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// Problem no 7.2,Page no.184
clc;clear;
close;
P=295 //KW //Power
N=100 //R.p.m
sigma_s=80 //MPa //shear stress
//Calculations
T_mean=((P*60000)*(2*%pi*N)**-1) //N*m
//T_max=T_mean=(%pi*D**3*sigma_s)*16**-1
D=((T_mean*16)*(%pi*sigma_s*10**6)**-1)**0.333 //m //Diameter of solid shaft
//For hollow shaft
//I_p_h=%pi*32**-1*(D_1**4-d_1**4) (equation 1)
//Now d_1=0.6*D_1
//substituting above value in equation 1,we get,
//I_p_h=0.0272*%pi*D_1**4
//For solid shaft
//I_p_s=%pi*32**-1*D**4
//T and sigma_s being the same then I_p*R**-1 will be the same for the two shafts
//Using relation I_p_h*R_1**-1=I_p_s*R**-1
//Substituting values and simplifying we get
D_1=(D**3*0.8704**-1)**0.3333333 //m //External diameter of hollow shaft
d_1=0.6*D_1 //cm //Internal diameter of hollow shaft
A_s=%pi*4**-1*(D*10**2)**2 //cm**2 //Area of solid shaft
A_h=%pi*4**-1*(((D_1*10**2)**2)-((d_1*10**2)**2))
W=(A_s-A_h)*A_s**-1*100 //Percentage //Percentage saving in weight
//Result
printf("Diameter of solid shaft is %.5f m",D)
printf("\n Percentage saving in weight is %.2f",W);printf(" %%")
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