A hollow shaft of diameter ratio 3/8 has to transfer 400KW power at 100 rpm. The maximum torque being 24% greater than mean torque. The maximum shear stress not to exceed 60 N/mm^2 and angle of twist in a length of 4m should not exceed 2 degree. Calculate external and internal diameter of shaft. Take G= 84 KN/mm^2.

Note:

If Power formula used for shaft then Torque will be Mean Torque i.e Tavg.

In Torsional Equation, Torque is Maximum Torque i.e Tmax.

For Hollow shaft in torsional equation radius will be half of the externl diameter.

Data: A hollow shaft of diameter ratio $\frac{3}{8}$

i.e $di=\frac{3}{8} d_o$ ($d_o$ external diameter, di=internal diameter)

$Power (P)=400KW=400*10^3W$

$N=100rpm$

Maximum torque ($T_{max}$)=24% greater than mean torque ($T_{avg})$

i.e $T_{max}$=1.24 $T_{avg}$)

Shear stress \tau_{max}=60 N/mm^2

Length $l=4m=4*10^3 mm$

Angle of twist $(\theta)$= $2^{\circ}$=\frac{2*\pi}{180} (radian)

Modulus of rigidity $(G)=84 KN/mm^2=84*10^3 N/mm^2$

For solid shaft, polar moment of inertia

$J_p=\frac{\pi d^4}{32} mm^4$

For hollow shaft, polar moment of inertia

$J_p=\frac{\pi (d_o^4-di^4)}{32} mm^4$

1. Power transmitted by shaft is,

$P=\frac{2 \pi N T_{avg}}{60}$ W

$400*10^3=\frac{2 \pi*100*T_{avg}}{60}$ W

$T_{avg}=38197.18 Nm$

$T_{avg}=38197.18*10^3 Nmm$

$T_{max}=$(1+24%)$ T_{avg}=1.24 T_{avg}=1.24-38197.18*10^3$

$T_{max}=47364.51*10^3 Nmm$

1. For strength:

$\frac{T}{J_p}=\frac{\tau}{r}$

But for hollow shaft,

$J_p=\frac{\pi(d0^4-di^4)}{32}=\frac{\pi (d0^4-(\frac{3*d0}{4}^4}{32})$

$=0.096 d^4$

and radius $(r)=\frac{do}{2}$ (always)

$\frac{47364.51*10^3}{0.096*do^4}=\frac{60}{\frac{do}{2}}$

$do=160.20mm$

$d=\frac{3}{8}*160.20=60.08mm$ (1)

1. For stiffness:

$\frac{T}{J_p}=\frac{G \theta}{l}$

$\frac{47364.51*10^3}{0.096*d^4}=\frac{84*10^3*(2*\frac{\pi}{180}}{4*10^3}$

$do^4=\frac{10^3*47364.5*4*10^3*180}{0.096*2*\pi*84*10^3}$

$do^4=161.06mm$

$di=\frac{3}{8}d0$

$di=60.39mm$ (2)

Values of di and do is greater of (1) and (2).

do=161.06mm and di=60.39mm