Question

A solid 1.25 in. diameter shaft is used in an aircraft engine to transmit 140 hp at 2930 rpm to a propeller that develops a thrust of 1700 lb. Determine the magnitudes of the principal stresses and the maximum shear stress at any point on the outside surface of the shaft. Enter the principal stresses at any point on the outside surface of the shaft. Use the sign convention that a tension normal stress is positive and a compression normal stress is negative. Also, use the naming convention that op 1 is larger algebraically than o p 2. Op 1 - 1 psi op 2 - 1 psi Enter the magnitude of the maximum shear stress at any point on the outside surface of the shaft. Enter a positive value for the magnitude Tmax= psi

Answer 1

Elnswer 1700 1700 Normal stress along s-direction = 1385.987 psi x d) 3.14(1-2637 Since no-load is acting on shaft in y-direction 4/4 - A Oy = 0 Torque on shaft is P B (270 Here, p=140 hp T = - 12 140x550 H-eboks 77000 ft-lb/s N= 2930 rpm 77000 = 251.081 lb-ft → T = 2(3-14)(2930) 60 = 251.081 X12 lb-in = 3012.977 lb-in -> Shear stress is exy TC J T을 7 / 22 CdA) 3012.977( 25 ) - 3014 (1025) 4 32 = 7860.61 psi

Now, Principle stresses are given by aty + T2= Log) + tay 1385-987-0,5 +67860.61) 1385.987 to + 2 2. = 692.993 + 7891.095 = 692.993 + 7891.095 = 18584.088 psi Ang O2= 692.993 7891.095 798.102 psi Any Zmay = 2007 (Exyde 1385.987- 978860-61) 2 = 7891.095 psi Imax = 7891.095 psilany Good Luck