PROBLEM 6.43

Determine the force in members CD and DF of the truss shown.

SOLUTION

Reactions:

0: (12 kN)(4.8 m) (12 kN)(2.4 m) (9.6 m) 0JM BΣ = + − =

9.00 kNB =

0: 9.00 kN 12.00 kN 12.00 kN 0yF JΣ = − − + =

15.00 kNJ =

Member CD:

0: 9.00 kN 0y CDF FΣ = + = 9.00 kNCDF C=

Member DF:

0: (1.8 m) (9.00 kN)(2.4 m) 0C DFM FΣ = − = 12.00 kNDFF T= static

s

PROBLEM 6.44

Determine the force in members FG and FH of the truss shown.

SOLUTION

Reactions:

0: (12 kN)(4.8 m) (12 kN)(2.4 m) (9.6 m) 0JM BΣ = + − =

9.00 kNB =

0: 9.00 kN 12.00 kN 12.00 kN 0yF JΣ = − − + =

15.00 kNJ =

Member FG:

30: 12.00 kN 15.00 kN 0

5y FGF FΣ = − − + =

5.00 kNFGF T=

Member FH:

0: (15.00 kN)(2.4 m) (1.8 m) 0G FHM FΣ = − =

20.0 kNFHF T= static

s

PROBLEM 6.47

Determine the force in members DF, EF, and EG of the truss shown.

SOLUTION

3tan

4β =

Reactions:

0= =A N

Member DF: 3

0: (16 kN)(6 m) (4 m) 05E DFM FΣ = + − =

40 kNDFF = + 40.0 kNDFF T=

Member EF: + 0: (16 kN)sin cos 0EFF Fβ βΣ = − =

16 tan 16(0.75) 12 kNEFF β= = = 12.00 kNEFF T=

Member EG: 4

0: (16 kN)(9 m) (3 m) 05F EGM FΣ = + =

60 kNEGF = − 60.0 kNEGF C= static

s

Determine the force in members GI, GJ, and HI of the

truss shown.

SOLUTION

Reactions:

0= =A N

Member GI: + 0: (16 kN)sin sin 0GIF Fβ βΣ = + =

16 kNGIF = − 16.00 kNGIF C=

Member GJ: 4

0: (16 kN)(9 m) (3 m) 05I GJM FΣ = − − =

60 kNGJF = − 60.0 kNGJF C=

Member HI: 3

0: (16 kN)(9 m) (4 m) 05G HIM FΣ = − + =

60 kNHIF = + 60.0 kNHIF T=

static

s

Determine the force in members AD, CD, and CE of the

truss shown.

SOLUTION

Reactions:

0: 36(2.4) (13.5) 20(9) 20(4.5) 0kM BΣ = − + + = 26.4 kN=B

0: 36 0 36 kNx x xF KΣ = − + = =K

0: 26.4 20 20 0 13.6 kNy y yF KΣ = − − + = =K

0: 36(1.2) 26.4(2.25) (1.2) 0C ADM FΣ = − − =

13.5 kNADF = − 13.5 kNADF C=

80: (4.5) 0

17A CDM F Σ = =

0CDF =

150: (2.4) 26.4(4.5) 0

17D CEM F Σ = − =

56.1 kNCEF = + 56.1 kNCEF T= static

s

Determine the force in members DG, FG, and FH of the

truss shown.

SOLUTION

See the solution to Problem 6.49 for free-body diagram and analysis to determine the reactions at the supports B and K.

26.4 kN=B ; 36.0 kNx =K ; 13.60 kNy =K

0: 36(1.2) 26.4(6.75) 20(2.25) (1.2) 0F DGM FΣ = − + − =

75 kNDGF = − 75.0 kNDGF C=

80: 26.4(4.5) (4.5) 0

17D FGM F Σ = − + =

56.1 kNFGF = + 56.1 kNFGF T=

15

0: 20(4.5) 26.4(9) (2.4) 017G FHM F Σ = − + =

69.7 kNFHF = + 69.7 kNFHF T= static

s

PROBLEM 6.51

A stadium roof truss is loaded as shown. Determine the force in members AB, AG, and FG.

SOLUTION

We pass a section through members AB, AG, and FG, and use the free body shown.

400: (6.3 ft) (1.8 kips)(14 ft)

41

(0.9 kips)(28 ft) 0

G ABM F Σ = − − =

8.20 kipsABF = + 8.20 kipsABF T=

30: (28 ft) (1.8 kips)(28 ft)

5

(1.8 kips)(14 ft) 0

D AGM F Σ = − +

+ =4.50 kipsAGF = + 4.50 kipsAGF T=

0: (9 ft) (1.8 kips)(12 ft) (1.8 kips)(26 ft)

(0.9 kips)(40 ft) 0A FGM FΣ = − − −

− =11.60 kipsFGF = − 11.60 kipsFGF C= 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.

828

PROBLEM 6.52

A stadium roof truss is loaded as shown. Determine the force in members AE, EF, and FJ.

SOLUTION

We pass a section through members AE, EF, and FJ, and use the free body shown.

2 2

80: (9 ft) (1.8 kips)(12 ft) (1.8 kips)(26 ft) (0.9 kips)(40 ft) 0

8 9F AEM F

Σ = − − − = +

17.46 kipsAEF = + 17.46 kipsAEF T=

0: (9 ft) (1.8 kips)(12 ft) (1.8 kips)(26 ft) (0.9 kips)(40 ft) 0A EFM FΣ = − − − − =

11.60 kipsEFF = − 11.60 kipsEFF C=

0: (8 ft) (0.9 kips)(8 ft) (1.8 kips)(20 ft) (1.8 kips)(34 ft) (0.9 kips)(48 ft) 0E FJM FΣ = − − − − − =

18.45 kipsFJF = − 18.45 kipsFJF C= static

s

PROBLEM 6.53

Determine the force in members CD and DF of the truss shown.

SOLUTION

5tan 22.62

12α α= = °

5 12sin cos

13 13α α= =

Member CD:

0: (9 m) (10 kN)(9 m) (10 kN)(6 m) (10 kN)(3 m) 0I CDM FΣ = + + + =

20 kNCDF = − 20.0 kNCDF C=

Member DF:

0: ( cos )(3.75 m) (10 kN)(3 m) (10 kN)(6 m) (10 kN)(9 m) 0C DFM F αΣ = + + + =

cos 48 kNDFF α = −

1248 kN 52.0 kN

13DF DFF F = − = −

52.0 kNDFF C= static

s

PROBLEM 6.54

Determine the force in members CE and EF of the truss shown.

SOLUTION

Member CE:

0: (2.5 m) (10 kN)(3 m) (10 kN)(6 m) 0F CEM FΣ = − − =

36 kNCEF = + 36.0 kNCEF T=

Member EF:

0: (6 m) (10 kN)(6 m) (10 kN)(3 m) 0I EFM FΣ = + + =

15 kNEFF = − 15.00 kNEFF C=

static

s

The truss shown was designed to support the roof of a food market. For the given loading, determine the force in members FG, EG, and EH.

SOLUTION

Reactions at supports. Because of the symmetry of the loading,

10,

2x yA A O= = = total load 4.48 kN= =A O

We pass a section through members FG, EG, and EH, and use the free body shown.

1.75 mSlope Slope

6 mFG FI= =

5.50 mSlope

2.4 mEG =

0: (0.6 kN)(7.44 m) (1.24 kN)(3.84 m)EMΣ = +

(4.48 kN)(7.44 m)

6(4.80 m) 0

6.25 FGF

−

− =

5.231 kNFGF = − 5.23 kNFGF C=

0: (5.50 m) (0.6 kN)(9.84 m)G EHM FΣ = +

(1.24 kN)(6.24 m) (1.04 kN)(2.4 m)

(4.48 kN)(9.84 m) 0

+ +− = 5.08 kNEHF T=

5.50 1.750: ( 5.231 kN) 4.48 kN 0.6 kN 1.24 kN 1.04 kN 0

6.001 6.25y EGF FΣ = + − + − − − =

0.1476 kNEGF = − 0.1476 kNEGF C= static

s

PROBLEM 6.56

The truss shown was designed to support the roof of a food market. For the given loading, determine the force in members KM, LM, and LN.

SOLUTION

Because of symmetry of loading,1

load2

=O 4.48 kN=O

We pass a section through KM, LM, LN, and use free body shown.

3.840: (3.68 m)

4

(4.48 kN 0.6 kN)(3.6 m) 0

M LNM F Σ = + − =

3.954 kNLNF = − 3.95 kNLNF C=

0: (4.80 m) (1.24 kN)(3.84 m)

(4.48 kN 0.6 kN)(7.44 m) 0L KMM FΣ = − −

+ − =

5.022 kNKMF = + 5.02 kNKMF T=

4.80 1.120: ( 3.954 kN) 1.24 kN 0.6 kN 4.48 kN 0

6.147 4y LMF FΣ = + − − − + =

1.963 kNLMF = − 1.963 kNLMF C=

static

s

PROBLEM 6.57

A Polynesian, or duopitch, roof truss is loaded as shown. Determine the force in members DF, EF, and EG.

SOLUTION

Free body: Truss:

0: 0x xF NΣ = =

0: (200 lb)(8 ) (400 lb)(7 6 5 )+(350 lb)(4 ) (300 lb)(3 2 ) (8 ) 0NM a a a a a a a a A aΣ = + + + + + + − =

1500 lb=A

0: 1500 lb 200 lb 3(400 lb) 350 lb 3(300 lb) 150 lb 0y yF NΣ = − − − − − + =

1300 lb 1300 lbyN = =N

We pass a section through DF, EF, and EG, and use the free body shown.

(We apply DFF at F.)

2 2

0: (200 lb)(18 ft) (400 lb)(12 ft) (400 lb)(6 ft) (1500 lb)(18 ft)

18(4.5 ft) 0

18 4.5

E

DF

M

F

Σ = + + −

− = +

3711 lb 3710 lbDF DFF F C= − =

0: (18 ft) (400 lb)(6 ft) (400 lb)(12 ft) 0A EFM FΣ = − − =

400 lbEFF = + 400 lbEFF T=

0: (4.5 ft) (1500 lb)(18 ft)+(200 lb)(18 ft) (400 lb)(12 ft) (400 lb)(6 ft) 0F EGM FΣ = − + + =

3600 lb 3600 lbEG EGF F T= + = static

s

PROBLEM 6.58

A Polynesian, or duopitch, roof truss is loaded as shown. Determine the force in members HI, GI, and GJ.

SOLUTION

See solution of Problem 6.57 for reactions:

1500 lb=A , 1300 lb=N

We pass a section through HI, GI, and GJ, and use the free body shown.

(We apply HIF at H.)

2 2

60: (8.5 ft) (1300 lb)(24 ft) (300 lb)(6 ft)

6 4

(300 lb)(12 ft) (300 lb)(18 ft) (150 lb)(24 ft) 0

G HIM F

Σ = + − + − − − =

2375.4 lb 2375 lbHI HIF F C= − =

0: (1300 lb)(18 ft) (300 lb)(6 ft) (300 lb)(12 ft)

(150 lb)(18 ft) (4.5 ft) 0I

GJ

M

F

Σ = − −− − =

3400 lb 3400 lbGJ GJF F T= + =

2 2

4 60: ( 2375.4 lb) 3400 lb 0

5 6 4x GIF FΣ = − − − − =

+

1779.4 lbGIF = − 1779 lbGIF C= static

s

Determine the force in members DE and DF of the truss shown

when P = 20 kips.

SOLUTION

Reactions:

2.5P= =C K

7.5tan

1822.62

β

β

=

= °

Member DE:

0: (2.5 )(6 ft) (12 ft) 0A DEM P FΣ = − =

1.25DEF P= +

For 20 kips,P = 1.25(20) 25 kipsDEF = + = + 25.0 kipsDEF T=

Member DF:

0: (12 ft) (2.5 )(6 ft) cos (5 ft) 0E DFM P P F βΣ = − − =

12 15 cos 22.62 (5 ft) 0DFP P F− − ° =

0.65DFF P= −

For 20 kips,P = 0.65(20) 13 kipsDFF = − = −

13.00 kipsDFF C=

static

s

Determine the force in members EG and EF of the truss shown

when P = 20 kips.

SOLUTION

Reactions:

2.5

7.5tan

651.34

P

α

α

= =

=

= °

C K

Member EG:

0: (18 ft) 2.5 (12 ft) (6 ft) (7.5 ft) 0F EGM P P P FΣ = − − + =

0.8 ;EGF P= +

For 20 kips,P = 0.8(20) 16 kipsEGF = = + 16.00 kipsEGF T=

Member EF:

0: 2.5 (6 ft) (12 ft) sin 51.34 (12 ft) 0A EFM P P FΣ = − + ° =

0.320 ;EFF P= −

For 20 kips,P = 0.320(20) 6.4 kipsEFF = − = −

6.40 kipsEFF C=

static

s

Determine the force in members EH and GI of the

truss shown. (Hint: Use section aa.)

SOLUTION

Reactions:

0: 0x xF AΣ = =

0: 12(45) 12(30) 12(15) (90) 0P yM AΣ = + + − =

12 kipsy =A

0: 12 12 12 12 0yF PΣ = − − − + = 24 kips=P

0: (12 kips)(30 ft) (16 ft) 0G EHM FΣ = − − =

22.5 kipsEHF = − 22.5 kipsEHF C=

0: 22.5 kips 0x GIF FΣ = − = 22.5 kipsGIF T=

static

s

Determine the force in members HJ and IL of the truss

shown. (Hint: Use section bb.)

SOLUTION

See the solution to Problem 6.61 for free body diagram and analysis to determine the reactions at supports A and P.

0; 12.00 kipsx y= =A A ; 24.0 kips=P

0: (16 ft) (12 kips)(15 ft) (24 kips)(30 ft) 0L HJM FΣ = − + =

33.75 kipsHJF = − 33.8 kipsHJF C=

0: 33.75 kips 0x ILF FΣ = − =

33.75 kipsILF = + 33.8 kipsILF T=

static

s

The diagonal members in the center panels of the truss shown are very slender and can act only in tension; such members are known as counters. Determine the forces in the counters that are acting under the given loading.

SOLUTION

Free body: Truss:

0: 0x xF FΣ = =

0: 4.8(3 ) 4.8(2 ) 4.8 2.4 (2 ) 0H yM a a a a F aΣ = + + − − =

13.20 kipsyF = + 13.20 kips=F

0: 13.20 kips 3(4.8 kips) 2(2.4 kips) 0yF HΣ = + − − =

6.00 kipsH = + 6.00 kips=H

Free body: ABF:

We assume that counter BG is acting.

9.60: 13.20 2(4.8) 0

14.6y BGF FΣ = − + − =

5.475BGF = + 5.48 kipsBGF T=

Since BG is in tension, our assumption was correct.

Free body: DEH:

We assume that counter DG is acting.

9.60: 6.00 2(2.4) 0

14.6y DGF FΣ = − + − =

1.825DGF = + 1.825 kipsDGF T=

Since DG is in tension, O.K. static

s

The diagonal members in the center panels of the truss shown are very slender and can act only in tension; such members are known as counters. Determine the forces in the counters that are acting under the given loading.

SOLUTION

Free body: Truss:

0: 0x xF FΣ = =

0: 4.8(2 ) 4.8 2.4 2.4(2 ) 0G yM F a a a a aΣ = − + + − − =

7.20yF = 7.20 kipsF =

Free body: ABF:

We assume that counter CF is acting.

9.60: 7.20 2(4.8) 0

14.6y CFF FΣ = + − =

3.65CFF = + 3.65 kipsCFF T=

Since CF is in tension, O.K.

Free body: DEH:

We assume that counter CH is acting.

9.60: 2(2.4 kips) 0

14.6y CHF FΣ = − =

7.30CHF = + 7.30 kipsCHF T=

Since CH is in tension, O.K. static

s

Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.)

SOLUTION

Structure (a):

Nonsimple truss with 4,r = 12,m = 8n = so 16 2 .r m n+ = = Check for determinacy:

One can solve joint F for forces in EF, FG and then solve joint E for yE and force in DE.

This leaves a simple truss ABCDGH with

3, 9, 6 so 12 2r m n r m n= = = + = = Structure is completely constrained and determinate.

Structure (b):

Simple truss (start with ABC and add joints alphabetically to complete truss) with 4,r = 13,m = 8n =

so 17 2 16r m n+ = > = Constrained but indeterminate static

s

PROBLEM 6.71 (Continued)

Structure (c):

Nonsimple truss with 3,r = 13,m = 8n = so 16 2 .+ = =r m n To further examine, follow procedure in part (a) above to get truss at left.

Since 1 0≠F (from solution of joint F),

1AM aFΣ = 0≠ and there is no equilibrium.

Structure is improperly constrained.

static

s