46-53

PROBLEM 6.46 A floor truss is loaded as shown. Determine the force in members CF, EF, and EG.

SOLUTION FBD Truss:

ΣM K = 0: (1 m ) 1 kN + 2 (1 kN ) + 3 (1.5 kN ) + 4 ( 2 kN ) + 5 ( 2 kN ) + 6 (1 kN ) − 6 Ay  = 0

A y = 5.25 kN ΣM E = 0: (1 m ) 1( 2 kN ) + 2 (1 kN − 5.25 kN )  + ( 0.5 m ) FCE = 0

FCF = 13.0 kN

FBD Section:

FCF = 13.00 kN T W

ΣM F = 0: (1 m ) 1( 2 kN ) + 2 ( 2 kN ) + 3 (1 kN − 5.25 kN )  − ( 0.5 m ) FEG = 0 FEG = −13.5 kN

FEG = 13.50 kN C W

ΣFy = 0: 5.25 kN − 1 kN − 2 kN − 2 kN −

FEF =

5 = 0.5590 kN 4

1 FEF = 0 5

FEF = 559 N T W

PROBLEM 6.47 A floor truss is loaded as shown. Determine the force in members FI, HI, and HJ.

SOLUTION FBD Truss:

ΣFx = 0: K x = 0

(

)

ΣM A = 0: (1m ) 6 K y − 0.5 kN − 5 (1 kN ) − 4 (1 kN ) − 3 (1.5 kN ) − 2 ( 2 kN ) − 1( 2 kN )  = 0  

K y = 3.75 kN FBD Section:

ΣM I = 0:

(1 m )( 3.75 kN − .5 kN ) − (.5 m ) FHJ

=0

FHJ = 6.5 kN ΣFy = 0:

FHJ = 6.50 kN C W

1 FHI − 1 kN − 0.5 kN + 3.75 kN = 0 5

FHI = −2.25 5 kN ΣFy = 0: −

FHI = 5.03 kN C W

2 FHI − FFI + FHJ = 0 5

FFI = 2 ( 2.25 kN ) + 6.50 kN

FFI = 11.00 kN T W

PROBLEM 6.48 A pitched flat roof truss is loaded as shown. Determine the force in members CE, DE, and DF.

SOLUTION FBD Truss:

ΣFx = 0: A x = 0

By load symmetry: A y = I y = 8 kips

FBD Section:

ΣM D = 0: ( 7 ft )( 2 kips − 8 kips ) + ( 3 ft ) ( FCE ) = 0 FCE = 14 kips

FCE = 14.00 kips T W

ΣM E = 0: ( 7 ft ) 1( 4 kips ) + 2 ( 2 kips − 8 kips ) 

( 4.5 ft ) FDF =

7 FDF = 0 51.25

8 51.25 kips 4.5

ΣFy = 0: 8 kips − 2 kips − 4 kips +

FDE = −1.692 kips

FDF = 12.73 kips C W

1.5 8 51.25 kips − 51.25 4.5

3 FDE = 0 58

FDE = 1.692 kips C W

PROBLEM 6.49 A pitched flat roof truss is loaded as shown. Determine the force in members EG, GH, and HJ.

SOLUTION FBD Truss:

By load symmetry: A y = I y = 8 kips

FBD Section:

ΣM = 0: ( 7 ft )( 8 kips − 2 kips ) − ( 6 ft ) FEG = 0 FEG = 7.00 kips T W ΣFx = 0:

FHJ = ΣFy = 0:

7 FHJ − 7.00 kips = 0 51.25

FHJ = 7.16 kips C W

51.25 kips

1.5 51.25

(

)

51.25 kips − FHG + ( 8 − 2 ) kips = 0

FHG = 7.50 kips C W

PROBLEM 6.50 A Howe scissors roof truss is loaded as shown. Determine the force in members DF, DG, and EG.

SOLUTION FBD Truss:

Σ FX = 0: A x = 0 By symmetry: A y = L y = 6 kN

FBD Section:

Notes:

15 m 6 2 5 5 yD = ⋅ = m 3 2 3 2 2 yE = ⋅ 1 = m 3 3 5 yF − yD = m 6 yG = 1 m yF =

yD − yG =

2 m 3

PROBLEM 6.50 CONTINUED ΣM D = 0: (1 m )

6 FEG + ( 2 m )( 2 kN ) + ( 4 m )(1 kN − 6 kN ) = 0 37

FEG =

8 37 kN 3

FEG = 16.22 kN T W

 1   3  ΣM A = 0: ( 2 m )( 2 kN ) + ( 4 m )( 2 kN ) − ( 6 m )  FDG  − (1 m )  FDG  = 0  10   10 

FDG = ΣFx = 0:

4 10 kN 3

6 3 12 FEG − FDG − FDF = 0 13 37 10

FDF = 13 kN

FDG = 4.22 kN C W 16 − 4 −

12 FDF = 0 13

FDF = 13.00 kN C W

PROBLEM 6.51 A Howe scissors roof truss is loaded as shown. Determine the force in members GI, HI, and HJ.

SOLUTION FBD Truss:

Σ Fx = 0: A x = 0 By symmetry: A y = L y = 6 kN

FBD Section:

Notes:

so

2 3 2 = 3

yI =

m

yH

⋅

5 5 = m 2 3

yH − yI = 1 m

PROBLEM 6.51 CONTINUED  12  ΣM I = 0: ( 4 m )( 6 kN − 1 kN ) − ( 2 m )( 2 kN ) − (1 m )  FHJ  = 0 13   FHJ =

52 kN 3

FHJ = 17.33 kN C W

 6  ΣM H = 0: ( 4 m )( 6 kN − 1 kN ) − ( 2 m )( 2 kN ) − (1 m )  FGI  = 0  37 

FGI =

8 37 kN 3

ΣM L = 0: ( 2 m )( 2 kN ) − ( 4 m ) FHI = 0

60

FGI = 16.22 kN T W FHI = 1.000 kN T W

PROBLEM 6.52 A Fink roof truss is loaded as shown. Determine the force in members BD, CD, and CE.

SOLUTION FBD Truss:

Σ Fx = 0: A x = 0

By symmetry: A y = K y = 3.6 kips FBD Section:

 16  ft  FCE + ( 5 ft )(1.2 kips ) + (10 ft )( 0.6 kips ) − (10 ft )( 3.6 kips ) = 0 ΣM D = 0:   3  FCE = 4.50 kips T W 4  ΣM A = 0: ( 6 ft )  FCD  − ( 5 ft )(1.2 kips ) = 0 5  FCD = 1.250 kips T W ΣFy = 0: ( 3.6 − 0.6 ) kips − 1.2 kips + FBD = 5.95 kips

4 8 (1.25 kips ) − FBD = 0 5 17 FBD = 5.95 kips C W

PROBLEM 6.53 A Fink roof truss is loaded as shown. Determine the force in members FH, FG, and EG.

SOLUTION FBD Truss:

Σ Fx = 0: A x = 0

By symmetry: A y = K y = 3.6 kips FBD Section:

ΣM F = 0: (15 ft )( 3.6 − .6 ) kips − (10 ft )(1.2 kips ) − ( 5 ft ) (1.2 kips ) − ( 8ft ) FEG = 0 FEG = 3.375 kips

FEG = 3.38 kips T W

 8  ΣM K = 0: ( 5 ft )(1.2 kips ) + (10 ft )(1.2 kips ) − (12 ft )  FFG  = 0  73 

FFG = ΣFy = 0

3 73 kips 16

8  3  8 73 kips  − FFH − 1.2 kips − 1.2 kips − 0.6 kip + 3.6 kips = 0  16 73   17 FFH = 4.4625 kips

62

FFG = 1.602 kips T W

FFH = 4.46 kips C W