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+clc
+//initialization of variables
+clear
+P=3 //tonne/m
+s=6 //m
+l=50 //cm
+b=20 //cm
+k=0.5 //m
+//calculations
+R=P*s/2
+sf=R-k*P
+bm=R*k-P*k^2/2
+tau_xy=1.5*sf*1000/(l*b)
+tau_max=tau_xy
+str=bm*s*10^5/(b*l*l)
+
+// consider the line a-a
+
+sigma_x=str*12.5/25
+sigma_y=0
+tau_xy=tau_xy*(1-(12.5/25)^2)
+
+sigma_1=(sigma_x+sigma_y)/2+sqrt((1/2*(sigma_x-sigma_y))^2+tau_xy^2)
+sigma_2=(sigma_x+sigma_y)/2-sqrt((1/2*(sigma_x-sigma_y))^2+tau_xy^2)
+
+printf('For the line a-a the bending stress and shearing stress are \n respectively %.2f kg/cm^2, %.2f kg/cm^2 ',sigma_x,tau_xy)
+printf('\n The principal stresses are %.2f kg/cm^2 (tension) %.2f kg/cm^2 (compression) ',sigma_1,sigma_2)
+
+//consider the line c-c
+printf('\n For the line c-c the bending stress and shearing stress are \n respectively %.2f kg/cm^2, %.2f kg/cm^2 ',sigma_x,tau_xy)
+printf('\n The principal stresses are %.2f kg/cm^2 (compression) %.2f kg/cm^2 (tension) ',sigma_2,sigma_1)
+
+//for the line b-b
+tau_xy=tau_max
+sigma_x=0
+sigma_y=0
+sigma_1=(sigma_x+sigma_y)/2+sqrt((1/2*(sigma_x-sigma_y))^2+tau_xy^2)
+sigma_2=(sigma_x+sigma_y)/2-sqrt((1/2*(sigma_x-sigma_y))^2+tau_xy^2)
+// results
+printf('\n For the line b-b the bending stress and shearing stress are \n respectively %.2f kg/cm^2, %.2f kg/cm^2 ',sigma_x,tau_xy)
+printf('\n The principal stresses are %.2f kg/cm^2 (tension) %.2f kg/cm^2 (compression) ',sigma_1,sigma_2)