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// Problem no 14.1,Page No.325
clc;clear;
close;
b=2 //m //width
FOS=1.5 //Factor of safety
//rho_mason=2.5*rho_w
mu=0.5 //coeffeicient of friction
//Calculations
//Let L=1 m (length of dam)
L=1
//W=b*H*L*rho
//After substituting values and Further simplifying we get
//W=2*H*rho
//Total Pressure
//P=W*H**2*2**-1
x_bar=b*2**-1 //Distance of Line of action of W from waterface
//Distance of pt where resultant cuts the base measured from Line of action
//x=P*W**-1*H*3**-1
//After substituting values and Further simplifying we get
//x=H**2*30**-1
//x_bar+x=2*b*3**-1
//After substituting values and Further simplifying we get
//1+H**2*30**-1=2*b*3**-1
H=(30*(2*b*3**-1-1))**0.5 //height of dam
//Frictional Resistance offered at the base
//F=mu*W
//After substituting values and Further simplifying we get
//F=3.16*rho
//Total Lateral Pressure
//P=W*H**2*2**-1
//P=4.99*W
//Factor of safety against sliding
//FOS1=F*P**-1=3.16*4.99**-1*rho_mason*rho_w**-1
FOS1=3.16*4.99**-1*2.5
//FOS1>FOS
//Result
printf("Dam is safe against sliding = %.2f m",FOS1)
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