A circular shaft has to transmit 550kW power at 115 RPM. Allowable shear stress=78 MPa. Find i) The required diameter of solid shaft ii)The diameters of hollow section section such that internal diameter=0.75 x external diameter.

given data : power transmitted P = 550KW Number of rotations N = 115rpm

i) Here what to be found out is the diameter of the solid shaft. This can be found by using torsion equation $${T\over J}= {\tau\over R}$$ torque can be calculated by the formulae $$P= {2\pi\ NT\over60}$$
P is given as 550KW so T = 45.67 KN-m

J is the polar moment of inertia it is calculated by the formulae $$J={\pi\ r^4\over2}$$ by substituting the value of J in the torsion equation we get

1. $${T\over\tau} = {\pi\ r^3\over 2}$$

the radius can be calculated by the above equation and shear stress is given as 78MPa by substituting all the above equations we get the radius of the shaft r =0.07196 meters or 71.96 mm

ii) for the hollow shaft R1 = internal diameter R2= external diameter R1 = 0.75* R2

$${T\over\tau} = {\pi\ (R2^4-R1^4)\over (2(R2))}$$ we get R2 = 0.114meters and R1=0.085