Engineering Mechanics - Statics Episode 2 Part 3 ppsx

Engineering Mechanics - Statics Chapter 6 Problem 6-27 Determine the force in each member of the truss and state if the members are in tension or compression Units Used: 3 kN = 10 N Given: P 1 = 4 kN P 2 = 0 kN a = 2m θ = 15 deg Solution: Take advantage of the symetry Initial Guesses: F BD = 1 kN F CD = 1 kN F CA = 1 kN F BC = 1 kN Given Joint D Joint B −P 1 2 F AB = 1 kN − F BD sin ( 2 θ ) − FCD sin ( 3 θ ) = 0 −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 (FCD + FCA)sin(θ ) + FBC = 0 ⎛ FBD ⎞ ⎜ ⎟ ⎜ FCD ⎟ ⎜ ⎟ F AB ⎟ = Find ( F ⎜ BD , F CD , F AB , F CA , FBC ) ⎜F ⎟ ⎜ CA ⎟ ⎜ FBC ⎟ ⎝ ⎠ ⎛ FFD ⎞ ⎛ FBD ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FED ⎟ ⎜ FCD ⎟ ⎜ ⎟ ⎜ ⎟ FGF ⎟ = ⎜ F AB ⎟ ⎜ ⎜F ⎟ ⎜F ⎟ ⎜ EG ⎟ ⎜ CA ⎟ ⎜ FFE ⎟ ⎜ FBC ⎟ ⎝ ⎠ ⎝ ⎠ 481 © 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 ⎛ FBD ⎞ ⎜ ⎟ ⎛ −4 ⎞ ⎜ FCD ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ FAB ⎟ = ⎜ −4 ⎟ kN ⎜F ⎟ ⎜ 0 ⎟ ⎜ CA ⎟ ⎜ ⎟ ⎜ FBC ⎟ ⎝ 0 ⎠ ⎝ ⎠ Chapter 6 ⎛ FFD ⎞ ⎜ ⎟ ⎛ −4 ⎞ ⎜ FED ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ FGF ⎟ = ⎜ −4 ⎟ kN ⎜F ⎟ ⎜ 0 ⎟ ⎜ EG ⎟ ⎜ ⎟ ⎜ FFE ⎟ ⎝ 0 ⎠ ⎝ ⎠ Positvive means Tension, Negative means Compression Problem 6-28 Determine the force in each member of the truss and state if the members are in tension or compression Units Used: 3 kN = 10 N Given: P 1 = 2 kN P 2 = 4 kN a = 2m θ = 15 deg Solution: Take advantage of the symmetry Initial Guesses: F BD = 1 kN F CD = 1 kN F CA = 1 kN F BC = 1 kN Given Joint D Joint B −P 1 2 F AB = 1 kN − F BD sin ( 2 θ ) − FCD sin ( 3 θ ) = 0 −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 (FCD + FCA)sin(θ ) + FBC = 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 ⎛ FBD ⎞ ⎜ ⎟ ⎜ FCD ⎟ ⎜ ⎟ ⎜ FAB ⎟ = Find ( FBD , FCD , FAB , FCA , FBC ) ⎜F ⎟ ⎜ CA ⎟ ⎜ FBC ⎟ ⎝ ⎠ ⎛ FBD ⎞ ⎜ ⎟ ⎛ −11.46 ⎞ ⎜ FCD ⎟ ⎜ 6.69 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ FAB ⎟ = ⎜ −13.46 ⎟ kN ⎜ F ⎟ ⎜ 6.69 ⎟ ⎜ CA ⎟ ⎜ ⎟ ⎜ FBC ⎟ ⎝ −3.46 ⎠ ⎝ ⎠ ⎛ FFD ⎞ ⎛ FBD ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FED ⎟ ⎜ FCD ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ FGF ⎟ = ⎜ FAB ⎟ ⎜F ⎟ ⎜F ⎟ ⎜ EG ⎟ ⎜ CA ⎟ ⎜ FFE ⎟ ⎜ FBC ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ FFD ⎞ ⎜ ⎟ ⎛ −11.46 ⎞ ⎜ FED ⎟ ⎜ 6.69 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ FGF ⎟ = ⎜ −13.46 ⎟ kN ⎜ F ⎟ ⎜ 6.69 ⎟ ⎜ EG ⎟ ⎜ ⎟ ⎜ FFE ⎟ ⎝ −3.46 ⎠ ⎝ ⎠ Positvive means Tension, Negative means Compression Problem 6-29 Determine the force in each member of the truss and state if the members are in tension or compression Units Used: 3 kip = 10 lb Given: F 1 = 2 kip F 2 = 1.5 kip F 3 = 3 kip F 4 = 3 kip a = 4 ft b = 10 ft Solution: a⎞ θ = atan ⎛ ⎟ ⎜ ⎝ 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 Engineering Mechanics - Statics Chapter 6 F AB = 1 lb F BC = 1 lb F CD = 1 lb F DE = 1 lb F AI = 1 lb F BI = 1 lb F CI = 1 lb F CG = 1 lb F CF = 1 lb F DF = 1 lb F EF = 1 lb F HI = 1 lb F GI = 1 lb F GH = 1 lb F FG = 1 lb Given Joint A F AI cos ( θ ) + FAB = 0 Joint B F BC − F AB = 0 F BI = 0 Joint C ( ) F CD − F BC + FCF − F CI cos ( θ ) = 0 ( ) F CG + FCF + F CI sin ( θ ) = 0 Joint D F DE − F CD = 0 F DF = 0 Joint I ( ) F 2 + FGI + F CI − F AI cos ( θ ) = 0 ( ) F HI − FBI + FGI − F AI − FCI sin ( θ ) = 0 Joint H F GH cos ( θ ) + F 1 = 0 −F GH sin ( θ ) − FHI = 0 Joint G Joint F (FFG − FGH − FGI )cos (θ ) = 0 −F 3 − F CG + ( FGH − FFG − FGI ) sin ( θ ) = 0 (FEF − FFG − FCF)cos (θ ) = 0 (FFG − FCF − FEF)sin(θ ) − F4 − FDF = 0 484 © 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 ⎛ FAB ⎞ ⎜ ⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FCD ⎟ ⎜F ⎟ ⎜ DE ⎟ ⎜ FAI ⎟ ⎜ ⎟ ⎜ FBI ⎟ ⎜ ⎟ ⎜ FCI ⎟ ⎜ F ⎟ = Find F , F , F , F , F , F , F , F , F , F , F , F , F , F , F ( AB BC CD DE AI BI CI CG CF DF EF HI GI GH ⎜ CG ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FDF ⎟ ⎜F ⎟ ⎜ EF ⎟ ⎜ FHI ⎟ ⎜ ⎟ ⎜ FGI ⎟ ⎜F ⎟ ⎜ GH ⎟ ⎜ FFG ⎟ ⎝ ⎠ ⎛ FAB ⎞ ⎜ ⎟ ⎛ 3.75 ⎞ ⎜ FBC ⎟ ⎜ 3.75 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FCD ⎟ = ⎜ 7.75 ⎟ kip ⎜ F ⎟ ⎜ 7.75 ⎟ ⎜ DE ⎟ ⎜ ⎟ ⎜ FAI ⎟ ⎝ −4.04 ⎠ ⎝ ⎠ ⎛ FBI ⎞ ⎜ ⎟ ⎛ 0 ⎞ ⎜ FCI ⎟ ⎜ 0.27 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FCG ⎟ = ⎜ 1.4 ⎟ kip ⎜ F ⎟ ⎜ −4.04 ⎟ ⎜ CF ⎟ ⎜ ⎟ ⎜ FDF ⎟ ⎝ 0 ⎠ ⎝ ⎠ ⎞ ⎛ FEF ⎟ ⎜ ⎛ −12.12 ⎞ ⎜ FHI ⎟ ⎜ 0.8 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FGI ⎟ = ⎜ −5.92 ⎟ kip ⎜ F ⎟ ⎜ −2.15 ⎟ ⎜ GH ⎟ ⎜ ⎟ ⎜ FFG ⎟ ⎝ −8.08 ⎠ ⎝ ⎠ Positive means Tension, Negative means Compression Problem 6-30 The Howe bridge truss is subjected to the loading shown Determine the force in members DE, EH, and HG, and state if the members are in tension or compression Units Used: 3 kN = 10 N 485 © 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 = 30 kN F 2 = 20 kN F 3 = 20 kN F 4 = 40 kN a = 4m b = 4m Solution: −F 2 a − F 3( 2a) − F 4( 3a) + Gy( 4a) = 0 Gy = F2 + 2F 3 + 3F4 4 Gy = 45 kN Guesses F DE = 1 kN F EH = 1 kN F HG = 1 kN Given −F DE − F HG = 0 Gy − F 4 − F EH = 0 F DE b + Gy a = 0 ⎛ FDE ⎞ ⎜ ⎟ FEH ⎟ = Find ( FDE , FEH , F HG) ⎜ ⎜F ⎟ ⎝ HG ⎠ ⎛ FDE ⎞ ⎛ −45 ⎞ ⎜ ⎟ FEH ⎟ = ⎜ 5 ⎟ kN ⎜ ⎜ ⎟ ⎜ F ⎟ ⎝ 45 ⎠ ⎝ HG ⎠ Positive (T) Negative (C) Problem 6-31 The Pratt bridge truss is subjected to the loading shown Determine the force in members LD, LK, CD, and KD, and state if the members are in tension or compression Units Used: 3 kN = 10 N 486 © 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 = 50 kN F 2 = 50 kN F 3 = 50 kN a = 4m b = 3m Solution: Ax = 0 Ay = 3F 3 + 4F2 + 5F 1 6 Guesses F LD = 1 kN F LK = 1 kN F CD = 1 kN F KD = 1 kN Given F 2 b + F 1( 2b) − Ay( 3b) − F LK a = 0 F CD a + F1 b − Ay( 2b) = 0 Ay − F1 − F2 − a ⎞ ⎛ ⎜ 2 2 ⎟ FLD = 0 ⎝ a +b ⎠ −F 3 − F KD = 0 ⎛ FLD ⎞ ⎜ ⎟ ⎜ FLK ⎟ ⎜ ⎟ = Find ( FLD , FLK , FCD , FKD) ⎜ FCD ⎟ ⎜F ⎟ ⎝ KD ⎠ ⎛ FLD ⎞ ⎞ ⎜ ⎟ ⎛ 0 ⎟ ⎜ FLK ⎟ ⎜ ⎜ −112.5 ⎟ kN ⎜ ⎟=⎜ ⎟ ⎜ FCD ⎟ ⎜ 112.5 ⎟ ⎜ F ⎟ ⎝ −50 ⎠ ⎝ KD ⎠ Positive (T) Negative (C) 487 © 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 6-32 The Pratt bridge truss is subjected to the loading shown Determine the force in members JI, JE, and DE, and state if the members are in tension or compression Units Used: 3 kN = 10 N Given: F 1 = 50 kN F 2 = 50 kN F 3 = 50 kN a = 4m b = 3m Solution: Initial Guesses Gy = 1 kN F JI = 1 kN F JE = 1 kN F DE = 1 kN Given Entire Truss −F 1 b − F2( 2b) − F3( 3b) + Gy( 6b) = 0 Section −F DE − F JI = 0 F JE + Gy = 0 Gy( 2b) − F DE a = 0 488 © 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 ⎛ Gy ⎞ ⎜ ⎟ ⎜ FJI ⎟ ⎜ ⎟ = Find ( Gy , FJI , FJE , FDE) Gy = 50 kN FJE ⎟ ⎜ ⎜F ⎟ ⎝ DE ⎠ ⎛ FJI ⎞ ⎛ −75 ⎞ ⎜ ⎟ FJE ⎟ = ⎜ −50 ⎟ kN ⎜ ⎜ ⎟ ⎜ F ⎟ ⎝ 75 ⎠ ⎝ DE ⎠ Positive means Tension, Negative means Compression Problem 6-33 The roof truss supports the vertical loading shown Determine the force in members BC, CK, and KJ and state if these members are in tension or compression Units Used: 3 kN = 10 N Given: F 1 = 4 kN F 2 = 8 kN a = 2m b = 3m 489 © 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 Ax = 1 kN Ay = 1 kN F BC = 1 kN F CK = 1 kN F KJ = 1 kN Given Ax = 0 F 2( 3a) + F 1( 4a) − Ay( 6a) = 0 ⎛ 2b ⎞ ⎟ + Ax⎜ ⎟ − Ay( 2a) = 0 ⎝3⎠ ⎝3⎠ F KJ ⎛ ⎜ 2b ⎞ F KJ + A x + 3a ⎞ ⎛ ⎜ 2 ⎟ FBC = 0 2 ⎝ b + 9a ⎠ F CK + A y + b ⎞F = 0 ⎛ BC ⎜ 2 2⎟ ⎝ b + 9a ⎠ ⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ ⎟ F KJ ⎟ = Find ( A , A , F , F ⎜ x y KJ CK , FBC ) ⎜F ⎟ ⎜ CK ⎟ ⎜ FBC ⎟ ⎝ ⎠ ⎛ Ax ⎞ ⎜ ⎟ ⎛ 0 ⎞ ⎜ Ay ⎟ ⎜ 6.667 ⎟ ⎜ ⎟ ⎜ ⎟ F KJ ⎟ = ⎜ 13.333 ⎟ kN Positive (T) ⎜ Negative (C) ⎜F ⎟ ⎜ 0 ⎟ CK ⎟ ⎜ ⎜ ⎟ ⎜ FBC ⎟ ⎝ −14.907 ⎠ ⎝ ⎠ Problem 6-34 Determine the force in members CD, CJ, KJ, and DJ of the truss which serves to support the deck of a bridge State if these members are in tension or compression Units Used: 3 kip = 10 lb Given: F 1 = 4000 lb F 2 = 8000 lb 490 © 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 BG = 1 kN Ax = 1 kN F HG = 1 kN Ay = 1 kN F BC = 1 kN Given − Ax = 0 − Ay( a) − a ⎡ ⎤ F ( b) = 0 HG ⎢ 2 2⎥ ⎣ ( c − b) + a ⎦ F 3( a) + F2( 2a) + F1( 3a) − A y( 4a) = 0 F BC + a ⎡ ⎤F + ⎛ a ⎞F − A = 0 HG ⎜ BG x ⎢ 2 2⎥ 2 2⎟ ⎣ ( c − b) + a ⎦ ⎝ a +c ⎠ Ay − F1 + c−b c ⎡ ⎤F + ⎛ ⎞ ⎢ ⎥ HG ⎜ 2 2 ⎟ FBG = 0 2 2 ⎣ ( c − b) + a ⎦ ⎝ a +c ⎠ 506 © 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 ⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ ⎟ ⎜ FHG ⎟ = Find ( Ax , Ay , FHG , FBG , FBC ) ⎜F ⎟ ⎜ BG ⎟ ⎜ FBC ⎟ ⎝ ⎠ ⎛ Ax ⎞ ⎜ ⎟ ⎛ 0 ⎞ ⎜ Ay ⎟ ⎜ 9 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ FHG ⎟ = ⎜ −10.062 ⎟ kN ⎜ F ⎟ ⎜ 1.803 ⎟ ⎜ BG ⎟ ⎜ ⎟ ⎝ 8 ⎠ ⎜ FBC ⎟ ⎝ ⎠ Positive (T) Negative (C) Problem 6-49 The skewed truss carries the load shown Determine the force in members CB, BE, and EF and state if these members are in tension or compression Assume that all joints are pinned Solution: ΣMB = 0; −P d − F EF d = 0 ΣME = 0; −P d + + → Σ Fx = 0; 2 F CB d = 0 F CB = FCB − F BE = 0 P BE = 5 P− 1 5 F EF = −P 5 P 2 P 2 F EF = P ( C) F CB = 1.12P ( T) F BE = 0.5P ( T) Problem 6-50 The skewed truss carries the load shown Determine the force in members AB, BF, and EF and state if these members are in tension or compression Assume that all joints are pinned 507 © 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: ΣMF = 0; −P 2d + P d + FAB d = 0 F AB = P F AB = P ( T) ΣMB = 0; −P d − F EF d = 0 F EF = −P F EF = P ( C) F BE = − 2 P F BF = 1.41P ( C) + → Σ Fx = 0; P + FBF 1 =0 2 Problem 6-51 Determine the force developed in members BC and CH of the roof truss and state if the members are in tension or compression Units Used: 3 kN = 10 N Given: F 1 = 1.5 kN F 2 = 2 kN 508 © 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 = 1.5 m b = 1m c = 2m d = 0.8 m Solution: a⎞ θ = atan ⎛ ⎟ ⎜ ⎝c⎠ a ⎞ ⎟ ⎝ c − b⎠ φ = atan ⎛ ⎜ Initial Guesses: E y = 1 kN F BC = 1 kN F CH = 1 kN Given −F 2( d) − F1( c) + Ey( 2c) = 0 F BC sin ( θ ) ( c) + FCH sin ( φ ) ( c − b) + E y( c) = 0 −F BC sin ( θ ) − FCH sin ( φ ) − F1 + Ey = 0 ⎛ Ey ⎞ ⎜ ⎟ ⎜ FBC ⎟ = Find ( Ey , FBC , FCH ) Ey = 1.15 kN ⎜F ⎟ ⎝ CH ⎠ ⎛ FBC ⎞ ⎛ −3.25 ⎞ ⎜ ⎟= kN ⎜ FCH ⎟ ⎜ 1.923 ⎟ ⎠ ⎝ ⎠ ⎝ Positive (T) Negative (C) Problem 6-52 Determine the force in members CD and GF of the truss and state if the members are in tension or compression Also indicate all zero-force members Units Used: 3 kN = 10 N Given: F 1 = 1.5 kN F 2 = 2 kN 509 © 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 = 1.5 m b = 1m c = 2m d = 0.8 m Solution: a⎞ θ = atan ⎛ ⎟ ⎜ ⎝c⎠ ⎞ ⎟ ⎝ c − b⎠ φ = atan ⎛ ⎜ a Initial Guesses: E y = 1 kN F CD = 1 kN F GF = 1 kN Given −F 2( d) − F1( c) + Ey( 2c) = 0 E y( b) + F CD sin ( θ ) ( b) = 0 E y( c) − FGF( a) = 0 ⎛ Ey ⎞ ⎜ ⎟ FCD ⎟ = Find ( E , F ⎜ y CD , FGF) E y = 1.15 kN ⎜F ⎟ ⎝ GF ⎠ ⎛ FCD ⎞ ⎛ −1.917 ⎞ Positive (T) ⎜ ⎟= kN Negative (C) ⎜ FGF ⎟ ⎜ 1.533 ⎟ ⎠ ⎝ ⎠ ⎝ DF and CF are zero force members Problem 6-53 Determine the force in members DE, DL, and ML of the roof truss and state if the members are in tension or compression Units Used: 3 kN = 10 N Given: F 1 = 6 kN F 2 = 12 kN F 3 = 12 kN 510 © 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 4 = 12 kN a = 4m b = 3m c = 6m Solution: θ = atan ⎛ ⎜ c − b⎞ ⎟ ⎝ 3a ⎠ ⎡b + ⎢ φ = atan ⎢ ⎣ 2 ⎤ ( c − b) ⎥ 3 a ⎥ ⎦ Initial Guesses: Ay = 1 kN F ML = 1 kN F DL = 1 kN F DE = 1 kN Given F 2( a) + F3( 2a) + F4( 3a) + F3( 4a) + F2( 5a) + F1( 6a) − A y( 6a) = 0 2 F 1( 2a) + F 2( a) − A y( 2a) + FML⎡b + ( c − b)⎤ = 0 ⎢ ⎥ ⎣ 3 ⎦ Ay − F 1 − F 2 − F 3 + F DE sin ( θ ) − FDL sin ( φ ) = 0 F ML + F DL cos ( φ ) + FDE cos ( θ ) = 0 ⎛ Ay ⎞ ⎜ ⎟ ⎜ FML ⎟ ⎜ ⎟ = Find ( Ay , FML , FDE , FDL) ⎜ FDE ⎟ ⎜F ⎟ ⎝ DL ⎠ Ay = 36 kN ⎛ FML ⎞ ⎛ 38.4 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ FDE ⎟ = ⎜ −37.1 ⎟ kN ⎜ F ⎟ ⎝ −3.8 ⎠ ⎝ DL ⎠ Positive (T), Negative (C) Problem 6-54 Determine the force in members EF and EL of the roof truss and state if the members are in 511 © 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 tension or compression Units Used: 3 kN = 10 N Given: F 1 = 6 kN F 2 = 12 kN F 3 = 12 kN F 4 = 12 kN a = 4m b = 3m c = 6m Solution: θ = atan ⎛ ⎜ c − b⎞ ⎟ ⎝ 3a ⎠ Initial Guesses: Iy = 1 kN F EF = 1 kN F EL = 1 kN Given −F 2( a) − F3( 2a) − F4( 3a) − F3( 4a) − F2( 5a) − F1( 6a) + Iy( 6a) = 0 −F 3( a) − F2( 2a) − F1( 3a) + Iy( 3a) + F EF cos ( θ ) ( c) = 0 −F 4 − F EL − 2F EF sin ( θ ) = 0 ⎛ Iy ⎞ ⎜ ⎟ ⎜ FEF ⎟ = Find ( Iy , FEF , FEL) ⎜F ⎟ ⎝ EL ⎠ Iy = 36 kN ⎛ FEF ⎞ ⎛ −37.108 ⎞ ⎜ ⎟= kN ⎜ FEL ⎟ ⎜ 6 ⎟ ⎠ ⎝ ⎠ ⎝ Positive (T) Negative (C) Problem 6-55 Two space trusses are used to equally support the uniform sign of mass M Determine the force developed in members AB, AC, and BC of truss ABCD and state if the members are in tension or compression Horizontal short links support the truss at joints B and D and there is a ball-and512 © 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 p socket joint at C pp j Given: M = 50 kg g = 9.81 m 2 s a = 0.25 m b = 0.5 m c = 2m Solution: h = 2 2 b −a ⎛ −a ⎞ AB = ⎜ −c ⎟ ⎜ ⎟ ⎝h⎠ ⎛a⎞ AD = ⎜ −c ⎟ ⎜ ⎟ ⎝h⎠ ⎛ 2a ⎞ BD = ⎜ 0 ⎟ ⎜ ⎟ ⎝0⎠ ⎛a⎞ BC = ⎜ 0 ⎟ ⎜ ⎟ ⎝ −h ⎠ ⎛0⎞ AC = ⎜ −c ⎟ ⎜ ⎟ ⎝0⎠ Guesses F AB = 1 N F AD = 1 N F AC = 1 N F BC = 1 N F BD = 1 N By = 1 N Given ⎛ 0 ⎟ ⎞ ⎜ 0 ⎟ AB AD AC F AB + FAD + FAC +⎜ =0 ⎜ −M g ⎟ AB AD AC ⎜ 2 ⎟ ⎝ ⎠ ⎞ ⎛ 0 ⎟ ⎜ F AB + FBD + F BC + ⎜ −B y ⎟ = 0 AB BD BC ⎜ 0 ⎟ ⎝ ⎠ −AB BD BC 513 © 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 ⎛ FAB ⎞ ⎜ ⎟ FAD ⎟ ⎜ ⎜F ⎟ ⎜ AC ⎟ = Find F , F , F , F , F , B ( AB AD AC BC BD y) ⎜ FBC ⎟ ⎜ ⎟ ⎜ FBD ⎟ ⎜ ⎟ ⎝ By ⎠ ⎛ By ⎞ ⎛ 566 ⎞ ⎜ ⎟ FAD ⎟ = ⎜ 584 ⎟ N ⎜ ⎜ ⎟ ⎜F ⎟ ⎝ 0 ⎠ ⎝ BD ⎠ ⎛ FAB ⎞ ⎛ 584 ⎞ ⎜ ⎟ FAC ⎟ = ⎜ −1133 ⎟ N ⎜ ⎜ ⎟ ⎜ F ⎟ ⎝ −142 ⎠ ⎝ BC ⎠ Positive (T), Negative (C) Problem 6-56 Determine the force in each member of the space truss and state if the members are in tension or compression.The truss is supported by short links at B, C, and D Given: F = 600 N a = 3m b = 1m c = 1.5 m Solution: ⎛b⎞ ⎜ ⎟ AB = −c ⎜ ⎟ ⎝ −a ⎠ ⎛0⎞ ⎜ ⎟ AC = c ⎜ ⎟ ⎝ −a ⎠ ⎛ −b ⎞ ⎜ ⎟ AD = −c ⎜ ⎟ ⎝ −a ⎠ ⎛ −b ⎞ ⎜ ⎟ CD = −2c ⎜ ⎟ ⎝ 0 ⎠ 514 © 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 ⎛ b ⎞ CB = ⎜ −2c ⎟ ⎜ ⎟ ⎝ 0 ⎠ Chapter 6 ⎛ −2b ⎞ BD = ⎜ 0 ⎟ ⎜ ⎟ ⎝ 0 ⎠ Guesses F BA = 1 N F BC = 1 N F CA = 1 N F DA = 1 N F BD = 1 N F DC = 1 N By = 1 N Bz = 1 N Cz = 1 N Given ⎛ 0 ⎞ F BA + FCA + F DA +⎜ 0 ⎟ =0 ⎜ ⎟ AB AC AD ⎝ −F ⎠ AB AC AD ⎛0⎟ ⎜ ⎞ F CA + FDC + F BC +⎜0 ⎟ =0 AC CD CB ⎜ Cz ⎟ ⎝ ⎠ −AC CD CB ⎛0⎞ ⎜ ⎟ F BC + FBD + F BA + ⎜ By ⎟ = 0 CB BD AB ⎜B ⎟ ⎝ z⎠ −CB BD −AB ⎛ FBA ⎞ ⎜ ⎟ FBC ⎟ ⎜ ⎜F ⎟ ⎜ CA ⎟ ⎜ FDA ⎟ ⎜ ⎟ FBD ⎟ = Find ( F , F ⎜ BA BC , FCA , F DA , F BD , F DC , By , B z , Cz) ⎜F ⎟ ⎜ DC ⎟ ⎜ By ⎟ ⎜ ⎟ ⎜ Bz ⎟ ⎟ ⎜ ⎝ Cz ⎠ 515 © 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 ⎛ By ⎞ ⎛ 1.421 × 10− 14 ⎞ ⎜ ⎟ ⎜ ⎟ Bz ⎟ = ⎜ ⎜ ⎟N 150 ⎜C ⎟ ⎜ ⎟ 300 ⎠ ⎝ z⎠ ⎝ Chapter 6 ⎛ FBA ⎞ ⎜ ⎟ ⎛ −175 ⎞ ⎟ FBC ⎟ ⎜ ⎜ ⎜ 79.1 ⎟ ⎜F ⎟ ⎜ CA ⎟ = ⎜ −335.4 ⎟ N ⎜ FDA ⎟ ⎜ −175 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ 25 ⎟ ⎜ FBD ⎟ ⎜ ⎟ ⎜ 79.1 ⎟ ⎝ ⎠ FDC ⎠ ⎝ Positive (T), Negative (C) Problem 6-57 Determine the force in each member of the space truss and state if the members are in tension or compression.The truss is supported by short links at A, B, and C Given: a = 4 ft b = 2 ft c = 3 ft d = 2 ft e = 8 ft ⎛ 0 ⎞ ⎜ ⎟ F = 500 lb ⎜ ⎟ ⎝ 0 ⎠ Solution: ⎛ −c ⎞ ⎜ ⎟ AD = a ⎜ ⎟ ⎝e⎠ ⎛ −c ⎞ ⎜ ⎟ BD = −b ⎜ ⎟ ⎝e ⎠ ⎛d⎞ ⎜ ⎟ CD = −b ⎜ ⎟ ⎝e ⎠ ⎛ 0 ⎞ ⎜ ⎟ AB = a + b ⎜ ⎟ ⎝ 0 ⎠ ⎛ −c − d ⎞ ⎛ −c − d ⎞ ⎜ a + b ⎟ BC = ⎜ 0 ⎟ AC = ⎜ ⎟ ⎜ ⎟ ⎝ 0 ⎠ ⎝ 0 ⎠ Guesses 516 © 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 BA = 1 lb F BC = 1 lb F BD = 1 lb F AD = 1 lb F AC = 1 lb F CD = 1 lb Ay = 1 lb Az = 1 lb B x = 1 lb B z = 1 lb Cy = 1 lb Cz = 1 lb Given F + FAD −AD AD + FBD −BD BD + F CD −CD CD =0 ⎛0 ⎞ ⎜ ⎟ F AD + F BA + FAC + ⎜ Ay ⎟ = 0 AD AB AC ⎜A ⎟ ⎝ z⎠ AD AB AC ⎛ Bx ⎞ ⎜ ⎟ −AB BC BD F BA + FBC + F BD +⎜ 0 ⎟ =0 AB BC BD ⎜B ⎟ ⎝ z⎠ ⎛0⎞ CD −AC −BC ⎜ C ⎟ F CD + F AC + F BC + ⎜ y⎟ = 0 CD AC BC ⎜C ⎟ ⎝ z⎠ ⎛ FBA ⎞ ⎜ ⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FBD ⎟ ⎜F ⎟ ⎜ AD ⎟ ⎜ FAC ⎟ ⎜ ⎟ ⎜ FCD ⎟ ⎜ ⎟ = Find ( FBA , FBC , FBD , FAD , FAC , FCD , Ay , Az , Bx , Bz , Cy , Cz) ⎜ Ay ⎟ ⎜ A ⎟ ⎜ z ⎟ ⎜ Bx ⎟ ⎜ ⎟ ⎜ Bz ⎟ ⎜ C ⎟ ⎜ y ⎟ ⎜ Cz ⎟ ⎝ ⎠ 517 © 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 ⎛ Ay ⎞ ⎜ ⎟ ⎛ −200 ⎞ ⎜ Az ⎟ ⎜ −667 ⎟ ⎟ ⎜B ⎟ ⎜ ⎜ 0 ⎟ ⎜ x⎟ = lb ⎜ Bz ⎟ ⎜ 667 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ −300 ⎟ ⎜ Cy ⎟ ⎜ ⎟ ⎜ 0 ⎟ ⎝ ⎠ ⎝ Cz ⎠ Chapter 6 ⎛ FBA ⎞ ⎜ ⎟ ⎛ 167 ⎞ ⎟ FBC ⎟ ⎜ ⎜ ⎜ 250 ⎟ ⎜F ⎟ ⎜ BD ⎟ = ⎜ −731 ⎟ lb ⎜ FAD ⎟ ⎜ 786 ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ −391 ⎟ ⎜ FAC ⎟ ⎜ ⎟ ⎜ 0 ⎟ ⎝ ⎠ FCD ⎠ ⎝ Positive (T) Negative (C) Problem 6-58 The space truss is supported by a ball-and-socket joint at D and short links at C and E Determine the force in each member and state if the members are in tension or compression Given: ⎛ 0 ⎞ ⎜ 0 ⎟ lb F1 = ⎜ ⎟ ⎝ −500 ⎠ ⎛ 0 ⎞ ⎜ ⎟ F 2 = 400 lb ⎜ ⎟ ⎝ 0 ⎠ a = 4 ft b = 3 ft c = 3 ft Solution: Find the external reactions Guesses E y = 1 lb Cy = 1 lb Cz = 1 lb Dx = 1 lb Dy = 1 lb Dz = 1 lb 518 © 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 ⎛ Dx ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ Dy ⎟ + ⎜ Ey ⎟ + ⎜ Cy ⎟ + F1 + F2 = 0 ⎜D ⎟ ⎜ 0 ⎟ ⎜C ⎟ ⎝ z⎠ ⎝ ⎠ ⎝ z⎠ D ⎞ ⎛0⎞ ⎛ −b ⎞ ⎛ 0 ⎞ ⎛ x ⎞ ⎛ −b ⎞ ⎛ 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎜ a ⎟ × F + ⎜ a ⎟ × F + ⎜ 0 ⎟ × D + 0 × Cy = 0 ⎜ ⎟ 1 ⎜ ⎟ 2 ⎜ ⎟ ⎜ y⎟ ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝0⎠ ⎝ c ⎠ ⎜ Dz ⎟ ⎝ c ⎠ ⎜ Cz ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ Ey ⎞ ⎜ ⎟ ⎜ Cy ⎟ ⎜C ⎟ ⎜ z ⎟ = Find E , C , C , D , D , D ( y y z x y z) ⎜ Dx ⎟ ⎜ ⎟ ⎜ Dy ⎟ ⎜ ⎟ ⎝ Dz ⎠ ⎛ Ey ⎞ ⎜ ⎟ ⎛ 266.667 ⎞ ⎜ Cy ⎟ ⎜ −400 ⎟ ⎟ ⎜C ⎟ ⎜ ⎜ ⎟ 0 ⎜ z⎟ = ⎟ lb ⎜ Dx ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎜ −266.667 ⎟ ⎜ Dy ⎟ ⎜ ⎟ ⎜ 500 ⎟ ⎝ ⎠ ⎝ Dz ⎠ Now find the force in each member ⎛ −b ⎞ ⎛ −b ⎞ ⎜ 0 ⎟ AC = ⎜ −a ⎟ AB = ⎜ ⎟ ⎜ ⎟ ⎝0⎠ ⎝c ⎠ ⎛0⎞ AD = ⎜ −a ⎟ ⎜ ⎟ ⎝c ⎠ ⎛0⎞ AE = ⎜ −a ⎟ ⎜ ⎟ ⎝0⎠ ⎛0⎞ ⎛b⎞ ⎜ −a ⎟ BE = ⎜ −a ⎟ BC = ⎜ ⎟ ⎜ ⎟ c ⎠ ⎝ ⎝0⎠ ⎛0⎞ BF = ⎜ −a ⎟ ⎜ ⎟ ⎝0⎠ ⎛b⎞ CD = ⎜ 0 ⎟ ⎜ ⎟ ⎝0⎠ ⎛0⎞ ⎛0⎞ ⎜ 0 ⎟ DE = ⎜ 0 ⎟ CF = ⎜ ⎟ ⎜ ⎟ ⎝ −c ⎠ ⎝ −c ⎠ ⎛ −b ⎞ DF = ⎜ 0 ⎟ ⎜ ⎟ ⎝ −c ⎠ ⎛ −b ⎞ EF = ⎜ 0 ⎟ ⎜ ⎟ ⎝0⎠ Guesses F AB = 1 lb F AC = 1 lb F AD = 1 lb F AE = 1 lb F BC = 1 lb F BE = 1 lb F BF = 1 lb F CD = 1 lb 519 © 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 F CF = 1 lb Chapter 6 F DE = 1lb F DF = 1 lb F EF = 1 lb Given F 1 + F AB AB F 2 + F BC BC AB BC + FAC AC + FBF BF AC BF + FAD AD + FBE BE AD BE + FAE + F AB AE AE −AB AB =0 =0 ⎛0⎟ ⎜ ⎞ −AE −BE EF −DE + FBE + F EF + F DE =0 ⎜ Ey ⎟ + FAE AE BE EF DE ⎜0⎟ ⎝ ⎠ F BF −BF BF + F CF −CF CF + F DF −DF DF + FEF −EF EF =0 ⎛0⎞ ⎜ ⎟ −BC −AC CD CF + FAC + FCD + FCF =0 ⎜ Cy ⎟ + FBC BC AC CD CF ⎜C ⎟ ⎝ z⎠ ⎛ FAB ⎞ ⎜ ⎟ ⎜ FAC ⎟ ⎜ ⎟ ⎜ FAD ⎟ ⎜F ⎟ ⎜ AE ⎟ ⎜ FBC ⎟ ⎜ ⎟ ⎜ FBE ⎟ ⎜ ⎟ = Find ( FAB , FAC , FAD , FAE , FBC , FBE , FBF , FCD , FCF , FDE , FDF , FEF) ⎜ FBF ⎟ ⎜F ⎟ ⎜ CD ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FDE ⎟ ⎜F ⎟ ⎜ DF ⎟ ⎜ FEF ⎟ ⎝ ⎠ 520 © 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 ... the publisher Engineering Mechanics - Statics Chapter Given: F = 30 kN F = 20 kN F = 20 kN F = 40 kN a = 4m b = 4m Solution: −F a − F 3( 2a) − F 4( 3a) + Gy( 4a) = Gy = F2 + 2F + 3F4 Gy = 45 kN... = kN F EL = kN Given −F 2( a) − F3( 2a) − F4( 3a) − F3( 4a) − F2( 5a) − F1( 6a) + Iy( 6a) = −F 3( a) − F2( 2a) − F1( 3a) + Iy( 3a) + F EF cos ( θ ) ( c) = −F − F EL − 2F EF sin ( θ ) = ⎛ Iy ⎞... publisher = Chapter Engineering Mechanics - Statics Solution: Guess E y = lb F GJ = lb Given −F 2( a) − F3( 2a) − F4( 3a) + Ey( 4a) = −F 4( a) + Ey( 2a) + F GJ sin ( θ ) ( 2a) = ⎞ ⎛ Ey ⎟ ⎜ = Find