Question 7.18: A solid shaft rotating on two bearings 1.5 m apart is subjec......

Fatigue strength in rotating bending is obtained as

S_e^*=0.5 \times \sigma_{\mathrm{ult}}=0.5 \times 652=326 \mathrm{~MPa}

Corrected fatigue strength is

S_e=326 \times 0.58=189 \mathrm{~MPa}

M_m=\frac{1}{2}\left(M_{\max }+M_{\min }\right)=\frac{1}{2}(3-3)=0

M_a=\frac{1}{2}(3+3)

M_a=3 \mathrm{~kN} \cdot \mathrm{m}

T_m=1.5 \mathrm{~kN} \cdot \mathrm{m}

T_a= 0

d=\left[\frac{16 N}{\pi \times S_e}\left\{\left(\frac{S_e}{\sigma_{\mathrm{ult}}}\right) \sqrt{3} \times\left(k_t T_m\right)+2\left(k_m M_a\right)\right\}\right]^{1 / 3}

d=\left[\frac{16 \times 1.5}{\pi \times 189}\left\{\left(\frac{189}{652}\right)\left(\sqrt{3} \times 2 \times 1.5 \times 10^6\right)+2 \times 1.5 \times 3 \times 10^6\right\}\right]^{1 / 3}

d = 75.16 mm

d = 15 mm

Using the equivalent bending moment criterion, the equivalent bending moment is

M_{e q}=\frac{1}{2}\left[k_m M+\sqrt{\left(k_m M\right)^2+\left(k_t T\right)^2}\right]

=\frac{1}{2}\left[1.5 \times 3.0 \times 10^6+\sqrt{\left(1.5 \times 3 \times 10^6\right)^2+\left(2 \times 1.5 \times 10^6\right)^2}\right]

= 4954163.457 N·mm

Design equation is

\sigma_b=\sigma_{\mathrm{all}}=\frac{32 M_{e q}}{\pi d^3}

\frac{652}{1.5}=\frac{32 \times 4954163.457 \times 1000}{\pi \times d^3}

d=\left\{\frac{32 \times 4954163.457 \times 1.5}{\pi \times 652}\right\}^{1 / 3}=48.78 \mathrm{~mm}

d = 50 mm

Using the maximum shear stress criterion, the equivalent torque is

T_{e q}=\sqrt{\left(k_m M\right)^2+\left(k_t T\right)^2}

=\sqrt{\left(1.5 \times 3 \times 10^6\right)^2+\left(2 \times 1.5 \times 10^6\right)^2}

= 5408326.913 N·mm

Design equation is

\tau_{\text {all }}=\frac{16 T_{e q}}{\pi d^3}

Assuming \tau_{\text {all }}=0.5 \times \frac{\sigma_{\mathrm{ult}}}{N}

\frac{0.5 \times 652}{1.5}=\frac{16 \times 5408326.913}{\pi \times d^3}

d=\left\{\frac{16 \times 5408326.913 \times 1.5}{\pi \times 652 \times 0.5}\right\}^{1 / 3}=50.23 \mathrm{~mm}

d = 51 mm

The safe diameter is 75 mm as obtained with fatigue design aspect. This explains the use of fatigue aspect of shaft design in case of rotating shaft with stationary loads.