# Solucionario de Mecanica vectorial para ingenieros Beer Johnston Cap5 10-estatica

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- 1.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 AB = 32 + 1.252 = 3.25 m BC = 32 + 42 = 5 mReactions: M A = 0: (84 kN)(3 m) C (5.25 m) = 0C = 48 kN Fx = 0: Ax C = 0 A x = 48 kNFy = 0: Ay = 84 kN = 0 A y = 84 kNJoint A: Fx = 0: 48 kN 12 FAB = 0 13 FAB = +52 kNFy = 0: 84 kN FAB = 52 kN T W5 (52 kN) FAC = 0 13 FAC = +64.0 kNFAC = 64.0 kN T WJoint C:FBC 48 kN = 5 3FBC = 80.0 kN C WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 737

2. 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 Free body: Entire truss Fx = 0: Bx = 0 M B = 0: C (15.75 ft) (945 lb)(12 ft) = 0 C = 720 lbFy = 0: By + 720 lb 945 lb = 0 B y = 225 lbFree body: Joint B: FAB FBC 225 lb = = 5 4 3FAB = 375 lb C W FBC = 300 lb T WFree body: Joint C: FAC FBC 720 lb = = 9.75 3.75 9 FBC = 300 lb TFAC = 780 lb C W(Checks)PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 738 3. 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 Free body: Entire truss Fx = 0: C x = 0 C x = 0 M B = 0: (1.92 kN)(3 m) + C y (4.5 m) = 0 C y = 1.28 kN C y = 1.28 kN Fy = 0: B 1.92 kN 1.28 kN = 0 B = 3.20 kNFree body: Joint B: FAB FBC 3.20 kN = = 5 3 4 FAB = 4.00 kN C W FBC = 2.40 kN C WFree body: Joint C: Fx = 0: 7.5 FAC + 2.40 kN = 0 8.5 FAC = +2.72 kNFy =FAC = 2.72 kN T W4 (2.72 kN) 1.28 kN = 0 (Checks) 8.5PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 739 4. 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 Reactions:M D = 0: Fy (24) (4 + 2.4)(12) (1)(24) = 0 Fy = 4.2 kipsFx = 0: Fx = 0 Fy = 0: D (1 + 4 + 1 + 2.4) + 4.2 = 0 D = 4.2 kipsJoint A: Fx = 0: FAB = 0FAB = 0 WFy = 0 : 1 FAD = 0 FAD = 1 kipJoint D:Fy = 0: 1 + 4.2 +Fx = 0:8 FBD = 0 17 FBD = 6.8 kips15 (6.8) + FDE = 0 17 FDE = +6 kipsFAD = 1.000 kip C WFBD = 6.80 kips C WFDE = 6.00 kips T WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 740 5. PROBLEM 6.4 (Continued)Joint E:Fy = 0 : FBE 2.4 = 0 FBE = +2.4 kipsFBE = 2.40 kips T WL Truss and loading symmetrical about cPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 741 6. 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 Free body: Truss Fx = 0: A x = 0 M A = 0: D (22.5) (10.8 kips)(22.5) (10.8 kips)(57.5) = 0 D = 38.4 kipsFy = 0: A y = 16.8 kipsFree body: Joint A: FAB F 16.8 kips = AD = 22.5 25.5 12FAB = 31.5 kips T W FAD = 35.7 kips C WFree body: Joint B: Fx = 0:FBC = 31.5 kips T WFy = 0:FBD = 10.80 kips C WFree body: Joint C: FCD FBC 10.8 kips = = 37 35 12 FBC = 31.5 kips TFCD = 33.3 kips C W(Checks)PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 742 7. 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 Free Body: Truss M E = 0: F (3 m) (900 N)(2.25 m) (900 N)(4.5 m) = 0 F = 2025 NFx = 0: Ex + 900 N + 900 N = 0 Ex = 1800 N E x = 1800 N Fy = 0: E y + 2025 N = 0 E y = 2025 N E y = 2025 N FAB = FBD = 0 WWe note that AB and BD are zero-force members: Free body: Joint A: FAC FAD 900 N = = 2.25 3.75 3FAC = 675 N T W FAD = 1125 N C WFree body: Joint D: FCD FDE 1125 N = = 3 2.23 3.75FCD = 900 N T W FDF = 675 N C WFree body: Joint E: Fx = 0: FEF 1800 N = 0FEF = 1800 N T WFy = 0: FCE 2025 N = 0FCE = 2025 N T WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 8. PROBLEM 6.6 (Continued)Free body: Joint F: Fy = 0:2.25 FCF + 2025 N 675 N = 0 3.75 FCF = 2250 NFx = FCF = 2250 N C W3 ( 2250 N) 1800 N = 0 (Checks) 3.75PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 9. 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: Truss Fy = 0: B y = 0 M B = 0: D(4.5 m) + (8.4 kN)(4.5 m) = 0 D = 8.4 kND = 8.4 kNFx = 0: Bx 8.4 kN 8.4 kN 8.4 kN = 0 Bx = +25.2 kN B x = 25.2 kNFree body: Joint A: FAB FAC 8.4 kN = = 5.3 4.5 2.8FAB = 15.90 kN C W FAC = 13.50 kN T WFree body: Joint C: Fy = 0: 13.50 kN 4.5 FCD = 0 5.3 FCD = +15.90 kNFx = 0: FBC 8.4 kN FCD = 15.90 kN T W2.8 (15.90 kN) = 0 5.3 FBC = 16.80 kNFBC = 16.80 kN C WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 10. PROBLEM 6.7 (Continued)Free body: Joint D: FBD 8.4 kN = 4.5 2.8FBD = 13.50 kN C WWe can also write the proportion FBD 15.90 kN = 4.5 5.3FBD = 13.50 kN C (Checks)PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 11. 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.SOLUTIONAD = 52 + 122 = 13 ft BCD = 122 + 162 = 20 ftReactions:Fx = 0: Dx = 0 M E = 0: D y (21 ft) (693 lb)(5 ft) = 0 Fy = 0: 165 lb 693 lb + E = 0D y = 165 lb E = 528 lbFx = 0:5 4 FAD + FDC = 0 13 5(1)Fy = 0:Joint D:12 3 FAD + FDC + 165 lb = 0 13 5(2)Solving (1) and (2), simultaneously: FAD = 260 lbFAD = 260 lb C WFDC = +125 lbFDC = 125 lb T WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 12. PROBLEM 6.8 (Continued)Joint E: Fx = 0:5 4 FBE + FCE = 0 13 5(3)Fy = 0:12 3 FBE + FCE + 528 lb = 0 13 5(4)Solving (3) and (4), simultaneously: FBE = 832 lbFBE = 832 lb C WFCE = +400 lbFCE = 400 lb T WJoint C:Force polygon is a parallelogram (see Fig. 6.11 p. 209) FAC = 400 lb T W FBC = 125 lb T WJoint A:Fx = 0:5 4 (260 lb) + (400 lb) + FAB = 0 13 5 FAB = 420 lbFy = 0:FAB = 420 lb C W12 3 (260 lb) (400 lb) = 0 13 5 0 = 0 (Checks)PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 13. PROBLEM 6.9 Determine the force in each member of the Pratt roof truss shown. State whether each member is in tension or compression.SOLUTION Free body: TrussFx = 0: Ax = 0Due to symmetry of truss and load Ay = H =1 total load = 21 kN 2Free body: Joint A:FAB FAC 15.3 kN = = 37 35 12 FAB = 47.175 kN FAC = 44.625 kNFAB = 47.2 kN C W FAC = 44.6 kN T WFree body: Joint B:From force polygon:FBD = 47.175 kN, FBC = 10.5 kNFBC = 10.50 kN C W FBD = 47.2 kN C WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 14. PROBLEM 6.9 (Continued)Free body: Joint C:Fy = 0:3 FCD 10.5 = 0 5Fx = 0: FCE +FCD = 17.50 kN T W4 (17.50) 44.625 = 0 5 FCE = 30.625 kNFree body: Joint E:DE is a zero-force memberFCE = 30.6 kN T W FDE = 0 WTruss and loading symmetrical about c LPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 15. PROBLEM 6.10 Determine the force in each member of the fan roof truss shown. State whether each member is in tension or compression.SOLUTION Free body: Truss Fx = 0 : A x = 0From symmetry of truss and loading: Ay = I =1 2Total loadA y = I = 6 kNFree body: Joint A:F FAB 5 kN = AC = 9.849 9 4FAB = 12.31 kN C W FAC = 11.25 kN T WFree body: Joint B: Fx =9 (12.31 kN + FBD + FDC ) = 0 9.849FBD + FBC = 12.31 kNor Fy =(1)4 (12.31 kN + FBD FBC ) 2 kN = 0 9.849FBD FBC = 7.386 kN(2)Add (1) and (2):2 FBD = 19.70 kNFBD = 9.85 kN C WSubtract (2) from (1):2 FBC = 4.924 kNFBC = 2.46 kN C WorPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 16. PROBLEM 6.10 (Continued)Free body: Joint D:FCD = 2.00 kN C WFrom force polygon:FDE = 9.85 kN C WFree body: Joint C: Fy =4 4 FCE (2.46 kN) 2 kN = 0 5 9.849FCE = 3.75 kN T W3 9 Fx = 0 : FCG + (3.75 kN) + (2.46 kN) 11.25 kN = 0 5 9.849 FCG = +6.75 kNFCG = 6.75 kN T WFrom the symmetry of the truss and loading: FEF = FDEFEF = 9.85 kN C WFEG = FCEFEG = 3.75 kN T WFFG = FCDFFG = 2.00 kN C WFFH = FBDFFH = 9.85 kN C WFGH = FBCFGH = 2.46 kN C WFGI = FACFGI = 11.25 kN T WFHI = FABFHI = 12.31 kN C WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 17. PROBLEM 6.11 Determine the force in each member of the Howe roof truss shown. State whether each member is in tension or compression.SOLUTION Free body: TrussFx = 0: H x = 0Because of the symmetry of the truss and loading: A = Hy =1 2Total loadA = H y = 1200 lbFree body: Joint A: FAB FAC 900 lb = = 5 4 3FAB = 1500 lb C W FAC = 1200 lb T WFree body: Joint C: BC is a zero-force member FCE = 1200 lb T WFBC = 0PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 18. PROBLEM 6.11 (Continued)Free body: Joint B: Fx = 0:orFBD + FBE = 1500 lb Fy = 0:or Add Eqs. (1) and (2): Subtract (2) from (1):4 4 4 FBD + FBC + (1500 lb) = 0 5 5 5(1)3 3 3 FBD FBE + (1500 lb) 600 lb = 0 5 5 5FBD FBE = 500 lb 2 FBD = 2000 lb 2 FBE = 1000 lb(2) FBD = 1000 lb C W FBE = 500 lb C WFree Body: Joint D: 4 4 (1000 lb) + FDF = 0 5 5Fx = 0:FDF = 1000 lbFDF = 1000 lb C W3 3 (1000 lb) ( 1000 lb) 600 lb FDE = 0 5 5Fy = 0:FDE = +600 lbFDE = 600 lb T WBecause of the symmetry of the truss and loading, we deduce that FEF = FBEFEF = 500 lb C WFEG = FCEFEG = 1200 lb T WFFG = FBCFFG = 0 WFFH = FABFFH = 1500 lb C WFGH = FACFGH = 1200 lb T WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 19. PROBLEM 6.12 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 Fx = 0: H x = 0Because of the symmetry of the truss and loading A = Hy =1 2Total loadA = H y = 1200 lbFree body: Joint A: FAB FAC 900 lb = = 5 4 3FAB = 1500 lb C W FAC = 1200 lb T WFree body: Joint C: BC is a zero-force member FBC = 0FCE = 1200 lb T WFree body: Joint B:Fx = 0:24 4 4 FBD + FBE + (1500 lb) = 0 25 5 5 24 FBD + 20 FBE = 30, 000 lbor Fy = 0:(1)7 3 3 FBD FBE + (1500) 600 = 0 25 5 5 7 FBD 15FBE = 7,500 lbor(2)PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 20. PROBLEM 6.12 (Continued)Multiply (1) by (3), (2) by 4, and add: 100 FBD = 120, 000 lbFBD = 1200 lb C W500 FBE = 30,000 lbFBE = 60.0 lb C WMultiply (1) by 7, (2) by 24, and add:Free body: Joint D: 24 24 FDF = 0 (1200 lb) + 25 25Fx = 0:FDF = 1200 lb Fy = 0:FDF = 1200 lb C W7 7 (1200 lb) (1200 lb) 600 lb FDE = 0 25 25 FDE = 72.0 lbFDE = 72.0 lb T WBecause of the symmetry of the truss and loading, we deduce that FEF = FBEFEF = 60.0 lb C WFEG = FCEFEG = 1200 lb T WFFG = FBCFFG = 0 WFFH = FABFFH = 1500 lb C WFGH = FACFGH = 1200 lb T WNote: Compare results with those of Problem 6.9.PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 21. PROBLEM 6.13 Determine the force in each member of the truss shown.SOLUTION 12.5 kN FCD FDG = = 2.5 6 6.5Joint D:FCD = 30 kN T W FDG = 32.5 kN C WF = 0: FCG = 0Joint C:BF =WF = 0: FFG = 32.5 kN CJoint G:WBF 2 (2.5 m) = 1.6667 m = BCF = tan 1 = 39.81 3 2Fy = 0: 12.5 kN FCF sin = 0 12.5 kN FCF sin 39.81 = 0 FCF = 19.526 kNFCF = 19.53 kN C WFx = 0: 30 kN FBC FCF cos = 0 30 kN FBC (19.526 kN) cos 39.81 = 0 FBC = +45.0 kNJoint F:Fx = 0: FBC = 45.0 kN T W6 6 FEF (32.5 kN) FCF cos = 0 6.5 6.5 6.5 FEF = 32.5 kN (19.526 kN) cos 39.81 6 FEF = 48.75 kNFy = 0: FBF FEF = 48.8 kN C W2.5 2.5 (32.5 kN) (19.526 kN)sin 39.81 = 0 FEF 6.5 6.5 FBF 2.5 ( 48.75 kN) 12.5 kN 12.5 kN = 0 6.5FBF = +6.25 kNFBF = 6.25 kN T WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 22. PROBLEM 6.13 (Continued)Joint B:tan =2.5 m ; = 51.34 2mFy = 0: 12.5 kN 6.25 kN FBE sin 51.34 = 0 FBE = 24.0 kNFBE = 24.0 kN C WFx = 0: 45.0 kN FAB + (24.0 kN) cos 51.34 = 0 FAB = +60 kNFAB = 60.0 kN T WJoint E: = 51.34 Fy = 0: FAE (24 kN) sin 51.34 (48.75 kN) FAE = +37.5 kN2.5 =0 6.5 FAE = 37.5 kN T WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 23. PROBLEM 6.14 Determine the force in each member of the roof truss shown. State whether each member is in tension or compression.SOLUTION Free body: Truss Fx = 0: A x = 0From symmetry of loading: Ay = E =1 Total load 2A y = E = 3.6 kNWe note that DF is a zero-force member and that EF is aligned with the load. Thus FDF = 0 W FEF = 1.2 kN C WFree body: Joint A: FAB FAC 2.4 kN = = 13 12 5FAB = 6.24 kN C W FAC = 2.76 kN T WFree body: Joint B: Fx = 0:3 12 12 FBC + FBD + (6.24 kN) = 0 3.905 13 13Fy = 0:2.5 5 5 FBC + FBD + (6.24 kN) 2.4 kN = 0 3.905 13 13(1) (2)PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 24. PROBLEM 6.14 (Continued)Multiply (1) by 2.5, (2) by 3, and add: 45 45 FBD + (6.24 kN) 7.2 kN = 0, FBD = 4.16 kN, 13 13FBD = 4.16 kN C WMultiply (1) by 5, (2) by 12, and add: 45 FBC + 28.8 kN = 0, FBC = 2.50 kN, 3.905FBC = 2.50 kN C WFree body: Joint C: Fy = 0:5 2.5 (2.50 kN) = 0 FCD 5.831 3.905 FCD = 1.867 kN T WFx = 0: FCE 5.76 kN +3 3 (2.50 kN) + (1.867 kN) = 0 3.905 5.831 FCE = 2.88 kN TFree body: Joint E: Fy = 0:5 FDE + 3.6 kN 1.2 kN = 0 7.81 FDE = 3.75 kNFx = 0: FCE FDE = 3.75 kN C W6 ( 3.75 kN) = 0 7.81FCE = +2.88 kN FCE = 2.88 kN T(Checks)PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 25. PROBLEM 6.15 Determine the force in each member of the Warren bridge truss shown. State whether each member is in tension or compression.SOLUTION Fx = 0: Ax = 0Free body: Truss Due to symmetry of truss and loading Ay = G =Free body: Joint A:1 Total load = 6 kips 2FAB FAC 6 kips = = 5 3 4FAB = 7.50 kips C W FAC = 4.50 kips T WFree body: Joint B:FBC FBD 7.5 kips = = 5 6 5FBC = 7.50 kips T W FBD = 9.00 kips C WFree body: Joint C: Fy = 0:4 4 (7.5) + FCD 6 = 0 5 5 FCD = 0 W3 Fx = 0: FCE 4.5 (7.5) = 0 5 FCE = +9 kipsFCE = 9.00 kips T WTruss and loading symmetrical about c LPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 26. PROBLEM 6.16 Solve Problem 6.15 assuming that the load applied at E has been removed. PROBLEM 6.15 Determine the force in each member of the Warren bridge truss shown. State whether each member is in tension or compression.SOLUTION Free body: TrussFx = 0: Ax = 0 M G = 0: 6(36) Ay (54) = 0 A y = 4 kips Fy = 0: 4 6 + G = 0 G = 2 kipsFree body: Joint A:FAB FAC 4 kips = = 5 3 4FAB = 5.00 kips C W FAC = 3.00 kips T WFree body Joint B:FBC FBD 5 kips = = 5 6 5FBC = 5.00 kips T W FBD = 6.00 kips C WFree body Joint C: M y = 0:4 4 (5) + FCD 6 = 0 5 53 3 Fx = 0: FCE + (2.5) (5) 3 = 0 5 5FCD = 2.50 kips T W FCE = 4.50 kips T WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 27. PROBLEM 6.16 (Continued)Free body: Joint D: 4 4 Fy = 0: (2.5) FDE = 0 5 5 FDE = 2.5 kipsFDE = 2.50 kips C W3 3 Fx = 0: FDF + 6 (2.5) (2.5) = 0 5 5 FDF = 3 kipsFree body: Joint F:FDF = 3.00 kips C WFEF FFG 3 kips = = 5 5 6FEF = 2.50 kips T W FFG = 2.50 kips C WFree body: Joint G:FEG 2 kips = 3 4Also:FFG 2 kips = 5 4FEG = 1.500 kips T WFFG = 2.50 kips C (Checks)PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 28. PROBLEM 6.17 Determine the force in member DE and in each of the members located to the left of DE for the inverted Howe roof truss shown. State whether each member is in tension or compression.SOLUTION Free body: Truss:Fx = 0: A x = 0 M H = 0: (400 lb)(4d ) + (800 lb)(3d ) + (800 lb)(2d ) + (800 lb)d Ay (4d ) = 0 Ay = 1600 lbAngles:6.72 = 32.52 10.54 6.72 tan = = 16.26 23.04 tan =Free body: Joint A:FAC FAB 1200 lb = = sin 57.48 sin106.26 sin16.26 FAB = 3613.8 lb C FAC = 4114.3 lb TFAB = 3610 lb C , FAC = 4110 lb T WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 29. PROBLEM 6.17 (Continued)Free body: Joint B: Fy = 0: FBC (800 lb) cos16.26 = 0 FBC = 768.0 lbFBC = 768 lb C WFx = 0: FBD + 3613.8 lb + (800 lb) sin16.26 = 0 FBD = 3837.8 lbFBD = 3840 lb C WFree body: Joint C:Fy = 0: FCE sin 32.52 + (4114.3 lb) sin 32.52 (768 lb) cos16.26 = 0 FCE = 2742.9 lbFCE = 2740 lb T WFx = 0: FCD (4114.3 lb) cos 32.52 + (2742.9 lb) cos 32.52 (768 lb) sin16.26 = 0 FCD = 1371.4 lbFCD = 1371 lb T WFree body: Joint E:FDE 2742.9 lb = sin 32.52 sin73.74FDE = 1536 lb C WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 30. PROBLEM 6.18 Determine the force in each of the members located to the right of DE for the inverted Howe roof truss shown. State whether each member is in tension or compression.SOLUTION Free body: TrussM A = 0: H (4d ) (800 lb)d (800 lb)(2d ) (800 lb)(3d ) (400 lb)(4d ) = 0 H = 1600 lbAngles:6.72 = 32.52 10.54 6.72 tan = = 16.26 23.04 tan =Free body: Joint H:FGH = (1200 lb) cot16.26 FGH = 4114.3 lb T FFH =1200 lb = 4285.8 lb sin16.26FGH = 4110 lb T W FFH = 4290 lb C WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 31. PROBLEM 6.18 (Continued)Free body Joint F: Fy = 0: FFG (800 lb) cos16.26 = 0 FFG = 768.0 lbFFG = 768 lb C WFx = 0: FDF 4285.8 lb + (800 lb) sin 16.26 = 0 FDF = 4061.8 lbFDF = 4060 lb C WFree body: Joint G: Fy = 0: FDG sin 32.52 (768.0 lb) cos16.26 = 0 FDG = +1371.4 lb FDG = 1371 lb T WFx = 0: FEG + 4114.3 lb (768.0 lb)sin16.26 (1371.4 lb) cos 32.52 = 0 FEG = +2742.9 lbFEG = 2740 lb T WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 32. PROBLEM 6.19 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: TrussFx = 0: A x = 0 M 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 kNFBC = 0 WWe note that BC is a zero-force member: Also:FCE = FACFree body: Joint A:Fx = 0:Fy = 0:(1) 1 2 1 2FAB +FAB +2 5 1 5FAC = 0(2)FAC + 3.50 kN = 0(3)Multiply (3) by 2 and add (2): 1 2FAB 7 kN = 0FAB = 9.90 kN C WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 33. PROBLEM 6.19 (Continued)Subtract (3) from (2): 1 5FAC 3.50 kN = 0 FAC = 7.826 kN FCE = FAC = 7.826 kNFrom (1):FAC = 7.83 kN T W FCE = 7.83 kN T WFree body: Joint B: Fy = 0:1 2FBD +1 2(9.90 kN) 2 kN = 0 FBD = 7.071 kN1Fx = 0: FBE +2(9.90 7.071)kN = 0 FBE = 2.000 kNFree body: Joint E:Fx = 0:2 5FBD = 7.07 kN C WFBE = 2.00 kN C W( FEG 7.826 kN) + 2.00 kN = 0 FEG = 5.590 kNFy = 0: FDE 1 5FEG = 5.59 kN T W(7.826 5.590)kN = 0 FDE = 1.000 kNFDE = 1.000 kN T WFree body: Joint D: Fx = 0:2 51 2(7.071 kN)FDF + FDG = 5.590 kNor Fy = 0:or( FDF + FDG ) +1 5( FDF FDG ) +1 2(4)(7.071 kN) = 2 kN 1 kN = 0FDE FDG = 4.472(5)Add (4) and (5):2 FDF = 10.062 kNFDF = 5.03 kN C WSubtract (5) from (4):2 FDG = 1.1180 kNFDG = 0.559 kN C WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 34. PROBLEM 6.20 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: TrussM A = 0: L(12 m) (2 kN)(2 m) (2 kN)(4 m) (1 kN)(6 m) = 0 L = 1.500 kNAngles: tan = 1 = 451 tan = 2 = 26.57Zero-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 WThus: We also note that FGI = FIK = FKLand(1)FHJ = FJL(2)PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 35. PROBLEM 6.20 (Continued)Free body: Joint L:FJL F 1.500 kN = KL = sin116.57 sin 45 sin18.43 FJL = 4.2436 kNFJL = 4.24 kN C W FKL = 3.35 kN T WFrom Eq. (1):FGI = FIK = FKLFGI = FIK = 3.35 kN T WFrom Eq. (2):FHJ = FJL = 4.2436 kNFHJ = 4.24 kN T WFGH FFH 4.2436 = = sin108.43 sin18.43 sin 53.14FFH = 5.03 kN C WFree body: Joint H:FGH = 1.677 kN T WFree body: Joint F: Fx = 0: FDF cos 26.57 (5.03 kN) cos 26.57 = 0 FDF = 5.03 kNFy = 0: FFG 1 kN + (5.03 kN) sin 26.57 ( 5.03 kN)sin 26.57 = 0 FFG = 3.500 kNFFG = 3.50 kN T WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 36. PROBLEM 6.21 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: TrussFx = 0: A x = 0Because of symmetry of loading: Ay = L =1 Total load 2A y = L = 1200 lbZero-Force Members. Examining joints C and H, we conclude that BC, EH, and GH are zero-force members. Thus FBC = FEH = 0WAlso,FCE = FAC(1)Free body: Joint AFABFAC 1000 lb = 2 1 5 FAB = 2236 lb C =FAB = 2240 lb C W FAC = 2000 lb T W FCE = 2000 lb T WFrom Eq. (1): Free body: Joint B Fx = 0:2 5FBD +2 5FBE +2 5(2236 lb) = 0FBD + FBE = 2236 lbor Fy = 0:1 5FBD 1 5FBE +1FBD FBE = 1342 lbor(2)5(2236 lb) 400 lb = 0(3)PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 37. PROBLEM 6.21 (Continued)Add (2) and (3):2 FBD = 3578 lbFBD = 1789 lb C WSubtract (3) from (1):2 FBE = 894 lbFBE = 447 lb C WFree body: Joint E 2Fx = 0:5FEG +2 5(447 lb) 2000 lb = 0FEG = 1789 lb T W Fy = 0: FDE +1 51(1789 lb) 5FDE = 600 lb(447 lb) = 0FDE = 600 lb C WFree body: Joint D 2Fx = 0:5FDF +2 52FDG +5(1789 lb) = 0FDF + FDG = 1789 lbor Fy = 0:1FDF 1FDG +15 5 5 + 600 lb 400 lb = 0(4)(1789 lb)FDF FDG = 2236 lbor Add (4) and (5):2 FDF = 4025 lbSubtract (5) from (4):2 FDG = 447 lb(5)FDF = 2010 lb C W FDG = 224 lb T WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 38. PROBLEM 6.22 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, 2L = 1200 lb(See F, B, diagram to the left for more details) Free body: Joint L 9 = 26.57 18 3 = tan 1 = 9.46 18 FJL FKL 1000 lb = = sin 63.43 sin 99.46 sin17.11 = tan 1FKL = 3352.7 lb CFJL = 3040 lb T W FKL = 3350 lb C WFree body: Joint K Fx = 0: 2 5FIK 2 5FJK 2 5(3352.7 lb) = 0FIK + FJK = 3352.7 lbor:Fy = 0:or: Add (1) and (2):1 5FIK 1 5FJK +(1) 1 5FIK FJK = 2458.3 lb(2)2 FIK = 5811.0FJK = 2905.5 lbSubtract (2) from (1):(3352.7) 400 = 0FIK = 2910 lb C W2 FJK = 894.4FJK = 447.2 lbFJK = 447 lb C WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 39. PROBLEM 6.22 (Continued)Free body: Joint J 2Fx = 0: 13 3Fy = 0:FIJ +136FIJ 37 1FGJ 376FGJ +37 151(3040 lb) 372(3040 lb) 5(447.2) = 0(447.2) = 0(3) (4)Multiply (4) by 6 and add to (3): 16 13FIJ 8 5(447.2) = 0FIJ = 360.54 lbFIJ = 361 lb T WMultiply (3) by 3, (4) by 2, and add: 16 37( FGJ 3040) 8 5(447.2) = 0FGJ = 2431.7 lbFGJ = 2430 lb T WFree body: Joint I 2Fx = 0: 52FFI 52FGI 5(2905.5) +2 13(360.54) = 0FFI + FGI = 2681.9 lbor Fy = 0:1 5FFI 1 5FGI +1 5(2905.5) (5) 3 13(360.54) 400 = 0FFI FGI = 1340.3 lbor(6)2 FFI = 4022.2Add (5) and (6):FFI = 2011.1 lbSubtract (6) from (5):FFI = 2010 lb C W2 FGI = 1341.6 lbFGI = 671 lb C WFree body: Joint F FromFx = 0: FDF = FFI = 2011.1 lb C 1 2011.1 lb = 0 Fy = 0: FFG 400 lb + 2 5 FFG = + 1400 lbFFG = 1400 lb T WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 40. PROBLEM 6.23 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 Fx = 0: A x = 0 M 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 kNFree body: Joint A FABFAC 4.20 kN = 2 1 5 FAB = 9.3915 kN =FAB = 9.39 kN C W FAC = 8.40 kN T WFree body: Joint B 2Fx = 0:5 1Fy = 0:5FBD + FBD 1 2 1 2FBC + FBC +2 5 1 5(9.3915) = 0(1)(9.3915) 2.4 = 0(2)Add (1) and (2): 3 5FBD +3 5(9.3915 kN) 2.4 kN = 0FBD = 7.6026 kNFBD = 7.60 kN C WMultiply (2) by 2 and add (1): 3 2FB + 4.8 kN = 0FBC = 2.2627 kNFBC = 2.26 kN C WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 41. PROBLEM 6.23 (Continued)Free body: Joint C 1Fx = 0:5 2Fy = 0:54FCD +17 1FCD +17FCE + FCE 1 2 1 2(2.2627) 8.40 = 0(3)(2.2627) = 0(4)Multiply (4) by 4 and add (1): 7 5FCD +5(2.2627) 8.40 = 02FCD = 0.1278 kNFCD = 0.128 kN C WMultiply (1) by 2 and subtract (2): 7 17FCE +3 2(2.2627) 2(8.40) = 0FCE = 7.068 kNFCE = 7.07 kN T WFree body: Joint D Fx = 0:2 51+ Fy = 0:51 5 +FDF +(0.1278) = 0FDF 2 51 2 FDE + (7.6026) 1.524 5(5)1.15 1 FDE + (7.6026) 1.524 5(0.1278) 2.4 = 0(6)Multiply (5) by 1.15 and add (6): 3.30 5FDF +3.30 5(7.6026) +3.15 5(0.1278) 2.4 = 0FDF = 6.098 kNFDF = 6.10 kN C WMultiply (6) by 2 and add (5): 3.30 3 FDE (0.1278) + 4.8 = 0 1.524 5FDE = 2.138 kNFDE = 2.14 kN C WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 42. PROBLEM 6.23 (Continued)Free body: Joint E Fx = 0:0.6 4 1 FEF + ( FEH FCE ) + (2.138) = 0 2.04 1.524 17(7)Fy = 0:1.95 1 1.15 FEF + ( FEH FCE ) (2.138) = 0 2.04 1.524 17(8)Multiply (8) by 4 and subtract (7): 7.2 FEF 7.856 kN = 0 2.04FEF = 2.23 kN T WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 43. PROBLEM 6.24 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 1.2FAB = 2.29 kN T W FAC = 2.29 kN C WFree body: Joint F FDF F 1.2 kN = EF = 2.29 2.29 2.1FDF = 2.29 kN T W FEF = 2.29 kN C WFree body: Joint D FBD FDE 2.29 kN = = 2.21 0.6 2.29FBD = 2.21 kN T W FDE = 0.600 kN C WFree body: Joint B Fx = 0:4 2.21 (2.29 kN) = 0 FBE + 2.21 kN 5 2.29FBE = 0 W 3 0.6 (2.29 kN) = 0 Fy = 0: FBC (0) 5 2.29FBC = 0.600 kNFBC = 0.600 kN C WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 44. PROBLEM 6.24 (Continued)Free body: Joint C Fx = 0: FCE +2.21 (2.29 kN) = 0 2.29FCE = 2.21kN Fy = 0: FCH 0.600 kN FCE = 2.21 kN C W 0.6 (2.29 kN) = 0 2.29FCH = 1.200 kNFCH = 1.200 kN C WFree body: Joint E Fx = 0: 2.21 kN 2.21 4 (2.29 kN) FEH = 0 2.29 5FEH = 0 W Fy = 0: FEJ 0.600 kN FEJ = 1.200 kN0.6 (2.29 kN) 0 = 0 2.29FEJ = 1.200 kN C WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 45. PROBLEM 6.25 For the tower and loading of Problem 6.24 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 6.24 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 1.2FGH = 3.03 kN T W FGI = 3.03 kN C WFree body: Joint L FJL F 1.2 kN = KL = 3.03 3.03 1.2FJL = 3.03 kN T W FKL = 3.03 kN C WFree body: Joint J Fx = 0: FHJ +2.97 (3.03 kN) = 0 3.03 FHJ = 2.97 kN T W0.6 (3.03 kN) = 0 3.03 = 1.800 kN FJK = 1.800 kN C WFy = 0: FJK 1.2 kN FJKFree body: Joint H Fx = 0:4 2.97 (3.03 kN) = 0 FHK + 2.97 kN 5 3.03 FHK = 0 W 0.6 3 (3.03) kN (0) = 0 3.03 5 = 1.800 kN FHI = 1.800 kN C WFy = 0: FHI 1.2 kN FHIPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 46. PROBLEM 6.25 (Continued)Free body: Joint I Fx = 0: FIK +2.97 (3.03 kN) = 0 3.03FIK = 2.97 kN Fy = 0: FIN 1.800 kN FIN = 2.40 kNFIK = 2.97 kN C W 0.6 (3.03 kN) = 0 3.03FIN = 2.40 kN C WFree body: Joint K Fx = 0: 4 2.97 (3.03 kN) = 0 FKN + 2.97 kN 5 3.03FKN = 0 W Fy = 0: FKD 0.6 3 (3.03 kN) 1.800 kN (0) = 0 3.03 5FKD = 2.40 kNFKD = 2.40 kN C WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 47. PROBLEM 6.26 Solve Problem 6.24 assuming that the cables hanging from the right side of the tower have fallen to the ground. PROBLEM 6.24 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 WConsidering next joint D, we note that BD and DE are zero-force members: FBD = FDE = 0 WFree body: Joint A F FAB 1.2 kN = AC = 2.29 2.29 1.2FAB = 2.29 kN T W FAC = 2.29 kN C WFree body: Joint B Fx = 0:4 2.21 (2.29 kN) = 0 FBE 5 2.29FBE = 2.7625 kN Fy = 0: FBC FBE = 2.76 kN T W0.6 3 (2.29 kN) (2.7625 kN) = 0 2.29 5FBC = 2.2575 kNFBC = 2.26 kN C WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 48. PROBLEM 6.26 (Continued)Free body: Joint C Fx = 0: FCE +2.21 (2.29 kN) = 0 2.29FCE = 2.21 kN C W Fy = 0: FCH 2.2575 kN FCH = 2.8575 kN0.6 (2.29 kN) = 0 2.29FCH = 2.86 kN C WFree body: joint E Fx = 0: 4 4 FEH (2.7625 kN) + 2.21 kN = 0 5 5FEH = 0 W 3 3 Fy = 0: FEJ + (2.7625 kN) (0) = 0 5 5FEJ = +1.6575 kNFEJ = 1.658 kN T WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 49. 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 M F = 0: G (20 ft) (15 kips)(16 ft) (40 kips)(15 ft) = 0G = 42 kips Fx = 0: Fx + 15 kips = 0Fx = 15 kips Fy = 0: Fy 40 kips + 42 kips = 0Fy = 2 kipsFree body: Joint F Fx = 0:1 5FDF 15 kips = 0 FDF = 33.54 kipsFy = 0: FBF 2 kips +2 5FDF = 33.5 kips T W(33.54 kips) = 0FBF = 28.00 kipsFBF = 28.0 kips C WFree body: Joint B Fx = 0: Fy = 0:5 29 2 29FAB + FAB 5 61 6 61FBD + 15 kips = 0(1)FBD + 28 kips = 0(2)PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 50. PROBLEM 6.27 (Continued)Multiply (1) by 6, (2) by 5, and add: 40 29FAB + 230 kips = 0 FAB = 30.96 kipsFAB = 31.0 kips C WMultiply (1) by 2, (2) by 5, and add: 40FBD 110 kips = 061FBD = 21.48 kipsFBD = 21.5 kips T WFree body: joint D 2Fy = 0:52FAD 5(33.54) +6 61(21.48) = 0FAD = 15.09 kips T W 1Fx = 0: FDE +5(15.09 33.54) 5(21.48) = 061FDE = 22.0 kips T WFree body: joint A 5Fx = 0:29Fy = 0: 1FAC +2 295FAC FAE +2 55FAE +29(30.36) 2 291 5(15.09) = 02(30.96) 5(15.09) = 0(3) (4)Multiply (3) by 2 and add (4): 8 29FAC +12 29(30.96) 4 5FAC = 28.27 kips,(15.09) = 0FAC = 28.3 kips C WMultiply (3) by 2 (4) by 5 and add: 8 5FAE +20 29(30.96) 12 5(15.09) = 0FAE = 9.50 kips T WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 51. PROBLEM 6.27 (Continued)Free body: Joint C From force triangle FCE 61=FCG 28.27 kips = 8 29FCE = 41.0 kips T W FCG = 42.0 kips C WFree body: Joint G Fx = 0: Fy = 0: 42 kips 42 kips = 0FEG = 0 W (Checks)PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 52. PROBLEM 6.28 Determine the force in each member of the truss shown. State whether each member is in tension or compression.SOLUTION Free body: Truss Fx = 0: H x = 0 M H = 0: 48(16) G (4) = 0 G = 192 kN Fy = 0: 192 48 + H y = 0H y = 144 kN H y = 144 kNZero-Force Members: Examining successively joints C, B, E, D, and F, we note that the following are zero-force members: BC, BE, DE, DG, FG Thus,FBC = FBE = FDE = FDG = FFG = 0Also note:FAB = FBD = FDF = FFH(1)FAC = FCE = FEG(2)Free body: Joint A:FAB FAC 48 kN = = 8 3 73FAB = 128 kNFAB = 128.0 kN T WFAC = 136.704 kNFAC = 136.7 kN C WFrom Eq. (1):FBD = FDF = FFH = 128.0 kN T WFrom Eq. (2):FCE = FEG = 136.7 kN C WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 53. PROBLEM 6.28 (Continued)Free body: Joint HAlsoFGH 145=144 kN 9FGH = 192.7 kN C WFFH 144 kN = 8 9 FFH = 128.0 kN T(Checks)PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 54. 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 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 WTruss of Problem 6.32a 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 WTruss of Problem 6.33a Starting with triangle EHK and adding two members at a time, we obtain successively joints D, J, C, G, I, B, F, and A, thus completing the truss. Therefore, this truss is a simple truss WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 55. 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 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 WTruss of Problem 6.32b. Starting with triangle CGH 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 WTruss of Problem 6.33b. Starting with triangle GLM and adding two members at a time, we obtain joints K and I but cannot continue, starting instead with triangle BCH, we obtain joint D but cannot continue, thus, this truss is not a simple truss WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 56. 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: FBC = 0 FB: Joint C: FCD = 0 FB: Joint J : FIJ = 0 FB: Joint I : FIL = 0 FB: Joint N : FMN = 0(a)FB: Joint M : FLM = 0 BC , CD, IJ , IL, LM , MN WThe zero-force members, therefore, are Truss (b)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 BC , BE , DE , EF , FG , IJ , KN , MN WThe zero-force members, therefore, arePROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 57. 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: 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 = 0AI , BJ , CK , DI , EI , FK , GK WThe zero-force members, therefore, are Truss (b)FB: Joint K : FFK = 0 FB: Joint O: FIO = 0(b)FK and IO WThe zero-force members, therefore, are All other members are either in tension or compression.PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 58. PROBLEM 6.33 For the given loading, determine the zero-force members in each of the two trusses shown.SOLUTION Truss (a)FB: Joint F : FBF = 0 FB: Joint B: FBG = 0 FB: Joint G: FGJ = 0 FB: Joint D: FDH = 0 (a)FB: Joint J : FHJ = 0 FB: Joint H : FEH = 0BF , BG , DH , EH , GJ , HJ WThe zero-force members, therefore, are Truss (b)FB: Joint A: FAB = FAF = 0 FB: Joint C: FCH = 0 FB: Joint E: FDE = FEJ = 0 FB: Joint L: FGL = 0 FB: Joint N : FIN = 0(b)AB, AF , CH , DE , EJ , GL, IN WThe zero-force members, therefore, arePROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 59. PROBLEM 6.34 Determine the zero-force members in the truss of (a) Problem 6.23, (b) Problem 6.28.SOLUTION (a)Truss of Problem 6.23 FB : Joint I : FIJ = 0 FB : Joint J : FGJ = 0 FB : Joint G: FGH = 0GH , GJ , IJ WThe zero-force members, therefore, are (b)Truss of Problem 6.28 FB : Joint C: FBC = 0 FB : Joint B: FBE = 0 FB : Joint E: FDE = 0 FB : Joint D: FDG = 0 FB : Joint F : FFG = 0 BC , BE , DE , DG, FG WThe zero-force members, therefore, arePROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 60. 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 = BxandDy = B yM z = 0: A(10 ft) (400 lb)(24 ft) = 0 A = 960 lbFx = 0: Bx + Dx + A = 0 2 Bx 960 lb = 0, Bx = 480 lbFy = 0: B y + Dy 400 lb = 02 By = 400 lb By = +200 lb B = (480 lb)i + (200 lb) j YThus Free body: C FCA = FAC FCB = FBC FCD = FCDJJJ G CA FAC = ( 24i + 10 j) CA 26 JJJ G CB FBC = (24i + 7k ) CB 25 JJJ G CD FCD = (24i 7k ) CD 25F = 0: FCA + FCB + FCD (400 lb) j = 0PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 61. 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 25i:j:10 FAC 400 lb = 0 26k:7 ( FBC FCD ) = 0 FCD = FBC 25(1) FAC = 1040 lb T WSubstitute for FAC and FCD in Eq. (1): 24 24 (10.40 lb) (2 FBC ) = 0 FBC = 500 lb 26 25FBC = FCD = 500 lb C WFree body: B FBC FBA FBDJJJ G CB = (500 lb) = (480 lb)i + (140 lb)k CB JJJ G BA FAB = FAB = (10 j 7k ) BA 12.21 = FBD kF = 0: FBA + FBD + FBC + (480 lb)i + (200 lb) j = 0Substituting for FBA , FBD , FBC and equating to zero the coefficients of j and k: j:10 FAB + 200 lb = 0 FAB = 244.2 lb 12.21k: 7 FAB FBD + 140 lb = 0 12.21 FBD = From symmetry:7 (244.2 lb) + 140 lb = +280 lb 12.21FAD = FABFAB = 244 lb C WFBD = 280 lb T W FAD = 244 lb C WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 62. 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 Fx = 0: 2 Bx = 0 Bx = Dx = 0 Fz = 0: Bz = 0 M cz = 0: 2 By (2.8 m) + (2184 N)(2 m) = 0 By = 780 N B = (780 N) j YThus Free body: A JJJ G AB FAB FAB = FAB = ( 0.8i 4.8 j + 2.1k ) AB 5.30 JJJG AC FAC = FAC = FAC (2i 4.8 j) AC 5.20 JJJG AD FAD = FAD = FAD (+0.8i 4.8 j 2.1k ) AD 5.30 F = 0: FAB + FAC + FAD (2184 N) j = 0PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 63. 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 5.30k:FAD = FABMultiply (1) by 6 and add (2): 16.8 FAC 2184 N = 0, FAC = 676 N 5.20 FAC = 676 N C WSubstitute for FAC and FAD in (1): 0.8 2 2 FAB + (676 N) = 0, FAB = 861.25 N 5.30 5.20 FAB = FAD = 861 N C WFree body: B JJJ G 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 F = 0: FAB + FBC + FBD + (780 N) j = 0Substituting for FAB , FBC , FBD and equating to zero the coefficients of i and k : i: k:130 N + 0.8 FBC = 0 FBC = +162.5 N 341.25 N 0.6 FBC FBD = 0 FBD = 341.25 0.6(162.5) = +243.75 NFrom symmetry:FBC = 162.5 N T WFCD = FBCFBD = 244 N T W FCD = 162.5 N T WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 64. 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 Fx = 0: 2 Bx + 2968 N = 0 Bx = Dx = 1484 N M cz = 0: 2 By (2.8 m) (2968 N)(4.8 m) = 0 By = 2544 N B = (1484 N)i (2544 N) j YThus Free body: A FAB = FABJJJ G AB ABFAB (0.8i 4.8 j + 2.1k ) 5.30 JJJG AC FAC = FAC = (2i 4.8 j) AC 5.20 JJJG AD = FAD AD FAD = (0.8i 4.8 j 2.1k ) 5.30 =FAC FADF = 0: FAB + FAC + FAD + (2968 N)i = 0PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 65. 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 5.30FAD = FABMultiply (1) by 6 and add (2): 16.8 FAC 6(2968 N) = 0, FAC = 5512 N 5.20 FAC = 5510 N C WSubstitute for FAC and FAD in (2): 4.8 4.8 2 FAB (5512 N) = 0, FAB = +2809 N 5.30 5.20 FAB = FAD = 2810 N T WFree body: B FAB FBC FBDJJJ G 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 kF = 0: FAB + FBC + FBD (1484 N)i (2544 N) j = 0Substituting for FAB , FBC , FBD and equating to zero the coefficients of i and k: i: k:+24 N + 0.8FBC 1484 N = 0, FBC = +1325 NFBC = 1325 N T W1113 N 0.6 FBC FBD = 0 FBD = 1113 N 0.6(1325 N) = 1908 N,From symmetry:FBD = 1908 N C WFCD = FBCFCD = 1325 N T WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 66. 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 Fx = 0: Ax = 0 M BC = 0: Ay (6 ft) + (1600 lb)(7.5 ft) = 0 A = (2000 lb)j YAy = 2000 lb By = CFrom symmetry:Fy = 0: 2 By 2000 lb 1600 lb = 0 B = (1800 lb) j YBy = 1800 lb F = 0: FAB + FAC + FAD (2000 lb) j = 0Free body: AFABi+k+ FAC2ik 2+ FAD (0.6 i + 0.8 j) (2000 lb) j = 0Factoring i, j, k and equating their coefficient to zero: 11FAC + 0.6 FAD = 0(1)0.8FAD 2000 lb = 02FAB +FAD = 2500 lb T W21 2FAB 1 2FAC = 0FAC = FABPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 67. PROBLEM 6.38* (Continued)Substitute for FAD and FAC into (1): 2 2Free body: BFAB + 0.6(2500 lb) = 0, FAB = 1060.7 lb,FBA FBCFAB = FAC = 1061 lb C WJJJ G BA i+k = FAB = + (1060.7 lb) = (750 lb)(i + k ) BA 2 = FBC kFBD = FBD (0.8 j 0.6k ) JJJ G BE FBE FBE = FBE = (7.5i + 8 j 6k ) BE 12.5F = 0: FBA + FBC + FBD + FBE + (1800 lb) j = 0Substituting for FBA , FBC , FBD + FBE and equate to zero the coefficients of i, j, k : i: 7.5 750 lb + FBE = 0, FBE = 1250 lb 12.5 FBE = 1250 lb C Wj: 8 0.8 FBD + (1250 lb) + 1800 lb = 0 12.5 FBD = 1250 lb C Wk:750 lb FBC 0.6( 1250 lb) 6 ( 1250 lb) = 0 12.5FBC = 2100 lb T W FBD = FCD = 1250 lb C WFrom symmetry: Free body: D F = 0: FDA + FDB + FDC + FDE i = 0We 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 lbi:1500 lb + FDE = 0FDE = 1500 lb T WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 68. 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 M 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 = 0Equate to zero the coefficients of i, j, k : i:3Dy 900 = 0, D y = 300 Nj:Dz = 0,D = (300 N) j Yk:1.8C + 1.8(300) 720 = 0C = (100 N) j YF = 0: B + 300 j + 100 j 1200 j = 0B = (800 N) j YFree body: B F = 0: FBA + FBC + FBE + (800 N) j = 0, with JJJ G BA FAB FBA = FAB (0.6i + 3j 0.75k ) = BA 3.15 FBE = FBE kFBC = FBC iSubstitute and equate to zero the coefficient of j, i, k : j: 3 FAB + 800 N = 0, FAB = 840 N, 3.315 i: 0.6 (840 N) + FBC = 0 3.15 FBC = 160.0 N T Wk: 0.75 (840 N) FBE = 0 3.15 FBE = 200 N T WFAB = 840 N C WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 69. PROBLEM 6.39* (Continued)Free body C: F = 0: FCA + FCB + FCD + FCE + (100 N) j = 0, with JJJ G F CA FCA = FAC = AC (1.2i + 3j 0.75 k ) CA 3.317FCB = (160 N) i FCD = FCD k FCE = FCEJJJ G F CE = CE (1.8i 3k ) CE 3, 499Substitute 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 W1.2 1.8 (110.57) 160 FCE = 0, FCE = 233.3 3.317 3.499FCE = 233 N C W0.75 3 (110.57) FCD (233.3) = 0 3.317 3.499FCD = 225 N T WFree body: D F = 0: FDA + FDC + FDE + (300 N) j = 0, with JJJ G F DA FDA = FAD = AD (1.2i + 3j + 2.25k ) DA 3.937 FDC = FCD k = (225 N)kFDE = FDE iSubstitute 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 W FDE = 1200 N T W 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 WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 70. 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 M 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 = 0Equate to zero the coefficient of i, j, k : 3Dy 2700 = 0 Dy = 900 N1.8Dz + 540 = 0 Dz = 300 N 1.8C + 1.8D y = 0 C = Dy = 900 NThus:C = (900 N) j D = (900 N) j + (300 N)kYF = 0: B 900 j + 900 j + 300k 900k = 0B = (600 N)k YFree body: BSince B is aligned with member BE: FAB = FBC = 0, FBE = 600 N T WFree body: C F = 0: FCA + FCD + FCE (900 N) j = 0,withPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 71. PROBLEM 6.40* (Continued)FCA FCDJJJ G F CA = FAC = AC ( 1.2i + 3j 0.75k ) CA 3.317 JJJ G F CE = FCD k FCE = FCE = CE (1.8i 3k ) CE 3.499Substitute and equate to zero the coefficient of j, i, k: j: 3 FAC 900 N = 0, FAC = 995.1 N 3.317 FAC = 995 N T Wi:1.2 1.8 (995.1) FCE = 0, FCE = 699.8 N 3.317 3.499FCE = 700 N C Wk:0.75 3 (995.1) FCD ( 699.8) = 0 3.317 3.499FCD = 375 N T WFree body: D F = 0: FDA + FDE + (375 N)k +(900 N)j + (300 N) k = 0 JJJ G F DA FDA = FAD = AD (1.2i + 3j + 2.25k ) DA 3.937withFDE = FDE iandSubstitute and equate to zero the coefficient j, i, k: j: 3 FAD + 900 N = 0, FAD = 1181.1 N 3.937 FAD = 1181N C Wi: 1.2 (1181.1 N) FDE = 0 3.937 FDE = 360 N T Wk: 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 WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 72. 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: M 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 = 0Equate to zero the coefficient of i, j, k: i : 9.6G + (10.08)(240) = 0 G = 252 lbG = (252 lb) j Yj: 11Bz (9.6)(275) = 0 Bz = 240 lb k : 11By + 11(252) (10.08)(275) = 0, By = 504 lbB = (504 lb) j (240 lb)k YPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 73. PROBLEM 6.41* (Continued)F = 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 YZero-force members The determination of these members will facilitate our solution. FB: C. WritingFx = 0, Fy = 0, Fz = 0Yields FBC = FCD = FCG = 0 YFB: F. WritingFx = 0, Fy = 0, Fz = 0Yields FBF = FEF = FFG = 0 YFB: A: Since FB: H: WritingAz = 0, writing Fz = 0 Fy = 0Yields FAD = 0 Y Yields FDH = 0 YFB: 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 YThe 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 WWe already found that Free body: AFy = 0Yields FAE = 252 lb T WFree body: HFz = 0Yields FEH = 240 lb C WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 74. PROBLEM 6.41* (Continued)F = 0: FEB + FEG + (240 lb)k (252 lb) j = 0Free body: EF FBE (11i 10.08 j) + EG (11i 9.6k ) + 240k 252 j = 0 14.92 14.6Equate to zero the coefficient of y and k: 10.08 j: FBE 252 = 0 14.92 FBE = 373 lb C W 9.6 FEG + 240 = 0 14.6 FEG = 365 lb T Wk:PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 75. 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 FCG = FDG = FFG = 0 WFree body: HFx = 0Yields FGH = 275 lb C WFree body: GF = 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 14.6Equate to zero the coefficient of i, j, k: 11 i: FEG + 275 = 0 14.6 FEG = 365 lb T W 10.08 j: FBG 252 = 0 13.92 FBG = 348 lb C W 9.6 9.6 k: (348) + (365) = 0 13.92 14.6 (Checks)PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 76. PROBLEM 6.43 A Warren bridge truss is loaded as shown. Determine the force in members CE, DE, and DF.SOLUTION Free body: Truss Fx = 0: k x = 0 M A = 0: k y (62.5 ft) (6000 lb)(12.5 ft) (6000 lb)(25 ft) = 0 k = k y = 3600 lb YFy = 0: A + 3600 lb 6000 lb 6000 lb = 0 A = 8400 lb YWe pass a section through members CE, DE, and DF and use the free body shown. M D = 0: FCE (15 ft) (8400 lb)(18.75 ft) + (6000 lb)(6.25 ft) = 0 FCE = +8000 lb Fy = 0: 8400 lb 6000 lb FCE = 8000 lb T W 15 FDE = 0 16.25FDE = +2600 lbFDE = 2600 lb T WM E = 0: 6000 lb (12.5 ft) (8400 lb)(25 ft) FDF (15 ft) = 0 FDF = 9000 lbFDF = 9000 lb C WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 77. PROBLEM 6.44 A Warren bridge truss is loaded as shown. Determine the force in members EG, FG, and FH.SOLUTION See solution of Problem 6.43 for free-body diagram of truss and determination of reactions: A = 8400 lbk = 3600 lb YWe pass a section through members EG, FG, and FH, and use the free body shown. M F = 0: (3600 lb)(31.25 ft) FEG (15 ft) = 0 FEG = +7500 lb +Fy = 0:FEG = 7500 lb T W15 FFG + 3600 lb = 0 16.25 FFG = 3900 lbFFG = 3900 lb C WM G = 0: FFH (15 ft) + (3600 lb)(25 ft) = 0 FFH = 6000 lbFFH = 6000 lb C WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 78. PROBLEM 6.45 Determine the force in members BD and DE of the truss shown.SOLUTION Member BD: M E = 0: FBD (4.5 m) (135 kN)(4.8 m) (135 kN)(2.4 m) = 0 FBD = +216 kNFBD = 216 kN T WMember DE: Fx = 0: 135 kN + 135 kN FDE = 0 FDE = +270 kNFDE = 270 kN T WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 79. PROBLEM 6.46 Determine the force in members DG and EG of the truss shown.SOLUTION Member DG: Fx = 0: 135 kN + 135 kN + 135 kN + FDG = 459 kN15 FDG = 0 17 FDG = 459 kN C WMember EG: M D = 0: (135 kN)(4.8 m) + (135 kN)(2.4 m) + FEG (4.5 m) = 0 FEG = 216 kNFEG = 216 kN C WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 80. PROBLEM 6.47 A floor truss is loaded as shown. Determine the force in members CF, EF, and EG.SOLUTION Free body: Truss Fx = 0: k x = 0 M A = 0: k y (4.8 m) (4 kN)(0.8 m) (4 kN)(1.6 m) (3 kN)(2.4 m) (2 kN)(3.2 m) (2 kN)(4 m) (1 kN)(4.8 m) = 0 k y = 7.5 kN k = 7.5 kN YThus: Fy = 0: A + 7.5 kN 18 kN = 0A = 10.5 kNA = 10.5 kN YWe pass a section through members CF, EF, and EG and use the free body shown. M E = 0: FCF (0.4 m) (10.5 kN)(1.6 m) + (2 kN)(1.6 m) + (4 kN)(0.8 m) = 0 FCF = +26.0 kN Fy = 0: 10.5 kN 2 kN 4 kN 4 kN FCF = 26.0 kN T W 1 5FEF = +1.118 kNFEF = 0 FEF = 1.118 kN T WM F = 0: (2 kN)(2.4 m) + (4 kN)(1.6 m) + (4 kN)(0.8 m) (10.5 kN)(2.4 m) FEG (0.4 m) = 0 FEG = 27.0 kNFEG = 27.0 kN C WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 81. PROBLEM 6.48 A floor truss is loaded as shown. Determine the force in members FI, HI, and HJ.SOLUTION See solution of Problem 6.47 for free-body diagram of truss and determination of reactions: A = 10.5 kN , k = 7.5 kN YWe pass a section through members FI, HI, and HJ, and use the free body shown. M H = 0: (7.5 kN)(1.6 m) (2 kN)(0.8 m) (1 kN)(1.6 m) FFI (0.4 m) = 0 FFI = +22.0 kN Fy = 0:1 5FFI = 22.0 kN T WFHI 2 kN 1 kN + 7.5 kN = 0 FHI = 10.06 kNFHI = 10.06 kN C WM I = 0: FHJ (0.4 m) + (7.5 kN)(0.8 m) (1 kN)(0.8 m) = 0 FHJ = 13.00 kNFHJ = 13.00 kN C WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 82. PROBLEM 6.49 A pitched flat roof truss is loaded as shown. Determine the force in members CE, DE, and DF.SOLUTION Reactions at supports: Because of the symmetry of the loading, Ax = 0 Ay = I =1 1 (Total load) = (8 kN) 2 2A = I = 4 kN YWe pass a section through members CD, DE, and DF, and use the free body shown. (We moved FDE to E and FDF to F) SlopeBJ =Slope2.16 m 9 = 9.6 m 40DE =1 m 5 = 2.4 m 12 0.46 m 0.46 m a= = = 2.0444 m 9 Slope BJ 40M D = 0: FCE (1 m) + (1 kN)(2.4 m) (4 kN)(2.4 m) = 0 FCE = 7.20 kN T WFCE = +7.20 kN M K = 0: (4 kN)(2.0444 m) (1 kN)(2.0444 m) 5 (2 kN)(4.4444 m) FDE (6.8444 m) = 0 13 FDE = 1.047 kNFDE = 1.047 kN C WM E = 0: (1 kN)(4.8 m) + (2kN)(2.4 m) (4 kN)(4.8 m) 40 FDF (1.54 m) = 0 41 FDF = 6.39 kNFDF = 6.39 kN C WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 83. PROBLEM 6.50 A pitched flat roof truss is loaded as shown. Determine the force in members EG, GH, and HJ.SOLUTION Reactions at supports: Because of the symmetry of the loading, Ax = 0 Ay = I =1 1 (Total load) = (8 kN) 2 2A = I = 4 kN YWe pass a section through members EG, GH, and HJ, and use the free body shown. M H = 0: (4 kN)(2.4 m) (1 kN)(2.4 m) FEG (2.08 m) = 0 FEG = +3.4615 kNFEG = 3.46 kN T WM J = 0: FGH (2.4 m) FEG (2.62 m) = 0 FGH = 2.62 (3.4615 kN) 2.4FGH = 3.7788 kN Fx = 0: FEG FGH = 3.78 kN C W2.4 FHJ = 0 2.46FHJ = 2.46 2.46 (3.4615 kN) FEG = 2.4 2.4FHJ = 3.548 kNFHJ = 3.55 kN C WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 84. PROBLEM 6.51 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 2A = L = 4.80 kips WWe pass a section through members DF, DG, and EG, and use the free body shown. We slide FDF to apply it at F: M G = 0: (0.8 kip)(24 ft) + (1.6 kips)(16 ft) + (1.6 kips)(8 ft) 8FDF (4.8 kips)(24 ft) 82 + 3.52(6 ft) = 0FDF = 10.48 kips, FDF = 10.48 kips C W M A = 0: (1.6 kips)(8 ft) (1.6 kips)(16 ft) 2.5 FDG 28 + 2.52(16 ft) 8 FDG 82 + 2.52(7 ft) = 0FDG = 3.35 kips, FDG = 3.35 kips C W M D = 0: (0.8 kips)(16 ft) + (1.6 kips)(8 ft) (4.8 kips)(16 ft) 8 FEG 82 + 1.52(4 ft) = 0FEG = +13.02 kips, FEG = 13.02 kips T WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 85. PROBLEM 6.52 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) 2Ay = L =Ax = 0,= 4.80 kipsA = L = 4.80 kips WWe pass a section through members GI, HI, and HJ, and use the free body shown.M 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 kipsFGI = 13.02 kips T WM L = 0: (1.6 kips)(8 ft) FHI (16 ft) = 0 FHI = +0.800 kips FHI = 0.800 kips T WWe slide FHG to apply it at H. M 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 kipsFHJ = 13.97 kips C WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 86. PROBLEM 6.53 A Pratt roof truss is loaded as shown. Determine the force in members CE, DE, and DF.SOLUTION Free body: Entire truss Fx = 0: Ax = 0Total load = 5(3 kN) + 2(1.5 kN) = 18 kN By symmetry: Ay = L =1 (18 kN) 2A = L = 9 kN WFree body: Portion ACD Note:Slope of ABDF is 6.75 3 = 9.00 4Force in CE: 2 M D = 0: FCE 6.75 m (9 kN)(6 m) + (1.5 kN)(6 m) + (3 kN)(3 m) = 0 3 FCE (4.5 m) 36 kN m = 0 FCE = +8 kNFCE = 8 kN T WForce in DE: M A = 0: FDE (6 m) + (3 kN)(6 m) + (3 kN)(3 m) = 0 FDE = 4.5 kNFDE = 4.5 kN C WPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 87. PROBLEM 6.53 (Continued)Force in DF: Sum moments about E where FCE and FDE intersect. M E = 0: (1.5 kN)(6 m) (9 kN)(6 m) + (3 kN)(3 m) +4 2 FCE 6.75 m = 0 5 3 4 FCE (4.5 m) 36 kN m 5 FCE = 10.00 kN C WFCE = 10.00 kNPROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 88. PROBLEM 6.54 A Pratt roof truss is loaded as shown. Determine the force in members FH, FI, and GI.SOLUTION Free body: Entire truss Fx = 0: Ax = 0Total load = 5(3 kN) + 2(1.5 kN) = 18 kN By symmetry: Ay = L =1 (18) = 9 kN 2Free body: Portion HIL Slope