PROBLEM 6.39

The portion of a power line transmission tower shown consists of nine members and is supported by a ball and socket at B and short links at C, D, and E. Determine the force in each of the members for the given loading.

SOLUTION

FBD Truss:

Joint FBDs:

( )( ) ( )0: 1 m 200 N 0 200 NBD z zM EΣ = − = =E k

( )( )0: 1 m 1200 N 0x z zM E DΣ = − − =

( )1200 N 200 N 1000 N 1000 Nz ZD = − = =D k

( )0: 1000 N 200 N 0 1200 Nz z zF BΣ = − − = =B k

( )( ) ( ) ( )0: .5 m 1200 N 1 m 0 600 NBz y yM CΣ = − = =C j

( )0: 200 N 0 200 Nx x xF BΣ = − = =B i

0: 1200 N 0y y yF B CΣ = + − =

( )1200 N 600 N 600 Ny yB = − =B j

0: 200 N 01.5

EAz

FFΣ = − = 300 N T EAF =

.50: 0 100 N

1.5x ED EA EDF F F FΣ = + = = −

100 N C EDF =

0: 0 200 N1.5

EAy EC EC

FF F FΣ = − − = = −

200 N C ECF =

0: 1000 N 01.5

ADz

FFΣ = − = 1500 N T ADF =

0.5 10: 100 N 01.5 2x AD CDF F FΣ = − − =

1500 N2 100 N 400 2

3CDF = − = −

566 N C CDF =

0: 01.52

CD ADy BD

F FF FΣ = − − − =

400 1000 600 NBDF = + − = − 600 N C BDF =

PROBLEM 6.39 CONTINUED

0: 600 N 200 N 400 N 0yFΣ = − − =

0: 01.25

ACz

FFΣ = = 0 ACF =

.50: 400 N1.25x BCF FΣ = − + 0 0ACF =

400 N T BCF =

.50: 200 N 400 N 01.25x ABF FΣ = − − − =

1200 1.25 NABF = 1342 N C ABF =

PROBLEM 6.40

The truss shown consists of 18 members and is supported by a ball and socket at A, two short links at B, and one short link at G. (a) Check that this truss is a simple truss, that it is completely constrained, and that the reactions at its supports are statically determinate. (b) For the given loading, determine the force in each of the six members joined at E.

SOLUTION

(a) To check for simple truss, start with ABDE and add three members at a time which meet at a single joint, thus successively adding joints F, G, H, and C, to complete the truss. This is, therefore, a simple truss.

There are six reaction force components, none of which are in-line, so they are determined by the six equilibrium equations. Constraints prevent motion. Truss is completely constrained and

statically determinate (b) FBD Truss:

( )( ) ( )( ) ( )0: 5.04 m 550 N 5.5 m 0 504 NBC y yM AΣ = + = = −A j

( )( ) ( )( )0: 5.5 m 480 N 4.8 m 550 N 0 0BF z zM AΣ = + − = =A

By inspection of joint : C 0DC BC GCF F F= = =

By inspection of joint :A 504 N T AE yF A= − =

0AD zF A= =

By inspection of joint : H 0DHF =

480 N C EHF =

By inspection of joint :F 0 EFF =

PROBLEM 6.40 CONTINUED

Then, since ED is the only non-zero member at D, not in the plane BDG, 0 DEF =

Joint E: 4.80: 480 N 07.3z EGF FΣ = − = 730 N T EGF =

5.040: 504 N 0 746 N7.46y BE BEF F FΣ = − − = = − 746 N C BEF =

PROBLEM 6.41

The truss shown consists of 18 members and is supported by a ball and socket at A, two short links at B, and one short link at G. (a) Check that this truss is a simple truss, that it is completely constrained, and that the reactions at its supports are statically determinate. (b) For the given loading, determine the force in each of the six members joined at G.

SOLUTION

(a) See part (a) solution 6.40 above (b) FBD Truss:

( ) ( )( )0: 4.8 m 5.04 m 480 N 0 504 NAB y yM GΣ = + = = −G

By inspection of joint : C 0 CGF =

By inspection of joint : H 550 N C HGF =

By inspection of joint :F 0 FGF =

Joint G:

5.5 5.50: 550 N 0

7.46 7.30x DG EGF F FΣ = − − =

PROBLEM 6.41 CONTINUED

5.04 5.040: 504 N 07.46 6.96y DG BGF F FΣ = − − − =

4.8 4.80: 07.30 6.96z EG BGF F FΣ = + =

Solving: 696 NBGF = − 696 N C BGF =

0DGF = 0 DGF =

730 NEGF = 730 N T EGF =

PROBLEM 6.42

A Warren bridge truss is loaded as shown. Determine the force in members CE, DE, and DF.

SOLUTION

FBD Truss:

Section ABDC:

( )( ) ( )( ) ( )0: 15 m 3 kN 20 m 3 kN 25 m 0K yM AΣ = + − =

4.2 kNy =A

( ) ( )( ) ( )( )0: 6 m 2.5 m 3 kN 7.5 m 4.2 kN 0D CEM FΣ = + − =

4 kNCEF = 4.00 kN T CEF =

120: 4.2 kN 3 kN 013y DEF FΣ = − − =

1.3 kNDEF = 1.300 kN T DEF =

50: 0

13x DF DE CEF F F FΣ = + + =

( ) ( )5 1.3 kN 4 kN 4.5 kN13DFF = − − = −

4.50 kN C DFF =

PROBLEM 6.43

A Warren bridge truss is loaded as shown. Determine the force in members EG, FG, and FH.

SOLUTION

FBD Truss:

Section FBD:

0: 0x xFΣ = =K

( ) ( )( ) ( )( )0: 25 m 10 m 3 kN 5 m 3 kN 0A yM KΣ = − − =

1.8 kNy =K

( )( ) ( )0: 10 m 1.8 kN 6 m 0G FHM FΣ = + =

3 kNFHF = − 3.00 kN C FHF =

( )( ) ( )( )0: 12.5 m 1.8 kN 6 m 0F EGM FΣ = − =

3.75 kNEGF = 3.75 kN T EGF =

120: 1.8 kN 013y FGF FΣ = + =

1.95 kNFGF = − 1.950 kN C FGF =

PROBLEM 6.44

A parallel chord Howe truss is loaded as shown. Determine the force in members CE, DE, and DF.

SOLUTION

FBD Truss:

FBD Section:

0: 0x xFΣ = =A

( ) ( ) ( ) ( )

( ) ( ) ( )

0: 4 ft 1 0.3 kip 2 0.6 kip 3 1 kip

4 0.6 kip 5 0.3 kip 6 0.3 kip 6 0

M

y

M

A

Σ = + +

+ + + − =

1.7 kipsy =A

( )( ) ( ) 40: 4 ft 3 kips 1.7 kips 2 ft 04.1D CEM F Σ = − + =

2.87 kipsCEF = 2.87 kips T CEF =

( )( ) ( )( )0: 8 ft 3 kips 1.7 kips 4 ft 0.3 kipEMΣ = − +

( ) 42 ft 04.1 DFF − =

5.125 kipsDFF = − 5.13 kips C DFF =

( )4 40: 04.1 17.21x DF CE DEF F F FΣ = + + =

( )17.21 5.125 2.87 kips 2.28 kips4.1DE DEF F= − − + =

2.28 kips T DEF =

PROBLEM 6.45

A parallel chord Howe truss is loaded as shown. Determine the force in members GI, GJ, and HJ.

SOLUTION

FBD Truss:

FBD Section:

( ) ( ) ( ) ( )24 ft 4 ft 0.3 kip 8 ft 0.6 kip 12 ft 1 kipA yM MΣ = − − −

( ) ( ) ( )16 ft 0.6 kip 20 ft 0.3 kip 24 ft 0.3 kip 0− − − =

1.7 kipsy =M

( ) ( ) 48 ft 1.7 0.3 kips 4 ft 0.3 kip 2 ft 04.1J GIM F Σ = − − − =

5.125 kips 5.3 kips T GI GIF F= =

( ) ( )12 ft 1.7 0.3 kips 8 ft 0.3 kipGMΣ = − −

( ) 44 ft 0.6 kip 2 ft 04.1 HJF − + =

6.15 kips 6.15 kips C HJF = − =

( )4 46.15 5.125 kips4.1 4.94x GJF FΣ = − −

1.235 kips T GJF =