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+
+
+//example 8.19
+//calculate 
+//Factor of safety against overturning
+//Factor of safety against sliding
+//Shear friction factor
+clc;funcprot(0);
+//given
+c=1;
+H=10;             //heigth of dam
+hw=10;            //heigth of water in reservior
+wb=8.25;          //bottom width
+Bt=1;             //top width
+Hs1=0.1;          //slope on upstream side
+gamma_w=9.81;     //unit weigth of water
+gamma_m=22.4;     //unit weigth of masonary
+f=1400;           //permissible shear stress at joint
+miu=0.75;         //coefficient of friction
+fi=atan(0.625);
+theta=atan(0.1);
+
+W1=Bt*H*gamma_m;
+W2=H*H*Hs1*gamma_m/2;
+W3=H*6.25*gamma_m/2;
+W4=hw*gamma_w*H*Hs1/2;
+P=gamma_w*hw^2/2;
+U=wb*gamma_w*hw*c/2;
+SumV=W1+W2+W3+W4-U;
+L3=2*(wb-(Hs1*H)-Bt)/3;
+L1=(wb-(Hs1*H)-Bt)+Bt/2;
+L2=(wb-(Hs1*H)-Bt)+Bt+(Hs1*H/3);
+L4=(wb-(Hs1*H)-Bt)+Bt+(2*Hs1*H/3);
+L5=2*wb/3;L6=hw/3;
+M1=W1*L1;M2=W2*L2;M3=W3*L3;M4=W4*L4;
+M5=U*L5;M6=P*L6;
+SumM=M1+M2+M3+M4-M5-M6;
+Mplus=M1+M2+M3+M4;
+Mminus=M5+M6;
+FOS=miu*SumV/P;
+SFF=(miu*SumV+wb*1400)/P;
+FOO=Mplus/Mminus;
+FOS=round(FOS*100)/100;
+SFF=round(SFF*10)/10;
+FOO=round(FOO*100)/100;
+mprintf("Factor of safety against sliding=%f.",FOS);
+mprintf("\nShear friction factor=%f.",SFF);
+mprintf("\nFactor of safety against overturning=%f.",FOO);
+mprintf("\nDam is unsafe against overturning");