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// Problem no 13.2,Page No.301
clc;clear;
close;
//W=10*w //KN/m //u.d.l
sigma=805*10**6 //Pa //Bending stress
Tou=0.85*10**6 //Pa //Shear stress
//Calculations
//M=W*L**2*10**-4*8**-1 //Max B.M
//F=W*L*10**-2*2**-1 //Max S.F
//y=h*2**-1 //depth
//A-b*h //Area of c/s
//Now using relation we get
//sigma=M*h*(2*I)**-1 //Bending stress
//AFter substituitng values we get
//805*10**6=w*l**2*h*(16*10**5*I)**-1 //Equation 1
//Again using the relation we get
//tou=F*A*y_bar*(I*b)**-1 //shear atress
//AFter substituitng values we get
//0.85*10**6=w*L*h**2*(16*10**5*I)**-1 //Equation 2
//Dividing equation 1 & 2 we get
//L*h**-1=10
//Let L*h**-1=Z
z=10
//Result
printf("The Ratio of span to depth ratio is %.2f",z)
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