Solution manual engineering mechanics statics 13th edition by r c hibbeler13e chap 06

6–1 Determine the force in each member of the truss and state if the members are in tension or compression Set P1 = 800 lb and P2 = 400 lb ft ft C A ft SOLUTION Method of Joints: In this case, the support reactions are not required for determining the member forces P2 Joint B: B P1 + ©F = 0; : x FBC cos 45° - FBA a b - 400 = (1) + c ©Fy = 0; FBC sin 45° + FBA a b - 800 = (2) T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) Solving Eqs (1) and (2) yields FBA = 285.71 lb (T) = 286 lb (T) Ans FBC = 808.12 lb (T) = 808 lb (T) Ans Joint C: + ©F = 0; : x FCA - 808.12 cos 45° = FCA = 571 lb (C) + c ©Fy = 0; Ans Cy - 808.12 sin 45° = Cy = 571 lb Note: The support reactions Ax and Ay can be determined by analyzing Joint A using the results obtained above © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 6–2 Determine the force on each member of the truss and state if the members are in tension or compression Set P1 = 500 lb and P2 = 100 lb ft ft C A ft SOLUTION Method of Joints: In this case, the support reactions are not required for determining the member forces P2 Joint B: B P1 + ©F = 0; : x FBC cos 45° - FBA a b - 100 = (1) + c ©Fy = 0; FBC sin 45° + FBA a b - 500 = (2) T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) Solving Eqs (1) and (2) yields FBA = 285.71 lb (T) = 286 lb (T) Ans FBC = 383.86 lb (T) = 384 lb (T) Ans Joint C: + ©F = 0; : x FCA - 383.86 cos 45° = FCA = 271 lb (C) + c ©Fy = 0; Ans Cy - 383.86 sin 45° = Cy = 271.43 lb Note: The support reactions Ax and Ay can be determined by analyzing Joint A using the results obtained above © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 6–3 Determine the force in each member of the truss, and state if the members are in tension or compression Set u = 0° D kN 1.5 m A SOLUTION B Support Reactions: Applying the equations of equilibrium to the free-body diagram of the entire truss,Fig.a, we have a + ©MA = 0; C 2m 2m kN NC (2 + 2) - 4(2) - 3(1.5) = NC = 3.125 kN + ©F = 0; : x - Ax = A x = kN + c ©Fy = 0; A y + 3.125 - = A y = 0.875 kN T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) Method of Joints: We will use the above result to analyze the equilibrium of joints C and A, and then proceed to analyze of joint B Joint C: From the free-body diagram in Fig b, we can write + c ©Fy = 0; 3.125 - FCD a b = FCD = 5.208 kN = 5.21 kN (C) + ©F = 0; : x Ans 5.208 a b - FCB = FCB = 4.167 kN = 4.17 kN (T) Ans Joint A: From the free-body diagram in Fig c, we can write + c ©Fy = 0; 0.875 - FAD a b = FAD = 1.458 kN = 1.46 kN (C) + ©F = 0; : x Ans FAB - - 1.458a b = FAB = 4.167 kN = 4.17 kN (T) Ans Joint B: From the free-body diagram in Fig d, we can write + c ©Fy = 0; FBD - = FBD = kN (T) + ©F = 0; : x 4.167 - 4.167 = Ans (check!) Note: The equilibrium analysis of joint D can be used to check the accuracy of the solution obtained above © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 u *6–4 Determine the force in each member of the truss, and state if the members are in tension or compression Set u = 30° D kN 1.5 m A SOLUTION B Support Reactions: From the free-body diagram of the truss, Fig a, and applying the equations of equilibrium, we have a + ©MA = 0; C 2m 2m kN NC cos 30°(2 + 2) - 3(1.5) - 4(2) = NC = 3.608 kN + ©F = 0; : x - 3.608 sin 30° - A x = A x = 1.196 kN + c ©Fy = 0; A y + 3.608 cos 30° - = A y = 0.875 kN T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) Method of Joints: We will use the above result to analyze the equilibrium of joints C and A, and then proceed to analyze of joint B Joint C: From the free-body diagram in Fig b, we can write + c ©Fy = 0; 3.608 cos 30° - FCD a b = FCD = 5.208 kN = 5.21 kN (C) + ©F = 0; : x Ans 5.208 a b - 3.608 sin 30° - FCB = FCB = 2.362 kN = 2.36 kN (T) Ans Joint A: From the free-body diagram in Fig c, we can write + c ©Fy = 0; 0.875 - FAD a b = FAD = 1.458 kN = 1.46 kN (C) + ©F = 0; : x Ans FAB - 1.458 a b - 1.196 = FAB = 2.362 kN = 2.36 kN (T) Ans Joint B: From the free-body diagram in Fig d, we can write + c ©Fy = 0; FBD - = FBD = kN (T) + ©F = 0; : x 2.362 - 2.362 = Ans (check!) Note: The equilibrium analysis of joint D can be used to check the accuracy of the solution obtained above © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 u 6–5 Determine the force in each member of the truss, and state if the members are in tension or compression 300 N 400 N D SOLUTION C 2m Method of Joints: Here, the support reactions A and C not need to be determined We will first analyze the equilibrium of joints D and B, and then proceed to analyze joint C 250 N A B 2m Joint D: From the free-body diagram in Fig a, we can write + ©F = 0; : x 400 - FDC = FDC = 400 N (C) + c ©Fy = 0; 200 N Ans FDA - 300 = FDA = 300 N (C) Ans + ©F = 0; : x T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) Joint B: From the free-body diagram in Fig b, we can write 250 - FBA = FBA = 250 N (T) + c ©Fy = 0; Ans FBC - 200 = FBC = 200 N (T) Ans Joint C: From the free-body diagram in Fig c, we can write + c ©Fy = 0; FCA sin 45° - 200 = FCA = 282.84 N = 283 N (C) + ©F = 0; : x Ans 400 + 282.84 cos 45° - NC = NC = 600 N Note: The equilibrium analysis of joint A can be used to determine the components of support reaction at A © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 6–6 Determine the force in each member of the truss, and state if the members are in tension or compression 600 N D 4m SOLUTION C Method of Joints: We will begin by analyzing the equilibrium of joint D, and then proceed to analyze joints C and E Joint D: From the free-body diagram in Fig a, + ©F = 0; : x 4m B A FDE a b - 600 = FDE = 1000 N = 1.00 kN (C) + c ©Fy = 0; 900 N E 6m Ans 1000 a b - FDC = Ans T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) FDC = 800 N (T) Joint C: From the free-body diagram in Fig b, + ©F = 0; : x FCE - 900 = FCE = 900 N (C) + c ©Fy = 0; Ans 800 - FCB = FCB = 800 N (T) Ans Joint E: From the free-body diagram in Fig c, R+ ©Fx ¿ = 0; - 900 cos 36.87° + FEB sin 73.74° = FEB = 750 N (T) Q+ ©Fy ¿ = 0; Ans FEA - 1000 - 900 sin 36.87° - 750 cos 73.74° = FEA = 1750 N = 1.75 kN (C) Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 6–7 Determine the force in each member of the Pratt truss, and state if the members are in tension or compression J 2m K 2m SOLUTION 2m Joint A: + c ©Fy = 0; L H A B D C E F 2m 2m 2m 2m 2m 2m 20 - FAL sin 45° = 10 kN FAL = 28.28 kN (C) + ©F = 0; : x I 20 kN 10 kN FAB - 28.28 cos 45° = FAB = 20 kN (T) Joint B: + ©F = 0; : x FBC - 20 = + c ©Fy = 0; T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) FBC = 20 kN (T) FBL = Joint L: R+ ©Fx = 0; FLC = +Q©Fy = 0; 28.28 - FLK = FLK = 28.28 kN (C) Joint C: + ©F = 0; : x FCD - 20 = FCD = 20 kN (T) + c ©Fy = 0; FCK - 10 = FCK = 10 kN (T) Joint K: R+ ©Fx - 0; 10 sin 45° - FKD cos (45° - 26.57°) = FKD = 7.454 kN (L) +Q©Fy = 0; 28.28 - 10 cos 45° + 7.454 sin (45° - 26.57°) - FKJ = FKJ = 23.57 kN (C) Joint J: + ©F = 0; : x 23.57 sin 45° - FJI sin 45° = FJI = 23.57 kN (L) + c ©Fy = 0; (23.57 cos 45°) - FJD = FJD = 33.3 kN (T) Ans FAL = FGH = FLK = FHI = 28.3 kN (C) Ans FAB = FGF = FBC = FFE = FCD = FED = 20 kN (T) Ans FBL = FFH = FLC = FHE = Ans FCK = FEI = 10 kN (T) Ans FKJ = FIJ = 23.6 kN (C) Ans Due to Symmetry © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by = FID = 7.45 kN (C) FKD Copyright and written permission should be obtained from the publisher prior to anyAns prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 G *6–8 Determine the force in each member of the truss and state if the members are in tension or compression Hint: The horizontal force component at A must be zero Why? 800 lb 600 lb B 60 C ft SOLUTION Joint C: + ©F = 0; : x FCB - 800 cos 60° = FCB = 400 lb (C) + c ©Fy = 0; Ans F - 400 = BD FBD = 666.7 = 667 lb (T) + c ©Fy = 0; ft T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) + ©F = 0; : x D FCD - 800 sin 60° = FCD = 693 lb (C) Joint B: A Ans FBA - Ans (666.7) - 600 = FBA = 1133 lb = 1.13 kip (C) Ans Member AB is a two-force member and exerts only a vertical force along AB at A © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 6–9 Determine the force in each member of the truss and state if the members are in tension or compression Hint: The vertical component of force at C must equal zero Why? B C 2m D SOLUTION A Joint A: + c ©Fy = 0; 1.5 m kN Ans - FAE + 7.5 a b = Ans + ©F = 0; : x FED = 4.5 kN(C) Ans + c ©Fy = 0; FEB = kN (T) Ans Joint E: Joint B: T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) FAE = 4.5 kN (C) + c ©Fy = 0; 22 (FBD) - - (7.5) = FBD = 19.8 kN (C) + ©F = 0; : x 2m kN FAB - = FAB = 7.5 kN (T) + ©F = 0; : x E FBC - Ans (7.5) (19.8) = 22 FBC = 18.5 kN (T) Ans Cy is zero because BC is a two-force member © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 6–10 Each member of the truss is uniform and has a mass of kg>m Remove the external loads of kN and kN and determine the approximate force in each member due to the weight of the truss State if the members are in tension or compression Solve the problem by assuming the weight of each member can be represented as a vertical force, half of which is applied at each end of the member B 2m D SOLUTION A 1.5 m Joint A: + c ©Fy = 0; Ans - FAE + 196.2a b = + ©F = 0; : x FED = 117.7 = 118 N (C) Ans + c ©Fy = 0; FEB = 215.8 = 216 N (T) Ans Joint E: Joint B: T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) Ans 22 (FBD) - 366.0 - 215.8 - (196.2) = FBD = 1045 = 1.04 kN (C) + ©F = 0; : x 2m kN FAE = 117.7 = 118 N (C) + c ©Fy = 0; E kN - 157.0 = F AB FAB = 196.2 = 196 N (T) + ©F = 0; : y C FBC - Ans (196.2) (1045) = 22 FBC = 857 N (T) Ans © 2013 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 ... prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s),... publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information... publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise For information