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clear
//
//
//Initilization of Variables
b1=200 //mm //Width at base
b2=100 //mm //Width at top
L=8 //m Length
P=500 //N //Load
//Calculations
//Consider a section at y metres from top
//At this section diameter d is
//d=b2+y*L**-1*(b1-b2)
//After Further simplifying we get
//d=b2+12.5*y //mm
//Moment of Inertia
//I=%pi*64**-1*d**4
//Section Modulus
//Z=%pi*32**-1*(b1+12.5*y)**3
//Moment
//M=5*10**5*y //N-mm
//Let sigma be the fibre stress at this section then
//M=sigma*Z
//After sub values in above equation and further simplifying we get
//sigma=5*10**5*32*%pi**-1*y*((b2+12.5*y)**3)**-1
//For sigma to be Max,d(sigma)*(dy)**-1=0
//16*10**6*%pi**-1*((b2+12.5*y)**-3+y*(-3)*(b2+12.5*y)**-4*12.5)
//After Further simplifying we get
//b2+12.5*y=37.5*y
//After Further simplifying we get
y=b2*25**-1 //m
//Stress at this section
sigma=5*10**5*32*%pi**-1*y*((b2+12.5*y)**3)**-1
//Result
printf("\n Stress at Extreme Fibre is max %0.2f m",y)
printf("\n Max stress is %0.2f N/mm**2",sigma)
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