The maximum torque is 20% more than the average torque. The permissible shear stress is 90 N/mm^2 and twist is 1 degree per 3 m length. Calculate the power transmitted if G=84000 N/mm^2.

Data: do = 120 mm, di = 100 mm

N = 200 rpm

$T_{max} = \frac{120}{100} \ T_{avg}$

$\zeta = 90 \ N/mm^2 $ $\theta = 1°$ l = 3000 mm

G = 84000 $N/mm^2$

Solution:

$J_p = \frac{\pi}{32} (120^4 – 100^4)$

$Jp = 10.54 \times 10^6 mm^4$

[1] $\frac{T_{max}}{Jp} = \frac{\zeta}{r}$

$\therefore$ $\frac{T_{max}}{10.54 \times 10^6} = \frac{90}{60}$

$T_{max} = 15.81 \times 10^6 \ Nmm$

[2] $\frac{T_{max}}{J_p} = \frac{G \ \theta}{l}$

$\therefore$ $T_{max} = 5.151 \times 10^6 \ Nmm$

$\therefore$ safe value of $T_{max} = 5.151 \times 10^6 \ Nmm$

$\therefore$ $T_{avg} = \frac{T_{max}}{1.2} = 4.293 \times 10^6 \ Nmm$

$P = \frac{2 \ \pi \ N \ T_{avg}}{60000}$

$\therefore$ $P = \frac{2 \pi \times 200 \times 4.293 \times 10^6}{60000}$

P = 89.91 kw.