Question

If the design of a solid shaft based on the maximum-distortion-energy leads to c (AMP® +37° where is the radius of the shaft, Mis the applied moment and is the applied torque. Calculate the minimum diameter of the shaft below in in to 2 decimal places, i allow77ksi. B -0.5 ft 3 ft -0.5 ft -2 ft 300 lb 120 lb 300 lb 120 lb
as soon as possible

Answer 1

Reactions at A and B in 6 RA+ RB = 420 +420 840 ebf. Taking moment about A; 420 x 1 + 420X 4 = RoxG Ro 350 lbf RA 840 - 350 agoebt 0.5 ft Radius q pulley Torque at Da (300 - 120 (0.5): goldfht Clock wise when viewed from А Torque at c: (309–120) (0-5)= 90 lbf be

120 + 300 420lbf 300+120= 4201bf D 바이터 3 ft ya aft RB = 350 lbf 700 efft RA= 490 lbf - 490 lbfft Bending moment diagram B 700 lbf.ft : M Maximum bending momento 3 Total, torque go + go lbf.ft T = 180 lbf.ft Using formula which is given for radius % 4 (4m?+ 372) ( 2 2 *? بله

For unit conversion all = 77 ksi 778 6.895 MPa = 530.896 MPa 700 lbf.ft - 700x 1.36 Nm a g49.07 Nm Nm T= 180 lbf ft 180 X1.36 244. OS Nm 4 (4x 949.08%+ 3x244,052) C= 2. 5301896 x ký loé)*xA 0.0132 6 m 13.26mm 11 Minimum diameth 2 2c= 26.52 mm