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+// Problem 5.14,Page no.136
+
+clc;clear;
+close;
+
+B=10 //cm //width of timber section
+D=15 //cm //depth of timber section
+b=10 //cm //width of steel plate
+t=12 //mm //thickness 
+w=3 //KN/m //Uniformly distributed Load
+L=4 //m //Span of beam
+m=20 //Ratio of modulus of elasticity of steel to timber 
+W=3 //KN/m //Load
+
+//Calculations
+
+y_1=15*2**-1 //C.G of timber 
+y_2=1.2*2**-1 //C.G of steel plate
+b_s=10*m**-1 //cm //Equivalent width of steel
+Y_bar=(10*1.2*0.6+15*0.5*8.7)*(10*1.2+15*0.5)**-1 //cm //distance of C.G from bottom edge
+
+I=1*12**-1*10*(1.2)**3+10*1.2*(3.72-0.6)**2+1*12**-1*0.5*(15)**3+0.5*15*(7.5-3.72)**2
+M=W*10**5*L**2*8**-1 //N*m
+
+Y_bar_1=3.72 //cm //C.G from bottom edge
+Y_bar_2=16.2-Y_bar //cm //C.G from top edge
+
+sigma_1=(M*I**-1*Y_bar_1)*10**-2 //N/mm**2 //stress at bottom
+
+sigma_2=(M*I**-1*Y_bar_2)*10**-2 //N/mm**2 //stress at top
+
+sigma_max=sigma_2*m**-1 
+
+//The Answers in book for Moment of Inertia about x-x axis onwards are incorrect
+
+//Result
+printf("Moment of Inertia = %.f N-m",M)
+printf("\n The Max Stress in steel is %.2f N/mm^2",sigma_1)
+printf("\n The Max Stress in timber is %.2f N/mm**2",sigma_max)