Engineering Mechanics - Statics Episode 2 Part 4 pdf

Engineering Mechanics - Statics Chapter 6 Positive (T) Negative (C) F AB F AC F AD F AE F BC F BE ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 300− 583.095 333.333 666.667− 0 500 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ lb= F BF F CD F CF F DE F DF F EF ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 0 300− 300− 0 424.264 300− ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ lb= Problem 6-59 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: F 1 200 300 500− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= F 2 0 400 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= a 4ft= b 3ft= c 3ft= Solution: Find the external reactions Guesses E y 1lb= C y 1lb= C z 1lb= D x 1lb= D y 1lb= D z 1lb= 521 © 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 D x D y D z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 0 E y 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + 0 C y C z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + F 1 + F 2 + 0= 0 a 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 1 × b− a 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F 2 ×+ 0 0 c ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ D x D y D z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ×+ b− 0 c ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 C y C z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ×+ 0= E y C y C z D x D y D z ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find E y C y , C z , D x , D y , D z , () = E y C y C z D x D y D z ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 33.333− 666.667− 200 200− 1.253− 10 13− × 300 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ lb= Now find the force in each member. AB b− 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = AC b− a− c ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = AD 0 a− c ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = AE 0 a− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = BC 0 a− c ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = BE b a− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = BF 0 a− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = CD b 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = CF 0 0 c− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = DE 0 0 c− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = DF b− 0 c− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = EF b− 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = Guesses F AB 1lb= F AC 1lb= F AD 1lb= F AE 1lb= F BC 1lb= F BE 1lb= F BF 1lb= F CD 1lb= F CF 1lb= F DE 1lb= F DF 1lb= F EF 1lb= Given 522 © 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 1 F AB AB AB + F AC AC AC + F AD AD AD + F AE AE AE + 0= F 2 F BC BC BC + F BF BF BF + F BE BE BE + F AB AB− AB + 0= 0 E y 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ F AE AE− AE + F BE BE− BE + F EF EF EF + F DE DE− DE + 0= F BF BF− BF F CF CF− CF + F DF DF− DF + F EF EF− EF + 0= 0 C y C z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ F BC BC− BC + F AC AC− AC + F CD CD CD + F CF CF CF + 0= F AB F AC F AD F AE F BC F BE F BF F CD F CF F DE F DF F EF ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find F AB F AC , F AD , F AE , F BC , F BE , F BF , F CD , F CF , F DE , F DF , F EF , () = 523 © 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 (T) Negative (C) F AB F AC F AD F AE F BC F BE ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 300− 971.825 1.121 10 11− × 366.667− 0 500 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ lb= F BF F CD F CF F DE F DF F EF ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 0 500− 300− 0 424.264 300− ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ lb= Problem 6-60 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 a ball-and-socket joints at A, B, and E. Hint: The support reaction at E acts along member EC. Why? Given: a 2m= F 200− 400 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N= b 1.5 m= c 5m= d 1m= e 2m= Solution: AC 0 ab+ 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = AD d a e ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = BC c− d− 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = BD c− b− e ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = CD d b− e ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = Guesses F AC 1N= F AD 1N= F BC 1N= F CD 1N= F EC 1N= F BD 1N= Given F F AD AD− AD + F BD BD− BD + F CD CD− CD + 0= 524 © 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 CD CD CD F BC BC− BC + F AC AC− AC + 0 0 F EC − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + 0= Positive (T) Negative (C) F AC F AD F BC F BD F CD F EC ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find F AC F AD , F BC , F BD , F CD , F EC , () = F AC F AD F BC F BD F CD F EC ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 221 343 148 186 397− 295− ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ N= Problem 6-61 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 ball-and-socket joints at C, D, E, and G. Units Used: kN 10 3 N= Given: F 3kN= a 2m= b 1.5 m= c 2m= d 1m= e 1m= Solution: F v F b 2 c 2 + 0 b c− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = 525 © 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 u AG 1 a 2 e 2 + e− a− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = u AE 1 a 2 d 2 + d a− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = u BC u AE = u AB 0 0 1− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = u BD u AG = u BE 1 a 2 c 2 + d 2 + d a− c ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = u BG 1 a 2 e 2 + c 2 + e− a− c ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = Guesses F AB 1kN= F AE 1kN= F AG 1kN= F BC 1kN= F BD 1kN= F BE 1kN= F BG 1kN= Given c− c 2 b 2 + Fa() a a 2 d 2 + F BC c()− a a 2 e 2 + F BD c()− 0= e a 2 e 2 + F BD a() d a 2 d 2 + F BC a()− 0= F v F AE u AE + F AG u AG + F AB u AB + 0= F AB − u AB F BG u BG + F BE u BE + F BC u BC + F BD u BD + 0= 526 © 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 AG F BC F BD F BE F BG ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find F AB F AE , F AG , F BC , F BD , F BE , F BG , () = F AB F AE F AG F BC F BD F BE F BG ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 2.4− 1.006 1.006 1.342− 1.342− 1.8 1.8 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ kN= Positive (T) Negative (C) Problem 6-62 Determine the force in members BD, AD, and AF of the space truss and state if the members are in tension or compression. The truss is supported by short links at A, B, D, and F. Given: F 0 250 250− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= a 6ft= b 6ft= θ 60 deg= Solution: Find the external reactions hbsin θ () = Guesses A x 1lb= B y 1lb= B z 1lb= D y 1lb= F y 1lb= F z 1lb= Given ⎛ ⎞ ⎛ ⎞ 527 © 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 A x 0 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + 0 B y B z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + 0 D y 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + 0 F y F z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + 0= 0.5b a h ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F× b a 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ A x 0 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ×+ b 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 B y B z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ×+ 0.5b 0 h ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 D y 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ×+ 0 a 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 F y F z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ×+ 0= A x B y B z D y F y F z ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find A x B y , B z , D y , F y , F z , () = A x B y B z D y F y F z ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 0 72 125 394− 72 125 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ lb= Now find the forces in the members AB 0 a− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = AC b− a− 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = AD 0.5− b a− h ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = AE 0.5− b 0 h ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = AF b− 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = BC b− 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = BD 0.5− b 0 h ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = CD 0.5b 0 h ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = CF 0 a 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = DE 0 a 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = DF 0.5− b a h− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = EF 0.5− b 0 h− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = Guesses F AB 1lb= F AC 1lb= F AD 1lb= F AE 1lb= F AF 1lb= F BC 1lb= F BD 1lb= F CD 1lb= F CF 1lb= 528 © 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 DE 1lb= F DF 1lb= F EF 1lb = Given F F DE DE− DE + F AE AE− AE + F EF EF EF + 0= F CF CF CF F CD CD CD + F BC BC− BC + F AC AC− AC + 0= F DE DE DE F DF DF DF + F AD AD− AD + F BD BD− BD + F CD CD− CD + 0 D y 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + 0= F AB AB− AB F BC BC BC + F BD BD BD + 0 B y B z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + 0= F AB AB AB F AC AC AC + F AF AF AF + F AD AD AD + F AE AE AE + A x 0 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + 0= F AB F AC F AD F AE F AF F BC F BD F CD F CF F DE F DF F EF ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find F AB F AC , F AD , F AE , F AF , F BC , F BD , F CD , F CF , F DE , F DF , F EF , () = 529 © 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 (T) Negative (C) F BD F AD F AF ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 144.3− 204.1 72.2 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= Problem 6-63 Determine the force in members CF, EF, and DF of the space truss and state if the members are in tension or compression. The truss is supported by short links at A, B, D, and F. Given: F 0 250 250− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= a 6ft= b 6ft= θ 60 deg= Solution: Find the external reactions hbsin θ () = Guesses A x 1lb= B y 1lb= B z 1lb= D y 1lb= F y 1lb= F z 1lb= Given F A x 0 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + 0 B y B z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + 0 D y 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + 0 F y F z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ + 0= 0.5b a h ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F× b a 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ A x 0 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ×+ b 0 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 B y B z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ×+ 0.5b 0 h ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 D y 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ×+ 0 a 0 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 0 F y F z ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ×+ 0= 530 © 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. [...]... publisher Engineering Mechanics - Statics Chapter 6 b = 0.9 m e = 1 .2 m c = 0.5 m θ = 45 deg Solution: BC is a two-force member Member AB : ΣMA = 0; e −F c + FBC 2 2 a +e b + F BC 2 a 2 ( c + d) = 0 2 a +e 2 a +e F BC = F c eb+a c+a d F BC = 20 0.3 N Thus, Cx = F BC Cy = F BC ΣF x = 0; ΣF y = 0; Ax − F BC e 2 Cx = 167 N 2 a +e a 2 Cy = 111 N 2 a +e e 2 =0 Ax = FBC 2 a +e Ay − F + FBC a 2 =0 2 a +e e 2 Ax... Engineering Mechanics - Statics Chapter 6 Given Ay − w1 a − MA − w1 a a − d 2 b +d 2 F CD = 0 d 2 d +b 2 Ax − b 2 b +d b F CD2 a = 0 2 2 b +d d 2 FCD = 0 FCD − Bx = 0 c c B y c − w2 =0 2 3 c FCD − w2 + B y = 0 2 2 2 b +d ⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ B ⎟ ⎜ x ⎟ = Find A , A , B , B , F , M ( x y x y CD A) ⎜ By ⎟ ⎜ ⎟ ⎜ FCD ⎟ ⎜ ⎟ ⎝ MA ⎠ ⎛ Ax ⎞ ⎛ 2. 8 ⎞ ⎜ ⎟= kip ⎜ Ay ⎟ ⎜ 5.1 ⎟ ⎝ ⎠ ⎝ ⎠ MA = 43 .2 kip⋅ ft ⎛ Bx ⎟ ⎛ 2. 8... permission in writing from the publisher Engineering Mechanics - Statics b −F CD − 2 −b + 2 a +b 2 2 2 2 2 b +c −c 2 F AD − c −F 3 − F DE − + FBD 2 a +b +c 2 a +b +c Chapter 6 =0 b 2 2 b +c F DF F DF = 0 F AD Joint B a −F BC − 2 2 a +b FBD = 0 b 2 2 a +b FBD = 0 −F 1 − F BA = 0 Joint E a 2 2 a +b FAE = 0 −F EF − b 2 2 a +b FAE = 0 ⎛ FBC ⎞ ⎜ ⎟ ⎜ FCF ⎟ ⎜ ⎟ ⎜ FCD ⎟ ⎜F ⎟ ⎜ AD ⎟ ⎜ FDF ⎟ ⎜ ⎟ ⎜ FDE ⎟ = Find ( FBC... Engineering Mechanics - Statics Chapter 6 Problem 6-6 4 Determine the force developed in each member of the space truss and state if the members are in tension or compression The crate has weight W Given: W = 150 lb a = 6 ft b = 6 ft c = 6 ft Solution: h = 2 Unit Vectors c − ⎛ a⎞ ⎜ ⎟ ⎝ 2 2 uAD = ⎛ −a ⎟ ⎜ ⎞ 1 ⎜ 2 ⎟ 2 0 ⎟ 2 ⎛ a⎞ ⎜ ⎟ h +⎜ ⎟ ⎝ 2 ⎝ h ⎠ uBD = ⎛a⎟ ⎜ ⎞ 1 2 2 0 ⎟ a⎞ 2 h +⎛ ⎟ ⎜h⎟ ⎜ ⎝ 2 ⎝ ⎠... permission in writing from the publisher Engineering Mechanics - Statics Chapter 6 Units Used: 3 kN = 10 N Given: a = 2 m M = 48 kN⋅ m b = 4m kN w1 = 8 m c = 2m kN w2 = 6 d = 6m m e = 3m Solution: Guesses Ax = 1 N Ay = 1 N MA = 1 N m Dx = 1 N Dy = 1 N By = 1 N Ey = 1 N Ex = 1 N Cy = 1 N Given Ay − w2 a − Dy = 0 a MA − w2 a − Dy a = 0 2 −w1 ( b + c) 2 E y − w1 −w1⎛ ⎜ 2 d+e 2 + B y b − E y( b + c) = 0 + Cy =... means, without permission in writing from the publisher Engineering Mechanics - Statics Chapter 6 F 2 = 500 N θ = 45 deg a = 0 .2 m b = 0 .2 m c = 0 .4 m d = 0 .4 m Solution: Guesses Ax = 1 N Ay = 1 N Cx = 1 N Cy = 1 N Given Ax − Cx = 0 Ay + Cy − F1 − F2 = 0 F1 a − Ay 2 a + Ax d = 0 ⎛ Ax ⎞ ⎜ ⎟ ⎜ Ay ⎟ ⎜ ⎟ = Find ( Ax , Ay , Cx , Cy) ⎜ Cx ⎟ ⎜C ⎟ ⎝ y⎠ −F 2 b + Cy( b + c) − Cx d = 0 ⎛ Ax ⎞ ⎜ ⎟ ⎛ 500 ⎞ ⎜ ⎟ ⎜ Ay... publisher Engineering Mechanics - Statics Chapter 6 Given: a = 55 ft f = 12 ft b = 10 ft F 1 = 5 kip c = 48 ft F 2 = 4 kip d = 5 ft F 3 = 2 kip e = 2 ft Solution: Member CE: ΣMC = 0; −F 3 e + E y( e + c) = 0 e Ey = F3 e+c E y = 80 lb ΣF y = 0; Cy − F3 + Ey = 0 Cy = F 3 − E y Cy = 1 920 lb Member ABD: ΣMA = 0; −F 1 a − F2( d + a) − Cy( a + d + b) + By( a + d) = 0 By = ΣF y = 0; F 1 a + F2( d + a) +... = 0; 3R − P = 0 Thus, P = 9T Man and seat: ΣF y = 0; T + P − W1 − W2 = 0 10T = W1 + W2 T = W1 + W2 10 P = 9T T = 16 lb P = 144 lb Seat: ΣF y = 0; P − N − W2 = 0 N = P − W2 N = 1 34 lb Problem 6-7 1 Determine the horizontal and vertical components of force that pins A and C exert on the frame Given: F = 500 N a = 0.8 m d = 0 .4 m 541 © 20 07 R C Hibbeler Published by Pearson Education, Inc., Upper Saddle... D a 2 2 a +b FBD + a 2 2 2 a +b +c FAD = 0 535 © 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 b −F CD − 2 −b + 2 a... +⎜ ⎟ ⎝ 2 ⎝ h ⎠ uBD = ⎛a⎟ ⎜ ⎞ 1 2 2 0 ⎟ a⎞ 2 h +⎛ ⎟ ⎜h⎟ ⎜ ⎝ 2 ⎝ ⎠ uAC = ⎛−a ⎟ ⎜ ⎞ 1 ⎜ 2 2 b ⎟ a⎞ 2 2 h +b +⎛ ⎟ ⎜ h ⎟ ⎜ ⎝ 2 ⎝ ⎠ uBC = ⎛a⎟ ⎜ ⎞ 1 2 2 b ⎟ a⎞ 2 2 h +b +⎛ ⎟ ⎜h⎟ ⎜ ⎝ 2 ⎝ ⎠ Guesses F AB = 1 lb B y = 1 lb Ax = 1 lb F AC = 1 lb Ay = 1 lb F AD = 1 lb F BC = 1 lb F BD = 1 lb F CD = 1 lb 533 © 20 07 R C Hibbeler Published by Pearson Education, Inc., Upper Saddle River, NJ All rights reserved . the publisher. Engineering Mechanics - Statics Chapter 6 F CD − b a 2 b 2 + F BD − b− a 2 b 2 + c 2 + F AD b b 2 c 2 + F DF −+ 0= F 3 − F DE − c b 2 c 2 + F DF − c− a 2 b 2 + c 2 + F AD + 0= Joint. Vectors hc 2 a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 −= u AD 1 h 2 a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + a− 2 0 h ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ = u BD 1 h 2 a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + a 2 0 h ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ = u AC 1 h 2 b 2 + a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + a 2 − b h ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ = u BC 1 h 2 b 2 + a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + a 2 b h ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ = Guesses. from the publisher. Engineering Mechanics - Statics Chapter 6 Positive (T) Negative (C) F BD F AD F AF ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 144 .3− 20 4. 1 72. 2 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ lb= Problem 6-6 3 Determine the force