# 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) + (4