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Simple stress problem (Strength of Materials - I, Mid-term Exam-44-1) Problem : Each of the two vertical links CF connecting the two horizontal members AD and EG has a 10×40-mm uniform rectangular cross section and is made of a steel with an ultimate strength in tension of 400 MPa, while each of the pins at C and F has a 20-mm diameter and is made of a steel with an ultimate strength in shear of 150 MPa. Determine the overall factor of safety for the links CF and the pins connecting them to the horizontal members. Solution : The equilibrium equation X 0.4FCF − 0.65 × 24 = 0 ME = 0, FCF = 39 kN FBE = FCF − 24 = 39 − 24 FBE = 15 kN The links CF are in tension (critical) A = (b − d) × t = (40 − 20) × 10 A = 200 mm2 σu = Fu 2A Fu = 2A × σu Fu = 2 × 200 × 400 Fu = 160 × 103 N Prof. Dr. M. Kemal APALAK 1 The pins at F are in double shear π 202 = 314. 16 mm2 4 Fu = , Fu = 2A × τu 2A = 2 × 314. 16 × 150 = 94248 N A = τu Fu Fu The critical force in link CF Fcr = 94248 N Factor of safety Fcr 94248 = FCF 39 × 103 F.S = 2. 416 6 F.S = Prof. Dr. M. Kemal APALAK 2