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
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.
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.
S MA = 0: E(2.5 m) - (12.5 kN)(2 m) - (12.5 kN)(4 m) Problem 6.167 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) = 0 E = 60 kN + S Fy = 0: Ay - (4)(12.5 kN) = 0 Ay = 50 kN S Fx = 0: Ax - E = 0 Ax= 60 kN +
S Fy = 0: FGD - 12.5 kN = 0 FGD = 32.5 kN C S Fx = 0: FGD - FCD = 0 Problem 6.167 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 - 12.5 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
S F = 0: FCG = 0 S F = 0: FFG - 32.5 kN = 0 FFG = 32.5 kN C Problem 6.167 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 - 32.5 kN = 0 FFG = 32.5 kN C FFG
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 6.167 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) = 1.6667 m 12.5 kN BF 2 b = BCF = tan-1 = 39.81o FBC FCD = 30 kN b + S Fy = 0: - 12.5 kN - FCF sin b = 0 - 12.5 kN - FCF sin 39.81o = 0 FCF = 19.53 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 +
S Fx = 0: - FEF - (32.5 kN) - FCF cos b = 0 A B C D F G 2 m 12.5 kN 2.5 m E 60 kN 50 kN Problem 6.167 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 = -32.5 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) - 12.5 kN - 12.5 kN = 0 FBF = 6.25 kN T
S Fy = 0: -12.5 kN -6.25 kN - FBE sin 51.34o = 0 FBE = -24.0 kN A B C D F G 2 m 12.5 kN 2.5 m E 60 kN 50 kN Problem 6.167 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: -12.5 kN -6.25 kN - FBE sin 51.34o = 0 FBE = -24.0 kN + FBE = 24.0 kN C + S Fx = 0: 45.0 kN - FAB + (24.0 kN) cos 51.34o = 0 FAB = 60.0 kN FAB = 60.0 kN T
S Fy = 0: FAE - (24 kN) sin 51.34o - (48.75 kN) = 0 B C D F G 2 m 12.5 kN 2.5 m E 60 kN 50 kN Problem 6.167 Solution Joint E FBE = 24 kN FAE g FEF = 48.75 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