Recent content by rakeshmech

Sorry! the force on the coupling would be 8080/0.125 (consider the radius... not diameter. silly error :) ) follow the same methodology

speed and power helps you to find the torque... you can use the equation I specified above, or you can also use this Power =(2*Pi*N*T)/60 also, we design machines with factor of safety as a function of yield strength rather than UTS. :)

total torque to be transmitted => T=9550P/N (N-m) => 9550*550/650 = 8080N-m force on the coupling= 8080/0.250 =40400 N assuming its a mild steel of yield strength 240MPa and shear strength of 120 MPa,with a factor of safety of 4 the design shear strength is 30Mpa we have area of each bolt...