Mecanica

PROBLEM 6.19
Determine the force in each of the members located to the left of member FG for the scissor roof truss shown. State whether each member is in tension or compression.

SOLUTION
FBD Truss:

ΣM A = 0: ( 6 ft ) 6 Ly − 3 (.5 kip ) − 2 (1 kip ) − 1(1 kip )  = 0  
L y = 0.75 kip

ΣFy = 0: Ay − 0.5 kip − 1 kip − 1 kip − 0.5 kip + 0.75 kip = 0
A y = 2.25 kips

ΣFx = 0: A x = 0Joint FBDs:
Joint A:
1.75 kips F F = AC = AB 1 5 8

FAB = 4.95 kips C FAC = 3.91 kips T FBC = 0 so FCE = 3.91 kips T

FAC = 3.913 kips By inspection of joint C: and Joint B:
ΣFy = 0:

FCE = FAC

4.95 kips F − 1 kip − BD = 0 2 2

FBD = 3.536 kips
ΣFx = 0: ( 4.95 kips − 3.536 kips )

FBD = 3.54 kips C
1 − FBE = 0 2

FBE = 1.000 kip Joint E:
ΣFx = 0:

FBE = 1.000 kip C

2 [FEG − 3.913 kips] + 1 kip = 0 5

FEG = 2.795 kips
ΣFy = 0:

FEG = 2.80 kips T

1 ( 2.795 kips − 3.913 kips ) + FDE = 0 5

FDE = 0.500 kip T

PROBLEM 6.19 CONTINUED
Joint D:
ΣFx = 0: ΣFy = 0: 1 2 ( 3.536 kips ) − ( FDF + FDG ) = 0 2 5

1 1 ( 3.536 kips ) − 1.5 kips + ( FDG − FDF ) = 0 2 5

Solving:

FDF = 2.516 kips FDG = 0.280 kip

so

FDF = 2.52 kips C FDG = 0.280 kip CPROBLEM 6.20
Determine the force in member FG and in each of the members located to the right of member FG for the scissor roof truss shown. State whether each member is in tension or compression.

SOLUTION
FBD Truss:   ΣM A = 0: ( 6 ft )  6Ly − 3 ( 0.5 kip ) − 2 (1 kip ) − 1(1 kip )  = 0 L y = 0.75 kip Inspection of joints K , J , and I , in order, shows that

FJK = 0 FIJ = 0 FHI = 0and that
Joint FBDs:

FIK = FKL ; FHJ = FJL and FGI = FIK
FJL = 2.1213 kips

0.75 F F = JL = KL 1 8 5

FJL = 2.12 kips C FKL = 1.677 kips T FHJ = 2.12 kips C

FKL = 1.6771 kips
and, from above:

FGI = FIK = 1.677 kips T
ΣFx = 0: ΣFy = 0:
2 1 ( FFH + FGH ) − ( 2.1213 kips ) = 0 5 2 1 1 ( FGH + FFH ) + ( 2.1213 kips ) = 0 5 2

Solving:

FFH = 2.516 kips FGH = − 0.8383 kipsFFH = 2.52 kips C FGH = 0.838 kips T

PROBLEM 6.20 CONTINUED
ΣFx = 0:
2 ( FDF − 2.516 kips ) = 0 5

FDF = 2.52 kips C

ΣFy = 0: FFG − 0.5 kip +

1 ( 2 )( 2.516 kips ) = 0 5

FFG = 1.750 kips T

PROBLEM 6.21
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
Joint FBDs:

FDF F 1.5 kN = EF = 2.29 2.29 1.2 FDF = FEF = 2.8625 kN FDF = 2.86 kN T FEF = 2.86 kN C FBD F 2.8625 kN = DE = = 1.25 kN 2.21 0.6 2.29 FBD = 2.7625 kN FBD = 2.76 kN T FDE = 0.750 kN C By symmetry of joint A vs. joint F FAB = 2.86 kN T FAC = 2.86 kN T ΣFx = 0: 2.7625 kN − 2.21 4 ( 2.8625 kN ) + FBE = 0 2.29 5 FBE = 0ΣFy = 0: FBC − 0.6 ( 2.8625 kN ) = 0; 2.29 FBC = 0.750 kN C 2.21 ( 2.8625 kN ) − FCE = 0 2.29

ΣFx = 0:

FCE = 2.7625 kN FCE = 2.76 kN C

ΣFy = 0: FCH − 0.75 kN −

0.6 ( 2.8625 kN ) = 0 2.21 FCH = 1.500 kN C

PROBLEM 6.21 CONTINUED
ΣFx = 0: 2.7625 kN − 2.21 4 ( 2.8625 kN ) − FHE = 0 2.29 5 FHE = 0 ΣFy = 0: FEJ − 0.75 kN − 0.6 ( 2.8625 kN ) = 0 2.29 FEJ = 1.500 kN

PROBLEM 6.22
For thetower and loading of Prob. 6.21 and knowing that FCH = FEJ = 1.5 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. SOLUTION
1.5 kN F F = JL = KL 1.2 3.03 3.03
Joint FBDs:

FJL = FKL = 3.7875 kN

FJL = 3.79 kN T FKL = 3.79 kN C

By symmetry:

FGH = 3.79 kN T FGI = 3.79 kN C ΣFx= 0: 2.97 ( 3.7875 kN ) − FHJ = 0 3.03 FHJ = 3.71 kN T

FHJ = 3.7125 kN ΣFy = 0: FJK −

0.6 ( 3.7875 kN ) − 1.5 kN = 0 3.03 FJK = 2.25 kN C

Knowing FHE = 0; by symmetry

FHK = 0 FHI = 2.25 kN C

ΣFx = 0:

2.97 ( 3.7875 kN ) − FIK = 0 3.03

FIK = 3.7125 FIK = 3.71 kN C

ΣFy = 0: FIN − 2.25 kN −

0.6 ( 3.7875 kN ) = 0 3.03 FIN = 3.00 kN C

Knowing that FHK = 0, by symmetry...