Problem 6-2

Determine the force in each member of the truss and state if the members are in tension orcompression.

Units Used:

kN 103 N=

Given:

P1 8 kN=

P2 10 kN=

Solution:

θ 45 deg=

Initial Guesses:

FAB 1 kN= FAD 1 kN= FDB 1 kN=

FDC 1 kN= FCB 1 kN=

Given

Joint A: FAB FAD cos θ( )+ 0=

P1− FAD sin θ( )− 0=

Joint D: FDB cos θ( ) FAD cos θ( )− FDC cos θ( )+ 0=

FAD FDB+ FDC−( )sin θ( ) P2− 0=

Joint C: FCB FDC sin θ( )+ 0=

PositivemeansTension,NegativemeansCompression

FAB

FAD

FDB

FDC

FCB

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

Find FAB FAD, FDB, FDC, FCB,( )=

FAB

FAD

FDB

FDC

FCB

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

8

11.31−

7.07

18.38−

13

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

kN=

© 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.

For the exclusive use of adopters of the Hibbeler series of books.

Problem 6-4

The truss, used to support a balcony, is subjected to the loading shown. Approximate each jointas a pin and determine the force in each member. State whether the members are in tension orcompression.

Units Used:

kip 103 lb=

Given:

P1 800 lb=

P2 0 lb=

a 4 ft=

θ 45 deg=

Solution:

Initial Guesses

FAB 1 lb= FAD 1 lb= FDC 1 lb=

FBC 1 lb= FBD 1 lb= FDE 1 lb=

Given

Joint A: FAB FAD cos θ( )+ 0=

P1− FAD sin θ( )− 0=

Joint B: FBC FAB− 0=

P2− FBD− 0=

Joint D : FDC FAD−( )cos θ( ) FDE+ 0=

FDC FAD+( )sin θ( ) FBD+ 0=

FAB

FAD

FBC

FBD

FDC

FDE

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

Find FAB FAD, FBC, FBD, FDC, FDE,( )=

© 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.

For the exclusive use of adopters of the Hibbeler series of books.

Positive means Tension,Negative means Compression

FAB

FAD

FBC

FBD

FDC

FDE

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

800

1131−

800

0

1131

1600−

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

lb=

Problem 6-7

Determine the force in each member of the truss and state if the members are in tension orcompression.

Units Used:

kN 103 N=

Given:

F1 3 kN=

F2 8 kN=

F3 4 kN=

F4 10 kN=

a 2 m=

b 1.5 m=

Solution: θ atanba⎛⎜⎝⎞⎟⎠

=

Initial Guesses

FBA 1 kN= FBC 1 kN= FAC 1 kN=

FAF 1 kN= FCD 1 kN= FCF 1 kN=

FDF 1 kN= FED 1 kN= FEF 1 kN=

Given

Joint B F1 FBC+ 0=

F2− FBA− 0=

Joint C FCD FBC− FAC cos θ( )− 0=

F3− FAC sin θ( )− FCF− 0=

Joint E FEF− 0=

Joint D FCD− FDF cos θ( )− 0=

F4− FDF sin θ( )− FED− 0=

Joint F FAF− FEF+ FDF cos θ( )+ 0=

© 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.

For the exclusive use of adopters of the Hibbeler series of books.

FCF FDF sin θ( )+ 0=

FBA

FAF

FDF

FBC

FCD

FED

FAC

FCF

FEF

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

Find FBA FAF, FDF, FBC, FCD, FED, FAC, FCF, FEF,( )=

Positive means tension,Negative means compression.

FBA

FAF

FDF

FBC

FCD

FED

FAC

FCF

FEF

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

8.00−

4.17

5.21

3.00−

4.17−

13.13−

1.46−

3.13−

0.00

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

kN=

Problem 6-9

The maximum allowable tensile force in the members of the truss is Tmax, and the maximumallowable compressive force is Cmax. Determine the maximum magnitude P of the two loadsthat can be applied to the truss.

Given:

Tmax 1500 lb=

Cmax 800 lb=

Solution:

Set P 1 lb=

Initial Guesses

FAB 1 lb= FAD 1 lb= FBD 1 lb=

FBC 1 lb= FCD 1 lb=

Given

Joint B FBC FAB−( ) 1

2P+ 0=

FBD− FAB FBC+( ) 1

2− 0=

Joint D FCD FAD−( ) 4

170=

FBD P− FAD FCD+( ) 1

17− 0=

Joint C FBC−1

2FCD

4

17− 0=

FAB

FBC

FAD

FCD

FBD

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

Find FAB FBC, FAD, FCD, FBD,( )=

FAB

FBC

FAD

FCD

FBD

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

0.471−

1.886−

1.374

1.374

1.667

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

lb=

© 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.

For the exclusive use of adopters of the Hibbeler series of books.

Now find the critical load

P1 PTmax

max FAB FBC, FAD, FCD, FBD,( )= P1 900 lb=

P2 PCmax

min FAB FBC, FAD, FCD, FBD,( )= P2 424.264 lb=

P min P1 P2,( )= P 424.3 lb=

Problem 6-10

Determine the force in each member of the truss and state if the members are in tension orcompression.

Given:

P1 0 lb=

P2 1000 lb=

a 10 ft=

b 10 ft=

Solution:

θ atanba⎛⎜⎝⎞⎟⎠

=

Initial Guesses:

FAB 1 lb= FAG 1 lb= FBG 1 lb=

FBC 1 lb= FDC 1 lb= FDE 1 lb=

FEG 1 lb= FEC 1 lb= FCG 1 lb=

Given

Joint B FBC FAB− 0=

FBG P1− 0=

Joint G FCG FAG−( )cos θ( ) FEG+ 0=

FCG FAG+( )− sin θ( ) FBG− 0=

Joint C FDC FBC− FCGcos θ( )− 0=

FEC FCGsin θ( )+ P2− 0=

Joint E FDE cos θ( ) FEG− 0=

FEC− FDE sin θ( )− 0=

Joint D FDE− cos θ( ) FDC− 0=

© 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.

For the exclusive use of adopters of the Hibbeler series of books.

FAB

FBC

FEG

FAG

FDC

FEC

FBG

FDE

FCG

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

Find FAB FBC, FEG, FAG, FDC, FEC, FBG, FDE, FCG,( )=

FAB

FBC

FEG

FAG

FDC

FEC

FBG

FDE

FCG

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

333

333

667−

471−

667

667

0

943−

471

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

lb= Positive means tension,Negative means compression.

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:

kN 103 N=

Given:

F1 50 kN=

F2 50 kN=

F3 50 kN=

a 4 m=

b 3 m=

Solution:

Initial Guesses

Gy 1 kN= FJI 1 kN=

FJE 1 kN= FDE 1 kN=

Given

Entire Truss

F1− b F2 2b( )− F3 3b( )− Gy 6b( )+ 0=

Section

FDE− FJI− 0= FJE Gy+ 0=

Gy 2b( ) FDE a− 0=

Gy

FJI

FJE

FDE

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

Find Gy FJI, FJE, FDE,( )= Gy 50 kN=

FJI

FJE

FDE

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

75−

50−

75

⎛⎜⎜⎝

⎞⎟⎟⎠

kN=

Positive means Tension,Negative means Compression

© 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.

For the exclusive use of adopters of the Hibbeler series of books.

Problem 6-33

The roof truss supports the vertical loading shown. Determine the force in members BC, CK, andKJ and state if these members are in tension or compression.

Units Used:

kN 103 N=

Given:

F1 4 kN=

F2 8 kN=

a 2 m=

b 3 m=

Solution:

Initial Guesses

Ax 1 kN= Ay 1 kN=

FBC 1 kN= FCK 1 kN=

FKJ 1 kN=

Given

Ax 0=

F2 3a( ) F1 4a( )+ Ay 6a( )− 0=

FKJ2b3

⎛⎜⎝

⎞⎟⎠

Ax2b3

⎛⎜⎝

⎞⎟⎠

+ Ay 2a( )− 0=

FKJ Ax+3a

b2 9a2+

⎛⎜⎝

⎞⎟⎠FBC+ 0=

FCK Ay+b

b2 9a2+

⎛⎜⎝

⎞⎟⎠FBC+ 0=

Ax

Ay

FKJ

FCK

FBC

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

Find Ax Ay, FKJ, FCK, FBC,( )=

Ax

Ay

FKJ

FCK

FBC

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

0.00

6.67

13.33

0.00

14.91−

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

kN= Positive (T)Negative (C)

© 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.

For the exclusive use of adopters of the Hibbeler series of books.

Problem 6-37

Determine the force in members BF, BG, and AB, and state if the members are in tension orcompression.

Units Used:

kN 103 N=

Given:

F1 5 kN= F4 10 kN=

F2 10 kN= a 4 m=

F3 5 kN= b 4 m=

Solution: θ atanab⎛⎜⎝⎞⎟⎠

=

Inital Guesses

FAB 1 kN= FBG 1 kN= FBF 1 kN=

Given

F1 F2+ F4+ FBG cos θ( )+ 0=

F1− 3a F22a− F4 a− FAB b+ 0=

FBF− 0=

FAB

FBG

FBF

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

Find FAB FBG, FBF,( )=

FAB

FBG

FBF

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

45

35.4−

0

⎛⎜⎜⎝

⎞⎟⎟⎠

kN= Positive (T)Negative (C)

© 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.

For the exclusive use of adopters of the Hibbeler series of books.