Ex 8

Statics – Exercise No. 8 6.10 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. Answer: FAB = FBC = FCD = 120 kN T; FAE = 192 kN T; FAF = 150 kN C; FBF = FBG = FCG = FCH = 0; FDH = FGH = FFG = 130 kN C; FEF = 120 kN C.

6.16 Determine the zero-force members in the truss shown for the given loading. Answer: BC, CD, IJ, IL, KL, LM, MN.

6.25 Determine the force in members EF and EG of the truss shown when P = 35 kN. Answer: FEF = 1.9846P; FEG = -7.143 P.

6.29 Determine the force in members CD and CE of the truss shown.

Answer: FCD = 64.3 kN T; FCE = 92.1 kN C.

6.33 Determine the force in members CE, DE and DF of the truss shown. Answer: FCE = 40 kN C; FDE = 16 kN C; FDF = 40 kN T.

6.42 Determine the force in members AB and AD of the truss shown. Answer: FAB = 161.6 kN C; FAD = 190.1 kN T.

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All the questions are taken from Ferdinand P. Beer and E. Russell Johnston Jr. “Vector Mechanics for Engineers, Statics” Second SI Metric Edition, McGraw-Hill Book Company.