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Engineering Mechanics - Statics Chapter 6 ΣF x = 0; A x E x − 0= E x 2P 3 = ΣF y = 0; E y P− 0= E y P= Joint E: ΣF x = 0; F EC 2 13 E x − 0= F EC 13 3 P= 1.20P= (T) ΣF y = 0; PF ED − F EC 3 13 − 0= F ED 0= Joint A: ΣF y = 0; F AB 1 5 F AD 1 5 − 0= F AB F AD = ΣF x = 0; A x 2F AB 2 5 − 0= F AB F AD = 5 6 P= 0.373P= (C) Joint D: ΣF x = 0; F AD 2 5 F DC 2 5 − 0= F DC 5 6 P= 0.373P= (C) ΣF y = 0; 2F AD 1 5 F DB − 0= F DB P 3 = (T) Joint B: ΣF x = 0; F AB 1 5 F BC 1 5 − 0= F BC 5 6 P= 0.373P= (C) 463 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 6 The maximum allowable tensile force in the members of the truss is T max and the maximum allowable compressive force is C max . Determine the maximum magnitude of the load P that can be applied to the truss. Units Used: kN 10 3 N= Given: T max 5kN= C max 3kN= d 2m= Solution: Set P 1kN= Initial Guesses: F AD 1kN= F AB 1kN= F BC 1kN= F BD 1kN= F CD 1kN= F CE 1kN= F DE 1kN= Given Joint A F AD 1 5 F AB 1 5 − 0= Joint B F BC 2 5 F AB 2 5 − 0= F BC F AB + () 1 5 F BD + 0= Joint D F CD F AD − () 2 5 0= F DE F BD − F AD F CD + () 1 5 − 0= Joint C F CD F BC + () − 2 5 F CE 2 13 − 0= F CD F BC − () 1 5 F CE 3 13 + P− 0= 464 Problem 6-17 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 6 F AD F AB F BC F BD F CD F CE F DE ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find F AD F AB , F BC , F BD , F CD , F CE , F DE , () = F AD F AB F BC F BD F CD F CE F DE ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 0.373− 0.373− 0.373− 0.333 0.373− 1.202 0 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ kN= Now Scale the answer P 1 P T max max F AD F AB , F BC , F BD , F CD , F CE , F DE , () = P 2 P C max min F AD F AB , F BC , F BD , F CD , F CE , F DE , = P min P 1 P 2 , () = P 4.16 kN= Problem 6-18 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? 465 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 6 kip 10 3 lb= Given: F 1 600 lb= F 2 800 lb= a 4ft= b 3ft= θ 60 deg= Solution: Initial Guesses F BA 1lb= F BD 1lb= F CB 1lb= F CD 1lb= Given Joint C F CB − F 2 cos θ()− 0= F CD − F 2 sin θ()− 0= Joint B F CB F BD b a 2 b 2 + + 0= F BA − F BD a a 2 b 2 + − F 1 − 0= Positive means Tension Negative means Compression F BA F BD F CB F CD ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find F BA F BD , F CB , F CD , () = F BA F BD F CB F CD ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 1.133− 10 3 × 666.667 400− 692.82− ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ lb= Problem 6-19 Determine the force in each member of the truss and state if the members are in tension or compression. Hint: The resultant force at the pin E acts along member ED. Why? Units Used: kN 10 3 N= 466 Problem 6-19 Units Used: © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 6 Given: F 1 3kN= F 2 2kN= a 3m= b 4m= Solution: Initial Guesses: F CB 1kN= F CD 1kN= F BA 1kN= F BD 1kN= F DA 1kN= F DE 1kN= Given Joint C F CB − F CD 2 a 2 a() 2 b 2 + − 0= F 2 − F CD b 2 a() 2 b 2 + − 0= Joint B F BA − F CB + 0= F 1 − F BD − 0= Joint D F CD F DA − F DE − () 2 a 2 a() 2 b 2 + 0= F BD F CD F DA + F DE − () b 2 a() 2 b 2 + + 0= 467 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 6 F CB F CD F BA F BD F DA F DE ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find F CB F CD , F BA , F BD , F DA , F DE , () = F CB F CD F BA F BD F DA F DE ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 3 3.606− 3 3− 2.704 6.31− ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ kN= Positive means Tension, Negative means Compression Problem 6-20 Each member of the truss is uniform and has a mass density ρ. 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. Given: ρ 8 kg m = g 9.81 m s 2 = F 1 0N= F 2 0N= a 3m= b 4m= 468 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 6 Solution: Initial Guesses: F CB 1N= F CD 1N= F BA 1N= F BD 1N= F DA 1N= F DE 1N= Given Joint C F CB − F CD 2 a 2 a() 2 b 2 + − 0= F 2 − F CD b 2 a() 2 b 2 + − ρg a 2 a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 b 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ++ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ − 0= Joint B F BA − F CB + 0= F 1 − F BD − ρga b 4 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 0= Joint D F BD F CD F DA + F DE − () b 2 a() 2 b 2 + + ρg b 4 3 a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 b 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ++ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ − 0= F CD F DA − F DE − () 2 a 2 a() 2 b 2 + 0= F CB F CD F BA F BD F DA F DE ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find F CB F CD , F BA , F BD , F DA , F DE , () = 469 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 6 F CB F CD F BA F BD F DA F DE ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 389 467− 389 314− 736 1204− ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ N= Positive means Tension, Negative means Compression Problem 6-21 Determine the force in each member of the truss in terms of the external loading and state if the members are in tension or compression. Solution: Joint B: + ↑ Σ F y = 0; F BA sin 2 θ()P− 0= F BA P csc 2 θ()= C() + → Σ F x = 0; F BA cos 2 θ()F BC − 0= F BC Pcot 2 θ()= C() Joint C: + → Σ F x = 0; P cot 2 θ()P+ F CD cos 2 θ()+ F CA cos θ()− 0= + ↑ Σ F y = 0; F CD sin 2 θ()F CA sin θ()− 0= 470 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 6 F CA cot 2 θ( ) 1+ cos θ() sin θ()cot 2 θ()− P= F CA cot θ()csc θ() sin θ()− 2 cos θ()+ P= T() F CD cot 2 θ()1+ P= C() Joint D: + → Σ F x = 0; F DA cot 2 θ()1+ ⎡⎣ ⎤⎦ cos 2 θ() ⎡⎣ ⎤⎦ P− 0= F DA cot 2 θ()1+ ⎡⎣ ⎤⎦ cos 2 θ() ⎡⎣ ⎤⎦ P= C() Problem 6-22 The maximum allowable tensile force in the members of the truss is T max , and the maximum allowable compressive force is C max . Determine the maximum magnitude P of the two loads that can be applied to the truss. Units Used: kN 10 3 N= Given: T max 2kN= C max 1.2 kN= L 2m= θ 30 deg= Solution: Initial guesses (assume all bars are in tension). Use a unit load for P and then scale the answer later. F BA 1kN= F BC 1kN= F CA 1kN= F CD 1kN= F DA 1kN= P 1kN= 471 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 6 Joint B + ↑ Σ F y = 0; F BA − sin 2 θ()P− 0= + → Σ F x = 0; F BA − cos 2 θ()F BC + 0= Joint C + ↑ Σ F y = 0; F CA − sin θ() F CD sin 2 θ()− 0= + → Σ F x = 0; F BC − P+ F CD cos 2 θ()− F CA cos θ()− 0= Joint D + → Σ F x = 0; F DA − F CD cos 2 θ()+ 0= F BA F BC F CA F CD F DA ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find F BA F BC , F CA , F CD , F DA , () = ans F BA F BC F CA F CD F DA ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = ans 1.155− 0.577− 2.732 1.577− 0.789− ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ kN= Now find the biggest tension and the biggest compression. T max ans()= T 2.732 kN= C min ans()= C 1.577− kN= Decide which is more important and scale the answer P min T max T C max − C ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ P ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ = P 732.051 N= Problem 6-23 472 Given © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 6 The Fink truss supports the loads shown. Determine the force in each member and state if the members are in tension or compression. Approximate each joint as a pin. Units Used: kip 10 3 lb= Given: F 1 500 lb= a 2.5 ft= F 2 1 kip= θ 30 deg= F 3 1 kip= Solution: Entire truss: ΣF x = 0; E x F 1 F 2 + F 3 + F 2 + F 1 + () sin θ()= E x 2000 lb= ΣM E = 0; A y − 4a cos θ() F 1 4a+ F 2 3a+ F 3 2a+ F 2 a+ 0= A y 2 F 1 2 F 2 + F 3 + 2 cos θ() = A y 2309.4 lb= ΣF y = 0; E y A y − 2 cos θ()F 1 + 2 cos θ()F 2 + cos θ()F 3 += E y 1154.7 lb= Joint A: ΣF y = 0; F AB cos θ()− F 1 A y + sin θ() = F AB 3.75 kip= (C) ΣF x = 0; F AH sin θ()− F 1 F AB cos θ()+= F AH 3 kip= (T) 473 Problem 6-24 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 6 Joint B: ΣF x = 0; F BC F AB = F BC 3.75 kip= (C) ΣF y = 0; F BH F 2 = F BH 1 kip= (C) Joint H: ΣF y = 0; F HC F 2 = F HC 1 kip= (T) ΣF x = 0; F GH F 2 − cos 90− deg θ+()F HC cos 90− deg θ+()− F AH += F GH 2 kip= (T) Joint E: ΣF y = 0; F EF F 1 E x sin θ()− E y cos θ()− () − sin θ() = F EF 3 kip= (T) ΣF x = 0; F ED E y − sin θ() E x cos θ()+ F EF cos θ()+= F ED 3.75 kip= (C) Joint D: ΣF x = 0; F DC F ED = F DC 3.75 kip= (C) ΣF y = 0; F DF F 2 = F DF 1 kip= (C) Joint C: ΣF x = 0; F CF F HC = F CF 1 kip= (T) ΣF y = 0; F CG F 3 F HC cos 90 deg θ−()2()+= F CG 2 kip= (C) Joint F: ΣF x = 0; F FG F EF F CF cos 90 deg θ−()2()−= F FG 2 kip= (T) 474 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 6 Problem 6-24 Determine the force in each member of the double scissors truss in terms of the load P and state if the members are in tension or compression. Solution: ΣΜ A = 0; P L 3 P 2 L 3 + D y L− 0= + ↑ ΣF y = 0; A y D y + 2 P− 0= Joint F: + ↑ ΣF y = 0; F FB 1 2 P− 0= + → ΣF x = 0; F FD F FE − F FB 1 2 − 0= 475 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 6 Joint E: + ↑ ΣF y = 0; F EC 1 2 P− 0= + → ΣF x = 0; F EF F EA − F EC 1 2 + 0= Joint B: + ↑ ΣF y = 0; F BA 1 2 F BD 1 5 + F FB 1 2 − 0= + → ΣF x = 0; F BA 1 2 F FB 1 2 + F BD 2 5 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 0= Joint C: + ↑ ΣF y = 0; F CA 1 5 F CD 1 2 + F EC 1 2 − 0= + → ΣF x = 0; F CA 2 5 F EC 1 2 − F CD 1 2 − 0= 476 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 6 Joint A: + → ΣF x = 0; F AE F BA 1 2 − F CA 2 5 − 0= Solving we find F EF 0.667 P T()= F FD 1.67 P T()= F AB 0.471 P C()= F AE 1.67 P T()= F AC 1.49 P C()= F BF 1.41 P T()= F BD 1.49 P C()= F EC 1.41 P T()= F CD 0.471 P C()= Problem 6-25 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? Units Used: kN 10 3 N= Given: F 1 6kN= 477 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 6 F 2 8kN= a 1.5 m= b 2m= c 2m= Solution: Initial Guesses: F AB 1kN= F AE 1kN= F EB 1kN= F BC 1kN= F BD 1kN= F ED 1kN= Given Joint A F AB a a 2 c 2 + F AE + 0= F AB c a 2 c 2 + F 1 − 0= Joint E F ED F AE − 0= F EB F 2 − 0= Joint B F BC F BD b b 2 c 2 + + F AB a a 2 c 2 + − 0= F EB − F BD c b 2 c 2 + − F AB c a 2 c 2 + − 0= F AB F AE F EB F BC F BD F ED ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find F AB F AE , F EB , F BC , F BD , F ED , () = 478 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 6 F AB F AE F EB F BC F BD F ED ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 7.5 4.5− 8 18.5 19.799− 4.5− ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ kN= Positive means Tension, Negative means Compresson. Problem 6-26 Each member of the truss is uniform and has a mass density ρ. Remove the external loads F 1 and F 2 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. Given: F 1 0= F 2 0= ρ 8 kg m = a 1.5 m= b 2m= c 2m= g 9.81 m s 2 = Solution: Find the weights of each bar. W AB ρga 2 c 2 += W BC ρgb= W BE ρgc= W AE ρga= W BD ρgb 2 c 2 += W DE ρgb= 479 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 6 Guesses F AB 1N= F AE 1N= F BE 1N= F BC 1N= F BD 1N= F DE 1N= Given Joint A F AE a a 2 c 2 + F AB + 0= c a 2 c 2 + F AB W AB W AE + 2 − 0= Joint E F DE F AE − 0= F BE W AE W BE + W DE + 2 − 0= Joint B F BC b b 2 c 2 + F BD + a a 2 c 2 + F AB − 0= c− a 2 c 2 + F AB F BE − c b 2 c 2 + F BD − W AB W BE + W BD + W BC + 2 − 0= F AB F AE F BC F BD F BE F DE ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find F AB F AE , F BC , F BD , F BE , F DE , () = Positive means tension, Negative means Compression. F AB F AE F BC F BD F BE F DE ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 196 118− 857 1045− 216 118− ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ N= 480 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 6 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: kN 10 3 N= Given: P 1 4kN= P 2 0kN= a 2m= θ 15 deg= Solution: Take advantage of the symetry. Initial Guesses: F BD 1kN= F CD 1kN= F AB 1kN= F CA 1kN= F BC 1kN= Given Joint D P 1 − 2 F BD sin 2 θ()− F CD sin 3 θ()− 0= Joint B P 2 − cos 2 θ()F BC − 0= F BD F AB − P 2 sin 2 θ()− 0= Joint C F CD cos θ() F CA cos θ()− 0= F CD F CA + () sin θ() F BC + 0= F BD F CD F AB F CA F BC ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find F BD F CD , F AB , F CA , F BC , () = F FD F ED F GF F EG F FE ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ F BD F CD F AB F CA F BC ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = 481 Problem 6-27 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 6 Positvive means Tension, Negative means Compression F BD F CD F AB F CA F BC ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 4− 0 4− 0 0 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ kN= F FD F ED F GF F EG F FE ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 4− 0 4− 0 0 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ kN= Problem 6-28 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: kN 10 3 N= Given: P 1 2kN= P 2 4kN= a 2m= θ 15 deg= Solution: Take advantage of the symmetry. Initial Guesses: F BD 1kN= F CD 1kN= F AB 1kN= F CA 1kN= F BC 1kN= Given Joint D P 1 − 2 F BD sin 2 θ()− F CD sin 3 θ()− 0= Joint B P 2 − cos 2 θ()F BC − 0= F BD F AB − P 2 sin 2θ()− 0= Joint C F CD cos θ() F CA cos θ()− 0= F CD F CA + () sin θ() F BC + 0= 482 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 6 F BD F CD F AB F CA F BC ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find F BD F CD , F AB , F CA , F BC , () = F FD F ED F GF F EG F FE ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ F BD F CD F AB F CA F BC ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = Positvive means Tension, Negative means Compression F BD F CD F AB F CA F BC ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 11.46− 6.69 13.46− 6.69 3.46− ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ kN= F FD F ED F GF F EG F FE ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 11.46− 6.69 13.46− 6.69 3.46− ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ kN= Problem 6-29 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: kip 10 3 lb= Given: F 1 2 kip= F 2 1.5 kip= F 3 3 kip= F 4 3 kip= a 4ft= b 10 ft= Solution: θ atan a b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Initial Guesses 483 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. dhanesh_h Mathcad - CombinedMathcads

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