Question

Part of a control linkage for an airplane consists of a rigid member CB and a flexible cable AB Originally the cable is unstretched

Answer 1

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Solution | LAB = /Bc? + Ac? 800mm (800) + (600) ber LAB = 1000mm A 1. LAB = 1 K 600mm Consider DACB ; - (:O = 0.61 (given) ) Pc = 90°to c = 90° +0.610 pc = 90.610 LABI = 19² +62_2ab cosde putting values, a= 800mm b= 600mm pc = 90.61 Lae' = 11800) + (600)2 2X 800x600 COS GOBI" LAR' = 994.877mm

Normal Strain in the Cable (EAB; EAR = 1 LABI - LABI LAB putting values LAB'=994.877mm LAB = 1000 mm EAB = 994.877 - 1000 1 1000 - 5.123X10-31 = EAB = 0.005123 mm/mm Therefore, Normal Shain in cable is 0.005123 mm/mm.