The A-36 hollow steel shaft is 2 m long and has an outer diameter of 40 mm. When it is rotating at 77 rad/s , it transmits 32 kW of power from the engine E to the generator G. The shear modulus of elasticity for A-36 steel is 75 GPa.

Determine the smallest thickness of the shaft if the allowable shear stress is Tallow 140 MPa and the shaft is restricted not to twist more than 0.05 rad

Answer:

The value is [tex]t = 8.28 *10^{-3} \ m[/tex]

Explanation:

From the question we are told that

The length is [tex]l = 2 \ m[/tex]

The outer diameter is [tex]d_o = 40 \ mm = 0.04 \ m[/tex]

The angular speed is [tex]w = 77 \ rad/s[/tex]

The amount of power transmitted is [tex]P = 32 \ kW = 32*10^{3} \ W[/tex]

The shear modulus of elasticity is [tex]G = 75 GPa = 75 *10^{9} \ Pa[/tex]

The allowable shear stress is [tex]\tau_{allow} = 140 MPa = 140 *10^{6}\ Pa[/tex]

The allowable angular displacement is [tex]\theta = 0.05 \ rad[/tex]

Generally the power transmitted is mathematically represented as

[tex]P = \tau * w[/tex]

Here [tex]\tau[/tex] is the amount of torque generated by the hollow steel shaft

[tex]32*10^{3} = \tau * 77[/tex]

=> [tex]\tau = 415.6 \ N\cdot m[/tex]

Generally the polar moment of inertia of the hollow steel shaft is mathematically represented as

[tex]J = \frac{\pi}{32 } ( d_o^4 - d_i^4 )[/tex]

Generally the allowable displacement of the hollow steel shaft is mathematically represented as

[tex]\theta = \frac{\tau * l }{ G J }[/tex]

=> [tex]\theta = \frac{ 415.6 * 2 }{ 75*10^{9} * \frac{\pi }{ 32} * ( d_o^4 - d_i^4 ) }[/tex]

=> [tex]0.05 = \frac{ 415.6 * 2 }{ 75*10^{9} * \frac{\pi }{ 32} * ( 0.04^4 - d_i^4 ) }[/tex]

=> [tex]d_i = 0.0235 \ m[/tex]

Generally the thickness of the shaft is mathematically represented as

[tex]t = \frac{d_o - d_i }{2}[/tex]

=> [tex]t = \frac{0.04 - 0.02345 }{2}[/tex]

=> [tex]t = 8.28 *10^{-3} \ m[/tex]