3.75 kips T FFH = 3.75 kips C Free body: Joint H: FGH 3.00 kips = 3 4 FGH = 2.25 kips T Also, FFH 3.00 kips = 5 4 FFH = 3.75 kips C (checks) From Eq. (2): FEG = FGH FEG = 2.25 kips T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 781 PROBLEM 6.23 The portion of truss shown represents the upper part of a power transmission line tower. For the given loading, determine the force in each of the members located above HJ. State whether each member is in tension or compression. SOLUTION Free body: Joint A: F FAB 1.2 kN = AC = 2.29 2.29 1.2 FAB = 2.29 kN T FAC = 2.29 kN C Free body: Joint F: FDF F 1.2 kN = EF = 2.29 2.29 2.1 FDF = 2.29 kN T FEF = 2.29 kN C Free body: Joint D: FBD FDE 2.29 kN = = 2.21 0.6 2.29 FBD = 2.21 kN T FDE = 0.600 kN C Free body: Joint B: ΣFx = 0: 4 2.21 (2.29 kN) = 0 FBE + 2.21 kN − 5 2.29 FBE = 0 3 0.6 (2.29 kN) = 0 ΣFy = 0: − FBC − (0) − 5 2.29 FBC = − 0.600 kN FBC = 0.600 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 782 PROBLEM 6.23 (Continued) Free body: Joint C: ΣFx = 0: FCE + 2.21 (2.29 kN) = 0 2.29 FCE = − 2.21 kN ΣFy = 0: − FCH − 0.600 kN − FCH = −1.200 kN FCE = 2.21 kN C 0.6 (2.29 kN) = 0 2.29 FCH = 1.200 kN C Free body: Joint E: ΣFx = 0: 2.21 kN − 2.21 4 (2.29 kN) − FEH = 0 2.29 5 FEH = 0 ΣFy = 0: − FEJ − 0.600 kN − FEJ = −1.200 kN 0.6 (2.29 kN) − 0 = 0 2.29 FEJ = 1.200 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 783 PROBLEM 6.24 For the tower and loading of Problem 6.23 and knowing that FCH = FEJ = 1.2 kN C and FEH = 0, determine the force in member HJ and in each of the members located between HJ and NO. State whether each member is in tension or compression. PROBLEM 6.23 The portion of truss shown represents the upper part of a power transmission line tower. For the given loading, determine the force in each of the members located above HJ. State whether each member is in tension or compression. SOLUTION Free body: Joint G: FGH F 1.2 kN = GI = 3.03 3.03 1.2 FGH = 3.03 kN T FGI = 3.03 kN C Free body: Joint L: FJL F 1.2 kN = KL = 3.03 3.03 1.2 FJL = 3.03 kN T FKL = 3.03 kN C Free body: Joint J: ΣFx = 0: − FHJ + 2.97 (3.03 kN) = 0 3.03 FHJ = 2.97 kN T 0.6 (3.03 kN) = 0 3.03 = −1.800 kN FJK = 1.800 kN C Fy = 0: − FJK − 1.2 kN − FJK Free body: Joint H: ΣFx = 0: 4 2.97 (3.03 kN) = 0 FHK + 2.97 kN − 5 3.03 FHK = 0 0.6 3 (3.03) kN − (0) = 0 3.03 5 = −1.800 kN FHI = 1.800 kN C ΣFy = 0: − FHI − 1.2 kN − FHI PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 784 PROBLEM 6.24 (Continued) Free body: Joint I: ΣFx = 0: FIK + 2.97 (3.03 kN) = 0 3.03 FIK = −2.97 kN ΣFy = 0: − FIN − 1.800 kN − FIN = −2.40 kN FIK = 2.97 kN C 0.6 (3.03 kN) = 0 3.03 FIN = 2.40 kN C Free body: Joint K: ΣFx = 0: − 4 2.97 (3.03 kN) = 0 FKN + 2.97 kN − 5 3.03 ΣFy = 0: − FKO − FKN = 0 0.6 3 (3.03 kN) − 1.800 kN − (0) = 0 3.03 5 FKO = −2.40 kN FKO = 2.40 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 785 PROBLEM 6.25 Solve Problem 6.23 assuming that the cables hanging from the right side of the tower have fallen to the ground. PROBLEM 6.23 The portion of truss shown represents the upper part of a power transmission line tower. For the given loading, determine the force in each of the members located above HJ. State whether each member is in tension or compression. SOLUTION Zero-Force Members: Considering joint F, we note that DF and EF are zero-force members: FDF = FEF = 0 Considering next joint D, we note that BD and DE are zero-force members: FBD = FDE = 0 Free body: Joint A: F FAB 1.2 kN = AC = 2.29 2.29 1.2 FAB = 2.29 kN T FAC = 2.29 kN C Free body: Joint B: ΣFx = 0: 4 2.21 (2.29 kN) = 0 FBE − 5 2.29 FBE = 2.7625 kN ΣFy = 0: − FBC − FBE = 2.76 kN T 0.6 3 (2.29 kN) − (2.7625 kN) = 0 2.29 5 FBC = −2.2575 kN FBC = 2.26 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 786 PROBLEM 6.25 (Continued) Free body: Joint C: ΣFx = 0: FCE + 2.21 (2.29 kN) = 0 2.29 ΣFy = 0: − FCH − 2.2575 kN − FCH = −2.8575 kN FCE = 2.21 kN C 0.6 (2.29 kN) = 0 2.29 FCH = 2.86 kN C Free body: Joint E: ΣFx = 0: − 4 4 FEH − (2.7625 kN) + 2.21 kN = 0 5 5 FEH = 0 3 3 ΣFy = 0: − FEJ + (2.7625 kN) − (0) = 0 5 5 FEJ = +1.6575 kN FEJ = 1.658 kN T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 787 PROBLEM 6.26 Determine the force in each of the members connecting joints A through F of the vaulted roof truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: ΣFx = 0: A x = 0 ΣM K = 0: (1.2 kN)6a + (2.4 kN)5a + (2.4 kN)4a + (1.2 kN)3a − Ay (6a) = 0 A y = 5.40 kN Free body: Joint A: FAB FAC 4.20 kN = 2 1 5 FAB = 9.3915 kN = FAB = 9.39 kN C FAC = 8.40 kN T Free body: Joint B: 2 ΣFx = 0: 5 1 ΣFy = 0: 5 FBD + FBD − 1 2 1 2 FBC + FBC + 2 5 1 5 (9.3915) = 0 (1) (9.3915) − 2.4 = 0 (2) Add Eqs. (1) and (2): 3 5 FBD + 3 5 (9.3915 kN) − 2.4 kN = 0 FBD = − 7.6026 kN FBD = 7.60 kN C Multiply Eq. (2) by –2 and add Eq. (1): 3 2 FB + 4.8 kN = 0 FBC = −2.2627 kN FBC = 2.26 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 788 PROBLEM 6.26 (Continued) Free body: Joint C: 1 ΣFx = 0: 5 2 ΣFy = 0: 5 4 FCD + 17 1 FCD + 17 FCE + FCE − 1 2 1 2 (2.2627) − 8.40 = 0 (3) (2.2627) = 0 (4) Multiply Eq. (4) by −4 and add Eq. (1): − 7 5 FCD + 5 (2.2627) − 8.40 = 0 2 FCD = 0.128 kN C FCD = − 0.1278 kN Multiply Eq. (1) by 2 and subtract Eq. (2): 7 17 FCE + 3 2 (2.2627) − 2(8.40) = 0 FCE = 7.07 kN T FCE = 7.068 kN Free body: Joint D: ΣFx = 0: 2 1 + ΣFy = 0: FDF + 5 5 1 5 + (0.1278) = 0 FDF − 2 5 1 2 FDE + (7.6026) 1.524 5 (5) 1.15 1 FDE + (7.6026) 1.524 5 (0.1278) − 2.4 = 0 (6) Multiply Eq. (5) by 1.15 and add Eq. (6): 3.30 5 FDF + 3.30 5 (7.6026) + 3.15 5 (0.1278) − 2.4 = 0 FDF = − 6.098 kN FDF = 6.10 kN C Multiply Eq. (6) by –2 and add Eq. (5): 3.30 3 FDE − (0.1278) + 4.8 = 0 1.524 5 FDE = − 2.138 kN FDE = 2.14 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 789 PROBLEM 6.26 (Continued) Free body: Joint E: ΣFx = 0: 0.6 4 1 FEF + ( FEH − FCE ) + (2.138) = 0 2.04 1.524 17 (7) ΣFy = 0: 1.95 1 1.15 FEF + ( FEH − FCE ) − (2.138) = 0 2.04 1.524 17 (8) Multiply Eq. (8) by 4 and subtract Eq. (7): 7.2 FEF − 7.856 kN = 0 2.04 FEF = 2.23 kN T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 790 PROBLEM 6.27 Determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: ΣM F = 0: G (20 ft) − (15 kips)(16 ft) − (40 kips)(15 ft) = 0 G = 42 kips ΣFx = 0: Fx + 15 kips = 0 Fx = 15 kips ΣFy = 0: Fy − 40 kips + 42 kips = 0 Fy = 2 kips Free body: Joint F: ΣFx = 0: 1 5 FDF − 15 kips = 0 FDF = 33.54 kips ΣFy = 0: FBF − 2 kips + 2 5 FDF = 33.5 kips T (33.54 kips) = 0 FBF = − 28.00 kips FBF = 28.0 kips C Free body: Joint B: ΣFx = 0: ΣFy = 0: 5 29 2 29 FAB + FAB − 5 61 6 61 FBD + 15 kips = 0 (1) FBD + 28 kips = 0 (2) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 791 PROBLEM 6.27 (Continued) Multiply Eq. (1) by 6, Eq. (2) by 5, and add: 40 29 FAB + 230 kips = 0 FAB = 31.0 kips C FAB = −30.96 kips Multiply Eq. (1) by 2, Eq. (2) by –5, and add: 40 61 FBD − 110 kips = 0 FBD = 21.5 kips T FBD = 21.48 kips Free body: Joint D: 2 ΣFy = 0: 5 2 FAD − 1 ΣFx = 0: FDE + 5 5 (33.54) + 6 61 (15.09 − 33.54) − (21.48) = 0 FAD = 15.09 kips T 5 (21.48) = 0 61 FDE = 22.0 kips T Free body: Joint A: 5 ΣFx = 0: 29 ΣFy = 0: − 1 FAC + 2 29 5 FAC − FAE + 2 5 5 FAE + 29 (30.36) − 2 29 1 5 (15.09) = 0 2 (30.96) − 5 (15.09) = 0 (3) (4) Multiply Eq. (3) by 2 and add Eq. (4): 8 29 FAC + 12 29 (30.96) − 4 5 FAC = −28.27 kips, (15.09) = 0 FAC = 28.3 kips C Multiply Eq. (3) by 2, Eq. (4) by 5, and add: − 8 5 FAE + 20 29 (30.96) − 12 5 (15.09) = 0 FAE = 9.50 kips T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 792 PROBLEM 6.27 (Continued) Free body: Joint C: From force triangle: FCE 61 = FCG 28.27 kips = 8 29 FCE = 41.0 kips T FCG = 42.0 kips C Free body: Joint G: ΣFx = 0: ΣFy = 0: 42 kips − 42 kips = 0 (Checks) FEG = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 793 PROBLEM 6.28 Determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Reactions: ΣFx = 0: Ex = 0 ΣM F = 0: E y = 45 kips ΣFy = 0: F = 60 kips Joint D: FCD FDH 15 kips = = 12 13 5 FCD = 36.0 kips T FDH = 39.0 kips C Joint H: ΣF = 0: FCH = 0 ΣF = 0: FGH = 39.0 kips C ΣF = 0: FCG = 0 Joint C: ΣF = 0: FBC = 36.0 kips T ΣF = 0: FBG = 0 Joint G: ΣF = 0: FFG = 39.0 kips C ΣF = 0: FBF = 0 Joint B: ΣF = 0: FAB = 36.0 kips T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 794 PROBLEM 6.28 (Continued) Joint A: AE = 122 + 152 = 19.21 ft tan 38.7° = 36 kips FAF sin 38.7° = 36 kips FAE FAF = 45.0 kips C FAE = 57.6 kips T Joint E: ΣFx = 0: + (57.6 kips) sin 38.7° + FEF = 0 FEF = −36.0 kips FEF = 36.0 kips C ΣFy = 0: (57.6 kips) cos 38.7° − 45 kips = 0 (Checks) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 795 PROBLEM 6.29 Determine whether the trusses of Problems 6.31a, 6.32a, and 6.33a are simple trusses. SOLUTION Truss of Problem 6.31a: Starting with triangle HDI and adding two members at a time, we obtain successively joints A, E, J, and B, but cannot go further. Thus, this truss is not a simple truss. Truss of Problem 6.32a: Starting with triangle ABC and adding two members at a time, we obtain joints D, E, G, F, and H, but cannot go further. Thus, this truss is not a simple truss. Truss of Problem 6.33a: Starting with triangle ABD and adding two members at a time, we obtain successively joints H, G, F, E, I, C, and J, thus completing the truss. Therefore, this is a simple truss. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 796 PROBLEM 6.30 Determine whether the trusses of Problems 6.31b, 6.32b, and 6.33b are simple trusses. SOLUTION Truss of Problem 6.31b: Starting with triangle CGM and adding two members at a time, we obtain successively joints B, L, F, A, K, J, then H, D, N, I, E, O, and P, thus completing the truss. Therefore, this truss is a simple truss. Truss of Problem 6.32b: Starting with triangle ABC and adding two members at a time, we obtain successively joints E, D, F, G, and H, but cannot go further. Thus, this truss is not a simple truss. Truss of Problem 6.33b: Starting with triangle GFH and adding two members at a time, we obtain successively joints D, E, C, A, and B, thus completing the truss. Therefore, this is a simple truss. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 797 PROBLEM 6.31 For the given loading, determine the zero-force members in each of the two trusses shown. SOLUTION Truss (a): FB: Joint B: FBJ = 0 FB: Joint D: FDI = 0 FB: Joint E: FEI = 0 FB: Joint I : FAI = 0 (a) FB: Joint F : FFK = 0 FB: Joint G: FGK = 0 FB: Joint K : FCK = 0 AI , BJ , CK , DI , EI , FK , GK The zero-force members, therefore, are Truss (b): FB: Joint K : FFK = 0 FB: Joint O : FIO = 0 (b) The zero-force members, therefore, are FK and IO All other members are either in tension or compression. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 798 PROBLEM 6.32 For the given loading, determine the zero-force members in each of the two trusses shown. SOLUTION Truss (a): FB: Joint B: FBC = 0 FB: Joint C: FCD = 0 FB: Joint J : FIJ = 0 FB: Joint I : FIL = 0 FB: Joint N : FMN = 0 FB: Joint M : FLM = 0 The zero-force members, therefore, are Truss (b): (a) BC , CD, IJ , IL, LM , MN FB: Joint C: FBC = 0 FB: Joint B: FBE = 0 FB: Joint G: FFG = 0 FB: Joint F : FEF = 0 FB: Joint E: FDE = 0 FB: Joint I : FIJ = 0 (b) FB: Joint M : FMN = 0 FB: Joint N : FKN = 0 The zero-force members, therefore, are BC , BE , DE , EF , FG , IJ , KN , MN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 799 PROBLEM 6.33 For the given loading, determine the zero-force members in each of the two trusses shown. SOLUTION Truss (a): Note: Reaction at F is vertical ( Fx = 0). Joint G: ΣF = 0, Joint D: ΣF = 0, Joint F : ΣF = 0, Joint G: ΣF = 0, Joint J : ΣF = 0, Joint I : ΣF = 0, FDG = 0 FDB = 0 FFG = 0 FGH = 0 FIJ = 0 FHI = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 800 PROBLEM 6.33 (Continued) Truss (b): Joint A: ΣF = 0, Joint C: ΣF = 0, Joint E: ΣF = 0, Joint F : ΣF = 0, Joint G: ΣF = 0, FAC = 0 FCE = 0 FEF = 0 FFG = 0 FGH = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 801 PROBLEM 6.34 Determine the zero-force members in the truss of (a) Problem 6.26, (b) Problem 6.28. SOLUTION (a) Truss of Problem 6.26: FB : Joint I : FIJ = 0 FB : Joint J : FGJ = 0 FB : Joint G: FGH = 0 GH , GJ , IJ The zero-force members, therefore, are (b) Truss of Problem 6.28: FB : Joint B: FBF = 0 FB : Joint B: FBG = 0 FB : Joint C: FCG = 0 FB : Joint C: FCH = 0 The zero-force members, therefore, are BF , BG, CG, CH PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 802 PROBLEM 6.35* The truss shown consists of six members and is supported by a short link at A, two short links at B, and a ball and socket at D. Determine the force in each of the members for the given loading. SOLUTION Free body: Truss: From symmetry: Dx = Bx and Dy = B y ΣM z = 0: − A(10 ft) − (400 lb)(24 ft) = 0 A = −960 lb ΣFx = 0: Bx + Dx + A = 0 2 Bx − 960 lb = 0, Bx = 480 lb ΣFy = 0: B y + Dy − 400 lb = 0 2 By = 400 lb By = +200 lb Thus, Free body: C: FCA = FAC FCB = FBC FCD = FCD B = (480 lb)i + (200 lb) j CA FAC = ( −24i + 10 j) CA 26 CB FBC = (−24i + 7k ) CB 25 CD FCD = (−24i − 7k ) CD 25 ΣF = 0: FCA + FCB + FCD − (400 lb) j = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 803 PROBLEM 6.35* (Continued) Substituting for FCA , FCB , FCD , and equating to zero the coefficients of i, j, k : 24 24 FAC − ( FBC + FCD ) = 0 26 25 i: − j: 10 FAC − 400 lb = 0 26 k: 7 ( FBC − FCD ) = 0 FCD = FBC 25 (1) FAC = 1040 lb T Substitute for FAC and FCD in Eq. (1): − 24 24 (10.40 lb) − (2 FBC ) = 0 FBC = −500 lb 26 25 Free body: B: FBC FBA FBD FBC = FCD = 500 lb C CB = (500 lb) = −(480 lb)i + (140 lb)k CB BA FAB = FAB = (10 j − 7k ) BA 12.21 = − FBD k ΣF = 0: FBA + FBD + FBC + (480 lb)i + (200 lb) j = 0 Substituting for FBA , FBD , FBC and equating to zero the coefficients of j and k: j: 10 FAB + 200 lb = 0 FAB = −244.2 lb 12.21 k: − 7 FAB − FBD + 140 lb = 0 12.21 FBD = − From symmetry: 7 (−244.2 lb) + 140 lb = +280 lb 12.21 FAD = FAB FAB = 244 lb C FBD = 280 lb T FAD = 244 lb C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 804 PROBLEM 6.36* The truss shown consists of six members and is supported by a ball and socket at B, a short link at C, and two short links at D. Determine the force in each of the members for P = (−2184 N)j and Q = 0. SOLUTION Free body: Truss: From symmetry: Dx = Bx and Dy = By ΣFx = 0: 2 Bx = 0 Bx = Dx = 0 ΣFz = 0: Bz = 0 ΣM c z = 0: − 2 By (2.8 m) + (2184 N)(2 m) = 0 By = 780 N Thus, Free body: A: B = (780 N) j AB FAB FAB = FAB = ( −0.8i − 4.8 j + 2.1k ) AB 5.30 AC FAC = FAC = FAC (2i − 4.8 j) AC 5.20 AD FAD = FAD = FAD (+0.8i − 4.8 j − 2.1k ) AD 5.30 ΣF = 0: FAB + FAC + FAD − (2184 N) j = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 805 PROBLEM 6.36* (Continued) Substituting for FAB , FAC , FAD , and equating to zero the coefficients of i, j, k : i: − 0.8 2 ( FAB + FAD ) + FAC = 0 5.30 5.20 (1) j: − 4.8 4.8 ( FAB + FAD ) − FAC − 2184 N = 0 5.30 5.20 (2) 2.1 ( FAB − FAD ) = 0 5.30 k: FAD = FAB Multiply Eq. (1) by –6 and add Eq. (2): 16.8 − FAC − 2184 N = 0, FAC = −676 N 5.20 FAC = 676 N C Substitute for FAC and FAD in Eq. (1): 0.8 2 − 2 FAB + (−676 N) = 0, FAB = −861.25 N 5.30 5.20 Free body: B: FAB = FAD = 861 N C AB = −(130 N)i − (780 N) j + (341.25 N)k FAB = (861.25 N) AB 2.8i − 2.1k FBC = FBC = FBC (0.8i − 0.6k ) 3.5 FBD = − FBD k ΣF = 0: FAB + FBC + FBD + (780 N) j = 0 Substituting for FAB , FBC , FBD and equating to zero the coefficients of i and k , i: k: −130 N + 0.8FBC = 0 FBC = +162.5 N 341.25 N − 0.6 FBC − FBD = 0 FBD = 341.25 − 0.6(162.5) = +243.75 N From symmetry: FCD = FBC FBC = 162.5 N T FBD = 244 N T FCD = 162.5 N T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 806 PROBLEM 6.37* The truss shown consists of six members and is supported by a ball and socket at B, a short link at C, and two short links at D. Determine the force in each of the members for P = 0 and Q = (2968 N)i. SOLUTION Free body: Truss: From symmetry: Dx = Bx and Dy = By ΣFx = 0: 2 Bx + 2968 N = 0 Bx = Dx = −1484 N ΣM cz ′ = 0: − 2 By (2.8 m) − (2968 N)(4.8 m) = 0 By = −2544 N Thus, Free body: A: FAB = FAB B = −(1484 N)i − (2544 N) j AB AB FAB (−0.8i − 4.8 j + 2.1k ) 5.30 AC FAC = FAC = (2i − 4.8 j) AC 5.20 AD = FAD AD FAD = (−0.8i − 4.8 j − 2.1k ) 5.30 = FAC FAD ΣF = 0: FAB + FAC + FAD + (2968 N)i = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 807 PROBLEM 6.37* (Continued) Substituting for FAB , FAC , FAD , and equating to zero the coefficients of i, j, k , 0.8 2 FAC + 2968 N = 0 ( FAB + FAD ) + 5.30 5.20 (1) 4.8 4.8 FAC = 0 ( FAB + FAD ) − 5.30 5.20 (2) i: − j: − k: 2.1 ( FAB − FAD ) = 0 5.30 FAD = FAB Multiply Eq. (1) by –6 and add Eq. (2): 16.8 − FAC − 6(2968 N) = 0, FAC = −5512 N 5.20 FAC = 5510 N C Substitute for FAC and FAD in Eq. (2): 4.8 4.8 − 2 FAB − (−5512 N) = 0, FAB = +2809 N 5.30 5.20 Free body: B: FAB FBC FBD FAB = FAD = 2810 N T BA = (2809 N) = (424 N)i + (2544 N) j − (1113 N)k BA 2.8 i − 2.1k = FBC = FBC (0.8i − 0.6k ) 3.5 = − FBD k ΣF = 0: FAB + FBC + FBD − (1484 N)i − (2544 N) j = 0 Substituting for FAB , FBC , FBD and equating to zero the coefficients of i and k , i: +24 N + 0.8FBC − 1484 N = 0, FBC = +1325 N k: −1113 N − 0.6 FBC − FBD = 0 FBD = −1113 N − 0.6(1325 N) = −1908 N, From symmetry: FCD = FBC FBC = 1325 N T FBD = 1908 N C FCD = 1325 N T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 808 PROBLEM 6.38* The truss shown consists of nine members and is supported by a ball and socket at A, two short links at B, and a short link at C. Determine the force in each of the members for the given loading. SOLUTION Free body: Truss: From symmetry: Az = Bz = 0 ΣFx = 0: Ax = 0 ΣM BC = 0: Ay (6 ft) + (1600 lb)(7.5 ft) = 0 Ay = −2000 lb A = −(2000 lb)j By = C From symmetry: ΣFy = 0: 2 By − 2000 lb − 1600 lb = 0 By = 1800 lb B = (1800 lb) j ΣF = 0: FAB + FAC + FAD − (2000 lb) j = 0 Free body: A: FAB i+k + FAC 2 i−k 2 + FAD (0.6 i + 0.8 j) − (2000 lb) j = 0 Factoring i, j, k and equating their coefficient to zero, 1 2 FAB + 1 FAC + 0.6 FAD = 0 (1) 0.8FAD − 2000 lb = 0 FAD = 2500 lb T 2 1 2 FAB − 1 2 FAC = 0 FAC = FAB PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 809 PROBLEM 6.38* (Continued) Substitute for FAD and FAC into Eq. (1): 2 2 Free body: B: FAB + 0.6(2500 lb) = 0, FAB = −1060.7 lb, FBA FBC FAB = FAC = 1061 lb C BA i+k = FAB = + (1060.7 lb) = (750 lb)(i + k ) BA 2 = − FBC k FBD = FBD (0.8 j − 0.6k ) BE FBE (7.5i + 8 j − 6k ) FBE = FBE = BE 12.5 ΣF = 0: FBA + FBC + FBD + FBE + (1800 lb) j = 0 Substituting for FBA , FBC , FBD , and FBE and equating to zero the coefficients of i, j, k , i: 7.5 750 lb + FBE = 0, FBE = −1250 lb 12.5 j: 8 0.8 FBD + (−1250 lb) + 1800 lb = 0 12.5 k: 750 lb − FBC − 0.6( −1250 lb) − FBE = 1250 lb C FBD = 1250 lb C 6 ( −1250 lb) = 0 12.5 FBC = 2100 lb T FBD = FCD = 1250 lb C From symmetry: Free body: D: ΣF = 0: FDA + FDB + FDC + FDE i = 0 We now substitute for FDA , FDB , FDC and equate to zero the coefficient of i. Only FDA contains i and its coefficient is −0.6 FAD = −0.6(2500 lb) = −1500 lb i: −1500 lb + FDE = 0 FDE = 1500 lb T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 810 PROBLEM 6.39* The truss shown consists of nine members and is supported by a ball and socket at B, a short link at C, and two short links at D. (a) Check that this truss is a simple truss, that it is completely constrained, and that the reactions at its supports are statically determinate. (b) Determine the force in each member for P = (−1200 N)j and Q = 0. SOLUTION Free body: Truss: ΣM B = 0: 1.8i × Cj + (1.8i − 3k ) × ( D y j + Dk ) + (0.6i − 0.75k ) × (−1200 j) = 0 −1.8 Ck + 1.8 D y k − 1.8 Dz j + 3D y i − 720k − 900i = 0 Equate to zero the coefficients of i, j, k : i: 3Dy − 900 = 0, D y = 300 N j: Dz = 0, k: 1.8C + 1.8(300) − 720 = 0 D = (300 N) j C = (100 N) j ΣF = 0: B + 300 j + 100 j − 1200 j = 0 B = (800 N) j Free body: B: ΣF = 0: FBA + FBC + FBE + (800 N) j = 0, with BA FAB (0.6i + 3j − 0.75k ) FBA = FAB = BA 3.15 FBC = FBC i FBE = − FBE k Substitute and equate to zero the coefficient of j, i, k : j: i: k: 3 FAB + 800 N = 0, FAB = −840 N, 3.315 0.6 (−840 N) + FBC = 0 3.15 0.75 − (−840 N) − FBE = 0 3.15 FAB = 840 N C FBC = 160.0 N T FBE = 200 N T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 811 PROBLEM 6.39* (Continued) Free body: C: ΣF = 0: FCA + FCB + FCD + FCE + (100 N) j = 0, with F CA FCA = FAC = AC (−1.2i + 3j − 0.75 k ) CA 3.317 FCB = −(160 N) i FCD = − FCD k FCE = FCE F CE = CE (−1.8i − 3k ) CE 3, 499 Substitute and equate to zero the coefficient of j, i, k : 3 FAC + 100 N = 0, FAC = −110.57 N 3.317 j: − i: k: − FAC = 110.6 N C FCE = 233 N C 1.2 1.8 FCE = 0, FCE = −233.3 (−110.57) − 160 − 3.317 3.499 FCD = 225 N T 0.75 3 (−110.57) − FCD − (−233.3) = 0 3.317 3.499 Free body: D: ΣF = 0: FDA + FDC + FDE + (300 N) j = 0, with F DA FDA = FAD = AD (−1.2i + 3j + 2.25k ) DA 3.937 FDC = FCD k = (225 N)k FDE = − FDE i Substitute and equate to zero the coefficient of j, i, k : j: i: k: 3 FAD + 300 N = 0, 3.937 FAD = −393.7 N, 1.2 − ( −393.7 N) − FDE = 0 3.937 FAD = 394 N C FDE = 120.0 N T 2.25 (−393.7 N) + 225 N = 0 (Checks) 3.937 Free body: E: Member AE is the only member at E which does not lie in the xz plane. Therefore, it is a zero-force member. FAE = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 812 PROBLEM 6.40* Solve Problem 6.39 for P = 0 and Q = (−900 N)k. PROBLEM 6.39* The truss shown consists of nine members and is supported by a ball and socket at B, a short link at C, and two short links at D. (a) Check that this truss is a simple truss, that it is completely constrained, and that the reactions at its supports are statically determinate. (b) Determine the force in each member for P = (−1200 N)j and Q = 0. SOLUTION Free body: Truss: ΣM B = 0: 1.8i × Cj + (1.8i − 3k ) × ( D y j + Dz k ) + (0.6i + 3j − 0.75k ) × (−900N)k = 0 1.8 Ck + 1.8D y k − 1.8Dz j + 3D y i + 540 j − 2700i = 0 Equate to zero the coefficient of i, j, k : 3Dy − 2700 = 0 Dy = 900 N −1.8Dz + 540 = 0 Dz = 300 N 1.8C + 1.8 Dy = 0 Thus, C = − D y = −900 N C = −(900 N) j D = (900 N) j + (300 N)k ΣF = 0: B − 900 j + 900 j + 300k − 900k = 0 B = (600 N)k Free body: B: Since B is aligned with member BE, FAB = FBC = 0, Free body: C: FBE = 600 N T ΣF = 0: FCA + FCD + FCE − (900 N) j = 0, with PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 813 PROBLEM 6.40* (Continued) FCA FCD F CA = FAC = AC ( −1.2i + 3j − 0.75k ) CA 3.317 F CE = − FCD k FCE = FCE = CE (−1.8i − 3k ) CE 3.499 Substitute and equate to zero the coefficient of j, i, k: j: 3 FAC − 900 N = 0, 3.317 FAC = 995.1 N FAC = 995 N T FCE = −699.8 N FCE = 700 N C i: − 1.2 1.8 FCE = 0, (995.1) − 3.317 3.499 k: − 0.75 3 (995.1) − FCD − ( −699.8) = 0 3.317 3.499 FCD = 375 N T Free body: D: ΣF = 0: FDA + FDE + (375 N)k +(900 N)j + (300 N) k = 0 F DA FDA = FAD = AD (−1.2i + 3j + 2.25k ) DA 3.937 with FDE = − FDE i and Substitute and equate to zero the coefficient j, i, k: j: i: k: 3 FAD + 900 N = 0, FAD = −1181.1 N 3.937 FAD = 1181 N C 1.2 − (−1181.1 N) − FDE = 0 3.937 FDE = 360 N T 2.25 (−1181.1 N + 375 N + 300 N = 0) 3.937 (Checks) Free body: E: Member AE is the only member at E which does not lie in the xz plane. Therefore, it is a zero-force member. FAE = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 814 PROBLEM 6.41* The truss shown consists of 18 members and is supported by a ball and socket at A, two short links at B, and one short link at G. (a) Check that this truss is a simple truss, that it is completely constrained, and that the reactions at its supports are statically determinate. (b) For the given loading, determine the force in each of the six members joined at E. SOLUTION (a) Check simple truss. (1) Start with tetrahedron BEFG. (2) Add members BD, ED, GD joining at D. (3) Add members BA, DA, EA joining at A. (4) Add members DH, EH, GH joining at H. (5) Add members BC, DC, GC joining at C. Truss has been completed: It is a simple truss. Free body: Truss: Check constraints and reactions. Six unknown reactions—ok; however, supports at A and B constrain truss to rotate about AB and support at G prevents such a rotation. Thus, Truss is completely constrained and reactions are statically determinate. Determination of reactions: ΣM A = 0: 11i × ( By j + Bz k ) + (11i − 9.6k ) × G j + (10.08 j − 9.6k ) × (275i + 240k ) = 0 11By k − 11By j + 11Gk + 9.6Gi − (10.08)(275)k + (10.08)(240)i − (9.6)(275) j = 0 Equate to zero the coefficient of i, j, k: i : 9.6G + (10.08)(240) = 0 G = −252 lb j: − 11Bz − (9.6)(275) = 0 Bz = −240 lb k : 11By + 11(−252) − (10.08)(275) = 0, By = 504 lb G = (−252 lb) j B = (504 lb) j − (240 lb)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 815 PROBLEM 6.41* (Continued) ΣF = 0: A + (504 lb) j − (240 lb)k − (252 lb) j + (275 lb)i + (240 lb)k = 0 A = −(275 lb)i − (252 lb) j Zero-force members. The determination of these members will facilitate our solution. FB: C: Writing ΣFx = 0, ΣFy = 0, ΣFz = 0 FB: F: Writing ΣFx = 0, ΣFy = 0, ΣFz = 0 FB: A: Since FB: H: Writing Az = 0, writing ΣFz = 0 ΣFy = 0 yields FBC = FCD = FCG = 0 yields FBF = FEF = FFG = 0 yields FAD = 0 yields FDH = 0 FB: D: Since FAD = FCD = FDH = 0, we need consider only members DB, DE, and DG. Since FDE is the only force not contained in plane BDG, it must be zero. Simple reasonings show that the other two forces are also zero. FBD = FDE = FDG = 0 The results obtained for the reactions at the supports and for the zero-force members are shown on the figure below. Zero-force members are indicated by a zero (“0”). (b) Force in each of the members joined at E. FDE = FEF = 0 We already found that Free body: A: ΣFy = 0 Free body: H: ΣFz = 0 yields FAE = 252 lb T yields FEH = 240 lb C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 816 PROBLEM 6.41* (Continued) ΣF = 0: FEB + FEG + (240 lb)k − (252 lb) j = 0 Free body: E: F FBE (11i − 10.08 j) + EG (11i − 9.6k ) + 240k − 252 j = 0 14.92 14.6 Equate to zero the coefficient of y and k: 10.08 j: − FBE − 252 = 0 14.92 k: FBE = 373 lb C 9.6 − FEG + 240 = 0 14.6 FEG = 365 lb T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 817 PROBLEM 6.42* The truss shown consists of 18 members and is supported by a ball and socket at A, two short links at B, and one short link at G. (a) Check that this truss is a simple truss, that it is completely constrained, and that the reactions at its supports are statically determinate. (b) For the given loading, determine the force in each of the six members joined at G. SOLUTION See solution to Problem 6.41 for part (a) and for reactions and zero-force members. (b) Force in each of the members joined at G. We already know that Free body: H: Free body: G: FCG = FDG = FFG = 0 yields FGH = 275 lb C ΣFx = 0 ΣF = 0: FGB + FGE + (275 lb)i − (252 lb) j = 0 FBG F (−10.08 j + 9.6k ) + EG (−11i + 9.6k ) + 275i − 252 j = 0 13.92 14.6 Equate to zero the coefficient of i, j, k: 11 i: − FEG + 275 = 0 14.6 FEG = 365 lb T 10.08 j: − FBG − 252 = 0 13.92 9.6 9.6 k: (−348) + (365) = 0 13.92 14.6 FBG = 348 lb C (Checks) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 818 PROBLEM 6.43 Determine the force in members CD and DF of the truss shown. SOLUTION Reactions: ΣM J = 0: (12 kN)(4.8 m) + (12 kN)(2.4 m) − B(9.6 m) = 0 B = 9.00 kN ΣFy = 0: 9.00 kN − 12.00 kN − 12.00 kN + J = 0 J = 15.00 kN Member CD: ΣFy = 0: 9.00 kN + FCD = 0 Member DF: ΣM C = 0: FDF (1.8 m) − (9.00 kN)(2.4 m) = 0 FCD = 9.00 kN C FDF = 12.00 kN T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 819 PROBLEM 6.44 Determine the force in members FG and FH of the truss shown. SOLUTION Reactions: ΣM J = 0: (12 kN)(4.8 m) + (12 kN)(2.4 m) − B (9.6 m) = 0 B = 9.00 kN ΣFy = 0: 9.00 kN − 12.00 kN − 12.00 kN + J = 0 J = 15.00 kN Member FG: 3 ΣFy = 0: − FFG − 12.00 kN + 15.00 kN = 0 5 FFG = 5.00 kN T Member FH: ΣM G = 0: (15.00 kN)(2.4 m) − FFH (1.8 m) = 0 FFH = 20.0 kN T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 820 PROBLEM 6.45 A Warren bridge truss is loaded as shown. Determine the force in members CE, DE, and DF. SOLUTION Free body: Truss: ΣFx = 0: k x = 0 ΣM A = 0: k y (62.5 ft) − (6000 lb)(12.5 ft) − (6000 lb)(25 ft) = 0 k = k y = 3600 lb ΣFy = 0: A + 3600 lb − 6000 lb − 6000 lb = 0 A = 8400 lb We pass a section through members CE, DE, and DF and use the free body shown. ΣM D = 0: FCE (15 ft) − (8400 lb)(18.75 ft) + (6000 lb)(6.25 ft) = 0 FCE = +8000 lb ΣFy = 0: 8400 lb − 6000 lb − FCE = 8000 lb T 15 FDE = 0 16.25 FDE = +2600 lb FDE = 2600 lb T ΣM E = 0: 6000 lb (12.5 ft) − (8400 lb)(25 ft) − FDF (15 ft) = 0 FDF = −9000 lb FDF = 9000 lb C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 821 PROBLEM 6.46 A Warren bridge truss is loaded as shown. Determine the force in members EG, FG, and FH. SOLUTION See solution of Problem 6.45 for free-body diagram of truss and determination of reactions: A = 8400 lb k = 3600 lb We pass a section through members EG, FG, and FH, and use the free body shown. ΣM F = 0: (3600 lb)(31.25 ft) − FEG (15 ft) = 0 FEG = +7500 lb ΣFy = 0: 15 FFG + 3600 lb = 0 16.25 FFG = −3900 lb ΣM G = 0: FFH (15 ft) + (3600 lb)(25 ft) = 0 FFH = −6000 lb FEG = 7500 lb T FFG = 3900 lb C FFH = 6000 lb C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 822 PROBLEM 6.47 Determine the force in members DF, EF, and EG of the truss shown. SOLUTION tan β = 3 4 Reactions: A=N=0 Member DF: 3 ΣM E = 0: + (16 kN)(6 m) − FDF (4 m) = 0 5 FDF = + 40 kN Member EF: + ΣF = 0: (16 kN) sin β − FEF cos β = 0 FEF = 16 tan β = 16(0.75) = 12 kN Member EG: ΣM F = 0: (16 kN)(9 m) + 4 FEG (3 m) = 0 5 FEG = −60 kN FDF = 40.0 kN T FEF = 12.00 kN T FEG = 60.0 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 823 PROBLEM 6.48 Determine the force in members GI, GJ, and HI of the truss shown. SOLUTION Reactions: A=N=0 Member GI: + ΣF = 0: (16 kN)sin β + FGI sin β = 0 FGI = −16 kN Member GJ: ΣM I = 0: − (16 kN)(9 m) − 4 FGJ (3 m) = 0 5 FGJ = −60 kN Member HI: ΣM G = 0: − (16 kN)(9 m) + 3 FHI (4 m) = 0 5 FHI = +60 kN FGI = 16.00 kN C FGJ = 60.0 kN C FHI = 60.0 kN T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 824 PROBLEM 6.49 Determine the force in members AD, CD, and CE of the truss shown. SOLUTION Reactions: ΣM k = 0: 36(2.4) − B(13.5) + 20(9) + 20(4.5) = 0 ΣFx = 0: −36 + K x = 0 B = 26.4 kN K x = 36 kN ΣFy = 0: 26.4 − 20 − 20 + K y = 0 K y = 13.6 kN ΣM C = 0: 36(1.2) − 26.4(2.25) − FAD (1.2) = 0 8 ΣM A = 0: FCD (4.5) = 0 17 FAD = −13.5 kN FAD = 13.5 kN C FCD = 0 15 ΣM D = 0: FCE (2.4) − 26.4(4.5) = 0 17 FCE = +56.1 kN FCE = 56.1 kN T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 825 PROBLEM 6.50 Determine the force in members DG, FG, and FH of the truss shown. SOLUTION See the solution to Problem 6.49 for free-body diagram and analysis to determine the reactions at the supports B and K. B = 26.4 kN ; K x = 36.0 kN ; K y = 13.60 kN ΣM F = 0: 36(1.2) − 26.4(6.75) + 20(2.25) − FDG (1.2) = 0 8 ΣM D = 0: − 26.4(4.5) + FFG (4.5) = 0 17 FDG = −75 kN FFG = +56.1 kN 15 ΣM G = 0: 20(4.5) − 26.4(9) + FFH 17 (2.4) = 0 FFH = +69.7 kN FDG = 75.0 kN C FFG = 56.1 kN T FFH = 69.7 kN T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 826 PROBLEM 6.51 A stadium roof truss is loaded as shown. Determine the force in members AB, AG, and FG. SOLUTION We pass a section through members AB, AG, and FG, and use the free body shown. ΣM G = 0: 40 41 FAB (6.3 ft) − (1.8 kips)(14 ft) − (0.9 kips)(28 ft) = 0 FAB = +8.20 kips FAB = 8.20 kips T FAG = +4.50 kips FAG = 4.50 kips T 3 ΣM D = 0: − FAG (28 ft) + (1.8 kips)(28 ft) 5 + (1.8 kips)(14 ft) = 0 ΣM A = 0: − FFG (9 ft) − (1.8 kips)(12 ft) − (1.8 kips)(26 ft) − (0.9 kips)(40 ft) = 0 FFG = 11.60 kips C FFG = −11.60 kips PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 827 PROBLEM 6.52 A stadium roof truss is loaded as shown. Determine the force in members AE, EF, and FJ. SOLUTION We pass a section through members AE, EF, and FJ, and use the free body shown. 8 ΣM F = 0: FAE (9 ft) − (1.8 kips)(12 ft) − (1.8 kips)(26 ft) − (0.9 kips)(40 ft) = 0 82 + 92 FAE = +17.46 kips FAE = 17.46 kips T ΣM A = 0: − FEF (9 ft) − (1.8 kips)(12 ft) − (1.8 kips)(26 ft) − (0.9 kips)(40 ft) = 0 FEF = −11.60 kips FEF = 11.60 kips C ΣM E = 0: − FFJ (8 ft) − (0.9 kips)(8 ft) − (1.8 kips)(20 ft) − (1.8 kips)(34 ft) − (0.9 kips)(48 ft) = 0 FFJ = −18.45 kips FFJ = 18.45 kips C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 828 PROBLEM 6.53 Determine the force in members CD and DF of the truss shown. SOLUTION tan α = 5 α = 22.62° 12 sin α = 5 12 cos α = 13 13 Member CD: ΣM I = 0: FCD (9 m) + (10 kN)(9 m) + (10 kN)(6 m) + (10 kN)(3 m) = 0 FCD = 20.0 kN C FCD = −20 kN Member DF: ΣM C = 0: ( FDF cos α )(3.75 m) + (10 kN)(3 m) + (10 kN)(6 m) + (10 kN)(9 m) = 0 FDF cos α = −48 kN 12 FDF = −48 kN 13 FDF = −52.0 kN FDF = 52.0 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 829 PROBLEM 6.54 Determine the force in members CE and EF of the truss shown. SOLUTION Member CE: ΣM F = 0: FCE (2.5 m) − (10 kN)(3 m) − (10 kN)(6 m) = 0 FCE = +36 kN FCE = 36.0 kN T Member EF: ΣM I = 0: FEF (6 m) + (10 kN)(6 m) + (10 kN)(3 m) = 0 FEF = −15 kN FEF = 15.00 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 830 PROBLEM 6.55 The truss shown was designed to support the roof of a food market. For the given loading, determine the force in members FG, EG, and EH. SOLUTION Reactions at supports. Because of the symmetry of the loading, Ax = 0, Ay = O = 1 total load 2 A = O = 4.48 kN We pass a section through members FG, EG, and EH, and use the free body shown. Slope FG = Slope FI = Slope EG = 1.75 m 6m 5.50 m 2.4 m ΣM E = 0: (0.6 kN)(7.44 m) + (1.24 kN)(3.84 m) − (4.48 kN)(7.44 m) 6 − FFG (4.80 m) = 0 6.25 FFG = −5.231 kN FFG = 5.23 kN C ΣM G = 0: FEH (5.50 m) + (0.6 kN)(9.84 m) + (1.24 kN)(6.24 m) + (1.04 kN)(2.4 m) −(4.48 kN)(9.84 m) = 0 ΣFy = 0: FEH = 5.08 kN T 5.50 1.75 (−5.231 kN) + 4.48 kN − 0.6 kN − 1.24 kN − 1.04 kN = 0 FEG + 6.001 6.25 FEG = −0.1476 kN FEG = 0.1476 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 831 PROBLEM 6.56 The truss shown was designed to support the roof of a food market. For the given loading, determine the force in members KM, LM, and LN. SOLUTION Because of symmetry of loading, O= O = 4.48 kN 1 load 2 We pass a section through KM, LM, LN, and use free body shown. 3.84 ΣM M = 0: FLN (3.68 m) 4 + (4.48 kN − 0.6 kN)(3.6 m) = 0 FLN = −3.954 kN FLN = 3.95 kN C ΣM L = 0: − FKM (4.80 m) − (1.24 kN)(3.84 m) + (4.48 kN − 0.6 kN)(7.44 m) = 0 FKM = +5.022 kN ΣFy = 0: FKM = 5.02 kN T 4.80 1.12 (−3.954 kN) − 1.24 kN − 0.6 kN + 4.48 kN = 0 FLM + 6.147 4 FLM = −1.963 kN FLM = 1.963 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 832 PROBLEM 6.57 A Polynesian, or duopitch, roof truss is loaded as shown. Determine the force in members DF, EF, and EG. SOLUTION Free body: Truss: ΣFx = 0: N x = 0 ΣM N = 0: (200 lb)(8a ) + (400 lb)(7a + 6a + 5a )+(350 lb)(4a ) + (300 lb)(3a + 2a + a ) − A(8a ) = 0 A = 1500 lb ΣFy = 0: 1500 lb − 200 lb − 3(400 lb) − 350 lb − 3(300 lb) − 150 lb + N y = 0 N y = 1300 lb N = 1300 lb We pass a section through DF, EF, and EG, and use the free body shown. (We apply FDF at F.) ΣM E = 0: (200 lb)(18 ft) + (400 lb)(12 ft) + (400 lb)(6 ft) − (1500 lb)(18 ft) 18 FDF (4.5 ft) = 0 − 182 + 4.52 FDF = −3711 lb ΣM A = 0: FEF (18 ft) − (400 lb)(6 ft) − (400 lb)(12 ft) = 0 FDF = 3710 lb C FEF = 400 lb T FEF = +400 lb ΣM F = 0: FEG (4.5 ft) − (1500 lb)(18 ft)+(200 lb)(18 ft) + (400 lb)(12 ft) + (400 lb)(6 ft) = 0 FEG = +3600 lb FEG = 3600 lb T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 833 PROBLEM 6.58 A Polynesian, or duopitch, roof truss is loaded as shown. Determine the force in members HI, GI, and GJ. SOLUTION See solution of Problem 6.57 for reactions: A = 1500 lb , N = 1300 lb We pass a section through HI, GI, and GJ, and use the free body shown. (We apply FHI at H.) 6 ΣM G = 0: FHI (8.5 ft) + (1300 lb)(24 ft) − (300 lb)(6 ft) 2 2 6 +4 − (300 lb)(12 ft) − (300 lb)(18 ft) − (150 lb)(24 ft) = 0 FHI = −2375.4 lb FHI = 2375 lb C ΣM I = 0: (1300 lb)(18 ft) − (300 lb)(6 ft) − (300 lb)(12 ft) − (150 lb)(18 ft) − FGJ (4.5 ft) = 0 FGJ = +3400 lb ΣFx = 0: − 4 6 FGI − (−2375.4 lb) − 3400 lb = 0 2 5 6 + 42 FGI = −1779.4 lb FGJ = 3400 lb T FGI = 1779 lb C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 834 PROBLEM 6.59 Determine the force in members DE and DF of the truss shown when P = 20 kips. SOLUTION Reactions: C = K = 2.5P 7.5 18 β = 22.62° tan β = Member DE: ΣM A = 0: (2.5 P )(6 ft) − FDE (12 ft) = 0 FDE = +1.25 P For P = 20 kips, FDE = +1.25(20) = +25 kips FDE = 25.0 kips T Member DF: ΣM E = 0: P (12 ft) − (2.5 P )(6 ft) − FDF cos β (5 ft) = 0 12 P − 15 P − FDF cos 22.62°(5 ft) = 0 FDF = −0.65 P For P = 20 kips, FDF = −0.65(20) = −13 kips FDF = 13.00 kips C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 835 PROBLEM 6.60 Determine the force in members EG and EF of the truss shown when P = 20 kips. SOLUTION Reactions: C = K = 2.5P 7.5 6 α = 51.34° tan α = Member EG: ΣM F = 0: P (18 ft) − 2.5 P (12 ft) − P (6 ft) + FEG (7.5 ft) = 0 FEG = +0.8 P; For P = 20 kips, FEG = 16.00 kips T FEG = 0.8(20) = +16 kips Member EF: ΣM A = 0: 2.5 P (6 ft) − P (12 ft) + FEF sin 51.34°(12 ft) = 0 FEF = −0.320 P; For P = 20 kips, FEF = −0.320(20) = −6.4 kips FEF = 6.40 kips C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 836 PROBLEM 6.61 Determine the force in members EH and GI of the truss shown. (Hint: Use section aa.) SOLUTION Reactions: ΣFx = 0: Ax = 0 ΣM P = 0: 12(45) + 12(30) + 12(15) − Ay (90) = 0 A y = 12 kips ΣFy = 0: 12 − 12 − 12 − 12 + P = 0 P = 24 kips ΣM G = 0: − (12 kips)(30 ft) − FEH (16 ft) = 0 FEH = −22.5 kips ΣFx = 0: FGI − 22.5 kips = 0 FEH = 22.5 kips C FGI = 22.5 kips T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 837 PROBLEM 6.62 Determine the force in members HJ and IL of the truss shown. (Hint: Use section bb.) SOLUTION See the solution to Problem 6.61 for free body diagram and analysis to determine the reactions at supports A and P. A x = 0; A y = 12.00 kips ; P = 24.0 kips ΣM L = 0: FHJ (16 ft) − (12 kips)(15 ft) + (24 kips)(30 ft) = 0 FHJ = −33.75 kips ΣFx = 0: 33.75 kips − FIL = 0 FIL = +33.75 kips FHJ = 33.8 kips C FIL = 33.8 kips T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 838 PROBLEM 6.63 Determine the force in members DG and FI of the truss shown. (Hint: Use section aa.) SOLUTION ΣM F = 0: FDG (4 m) − (5 kN)(3 m) = 0 FDG = +3.75 kN ΣFy = 0: − 3.75 kN − FFI = 0 FFI = −3.75 kN FDG = 3.75 kN T FFI = 3.75 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 839 PROBLEM 6.64 Determine the force in members GJ and IK of the truss shown. (Hint: Use section bb.) SOLUTION ΣM I = 0: FGJ (4 m) − (5 kN)(6 m) − (5 kN)(3 m) = 0 FGJ = +11.25 kN ΣFy = 0: − 11.25 kN − FIK = 0 FIK = −11.25 kN FGJ = 11.25 kN T FIK = 11.25 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 840 PROBLEM 6.65 The diagonal members in the center panels of the truss shown are very slender and can act only in tension; such members are known as counters. Determine the forces in the counters that are acting under the given loading. SOLUTION Free body: Truss: ΣFx = 0: Fx = 0 ΣM H = 0: 4.8(3a) + 4.8(2a) + 4.8a − 2.4a − Fy (2a) = 0 F = 13.20 kips Fy = +13.20 kips ΣFy = 0: H + 13.20 kips − 3(4.8 kips) − 2(2.4 kips) = 0 H = +6.00 kips H = 6.00 kips Free body: ABF: We assume that counter BG is acting. ΣFy = 0: − 9.6 FBG + 13.20 − 2(4.8) = 0 14.6 FBG = +5.475 FBG = 5.48 kips T Since BG is in tension, our assumption was correct. Free body: DEH: We assume that counter DG is acting. ΣFy = 0: − 9.6 FDG + 6.00 − 2(2.4) = 0 14.6 FDG = +1.825 FDG = 1.825 kips T Since DG is in tension, O.K. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 841 PROBLEM 6.66 The diagonal members in the center panels of the truss shown are very slender and can act only in tension; such members are known as counters. Determine the forces in the counters that are acting under the given loading. SOLUTION Free body: Truss: ΣFx = 0: Fx = 0 ΣM G = 0: − Fy a + 4.8(2a) + 4.8a − 2.4a − 2.4(2a) = 0 Fy = 7.20 F = 7.20 kips Free body: ABF: We assume that counter CF is acting. ΣFy = 0: 9.6 FCF + 7.20 − 2(4.8) = 0 14.6 FCF = +3.65 FCF = 3.65 kips T Since CF is in tension, O.K. Free body: DEH: We assume that counter CH is acting. ΣFy = 0: 9.6 FCH − 2(2.4 kips) = 0 14.6 FCH = +7.30 FCH = 7.30 kips T Since CH is in tension, O.K. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 842 PROBLEM 6.67 The diagonal members in the center panels of the power transmission line tower shown are very slender and can act only in tension; such members are known as counters. For the given loading, determine (a) which of the two counters listed below is acting, (b) the force in that counter. Counters CJ and HE SOLUTION Free body: Portion ABDFEC of tower. We assume that counter CJ is acting and show the forces exerted by that counter and by members CH and EJ. ΣFx = 0: 4 FCJ − 2(1.2 kN)sin 20° = 0 FCJ = +1.026 kN 5 Since CJ is found to be in tension, our assumption was correct. Thus, the answers are (a) CJ (b) 1.026 kN T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 843 PROBLEM 6.68 The diagonal members in the center panels of the power transmission line tower shown are very slender and can act only in tension; such members are known as counters. For the given loading, determine (a) which of the two counters listed below is acting, (b) the force in that counter. Counters IO and KN SOLUTION Free body: Portion of tower shown. We assume that counter IO is acting and show the forces exerted by that counter and by members IN and KO. ΣFx = 0 : 4 FIO − 4(1.2 kN)sin 20° = 0 FIO = +2.05 kN 5 Since IO is found to be in tension, our assumption was correct. Thus, the answers are (a) IO (b) 2.05 kN T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 844 PROBLEM 6.69 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.) SOLUTION Structure (a) Number of members: m = 16 Number of joints: n = 10 Reaction components: r=4 m + r = 20, 2 n = 20 (a) m + r = 2n Thus, To determine whether the structure is actually completely constrained and determinate, we must try to find the reactions at the supports. We divide the structure into two simple trusses and draw the free-body diagram of each truss. This is a properly supported simple truss – O.K. This is an improperly supported simple truss. (Reaction at C passes through B. Thus, Eq. ΣM B = 0 cannot be satisfied.) Structure is improperly constrained. Structure (b) m = 16 n = 10 r=4 m + r = 20, 2n = 20 (b) m + r = 2n Thus, PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 845 PROBLEM 6.69 (Continued) We must again try to find the reactions at the supports dividing the structure as shown. Both portions are simply supported simple trusses. Structure is completely constrained and determinate. Structure (c) m = 17 n = 10 r=4 m + r = 21, 2n = 20 (c) m + r > 2n Thus, This is a simple truss with an extra support which causes reactions (and forces in members) to be indeterminate. Structure is completely constrained and indeterminate. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 846 PROBLEM 6.70 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.) SOLUTION Structure (a): Nonsimple truss with r = 4, m = 16, n = 10, so m + r = 20 = 2n, but we must examine further. FBD Sections: FBD I: ΣM A = 0 T1 II: ΣFx = 0 I: ΣFx = 0 T2 I: ΣFy = 0 Ax ΣM E = 0 Ay ΣFy = 0 Cy II: II: Ey Since each section is a simple truss with reactions determined, structure is completely constrained and determinate. Structure (b): Nonsimple truss with r = 3, m = 16, n = 10, so Structure is partially constrained. m + r = 19 < 2n = 20 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 847 PROBLEM 6.70 (Continued) Structure (c): Simple truss with r = 3, m = 17, n = 10, m + r = 20 = 2n, but the horizontal reaction forces Ax and E x are collinear and no equilibrium equation will resolve them, so the structure is improperly constrained and indeterminate. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 848 PROBLEM 6.71 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.) SOLUTION Structure (a): Nonsimple truss with r = 4, m = 12, n = 8 so r + m = 16 = 2n. Check for determinacy: One can solve joint F for forces in EF, FG and then solve joint E for E y and force in DE. This leaves a simple truss ABCDGH with r = 3, m = 9, n = 6 so r + m = 12 = 2n Structure is completely constrained and determinate. Structure (b): Simple truss (start with ABC and add joints alphabetically to complete truss) with r = 4, m = 13, n = 8 so r + m = 17 > 2n = 16 Constrained but indeterminate PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 849 PROBLEM 6.71 (Continued) Structure (c): Nonsimple truss with r = 3, m = 13, n = 8 so r + m = 16 = 2n. To further examine, follow procedure in part (a) above to get truss at left. Since F1 ≠ 0 (from solution of joint F), ΣM A = aF1 ≠ 0 and there is no equilibrium. Structure is improperly constrained. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 850 PROBLEM 6.72 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.) SOLUTION Structure (a) Number of members: m = 12 Number of joints: n=8 Reaction components: r =3 m + r = 15, Thus, 2 n = 16 m + r < 2n Structure is partially constrained. Structure (b) m = 13, n = 8 r =3 m + r = 16, 2n = 16 Thus, m + r = 2n To verify that the structure is actually completely constrained and determinate, we observe that it is a simple truss (follow lettering to check this) and that it is simply supported by a pin-and-bracket and a roller. Thus, structure is completely constrained and determinate. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 851 PROBLEM 6.72 (Continued) Structure (c) m = 13, n = 8 r=4 m + r = 17, 2n = 16 Thus, m + r > 2n Structure is completely constrained and indeterminate. This result can be verified by observing that the structure is a simple truss (follow lettering to check this), therefore it is rigid, and that its supports involve four unknowns. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 852 PROBLEM 6.73 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.) SOLUTION Structure (a): Rigid truss with r = 3, m = 14, n = 8, so Structure (b): r + m = 17 > 2n = 16 so completely constrained but indeterminate Simple truss (start with ABC and add joints alphabetically), with r = 3, m = 13, n = 8, so r + m = 16 = 2n so completely constrained and determinate Structure (c): Simple truss with r = 3, m = 13, n = 8, so r + m = 16 = 2n, but horizontal reactions ( Ax and Dx ) are collinear, so cannot be resolved by any equilibrium equation. Structure is improperly constrained. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 853 PROBLEM 6.74 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.) SOLUTION Structure (a): m = 12 No. of members No. of joints n = 8 m + r = 16 = 2n No. of reaction components r = 4 unknows = equations FBD of EH: ΣM H = 0 FDE ; ΣFx = 0 FGH ; ΣFy = 0 Hy Then ABCDGF is a simple truss and all forces can be determined. This example is completely constrained and determinate. Structure (b): No. of members m = 12 No. of joints n = 8 m + r = 15 < 2n = 16 No. of reaction components r = 3 unknows < equations partially constrained Note: Quadrilateral DEHG can collapse with joint D moving downward; in (a), the roller at F prevents this action. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 854 PROBLEM 6.74 (Continued) Structure (c): No. of members m = 13 No. of joints n = 8 m + r = 17 > 2n = 16 No. of reaction components r = 4 unknows > equations completely constrained but indeterminate PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 855 PROBLEM 6.75 Determine the force in member BD and the components of the reaction at C. SOLUTION We note that BD is a two-force member. The force it exerts on ABC, therefore, is directed along line BD. Free body: ABC: BD = (24) 2 + (10) 2 = 26 in. 10 ΣM C = 0: (160 lb)(30 in.) − FBD (16 in.) = 0 26 FBD = +780 lb ΣM x = 0: C x + FBD = 780 lb T 24 (780 lb) = 0 26 C x = −720 lb ΣFy = 0: C y − 160 lb + 10 (780 lb) = 0 26 C y = −140.0 lb C x = 720 lb C y = 140.0 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 856 PROBLEM 6.76 Determine the force in member BD and the components of the reaction at C. SOLUTION We note that BD is a two-force member. The force it exerts on ABC, therefore, is directed along line BD. Free body: ABC: Attaching FBD at D and resolving it into components, we write ΣM C = 0: 450 FBD (240 mm) = 0 (400 N)(135 mm) + 510 FBD = −255 N ΣFx = 0: C x + FBD = 255 N C 240 (−255 N) = 0 510 C x = +120.0 N ΣFy = 0: C y − 400 N + 450 (−255 N) = 0 510 C y = +625 N C x = 120.0 N C y = 625 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 857 PROBLEM 6.77 Determine the components of all forces acting on member ABCD of the assembly shown. SOLUTION Free body: Entire assembly: ΣM B = 0: D (120 mm) − (480 N)(80 mm) = 0 D = 320 N ΣFx = 0: Bx + 480 N = 0 B x = 480 N ΣFy = 0: By + 320 N = 0 B y = 320 N Free body: Member ABCD: ΣM A = 0: (320 N)(200 mm) − C (160 mm) − (320 N)(80 mm) − (480 N)(40 mm) = 0 C = 120.0 N ΣFx = 0: Ax − 480 N = 0 A x = 480 N ΣFy = 0: Ay − 320 N − 120 N + 320 N = 0 A y = 120.0 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 858 PROBLEM 6.78 Determine the components of all forces acting on member ABD of the frame shown. SOLUTION Free body: Entire frame: ΣM D = 0: (300 lb) − (12 ft) + (450 lb)(4 ft) − E (6 ft) = 0 E = +900 lb E = 900 lb D x = 900 lb ΣFx = 0: Dx − 900 lb = 0 ΣFy = 0: Dy − 300 lb − 450 lb = 0 D y = 750 lb Free body: Member ABD: We note that BC is a two-force member and that B is directed along BC. ΣM A = 0: (750 lb)(16 ft) − (900 lb)(6 ft) − B (8 ft) = 0 B = +825 lb B = 825 lb ΣFx = 0: Ax + 900 lb = 0 Ax = −900 lb ΣFy = 0: Ay + 750 lb − 825 lb = 0 Ay = +75 lb A x = 900 lb A y = 75.0 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 859 PROBLEM 6.79 For the frame and loading shown, determine the components of all forces acting on member ABC. SOLUTION Free body: Entire frame: ΣM E = 0: − Ax (4) − (20 kips)(5) = 0 Ax = −25 kips, A x = 25.0 kips ΣFy = 0: Ay − 20 kips = 0 A y = 20.0 kips Ay = 20 kips Free body: Member ABC: Note: BE is a two-force member, thus B is directed along line BE and By = 2 Bx . 5 ΣM C = 0: (25 kips)(4 ft) − (20 kips)(10 ft) + Bx (2 ft) + By (5 ft) = 0 −100 kip ⋅ ft + Bx (2 ft) + 2 Bx (5 ft) = 0 5 Bx = 25 kips By = B x = 25.0 kips 2 2 ( Bx ) = (25) = 10 kips 5 5 B y = 10.00 kips ΣFx = 0: C x − 25 kips − 25 kips = 0 C x = 50 kips ΣFy = 0: C y + 20 kips − 10 kips = 0 C y = −10 kips C x = 50.0 kips C y = 10.00 kips PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 860 PROBLEM 6.80 Solve Problem 6.79 assuming that the 20-kip load is replaced by a clockwise couple of magnitude 100 kip ⋅ ft applied to member EDC at Point D. PROBLEM 6.79 For the frame and loading shown, determine the components of all forces acting on member ABC. SOLUTION Free body: Entire frame: ΣFy = 0: Ay = 0 ΣM E = 0: − Ax (4 ft) − 100 kip ⋅ ft = 0 Ax = −25 kips A x = 25.0 kips A = 25.0 kips B x = 25.0 kips Free body: Member ABC: Note: BE is a two-force member, thus B is directed along line BE and By = 2 Bx . 5 ΣM C = 0: (25 kips)(4 ft) + Bx (2 ft) + By (5 ft) = 0 100 kip ⋅ ft + Bx (2 ft) + 2 Bx (5 ft) = 0 5 Bx = −25 kips By = B y = 10.00 kips 2 2 Bx = ( −25) = −10 kips; 5 5 ΣFx = 0: − 25 kips + 25 kips + C x = 0 Cx = 0 ΣFy = 0: +10 kips + C y = 0 C y = −10 kips C y = 10 kips C = 10.00 kips PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 861 PROBLEM 6.81 Determine the components of all forces acting on member ABCD when θ = 0. SOLUTION Free body: Entire assembly: ΣM B = 0: A(200) − (150 N)(500) = 0 A = +375 N A = 375 N B x = 375 N ΣFx = 0: Bx + 375 N = 0 Bx = −375 N ΣFy = 0: By − 150 N = 0 By = +150 N B y = 150 N Free body: Member ABCD: We note that D is directed along DE, since DE is a two-force member. ΣM C = 0: D (300) − (150 N)(100) + (375 N)(200) = 0 D = −200 N D = 200 N ΣFx = 0: C x + 375 − 375 = 0 Cx = 0 ΣFy = 0: C y + 150 − 200 = 0 C y = +50.0 N C = 50.0 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 862 PROBLEM 6.82 Determine the components of all forces acting on member ABCD when θ = 90°. SOLUTION Free body: Entire assembly: ΣM B = 0: A(200) − (150 N)(200) = 0 A = +150.0 N A = 150.0 N ΣFx = 0: Bx + 150 − 150 = 0 Bx = 0 B = 0 ΣFy = 0: By = 0 Free body: Member ABCD: We note that D is directed along DE, since DE is a two-force member. ΣM C = 0: D (300) + (150 N)(200) = 0 D = −100.0 N D = 100.0 N ΣFx = 0: C x + 150 N = 0 C x = −150 N ΣFy = 0: C y − 100 N = 0 C y = +100.0 N Cx = 150.0 N C y = 100.0 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 863 PROBLEM 6.83 Determine the components of the reactions at A and E if a 750-N force directed vertically downward is applied (a) at B, (b) at D. SOLUTION Free body: Entire frame: The following analysis is valid for both parts (a) and (b) since position of load on its line of action is immaterial. ΣM E = 0: − (750 N)(80 mm) − Ax (200 mm) = 0 Ax = − 300 N A x = 300 N ΣFx = 0: E x − 300 N = 0 E x = 300 N E x = 300 N ΣFy = 0: Ay + E y − 750 N = 0 (a) (1) Load applied at B. Free body: Member CE: CE is a two-force member. Thus, the reaction at E must be directed along CE. Ey 300 N From Eq. (1): = 75 mm 250 mm E y = 90 N Ay + 90 N − 750 N = 0 Ay = 660 N Thus, reactions are (b) A x = 300 N , E x = 300 N , A y = 660 N E y = 90.0 N Load applied at D. Free body: Member AC: AC is a two-force member. Thus, the reaction at A must be directed along AC. Ay 300 N = 125 mm 250 mm Ay = 150 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 864 PROBLEM 6.83 (Continued) Ay + E y − 750 N = 0 From Eq. (1): 150 N + E y − 750 N = 0 E y = 600 N E y = 600 N Thus, reactions are A x = 300 N , E x = 300 N , A y = 150.0 N E y = 600 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 865 PROBLEM 6.84 Determine the components of the reactions at A and E if a 750-N force directed vertically downward is applied (a) at B, (b) at D. SOLUTION Free-body: Entire frame: The following analysis is valid for both parts (a) and (b) since position of load on its line of action is immaterial. ΣM E = 0: − (750 N)(240 mm) − Ax (400 mm) = 0 Ax = −450 N A x = 450 N ΣFx = 0: E x − 450 N = 0 E x = 450 N E x = 450 N ΣFy = 0: Ay + E y − 750 N = 0 (a) (1) Load applied at B. Free body: Member CE: CE is a two-force member. Thus, the reaction at E must be directed along CE. Ey 450 N From Eq. (1): = 240 mm ; E y = 225 N 480 mm Ay + 225 − 750 = 0; A y = 525 N Thus, reactions are A x = 450 N , , E y = 225 N A x = 450 N , A y = 150.0 N E x = 450 N , E x = 450N (b) A y = 525 N Load applied at D. Free body: Member AC: AC is a two-force member. Thus, the reaction at A must be directed along AC. Ay 450 N From Eq. (1): = 160 mm 480 mm A y = 150.0 N Ay + E y − 750 N = 0 150 N + E y − 750 N = 0 E y = 600 N E y = 600 N Thus, reactions are E y = 600 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 866 PROBLEM 6.85 Determine the components of the reactions at A and E if the frame is loaded by a clockwise couple of magnitude 36 N · m applied (a) at B, (b) at D. SOLUTION Free body: Entire frame: The following analysis is valid for both parts (a) and (b) since the point of application of the couple is immaterial. ΣM E = 0: −36 N ⋅ m − Ax (0.2 m) = 0 Ax = − 180 N A x = 180 N ΣFx = 0: − 180 N + E x = 0 E x = 180 N E x = 180 N ΣFy = 0: Ay + E y = 0 (a) (1) Couple applied at B. Free body: Member CE: AC is a two-force member. Thus, the reaction at E must be directed along EC. Ey 180 N From Eq. (1): = 0.075 m 0.25 m E y = 54 N Ay + 54 N = 0 Ay = − 54 N A y = 54.0 N Thus, reactions are (b) A x = 180.0 N , E x = 180.0 N , A y = 54.0 N E y = 54.0 N Couple applied at D. Free body: Member AC: AC is a two-force member. Thus, the reaction at A must be directed along EC. Ay 180 N = 0.125 m 0.25 m Ay = 90 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 867 PROBLEM 6.85 (Continued) From Eq. (1): Ay + E y = 0 90 N + E y = 0 E y = − 90 N E y = 90 N Thus, reactions are A x = 180.0 N E x = −180.0 N A y = 90.0 N , , E y = 90.0 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 868 PROBLEM 6.86 Determine the components of the reactions at A and E if the frame is loaded by a clockwise couple of magnitude 36 N · m applied (a) at B, (b) at D. SOLUTION Free body: Entire frame: The following analysis is valid for both parts (a) and (b) since the point of application of the couple is immaterial. ΣM E = 0: −36 N ⋅ m − Ax (0.4 m) = 0 Ax = − 90 N A x = 90.0 N ΣFx = 0: −90 +E x = 0 E x = 90 N E x = 90.0 N ΣFy = 0: Ay + E y = 0 (a) (1) Couple applied at B. Free body: Member CE: AC is a two-force member. Thus, the reaction at E must be directed along EC. Ey 90 N From Eq. (1): = 0.24 m ; E y = 45.0 N 0.48 m Ay + 45 N = 0 Ay = − 45 N A y = 45.0 N Thus, reactions are (b) A x = 90.0 N , E x = 90.0 N , A y = 45.0 N E y = 45.0 N Couple applied at D. Free body: Member AC: AC is a two-force member. Thus, the reaction at A must be directed along AC. Ay 90 N = 0.16 m ; A y = 30 N 0.48 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 869 PROBLEM 6.86 (Continued) From Eq. (1): Ay + E y = 0 30 N + E y = 0 E y = − 30 N E y = 30 N A x = 90.0 N Thus, reactions are E x = − 90.0 N A y = 30.0 N , , E y = 30.0 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 870 PROBLEM 6.87 Determine all the forces exerted on member AI if the frame is loaded by a clockwise couple of magnitude 1200 lb · in. applied (a) at Point D, (b) at Point E. SOLUTION Free body: Entire frame: Location of couple is immaterial. ΣM H = 0: I (48 in.) − 1200 lb ⋅ in. = 0 I = +25.0 lb (a) and (b) I = 25.0 lb We note that AB, BC, and FG are two-force members. Free body: Member AI: tan α = (a) 20 5 = 48 12 α = 22.6° Couple applied at D. ΣFy = 0: − ΣM G = 0: 5 A + 25 lb = 0 13 A = +65.0 lb A = 65.0 lb 12 (65 lb)(40 in.) − C (20 in.) = 0 13 C = +120 lb ΣFx = 0: − 12 (65 lb) + 120 lb + G = 0 13 G = −60.0 lb 22.6° C = 120 lb G = 60 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 871 PROBLEM 6.87 (Continued) (b) Couple applied at E. ΣFy = 0: − ΣM G = 0: 5 A + 25 lb = 0 13 A = +65.0 lb A = 65.0 lb 22.6° 12 (65 lb) + (40 in.) − C (20 in.) − 1200 lb ⋅ in. = 0 13 C = +60.0 lb C = 60.0 lb ΣFx = 0: − 12 (65 lb) + 60 lb + G = 0 13 G=0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 872 PROBLEM 6.88 Determine all the forces exerted on member AI if the frame is loaded by a 40-lb force directed horizontally to the right and applied (a) at Point D, (b) at Point E. SOLUTION Free body: Entire frame: Location of 40-lb force on its line of action DE is immaterial. ΣM H = 0: I (48 in.) − (40 lb)(30 in.) = 0 I = +25.0 lb (a) and (b) I = 25.0 lb We note that AB, BC, and FG are two-force members. Free body: Member AI: tan α = (a) 20 5 = 48 12 α = 22.6° Force applied at D. ΣFy = 0: − ΣM G = 0: 5 A + 25 lb = 0 13 A = +65.0 lb 12 (65 lb)(40 in.) − C (20 in.) = 0 13 C = +120.0 lb ΣFx = 0: − 12 (65 lb) + 120 lb + G = 0 13 G = −60.0 lb A = 65.0 lb 22.6° C = 120.0 lb G = 60.0 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 873 PROBLEM 6.88 (Continued) (b) Force applied at E. ΣFy = 0: − ΣM G = 0: 5 A + 25 lb = 0 13 A = +65.0 lb A = 65.0 lb 22.6° 12 (65 lb)(40 in.) − C (20 in.) − (40 lb)(10 in.) = 0 13 C = +100.0 lb C = 100.0 lb ΣFx = 0: − 12 (65 lb) + 100 lb + 40 lb + G = 0 13 G = −80.0 lb G = 80.0 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 874 PROBLEM 6.89 Determine the components of the reactions at A and B, (a) if the 500-N load is applied as shown, (b) if the 500-N load is moved along its line of action and is applied at Point F. SOLUTION Free body: Entire frame: Analysis is valid for either parts (a) or (b), since position of 100-lb load on its line of action is immaterial. ΣM A = 0: By (10) − (100 lb)(6) = 0 By = + 60 lb ΣFy = 0: Ay + 60 − 100 = 0 Ay = + 40 lb ΣFx = 0: Ax + Bx = 0 (a) (1) Load applied at E. Free body: Member AC: Since AC is a two-force member, the reaction at A must be directed along CA. We have Ax 40 lb = 10 in. 5 in. From Eq. (1): A x = 80.0 lb , A y = 40.0 lb B x = 80.0 lb , B y = 60.0 lb − 80 + Bx = 0 Bx = + 80 lb Thus, PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 875 PROBLEM 6.89 (Continued) (b) Load applied at F. Free body: Member BCD: Since BCD is a two-force member (with forces applied at B and C only), the reaction at B must be directed along CB. We have, therefore, Bx = 0 The reaction at B is From Eq. (1): The reaction at A is B y = 60.0 lb Bx = 0 Ax + 0 = 0 Ax = 0 Ax = 0 A y = 40.0 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 876 PROBLEM 6.90 (a) Show that when a frame supports a pulley at A, an equivalent loading of the frame and of each of its component parts can be obtained by removing the pulley and applying at A two forces equal and parallel to the forces that the cable exerted on the pulley. (b) Show that if one end of the cable is attached to the frame at a Point B, a force of magnitude equal to the tension in the cable should also be applied at B. SOLUTION First note that, when a cable or cord passes over a frictionless, motionless pulley, the tension is unchanged. ΣM C = 0: rT1 − rT2 = 0 T1 = T2 (a) Replace each force with an equivalent force-couple. (b) Cut the cable and replace the forces on pulley with equivalent pair of forces at A as above. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 877 PROBLEM 6.91 A 3-ft-diameter pipe is supported every 16 ft by a small frame like that shown. Knowing that the combined weight of the pipe and its contents is 500 lb/ft and assuming frictionless surfaces, determine the components (a) of the reaction at E, (b) of the force exerted at C on member CDE. SOLUTION Free body: 16-ft length of pipe: W = (500 lb/ft)(16 ft) = 8 kips Force Triangle B D 8 kips = = 3 5 4 B = 6 kips D = 10 kips Determination of CB = CD. We note that horizontal projection of BO + OD = horizontal projection of CD Thus, 3 4 r + r = (CD ) 5 5 8 CB = CD = r = 2(1.5 ft) = 3 ft 4 Free body: Member ABC: ΣM A = 0: Cx h − (6 kips)(h − 3) Cx = h−3 (6 kips) h Cx = 9−3 (6 kips) = 4 kips 9 (1) For h = 9 ft, PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 878 PROBLEM 6.91 (Continued) Free body: Member CDE: From above, we have Cx = 4.00 kips ΣM E = 0: (10 kips)(7 ft) − (4 kips)(6 ft) − C y (8 ft) = 0 C y = +5.75 kips, 3 ΣFx = 0: − 4 kips + (10 kips) + Ex = 0 5 Ex = −2 kips, 4 ΣFy = 0: 5.75 kips − (10 kips) + E y = 0, 5 C y = 5.75 kips E x = 2.00 kips E y = 2.25 kips PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 879 PROBLEM 6.92 Solve Problem 6.91 for a frame where h = 6 ft. PROBLEM 6.91 A 3-ft-diameter pipe is supported every 16 ft by a small frame like that shown. Knowing that the combined weight of the pipe and its contents is 500 lb/ft and assuming frictionless surfaces, determine the components (a) of the reaction at E, (b) of the force exerted at C on member CDE. SOLUTION See solution of Problem 6.91 for derivation of Eq. (1). For h = 6 ft, C x = h−3 6−3 = 3 kips (6 kips) = h 6 Free body: Member CDE: From above, we have Cx = 3.00 kips ΣM E = 0: (10 kips)(7 ft) − (3 kips)(6 ft) − C y (8 ft) = 0 C y = +6.50 kips, 3 ΣFx = 0: −3 kips + (10 kips) + Ex = 0 5 Ex = −3.00 kips, 4 ΣFy = 0: 6.5 kips − (10 kips) + E y = 0 5 E y = 1.500 kips C y = 6.50 kips E x = 3.00 kips E y = 1.500 kips PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 880 PROBLEM 6.93 Knowing that the pulley has a radius of 0.5 m, determine the components of the reactions at A and E. SOLUTION FBD Frame: ΣM A = 0: (7 m) E y − (4.5 m)(700 N) = 0 ΣFy = 0: Ay − 700 N + 450 N = 0 ΣFx = 0: Ax − E x = 0 E y = 450 N A y = 250 N Ax = E x Dimensions in m FBD Member ABC: ΣM C = 0: (1 m)(700 N) − (1 m)(250 N) − (3 m) Ax = 0 A x = 150.0 N so E x = 150.0 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 881 PROBLEM 6.94 Knowing that the pulley has a radius of 50 mm, determine the components of the reactions at B and E. SOLUTION Free body: Entire assembly: ΣM E = 0 : − (300 N)(350 mm) − Bx (150 mm) = 0 Bx = −700 N B x = 700 N ΣFx = 0: − 700 N + E x = 0 E x = 700 N E x = 700 N ΣFy = 0: By + E y − 300 N = 0 (1) Free body: Member ACE: ΣM C = 0: (700 N)(150 mm) − (300 N)(50 mm) − E y (180 mm) = 0 E y = 500 N From Eq. (1): E y = 500 N By + 500 N − 300 N = 0 By = −200 N B y = 200 N Thus, reactions are B x = 700 N , E x = 700 N , B y = 200 N E y = 500 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 882 PROBLEM 6.95 A trailer weighing 2400 lb is attached to a 2900-lb pickup truck by a ball-and-socket truck hitch at D. Determine (a) the reactions at each of the six wheels when the truck and trailer are at rest, (b) the additional load on each of the truck wheels due to the trailer. SOLUTION (a) Free body: Trailer: (We shall denote by A, B, C the reaction at one wheel.) ΣM A = 0: − (2400 lb)(2 ft) + D (11 ft) = 0 D = 436.36 lb ΣFy = 0: 2 A − 2400 lb + 436.36 lb = 0 A = 982 lb A = 981.82 lb Free body: Truck. ΣM B = 0: (436.36 lb)(3 ft) − (2900 lb)(5 ft) + 2C (9 ft) = 0 C = 732.83 lb C = 733 lb ΣFy = 0: 2B − 436.36 lb − 2900 lb + 2(732.83 lb) = 0 B = 935.35 lb (b) B = 935 lb Additional load on truck wheels. Use free body diagram of truck without 2900 lb. ΣM B = 0: (436.36 lb)(3 ft) + 2C (9 ft) = 0 C = −72.73 lb ΣFy = 0: 2B − 436.36 lb − 2(72.73 lb) = 0 B = 290.9 lb ΔC = −72.7 lb ΔB = +291 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 883 PROBLEM 6.96 In order to obtain a better weight distribution over the four wheels of the pickup truck of Problem 6.95, a compensating hitch of the type shown is used to attach the trailer to the truck. The hitch consists of two bar springs (only one is shown in the figure) that fit into bearings inside a support rigidly attached to the truck. The springs are also connected by chains to the trailer frame, and specially designed hooks make it possible to place both chains in tension. (a) Determine the tension T required in each of the two chains if the additional load due to the trailer is to be evenly distributed over the four wheels of the truck. (b) What are the resulting reactions at each of the six wheels of the trailer-truck combination? PROBLEM 6.95 A trailer weighing 2400 lb is attached to a 2900-lb pickup truck by a ball-and-socket truck hitch at D. Determine (a) the reactions at each of the six wheels when the truck and trailer are at rest, (b) the additional load on each of the truck wheels due to the trailer. SOLUTION (a) We small first find the additional reaction Δ at each wheel due to the trailer. Free body diagram: (Same Δ at each truck wheel) ΣM A = 0: − (2400 lb)(2 ft) + 2 Δ (14 ft) + 2 Δ (23 ft) = 0 Δ = 64.86 lb ΣFy = 0: 2 A − 2400 lb + 4(64.86 lb) = 0; A = 1070 lb; A = 1070 lb Free body: Truck: (Trailer loading only) ΣM D = 0: 2 Δ (12 ft) + 2 Δ (3 ft) − 2T (1.7 ft) = 0 T = 8.824Δ = 8.824(64.86 lb) T = 572.3 lb T = 572 lb Free body: Truck: (Truck weight only) ΣM B = 0: − (2900 lb)(5 ft) + 2C ′(9 ft) = 0 C ′ = 805.6 lb C′ = 805.6 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 884 PROBLEM 6.96 (Continued) ΣFy = 0: 2B′ − 2900 lb + 2(805.6 lb) = 0 B′ = 644.4 lb Actual reactions: B = B′ + Δ = 644.4 lb + 64.86 = 709.2 lb C = C ′ + Δ = 805.6 lb + 64.86 = 870.46 lb From part a: B′ = 644.4 lb B = 709 lb C = 870 lb A = 1070 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 885 PROBLEM 6.97 The cab and motor units of the front-end loader shown are connected by a vertical pin located 2 m behind the cab wheels. The distance from C to D is 1 m. The center of gravity of the 300-kN motor unit is located at Gm, while the centers of gravity of the 100-kN cab and 75-kN load are located, respectively, at Gc and Gl. Knowing that the machine is at rest with its brakes released, determine (a) the reactions at each of the four wheels, (b) the forces exerted on the motor unit at C and D. SOLUTION (a) Free body: Entire machine: A = Reaction at each front wheel B = Reaction at each rear wheel ΣM A = 0: 75(3.2 m) − 100(1.2 m) + 2 B (4.8 m) − 300(5.6 m) = 0 2 B = 325 kN B = 162.5 kN ΣFy = 0: 2 A + 325 − 75 − 100 − 300 = 0 2 A = 150 kN A = 75.0 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 886 PROBLEM 6.97 (Continued) (b) Free body: Motor unit: ΣM D = 0: C (1 m) + 2 B (2.8 m) − 300(3.6 m) = 0 C = 1080 − 5.6 B Recalling (1) B = 162.5 kN, C = 1080 − 5.6(162.5) = 170 kN C = 170.0 kN ΣFx = 0: Dx − 170 = 0 D x = 170.0 kN ΣFy = 0: 2(162.5) − Dy − 300 = 0 D y = 25.0 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 887 PROBLEM 6.98 Solve Problem 6.97 assuming that the 75-kN load has been removed. PROBLEM 6.97 The cab and motor units of the front-end loader shown are connected by a vertical pin located 2 m behind the cab wheels. The distance from C to D is 1 m. The center of gravity of the 300-kN motor unit is located at Gm, while the centers of gravity of the 100-kN cab and 75-kN load are located, respectively, at Gc and Gl. Knowing that the machine is at rest with its brakes released, determine (a) the reactions at each of the four wheels, (b) the forces exerted on the motor unit at C and D. SOLUTION (a) Free body: Entire machine: A = Reaction at each front wheel B = Reaction at each rear wheel ΣM A = 0: 2 B (4.8 m) − 100(1.2 m) − 300(5.6 m) = 0 2 B = 375 kN B = 187.5 kN ΣFy = 0: 2 A + 375 − 100 − 300 = 0 2 A = 25 kN (b) A = 12.50 kN Free body: Motor unit: See solution of Problem 6.97 for free body diagram and derivation of Eq. (1). With B = 187.5 kN, we have C = 1080 − 5.6(187.5) = 30 kN C = 30.0 kN ΣFx = 0: Dx − 30 = 0 D x = 30.0 kN ΣFy = 0: 2(187.5) − Dy − 300 = 0 D y = 75.0 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 888 PROBLEM 6.99 For the frame and loading shown, determine the components of all forces acting on member ABE. SOLUTION FBD Frame: ΣM E = 0: (1.8 m) Fy − (2.1 m)(12 kN) = 0 Fy = 14.00 kN ΣFy = 0: − E y + 14.00 kN − 12 kN = 0 E y = 2 kN E y = 2.00 kN FBD member BCD: ΣM B = 0: (1.2 m)C y − (12 kN)(1.8 m) = 0 C y = 18.00 kN But C is ⊥ ACF, so Cx = 2C y ; Cx = 36.0 kN ΣFx = 0: − Bx + C x = 0 Bx = C x = 36.0 kN Bx = 36.0 kN on BCD ΣFy = 0: − By + 18.00 kN − 12 kN = 0 By = 6.00 kN on BCD On ABE: B x = 36.0 kN E x = 23.0 kN B y = 6.00 kN FBD member ABE: ΣM A = 0: (1.2 m)(36.0 kN) − (0.6 m)(6.00 kN) + (0.9 m)(2.00 kN) − (1.8 m)( E x ) = 0 ΣFx = 0: − 23.0 kN + 36.0 kN − Ax = 0 ΣFy = 0: − 2.00 kN + 6.00 kN − Ay = 0 A x = 13.00 kN A y = 4.00 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 889 PROBLEM 6.100 For the frame and loading shown, determine the components of all forces acting on member ABE. PROBLEM 6.99 For the frame and loading shown, determine the components of all forces acting on member ABE. SOLUTION FBD Frame: ΣM F = 0: (1.2 m)(2400 N) − (4.8 m) E y = 0 E y = 600 N FBD member BC: Cy = 4.8 8 Cx = Cx 5.4 9 ΣM C = 0: (2.4 m) By − (1.2 m)(2400 N) = 0 By = 1200 N B y = 1200 N On ABE: ΣFy = 0: −1200 N + C y − 2400 N = 0 C y = 3600 N 9 Cx = C y 8 so ΣFx = 0: − Bx + C x = 0 Bx = 4050 N C x = 4050 N on BC B x = 4050 N On ABE: PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 890 PROBLEM 6.100 (Continued) FBD member AB0E: ΣM A = 0: a (4050 N) − 2 aE x = 0 E x = 2025 N ΣFx = 0 : − Ax + (4050 − 2025) N = 0 ΣFy = 0: 600 N + 1200 N − Ay = 0 E x = 2025 N A x = 2025 N A y = 1800 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 891 PROBLEM 6.101 For the frame and loading shown, determine the components of all forces acting on member ABD. SOLUTION Free body: Entire frame: Σ M A = 0: E (12 in.) − (360 lb)(15 in.) − (240 lb)(33 in.) = 0 E = +1110 lb E = +1110 lb ΣFx = 0: Ax + 1110 lb = 0 Ax = −1110 lb A x = 1110 lb ΣFy = 0: Ay − 360 lb − 240 lb = 0 A y = 600 lb Ay = +600 lb Free body: Member CDE: Σ M C = 0: (1110 lb)(24 in.) − Dx (12 in.) = 0 Dx = +2220 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 892 PROBLEM 6.101 (Continued) Free body: Member ABD: D x = 2220 lb From above: ΣM B = 0: Dy (18 in.) − (600 lb)(6 in.) = 0 Dy = +200 lb D y = 200 lb ΣFx = 0: Bx + 2220 lb − 1110 lb = 0 Bx = −1110 lb ΣFy = 0: By + 200 lb + 600 lb = 0 By = −800 lb B x = 1110 lb B y = 800 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 893 PROBLEM 6.102 Solve Problem 6.101 assuming that the 360-lb load has been removed. PROBLEM 6.101 For the frame and loading shown, determine the components of all forces acting on member ABD. SOLUTION Free body diagram of entire frame. ΣM A = 0: E (12 in.) − (240 lb)(33 in.) = 0 E = +660 lb E = 660 lb ΣFx = 0: Ax + 660 lb = 0 Ax = −660 lb A x = 660 lb ΣFy = 0: Ay − 240 lb = 0 A y = 240 lb Ay = +240 lb Free body: Member CDE: ΣM C = 0: (660 lb)(24 in.) − Dx (12 in.) = 0 Dx = +1320 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 894 PROBLEM 6.102 (Continued) Free body: Member ABD: D x = 1320 lb From above: ΣM B = 0: Dy (18 in.) − (240 lb)(6 in.) = 0 Dy = +80 lb D y = 80.0 lb ΣFx = 0: Bx + 1320 lb − 660 lb = 0 Bx = −660 lb ΣFy = 0: By + 80 lb + 240 lb = 0 By = −320 lb B x = 660 lb B y = 320 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 895 PROBLEM 6.103 For the frame and loading shown, determine the components of the forces acting on member CDE at C and D. SOLUTION Free body: Entire frame: ΣM y = 0: Ay − 25 lb = 0 A y = 25 lb Ay = 25 lb ΣM F = 0: Ax (6.928 + 2 × 3.464) − (25 lb)(12 in.) = 0 Ax = 21.651 lb A y = 21.65 lb ΣFx = 0: F − 21.651 lb = 0 F = 21.651 lb F = 21.65 lb Free body: Member CDE: ΣM C = 0: Dy (4 in.) − (25 lb)(10 in.) = 0 Dy = +62.5 lb ΣFy = 0: −C y + 62.5 lb − 25 lb = 0 C y = +37.5 lb D y = 62.5 lb C y = 37.5 lb Free body: Member ABD: ΣM B = 0: Dx (3.464 in.) + (21.65 lb)(6.928 in.) −(25 lb)(4 in.) − (62.5 lb)(2 in.) = 0 Dx = +21.65 lb Return to free body: Member CDE: From above: Dx = +21.65 lb D x = 21.7 lb C x = 21.7 lb ΣFx = 0: C x − 21.65 lb C x = +21.65 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 896 PROBLEM 6.104 For the frame and loading shown, determine the components of the forces acting on member CFE at C and F. SOLUTION Free body: Entire frame: ΣM D = 0: (40 lb)(13 in.) + Ax (10 in.) = 0 Ax = −52 lb, A x = 52 lb Free body: Member ABF: ΣM B = 0: − (52 lb)(6 in.) + Fx (4 in.) = 0 Fx = +78 lb Free body: Member CFE: Fx = 78.0 lb From above: ΣM C = 0: (40 lb)(9 in.) − (78 lb)(4 in.) − Fy (4 in.) = 0 Fy = +12 lb Fy = 12.00 lb ΣFx = 0: C x − 78 lb = 0 C x = +78 lb ΣFy = 0: − 40 lb + 12 lb + C y = 0; C y = +28 lb C x = 78.0 lb C y = 28.0 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 897 PROBLEM 6.105 For the frame and loading shown, determine the components of all forces acting on member ABD. SOLUTION Free body: Entire frame: Dimensions in mm ΣM A = 0: F (500) − (3 kN)(600) − (2 kN)(1000) = 0 F = +7.60 kN F = 7.60 kN ΣFx = 0: Ax + 7.60 N = 0, Ax = −7.60 N ΣFy = 0: Ay − 3 kN − 2 kN = 0 Ay = +5 kN A x = 7.60 kN A y = 5.00 kN Free body: Member CDE: ΣM C = 0: D y (400) − (3 kN)(200) − (2 kN)(600) = 0 Dy = +4.50 kN Free body: Member ABD: D y = 4.50 kN From above: ΣM B = 0: Dx (200) − (4.50 kN)(400) − (5 kN)(400) = 0 Dx = +19 kN D x = 19.00 kN B x = 11.40 kN ΣFx = 0: Bx + 19 kN − 7.60 kN = 0 Bx = −11.40 kN ΣFy = 0: By + 5 kN − 4.50 kN = 0 By = −0.50 kN B y = 0.500 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 898 PROBLEM 6.106 Solve Problem 6.105 assuming that the 3-kN load has been removed. PROBLEM 6.105 For the frame and loading shown, determine the components of all forces acting on member ABD. SOLUTION Free body: Entire frame: Dimensions in mm ΣM A = 0: F (500) − (2 kN)(1000) = 0 F = +4 kN F = 4.00 kN ΣFx = 0: Ax + 4 kN = 0, Ax = − 4 kN A x = 4.00 kN ΣFy = 0: Ay − 2 kN = 0, A y = 2.00 kN Ay = +2 kN Free body: Member CDE: ΣM C = 0: D y (400) − (2 kN)(600) = 0 Dy = +3.00 kN Free body: Member ABD: D y = 3.00 kN From above: ΣM B = 0: Dx (200) − (3 kN)(400) − (2 kN)(400) = 0 Dx = +10.00 kN D x = 10.00 kN ΣFx = 0: Bx + 10 kN − 4 kN = 0 Bx = −6 kN ΣFy = 0: By + 2 kN − 3 kN = 0 By = +1 kN B x = 6.00 kN B y = 1.000 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 899 PROBLEM 6.107 Determine the reaction at F and the force in members AE and BD. SOLUTION Free body: Entire frame: ΣM C = 0: Fy (9 in.) − (450 lb)(24 in.) = 0 Fy = 1200 lb Fy = 1200 lb Free body: Member DEF: ΣM J = 0: (1200 lb)(4.5 in.) − Fx (18 in.) = 0 Fx = 300 lb 3 ΣM D = 0: − Fx (24 in.) − FAE (12 in.) = 0 5 10 10 FAE = − Fx = − (300 lb) 3 3 FAE = −1000 lb ΣFy = 0: 1200 lb + FBD = 4 4 (−1000 lb) − FBD = 0 5 5 5 (1200 lb) − 1000 lb = +500 lb 4 Fx = 300 lb FAE = 1000 lb C FBD = 500 lb T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 900 PROBLEM 6.108 For the frame and loading shown, determine the reactions at A, B, D, and E. Assume that the surface at each support is frictionless. SOLUTION Free body: Entire frame: ΣFx = 0: A − B + (1000 lb)sin 30° = 0 A − B + 500 = 0 (1) ΣFy = 0: D + E − (1000 lb) cos 30° = 0 D + E − 866.03 = 0 (2) Free body: Member ACE: ΣM C = 0: − A(6 in.) + E (8 in.) = 0 E= 3 A 4 (3) Free body: Member BCD: ΣM C = 0: − D(8 in.) + B (6 in.) = 0 D= 3 B 4 (4) Substitute E and D from Eqs. (3) and (4) into Eq. (2): − 3 3 A + B − 866.06 = 0 4 4 A + B − 1154.71 = 0 (5) From Eq. (1): A − B + 500 = 0 (6) Eqs. (5) + (6): 2 A − 654.71 = 0 A = 327.4 lb A = 327 lb Eqs. (5) − (6): 2 B − 1654.71 = 0 B = 827.4 lb B = 827 lb From Eq. (4): D = 3 (827.4) 4 D = 620.5 lb D = 621 lb From Eq. (3): E = 3 (327.4) 4 E = 245.5 lb E = 246 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 901 PROBLEM 6.109 The axis of the three-hinge arch ABC is a parabola with the vertex at B. Knowing that P = 112 kN and Q = 140 kN, determine (a) the components of the reaction at A, (b) the components of the force exerted at B on segment AB. SOLUTION Free body: Segment AB: ΣM A = 0: Bx (3.2 m) − B y (8 m) − P(5 m) = 0 (1) Bx (2.4 m) − By (6 m) − P (3.75 m) = 0 (2) 0.75 (Eq. 1): Free body: Segment BC: ΣM C = 0: Bx (1.8 m) + B y (6 m) − Q(3 m) = 0 Add Eqs. (2) and (3): (3) 4.2 Bx − 3.75P − 3Q = 0 Bx = (3.75 P + 3Q)/4.2 From Eq. (1): (3.75P + 3Q) (4) 3.2 − 8B y − 5 P = 0 4.2 B y = (− 9P + 9.6Q )/33.6 (5) given that P = 112 kN and Q = 140 kN. (a) Reaction at A. Considering again AB as a free body, ΣFx = 0: Ax − Bx = 0; Ax = Bx = 200 kN A x = 200 kN ΣFy = 0: Ay − P − B y = 0 Ay − 112 kN − 10 kN = 0 Ay = + 122 kN A y = 122.0 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 902 PROBLEM 6.109 (Continued) (b) Force exerted at B on AB. From Eq. (4): Bx = (3.75 × 112 + 3 × 140)/4.2 = 200 kN B x = 200 kN From Eq. (5): B y = (− 9 × 112 + 9.6 × 140)/33.6 = + 10 kN B y = 10.00 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 903 PROBLEM 6.110 The axis of the three-hinge arch ABC is a parabola with the vertex at B. Knowing that P = 140 kN and Q = 112 kN, determine (a) the components of the reaction at A, (b) the components of the force exerted at B on segment AB. SOLUTION Free body: Segment AB: ΣM A = 0: Bx (3.2 m) − B y (8 m) − P(5 m) = 0 (1) Bx (2.4 m) − By (6 m) − P (3.75 m) = 0 (2) 0.75 (Eq. 1): Free body: Segment BC: ΣM C = 0: Bx (1.8 m) + B y (6 m) − Q(3 m) = 0 Add Eqs. (2) and (3): (3) 4.2 Bx − 3.75P − 3Q = 0 Bx = (3.75 P + 3Q)/4.2 (3.75P + 3Q) From Eq. (1): (4) 3.2 − 8B y − 5 P = 0 4.2 B y = (− 9P + 9.6Q )/33.6 (5) given that P = 140 kN and Q = 112 kN. (a) Reaction at A. ΣFx = 0: Ax − Bx = 0; Ax = Bx = 205 kN A x = 205 kN ΣFy = 0: Ay − P − B y = 0 Ay − 140 kN − (− 5.5 kN) = 0 A y = 134.5 kN Ay = 134.5 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 904 PROBLEM 6.110 (Continued) (b) Force exerted at B on AB. From Eq. (4): Bx = (3.75 × 140 + 3 × 112)/4.2 = 205 kN B x = 205 kN From Eq. (5): B y = (− 9 × 140 + 9.6 × 112)/33.6 = − 5.5 kN B y = 5.50 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 905 PROBLEM 6.111 Members ABC and CDE are pin-connected at C and supported by four links. For the loading shown, determine the force in each link. SOLUTION Member FBDs: I FBD I: FBD II: FBDs combined: ΣM B = 0: II 1 aC y − a 1 M D = 0: aC y − a ΣM G = 0: aP − a FAF = 0 FAF = 2C y 2 2 1 2 FEH = 0 FEH = 2C y 1 FAF − a 2 FEH = 0 P = Cy = P 2 1 2 2C y + 1 2C y 2 so FAF = 2 P C 2 FEH = 2 P T 2 FBD I: FBD II: ΣFy = 0: 1 2 FAF + ΣFy = 0: − C y + 1 2 1 2 FBG − P + C y = 0 FDG − 1 2 FEH = 0 − 1 P P + FBG − P + = 0 2 2 2 FBG = 0 1 P P + FDG − = 0 2 2 2 FDG = 2 P C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 906 PROBLEM 6.112 Members ABC and CDE are pin-connected at C and supported by four links. For the loading shown, determine the force in each link. SOLUTION Member FBDs: I II FBD I: ΣM I = 0: 2aC y + aC x − aP = 0 2C y + C x = P FBD II: ΣM J = 0: 2aC y − aC x = 0 2C y − C x = 0 Solving, FBD I: Cx = P P ; C y = as shown. 2 4 ΣFx = 0: − 1 2 FBG + C x = 0 FBG = C x 2 ΣFy = 0: FAF − P + FBD II: ΣFx = 0: − Cx + ΣFy = 0: − C y + 1 2 1 2 P P + = 0 2 2 4 FDG = 0 FDG = Cx 2 1 2 P P P P + FEH = 0 FEH = − = − 4 2 4 2 2 FBG = P 2 P 4 FAF = FDG = P FEH = 2 C C C P T 4 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 907 PROBLEM 6.113 Members ABC and CDE are pin-connected at C and supported by four links. For the loading shown, determine the force in each link. SOLUTION Member FBDs: I II From FBD I: ΣM J = 0: a a 3a Cx + C y − P = 0 Cx + 3C y = P 2 2 2 FBD II: ΣM K = 0: a 3a C x − C y = 0 Cx − 3C y = 0 2 2 Solving, Cx = FBD I: P P ; C y = as drawn. 2 6 ΣM B = 0: aC y − a ΣFx = 0: − 1 2 FAG = 0 FAG = 2C y = P 6 2 FAG = 1 1 2 2 FAG + FBF − Cx = 0 FBF = FAG + Cx 2 = P+ P 6 2 2 2 FBF = FBD II: ΣM D = 0: a 2 P C 6 1 2 FEH + aC y = 0 FEH = − 2C y = − P 6 2 ΣFx = 0: C x − 2 2 P C 3 FEH = 2 P T 6 1 1 2 2 FDI + FEH = 0 FDI = FEH + C x 2 = − P+ P 6 2 2 2 FDI = 2 P C 3 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 908 PROBLEM 6.114 Members ABC and CDE are pin-connected at C and supported by the four links AF, BG, DG, and EH. For the loading shown, determine the force in each link. SOLUTION We consider members ABC and CDE: Free body: CDE: ΣM J = 0: C x (4a) + C y (2a) = 0 C y = −2Cx (1) Free body: ABC: ΣM K = 0: Cx (2a) + C y (4a) − P(3a) = 0 Substituting for Cy from Eq. (1): Cx (2a ) − 2Cx (4a) − P(3a ) = 0 1 Cx = − P 2 ΣFx = 0: 1 2 1 C y = −2 − P = + P 2 FBG + C x = 0 P 1 , FBG = − 2Cx = − 2 − P = + 2 2 FBG = P 2 T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 909 PROBLEM 6.114 (Continued) ΣFy = 0: − FAF − 1 2 FBG − P + C y = 0 FAF = − 1 2 P 2 −P+P=− P 2 FAF = P 2 C Free body: CDE: ΣFy = 0: 1 2 FDG − C y = 0 FDG = 2C y = + 2 P ΣFx = 0: − FEH − Cx − 1 2 FDG = 2 P T FDG = 0 P 1 1 2P = − FEH = − − P − 2 2 2 FEH = P 2 C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 910 PROBLEM 6.115 Solve Problem 6.112 assuming that the force P is replaced by a clockwise couple of moment M0 applied to member CDE at D. PROBLEM 6.112 Members ABC and CDE are pin-connected at C and supported by four links. For the loading shown, determine the force in each link. SOLUTION Free body: Member ABC: ΣM J = 0: C y (2a) + Cx ( a) = 0 Cx = − 2C y Free body: Member CDE: ΣM K = 0: C y (2a ) − Cx (a ) − M 0 = 0 C y (2a) − (− 2C y )(a) − M 0 = 0 Cx = − 2C y : D ΣFx = 0: 2 D + Cx = 0; 2 − M0 D= M0 䉰 4a M Cx = − 0 䉰 2a Cy = M0 =0 2a FDG = 2a M0 2a T ΣM D = 0: E (a) − C y ( a) + M 0 = 0 M E (a) − 0 4a (a) + M 0 = 0 E=− 3 M0 4 a FEH = 3 M0 4 a C Return to free body of ABC: ΣFx = 0: B 2 + Cx = 0; B= B 2 − M0 =0 2a M0 FBG = 2a ΣM B = 0: A( a) + C y (a); A(a) + M0 2a M0 (a ) = 0 4a A=− M0 4a FAF = M0 4a T C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 911 PROBLEM 6.116 Solve Problem 6.114 assuming that the force P is replaced by a clockwise couple of moment M0 applied at the same point. PROBLEM 6.114 Members ABC and CDE are pin-connected at C and supported by the four links AF, BG, DG, and EH. For the loading shown, determine the force in each link. SOLUTION Free body: CDE: (same as for Problem 6.114) ΣM J = 0: C x (4a) + C y (2a) = 0 C y = −2Cx (1) Free body: ABC: ΣM K = 0: C x (2a) + C y (4a) − M 0 = 0 Substituting for Cy from Eq. (1): Cx (2a ) − 2Cx (4a) − M 0 = 0 M0 6a M C y = −2 − 0 6a Cx = − ΣFx = 0: 1 2 M0 = + 3a FBG + C x = 0 2M 0 6a FBG = − 2C x = + ΣFy = 0: − FAF − 1 2 FBG = 2M 0 6a T FBG + C y = 0 FAF = − 1 2 M 2M 0 M 0 + =+ 0 6a 3a 6a FAF = M0 6a T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 912 PROBLEM 6.116 (Continued) Free body: CDE: (Use F.B. diagram of Problem 6.114.) ΣFy = 0: 1 2 FDG − C y = 0 FDG = 2C y = + ΣFx = 0: − FEH − C x − 1 2 2M 0 3a FDG = 2M 0 3a T FDG = 0 M FEH = − − 0 6a 1 2M 0 − 2 3a M = − 0 , 6a FEH = M0 6a C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 913 PROBLEM 6.117 Four beams, each of length 3a, are held together by single nails at A, B, C, and D. Each beam is attached to a support located at a distance a from an end of the beam as shown. Assuming that only vertical forces are exerted at the connections, determine the vertical reactions at E, F, G, and H. SOLUTION We shall draw the free body of each member. Force P will be applied to member AFB. Starting with member AED, we shall express all forces in terms of reaction E. Member AFB: ΣM D = 0: A(3a) + E (a ) = 0 E 3 ΣM A = 0: − D (3a ) − E (2a ) = 0 A=− D=− Member DHC: 2E 3 2E ΣM C = 0: − (3a) − H (a) = 0 3 H = − 2E 2E ΣM H = 0: − (2a) + C (a ) = 0 3 C=+ Member CGB: 4E 3 4E ΣM B = 0: + (3a ) − G (a ) = 0 3 G = + 4E 4E ΣM G = 0: + (2a) + B(a ) = 0 3 B=− (1) (2) 8E 3 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 914 PROBLEM 6.117 (Continued) Member AFB: ΣFy = 0: F − A − B − P = 0 E 8E F − − − − −P =0 3 3 F = P − 3E (3) ΣM A = 0: F ( a) − B(3a) = 0 8E ( P − 3E )(a) − − (3a ) = 0 3 P − 3E + 8 E = 0; E = − P 5 E= P 5 Substitute E = − P5 into Eqs. (1), (2), and (3). P H = −2 E = − 2 − 5 P G = + 4E = 4 − 5 P F = P − 3E = P − 3 − 5 H =+ 2P 5 H= 2P 5 G=− 4P 5 G= 4P 5 F =+ 8P 5 F= 8P 5 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 915 PROBLEM 6.118 Four beams, each of length 2a, are nailed together at their midpoints to form the support system shown. Assuming that only vertical forces are exerted at the connections, determine the vertical reactions at A, D, E, and H. SOLUTION Note that, if we assume P is applied to EG, each individual member FBD looks like so 2 Fleft = 2 Fright = Fmiddle Labeling each interaction force with the letter corresponding to the joint of its application, we see that B = 2 A = 2F C = 2B = 2D G = 2C = 2 H P + F = 2G ( = 4C = 8 B = 16 F ) = 2 E From P + F = 16 F , F= P 15 so A = P 15 D= 2P 15 H= 4P 15 E= 8P 15 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 916 PROBLEM 6.119 Each of the frames shown consists of two L-shaped members connected by two rigid links. For each frame, determine the reactions at the supports and indicate whether the frame is rigid. SOLUTION (a) Member FBDs: I II FBD I: ΣM A = 0: aF1 − 2aP = 0 F1 = 2 P; FBD II: ΣM B = 0: − aF2 = 0 F2 = 0 ΣFy = 0: Ay − P = 0 A y = P ΣFx = 0: Bx + F1 = 0, Bx = − F1 = −2 P B x = 2P ΣFy = 0: B y = 0 FBD I: ΣFx = 0: − Ax − F1 + F2 = 0 so B = 2P Ax = F2 − F1 = 0 − 2 P A x = 2P so A = 2.24P 26.6° Frame is rigid. (b) FBD left: FBD whole: I FBD I: ΣM E = 0: FBD II: ΣM B = 0: II a a 5a P + Ax − Ay = 0 Ax − 5 Ay = − P 2 2 2 3aP + aAx − 5aAy = 0 Ax − 5 Ay = −3P This is impossible unless P = 0. not rigid PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 917 PROBLEM 6.119 (Continued) (c) Member FBDs: I FBD I: II ΣFy = 0: A − P = 0 A = P ΣM D = 0: aF1 − 2aA = 0 F1 = 2 P ΣFx = 0: F2 − F1 = 0 F2 = 2 P FBD II: ΣM B = 0: 2aC − aF1 = 0 C = F1 =P 2 ΣFx = 0: F1 − F2 + Bx = 0 Bx = P − P = 0 ΣFx = 0: By + C = 0 B y = − C = − P C = P B=P Frame is rigid. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 918 PROBLEM 6.120 Each of the frames shown consists of two L-shaped members connected by two rigid links. For each frame, determine the reactions at the supports and indicate whether the frame is rigid. SOLUTION (a) Member FBDs: I FBD II: FBD I: ΣFy = 0: B y = 0 II ΣM B = 0: aF2 = 0 F2 = 0 ΣM A = 0: aF2 − 2aP = 0, but F2 = 0 so P = 0 (b) not rigid for P ≠ 0 Member FBDs: Note: Seven unknowns ( Ax , Ay , Bx , B y , F1 , F2 , C ), but only six independent equations System is statically indeterminate. System is, however, rigid. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 919 PROBLEM 6.120 (Continued) (c) FBD whole: FBD right: I FBD I: ΣM A = 0: 5aBy − 2aP = 0 ΣFy = 0: Ay − P + FBD II: FBD I: II ΣM C = 0: 2 P=0 5 a 5a Bx − B y = 0 Bx = 5 B y 2 2 ΣFx = 0: Ax + Bx = 0 Ax = − Bx By = 2 P 5 Ay = 3 P 5 B x = 2P A x = 2P A = 2.09P B = 2.04P 16.70° 11.31° System is rigid. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 920 PROBLEM 6.121 Each of the frames shown consists of two L-shaped members connected by two rigid links. For each frame, determine the reactions at the supports and indicate whether the frame is rigid. SOLUTION Note: In all three cases, the right member has only three forces acting, two of which are parallel. Thus, the third force, at B, must be parallel to the link forces. (a) FBD whole: ΣM A = 0: − 2aP − ΣFx = 0: Ax − ΣFy = 0: (b) a 4 1 B + 5a B = 0 B = 2.06 P 4 17 17 4 17 Ay − P + B=0 1 17 B = 2.06P 14.04° A = 2.06P 14.04° A x = 2P B = 0 Ay = P 2 rigid FBD whole: Since B passes through A, ΣM A = 2aP = 0 only if P = 0. no equilibrium if P ≠ 0 not rigid PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 921 PROBLEM 6.121 (Continued) (c) FBD whole: ΣM A = 0: 5a 1 17 ΣFx = 0: Ax + B+ 4 17 ΣFy = 0: Ay − P + 3a 4 17 B − 2aP = 0 B = P 4 17 4 B=0 1 17 B = 1.031P 14.04° A = 1.250P 36.9° Ax = − P B=0 Ay = P − P 3P = 4 4 System is rigid. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 922 PROBLEM 6.122 The shear shown is used to cut and trim electronic circuit board laminates. For the position shown, determine (a) the vertical component of the force exerted on the shearing blade at D, (b) the reaction at C. SOLUTION We note that BD is a two-force member. Free body: Member ABC: We have the components: Px = (400 N) sin 30° = 200 N Py = (400 N) cos 30° = 346.41 N (FBD ) x = 25 FBD 65 (FBD ) y = 60 FBD 65 ΣM C = 0: ( FBD ) x (45) + ( FBD ) y (30) − Px (45 + 300sin 30°) − Py (30 + 300cos 30°) = 0 25 60 FBD (45) + FBD (30) = (200)(195) + (346.41)(289.81) 65 65 45 FBD = 139.39 × 103 FBD = 3097.6 N (a) Vertical component of force exerted on shearing blade at D. ( FBD ) y = 60 60 (3097.6 N) = 2859.3 N FBD = 65 65 (FBD ) y = 2860 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 923 PROBLEM 6.122 (Continued) (b) Returning to FB diagram of member ABC, ΣFx = 0: ( FBD ) x − Px − C x = 0 Cx = ( FBD ) x − Px = 25 FBD − Px 65 25 (3097.6) − 200 65 Cx = +991.39 = C x = 991.39 N ΣFy = 0: ( FBD ) y − Py − C y = 0 C y = ( FBD ) y − Py = C y = +2512.9 N C = 2295 N 60 60 (3097.6) − 346.41 FBD − Py = 65 65 C y = 2512.9 C = 2700 N 68.5° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 924 PROBLEM 6.123 The press shown is used to emboss a small seal at E. Knowing that P = 250 N, determine (a) the vertical component of the force exerted on the seal, (b) the reaction at A. SOLUTION FBD Stamp D: ΣFy = 0: E − FBD cos 20° = 0, E = FBD cos 20° FBD ABC: ΣM A = 0: (0.2 m)(sin 30°)( FBD cos 20°) + (0.2 m)(cos 30°)( FBD sin 20°) − [(0.2 m)sin 30° + (0.4 m) cos15°](250 N) = 0 FBD = 793.64 N C and, from above, E = (793.64 N) cos 20° (a) E = 746 N ΣFx = 0: Ax − (793.64 N)sin 20° = 0 A x = 271.44 N ΣFy = 0: Ay + (793.64 N) cos 20° − 250 N = 0 A y = 495.78 N so (b) A = 565 N 61.3° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 925 PROBLEM 6.124 The press shown is used to emboss a small seal at E. Knowing that the vertical component of the force exerted on the seal must be 900 N, determine (a) the required vertical force P, (b) the corresponding reaction at A. SOLUTION FBD Stamp D: ΣFy = 0: 900 N − FBD cos 20° = 0, FBD = 957.76 N C (a) FBD ABC: ΣM A = 0: [(0.2 m)(sin 30°)](957.76 N) cos 20° + [(0.2 m)(cos 30°)](957.76 N) sin 20° − [(0.2 m)sin 30° + (0.4 m) cos15°]P = 0 P = 301.70 N, (b) P = 302 N ΣFx = 0: Ax − (957.76 N)sin 20° = 0 A x = 327.57 N ΣFy = 0: −Ay + (957.76 N) cos 20° − 301.70 N = 0 A y = 598.30 N so A = 682 N 61.3° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 926 PROBLEM 6.125 Water pressure in the supply system exerts a downward force of 135 N on the vertical plug at A. Determine the tension in the fusible link DE and the force exerted on member BCE at B. SOLUTION Free body: Entire linkage: + ΣFy = 0: C − 135 = 0 C = + 135 N Free body: Member BCE: ΣFx = 0: Bx = 0 ΣM B = 0: (135 N)(6 mm) − TDE (10 mm) = 0 TDE = 81.0 N ΣFy = 0: 135 + 81 − By = 0 By = + 216 N B = 216 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 927 PROBLEM 6.126 An 84-lb force is applied to the toggle vise at C. Knowing that θ = 90°, determine (a) the vertical force exerted on the block at D, (b) the force exerted on member ABC at B. SOLUTION We note that BD is a two-force member. Free body: Member ABC: BD = (7) 2 + (24)2 = 25 in. We have ( FBD ) x = 7 24 FBD , ( FBD ) y = FBD 25 25 ΣM A = 0: ( FBD ) x (24) + ( FBD ) y (7) − 84(16) = 0 7 24 FBD (24) + FBD (7) = 84(16) 25 25 336 FBD = 1344 25 FBD = 100 lb tan α = 24 α = 73.7° 7 FBD = 100.0 lb (b) Force exerted at B. (a) Vertical force exerted on block. ( FBD ) y = 24 24 FBD = (100 lb) = 96 lb 25 25 73.7° (FBD ) y = 96.0 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 928 PROBLEM 6.127 Solve Problem 6.126 when θ = 0. PROBLEM 6.126 An 84-lb force is applied to the toggle vise at C. Knowing that θ = 90°, determine (a) the vertical force exerted on the block at D, (b) the force exerted on member ABC at B. SOLUTION We note that BD is a two-force member. Free body: Member ABC: BD = (7) 2 + (24)2 = 25 in. We have ( FBD ) x = 7 24 FBD , ( FBD ) y = FBD 25 25 ΣM A = 0: ( FBD ) x (24) + ( FBD ) y (7) − 84(40) = 0 7 24 FBD (24) + FBD (7) = 84(40) 25 25 336 FBD = 3360 25 FBD = 250 lb tan α = 24 α = 73.7° 7 FBD = 250.0 lb (b) Force exerted at B. (a) Vertical force exerted on block. ( FBD ) y = 24 24 FBD = (250 lb) = 240 lb 25 25 73.7° (FBD ) y = 240 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 929 PROBLEM 6.128 For the system and loading shown, determine (a) the force P required for equilibrium, (b) the corresponding force in member BD, (c) the corresponding reaction at C. SOLUTION Member FBDs: FBD I: I: ΣM C = 0: R( FBD sin 30°) − [ R(1 − cos 30°)](100 N) − R(50 N) = 0 FBD = 126.795 N ΣFx = 0: −C x + (126.795 N) cos 30° = 0 (b) FBD = 126.8 N T C x = 109.808 N ΣFy = 0: C y + (126.795 N) sin 30° − 100 N − 50 N = 0 C y = 86.603 N II: so (c) C = 139.8 N 38.3° FBD II: ΣM A = 0: aP − a[(126.795 N) cos 30°] = 0 (a ) P = 109.8 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 930 PROBLEM 6.129 The Whitworth mechanism shown is used to produce a quick-return motion of Point D. The block at B is pinned to the crank AB and is free to slide in a slot cut in member CD. Determine the couple M that must be applied to the crank AB to hold the mechanism in equilibrium when (a) α = 0, (b) α = 30°. SOLUTION (a) Free body: Member CD: ΣM C = 0: B(0.5 m) − (1200 N)(0.7 m) = 0 B = 1680 N Free body: Crank AB: ΣM A = 0: M − (1680 N)(0.1 m) = 0 M = 168.0 N ⋅ m (b) M = 168.0 N ⋅ m Geometry: AB = 100 mm, α = 30° tan θ = 50 θ = 5.87° 486.6 BC = (50)2 + (486.6) 2 = 489.16 mm Free body: Member CD: ΣM C = 0: B (0.48916) − (1200 N)(0.7) cos 5.87° = 0 B = 1708.2 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 931 PROBLEM 6.129 (Continued) Free body: Crank AB: ΣM A = 0: M − ( B sin 65.87°)(0.1 m) = 0 M = (1708.2 N)(0.1 m)sin 65.87° M = 155.90 N ⋅ m M = 155.9 N ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 932 PROBLEM 6.130 Solve Problem 6.129 when (a) α = 60°, (b) α = 90°. PROBLEM 6.129 The Whitworth mechanism shown is used to produce a quick-return motion of Point D. The block at B is pinned to the crank AB and is free to slide in a slot cut in member CD. Determine the couple M that must be applied to the crank AB to hold the mechanism in equilibrium when (a) α = 0, (b) α = 30°. SOLUTION (a) Geometry: AB = 100 mm, α = 60° BE = 100 cos 60° = 86.60 CE = 400 + 100sin 60° = 450 86.60 tan θ = θ = 10.89° 450 BC = (86.60)2 + (450) 2 = 458.26 mm Free body: Member CD: ΣM C = 0: B(0.45826 m) − (1200 N)(0.7 m) cos10.89° = 0 B = 1800.0 N Free body: Crank AB: ΣM A = 0: M − ( B sin 40.89°)(0.1 m) = 0 M = (1800.0 N)(0.1 m)sin 40.89° M = 117.83 N ⋅ m M = 117.8 N ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 933 PROBLEM 6.130 (Continued) (b) Free body: Member CD: 0.1 m 0.4 m θ = 14.04° tan θ = BC = (0.1) 2 + (0.4) 2 = 0.41231 m ΣM C = 0: B(0.41231) − (1200 N)(0.7 cos14.04°) = 0 B = 1976.4 N Free body: Crank AB: ΣM A = 0: M − ( B sin14.04°)(0.1 m) = 0 M = (1976.4 N)(0.1 m) sin14.04° M = 47.948 N ⋅ m M = 47.9 N ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 934 PROBLEM 6.131 A couple M of magnitude 1.5 kN ⋅ m is applied to the crank of the engine system shown. For each of the two positions shown, determine the force P required to hold the system in equilibrium. SOLUTION (a) FBDs: 50 mm 175 mm 2 = 7 Dimensions in mm Note: tan θ = FBD whole: ΣM A = 0: (0.250 m)C y − 1.5 kN ⋅ m = 0 C y = 6.00 kN FBD piston: ΣFy = 0: C y − FBC sin θ = 0 FBC = Cy sin θ = 6.00 kN sinθ ΣFx = 0: FBC cos θ − P = 0 P = FBC cos θ = 6.00 kN = 7 kips tan θ P = 21.0 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 935 PROBLEM 6.131 (Continued) (b) FBDs: Dimensions in mm 2 as above 7 Note: tan θ = FBD whole: ΣM A = 0: (0.100 m)C y − 1.5 kN ⋅ m = 0 C y = 15 kN ΣFy = 0: C y − FBC sin θ = 0 FBC = ΣFx = 0: Cy sin θ FBC cos θ − P = 0 P = FBC cos θ = Cy tan θ = 15 kN 2/7 P = 52.5 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 936 PROBLEM 6.132 A force P of magnitude 16 kN is applied to the piston of the engine system shown. For each of the two positions shown, determine the couple M required to hold the system in equilibrium. SOLUTION (a) FBDs: Note: FBD piston: 50 mm 175 mm 2 = 7 Dimensions in mm tan θ = ΣFx = 0: FBC cos θ − P = 0 FBC = P cos θ ΣFy = 0: C y − FBC sin θ = 0 C y = FBC sin θ = P tan θ = FBD whole: 2 P 7 ΣM A = 0: (0.250 m)C y − M = 0 2 M = (0.250 m) (16 kN) 7 = 1.14286 kN ⋅ m M = 1143 N ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 937 PROBLEM 6.132 (Continued) (b) FBDs: Dimensions in mm tan θ = Note: FBD piston, as above: FBD whole: 2 as above 7 C y = P tan θ = 2 P 7 2 ΣM A = 0: (0.100 m)C y − M = 0 M = (0.100 m) (16 kN) 7 M = 0.45714 kN ⋅ m M = 457 N ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 938 PROBLEM 6.133 The pin at B is attached to member ABC and can slide freely along the slot cut in the fixed plate. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium when θ = 30°. SOLUTION Free body: Member ABC: ΣM C = 0: (25 lb)(13.856 in.) − B(3 in.) = 0 B = +115.47 lb ΣFy = 0: − 25 lb + C y = 0 C y = +25 lb ΣFx = 0: 115.47 lb − Cx = 0 Cx = +115.47 lb Free body: Member CD: β = sin −1 5.196 ; β = 40.505° 8 CD cos β = (8 in.) cos 40.505° = 6.083 in. ΣM D = 0: M − (25 lb)(5.196 in.) − (115.47 lb)(6.083 in.) = 0 M = +832.3 lb ⋅ in. M = 832 lb ⋅ in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 939 PROBLEM 6.134 The pin at B is attached to member ABC and can slide freely along the slot cut in the fixed plate. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium when θ = 60°. SOLUTION Free body: Member ABC: ΣM C = 0: (25 lb)(8 in.) − B(5.196 in.) = 0 B = +38.49 lb ΣFx = 0: 38.49 lb − C x = 0 Cx = +38.49 lb ΣFy = 0: − 25 lb + C y = 0 C y = +25 lb Free body: Member CD: 3 8 β = sin −1 ; β = 22.024° CD cos β = (8 in.) cos 22.024° = 7.416 in. ΣM D = 0: M − (25 lb)(3 in.) − (38.49 lb)(7.416 in.) = 0 M = +360.4 lb ⋅ in. M = 360 lb ⋅ in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 940 PROBLEM 6.135 Rod CD is attached to the collar D and passes through a collar welded to end B of lever AB. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium when θ = 30°. SOLUTION FBD DC: ΣFx′ = 0: D y sin 30° − (150 N) cos 30° = 0 Dy = (150 N) ctn 30° = 259.81 N FBD machine: ΣM A = 0: (0.100 m)(150 N) + d (259.81 N) − M = 0 d = b − 0.040 m b= so 0.030718 m tan 30 b = 0.053210 m d = 0.0132100 m M = 18.4321 N ⋅ m M = 18.43 N ⋅ m. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 941 PROBLEM 6.136 Rod CD is attached to the collar D and passes through a collar welded to end B of lever AB. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium when θ = 30°. SOLUTION B ⊥ CD Note: FBD DC: ΣFx′ = 0: D y sin 30° − (300 N) cos 30° = 0 Dy = 300 N = 519.62 N tan 30° FBD machine: ΣM A = 0: 0.200 m 519.62 N − M = 0 sin 30° M = 207.85 N ⋅ m M = 208 N ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 942 PROBLEM 6.137 Two rods are connected by a frictionless collar B. Knowing that the magnitude of the couple MA is 500 lb · in., determine (a) the couple MC required for equilibrium, (b) the corresponding components of the reaction at C. SOLUTION (a) Free body: Rod AB & collar: ΣM A = 0: ( B cos α )(6 in.) + ( B sin α )(8 in.) − M A = 0 B = (6cos 21.8° + 8sin 21.8°) − 500 = 0 B = 58.535 lb Free body: Rod BC: +ΣM C = 0: M C − B = 0 M C = B = (58.535 lb)(21.541 in.) = 1260.9 lb ⋅ in. M C = 1261 lb ⋅ in. (b) ΣFx = 0: Cx + B cos α = 0 Cx = − B cos α = −(58.535 lb) cos 21.8° = −54.3 lb C x = 54.3 lb ΣFy = 0: C y − B sin α = 0 C y = B sin α = (58.535 lb) sin 21.8° = +21.7 lb C y = 21.7 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 943 PROBLEM 6.138 Two rods are connected by a frictionless collar B. Knowing that the magnitude of the couple MA is 500 lb · in., determine (a) the couple MC required for equilibrium, (b) the corresponding components of the reaction at C. SOLUTION (a) Free body: Rod AB: ΣM A = 0: B(10 in.) − 500 lb ⋅ in. = 0 B = 50.0 Free body: Rod BC & collar: ΣM C = 0: M C − (0.6 B)(20 in.) − (0.8 B)(8 in.) = 0 M C − (30 lb)(20 in.) − (40 lb)(8 in.) = 0 M C = 920 lb ⋅ in. (b) M C = 920 lb ⋅ in. C x = 30.0 lb ΣFx = 0: Cx + 0.6 B = 0 Cx = −0.6 B = −0.6(50.0 lb) = −30.0 lb ΣFy = 0: C y − 0.8B = 0 C y = 0.8 B = 0.8(50.0 lb) = +40.0 lb C y = 40.0 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 944 PROBLEM 6.139 Two hydraulic cylinders control the position of the robotic arm ABC. Knowing that in the position shown the cylinders are parallel, determine the force exerted by each cylinder when P = 160 N and Q = 80 N. SOLUTION Free body: Member ABC: ΣM B = 0: 4 FAE (150 mm) − (160 N)(600 mm) = 0 5 FAE = +800 N FAE = 800 N T 3 ΣFx = 0: − (800 N) + Bx − 80 N = 0 5 Bx = +560 N 4 ΣFy = 0: − (800 N) + By − 160 N = 0 5 By = +800 N Free body: Member BDF: ΣM F = 0: (560 N)(400 mm) − (800 N)(300 mm) − FDG = −100 N 4 FDG (200 mm) = 0 5 FDG = 100.0 N C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 945 PROBLEM 6.140 Two hydraulic cylinders control the position of the robotic arm ABC. In the position shown, the cylinders are parallel and both are in tension. Knowing the FAE = 600 N and FDG = 50 N, determine the forces P and Q applied at C to arm ABC. SOLUTION Free body: Member ABC: ΣM B = 0: 4 (600 N)(150 mm) − P(600 mm) = 0 5 P = 120.0 N P = +120 N ΣM C = 0: 4 (600)(750 mm) − By (600 mm) = 0 5 By = +600 N Free body: Member BDF: ΣM F = 0: Bx (400 mm) − (600 N)(300 mm) 4 − (50 N)(200 mm) = 0 5 Bx = +470 N Return to free body: Member ABC: 3 ΣFx = 0: − (600 N) + 470 N − Q = 0 5 Q = +110 N Q = 110.0 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 946 PROBLEM 6.141 The tongs shown are used to apply a total upward force of 45 kN on a pipe cap. Determine the forces exerted at D and F on tong ADF. SOLUTION FBD whole: By symmetry, FBD ADF: A = B = 22.5 kN ΣM F = 0: (75 mm) D − (100 mm)(22.5 kN) = 0 D = 30.0 kN ΣFx = 0: Fx − D = 0 Fx = D = 30 kN ΣFy = 0: 22.5 kN − Fy = 0 Fy = 22.5 kN F = 37.5 kN so 36.9° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 947 PROBLEM 6.142 If the toggle shown is added to the tongs of Problem 6.141 and a single vertical force is applied at G, determine the forces exerted at D and F on tong ADF. SOLUTION Free body: Toggle: By symmetry, Ay = 1 (45 kN) = 22.5 kN 2 AG is a two-force member. Ax 22.5 kN = 22 mm 55 mm Ax = 56.25 kN Free body: Tong ADF: ΣFy = 0: 22.5 kN − Fy = 0 Fy = +22.5 kN ΣM F = 0 : D (75 mm) − (22.5 kN)(100 mm) − (56.25 kN)(160 mm) = 0 D = +150 kN D = 150.0 kN ΣFx = 0: 56.25 kN − 150 kN + Fx = 0 Fx = 93.75 kN F = 96.4 kN 13.50° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 948 PROBLEM 6.143 A small barrel weighing 60 lb is lifted by a pair of tongs as shown. Knowing that a = 5 in., determine the forces exerted at B and D on tong ABD. SOLUTION We note that BC is a two-force member. Free body: Tong ABD: Bx By = 15 5 Bx = 3By ΣM D = 0: By (3 in.) + 3By (5 in.) − (60 lb)(9 in.) = 0 By = 30 lb Bx = 3By : Bx = 90 lb ΣFx = 0: −90 lb + Dx = 0 D x = 90 lb ΣFy = 0: 60 lb − 30 lb − D y = 0 D y = 30 lb B = 94.9 lb D = 94.9 lb 18.43° 18.43° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 949 PROBLEM 6.144 A 39-ft length of railroad rail of weight 44 lb/ft is lifted by the tongs shown. Determine the forces exerted at D and F on tong BDF. SOLUTION Free body: Rail: W = (39 ft)(44 lb/ft) = 1716 lb Free body: Upper link: By symmetry, 1 E y = Fy = W = 858 lb 2 By symmetry, 1 ( FAB ) y = ( FAC ) y = W = 858 lb 2 Since AB is a two-force member, ( FAB ) x ( FAB ) y = 9.6 6 Free Body: Tong BDF: ( FAB ) x = 9.6 (858) = 1372.8 lb 6 ΣM D = 0: (Attach FAB at A.) Fx (8) − ( FAB ) x (18) − Fy (0.8) = 0 Fx (8) − (1372.8 lb)(18) − (858 lb)(0.8) = 0 Fx = +3174.6 lb F = 3290 lb 15.12° ΣFx = 0: − Dx + ( FAB ) x + Fx = 0 Dx = ( FAB ) x + Fx = 1372.8 + 3174.6 = 4547.4 lb ΣFy = 0: Dy + ( FAB ) y − Fy = 0 Dy = 0 D = 4550 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 950 PROBLEM 6.145 Determine the magnitude of the gripping forces produced when two 300-N forces are applied as shown. SOLUTION We note that AC is a two-force member. FBD handle CD: ΣM D = 0: − (126 mm)(300 N) − (6 mm) 1 + (30 mm) A = 0 8.84 2.8 8.84 A A = 2863.6 8.84 N Dimensions in mm FBD handle AB: ΣM B = 0: (132 mm)(300 N) − (120 mm) + (36 mm) F = 0 1 8.84 (2863.6 8.84 N) F = 8.45 kN Dimensions in mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 951 PROBLEM 6.146 The compound-lever pruning shears shown can be adjusted by placing pin A at various ratchet positions on blade ACE. Knowing that 300-lb vertical forces are required to complete the pruning of a small branch, determine the magnitude P of the forces that must be applied to the handles when the shears are adjusted as shown. SOLUTION We note that AB is a two-force member. ( FAB ) x ( FAB ) y = 0.65 in. 0.55 in. 11 ( FAB ) y = ( FAB ) x 13 (1) Free body: Blade ACE: ΣM C = 0: (300 lb)(1.6 in.) − ( FAB ) x (0.5 in.) − ( FAB ) y (1.4 in.) = 0 Use Eq. (1): ( FAB ) x (0.5 in.) + 11 ( FAB ) x (1.4 in.) = 480 lb ⋅ in. 13 1.6846( FAB ) x = 480 ( FAB ) y = 11 (284.9 lb) 13 ( FAB ) x = 284.9 lb ( FAB ) y = 241.1 lb Free body: Lower handle: ΣM D = 0: (241.1 lb)(0.75 in.) − (284.9 lb)(0.25 in.) − P(3.5 in.) = 0 P = 31.3 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 952 PROBLEM 6.147 The pliers shown are used to grip a 0.3-in.-diameter rod. Knowing that two 60-lb forces are applied to the handles, determine (a) the magnitude of the forces exerted on the rod, (b) the force exerted by the pin at A on portion AB of the pliers. SOLUTION Free body: Portion AB: (a) (b) ΣM A = 0: Q (1.2 in.) − (60 lb)(9.5 in.) = 0 Q = 475 lb ΣFx = 0: Q(sin 30°) + Ax = 0 (475 lb)(sin 30°) + Ax = 0 Ax = −237.5 lb A x = 237.5 lb ΣFy = 0: − Q (cos 30°) + Ay − 60 lb = 0 −(475 lb)(cos 30°) + Ay − 60 lb = 0 Ay = +471.4 lb A y = 471.4 lb A = 528 lb 63.3° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 953 PROBLEM 6.148 In using the bolt cutter shown, a worker applies two 300-N forces to the handles. Determine the magnitude of the forces exerted by the cutter on the bolt. SOLUTION FBD cutter AB: I FBD I: FBD handle BC: II Dimensions in mm ΣFx = 0: Bx = 0 FBD II: ΣM C = 0: (12 mm)By − (448 mm)300 N = 0 By = 11, 200.0 N Then FBD I: ΣM A = 0: (96 mm) By − (24 mm) F = 0 F = 4 By F = 44,800 N = 44.8 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 954 PROBLEM 6.149 The specialized plumbing wrench shown is used in confined areas (e.g., under a basin or sink). It consists essentially of a jaw BC pinned at B to a long rod. Knowing that the forces exerted on the nut are equivalent to a clockwise (when viewed from above) couple of magnitude 135 lb ⋅ in., determine (a) the magnitude of the force exerted by pin B on jaw BC, (b) the couple M0 that is applied to the wrench. SOLUTION Free body: Jaw BC: This is a two-force member. Cy 1.5 in. Free body: Nut: = Cx C y = 2.4 C x in. 5 8 ΣFx = 0: Bx = C x (1) ΣFy = 0: By = C y = 2.4 C x (2) ΣFx = 0: C x = Dx ΣM = 135 lb ⋅ in. C x (1.125 in.) = 135 lb ⋅ in. C x = 120 lb (a) Eq. (1): Bx = C x = 120 lb Eq. (2): By = C y = 2.4(120 lb) = 288 lb ( B = Bx2 + By2 (b) 1/ 2 ) B = 312 lb = (1202 + 2882 )1/2 Free body: Rod: ΣM D = 0: − M 0 + By (0.625 in.) − Bx (0.375 in.) = 0 − M 0 + (288)(0.625) − (120)(0.375) = 0 M 0 = 135.0 lb ⋅ in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 955 PROBLEM 6.150 Determine the force P that must be applied to the toggle CDE to maintain bracket ABC in the position shown. SOLUTION We note that CD and DE are two-force members. Free body: Joint D: ( FCD ) x ( FCD ) y = 30 150 Similarly, ( FCD ) y = 5( FCD ) x Dimensions in mm ( FDE ) y = 5( FDE ) x ΣFy = 0: ( FDE ) y = ( FCD ) y It follows that ( FDE ) x = ( FCD ) x ΣFx = 0: P − ( FDE ) x − ( FCD ) x = 0 ( FDE ) x = ( FCD ) x = Also, 1 P 2 1 ( FDE ) y = ( FCD ) y = 5 P = 2.5P 2 Free body: Member ABC: 1 ΣM A = 0: (2.5 P )(300) + P (450) − (910 N)(150) = 0 2 (750 + 225) P = (910 N)(150) P = 140.0 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 956 PROBLEM 6.151 Determine the force P that must be applied to the toggle CDE to maintain bracket ABC in the position shown. SOLUTION We note that CD and DE are two-force members. Free body: Joint D: ( FCD ) x ( FCD ) y = 30 150 Similarly, ( FCD ) y = 5( FCD ) x ( FDE ) y = 5( FDE ) x Dimensions in mm ΣFy = 0: ( FDE ) y = ( FCD ) y It follows that ( FDE ) x = ( FCD ) x Σ Fx = 0: ( FDE ) x + ( FCD ) x − P = 0 ( FDE ) x = ( FCD ) x = Also, 1 P 2 1 ( FDE ) y = ( FCD ) y = 5 P = 2.5P 2 Free body: Member ABC: 1 ΣM A = 0: (2.5P)(300) − P (450) − (910 N)(150) = 0 2 (750 − 225) P = (910 N)(150) P = 260 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 957 PROBLEM 6.152 A 45-lb shelf is held horizontally by a self-locking brace that consists of two parts EDC and CDB hinged at C and bearing against each other at D. Determine the force P required to release the brace. SOLUTION Free body: Shelf: ΣM A = 0: By (10 in.) − (45 lb)(6.25 in.) = 0 By = 28.125 lb Free body: Portion ECB: ΣM E = 0: − Bx (7.5 in.) − P(7.5 in.) − (28.125 lb)(10 in.) = 0 Bx = −37.5 − P Free body: Portion CDB: ΣM C = 0: − (28.125 lb)(4.6 in.) − Bx (2.2 in.) = 0 −(28.125 lb)(4.6 in.) − ( −37.5 − P)(2.2 in.) = 0 P = 21.3 lb P = 21.3 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 958 PROBLEM 6.153 The telescoping arm ABC is used to provide an elevated platform for construction workers. The workers and the platform together have a mass of 200 kg and have a combined center of gravity located directly above C. For the position when θ = 20°, determine (a) the force exerted at B by the single hydraulic cylinder BD, (b) the force exerted on the supporting carriage at A. SOLUTION a = (5 m) cos 20° = 4.6985 m Geometry: b = (2.4 m) cos 20° = 2.2553 m c = (2.4 m) sin 20° = 0.8208 m d = b − 0.5 = 1.7553 m e = c + 0.9 = 1.7208 m tan β = e 1.7208 = ; β = 44.43° d 1.7553 Free body: Arm ABC: We note that BD is a two-force member. W = (200 kg)(9.81 m/s 2 ) = 1.962 kN (a) ΣM A = 0: (1.962 kN)(4.6985 m) − FBD sin 44.43°(2.2553 m) + FBD cos 44.43(0.8208 m) = 0 9.2185 − FBD (0.9927) = 0: FBD = 9.2867 kN FBD = 9.29 kN (b) 44.4° ΣFx = 0: Ax − FBD cos β = 0 Ax = (9.2867 kN) cos 44.43° = 6.632 kN A x = 6.632 kN ΣFy = 0: Ay − 1.962 kN + FBD sin β = 0 Ay = 1.962 kN − (9.2867 kN)sin 44.43° = −4.539 kN A y = 4.539 kN A = 8.04 kN 34.4° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 959 PROBLEM 6.154 The telescoping arm ABC can be lowered until end C is close to the ground, so that workers can easily board the platform. For the position when θ = −20°, determine (a) the force exerted at B by the single hydraulic cylinder BD, (b) the force exerted on the supporting carriage at A. SOLUTION a = (5 m) cos 20° = 4.6985 m Geometry: b = (2.4 m) cos 20° = 2.2552 m c = (2.4 m) sin 20° = 0.8208 m d = b − 0.5 = 1.7553 m e = 0.9 − c = 0.0792 m tan β = e 0.0792 = ; β = 2.584° d 1.7552 Free body: Arm ABC: We note that BD is a two-force member. W = (200 kg)(9.81 m/s 2 ) W = 1962 N = 1.962 kN (a) ΣM A = 0: (1.962 kN)(4.6985 m) − FBD sin 2.584°(2.2553 m) − FBD cos 2.584°(0.8208 m) = 0 9.2185 − FBD (0.9216) = 0 FBD = 10.003 kN FBD = 10.00 kN (b) 2.58° ΣFx = 0: Ax − FBD cos β = 0 Ax = (10.003 kN)cos 2.583° = 9.993 kN A x = 9.993 kN ΣFy = 0: Ay − 1.962 kN + FBD sin β = 0 Ay = 1.962 kN − (10.003 kN)sin 2.583° = −1.5112 kN A y = 1.5112 kN A = 10.11 kN 8.60° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 960 PROBLEM 6.155 The bucket of the front-end loader shown carries a 3200-lb load. The motion of the bucket is controlled by two identical mechanisms, only one of which is shown. Knowing that the mechanism shown supports one-half of the 3200-lb load, determine the force exerted (a) by cylinder CD, (b) by cylinder FH. SOLUTION Free body: Bucket: (one mechanism) ΣM D = 0: (1600 lb)(15 in.) − FAB (16 in.) = 0 FAB = 1500 lb Note: There are two identical support mechanisms. Free body: One arm BCE: 8 20 β = 21.8° tan β = ΣM E = 0: (1500 lb)(23 in.) + FCD cos 21.8°(15 in.) − FCD sin 21.8°(5 in.) = 0 FCD = −2858 lb FCD = 2.86 kips C Free body: Arm DFG: ΣM G = 0: (1600 lb)(75 in.) + FFH sin 45°(24 in.) − FFH cos 45°(6 in.) = 0 FFH = −9.428 kips FFH = 9.43 kips C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 961 PROBLEM 6.156 The motion of the bucket of the front-end loader shown is controlled by two arms and a linkage that are pin-connected at D. The arms are located symmetrically with respect to the central, vertical, and longitudinal plane of the loader; one arm AFJ and its control cylinder EF are shown. The single linkage GHDB and its control cylinder BC are located in the plane of symmetry. For the position and loading shown, determine the force exerted (a) by cylinder BC, (b) by cylinder EF. SOLUTION Free body: Bucket ΣM J = 0: (4500 lb)(20 in.) − FGH (22 in.) = 0 FGH = 4091 lb Free body: Arm BDH ΣM D = 0: − (4091 lb)(24 in.) − FBC (20 in.) = 0 FBC = −4909 lb FBC = 4.91 kips C Free body: Entire mechanism (Two arms and cylinders AFJE) Note: Two arms thus 2 FEF 18 in. 65 in. β = 15.48° tan β = ΣM A = 0: (4500 lb)(123 in.) + FBC (12 in.) + 2 FEF cos β (24 in.) = 0 (4500 lb)(123 in.) − (4909 lb)(12 in.) + 2 FEF cos 15.48°(24 in.) = 0 FEF = 10.69 kips C FEF = −10.690 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 962 PROBLEM 6.157 The motion of the backhoe bucket shown is controlled by the hydraulic cylinders AD, CG, and EF. As a result of an attempt to dislodge a portion of a slab, a 2-kip force P is exerted on the bucket teeth at J. Knowing that θ = 45°, determine the force exerted by each cylinder. SOLUTION Free body: Bucket: ΣM H = 0: (Dimensions in inches) 4 3 FCG (10) + FCG (10) + P cos θ (16) + P sin θ (8) = 0 5 5 FCG = − P (16 cos θ + 8 sin θ ) 14 (1) Free body: Arm ABH and bucket: (Dimensions in inches) ΣM B = 0: 4 3 FAD (12) + FAD (10) + P cos θ (86) − P sin θ (42) = 0 5 5 FAD = − P (86 cos θ − 42 sin θ ) 15.6 (2) Free body: Bucket and arms IEB + ABH: Geometry of cylinder EF: 16 in. 40 in. β = 21.801° tan β = PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 963 PROBLEM 6.157 (Continued) ΣM I = 0: FEF cos β (18 in.) + P cos θ (28 in.) − P sin θ (120 in.) = 0 P(120 sin θ − 28 cos θ ) cos 21.8°(18) P = (120 sin θ − 28 cos θ ) 16.7126 FEF = For P = 2 kips, (3) θ = 45° From Eq. (1): FCG = − 2 (16 cos 45° + 8 sin 45°) = −2.42 kips 14 From Eq. (2): FAD = − 2 (86 cos 45° − 42 sin 45°) = −3.99 kips 15.6 From Eq. (3): FEF = 2 (120 sin 45° − 28 cos 45°) = +7.79 kips 16.7126 FCG = 2.42 kips C FAD = 3.99 kips C FEF = 7.79 kips T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 964 PROBLEM 6.158 Solve Problem 6.157 assuming that the 2-kip force P acts horizontally to the right (θ = 0). PROBLEM 6.157 The motion of the backhoe bucket shown is controlled by the hydraulic cylinders AD, CG, and EF. As a result of an attempt to dislodge a portion of a slab, a 2-kip force P is exerted on the bucket teeth at J. Knowing that θ = 45°, determine the force exerted by each cylinder. SOLUTION Free body: Bucket: ΣM H = 0: (Dimensions in inches) 4 3 FCG (10) + FCG (10) + P cos θ (16) + P sin θ (8) = 0 5 5 FCG = − P (16 cos θ + 8 sin θ ) 14 (1) Free body: Arm ABH and bucket: (Dimensions in inches) ΣM B = 0: 4 3 FAD (12) + FAD (10) + P cos θ (86) − P sin θ (42) = 0 5 5 FAD = − P (86 cos θ − 42 sin θ ) 15.6 (2) Free body: Bucket and arms IEB + ABH: Geometry of cylinder EF: 16 in. 40 in. β = 21.801° tan β = PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 965 PROBLEM 6.158 (Continued) ΣM I = 0: FEF cos β (18 in.) + P cos θ (28 in.) − P sin θ (120 in.) = 0 P(120 sin θ − 28 cos θ ) cos 21.8°(18) P (120 sin θ − 28 cos θ ) = 16.7126 FEF = For P = 2 kips, (3) θ =0 From Eq. (1): FCG = − 2 (16 cos 0 + 8 sin 0) = −2.29 kips 14 From Eq. (2): FAD = − 2 (86 cos 0 − 42 sin 0) = −11.03 kips 15.6 From Eq. (3): FEF = 2 (120 sin 0 − 28 cos 0) = −3.35 kips 16.7126 FCG = 2.29 kips C FAD = 11.03 kips C FEF = 3.35 kips C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 966 PROBLEM 6.159 In the planetary gear system shown, the radius of the central gear A is a = 18 mm, the radius of each planetary gear is b, and the radius of the outer gear E is (a + 2b). A clockwise couple of magnitude MA = 10 N ⋅ m is applied to the central gear A and a counterclockwise couple of magnitude MS = 50 N ⋅ m is applied to the spider BCD. If the system is to be in equilibrium, determine (a) the required radius b of the planetary gears, (b) the magnitude ME of the couple that must be applied to the outer gear E. SOLUTION FBD Central Gear: By symmetry, F1 = F2 = F3 = F ΣM A = 0: 3(rA F ) − 10 N ⋅ m = 0, ΣM C = 0: rB ( F − F4 ) = 0, FBD Gear C: F= 10 N⋅m 3rA F4 = F ΣFx′ = 0: Cx′ = 0 ΣFy′ = 0: C y′ − 2 F = 0, C y′ = 2 F Gears B and D are analogous, each having a central force of 2F. ΣM A = 0: 50 N ⋅ m − 3(rA + rB )2F = 0 FBD Spider: 50 N ⋅ m − 3(rA + rB ) 20 N⋅m = 0 rA rA + rB r = 2.5 = 1 + B , rA rA Since rA = 18 mm, rB = 1.5rA rB = 27.0 mm FBD Outer Gear: (a) ΣM A = 0: 3(rA + 2rB ) F − M E = 0 3(18 mm + 54 mm) 10 N ⋅ m − ME = 0 54 mm M E = 40.0 N ⋅ m (b) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 967 PROBLEM 6.160 The gears D and G are rigidly attached to shafts that are held by frictionless bearings. If rD = 90 mm and rG = 30 mm, determine (a) the couple M0 that must be applied for equilibrium, (b) the reactions at A and B. SOLUTION (a) Projections on yz plane. Free body: Gear G: ΣM G = 0: 30 N ⋅ m − J (0.03 m) = 0; J = 1000 N Free body: Gear D: ΣM D = 0: M 0 − (1000 N)(0.09 m) = 0 M 0 = (90.0 N ⋅ m)i M 0 = 90 N ⋅ m (b) Gear G and axle FH: ΣM F = 0: H (0.3 m) − (1000 N)(0.18 m) = 0 H = 600 N ΣFy = 0: F + 600 − 1000 = 0 F = 400 N Gear D and axle CE: ΣM C = 0: (1000 N)(0.18 m) − E (0.3 m) = 0 E = 600 N ΣFy = 0: 1000 − C − 600 = 0 C = 400 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 968 PROBLEM 6.160 (Continued) Free body: Bracket AE: ΣFy = 0: A − 400 + 400 = 0 A = 0 ΣM A = 0: M A + (400 N)(0.32 m) − (400 N)(0.2 m) = 0 M A = −48 N ⋅ m Free body: Bracket BH: ΣFy = 0: B − 600 + 600 = 0 M A = −(48.0 N ⋅ m)i B=0 ΣM B = 0: M B + (600 N)(0.32 m) − (600 N)(0.2 m) = 0 M B = −72 N ⋅ m M B = −(72.0 N ⋅ m)i PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 969 PROBLEM 6.161* Two shafts AC and CF, which lie in the vertical xy plane, are connected by a universal joint at C. The bearings at B and D do not exert any axial force. A couple of magnitude 500 lb ⋅ in. (clockwise when viewed from the positive x-axis) is applied to shaft CF at F. At a time when the arm of the crosspiece attached to shaft CF is horizontal, determine (a) the magnitude of the couple that must be applied to shaft AC at A to maintain equilibrium, (b) the reactions at B, D, and E. (Hint: The sum of the couples exerted on the crosspiece must be zero.) SOLUTION We recall from Figure 4.10 that a universal joint exerts on members it connects a force of unknown direction and a couple about an axis perpendicular to the crosspiece. Free body: Shaft DF: ΣM x = 0: M C cos30° − 500 lb ⋅ in. = 0 M C = 577.35 lb ⋅ in. Free body: Shaft BC: We use here x′, y ′, z with x′ along BC . ΣM C = 0: − M R i′ − (577.35 lb ⋅ in.)i′ + (−5 in.)i′ × ( By j ′ + Bz k ) = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 970 PROBLEM 6.161* (Continued) Equate coefficients of unit vectors to zero: i: M A − 577.35 lb ⋅ in. = 0 j: Bz = 0 k: By = 0 M A = 577.35 lb ⋅ in. M A = 577 lb ⋅ in. B=0 ΣF = 0: B + C = 0, B=0 since B = 0, C=0 Return to free body of shaft DF. ΣM D = 0 (Note that C = 0 and M C = 577.35 lb ⋅ in. ) (577.35 lb ⋅ in.)(cos 30°i + sin 30° j) − (500 lb ⋅ in.)i +(6 in.)i × ( Ex i + E y j + Ez k ) = 0 (500 lb ⋅ in.)i + (288.68 lb ⋅ in.) j − (500 lb ⋅ in.)i + (6 in.) E y k − (6 in.) Ez j = 0 Equate coefficients of unit vectors to zero: j: 288.68 lb ⋅ in. − (6 in.)Ez = 0 Ez = 48.1 lb k: Ey = 0 ΣF = 0: C + D + E = 0 0 + Dy j + Dz k + Ex i + (48.1 lb)k = 0 i: Ex = 0 j: Dy = 0 k: Dz + 48.1 lb = 0 Dz = −48.1 lb Reactions are: B=0 D = −(48.1 lb)k E = (48.1 lb)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 971 PROBLEM 6.162* Solve Problem 6.161 assuming that the arm of the crosspiece attached to shaft CF is vertical. PROBLEM 6.161 Two shafts AC and CF, which lie in the vertical xy plane, are connected by a universal joint at C. The bearings at B and D do not exert any axial force. A couple of magnitude 500 lb · in. (clockwise when viewed from the positive x-axis) is applied to shaft CF at F. At a time when the arm of the crosspiece attached to shaft CF is horizontal, determine (a) the magnitude of the couple that must be applied to shaft AC at A to maintain equilibrium, (b) the reactions at B, D, and E. (Hint: The sum of the couples exerted on the crosspiece must be zero.) SOLUTION Free body: Shaft DF. Σ M x = 0: M C − 500 lb ⋅ in. = 0 M C = 500 lb ⋅ in. Free body: Shaft BC: We resolve −(520 lb ⋅ in.)i into components along x′ and y ′ axes: −MC = −(500 lb ⋅ in.)(cos30°i′ + sin 30° j′) ΣMC = 0: M Ai′ − (500 lb ⋅ in.)(cos30°i′ + sin 30° j′) + (5 in.)i′ × ( By′ j′ + Bz k ) = 0 M A i′ − (433 lb ⋅ in.)i ′ − (250 lb ⋅ in.) j + (5 in.) By′k − (5 in.) Bz j′ = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 972 PROBLEM 6.162* (Continued) Equate to zero coefficients of unit vectors: M A = 433 lb ⋅ in. i′: M A − 433 lb ⋅ in. = 0 j′: − 250 lb ⋅ in. − (5 in.) Bz = 0 Bz = −50 lb k : B y′ = 0 B = −(50 lb)k Reactions at B. ΣF = 0: B − C = 0 −(50 lb)k − C = 0 C = −(50 lb)k Return to free body of shaft DF. ΣM D = 0: (6 in.)i × ( Ex i + E y j + Ez k ) − (4 in.)i × (−50 lb)k − (500 lb ⋅ in.)i + (500 lb ⋅ in.)i = 0 (6 in.) E y k − (6 in.) Ez j − (200 lb ⋅ in.) j = 0 k : Ey = 0 j: −(6 in.) Ez − 200 lb ⋅ in. = 0 Ez = −33.3 lb ΣF = 0: C + D + E = 0 −(50 lb)k + Dy j + Dz k + Ex i − (33.3 lb)k = 0 i : Ex = 0 k : −50 lb − 33.3 lb + Dz = 0 Dz = 83.3 lb Reactions are B = −(50 lb)k D = (83.3 lb)k E = −(33.3 lb)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 973 PROBLEM 6.163* The large mechanical tongs shown are used to grab and lift a thick 7500-kg steel slab HJ. Knowing that slipping does not occur between the tong grips and the slab at H and J, determine the components of all forces acting on member EFH. (Hint: Consider the symmetry of the tongs to establish relationships between the components of the force acting at E on EFH and the components of the force acting at D on DGJ.) SOLUTION Free body: Pin A: T = W = mg = (7500 kg)(9.81 m/s 2 ) = 73.575 kN ΣFx = 0: ( FAB ) x = ( FAC ) x 1 ΣFy = 0: ( FAB ) y = ( FAC ) y = W 2 Also, ( FAC ) x = 2( FAC ) y = W Free body: Member CDF: 1 Σ M D = 0: W (0.3) + W (2.3) − Fx (1.8) − Fy (0.5 m) = 0 2 or 1.8Fx + 0.5Fy = 1.45W (1) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 974 PROBLEM 6.163* (Continued) ΣFx = 0: Dx − Fx − W = 0 Ex − Fx = W or (2) 1 ΣFy = 0: Fy − Dy + W = 0 2 1 E y − Fy = W 2 or (3) Free body: Member EFH: 1 ΣM E = 0: Fx (1.8) + Fy (1.5) − H x (2.3) + W (1.8 m) = 0 2 1.8Fx + 1.5Fy = 2.3H x − 0.9W or (4) ΣFx = 0: Ex + Fx − H x = 0 Ex + Fx = H x or 2 Fx = H x − W Subtract Eq. (2) from Eq. (5): Subtract Eq. (4) from 3 × (1): 3.6Fx = 5.25W − 2.3H x Add Eq. (7) to 2.3 × Eq. (6): 8.2Fx = 2.95W Fx = 0.35976W (5) (6) (7) (8) Substitute from Eq. (8) into Eq. (1): (1.8)(0.35976W ) + 0.5Fy = 1.45W 0.5Fy = 1.45W − 0.64756W = 0.80244W Fy = 1.6049W (9) Substitute from Eq. (8) into Eq. (2): Ex − 0.35976W = W ; Ex = 1.35976W Substitute from Eq. (9) into Eq. (3): 1 E y − 1.6049W = W 2 From Eq. (5): H x = Ex + Fx = 1.35976W + 0.35976W = 1.71952W Recall that 1 Hy = W 2 E y = 2.1049W PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 975 PROBLEM 6.163* (Continued) Since all expressions obtained are positive, all forces are directed as shown on the free-body diagrams. Substitute W = 73.575 kN E y = 154.9 kN E x = 100.0 kN Fx = 26.5 kN Fy = 118.1 kN H y = 36.8 kN H x = 126.5 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 976 PROBLEM 6.164 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Free Body: Truss: ΣM E = 0: F (3 m) − (900 N)(2.25 m) − (900 N)(4.5 m) = 0 F = 2025 N ΣFx = 0: Ex + 900 N + 900 N = 0 Ex = −1800 N E x = 1800 N ΣFy = 0: E y + 2025 N = 0 E y = −2025 N E y = 2025 N FAB = FBD = 0 We note that AB and BD are zero-force members. Free body: Joint A: FAC FAD 900 N = = 2.25 3.75 3 FAC = 675 N T FAD = 1125 N C Free body: Joint D: FCD FDE 1125 N = = 3 2.23 3.75 Free body: Joint E: ΣFx = 0: FEF − 1800 N = 0 ΣFy = 0: FCE − 2025 N = 0 FCD = 900 N T FDF = 675 N C FEF = 1800 N T FCE = 2025 N T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 977 PROBLEM 6.164 (Continued) Free body: Joint F: ΣFy = 0: 2.25 FCF + 2025 N − 675 N = 0 3.75 FCF = −2250 N ΣFx = − FCF = 2250 N C 3 (−2250 N) − 1800 N = 0 (Checks) 3.75 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 978 PROBLEM 6.165 Using the method of joints, determine the force in each member of the roof truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: ΣFx = 0: A x = 0 From symmetry of loading: Ay = E = 1 2 total load A y = E = 3.6 kN We note that DF is a zero-force member and that EF is aligned with the load. Thus, FDF = 0 FEF = 1.2 kN C Free body: Joint A: FAB FAC 2.4 kN = = 13 12 5 FAB = 6.24 kN C FAC = 2.76 kN T Free body: Joint B: ΣFx = 0: Σ Fy = 0: 3 12 12 FBC + FBD + (6.24 kN) = 0 3.905 13 13 − 2.5 5 5 FBC + FBD + (6.24 kN) − 2.4 kN = 0 3.905 13 13 (1) (2) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 979 PROBLEM 6.165 (Continued) Multiply Eq. (1) by 2.5, Eq. (2) by 3, and add: 45 45 FBD + (6.24 kN) − 7.2 kN = 0, FBD = −4.16 kN, 13 13 FBD = 4.16 kN C Multiply Eq. (1) by 5, Eq. (2) by –12, and add: 45 FBC + 28.8 kN = 0, FBC = −2.50 kN, 3.905 FBC = 2.50 kN C Free body: Joint C: ΣFy = 0: 5 2.5 (2.50 kN) = 0 FCD − 5.831 3.905 ΣFx = 0: FCE − 5.76 kN + FCD = 1.867 kN T 3 3 (2.50 kN) + (1.867 kN) = 0 3.905 5.831 FCE = 2.88 kN T Free body: Joint E: ΣFy = 0: 5 FDE + 3.6 kN − 1.2 kN = 0 7.81 FDE = −3.75 kN ΣFx = 0: − FCE − FDE = 3.75 kN C 6 (−3.75 kN) = 0 7.81 FCE = +2.88 kN FCE = 2.88 kN T (Checks) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 980 PROBLEM 6.166 A Howe scissors roof truss is loaded as shown. Determine the force in members DF, DG, and EG. SOLUTION Reactions at supports. Because of symmetry of loading, Ax = 0, Ay = L = 1 1 (total load) = (9.60 kips) = 4.80 kips 2 2 A = L = 4.80 kips We pass a section through members DF, DG, and EG, and use the free body shown. We slide FDF to apply it at F: ΣM G = 0: (0.8 kip)(24 ft) + (1.6 kips)(16 ft) + (1.6 kips)(8 ft) 8 FDF − (4.8 kips)(24 ft) − 82 + 3.52 (6 ft) = 0 FDF = −10.48 kips, FDF = 10.48 kips C ΣM A = 0: − (1.6 kips)(8 ft) − (1.6 kips)(16 ft) − 2.5FDG 2 8 + 2.5 2 (16 ft) − 8FDG 82 + 2.52 FDG = −3.35 kips, FDG = 3.35 kips C ΣM D = 0: (0.8 kips)(16 ft) + (1.6 kips)(8 ft) − (4.8 kips)(16 ft) − (7 ft) = 0 8 FEG 82 + 1.52 (4 ft) = 0 FEG = +13.02 kips, FEG = 13.02 kips T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 981 PROBLEM 6.167 A Howe scissors roof truss is loaded as shown. Determine the force in members GI, HI, and HJ. SOLUTION Reactions at supports. Because of symmetry of loading, 1 (Total load) 2 1 = (9.60 kips) 2 Ax = 0, Ay = L = = 4.80 kips A = L = 4.80 kips We pass a section through members GI, HI, and HJ, and use the free body shown. ΣM H = 0: − 16 FGI 162 + 32 (4 ft) + (4.8 kips)(16 ft) − (0.8 kip)(16 ft) − (1.6 kips)(8 ft) = 0 FGI = +13.02 kips FGI = 13.02 kips T ΣM L = 0: (1.6 kips)(8 ft) − FHI (16 ft) = 0 FHI = +0.800 kips FHI = 0.800 kips T We slide FHG to apply it at H. ΣM I = 0: 8FHJ 82 + 3.52 (4 ft) + (4.8 kips)(16 ft) − (1.6 kips)(8 ft) − (0.8 kip)(16 ft) = 0 FHJ = −13.97 kips FHJ = 13.97 kips C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 982 PROBLEM 6.168 Rod CD is fitted with a collar at D that can be moved along rod AB, which is bent in the shape of an arc of circle. For the position when θ = 30°, determine (a) the force in rod CD, (b) the reaction at B. SOLUTION FBD: (a) ΣM C = 0: (15 in.)(20 lb − By ) = 0 ΣFy = 0: − 20 lb + FCD sin 30° − 20 lb = 0 (b) B y = 20 lb FCD = 80.0 lb T ΣFx = 0: (80 lb) cos30° − Bx = 0 B x = 69.282 lb so B = 72.1 lb 16.10° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 983 PROBLEM 6.169 For the frame and loading shown, determine the components of all forces acting on member ABC. SOLUTION Free body: Entire frame: ΣFx = 0: Ax + 18 kN = 0 Ax = −18 kN A x = 18.00 kN ΣM E = 0: −(18 kN)(4 m) − Ay (3.6 m) = 0 Ay = −20 kN A y = 20.0 kN ΣFy = 0: − 20 kN + F = 0 F = +20 kN F = 20 kN Free body: Member ABC Note: BE is a two-force member, thus B is directed along line BE. ΣM C = 0: B(4 m) − (18 kN)(6 m) + (20 kN)(3.6 m) = 0 B = 9 kN B = 9.00 kN Cx = 9.00 kN ΣFx = 0: Cx − 18 kN + 9 kN = 0 Cx = 9 kN ΣFy = 0: C y − 20 kN = 0 C y = 20 kN C y = 20.0 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 984 PROBLEM 6.170 Knowing that each pulley has a radius of 250 mm, determine the components of the reactions at D and E. SOLUTION Free body: Entire assembly: ΣM E = 0: (4.8 kN)(4.25 m) − Dx (1.5 m) = 0 Dx = +13.60 kN D x = 13.60 kN E x = 13.60 kN ΣFx = 0: Ex + 13.60 kN = 0 Ex = −13.60 kN Σ Fy = 0: Dy + E y − 4.8 kN = 0 (1) Free body: Member ACE: ΣM A = 0 : (4.8 kN)(2.25 m) + E y (4 m) = 0 E y = −2.70 kN From Eq. (1): E y = 2.70 kN Dy − 2.70 kN − 4.80 kN = 0 Dy = +7.50 kN D y = 7.50 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 985 PROBLEM 6.171 For the frame and loading shown, determine the components of the forces acting on member DABC at B and D. SOLUTION Free body: Entire frame: ΣM G = 0: H (0.6 m) − (12 kN)(1 m) − (6 kN)(0.5 m) = 0 H = 25 kN H = 25 kN Free body: Member BEH: ΣM F = 0: Bx (0.5 m) − (25 kN)(0.2 m) = 0 Bx = +10 kN Free body: Member DABC: B x = 10.00 kN From above: ΣM D = 0: − By (0.8 m) + (10 kN + 12 kN)(0.5 m) = 0 By = +13.75 kN B y = 13.75 kN ΣFx = 0: − Dx + 10 kN + 12 kN = 0 Dx = + 22 kN D x = 22.0 kN ΣFy = 0: − Dy + 13.75 kN = 0 Dy = +13.75 kN D y = 13.75 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 986 PROBLEM 6.172 For the frame and loading shown, determine (a) the reaction at C, (b) the force in member AD. SOLUTION Free body: Member ABC: ΣM C = 0: + (100 lb)(45 in.) + 4 4 FAD (45 in.) + FBE (30 in.) = 0 5 5 3FAD + 2 FBF = − 375 lb (1) Free Body: Member DEF: ΣM F = 0: 4 4 FAD (30 in.) + FBF (15 in.) = 0 5 5 FBE = −2FAD (b) Substitute from Eq. (2) into Eq. (1): 3FAD + 2(− 2FAD ) = − 375 lb FAD = + 375 lb From Eq. (2): FAD = 375 lb T FBE = − 2 FAD = − 2(375 lb) FBE = −750 lb (a) (2) FBE = 750 lb C. Return to free body of member ABC. 4 4 ΣFx = 0: C x + 100 lb + FAD + FBE = 0 5 5 C x + 100 + 4 4 (375) + (−750) = 0 5 5 Cx = + 200 lb Cx = 200 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 987 PROBLEM 6.172 (Continued) 3 3 ΣFy = 0: C y − FAD − FBF = 0 5 5 3 3 C y − (375) − ( − 750) = 0 5 5 C y = − 225 lb C y = 225 lb α = 48.37° C = 301.0 lb C = 301 lb 48.4° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 988 PROBLEM 6.173 The control rod CE passes through a horizontal hole in the body of the toggle system shown. Knowing that link BD is 250 mm long, determine the force Q required to hold the system in equilibrium when β = 20°. SOLUTION We note that BD is a two-force member. Free body: Member ABC: Dimensions in mm Since BD = 250, θ = sin −1 33.404 ; θ = 7.679° 250 Σ M C = 0: ( FBD sin θ )187.94 − ( FBD cos θ )68.404 + (100 N)328.89 = 0 FBD [187.94sin 7.679° − 68.404 cos 7.679°] = 32,889 FBD = 770.6 N ΣFx = 0: (770.6 N) cos 7.679° = Cx = 0 Cx = + 763.7 N Member CQ: ΣFx = 0: Q = Cx = 763.7 N Q = 764 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 989 PROBLEM 6.174 Determine the magnitude of the gripping forces exerted along line aa on the nut when two 50-lb forces are applied to the handles as shown. Assume that pins A and D slide freely in slots cut in the jaws. SOLUTION FBD jaw AB: ΣFx = 0: Bx = 0 ΣM B = 0: (0.5 in.)Q − (1.5 in.) A = 0 A= Q 3 ΣFy = 0: A + Q − By = 0 By = A + Q = FBD handle ACE: 4Q 3 By symmetry and FBD jaw DE, Q 3 Ex = Bx = 0 D= A= E y = By = 4Q 3 ΣM C = 0: (5.25 in.)(50 lb) + (0.75 in.) Q 4Q − (0.75 in.) =0 3 3 Q = 350 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 990 PROBLEM 6.175 Knowing that the frame shown has a sag at B of a = 1 in., determine the force P required to maintain equilibrium in the position shown. SOLUTION We note that AB and BC are two-force members. Free body: Toggle: Cy = By symmetry, P 2 Cy Cx = 10 in. a Cx = 10 10 P 5 P Cy = ⋅ = a a 2 a Free body: Member CDE: ΣM E = 0: Cx (6 in.) − C y (20 in.) − (50 lb)(10 in.) = 0 5P P (b) − (20) = 500 2 a 30 P − 10 = 500 a For (1) a = 1.0 in. 30 P − 10 = 500 1 20 P = 500 P = 25.0 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 991 PROBLEM 6.F1 For the frame and loading shown, draw the free-body diagram(s) needed to determine the forces acting on member ABC at B and C. SOLUTION We note that BD is a two-force member. Free body: ABC PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 992 PROBLEM 6.F2 For the frame and loading shown, draw the free-body diagram(s) needed to determine all forces acting on member GBEH. SOLUTION We note that AG, DEF, and DH are two-force members. Free body: ABC Note: Sum moments about A to determine By. Note: Sum moments about E to determine FDH (which becomes zero). Free body: DEF Sum forces in y to determine Ey (which also becomes zero). Free body: GBEH Note: With By, Ey, and FDH previously determined, apply three equilibrium equations to determine three remaining forces on GBEH. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 993 PROBLEM 6.F3 For the frame and loading shown, draw the free-body diagram(s) needed to determine the reactions at B and F. SOLUTION We note that BC is two-force member. Free body: ACF PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 994 PROBLEM 6.F4 Knowing that the surfaces at A and D are frictionless, draw the free-body diagram(s) needed to determine the forces exerted at B and C on member BCE. SOLUTION Free body: ACD Note: Sum moments about H to relate Cx and Cy. Free body: BCE Note: Sum moments about B to relate Cx and Cy ; combine with relation from free body ACD to determine Cx and Cy ; finally, sum forces in x and y to determine Bx and By. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 995 PROBLEM 6.F5 The position of member ABC is controlled by the hydraulic cylinder CD. Knowing that θ = 30°, draw the free-body diagram(s) needed to determine the force exerted by the hydraulic cylinder on pin C, and the reaction at B. SOLUTION We note that CD is a two-force member. Free body: ABC Note: To find θ, consider geometry of triangle BCD: Law of cosines: (CD)2 = (0.5)2 + (1.5)2 − 2(0.5)(1.5) cos 60° CD = 1.32288 m Law of sines: sin q sin 60° = 0.5 m 1.32288 m θ = 19.1066° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 996 PROBLEM 6.F6 Arm ABC is connected by pins to a collar at B and to crank CD at C. Neglecting the effect of friction, draw the free-body diagram(s) needed to determine the couple M to hold the system in equilibrium when θ = 30°. SOLUTION Free body: ABC Note: Sum forces in x to determine Cx ; sum moments about B to determine Cy. Free body: CD Note: Sum moments about D to determine M. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 997 PROBLEM 6.F7 Since the brace shown must remain in position even when the magnitude of P is very small, a single safety spring is attached at D and E. The spring DE has a constant of 50 lb/in. and an unstretched length of 7 in. Knowing that l = 10 in. and that the magnitude of P is 800 lb, draw the free-body diagram(s) needed to determine the force Q required to release the brace. SOLUTION Free body: ABC Note: Sum moments about C to relate Q to A. Spring force. Unstretched length: Stretched length: 0 = 7 in. = 10 in. k = 50 lb/in. F = kx = k ( − 0 ) = 50(10 − 7) = 150 lb Free body: ADB Note: Sum moments about B to determine A; use relation from free body ABC to determine Q. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 998 PROBLEM 6.F8 A log weighing 800 lb is lifted by a pair of tongs as shown. Draw the free-body diagram(s) needed to determine the forces exerted at E and F on tong DEF. SOLUTION We note that AC and BD are two-force members. Free body: AB Note: Free body: LOG Sum moments about A to determine FBD. Free body: DEF Note: Sum moments about E to find Fx; sum forces in x and y to find Ex and Ey. Note: Sum moments about G to determine Fy. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 999