1. Problem 6.167
12.5 kN
12.5 kN
12.5 kN
12.5 kN
Using the method of joints, determine the force in each member of the truss shown.
2 m
2 m
2 m A B C D
2.5 m G F E

2. Solving Problems on Your Own
12.5 kN
12.5 kN
12.5 kN
12.5 kN
Using the method of joints, determine the force in each member of the truss shown.
2 m
2 m
2 m A B C D
1. Draw the free-body diagram of
the entire truss, and use this
diagram to determine the
reactions at the supports.
2.5 m G F E
2. Locate a joint connecting only two members, and draw the
free-body diagram for its pin. Use this free-body diagram to
determine the unknown forces in each of the two members.
Assuming all members are in tension, if the answer obtained
from SFx = 0 and SFy = 0 is positive, the member is in
tension. A negative answer means the member is in
compression.

3. Solving Problems on Your Own
12.5 kN
12.5 kN
12.5 kN
12.5 kN
Using the method of joints, determine the force in each member of the truss shown.
2 m
2 m
2 m A B C D
2.5 m G F E
3. Next, locate a joint where the forces in only two of the
connected members are still unknown. Draw the free-body
diagram of the pin and use it as indicated in step 2 to
determine the two unknown forces.
4. Repeat this procedure until the forces in all the members of
the truss have been determined.

4. S MA = 0: E(2.5 m) - (12.5 kN)(2 m) - (12.5 kN)(4 m)
Problem Solution
12.5 kN
12.5 kN
12.5 kN
12.5 kN
Draw the free-body diagram of the entire truss, and use it to
determine reactions at the supports.
2 m
2 m
2 m A Ax B C D Ay
2.5 m G F E E +
S MA = 0: E(2.5 m) - (12.5 kN)(2 m) - (12.5 kN)(4 m)
- (12.5 kN)(6 m) = E = 60 kN +
S Fy = 0: Ay - (4)(12.5 kN) = Ay = 50 kN
S Fx = 0: Ax - E = Ax= 60 kN +

5. S Fy = 0: FGD - 12.5 kN = 0 FGD = 32.5 kN C S Fx = 0: FGD - FCD = 0
Problem Solution A B C D F G
2 m
12.5 kN
2.5 m E
60 kN
50 kN
Locate a joint connecting only two
members, and draw the free-body diagram for its pin. Use the free-body diagram to determine the unknown forces in each of the two members.
Joint D
12.5 kN
2.5
6.5 +
S Fy = 0: FGD kN = 0
FCD
FGD = 32.5 kN C
6.5
2.5 6
6.5 6
S Fx = 0: FGD - FCD = 0 +
FGD
FCD = 30 kN T

6. S F = 0: FCG = 0 S F = 0: FFG - 32.5 kN = 0 FFG = 32.5 kN C
Problem Solution A B C D F G
2 m
12.5 kN
2.5 m E
60 kN
50 kN
Next, locate a joint where the forces in only two of the connected members are still unknown. Draw the free-body diagram of the pin and use it to determine the two unknown forces.
Joint G
FCG
S F = 0: FCG = 0
32.5 kN
S F = 0: FFG kN = 0
FFG = 32.5 kN C
FFG

7. S Fx = 0: 30 kN - FCF cos b - FBC = 0 A B C D F G
2 m
12.5 kN
2.5 m E
60 kN
50 kN
Problem Solution
Repeat this procedure until the forces in all the members of the truss have been determined.
Joint C 2 3
BF = (2.5 m) = m
12.5 kN BF 2
b = BCF = tan = 39.81o
FBC
FCD = 30 kN b +
S Fy = 0: kN - FCF sin b = 0
kN - FCF sin 39.81o = 0
FCF = kN C
FCF
S Fx = 0: 30 kN - FCF cos b - FBC = 0
30 kN - (-19.53) cos 39.81o - FBC = 0
FBC = 45.0 kN T +

8. S Fx = 0: - FEF - (32.5 kN) - FCF cos b = 0A B C D F G
2 m
12.5 kN
2.5 m E
60 kN
50 kN
Problem Solution
FCF = 19.53kN
Joint F
FBF
b=39.81o
6.5
FEF
FFG = 32.5 kN
2.5 6 + 6
6.5 6
6.5
S Fx = 0: FEF (32.5 kN) - FCF cos b = 0
6.5 6
FEF = kN - ( ) (19.53) cos 39.81o FEF = 48.8 kN C
2.5
6.5
2.5
6.5 +
S Fy = 0: FBF FEF (32.5 kN) - (19.53) sin b = 0
2.5
6.5
FBF (-48.8 kN) kN kN = 0
FBF = 6.25 kN T

9. S Fy = 0: -12.5 kN -6.25 kN - FBE sin 51.34o = 0 FBE = -24.0 kNA B C D F G
2 m
12.5 kN
2.5 m E
60 kN
50 kN
Problem Solution
12.5 kN
Joint B
FAB
FBC = 45.0 kN g
FBF = 6.25kN
FBE
2.5 m
2 m
tan g = ; g = 51.34o
S Fy = 0: kN kN - FBE sin 51.34o = 0
FBE = kN +
FBE = 24.0 kN C +
S Fx = 0: kN - FAB + (24.0 kN) cos 51.34o = 0
FAB = 60.0 kN
FAB = 60.0 kN T

10. S Fy = 0: FAE - (24 kN) sin 51.34o - (48.75 kN) = 0B C D F G
2 m
12.5 kN
2.5 m E
60 kN
50 kN
Problem Solution
Joint E
FBE = 24 kN
FAE g
FEF = kN
6.5
2.5
60 kN 6
g = 51.34o
2.5
6.5 +
S Fy = 0: FAE - (24 kN) sin 51.34o - (48.75 kN) = 0
FAE = 37.5 kN FAE = 37.5 kN T