Using thethod of joints, me determine the force in each ... · Using the method of joints, determine the force inh member of the eac truss shown. State whether each member is in tension

Using the method of joints, determine the force in each member of the truss

shown. State whether each member is in tension or compression.

SOLUTION

Free body: Entire truss:

0: 0 0y y yF BΣ = = =B

0: (3.2 m) (48 kN)(7.2 m) 0C xM BΣ = − − = 108 kN 108 kNx xB = − =B

0: 108 kN 48 kN 0xF CΣ = − + = 60 kNC = 60 kN=C

Free body: Joint B:

108 kN

5 4 3BCAB FF = =

180.0 kNABF T=

144.0 kNBCF T=

Free body: Joint C:

60 kN

13 12 5AC BCF F

= =

156.0 kNACF C=

144 kN (checks)BCF =static

s

Using the method of joints, determine the force in each member of the

truss shown. State whether each member is in tension or compression.

SOLUTION

Reactions:

0: 1260 lbAMΣ = =C

0: 0x xFΣ = =A

0: 960 lby yFΣ = =A

Joint B:

300 lb

12 13 5BCAB FF = =

720 lbABF T=

780 lbBCF C=

Joint A:

40: 960 lb 0

5y ACF FΣ = − − =

1200 lbACF = 1200 lbACF C=

30: 720 lb (1200 lb) 0 (checks)

5xFΣ = − =

static

s

Using the method of joints, determine the force in each member of the truss

shown. State whether each member is in tension or compression.

SOLUTION

2 2

2 2

3 1.25 3.25 m

3 4 5 m

AB

BC

= + =

= + =

Reactions:

0: (84 kN)(3 m) (5.25 m) 0AM CΣ = − =

48 kN=C

0: 0x xF A CΣ = − =

48 kNx =A

0: 84 kN 0y yF AΣ = = =

84 kNy =A

Joint A:

120: 48 kN 0

13x ABF FΣ = − =

52 kNABF = + 52.0 kNABF T=

50: 84 kN (52 kN) 0

13y ACF FΣ = − − =

64.0 kNACF = + 64.0 kNACF T=

Joint C:

48 kN

5 3BCF

= 80.0 kNBCF C= static

s

Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression.

SOLUTION

Free body: Truss:

From the symmetry of the truss and loading, we find 600 lb= =C D

Free body: Joint B: 300 lb

2 15BCAB FF

= =

671 lbABF T= 600 lbBCF C=

Free body: Joint C:

30: 600 lb 0

5y ACF FΣ = + =

1000 lbACF = − 1000 lbACF C=

40: ( 1000 lb) 600 lb 0

5x CDF FΣ = − + + = 200 lbCDF T=

From symmetry:

1000 lb , 671 lb ,AD AC AE ABF F C F F T= = = = 600 lbDE BCF F C= =

static

s


SOLUTION

Reactions:

0: (24) (4 2.4)(12) (1)(24) 0D yM FΣ = − + − =

4.2 kipsy =F

0: 0x xFΣ = =F

0: (1 4 1 2.4) 4.2 0yF DΣ = − + + + + =

4.2 kips=D

Joint A:

0: 0x ABF FΣ = = 0ABF =

0 : 1 0y ADF FΣ = − − =

1 kipADF = − 1.000 kipADF C=

Joint D: 8

0: 1 4.2 017y BDF FΣ = − + + =

6.8 kipsBDF = − 6.80 kipsBDF C=

150: ( 6.8) 0

17x DEF FΣ = − + =

6 kipsDEF = + 6.00 kipsDEF T= static

s

PROBLEM (Continued)

Joint E:

0 : 2.4 0y BEF FΣ = − =

2.4 kipsBEF = + 2.40 kipsBEF T=

Truss and loading symmetrical about c .L

static

s


SOLUTION

2 2

2 2

5 12 13 ft

12 16 20 ft

AD

BCD

= + =

= + =

Reactions: 0: 0x xF DΣ = =

0: (21 ft) (693 lb)(5 ft) 0E yM DΣ = − = 165 lby =D

0: 165 lb 693 lb 0yF EΣ = − + = 528 lb=E

Joint D: 5 4

0: 013 5x AD DCF F FΣ = + = (1)

12 30: 165 lb 0

13 5y AD DCF F FΣ = + + = (2)

Solving Eqs. (1) and (2) simultaneously,

260 lbADF = − 260 lbADF C=

125 lbDCF = + 125 lbDCF T= static

s

PROBLEM (Continued)

Joint E:

5 40: 0

13 5x BE CEF F FΣ = + = (3)

12 30: 528 lb 0

13 5y BE CEF F FΣ = + + = (4)

Solving Eqs. (3) and (4) simultaneously,

832 lbBEF = − 832 lbBEF C=

400 lbCEF = + 400 lbCEF T=

Joint C:

Force polygon is a parallelogram (see Fig. 6.11, p. 209).

400 lbACF T=

125.0 lbBCF T=

Joint A: 5 4

0: (260 lb) (400 lb) 013 5x ABF FΣ = + + =

420 lbABF = − 420 lbABF C=

12 30: (260 lb) (400 lb) 0

13 50 0 (Checks)

yFΣ = − =

=

static

s

Using the method of joints, determine the force in each member of the truss shown.

State whether each member is in tension or compression.

SOLUTION


0: 2(5 kN) 0x xF CΣ = + =

10 kN 10 kNxxC = − =C

0: (2 m) (5 kN)(8 m) (5 kN)(4 m) 0CM DΣ = − − =

30 kN 30 kND = + =D

0: 30 kN 0 30 kN 30 kNyy y yF C CΣ = + = = − =C

Free body: Joint A:

5 kN

4 117AB ADF F

= =

20.0 kNABF T=

20.6 kNADF C=

Free body: Joint B:

10: 5 kN 0

5x BDF FΣ = + =

5 5 kNBDF = − 11.18 kNBDF C=

20: 20 kN ( 5 5 kN) 0

5y BCF FΣ = − − − =

30 kNBCF = + 30.0 kNBCF T=

static

s

PROBLEM(Continued)

Free body: Joint C:

0: 10 kN 0x CDF FΣ = − =

10 kNCDF = + 10.00 kNCDF T=

0: 30 kN 30 kN 0 (checks)yFΣ = − =

static

s

PROBLEM 6.8


SOLUTION

Reactions:

0: 16 kNC xMΣ = =A

0: 9 kNy yFΣ = =A

0: 16 kNxFΣ = =C

Joint E:

3 kN

5 4 3BE DEF F= =

5.00 kNBEF T=

4.00 kNDEF C=

Joint B:

40: (5 kN) 0

5x ABF FΣ = − =

4 kNABF = + 4.00 kNABF T=

30: 6 kN (5 kN) 0

5y BDF FΣ = − − − =

9 kNBDF = − 9.00 kNBDF C=

Joint D:

30: 9 kN 0

5y ADF FΣ = − + =

15 kNADF = + 15.00 kNADF T=

40: 4 kN (15 kN) 0

5x CDF FΣ = − − − =

16 kNCDF = − 16.00 kNCDF C= static

s

Determine the force in each member of the Gambrel roof truss shown. State whether each member is in tension or compression.

SOLUTION

Free body: Truss:

0: 0x xFΣ = =H

Because of the symmetry of the truss and loading,

1total load

2y= =A H

1200 lby= =A H

Free body: Joint A:

900 lb

5 4 3ACAB FF = = 1500 lbAB C=F

1200 lbAC T=F

Free body: Joint C:

BC is a zero-force member.

0BC =F 1200 lbCEF T=

Free body: Joint B:

24 4 40: (1500 lb) 0

25 5 5x BD BEF F FΣ = + + =

or 24 20 30,000 lbBD BEF F+ = − (1)

7 3 3

0: (1500) 600 025 5 5y BD BEF F FΣ = − + − =

or 7 15 7,500 lbBD BEF F− = − (2)static

s

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

755

PROBLEM 6.9 (Continued)

Multiply Eq. (1) by 3, Eq. (2) by 4, and add:

100 120,000 lbBDF = − 1200 lbBDF C=

Multiply Eq. (1) by 7, Eq. (2) by –24, and add:

500 30,000 lbBEF = − 60.0 lbBEF C=

Free body: Joint D:

24 24

0: (1200 lb) 025 25x DFF FΣ = + =

1200 lbDFF = − 1200 lbDFF C=

7 7

0: (1200 lb) ( 1200 lb) 600 lb 025 25y DEF FΣ = − − − − =

72.0 lbDEF = 72.0 lbDEF T=

Because of the symmetry of the truss and loading, we deduce that

=EF BEF F 60.0 lbEFF C=

EG CEF F= 1200 lbEGF T=

FG BCF F= 0FGF =

FH ABF F= 1500 lbFHF C=

GH ACF F= 1200 lbGHF T=

Note: Compare results with those of Problem 6.11.

static

s

Determine the force in each member of the Howe roof truss shown. State whether each member is in tension or compression.

SOLUTION

Free body: Truss:

0: 0x xFΣ = =H


1total load

2yA H= =

1200 lby= =A H

Free body: Joint A:

900 lb

5 4 3ACAB FF = = 1500 lbAB C=F

1200 lbAC T=F

Free body: Joint C:

BC is a zero-force member.

0BC =F 1200 lbCE T=F static

s


Free body: Joint B:

4 4 40: (1500 lb) 0

5 5 5x BD BCF F FΣ = + + =

or 1500 lbBD BEF F+ = − (1)

3 3 30: (1500 lb) 600 lb 0

5 5 5y BD BEF F FΣ = − + − =

or 500 lbBD BEF F− = − (2)

Add Eqs. (1) and (2): 2 2000 lbBDF = − 1000 lbBDF C=

Subtract Eq. (2) from Eq. (1): 2 1000 lbBEF = − 500 lbBEF C=

Free Body: Joint D:

4 40: (1000 lb) 0

5 5x DFF FΣ = + =

1000 lbDFF = − 1000 lbDFF C=

3 30: (1000 lb) ( 1000 lb) 600 lb 0

5 5y DEF FΣ = − − − − =

600 lbDEF = + 600 lbDEF T=


=EF BEF F 500 lbEFF C=

EG CEF F= 1200 lbEGF T=

FG BCF F= 0FGF =

FH ABF F= 1500 lbFHF C=

GH ACF F= 1200 lbGHF T= static

s

Determine the force in each member of the Pratt roof truss shown. State whether each member is in tension or compression.

SOLUTION

Free body: Truss:

0: 0x xF AΣ = =

Due to symmetry of truss and load,

1 total load 21 kN

2yA H= = =

Free body: Joint A:

15.3 kN

37 35 12ACAB FF = =

47.175 kN 44.625 kNAB ACF F= = 47.2 kNABF C=

44.6 kNACF T=

Free body: Joint B:

From force polygon: 47.175 kN, 10.5 kNBD BCF F= = 10.50 kNBCF C=

47.2 kNBDF C= static

s


759


Free body: Joint C: 3

0: 10.5 05y CDF FΣ = − = 17.50 kNCDF T=

4

0: (17.50) 44.625 05x CEF FΣ = + − =

30.625 kNCEF = 30.6 kNCEF T=

Free body: Joint E: DE is a zero-force member. 0DEF =

Truss and loading symmetrical about .cL

static

s

Determine the force in each member of the Fink roof truss shown.

State whether each member is in tension or compression.

SOLUTION

Free body: Truss:

0: 0x xFΣ = =A


1 total load

2y = =A G

6.00 kNy = =A G

Free body: Joint A:

4.50 kN

2.462 2.25 1ACAB FF = =

11.08 kNABF C=

10.125 kNACF = 10.13 kNACF T=

Free body: Joint B:

3 2.25 2.250: (11.08 kN) 0

5 2.462 2.462x BC BDF F FΣ = + + = (1)

4 11.08 kN0: 3 kN 0

5 2.462 2.462BD

y BCF

F FΣ = − + + − = (2)

Multiply Eq. (2) by –2.25 and add to Eq. (1):

126.75 kN 0 2.8125

5 BC BCF F+ = = − 2.81 kNBCF C=

Multiply Eq. (1) by 4, Eq. (2) by 3, and add:

12 12(11.08 kN) 9 kN 0

2.462 2.462BDF + − =

9.2335 kNBDF = − 9.23 kNBDF C= static

s

PROBLEM (Continued)

Free body: Joint C:4 4

0: (2.8125 kN) 05 5y CDF FΣ = − =

2.8125 kN,CDF = 2.81 kNCDF T=

3 30: 10.125 kN (2.8125 kN) (2.8125 kN) 0

5 5x CEF FΣ = − + + =

6.7500 kNCEF = + 6.75 kNCEF T=


DE CDF F= 2.81 kNCDF T=

DF BDF F= 9.23 kNDFF C=

EF BCF F= 2.81 kNEFF C=

EG ACF F= 10.13 kNEGF T=

FG ABF F= 11.08 kNFGF C=

static

s

PROBLEM 6.13

Using the method of joints, determine the force in each member of the double-pitch roof truss shown. State whether each member is in tension or compression.

SOLUTION

Free body: Truss:

0: (18 m) (2 kN)(4 m) (2 kN)(8 m) (1.75 kN)(12 m)

(1.5 kN)(15 m) (0.75 kN)(18 m) 0AM HΣ = − − −

− − =

4.50 kN=H

0: 0

0: 9 0

9 4.50

x x

y y

y

F A

F A H

A

Σ = =Σ = + − =

= − 4.50 kNy =A

Free body: Joint A:

3.50 kN

2 157.8262 kN

ACAB

AB

FF

F C

= =

= 7.83 kNABF C=

7.00 kNACF T= static

s


Free body: Joint B:

2 2 10: (7.8262 kN) 0

5 5 2x BD BCF F FΣ = + + =

or 0.79057 7.8262 kNBD BCF F+ = − (1)

1 1 10: (7.8262 kN) 2 kN 0

5 5 2y BD BCF F FΣ = + − − =

or 1.58114 3.3541BD BCF F− = − (2)

Multiply Eq. (1) by 2 and add Eq. (2):

3 19.0065

6.3355 kNBD

BD

F

F

= −= − 6.34 kNBDF C=

Subtract Eq. (2) from Eq. (1):

2.37111 4.4721

1.8861 kNBC

BC

F

F

= −= − 1.886 kNBCF C=

Free body: Joint C:

2 10: (1.8861 kN) 0

5 2y CDF FΣ = − =

1.4911 kNCDF = + 1.491 kNCDF T=

1 10: 7.00 kN (1.8861 kN) (1.4911 kN) 0

2 5x CEF FΣ = − + + =

5.000 kNCEF = + 5.00 kNCEF T=

Free body: Joint D:

2 1 2 10: (6.3355 kN) (1.4911 kN) 0

5 2 5 5x DF DEF F FΣ = + + − =

or 0.79057 5.5900 kNDF DEF F+ = − (1)

1 1 1 20: (6.3355 kN) (1.4911 kN) 2 kN 0

5 2 5 5y DF DEF F FΣ = − + − − =

or 0.79057 1.1188 kNDF DEF F− = − (2)

Add Eqs. (1) and (2): 2 6.7088 kNDFF = −3.3544 kNDFF = − 3.35 kNDFF C=

Subtract Eq. (2) from Eq. (1): 1.58114 4.4712 kNDEF = −2.8278 kNDEF = − 2.83 kNDEF C= static

s


Free body: Joint F:

1 20: (3.3544 kN) 0

2 5x FGF FΣ = + =

4.243 kNFGF = − 4.24 kNFGF C=

1 10: 1.75 kN (3.3544 kN) ( 4.243 kN) 0

5 2y EFF FΣ = − − + − − =

2.750 kNEFF = 2.75 kNEFF T=

Free body: Joint G:

1 1 10: (4.243 kN) 0

2 2 2x GH EGF F FΣ = − + =

or 4.243 kNGH EGF F− = − (1)

1 1 10: (4.243 kN) 1.5 kN 0

2 2 2y GH EGF F FΣ = − − − − =

or 6.364 kNGH EGF F+ = − (2)

Add Eqs. (1) and (2): 2 10.607GHF = −

5.303GHF = − 5.30 kNGHF C=

Subtract Eq. (1) from Eq. (2): 2 2.121 kNEGF = −

1.0605 kNEGF = − 1.061 kNEGF C=

Free body: Joint H:

3.75 kN

1 1EHF = 3.75 kNEHF T=

We can also write 3.75 kN

12GHF

= 5.30 kN (Checks)GHF C=static

s

PROBLEM 6.14

The truss shown is one of several supporting an advertising panel. Determine the force in each member of the truss for a wind load equivalent to the two forces shown. State whether each member is in tension or compression.

SOLUTION


0: (800 N)(7.5 m) (800 N)(3.75 m) (2 m) 0FM AΣ = + − =

2250 NA = + 2250 N=A

0: 2250 N 0y yF FΣ = + =

2250 N 2250 Ny yF = − =F

0: 800 N 800 N 0x xF FΣ = − − + = 1600 N 1600 Nx xF = + =F

Joint D:

800 N

8 15 17DE BDF F= =

1700 NBDF C=

1500 NDEF T=

Joint A:

2250 N

15 17 8ACAB FF= =

2250 NABF C=

1200 NACF T=

Joint F:

0: 1600 N 0x CFF FΣ = − =

1600 NCFF = + 1600 NCFF T=

0: 2250 N 0y EFF FΣ = − =

2250 NEFF = + 2250 NEFF T= static

s


Joint C: 8

0: 1200 N 1600 N 017x CEF FΣ = − + =

850 NCEF = − 850 NCEF C=

150: 0

17y BC CEF F FΣ = + =

15 15( 850 N)

17 17BC CEF F= − = − −

750 NBCF = + 750 NBCF T=

Joint E: 8

0: 800 N (850 N) 017x BEF FΣ = − − + =

400 NBEF = − 400 NBEF C=

150: 1500 N 2250 N (850 N) 0

170 0 (checks)

yFΣ = − + =

=

static

s

PROBLEM 6.23

The portion of truss shown represents the upper part of a power transmission line tower. For the given loading, determine the force in each of the members located above HJ. State whether each member is in tension or compression.

SOLUTION

Free body: Joint A:

1.2 kN

2.29 2.29 1.2ACAB FF = = 2.29 kNABF T=

2.29 kNACF C=

Free body: Joint F:

1.2 kN

2.29 2.29 2.1= =DF EFF F

2.29 kNDFF T=

2.29 kNEFF C=

Free body: Joint D:

2.29 kN

2.21 0.6 2.29BD DEF F= = 2.21 kNBDF T=

0.600 kNDEF C=

Free body: Joint B:

4 2.210: 2.21 kN (2.29 kN) 0

5 2.29x BEF FΣ = + − =

0BEF =

3 0.60: (0) (2.29 kN) 0

5 2.29y BCF FΣ = − − − =

0.600 kNBCF = − 0.600 kNBCF C= static

s


783


Free body: Joint C:

2.21

0: (2.29 kN) 02.29x CEF FΣ = + =

2.21 kNCEF = − 2.21 kNCEF C=

0.6

0: 0.600 kN (2.29 kN) 02.29y CHF FΣ = − − − =

1.200 kNCHF = − 1.200 kNCHF C=

Free body: Joint E:

2.21 4

0: 2.21 kN (2.29 kN) 02.29 5

Σ = − − =x EHF F

0EHF =

0.6

0: 0.600 kN (2.29 kN) 0 02.29y EJF FΣ = − − − − =

1.200 kNEJF = − 1.200 kNEJF C=

static

s

PROBLEM 6.24

For the tower and loading of Problem 6.23 and knowing that FCH = FEJ = 1.2 kN C and FEH = 0, determine the force in member HJ and in each of the members located between HJ and NO. State whether each member is in tension or compression.

PROBLEM 6.23 The portion of truss shown represents the upper part of a power transmission line tower. For the given loading, determine the force in each of the members located above HJ. State whether each member is in tension or compression.

SOLUTION

Free body: Joint G:

1.2 kN

3.03 3.03 1.2GH GIF F

= = 3.03 kNGHF T=

3.03 kNGIF C=

Free body: Joint L:

1.2 kN

3.03 3.03 1.2= =JL KLF F

3.03 kNJLF T=

3.03 kNKLF C=

Free body: Joint J:

2.970: (3.03 kN) 0

3.03Σ = − + =x HJF F

2.97 kNHJF T=

0.60: 1.2 kN (3.03 kN) 0

3.03y JKF F= − − − =

1.800 kNJKF = − 1.800 kNJKF C=

Free body: Joint H:

4 2.970: 2.97 kN (3.03 kN) 0

5 3.03x HKF FΣ = + − =

0HKF =

0.6 30: 1.2 kN (3.03) kN (0) 0

3.03 5y HIF FΣ = − − − − =

1.800 kNHIF = − 1.800 kNHIF C= static

s


Free body: Joint I:

2.970: (3.03 kN) 0

3.03x IKF FΣ = + =

2.97 kNIKF = − 2.97 kNIKF C=

0.60: 1.800 kN (3.03 kN) 0

3.03y INF FΣ = − − − =

2.40 kNINF = − 2.40 kNINF C=

Free body: Joint K:

4 2.970: 2.97 kN (3.03 kN) 0

5 3.03x KNF FΣ = − + − =

0KNF =

0.6 30: (3.03 kN) 1.800 kN (0) 0

3.03 5y KOF FΣ = − − − − =

2.40 kNKOF = − 2.40 kNKOF C=

static

s

PROBLEM 6.25

Solve Problem 6.23 assuming that the cables hanging from the right side of the tower have fallen to the ground.

PROBLEM 6.23 The portion of truss shown represents the upper part of a power transmission line tower. For the given loading, determine the force in each of the members located above HJ. State whether each member is in tension or compression.

SOLUTION

Zero-Force Members:

Considering joint F, we note that DF and EF are zero-force members:

0= =DF EFF F

Considering next joint D, we note that BD and DE are zero-force members:

0BD DEF F= =

Free body: Joint A:

1.2 kN

2.29 2.29 1.2ACAB FF = = 2.29 kNABF T=

2.29 kNACF C=

Free body: Joint B:

4 2.210: (2.29 kN) 0

5 2.29Σ = − =x BEF F

2.7625 kNBEF = 2.76 kNBEF T=

0.6 30: (2.29 kN) (2.7625 kN) 0

2.29 5y BCF FΣ = − − − =

2.2575 kNBCF = − 2.26 kNBCF C= static

s


Free body: Joint C:

2.210: (2.29 kN) 0

2.29Σ = + =x CEF F

2.21 kNCEF C=

0.60: 2.2575 kN (2.29 kN) 0

2.29y CHF FΣ = − − − =

2.8575 kNCHF = − 2.86 kNCHF C=

Free body: Joint E:

4 40: (2.7625 kN) 2.21 kN 0

5 5x EHF FΣ = − − + =

0EHF =

3 30: (2.7625 kN) (0) 0

5 5y EJF FΣ = − + − =

1.6575 kNEJF = + 1.658 kNEJF T=

static

s

Documents

Using thethod of joints, me determine the force in each ... · Using the method of joints, determine the force inh member of the eac truss shown. State whether each member is in tension