- #1
manal950
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A solid steel shaft is subjected to a torque of 45 KN/m. If the angle of twist is 0.5 degree per meter length of the shaft and the shear stress is not exceed 90 MN/m^2
(I) find the suitable diameter of the shaft . Take C = 80 GN/m^2
(II) Maximum shear strain
we can solve the question by torsion equation
T/Ip = e/R = CQ/l
T maximum twisting torque in Nm
Ip polar moment of inertia = pid^4/32 in m^4
e shear stress in N/m^2
R radius of the shaft
Q the angle of the shaft
l length if the shaft in m
my question know why in solving fined two diameter ?
(I) find the suitable diameter of the shaft . Take C = 80 GN/m^2
(II) Maximum shear strain
we can solve the question by torsion equation
T/Ip = e/R = CQ/l
T maximum twisting torque in Nm
Ip polar moment of inertia = pid^4/32 in m^4
e shear stress in N/m^2
R radius of the shaft
Q the angle of the shaft
l length if the shaft in m
my question know why in solving fined two diameter ?