Answer:
a
[tex]d = 0.0223 \ m[/tex]
b
[tex]d = 0.0481 \ m[/tex]
Explanation:
From the question we are told that
The maximum allowable shear stress is [tex]\sigma = 70 MPa = 70 *10^{6} \ Pa[/tex]
The power is [tex]P = 40 \ kW = 40 *10^{3} \ W[/tex]
considering question a
The shaft speed is given as [tex]v = 2500\ rev/min[/tex]
Generally the torque experienced by the shaft is mathematically represented as
[tex]\tau = \frac{ 9.55 * P}{v}[/tex]
=> [tex]\tau = \frac{ 9.55 * 40 *10^{3}}{ 2500}[/tex]
=> [tex]\tau = 152.8 \ N \cdot m[/tex]
Generally the maximum torque experienced by the shaft is mathematically represented as
[tex]\tau_m = \frac{2 \tau }{ \pi r^2 }[/tex]
Generally diameter = 2 * radius (r)
So
[tex]\tau_m = \frac{2 \tau }{ \pi 4 d^2 }[/tex]
Generally the maximum allowable shear stress is mathematically represented as
[tex]\sigma = \frac{2 \tau }{ \pi 4 d^2 } * \frac{32}{d}[/tex]
=> [tex]\sigma = \frac{16 \tau }{ \pi d^3}[/tex]
=> [tex]d = \sqrt[3]{\frac{16 * \tau }{ \pi \sigma } }[/tex]
=> [tex]d = \sqrt[3]{\frac{16 * 152.8 }{ \pi * 70 *10^{6} } }[/tex]
=> [tex]d = 0.0223 \ m[/tex]
considering question b
The shaft speed is given as [tex]v = 250\ rev/min[/tex]
Generally the torque experienced by the shaft is mathematically represented as
[tex]\tau = \frac{ 9.55 * 40 *10^{3}}{250 }[/tex]
=> [tex]\tau = 1528 \ N \cdot m[/tex]
Generally the shaft diameter is mathematically represented as
[tex]d = \sqrt[3]{\frac{16 * \tau }{ \pi \sigma } }[/tex]
=>[tex]d = \sqrt[3]{\frac{16 * 1528 }{ 3.142 * 70 *10^{6} } }[/tex]
=>[tex]d = 0.0481 \ m[/tex]
