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  • Publicado : 7 de septiembre de 2010
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PROBLEM 6.1
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
FBD Truss:

ΣM B = 0: ( 6.25 m ) C y − ( 4 m )( 315 N ) = 0
ΣFy = 0: By − 315 N + C y = 0

C y = 240 N

B y = 75 N

ΣFx = 0:

Bx = 0

Joint FBDs: Joint B:
N

FAB F 75 N = BC = 5 4 3

FAB = 125.0 N C FBC = 100.0 N TJoint C:
N

By inspection:

FAC = 260 N C

PROBLEM 6.2
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
FBD Truss:

ΣM A = 0: (14 ft ) Cx − ( 7.5 ft )( 5.6 kips ) = 0 ΣFx = 0: − Ax + Cx = 0 ΣFy = 0: Ay − 5.6 kips = 0

C x = 3 kips

A x = 3 kips A y = 5.6 kips

Joint FBDs:Joint C: FBC F 3 kips = AC = 5 4 3 FBC = 5.00 kips C FAC = 4.00 kips T Joint A: FAB 1.6 kips = 8.5 4 FAB = 3.40 kips T

PROBLEM 6.3
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
FBD Truss:
ΣM B = 0: ( 6 ft )( 6 kips ) − ( 9 ft ) C y = 0 ΣFy = 0: By − 6 kips − C y = 0 ΣFx = 0: C x = 0

C y =4 kips

B y = 10 kips

Joint FBDs:
Joint C: FAC F 4 kips = BC = 17 15 8 FAC = 8.50 kips T FBC = 7.50 kips C

Joint B:

By inspection:

FAB = 12.50 kips C

FAB 10 kips = 5 4

PROBLEM 6.4
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
FBD Truss:
ΣM B = 0: (1.5 m ) C y + ( 2 m)(1.8 kN ) − 3.6 m ( 2.4 kN ) = 0

C y = 3.36 kN
ΣFy = 0: By + 3.36 kN − 2.4 kN = 0

B y = 0.96 kN

Joint FBDs:
Joint D: ΣFy = 0:

2 FAD − 2.4 kN = 0 2.9 ΣFx = 0: FCD − 2.1 FAD = 0 2.9

FAD = 3.48 kN T

FCD = Joint C:

2.1 (3.48 kN) 2.9

FCD = 2.52 kN C

By inspection:

FAC = 3.36 kN C FBC = 2.52 kN C

Joint B:

ΣFy = 0:

4 FAB − 0.9 kN = 0 5

FAB = 1.200 kN T PROBLEM 6.5
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
FBD Truss:
ΣFx = 0 : C x = 0
By symmetry: C y = D y = 6 kN

Joint FBDs:
Joint B:

ΣFy = 0: − 3 kN +

1 FAB = 0 5

FAB = 3 5 = 6.71 kN T
ΣFx = 0: 2 FAB − FBC = 0 5

FBC = 6.00 kN C

Joint C:

ΣFy = 0: 6 kN − ΣFx = 0: 6 kN− Joint A:

3 FAC = 0 5

FAC = 10.00 kN C FCD = 2.00 kN T

4 FAC + FCD = 0 5

 1  3  ΣFy = 0: − 2  3 5 kN  + 2  10 kN  − 6 kN = 0 check 5 5    

By symmetry:

FAE = FAB = 6.71 kN T FAD = FAC = 10.00 kN C FDE = FBC = 6.00 kN C

PROBLEM 6.6
Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension orcompression.

SOLUTION
FBD Truss:
ΣM A = 0: ( 25.5 ft ) C y + ( 6 ft )( 3 kips ) − ( 8 ft )( 9.9 kips ) = 0

C y = 2.4 kips
ΣFy = 0: Ay + 2.4 kips − 9.9 kips = 0

A y = 7.4 kips
ΣFx = 0: − Ax + 3 kips = 0

A x = 3 kips Joint FBDs:
Joint C: 2.4 kips F F = CD = BC 12 18.5 18.5 FCD = 3.70 kips T FBC = 3.70 kips C or: ΣFx = 0: FBC = FCD ΣFy = 0: 2.4 kips − 2 6 FBC = 0 18.5

same answers

JointD: ΣFx = 0: 3 kips +

17.5 4 ( 3.70 kips ) − FAD = 0 18.5 5
FAD = 8.13 kips T

FAD = 8.125 kips ΣFy = 0:

6 3 ( 3.7 kips ) + (8.125 kips ) − FBD = 0 18.5 5
FBD = 6.075 kips FBD = 6.08 kips C

PROBLEM 6.6 CONTINUED
Joint A:
ΣFx = 0: −3 kips + FAB = 4.375 kips

4 4 (8.125 kips ) − FAB = 0 5 5
FAB = 4.38 kips C

PROBLEM 6.7
Using the method of joints, determine the force in eachmember of the truss shown. State whether each member is in tension or compression.

SOLUTION
FBD Truss:

ΣFy = 0: Ay − 480 N = 0

A y = 480 N Dx = 0 Ax = 0

ΣM A = 0: ( 6 m ) Dx = 0 ΣFx = 0: − Ax = 0

Joint FBDs:

Joint A: 480 N F F = AB = AC 6 2.5 6.5
FAB = 200 N C FAC = 520 N T

Joint B:

200 N F F = BE = BC 2.5 6 6.5

FBE = 480 N C FBC = 520 N T

PROBLEM 6.7 CONTINUED...
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