Resistencia de materiales, Apuntes de Elasticidad y Resistencia de materiales

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5-139. The engine of the helicopter is delivering 800 hp to the rotor shaft AB when the blade is rotating at 1500 rev/min. Determine to the nearest ¿ in. the diameter of the shaft AB if the allowable shear stress is Tallow = 8 ksi and the vibrations limit the angle of twist of the shaft to 0.05 rad. The shaft is 2 ft long and made of L2 tool steel. Internal Torque: rev (27 rady 1 min = 1500 —| 2 = 50.0 rad: o min cr dl 550 ft-1b/s P = 800» ( 15 )- «40000615 P _ 440.000 T=-= w S00% =2801.13 lb. ft Allowable Shear Stress: Assume failure due to shear stress Te Tau == 7 2801.13(12) ($) ON d =2.716in. Ans 8(10%) = Angle of Twist : Assume failure due to angle of twist limitation qalE 2801.13(12(2)(12) ¿(9 10109) d=1.966 in. Shear stress failure controls the design. Hence, Used = E in. diameter shaft. Ans The two circular rod segments, one of aluminum and of copper, are fixed to the rigid walls such that is a gap of 0.008 in. between them when 7, = 60%F. ger temperature T, is required in order to just the 820? Each rod has a diameter of 1.25 in., 13(109)/*F, E, = 10(10%) ksi, 24 = 9.4(10%)/F, 18(10?) ksi. Determine the average normal stress in if T, = 200"F. Thermal Expansion : To close the gap 87 = 057 L,¿+0AT Lo, 0.008 =13(10"%) (7, -60)(8) +9.4( 107%) (7; -60)(4) T=116%F Ans Compatibility : 0,008 = (S7)cu (Epa + (Era (8 A A 0008 =9:4(10%) 2 o di Ñ F8) A O O F = 14.195 kip Average Normal Stress : = 11.6 ksi Ans León. al (¿a 7 EE TT