Problem 6.167 12.5 kN 12.5 kN 12.5 kN 12.5 kN Using the method of joints, determine the force in...
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Problem 6.167 Using the method of joints, determine the force in each member of the truss shown. A B C D E F G 2 m 12.5 kN 2.5 m 2 m 2 m 12.5 kN 12.5 kN 12.5 kN
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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.
Transcript of Problem 6.167 12.5 kN 12.5 kN 12.5 kN 12.5 kN Using the method of joints, determine the force in...
Problem 6.167
Using the method of joints, determine the force in each member of the truss shown.
AB C
D
E
F
G
2 m12.5 kN
2.5 m
2 m 2 m12.5 kN 12.5 kN 12.5 kN
Solving Problems on Your Own
1. Draw the free-body diagram of the entire truss, and use this diagram to determine the reactions at the supports.
Using the method of joints, determine the force in each member of the truss shown.
AB C
D
E
F
G
2 m12.5 kN
2.5 m
2 m 2 m12.5 kN 12.5 kN 12.5 kN
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 Fx = 0 and Fy = 0 is positive, the member is in tension. A negative answer means the member is in compression.
Problem 6.167
Solving Problems on Your Own
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.
Using the method of joints, determine the force in each member of the truss shown.
AB C
D
E
F
G
2 m12.5 kN
2.5 m
2 m 2 m12.5 kN 12.5 kN 12.5 kN
4. Repeat this procedure until the forces in all the members of the truss have been determined.
Problem 6.167
+
Problem 6.167 Solution
Draw the free-body diagram of the entire truss, and use it todetermine reactions at the supports.
AB C
D
E
F
G
2 m12.5 kN
2 m 2 m12.5 kN 12.5 kN 12.5 kN
2.5 m
Ax
Ay
E
MA = 0: E(2.5 m) - (12.5 kN)(2 m) - (12.5 kN)(4 m) - (12.5 kN)(6 m) = 0 E = 60 kN
Fy = 0: Ay - (4)(12.5 kN) = 0 Ay = 50 kNFx = 0: Ax - E = 0 Ax= 60 kN
+
+
Problem 6.167 Solution
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.
12.5 kN
FCD
FGD
2.56
6.5
Fy = 0: FGD - 12.5 kN = 0
+
+
2.56.5
66.5
FGD = 32.5 kN C
Fx = 0: FGD - FCD = 0
FCD = 30 kN T
Joint D
A B CD
FG
2 m
12.5 kN
2 m 2 m
12.5 kN 12.5 kN 12.5 kN
2.5 m
E60 kN
60 kN
50 kN
Problem 6.167 Solution
A B CD
FG
2 m
12.5 kN
2 m 2 m
12.5 kN 12.5 kN 12.5 kN
2.5 m
E60 kN
60 kN
50 kN
FCG
FFG
F = 0: FFG - 32.5 kN = 0
FFG = 32.5 kN C
Joint G
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.
32.5 kN
F = 0: FCG = 0
Problem 6.167 Solution
FBC
FCF
Fy = 0: - 12.5 kN - FCF sin= 0 - 12.5 kN - FCF sin39.81o= 0 FCF = 19.53 kN C
Joint C
FCD = 30 kN
Repeat this procedure until the forces in all the members of the truss have been determined.
12.5 kN
A B CD
FG
2 m
12.5 kN
2 m 2 m
12.5 kN 12.5 kN 12.5 kN
2.5 m
E60 kN
60 kN
50 kN
= BCF = tan-1 = 39.81o
23
BF2
BF = (2.5 m) = 1.6667 m
Fx = 0: 30 kN - FCF cos -FBC = 0 30 kN - (-19.53) cos39.81o-FBC = 0 FBC = 45.0 kN T
+
+
Problem 6.167 Solution
FEF
Fy = 0: FBF - FEF - (32.5 kN) - (19.53) sin = 0
Joint F=39.81oA B C
D
FG
2 m
12.5 kN
2 m 2 m
12.5 kN 12.5 kN 12.5 kN
2.5 m
E60 kN
60 kN
50 kN
Fx = 0: - FEF - (32.5 kN) - FCF cos = 0+
+
FBF
2.56
6.5 FFG = 32.5 kN
FCF = 19.53kN
66.5
66.5
FEF = -32.5 kN - ( ) (19.53) cos39.81o FEF = 48.8 kN C6.56
2.56.5
2.56.5
FBF - (-48.8 kN) - 12.5 kN - 12.5 kN = 0 2.56.5
FBF = 6.25 kN T
Problem 6.167 Solution
FBE
Fy = 0: -12.5 kN -6.25 kN - FBE sin 51.34o = 0 FBE = -24.0 kN
Joint B
A B CD
FG
2 m
12.5 kN
2 m 2 m
12.5 kN 12.5 kN 12.5 kN
2.5 m
E60 kN
60 kN
50 kN
+
+
FAB FBC = 45.0 kN
FBF = 6.25kN
12.5 kN
tan = ; = 51.34o2.5 m2 m
Fx = 0: 45.0 kN - FAB + (24.0 kN) cos 51.34o = 0 FAB = 60.0 kN
FBE = 24.0 kN C
FAB = 60.0 kN T
Problem 6.167 Solution
Fy = 0: FAE - (24 kN) sin 51.34o - (48.75 kN) = 0
Joint E
A B CD
FG
2 m
12.5 kN
2 m 2 m
12.5 kN 12.5 kN 12.5 kN
2.5 m
E60 kN
60 kN
50 kN
+
FAEFEF = 48.75 kN
FBE = 24 kN
2.56.5
FAE = 37.5 kN FAE = 37.5 kN T
2.56
6.560 kN
= 51.34o
