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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

ΣFx = 0:

C y = 240 N

B y = 75 N

Bx = 0

Joint FBDs: Joint B:

FAB F 75 N = BC = 5 4 3 N

Joint C:

FBC = 100.0 N T W

By inspection: N

FAB = 125.0 N C W

FAC = 260 N C W

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 W FAC = 4.00 kips T W Joint A: FAB 1.6 kips = 8.5 4 FAB = 3.40 kips T W

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

C y = 4 kips

B y = 10 kips

ΣFx = 0: C x = 0

Joint FBDs: Joint C:

FAC F 4 kips = BC = 17 15 8 FAC = 8.50 kips T W FBC = 7.50 kips C W

Joint B:

By inspection:

FAB 10 kips = 5 4

FAB = 12.50 kips C W

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 −

FCD =

FAD = 3.48 kN T W

2.1 FAD = 0 2.9

2.1 (3.48 kN) 2.9

FCD = 2.52 kN C W

Joint C:

By inspection:

FAC = 3.36 kN C W FBC = 2.52 kN C W

Joint B:

ΣFy = 0:

4 FAB − 0.9 kN = 0 5

FAB = 1.200 kN T W

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 ΣFx = 0 : C x = 0

FBD Truss:

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 W ΣFx = 0:

Joint C:

2 FAB − FBC = 0 5

ΣFy = 0: 6 kN − ΣFx = 0: 6 kN −

FBC = 6.00 kN C W

3 FAC = 0 5

FAC = 10.00 kN C W

4 FAC + FCD = 0 5

FCD = 2.00 kN T W

Joint A:  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 W FAD = FAC = 10.00 kN C W FDE = FBC = 6.00 kN C W

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 or compression.

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 2.4 kips F F = CD = BC 12 18.5 18.5

Joint FBDs: Joint C:

FCD = 3.70 kips T W FBC = 3.70 kips C W or: ΣFx = 0: FBC = FCD

ΣFy = 0: 2.4 kips − 2

6 FBC = 0 18.5

same answers

Joint D: ΣFx = 0: 3 kips +

17.5 4 ( 3.70 kips ) − FAD = 0 18.5 5

FAD = 8.125 kips ΣFy = 0:

FAD = 8.13 kips T W

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

FBD = 6.08 kips C W

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 W

PROBLEM 6.7 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:

ΣFy = 0: Ay − 480 N = 0

A y = 480 N

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

Dx = 0 Ax = 0

Joint FBDs:

Joint A: 480 N F F = AB = AC 6 2.5 6.5

FAB = 200 N C W FAC = 520 N T W

Joint B:

200 N F F = BE = BC 2.5 6 6.5

FBE = 480 N C W FBC = 520 N T W

PROBLEM 6.7 CONTINUED Joint C: FCD = FCE = 520 N T W

By inspection:

ΣFx = 0:

Joint D:

2.5 ( 520 N ) − FDE = 0 6.5

FDE = 200 N C W

PROBLEM 6.8 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 E = 0: ( 9 ft ) Fy − ( 6.75 ft )( 4 kips ) − (13.5 ft )( 4 kips ) = 0

Fy = 9 kips ΣFy = 0: − E y + 9 kips = 0

E y = 9 kips

ΣFx = 0: − Ex + 4 kips + 4 kips = 0

E x = 8 kips FEC = 9.00 kips T W

By inspection of joint E:

FEF = 8.00 kips T W FAB = 0 W

By inspection of joint B:

FBD = 0 W

Joint FBDs:

Joint F: ΣFx = 0:

4 FCF − 8 kips = 0 5

FCF = 10.00 kips C W

3 (10 kips) = 0 5

FDF = 6.00 kips T W

ΣFy = 0: FDF −

Joint C: ΣFx = 0: 4 kips −

4 (10 kips ) + FCD = 0 5 FCD = 4.00 kips T W

ΣFy = 0: FAC − 9 kips +

3 (10 kips ) = 0 5 FAC = 3.00 kips T W

PROBLEM 6.8 CONTINUED

Joint A:

ΣFx = 0: 4 kips −

4 FAD = 0 5

FAD = 5.00 kips C W