Design protective type of cast iron flange coupling for a steel shaft transmitting 15 kW at 210 r.p.m. and with an allowable shear stress 40 MPa. The working stress in the bolts should not exceed 35 MPa. Assume that the same material is used for shaft and key and that the crushing stress is twice the value of its shear stress. The maximum torque is 24.5% greater than the full load torque. The shear stress for cast iron is 15 MPa. Also draw an appropriate sketch of protective type flange coupling.

Subject: Machine Design -I

Topic: Shaft, Keys and Coupling

Difficulty: Medium

Design for hub:

Mean torque transmitted by the shaft,

$T_mean=\frac{P*60}{2 \pi N}=\frac{15*a0^3*60}{2 \pi 210}=682.09N-m=682.09*10^3 N-mm$

Maximum torque transmitted ,

$T_max=1.24*T_mean=845.79*10^3 N-mm$

$T_max=\frac{pi}{16}8\tau_s8d^2$

$d=47.57=50mm$

Outer diameter and length of hub,

D=2d=2*50=100mm

L-1.5d=1.5850=75mm

Check the induced shear stress for the hub material which is cast iron, by considering it as a hollow shaft.

$T_max=\frac{\pi}{16}*\tau_c*\frac{D^4-d^4}{D}$

$845.79*10^3=\frac{\pi}{16}*\tau_c*\frac{100^4-50^4}{100}$

$\tau_c=4.59MPa\lt15MPa$

Design for key:

Width and thickness are equal i.e

$w=t=d/4=50/4=12.5 or 14mm$

Take length of key equal to length of hub

$l=L=75mm$

$T_max=l*w*\tau_k*\frac{d}{2}$

$845.79*10^3=75*14*\tau_k*\frac{50}{2}$

$\tau_k=32.20 MPa$

$\sigma_c=2\tau_k=2*32.20=64.40MPa$

Design for flange:

The thickness for the flange is taken as 0.5*d

Therefore,

$t_f=0.5*50=25mm$

$T_max=\frac{\pi D^2}{16}*\tau_c*\tau_f$

$845.79*10^3=\frac{\pi 100^2}{16}*\tau_c*25$

$t_c=17.23\lt15MPa$

Design for bolts:

Let $d_1$=Nominal diameter of bolts

Number of bolts, n=4........ d=50mm

Pitch circle diameter of bolts,

D_1=3d=3*50=150mm

The bolts are subjected to sheer stress due to the torque transmitted.

$T_max=\frac{\pi d_1^2}{16}*\tau_b*n*\frac{d_1}{2}$

$845.79*10^3=\frac{\pi d_1^2}{16}*35*4*\frac{150}{2}$

$d_1=20.25mm or 22mm$

Outer diameter of the flange,

$D_2=4d=4*50=200mm$

Thickness of the protective circumferential flange,

$t_f$=$0.25d=0.25*50=12.5mm$