Engineering Mechanics - Statics Chapter 6 Positive means Tension, Negative means Compression F AB F AD F BC F BD F DC F DE ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 600 849− 600 400− 1414 1600− ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ lb= Problem 6-4 The truss, used to support a balcony, is subjected to the loading shown. Approximate each joint as a pin and determine the force in each member. State whether the members are in tension or compression. Units Used: kip 10 3 lb= Given: P 1 800 lb= P 2 0lb= a 4ft= θ 45 deg= Solution: Initial Guesses F AB 1lb= F AD 1lb= F DC 1lb= F BC 1lb= F BD 1lb= F DE 1lb= Given Joint A: F AB F AD cos θ () + 0= P 1 − F AD sin θ () − 0= Joint B: F BC F AB − 0= P 2 − F BD − 0= 441 © 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 D: F DC F AD − () cos θ () F DE + 0= F DC F AD + () sin θ () F BD + 0= F AB F AD F BC F BD F DC F DE ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find F AB F AD , F BC , F BD , F DC , F DE , () = Positive means Tension, Negative means Compression F AB F AD F BC F BD F DC F DE ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 800 1131− 800 0 1131 1600− ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ lb= Problem 6-5 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 20 kN= P 2 10 kN= a 1.5 m= e 2m= Solution: θ atan e a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = 442 © 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 Initial Guesses: F AB 1kN= F AG 1kN= F CF 1kN= F BC 1kN= F BG 1kN= F DE 1kN= F CG 1kN= F FG 1kN= F EF 1kN= F CD 1kN= F DF 1kN= Given Joint B F BC F AB cos θ () − 0= F BG − F AB sin θ () − 0= Joint G F FG F CG cos θ () + F AG − 0= F CG sin θ () F BG + P 1 − 0= Joint C F BC − F CD + F CF F CG − () cos θ () + 0= F CG F CF + () − sin θ () 0= Joint D F CD − F DE cos θ () + 0= F DF − F DE sin θ () − 0= Joint F F EF F FG − F CF cos θ () − 0= F DF F CF sin θ () + P 2 − 0= Joint E F DE − cos θ () F EF − 0= 443 © 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 BC F CG F CD F AG F BG F FG F DF F CF F DE F EF ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find F AB F BC , F CG , F CD , F AG , F BG , F FG , F DF , F CF , F DE , F EF , () = F AB F BC F CG F CD ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 21.88− 13.13− 3.13 9.37− ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ kN= F AG F BG F FG F DF ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 13.13 17.5 11.25 12.5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ kN= F CF F DE F EF ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 3.13− 15.62− 9.37 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN= Positive means Tension, Negative means Compression Problem 6-6 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 40 kN= P 2 20 kN= a 1.5 m= e 2m= 444 © 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: θ atan e a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Initial Guesses: F AB 1kN= F AG 1kN= F CF 1kN= F BC 1kN= F BG 1kN= F DE 1kN= F CG 1kN= F FG 1kN= F EF 1kN= F CD 1kN= F DF 1kN= Given Joint B F BC F AB cos θ () − 0= F BG − F AB sin θ () − 0= Joint G F FG F CG cos θ () + F AG − 0= F CG sin θ () F BG + P 1 − 0= Joint C F BC − F CD + F CF F CG − () cos θ () + 0= F CG F CF + () − sin θ () 0= Joint D F CD − F DE cos θ () + 0= F DF − F DE sin θ () − 0= Joint F F EF F FG − F CF cos θ () − 0= F DF F CF sin θ () + P 2 − 0= Joint E F DE − cos θ () F EF − 0= 445 © 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 BC F CG F CD F AG F BG F FG F DF F CF F DE F EF ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find F AB F BC , F CG , F CD , F AG , F BG , F FG , F DF , F CF , F DE , F EF , () = F AB F BC F CG F CD ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 43.75− 26.25− 6.25 18.75− ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ kN= F AG F BG F FG F DF ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 26.25 35 22.5 25 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ kN= F CF F DE F EF ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 6.25− 31.25− 18.75 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN= Positive means Tension, Negative means Compression Problem 6-7 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: F 1 3kN= F 2 8kN= F 3 4kN= F 4 10 kN= 446 © 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 a 2m= b 1.5 m= Solution: θ atan b a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Initial Guesses F BA 1kN= F BC 1kN= F AC 1kN= F AF 1kN= F CD 1kN= F CF 1kN= F DF 1kN= F ED 1kN= F EF 1kN= Given Joint B F 1 F BC + 0= F 2 − F BA − 0= Joint C F CD F BC − F AC cos θ () − 0= F 3 − F AC sin θ () − F CF − 0= Joint E F EF − 0= Joint D F CD − F DF cos θ () − 0= F 4 − F DF sin θ () − F ED − 0= Joint F F AF − F EF + F DF cos θ () + 0= F CF F DF sin θ () + 0= F BA F AF F DF F BC F CD F ED F AC F CF F EF ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find F BA F AF , F DF , F BC , F CD , F ED , F AC , F CF , F EF , () = 447 © 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 Positive means tension, Negative means compression. F BA F AF F DF F BC F CD F ED F AC F CF F EF ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 8− 4.167 5.208 3− 4.167− 13.125− 1.458− 3.125− 0 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ kN= Problem 6-8 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: Σ M A = 0; P− aC y 2a+ Pa− 0= C y P= Joint C: Σ F x = 0; 1 2 F BC 4 17 F CD − 0= Σ F y = 0; P 1 17 F CD + 1 2 F BC − 0= 448 © 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 BC 42 P 3 = 1.886 P= (C) F CD 17 P 3 = 1.374 P= (T) Joint B: Σ F x = 0; P 1 2 F CD − 1 2 F AB + 0= Σ F y = 0; 1 2 F CD 1 2 F AB + F BD − 0= F AB 2 P 3 = 0.471P= C() F BD 5P 3 = 1.667P= T() Joint D: Σ F x = 0; F DA F CD = 1.374P= (T) Problem 6-9 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. Given: T max 1500 lb= C max 800 lb= Solution: Set P 1lb= Initial Guesses F AB 1lb= F AD 1lb= F BD 1lb= F BC 1lb= F CD 1lb= 449 © 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 F BC F AB − () 1 2 P+ 0= F BD − F AB F BC + () 1 2 − 0= Joint D F CD F AD − () 4 17 0= F BD P− F AD F CD + () 1 17 − 0= Joint C F BC − 1 2 F CD 4 17 − 0= F AB F BC F AD F CD F BD ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find F AB F BC , F AD , F CD , F BD , () = F AB F BC F AD F CD F BD ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 0.471− 1.886− 1.374 1.374 1.667 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ lb= Now find the critical load P 1 P T max max F AB F BC , F AD , F CD , F BD , () = P 1 900lb= P 2 P C max min F AB F BC , F AD , F CD , F BD , () = P 2 424.264 lb= P min P 1 P 2 , () = P 424.3 lb= Problem 6-10 Determine the force in each member of the truss and state if the members are in tension or compression. Given: P 1 0lb= 450 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. [...]... = 1 N F DE = 1 N Given Joint C −F CB − FCD −F 2 − F CD Joint B 2a 2 ( 2 a) + b =0 2 −F BA + F CB = 0 ⎛ b⎞ −F 1 − F BD − ρ g⎜ a + ⎟ = 0 ⎝ 4⎠ Joint D 2 2⎤ ⎛ a⎞ + ⎛ b⎞ ⎥ = 0 ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎝ 4⎠ ⎦ ⎡a − ρ g⎢ + 2 2 2 ( 2 a) + b b ( F BD + FCD + FDA − FDE (FCD − FDA − FDE) ⎡b − ρ g⎢ + 3 2 2 ⎣4 ( 2 a) + b ) b 2a 2 2 2 ⎛ a⎞ + ⎛ b⎞ ⎥ = 0 ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎝ 4⎠ ⎦ =0 2 ( 2 a) + b ⎛ FCB ⎞ ⎜ ⎟ FCD ⎟ ⎜ ⎜F ⎟ ⎜ BA ⎟ = Find... publisher Engineering Mechanics - Statics Chapter 6 Given: F 1 = 3 kN F 2 = 2 kN a = 3m b = 4m Solution: Initial Guesses: F CB = 1 kN F CD = 1 kN F BA = 1 kN F BD = 1 kN F DA = 1 kN F DE = 1 kN Given Joint C −F 2 − F CD Joint B 2a −F CB − FCD =0 2 ( 2 a) + b b 2 2 =0 2 ( 2 a) + b −F BA + F CB = 0 −F 1 − F BD = 0 Joint D (FCD − FDA − FDE) ( 2a 2 =0 2 ( 2 a) + b F BD + FCD + FDA − FDE ) b 2 =0 2 ( 2 a) +... in writing from the publisher Engineering Mechanics - Statics Chapter 6 Joint E: + ↑ ΣF y = 0; + → ΣFx = 0; 1 F EC −P=0 2 1 F EF − F EA + F EC =0 2 Joint B: + ↑ ΣF y = 0; F BA + → ΣFx = 0; F BA 1 + F BD + F FB 2 1 2 1 1 1 − F FB 5 =0 2 ⎛ 2 ⎞=0 ⎟ ⎝ 5⎠ − F BD ⎜ 2 Joint C: + ↑ ΣF y = 0; F CA + → ΣFx = 0; F CA 1 5 2 5 + FCD 1 − FEC 1 2 2 − FEC 1 − FCD 1 =0 2 =0 2 476 © 20 07 R C Hibbeler Published by Pearson... permission in writing from the publisher Engineering Mechanics - Statics F CA = Chapter 6 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 Fx = 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 -2 2 The maximum allowable tensile force... ΣME = 0; − Ay 4a cos ( θ ) + F 14a + F 23 a + F 32a + F 2 a = 0 Ay = ΣF y = 0; 2 F1 + 2 F2 + F3 E x = 20 00 lb Ay = 23 09.4 lb 2 cos ( θ ) E y = − Ay + 2 cos ( θ ) F1 + 2 cos ( θ ) F 2 + cos ( θ ) F3 E y = 1154.7 lb Joint A: ΣF y = 0; F AB = −cos ( θ ) F 1 + Ay sin ( θ ) F AB = 3.75 kip ΣF x = 0; (C) F AH = −sin ( θ ) F 1 + F AB cos ( θ ) F AH = 3 kip (T) 473 © 20 07 R C Hibbeler Published by Pearson Education,... permission in writing from the publisher Engineering Mechanics - Statics Chapter 6 ⎛ FAB ⎞ ⎜ ⎟ ⎛ 22 .361 ⎞ ⎜ FBF ⎟ ⎜ 20 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FBC ⎟ = ⎜ 20 ⎟ kN ⎜F ⎟ ⎜ 0 ⎟ ⎜ ED ⎟ ⎜ ⎟ ⎝ 20 ⎠ ⎜ FAF ⎟ ⎝ ⎠ ⎛ FFC ⎞ ⎜ ⎟ ⎛ 28 .28 4 ⎞ ⎜ FEC ⎟ ⎜ 20 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FGB ⎟ = ⎜ 30 ⎟ kN ⎜F ⎟ ⎜ 0 ⎟ ⎜ FE ⎟ ⎜ ⎟ ⎝ 0 ⎠ ⎜ FCD ⎟ ⎝ ⎠ Positive means Tension, Negative means Compression Problem 6-1 4 Determine the force in each member of... writing from the publisher Engineering Mechanics - Statics Chapter 6 Problem 6 -2 4 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; + ↑ ΣF y = 0; P L 2L +P − Dy L = 0 3 3 Ay + Dy − 2 P = 0 Joint F: + 1 ↑ ΣF y = 0; F FB + → ΣFx = 0; F FD − F FE − F FB −P=0 2 1 =0 2 475 © 20 07 R C Hibbeler Published... publisher Engineering Mechanics - Statics Chapter 6 Units Used: 3 kip = 10 lb Given: F 1 = 600 lb F 2 = 800 lb a = 4 ft b = 3 ft θ = 60 deg Solution: Initial Guesses F BA = 1 lb F BD = 1 lb F CB = 1 lb F CD = 1 lb Given Joint C Joint B −F CB − F2 cos ( θ ) = 0 F CB + FBD b 2 a +b −F CD − F 2 sin ( θ ) = 0 =0 2 ⎛ FBA ⎞ ⎜ ⎟ ⎜ FBD ⎟ ⎜ ⎟ = Find ( FBA , FBD , FCB , FCD) ⎜ FCB ⎟ ⎜F ⎟ ⎝ CD ⎠ −F BA − F BD a 2 2 a... 0; 3 Ax d − P d = 0 2 Ax = 2P 3 4 62 © 20 07 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 2P ΣF x = 0; Ax − Ex... without permission in writing from the publisher Engineering Mechanics - Statics Chapter 6 Given Joint B + ↑ Σ Fy = 0; + → Σ Fx = 0; −F BA sin ( 2 θ ) − P = 0 −F BA cos ( 2 θ ) + F BC = 0 Joint C + ↑Σ Fy = 0; −F CA sin ( θ ) − F CD sin ( 2 θ ) = 0 + → Σ Fx = 0; Joint D −F BC + P − FCD cos ( 2 θ ) − FCA cos ( θ ) = 0 + → Σ Fx = 0; −F DA + F CD cos ( 2 θ ) = 0 ⎛ FBA ⎞ ⎜ ⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FCA ⎟ = Find ( . F EF , () = F AB F BC F CG F CD ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 43.75− 26 .25 − 6 .25 18.75− ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ kN= F AG F BG F FG F DF ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 26 .25 35 22 .5 25 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ kN= F CF F DE F EF ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 6 .25 − 31 .25 − 18.75 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ kN= Positive. the publisher. Engineering Mechanics - Statics Chapter 6 F BC 42 P 3 = 1.886 P= (C) F CD 17 P 3 = 1.374 P= (T) Joint B: Σ F x = 0; P 1 2 F CD − 1 2 F AB + 0= Σ F y = 0; 1 2 F CD 1 2 F AB + F BD −. permission in writing from the publisher. Engineering Mechanics - Statics Chapter 6 F AB F BF F BC F ED F AF ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 22 .361− 20 20 0 20 − ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ kN= F FC F EC F GB F FE F CD ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 28 .28 4− 20 30 0 0 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ kN= Positive