Answer to Question #182781 in Mechanical Engineering for Priya Bapuji Zade

Question

Design a bushed pin type flexible coupling is usedto connect two shafts and transmit 5 

kWpower at 720 rpm Shafts, keys and pins aremade of commercial steel (Syt = Syc = 240 

N/mm2

) and the factor of safety is 3. The flanges are made of grey cast iron FG 200(Sut= 

200 N/mm2

) and the factor of safety is6. Assume,Sys = 0.5Syt and Sus = 0.5Sut.

Answer 1

Transmit power = 5 kW, RPM of shaft= 720 ,Syt = Syc = 240 N/mm2, FOS=3 and

flanges are made of grey cast iron FG 200(Sut= 200 N/mm2),FOS=6 and Sys = 0.5Syt and Sus = 0.5Sut.

(i) Power tansmitted,

$P= \frac{2\pi N T}{60}$

$5000= \frac{2\pi \times 720 \times T}{60}$

T= 66.31 N-m, this is the torque transmitted by shaft.

(ii) Design of Key: - Some data are assumed from design databook

for key following relation is given as

Syt = Syc = 240 N/mm2, FOS=3

$\tau_k= \frac{S_{yt}}{FOS}= \frac{240}{3}= 80 MPa$

width of key = 16 mm whereas thickness of key 10 mm

$T= L w \tau_k d/2$

d = 100mm

(iii) DEsign of flange

FOS=6 and Sys = 0.5Syt and Sus = 0.5Sut.

$\tau_c=40 MPa$

$T= (\pi D^2/2) \times \tau_c \times t_f$

t= 25 mm