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

Question #182781

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.



1
Expert's answer
2021-04-23T08:25:36-0400

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



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